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http://www.avabodh.com/cin/pointer.html
[ "# Pointer Dereferencing\n\nTake this C code as example:\n```// this is defined globally.\nint globalVar;\n// These lines of code are part of some function\nint b;\nint * ptr;\nptr = &globalVar;\n*ptr = 100;\nb = *ptr;\n```\n\nGenerated assembly code:\n```movl \\$globalVar, -4(%ebp)\nmovl -4(%ebp), %eax\nmovl \\$100, (%eax)\nmovl -4(%ebp), %eax\nmovl (%eax), %eax\nmovl %eax, -8(%ebp)\n```\nLocation of local variables of the stack (local variables are explained here)\n```ptr => -4(%ebp)\nb => -8(%ebp)\n```\nThe use of registers as temporary memory is described here\n```# ptr = &globarVar\nmovl \\$globalVar, -4(%ebp)\n\n# tmp = ptr\nmovl -4(%ebp), %eax\n\n# *tmp = 100\nmovl \\$100, (%eax)\n\n# tmp = ptr\nmovl -4(%ebp), %eax\n\n# tmp = *tmp\nmovl (%eax), %eax\n\n# b = tmp\nmovl %eax, -8(%ebp)\n```" ]
[ null ]
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https://www.tl6.net/que8shu/10392.html
[ "", null, "1,1, 2, 3, 5, 8, 13, 21, 34, 55, ···\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ···\n\nLn =Fn-1 +Fn+1,  其中 n=1,2,3, ···\n\nL1 =F0 +F2 = 0+ 1=1\n\nL2 =F1 +F3 = 1+ 2=3\n\nL3 =F2 +F4 = 1+ 3=4\n\nL4 =F3 +F5 = 2+ 5=7\n\nL5 =F4 +F6 = 3+ 8=11\n\nL6 =F5 +F7 = 5+ 13=18\n\nL7 =F6 +F8 = 8+ 21=29\n\nL8 =F7 +F9 =13+ 34=47\n\nL9 = F8 +F10 =21+ 55=76\n\nL10 =F9 + F11 =34+ 89=123\n\nL11 =F10 +F12 =55+144=199\n\nL12 =F11 +F13 =89+233=322\n\nL13 =F12 +F14 =144+377=521\n\n............\n\n1,3,4,7,11,18,29,47,76,123,199,332,521,···\n\nLn =Fn-1 +Fn+1\n\n=(Fn-3 +Fn-2 ) +(Fn-1 +Fn )\n\n=(Fn-3 +Fn-1 ) +( Fn-2 + Fn )\n\n=Ln-2+Ln-1\n\nL2-1=3-1=2=1×2\n\nL3-1= 4-1 = 3 = 1×3\n\nL5-1= 11-1 =10=2×5\n\nL7-1= 29-1 =28=4×7\n\nL11-1= 199-1 =198=18×11\n\nL13-1 = 521-1 =520 =40×13" ]
[ null, "https://www.tl6.net/uploads/allimg/211004/1-211004125034338.jpg", null ]
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https://cantera.org/documentation/docs-2.0/doxygen/html/classCantera_1_1AxiStagnFlow.html
[ "Cantera  2.0\nAxiStagnFlow Class Reference\n\nA class for axisymmetric stagnation flows. More...\n\n#include <StFlow.h>\n\nInheritance diagram for AxiStagnFlow:\n[legend]\nCollaboration diagram for AxiStagnFlow:\n[legend]\n\n## Public Member Functions\n\nAxiStagnFlow (IdealGasPhase *ph=0, size_t nsp=1, size_t points=1)\n\nvirtual void eval (size_t j, doublereal *x, doublereal *r, integer *mask, doublereal rdt)\nEvaluate the residual function for axisymmetric stagnation flow.\n\nvirtual std::string flowType ()\n\nint domainType ()\nDomain type flag.\n\nsize_t domainIndex ()\nThe left-to-right location of this domain.\n\nbool isConnector ()\nTrue if the domain is a connector domain.\n\nconst OneDimcontainer () const\nThe container holding this domain.\n\nvoid setContainer (OneDim *c, size_t index)\nSpecify the container object for this domain, and the position of this domain in the list.\n\nvoid setBandwidth (int bw=-1)\n\nsize_t bandwidth ()\nSet the Jacobian bandwidth for this domain.\n\nvirtual void setInitialState (doublereal *xlocal=0)\n\nRefiner & refiner ()\nReturn a reference to the grid refiner.\n\nsize_t nComponents () const\nNumber of components at each grid point.\n\nvoid checkComponentIndex (size_t n) const\nCheck that the specified component index is in range Throws an exception if n is greater than nComponents()-1.\n\nvoid checkComponentArraySize (size_t nn) const\nCheck that an array size is at least nComponents() Throws an exception if nn is less than nComponents().\n\nsize_t nPoints () const\nNumber of grid points in this domain.\n\nvoid checkPointIndex (size_t n) const\nCheck that the specified point index is in range Throws an exception if n is greater than nPoints()-1.\n\nvoid checkPointArraySize (size_t nn) const\nCheck that an array size is at least nPoints() Throws an exception if nn is less than nPoints().\n\nvoid setComponentName (size_t n, std::string name)\n\nvoid setComponentType (size_t n, int ctype)\n\nvoid setBounds (size_t nl, const doublereal *lower, size_t nu, const doublereal *upper)\nSet the lower and upper bounds for each solution component.\n\nvoid setBounds (size_t n, doublereal lower, doublereal upper)\n\nvoid setTolerances (size_t nr, const doublereal *rtol, size_t na, const doublereal *atol, int ts=0)\nset the error tolerances for all solution components.\n\nvoid setTolerances (size_t n, doublereal rtol, doublereal atol, int ts=0)\nset the error tolerances for solution component n.\n\nvoid setTolerances (doublereal rtol, doublereal atol, int ts=0)\nset scalar error tolerances.\n\nvoid setTolerancesTS (doublereal rtol, doublereal atol)\n\nvoid setTolerancesSS (doublereal rtol, doublereal atol)\n\ndoublereal rtol (size_t n)\nRelative tolerance of the nth component.\n\ndoublereal atol (size_t n)\nAbsolute tolerance of the nth component.\n\ndoublereal upperBound (size_t n) const\nUpper bound on the nth component.\n\ndoublereal lowerBound (size_t n) const\nLower bound on the nth component.\n\nvoid initTimeInteg (doublereal dt, const doublereal *x0)\nPrepare to do time stepping with time step dt.\n\nPrepare to solve the steady-state problem.\n\nbool transient ()\nTrue if not in steady-state mode.\n\nvoid needJacUpdate ()\nSet this if something has changed in the governing equations (e.g.\n\nvoid evalss (doublereal *x, doublereal *r, integer *mask)\nEvaluate the steady-state residual at all points, even if in transient mode.\n\nvirtual doublereal residual (doublereal *x, size_t n, size_t j)\n\nint timeDerivativeFlag (size_t n)\n\nvoid setAlgebraic (size_t n)\n\nvirtual void update (doublereal *x)\nDoes nothing.\n\ndoublereal time () const\n\nvoid incrementTime (doublereal dt)\n\nsize_t index (size_t n, size_t j) const\n\ndoublereal value (const doublereal *x, size_t n, size_t j) const\n\nsize_t size () const\n\nvoid locate ()\nFind the index of the first grid point in this domain, and the start of its variables in the global solution vector.\n\nvirtual size_t loc (size_t j=0) const\nLocation of the start of the local solution vector in the global solution vector,.\n\nsize_t firstPoint () const\nThe index of the first (i.e., left-most) grid point belonging to this domain.\n\nsize_t lastPoint () const\nThe index of the last (i.e., right-most) grid point belonging to this domain.\n\nSet the left neighbor to domain 'left.\n\nSet the right neighbor to domain 'right.\n\nvoid append (Domain1D *right)\nAppend domain 'right' to this one, and update all links.\n\nDomain1Dleft () const\nReturn a pointer to the left neighbor.\n\nDomain1Dright () const\nReturn a pointer to the right neighbor.\n\ndouble prevSoln (size_t n, size_t j) const\nValue of component n at point j in the previous solution.\n\nvoid setID (const std::string &s)\nSpecify an identifying tag for this domain.\n\nstd::string id () const\n\nvoid setDesc (const std::string &s)\nSpecify descriptive text for this domain.\n\nconst std::string & desc ()\n\nvirtual void showSolution_s (std::ostream &s, const doublereal *x)\n\ndoublereal z (size_t jlocal) const\n\ndoublereal zmin () const\n\ndoublereal zmax () const\n\nvoid setProfile (std::string name, doublereal *values, doublereal *soln)\n\nvector_fpgrid ()\n\nconst vector_fpgrid () const\n\ndoublereal grid (size_t point)\n\nvoid setGrid (size_t n, const doublereal *z)\n\nvirtual doublereal initialValue (size_t n, size_t j)\nInitial value of solution component n at grid point j.\n\n## Public Attributes\n\ndoublereal m_zfixed\n\ndoublereal m_tfixed\n\n## Protected Attributes\n\ndoublereal m_rdt\n\nsize_t m_nv\n\nsize_t m_points\n\nvector_fp m_slast\n\ndoublereal m_time\n\nvector_fp m_max\n\nvector_fp m_min\n\nvector_fp m_rtol_ss\n\nvector_fp m_rtol_ts\n\nvector_fp m_atol_ss\n\nvector_fp m_atol_ts\n\nvector_fp m_z\n\nOneDimm_container\n\nsize_t m_index\n\nint m_type\n\nsize_t m_iloc\nStarting location within the solution vector for unknowns that correspond to this domain.\n\nsize_t m_jstart\n\nDomain1Dm_left\n\nDomain1Dm_right\n\nstd::string m_id\nIdentity tag for the domain.\n\nstd::string m_desc\n\nRefiner * m_refiner\n\nvector_int m_td\n\nstd::vector< std::string > m_name\n\nint m_bw\n\n## Problem Specification\n\nvirtual void setupGrid (size_t n, const doublereal *z)\n\nthermo_tphase ()\n\nKineticskinetics ()\n\nvirtual void init ()\nInitialize.\n\nvoid setThermo (IdealGasPhase &th)\nSet the thermo manager.\n\nvoid setKinetics (Kinetics &kin)\nSet the kinetics manager. The kinetics manager must.\n\nvoid setTransport (Transport &trans, bool withSoret=false)\nset the transport manager\n\nvoid enableSoret (bool withSoret)\n\nbool withSoret () const\n\nvoid setPressure (doublereal p)\nSet the pressure.\n\nvirtual void setState (size_t point, const doublereal *state, doublereal *x)\n\nvirtual void _getInitialSoln (doublereal *x)\nWrite the initial solution estimate into array x.\n\nvirtual void _finalize (const doublereal *x)\nIn some cases, a domain may need to set parameters that depend on the initial solution estimate.\n\nvoid setFixedTempProfile (vector_fp &zfixed, vector_fp &tfixed)\nSometimes it is desired to carry out the simulation using a specified temperature profile, rather than computing it by solving the energy equation.\n\nvoid setTemperature (size_t j, doublereal t)\nSet the temperature fixed point at grid point j, and disable the energy equation so that the solution will be held to this value.\n\nvoid setMassFraction (size_t j, size_t k, doublereal y)\nSet the mass fraction fixed point for species k at grid point j, and disable the species equation so that the solution will be held to this value.\n\ndoublereal T_fixed (size_t j) const\nThe fixed temperature value at point j.\n\ndoublereal Y_fixed (size_t k, size_t j) const\nThe fixed mass fraction value of species k at point j.\n\nvirtual std::string componentName (size_t n) const\nName of the nth component. May be overloaded.\n\nsize_t componentIndex (std::string name) const\n\nvirtual void showSolution (const doublereal *x)\nPrint the solution.\n\nvirtual void save (XML_Node &o, const doublereal *const sol)\nSave the current solution for this domain into an XML_Node.\n\nvirtual void restore (const XML_Node &dom, doublereal *soln)\n\nvoid solveEnergyEqn (size_t j=npos)\n\nvoid fixTemperature (size_t j=npos)\n\nbool doSpecies (size_t k)\n\nbool doEnergy (size_t j)\n\nvoid solveSpecies (size_t k=npos)\n\nvoid fixSpecies (size_t k=npos)\n\nvoid integrateChem (doublereal *x, doublereal dt)\n\nvoid resize (size_t components, size_t points)\nChange the grid size.\n\nvirtual void setFixedPoint (int j0, doublereal t0)\n\nvoid setJac (MultiJac *jac)\n\nvoid setGas (const doublereal *x, size_t j)\nSet the gas object state to be consistent with the solution at point j.\n\nvoid setGasAtMidpoint (const doublereal *x, size_t j)\nSet the gas state to be consistent with the solution at the midpoint between j and j + 1.\n\ndoublereal density (size_t j) const\n\nvirtual bool fixed_mdot ()\n\nvoid setViscosityFlag (bool dovisc)\n\ndoublereal m_inlet_u\n\ndoublereal m_inlet_V\n\ndoublereal m_inlet_T\n\ndoublereal m_rho_inlet\n\nvector_fp m_yin\n\ndoublereal m_surface_T\n\ndoublereal m_press\n\nvector_fp m_dz\n\nvector_fp m_rho\n\nvector_fp m_wtm\n\nvector_fp m_wt\n\nvector_fp m_cp\n\nvector_fp m_enth\n\nvector_fp m_visc\n\nvector_fp m_tcon\n\nvector_fp m_diff\n\nvector_fp m_multidiff\n\nArray2D m_dthermal\n\nArray2D m_flux\n\nArray2D m_wdot\n\nvector_fp m_surfdot\n\nsize_t m_nsp\n\nIdealGasPhasem_thermo\n\nKineticsm_kin\n\nTransportm_trans\n\nMultiJacm_jac\n\nbool m_ok\n\nstd::vector< bool > m_do_energy\n\nbool m_do_soret\n\nstd::vector< bool > m_do_species\n\nint m_transport_option\n\nArray2D m_fixedy\n\nvector_fp m_fixedtemp\n\nvector_fp m_zfix\n\nvector_fp m_tfix\n\nbool m_dovisc\n\ndoublereal component (const doublereal *x, size_t i, size_t j) const\n\ndoublereal conc (const doublereal *x, size_t k, size_t j) const\n\ndoublereal cbar (const doublereal *x, size_t k, size_t j) const\n\ndoublereal wdot (size_t k, size_t j) const\n\nvoid getWdot (doublereal *x, size_t j)\nwrite the net production rates at point j into array m_wdot\n\nvoid updateThermo (const doublereal *x, size_t j0, size_t j1)\nupdate the thermodynamic properties from point j0 to point j1 (inclusive), based on solution x.\n\ndoublereal cdif2 (const doublereal *x, size_t n, size_t j, const doublereal *f) const\n\ndoublereal T (const doublereal *x, size_t j) const\n\ndoublereal & T (doublereal *x, size_t j)\n\ndoublereal T_prev (size_t j) const\n\ndoublereal rho_u (const doublereal *x, size_t j) const\n\ndoublereal u (const doublereal *x, size_t j) const\n\ndoublereal V (const doublereal *x, size_t j) const\n\ndoublereal V_prev (size_t j) const\n\ndoublereal lambda (const doublereal *x, size_t j) const\n\ndoublereal Y (const doublereal *x, size_t k, size_t j) const\n\ndoublereal & Y (doublereal *x, size_t k, size_t j)\n\ndoublereal Y_prev (size_t k, size_t j) const\n\ndoublereal X (const doublereal *x, size_t k, size_t j) const\n\ndoublereal flux (size_t k, size_t j) const\n\ndoublereal dVdz (const doublereal *x, size_t j) const\n\ndoublereal dYdz (const doublereal *x, size_t k, size_t j) const\n\ndoublereal dTdz (const doublereal *x, size_t j) const\n\ndoublereal shear (const doublereal *x, size_t j) const\n\ndoublereal divHeatFlux (const doublereal *x, size_t j) const\n\nsize_t mindex (size_t k, size_t j, size_t m)\n\nvoid updateDiffFluxes (const doublereal *x, size_t j0, size_t j1)\nUpdate the diffusive mass fluxes.\n\nvoid updateTransport (doublereal *x, size_t j0, size_t j1)\nUpdate the transport properties at grid points in the range from j0 to j1, based on solution x.\n\n## Detailed Description\n\nA class for axisymmetric stagnation flows.\n\nDefinition at line 507 of file StFlow.h.\n\n## Member Function Documentation\n\n void eval ( size_t jg, doublereal * xg, doublereal * rg, integer * diagg, doublereal rdt )\nvirtual\n\nEvaluate the residual function for axisymmetric stagnation flow.\n\nIf jpt is less than zero, the residual function is evaluated at all grid points. If jpt >= 0, then the residual function is only evaluated at grid points jpt-1, jpt, and jpt+1. This option is used to efficiently evaluate the Jacobian numerically.\n\nReimplemented from Domain1D.\n\nDefinition at line 342 of file StFlow.cpp.\n\n virtual void init ( )\ninlinevirtualinherited\n\nInitialize.\n\nThis method is called by OneDim::init() for each domain once at the beginning of a simulation. Base class method does nothing, but may be overloaded.\n\nReimplemented from Domain1D.\n\nDefinition at line 82 of file StFlow.h.\n\n void setThermo ( IdealGasPhase & th )\ninlineinherited\n\nSet the thermo manager.\n\nNote that the flow equations assume the ideal gas equation.\n\nDefinition at line 89 of file StFlow.h.\n\n void setKinetics ( Kinetics & kin )\ninlineinherited\n\nSet the kinetics manager. The kinetics manager must.\n\nDefinition at line 94 of file StFlow.h.\n\n void setTransport ( Transport & trans, bool withSoret = false )\ninherited\n\nset the transport manager\n\nInstall a transport manager.\n\nDefinition at line 242 of file StFlow.cpp.\n\nReferences Transport::model(), and Array2D::resize().\n\n void setPressure ( doublereal p )\ninlineinherited\n\nSet the pressure.\n\nSince the flow equations are for the limit of small Mach number, the pressure is very nearly constant throughout the flow.\n\nDefinition at line 108 of file StFlow.h.\n\n virtual void setState ( size_t point, const doublereal * state, doublereal * x )\ninlinevirtualinherited\nTodo:\nremove? may be unused\n\nReimplemented from Domain1D.\n\nDefinition at line 114 of file StFlow.h.\n\nReferences StFlow::setMassFraction(), and StFlow::setTemperature().\n\n virtual void _getInitialSoln ( doublereal * x )\ninlinevirtualinherited\n\nWrite the initial solution estimate into array x.\n\nReimplemented from Domain1D.\n\nDefinition at line 125 of file StFlow.h.\n\nReferences StFlow::T_fixed(), and StFlow::Y_fixed().\n\n void _finalize ( const doublereal * x )\nvirtualinherited\n\nIn some cases, a domain may need to set parameters that depend on the initial solution estimate.\n\nIn such cases, the parameters may be set in method _finalize. This method is called just before the Newton solver is called, and the x array is guaranteed to be the local solution vector for this domain that will be used as the initial guess. If no such parameters need to be set, then method _finalize does not need to be overloaded.\n\nReimplemented from Domain1D.\n\nDefinition at line 306 of file StFlow.cpp.\n\n void setFixedTempProfile ( vector_fp & zfixed, vector_fp & tfixed )\ninlineinherited\n\nSometimes it is desired to carry out the simulation using a specified temperature profile, rather than computing it by solving the energy equation.\n\nThis method specifies this profile.\n\nDefinition at line 141 of file StFlow.h.\n\n void setTemperature ( size_t j, doublereal t )\ninlineinherited\n\nSet the temperature fixed point at grid point j, and disable the energy equation so that the solution will be held to this value.\n\nDefinition at line 151 of file StFlow.h.\n\nReferenced by StFlow::_finalize(), and StFlow::setState().\n\n void setMassFraction ( size_t j, size_t k, doublereal y )\ninlineinherited\n\nSet the mass fraction fixed point for species k at grid point j, and disable the species equation so that the solution will be held to this value.\n\nnote: in practice, the species are hardly ever held fixed.\n\nDefinition at line 162 of file StFlow.h.\n\nReferenced by StFlow::_finalize(), and StFlow::setState().\n\n doublereal T_fixed ( size_t j ) const\ninlineinherited\n\nThe fixed temperature value at point j.\n\nDefinition at line 169 of file StFlow.h.\n\nReferenced by StFlow::_getInitialSoln(), AxiStagnFlow::eval(), and FreeFlame::eval().\n\n doublereal Y_fixed ( size_t k, size_t j ) const\ninlineinherited\n\nThe fixed mass fraction value of species k at point j.\n\nDefinition at line 175 of file StFlow.h.\n\nReferenced by StFlow::_getInitialSoln().\n\n string componentName ( size_t n ) const\nvirtualinherited\n\nName of the nth component. May be overloaded.\n\nReimplemented from Domain1D.\n\nDefinition at line 985 of file StFlow.cpp.\n\nReferences Phase::speciesName().\n\nReferenced by StFlow::showSolution().\n\n void showSolution ( const doublereal * x )\nvirtualinherited\n\nPrint the solution.\n\nReimplemented from Domain1D.\n\nDefinition at line 887 of file StFlow.cpp.\n\nReferences StFlow::componentName(), StFlow::updateThermo(), and Cantera::writelog().\n\n void save ( XML_Node & o, const doublereal *const sol )\nvirtualinherited\n\nSave the current solution for this domain into an XML_Node.\n\nParameters\n o XML_Node to save the solution to. sol Current value of the solution vector. The object will pick out which part of the solution vector pertains to this object.\n\nReimplemented from Domain1D.\n\nDefinition at line 1170 of file StFlow.cpp.\n\n void resize ( size_t ncomponents, size_t points )\nvirtualinherited\n\nChange the grid size.\n\nCalled after grid refinement.\n\nReimplemented from Domain1D.\n\nDefinition at line 197 of file StFlow.cpp.\n\nReferences Array2D::resize(), and Domain1D::resize().\n\n void setGas ( const doublereal * x, size_t j )\ninherited\n\nSet the gas object state to be consistent with the solution at point j.\n\nDefinition at line 280 of file StFlow.cpp.\n\n void setGasAtMidpoint ( const doublereal * x, size_t j )\ninherited\n\nSet the gas state to be consistent with the solution at the midpoint between j and j + 1.\n\nDefinition at line 293 of file StFlow.cpp.\n\nReferenced by StFlow::updateTransport().\n\n void getWdot ( doublereal * x, size_t j )\ninlineprotectedinherited\n\nwrite the net production rates at point j into array m_wdot\n\nDefinition at line 303 of file StFlow.h.\n\nReferences Kinetics::getNetProductionRates(), and StFlow::setGas().\n\nReferenced by AxiStagnFlow::eval(), and FreeFlame::eval().\n\n void updateThermo ( const doublereal * x, size_t j0, size_t j1 )\ninlineprotectedinherited\n\nupdate the thermodynamic properties from point j0 to point j1 (inclusive), based on solution x.\n\nDefinition at line 312 of file StFlow.h.\n\nReferenced by AxiStagnFlow::eval(), FreeFlame::eval(), and StFlow::showSolution().\n\n void updateDiffFluxes ( const doublereal * x, size_t j0, size_t j1 )\nprotectedinherited\n\nUpdate the diffusive mass fluxes.\n\nDefinition at line 942 of file StFlow.cpp.\n\nReferenced by AxiStagnFlow::eval(), and FreeFlame::eval().\n\n void updateTransport ( doublereal * x, size_t j0, size_t j1 )\nprotectedinherited\n\nUpdate the transport properties at grid points in the range from j0 to j1, based on solution x.\n\nDefinition at line 590 of file StFlow.cpp.\n\nReferenced by AxiStagnFlow::eval(), and FreeFlame::eval().\n\n int domainType ( )\ninlineinherited\n\nDomain type flag.\n\nDefinition at line 71 of file Domain1D.h.\n\n size_t domainIndex ( )\ninlineinherited\n\nThe left-to-right location of this domain.\n\nDefinition at line 78 of file Domain1D.h.\n\nReferenced by Cantera::bound_step().\n\n bool isConnector ( )\ninlineinherited\n\nTrue if the domain is a connector domain.\n\nDefinition at line 85 of file Domain1D.h.\n\n const OneDim& container ( ) const\ninlineinherited\n\nThe container holding this domain.\n\nDefinition at line 92 of file Domain1D.h.\n\n void setContainer ( OneDim * c, size_t index )\ninlineinherited\n\nSpecify the container object for this domain, and the position of this domain in the list.\n\nDefinition at line 100 of file Domain1D.h.\n\n size_t bandwidth ( )\ninlineinherited\n\nSet the Jacobian bandwidth for this domain.\n\nWhen class OneDim computes the bandwidth of the overall multi-domain problem (in OneDim::resize()), it calls this method for the bandwidth of each domain. If setBandwidth has not been called, then a negative bandwidth is returned, in which case OneDim assumes that this domain is dense – that is, at each point, all components depend on the value of all other components at that point. In this case, the bandwidth is bw = 2*nComponents() - 1. However, if this domain contains some components that are uncoupled from other components at the same point, then this default bandwidth may greatly overestimate the true bandwidth, with a substantial penalty in performance. For such domains, use method setBandwidth to specify the bandwidth before passing this domain to the Sim1D or OneDim constructor.\n\nDefinition at line 129 of file Domain1D.h.\n\nReferenced by OneDim::resize().\n\n Refiner& refiner ( )\ninlineinherited\n\nReturn a reference to the grid refiner.\n\nDefinition at line 172 of file Domain1D.h.\n\nReferenced by Sim1D::refine(), Sim1D::setGridMin(), and Sim1D::setRefineCriteria().\n\n size_t nComponents ( ) const\ninlineinherited\n\nNumber of components at each grid point.\n\nDefinition at line 177 of file Domain1D.h.\n\n void checkComponentIndex ( size_t n ) const\ninlineinherited\n\nCheck that the specified component index is in range Throws an exception if n is greater than nComponents()-1.\n\nDefinition at line 183 of file Domain1D.h.\n\n void checkComponentArraySize ( size_t nn ) const\ninlineinherited\n\nCheck that an array size is at least nComponents() Throws an exception if nn is less than nComponents().\n\nUsed before calls which take an array pointer.\n\nDefinition at line 192 of file Domain1D.h.\n\n size_t nPoints ( ) const\ninlineinherited\n\nNumber of grid points in this domain.\n\nDefinition at line 199 of file Domain1D.h.\n\n void checkPointIndex ( size_t n ) const\ninlineinherited\n\nCheck that the specified point index is in range Throws an exception if n is greater than nPoints()-1.\n\nDefinition at line 205 of file Domain1D.h.\n\n void checkPointArraySize ( size_t nn ) const\ninlineinherited\n\nCheck that an array size is at least nPoints() Throws an exception if nn is less than nPoints().\n\nUsed before calls which take an array pointer.\n\nDefinition at line 214 of file Domain1D.h.\n\n void setBounds ( size_t nl, const doublereal * lower, size_t nu, const doublereal * upper )\ninlineinherited\n\nSet the lower and upper bounds for each solution component.\n\nDefinition at line 254 of file Domain1D.h.\n\nReferences Cantera::int2str().\n\n void setTolerances ( size_t nr, const doublereal * rtol, size_t na, const doublereal * atol, int ts = 0 )\ninherited\n\nset the error tolerances for all solution components.\n\nDefinition at line 16 of file Domain1D.cpp.\n\nReferences Cantera::int2str().\n\n void setTolerances ( size_t n, doublereal rtol, doublereal atol, int ts = 0 )\ninherited\n\nset the error tolerances for solution component n.\n\nDefinition at line 34 of file Domain1D.cpp.\n\n void setTolerances ( doublereal rtol, doublereal atol, int ts = 0 )\ninherited\n\nset scalar error tolerances.\n\nAll solution components will have the same relative and absolute error tolerances.\n\nDefinition at line 47 of file Domain1D.cpp.\n\n doublereal rtol ( size_t n )\ninlineinherited\n\nRelative tolerance of the nth component.\n\nDefinition at line 285 of file Domain1D.h.\n\n doublereal atol ( size_t n )\ninlineinherited\n\nAbsolute tolerance of the nth component.\n\nDefinition at line 290 of file Domain1D.h.\n\n doublereal upperBound ( size_t n ) const\ninlineinherited\n\nUpper bound on the nth component.\n\nDefinition at line 295 of file Domain1D.h.\n\nReferenced by Cantera::bound_step(), and Inlet1D::save().\n\n doublereal lowerBound ( size_t n ) const\ninlineinherited\n\nLower bound on the nth component.\n\nDefinition at line 300 of file Domain1D.h.\n\nReferenced by Cantera::bound_step(), and Inlet1D::save().\n\n void initTimeInteg ( doublereal dt, const doublereal * x0 )\ninlineinherited\n\nPrepare to do time stepping with time step dt.\n\nCopy the internally-stored solution at the last time step to array x0.\n\nDefinition at line 310 of file Domain1D.h.\n\nReferences Domain1D::loc().\n\nReferenced by OneDim::initTimeInteg().\n\ninlineinherited\n\nPrepare to solve the steady-state problem.\n\nSet the internally-stored reciprocal of the time step to 0,0\n\nDefinition at line 319 of file Domain1D.h.\n\ninlineinherited\n\nDefinition at line 324 of file Domain1D.h.\n\n bool transient ( )\ninlineinherited\n\nTrue if not in steady-state mode.\n\nDefinition at line 329 of file Domain1D.h.\n\n void needJacUpdate ( )\ninherited\n\nSet this if something has changed in the governing equations (e.g.\n\nSignal that the current Jacobian is no longer valid.\n\nthe value of a constant has been changed, so that the last-computed Jacobian is no longer valid. Note: see file OneDim.cpp for the implementation of this method.\n\nDefinition at line 391 of file OneDim.cpp.\n\nReferences OneDim::jacobian(), OneDim::saveStats(), and MultiJac::setAge().\n\n void evalss ( doublereal * x, doublereal * r, integer * mask )\ninlineinherited\n\nEvaluate the steady-state residual at all points, even if in transient mode.\n\nUsed only to print diagnostic output.\n\nDefinition at line 345 of file Domain1D.h.\n\nReferences Domain1D::eval(), and Cantera::npos.\n\n virtual void update ( doublereal * x )\ninlinevirtualinherited\n\nDoes nothing.\n\nDefinition at line 373 of file Domain1D.h.\n\n void locate ( )\ninlineinherited\n\nFind the index of the first grid point in this domain, and the start of its variables in the global solution vector.\n\nDefinition at line 413 of file Domain1D.h.\n\nReferences Domain1D::lastPoint(), Domain1D::loc(), Domain1D::locate(), and Domain1D::m_iloc.\n\nReferenced by Domain1D::linkLeft(), Domain1D::locate(), and Domain1D::resize().\n\n virtual size_t loc ( size_t j = 0 ) const\ninlinevirtualinherited\n\nLocation of the start of the local solution vector in the global solution vector,.\n\nDefinition at line 438 of file Domain1D.h.\n\nReferences Domain1D::m_iloc.\n\n size_t firstPoint ( ) const\ninlineinherited\n\nThe index of the first (i.e., left-most) grid point belonging to this domain.\n\nDefinition at line 446 of file Domain1D.h.\n\n size_t lastPoint ( ) const\ninlineinherited\n\nThe index of the last (i.e., right-most) grid point belonging to this domain.\n\nDefinition at line 454 of file Domain1D.h.\n\n void linkLeft ( Domain1D * left )\ninlineinherited\n\nSet the left neighbor to domain 'left.\n\n' Method 'locate' is called to update the global positions of this domain and all those to its right.\n\nDefinition at line 463 of file Domain1D.h.\n\nReferences Domain1D::left(), and Domain1D::locate().\n\nReferenced by Domain1D::append().\n\n void linkRight ( Domain1D * right )\ninlineinherited\n\nSet the right neighbor to domain 'right.\n\n'\n\nDefinition at line 471 of file Domain1D.h.\n\nReferences Domain1D::right().\n\nReferenced by Domain1D::append().\n\n void append ( Domain1D * right )\ninlineinherited\n\nAppend domain 'right' to this one, and update all links.\n\nDefinition at line 478 of file Domain1D.h.\n\n Domain1D* left ( ) const\ninlineinherited\n\nReturn a pointer to the left neighbor.\n\nDefinition at line 486 of file Domain1D.h.\n\n Domain1D* right ( ) const\ninlineinherited\n\nReturn a pointer to the right neighbor.\n\nDefinition at line 493 of file Domain1D.h.\n\n double prevSoln ( size_t n, size_t j ) const\ninlineinherited\n\nValue of component n at point j in the previous solution.\n\nDefinition at line 500 of file Domain1D.h.\n\nReferenced by ReactingSurf1D::eval().\n\n void setID ( const std::string & s )\ninlineinherited\n\nSpecify an identifying tag for this domain.\n\nDefinition at line 507 of file Domain1D.h.\n\nReferences Domain1D::m_id.\n\nReferenced by StFlow::StFlow().\n\n void setDesc ( const std::string & s )\ninlineinherited\n\nSpecify descriptive text for this domain.\n\nDefinition at line 522 of file Domain1D.h.\n\n doublereal initialValue ( size_t n, size_t j )\nvirtualinherited\n\nInitial value of solution component n at grid point j.\n\nDefinition at line 207 of file Domain1D.cpp.\n\n## Member Data Documentation\n\n size_t m_iloc\nprotectedinherited\n\nStarting location within the solution vector for unknowns that correspond to this domain.\n\nRemember there may be multiple domains associated with this problem\n\nDefinition at line 628 of file Domain1D.h.\n\nReferenced by Domain1D::loc(), and Domain1D::locate().\n\n std::string m_id\nprotectedinherited\n\nIdentity tag for the domain.\n\nDefinition at line 635 of file Domain1D.h.\n\nReferenced by StFlow::save(), and Domain1D::setID().\n\nThe documentation for this class was generated from the following files:" ]
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https://www.gcflcm.com/gcf-of-84-and-91
[ "# What is the Greatest Common Factor of 84 and 91?\n\nGreatest common factor (GCF) of 84 and 91 is 7.\n\nGCF(84,91) = 7\n\nWe will now calculate the prime factors of 84 and 91, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 84 and 91.\n\nGCF Calculator and\nand\n\n## How to find the GCF of 84 and 91?\n\nWe will first find the prime factorization of 84 and 91. After we will calculate the factors of 84 and 91 and find the biggest common factor number .\n\n### Step-1: Prime Factorization of 84\n\nPrime factors of 84 are 2, 3, 7. Prime factorization of 84 in exponential form is:\n\n84 = 22 × 31 × 71\n\n### Step-2: Prime Factorization of 91\n\nPrime factors of 91 are 7, 13. Prime factorization of 91 in exponential form is:\n\n91 = 71 × 131\n\n### Step-3: Factors of 84\n\nList of positive integer factors of 84 that divides 84 without a remainder.\n\n1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42\n\n### Step-4: Factors of 91\n\nList of positive integer factors of 91 that divides 84 without a remainder.\n\n1, 7, 13\n\n#### Final Step: Biggest Common Factor Number\n\nWe found the factors and prime factorization of 84 and 91. The biggest common factor number is the GCF number.\nSo the greatest common factor 84 and 91 is 7.\n\nAlso check out the Least Common Multiple of 84 and 91" ]
[ null ]
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https://www.colorhexa.com/222f2e
[ "# #222f2e Color Information\n\nIn a RGB color space, hex #222f2e is composed of 13.3% red, 18.4% green and 18% blue. Whereas in a CMYK color space, it is composed of 27.7% cyan, 0% magenta, 2.1% yellow and 81.6% black. It has a hue angle of 175.4 degrees, a saturation of 16% and a lightness of 15.9%. #222f2e color hex could be obtained by blending #445e5c with #000000. Closest websafe color is: #333333.\n\n• R 13\n• G 18\n• B 18\nRGB color chart\n• C 28\n• M 0\n• Y 2\n• K 82\nCMYK color chart\n\n#222f2e color description : Very dark (mostly black) cyan.\n\n# #222f2e Color Conversion\n\nThe hexadecimal color #222f2e has RGB values of R:34, G:47, B:46 and CMYK values of C:0.28, M:0, Y:0.02, K:0.82. Its decimal value is 2240302.\n\nHex triplet RGB Decimal 222f2e `#222f2e` 34, 47, 46 `rgb(34,47,46)` 13.3, 18.4, 18 `rgb(13.3%,18.4%,18%)` 28, 0, 2, 82 175.4°, 16, 15.9 `hsl(175.4,16%,15.9%)` 175.4°, 27.7, 18.4 333333 `#333333`\nCIE-LAB 18.234, -5.732, -1.157 2.169, 2.57, 2.966 0.281, 0.334, 2.57 18.234, 5.847, 191.412 18.234, -5.448, -0.514 16.032, -3.904, 0.252 00100010, 00101111, 00101110\n\n# Color Schemes with #222f2e\n\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #2f2223\n``#2f2223` `rgb(47,34,35)``\nComplementary Color\n• #222f28\n``#222f28` `rgb(34,47,40)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #222a2f\n``#222a2f` `rgb(34,42,47)``\nAnalogous Color\n• #2f2822\n``#2f2822` `rgb(47,40,34)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #2f222a\n``#2f222a` `rgb(47,34,42)``\nSplit Complementary Color\n• #2f2e22\n``#2f2e22` `rgb(47,46,34)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #2e222f\n``#2e222f` `rgb(46,34,47)``\n• #232f22\n``#232f22` `rgb(35,47,34)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #2e222f\n``#2e222f` `rgb(46,34,47)``\n• #2f2223\n``#2f2223` `rgb(47,34,35)``\n• #020303\n``#020303` `rgb(2,3,3)``\n• #0d1111\n``#0d1111` `rgb(13,17,17)``\n• #172020\n``#172020` `rgb(23,32,32)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #2d3e3c\n``#2d3e3c` `rgb(45,62,60)``\n• #374d4b\n``#374d4b` `rgb(55,77,75)``\n• #425b59\n``#425b59` `rgb(66,91,89)``\nMonochromatic Color\n\n# Alternatives to #222f2e\n\nBelow, you can see some colors close to #222f2e. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #222f2b\n``#222f2b` `rgb(34,47,43)``\n• #222f2c\n``#222f2c` `rgb(34,47,44)``\n• #222f2d\n``#222f2d` `rgb(34,47,45)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #222f2f\n``#222f2f` `rgb(34,47,47)``\n• #222e2f\n``#222e2f` `rgb(34,46,47)``\n• #222d2f\n``#222d2f` `rgb(34,45,47)``\nSimilar Colors\n\n# #222f2e Preview\n\nThis text has a font color of #222f2e.\n\n``<span style=\"color:#222f2e;\">Text here</span>``\n#222f2e background color\n\nThis paragraph has a background color of #222f2e.\n\n``<p style=\"background-color:#222f2e;\">Content here</p>``\n#222f2e border color\n\nThis element has a border color of #222f2e.\n\n``<div style=\"border:1px solid #222f2e;\">Content here</div>``\nCSS codes\n``.text {color:#222f2e;}``\n``.background {background-color:#222f2e;}``\n``.border {border:1px solid #222f2e;}``\n\n# Shades and Tints of #222f2e\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010101 is the darkest color, while #f5f8f8 is the lightest one.\n\n• #010101\n``#010101` `rgb(1,1,1)``\n• #090d0d\n``#090d0d` `rgb(9,13,13)``\n• #121818\n``#121818` `rgb(18,24,24)``\n• #1a2423\n``#1a2423` `rgb(26,36,35)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #2a3a39\n``#2a3a39` `rgb(42,58,57)``\n• #324644\n``#324644` `rgb(50,70,68)``\n• #3b514f\n``#3b514f` `rgb(59,81,79)``\n• #435d5b\n``#435d5b` `rgb(67,93,91)``\n• #4b6866\n``#4b6866` `rgb(75,104,102)``\n• #537371\n``#537371` `rgb(83,115,113)``\n• #5c7f7c\n``#5c7f7c` `rgb(92,127,124)``\n• #648a87\n``#648a87` `rgb(100,138,135)``\n• #6d9592\n``#6d9592` `rgb(109,149,146)``\n• #789d9a\n``#789d9a` `rgb(120,157,154)``\n• #83a5a3\n``#83a5a3` `rgb(131,165,163)``\n• #8faeab\n``#8faeab` `rgb(143,174,171)``\n• #9ab6b4\n``#9ab6b4` `rgb(154,182,180)``\n• #a5bebc\n``#a5bebc` `rgb(165,190,188)``\n• #b1c6c5\n``#b1c6c5` `rgb(177,198,197)``\n• #bccfcd\n``#bccfcd` `rgb(188,207,205)``\n• #c8d7d6\n``#c8d7d6` `rgb(200,215,214)``\n• #d3dfde\n``#d3dfde` `rgb(211,223,222)``\n• #dee7e7\n``#dee7e7` `rgb(222,231,231)``\n• #eaf0ef\n``#eaf0ef` `rgb(234,240,239)``\n• #f5f8f8\n``#f5f8f8` `rgb(245,248,248)``\nTint Color Variation\n\n# Tones of #222f2e\n\nA tone is produced by adding gray to any pure hue. In this case, #282929 is the less saturated color, while #034e48 is the most saturated one.\n\n• #282929\n``#282929` `rgb(40,41,41)``\n• #252c2b\n``#252c2b` `rgb(37,44,43)``\n• #222f2e\n``#222f2e` `rgb(34,47,46)``\n• #1f3231\n``#1f3231` `rgb(31,50,49)``\n• #1c3533\n``#1c3533` `rgb(28,53,51)``\n• #193836\n``#193836` `rgb(25,56,54)``\n• #163b39\n``#163b39` `rgb(22,59,57)``\n• #123f3b\n``#123f3b` `rgb(18,63,59)``\n• #0f423e\n``#0f423e` `rgb(15,66,62)``\n• #0c4540\n``#0c4540` `rgb(12,69,64)``\n• #094843\n``#094843` `rgb(9,72,67)``\n• #064b46\n``#064b46` `rgb(6,75,70)``\n• #034e48\n``#034e48` `rgb(3,78,72)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #222f2e is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
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https://www.nagwa.com/en/videos/353154378367/
[ "# Video: Finding the Resistance of Components in Series\n\nFour identical resistors are placed in series in a circuit. The resistance of all four resistors together is 36 Ω. What is the resistance of each resistor individually?\n\n03:22\n\n### Video Transcript\n\nFour identical resistors are placed in series in a circuit. The resistance of all four resistors together is 36 ohms. What is the resistance of each resistor individually?\n\nAlright, so we’ve got a question about resistors here. Let’s see what we know about these resistors. First of all, we know that there’s four of them. And we also know that they’re all identical. Also, we know that they’re placed in series in a circuit. The resistance of all four resistors together is 36 ohms. We need to find the resistance of each resistor individually.\n\nNow in order to help us do this, we can draw a partial circuit diagram. And that looks something like this. We’ve got the four resistors, which we’ll label 𝑅 one, 𝑅 two, 𝑅 three, and 𝑅 four. And we’ve got dots on either side because we don’t know what happens in the rest of the circuit. We don’t know how it’s wired up.\n\nHowever, our main point of focus is the four resistors wired up in series. And that is quite important, the fact that they’re in series. Now, one of the most important things that we know from the question is that these four resistors are identical. Therefore, we can confidently say that the resistances of all of these resistors are equal. 𝑅 one is equal to 𝑅 two is equal to 𝑅 three is equal to 𝑅 four. So for convenience, let’s throw away the subscripts. Let’s just call each one of these resistances 𝑅.\n\nNow we know the total resistance of the four resistors together. We know that is 36 ohms. And this is going to be the key to calculating the resistance of each one of these resistors. However, before we can do that, we need to know how the total resistance of resistors in series is calculated.\n\nSo the total resistance of a bunch of resistors in series is given the following way. If we’ve got 𝑛 resistors lined up in series, then the total resistance, 𝑅 sub tot, standing for total, is equal to the first resistance plus the second resistance plus so on and so forth until we get to 𝑅 sub 𝑛. And that is the resistance of the 𝑛th resistor, the final resistor in the series.\n\nBasically, in other words, you just add up all the resistances together if the resistors are in series. And that ends up being the total resistance of the system. In our situation, we’ve got four resistors. Therefore, the value of 𝑛 is four. And so our total resistance is equal to 𝑅 one plus 𝑅 two plus 𝑅 three plus 𝑅 four.\n\nNow there are two modifications to this equation that we can make. First of all, as we’ve already said, the resistances of each one of these resistors is the same. And we’ve called this 𝑅. So once again, we can get rid of the subscripts. And in fact, we can simplify this down further because 𝑅 plus 𝑅 plus 𝑅 plus 𝑅 is four 𝑅.\n\nSecondly, we know the total resistance of all four resistors together. We’ve been given this in the question. That resistance is 36 ohms. So we’ll replace 𝑅 sub tot with 36 ohms. And at that point, we are very close to finding the resistance of one of these resistors. All we need to do is divide both sides of the equation by four. And so the fours on the right-hand side cancel, which means that all we need to do on the left-hand side is divide 36 by four, which leaves us with our final answer. The resistance of each resistor individually is nine ohms." ]
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https://www.javaguides.net/2023/09/r-program-to-drop-rows-from-dataframe.html
[ "# 1. Introduction\n\nWhen working with dataframes in R, there might be situations where we need to drop specific rows, either based on their index or based on some conditions. In this guide, we will explore how to achieve this using R.\n\n# 2. Program Overview\n\n1. Create an initial dataframe.\n\n2. Drop rows based on their index.\n\n3. Drop rows based on a condition.\n\n# 3. Code Program\n\n``````# Load necessary libraries\nlibrary(dplyr)\n\n# Create an initial dataframe\ndf <- data.frame(\nName = c('Alice', 'Bob', 'Charlie', 'David', 'Eve'),\nAge = c(25, 30, 28, 35, 29),\nOccupation = c('Engineer', 'Doctor', 'Lawyer', 'Artist', 'Engineer')\n)\n\n# Print the original dataframe\nprint(\"Original Dataframe:\")\nprint(df)\n\n# Drop the 2nd and 4th rows\ndf <- df[-c(2,4), ]\n\n# Print the dataframe after dropping rows by index\nprint(\"Dataframe after Dropping 2nd and 4th Rows:\")\nprint(df)\n\n# Drop rows where Occupation is 'Engineer'\ndf <- df %>% filter(Occupation != \"Engineer\")\n\n# Print the dataframe after dropping rows by condition\nprint(\"Dataframe after Dropping Rows with Occupation as 'Engineer':\")\nprint(df)\n``````\n\n### Output:\n\n``` \"Original Dataframe:\"\nName Age Occupation\n1 Alice 25 Engineer\n2 Bob 30 Doctor\n3 Charlie 28 Lawyer\n4 David 35 Artist\n5 Eve 29 Engineer\n\n \"Dataframe after Dropping 2nd and 4th Rows:\"\nName Age Occupation\n1 Alice 25 Engineer\n3 Charlie 28 Lawyer\n5 Eve 29 Engineer\n\n \"Dataframe after Dropping Rows with Occupation as 'Engineer':\"\nName Age Occupation\n3 Charlie 28 Lawyer\n```\n\n# 4. Step By Step Explanation\n\n- We start by creating a dataframe df with columns Name, Age, and Occupation.- To drop rows based on their index, we use negative indexing. For instance, to drop the 2nd and 4th rows, we use df <- df[-c(2,4), ].- To drop rows based on a specific condition, we use the filter function from the dplyr package. In our example, we drop rows where the Occupation column has the value 'Engineer'." ]
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https://brainmass.com/business/capital-budgeting/new-project-evaluation-initial-investment-cash-flows-npv-irr-42667
[ "Explore BrainMass\n\n# New Project evaluation: initial investment, cash flows, NPV, IRR\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nProject Evaluation:\n\nRevenues generated by a new fad product are forecast as follows:\nYear Revenues\n1 \\$40,000.00\n2 30,000.00\n3 20,000.00\n4 10,000.00\nthereafter 0\n\nExpenses are expected to be 40% of revenues, and working capital required in each year is expected to be 20% of revenues in the following year. The product requires an immediate investment of \\$45,000.00 in plant equipment.\n\na) What is the initial investment in the product? Remember working capital.\nb) If the plant and equipment are depreciated over 4 years to a salvage value of zero using straight-line depreciation, and the firm's tax rate is 40%, what are the project cash flows in each year?\nc) If the opportunity cost of capital is 12%, what is the project Net Present Value?\nd) What is the project Internal Rate of Return?\n\nThanks!\n\n#### Solution Preview\n\na) What is the initial investment in the product? Remember working capital.\nInitial investment is simply the sum of immediate investment (\\$45,000) and the working capital to start the operation, which is the 20% of first year's revenue (\\$40,000). Thus:\n\nI= 45,000 + 40,000 x 0.2= \\$53,000\n\nb) If the plant and equipment are depreciated over 4 years to a salvage value of zero using straight-line depreciation, and the firm's tax rate is 40%, what are the project cash flows in each year?\nLet's write the formula of the cash flow (CF) first:\n\nCF = Net Operating Profit - Taxes - Net Change in Working Capital\n\nNow we need to find the net operating profit (NOP) first in order to use in our formula. NOP is the difference between the revenues (R) and the expenses ...\n\n#### Solution Summary\n\nThe solution is a detailed set of facts and calculations to make the answers easy to understand.\n\n\\$2.49" ]
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https://www.unitconverters.net/energy/btu-th-to-watt-second.htm
[ "", null, "Home / Energy Conversion / Convert Btu (th) to Watt-second\n\n# Convert Btu (th) to Watt-second\n\nPlease provide values below to convert Btu (th) [Btu (th)] to watt-second [W*s], or vice versa.\n\n From: Btu (th)", null, "To: watt-second\n\n### Btu (th) to Watt-second Conversion Table\n\nBtu (th) [Btu (th)]Watt-second [W*s]\n0.01 Btu (th)10.5434999997 W*s\n0.1 Btu (th)105.4349999974 W*s\n1 Btu (th)1054.3499999744 W*s\n2 Btu (th)2108.6999999489 W*s\n3 Btu (th)3163.0499999233 W*s\n5 Btu (th)5271.7499998722 W*s\n10 Btu (th)10543.499999744 W*s\n20 Btu (th)21086.999999489 W*s\n50 Btu (th)52717.499998722 W*s\n100 Btu (th)105434.99999744 W*s\n1000 Btu (th)1054349.9999744 W*s\n\n### How to Convert Btu (th) to Watt-second\n\n1 Btu (th) = 1054.3499999744 W*s\n1 W*s = 0.0009484517 Btu (th)\n\nExample: convert 15 Btu (th) to W*s:\n15 Btu (th) = 15 × 1054.3499999744 W*s = 15815.249999617 W*s" ]
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http://www.thermopedia.com/content/137/
[ "# Mie solution for spherical particles\n\n## THE MIE SOLUTION FOR SPHERICAL PARTICLES\n\nLeading to: Limiting cases of the general Mie theory\n\nThe scheme of the problem is shown in Fig. 1.", null, "Figure 1. Scheme of the scattering problem for a spherical particle\n\nThe amount of scattering and absorption by a particle is usually expressed in terms of the scattering cross section Cs and absorption cross section Ca. The total amount of absorption and scattering, or extinction, is expressed in term of the extinction cross section Ct. The dimensionless efficiency factors are often used instead of cross sections,\n\n(1)", null, "Absorption and scattering of radiation by a two-layer spherical particle depend on the complex index of refraction of the core and mantle substances (m' = n'-iκ' and m'' = n''- iκ'', correspondingly) and also on diffraction parameters x' = 2πa'/λ, x'' = 2πa''/λ, where a' is the core radius, and a'' is the external radius of the particle.\n\nAccording to Mie theory, the efficiency factors of scattering and extinction are expressed in the form of the following series:\n\n(2)", null, "where complex coefficients ak, bk are called the Mie coefficients. The efficiency factor of absorption is determined as a difference, Qa = Qt - Qs. The terms in the Mie series correspond to the partial waves of different orders. The number of terms, which should be taken into account, increases with increasing the diffraction parameter. As a result, the calculations for particles of radius much greater than the wavelength are more complicated.\n\nThe angular characteristics of scattering are expressed by the following complex amplitude functions corresponding to perpendicular polarizations:\n\n(3a)", null, "(3b)", null, "Here, μ = cosθ, where θ is the angle of scattering measured from the direction of the incident radiation (see Fig. 1), and πk and τk are special angular functions defined later by Eq. (22). The scattering (phase) function for linear polarized incident radiation is\n\n(4)", null, "where\n\n(5)", null, "Angle φ is measured from the polarization plane of the incident radiation. For unpolarized (randomly polarized) incident radiation, the following relations hold true:\n\n(6)", null, "and the polarization degree of scattered radiation is\n\n(7)", null, "For calculating the asymmetry factor of scattering, one can use the Debye equation,\n\n(8)", null, "where the asterisk denotes a complex conjugate quantity. Remember that the asymmetry factor of scattering is defined as\n\n(9)", null, "According to Eq. (8), the value of μ does not depend on polarization of the incident radiation.\n\nThe Mie coefficients for two-layer spherical particles are determined by\n\n(10)", null, "where", null, "", null, "", null, "(11)", null, "and αk, βk, γk expressions will be given later [see Eqs. (17) and (18)].\n\nFor hollow particles, Eqs. (10) and (11) become\n\n(12)", null, "(13a)", null, "(13b)", null, "(13c)", null, "(13d)", null, "For homogeneous particles, the simplifications of the Mie coefficients are considerable,\n\n(14)", null, "For perfectly reflecting homogeneous particles (|m| → ∞),\n\n(15)", null, "and for two-layer particles with perfectly reflecting central core,\n\n(16a)", null, "(16b)", null, "(16c)", null, "(16d)", null, "Functions αk, βk, γk are expressed by the Riccati-Bessel functions ψk, ζk (Abramowitz and Stegun, 1965),\n\n(17)", null, "The Riccati-Bessel functions are related to the Bessel functions and the Hankel second-kind functions in a simple manner,\n\n(18)", null, "Two recursion relations,\n\n(19)", null, "and formulas for derivatives,\n\n(20)", null, "can be used in calculations of the Mie coefficients. As a result, we have the following expressions for logarithmic derivatives αk and βk:\n\n(21)", null, "Special angular functions πk, τk are expressed by associated Legendre polynomials,\n\n(22)", null, "and can be calculated following the recursion relations derived by Hosemann (1971),\n\n(23)", null, "For calculations employing Eqs. (19) and (23), it is necessary to know the following initial functions:\n\n(24a)", null, "(24b)", null, "(24c)", null, "We have now all the expressions for calculations of absorption and scattering of radiation by homogeneous, hollow, or two-layer spherical particles.\n\nThe calculations of the Riccati-Bessel functions based on recursion relations (19) starting from the first functions [(24a) and (24b)] up to higher-order functions (the so-called upward recursion) may lead to significant computational errors. The latter limitation is important in the case of large values of x and |m|x when the functions are of the order close to the argument are calculated with great error. This problem was overcome by Kattawar and Plass (1967), who showed that some components of the Riccati-Bessel functions are well calculated by using downward recursion. The detailed analysis of the accuracy and stability of several algorithms for Mie scattering calculations have been performed by Wiscombe (1980). A comparison of Mie scattering subroutines can be found in the paper by Felske et al. (1983). The reader can find the early FORTRAN codes for homogeneous and various two-layer particles in appendices of the books by Bohren and Huffman (1983) and Dombrovsky (1996). The algorithms for the general case of stratified spheres were presented by Toon and Ackerman (1981) and Bhandari (1985). Additional codes for multilayered spheres are listed by Flatau (2000) and Wriedt (2000). Note that the work on improving the Mie scattering algorithms continuesd. One can find some new results in recent papers by Yang (2003), Du (2004), Li et al. (2006), and Cai et al. (2008). A more detailed bibliography on this subject can be found in the recent monograph by Dombrovsky and Baillis (2010).\n\nFortunately, this advanced technique of the Mie calculations should only be used mainly in the limiting cases when the general solution is degenerated and reliable estimates can be obtained on the basis of known approximations. The geometrical optics approximation for very large particles is a good illustration of the latter statement, which is a particular case of the general observation, namely, one should find an alternative physical approach in the range when the ordinary procedure leads to more and more complicated mathematics.\n\n#### REFERENCES\n\nAbramowitz, M. and Stegun, I. A. (Eds.), Handbook of Mathematical Functions, Dover, New York, 1965.\n\nBhandari, R., Scattering coefficients for a multilayered sphere: analytic expressions and algorithms, Appl. Opt., vol. 24, no, 13, pp. 1960-1967, 1985.\n\nBohren, C. F. and Huffman, D. R., Absorption and Scattering of Light by Small Particles, Wiley, Hoboken, NJ, 1983.\n\nCai, W., Zhao, Y., and Ma, L., Direct recursion of the ratio of Bessel functions with applications to Mie scattering calculations, J. Quant. Spectrosc. Radiat. Transfer, vol. 109, no. 16, pp. 2673-2678, 2008.\n\nDombrovsky, L. A., Radiation Heat Transfer in Disperse Systems, Begell House, Redding, CT, and New York, 1996.\n\nDombrovsky, L. A. and Baillis, D., Thermal Radiation in Disperse Systems: An Engineering Approach, Begell House, Redding, CT, and New York, 2010.\n\nDu, H., Mie-scattering calculation, Appl. Opt., vol. 43, no. 9, pp. 1951-1956, 2004.\n\nFelske, J. D., Chu, Z. Z., and Ku, J. C., Mie scattering subroutines (DBMIE and MIEV0): A comparison of computational times, Appl. Opt., vol. 22, no. 15, pp, 2240-2241, 1983.\n\nFlatau, P. J., SCATTERLIB: Light Scattering Codes Library, atol.uscd.edu/~pflatau/scatlib/, 2000.\n\nHosemann, J. P., Computation of angular functions πn and τn occurring the Mie theory, Appl. Opt., vol. 10, no. 6, pp. 1452-53, 1971.\n\nKattawar, G. W. and Plass, G. N., Electromagnetic scattering from absorbing spheres, Appl. Opt., vol. 6, no. 8, pp. 1377-1382, 1967.\n\nLi, R., Han, X., Jiang, H., and Ren, K. F., Debye series for light scattering by a multilayered sphere, Appl. Opt., vol. 45, no. 6, pp. 1260-1270, 2006.\n\nToon, O. B. and Ackerman, T. P., Algorithms for the calculation of scattering by stratified spheres, Appl. Opt., vol. 20, no. 20, pp. 3657-3660, 1981.\n\nWiscombe, W. J., Improved Mie scattering algorithms, Appl. Opt., vol. 19, no. 9 pp. 1505-1509, 1980.\n\nWriedt, T., Electromagnetic scattering programs, www.t-matrix.de, 2000.\n\nYang, W., Improved recursive algorithm for light scattering by a multilayered sphere, Appl. Opt., vol. 42, no. 9, pp. 1710-1720, 2003.\n\n#### References\n\n1. Abramowitz, M. and Stegun, I. A. (Eds.), Handbook of Mathematical Functions, Dover, New York, 1965.\n2. Bhandari, R., Scattering coefficients for a multilayered sphere: analytic expressions and algorithms, Appl. Opt., vol. 24, no, 13, pp. 1960-1967, 1985.\n3. Bohren, C. F. and Huffman, D. R., Absorption and Scattering of Light by Small Particles, Wiley, Hoboken, NJ, 1983.\n4. Cai, W., Zhao, Y., and Ma, L., Direct recursion of the ratio of Bessel functions with applications to Mie scattering calculations, J. Quant. Spectrosc. Radiat. Transfer, vol. 109, no. 16, pp. 2673-2678, 2008.\n5. Dombrovsky, L. A., Radiation Heat Transfer in Disperse Systems, Begell House, Redding, CT, and New York, 1996.\n6. Dombrovsky, L. A. and Baillis, D., Thermal Radiation in Disperse Systems: An Engineering Approach, Begell House, Redding, CT, and New York, 2010.\n7. Du, H., Mie-scattering calculation, Appl. Opt., vol. 43, no. 9, pp. 1951-1956, 2004.\n8. Felske, J. D., Chu, Z. Z., and Ku, J. C., Mie scattering subroutines (DBMIE and MIEV0): A comparison of computational times, Appl. Opt., vol. 22, no. 15, pp, 2240-2241, 1983.\n9. Flatau, P. J., SCATTERLIB: Light Scattering Codes Library, atol.uscd.edu/~pflatau/scatlib/, 2000.\n10. Hosemann, J. P., Computation of angular functions Ï€n and Ï„n occurring the Mie theory, Appl. Opt., vol. 10, no. 6, pp. 1452-53, 1971.\n11. Kattawar, G. W. and Plass, G. N., Electromagnetic scattering from absorbing spheres, Appl. Opt., vol. 6, no. 8, pp. 1377-1382, 1967.\n12. Li, R., Han, X., Jiang, H., and Ren, K. F., Debye series for light scattering by a multilayered sphere, Appl. Opt., vol. 45, no. 6, pp. 1260-1270, 2006.\n13. Toon, O. B. and Ackerman, T. P., Algorithms for the calculation of scattering by stratified spheres, Appl. Opt., vol. 20, no. 20, pp. 3657-3660, 1981.\n14. Wiscombe, W. J., Improved Mie scattering algorithms, Appl. Opt., vol. 19, no. 9 pp. 1505-1509, 1980.\n15. Wriedt, T., Electromagnetic scattering programs, www.t-matrix.de, 2000.\n16. Yang, W., Improved recursive algorithm for light scattering by a multilayered sphere, Appl. Opt., vol. 42, no. 9, pp. 1710-1720, 2003." ]
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http://coin-lab.org/content/research/single_obj.html
[ "# Evolutionary Single-objective Optimization\n\nCOIN lab also addresses fundamental aspects of single-objective evolutionary algorithms and attempts to develop efficient methodologies. Some of Prof. Deb’s earlier studies are now used as default methodologies. The penalty parameter-less constraint handling approach (CMAME, 2000) is one such contribution. The simulated binary recombination (SBX) operator proposed in 1995 is another real-parameter recombination operator which is routinely used in EC studies and has also been adopted in commercial softwares. Similarly, polynomial mutation operator (suggested in Deb’s (2001) book) is another commonly-used EC operator. Currently, COIN lab is investigating a non-uniform mapping binary-coded genetic algorithm and its extension for a better application. Another study on creating an efficient initial population for limited budget application is interesting.", null, "## Parameter-less Constraint Handling\n\nConstraints are inevitable in practice. However, there has been a lukewarm interest shown by EMO researchers in making their algorithms applied to constrained problems. Research at COIN lab has always focused on unconstrained and constrained problems alike. In 2000, we developed a parameter-less constraint handling approach, which is now known as 'Deb's Parameter-less Approach'. Without using any parameter the methodology compares two solutions by emphasizing feasible over infeasible, less violated infeasible, and better objective-based feasible solutions. The idea can be implemented wither by changing the binary tournament selection or by defining a Fitness function using objective and constraint violations. The idea has also been extended for multi-objective optmization by redefining domination principle using 'constrained domination principle'.\n\n#### References\n\nDeb, K. (2000). An efficient constraint handling method for genetic algorithms. Computer Methods in Applied Mechanics and Engineering, 186, 311–338.\n\nDeb, K., Pratap. A, Agarwal, S., and Meyarivan, T. (2002). A fast and elitist multiobjective genetic algorithm: NSGA-II. IEEE Transaction on Evolutionary Computation, 6(2), 181–197.\n\n## Real-parameter Recombination Operator - Simulated Binary Crossover\n\nWhile recombining two real-parameter vectors to create one or more real-parameter vectors, EC researchers have proposed ad-hoc procedures by restricting offsprings vectors to be picked uniformly from the region enclosed (or extended) by two parent vectors. In 1995, we have shown that if real variables are coded in long binary strings, the resulting single-point crossover operators will create offsprings which are not uniformly distributed, rather has a biased distribution closer to the parent vectors. This is due to the fact that there are more crossovers happen at the least significant bits resulting in near parent points. Based on this observation, we have proposed Simulated Binary Crossover (SBX) in which a non-dimensionalized probability distribution is used to create offsprings near parents variable wise. The distribution index eta_c controls the relative spread of offspring from parent. Higher eta_c means lesser spread. The idea was later extended to have Parent-centric crossover (PCX) operated vector-wise.\n\n#### References\n\nDeb, K. and Agrawal, R. B. (1995). Simulated binary crossover for continuous search space. Complex Systems, 9, 115–148.\n\nDeb, K. and Kumar, A. (1995). Real-coded genetic algorithms with simulated binary crossover: Studies on multimodal and multiobjective problems. Complex Systems, 9(6), 431–454.\n\nDeb, K. and Beyer, H.-G. (2001). Self-adaptive genetic algorithms with simulated binary crossover. Evolutionary Computation Journal, 9(2), 197–221.\n\nDeb, K., Joshi, D., and Anand, A. (2002). Real-coded evolutionary algorithms with parentcentric recombination. Proceedings of the Congress on Evolutionary Computation (CEC2002). (Honolulu, USA). (pp. 61–66).\n\n## Real-parameter Mutation Operator - Polynomial Mutation\n\nReal-parameter mutation means a perturbation of an existing parent vector to a neighboring vector. Gaussian probabilities are often used for such purposes with a user-defined variance. In polynomial mutation, a polynomial probability distribution was proposed which is similar in principle to Gaussian distribution. The distribution index eta_m controls the spread of mutated point from parent. Higher eta_m produces smaller spread. A comparison of polynomial mutation with Gaussian mutation can be found at the second reference given below.\n\n#### References\n\nDeb, K. (2001). Multi-objective optimization using evolutionary algorithms. Chichester, London: Wiley.\n\nDeb, K. and Deb, D. (2014). Analysing mutation schemes for real-parameter genetic algorithms. Int. J. Artificial Intelligence and Soft Computing, 4(1), Inderscience Enterprises Ltd., 1–28 (DOI: 10.1504/IJAISC.2014.059280)" ]
[ null, "http://coin-lab.org/images/research/constraint_handling.png", null ]
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https://docs.trifacta.com/plugins/viewsource/viewpagesrc.action?pageId=177669551
[ "This example illustrates how you can apply the following percentile-related functions to your transformations:\n\n• MEDIAN - Calculate the median value from a column of values. See MEDIAN Function.\n• PERCENTILE - Calculate a specified percentile for a column of values. See PERCENTILE Function.\n• QUARTILE - Calculate a specified quartile for a column of values. See QUARTILE Function.\n\nThe following functions use an approximation technique for calculating median, percentile, and quartiles. In some cases, these calculations can be computed faster across large datasets.\n\nSource:\n\nThe following table lists each student's height in inches:\n\nStudentHeight\n164\n265\n363\n464\n562\n666\n766\n865\n969\n1066\n1173\n1269\n1369\n1461\n1564\n1661\n1771\n1867\n1973\n2066\n\nTransformation:\n\nUse the following transformations to calculate the median height in inches, a specified percentile and the first quartile.\n\n• The first function uses a precise algorithm which can be slow to execute across large datasets.\n• The second function uses an appropriate approximation algorithm, which is much faster to execute across large datasets.\n• These approximate functions can use an error boundary parameter, which is set to 0.4 (0.4%) across all functions.\n\nMedian: This transformation calculates the median value, which corresponds to the 50th percentile.", null, "", null, "Percentile: This transformation calculates the 68th percentile.", null, "", null, "Quartile: This transformation calculates the first quartile, which corresponds to the 25th percentile.", null, "", null, "Results:\n\nstudentIdheightInapproxPercentile25Inpercentile25InapproxPercentile68Inpercentile68InapproxMedianInmedianIn\n164646467.166.926666\n265646467.166.926666\n363646467.166.926666\n464646467.166.926666\n562646467.166.926666\n666646467.166.926666\n766646467.166.926666\n865646467.166.926666\n969646467.166.926666\n1066646467.166.926666\n1173646467.166.926666\n1269646467.166.926666\n1369646467.166.926666\n1461646467.166.926666\n1564646467.166.926666\n1661646467.166.926666\n1771646467.166.926666\n1867646467.166.926666\n1973646467.166.926666\n2066646467.166.926666" ]
[ null, "https://docs.trifacta.com/plugins/servlet/confluence/placeholder/macro", null, "https://docs.trifacta.com/plugins/servlet/confluence/placeholder/macro", null, "https://docs.trifacta.com/plugins/servlet/confluence/placeholder/macro", null, "https://docs.trifacta.com/plugins/servlet/confluence/placeholder/macro", null, "https://docs.trifacta.com/plugins/servlet/confluence/placeholder/macro", null, "https://docs.trifacta.com/plugins/servlet/confluence/placeholder/macro", null ]
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https://math.stackexchange.com/questions/570635/the-continuity-of-the-expectation-of-a-continuous-stochastic-procees
[ "# The continuity of the expectation of a continuous stochastic procees\n\nLet $X_t$ be a continuous stochastic process on a filtered space $(\\Omega, \\mathcal F, \\mathcal F_t, \\mathbb P)$.\n\nIs $\\mathbb E[X_t]$ necessarily a continuous function?\n\nMy first answer would be no. For example if $X_t$ admits densities $f(t,x)$, the first equality in:\n\n$$\\lim_{t \\rightarrow t_0} \\int_{\\mathbb R} x f(t, x) dx=\\int_{\\mathbb R} \\lim_{t \\rightarrow t_0} x f(t,x)=\\int_{\\mathbb R} x f(t_0,x)$$\n\nrequires $f$ to be continuous in $t$ uniformly in $x$ to hold.\n\nExamples where $\\mathbb E[X_t]$ is indeed continuous are abundant. Counterexamples where it is not? Thanks.\n\nOne counterexample is the following:\n\nLet $$B_s$$ be a standard Brownian motion, and let $$S = \\inf\\{s : B_s = 1\\}$$ be the first time it hits 1. Since Brownian motion is recurrent, $$S < \\infty$$ almost surely. Let $$Y_s = B_{s \\wedge S}$$; then $$Y_s$$ is a continuous martingale and $$\\lim_{s \\to \\infty} Y_s = 1$$ almost surely. Finally, let $$X_t = \\begin{cases} Y_{t/(1-t)}, & 0 \\le t < 1 \\\\ 1, & t \\ge 1.\\end{cases}$$ $$X_t$$ is continuous since $$\\lim_{t \\to 1^-} X_t = \\lim_{s \\to +\\infty} Y_s = 1$$. But for $$t < 1$$, we have $$E X_t = E Y_{t/(1-t)} = 0$$ since $$Y_s$$ is a martingale, and for $$t \\ge 1$$, $$E X_t = 1$$. So $$E X_t$$ is discontinuous.\n\n$$X_t$$ is a useful example to keep in mind; for instance, it is a local martingale with respect to its natural filtration, but not a martingale.\n\nEdit: Hans asks in comments whether we can have an example where the second moments are finite but discontinuous. The answer is yes. Let $$X_t$$ be as above and set $$Z_t = \\sqrt{1-X_t}$$. Note that $$X_t \\le 1$$ everywhere so the square root makes sense, and $$Z_t$$ is a continuous process. Then we have $$E[Z_t^2] = \\begin{cases} 1, & t < 1 \\\\ 0, & t \\ge 1 \\end{cases}$$ which is discontinuous. The second moments $$E[Z_t^2]$$ are not only finite but uniformly bounded.\n\nMoreover, if we wish to consider variance (centered second moment) instead, notice that by continuity of $$Z_t$$, as $$t \\uparrow 1$$ we have $$Z_t \\to Z_1 = 0$$ almost surely. Since we showed $$E[Z_t^2]$$ is uniformly bounded, we have that $$\\{Z_t\\}$$ is uniformly integrable (see https://math.stackexchange.com/a/184484/822), hence $$E Z_t \\to E Z_1 = 0$$. So as $$t \\uparrow 1$$, $$\\operatorname{Var}(Z_t) = E[Z_t^2] - (E Z_t)^2 \\to 1$$, whereas $$\\operatorname{Var}(Z_1) = 0$$.\n\n• Thank you very much Nate. It is staggering how stopping/time reversing can produce so many counterexamples Nov 17, 2013 at 18:20\n• What about the continuity of the second moment if the second moments are all finite?\n– Hans\nFeb 3, 2015 at 4:19\n• @Hans: Consider $\\sqrt{1-X_t}$. Feb 3, 2015 at 5:31\n• Thank you, Nate. I actually meant the variance. What is a counterexample in this case?\n– Hans\nFeb 3, 2015 at 5:58\n• @Hans: $Z_t = \\sqrt{1-X_t}$ still works. Of course we have $E Z_t^2 = 1$ for $t < 1$ and $0$ for $t \\ge 1$. But since $Z_t$ is $L^2$-bounded, it is uniformly integrable, so as $t \\uparrow 1$ we have $E Z_t \\to E Z_1 = 0$. So as $t \\uparrow 1$ we have $\\operatorname{Var}(Z_t) = E Z_t^2 - (E Z_t)^2 = 1 - (E Z_t)^2 \\to 1$, while $\\operatorname{Var}(Z_1) = 0$. Feb 3, 2015 at 7:26" ]
[ null ]
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https://cubinforcongress.com/what-is-the-distinction-between-polynomial-operate-polynomial-expression-and-polynomial-equation.html
[ "# What Is the Distinction Between Polynomial Operate, Polynomial Expression and Polynomial Equation?\n\nMore often than not, you may match a polynomial (or a polynomial expression) to the…", null, "More often than not, you may match a polynomial (or a polynomial expression) to the perform it depicts. There are conditions, although, if you’ll want to distinguish between the 2. In case your coefficients are in a subject of two parts (the place 1+1=0), as an example, the polynomials x+1 and x2+1 mirror the exact same perform as a result of whose values stay the identical as each member within the subject. In consequence, distinguishing between a polynomial and the perform it represents could be important in some disciplines of arithmetic, equivalent to up to date algebra and arithmetic foundations. More often than not, this isn’t an element to think about. 2 issues are equal, in response to an equation. The polynomial equation x3–x+2=x4+1, for instance, states that the 2 polynomials x3–3x+2 and x4+1 are equal. All sides of the equation is often zero. The equation 0=x4x2+x1 is logically equal to that closing equation. A solution of the equal polynomial equation f(x)=0 is a root of a polynomial f(x). For instance, as a result of 1 is a options of the issue x4x2+x1=0, it’s a root of one thing just like the polynomial x4x2+x1. Allow us to undergo the overview of all three of the matters to know this just a little extra clearly.\n\n## Polynomial Operate\n\nThe essential, most frequently used, and most important mathematical perform is the polynomial perform. These capabilities are used to characterize algebraic expressions that meet explicit standards. They’re additionally able to performing a variety of duties. Due to their big selection of purposes, it is very important be taught and comprehend polynomial capabilities. Polynomial is fashioned by combining the phrases poly and nomial. As a result of “poly” implies “many” in addition to “nomial” implies “time period,” we might say that polynomials comprise “algebraic expressions containing many parts.” Let’s get began with the definitions and sorts of polynomial capabilities. Polynomial capabilities are equations that embrace variables of differing levels, elements, constructive exponents, in addition to constants, amongst different issues. Listed below are a number of polynomial perform situations:\n\n• f(x) = 3x2– 5\n• g(x) = -7x3+ (1/2) x – 7\n• h(x) = 3x4+ 7x3 – 12x2\n\nPolynomial Operate Varieties: A polynomial’s id is measured because the variety of parts it comprises. Monomials, binomials, in addition to trinomials are certainly the three most prevalent polynomials we expertise.\n\n• Polynomials with just one time period are referred to as monomials.\n• Polynomials with solely two phrases are generally known as binomials.\n• Polynomials with solely three phrases are referred to as trinomials.\n\nPolynomials are certainly classed in response to their levels. Linear polynomial perform, quadratic polynomial perform, cubic polynomial perform, in addition to zero polynomial perform are the 4 hottest types of polynomials used throughout precalculus in addition to algebra.\n\n## Polynomial Expressions\n\nThese phrases “polynomial” in addition to “nomial” are fashioned up of two phrases that indicate “many phrases.” It’s a mathematical image that’s used to indicate expressions. A polynomial expression is an announcement that comprises no less than two integers and one arithmetic operation. Polynomial Expressions: What Are They? A polynomial consists of phrases, every of which has a coefficient, whereas an expression is an announcement together with no less than two numbers or one mathematical operation. Polynomial expressions are expressions that meet the polynomial requirement. Let’s take a look on the following situations to find out in the event that they’re polynomial expressions:\n\n• x2 + 3√x + 1 – No\n• x2 + √3x + 1 – Sure\n\nThis similar expression x2 + 3x + 1 isn’t a polynomial expression inside the two conditions talked about above because the variable does have a fractional exponent, particularly, 1/2, which might be a non-integer worth; nevertheless, the expression x2 + 3 x + 1 is certainly a polynomial expression because the fractional energy 1/2 has been on the fixed, though on this case is 3. Customary Type of Polynomial Expressions: Every time the phrases of a polynomial expression are organized from best to lowest diploma, the usual type is obtained.\n\n## Have you learnt the reply of the beneath query?\n\n1. The variety of polynomials having zeroes as -2 and 5 is\n2. 1\n3. 2\n4. 3\n5. greater than 3\n\n## Polynomial Equation\n\nA polynomial equation can be one with a number of phrases and normally comprises variables equivalent to coefficient in addition to exponent. Polynomials can have a variety of exponents. The largest exponent of a polynomial is termed its diploma. A polynomial’s diploma specifies the utmost root system that actually can exist in a polynomial equation. A polynomial equation is a polynomial expression that has been set to equal zero. As an example, if the best exponent being 3, the polynomial equation may have three roots. If we all know what number of roots there are, we are able to simply write the polynomial equation. Factoring polynomials by way of levels in addition to variables within the polynomial equation permits them to be resolved. Occasion of a Polynomial Equation: 4x2 + 6x + 21=0 whereby 4x2+6x+21 is a polynomial expression printed on the left facet that has been corrected to deliver the polynomial expression equal to 0, leading to a full polynomial equation. Technique for Polynomial Equations Expression: This polynomial equation formulation is normally written as anxn, the place a represents a coefficient, x represents a variable, in addition to n represents an exponent." ]
[ null, "https://cubinforcongress.com/wp-content/uploads/2022/02/What-Is-the-Difference-Between-Polynomial-Function-Polynomial-Expression-and-Polynomial-Equation.jpg", null ]
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https://blog.enterprisedna.co/how-to-use-iterating-functions-to-fix-total-issues-in-power-bi-dax-concepts/
[ "", null, "# How To Use Iterating Functions To Fix Total Issues In Power BI – DAX Concepts\n\nI’m going to show you today how to use iterating functions when you’re trying to fix Total issues in Power BI. You may watch the full video of this tutorial at the bottom of this blog.\n\nWhen I go through the questions posted on the Enterprise DNA support forum for members, I see a lot of people having issues with the Totals they get on their report regardless if they’re using advanced logic or not.\n\nHere, I’ll show you why that happens and how you can get around those issues.\n\n## Computing Min Revenue Totals\n\nLet’s use a simple example on computing Min Revenue Totals.\n\nThe data below shows that I have my Total Revenue as well as the Total Revenue Last Year.", null, "Getting last year’s revenue is simple. I just used the DATEADD function.", null, "The Total Revenue shows data for the year 2018. Then you’ll see that we subtracted one year under the DATEADD function, which gives us the 2017 data.\n\nYou’ll see that we also have another column after Revenue Last Year, which is the Min Revenue Year. Basically, this column should show the lowest numbers out of both 2018 and 2017.", null, "Alternatively, you can also click on the legend and choose a different year. So you can use the revenue for 2016 and 2017, for example.", null, "But for this example, let’s stick to the 2017 and 2018 data.\n\nIf you’ve done some work using Excel, you’ll see that the formula I’m using here is the same.", null, "Once you hit enter, you can go through the numbers and see that as you work your way through the rows, the numbers make sense. That is, until you reach the Total at the very bottom.", null, "You’ll notice that it shows the same Total as the Total Revenue column even if the data throughout the rows are different. This is definitely not the result we’re looking for.\n\nTo understand why this happened, you also have to understand the context. Here, the context only lies on the rows preceding the Total where it chooses the lower amount between the two columns. But at the end, it simply compared the two totals from 2017 and 2018, picked which one was the minimum amount, and took that as the result.", null, "It doesn’t understand that you want the total of all the Minimum Revenue listed throughout that column.\n\n## Using SUMX To Fix The Total\n\nThis is where iterating functions come in. For this example, I’m going to use SUMX.", null, "Using SUMX as the iterating function, I can start working through the table virtually.\n\nI’ll start by using the SUMMARIZE function for the Customers. Then, I’m using the MIN function to get the minimum value for the Total Revenue for 2017 and 2018.", null, "When you use iterating functions, the calculation happening at every single row happens virtually.\n\nAll of these calculations are stored into memory, after which the SUMX function adds all of that stored data to come up with the right total at the bottom of that column.\n\nIn other words, it finally adds more context to make the system understand what kind of result you’re looking for.\n\n## Conclusion\n\nWith this tutorial, hopefully, you were able to see how to fix Total issues you’ve been encountering with Power BI. The key here is understanding how iterating functions like SUMX work. Remember that these functions can be used to virtually work through each row of data, and you can get the right results.\n\nSo, if you get stuck on a Total in Power BI, you can play around with techniques like this. See if you can work in some parameters that follow the same context we applied to retrieve the value that you’re looking for.\n\nAll the best.\n\nSam\n\n***** Related Course Modules *****\nLearning Summit Series\nUnique Analytical Scenarios\nDAX Formula Deep Dives\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]
[ null, "https://blog.enterprisedna.co/wp-content/uploads/2020/04/How-To-Use-Iterating-Functions-To-Solve-Total-Issues-In-Power-BI-DAX-Concepts-1084x584.png", null, "https://lh6.googleusercontent.com/kfLJmLVYyj6ceH36V9JJ_WycWkZQNlF0dbBxhLQgkcRoaeDr_E_OwIDy6izVDZP36NMxZTWcG1hsiPk5O8PdMkmymfKYfOw0XCPHvyRxWjv32CjD6XSpc4F7WukczAbxnMB85lB-", null, "https://lh6.googleusercontent.com/lg7O5Dx4qUVO5O2qhZR4IQ5zOMPMY-Oe5Cu5eMKZrz7EDtMGlcKR4eSHbQW0MZYc-zFy2_by1HT3leN-Izmffu8lEVraa3LVVzBMNGumgDyT9LFTiFBM_kaUkKGOFSyGxN_34tuL", null, "https://lh6.googleusercontent.com/qGi4fqJq88eR4kU8ksVD87PCdwoxr-oUz8fzkNgj9qHyiGlf4PJ0XWeulZRWkgJ9OepVDpNt_aEVAEBVdvmnI3LZ9MGxrgT8MRTHze6C0Y7MmoPNX6Ccg4_wxi15_KYhxhVqwnLA", null, "https://lh3.googleusercontent.com/F7te6yApJj19pVHv_WpZxOmA009MWFfssZqWEgCtZLRqIlrS2X6PwjfvByaQ7UYneH4esZ6rofiAxNc2xFkHqf51CsdFftruIoqdZ5wYfSRTSmxmlKafaABcwBvNI2pcscfbVm68", null, "https://lh4.googleusercontent.com/_SZ92eMk0i5j4lM1lszbVJNo5gp8MKrrWXNe70TB5dj_IXhRQ386LQeUfm00o8R0RQ2j2Xz-7htfEsRfT81xo6QNbI_M5kCvAXgC8f0AKV3YqIF6Go1KZpr9PKIHmbWcQSHW55G2", null, "https://lh3.googleusercontent.com/2qVXvBaDHJ2dvsjGfv3eQSAkLEY8ItZNrm0NFVeeSj7uqkga6ZGNSpLSEBnH6uF-vK3_dqJeoQ0wyWLFzHOgF7ry_xZXqazuOemH7_4ND-0Rda_w4O1HK6FbfnoYsX1JIdaKR5Lb", null, "https://lh6.googleusercontent.com/ih1ol27srvcCcCujlutblBseOyQPZPfnFtdqj5Ux95wbx63jjKFPwS4tLB8k8fwcXBkNp0USiErXzvNAfGBPJ0U9PUExHK_ahn_r0HIE9kgbZ2bwpRTITD2lvcFBwAWoN4xH5TEt", null, "https://lh6.googleusercontent.com/e1HKDkUTzbA9E_O5vnvbAdmxLTXVnQ1KQOKzKQItqZ_1AaUPYyMlWmOsn4o1UEc9tdBJ0jNtxorq2QVnJ0rLfeNQqNsrpKaePhHBJGylxNUFhCtgyFBqSdCExN56-7WmU3edKuLd", null, "https://lh5.googleusercontent.com/4064Ei6cyT_HSd8DTTpVYPfDq4Q1SuZiUClLJem-pgz-mygqCFrc8JPqeSsqTcnnCz8iWaZQeel2NIFd0ehq2VI5UqImKhr44vuGEmOB0PvwsNfv1bDSKT7OMQ_OCBHJPzPWcHEk", null ]
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http://wiki.xentax.com/index.php/Lemmings_2
[ "# Lemmings 2\n\nChoose archive extension:\n\n## VOC\n\n### Format Specifications\n\n```// MexCom - Recreation of formats\\Lemmings 2 (VOC)\\VOC.bms\n// THIS IS AN AUTOMATED SPECIFICATION\n// ----------------------------\n// LEGEND\n// ----------------------------\n// #DECLARE (Set variable to value)\n// \\$\\$ CALCULATE (Calculate a new value)\n// ** (Section that repeats itself on condition)\n// // (Comment)\n// uint32{4} (Unsigned 32-bit value, 4 bytes)\n// uint16{2} (Unsigned 16-bit value, 2 bytes)\n// ubyte{1} (Unsigned 8-bit value, 1 byte)\n// char{n} (String value, n bytes in length\n// ----------------------------\n\n// Format Specification\n\nuint32{4}\t\tD\n#DECLARE FST = CURRENT OFFSET\nuint32{4}\t\tFS\n#DECLARE FO = 512\n\n// Resources have no name, are located at FO and have a size of FS\n#DECLARE FN = 512\n\\$\\$ CALCULATE FN / 4\n\\$\\$ CALCULATE FN - 1\n\n** Start Repeated entry (T) {FN}\n\n==>FST\nuint32{4}\t\tOffset of resource (FO)\n\\$\\$ CALCULATE FO + 512\n#DECLARE FST = CURRENT OFFSET\nuint32{4}\t\tFS\n\n** IF FS = 0\n#DECLARE FS = CURRENT OFFSET\n\\$\\$ CALCULATE FS + 1\n\\$\\$ CALCULATE FS - FO\n\n// Resources have no name, are located at FO and have a size of FS\n\n** ELSE\n\\$\\$ CALCULATE FS - FO\n\n// Resources have no name, are located at FO and have a size of FS\n** END IF\n** End Repeated entry (T)\n// ----------------------------\n// MexCom - Recreation complete\n```\n\n### MultiEx BMS\n\n```Get D Long 0 ;\nSavePos FST 0 ;\nGet FS Long 0 ;\nSet FO Long 512 ;\nLog \"\" FO FS 0 0 ;\nSet FN Long 512 ;\nMath FN /= 4 ;\nMath FN -= 1 ;\nFor T = 1 To FN ;\nGoTo FST 0 ;\nGet FO Long 0 ;\nMath FO += 512 ;\nSavePos FST 0 ;\nGet FS Long 0 ;\nIf FS = 0 ;\nGoTo EOF 0 ;\nSavePos FS 0 ;\nMath FS += 1 ;\nMath FS -= FO ;\nLog \"\" FO FS 0 0 ;\nSet T Long FN ;\nElse ;\nMath FS -= FO ;\nLog \"\" FO FS 0 0 ;\nEndIf ;\nNext T ;\n```" ]
[ null ]
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https://www.cse.chalmers.se/research/group/logic/TypesSS05/resources/coq/CoqArt/induc-fond/use_le_ind_max.html
[ "# Using maximal induction principles (le)\n\nUse the Scheme command to generate the maximal induction principle for le.\n\nThe following text defines a function with a pre-condition based on le and proves a small theorem about this function.\n\n```Definition pred_partial: forall (n : nat), n <> 0 -> nat.\nintros n; case n.\nintros h; elim h; reflexivity.\nintros p h'; exact p.\nDefined.\n\nTheorem le_2_n_not_zero: forall (n : nat), 2 <= n -> (n <> 0).\nProof.\nintros n Hle; elim Hle; intros; discriminate.\nQed.\n```\n\nProve the following theorem, using the maximal induction principle.\n\n```Theorem le_2_n_pred:\nforall (n : nat) (h : 2 <= n), (pred_partial n (le_2_n_not_zero n h) <> 0).\n```\n\n## Solution\n\nLook at this file\n\nGoing home\nPierre Castťran" ]
[ null ]
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https://www.physicsoverflow.org/40097/please-mathematical-freezing-process-first-studied-harlan
[ "#", null, "Could you help me please with a mathematical model of the soil freezing process first studied by Harlan?\n\n+ 0 like - 0 dislike\n222 views\n\nHello, dear colleague. Now I'm dealing with issues of modeling heat and mass transfer in frozen and thawed soils. I am solving this problem numerically using the finite volume method. Below I give an equation from Harlan for mass transfer,\n\n∂/∂x(K(∂ψ/∂x)) = (∂θu/∂t) + (ρiw)(∂θi/∂t),\n\nwhere K is the hydraulic conductivity of soil, [m/s]; θi is the volumetric ice fraction, i.e., the volume of the ice in per unit volume of frozen soil - dimensionless quantity; ψ is the soil suction potential, which controls the flow of the soil water [m]; T is the temperature, [K]; x is the position coordinate, [m]; t is the time, [s], θu is the volumetric unfrozen water fraction - dimensionless quantity, ρi - ice density [kg/m3], ρw - water density [kg/m3].\n\nGive me please the most thorough explanation for these four questions:\n\nWhat is the physical meaning of this equation?\nWhat is the physical meaning of its left and right side?\nIn the right side of this equation, why has the second term a factor (ρiw)?\nIn which program can I put this equation to see its physical meaning?\n\nP.S. Do you know quality and intuitive (with detailed explanations) articles, books, theses (in English language) on the subject of modeling of heat and mass transfer processes in frozen and thawed soils by the control (finite) volume method. This subject is very interesting. I am looking for the treatment one, two and three-dimensional problems, as well as software environments where you can realize the solution of these problems by “my” formulas (instead of formulas 'built into' these systems).\n\nPlease add a reference to the context from which you obtained the equation.\n\n@ArnoldNeumaier\n\nArticle:\n\nThe equation you cite is a continuity equation describing mass transport. See Harlan's paper\n\nAnalysis of coupled heat-fluid transport in partially frozen soil\n\nwhere you can find details. There is plenty of software for solving partial differential equations, see, e.g. here, but you need to find out for yourself which package is suitable for your problem.\n\n Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the \"link\" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\\varnothing$ in the following word:$\\varnothing\\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register." ]
[ null, "https://www.physicsoverflow.org/qa-plugin/po-printer-friendly/print_on.png", null ]
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https://gitlab.mpi-sws.org/gpirlea/iris/-/commit/1e249d7541fee56a8b49804b2e2fa4f0eb3a22d1
[ "### General invariant tactic.\n\nparent 040db00a\n ... ... @@ -90,6 +90,18 @@ Section proofs. + iIntros \"HP\". iApply \"Hclose\". iLeft. iNext. by iApply \"HP'\". - iDestruct (cinv_own_1_l with \"Hγ' Hγ\") as %[]. Qed. Global Instance into_inv_cinv N γ P : IntoInv (cinv N γ P) N. Global Instance elim_inv_cinv p γ E N P P' Q Q' : ElimModal True (|={E,E∖↑N}=> (▷ P ∗ cinv_own γ p) ∗ (▷ P ={E∖↑N,E}=∗ True))%I P' Q Q' → ElimInv (↑N ⊆ E) N (cinv N γ P) [cinv_own γ p] P' Q Q'. Proof. rewrite /ElimInv/ElimModal. iIntros (Helim ?) \"(#H1&(Hown&_)&H2)\". iApply Helim; auto. iFrame \"H2\". iMod (cinv_open E N γ p P with \"[#] [Hown]\") as \"(HP&Hown&Hclose)\"; auto. by iFrame. Qed. End proofs. Typeclasses Opaque cinv_own cinv.\n From iris.base_logic.lib Require Export fancy_updates. From stdpp Require Export namespaces. From stdpp Require Export namespaces. From iris.base_logic.lib Require Import wsat. From iris.algebra Require Import gmap. From iris.proofmode Require Import tactics coq_tactics intro_patterns. From iris.proofmode Require Import tactics. Set Default Proof Using \"Type\". Import uPred. ... ... @@ -94,6 +94,19 @@ Proof. iApply \"HP'\". iFrame. Qed. Global Instance into_inv_inv N P : IntoInv (inv N P) N. Global Instance elim_inv_inv E N P P' Q Q' : ElimModal True (|={E,E∖↑N}=> ▷ P ∗ (▷ P ={E∖↑N,E}=∗ True))%I P' Q Q' → ElimInv (↑N ⊆ E) N (inv N P) [] P' Q Q'. Proof. rewrite /ElimInv/ElimModal. iIntros (Helim ?) \"(#H1&_&H2)\". iApply Helim; auto; iFrame. iMod (inv_open _ N with \"[#]\") as \"(HP&Hclose)\"; auto. iFrame. by iModIntro. Qed. Lemma inv_open_timeless E N P `{!Timeless P} : ↑N ⊆ E → inv N P ={E,E∖↑N}=∗ P ∗ (P ={E∖↑N,E}=∗ True). Proof. ... ... @@ -101,29 +114,3 @@ Proof. iIntros \"!> {\\$HP} HP\". iApply \"Hclose\"; auto. Qed. End inv. Tactic Notation \"iInvCore\" constr(N) \"as\" tactic(tac) constr(Hclose) := let Htmp := iFresh in let patback := intro_pat.parse_one Hclose in let pat := constr:(IList [[IIdent Htmp; patback]]) in iMod (inv_open _ N with \"[#]\") as pat; [idtac|iAssumption || fail \"iInv: invariant\" N \"not found\"|idtac]; [solve_ndisj || match goal with |- ?P => fail \"iInv: cannot solve\" P end |tac Htmp]. Tactic Notation \"iInv\" constr(N) \"as\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1 x2) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) simple_intropattern(x3) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1 x2 x3) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) simple_intropattern(x3) simple_intropattern(x4) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1 x2 x3 x4) pat) Hclose.\n ... ... @@ -110,4 +110,17 @@ Section proofs. + iFrame. iApply \"Hclose\". iNext. iLeft. by iFrame. - iDestruct (na_own_disjoint with \"Htoki Htoki2\") as %?. set_solver. Qed. Global Instance into_inv_na p N P : IntoInv (na_inv p N P) N. Global Instance elim_inv_na p F E N P P' Q Q': ElimModal True (|={E}=> (▷ P ∗ na_own p (F∖↑N)) ∗ (▷ P ∗ na_own p (F∖↑N) ={E}=∗ na_own p F))%I P' Q Q' → ElimInv (↑N ⊆ E ∧ ↑N ⊆ F) N (na_inv p N P) [na_own p F] P' Q Q'. Proof. rewrite /ElimInv/ElimModal. iIntros (Helim (?&?)) \"(#H1&(Hown&_)&H2)\". iApply Helim; auto. iFrame \"H2\". iMod (na_inv_open p E F N P with \"[#] [Hown]\") as \"(HP&Hown&Hclose)\"; auto. by iFrame. Qed. End proofs.\n From iris.bi Require Export bi. From stdpp Require Import namespaces. Set Default Proof Using \"Type\". Import bi. ... ... @@ -467,6 +468,31 @@ Proof. by apply as_valid. Qed. Lemma as_valid_2 (φ : Prop) {PROP : bi} (P : PROP) `{!AsValid φ P} : P → φ. Proof. by apply as_valid. Qed. (* Input: `P`; Outputs: `N`, Extracts the namespace associated with an invariant assertion. Used for `iInv`. *) Class IntoInv {PROP : bi} (P: PROP) (N: namespace). Arguments IntoInv {_} _%I _. Hint Mode IntoInv + ! - : typeclass_instances. (* Input: `Pinv`; - `Pinv`, an invariant assertion - `Ps_aux` is a list of additional assertions needed for opening an invariant; - `Pout` is the assertion obtained by opening the invariant; - `Q` is a goal on which iInv may be invoked; - `Q'` is the transformed goal that must be proved after opening the invariant. There are similarities to the definition of ElimModal, however we want to be general enough to support uses in settings where there is not a clearly associated instance of ElimModal of the right form (e.g. to handle Iris 2.0 usage of iInv). *) Class ElimInv {PROP : bi} (φ: Prop) (N: namespace) (Pinv : PROP) (Ps_aux: list PROP) (Pout Q Q': PROP) := elim_inv : φ → Pinv ∗ [∗] Ps_aux ∗ (Pout -∗ Q') ⊢ Q. Arguments ElimInv {_} _ _ _ _%I _%I _%I _%I : simpl never. Arguments elim_inv {_} _ _ _%I _%I _%I _%I _%I _%I. Hint Mode ElimInv + - - ! - - - - : typeclass_instances. (* We make sure that tactics that perform actions on *specific* hypotheses or parts of the goal look through the [tc_opaque] connective, which is used to make definitions opaque for type class search. For example, when using `iDestruct`, ... ...\n ... ... @@ -209,6 +209,16 @@ Proof. repeat apply sep_mono=>//; apply affinely_persistently_if_flag_mono; by destruct q1. Qed. Lemma envs_lookup_delete_list_cons Δ Δ' Δ'' rp j js p1 p2 P Ps : envs_lookup_delete rp j Δ = Some (p1, P, Δ') → envs_lookup_delete_list rp js Δ' = Some (p2, Ps, Δ'') → envs_lookup_delete_list rp (j :: js) Δ = Some (p1 && p2, (P :: Ps), Δ''). Proof. rewrite //= => -> //= -> //=. Qed. Lemma envs_lookup_delete_list_nil Δ rp : envs_lookup_delete_list rp [] Δ = Some (true, [], Δ). Proof. done. Qed. Lemma envs_lookup_snoc Δ i p P : envs_lookup i Δ = None → envs_lookup i (envs_snoc Δ p i P) = Some (p, P). Proof. ... ... @@ -1159,6 +1169,20 @@ Proof. rewrite envs_entails_eq => ???? HΔ. rewrite envs_replace_singleton_sound //=. rewrite HΔ affinely_persistently_if_elim. by eapply elim_modal. Qed. (** * Invariants *) Lemma tac_inv_elim Δ1 Δ2 Δ3 js j p φ N P' P Ps Q Q' : envs_lookup_delete_list false js Δ1 = Some (p, P :: Ps, Δ2) → ElimInv φ N P Ps P' Q Q' → φ → envs_app false (Esnoc Enil j P') Δ2 = Some Δ3 → envs_entails Δ3 Q' → envs_entails Δ1 Q. Proof. rewrite envs_entails_eq => ???? HΔ. rewrite envs_lookup_delete_list_sound //. rewrite envs_app_singleton_sound //=. rewrite HΔ //= affinely_persistently_if_elim //=. rewrite -sep_assoc. by eapply elim_inv. Qed. End bi_tactics. Section sbi_tactics. ... ...\n From iris.proofmode Require Import coq_tactics. From iris.proofmode Require Import base intro_patterns spec_patterns sel_patterns. From iris.bi Require Export bi big_op. From stdpp Require Import namespaces. From iris.proofmode Require Export classes notation. From iris.proofmode Require Import class_instances. From stdpp Require Import hlist pretty. ... ... @@ -214,6 +215,14 @@ Tactic Notation \"iAssumption\" := |fail \"iAssumption:\" Q \"not found\"] end. Tactic Notation \"iAssumptionListCore\" := repeat match goal with | |- envs_lookup_delete_list _ ?ils ?p = Some (_, ?P :: ?Ps, _) => eapply envs_lookup_delete_list_cons; [by iAssumptionCore |] | |- envs_lookup_delete_list _ ?ils ?p = Some (_, [], _) => eapply envs_lookup_delete_list_nil end. (** * False *) Tactic Notation \"iExFalso\" := apply tac_ex_falso. ... ... @@ -1864,6 +1873,80 @@ Tactic Notation \"iMod\" open_constr(lem) \"as\" \"(\" simple_intropattern(x1) Tactic Notation \"iMod\" open_constr(lem) \"as\" \"%\" simple_intropattern(pat) := iDestructCore lem as false (fun H => iModCore H; iPure H as pat). Tactic Notation \"iAssumptionInv\" constr(N) := let rec find Γ i P := lazymatch Γ with | Esnoc ?Γ ?j ?P' => first [assert (IntoInv P' N) by apply _; unify P P'; unify i j|find Γ i P] end in match goal with | |- envs_lookup_delete _ ?i (Envs ?Γp ?Γs) = Some (_, ?P, _) => first [is_evar i; fail 1 | env_reflexivity] | |- envs_lookup_delete _ ?i (Envs ?Γp ?Γs) = Some (_, ?P, _) => is_evar i; first [find Γp i P | find Γs i P]; env_reflexivity end. Tactic Notation \"iInvCore\" constr(N) \"with\" constr(Hs) \"as\" tactic(tac) constr(Hclose) := let hd_id := fresh \"hd_id\" in evar (hd_id: ident); let hd_id := eval unfold hd_id in hd_id in let Htmp1 := iFresh in let Htmp2 := iFresh in let patback := intro_pat.parse_one Hclose in eapply tac_inv_elim with _ _ (hd_id :: Hs) Htmp1 _ _ N _ _ _ _; first eapply envs_lookup_delete_list_cons; swap 2 3; [ iAssumptionInv N || fail \"iInv: invariant\" N \"not found\" | apply _ || let I := match goal with |- ElimInv _ ?N ?I _ _ _ _ => I end in fail \"iInv: cannot eliminate invariant \" I \" with namespace \" N | iAssumptionListCore || fail \"iInv: other assumptions not found\" | try (split_and?; solve [ fast_done | solve_ndisj ]) | env_reflexivity |]; let pat := constr:(IList [[IIdent Htmp2; patback]]) in iDestruct Htmp1 as pat; tac Htmp2. Tactic Notation \"iInvCore\" constr(N) \"as\" tactic(tac) constr(Hclose) := let tl_ids := fresh \"tl_ids\" in evar (tl_ids: list ident); let tl_ids := eval unfold tl_ids in tl_ids in iInvCore N with tl_ids as (fun H => tac H) Hclose. Tactic Notation \"iInvCoreParse\" constr(N) \"with\" constr(Hs) \"as\" tactic(tac) constr(Hclose) := let Hs := words Hs in let Hs := eval vm_compute in (INamed <\\$> Hs) in iInvCore N with Hs as (fun H => tac H) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1 x2) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) simple_intropattern(x3) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1 x2 x3) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) simple_intropattern(x3) simple_intropattern(x4) \")\" constr(pat) constr(Hclose) := iInvCore N as (fun H => iDestructHyp H as (x1 x2 x3 x4) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"with\" constr(Hs) \"as\" constr(pat) constr(Hclose) := iInvCoreParse N with Hs as (fun H => iDestructHyp H as pat) Hclose. Tactic Notation \"iInv\" constr(N) \"with\" constr(Hs) \"as\" \"(\" simple_intropattern(x1) \")\" constr(pat) constr(Hclose) := iInvCoreParse N with Hs as (fun H => iDestructHyp H as (x1) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"with\" constr(Hs) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) \")\" constr(pat) constr(Hclose) := iInvCoreParse N with Hs as (fun H => iDestructHyp H as (x1 x2) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"with\" constr(Hs) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) simple_intropattern(x3) \")\" constr(pat) constr(Hclose) := iInvCoreParse N with Hs as (fun H => iDestructHyp H as (x1 x2 x3) pat) Hclose. Tactic Notation \"iInv\" constr(N) \"with\" constr(Hs) \"as\" \"(\" simple_intropattern(x1) simple_intropattern(x2) simple_intropattern(x3) simple_intropattern(x4) \")\" constr(pat) constr(Hclose) := iInvCoreParse N with Hs as (fun H => iDestructHyp H as (x1 x2 x3 x4) pat) Hclose. Hint Extern 0 (_ ⊢ _) => iStartProof. (* Make sure that by and done solve trivial things in proof mode *) ... ...\n From iris.proofmode Require Import tactics. From iris.base_logic Require Import base_logic lib.invariants. From iris.base_logic Require Import base_logic. From iris.base_logic.lib Require Import invariants cancelable_invariants na_invariants. Section base_logic_tests. Context {M : ucmraT}. ... ... @@ -48,7 +49,7 @@ Section base_logic_tests. End base_logic_tests. Section iris_tests. Context `{invG Σ}. Context `{invG Σ, cinvG Σ, na_invG Σ}. Lemma test_masks N E P Q R : ↑N ⊆ E → ... ... @@ -58,4 +59,89 @@ Section iris_tests. iApply (\"H\" with \"[% //] [\\$] [> HQ] [> //]\"). by iApply inv_alloc. Qed. Lemma test_iInv_1 N P: inv N (bi_persistently P) ={⊤}=∗ ▷ P. Proof. iIntros \"#H\". iInv N as \"#H2\" \"Hclose\". iMod (\"Hclose\" with \"H2\"). iModIntro. by iNext. Qed. Lemma test_iInv_2 γ p N P: cinv N γ (bi_persistently P) ∗ cinv_own γ p ={⊤}=∗ cinv_own γ p ∗ ▷ P. Proof. iIntros \"(#?&?)\". iInv N as \"(#HP&Hown)\" \"Hclose\". iMod (\"Hclose\" with \"HP\"). iModIntro. iFrame. by iNext. Qed. Lemma test_iInv_3 γ p1 p2 N P: cinv N γ (bi_persistently P) ∗ cinv_own γ p1 ∗ cinv_own γ p2 ={⊤}=∗ cinv_own γ p1 ∗ cinv_own γ p2 ∗ ▷ P. Proof. iIntros \"(#?&Hown1&Hown2)\". iInv N as \"(#HP&Hown2)\" \"Hclose\". iMod (\"Hclose\" with \"HP\"). iModIntro. iFrame. by iNext. Qed. Lemma test_iInv_4 t N E1 E2 P: ↑N ⊆ E2 → na_inv t N (bi_persistently P) ∗ na_own t E1 ∗ na_own t E2 ⊢ |={⊤}=> na_own t E1 ∗ na_own t E2 ∗ ▷ P. Proof. iIntros (?) \"(#?&Hown1&Hown2)\". iInv N as \"(#HP&Hown2)\" \"Hclose\". iMod (\"Hclose\" with \"[HP Hown2]\"). { iFrame. done. } iModIntro. iFrame. by iNext. Qed. (* test named selection of which na_own to use *) Lemma test_iInv_5 t N E1 E2 P: ↑N ⊆ E2 → na_inv t N (bi_persistently P) ∗ na_own t E1 ∗ na_own t E2 ={⊤}=∗ na_own t E1 ∗ na_own t E2 ∗ ▷ P. Proof. iIntros (?) \"(#?&Hown1&Hown2)\". iInv N with \"Hown2\" as \"(#HP&Hown2)\" \"Hclose\". iMod (\"Hclose\" with \"[HP Hown2]\"). { iFrame. done. } iModIntro. iFrame. by iNext. Qed. Lemma test_iInv_6 t N E1 E2 P: ↑N ⊆ E1 → na_inv t N (bi_persistently P) ∗ na_own t E1 ∗ na_own t E2 ={⊤}=∗ na_own t E1 ∗ na_own t E2 ∗ ▷ P. Proof. iIntros (?) \"(#?&Hown1&Hown2)\". iInv N with \"Hown1\" as \"(#HP&Hown1)\" \"Hclose\". iMod (\"Hclose\" with \"[HP Hown1]\"). { iFrame. done. } iModIntro. iFrame. by iNext. Qed. (* test robustness in presence of other invariants *) Lemma test_iInv_7 t N1 N2 N3 E1 E2 P: ↑N3 ⊆ E1 → inv N1 P ∗ na_inv t N3 (bi_persistently P) ∗ inv N2 P ∗ na_own t E1 ∗ na_own t E2 ={⊤}=∗ na_own t E1 ∗ na_own t E2 ∗ ▷ P. Proof. iIntros (?) \"(#?&#?&#?&Hown1&Hown2)\". iInv N3 with \"Hown1\" as \"(#HP&Hown1)\" \"Hclose\". iMod (\"Hclose\" with \"[HP Hown1]\"). { iFrame. done. } iModIntro. iFrame. by iNext. Qed. (* iInv should work even where we have \"inv N P\" in which P contains an evar *) Lemma test_iInv_8 N : ∃ P, inv N P ={⊤}=∗ P ≡ True ∧ inv N P. Proof. eexists. iIntros \"#H\". iInv N as \"HP\" \"Hclose\". iMod (\"Hclose\" with \"[\\$HP]\"). auto. Qed. End iris_tests.\nMarkdown is supported\n0% or\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]
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https://math.stackexchange.com/questions/301163/the-time-complexity-of-finding-a-neighborhood-graph-provided-an-unordered-adjace
[ "The time complexity of finding a neighborhood graph provided an unordered adjacency matrix\n\nImagine I have an unordered adjacency matrix for some graph $G$ with a set of vertices $V$ and a set of edges $E$. I would like to find a subset of edges that determines a $k$-hop neighborhood graph about a chosen vertex $v_i \\in V$ (i.e. a graph consisting of the vertices and edges \"reachable\" in $k$ edge-wise hops from $v_i$ and the set of all edges between these vertices that exist in $G$).\n\nWe allow any standard data structure for $G$. Can I achieve a linear or near-linear time complexity for returning an adjacency matrix for this neighborhood graph as a function of the number of vertices in the neighborhood graph $||V_{neighborhood}||$? What is the best known algorithm for finding these neighborhood graphs?\n\nWhat if we know something about the structure of the graph, perhaps that we're on a rectangular or hexagonal lattice?\n\n• You could use Dijkstra's algorithm to solve the shortest path problem from $v_i$ and then pick those vertices with path length at most $k$. Complexity will be $O(|E|+|V|\\log |V|)$ – polkjh Feb 12 '13 at 12:56\n• @polkjh That should work, so a \"yes\" for the \"almost linear time\". – R.H. Feb 12 '13 at 13:10\n• Is $v_i$ fixed? Or you want these $k$-hop neighbourhoods for all vertices? – Paresh Feb 12 '13 at 13:11\n• If it is fixed, you can do it in linear time. – Paresh Feb 12 '13 at 13:11\n• @Paresh Thanks - Do you know if the linear time algorithm has a name? And if we want them for all vertices, can we do better than to repeatedly apply the linear time algorithm for a single vertex? – R.H. Feb 12 '13 at 13:13\n\nYou can do a simple Breadth First Search from the start node. It starts with the first node, and adds all its neighbours to a queue. Then, it de-queues each node, finds its unvisited neighbors to the queue and marks the current node visited. The key observation here is that the all the vertices that are at some distance $d$ from the start vertex, are in one contiguous block in the FIFO queue. If you can keep track of how deep you are in the queue, you can stop after $k$-levels. More specifically, the moment you de-queue a vertex that you know is at a distance of $k$ from the start node, you don't have to enqueue any more, just go on till the queue is empty, and you are done. Using BFS in this way will give you vertices in increasing order of hops/distance from the start.\n\nYou can also adapt the Depth First Search algorithm for this, by limiting the depth to $k$. This is easier to code, but won't give vertices in order of distances from the start. So, for some of the discovered vertices, you may later find a shorter path. But no vertex will have to be discarded after being discovered in either methods.\n\nBoth these algorithms are linear in the size of the graph - $O(|V|+ |E|)$ even if they traverse the whole graph. But since you are stopping much sooner, they would be very efficient. More specifically, they will visit only those edges and vertices that are actually in the $k$-hop neighborhood, and nothing more -- $O(|V_s|+|E_s|)$, where $V_s$ and $E_s$ are the number of vertices and edges in the $k$-neighborhood subgraph. Another consideration can be that if $k$ is small but vertex degrees are very large, you might prefer DFS.\n\nk-hops for all vertices:\nIf the graph is sparse, that is, $|E| = O(|V|)$, then running a BFS/DFS from each node should be the best option asymptotically. Rectangular or hexagonal lattice graphs are sparse graphs. So, complexity would be $O(|V|^2)$. I do not think that there should be anything asymptotically faster than this for sparse graphs, though there may be other solutions that are more efficient (but still quadratic).\n\nIf the graph is not sparse, you can compute the $k$-th power of the adjacency matrix in $O(n^{\\omega}\\log k)$ time, where $\\omega$ is the constant of matrix multiplication. The current best known algorithm has $\\omega = 2.3727$. But in practice, the simple cubic matrix multiplication ($\\omega = 3$) is the most efficient. You could also use the all-pair shortest path algorithm by Floyd-Warshall and find those vertices that have distance atmost $k$. This is $O(|V|^3)$\n\n• Wouldn't we expect that the time for find each subgraph with a BFS or DFS with bounded depth $k$, would be $O(||E_s||+||V_s||)$ where $E_s$ and $E_s$ are the ultimate number of edges and vertices in the subgraph (and not the full graph, $G$)? – R.H. Feb 12 '13 at 15:48\n• You are right. This is exactly what I meant when I said \"More specifically, they will visit only those edges and vertices that are actually in the k-hop neighborhood, and nothing more.\" I'll add your description to this line as it is more formal. – Paresh Feb 12 '13 at 15:53\n• Thus, our final time complexity for computing the $k$-hop subgraphs for all vertices would be $\\approx O(||V_G||)$ for fixed $k$, right? – R.H. Feb 12 '13 at 15:54\n• I think this is a bit more involved. In the graphs you mention - rectangular or hexagonal lattices - the time complexity would indeed be $O(|V_G|)$. However, for general graphs, it could be quadratic - it will depend on the degree distribution of the graphs and maybe other things. The same edges/vertices may be visited multiple times when running BFS/DFS from different vertices. – Paresh Feb 12 '13 at 19:18" ]
[ null ]
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https://www.cs.carleton.edu/cs_comps/0405/shape/marching_cubes.html
[ "# An Implementation of the Marching Cubes Algorithm\n\nBy Ben Anderson\n\n#### Abstract:\n\nThe Marching cubes algorithm can be described as follows:\nGiven an object, a test to determine whether an arbitrary point is within the object, and bounds within which the object exists:\nDivide the space within the bounds into an arbitrary number of cubes.  Test the corners of every cube for whether they are inside the object.  For every cube where some corners are inside and some corners are outside the object, the surface must pass through that cube, intersecting the edges of the cube in between corners of opposite classification.  Draw a surface within each cube connecting these intersections.  You have your object.\n\n#### Full Text:\n\nOnce we have masked images of every slice of the object we are to reconstruct (ie, a black and white image where black is inside and white is outside the object), we can go about reconstructing the original surface.  One way of doing this is to use the Marching Cubes algorithm.\n\n### In 2D:\n\nTo understand how the Marching Cubes algorithm works, let's take a 2D case and what might be called the \"Marching Squares\" algorithm.  Here's our object:", null, "Now, let's divide it into squares:", null, "We can tell by looking at it which vertices are in, and which are outside of the object, so let's label them, red for inside, blue for out:", null, "Now we know that somewhere on each edge between an inside corner and an outside one the original surface must intersect our squares (purple dots):", null, "Within each square, let's connect the purple dots.  Now we have an approximation of the original surface (the purple lines):", null, "## In 3D:\n\nImplementing the algorithm in 3D works much the same as it did in 2D.  For slice data like the Visible Human Male dataset, you stack the slices in 3D, knowing each slice is 1mm or 3 pixels appart.  In order to be able to test the vertices of each cube, you must choose your cube size to align with the slices, either using 1mm cubes (or rectangles 1mm high and 1/3rd of a mm thick since pixels are only 1/3rd of a mm wide), or some multiple of 1mm cubes (ie 2mm or 3 mm) so that each horizontal side of your cubes falls on the plane of a slice.  You then can test each vertex by going to the masked slices corresponding to each cube's top and bottom z values.  You now have a bunch of cubes with labeled corners.  For each cube, you know the surface intersects the cube along the edges in between corners of opposing classifications.  Each cube should look something like this:", null, "In 2D to approximate the surface we simply had to draw lines between each purple dot.  In 3D this would give us a strange line drawing like this:", null, "Which doesn't look very much like a piece of a circle.  Instead, what we want to do is triangulate the cube where filled triangles will represent the surface passing through the cube.  Something like\nthis:", null, "However, you'll note that this isn't as easy as drawing a line connecting all the purple dots in a square.  For one thing, a triangle connects 3 dots, and for another, it's hard to tell which three dots to connect in which triangle.  If you just randomly connected dots in a triangle, you might get something like:", null, "Which really doesn't look any good.  The other issue is that since you're going to be doing this to a lot of cubes, you probably want whatever solution you find to be fast.  And what is really fast?  A lookup table.  Consider a cube.  Each corner can be either inside or outside the object.  There are eight corners.  That means you're working with 28, or 256 different possibilities for what a cube might look like, which is definitely feasible.  Triangulating each of those possibilities might be a pain, but luckily, given identical triangulations for cubes whose vertex classification is opposite of eachother, rotations and mirroring, there are in fact only 14 unique triangulations:", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Each triangulation contains zero to four triangles and can be stored as a list of triangles where each triangle is a list of 3 numbers which are indexes to the edges on which each triangles vertices lie. for an example, here's some ASCII art taken from the code labeling each edge and each corner with an index:\n\n//                 v7_______e6_____________v6\n//                  /|                    /|\n//                 / |                   / |\n//              e7/  |                e5/  |\n//               /___|______e4_________/   |\n//            v4|    |                 |v5 |e10\n//              |    |                 |   |\n//              |    |e11              |e9 |\n//            e8|    |                 |   |\n//              |    |_________________|___|\n//              |   / v3      e2       |   /v2\n//              |  /                   |  /\n//              | /e3                  | /e1\n//              |/_____________________|/\n//              v0         e0          v1\n\nAs you can see, the triangle in the second example might be defined as [e3,e2,e11].  A cube can be looked up in the table by taking the classification of each vertex and converting it to 0's and 1's and then forming a binary number.  Again, using the same cube as an example, v3 is inside the shape and every other vertex is out.  This results in a binary number of 00010000 or 23 or 8, so the entry for 8 in the lookup table might be:\n\n8: [e1,e2,e11]\n\nIn the actual code, it doesn't look quite like that, but you get the idea.\n\nAnd so, in order to compute the triangulation for the entire image, you do this for every cube, offset the triangulations to their appropriate locations in the space and voila!", null, "As you can tell, this is a blocky rendering of a foot.  You were probably hoping for somthing better.  Well, luckily, this is only rendered using 1/3rd of the slices, (fairly big cubes).  To overcome this, the first thing you might think of is to make the cubes smaller resulting in a more accurate rendering of a foot:", null, "Well, that's better, but as you can tell, it's still fairly blocky.  Well, we can do better.  To understand how, let's go back to 2D.\n\nHere's our approximation in 2D from earlier:", null, "Now, without even using smaller cubes, we can come up with a much better approximation of this surface, and one that's much smoother (the surface right now is made up of facets meeting at 45° angles).  How can we do that?  Like this:", null, "The only difference here is that the purple dots have been moved.  Instead of placing the purple dots midway between the red and the blue ones, they have been moved to meet up with the actual intersection of the surface with the edges in question.  This results in a more accurate representation of the surface, as well as a smoother one using the same number of cubes as before.  Let's see what that looks like in 3D:", null, "Now, that is smoother, and you can almost make out the toenail ont he big toe, but if you're saying to yourself, \"That's still pretty blocky,\" you'd be right.\n\nThe main problem is that while in 2D we can compute exact intersections for every edge of our shape, given only slice data, in 3D, we cannot do this.  Take this picture for example:", null, "This is a side view of a row of cubes.  The top line is a slice, as is the bottom.  The problem is that while we can compute exact intersections on a given slice, in between, we have no idea where the\nobject lies.  While the actual surface is represented by the dotted red line, in between layers, we have no information as to where that line intersects the cubes.  The only choice is to guess and place it midway between slices given that in any single cube, the only information you have between layers is that on top of the cube every point is inside and on the bottom, every point is out.  This results in a stairstepping effect particularly apparent when the surface becomes more horizontal (closer to parallel with the slices) and there is a greater horizontal distance (a larger number of cubes) between intersections with the cutting planes.   What we get is an effect like this:", null, "Luckily, there's still more we can do.\n\nOne step would be to take a trick from CS 317 (Graphics) and use Gouraud Shading to render the object.  For those of you who don't know what that is, you'll have to take Jack Goldfeather's course next year.  A basic explanation is that for each vertex, you can compute an average surface normal at that point by averaging the normals of all triangles sharing that vertex.  At each vertex then, you can use that normal to compute the lighting at that point.  Given the lighting at each vertex of a triangle, you can interpolate for every point between to determine how bright the surface is at that point.  The overall effect is to blur the appearance of edges without changing the actual geometry of the object.  Doing this we get:", null, "Which is smoother still, but still not quite what we'd like.  To do better we're going to have to work with a little post-processing.\n\n## Coloring the Visible Human\n\nSo far all the images we've seen have been red.  As you know, feet are not red.  There are two basic options for coloring reconstructions.  One is to simply use a more appropriate color:", null, "As you can see, this is significantly more foot like.  The second option is to use the actual color of the images just inside the point of intersection:", null, "This also looks footlike.  The bronze color from the first image was actually taken from one of the rgb values in the back of the foot.  However, as you can see from the blue tint on the front of the foot, taking the actual rgb values has the unfortunate side effect of grabbing values that contain some of the blue from the ice.  This is partly because of the fact that the threshold for being outside the object is 60% blue (rather than 50%) and partly because the rgb values for vertices in between slices are arbitrarily taken from one of the slices and sometimes that slice at that xy value is blue.\n\n## Analysis:\n\nThe marching cubes algorithm is very well suited to surface reconstruction.  Given a surface for which you can test arbitrary points for whether they fall inside or outside the object, it's only weakness is occasional extraneous triangles.  It is fast (linear increases in time as area increases), accurate and works with arbitrarily shaped objects.  With slice data it's only additional weakness is a stairstepping effect when the surface approaches parallel with the slices.  Both of these weaknesses can be overcome with post processing.\n\nFor much cooler pictures, check out our results.\n\nReferences:\n1. Lorensen, W. E. and Cline, H. E., \"Marching Cubes: A High Resolution 3D Surface Construction Algorithm,\" Computer Graphics, vol. 21, no. 3, pp. 163-169, July 1987.\n2. Lorensen, W. E., \"Marching Through the Visible Man,\" IEEE Visualization, Proceedings of the 6th conference on Visualization '95, pp. 368-373. 1994.\n3. Lorensen, W. E., Schroeder, W. J. and Zarge, Jonathan A., \"Decimation of Triangle Meshes,\" Internationa Conference on Computer Graphics and Interactive Techniques, Proceedings of the 19th annual conference on Computer Graphics and Interactive Techniques, pp. 65-70. 1992.\n4. Taubin, G., \"Curve and Surface Smoothing without Shrinkage,\" IBM Research Report RC-19536, September 1994." ]
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https://au.mathworks.com/matlabcentral/answers/389054-i-want-to-convert-from-cartesian-to-spherical-coordinates-in-matlab-i-have-tried-to-use-cart2sph-in
[ "I want to convert from Cartesian to spherical coordinates in MATLAB. I have tried to use cart2sph in MATLAB but it is showing some error\n\nAhmad Bilal (view profile)\n\non 18 Mar 2018\nLatest activity Answered by Ahmet Cecen\n\nAhmet Cecen (view profile)\n\non 18 Mar 2018\nAccepted Answer by Ahmet Cecen\n\nAhmet Cecen (view profile)\n\nMy MATLAB Code is as follows :\nfunction [phi,theta] = calc_phitheta_geom(xyz_source,xyz_mic1,xyz_mic2)\nmid_p =(xyz_mic1 + xyz_mic2)/2 %midpoint calculation of xyz_mic1 and xyz_mic2\nV = mid_p - xyz_source %Vector from midpoint till xyz_source\n[phi,theta] = cart2sph(V) %converting Vector V from cartesian to spherical\nend\nNow actually Matlab command requires parameter for cart2sph as follows:\n[phi,theta] = cart2sph(x,y,z) % here x y z are cartesian coordinates.\nMy question is that when I am using [phi,theta] = cart2sph(V) it is showing not enough input arguments.\nHow can I resolve this issue? Can anybody help me in this regard?\nThank you.\n\nTags\n\nAnswer by Ahmet Cecen\n\non 18 Mar 2018" ]
[ null ]
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https://quant.stackexchange.com/questions/tagged/modelling?tab=Newest
[ "# Questions tagged [modelling]\n\nThe tag has no usage guidance.\n\n27 questions\nFilter by\nSorted by\nTagged with\n50 views\n\n### Predicting Stock Returns vs Stock Price and then computing returns\n\nI am working on building a model to predict the beta-adjusted returns of a stock using a set of features. At the moment, when I try to predict the stock price directly, instead of the returns, the ...\n73 views\n\n### What is the state of the art govie bond term structure recently\n\nSpecifically, the US govt bond market is segmented and the shape is difficult to model in a structural model, because recently there is a maturity gap from 12 to 20 year, and the front and back ends ...\n34 views\n\n### GARCH(1,1) forecast plot in R with training data\n\nI've fit a GARCH(1,1) model in R and would like to create a plot similar to the one in this question: Is this the correct way to forecast stock price volatility using GARCH Could someone direct me to ...\n31 views\n\n### Calibrating g2++ in negative interest rate environment\n\nI am working on a g2++ model in a dualcurve setup for both Euribor and EONIA. I have the model built, but have some issues in calibrating it - I get a perfect fit with a Nelder-Mead algorithm, but it ...\n167 views\n\n### Risk neutral modelling of a stock\n\nSuppose a stock $S$ follows $$dS(t) = \\alpha(t)S(t)dt + \\sigma(t)S(t)dW(t),$$ where $W(t)$ is a Brownian motion under $P$. Also suppose there is a short rate process $r(t)$. My question would be is ...\n93 views\n\n### Which methods are there to determine the price of futures contracts?\n\nWhich method apart from the cost of carry model exists, and which works best in real life? How does the market expectations impact on the futures price?\n127 views\n\n### Modeling independent variables that have an asymmetric impact on the dependent variable\n\nI'm trying to regress a dependent variable on an independent variable that has an asymmetric impact. E.g., the dependent variable is much more responsive to an increase in the independent variable ...\n66 views\n\n### Forecasting default rates using a macroeconomic model\n\nI am trying to forecast corporate default rates using macroeconomic data. I have a few explanatory variables (all the variables are explained in figure 2), which range from 2000 to 2017. On this ...\n46 views\n\n### goodness of fit metric\n\nI am trying to approximate the returns of asset A by means of a linear combination of other assets A'=aB0+bB1+c*B2.... I have this quite figured out but I'm not sure what a good metric for goodness ...\n159 views\n\n### Are processes with independent increments (which are not Lévy) used in finance?\n\nFrom Jacod and Shiryaev's Limit Theorems for Stochastic Processes, we get the following definitions. Definitions: A process with independent increments (abbreviated PII) $X = (X_t)_{t \\geq 0}$ on a ...\n558 views\n\n### Margin modelling to backtest futures investment strategy\n\nLet say that I have access to continuous daily time series for 20+ years of data for E-mini S&P 500 Index Futures. I have a long/short strategy to backtest that places orders either on open or ...\n39 views\n\n### Passage from dates ranges to real numbers in modelling : which market practice?\n\nLet's say I model a 6M forward Libor rate as a process $(L^1_t)_t$ that's a diffusion, with in view a Monte-Carlo (MC) pricing of some product. At some point I will have real life dates $T_i$'s that I ...\n11k views\n\n### What is a regime switch?\n\nI've come across the term regime switch in volatilities when reading about the modelling of interest rates but could not find a definition for a regime switch and what a regime is. Can somebody give ...\n541 views\n\n### LIBOR 3M and 1M from Vasicek model\n\nI would like to discuss my approach toward modelling of interest rates with respect to its downsides and advantages. My problem is to forecast daily LIBOR 3M and LIBOR 1M over a particular time ...\n233 views\n\n### Extensions of CIR\n\nI could need some advice on extensions of the CIR model. The standard CIR reads $dr(t)=\\kappa(\\theta-r(t))dt + \\sigma \\sqrt{r(t)} dW(t)$. A possible extension, if we would like the short-rate to ...\n782 views\n\n### CIR model and calibration\n\nI am new to quantitative finance. We know that in the CIR model the short rate can't go negative. My question then concerns calibration of CIR to a ZCB yield curve. Is it (and why?) possible to ...\n1k views\n\n### How to backtest Value at Risk Models using Conditional and Unconditional tests?\n\nI am trying to carry out backtesting on a number of Value at Risk figures i obtained using var/covar, historical, and monte carlo simulation. The two methods im using are the Kupiec test (...\n165 views\n\n### Local volatility parametrization using the spot\n\nIs it possible to estimate the local volatility using the spot price S at time t instead of the strike price K and the expiry date T ? Any help would be appreciated.\n5k views\n\n### Option Pricing Model Calibration In Practice\n\nI'm curious how an option pricing model like the Heston model is calibrated in practice. Here's how I imagine it happens: Let's say I have access to the most recent option prices on a given stock ...\n619 views\n\n### Why do people always seek finite-variance models for option pricing\n\nFor the purpose of getting fatter tails than the Guassian, I have seen people for example use $\\alpha$-stable processes to model the stock. But in that case they end up using 'tempered' versions of ...\n96 views\n\n### EGARCH formulation\n\nI am a bit confused about the formulation of the EGARCH(1,1) model. First, we have the error term: $\\epsilon_t=\\sigma_t*\\zeta_t$, where $\\zeta_t$ is white noise. Now the EGARCH(1,1) should be: log(...\n689 views\n\n### GARCH modelling and forecasting\n\nI have a few questions regarding GARCH modelling and forecasting and it would be great if someone could help me. I am modelling the log return of oil spot prices using various GARCH models: GARCH, ...\n2k views\n\n### Is it too important that my residuals be normal? I am Using an ARMA/GARCH model\n\nI am trying to fit an ARMA/GARCH model to a time series. I found that the best candidate is an ARMA(1,0) + GARCH(1,1) with gaussian white noise It has coefficients with p-values near cero and the ...\n9k views\n\n### Null and Alternative hypothesis for multiple linear regression\n\nI have 1 dependent variable and 3 independent variables. I run multiple regression, and find that the p value for one of the independent variables is higher than 0.05 (95% is my confidence level). I ...\n4k views\n\n### Models for simulating FX movements\n\nMy goal is to develop a model to simulate long term FX movements. (I am not sure if long term makes any difference, but if it does I am more interested in long term fx movements) These Monte Carlo ..." ]
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https://uk.mathworks.com/matlabcentral/cody/problems/1988-remove-the-middle-element-from-a-vector/solutions/533579
[ "Cody\n\n# Problem 1988. Remove the middle element from a vector\n\nSolution 533579\n\nSubmitted on 23 Nov 2014 by Abdullah Caliskan\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% x = [1,2,3]; y_correct = 2; assert(isequal(remove_middle(x),y_correct))\n\nans = 2\n\n2   Pass\n%% x = [1,2,3,4]; y_correct = 2; assert(isequal(remove_middle(x),y_correct))\n\nans = 2\n\n3   Pass\n%% x = []; y_correct = []; assert(isequal(remove_middle(x),y_correct))\n\nans = []" ]
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https://wiki.portal.chalmers.se/agda/Main/TypesSummerSchool2007
[ "# TypesSummerSchool2007\n\nThe Types Summer School 2007 is held in Bertinoro, Italy, August 19 - 31.\n\nThe examples from the lectures can be found here: examples/SummerSchool07/Lecture\n\n## Exercises\n\n### 1. Natural numbers\n\n1.1. Define a datatype of `Nat` natural numbers with constructors `zero` and `suc`.\n\n1.2. Define an addition function `_+_` on natural numbers, by recursion over the first argument.\n\n1.3. Define multiplication.\n\n1.4. Given the identity type\n\n``` data _==_ {A : Set}(x : A) : A -> Set where\nrefl : x == x\n```\n\nprove that _+_ is associative, i.e. give a function\n\n``` assoc : (x y z : Nat) -> (x + (y + z)) == ((x + y) + z)\n```\n\n### 2. Vectors\n\nVectors are lists of a fixed length:\n\n``` data Vec (A : Set) : Nat -> Set where\nε : Vec A zero\n_►_ : {n : Nat} -> A -> Vec A n -> Vec A (suc n)\n```\n\n2.1. Define a function `vec` which computes a vector where all the elements are the same.\n\n``` vec : {A : Set}{n : Nat} -> A -> Vec A n\n```\n\nHint: you need to do recursion over n\n\n2.2. Define a function `_<*>_` to point-wise apply a vector of functions to a vector of arguments.\n\n``` _<*>_ : {A B : Set}{n : Nat} -> Vec (A -> B) n -> Vec A n -> Vec B n\n```\n\n2.3. Use `vec` and `_<*>_` to define a function\n\n``` map : {A B : Set}{n : Nat} -> (A -> B) -> Vec A n -> Vec B n\n```\n\n2.4. Use `vec` and `_<*>_` to define a function\n\n``` zip : {A B C : Set}{n : Nat} -> (A -> B -> C) -> Vec A n -> Vec B n -> Vec C n\n```\n\n### 3. Finite sets\n\nThe family of finite sets indexed by size is defined by:\n\n``` data Fin : Nat -> Set where\nfzero : {n : Nat} -> Fin (suc n)\nfsuc : {n : Nat} -> Fin n -> Fin (suc n)\n```\n\nThe elements of `Fin n` are the natural numbers less than `n` (but with different names for the constructors).\n\n3.1. Show that there are no elements in `Fin zero`, i.e. give a function\n\n``` empty : Fin zero -> False\n```\n\nwhere\n\n``` data False : Set where\n```\n\nis the empty datatype. Hint: use an absurd pattern `()`.\n\n3.2. We can use elements of finite sets as positions in vectors. Define a look up function\n\n``` _!_ : {A : Set}{n : Nat} -> Vec A n -> Fin n -> A\n```\n\n3.3. In fact, `_!_` is an isomorphism between `Vec A n` and `Fin n -> A`. Define its inverse\n\n``` tabulate : {A : Set}{n : Nat} -> (Fin n -> A) -> Vec A n\n```\n\n### 4. Predicates over lists\n\nThe list datatype is given by\n\n``` data List (A : Set) : Set where\n[] : List A\n_::_ : A -> List A -> List A\n```\n\n4.1. Define `map` and `_++_` for lists.\n\n``` map : {A B : Set} -> (A -> B) -> List A -> List B\n_++_ : {A : Set} -> List A -> List A -> List A\n```\n\n4.2. Define a datatype family `All P` which represents proofs that `P` holds for all elements of a list\n\n``` data All {A : Set}(P : A -> Set) : List A -> Set where\n```\n\n4.3. Define a datatype family `Some P` which represents proofs that `P` holds for some element of a list\n\n``` data Some {A : Set}(P : A -> Set) : List A -> Set where\n```\n\n4.4. State and prove some interesting lemmas about the interaction between `All` and `Some` and the list functions `map` and `_++_`.\n\n4.5. Define list membership in terms of `Some`.\n\n``` _∈_ : {A : Set} -> A -> List A -> Set\n```\n\nHint: you need to use the identity type from exercise 1.4.\n\n4.6. We can define natural numbers as lists:\n\n``` record True : Set where\ntt : True\ntt = record {} -- the unique element of True\n\nNat = List True\nzero : Nat\nzero = []\nsuc : Nat -> Nat\nsuc n = tt :: n\n```\n\nNow define `Vec` and `Fin` using `All` and `Some`.\n\n4.7. Dependent pairs can be defined by\n\n``` data _×_ (A : Set)(B : A -> Set) : Set where\n_,_ : (x : A) -> B x -> A × B\n```\n\nWe can recover the simply typed pairs:\n\n``` _∧_ : (A B : Set) -> Set\nA ∧ B = A × \\_ -> B\n```\n\nNow if you know `All P xs` and `Some Q xs` then you can get an `A` satisfying both `P` and `Q`. Prove this by defining the following look up function:\n\n``` _!_ : {A : Set}{P : A -> Set}{Q : A -> Set}{xs : List A} ->\nAll P xs -> Some Q xs -> A × (\\z -> P z ∧ Q z)\n```\n\n4.8. (Harder) For a decidable predicate it is the case that it either holds for each element in a list, or there is an element for which it doesn't hold. Given\n\n``` ¬_ : Set -> Set\n¬ P = P -> False\n\ndata _∨_ (A B : Set) : Set where\ninl : A -> A ∨ B\ninr : B -> A ∨ B\n\ndata Bool : Set where\nfalse : Bool\ntrue : Bool\n\ndata IsTrue : Bool -> Set where\nisTrue : IsTrue true\n\nHolds : {A : Set} -> (A -> Bool) -> A -> Set\nHolds p x = IsTrue (p x)\n```\n\ndefine the function `all`:\n\n``` all : {A : Set}(p : A -> Bool)(xs : List A) ->\nAll (Holds p) xs ∨ Some (\\x -> ¬ Holds p x) xs\n```\n\nHint: you might find the function `decide` helpful\n\n``` decide : {A : Set}(p : A -> Bool)(x : A) -> Holds p x ∨ ¬ Holds p x\n```" ]
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https://tipcalc.net/how-much-is-a-25-percent-tip-on-5
[ "# Tip Calculator\n\nHow much is a 25 percent tip on \\$5?\n\nTIP:\n\\$ 0\nTOTAL:\n\\$ 0\n\nTIP PER PERSON:\n\\$ 0\nTOTAL PER PERSON:\n\\$ 0\n\n## How much is a 25 percent tip on \\$5? How to calculate this tip?\n\nAre you looking for the answer to this question: How much is a 25 percent tip on \\$5? Here is the answer.\n\nLet's see how to calculate a 25 percent tip when the amount to be paid is 5. Tip is a percentage, and a percentage is a number or ratio expressed as a fraction of 100. This means that a 25 percent tip can also be expressed as follows: 25/100 = 0.25 . To get the tip value for a \\$5 bill, the amount of the bill must be multiplied by 0.25, so the calculation is as follows:\n\n1. TIP = 5*25% = 5*0.25 = 1.25\n\n2. TOTAL = 5+1.25 = 6.25\n\n3. Rounded to the nearest whole number: 6\n\nIf you want to know how to calculate the tip in your head in a few seconds, visit the Tip Calculator Home.\n\n## So what is a 25 percent tip on a \\$5? The answer is 1.25!\n\nOf course, it may happen that you do not pay the bill or the tip alone. A typical case is when you order a pizza with your friends and you want to split the amount of the order. For example, if you are three, you simply need to split the tip and the amount into three. In this example it means:\n\n1. Total amount rounded to the nearest whole number: 6\n\n2. Split into 3: 2\n\nSo in the example above, if the pizza order is to be split into 3, you’ll have to pay \\$2 . Of course, you can do these settings in Tip Calculator. You can split the tip and the total amount payable among the members of the company as you wish. So the TipCalc.net page basically serves as a Pizza Tip Calculator, as well.\n\n## Tip Calculator Examples (BILL: \\$5)\n\nHow much is a 5% tip on \\$5?\nHow much is a 10% tip on \\$5?\nHow much is a 15% tip on \\$5?\nHow much is a 20% tip on \\$5?\nHow much is a 25% tip on \\$5?\nHow much is a 30% tip on \\$5?" ]
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https://www.mis.mpg.de/calendar/conferences/2020/cominds/abstracts.html
[ "# GAMM AG Workshop Computational and Mathematical Methods in Data Science\n\n## Abstracts for the talks\n\nMarco Cuturi\nGoogle Brain & Institut Polytechnique de Paris\nSupervised Quantile Normalization for Matrix Factorization using Optimal Transport\nWe present in this recent work (https://arxiv.org/pdf/2002.03229.pdf) a recent application of our framework to carry out \"soft\" sorting and ranking using regularized OT. We expand this framework to include \"soft\" quantile normalization operators that can be differentiated efficiently, and apply it to the problem of dimensionality reduction: we ask how features can be normalized with a target distribution of quantiles to recover \"easy to factorize\" matrices. We provide algorithms to do this, as well as empirical evidence of recovery.\n\nAxel Klawonn\nUniversity of Cologne\nMachine learning in adaptive domain decomposition methods\nThe convergence rate of domain decomposition methods is generally determined by the eigenvalues of the preconditioned system. For second-order elliptic partial differential equations, coefficient discontinuities with a large contrast can lead to a deterioration of the convergence rate. A remedy can be obtained by enhancing the coarse space with elements, which are often called constraints, that are computed by solving small eigenvalue problems on portions of the interface of the domain decomposition, i.e., edges in two dimensions or faces and edges in three dimensions. In the present work, without restriction of generality, the focus is on two dimensions. In general, it is difficult to predict where these constraints have to be computed, i.e., on which edges. Here, a machine learning based strategy using neural networks is suggested to predict the geometric location of these edges in a preprocessing step. This reduces the number of eigenvalue problems that have to be solved during the iteration. Numerical experiments for model problems and realistic microsections using regular decompositions as well as those from graph partitioners are provided, showing very promising results.\n\nThis is joint work with Alexander Heinlein, Martin Lanser, and Janine Weber.\n\nKathlén Kohn\nKTH Stockholm\nInvariant theory and scaling algorithms for maximum likelihood estimation\nThe task of fitting data to a model is fundamental in statistics. For this, a widespread approach is finding a maximum likelihood estimate (MLE), where one maximizes the likelihood of observing the data as we range over the model. For two common statistical settings (log-linear models and Gaussian transformation families), we show that this approach is equivalent to a capacity problem in invariant theory: finding a point of minimal norm in an orbit under a corresponding group action. The existence of the MLE can then be characterized by stability notions under the action. This dictionary between invariant theory and statistics has already led to the solution of long-standing questions concerning the MLE of matrix normal models. Moreover, algorithms from statistics can be used in invariant theory, and vice versa.\n\nThis talk is based on joint work with Carlos Améndola, Philipp Reichenbach and Anna Seigal.\n\nDaniel Kressner\nEPF Lausanne\nRandomized trace estimates for indefinite matrices with an application to determinants\nThe need for estimating the determinant of a large-scale symmetric positive definite (SPD) matrix A features prominently in several machine learning tasks, most notably in maximum likelihood estimation for Gaussian process regression. Because of log(det(A)) = trace( log(A) ), estimating the log-determinant is equivalent to estimating the trace of the matrix logarithm. This simple relation allows for the application of Hutchinson's randomized trace estimator, which approximates the trace of a symmetric matrix B by computing the average of x'*B*x for many samples of a random vector x. When estimating determinants, two major obstacles arise: (1) The matrix B=log(A) can be indefinite even when A is SPD. This complicates the analysis; nearly all existing tail bounds only apply to SPD matrices. (2) The exact evaluation of x'*log(A)*x requires the computation of the matrix logarithm, which is expensive. Recent work by Ubaru, Chen, and Saad considers the use of Rademacher random vectors and addresses (1) by shifting the matrix and (2) by using Lanczos quadrature. In this talk, we will explain why the tail bounds obtained from shifting are overly pessimistic. A new bound is presented, which reduces the required number of samples by up to a factor n, where n is the size of the matrix. We will also extend these results to Gaussian random vectors, which requires some careful consideration for the Lanczos quadrature because of the unboundedness of such vectors.\n\nThis talk is based on joint work with Alice Cortinovis.\n\nGitta Kutyniok\nTU Berlin\nThe Mathematics of Deep Learning: Can we Open the Black Box of Deep Neural Networks?\nDespite the outstanding success of deep neural networks in real-world applications, most of the related research is empirically driven and a comprehensive mathematical foundation is still missing. Regarding deep learning as a statistical learning problem, the necessary theory can be divided into the research directions of expressivity, learning, and generalization. Recently, the new direction of interpretability became important as well.\n\nIn this talk, we will provide an introduction into those four research foci. We will then delve a bit deeper into the area of expressivity, namely the approximation capacity of neural network architectures as one of the most developed mathematical theories so far, and discuss some recent work. Finally, we will provide a survey about the novel and highly relevant area of interpretability, which aims at developing an understanding how a given network reaches decisions, and discuss the very first mathematically founded approach to this problem.\n\nGuido Montúfar\nMax Planck Institute for Mathematics in the Sciences\nComputing the Unique Information\nGiven a pair of predictor variables and a response variable, how much information do the predictors have about the response, and how is this information distributed between unique, redundant, and synergistic components? Recent work has proposed to quantify the unique component of the decomposition as the minimum value of the conditional mutual information over a constrained set of information channels. We present an efficient iterative divergence minimization algorithm to solve this optimization problem with convergence guarantees and evaluate its performance against other techniques.\n\nThis is joint work with Pradeep Kr. Banerjee and Johannes Rauh.\n\nChristoph Schnörr\nHeidelberg University\nAssignment Flows for Data Labeling and Pattern Formation on Graphs\nAssignment flows are smooth dynamical systems for labeling data on graphs. After introducing the basic approach, I explain a recent extension to unsupervised scenarios, self-assignment flows, that combine computation of prototypes for data coding and data labeling using these prototypes. A brief outlook on parameter learning and the problem to understand their function conclude the talk.\n\nIngo Steinwart\nUniversity of Stuttgart\nDensity-Based Cluster Analysis\nA central, initial task in data science is cluster analysis, where the goal is to find clusters in unlabeled data. One widely accepted definition of clusters has its roots in a paper by Carmichael et al., where clusters are described to be densely populated areas in the input space that are separated by less populated areas. The mathematical translation of this idea usually assumes that the data is generated by some unknown probability measure that has a density with respect to the Lebesgue measure. Given a threshold level, the clusters are then defined to be the connected components of the density level set. However, choosing this threshold and possible width parameters of a density estimator, which is left to the user, is a notoriously difficult problem, typically only addressed by heuristics.\n\nIn this talk, I show how a simple algorithm based on a density estimator can find the smallest level for which there are more than one connected component in the level set. Unlike other cluster algorithms this approach is fully adaptive in the sense that it does not require the user to guess crucial hyper-parameters. Finally, I present some numerical illustrations.\n\nJoel Tropp\nCalifornia Institute of Technology\nSketchySVD\nThis talk asserts that randomized linear algebra is a natural tool for on-the-fly compression of data matrices that arise from large-scale scientific simulations and data collection. The technical contribution consists in a new algorithm for constructing an accurate low-rank approximation of a huge matrix from streaming data. Among other applications, we show how the SVD of a large-scale sea surface temperature dataset exposes features of the global climate.\n\n## Date and Location\n\nSeptember 10 - 11, 2020\nMax Planck Institute for Mathematics in the Sciences\n\n## Scientific Organizers\n\nMax von Renesse\nLeipzig University\n\nAndré Uschmajew\nMPI for Mathematics in the Sciences" ]
[ null ]
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http://wims.espe-nice-toulon.fr/wims/wims.cgi?lang=en&+module=H6%2Fgeometry%2Fellchoice.en
[ "!! used as default html header if there is none in the selected theme. Choice of ellipses\n\n# Choice of ellipses --- Introduction ---\n\nAn ellipse (oriented horizontally/vertically) is described by an equation of the form\n\na(x-x0)2 + b(y-y0)2 = c,\n\nor of the form\n\na x2+ b y2+ c x+ d y+ e = 0,\n\nwhere (x0,y0) is the center of the ellipse and the ratio between the two axes of the ellipse is described by the ratio between a and b (how?).\n\nThis online exercise can either plot an ellipse and ask you to recognize its equation among a list to choose from, or give an equation and ask you to recognize the plotted ellipse among a list to choose from.\n\nConfiguration of the exercise :\n• .\n• .\n• And a session will be composed of ( with a score assigned at the end of each session).\nOther exercises on:\nIn order to access WIMS services, you need a browser supporting forms. In order to test the browser you are using, please type the word wims here: and press Enter''." ]
[ null ]
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https://swmath.org/software/14162
[ "# AS 307\n\nAlgorithm AS 307: Bivariate location depth. he half-space depth of a point θ relative to a bivariate data set {x 1 ,⋯,x n } is given by the smallest number of data points contained in a closed half-plane of which the boundary line passes through θ. A straightforward algorithm for the half-space depth needs O(n 2 ) steps. The simplicial depth of θ relative to {x 1 ,⋯,x n } is given by the number of data triangles Δ(x i ,x j ,x k ) that contain θ; this appears to require O(n 3 ) steps. The algorithm proposed here computes both depths in O(nlogn) time, by combining geometric properties with certain sorting and updating mechanisms. Both types of depth can be used for data description, bivariate confidence regions, p-values, quality indices and control charts. Moreover, the algorithm can be extended to the computation of depth contours and bivariate sign test statistics.\n\n##", null, "Keywords for this software\n\nAnything in here will be replaced on browsers that support the canvas element\n\n## References in zbMATH (referenced in 61 articles , 1 standard article )\n\nShowing results 1 to 20 of 61.\nSorted by year (citations)" ]
[ null, "https://swmath.org/media/img/minus.png", null ]
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https://en.m.wikipedia.org/wiki/Statistical_mechanics
[ "# Statistical mechanics\n\nIn physics, statistical mechanics is a mathematical framework that applies statistical methods and probability theory to large assemblies of microscopic entities. It does not assume or postulate any natural laws, but explains the macroscopic behavior of nature from the behavior of such ensembles.\n\nSometimes called statistical physics or statistical thermodynamics, its applications include many problems in the fields of physics, biology, chemistry, and neuroscience. Its main purpose is to clarify the properties of matter in aggregate, in terms of physical laws governing atomic motion.\n\nStatistical mechanics arose out of the development of classical thermodynamics, a field for which it was successful in explaining macroscopic physical properties—such as temperature, pressure, and heat capacity—in terms of microscopic parameters that fluctuate about average values and are characterized by probability distributions.\n\nThe founding of the field of statistical mechanics is generally credited to three physicists:\n\nWhile classical thermodynamics is primarily concerned with thermodynamic equilibrium, statistical mechanics has been applied in non-equilibrium statistical mechanics to the issues of microscopically modeling the speed of irreversible processes that are driven by imbalances. Examples of such processes include chemical reactions and flows of particles and heat. The fluctuation–dissipation theorem is the basic knowledge obtained from applying non-equilibrium statistical mechanics to study the simplest non-equilibrium situation of a steady state current flow in a system of many particles.\n\n## History\n\nIn 1738, Swiss physicist and mathematician Daniel Bernoulli published Hydrodynamica which laid the basis for the kinetic theory of gases. In this work, Bernoulli posited the argument, still used to this day, that gases consist of great numbers of molecules moving in all directions, that their impact on a surface causes the gas pressure that we feel, and that what we experience as heat is simply the kinetic energy of their motion.\n\nIn 1859, after reading a paper on the diffusion of molecules by Rudolf Clausius, Scottish physicist James Clerk Maxwell formulated the Maxwell distribution of molecular velocities, which gave the proportion of molecules having a certain velocity in a specific range. This was the first-ever statistical law in physics. Maxwell also gave the first mechanical argument that molecular collisions entail an equalization of temperatures and hence a tendency towards equilibrium. Five years later, in 1864, Ludwig Boltzmann, a young student in Vienna, came across Maxwell's paper and spent much of his life developing the subject further.\n\nStatistical mechanics was initiated in the 1870s with the work of Boltzmann, much of which was collectively published in his 1896 Lectures on Gas Theory. Boltzmann's original papers on the statistical interpretation of thermodynamics, the H-theorem, transport theory, thermal equilibrium, the equation of state of gases, and similar subjects, occupy about 2,000 pages in the proceedings of the Vienna Academy and other societies. Boltzmann introduced the concept of an equilibrium statistical ensemble and also investigated for the first time non-equilibrium statistical mechanics, with his H-theorem.\n\nThe term \"statistical mechanics\" was coined by the American mathematical physicist J. Willard Gibbs in 1884.[note 1] \"Probabilistic mechanics\" might today seem a more appropriate term, but \"statistical mechanics\" is firmly entrenched. Shortly before his death, Gibbs published in 1902 Elementary Principles in Statistical Mechanics, a book which formalized statistical mechanics as a fully general approach to address all mechanical systems—macroscopic or microscopic, gaseous or non-gaseous. Gibbs' methods were initially derived in the framework classical mechanics, however they were of such generality that they were found to adapt easily to the later quantum mechanics, and still form the foundation of statistical mechanics to this day.\n\n## Principles: mechanics and ensembles\n\nIn physics, two types of mechanics are usually examined: classical mechanics and quantum mechanics. For both types of mechanics, the standard mathematical approach is to consider two concepts:\n\nUsing these two concepts, the state at any other time, past or future, can in principle be calculated. There is however a disconnect between these laws and everyday life experiences, as we do not find it necessary (nor even theoretically possible) to know exactly at a microscopic level the simultaneous positions and velocities of each molecule while carrying out processes at the human scale (for example, when performing a chemical reaction). Statistical mechanics fills this disconnection between the laws of mechanics and the practical experience of incomplete knowledge, by adding some uncertainty about which state the system is in.\n\nWhereas ordinary mechanics only considers the behaviour of a single state, statistical mechanics introduces the statistical ensemble, which is a large collection of virtual, independent copies of the system in various states. The statistical ensemble is a probability distribution over all possible states of the system. In classical statistical mechanics, the ensemble is a probability distribution over phase points (as opposed to a single phase point in ordinary mechanics), usually represented as a distribution in a phase space with canonical coordinate axes. In quantum statistical mechanics, the ensemble is a probability distribution over pure states,[note 2] and can be compactly summarized as a density matrix.\n\nAs is usual for probabilities, the ensemble can be interpreted in different ways:\n\n• an ensemble can be taken to represent the various possible states that a single system could be in (epistemic probability, a form of knowledge), or\n• the members of the ensemble can be understood as the states of the systems in experiments repeated on independent systems which have been prepared in a similar but imperfectly controlled manner (empirical probability), in the limit of an infinite number of trials.\n\nThese two meanings are equivalent for many purposes, and will be used interchangeably in this article.\n\nHowever the probability is interpreted, each state in the ensemble evolves over time according to the equation of motion. Thus, the ensemble itself (the probability distribution over states) also evolves, as the virtual systems in the ensemble continually leave one state and enter another. The ensemble evolution is given by the Liouville equation (classical mechanics) or the von Neumann equation (quantum mechanics). These equations are simply derived by the application of the mechanical equation of motion separately to each virtual system contained in the ensemble, with the probability of the virtual system being conserved over time as it evolves from state to state.\n\nOne special class of ensemble is those ensembles that do not evolve over time. These ensembles are known as equilibrium ensembles and their condition is known as statistical equilibrium. Statistical equilibrium occurs if, for each state in the ensemble, the ensemble also contains all of its future and past states with probabilities equal to the probability of being in that state.[note 3] The study of equilibrium ensembles of isolated systems is the focus of statistical thermodynamics. Non-equilibrium statistical mechanics addresses the more general case of ensembles that change over time, and/or ensembles of non-isolated systems.\n\n## Statistical thermodynamics\n\nThe primary goal of statistical thermodynamics (also known as equilibrium statistical mechanics) is to derive the classical thermodynamics of materials in terms of the properties of their constituent particles and the interactions between them. In other words, statistical thermodynamics provides a connection between the macroscopic properties of materials in thermodynamic equilibrium, and the microscopic behaviours and motions occurring inside the material.\n\nWhereas statistical mechanics proper involves dynamics, here the attention is focussed on statistical equilibrium (steady state). Statistical equilibrium does not mean that the particles have stopped moving (mechanical equilibrium), rather, only that the ensemble is not evolving.\n\n### Fundamental postulate\n\nA sufficient (but not necessary) condition for statistical equilibrium with an isolated system is that the probability distribution is a function only of conserved properties (total energy, total particle numbers, etc.). There are many different equilibrium ensembles that can be considered, and only some of them correspond to thermodynamics. Additional postulates are necessary to motivate why the ensemble for a given system should have one form or another.\n\nA common approach found in many textbooks is to take the equal a priori probability postulate. This postulate states that\n\nFor an isolated system with an exactly known energy and exactly known composition, the system can be found with equal probability in any microstate consistent with that knowledge.\n\nThe equal a priori probability postulate therefore provides a motivation for the microcanonical ensemble described below. There are various arguments in favour of the equal a priori probability postulate:\n\n• Ergodic hypothesis: An ergodic system is one that evolves over time to explore \"all accessible\" states: all those with the same energy and composition. In an ergodic system, the microcanonical ensemble is the only possible equilibrium ensemble with fixed energy. This approach has limited applicability, since most systems are not ergodic.\n• Principle of indifference: In the absence of any further information, we can only assign equal probabilities to each compatible situation.\n• Maximum information entropy: A more elaborate version of the principle of indifference states that the correct ensemble is the ensemble that is compatible with the known information and that has the largest Gibbs entropy (information entropy).\n\nOther fundamental postulates for statistical mechanics have also been proposed. For example, recent studies shows that the theory of statistical mechanics can be built without the equal a priori probability postulate. One such formalism is based on the fundamental thermodynamic relation together with the following set of postulates:\n\n1. The probability density function is proportional to some function of the ensemble parameters and random variables.\n2. Thermodynamic state functions are described by ensemble averages of random variables.\n3. The entropy as defined by Gibbs entropy formula matches with the entropy as defined in classical thermodynamics.\n\nwhere the third postulate can be replaced by the following:\n\n1. At infinite temperature, all the microstates have the same probability.\n\n### Three thermodynamic ensembles\n\nThere are three equilibrium ensembles with a simple form that can be defined for any isolated system bounded inside a finite volume. These are the most often discussed ensembles in statistical thermodynamics. In the macroscopic limit (defined below) they all correspond to classical thermodynamics.\n\nMicrocanonical ensemble\ndescribes a system with a precisely given energy and fixed composition (precise number of particles). The microcanonical ensemble contains with equal probability each possible state that is consistent with that energy and composition.\nCanonical ensemble\ndescribes a system of fixed composition that is in thermal equilibrium[note 4] with a heat bath of a precise temperature. The canonical ensemble contains states of varying energy but identical composition; the different states in the ensemble are accorded different probabilities depending on their total energy.\nGrand canonical ensemble\ndescribes a system with non-fixed composition (uncertain particle numbers) that is in thermal and chemical equilibrium with a thermodynamic reservoir. The reservoir has a precise temperature, and precise chemical potentials for various types of particle. The grand canonical ensemble contains states of varying energy and varying numbers of particles; the different states in the ensemble are accorded different probabilities depending on their total energy and total particle numbers.\n\nFor systems containing many particles (the thermodynamic limit), all three of the ensembles listed above tend to give identical behaviour. It is then simply a matter of mathematical convenience which ensemble is used. The Gibbs theorem about equivalence of ensembles was developed into the theory of concentration of measure phenomenon, which has applications in many areas of science, from functional analysis to methods of artificial intelligence and big data technology.\n\nImportant cases where the thermodynamic ensembles do not give identical results include:\n\n• Microscopic systems.\n• Large systems at a phase transition.\n• Large systems with long-range interactions.\n\nIn these cases the correct thermodynamic ensemble must be chosen as there are observable differences between these ensembles not just in the size of fluctuations, but also in average quantities such as the distribution of particles. The correct ensemble is that which corresponds to the way the system has been prepared and characterized—in other words, the ensemble that reflects the knowledge about that system.\n\nThermodynamic ensembles\nMicrocanonical Canonical Grand canonical\nFixed variables $E,N,V$  $T,N,V$  $T,\\mu ,V$\nMicroscopic features Number of microstates Canonical partition function Grand partition function\n$W$  $Z=\\sum _{k}e^{-E_{k}/k_{B}T}$  ${\\mathcal {Z}}=\\sum _{k}e^{-(E_{k}-\\mu N_{k})/k_{B}T}$\nMacroscopic function Boltzmann entropy Helmholtz free energy Grand potential\n$S=k_{B}\\log W$  $F=-k_{B}T\\log Z$  $\\Omega =-k_{B}T\\log {\\mathcal {Z}}$\n\n### Calculation methods\n\nOnce the characteristic state function for an ensemble has been calculated for a given system, that system is 'solved' (macroscopic observables can be extracted from the characteristic state function). Calculating the characteristic state function of a thermodynamic ensemble is not necessarily a simple task, however, since it involves considering every possible state of the system. While some hypothetical systems have been exactly solved, the most general (and realistic) case is too complex for an exact solution. Various approaches exist to approximate the true ensemble and allow calculation of average quantities.\n\n#### Exact\n\nThere are some cases which allow exact solutions.\n\n#### Monte Carlo\n\nAlthough some problems in statistical physics can be solved analytically using approximations and expansions, most current research utilizes the large processing power of modern computers to simulate or approximate solutions. A common approach to statistical problems is to use a Monte Carlo simulation to yield insight into the properties of a complex system. Monte Carlo methods are important in computational physics, physical chemistry, and related fields, and have diverse applications including medical physics, where they are used to model radiation transport for radiation dosimetry calculations.\n\nThe Monte Carlo method examines just a few of the possible states of the system, with the states chosen randomly (with a fair weight). As long as these states form a representative sample of the whole set of states of the system, the approximate characteristic function is obtained. As more and more random samples are included, the errors are reduced to an arbitrarily low level.\n\n#### Other\n\n• For rarefied non-ideal gases, approaches such as the cluster expansion use perturbation theory to include the effect of weak interactions, leading to a virial expansion.\n• For dense fluids, another approximate approach is based on reduced distribution functions, in particular the radial distribution function.\n• Molecular dynamics computer simulations can be used to calculate microcanonical ensemble averages, in ergodic systems. With the inclusion of a connection to a stochastic heat bath, they can also model canonical and grand canonical conditions.\n• Mixed methods involving non-equilibrium statistical mechanical results (see below) may be useful.\n\n## Non-equilibrium statistical mechanics\n\nMany physical phenomena involve quasi-thermodynamic processes out of equilibrium, for example:\n\nAll of these processes occur over time with characteristic rates. These rates are important in engineering. The field of non-equilibrium statistical mechanics is concerned with understanding these non-equilibrium processes at the microscopic level. (Statistical thermodynamics can only be used to calculate the final result, after the external imbalances have been removed and the ensemble has settled back down to equilibrium.)\n\nIn principle, non-equilibrium statistical mechanics could be mathematically exact: ensembles for an isolated system evolve over time according to deterministic equations such as Liouville's equation or its quantum equivalent, the von Neumann equation. These equations are the result of applying the mechanical equations of motion independently to each state in the ensemble. These ensemble evolution equations inherit much of the complexity of the underlying mechanical motion, and so exact solutions are very difficult to obtain. Moreover, the ensemble evolution equations are fully reversible and do not destroy information (the ensemble's Gibbs entropy is preserved). In order to make headway in modelling irreversible processes, it is necessary to consider additional factors besides probability and reversible mechanics.\n\nNon-equilibrium mechanics is therefore an active area of theoretical research as the range of validity of these additional assumptions continues to be explored. A few approaches are described in the following subsections.\n\n### Stochastic methods\n\nOne approach to non-equilibrium statistical mechanics is to incorporate stochastic (random) behaviour into the system. Stochastic behaviour destroys information contained in the ensemble. While this is technically inaccurate (aside from hypothetical situations involving black holes, a system cannot in itself cause loss of information), the randomness is added to reflect that information of interest becomes converted over time into subtle correlations within the system, or to correlations between the system and environment. These correlations appear as chaotic or pseudorandom influences on the variables of interest. By replacing these correlations with randomness proper, the calculations can be made much easier.\n\n• Boltzmann transport equation: An early form of stochastic mechanics appeared even before the term \"statistical mechanics\" had been coined, in studies of kinetic theory. James Clerk Maxwell had demonstrated that molecular collisions would lead to apparently chaotic motion inside a gas. Ludwig Boltzmann subsequently showed that, by taking this molecular chaos for granted as a complete randomization, the motions of particles in a gas would follow a simple Boltzmann transport equation that would rapidly restore a gas to an equilibrium state (see H-theorem).\n\nThe Boltzmann transport equation and related approaches are important tools in non-equilibrium statistical mechanics due to their extreme simplicity. These approximations work well in systems where the \"interesting\" information is immediately (after just one collision) scrambled up into subtle correlations, which essentially restricts them to rarefied gases. The Boltzmann transport equation has been found to be very useful in simulations of electron transport in lightly doped semiconductors (in transistors), where the electrons are indeed analogous to a rarefied gas.\n\nA quantum technique related in theme is the random phase approximation.\n• BBGKY hierarchy: In liquids and dense gases, it is not valid to immediately discard the correlations between particles after one collision. The BBGKY hierarchy (Bogoliubov–Born–Green–Kirkwood–Yvon hierarchy) gives a method for deriving Boltzmann-type equations but also extending them beyond the dilute gas case, to include correlations after a few collisions.\n• Keldysh formalism (a.k.a. NEGF—non-equilibrium Green functions): A quantum approach to including stochastic dynamics is found in the Keldysh formalism. This approach is often used in electronic quantum transport calculations.\n• Stochastic Liouville equation.\n\n### Near-equilibrium methods\n\nAnother important class of non-equilibrium statistical mechanical models deals with systems that are only very slightly perturbed from equilibrium. With very small perturbations, the response can be analysed in linear response theory. A remarkable result, as formalized by the fluctuation–dissipation theorem, is that the response of a system when near equilibrium is precisely related to the fluctuations that occur when the system is in total equilibrium. Essentially, a system that is slightly away from equilibrium—whether put there by external forces or by fluctuations—relaxes towards equilibrium in the same way, since the system cannot tell the difference or \"know\" how it came to be away from equilibrium.: 664\n\nThis provides an indirect avenue for obtaining numbers such as ohmic conductivity and thermal conductivity by extracting results from equilibrium statistical mechanics. Since equilibrium statistical mechanics is mathematically well defined and (in some cases) more amenable for calculations, the fluctuation–dissipation connection can be a convenient shortcut for calculations in near-equilibrium statistical mechanics.\n\nA few of the theoretical tools used to make this connection include:\n\n### Hybrid methods\n\nAn advanced approach uses a combination of stochastic methods and linear response theory. As an example, one approach to compute quantum coherence effects (weak localization, conductance fluctuations) in the conductance of an electronic system is the use of the Green–Kubo relations, with the inclusion of stochastic dephasing by interactions between various electrons by use of the Keldysh method.\n\n## Applications\n\nThe ensemble formalism can be used to analyze general mechanical systems with uncertainty in knowledge about the state of a system. Ensembles are also used in:\n\nStatistical physics explains and quantitatively describes superconductivity, superfluidity, turbulence, collective phenomena in solids and plasma, and the structural features of liquid. It underlies the modern astrophysics. In solid state physics, statistical physics aids the study of liquid crystals, phase transitions, and critical phenomena. Many experimental studies of matter are entirely based on the statistical description of a system. These include the scattering of cold neutrons, X-ray, visible light, and more. Statistical physics also plays a role in materials science, nuclear physics, astrophysics, chemistry, biology and medicine (e.g. study of the spread of infectious diseases).\n\nAnalytical and computational techniques derived from statistical physics of disordered systems, can be extended to large-scale problems, including machine learning, e.g., to analyze the weight space of deep neural networks. Statistical physics is thus finding applications in the area of medical diagnostics.\n\n### Quantum statistical mechanics\n\nQuantum statistical mechanics is statistical mechanics applied to quantum mechanical systems. In quantum mechanics, a statistical ensemble (probability distribution over possible quantum states) is described by a density operator S, which is a non-negative, self-adjoint, trace-class operator of trace 1 on the Hilbert space H describing the quantum system. This can be shown under various mathematical formalisms for quantum mechanics. One such formalism is provided by quantum logic." ]
[ null ]
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https://www.mdpi.com/2072-666X/10/9/597/htm
[ "Next Article in Journal\nA Gas Mixture Prediction Model Based on the Dynamic Response of a Metal-Oxide Sensor\n\nMicromachines 2019, 10(9), 597; https://doi.org/10.3390/mi10090597\n\nArticle\nTrajectory Definition with High Relative Accuracy (HRA) by Parametric Representation of Curves in Nano-Positioning Systems\nby", null, "Lucía Díaz Pérez 1,*", null, ",", null, "Beatriz Rubio Serrano 2,", null, "Esmeralda Mainar Maza 2 and", null, "Marta Torralba Gracia 3", null, "1\nAragon Institute of Engineering Research (I3A), University of Zaragoza, C/María de Luna 3, 50018 Zaragoza, Spain\n2\nDepartment of Applied Mathematics, University Research Institute of Mathematics and its Applications(IUMA), University of Zaragoza, C/María de Luna, 3, 50018 Zaragoza, Spain\n3\nCentro Universitario de la Defensa, Ctra. Huesca s/n, 50090 Zaragoza, Spain\n*\nAuthor to whom correspondence should be addressed.\nReceived: 8 July 2019 / Accepted: 7 September 2019 / Published: 10 September 2019\n\nAbstract\n\n:\nNanotechnology applications demand high accuracy positioning systems. Therefore, in order to achieve sub-micrometer accuracy, positioning uncertainty contributions must be minimized by implementing precision positioning control strategies. The positioning control system accuracy must be analyzed and optimized, especially when the system is required to follow a predefined trajectory. In this line of research, this work studies the contribution of the trajectory definition errors to the final positioning uncertainty of a large-range 2D nanopositioning stage. The curve trajectory is defined by curve fitting using two methods: traditional CAD/CAM systems and novel algorithms for accurate curve fitting. This novel method has an interest in computer-aided geometric design and approximation theory, and allows high relative accuracy (HRA) in the computation of the representations of parametric curves while minimizing the numerical errors. It is verified that the HRA method offers better positioning accuracy than commonly used CAD/CAM methods when defining a trajectory by curve fitting: When fitting a curve by interpolation with the HRA method, fewer data points are required to achieve the precision requirements. Similarly, when fitting a curve by a least-squares approximation, for the same set of given data points, the HRA method is capable of obtaining an accurate approximation curve with fewer control points.\nKeywords:\nnano-positioning; CAD/CAM systems; high relative accuracy (HRA)\n\n1. Introduction\n\nCurrent trends in precision engineering demand high accuracy positioning systems that can be used for measuring and manufacturing applications . For these, the meteorological challenge is an obstacle to the achievement of high accuracy, repeatability, and stability at sub-micrometer and nanometre scales over long travel ranges . In these systems, the positioning error is minimized by implementing precision engineering design principles. In addition, the control system performance must be optimized to reduce the positioning uncertainty. In this line of research, at the University of Zaragoza, a novel nanopositioning platform (NanoPla) has been designed, built, and implemented . The NanoPla requirements imply sub-micrometer accuracy in a large working range of 50 mm × 50 mm.\nIn some nanomanufacturing applications, nanopositioning stages are required to follow a predefined trajectory with a sub-micrometer precision . The definition of a trajectory in traditional manufacturing systems, such as machine tools, proceeds as follows: First, the geometry is defined in a CAD program; then, this geometry is exported to a CAM program, where the machine and the toolpath generation conditions are set. However, simpler geometries can be directly designed in the CAM program. In addition, some programs include CAM and CAD functions within the same software. Finally, by means of a post-processor, a specific numerical control (NC) program is created . Nevertheless, creating a complex curve in a CAD/CAM system is not always straightforward [6,7,8]. In CAD/CAM systems, complex curve trajectories are usually defined by curve fitting, having as input a set of points belonging to the nominal curve trajectory. The resultant fitting curve produces an equation that can compute points anywhere along the trajectory. This equation is determined by a set of control points together with the fitting method basis. However, trajectory definition by curve fitting in CAD/CAM systems results in fitting errors that, even though in traditional manufacturing systems are negligible, in nanopositioning stages can contribute greatly to the final positioning uncertainty of the stage.\nIn CAD/CAM systems, curve fitting is commonly performed by interpolation or by least-squares approximation. There are many types of interpolation methods, such as linear, circular, and polynomial interpolation . Linear interpolation results in polylines, inducing large geometrical deviations that can only be reduced at the expense of the amount of data. On the contrary, spline interpolation can transform a large number of linear blocks into just a few spline blocks. In spline interpolation, the resultant curve depends on the control points obtained. When the trajectory is obtained by spline interpolation of a great number of interpolation points, a great number of control points is needed and the computational complexity increases. On the other hand, the curve obtained by a spline least-squares approximation does not necessarily contain all the given data points, but only approximately fits them. The number of control points that define the approximation curve is related to how close the resultant curve fits the given data points. Nevertheless, control points may be redundant or inadequate. Redundant control points unnecessarily increase the computational complexity without decreasing the fitting errors, whereas inadequate control points increase the fitting errors and, thus, the curve fails to satisfy the precision requirements . Thus, the challenge of curve fitting, either by interpolation or approximation, lies in obtaining the most accurate trajectory, without compromising the computational efficiency.\nOther works have focused on optimizing trajectory generation for a smooth interpolating motion, in order to respect the machine kinematic feed-rate and acceleration limits, as well as to avoid fluctuations due to discontinuities in the first derivatives along the tool path. Spline-type interpolation has been applied in high-speed machining to limit the speed discontinuities while respecting the contour tolerance [10,12]. The target of these works is to achieve a confined error and minimal machining time; however, this task becomes harder for more accurate tolerances.\nIn the NanoPla application, speed and acceleration control is not as critical as accurately following a predefined trajectory. Therefore, an accurate curve fitting method capable of offering high accuracy in position definition along the whole trajectory is required. This work presents a novel method for accurate curve fitting with shape-preserving representations, allowing high relative accuracy (HRA) in the computations. Performing an algorithm with HRA is a very desirable goal, because it implies that the relative errors of the computations will be of the order of the machine precision, independent of the size of the problem. The proposed method, which will be called the HRA method, is based on recent advances in numerical linear algebra (see and the references therein). This method considerably improves the precision in curve fitting and, thus, minimizes the contribution of the trajectory fitting errors in the total positioning uncertainty of a nanopositioning stage, such as the NanoPla. The obtained fitting trajectories have continuous successive derivatives, which can also be efficiently evaluated. In this work, curve fitting is performed by interpolation and by least-squares approximation. The HRA method has been developed taking into account both cases. In this article, the contribution to the final positioning error of the NanoPla in terms of the fitting errors of the novel HRA method is compared to the contribution of the fitting errors obtained by traditional CAD/CAM systems, considering also the number of data points and control points required to define the resultant curve, in each case. In addition, the relevance of the resultant fitting errors of each method, in terms of the final positioning error of the NanoPla, is experimentally analyzed.\n\n2. Curve Fitting in Computer Aided Geometric Design (CAGD)\n\nDue to the importance of curve definition in this work, this section briefly explains the procedures for defining a curve by interpolation or by least squares approximation.\nInterpolation is a fundamental tool in CAGD. Given a basis $( u 0 , … , u n )$ of functions defined on I and a set of data points, $p 1 , … , p l + 1$, corresponding to parameters $t 1 , … , t l + 1$, the objective of interpolation is to find a curve,\n$γ ( t ) : = ∑ i = 0 n P i + 1 u i ( t ) , t ∈ I ,$\nwhich passes through the given data points; that is, $γ ( t i ) = p i$, $i = 1 , … , l + 1$ (see Figure 1). The points $P 1 , … , P n + 1$ are called control points. The polygon $P 1 ⋯ P n + 1$, whose vertices are the control points, is called the control polygon.\nIt is worth noting that the number of control points $P 1 , … , P n + 1$ necessary to define the curve is equal to the number of interpolated data points ($l = n$). A better approximation to the curve to be fitted would be expected by increasing the number of given data points. Nevertheless, increasing the number of data points increases the number of control points and, thus, the complexity of the curve and computation.\nOn the other hand, an approximation curve is a curve that approximately fits the given data points but does not necessarily include them (see Figure 2). The most common technique for finding such curves is known as least squares approximation. It should be noted that, in an approximation, the number of control points that define the approximation curve are fewer than the number of given data points ($l > n$). Therefore, when fitting the same set of given data points $p 1 , … , p l + 1$, an approximation curve requires less computation than an interpolating curve representation (see Figure 2).\n\n3. Materials and Methods\n\nNow that the basis of curve definition has been explained, we first present an overview of the NanoPla. The NanoPla is a nanopositioning system, which is used to perform the experimental analysis of trajectory definition errors. Then, we define the analysis procedure which is utilized in this work for the comparison of the novel curve fitting method with traditional methods.\n\n3.1. Nanopositioning Platform (NanoPla)\n\nThe NanoPla consists of three stages: a superior and an inferior base, both of which are fixed, and a moving platform placed in the middle (Figure 3). Three air-bearings levitate the moving platform, while Halbach linear motors generate the required forces for planar motion. A 2D laser system works as a positioning sensor, which provides positioning feedback in the X- and Y-axes and rotation around the Z-axis . The control strategy presented in was adapted and implemented in the NanoPla, in order to minimize the errors introduced by the control hardware and software. The resultant positioning control system of the NanoPla performs the motion and allows accurate positioning over a large range, up to 50 mm × 50 mm.\nThe NanoPla has been designed to work together with different kinds of tools and probes in various applications, such as metrology or nanomanufacturing. Some of these applications require the NanoPla to follow a specific trajectory; for example, one previously designed in a CAD/CAM system. Nevertheless, defining complex curve trajectories in traditional CAD/CAM systems results in trajectory definition errors, which can highly increase the final positioning error of the NanoPla. In Section 5, the positioning uncertainty of the NanoPla positioning system is calculated, in order to confine the trajectory definition errors such that they do not significantly affect the final positioning error.\n\n3.2. Analysis Procedure\n\nIn this work, we analyze the errors in the definition of a trajectory determined by fitting a set of given data points $p 1 , … , p l + 1$. These given data points belong to a curve which was chosen to be a parametric curve, such that the resultant trajectory can be compared with the nominal values of the curve. The selected parametric curve is a cycloid, where the components $x ( t )$, $y ( t )$ take the form:\n$x ( t ) : = r ( t − sin t ) , t ∈ [ 0 , 2 π ] y ( t ) : = r ( 1 − cos t ) .$\nCycloids are commonly used geometric figures in manufacturing applications (e.g., as gear tooth geometry). The cycloid is a transcendental curve and, thus, cannot be expressed by polynomials exactly. For this reason, cycloids cannot be incorporated into most commercial CAD systems .\nThe rest of the paper is organized as follows: first, the HRA method is presented in Section 4. Then, the positioning uncertainty of the NanoPla control system is assessed in Section 5. Knowing the positioning uncertainty of the system allows us to confine the error in the trajectory definition (i.e., fitting error), such that it would not have a significant influence on the total NanoPla positioning error. Then, in Section 6, curve fitting is applied to a determined set of data points, $p 1 , ⋯ , p l + 1$, belonging to a cycloid of radius 1 mm, obtained for equidistant parameters on the interval $t ∈ ( 0 , 2 π )$. Curve fitting is performed by interpolation and least-squares approximation using two different methods: the HRA method, as described in Section 4, and in commonly used commercial CAD/CAM software. The resultant curve obtained by linear interpolation, the most commonly used interpolation method in numerical control (NC), is also included in the comparison. In addition, the resultant fitting errors are related to the number of control points required for the curve construction in each case, due to their relation to the final computational complexity. The target of Section 6 is to confirm that the HRA method is capable of satisfying the precision requirements with a minimum quantity of data points in the interpolation case, and with a minimum number of control points in the case of the least-squares approximation, in contrast to CAD/CAM methods. Finally, in Section 7, the relevance of the curve fitting errors in the total positioning error of the NanoPla is experimentally studied. For the experiments, the NanoPla is set to stop at certain positions on the trajectories defined by interpolating a set of points, either with the HRA method or by using CAM software. The contribution of the fitting errors to the final position of the NanoPla in each case is assessed. Figure 4 presents a diagram of the procedure followed in this work.\n\n4. Accurate Curve Fitting with Shape-Preserving Representations: HRA Method\n\nIn this section, the HRA method for curve fitting is presented. Section 4.1 introduces HRA and a factorization of square strictly totally positive (STP) matrices, which allows for solving HRA linear systems with these matrices. Section 4.2 introduces the class of fg-Bernstein bases and the mentioned bidiagonal factorization of their STP collocation matrices. Section 4.3 presents a procedure for accurate curve fitting by means of interpolation and least-squares approximation using algorithms for solving linear systems with HRA.\n\n4.1. High Relative Accuracy and Bidiagonal Factorizations\n\nLet us recall that a quantity X can be obtained with HRA if the relative error of the computed value $X ˜$ can be bounded, as follows:\n$| | X − X ˜ | | | | X | | < C u ,$\nwhere C is a positive constant independent of the arithmetic precision and u is the unit round-off. An algorithm can be computed with HRA when it only uses products, quotients, the addition of numbers with the same sign, or subtraction of initial data (see ). Performing an algorithm with HRA is a very desirable goal. HRA implies that the relative errors of the computations are of the order of the machine precision, independent of the size of the problem. This goal is difficult to assure, although, in recent years, there have been some advances; in particular, in the field of numerical linear algebra. Let us recall that a matrix is totally positive (TP) if all of its minors are nonnegative and it is STP if all of its minors are positive (see ). Up until now, computations with HRA have been guaranteed only for a few classes of STP matrices. Previously, a reparametrization of the matrices was needed.\nNeville elimination is a particularly important procedure when studying TP matrices. Neville elimination is an alternative to Gaussian elimination and makes zeros in a column of a matrix by adding to a given row the previous one multiplied by an appropriate coefficient called a multiplier (see [18,19,20]). Neville elimination has been used to characterize TP and STP matrices (see [18,19]). From Theorem 4.1 of and p. 116 of , a given matrix A is STP if and only if the Neville elimination of A and $A T$ can be performed without row exchanges, all the multipliers of the Neville elimination of A and $A T$ are positive, and all the diagonal pivots of the Neville elimination of A are positive.\nAccording to , together with the arguments of p. 116 of , an STP matrix $A ∈ R ( n + 1 ) × ( n + 1 )$ can be factorized into the form\n$A = F n F n − 1 ⋯ F 1 D G 1 ⋯ G n − 1 G n ,$\nwhere $F i$ and $G i$ are the lower and upper triangular bidiagonal matrices of the form\n$F i = 1 0 1 ⋱ ⋱ 0 1 m i + 1 , 1 1 ⋱ ⋱ m n + 1 , n + 1 − i 1 , G i T = 1 0 1 ⋱ ⋱ 0 1 m ^ i + 1 , 1 1 ⋱ ⋱ m ^ n + 1 , n + 1 − i 1$\nand $D = diag p 1 , 1 , … , p n + 1 , n + 1$. The entries $m i , j$ and $m ^ i , j$ are the multipliers of the Neville elimination of A and $A T$, respectively, and the diagonal entries $p i , i$ are the diagonal pivots of the Neville elimination of A. In fact, a unique bidiagonal factorization can be obtained for nonsingular TP matrices (see [19,20]).\nBidiagonal factorizations have played a crucial role in deriving algorithms with HRA for STP matrices. When the bidiagonal factorization of the considered matrix is obtained with HRA, the computation of the inverse matrix, its eigenvalues and singular values, the solutions of some linear systems, and the computation of its $Q R$ factorization can be also performed with HRA, using the algorithms presented by Koev in [16,21].\nTable 1 illustrates the accuracy of the computed solutions of $A x = c$ when using Koev’s algorithm with the bidiagonal factorization of A. For different values of n, collocation matrices $A n$ of Bernstein polynomials of degree n on $[ 0 , 1 ]$ corresponding to\n$A n : = n − 1 j − 1 t i j − 1 ( 1 − t i ) n − j 1 ≤ i , j ≤ n , 0 < t 1 < ⋯ < t n < 1 ,$\nhave been considered at equidistant parameters. The solution of the systems $A n x = c n$, where $c n$ is a vector formed by random numbers, have been obtained. The linear systems have also been solved by using Mathematica with a precision of 100 digits; this solution was considered to be exact. Furthermore, two approximations have been computed with Matlab, the first one using the Matlab command \\ and the second one using Koev’s algorithm with the bidiagonal factorization of $A n$.\nAs shown in Table 1, the accuracy of the obtained results when using the bidiagonal factorization of the coefficient matrix and Koev’s algorithm does not depend on the dimension of the considered problem. It is much better than the precision obtained when using the usual resolution with the Matlab command \\.\n\n4.2. The Class of Fg-Bernstein Bases\n\nIn CAGD, polynomial curves are usually represented in terms of the Bernstein basis of the space of polynomials of degree less than or equal to n, defined on $I = [ a , b ]$ by\n$( B 0 n , … , B n n ) , B k n ( t ) : = n k t − a b − a k b − t b − a n − k , t ∈ I , k = 0 , … , n .$\nIn , this basis was generalized as follows: let us suppose that $f , g : I → R$ are nonnegative continuous functions. For a given $n ∈ N$, the corresponding fg-Bernstein basis of order n is defined as\n$( u 0 n , … , u n n ) , u k n ( t ) : = n k f k ( t ) g n − k ( t ) , t ∈ I , k = 0 , … , n .$\nTheorem 2 of proves that, given nonnegative functions $f , g : I → R$ such that $f ( t ) ≠ 0$, $g ( t ) ≠ 0$, $∀ t ∈ ( a , b )$, and $f / g$ is a strictly increasing function, then\n$A : = n j − 1 f j − 1 ( t i ) g n − j + 1 ( t i ) 1 ≤ i , j ≤ n + 1 , a < t 1 < ⋯ < t n + 1 < b$\nis STP. Moreover, in Theorem 3 of , the following bidiagonal decomposition (3) of the collocation matrices (8) was deduced\n$A = F n F n − 1 ⋯ F 1 D G 1 ⋯ G n − 1 G n ,$\nwhere $F i$ and $G i$, $1 ≤ i ≤ n$, are lower and upper triangular bidiagonal matrices of the form (4) and $D = diag p 1 , 1 , … , p n + 1 , n + 1$. The entries $m i , j , m ^ i , j$, and $p i , i$ are given by\n$m i , j = g n − j + 1 ( t i ) g ( t i − j ) g n − j + 2 ( t i − 1 ) ∏ k = 1 j − 1 f ( t i ) g ( t i − k ) − f ( t i − k ) g ( t i ) ∏ k = 2 j f ( t i − 1 ) g ( t i − k ) − f ( t i − k ) g ( t i − 1 ) , m ^ i , j = n − i + 2 i − 1 f ( t j ) g ( t j ) , 1 ≤ j < i ≤ n + 1 , p i , i = n i − 1 g n − i + 1 ( t i ) ∏ k = 1 i − 1 g ( t k ) ∏ k = 1 i − 1 f ( t i ) g ( t k ) − f ( t k ) g ( t i ) , 1 ≤ i ≤ n + 1 .$\nLet us observe that a sufficient condition to obtain the bidiagonal decomposition of A with HRA is that the expressions $f ( t i )$, $g ( t i )$, and $f ( t i ) g ( t k ) − f ( t k ) g ( t i )$, for all $k < i$, can be computed with HRA.\nThere are many interesting choices of functions f and g allowing the definition of fg-Bernstein bases whose collocation matrices can be factorized as in (9) (see, e.g., Section 3 of ).\n\n4.3. Curve Fitting with Fg-Bernstein Bases\n\nThis section presents an accurate method to solve interpolation and approximation problems using the fg-Bernstein bases introduced in Section 4.2.\nThe approach of the problem is as follows. Given $I = [ a , b ]$ and a set of parameters $a < t 1 < ⋯ < t l + 1 < b$, it is desirable to compute a function\n$p ( t ) : = ∑ j = 1 n + 1 c j n j − 1 f j − 1 ( t ) g n − j + 1 ( t ) , t ∈ I ,$\nexpressed in terms of a fg-Bernstein basis of order n, as defined in (7), such that\n$p ( t i ) = p i , 1 ≤ i ≤ l + 1 .$\nIn order to compute the coefficients of $p ( t )$ with respect to the considered fg-Bernstein basis, the linear system $A c = p$ has to be solved, where\n$A : = n j − 1 f j − 1 ( t i ) g n − j + 1 ( t i ) 1 ≤ i ≤ l + 1 ; 1 ≤ j ≤ n + 1$\nis the collocation matrix of the fg-Bernstein basis corresponding to the nodes $t 1 < ⋯ < t l + 1$, $p = ( p 1 , … , p l + 1 ) T$ is the data vector, and $c = ( c 1 , … , c n + 1 ) T$ is the vector with the coefficients to be calculated. Using Theorem 2 of , we can easily deduce that A is STP.\nLet us observe that, in an interpolation problem, $l = n$; that is, the number of data points $( p 1 , … , p l + 1 )$ is equal to the number of the coefficients $( c 1 , … , c n + 1 )$. Moreover, the coefficient matrix is nonsingular and the linear system has a unique solution.\nNow, let us describe how to solve this interpolation problem by means of the bidiagonal decomposition (8) of the collocation matrix of the fg-Bernstein basis.\nIn , an efficient algorithm for computing the solution of systems $A c = p$ was presented. In , the library “TNSolve” was made available, which contains an implementation of the mentioned algorithm for use with Matlab (or Octave). Its computational cost is $O ( n 2 )$ elementary operations and requires, as input arguments, the bidiagonal factorization (3) of the matrix A and the vector p of the linear system $A c = p$. Furthermore, a Matlab (or Octave) function, named “TNBDA”, has been implemented, which takes as input arguments a sequence of parameters $a < t 1 < t 2 < ⋯ < t l + 1 < b$ and computes the bidiagonal factorization (9) of the collocation matrix A. Section 6.1 shows the accuracy of the computed solutions of $A c = p$ when using the function “TNSolve” with the bidiagonal factorization of the collocation matrix of the fg-Bernstein basis given by “TNBDA”.\nOn the other hand, let us observe that, in an approximation problem, $l > n$; that is, the number of data points $( p 1 , … , p l + 1 )$ is greater than the number of the coefficients $( c 1 , … , c n + 1 )$. In this case, the polynomial, $p ( t )$, given in (11), is the solution to the least squares problem if it minimizes the sum of the squares of the deviations from the data\n$∑ i = 1 l + 1 | p i − p ( t i ) | 2 .$\nComputing the coefficients $c j$ of this polynomial $p ( t )$ is equivalent to solving, in the least squares sense, the overdetermined linear system $A c = p$, where A is defined in (12) and $p = ( p 1 , … , p l + 1 ) T$ is the data vector.\nThe unique solution of the problem is given by\n$A T A c = A T p .$\nSolving the previous normal Equation (13) is a worse-conditioned problem than computing the solution through the QR decomposition of the coefficient matrix A, which is the usual approach. Now, following the approach of , let us describe how to solve the approximation problem by means of the bidiagonal decomposition of the collocation matrix of the fg-Bernstein basis.\nIn , an efficient algorithm for computing the $Q R$ decomposition of an STP matrix A was presented. In , the Matlab (or Octave) library “TNQR”, containing an implementation of the above-mentioned algorithm, is available. Assuming that the bidiagonal factorization of A is known, “TNQR” computes the matrix Q and the bidiagonal factorization of the matrix R with HRA. Now, following the approach of , let us describe how to solve the least squares problem by means of a bidiagonal decomposition for rectangular matrices that generalizes the bidiagonal factorization described before when $l = n$ (i.e., in the square case) and the $Q R$ decomposition provided by “TNQR”.\nIn order to compute the solution of the least squares problem, let us define the $R ( l + 1 ) × ( n + 1 )$ matrix M such that\n$M i , i : = p i , i , i = 1 , … , n + 1 , M i , j : = m i , j , j = 1 , … , n + 1 ; i = j + 1 , … , l + 1 , M i , j : = m ^ i , j , i = 1 , … , n ; j = i + 1 , … , n + 1 ,$\nwhere the $m i , j$, $m ^ i , j$, and $p i , i$ are obtained as in (10). Then, using “TNQR”, the $Q R$ decomposition of A can be obtained, such that\n$A = Q R 0 ,$\nwhere $Q ∈ R ( l + 1 ) × ( l + 1 )$ is an orthogonal matrix and $R ∈ R ( n + 1 ) × ( n + 1 )$ is an upper triangular matrix with positive diagonal entries. According to Section 1.3.1 in , the solution of the least squares problem is obtained from\n$d 1 d 2 = Q T p , R c = d 1 ,$\nwhere $d 1 ∈ R n + 1$, $d 2 ∈ R l − n$. The matrices Q and R have a special structure, as described in . In particular, R is nonsingular and TP. In order to obtain the solution of the upper triangular system $R c = d 1$, the routine “TNSolve” of has been used, which performs the bidiagonal decomposition of the upper triangular TP matrix R. The numerical experiments in Section 6.2 show that the accuracy of the computed solutions does not depend on the size of the considered problem.\n\n5. Positioning uncertainty of the NanoPla\n\nThe total NanoPla positioning error at a certain point of the trajectory ($e p o s i t i o n$) results from the sum of the positioning error of the system ($e N a n o P l a$) and the error in the definition of the trajectory ($e t r a j e c t o r y$), as represented in Equation (15)\n$e p o s i t i o n = e N a n o P l a + e t r a j e c t o r y .$\nTherefore, it is convenient to analyze and calculate the positioning uncertainty of the NanoPla system, in order to confine the curve fitting error in the trajectory definition (Ce), such that it does not have a significant influence on the final positioning error.\nThe NanoPla control system inputs are the X and Y coordinates of the desired position and the laser system readouts of the actual NanoPla position. In addition, rotation of the moving platform around the Z-axis is controlled to avoid laser system misalignment. The design and control system of the NanoPla has been optimized to correct systematic errors and minimize random positioning errors. The control strategy is computed in Simulink (Matlab), which outputs the phase voltages to be generated for each linear motor. These instructions are sent by a serial communication interface (SCI) to four digital motor control (DMC) kits of Texas Instruments. Each kit drives the phase voltages of one linear motor by pulse width modulation signals. The DMC kit works with 32-bit data and the input to the high-resolution PWM (HRPWM) module is a fixed point of 19 bits ($11 + 8$) bits. When the current flows through the motor’s coils, each motor generates a horizontal force and a vertical force. The four motors produce planar motion and are placed in parallel pairs; the first pair generates a force in the X-axis and the second pair in the Y-axis. These forces produce a displacement of the NanoPla, ultimately to the desired position. The vertical forces generated by the four motors help the air-bearings to levitate the platform. During computation of the control task, real numbers are represented with finite precision . In Figure 5, a block diagram of the control system is represented, including the data type of the transmitted information, which affects the resolution of the system.\nIn Simulink, data are represented in double-precision floating-point format (64 bits) and, in this case, the rounding operation has no influence on the calculated results. The errors derived from the resolution of the data type at the SCI are also negligible. Nevertheless, the resolution of the HRPWM driving the phase currents contributes to the final positioning uncertainty. The other main contributors are the noise of the laser system, the noise of the generated currents, and the vibrations of the NanoPla. These errors cause root mean square (RMS) deviations of the positioning error in the open-loop, which can be experimentally measured. In addition, the geometrical errors of the 2D laser system have been characterized by a self-calibration procedure, as defined in , and corrected in the measurements. The uncertainty of the calibrated laser system also contributes to the final positioning uncertainty. Even though the laser system resolution is included inside the standard uncertainty of the laser system, its value is shown separately in the table, so it can be compared to the other contributor’s magnitudes. The calculation of the positioning uncertainty and its contributors are represented in Table 2, according to .\nThe resultant positioning uncertainty in the X–Y-plane, $U X Y$ ($k = 2$), was equal to 0.50 $μ$m over the working range of 50 mm × 50 mm. Additionally, it is worth mentioning that, when working in a closed-loop, the control system tends to correct the positioning error caused by the resolution of the system , which results in a difference between position feedback and position reference, generally smaller than 0.02 $μ$m, and a RMS positioning noise, of approximately ±0.20 $μ$m, whose exact values depend on the position and the controller configuration. The error introduced by the definition of the NanoPla trajectory should be confined, in order not to significantly increase the final positioning uncertainty. A confined fitting error of 0.05 $μ$m or lower increases the final positioning uncertainty by less than 0.01 $μ$m, which is considered acceptable for the requirements of the NanoPla.\n\n6. Analysis of the Curve Fitting Errors\n\nThe target of this section is to perform curve fitting on a set of given data points using the HRA method and CAD/CAM software, in order to confirm that the HRA method provides a better fit. In this study, the curves that defined the trajectory of the NanoPla were calculated by curve fitting a set of data points $p 1 , ⋯ , p l + 1$ with equidistant parameter values $t 1 , ⋯ , t l + 1$, belonging to a cycloid curve of radius 1 mm, as defined in (2).\nCurve fitting was performed using the HRA method and CAD/CAM systems with the requirement of obtaining a fitting error lower than 0.05 $μ$m along the whole trajectory, in order to fulfill the NanoPla precision requirements without compromising the number of required data points and control points of the fitting curve. In the first subsection, the curves are obtained by interpolating sets of data points with different dimensions; that is, for different values of $l + 1$. The smallest possible set of points required to fulfill the tolerance was obtained for each method. In the second subsection, the curves were obtained by performing least-squares approximations of sets of data points for different values of $l + 1$ and the minimum number of control points required to fulfill the tolerance is obtained for each method. In both cases, the results obtained with the HRA method and with the commercial CAD/CAM software were analyzed.\n\n6.1. Curve Fitting By Interpolation\n\nAs previously noted, interpolation consists of finding a curve through $l + 1$ data points $p 1 , … , p l + 1$ with parameter values $t 1 , … , t l + 1$. For this reason, when interpolating through a set of points using an exact method, the interpolation errors at these points were zero. Nevertheless, at non-data intervals (data between interpolation points), the interpolation errors can be calculated as the difference between the nominal parametric curve and the interpolated curve, when the equation of the nominal curve is known. In addition, as mentioned in Section 2, in an interpolation problem, the number of control points equals the number of given data points ($l = n$). Thus, in this subsection, the number of data points and control points will be indistinctly referred to as $n + 1$.\nFor comparison, the curve trajectory was generated by three exact interpolation methods: linear interpolation, CAD/CAM system interpolation, and the HRA method presented in Section 4. The interpolation results of some commonly used CAD and CAM systems have been compared and, even though their results slightly differed, the errors had similar magnitude; thus, the difference was not significant. Therefore, the results of CAM software have been used as a reference in this subsection. On the other hand, interpolation with the HRA method was performed using two different fg-Bernstein bases. One of them is the polynomial basis obtained with $f ( t ) = t$ and $g ( t ) = 1 − t$, $t ∈ [ 0 , 1 ]$ (HRA interpolation with this basis will be called HRA IB1) and the other one is the trigonometric basis obtained with $f ( t ) = sin ( Δ + t ) / 2$ and $g ( t ) = sin ( Δ − t ) / 2$ , $t ∈ [ − Δ , Δ ]$ and $Δ = 1.57$ (HRA interpolation with this basis will be called HRA IB2). Thus, interpolation with the HRA method was applied as follows: Firstly, for different values of $n + 1$, collocation matrices at equidistant parameters in the interior of the interval domain of the selected fg-Bernstein bases were considered. Secondly, in order to obtain the $n + 1$ control points, the linear system $A c = p$ was computed by applying the interpolation method explained in Section 4, when $l = n$. Knowing the control points, the parametric curve was evaluated and the deviations of the interpolated curve from the cycloidal nominal were obtained by computing the distance at equally distributed points along the profile curve.\nFigure 6a shows the resultant curves for the three interpolation methods, for $n + 1 = 7$; that is, seven interpolation points and seven control points. For simplification and due to symmetry, only half of the curves are represented. As can be seen, the greatest interpolation errors occurred when applying linear interpolation. With this method, the maximum interpolation errors appear at the middle zone of the intervals, whose interpolating points were further from each other (interval between $p 3$ and $p 4$). Nevertheless, when applying CAM/CAD system spline interpolation and the HRA method, the interpolation errors were reduced. In both cases, the maximum interpolation errors appeared in the middle zones of the first intervals. The first non-data interval is shown in Figure 6b.\nThe CAD and CAM systems spline bases are usually unknown for the user. In other works, such as , the interpolation algorithm had to be experimentally extracted to simulate the interpolation errors. In this work, the interpolation errors were calculated graphically in the CAM program. However, the interpolation errors of linear interpolation and the HRA method were calculated analytically. In both cases, the errors of the interpolated curve from the cycloidal nominal were obtained by computing the distance at equally distributed points along the profile curve. The interpolation errors obtained by the HRA method IB1, along the whole range of the cycloid, are represented in Figure 7, for $n + 1 = 7$, 9, and 11. As shown, the maximum interpolation errors occurred in the first interval, then decreased by more than half.\nIn Table 3, the fitting errors of the interpolated curves for the number of interpolated points $n + 1 = 7$, 9, 11, and 21 are represented. It was observed that, for this specific curve, the first and second non-data intervals were the most critical, because they contained the greatest fitting errors when interpolating with CAM system splines, as well as with the HRA method. For this reason and for simplification, only these intervals are represented in Table 3. As can be seen, linear interpolation always presented the worst results. In addition, for the same number $n + 1$ of interpolation points, the HRA method presented a considerably smaller error than the CAM system splines. When interpolating with HRA method IB1, for a number of interpolation points higher than 11, the resultant errors became negligible for the NanoPla application (Ce < 0.05 $μ$m); while, in the HRA method IB2, 16 interpolation points were required. In contrast, it has been verified that more than 49 interpolation points are necessary to achieve an error lower than 0.05 $μ$m at the non-data intervals, when interpolating with CAM/CAD system splines, and more than 250 interpolation points are needed when applying linear interpolation. Therefore, it can be inferred that the HRA method is capable of achieving accurate curve fitting while interpolating a smaller number of data points, in comparison with commonly used CAD/CAM systems.\n\n6.2. Curve Fitting by Least Squares Approximation\n\nLeast-squares approximation is not an exact method; that is, the resultant fitting curve does not necessarily contain the given data points, but only approximately fits them. In an approximation problem, the number of control points can be externally defined and is directly related to the fitting errors. As mentioned above, the number of control points defines the complexity of the fitting curve: typically, a smaller number of control points results in a simpler curve, but also in greater fitting errors. Nevertheless, an elevated number of control points does not assure accurate fitting, as some control points may be redundant or inadequate . In addition, in an approximation problem, the fitting tolerance is defined as the maximum permitted error between the given data points and the resultant approximation curve. Nevertheless, the error between the smooth curve and the nominal curve can be greater over non-data intervals. For this reason, the dimension $l + 1$ of the set of given data points has to be large enough such that the fitting error does not increase significantly over the non-data intervals.\nFor comparison, curve trajectories were generated by two least-squares approximation methods: CAD/CAM system approximation and the HRA method presented in Section 4. In CAD/CAM systems, when curve fitting by approximation, a tolerance for the fitting error must be defined by the designer. The smaller this tolerance is, the greater the number of control points the approximation curve requires. Similar to the interpolation operation, the approximation operation was performed using some of the most commonly used CAM/CAD systems and the results were compared. Even though the obtained results were similar, some programs presented more information to the user about the approximation function than others. For example, in some programs the approximation spline degree can be selected, the points in the curve where the maximum fitting error occurs are shown, and the magnitude of the consequent error, along with the magnitude of the average fitting error along the whole curve, is calculated. In this section, the results of CAD software which calculated the maximum fitting error (confined error, Ce) and graphically showed its location was used as a reference. On the other hand, the least squares approximation with the HRA method was performed using two different fg-Bernstein bases, the same as those in the previous subsection. One of them is the polynomial basis obtained with $f ( t ) = t$ and $g ( t ) = 1 − t$, $t ∈ [ 0 , 1 ]$ (HRA least-squares approximation with this basis will be called HRA LSB1), and the other one is the trigonometric basis obtained with $f ( t ) = sin ( Δ + t ) / 2$ and $g ( t ) = sin ( Δ − t ) / 2$ , $t ∈ [ − Δ , Δ ]$ and $Δ = 1.57$ (HRA least-squares approximation with this second basis will be called HRA LSB2). The least squares approximation with the HRA method was carried out as follows: first, for different values of $l + 1$, collocation matrices at equidistant parameters in the interior of the interval domain of the selected fg-Bernstein bases were considered. Second, in order to obtain the $n + 1$ control points, the linear system $A c = p$ was computed by applying the interpolation method explained in Section 4, with $l > n$. Knowing the control points, the parametric curve was evaluated and the maximum fitting error of the approximated curve from the cycloidal nominal was obtained as the confined error, by computing the maximum distance at equally distributed points along the profile curve.\nAccording to the positioning accuracy of the NanoPla, the fitting tolerance was set to be smaller than 0.05 $μ$m for the approximation operation in every case. Table 4 represents the number of control points that were required to generate a fitting curve from the given data points, $p 1 , … , p l + 1$, by least squares approximation, with the reference CAD method and the HRA method. The resultant confined error (Ce)—that is, the maximum fitting error—obtained for every method is also represented in the table. In the case of the CAD method, the confined error was calculated and provided by the software. For the HRA method, the confined error was analytically calculated. In addition, the results of the HRA method are shown for the two different bases, HRA LSB1 and HRA LSB2.\nIt can be observed that, in every case, with the same set of given data points, the HRA methods were capable of obtaining an approximation curve within a tolerance of 0.05 $μ$m with fewer control points, in comparison with the CAD method. In addition, it is worth noting that, for the HRA methods, when the number of given data points increased, the number of control points required to fulfill the tolerance constraint remained practically constant, in contrast to the CAD method. Thus, it can be inferred that the HRA method provided a better approximation, requiring fewer control points, and that it did not define redundant control points.\n\n7. Experimental Results\n\nThe previous section showed that the HRA method is capable of performing accurate curve fitting requiring fewer data points in the interpolation operation and with a minimum number of control points in a least-squares approximation, in comparison with CAD/CAM systems. The resultant fitting curve can be used to define the trajectory of a positioning system, such as the NanoPla. This section experimentally analyzes the relevance of the trajectory definition errors in the final positioning error of the NanoPla when the trajectory is defined by fitting a certain set of data points.\nThe NanoPla was required to displace itself to positions along a defined trajectory with minimum positioning error. As was assessed in Section 5, the positioning uncertainty $U X Y$ ($k = 2$) of the NanoPla was $0.50$ $μ$m. Therefore, trajectory definition errors lower than $0.05$ $μ$m were considered to be negligible in this application. In order to analyze the contribution of the trajectory definition errors to the final positioning error of the NanoPla, the trajectory was defined by interpolating a set of points using both the CAM/CAD system methods and the HRA method. Then, the position of the NanoPla at certain points of the trajectories was compared to the nominal curve trajectory position.\nIn Section 6, it was stated that the interpolating errors when calculating a cycloid (r = 1 mm) based on 11 points were lower than $0.05$ $μ$m, when using the HRA method. Nevertheless, the errors were up to $4.5$ $μ$m when interpolating with CAM system splines. In this experiment, the trajectory of the NanoPla was defined by the curves resulting from interpolating a set of 11 points of a parametric cycloid (r = 1 mm) separated by equal intervals of t ($△ t = 2 π / 10$ rad). The interpolation was performed by CAM system splines and by the HRA method, in order to verify the interpolating error effect when implementing the trajectories in the NanoPla. In order to observe the effect of the positioning noise of the NanoPla in a closed-loop, the control system of the NanoPla was set to keep the moving platform still at certain positions in the trajectories. The resultant X,Y positions of the NanoPla were experimentally recorded for 30 s, with a readout taken every $0.12$ seconds.\nFigure 8 represents the curve trajectories defined by interpolating 11 data points ($p 1 , … , p 11$) with the HRA method and with CAM splines; in addition, the parametric curve is also included. Due to the fact that the fitting errors occurred at a micrometer scale, they cannot be observed in Figure 8a. For clarification, Figure 8b,c show an amplification of the first and second non-data intervals, where the greatest fitting errors take place. Figure 8b,c also include the 250 experimental readouts of the laser system which represents the positions of the platform where the control system was set to stay still at points belonging to the trajectory of the curve, as defined by the HRA method ($T P 1 H R A$ , $T P 2 H R A$, $T P 3 H R A$, and $T P 4 H R A$; magenta in the figure) and by CAM system splines ($T P 1 C A M$, $T P 2 C A M$ , $T P 3 C A M$, and $T P 4 C A M$; blue in the figure). These trajectory points (TP) were selected in the non-data intervals of the curve, where the greatest fitting errors occurred, as can be seen in the figures. Therefore, they present a worst-case scenario.\nIn Figure 8b,c, it can be seen that the HRA method trajectory was almost coincident with the parametric curve, while the CAM trajectory presented a larger fitting error. At the trajectory points (TP) where the NanoPla was set to keep still, the positioning error was the result of the sum between the positioning error of the NanoPla and the errors in the definition of the trajectory (Equation (15)), which were the curve fitting errors. In Table 5, the NanoPla positioning error and the trajectory definition error at each trajectory point, as well as the total positioning error, are represented. The NanoPla positioning error at the TPs is expressed as the sum of a constant value and a standard deviation. The constant value is the difference between the laser system position feedback and the reference position in the controller, calculated by averaging all laser readouts at the TP. The standard deviation includes the RMS positioning noise and the calibrated laser system uncertainty.\nThe interpolating errors for $T P 1 H R A$ and $T P 2 H R A$ were 0.04 $μ$m and 0.03 $μ$m, respectively; as can be seen from Figure 8b, they did not significantly affect the final positioning accuracy. On the contrary, the interpolating errors of $T P 1 C A M$ and $T P 2 C A M$ were 3.1 $μ$m and 3.8 $μ$m, respectively, and they had a significant effect in the total positioning error of the NanoPla, being the greatest contributor. Similarly, in the second non-data interval (Figure 8c), the interpolating error for $T P 3 H R A$ and $T P 4 H R A$ was 0.007 $μ$m in both cases, which was negligible in comparison to the NanoPla positioning error. The interpolating errors of $T P 3 C A M$ and $T P 4 C A M$ were 4.5 $μ$m in both cases. As in the first interval, they were the greatest contributor to the final positioning error of the NanoPla.\n\n8. Conclusions\n\nIn nanopositioning systems, such as the NanoPla, the final positioning accuracy can be affected by errors in the definition of the trajectory. Hence, this work proposed the use of a novel method for the parametric representation of curves, allowing for curve fitting with high relative accuracy for the definition of curve trajectories. This method, which uses collocation matrices of fg-Bernstein bases, is applicable in CAGD and greatly improves the accuracy in trajectory definition by curve fitting, when compared to traditional CAD/CAM methods implemented in CAD/CAM software.\nIn this work, the positioning uncertainty of the NanoPla ($k = 2$) was calculated to be 0.50 $μ$m. Thus, the trajectory definition errors were required to be lower than 0.05 $μ$m, in order not to significantly affect the final positioning accuracy of the system. Sets of given data points were curve fitted by interpolation and least-squares approximation, using both the proposed HRA method and CAD/CAM software, with the requirement of fulfilling the tolerance of 0.05 $μ$m. Finally, it was experimentally verified that, when defining the trajectory by interpolating a given set of 11 data points with the HRA method, the fitting errors at non-data intervals were negligible, in comparison to the NanoPla positioning uncertainty. In contrast, when defining the trajectory with CAM splines with the same set of 11 interpolation points, the interpolation errors were the greatest contributors to the NanoPla final positioning error. This work focused on the contribution of the trajectory definition errors to the final positioning accuracy of the NanoPla. In future work, the dynamic response of the NanoPla when performing a trajectory of motion will be studied.\nThe importance in this work of the fg-Bernstein bases arises from the fact that their collocation matrices admit many algebraic computations with HRA. In particular, the proposed HRA method can be applied to perform interpolation and least-squares approximation with high precision and can be considered as a modeling tool for trajectories. The benefits of the proposed HRA method that have been shown in this paper are that it is capable of performing accurate interpolation with a minimum quantity of data points and, when performing a least-squares approximation, it is capable of obtaining an accurate approximation curve with a minimum number of control points. In addition, the number of obtained control points was independent of the size of the problem. On the contrary, in CAD/CAM systems, a higher number of data points was required for accurate curve fitting and the number of control points increased with the size of the problem, which may result in redundant or inadequate control points. As a result, the proposed HRA method offers a better positioning accuracy than commonly used CAD/CAM methods when defining a trajectory by curve fitting (either by interpolation or least-squares approximation). Hence, implementing the HRA method in CAD/CAM systems would be of great interest to nanomanufacturing applications in which a nanopositioning stage is required to follow a predefined trajectory.\n\nAuthor Contributions\n\nL.D.P., M.T.G., J.A.A.G. and J.A.Y.F. developed the positioning control system of the NanoPla.; B.R.S. and E.M.M. developed the HRA method. All authors contributed to the conceptualization of the project and definition of the methodology. L.C.D.P. and B.R.S. performed the experiments and wrote the manuscript draft. 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Curve fitting by least squares approximation, where $p 1 , p 2 , p 3 , p 4 , p 5 , p 6$ are the given data and $P 1 , P 2 , P 3$ are the control points.\nFigure 3. (a) Photograph of the NanoPla and the control system components; (b) front and exploded view of the NanoPla.\nFigure 3. (a) Photograph of the NanoPla and the control system components; (b) front and exploded view of the NanoPla.\nFigure 4. Diagram of the procedure for the analysis of the methods.\nFigure 4. Diagram of the procedure for the analysis of the methods.\nFigure 5. Scheme of the data flow in the NanoPla control system.\nFigure 5. Scheme of the data flow in the NanoPla control system.\nFigure 6. (a) Resultant interpolated curves obtained for $n + 1 = 7$, represented for $t ∈ ( 0 , π )$; (b) resultant interpolated curves obtained for $n + 1 = 7$, in the first non-data interval.\nFigure 6. (a) Resultant interpolated curves obtained for $n + 1 = 7$, represented for $t ∈ ( 0 , π )$; (b) resultant interpolated curves obtained for $n + 1 = 7$, in the first non-data interval.\nFigure 7. Interpolation errors obtained by the high relative accuracy (HRA) method IB1.\nFigure 7. Interpolation errors obtained by the high relative accuracy (HRA) method IB1.\nFigure 8. (a) Trajectories defined by interpolating the data points $p 1 , … , p 11$ with the HRA method and CAM splines; detail of the first non-data interval (b) and the second non-data interval (c), including the position of the NanoPla at certain trajectory points.\nFigure 8. (a) Trajectories defined by interpolating the data points $p 1 , … , p 11$ with the HRA method and CAM splines; detail of the first non-data interval (b) and the second non-data interval (c), including the position of the NanoPla at certain trajectory points.\nTable 1. Relative errors in solving $A n x = c n$ using the bidiagonal factorization of collocation matrices of Bernstein polynomials (5) and Koev’s algorithm.\nTable 1. Relative errors in solving $A n x = c n$ using the bidiagonal factorization of collocation matrices of Bernstein polynomials (5) and Koev’s algorithm.\nn$A n ∖ c n$HRA\n10$3.3579 × 10 − 13$$1.1191 × 10 − 15$\n20$1.2413 × 10 − 9$$6.2974 × 10 − 16$\n25$4.1424 × 10 − 7$$2.0843 × 10 − 15$\n50$1.0000$$7.5480 × 10 − 15$\nTable 2. NanoPla positioning uncertainty contributors and calculation.\nTable 2. NanoPla positioning uncertainty contributors and calculation.\nSourceJustificationValue\nResolution at the HRPWM $u H R P W M$Resolution of $2.62 × 10 − 5$ V $700 / 12$ nm\nLaser system resolution $u L r e s$Resolution of $1.58$ nm($1.58 / 12$ nm)\nCalibrated Laser system $u L c a l$Geometrical errors + measuring system calibration 99 nm\nRMS positioning error $u R M S$Laser system noise + phase currents noise + vibrations110 nm\nPositioning uncertainty $U X Y$ ($k = 2$)$U X Y ( k = 2 ) = k u H R P W M 2 + u L r e s 2 + u L c a l 2 + u R M S 2$501 nm\nTable 3. Maximum interpolating errors for a curve generated based on a set of points $p 1 , … , p n + 1$ in the first and second non-data intervals, for different interpolation methods.\nTable 3. Maximum interpolating errors for a curve generated based on a set of points $p 1 , … , p n + 1$ in the first and second non-data intervals, for different interpolation methods.\n$n + 1$IntervalsLinearCAMHRA IB1HRA IB2\n7$( 0 , 2 π / 6 )$$27.6$$μ$m18.0 $μ$m20.17 $μ$m17.84 $μ$m\n7$( 2 π / 6 , 4 π / 6 )$93.5 $μ$m17.1 $μ$m5.31 $μ$m5.60 $μ$m\n9$( 0 , 2 π / 8 )$11.8 $μ$m8.0 $μ$m1.25 $μ$m3.95 $μ$m\n9$( 2 π / 8 , 4 π / 8 )$41.9 $μ$m8.5 $μ$m0.22 $μ$m0.93 $μ$m\n11$( 0 , 2 π / 10 )$6.1 $μ$m4.0 $μ$m0.05 $μ$m0.94 $μ$m\n11$( 2 π / 10 , 4 π / 10 ) )$22.0 $μ$m4.5 $μ$m<0.01 $μ$m0.18 $μ$m\n21$( 0 , 2 π / 20 )$0.7 $μ$m0.5 $μ$m≪1 nm<0.01 $μ$m\n21$( 2 π / 20 , 4 π / 20 )$2.8 $μ$m0.6 $μ$m≪1 nm<1 nm\nTable 4. Number of control points ($P i$, $i = 1 , … , n + 1$) and maximum fitting errors (Ce) for a curve generated based on a set of points $p i$, $i = 1 , … , l + 1$, using different approximation methods.\nTable 4. Number of control points ($P i$, $i = 1 , … , n + 1$) and maximum fitting errors (Ce) for a curve generated based on a set of points $p i$, $i = 1 , … , l + 1$, using different approximation methods.\n$l + 1$CAD HRA LSB1 HRA LSB2\n$n + 1$Ce$n + 1$Ce$n + 1$Ce\n5121$0.045$$μ$m10$0.004$$μ$m12$0.046$$μ$m\n10138$0.006$$μ$m11$0.008$$μ$m13$0.037$$μ$m\n25144$0.023$$μ$m11$0.013$$μ$m14$0.017$$μ$m\n50134$0.046$$μ$m11$0.015$$μ$m14$0.021$$μ$m\n100138$0.036$$μ$m11$0.016$$μ$m14$0.023$$μ$m\nTable 5. NanoPla positioning system errors ($e N a n o P l a$), trajectory errors ($e t r a j e c t o r y$) and total positioning errors ($e p o s i t i o n$) at certain points of the trajectory (totally positive (TP)) defined by the HRA method and CAM system.\nTable 5. NanoPla positioning system errors ($e N a n o P l a$), trajectory errors ($e t r a j e c t o r y$) and total positioning errors ($e p o s i t i o n$) at certain points of the trajectory (totally positive (TP)) defined by the HRA method and CAM system.\nTrajectory Points$e NanoPla$$e trajectory$$e position$\n$T P 1 H R A$$0.01 ± 0.16$$μ$m$0.04$$μ$m$0.05 ± 0.16$$μ$m\n$T P 1 C A M$$0.02 ± 0.21$$μ$m$3.1$$μ$m$3.12 ± 0.21$$μ$m\n$T P 2 H R A$$0.01 ± 0.19$$μ$m$0.03$$μ$m$0.04 ± 0.19$$μ$m\n$T P 2 C A M$$0.00 ± 0.13$$μ$m$3.8$$μ$m$3.80 ± 0.13$$μ$m\n$T P 3 H R A$$0.02 ± 0.22$$μ$m$0.007$$μ$m$0.027 ± 0.22$$μ$m\n$T P 3 C A M$$0.01 ± 0.32$$μ$m$4.5$$μ$m$4.51 ± 0.32$$μ$m\n$T P 4 H R A$$0.02 ± 0.31$$μ$m$0.007$$μ$m$0.027 ± 0.31$$μ$m\n$T P 4 C A M$$0.01 ± 0.16$$μ$m$4.5$$μ$m$4.51 ± 0.16$$μ$m" ]
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https://www.daniweb.com/programming/software-development/threads/153327/hellow
[ "how can i draw a circle divided to many sectors ,each secteor area deppends on specific angle for each sector in the circle ,\n\nthanx for help ,\nregards\n\nyou will have to use some sort of gui libraries and/or compiler. Depends on your operating system.\n\nIf you don't know it already, start a course in trigonometry. It might help.\n\nTrig isn't all that difficult.. I put together a small example of how to create a bitmap, draw a circle and a sector in it (with a 50 deg angle). I tried to keep the code as short as possible, but it is still over 200 lines, so I will …\n\n## All 9 Replies\n\nyou will have to use some sort of gui libraries and/or compiler. Depends on your operating system.\n\nIf you don't know it already, start a course in trigonometry. It might help.\n\ntrig will help for the math part but is useless for the gui part. What you want is an easy way to draw pie charts in C or C++. Read some of these google links.\n\n(x - a) ^ 2 + (y - b) ^ 2 = r ^ 2\nWhere a and b are the center coordinates, and r is the radius. This should help you solve for x, and y, given one or the other.\nThis is useful for plotting, but sin, cos, and tan will be useful for splitting it up into sectors. (Remember a full rotation is 2 * pi)\n\nTrig isn't all that difficult.. I put together a small example of how to create a bitmap, draw a circle and a sector in it (with a 50 deg angle). I tried to keep the code as short as possible, but it is still over 200 lines, so I will post the important part and attach the whole code. It should draw on the bitmap and save it as circle.bmp in the project folder.\n\n``````const double PI = 3.141592;\nconst double deg_to_rad = PI / 180;\nconst double rad_to_deg = 180 / PI;\n\nvoid draw_sector(Bitmap &bmp, int deg_startangle, int deg_finangle, int radius) {\nint centre_x = bmp.width / 2;\nint centre_y = bmp.height / 2;\n\n// Start Angle\nbmp.line(\ncentre_x, // x1\ncentre_y, // y1\n);\n\n// Finish Angle\nbmp.line(\ncentre_x, // x1\ncentre_y, // y1\n);\n}\n\nint main() {\n// Create Bitmap object\nBitmap bmp(500, 500);\n\n// Centre of bitmap\nint centre_x = bmp.width / 2;\nint centre_y = bmp.height / 2;\n\nint x1, y1, x2, y2; // Temp location variables\n\n// Set default old coordient\ny2 = centre_y;\n\ncout << \"Drawing circle with a sector of 50 deg\\n\";\n\n// Draw a circle ( loop from 0 to (2*PI) ) with 1deg accuracy\nfor (double i = 0; i < (360 * deg_to_rad); i += (1 * deg_to_rad)) {\n// Get next coordients to draw a line to\nx1 = int(centre_x + cos(i) * 150);\ny1 = int(centre_y + sin(i) * 150);\n\n// Draw line from old coordients to new ones\nbmp.line(x1, y1, x2, y2);\n\n// Set old coordients\nx2 = x1;\ny2 = y1;\n}\n\n// Draw a sector between angle 0 and 50\n\nbmp.save(\"circle.bmp\");\n\ncout << \"\\nDone!\\n\\nPress <ENTER> to continue..\\n\";\n\ncin.ignore();\nreturn 0;\n}``````\n\nAnd here are some short tutorials to help :)\nTrigonometry\n\nHope all this helps :]\n\ncommented: Whoo! You're a genius kid XP +4\n\nthanks all my freinds , you response is realy helpful....\n\nBest regards\nextrov\n\nHello is spelled without the w. =)\nAlso, trigonometry is the way to go. You can probably find a tutorial or something: www.google.com/search?q=Trigonometry+Tutorial\n\nMy fingers get out of sync with my brain so often, every time I try to demo \"Hello world\" it comes out \"Hellow orld\"\n\nOr maybe it's just an emphatic variant of Hello?\n\nMy fingers get out of sync with my brain so often, every time I try to demo \"Hello world\" it comes out \"Hellow orld\"\n\nOr maybe it's just an emphatic variant of Hello?\n\nyes thats it , u gut it ...\nmy be sometimes when people miss something ,anothers add it ...\n\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge." ]
[ null ]
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https://eddmann.com/posts/mergesort-in-clojure-using-post-conditionals/
[ "# Mergesort in Clojure using Post Conditionals\n\nWhilst currently reading through The Joy of Clojure book I was introduced to the concept of pre and post-conditionals, similar to another language I have heard about called Eiffel. To experiment with this feature I decided to create a simple merge-sort algorithm implementation which provided the post invariant that its returned values were sorted by the provided predicate.\n\n``````(defn sorted? [pred? col]\n(every? #(apply pred? %) (partition 2 1 col)))\n\n(defn merge\n([pred? left right]\n{:post [(sorted? pred? %)]}\n(loop [[l & lr :as ls] left [r & rr :as rs] right acc []]\n(if (and l r)\n(if (pred? l r)\n(recur lr rs (conj acc l))\n(recur ls rr (conj acc r)))\n(concat acc ls rs)))))\n\n(defn half [col]\n(split-at (/ (count col) 2) col))\n\n(defn merge-sort\n([col] (merge-sort < col))\n([pred? col]\n{:post [(sorted? pred? %)]}\n(if (< (count col) 2)\ncol\n(let [[left right] (half col)]\n(merge pred? (merge-sort pred? left) (merge-sort pred? right))))))\n``````\n\nI really liked the manner in which I could use the ‘partition’ function to succinctly map the task of splitting on each co-located pair of values within a collection in code. We are able to then put this implementation into practice by optionally providing a predicate (else defaulting to ascending order).\n\n``````(def numbers (shuffle (range 1 11))) ; [1 9 7 2 10 3 6 5 4 8]\n\n(merge-sort numbers) ; (1 2 3 4 5 6 7 8 9 10)\n(merge-sort > numbers) ; (10 9 8 7 6 5 4 3 2 1)\n``````" ]
[ null ]
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https://docksci.com/latent-risk-and-trend-models-for-the-evolution-of-annual-fatality-numbers-in-30-_5acd969cd64ab2d5697b6725.html
[ "Accident Analysis and Prevention 71 (2014) 327–336\n\nContents lists available at ScienceDirect\n\nAccident Analysis and Prevention journal homepage: www.elsevier.com/locate/aap\n\nLatent risk and trend models for the evolution of annual fatality numbers in 30 European countries Emmanuelle Dupont a,∗ , Jacques J.F. Commandeur b , Sylvain Lassarre c , Frits Bijleveld b , Heike Martensen a , Constantinos Antoniou d , Eleonora Papadimitriou d , George Yannis d , f ˜ Elke Hermans e , Katherine Pérez f , Elena Santamarina-Rubio , Davide Shingo Usami g , Gabriele Giustiniani g a\n\nBRSI, Belgian Road Safety Institute, Belgium SWOV Institute for Road Safety Research and VU University Amsterdam, The Netherlands c IFSTTAR, French Institute of Science and Technology for Transports, Development, and Networks, France d NTUA, National Technical University of Athens, Greece e Hasselt University, Transportation Research Institute, Belgium f ASPB, Agència de Salut Pública de Barcelona, CIBER de Epidemiología y Salud Pública (CIBERESP), Institut d’Investigació Biomèdica Sant Pau (IIB Sant Pau), Barcelona, Spain g La Sapienza University of Rome, Research Centre for Transport and Logistics, Italy b\n\na r t i c l e\n\ni n f o\n\nArticle history: Received 18 January 2014 Received in revised form 10 May 2014 Accepted 12 June 2014 Available online 9 July 2014 Keywords: Risk Exposure Structural time series models Latent risk model Stochastic trend model Forecasting\n\na b s t r a c t In this paper a unified methodology is presented for the modelling of the evolution of road safety in 30 European countries. For each country, annual data of the best available exposure indicator and of the number of fatalities were simultaneously analysed with the bivariate latent risk time series model. This model is based on the assumption that the amount of exposure and the number of fatalities are intrinsically related. It captures the dynamic evolution in the fatalities as the product of the dynamic evolution in two latent trends: the trend in the fatality risk and the trend in the exposure to that risk. Before applying the latent risk model to the different countries it was first investigated and tested whether the exposure indicator at hand and the fatalities in each country were in fact related at all. If they were, the latent risk model was applied to that country; if not, a univariate local linear trend model was applied to the fatalities series only, unless the latent risk time series model was found to yield better forecasts than the univariate local linear trend model. In either case, the temporal structure of the unobserved components of the optimal model was established, and structural breaks in the trends related to external events were identified and captured by adding intervention variables to the appropriate components of the model. As a final step, for each country the optimally modelled developments were projected into the future, thus yielding forecasts for the number of fatalities up to and including 2020. © 2014 Elsevier Ltd. All rights reserved.\n\n1. Introduction The temporal evolution of the number of accidents and victims (fatalities, severely injured, injured) is a major topic of interest in many road safety studies (see, e.g., COST 329, 2004; Lassarre et al., 2012; Commandeur et al., 2013). These quantities are counted on a monthly or yearly basis in all European countries. Both basic and more sophisticated statistical models have been proposed to capture these evolutions. Many models assume only a (log) linear\n\nfunction of time for the modelling of the number of fatalities, for instance some of the models discussed in Elvik (2010). Accidents and their consequences in terms of victims are occurrences of failures in the road transportation system. Each time a road user makes a trip, he or she is exposed to harm with a certain probability of being involved in an accident or being killed. The number of traffic fatalities in a certain period is obviously dependent on the total exposure resulting from the amount of traffic in the traffic system in that same period. The amount of exposure determines the scale of the road safety problem and is therefore an essential factor in the assessment of road safety. Many models include some measure of traffic volume as an approximation of exposure. Frequently (e.g., Oppe, 1989, 1991) the total number of\n\n328\n\nE. Dupont et al. / Accident Analysis and Prevention 71 (2014) 327–336\n\nmotor vehicle kilometres travelled is used as a measure of traffic volume, which is only an approximation of exposure (because it is based on survey data, for example). It is then used to calculate the empirical risk as the number of fatalities divided by the number of kilometres travelled. This paper tries to overcome the limitations of previous models by considering three issues. First, it is acknowledged that the actual exposure is latent and can only be approximated by whichever measure of traffic volume is chosen. Consequent on the assumption of latent exposure, the risk derived analogous to that reported by Oppe (1989, 1991) now becomes a latent risk by definition. Second, latent exposure and the latent risk are simultaneously used for the modelling of the number of fatalities instead of in two separate steps as was done in Oppe (1989, 1991). When seeking to understand the evolution in the number of fatalities or accidents this approach makes it easier to decide whether a change is to be attributed to a change in exposure or to a change in the risk road users are exposed to. Finally, as indicated in Elvik (2010), the trends underlying the developments of the number of traffic fatalities need not be stable over time, with the consequence that modelling them as stable may yield poor in-sample predictions as well as poor forecasting results. Another aspect paramount to the analysis of data spanning an extended period of time, therefore, is that relations need not be invariant over the whole observed period. One way to handle such changes is to allow (model) parameters to be time varying as well, thus yielding improved in-sample predictions and forecasting results. The model described and applied in this paper – called the Latent Risk Time series model (Bijleveld et al., 2008; Bijleveld, 2008; COST 329, 2004) – is designed to accommodate all three of the above-mentioned properties. The aim of the analyses presented in this paper is to obtain forecasts for the number of traffic fatalities in each of the European countries in 2020 in a similar way by means of the structural time series approach, using comparable data as much as possible. As the purpose is to provide some comparable robust forecasts to help policy makers develop long-term targets and strategies for successful road safety policies, the models are focused to the analysis of two basic components: exposure and risk, and the introduction of idiosyncratic explanatory factors has been avoided whenever possible. In total, the results for some 30 countries are presented in this paper that are based on the work performed for the EC FP7 project DaCoTA (see Martensen and Dupont, 2010; Dupont and Martensen, 2012; Lassarre et al., 2012). The use of the homogeneous modelling technique allows to compare the past and future developments of the various countries and to address the following important questions: • Has there been a continuous, smooth development of road safety or were there abrupt changes in these developments? • If there were changes, can these be attributed to changes in the risk, or rather to changes in exposure? • Where does the past development tend to (if continued)? This last issue is particularly important for the setting of realistic road safety targets by policy makers. The European Commission has set the target to halve the number of road deaths in 2020 as compared to 2010. 2. Methodology One of the most important outcomes of road safety – quantified as the number of fatalities – is a joint function of the “level of dangerousness” of the traffic system, or road risk, and of the extent in which road users are confronted with this risk, here defined\n\nas the exposure to risk. This framework, where the fatality trend is decomposed into a risk and exposure trend, was made popular by Oppe (1989, 1991). This decomposition implies that two series of observations have to be analysed in parallel in order to model the development of road safety: one for the road safety indicator, the other for the exposure indicator. In the models presented here, the number of fatalities is the road safety indicator. The indicator for exposure is related to traffic volume and either the number of vehicle kilometres travelled or the size of the vehicle fleet can be used, depending on the availability of mobility data in the different European countries. The assumption that the development in traffic safety is the product of the respective developments in exposure and risk can be summarised as follows: Vehicle kilometres = Exposure Fatalities = Exposure × Risk\n\n(1)\n\nExcept for the time dependent specification, these two equations define the Latent Risk Time series model (LRT). In the LRT model both traffic volume and fatalities are treated as dependent variables. Traffic volume is modelled as a measure of “exposure” which can be subject to error. The number of fatalities, on the other hand, is defined as the product of “exposure” and “risk” and is also subject to random variation. Traffic volume and the number of fatalities are considered to be the manifest counterparts of “exposure” and “exposure times risk”, respectively, where “exposure” and “risk” are treated as latent (i.e., unobserved) variables. By taking the logarithm of the two equations in (1) (thus turning the multiplicative model into an additive one), and adding an error term (also known as a disturbance term) to the latent variables, we obtain: log(Traffic Volume) = log(Exposure) + error(Exposure) log(Fatalities) = log(Exposure) + log(Risk) + error(Fatalities)\n\n(2)\n\nThis implies that the disturbances in the original model formulation (1) should also be considered a multiplicative variable. This may seem a questionable assumption. One should note, however, that the additive Gaussian noise model (with constant variance) might not be appropriate for fatality count data, and possibly even not for traffic volume data. The implicit assumption of multiplicative errors is actually quite commonly applied, if only for practical reasons, as it substantially simplifies modelling. Because the equations in (2) define the way in which the latent variables exposure and risk can be inferred from the observations, they are called the measurement equations. When observed over time, these equations can therefore be interpreted as a decomposition of an observed time series (e.g., log(Traffic Volumet )) into a trend, which is the latent variable log(Exposuret ), and an error term, which is then also known as an irregular component (the error term error(Exposuret )). As can be seen in (2), the log(Exposure) is present in both measurement equations. There is a trend in this bivariate process, which depends in both equations on the exposure plus a specific trend related to the risk for the number of fatalities. In order to specify the dynamics of the model, two linear state equations are introduced for each of the latent variables log(Exposure) and log(Risk) in addition to the measurement equations in (2). One of these state equations is called the level equation, and the other the slope equation. The equations are linear, and define that the slope component at a certain time point is equal to the slope component at the previous time point plus some additive random disturbance, while the level at a certain time point is equal to the level at the previous time point plus the slope at the previous time point plus some additive random disturbance. In the absence of any random disturbance, this means that the level follows a straight\n\nE. Dupont et al. / Accident Analysis and Prevention 71 (2014) 327–336\n\nline, and the slope is constant. Introducing a positive random disturbance in the slope component means that the level component will start to increase faster (the slope will become steeper), and continue to do so until a new random disturbance changes the slope. Likewise, introducing a positive random disturbance in the level component means that the level component will stay at a higher level until a new random disturbance changes it. Both disturbances have a long-term effect, as opposed to the disturbances in the measurement equations, which have a short-term effect. This specification is identical to the specification of the local linear trend model (see Harvey, 1989; Commandeur and Koopman, 2007; Durbin and Koopman, 2012).1 We can now provide the complete formulation of the LRT model as it was presented in Bijleveld et al. (2008): log(Traffic Volumet ) = level(log(Exposuret )) + εet level(log(Exposuret+1 ) = level(log(Exposuret ) e + slope(log(Exposuret )) + \u0003t=1 e slope(log(Exposuret+1 )) = slope(log(Exposuret )) + \u0004t+1\n\n(3) f\n\nlog(Fatalitiest ) = level(log(Exposuret )) + level(log (Risk)t ) + εt f\n\nlevel(log(Riskt+1 )) = level(log(Riskt )) + sloope(log (Risk)t ) + \u0003t slope(log(Riskt+1 )) =\n\nf slope(log(Riskt )) + \u0004t+1\n\nwhere\n\n– εet is the irregular component (disturbance term) for the measurement equation of traffic volume, f – εt is the irregular component (disturbance term) for the measurement equation of the fatalities, e – \u0003t+1 is the disturbance term for the state equation of the level of the latent exposure, r – \u0003t+1 is the disturbance term for the state equation of the level of the latent risk, e – \u0004t+1 is the disturbance term for the state equation of the slope of the latent exposure, r is the disturbance term for the state equation of the slope of – \u0004t+1 the latent risk.\n\nThese terms constitute the random components of the model. Because the two dependent variables are both in logarithms, the two slope components in (3) can be interpreted as a yearly rate of change.\n\n329\n\nThe covariance matrix of the disturbances specifies the dynamics in and between the components of the models due to their randomness:\n\nA\n\n0\n\n0\n\n˝ = ⎣0\n\nB\n\n0⎦\n\n0\n\n0\n\nC\n\nwith\u0006\n\n\u0006\n\nA= \u0005\u00032e\n\n(4)\n\n\u0005ε2e cov(εe , εf )\n\ncov(\u0003 e , \u0003 r )\n\ncov(εe , εf ) \u0005 2f\n\ncov(\u0003 e , \u0003 r ) \u0005\u00032r\n\n\u0007\n\n,\n\nand\n\nC=\n\nB=\n\n\u0006\n\n\u0005\u00042e cov(\u0004 e , \u0004 r )\n\ncov(\u0004 e , \u0004 r ) \u0005\u00042r\n\n\u0007\n\nThe values of the variances of the disturbances on the diagonal of matrices B and C determine whether the corresponding level and slope components are stochastic or deterministic. If a disturbance variance is zero then the corresponding component is deterministic and fixed: in this case the slope component does not change over time. If the value of the variance is larger than zero, on the other hand, then the corresponding component is stochastic, and its value therefore changes over time. The tests conducted in order to determine whether components should be considered as deterministic or stochastic are described in the “Model selection” section below. 3. Investigation of correlations between components As indicated in (4), some of the random components in the model may be correlated, such as the disturbances of the slope components. When such a correlation happens to be (close to) plus or minus one, then we have the special situation that the level and/or slope components are – as is said – common. When unobserved components are common this means that the changes or “shocks” driving the dynamics of the (two or multiple) time series are perfectly linearly related. Stated differently, the level and/or slope components then change in the same way at the same points in time, and the corresponding time series therefore display common behaviour. The identification of common factors is important because it allows the improvement of models to make them more efficient, but also because factor components are informative in themselves of the dynamics governing the evolution of the trends considered. To explore this structure, a bivariate SUTSE (Seemingly Unrelated Time Series Equations) model (Harvey, 1989) of the traffic volume and the number of fatalities with a complete covariance structure of the disturbances of the levels, slopes, and irregulars can be applied: log(Traffic Volumet ) = level(log(Traffic Volumet )) + εt e level(log(Traffic Volumet+1 )) = level(log(Traffic Volumet )) e +slope(log(Traffic Volumet )) + \u0003t+1 slope(log(Traffic Volumet+1 )) = slope(log(Traffic Volumet )) + \u0004t+1 e (5)\n\n1\n\nARMA and ARIMA models or their extensions are frequently applied analysis techniques for the forecasting of fatality numbers. When it comes to linear Gaussian time series models, there are many equivalencies between the ARIMA and the structural time series framework. It can be shown, for example, that the local level model and the ARIMA (0 1 1) model yield identical forecasts; the same applies to the local linear trend model and the ARIMA (0 2 2) model. For a comprehensive overview of the equivalencies between ARIMA and structural time series models we refer to Appendix 1 in Harvey (1989). The reasons that we prefer the structural time series approach is that these models do not require the time series to be stationary, and that missing observations and multivariate time series are easily handled in the latter framework while this is relatively difficult in a pure ARIMA modelling context. Moreover, the structural time series framework yields interpretable components such as trends, cycles, and seasonal patterns, which moreover can be inspected for additional model validation; this is not the case in ARIMA models.\n\nlog(Fatalitiest ) = level(log(Fatalitiest )) + εt f level(log(Fatalitiest+1 )) = level(log(Fatalitiest )) + slope(log(Fatalitiest )) + \u0003t+1\n\nf\n\nslope(log(Fatalitiest+1 )) = slope(log(Fatalitiest )) + \u0004t+1 f\n\nNote that SUTSE model (5) is different from LRT model (3) and (4) because it no longer contains a latent risk variable. The disturbances of the unobserved level components of the fatalities and traffic volume can be correlated and the disturbances of the unobserved slope components of the fatalities and traffic volume can be correlated as well. There is an intimate relation between such common component models and what is known in the time series literature as cointegration.\n\n330\n\nE. Dupont et al. / Accident Analysis and Prevention 71 (2014) 327–336\n\nTable 1 Rules for the use of a LRT model depending on the correlation between the slope components of the bivariate SUTSE model. Type of correlation between the slope disturbances\n\nNo correlation (0)\n\nFull correlation (1)\n\nMedium correlation (0.1–0.9)\n\nConsequences for the model\n\nIndependence between fatalities and exposure\n\nStrong dependency: common components (same stochastic slope), cointegration Long-term linear relationship\n\nWeak dependency\n\nE(fat.|exp.) = E(fat.)\n\nlog(Fatt ) = b level(log(Expt )) + a + ct + εt\n\nUnivariate LLT\n\nBivariate LRT with deterministic risk trend\n\nModel\n\nWhen the correlation is negligible, there is no relationship between fatalities and exposure, meaning that knowledge of exposure does not bring any additional information to predict the number of fatalities. In that case, it can be proven that the trend for the exposure can be ignored altogether and that a univariate Local Linear Trend model (LLT) applied to the fatalities yields very similar results. When the correlation between the slope disturbances is equal to 1 and the level components are deterministic, both time series share the same stochastic slope and are called trend stationary with a deterministic linear trend for the risk. In that case, we obtain a LRT model by constraining the b coefficient for exposure to 1 (and modifying the risk component accordingly). Otherwise, there is a weak correlation between the two time series, and a LRT model provides a solution through the estimation of a covariance between the slope disturbances of risk and exposure. A systematic overview of all the possibilities is given in Table 1. 4. Model selection Irrespective of whether the LRT model (3) and (4) or the SUTSE model (5) is applied, various types of models can be obtained depending on whether the unobserved components – the level and the slope components – are treated stochastically or deterministically, i.e., as random or as fixed. In order to select the best model for the country data at hand, the analysis was started with a full model where all the components were treated stochastically. Then, for each disturbance variance the log-likelihood of the full model was compared with the log-likelihood of the model where the disturbance variance had been fixed on zero, and a likelihood ratio test was used to decide whether to treat the component stochastically or deterministically. When a stochastically treated component was thus found to yield a better model, it was next also explicitly tested whether its covariance term with the other corresponding component in the model (which would be either a level or a slope) should be fixed on zero or not. The software for all the analyses presented in this paper was written in R version 2, with the dlm package of Petris et al. (2009) as its central routine for filtering and smoothing. 5. Interventions Interventions are implemented using variables that model the effect of a known particular event on the level or on the slope of the time series (e.g., the introduction of a law, the beginning of a crisis, a change in counting methods, etc.). Interventions that are assumed to have an immediate and permanent effect are coded 0 for all time points prior to the event and 1 for the time point of the event and those following. These variables allow testing whether a significant change indeed took place at the assumed time of the intervention. Several types of interventions can be defined depending on the particular model equation they are inserted into. The intervention is included into the measurement equation when it is suspected that some change in the series reflects a change in the way it has been\n\nBivariate LRT with stochastic risk trend\n\nmeasured and not a change in the phenomenon itself. An example is the change in definition where fatalities were first defined as “victims who died within 24 h after the accident” and later defined as “victims who died within 30 days after the accident”. An intervention is included into the level equation if it is thought to have caused a permanent reduction (or increase) in either the fatality risk (e.g., the seat-belt law) or in the exposure (e.g., the introduction of taxes, an economic crisis). A level intervention considered here takes the form of a step: the fatality risk, for example, increases or decreases at the moment of the intervention and it remains at that level afterwards. An intervention is included into the slope equation when something is suspected to have caused a change in direction – or steepness – of the development of either fatality risk or exposure. This could, for example, be an increased commitment to road safety improvement in a country, due to which the fatality risk decreases at a faster rate than before. The selection of “candidates for interventions” should be based on actual knowledge about past developments (for example, the moments at which laws have been introduced or the method for the estimation of traffic volume has been adapted), as well as on the results of the analyses of the data (large, visible changes in the series). 6. Forecasts Once an appropriate model is identified, forecasts are obtained by projecting the trends that are observed in the past into the future. This does not necessarily mean that the forecasts will correctly predict what is going to happen. But when assessing the quality of the model, we can check the accuracy of the forecasts by forecasting a certain number of years (5–10) and then compare these forecasts with the observations. The implementation of interventions in the model at time points where the series displays extreme values or changes reduces the error variance and consequently the confidence interval around the forecasts. When one knows that a change occurred because of a particular event, one can also be confident that the change in the value of the relevant component is no part of the random variations in the series, and that such changes are thus unlikely to happen again in the future. However, when the reasons for the changes in the past are not really understood, one may expect that similar changes can happen in the future as well. In the latter case, correcting the models for past unexplained disturbances by introducing exclusively data driven interventions artificially reduces the confidence intervals for the forecasts; the implementation of such exclusively data driven interventions in the models should therefore be avoided. To summarise, time series models give us an estimate of the future development, under the assumption that the model is adequate and that the past development is continued. Moreover, they quantify the uncertainty of the forecasts: the weaker the predictive performance in the past, the larger the uncertainty in the forecasts will be as reflected in the corresponding confidence intervals.\n\nE. Dupont et al. / Accident Analysis and Prevention 71 (2014) 327–336\n\n331\n\nTable 2 Type of exposure indicator selected and length of the data series for the different countries, along with the types of correlations identified between the development of the exposure and fatality series. Unless otherwise specified, the exposure and fatality series are of identical length. Exposure indicator Vehicle kilometres 20 countries\n\nVehicle fleet seven countries\n\nFuel consumption one country\n\nNone available\n\nAustria (1990–2010) Belgium (1970–2010) Czech Republic (1990–2010, exp.: 1995–2010)) Denmark (1980–2010) Finland (1975–2010) France (1957–2010) Germany (1991–2010) Hungary (1993–2010) Iceland (1975–2010, exp.: 1980–2010) Ireland (1970–2010) Italy (1980–2010) Norway (1973–2009) Poland (1975–2010,exp.: 1996–2008) Romania (1990–2010) Slovenia (1970–2010) Spain (1961–2009) Sweden (1970–2009) Switzerland (1975–2010) The Netherlands (1950–2010) UK (1983–2010) 5 “common slopes” 4 correlated series\n\nBulgaria (2001–2010) Estonia (1991–2010, exp.: 1997–2008) Greece (1960–2010) Latvia (1975–2010, exp.: 1996–2009) Luxembourg (1975–2010) Portugal (1979–2008) Slovakia (1990–2010, exp.: 1991–2002)\n\nCyprus (1991–2010)\n\nLithuania (2001–2010) Malta (1990–2010)\n\n2 “common slopes”\n\nNo correlation\n\n7. Data Road safety fatalities – although by no means the only interesting measure – are the key measurement to analyse and compare the development of road safety across countries, because they are less susceptible to underreporting than other measures. The yearly number of road traffic fatalities in the different European countries is available in the CARE database. CARE includes harmonised fatality data, which conform to the European definition of fatalities within 30 days from the accident. The EU-15 Member States have fatality data that span the entire period 1991–2008. For countries that became a member of the European Union later on, however, data availability is limited to a smaller number of years (e.g., from 2005 onwards for Estonia and Slovakia, from 2003 onwards for Hungary, from 2000 onwards for Slovenia etc.). Various exposure indicators were investigated, including the number of vehicle kilometres travelled, the vehicle fleet, the fuel consumption and the road length. The analyses could not be carried out on the basis of vehicle kilometres data for all European countries and in these cases alternative exposure indicators have been used. In 20 countries, the most preferable exposure indicator was available, which is vehicle kilometres. In seven countries the size of the vehicle fleet was the only available exposure indicator. Fuel consumption has been used as exposure indicator in the case of Cyprus. Finally, for two countries (Lithuania and Malta) no exposure indicator was available at all.\n\nmodel (5).2 For each country, the length of the series of observations available and included in the analysis is also mentioned. It is clear from this table that a correlation was identified more often when vehicle kilometres could be used as exposure indicator than for other types of exposure indicators. In all instances where a non-zero correlation was established between the two series, this correlation applied to the slope components (not to the level components) and was positive. Given that the values of the slope component for the exposure tend to be positive (indicating that exposure is always increasing) while those for the risk are most often negative (indicating that the risk decreases over time), positively correlated slope components imply that the decrease in the annual fatality numbers slows down when the increase in the annual number of vehicle kilometres becomes stronger. Common slopes (i.e., a correlation close to 1) were observed in five of the nine countries where a relationship could be identified on the basis of vehicle kilometres (i.e., Denmark, Finland, France, the Netherlands, and the UK). Figs. 1–4 contain the results of the SUTSE model for the Netherlands. The common slopes identified for this country are illustrated in Figs. 2 and 4. Common slopes were also observed for Portugal and Estonia, where vehicle fleet was used as the exposure indicator. The identification of a satisfactory relation between the exposure indicator and fatality series was one condition that determined whether the LRT model or the LLT model was used to capture the past developments in the annual fatality numbers of a country, and to project these developments into the future. In all eleven countries where a relationship between the exposure indicator and the fatality series could be demonstrated with the SUTSE model (5),\n\n8. Results A relationship between the exposure and fatality series could not be identified in all countries. Table 2 summarises the different types of exposure indicators that have been used for each country, and also indicates the countries for which correlated series or even common slopes could be established on the basis of the SUTSE\n\n2 The correlations range in the interval [0.1, 0.9]. This interval is explained by the fact that the correlations between the slope and level disturbances of the two series have been tested in two steps: (1) a test of whether H0 : \u0006(\u0003e ,\u0003f ) = 0, \u0006(\u0004 e ,\u0004 f ) = \b \b 1. This testing 0 and H0 : \u0006(εe ,εf ) = 0 can be rejected and (2) a test of whether \u0006 ∼ = strategy consequently only informs whether correlation coefficient(s) are significantly different from 0 and from 1, without providing information of its (their) exact estimated value.\n\n332\n\nE. Dupont et al. / Accident Analysis and Prevention 71 (2014) 327–336\n\nFig. 1–4. Developments of the state components for the exposure (upper graphs) and the fatalities (lower graphs), as estimated on the basis of the SUTSE model for The Netherlands. The trend (level) developments are shown in the left-hand graphs, the slope developments in the right-hand graphs (both in the anti-logs, the scale 1.05 for the slope means an increase of the series with 100*ln(1.05) = 5% per year), including their 95% confidence intervals. Note the similarity between the slope components which have a correlation of 1 (and are therefore common).\n\nthe development of the annual fatality numbers was modelled and defined as the result of the combined developments of the risk and of the exposure according to the LRT model (3). When no significant relationship between the exposure indicator and fatality series could be detected, on the other hand, the forecasting accuracy for the fatalities obtained with the bivariate LRT model was compared with the forecasting accuracy of the univariate LLT model applied to the fatalities series only, thereby disregarding the development of the exposure indicator. For 5 of the 19 countries for which no significant relation between the exposure indicator and the fatality series was established, the LRT model still resulted in better forecasts than the LLT model and\n\nthese five countries were therefore analysed with the LRT model. The remaining fourteen countries with no significant correlation between exposure indicator and fatalities were all analysed with the LLT model. In addition, as already mentioned in the section on Model selection, various types of LRT and LLT models can be obtained depending on whether the unobserved components – the level and the slope components – are treated stochastically or deterministically, i.e., as random or as fixed. Table 3 provides an overview of the temporal structure of the optimal models that were identified for the 16 European countries to which the LRT model was applied. Whenever common – or highly correlated – slopes were observed\n\nE. Dupont et al. / Accident Analysis and Prevention 71 (2014) 327–336\n\n333\n\nTable 3 Overview of the optimal LRT model types for the different countries. Most frequently selected models\n\nOther models\n\nExposure trend Level fixed, slope random\n\nLevel fixed, slope random\n\nLevel fixed, slope random\n\nRisk trend Level random, slope fixed\n\nLevel fixed, slope fixed\n\nLevel fixed, slope random\n\nAustria (no component fixed)\n\nCyprus\n\nUK Italy\n\nFinland (only slope risk fixed) Slovenia (only level exposure fixed)\n\nDenmark France The Netherlands Spain Switzerland Norway Portugal Estonia Belgium Germany ⇒10/16 countries\n\nfor the fatality and exposure indicator series, the slope component of the risk was treated deterministically (i.e., as fixed). As the table indicates the most common type of LRT model is one where the slope component for exposure is random and its level is fixed, and where the level component of the risk is random and its slope is fixed. A typical example is given in Fig. 5-8, which contains the results of the LRT model obtained for the Netherlands. For the exposure, the slope changes reflect the fact that the rate of change is decreasing over the years, i.e., the exposure continuously increases, but not as fast as it did in the past. In contrast, the risk is characterised by ups and downs (due to random variations in the level component), but–as is illustrated in the bottom graph on the right of Fig. 8 – the general direction of the year-to-year changes in the number of fatalities per unit of exposure is constant. For the\n\nTable 4 Overview of the optimal univariate LLT model types identified for the different countries. Fatality trend Level random, slope fixed\n\nLevel and slope fixed\n\nLevel fixed\n\nBulgaria Greece Luxembourg Lithuania Ireland Poland Sweden Latvia Slovakia ⇒9/14 countries\n\nHungary Iceland Malta\n\nCzech Republic Romania\n\nTable 5 Overview of the intervention variables identified for the different countries, including their type and cause. Country\n\nYear\n\nType of intervention\n\nReason\n\nGreece\n\n1986 1991 1996 2002 2008 1991 1999 2009\n\nFatalities level Fatalities level Fatalities slope Fatalities level Fatalities level Exposure level Exposure level Risk level\n\n1989 1974 1974 2004 1989 1990 2008 1983 2003 1975 1982 1984 1989 1993 1994 2004 2007\n\nFatalities level Exposure level Risk level Risk level Fatalities level Exposure level Fatalities slope Risk level Exposure level Fatalities level Risk slope Exposure slope Risk slope Level risk Risk slope Risk slope Exposure slope\n\n2008\n\nRisk level\n\n1993 1991–1993 1991–1993 2008–2012 2008–2012\n\nExposure level Exposure slope Risk slope Exposure slope Risk slope\n\nEconomic recession Change in the vehicle fleet (car exchange scheme) 30-days definition introduced for fatalities Increase of motorway length by 19% Economic recession, large set of road safety measures introduced Fatalities registration form adapted 30-days definition introduced Change in the vehicle fleet (trailers and semitrailers\n\n## Latent risk and trend models for the evolution of annual fatality numbers in 30 European countries.\n\nIn this paper a unified methodology is presented for the modelling of the evolution of road safety in 30 European countries. For each country, annual ..." ]
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https://lists.defectivebydesign.org/archive/html/help-gsl/2021-08/msg00003.html
[ "help-gsl\n[Top][All Lists]\n\n## Question of polynomial multiplication.\n\n From: 김현성 Subject: Question of polynomial multiplication. Date: Tue, 17 Aug 2021 16:03:02 +0000\n\n```Dear GSL developers\n\nHello,\n\nWhile I am reading polynomial section of GSL manual, I realized that GSL does\nnot provide the function calculates two polynomial multiplication.\nI wonder it is not constructed yet or it had been considered but does not fit\nwith GSL design.\n\nI found 3 basic algorithms for polynomial multiplication.\n\nl Direct coefficient vector convolution.\n\nl Dived and Conquer approach algorithm.\n\nl FFT for fast convolution calculation. <- In my opinion, this algorithm will\nnot be able in GSL because of interdependence\nI think above two will be useful, if they are provided in GSL. How about you?\n\nThanks\n\nHyean Sung\n```" ]
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https://deepai.org/publication/a-5-2-approximation-algorithm-for-coloring-rooted-subtrees-of-a-degree-3-tree
[ "DeepAI\n\n# A 5/2-Approximation Algorithm for Coloring Rooted Subtrees of a Degree 3 Tree\n\nA rooted tree R⃗ is a rooted subtree of a tree T if the tree obtained by replacing the directed edges of R⃗ by undirected edges is a subtree of T. We study the problem of assigning minimum number of colors to a given set of rooted subtrees R of a given tree T such that if any two rooted subtrees share a directed edge, then they are assigned different colors. The problem is NP hard even in the case when the degree of T is restricted to 3. We present a 5/2-approximation algorithm for this problem. The motivation for studying this problem stems from the problem of assigning wavelengths to multicast traffic requests in all-optical WDM tree networks.\n\n07/09/2019\n\n### PTAS and Exact Algorithms for r-Gathering Problems on Tree\n\nr-gathering problem is a variant of facility location problems. In this ...\n04/09/2021\n\n### On the minimum spanning tree problem in imprecise set-up\n\nIn this article, we study the Euclidean minimum spanning tree problem in...\n03/14/2023\n\n### Coloring and Recognizing Directed Interval Graphs\n\nA mixed interval graph is an interval graph that has, for every pair of ...\n02/11/2019\n\n### Geometric Multicut\n\nWe study the following separation problem: Given a collection of colored...\n02/01/2023\n\n### Parameterized Algorithms for Colored Clustering\n\nIn the Colored Clustering problem, one is asked to cluster edge-colored ...\n06/18/2019\n\n### Rooting for phylogenetic networks\n\nThis paper studies the relationship between undirected (unrooted) and di...\n11/23/2018" ]
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https://zbmath.org/?q=an:0819.35051
[ "# zbMATH — the first resource for mathematics\n\nOn a class of nonlinear elliptic equations. (English) Zbl 0819.35051\nBojarski, Bogdan (ed.) et al., Partial differential equations. Part 1. The 36th semester held at the Stefan Banach International Mathematical Center in Warsaw, Poland, from September 17 to December 17, 1990. Warsaw: Polish Academy of Sciences, Inst. of Mathematics. Banach Cent. Publ. 27, Part 1, 75-80 (1992).\nLet $$\\Omega$$ be a bounded domain of $$\\mathbb{R}^ n$$ with boundary $$\\Gamma$$, $$n\\geq 1$$. The goal of this note is to summarize results regarding existence and number of solutions of the equation $\\Delta\\varphi- |\\nabla \\varphi|^ q+ \\lambda \\varphi^ p=0 \\quad \\text{in }\\Omega, \\qquad \\varphi>0 \\quad \\text{in }\\Omega, \\qquad \\varphi=0 \\quad \\text{on }\\Gamma, \\tag{1}$ $$\\lambda>0$$, $$p,q>1$$.\nFor the entire collection see [Zbl 0771.00021].\n\n##### MSC:\n 35J65 Nonlinear boundary value problems for linear elliptic equations\n##### Keywords:\nmultiplicity of solutions" ]
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http://algorist.com/algowiki/index.php?title=Data-structures-TADM2E&oldid=83
[ "(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\n# Data Structures\n\nStacks, Queues, and Lists\n\n<br>3-1. A common problem for compilers and text editors is determining whether the parentheses in a string are balanced and properly nested. For example, the string $((())())()$ contains properly nested pairs of parentheses, which the strings $)()($ and $())$ do not. Give an algorithm that returns true if a string contains properly nested and balanced parentheses, and false if otherwise. For full credit, identify the position of the first offending parenthesis if the string is not properly nested and balanced.\n\n<br>3-2. Write a program to reverse the direction of a given singly-linked list. In other words, after the reversal all pointers should now point backwards. Your algorithm should take linear time.\n\n<br>3-3. We have seen how dynamic arrays enable arrays to grow while still achieving constant-time amortized performance. This problem concerns extending dynamic arrays to let them both grow and shrink on demand.\n\n1. Consider an underflow strategy that cuts the array size in half whenever the array falls below half full. Give an example sequence of insertions and deletions where this strategy gives a bad amortized cost.\n2. Then, give a better underflow strategy than that suggested above, one that achieves constant amortized cost per deletion.\n\nTrees and Other Dictionary Structures\n\n<br>3-4. Design a dictionary data structure in which search, insertion, and deletion can all be processed in $O(1)$ time in the worst case. You may assume the set elements are integers drawn from a finite set ${1,2,..,n}$, and initialization can take $O(n)$ time.\n\n<br>3-5. Find the overhead fraction (the ratio of data space over total space) for each of the following binary tree implementations on $n$ nodes:\n\n1. All nodes store data, two child pointers, and a parent pointer. The data field requires four bytes and each pointer requires four bytes.\n2. Only leaf nodes store data; internal nodes store two child pointers. The data field requires four bytes and each pointer requires two bytes.\n\n<br>3-6. Describe how to modify any balanced tree data structure such that search, insert, delete, minimum, and maximum still take $O(\\log n)$ time each, but successor and predecessor now take $O(1)$ time each. Which operations have to be modified to support this?\n\n<br>3-7. Suppose you have access to a balanced dictionary data structure, which supports each of the operations search, insert, delete, minimum, maximum, successor, and predecessor in $O(\\log n)$ time. Explain how to modify the insert and delete operations so they still take $O( \\log n)$ but now minimum and maximum take $O(1)$ time. (Hint: think in terms of using the abstract dictionary operations, instead of mucking about with pointers and the like.)\n\n<br>3-8. Design a data structure to support the following operations:\n\n1. insert(x,T) -- Insert item $x$ into the set $T$.\n2. delete(k,T) -- Delete the $k$th smallest element from $T$.\n3. member(x,T) -- Return true iff $x \\in T$.\n\nAll operations must take $O(\\log n)$ time on an $n$-element set.\n\n<br>3-9. A concatenate operation takes two sets $S_1$ and $S_2$, where every key in $S_1$ is smaller than any key in $S_2$, and merges them together. Give an algorithm to concatenate two binary search trees into one binary search tree. The worst-case running time should be $O(h)$, where $h$ is the maximal height of the two trees.\n\nApplications of Tree Structures\n\n<br>3-10. In the bin-packing problem, we are given $n$ metal objects, each weighing between zero and one kilogram. Our goal is to find the smallest number of bins that will hold the $n$ objects, with each bin holding one kilogram at most.\n\n1. The best-fit heuristic for bin packing is as follows. Consider the objects in the order in which they are given. For each object, place it into the partially filled bin with the smallest amount of extra room after the object is inserted.. If no such bin exists, start a new bin. Design an algorithm that implements the best-fit heuristic (taking as input the $n$ weights $w_1,w_2,...,w_n$ and outputting the number of bins used) in $O(n \\log n)$ time.\n2. Repeat the above using the worst-fit heuristic, where we put the next object in the partially filled bin with the largest amount of extra room after the object is inserted.\n\n<br>3-11. Suppose that we are given a sequence of $n$ values $x_1,x_2,...,x_n$ and seek to quickly answer repeated queries of the form: given $i$ and $j$, find the smallest value in $x_i,\\ldots,x_j$.\n\n1. Design a data structure that uses $O(n^2)$ space and answers queries in $O(1)$ time.\n2. Design a data structure that uses $O(n)$ space and answers queries in $O(\\log n)$ time. For partial credit, your data structure can use $O(n \\log n)$ space and have $O(\\log n)$ query time.\n\n<br>3-12. Suppose you are given an input set $S$ of $n$ numbers, and a black box that if given any sequence of real numbers and an integer $k$ instantly and correctly answers whether there is a subset of input sequence whose sum is exactly $k$. Show how to use the black box $O(n)$ times to find a subset of $S$ that adds up to $k$.\n\n<br>3-13. Let $A[1..n]$ be an array of real numbers. Design an algorithm to perform any sequence of the following operations:\n\n1. Add(i,y) -- Add the value $y$ to the $i$th number.\n2. Partial-sum(i) -- Return the sum of the first $i$ numbers, i.e.\n\n$\\sum_{j=1}^i A[j]$. There are no insertions or deletions; the only change is to the values of the numbers. Each operation should take $O(\\log n)$ steps. You may use one additional array of size $n$ as a work space.\n\n<br>3-14. Extend the data structure of the previous problem to support insertions and deletions. Each element now has both a key and a value. An element is accessed by its key. The addition operation is applied to the values, but the elements are accessed by its key. The Partial-sum operation is different.\n\n1. Add(k,y) -- Add the value $y$ to the item with key $k$.\n2. Insert(k,y) -- Insert a new item with key $k$ and value $y$.\n3. Delete(k) -- Delete the item with key $k$.\n4. Partial-sum(k) --\n\nReturn the sum of all the elements currently in the set whose key is less than $y$, i.e. $\\sum_{x_j<y} x_i$. The worst case running time should still be $O(n \\log n)$ for any sequence of $O(n)$ operations.\n\n<br>3-15. Design a data structure that allows one to search, insert, and delete an integer $X$ in $O(1)$ time (i.e., constant time, independent of the total number of integers stored). Assume that $1 \\leq X \\leq n$ and that there are $m+n$ units of space available, where $m$ is the maximum number of integers that can be in the table at any one time. (Hint: use two arrays $A[1..n]$ and $B[1..m]$.) You are not allowed to initialize either $A$ or $B$, as that would take $O(m)$ or $O(n)$ operations. This means the arrays are full of random garbage to begin with, so you must be very careful.\n\nImplementation Projects\n\n<br>3-16. Implement versions of several different dictionary data structures, such as linked lists, binary trees, balanced binary search trees, and hash tables. Conduct experiments to assess the relative performance of these data structures in a simple application that reads a large text file and reports exactly one instance of each word that appears within it. This application can be efficiently implemented by maintaining a dictionary of all distinct words that have appeared thus far in the text and inserting/reporting each word that is not found. Write a brief report with your conclusions.\n\n<br>3-17. A Caesar shift (see cryptography) is a very simple class of ciphers for secret messages. Unfortunately, they can be broken using statistical properties of English. Develop a program capable of decrypting Caesar shifts of sufficiently long texts.\n\nInterview Problems\n\n<br>3-18. What method would you use to look up a word in a dictionary?\n\n<br>3-19. Imagine you have a closet full of shirts. What can you do to organize your shirts for easy retrieval?\n\n<br>3-20. Write a function to find the middle node of a singly-linked list.\n\n<br>3-21. Write a function to compare whether two binary trees are identical. Identical trees have the same key value at each position and the same structure.\n\n<br>3-22. Write a program to convert a binary search tree into a linked list.\n\n<br>3-23. Implement an algorithm to reverse a linked list. Now do it without recursion.\n\n<br>3-24. What is the best data structure for maintaining URLs that have been visited by a Web crawler? Give an algorithm to test whether a given URL has already been visited, optimizing both space and time.\n\n<br>3-25. You are given a search string and a magazine. You seek to generate all the characters in search string by cutting them out from the magazine. Give an algorithm to efficiently determine whether the magazine contains all the letters in the search string.\n\n<br>3-26. Reverse the words in a sentence---i.e., My name is Chris becomes Chris is name My. Optimize for time and space.\n\n<br>3-27. Determine whether a linked list contains a loop as quickly as possible without using any extra storage. Also, identify the location of the loop.\n\n<br>3-28. You have an unordered array $X$ of $n$ integers. Find the array $M$ containing $n$ elements where $M_i$ is the product of all integers in $X$ except for $X_i$. You may not use division. You can use extra memory. (Hint: There are solutions faster than $O(n^2)$.)\n\n<br>3-29. Give an algorithm for finding an ordered word pair (e.g., New York) occurring with the greatest frequency in a given webpage. Which data structures would you use? Optimize both time and space." ]
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https://www.colorhexa.com/548673
[ "# #548673 Color Information\n\nIn a RGB color space, hex #548673 is composed of 32.9% red, 52.5% green and 45.1% blue. Whereas in a CMYK color space, it is composed of 37.3% cyan, 0% magenta, 14.2% yellow and 47.5% black. It has a hue angle of 157.2 degrees, a saturation of 22.9% and a lightness of 42.7%. #548673 color hex could be obtained by blending #a8ffe6 with #000d00. Closest websafe color is: #669966.\n\n• R 33\n• G 53\n• B 45\nRGB color chart\n• C 37\n• M 0\n• Y 14\n• K 47\nCMYK color chart\n\n#548673 color description : Mostly desaturated dark cyan - lime green.\n\n# #548673 Color Conversion\n\nThe hexadecimal color #548673 has RGB values of R:84, G:134, B:115 and CMYK values of C:0.37, M:0, Y:0.14, K:0.47. Its decimal value is 5539443.\n\nHex triplet RGB Decimal 548673 `#548673` 84, 134, 115 `rgb(84,134,115)` 32.9, 52.5, 45.1 `rgb(32.9%,52.5%,45.1%)` 37, 0, 14, 47 157.2°, 22.9, 42.7 `hsl(157.2,22.9%,42.7%)` 157.2°, 37.3, 52.5 669966 `#669966`\nCIE-LAB 52.031, -21.396, 4.933 15.275, 20.172, 19.307 0.279, 0.368, 20.172 52.031, 21.958, 167.016 52.031, -23.84, 10.005 44.913, -17.891, 5.952 01010100, 10000110, 01110011\n\n# Color Schemes with #548673\n\n• #548673\n``#548673` `rgb(84,134,115)``\n• #865467\n``#865467` `rgb(134,84,103)``\nComplementary Color\n• #54865a\n``#54865a` `rgb(84,134,90)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #548086\n``#548086` `rgb(84,128,134)``\nAnalogous Color\n• #865a54\n``#865a54` `rgb(134,90,84)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #865480\n``#865480` `rgb(134,84,128)``\nSplit Complementary Color\n• #867354\n``#867354` `rgb(134,115,84)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #735486\n``#735486` `rgb(115,84,134)``\n• #678654\n``#678654` `rgb(103,134,84)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #735486\n``#735486` `rgb(115,84,134)``\n• #865467\n``#865467` `rgb(134,84,103)``\n• #37574b\n``#37574b` `rgb(55,87,75)``\n• #406758\n``#406758` `rgb(64,103,88)``\n• #4a7666\n``#4a7666` `rgb(74,118,102)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #5e9680\n``#5e9680` `rgb(94,150,128)``\n• #6ba28d\n``#6ba28d` `rgb(107,162,141)``\n• #7bac99\n``#7bac99` `rgb(123,172,153)``\nMonochromatic Color\n\n# Alternatives to #548673\n\nBelow, you can see some colors close to #548673. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #548667\n``#548667` `rgb(84,134,103)``\n• #54866b\n``#54866b` `rgb(84,134,107)``\n• #54866f\n``#54866f` `rgb(84,134,111)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #548677\n``#548677` `rgb(84,134,119)``\n• #54867b\n``#54867b` `rgb(84,134,123)``\n• #548680\n``#548680` `rgb(84,134,128)``\nSimilar Colors\n\n# #548673 Preview\n\nThis text has a font color of #548673.\n\n``<span style=\"color:#548673;\">Text here</span>``\n#548673 background color\n\nThis paragraph has a background color of #548673.\n\n``<p style=\"background-color:#548673;\">Content here</p>``\n#548673 border color\n\nThis element has a border color of #548673.\n\n``<div style=\"border:1px solid #548673;\">Content here</div>``\nCSS codes\n``.text {color:#548673;}``\n``.background {background-color:#548673;}``\n``.border {border:1px solid #548673;}``\n\n# Shades and Tints of #548673\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010101 is the darkest color, while #f4f8f7 is the lightest one.\n\n• #010101\n``#010101` `rgb(1,1,1)``\n• #080d0c\n``#080d0c` `rgb(8,13,12)``\n• #101916\n``#101916` `rgb(16,25,22)``\n• #182620\n``#182620` `rgb(24,38,32)``\n• #1f322b\n``#1f322b` `rgb(31,50,43)``\n• #273e35\n``#273e35` `rgb(39,62,53)``\n• #2e4a3f\n``#2e4a3f` `rgb(46,74,63)``\n• #36564a\n``#36564a` `rgb(54,86,74)``\n• #3d6254\n``#3d6254` `rgb(61,98,84)``\n• #456e5e\n``#456e5e` `rgb(69,110,94)``\n• #4c7a69\n``#4c7a69` `rgb(76,122,105)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #5c927d\n``#5c927d` `rgb(92,146,125)``\n• #649e88\n``#649e88` `rgb(100,158,136)``\n• #70a591\n``#70a591` `rgb(112,165,145)``\n``#7cad9a` `rgb(124,173,154)``\n• #88b4a3\n``#88b4a3` `rgb(136,180,163)``\n``#94bcad` `rgb(148,188,173)``\n• #a0c3b6\n``#a0c3b6` `rgb(160,195,182)``\n• #accbbf\n``#accbbf` `rgb(172,203,191)``\n• #b8d3c8\n``#b8d3c8` `rgb(184,211,200)``\n``#c4dad2` `rgb(196,218,210)``\n• #d0e2db\n``#d0e2db` `rgb(208,226,219)``\n• #dce9e4\n``#dce9e4` `rgb(220,233,228)``\n• #e8f1ee\n``#e8f1ee` `rgb(232,241,238)``\n• #f4f8f7\n``#f4f8f7` `rgb(244,248,247)``\nTint Color Variation\n\n# Tones of #548673\n\nA tone is produced by adding gray to any pure hue. In this case, #65756f is the less saturated color, while #00da87 is the most saturated one.\n\n• #65756f\n``#65756f` `rgb(101,117,111)``\n• #5c7e71\n``#5c7e71` `rgb(92,126,113)``\n• #548673\n``#548673` `rgb(84,134,115)``\n• #4c8e75\n``#4c8e75` `rgb(76,142,117)``\n• #439777\n``#439777` `rgb(67,151,119)``\n• #3b9f79\n``#3b9f79` `rgb(59,159,121)``\n• #32a87b\n``#32a87b` `rgb(50,168,123)``\n• #2ab07d\n``#2ab07d` `rgb(42,176,125)``\n• #22b87f\n``#22b87f` `rgb(34,184,127)``\n• #19c181\n``#19c181` `rgb(25,193,129)``\n• #11c983\n``#11c983` `rgb(17,201,131)``\n• #09d185\n``#09d185` `rgb(9,209,133)``\n• #00da87\n``#00da87` `rgb(0,218,135)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #548673 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
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https://codereview.stackexchange.com/questions/175366/french-republican-date-conversion
[ "French Republican date conversion\n\nThis program converts a date given from the French Republican Calendar to the corresponding date in our standard Gregorian Calendar. I'm looking for some improvements that I should start making in my code as well as some tips on how to write less lengthy code.\n\nFor example, the RepubtoGregDate function is very long and I believe there is probably a more effective way of implementing it.\n\nI'm very new to Python and have no prior knowledge in programming so it would nice to have some suggestions in how to improve my code.\n\n#Python Assignment - French Republican to Gregorian Date Convertor\n#Coded by - -\nimport sys\n\ndef Main():\nStartup()\nwhile True:\ndayN, monthN, yearN, date = Date()\nStopProgram(date)\nValidDate = DateVal(yearN, monthN, dayN)\nValLoop(ValidDate)\nmonthW = MonthNtoW(monthN)\nyearRN = YearNtoRN(yearN)\nRepubDate = FullRepubDate(dayN, monthW, yearRN)\nLeap = LeapCheck(yearN)\nGrDate = RepubtoGregDate(yearN, monthN, dayN, Leap)\n\ndef Startup():\nprint(\"Republican to Gregorian Date Converter\\n\")\nprint(\"This program will convert a date from the French Republican Calendar to the equivalent date in the Gregorian Calendar.\")\nprint(\"To stop calculating dates and close the program, simply input: 0 0 0.\\n\")\n\n# Asks for a Republican Date and returns day, month and year.\ndef Date():\nprint(\"Input a Republican Date in the format: dd mm yy.\")\ndate = input(\"Please enter a Republican Date: \")\nprint(\"\")\ndate = date.split()\ndayN = int(date)\nmonthN = int(date)\nyearN = int(date)\nreturn(dayN, monthN, yearN, date)\n\n# Ends the program if the sentinal is matched.\ndef StopProgram(date):\nif date == ['0', '0', '0']:\nsys.exit(print(\"Program Closed.\"))\n\n# Validates the date, making sure it fits the Republican format and that\n# the date is between the calendars real world use: 1793 - 1806.\ndef DateVal(yearN, monthN, dayN):\nif yearN >= 1 and yearN <= 14:\nValidYear = True\nelse:\nValidYear = False\nif monthN >= 1 and monthN <= 12:\nValidMonth = True\nelse:\nValidMonth = False\nif dayN >= 1 and dayN <= 30:\nValidDay = True\nelse:\nValidDay = False\nif yearN < 2 and monthN < 1 and dayN < 14:\nInvalidDate = True\nelif yearN > 14 and monthN > 4 and dayN > 11:\nInvalidDate = True\nelse:\nInvalidDate = False\nif ValidYear == True and ValidMonth == True and ValidDay == True and InvalidDate == False:\nValidDate = True\nelse:\nValidDate = False\nreturn(ValidDate)\n\n# Repeats the Date() Function if the Date is invalid.\ndef ValLoop(ValidDate):\nwhile ValidDate == False:\nprint(\"The entered date was not a valid gregorian date, please try again.\")\ndayN, monthN, yearN = Date()\n\n# Finds the month number's corresponding month word.\ndef MonthNtoW(monthN):\nMonthWDict = {1 : \"Vendémiaire\", 2 : \"Brumaire\", 3 : \"Frimaire\", 4 : \"Nivôse\", 5 : \"Pluviôse\", 6 : \"Ventôse\", 7 : \"Germinal\", 8 : \"Floréal\", 9 : \"Prairial\", 10 : \"Messidor\", 11 : \"Thermidor\", 12 : \"Fructidor\"}\nmonthW = MonthWDict[monthN]\nreturn monthW\n\n# Finds the year number's corresponding year in roman numerals.\ndef YearNtoRN(yearN):\nYearRNDict = {1 : \"I\", 2 : \"II\", 3 : \"III\", 4 : \"IV\", 5 : \"V\", 6 : \"VI\", 7 : \"VII\", 8 : \"VIII\", 9 : \"IX\", 10 : \"X\", 11 : \"XI\", 12 : \"XII\", 13 : \"XIII\", 14 : \"XIV\"}\nyearRN = YearRNDict[yearN]\nreturn(yearRN)\n\n# Outputs the Republican date in its written format.\ndef FullRepubDate(dayN, monthW, yearRN):\ndayN = str(dayN)\nRepubDate = dayN + \" \" + monthW + \" an \" + yearRN\nprint(\"The French Republican Date is: \", RepubDate)\ndayN = int(dayN)\nreturn(RepubDate)\n\n# Checks if the year is a leap year.\ndef LeapCheck(yearN):\nif yearN == 3 or yearN == 7 or yearN == 11:\nLeap = True\nelse:\nLeap = False\nreturn(Leap)\n\n# Converts the Republican date to a Gregorian date.\n# It does this by converting the date to days in the year and then adds\n# that to the day the Republican Calendar started in The Gregorian calendar.\ndef RepubtoGregDate(yearN, monthN, dayN, Leap):\nmonthL = 30\nStartDate = 266\nStartYear = 1792\nYDays = 365\nLYDays = 366\n\nYear = StartYear + (yearN - 1)\nDay = ((monthN - 1) * monthL) + dayN\nif Leap == True and Day >= 60:\nDay += 1\n\nPH = StartDate + Day\nif Leap == True:\nif PH > LYDays:\nYear += 1\nDay = PH - LYDays\nelse:\nDay = PH\nif Leap == False:\nif PH > YDays:\nYear += 1\nDay = PH - YDays\nelse:\nDay = PH\n\nif Leap == False:\nif Day >= 1 and Day <= 31:\nGregMonth = \"January\"\nif Day >= 32 and Day <= 59:\nDay -= 31\nGregMonth = \"February\"\nif Day >= 60 and Day <= 90:\nDay -= 59\nGregMonth = \"March\"\nif Day >= 91 and Day <= 120:\nDay -= 90\nGregMonth = \"April\"\nif Day >= 121 and Day <= 151:\nDay -= 120\nGregMonth = \"May\"\nif Day >= 152 and Day <= 181:\nDay -= 151\nGregMonth = \"June\"\nif Day >= 182 and Day <= 212:\nDay -= 181\nGregMonth = \"July\"\nif Day >= 213 and Day <= 243:\nDay -= 212\nGregMonth = \"August\"\nif Day >= 244 and Day <= 273:\nDay -= 243\nGregMonth = \"September\"\nif Day >= 274 and Day <= 304:\nDay -= 273\nGregMonth = \"October\"\nif Day >= 305 and Day <= 334:\nDay -= 304\nGregMonth = \"November\"\nif Day >= 335 and Day <= 365:\nDay -= 334\nGregMonth = \"December\"\n\nif Leap == True:\nif Day >= 1 and Day <= 31:\nGregMonth = \"January\"\nif Day >= 32 and Day <= 60:\nDay -= 32\nGregMonth = \"February\"\nif Day >= 61 and Day <= 91:\nDay -= 60\nGregMonth = \"March\"\nif Day >= 92 and Day <= 121:\nDay -= 91\nGregMonth = \"April\"\nif Day >= 122 and Day <= 152:\nDay -= 121\nGregMonth = \"May\"\nif Day >= 153 and Day <= 182:\nDay -= 152\nGregMonth = \"June\"\nif Day >= 183 and Day <= 213:\nDay -= 182\nGregMonth = \"July\"\nif Day >= 214 and Day <= 244:\nDay -= 213\nGregMonth = \"August\"\nif Day >= 245 and Day <= 274:\nDay -= 244\nGregMonth = \"September\"\nif Day >= 275 and Day <= 305:\nDay -= 274\nGregMonth = \"October\"\nif Day >= 306 and Day <= 335:\nDay -= 305\nGregMonth = \"November\"\nif Day >= 336 and Day <= 366:\nDay -= 335\nGregMonth = \"December\"\n\nDay = str(Day)\nYear = str (Year)\nGregDate = Day + \" \" + GregMonth + \" \" + Year\nDay = int(Day)\nYear = int(Year)\nprint(\"The Gregorian Date is: \", GregDate, \"\\n\")\nreturn(GregDate)\n\n#Main\nMain()\n• dayN, monthN, yearN = Date() ValueError: too many values to unpack (expected 3) When I try to run, it will give me an error. Also unclear to me what Dates are valid... – Ludisposed Sep 11 '17 at 14:33\n\nRunning your code through PEP8 will produce alot of errors, you can see it for yourself.\n\nJust to name a few PEP8 improvements:\n\n1. Use indentation of 4 spaces, it looks cleaner\n2. Functions should be named with lowerscore letters. Example: stop_program\n3. You use a lot of if var == False: which could be rewritten to if not var:\n4. Same goes for if var == True: which is just if var:\n5. You have some calculations like if val > 0 and val < 12: which can be rewritten to if 0 < val < 12:\n6. When using comments don't indent them! - #comment should be # comment\n\nOther improvements:\n\nWhen possible use libraries, be sure to check out datetime library\n\nI thought I'd have a go at this. I've done it in c# - yes I know it's a different language but hold on - you should be able to draw some useful general principles from it.\n\nI've tried to make it more readable and implemented an OOP approach. Although I would have liked to make inroads in simplifying the actual algorithm which does the conversion, I think you'll probably see what I've done.\n\nSee the link attached - i'm hoping it is a little more readable - in terms of having small methods which do a particular job. IsValid(), IsDayValid() IsYearValid() etc are all small methods which do a particular and improve readability.\n\nusing System;\nnamespace ConsoleApplication1\n{\ninternal class Program\n{\nprivate static void Main(string[] args)\n{\n// prints welcome\n// gets date input\nDateConverter dc = new DateConverter();\n\ndc.PrintConvertedDate();\ndc.PrintRepublicanDate();\n\n}\n}\n\npublic class DateConverter\n{\npublic RepublicanDate republicDate { get; set; }\n\npublic DateConverter()\n{\nPrintWelcome();\nthis.republicDate = GetDate();\n}\n\npublic void PrintWelcome()\n{\nConsole.WriteLine(\"\\nRepublican to Gregorian Date Converter\");\nConsole.WriteLine(\"\\nThis program will convert a date from the French Republican Calendar to the equivalent date in the Gregorian Calendar.\");\nConsole.WriteLine(\"\\nTo stop calculating dates and close the program, simply input: 0 0 0.\\n\");\n}\n\npublic DateTime ConvertDate()\n{\nreturn republicDate.ConvertToGregorianDate();\n}\n\npublic void PrintConvertedDate()\n{\n// prints converted date\nConsole.WriteLine(\"This converted gregorian date is: \" + this.ConvertDate());\n}\n\npublic void PrintRepublicanDate()\n{\n// prints the republic date in its original string format.\nConsole.WriteLine(republicDate.ToString());\n}\n\nprivate RepublicanDate GetDate()\n{\nRepublicanDate date;\n\n// repeat untill we get a valid date value\ndo\n{\ndate = new RepublicanDate();\n} while (date.IsInvalid());\n\nreturn date;\n}\n}\n\npublic class RepublicanDate\n{\nprivate DayMonthYear input;\n\npublic RepublicanDate()\n{\ninput = GetInputDate();\n}\n\nprivate DayMonthYear GetInputDate()\n{\nConsole.WriteLine(\"Input a Republican Date in the format: 'dd mm yy'. Please enter a republican date: \\n\");\nstring date = Console.ReadLine();\nstring[] dateArray = date.Split(' ');\n\nint day = int.Parse(dateArray);\nint month = int.Parse(dateArray);\nint year = int.Parse(dateArray);\n\nDayMonthYear dt = new DayMonthYear(day, month, year);\n\nreturn dt;\n}\n\n#region Validity methods\n\npublic bool IsInvalid()\n{\nif (IsYearInvalid(input))\n{\nreturn true;\n}\nelse if (IsMonthInvalid(input))\n{\nreturn true;\n}\nelse if (IsDayInvalid(input))\n{\nreturn true;\n}\nelse\n{\nreturn false;\n}\n}\n\nprivate bool IsDayInvalid(DayMonthYear input)\n{\nif (input.Day >= 1 && input.Day <= 30)\n{\nreturn false;\n}\nelse\n{\nreturn true;\n}\n}\n\nprivate bool IsMonthInvalid(DayMonthYear input)\n{\nif (input.Month >= 1 && input.Month <= 12)\n{\nreturn false;\n}\nelse\n{\nreturn true;\n}\n}\n\nprivate bool IsYearInvalid(DayMonthYear input)\n{\nif (input.Year >= 1 && input.Year <= 14)\n{\nreturn false;\n}\nelse\n{\nreturn true;\n}\n}\n\n#endregion Validity methods\n\npublic override string ToString()\n{\nreturn String.Format(\"The french Rpublic Date is: {0} {1} and {2}\", input.Day, GetFrenchMonth(input.Month), GetFrenchYear(input.Year));\n}\n\n#region To String Helper methods\n\nprivate string GetFrenchMonth(int monthNumber)\n{\nswitch (monthNumber)\n{\ncase 1:\nreturn \"Vendémiaire\";\n\ncase 2:\nreturn \"Brumaire\";\n\ncase 3:\nreturn \"Frimaire\";\n\ncase 4:\nreturn \"Nivôse\";\n\ncase 5:\nreturn \"Pluviôse\";\n\ncase 6:\nreturn \"Ventôse\";\n\ncase 7:\nreturn \"Germinal\";\n\ncase 8:\nreturn \"Floréal\";\n\ncase 9:\nreturn \"Prairial\";\n\ncase 10:\nreturn \"Messidor\";\n\ncase 11:\nreturn \"Thermidor\";\n\ncase 12:\nreturn \"Fructidor\";\n\ndefault:\nthrow new ArgumentOutOfRangeException(\"\\n The months should only be between 1 and 12\");\n}\n}\n\nprivate string GetFrenchYear(int YearNumber)\n{\nswitch (YearNumber)\n{\ncase 1:\nreturn \"I\";\n\ncase 2:\nreturn \"II\";\n\ncase 3:\nreturn \"III\";\n\ncase 4:\nreturn \"IV\";\n\ncase 5:\nreturn \"V\";\n\ncase 6:\nreturn \"VI\";\n\ncase 7:\nreturn \"VII\";\n\ncase 8:\nreturn \"VIII\";\n\ncase 9:\nreturn \"IX\";\n\ncase 10:\nreturn \"X\";\n\ncase 11:\nreturn \"XI\";\n\ncase 12:\nreturn \"XII\";\n\ncase 13:\nreturn \"XIII\";\n\ncase 14:\nreturn \"XIV\";\n\ndefault:\nthrow new ArgumentOutOfRangeException(\"\\n The year should only be between 1 and 14\");\n}\n}\n\n#endregion To String Helper methods\n\npublic DateTime ConvertToGregorianDate()\n{\n// just returned a dummy value, but all your conversion logic could go here:\nreturn new DateTime();\n\n// I can't comment too much on the logic seeing i don't understand it\n// but it looks like you can certain break it up\n// to make it more readable and understandable.\n\n// all the conversion logic should go here.\n//monthL = 30\n//StartDate = 266\n//StartYear = 1792\n//YDays = 365\n//LYDays = 366\n\n//Year = StartYear + (yearN - 1)\n//Day = ((monthN - 1) * monthL) + dayN\n//if Leap == True and Day >= 60:\n// Day += 1\n\n//PH = StartDate + Day\n//if Leap == True:\n// if PH > LYDays:\n// Year += 1\n// Day = PH - LYDays\n// else:\n// Day = PH\n//if Leap == False:\n// if PH > YDays:\n// Year += 1\n// Day = PH - YDays\n// else:\n// Day = PH\n\n//if Leap == False:\n// if Day >= 1 and Day <= 31:\n// GregMonth = \"January\"\n// if Day >= 32 and Day <= 59:\n// Day -= 31\n// GregMonth = \"February\"\n// if Day >= 60 and Day <= 90:\n// Day -= 59\n// GregMonth = \"March\"\n// if Day >= 91 and Day <= 120:\n// Day -= 90\n// GregMonth = \"April\"\n// if Day >= 121 and Day <= 151:\n// Day -= 120\n// GregMonth = \"May\"\n// if Day >= 152 and Day <= 181:\n// Day -= 151\n// GregMonth = \"June\"\n// if Day >= 182 and Day <= 212:\n// Day -= 181\n// GregMonth = \"July\"\n// if Day >= 213 and Day <= 243:\n// Day -= 212\n// GregMonth = \"August\"\n// if Day >= 244 and Day <= 273:\n// Day -= 243\n// GregMonth = \"September\"\n// if Day >= 274 and Day <= 304:\n// Day -= 273\n// GregMonth = \"October\"\n// if Day >= 305 and Day <= 334:\n// Day -= 304\n// GregMonth = \"November\"\n// if Day >= 335 and Day <= 365:\n// Day -= 334\n// GregMonth = \"December\"\n\n//if Leap == True:\n// if Day >= 1 and Day <= 31:\n// GregMonth = \"January\"\n// if Day >= 32 and Day <= 60:\n// Day -= 32\n// GregMonth = \"February\"\n// if Day >= 61 and Day <= 91:\n// Day -= 60\n// GregMonth = \"March\"\n// if Day >= 92 and Day <= 121:\n// Day -= 91\n// GregMonth = \"April\"\n// if Day >= 122 and Day <= 152:\n// Day -= 121\n// GregMonth = \"May\"\n// if Day >= 153 and Day <= 182:\n// Day -= 152\n// GregMonth = \"June\"\n// if Day >= 183 and Day <= 213:\n// Day -= 182\n// GregMonth = \"July\"\n// if Day >= 214 and Day <= 244:\n// Day -= 213\n// GregMonth = \"August\"\n// if Day >= 245 and Day <= 274:\n// Day -= 244\n// GregMonth = \"September\"\n// if Day >= 275 and Day <= 305:\n// Day -= 274\n// GregMonth = \"October\"\n// if Day >= 306 and Day <= 335:\n// Day -= 305\n// GregMonth = \"November\"\n// if Day >= 336 and Day <= 366:\n// Day -= 335\n// GregMonth = \"December\"\n\n//Day = str(Day)\n//Year = str (Year)\n//GregDate = Day + \" \" + GregMonth + \" \" + Year\n//Day = int(Day)\n//Year = int(Year)\n//print(\"The Gregorian Date is: \", GregDate, \"\\n\")\n//return(GregDate)\n}\n}\n\npublic struct DayMonthYear\n{\npublic int Day { get; set; }\npublic int Month { get; set; }\npublic int Year { get; set; }\n\npublic DayMonthYear(int day, int month, int year)\n: this()\n{\nthis.Day = day;\nthis.Month = month;\nthis.Year = year;\n}\n}\n}" ]
[ null ]
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https://www.convertunits.com/from/kilonewton/square+meter/to/millipascal
[ "## ››Convert kilonewton/square metre to millipascal\n\n kilonewton/square meter millipascal\n\nHow many kilonewton/square meter in 1 millipascal? The answer is 1.0E-6.\nWe assume you are converting between kilonewton/square metre and millipascal.\nYou can view more details on each measurement unit:\nkilonewton/square meter or millipascal\nThe SI derived unit for pressure is the pascal.\n1 pascal is equal to 0.001 kilonewton/square meter, or 1000 millipascal.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between kilonewtons/square meter and millipascals.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of kilonewton/square meter to millipascal\n\n1 kilonewton/square meter to millipascal = 1000000 millipascal\n\n2 kilonewton/square meter to millipascal = 2000000 millipascal\n\n3 kilonewton/square meter to millipascal = 3000000 millipascal\n\n4 kilonewton/square meter to millipascal = 4000000 millipascal\n\n5 kilonewton/square meter to millipascal = 5000000 millipascal\n\n6 kilonewton/square meter to millipascal = 6000000 millipascal\n\n7 kilonewton/square meter to millipascal = 7000000 millipascal\n\n8 kilonewton/square meter to millipascal = 8000000 millipascal\n\n9 kilonewton/square meter to millipascal = 9000000 millipascal\n\n10 kilonewton/square meter to millipascal = 10000000 millipascal\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from millipascal to kilonewton/square meter, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Millipascal\n\nThe SI prefix \"milli\" represents a factor of 10-3, or in exponential notation, 1E-3.\n\nSo 1 millipascal = 10-3 pascals.\n\nThe definition of a pascal is as follows:\n\nThe pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]
[ null ]
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https://inches-to-feet.appspot.com/925-inches-to-feet.html
[ "Inches To Feet\n\n# 925 in to ft925 Inches to Feet\n\nin\n=\nft\n\n## How to convert 925 inches to feet?\n\n 925 in * 0.0833333333 ft = 77.0833333333 ft 1 in\nA common question is How many inch in 925 foot? And the answer is 11100.0 in in 925 ft. Likewise the question how many foot in 925 inch has the answer of 77.0833333333 ft in 925 in.\n\n## How much are 925 inches in feet?\n\n925 inches equal 77.0833333333 feet (925in = 77.0833333333ft). Converting 925 in to ft is easy. Simply use our calculator above, or apply the formula to change the length 925 in to ft.\n\n## Convert 925 in to common lengths\n\nUnitLength\nNanometer23495000000.0 nm\nMicrometer23495000.0 µm\nMillimeter23495.0 mm\nCentimeter2349.5 cm\nInch925.0 in\nFoot77.0833333333 ft\nYard25.6944444444 yd\nMeter23.495 m\nKilometer0.023495 km\nMile0.0145991162 mi\nNautical mile0.0126862851 nmi\n\n## What is 925 inches in ft?\n\nTo convert 925 in to ft multiply the length in inches by 0.0833333333. The 925 in in ft formula is [ft] = 925 * 0.0833333333. Thus, for 925 inches in foot we get 77.0833333333 ft.\n\n## 925 Inch Conversion Table", null, "## Alternative spelling\n\n925 Inches to ft, 925 Inches in ft, 925 in to ft, 925 in in ft, 925 Inch to ft, 925 Inch in ft, 925 Inch to Feet, 925 Inch in Feet, 925 Inches to Feet, 925 Inches in Feet, 925 Inches to Foot, 925 Inches in Foot, 925 in to Foot, 925 in in Foot" ]
[ null, "https://inches-to-feet.appspot.com/image/925.png", null ]
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https://pythonexamples.org/python-list-methods/
[ "# Python List Methods\n\nContents\n\n## Python List Methods\n\nPython List data type provides a set of methods that we can call on the list objects.\n\nIn this tutorial, you will learn about all of the Python List methods with description for each of them, and a well detailed example. Also, dedicated tutorials are written for each of the List methods. Link to the respective tutorial is provided at the end of each method.\n\n### 1. List append() method\n\nPython List append() method appends given object to the end of this list.\n\nPython List append()\n\nIn the following program, we take a list of strings in my_list with some initial values. We have to append an object 'mango' to this list. Call append() method on the list my_list and pass the object 'list' as argument to the method.\n\nPython Program\n\n``````my_list = ['apple', 'banana', 'cherry']\nmy_list.append('mango')\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``['apple', 'banana', 'cherry', 'mango']``\n\n### 2. List clear() method\n\nPython List clear() method removes all the items from the list. The method makes the list empty.\n\nPython List clear()\n\nIn the following program, we take a list my_list which is initialized with [52, 41, 63]. We have to make this list empty. Call clear() method on the list my_list, and the items in the list shall be cleared (removed).\n\nPython Program\n\n``````my_list = [52, 41, 63]\nmy_list.clear()\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``[]``\n\n### 3. List copy() method\n\nPython List copy() method creates a shallow copy of the list, and returns the copied list.\n\nPython List copy()\n\nIn the following program, we take a list my_list with values [22, 44, 66]. We have to make a copy of this list. Call copy() method on the list my_list, and store the returned value in a variable.\n\nPython Program\n\n``````my_list = [22, 44, 66]\nmy_list_copy = my_list.copy()\nprint(f\"Original list : {my_list}\")\nprint(f\"Copied list : {my_list_copy}\")``````\nRun Code Copy\n\nOutput\n\n``````Original list : [22, 44, 66]\nCopied list : [22, 44, 66]``````\n\n### 4. List count() method\n\nPython List count() method counts the number of occurrences of specified value in the list, and returns that count.\n\nPython List count()\n\nIn the following program, we take a list my_list with some string values. We have to count the number of occurrences of the value 'apple' in this list. Call count() method on the list my_list, and pass the value 'apple' as argument to the method.\n\nPython Program\n\n``````my_list = ['apple', 'fig', 'apple', 'cherry']\n\ncount = my_list.count('apple')\nprint(f\"Count : {count}\")``````\nRun Code Copy\n\nOutput\n\n``Count : 2``\n\n### 5. List extend() method\n\nPython List extend() method extends the list with items from given iterable.\n\nPython List extend()\n\nIn the following program, we take a list my_list with some string values. We have to extend this list with items from an another list another_list. Call extend() method on the list my_list, and pass another_list as argument to the method.\n\nPython Program\n\n``````my_list = ['apple', 'banana', 'cherry']\nanother_list = ['mango', 'fig']\n\nmy_list.extend(another_list)\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``['apple', 'banana', 'cherry', 'mango', 'fig']``\n\n### 6. List index() method\n\nPython List index() method is used to find the index (position) of the specified value in the list.\n\nPython List index()\n\nIn the following program, we define a list my_list with four string values. We have to find the index of the value 'cherry' in this list. Call index() method on the list, and pass the value as argument.\n\nPython Program\n\n``````my_list = ['apple', 'banana', 'cherry', 'fig']\nvalue_index = my_list.index('cherry')\nprint(f\"Index : {value_index}\")``````\nRun Code Copy\n\nOutput\n\n``Index : 2``\n\n### 7. List insert() method\n\nPython List insert() method inserts given object at specified index in the list.\n\nPython List insert()\n\nIn the following program, we take a list my_list with some initial values. We have to insert the object 99 at index 3 in this list. Call insert() method on the list my_list, and pass the index and object as arguments to the method.\n\nPython Program\n\n``````my_list = [11, 22, 33, 44, 55, 66]\nmy_list.insert(3, 99)\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``[11, 22, 33, 99, 44, 55, 66]``\n\n### 8. List pop() method\n\nPython List pop() method removes the object at specified index in the list.\n\nPython List pop()\n\nIn the following program, we take a list my_list with some initial values. We have to remove the item at index 3 in this list. Call pop() method on the list my_list, and pass the index as argument to the method.\n\nPython Program\n\n``````my_list = [2, 4, 6, 8, 10, 12]\nmy_list.pop(3) # index=3\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``[2, 4, 6, 10, 12]``\n\n### 9. List remove() method\n\nPython List remove() method removes the first occurrence of specified item from the list.\n\nPython List remove()\n\nIn the following program, we take a list my_list with some initial values. We have to remove the item with value 8 in this list. Call remove() method on the list my_list, and pass the value 8 as argument to the method.\n\nPython Program\n\n``````my_list = [2, 4, 6, 8, 10, 12]\nmy_list.remove(8)\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``[2, 4, 6, 10, 12]``\n\n### 10. List reverse() method\n\nPython List reverse() method reverses the order of items in the list in place.\n\nPython List reverse()\n\nIn the following program, we take a list my_list with values [2, 4, 6, 8, 10]. We have to reverse this list. Call reverse() method on the list my_list.\n\nPython Program\n\n``````my_list = [2, 4, 6, 8, 10]\nmy_list.reverse()\nprint(my_list)``````\nRun Code Copy\n\nOutput\n\n``[10, 8, 6, 4, 2]``\n\n### 11. List sort() method\n\nPython List sort() method sorts the items in the list in place in ascending order.\n\nPython List sort()\n\nIn the following program, we take a list my_list with numeric values. We have to sort this list of numbers in ascending order. Call sort() method on the list my_list, and pass no arguments.\n\nPython Program\n\n``````my_list = [5, 3, 1, 6, 2, 4]\nprint(f\"Original list : {my_list}\")\n\nmy_list.sort()\nprint(f\"Sorted list : {my_list}\")``````\nRun Code Copy\n\nOutput\n\n``````Original list : [5, 3, 1, 6, 2, 4]\nSorted list : [1, 2, 3, 4, 5, 6]``````\n\n## Summary\n\nIn this Python Lists tutorial, we learned about different methods that we call on list objects.\n\nCode copied to clipboard successfully 👍" ]
[ null ]
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https://123dok.net/document/q7w4g4lk-sharp-asymptotics-of-the-lp.html
[ "# Sharp asymptotics of the Lp\n\n(1)\n\n## Sharp asymptotics of the L p approximation error for interpolation on block partitions\n\n### Yuliya Babenko, Tatyana Leskevich, Jean-Marie Mirebeau February 9, 2011\n\nAbstract\n\nAdaptive approximation (or interpolation) takes into account local variations in the behavior of the given function, adjusts the approximant depending on it, and hence yields the smaller error of approximation. The question of constructing optimal approximating spline for each functionproved to be very hard. In fact, no polynomial time algorithm of adaptive spline approximation can be designed and no exact formula for the optimal error of approximation can be given. Therefore, the next natural question would be to study the asymptotic behavior of the error and construct asymptotically optimal sequences of partitions.\n\nIn this paper we provide sharp asymptotic estimates for the error of interpolation by splines on block partitions in IRd. We consider various projection operators to define the interpolant and provide the analysis of the exact constant in the asymptotics as well as its explicit form in certain cases.\n\n(2)\n\nYuliya Babenko\n\nDepartment of Mathematics and Statistics Sam Houston State University\n\nBox 2206\n\nHuntsville, TX, USA 77340-2206 Phone: 936.294.4884\n\nFax: 936.294.1882\n\nEmail: babenko@shsu.edu Tatyana Leskevich\n\nDepartment of Mathematical Analysis Dnepropetrovsk National University pr. Gagarina, 72,\n\nDnepropetrovsk, UKRAINE, 49050 Email: tleskevich@gmail.com Jean-Marie Mirebeau\n\nUPMC Univ Paris 06, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France CNRS, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France\n\nEmail: mirebeau@ann.jussieu.fr\n\n(3)\n\n### 1 Introduction\n\nThe goal of this paper is to study the adaptive approximation by interpolating splines defined over block partitions in IRd. With the help of introduced projection operator we shall handle the general case, and then apply the obtained estimates to several different interpolating schemes most commonly used in practice.\n\nOur approach is to introduce the “error function” which reflects the interaction of approximation procedure with polynomials. Throughout the paper we shall study the asymptotic behavior of the ap- proximation error and, whenever possible, the explicit form of the error function which plays a major role in finding the constants in the formulae for exact asymptotics.\n\n### 1.1 The projection operator\n\nLet us first introduce the definitions that will be necessary to state the main problem and the results of this paper.\n\nWe consider a fixed integerd≥1 and we denote byx= (x1,· · ·, xd) the elements of Rd. A blockR is a subset ofRd of the form\n\nR= Y\n\n1≤i≤d\n\n[ai, bi]\n\nwhere ai < bi, for all 1≤i≤d. For any blockR⊂Rd, byLp(R), 1≤p≤ ∞, we denote the space of measurable functionsf :R→IR for which the value\n\nkfkp=kfkLp(R):=\n\n\n\n\n\n Z\n\nR\n\n|f(x)|pdx\n\n1 p\n\n, if 1≤p <∞, esssup{|f(x)|: x∈R}, if p=∞.\n\nis finite. We also consider the spaceC0(R) of continuous functions onRequipped with the uniform norm k · kL(R). We shall make a frequent use of the canonical blockId, whereIis the interval\n\nI:=\n\n−1 2,1\n\n2\n\n.\n\nNext we define the space V := C0(Id) and the norm k · kV := k · kL(Id). Throughout this paper we consider a linear and bounded (hence, continuous) operator I :V →V.This implies that there exists a constantCI such that\n\nkIukV ≤CIkukV for allu∈V. (1.1)\n\nWe assume furthermore that I is a projector, which means that it satisfies\n\nI◦I = I. (1.2)\n\nLetRbe an arbitrary block. It is easy to show that there exists a uniquex0∈Rd and a unique diagonal matrixD with positive diagonal coefficients such that the transformation\n\nφ(x) :=x0+Dx satisfies φ(Id) =R. (1.3) The volume ofR, denoted by|R|, is equal to det(D). For any functionf ∈C0(R) we then define\n\nIRf := I(f◦φ)◦φ−1. (1.4)\n\nNote that\n\nkf −IRfkLp(R)= (detD)1pkf◦φ−I(f ◦φ)kLp(Id). (1.5) A block partition Rof a block R0 is a finite collection of blocks such that their union covers R0 and which pairwise intersections have zero Lebesgue measure. If R is a block partition of a blockR0 and\n\n(4)\n\niff ∈C0(R0), by IRf ∈L(R0) we denote the (possibly discontinuous) function which coincides with IRf on the interior of each blockR∈ R.\n\nMain Question. The purpose of this paper is to understand the asymptotic behavior of the quantity kf −IRNfkLp(R0)\n\nfor each given function f onR0 from some class of smoothness, where (RN)N≥1 is a sequence of block partitions ofR0that are optimally adapted tof.\n\nNote that the exact value of this error can be explicitly computed only in trivial cases. Therefore, the natural question is to study the asymptotic behavior of the error function, i.e. the behavior of the error as the number of elements of the partitionRN tends to infinity.\n\nMost of our results hold with only assumptions (1.1) of continuity of the operator I , the projection axiom (1.2), and the definition of IR given by (1.4). Our analysis therefore applies to various projection operators I, such as the L2-orthogonal projection on a space of polynomials, or spline interpolating schemes described in§1.4.\n\n### 1.2 History\n\nThe main problem formulated above is interesting for functions of arbitrary smoothness as well as for various classes of splines (for instance, for splines of higher order, interpolating splines, best approximating splines, etc.). In the univariate case general questions of this type have been investigated by many authors.\n\nThe results are more or less complete and have numerous applications (see, for example, ).\n\nFewer results are known in the multivariate case. Most of them, starting with the works [1, 9], are for the case of approximation by splines on triangulations (for review of existing results see, for instance [2, 3, 6, 7, 10, 12]). However, in applications where preferred directions exist, box partitions are sometimes more convenient and efficient.\n\nThe first result on the error of interpolation on rectangular partitions by bivariate splines linear in each variable (or bilinear) is due to D’Azevedo who obtained local (on a single rectangle) error estimates.\n\nIn Babenko obtained the exact asymptotics for the error (inL1,L2, andLnorms) of interpolation ofC2(Id) functions by bilinear splines.\n\nIn Babenko generalized the result to interpolation and quasiinterpolation of a functionf ∈C2(Id) with arbitrary but fixed throughout the domain signature (number of positive and negative second-order partial derivatives). However, the norm used to measure the error of approximation was uniform.\n\nIn this paper we use a different, more abstract, approach which allows us to obtain the exact asymp- totics of the error in a more general framework which can be applied to many particular interpolation schemes by an appropriate choice of the interpolation operator. In general, the constant in the asymp- totics is implicit. However, imposing additional assumptions on the interpolation operator allows us to compute the constant explicitly.\n\nThe paper is organized as follows. Section 1.5 contains the statements of main approximation results.\n\nThe closer study of the error function, as well as its explicit formulas under some restrictions, can be found in Section 2. The proofs of the theorems about asymptotic behavior of the error are contained in Section 3.\n\n### 1.3 Polynomials and the error function\n\nIn order to obtain the asymptotic error estimates we need to study the interaction of the projection operator I with polynomials.\n\nThe notationαalways refers to ad-vector of non-negative integers α= (α1,· · ·, αd)∈ZZd+. For each αwe define the following quantities\n\n|α|:= X\n\n1≤i≤d\n\nαi, α! := Y\n\n1≤i≤d\n\nαi!, max(α) := max\n\n1≤i≤dαi.\n\n(5)\n\nWe also define the monomial\n\nXα:= Y\n\n1≤i≤d\n\nXiαi,\n\nwhere the variable isX = (X1, ..., Xd)∈IRd. Finally, for each integerk≥0 we define the following three vector spaces of polynomials\n\nIPk := Vect{Xα: |α| ≤k},\n\nIPk := Vect{Xα: max(α)≤kand|α| ≤k+ 1}, IP∗∗k := Vect{Xα: max(α)≤k}.\n\n(1.6) Note that clearly dim(IP∗∗k ) = (k+ 1)d. In addition, a classical combinatorial argument shows that\n\ndim IPk = k+d\n\nd\n\n. Furthermore,\n\ndim IPk = dim IPk+1−d=\n\nk+d+ 1 d\n\n−d.\n\nByVI we denote the image of I, which is a subspace ofV =C0(Id). Since I is a projector (1.2), we have\n\nVI={I(f) : f ∈V}={f ∈V : f = I(f)}. (1.7) From this point on, the integerkis fixed and defined as follows\n\nk=k(I) := max{k0≥0 : IPk0⊂VI} (1.8) Hence, the operator I reproduces polynomials of total degree less or equal than k. (If k =∞ then we obtain, using the density of polynomials in V and the continuity of I, that I(f) =f for all f ∈V. We exclude this case from now on.)\n\nIn what follows, bymwe denote the integer defined by\n\nm=m(I) :=k+ 1, (1.9)\n\nwherek=k(I) is defined in (1.8). By IHmwe denote the space of homogeneous polynomials of degreem IHm:= Vect{Xα: |α|=m}.\n\nWe now introduce a functionKI on IHm, further referred to as the “error function”.\n\nDefinition 1.1 (Error Function) For all π∈IHm\n\nKI(π) := inf\n\n|R|=1kπ−IRπkLp(R), (1.10) where the infimum is taken over all blocks Rof unitd-dimensional volume.\n\nThe error functionK plays a major role in our asymptotical error estimates developed in the next subsection. Hence, we dedicate§2 to its close study, and we provide its explicit form in various cases.\n\nThe optimization (1.10) among blocks can be rephrased into an optimization among diagonal matrices.\n\nIndeed, if |R| = 1, then there exists a unique x0 ∈ Rd and a unique diagonal matrix with positive coefficients such that R = φ(Id) with φ(x) =x0+Dx. Furthermore, the homogeneous component of degree m is the same in bothπ◦φ and π◦D, hence π◦φ−π◦D ∈ IPk (recal thatm = k+ 1) and therefore this polynomial is reproduced by the projection operator I. Using the linearity of I, we obtain\n\nπ◦φ−I(π◦φ) =π◦D−I(π◦D).\n\nCombining this with (1.5), we obtain that KI(π) = inf\n\ndetD=1 D≥0\n\nkπ◦D−I(π◦D)kLp(Id), (1.11) where the infimum is taken over the set of diagonal matrices with non-negative entries and unit determi- nant.\n\n(6)\n\n### 1.4 Examples of projection operators\n\nIn this section we define several possible choices for the projection operator I which are consistent with (1.8) and, in our opinion, are most useful for practical purposes. However, many other possibilities could be considered.\n\nDefinition 1.2 (L2(Id) orthogonal projection) We may defineI(f)as theL2(Id) orthogonal projec- tion of f onto one of the spaces of polynomials IPk, IPk or IP∗∗k defined in (1.6).\n\nIf the projection operator I is chosen as in Definition 1.2, then a simple change of variables shows that for any blockR, the operator IRdefined by (1.4) is theL2(R) orthogonal projection onto the same space of polynomials.\n\nTo introduce several possible interpolation schemes for which we obtain the estimates using our approach, we consider a setUk⊂Iof cardinality #(Uk) =k+ 1 (special cases are given below). For any u= (u1,· · ·ud)∈Ukd we define an element of IP∗∗k as follows\n\nµu(X) := Y\n\n1≤i≤d\n\n Y\n\nv∈Uk\n\nv6=ui\n\nXi−v ui−v\n\n∈IP∗∗k .\n\nClearly,µu(u) =µu(u1,· · · , ud) = 1 andµu(v) =µu(v1,· · · , vd) = 0 if v= (v1,· · ·, vd)∈ Ukd and v6=u.\n\nIt follows that the elements of B := (µu)u∈Ud\n\nk are linearly independent. Since #(B) = #(Ukd) = (k+ 1)d= dim(IP∗∗k ),B is a basis of IP∗∗k .\n\nTherefore, any element ofµ∈IP∗∗k can be written in the form µ(X) = X\n\nu∈Ukd\n\nλuµu(X).\n\nIt follows that there is a unique element of µ ∈IP∗∗k such that µ(u) = f(u) for all u∈ Ukd. We define If :=µ, namely\n\n(If)(X) := X\n\nu∈Ukd\n\nf(u)µu(X)∈IP∗∗k .\n\nWe may takeUk to be the set ofk+ 1 equi-spaced points on I Uk=\n\n−1 2+n\n\nk : 0≤n≤k\n\n. (1.12)\n\nWe obtain a different, but equally relevant, operator I by choosingUk to be the set of Tchebychev points onI\n\nUk= 1\n\n2cosnπ k\n\n: 0≤n≤k\n\n. (1.13)\n\nDifferent interpolation procedures can be used to construct I. Another convenient interpolation scheme is to take\n\nI(f)∈IPk\n\nand I(f) =f on a subset ofUkd. This subset contains dim IPk points, which are convenient to choose first on the boundary of Id and then (if needed) at some interior lattice points. Note that since dim IPk <\n\n#(Ukd) = (k+ 1)d, it is always possible to construct such an operator.\n\nIf the projection operator I is chosen as described above, then for any blockR and anyf ∈C0(R), IR(f) is the unique element of respective space of polynomials which coincides withf at the imageφ(p) of the pointspmentioned in the definition of I, by the transformationφdescribed in (1.3).\n\n(7)\n\n### 1.5 Main results\n\nIn order to obtain the approximation results we often impose a slight technical restriction (which can be removed, see for instance ) on sequences of block partitions, which is defined as follows.\n\nDefinition 1.3 (admissibility) We say that a sequence (RN)N≥1 of block partitions of a block R0 is admissible if#(RN)≤N for all N≥1, and\n\nsup\n\nN≥1\n\nN1d sup\n\nR∈RN\n\ndiam(R)\n\n<∞ (1.14)\n\nWe recall that the approximation error is measured inLp norm, where the exponent pis fixed and 1≤p≤ ∞. We defineτ∈(0,∞) by\n\n1 τ := m\n\nd +1\n\np. (1.15)\n\nIn the following estimates we identified dmf(x) with an element of IHm according to dmf(x)\n\nm! ∼ X\n\n|α|=m\n\nmf(x)\n\n∂xα Xα\n\nα! . (1.16)\n\nWe now state the asymptotically sharp lower bound for the approximation error of a function f on an admissible sequence of block partitions.\n\nTheorem 1.4 Let R0 be a block and let f ∈Cm(R0). For any admissible sequence of block partitions (RN)N≥1 ofR0\n\nlim inf\n\nN→∞ Nmdkf −IRNfkLp(R0)\n\nKI\n\ndmf m!\n\nL\n\nτ(R0)\n\n.\n\nThe next theorem provides an upper bound for the projection error of a functionf when an optimal sequence of block partitions is used. It confirms the sharpness of the previous theorem.\n\nTheorem 1.5 Let R0 be a block and let f ∈ Cm(R0). Then there exists a (perhaps non-admissible) sequence (RN)N≥1,#RN ≤N, of block partitions ofR0 satisfying\n\nlim sup\n\nN→∞\n\nNmdkf−IRNfkLp(R0)\n\nKI\n\ndmf m!\n\nL\n\nτ(R0)\n\n. (1.17)\n\nFurthermore, for all ε > 0 there exists an admissible sequence (RεN)N≥1 of block partitions of R0 satisfying\n\nlim sup\n\nN→∞\n\nNmdkf−IRε\n\nNfkLp(R0)\n\nKI dmf\n\nm!\n\nL\n\nτ(R0)\n\n+ε. (1.18)\n\nAn important feature of these estimates is the “lim sup”. Recall that the upper limit of a sequence (uN)N≥N0 is defined by\n\nlim sup\n\nN→∞\n\nuN := lim\n\nN→∞sup\n\nn≥N\n\nun, and is in general strictly smaller than the supremum supN≥N\n\n0uN. It is still an open question to find an appropriate upper estimate of supN≥N0Nmdkf−IRNfkLp(R0) when optimally adapted block partitions are used.\n\nIn order to have more control of the quality of approximation on various parts of the domain we introduce a positive weight function Ω∈C0(R0). For 1≤p≤ ∞ and for any u∈Lp(R0) as usual we define\n\nkukLp(R0,Ω):=kuΩkLp(R0). (1.19) Remark 1.6 Theorems 1.4, 1.5 and 1.7 below also hold when the normk · kLp(R0) (resp k · kLτ(R0)) is replaced with the weighted normk · kLp(R0,Ω) (respk · kLτ(R0,Ω)) defined in (1.19).\n\n(8)\n\nIn the following section we shall use some restrictive hypotheses on the interpolation operator in order to obtain an explicit formula for the shape function. In particular, Propositions 2.7, 2.8, and equation (2.20) show that, under some assumptions, there exists a constantC=C(I)>0 such that\n\n1 CKI\n\ndmf m!\n\nd v u u u t\n\nY\n\n1≤i≤d\n\nmf\n\n∂xmi\n\n≤CKI\n\ndmf m!\n\n.\n\nThese restrictive hypotheses also allow to improve slightly the estimate (1.18) as follows.\n\nTheorem 1.7 If the hypotheses of Proposition 2.7 or 2.8 hold, and if KI\n\ndmf m!\n\n> 0 everywhere on R0, then there exists an admissiblesequence of partitions(RN)N≥1 which satisfies the optimal estimate (1.17).\n\nThe proofs of the Theorems 1.4, 1.5 and 1.7 are given in §3. Each of these proofs can be adapted to weighted norms, hence establishing Remark 1.6. Some details on how to adapt proofs for the case of weighted norms are provided at the end of each proof.\n\n### 2 Study of the error function\n\nIn this section we perform a close study of the error function KI, since it plays a major role in our asymptotic error estimates. In the first subsection§2.1 we investigate general properties which are valid for any continuous projection operator I. However, we are not able to obtain an explicit form of KI\n\nunder such general assumptions. Recall that in§1.4 we presented several possible choices of projection operators I that seem more likely to be used in practice. In §2.2 we identify four important properties shared by these examples. These properties are used in§2.3 to obtain an explicit form ofKI.\n\n### 2.1 General properties\n\nThe error functionKobeys the following important invariance property with respect to diagonal changes of coordinates.\n\nProposition 2.1 For all π∈IHm and all diagonal matricesD with non-negative coefficients KI(π◦D) = (detD)mdKI(π).\n\nProof: We first assume that the diagonal matrix D has positive diagonal coefficients. Let D be a diagonal matrix with positive diagonal coefficient and which satisfies detD= 1. Let alsoπ∈IHm. Then\n\nπ◦(DD) =π◦((detD)1dD) = (det˜ D)mdπ◦D,˜\n\nwhere ˜D := (detD)1dDD satisfies det ˜D = detD = 1 and is uniquely determined byD. According to (1.11) we therefore have\n\nKI(π◦D) = inf\n\ndetD=1 D≥0\n\nkπ◦(DD)−I(π◦(DD))kLp(Id)\n\n= (detD)md inf\n\ndet ˜D=1 D≥0˜\n\nkπ◦D˜ −I(π◦D)k˜ Lp(Id)\n\n= (detD)mdKI(π),\n\nwhich concludes the proof in the case whereD has positive diagonal coefficients.\n\nLet us now assume that D is a diagonal matrix with non-negative diagonal coefficients and such that det(D) = 0. LetD0 be a diagonal matrix with positive diagonal coefficients, and such thatD=DD0and detD0 = 2. We obtain\n\nKI(π◦D) =KI(π◦(DD0)) = 2mdKI(π◦D),\n\n(9)\n\nwhich implies thatKI(π◦D) = 0 and concludes the proof.\n\nThe next proposition shows that the exponentpused for measuring the approximation error plays a rather minor role. ByKp we denote the error function associated with the exponent p.\n\nProposition 2.2 There exists a constant c >0 such that for all 1≤p1≤p2≤ ∞we have on IHm\n\ncK≤Kp1 ≤Kp2 ≤K.\n\nProof: For any functionf ∈V =C0(Id) and for any 1≤p1≤p2≤ ∞by a standard convexity argument we obtain that\n\nkfkL1(Id)≤ kfkLp\n\n1(Id)≤ kfkLp\n\n2(Id)≤ kfkL(Id). Using (1.11), it follows that\n\nK1≤Kp1 ≤Kp2 ≤K\n\non IHm. Furthermore, the following semi norms on IHm\n\n|π|1:=kπ−IπkL1(Id) and|π|:=kπ−IπkL(Id)\n\nvanish precisely on the same subspace of IHm, namely VI∩Hm = {π ∈ IHm : π = Iπ}. Since IHm\n\nhas finite dimension, it follows that they are equivalent. Hence, there exists a constant c >0 such that c| · |≤ | · |1 on IHm. Using (1.11), it follows thatcK≤K1, which concludes the proof.\n\n### 2.2 Desirable properties of the projection operator\n\nThe examples of projection operators presented in §1.4 share some important properties which allow to obtain the explicit expression of the error functionKI. These properties are defined below and calledH±, Hσ,H or H∗∗. They are satisfied when operator I is the interpolation at equispaced points (Definition 1.12), at Tchebychev points (Definition 1.13), and usually on the most interesting sets of other points.\n\nThey are also satisfied when I is theL2(Id) orthogonal projection onto IPk or IP∗∗k (Definition 1.2).\n\nThe first property reflects the fact that a coordinatexi onId can be changed to −xi, independently of the projection process.\n\nDefinition 2.3 (H± hypothesis) We say that the interpolation operator I satisfies the H± hypothesis if for any diagonal matrix D with entries in±1 we have for allf ∈V\n\nI(f◦D) = I(f)◦D.\n\nThe next property implies that the different coordinatesx1,· · ·, xdonId play symmetrical roles with respect to the projection operator.\n\nDefinition 2.4 (Hσ hypothesis) If Mσ is a permutation matrix, i.e. (Mσ)ij :=δiσ(j) for some per- mutation σof {1,· · · , d}, then for allf ∈V\n\nI(f◦Mσ) = I(f)◦Mσ.\n\nAccording to (1.8), the projection operator I reproduces the space of polynomials IPk. However, in many situations the spaceVI of functions reproduced by I is larger than IPk. In particularVI= IP∗∗k when I is the interpolation on equispaced or Tchebychev points, and VI = IPk (resp IPk, IP∗∗k ) when I is the L2(Id) orthogonal projection onto IPk (resp IPk, IP∗∗k ).\n\nIt is particularly useful to know whether the projection operator I reproduces the elements of IPk, and we therefore give a name to this property. Note that it clearly does not hold for the L2(Id) orthogonal projection onto IPk.\n\nDefinition 2.5 (H hypothesis) The following inclusion holds : Pk⊂VI.\n\n(10)\n\nOn the contrary it is useful to know that some polynomials, and in particular pure powers xmi , are not reproduced by I.\n\nDefinition 2.6 (H∗∗ hypothesis)\n\nIf X\n\n1≤i≤d\n\nλixmi ∈VI then (λ1,· · ·, λd) = (0,· · · ,0).\n\nThis condition obviously holds if I(f)∈IP∗∗k (polynomials of degree≤kin each variable) for allf. Hence, it holds for all the examples of projection operators given in the previous subsection§1.4.\n\n### 2.3 Explicit formulas\n\nIn this section we provide the explicit expression forKwhen some of the hypothesesH±,Hσ,HorH∗∗\n\nhold. Letπ∈IHmand letλi be the corresponding coefficient ofXimin π, for all 1≤i≤d. We define K(π) := d\n\ns Y\n\n1≤i≤d\n\ni|\n\nand\n\ns(π) := #{1≤i≤d: λi>0}.\n\nIf dmm!f(x) is identified by (1.16) to an element of IHm, then one has\n\nK\n\ndmf(x) m!\n\n= 1 m!\n\nd\n\nv u u u t\n\nY\n\n1≤i≤d\n\nmf\n\n∂xmi (x)\n\n. (2.20)\n\nProposition 2.7 If m is odd and ifH±,Hσ andH hold, then Kp(π) =C(p)K(π), where\n\nC(p) :=\n\nX\n\n1≤i≤d\n\nXim−I\n\n X\n\n1≤i≤d\n\nXim\n\nL\n\np(Id)\n\n>0.\n\nProposition 2.8 If m is even and ifHσ,H andH∗∗ hold then Kp(π) =C(p, s(π))K(π).\n\nFurthermore,\n\nC(p,0) =C(p, d) =\n\nX\n\n1≤i≤d\n\nXim−I\n\n X\n\n1≤i≤d\n\nXim\n\nL\n\np(Id)\n\n>0. (2.21)\n\nOther constantsC(p, s)are positive and obey C(p, s) =C(p, d−s).\n\nNext we turn to the proofs of Propositions 2.7 and 2.8.\n\n(11)\n\nProof of Proposition 2.7 Letπ∈IHmand letλi be the coefficient ofXimin π. Denote by π:= X\n\n1≤i≤d\n\nλiXim\n\nso thatπ−π∈IPk and, more generally,π◦D−π◦D∈IPk for any diagonal matrixD. The hypothesis Hstates that the projection operator I reproduces the elements of IPk, and therefore\n\nπ◦D−I(π◦D) =π◦D−I(π◦D).\n\nHence, KI(π) =KI) according to (1.11). If there exists i0, 1≤i0 ≤d, such that λi0 = 0, then we denote byDthe diagonal matrix of entriesDii = 1 ifi6=i0 and 0 ifi=i0. Applying Proposition 2.1 we find\n\nKI(π) =KI) =KI◦D) = (detD)mdKI) = 0.\n\nwhich concludes the proof. We now assume that all the coefficients λi, 1≤i≤d, are different from 0, and we denote byεibe the sign ofλi. Applying Proposition 2.1 to the diagonal matrixDof entries|λi|m1 we find that\n\nKI(π) =KI) = (detD)mdKI◦D−1) =K(π)KI\n\n X\n\n1≤i≤d\n\nεiXim\n\n.\n\nUsing theH±hypothesis with the diagonal matrixDof entriesDiii, and recalling thatmis odd, we find that\n\nKI\n\n X\n\n1≤i≤d\n\nεiXim\n\n=KI\n\n X\n\n1≤i≤d\n\nXim\n\n. We now define the functions\n\ngi:=Xim−I(Xim) for 1≤i≤d.\n\nIt follows from (1.11) that KI\n\n X\n\n1≤i≤d\n\nXim\n\n= inf\n\nQ\n\n1≤i≤dai=1\n\nX\n\n1≤i≤d\n\naigi L\n\np(Id)\n\n,\n\nwhere the infimum is taken over alld-vectors of positive reals of product 1. Let us consider such ad-vector (a1,· · · , ad), and a permutationσof the set{1,· · ·, d}. TheHσ hypothesis implies that the quantity\n\nX\n\n1≤i≤d\n\naσ(i)gi\n\nL\n\np(Id)\n\nis independent ofσ. Hence, summing over all permutations, we obtain\n\nX\n\n1≤i≤d\n\naigi L\n\np(Id)\n\n= 1 d!\n\nX\n\nσ\n\nX\n\n1≤i≤d\n\naσ(i)gi L\n\np(Id)\n\n≥ 1 d\n\nX\n\n1≤i≤d\n\ngi L\n\np(Id)\n\nX\n\n1≤i≤d\n\nai. (2.22)\n\nThe right-hand side is minimal when a1=· · ·=ad= 1, which shows that\n\nX\n\n1≤i≤d\n\naigi\n\nL\n\np(Id)\n\nX\n\n1≤i≤d\n\ngi\n\nL\n\np(Id)\n\n=C(p)\n\nwith equality whenai= 1 for alli. Note as a corollary that\n\nKIε) =kπε−I(πε)kLp(Id)=C(p) where πε= X\n\n1≤i≤d\n\nεiXim. (2.23)\n\n(12)\n\nIt remains to prove thatC(p)>0. Using the hypothesisH±, we find that for allµi∈ {±1}we have\n\nX\n\n1≤i≤d\n\nµigi\n\nL\n\np(Id)\n\n=C(p).\n\nIn particular, for any 1≤i0≤done has 2kgi0kLp(Id)\n\nX\n\n1≤i≤d\n\ngi L\n\np(Id)\n\n+\n\n2gi0− X\n\n1≤i≤d\n\ngi L\n\np(Id)\n\n≤2C(p).\n\nIf C(p) = 0, it follows that gi0 = 0 and therefore that Xim\n\n0 = I(Xim\n\n0), for any 1 ≤ i0 ≤d. Using the assumptionH, we find that the projection operator I reproduces all the polynomials of degreem=k+ 1, which contradicts the definition (1.8) of the integerk.\n\nProof of proposition 2.8 We define λi, π and εi ∈ {±1} as before and we find, using similar reasoning, that\n\nKI(π) =K(π)KI\n\n X\n\n1≤i≤d\n\nεiXim\n\n. For 1≤s≤dwe define\n\nC(p, s) :=KI\n\n X\n\n1≤i≤s\n\nXim− X\n\ns+1≤i≤d\n\nXim\n\n. From the hypothesisHσ it follows that KI(π) =K(π)C(p, s(π)).\n\nUsing againHσ and the fact thatKI(π) =KI(−π) for allπ∈IHm, we find that C(p, s) =KI\n\n X\n\n1≤i≤s\n\nXim− X\n\ns+1≤i≤d\n\nXim\n\n=KI\n\n−\n\n X\n\n1≤i≤d−s\n\nXim− X\n\nd−s+1≤i≤d\n\nXim\n\n=C(p, d−s).\n\nWe definegi:=Xim−I(Xim), as in the proof of Proposition 2.7. We obtain the expression forC(p,0) by summing over all permutations as in (2.22)\n\nC(p,0) =\n\nX\n\n1≤i≤d\n\ngi L\n\np(Id)\n\n.\n\nThis concludes the proof of the first part of Proposition 2.8. We now prove that C(p, s) > 0 for all 1≤p≤ ∞and alls∈ {0,· · ·, d}. To this end we define the following quantity onRd\n\nkakK:=\n\nX\n\n1≤i≤d\n\naigi L\n\np(Id)\n\n.\n\nNote thatkakK = 0 if and only if\n\nX\n\n1≤i≤d\n\naiXim= X\n\n1≤i≤d\n\naiI(Xim),\n\nand the hypothesisH∗∗ precisely states that this equality occurs if and only ifai= 0, for all 1≤i≤d.\n\nHence,k · kK is a norm onRd. Furthermore, let Es:=\n\na∈Rs+×Rd−s : Y\n\n1≤i≤d\n\n|ai|= 1\n\n(13)\n\nThen\n\nC(p, s) = inf\n\na∈Es\n\nkakK.\n\nSinceEs is a closed subset ofRd, which does not contain the origin, this infimum is attained. It follows that C(p, s)>0, and that there exists a rectangleRε of unit volume such that\n\nKIε) =kπε−IπεkLp(Rε)=C(p, s(πε)) where πε= X\n\n1≤i≤d\n\nεiXim. (2.24)\n\n### 3 Proof of the approximation results\n\nIn this section, let the block R0, the integer m, the functionf ∈Cm(R0) and the exponentpbe fixed.\n\nWe conduct our proofs for 1 ≤p <∞ and provide comments on how to adjust our arguments for the casep=∞.\n\nFor eachx∈R0byµx we denote them-th degree Taylor polynomial off at the pointx µxx(X) := X\n\n|α|≤m\n\nmf\n\n∂xα(x)(X−x)α\n\nα! , (3.25)\n\nand we defineπxto be the homogeneous component of degreemin µx, πxx(X) := X\n\n|α|=m\n\nmf\n\n∂xα(x)Xα\n\nα! . (3.26)\n\nSince πx and µx are polynomials of degree m, their m-th derivative is constant, and clearly dmπx = dmµx = dmf(x). In particular, for any x ∈ R0 the polynomial µx−πx belongs to IPk (recall that k=m−1) and is therefore reproduced by the projection operator I. It follows that for anyx∈R0and any blockR\n\nπx−IRπxx−IRµx. (3.27)\n\nIn addition, we introduce a measureρof the degeneracy of a blockR ρ(R) :=diam(R)d\n\n|R| .\n\nGiven any functiong∈Cm(R) and anyx∈R we can define, similarly to (3.26), a polynomial ˆπx∈IHm associated to gatx. We then define\n\nkdmgkL(R):= sup\n\nx∈R\n\nsup\n\n|u|=1\n\n|ˆπx(u)|\n\n!\n\n. (3.28)\n\nProposition 3.1 There exists a constant C=C(m, d)>0 such that for any block R and any function g∈Cm(R)\n\nkg−IRgkLp(R)≤C|R|τ1ρ(R)mdkdmgkL(R). (3.29) Proof: Letx0∈Rand letg0be the Taylor polynomial forgof degreem−1 at pointx0which is defined as follows\n\ng0(X) := X\n\n|α|≤m−1\n\nαf(x0)\n\n∂xα\n\n(X−x0)α α! . Letx∈Rand letx(t) =x0+t(x−x0). We have\n\ng(x) =g0(x) + Z 1\n\nt=0\n\ndmgx(t)(x−x0)\n\nm! (1−t)mdt.\n\n(14)\n\nHence,\n\n|g(x)−g0(x)| ≤ Z 1\n\nt=0\n\nkdmgkL(R)|x−x0|m(1−t)mdt≤ 1\n\nm+ 1kdmgkL(R)diam(R)m. (3.30) Sinceg0 is a polynomial of degree at mostm−1, we have g0= Ig0. Hence,\n\nkg−IRgkLp(R) ≤ |R|1pkg−IRgkL(R)\n\n= |R|1pk(g−g0)−IR(g−g0)kL(R)\n\n≤ (1 +CI)|R|1pkg−g0kL(R),\n\nwhereCI is the operator norm of I :V →V. Combining this estimate with (3.30), we obtain (3.29).\n\n### 3.1 Proof of Theorem 1.4 (Lower bound)\n\nThe following lemma allows us to bound the interpolation error off on the blockR from below.\n\nLemma 3.2 For any blockR⊂R0 and x∈R we have\n\nkf −IRfkLp(R)≥ |R|1τ KIx)−ω(diamR)ρ(R)md ,\n\nwhere the function ω is positive, depends only onf and m, and satisfiesω(δ)→0as δ→0.\n\nProof: Leth:=f−µx, where µxis defined in (3.25) Using (3.27), we obtain kf−IRfkLp(R) ≥ kπx−IRπxkLp(R)− kh−IRhkLp(R)\n\n≥ |R|τ1KIx)− kh−IRhkLp(R), and according to (3.29) we have\n\nkh−IRhkLp(R)≤C0|R|1τρ(R)mdkdmhkL(R). Observe that\n\nkdmhkL(R)=kdmf−dmπxkL(R)=kdmf−dmf(x)kL(R). We introduce the modulus of continuityω of them-th derivatives off.\n\nω(r) := sup\n\nx1,x2∈R0:\n\n|x1−x2|≤r\n\nkdmf(x1)−dmf(x2)k= sup\n\nx1,x2∈R0:\n\n|x1−x2|≤r\n\nsup\n\n|u|≤1\n\nx1(u)−πx2(u)|\n\n!\n\n(3.31)\n\nBy settingω=C0ωwe conclude the proof of this lemma.\n\nWe now consider an admissible sequence of block partitions (RN)N≥0. For all N ≥0,R∈ RN and x∈R, we define\n\nφN(x) :=|R| and ψN(x) := KIx)−ω(diam(R))ρ(R)md\n\n+, where λ+ := max{λ,0}. We now apply Holder’s inequality R\n\nR0f1f2 ≤ kf1kLp\n\n1(R0)kf2kLp\n\n2(R0) with the functions\n\nf1Nd ψNτ andf2Nd and the exponentsp1=pτ and p2= d .Note that p1\n\n1 +p1\n\n2\n\n1 p+md\n\n= 1. Hence, Z\n\nR0\n\nψNτ ≤ Z\n\nR0\n\nφ\n\nmp d\n\nN ψNp τpZ\n\nR0\n\nφ−1N d\n\n. (3.32)\n\n(15)\n\nNote thatR\n\nR0φ−1N = #(RN)≤N. Furthermore, if R∈ RN andx∈R then according to Lemma 3.2 φN(x)mdψN(x) =|R|1τ1pψN(x)≤ |R|1pkf −IRfkLp(R).\n\nHence,\n\nZ\n\nR0\n\nφ\n\nmp d\n\nN ψNp 1p\n\n\"\n\nX\n\nR∈RN\n\n1\n\n|R|\n\nZ\n\nR\n\nkf−IRfkpL\n\np(R)\n\n#1p\n\n=kf−IRfkLp(R0). (3.33) Inequality (3.32) therefore leads to\n\nNkLτ(R0)≤ kf−IRNfkLp(R0)Nmd. (3.34) Since the sequence (RN)N≥0 is admissible, there exists a constant CA >0 such that for all N and all R ∈ RN we have diam(R) ≤ CAN1d. We introduce a subset of R0N ⊂ RN which collects the most degenerate blocks\n\nR0N ={R∈ RN : ρ(R)≥ω(CAN1d)m1},\n\nwhereω is the function defined in Lemma 3.2. ByRN0 we denote the portion ofR0covered by R0N. For allx∈R0\\R0N we obtain\n\nψN(x)≥KIx)−ω(CANd1)1−d1.\n\nWe defineεN :=ω(CAN1d)1−1d and we notice that εN →0 asN → ∞. Hence, kψNkτL\n\nτ(R0)\n\n(KIx)−εN)+\n\nτ\n\nLτ(R0\\R0N)\n\n(KIx)−εN)+\n\nτ\n\nLτ(R0)−Cτ|R0N|,\n\nwhere C:= maxx∈R0KIx). Next we observe that |R0N| →0 asN →+∞: indeed for all R∈ R0N we have\n\nSince #(R0N)≤N, we obtain|R0N| ≤CAdω(CAN1d)m1, and the right-hand side tends to 0 asN → ∞.\n\nWe thus obtain lim inf\n\nN→∞NkLτ(R0)≥ lim\n\nN→∞\n\n(KIx)−εN)+ L\n\nτ(R0)=kKIx)kLτ(R0). Combining this result with (3.34), we conclude the proof of the announced estimate.\n\nNote that this proof also works with the exponentp=∞by changing Z\n\nR0\n\nφ\n\nmp d\n\nN ψNp τp\n\ninto kφNmdψNkτL(R\n\n0)\n\nin (3.32) and performing the standard modification in (3.33).\n\nRemark 3.3 As announced in Remark 1.6, this proof can be adapted to the weighted normk · kLp(R0,Ω)\n\nassociated to a positive weight function Ω ∈C0(R0) and defined in (1.19). For that purpose let rN :=\n\nsup{diam(R) : R∈ RN} and let\n\nN(x) := inf\n\nx0∈R0\n\n|x−x0|≤rN\n\nΩ(x0).\n\nThe sequence of functions ΩN increases with N and tends uniformly to Ωas N → ∞. If R ∈ RN and x∈R, then\n\nkf−IRfkLp(R,Ω)≥ΩN(x)kf−IRfkLp(R).\n\nThe main change in the proof is that the functionψN should be replaced withψN0 := ΩNψN. Other details are left to the reader.\n\n(16)\n\n### 3.2 Proof of the upper estimates\n\nThe proof of Theorems 1.5 (and 1.7) is based on the actual construction of an asymptotically optimal sequence of block partitions. To that end we introduce the notion of a local block specification.\n\nDefinition 3.4 (local block specification) A local block specification on a block R0 is a (possibly discontinuous) map x7→R(x)which associates to each point x∈R0 a block R(x), and such that\n\n• The volume |R(x)|is a positive continuous function of the variablex∈R0.\n\n• The diameter is bounded : sup{diam(R(x)) : x∈R0}<∞.\n\nThe following lemma shows that it is possible to build sequences of block partitions ofR0adapted in a certain sense to a local block specification.\n\nLemma 3.5 LetR0 be a block in IRd and letx7→R(x)be a local block specification on R0. Then there exists a sequence (Pn)n≥1 of block partitions ofR0,Pn =Pn1∪ Pn2,satisfying the following properties.\n\n• (The number of blocks in Pn is asymptotically controlled)\n\nn→∞lim\n\n#(Pn) n2d =\n\nZ\n\nR0\n\n|R(x)|−1dx. (3.35)\n\n• (The elements of Pn1 follow the block specifications) For eachR∈ Pn1 there existsy∈R0 such that R is a translate ofn−2R(y), and|x−y| ≤ diam(R0)\n\nn for allx∈R. (3.36)\n\n• (The elements of Pn2 have a small diameter)\n\nn→∞lim n2 sup\n\nR∈P2n\n\ndiam(R)\n\n!\n\n= 0. (3.37)\n\nProof: See Appendix.\n\nWe recall that the blockR0, the exponentpand the functionf ∈Cm(R0) are fixed, and that at each point x∈R0 the polynomial πx∈IHm is defined by (3.26). The sequence of block partitions described in the previous lemma is now used to obtain an asymptotical error estimate.\n\nLemma 3.6 Letx7→R(x)be a local block specification such that for allx∈R0\n\nx−IR(x)x)kLp(R(x)) ≤1. (3.38)\n\nLet(Pn)n≥1be a sequence of block partitions satisfying the properties of Lemma 3.5, and let for allN ≥0 n(N) := max{n≥1 : #(Pn)≤N}.\n\nThenRN :=Pn(N) is an admissible sequence of block partitions and lim sup\n\nN→∞\n\nNmdkf−IRNfkLp(R0)≤ Z\n\nR0\n\nR(x)−1dx τ1\n\n. (3.39)\n\nProof: Letn≥0 and letR∈ Pn. IfR∈ Pn1 then lety∈R0 be as in (3.36). Using (3.29) we find kf−IRfkLp(R) ≤ kπy−IRπykLp(R)+k(f−πy)−IR(f−πy)kLp(R)\n\n≤ n2dτy−IR(y)πykLp(R(y))+C|R|1pdiam(R)mkdmf−dmπykL∞(R)\n\n≤ n2dτ +Cn2dτ |R(y)|1pdiam(R(y))mkdmf−dmf(y)kL(R)\n\n≤ n2dτ (1 +C0ω(n−1diam(R0))),\n\nUpdating...\n\n## References\n\nRelated subjects :" ]
[ null ]
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https://gm-567.com/qa/how-do-you-add-60-percent-to-a-price.html
[ "", null, "# How Do You Add 60 Percent To A Price?\n\n## How do you add 30% to a price?\n\nWhen the cost is \\$5.00 you add 0.30 × \\$5.00 = \\$1.50 to obtain a selling price of \\$5.00 + \\$1.50 = \\$6.50.\n\nThis is what I would call a markup of 30%.\n\n0.70 × (selling price) = \\$5.00..\n\n## How do you add 5% to a price?\n\nTo convert 5 percent to a decimal number, so that you can multiply it with any number to add 5 percent to it, you divide 5 by 100 and then add 1. Here is the math to illustrate. Now you can multiply 1.05 by any number that you want to add 5 percent to.\n\n## What is a 10% increase of 50?\n\nTo increase 10 by 50 percent, you add the value of 50 percent, so you add 10 and 5. This gives you an answer of 15. This is what you get when you increase 10 by 50 percent.\n\n## How do you add 15% to a price?\n\nIf you have a starting amount and you want to add a percentage, simply multiply the percent by the original amount to find the amount that gets added.\n\n## How do you add 40 percent to a price?\n\nAn alternative to that is to designate the cost amount as 100% and add the markup percentage to it. For example if your cost is \\$10.00 and you wish to markup that price by 40%, 100% + 40% = 140%. Multiply the \\$10.00 cost by 140% and get the retail price of \\$14.00.\n\n## How do you add percentage to a price?\n\nIf your calculator does not have a percent key and you want to add a percentage to a number multiply that number by 1 plus the percentage fraction. For example 25000+9% = 25000 x 1.09 = 27250. To subtract 9 percent multiply the number by 1 minus the percentage fraction. Example: 25000 – 9% = 25000 x 0.91 = 22750.\n\n## How do you increase a number by 100 percent?\n\nTo calculate the percentage increase:First: work out the difference (increase) between the two numbers you are comparing.Increase = New Number – Original Number.Then: divide the increase by the original number and multiply the answer by 100.% increase = Increase ÷ Original Number × 100.More items…\n\n## What is cost plus 10 percent?\n\nCost plus percentage stands for a form of a contract, for example for construction work. According to such a contact, a contractor is paid for the costs needed to perform the work plus a percentage fee — 10 percent, for example.\n\n## How do you find 10% of a number?\n\nTo calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent.\n\n## How do you discount a price?\n\nFollow the steps below:Convert the percentage to a decimal. Represent the discount percentage in decimal form. … Multiply the original price by the decimal. Take the original price of the item and multiply it by the decimal determined in step one. … Subtract the discount from the original price.\n\n## How do you add 20% to a price?\n\nIf you know the wholesale price of an item and want to calculate how much you must add for a 20 percent markup, multiply the wholesale price by 0.2, which is 20 percent expressed in decimal form. The result is the amount of markup you should add.\n\n## What is a 20% markup?\n\nThe Markup percentage is the percentage of the selling price not represented in the cost of the goods. So if the markup is 20%, then 80% of the selling price is the cost. Your cost is \\$938, so the \\$938/80% = \\$1172.50 would be the cost for a product with a 20% markup.\n\n## How do you calculate a 5% increase?\n\nPercent increase formulaIdentify the original value and the new value.Input the values into the formula.Subtract the original value from the new value, then divide the result by the original value.Multiply the result by 100. … Check your answer using the percentage increase calculator.\n\n## What is a 100% increase?\n\nAn increase of 100% in a quantity means that the final amount is 200% of the initial amount (100% of initial + 100% of increase = 200% of initial). In other words, the quantity has doubled. An increase of 800% means the final amount is 9 times the original (100% + 800% = 900% = 9 times as large).\n\n## How do you add 10% to a price?\n\nTo increase a number by a percentage amount, multiply the original amount by 1+ the percent of increase. In the example shown, Product A is getting a 10 percent increase. So you first add 1 to the 10 percent, which gives you 110 percent. You then multiply the original price of 100 by 110 percent." ]
[ null, "https://mc.yandex.ru/watch/68551822", null ]
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https://manpages.org/nextafter
[ "", null, "nextafter(1) nexttowardl\n\n## SYNOPSIS\n\n#include <math.h>\n\ndouble nextafter(double x, double y);\nfloat nextafterf(float\nx, float y);\nlong double nextafterl(long double\nx, long double y);\ndouble nexttoward(double\nx, long double y);\nfloat nexttowardf(float\nx, long double y);\nlong double nexttowardl(long double\nx, long double y);\n\n## DESCRIPTION\n\nThe nextafter(), nextafterf(), and nextafterl() functions shall compute the next representable floating-point value following x in the direction of y. Thus, if y is less than x, nextafter() shall return the largest representable floating-point number less than x. The nextafter(), nextafterf(), and nextafterl() functions shall return y if x equals y.\n\nThe nexttoward(), nexttowardf(), and nexttowardl() functions shall be equivalent to the corresponding nextafter() functions, except that the second parameter shall have type long double and the functions shall return y converted to the type of the function if x equals y.\n\nAn application wishing to check for error situations should set errno to zero and call feclearexcept(FE_ALL_EXCEPT) before calling these functions. On return, if errno is non-zero or fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) is non-zero, an error has occurred.\n\n## RETURN VALUE\n\nUpon successful completion, these functions shall return the next representable floating-point value following x in the direction of y.\n\nIf x== y, y (of the type x) shall be returned.\n\nIf x is finite and the correct function value would overflow, a range error shall occur and ±HUGE_VAL, ±HUGE_VALF, and ±HUGE_VALL (with the same sign as x) shall be returned as appropriate for the return type of the function.\n\nIf x or y is NaN, a NaN shall be returned.\n\nIf x!= y and the correct function value is subnormal, zero, or underflows, a range error shall occur, and either the correct function value (if representable) or 0.0 shall be returned.\n\n## ERRORS\n\nThese functions shall fail if:\n\nRange Error\nThe correct value overflows.\n\nIf the integer expression (math_errhandling & MATH_ERRNO) is non-zero, then errno shall be set to [ERANGE]. If the integer expression (math_errhandling & MATH_ERREXCEPT) is non-zero, then the overflow floating-point exception shall be raised.\n\nRange Error\nThe correct value is subnormal or underflows.\n\nIf the integer expression (math_errhandling & MATH_ERRNO) is non-zero, then errno shall be set to [ERANGE]. If the integer expression (math_errhandling & MATH_ERREXCEPT) is non-zero, then the underflow floating-point exception shall be raised.\n\nThe following sections are informative.\n\nNone.\n\n## APPLICATION USAGE\n\nOn error, the expressions (math_errhandling & MATH_ERRNO) and (math_errhandling & MATH_ERREXCEPT) are independent of each other, but at least one of them must be non-zero.\n\nNone.\n\nNone." ]
[ null, "https://manpages.org/assets/menu-533ee2b7609b0e242c51d7f42ca18b7f.png", null ]
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https://istd.sutd.edu.sg/undergraduate/courses/50-039-theory-and-practice-of-deep-learning
[ "# 50.039 Theory and Practice of Deep Learning\n\nHome / Education / Undergraduate / Courses / 50.039 Theory and Practice of Deep Learning\n\n#### Course Description\n\nThe course goal is to familiarize students with deep learning for vision, text, reinforcement learning and multimodal combinations thereof. The meaning of deep learning for this course is the training and application of neural networks as prediction models for various setups of input and output modalities. The course will include coding a subset of approaches for vision (so to avoid overlap with computer vision), approaches for sequential data such as 1D‐CNNs, temporal causal networks and recurrent neural networks, multimodal approaches, attention models, explainable AI and generative adversarial neural nets. The coding will cover the whole chain from data loader to training to parameter tuning to performance evaluation. The course will also focus on important practical aspects which are required to make the training part of deep learning work on smaller datasets such as transfer learning, the various forms of data augmentation, different optimizers and learning rate tuning.\n\n#### Pre-requisite\n\n50.007 Machine Learning and ability to to code and debug in Python 3\n\n#### Learning Objective\n\nBy the end of the course, students will be able to\n\n1. Explain the concept of ﴾discriminative﴿ learning from data, generalization and overfitting.\n2. Be able to explain how the decision boundary of a linear neuron depends on its parameters in a general vector space.\n3. Be able to explain what backpropagation is used for.\n4. List for different prediction problems. Suitable outputs of predictors and suitable loss functions.\n5. Explain the basic ideas behind convolutional and recurrent neural networks.\n6. Be able to retell methods of data augmentation.\n7. Be able to use deep learning toolboxes for loading data, training and performance evaluation of deep neural networks.\n8. Construct data loaders for custom datasets of various types with various types of ground truth annotations, such as images, sequential data e.g. text and multi‐modal data.\n9. Being able to set up neural networks for vision, and apply training algorithms to them and evaluate the performance of transfer learning tasks with two state of the art deep learning toolboxes.\n10. Being able to set up neural networks for sequence classification, and apply training algorithms to them and evaluate the performance of transfer learning tasks with one state of the art deep learning toolboxes.\n11. Be able to explain the functioning principle of a generative adversarial neural network.\n\n#### Measurable Outcomes\n\n1. Evaluate deep learning models and loss functions by suitability based on the intended prediction outputs for several prediction task types (ad: LO suitable loss functions).\n2. Compute the terms used in backpropagation for a given neural network topology (ad: LO backprop)\n3. Sketch the set of points for which a linear unit has constant outputs, sketch the directions for which the function values change fastest for a linear unit (ad LO decision boundary of a linear unit).\n4. For a given model and a layer give arguments when to use a fully connected and when to use convolutional layers.\nfor a given problem give arguments when to use feedforward and when to use recurrent neural networks ad LO convolutional NNs, recurrent NNs.\n5. Create code for dataloaders for custom datasets with groundtruths for image and text data.\n6. Create code for neural networks for vision, which loads data, applies training algorithms and evaluates the prediction performance with two state of the art deep learning toolboxes.\n7. Create code for neural networks for sequence classification, which loads data, applies training algorithms and evaluates the prediction performance of transfer learning tasks with one state of the art deep learning toolbox.\n8. Compute train and test loss curves from a training process and be able to examine the amount of overfitting when comparing different pairs of train/test loss curves (ad LO generalization and overfitting).\n9. Employ data augmentation methods and transfer learning in code, evaluate their impact on performance (ad LO data augmentation).\n10. Produce example code for a generative deep neural network, be able to give possible use cases for GANs (ad LO generative adversarial neural network).\n\n#### Topics Covered\n\n• Introduction, Review of ML\n• Syntactic Tagging, Word Senses and Embeddings\n• Language Modeling\n• Chunking (Shallow Syntactic Parsing)\n• Information Extraction\n• Syntactic Parsing\n• Semantic Role Labeling (Shallow Semantic Parsing)\n• Semantic Parsing\n• Sentiment Analysis\n• Text Generation\n• Machine Translation\n\nImage Credit" ]
[ null ]
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https://incomeaftertax.com/23180-plus-vat-de
[ "# €23,180 Plus VAT DE\n\nBelow is the calculation of how much €23,180 Plus VAT DE is for the year 2023. On this page, you will also find all other related information to €23,180 Plus VAT in Germany calculation.\n\n## €23,180 Plus VAT in Germany\n\nValidated and accurate as of October 3rd, 2023 at 12:00 AM.\n\n## How Much is €23180 Plus VAT in the Germany?\n\nAssuming the VAT rate is 19%, the €23,180 plus VAT in the Germany equals €27,584.\n\nPlease check below to learn how €23,180 plus VAT in the DE is calculated.\n\n## How is €23,180 Plus VAT Germany Calculated?\n\nTo calculate the €23,180 Plus VAT in the Germany, you will need to:\n\n1. First, take the total amount of €23,180\n2. Then, divide that by 100 (%) and multiply by 19 (VAT rate %)\n3. €23,180 plus the number above (€4,404) is equal to €27,584\n\nAs you can see, the €23,180 Plus VAT in the Germany calculation is pretty simple. The final result is €27,584, which is how much you are left after adding the VAT.\n\nRemember, this €23,180 Plus VAT DE calculation is just for guidance. Always consult a professional accountant or tax advisor. You can also check a full-length article where Germany VAT is explained in detail.\n\n## How Much VAT Should I Add to The Sum of €23,180 in the DE?\n\nThe answer to this question depends on the value-added tax (VAT) rate that is in effect in the Germany. Currently, the standard VAT rate is 19%, so the amount of VAT added to €23,180 would be €4,404.\n\nWe think these three links are helpful and related to the €23,180 Plus VAT DE: German VAT Refund, VAT in Germany , and Taxation in Germany.", null, "" ]
[ null, "https://incomeaftertax.com/DannyMe.png", null ]
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https://link.springer.com/article/10.1007/BF01086236
[ "Ukrainian Mathematical Journal\n\n, Volume 24, Issue 3, pp 251–258\n\nCoefficient indicators of stability of Lyapunov exponents of two-dimensional linear systems\n\n• N. A. Izobov\nArticle\n\nKeywords\n\nLinear System Lyapunov Exponent Coefficient Indicator\nThese keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.\n\nLiterature cited\n\n1. 1.\nK. P. Persidskii “On eigenvalues of differential equations,” Izv. Akad. Nauk KazSSR,42 (1947).Google Scholar\n2. 2.\nB. F. Bylov, E. É. Vinograd, D. M. Grobman, and V. V. Nemytskii, Theory of Lyapunov Exponents [in Russian], Nauka, Moscow (1966).Google Scholar\n3. 3.\nN. P. Erugin, Reducible Systems [in Russian], Proceedings of the V. A. Steklov Mathematical Institute of the Academy of Sciences, SSSR, Izd-vo AN SSSR, Leningrad-Moscow (1946).Google Scholar\n4. 4.\nR. É. Vinograd, “A new proof of Perron's theorem and certain properties of regular systems,” Usp. Matem. Nauk,9, No. 2 (1954).Google Scholar\n5. 5.\nV. M. Millionshchikov, “Rough properties of linear systems of differential equations,” Different. Urav.,5, No. 10 (1969).Google Scholar\n6. 6.\nB. F. Bylov and N. A. Izobov, “Necessary and sufficient condition for stability of characteristic exponents of linear systems,” Different. Urav.,5, No. 10 (1969).Google Scholar\n7. 7.\nYu. S. Bogdanov, “Eigenvalues of systems of linear differential equations,” Matem. Sb.,41, No. 4 (1957).Google Scholar\n8. 8.\nD. M. Grobman, “Characteristic exponents of systems close to linear,” Matem. Sb.,30, No. 1 (1952).Google Scholar" ]
[ null ]
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https://deepai.org/publication/equivalence-test-in-multi-dimensional-space-with-applications-in-a-b-testing
[ "", null, "# Equivalence Test in Multi-dimensional Space with Applications in A/B Testing\n\nIn this paper, we provide a statistical testing framework to check whether a random sample splitting in a multi-dimensional space is carried out in a valid way, which could be directly applied to A/B testing and multivariate testing to ensure the online traffic split is truly random with respect to the covariates. We believe this is an important step of quality control that is missing in many real world online experiments. Here, we propose a randomized chi-square test method, compared with propensity score and distance components (DISCO) test methods, to test the hypothesis that the post-split categorical data sets have the same multi-dimensional distribution. The methods can be easily generalized to continuous data. We also propose a resampling procedure to adjust for multiplicity which in practice often has higher power than some existing method such as Holm's procedure. We try the three methods on both simulated and real data sets from Adobe Experience Cloud and show that each method has its own advantage while all of them establish promising power. To our knowledge, we are among the first ones to formulate the validity of A/B testing into a post-experiments statistical testing problem. Our methodology is non-parametric and requires minimum assumption on the data, so it can also have a wide range of application in other areas such as clinical trials, medicine, and recommendation system where random data splitting is needed.\n\n## Authors\n\n##### This week in AI\n\nGet the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.\n\n## 1. Introduction\n\nA/B testing (also known as split testing or bucket testing) is a method of comparing two versions of a webpage or app against each other to determine which one performs better. It is essentially a special case of multiple variants of a page which are shown to users at random, and statistical analysis is used to determine which variation performs better for a given key performance indicator (KPI). Sometimes, the results from A/B testing are used as a baseline to be compared with other digital marketing services, e.g. personalization. Thus, it is important that A/B testing or multi-variate testing results are truly valid so that we make the correct inference. One important aspect is to get rid of any confounding. In other words, we want to make sure that the randomly assigned groups have the same profile distributions. In this way any difference in the KPI actually results from the testing objects, instead of from some characteristics of the users such as age or gender. Despite the large volume of literature on A/B testing design, to our knowledge, there has not been any papers which aims to validate the balance of population distribution of different groups. We want to point out that our distribution testing methods are non-parametric by design and can have much more general applications beyond A/B testing, such as in randomized medical clinical trials, in design of experiments in manufacturing fault detection, and in bandits problem in personalized medicine, basically anywhere when uniformly random data splitting is required in a multi-dimensional space.\n\nThe rest of this paper has the following structure: Section 2 discusses past literature on theory and application of randomized controlled trials (RCT) and observational studies. Section 3 introduces some statistical tools, i.e., distance-covariance analysis, propensity score method, randomized chi-square test, and a resampling technique, to test equivalence of multi-dimensional categorical distribution. Section 4 compares and analyzes the above methods on simulated data. Section 5 applies our methods on real traffic data of online marketing from Adobe Experience Cloud. Section 6 discusses further generalization and implementations of our methodology.\n\n## 2. Past Work\n\nRandomized controlled trials are the simplest method to yield unbiased estimates of the treatment effects under a minimal set of assumptions. Researchers have been developing new methodology that boosts the efficiency of randomized trials when sample size is small. Sample size may not be a problem in some e-commerce settings as the number of online visitors can easily build up to hundreds of thousands. However, in other settings such as clinical trials, the number of samples (patients) is often small\n\n(bhat2017, ). (bhat2017, ) formulates online/offline A-B testing as a (computationally challenging) dynamic optimization problem and develops approximation and exact algorithms. In particular, the paper assumes that response is linear in the treatment and covariates: , where or according to whether subject is assigned to treatment or control group, and denotes covariates of subject\n\n. The objective is to maximize overall precision, which is the inverse of the standard error of\n\n. In online setting, is measurable. It can also incorporates other goals such as controlled selection bias and endogenous stopping. Despite clean theoretical guarantees and relatively easy implementation of this algorithm, the linear model is not applicable in many real cases, for example, if is 0 or 1 denoting whether a user convert or not. In this case, a generalized linear model makes more sense but we are unclear about any theoretical results in this scenario. Also the theory requires to have elliptical distribution, which is not the case in many real applications, though one real experiment with categorical data in the paper still performs well. More importantly, the number of covariates may not be clear at the beginning of A/B testing as online user profile may not be available at the time of experiments. Since the current paper is concerned with experiments in e-commerce, the sample size does not pause a big issue.\n\nThere are numerous works on the implementation, validity, and efficiency of A/B testing. (kohavi2009, )\n\ngives a detailed summary of the things data scientists need to pay attention to when carrying out A/B testing, and in general multivariate testing, such as determining the right sample size before testing and stopping the experiments when buggy or unintentionally poor results come up. When there are more than one variant to test, such as the choice of the picture or the color of the text on a website, online experimenters can either do multiple pairwise comparison of each individual variant, and apply methods of multiple testing adjustments, such as Bonferroni adjustment, if the number of variants is small, or do a multivariate test altogether. This paper mentions some possibilities that A/B testing split might not be truly random. One of the most common in practice is the effect of robots. Robots can introduce significant skew into estimates, enough to render the split invalid, and cause many metrics to be significant when they should not have been. For some websites robots are thought to provide up to half the pageviews on the site\n\n(kohavi2004, ). However, it is difficult to clearly delineate between the human users and robots (tan2002, ). Despite mentioning possible confounding factors in online experiments, this paper does not go into details on how to get rid of them, and how to test whether the splitting of traffic is truly random.\n\nWhen randomized trials are not possible, techniques in observational studies are needed to get valid inference. (gordon2016, )\n\nfocuses on the evidence of advertising measurements from big field experiments in Facebook. It uses the advertising effects measured from RCTs as benchmark, and compares the causal effects measured from observational studies to the benchmark. Two methods proposed by the paper to get rid of confounding in non randomized experiments are matching and regression adjustments. Exact matching and propensity matching are two common ways in practice for the former. Inverse-probability-weighed regression adjustment (IPWRA)\n\n(wooldridge2007, ) incorporates propensity information into regression adjustments to provide more robust results. This paper also mentions checking the equivalence of distribution in A/B split so as to create valid benchmark. Yet it simply checks each covariate independently without adjusting for multiplicity. This can incur false positives when the number of covariates is large and cannot detect difference in covariance structure. A/B testing is a design experiment so we do not need to use the methodology of observational studies in our paper.\n\nThere are also papers on the continuous monitoring of A/B testing and how to effectively run a large number of A/B testing. If we plot online experiments results continuously and declare significance the first time p-value is smaller than a pre-defined threshold, we will find that the actual type I error can be much larger than the threshold.\n\n(johari2015, ) introduces methods to continuously monitoring A/B while still controlling type I error or the false discovery rate. (tang2010, ) describes overlapping experiments infrastructure that helps run experiments faster and produce better decisions. (zhang2017, ) describes an adaptive sample size modification for cold start of A/B testing. Network effect can also come into play, especially for experiments conducted in companies like Facebook or LinkedIn. (gui2015, ) proposes an A/B testing effect analysis by taking into account network effect into a linear model.\n\nAll these works contribute to the validity or efficiency of online experiments. We emphasize that our ”randomness check” has a very general application, regardless of how traffic is split, the distribution of the covariates, or whether the experiments are done offline or monitored continuously.\n\n## 3. Methodology\n\nTo formulate the problem mathematically, assume the distributor has assigned incoming users to\n\ngroups. Each user is represented by a row vector of length\n\n, with each entry representing certain profile information, such as age, area, gender, or number of times of past visit. Hence, we have data sets, each of dimension . Our goal is to infer whether the split achieves desired randomness by testing whether the -dimensional data have the same distribution. Throughout our analysis, we assume each column is categorical and each row (user) is independent. These are reasonable assumptions because many important profile informations, such as gender, age, region, are categorical. We assume the number of user in each group is large, and the network effect is negligible compared to the scale of user numbers.\n\nIn the following subsections, we first state the method that is used currently by online experimenters to do the A/B split validity check (gordon2016, ) as baseline. Then we describe a method proposed by (rizzo2010, )\n\n, called distance components (DISCO), of measuring the total dispersion of the samples, which admits a partition of the total dispersion into components analogous to the variance components in ANOVA. We will also apply propensity method and introduce a randomized chi-square test. Each of these distribution test method has its own advantage as we show in Section\n\n4. Finally, we propose a resampling technique that controls family-wise error rate (FWER) under any condition while still maintaining high power compared to some other multiplicity adjustment methods, as again shown in Section 4. Our contribution is to apply and modify existing testing methods to the ubiquitous A/B-multi testing validity check.\n\n### 3.1. Baseline\n\n(gordon2016, ) and some other papers apply F-test 111see https://en.wikipedia.org/wiki/F-test for an introduction of F-test. to each of the column (covariate) or some key columns of the split data sets, and claim the split is truly random if none of the p-value is smaller than 0.05. This is perhaps the most straightforward and simple way of approaching the randomization check problem, but with some potential issues. First, F-test can only test difference in mean, but cannot detect other distributional differences, such as variance (we will see in the next subsection DISCO provides a solution to this). Second, even if each column has the same distribution among the data sets, the interaction among the columns can be different for each data set, which also contribution to multi-dimensional distribution heterogeneity. Third, multiplicity adjustment is needed if the dimension is large, which is often the case for online marketing data. Otherwise, even an A/A test can have very high false positive rate.\n\nTo address the third issue, we apply a resampling technique (introduced later) to adjust for multiplicity. It has the advantage of controlling FWER under any dependence structure of p-values, while maintaining high power compared to other p-value adjustment methods, such as Bonferroni or Holmes method.\n\n### 3.2. DISCO one-step test\n\n(rizzo2010, ) propose a new method, called distance components (DISCO), of measuring the total dispersion of the samples in multi-dimensions, which admits a partition of the total dispersion into components analogous to the variance components in ANOVA. They introduce a measure of dispersion based on Euclidean distances between all pairs of sample elements, for any power , which is called the index, of distances such that . The method is based on the following key definitions and theorem.\n\nSuppose that and are independent and identically distributed (i.i.d.), and and are i.i.d., independent of . If is a constant such that and , define the -distance (energy distance) between the distributions of and as\n\n (1) Eα(X,Y)=2E||X−Y||α−E||X−X′||α−E||Y−Y′||α.\n\nThen we have the following theorem.\n\n###### Theorem 3.1 ().\n\nSuppose that are i.i.d. with distribution , are i.i.d. with distribution , and is independent of . If is a constant such tha and , then the following statements hold:\n(i)\n(ii) If , then if and only if .\n(iii) If , then if and only if .\n\nThe proof of Theorem 3.1 can be read in (rizzo2010, ). Based on this theorem, DISCO statistics, which can be thought of as a variation of the ANOVA statistics, can be developed. Define the empirical distance between distributions as follows. For two -dimensional samples and , the -distance between and is defined as\n\n (2) dα(A,B)=n1n2n1+n2[2gα(A,B)−gα(A,A)−gα(B,B)],\n\nwhere\n\n (3) gα(A,B)=1n1n2n1∑i=1n2∑m=1||ai−bm||α.\n\nNote that is within-sample dispersion and is between-sample dispersion. Similar to ANOVA analysis, we can also write total dispersion as summation of between-sample and within-sample dispersion here. Let be -dimensional samples with sizes . The -sample -distance statistic that takes the role of ANOVA sum of squares for treatments is the weighted sum of dispersion statistics:\n\n (4) Sα(A1,...,AK)=∑1≤j\n\nSimilarly, the total dispersion of the observed response is\n\n (5) Tα(A1,...,AK)=N2gα(A,A),\n\nwhere is the pooled sample, and the within-sample dispersion is\n\n (6) Wα(A1,...,AK)=K∑j=1nj2gα(Aj,Aj).\n\nNote that we have , and when , the decomposition is exactly the ANOVA decomposition of the total squared error: . Hence, ANOVA is a special case of DISCO method.\n\nBased on the decomposition, the final statistics for testing equality of distribution is\n\n (7) Dn,α=Sα/(K−1)Wα/(NK),\n\nwith (if , the above statistics can only test equality of mean). The distribution of is complicated so (rizzo2010, ) uses permutation, which is a simplified version of our resampling technique, to obtain rejection thresholds. As we will see in Section 4\n\n, DISCO one-step test, though simple to implement, does not provide information about which dimensions are problematic when the null hypothesis of equal distribution is rejected.\n\n### 3.3. Propensity score method\n\nPropensity score comparison tests whether the -dimensional data sets have the same distribution by fitting a model to the data to obtain the likelihood of a data point being assigned to a particular data set. To be specific, combine the data sets to be one big data set , which is also the -dimensional covariate space. The response is a length- vector with each entry being the class label (1,2,…,or\n\n) of each data point. We can then fit logistic regression, tree-based model, support vector machine, or any class prediction model to the data to obtain a predicted label for each data point. Here we use multiple logistic regression. If the\n\noriginal data sets truly have the same distribution, the predicted labels should have about uniform distribution on\n\n. We use chi-square test on a contingency table to test the uniformity of distribution. However, we can show, for logistic regression, that only when the data sets are exactly the same is the distribution of the predicted labels truly uniform. Otherwise, the predicted label will tend to be the same as the actual label due to overfitting. To resolve this, we can randomly choose a proportion of rows in each data set to be training data, and the rest are test data. But this reduces the effective test sample size in the stage of chi-square test. For example, if we choose , then only 1/5 of the data are used to chi-square test. This can shrink power when the total number of data points are not too large. Here, like in DISCO one-step test introduced in the previous subsection, we use permutation method to get the rejection threshold for the p-value obtained from the chi-square test.\n\nThe propensity score method, in some sense, takes into account both marginal distributions and interactions because all the covariates appear together on the right hand side of the link function. Interaction terms can also be added, but it is unclear how many of the interaction terms are enough since adding all the interactions are quite unfeasible when the number of covariates is large. However, as we will show in the simulation section, the propensity score method is not very sensitive to non-mean difference (such as variance difference) in distribution. Next, we will give a more direct approach to testing marginal and interaction homogeneity.\n\n### 3.4. Randomized chi-square test\n\nSince the data are categorical, the distribution is determined by the multinomial distribution parameters , where each denotes the probability of seeing a particular combination of categories from the columns, with , where is the total number of categories. With columns in total, the total number of categories can be huge ( is most likely a lower bound). Theoretically, we can create a contingency table and apply chi-square test to test whether the multinomial distribution is independent of data sets (equivalent to the data sets having the same distribution). Yet this is not feasible under computation time constraints. Nor can we reduce the dimensions of the table without losing information. For example, two distributions can have the same two-way interaction but different three-way interaction. However, if we have further information about the number of categories of each column, we can reduce accordingly. A simple example is that all columns are binary (two categories). The distribution in this case is determined by marginal distribution and two-way interaction. Without further information, we compromise by introducing randomization: choosing a small subset of columns each time, say columns, to do chi-square test, and then repeat the process times. The argument is that even if the non-null columns are sparse, by repeatedly choosing times for relatively large , the non-null columns still have large probability of being picked at least once. Assume there are columns in total, the probability that one single column is not picked even once is , which is approximately when is small. Thus, and have equal impacts on the selection probability. In practice, increasing costs less computation than increasing .\n\nNote that randomized chi-square only applies to categorical data among the three methods we introduce. In practice, we can “categorize” continuous data into buckets, which sometimes gives us higher heterogeneous distribution detection power than applying the other methods directly on the continuous data.\n\n### 3.5. A resampling technique\n\nResampling\n\nis often used when the distribution of the test statistics is not known or hard to derive. In our randomized chi-square test, for example, the distribution of P-values depends on the unknown distribution of the original data sets. Thus, we can use resampling to set the threshold of rejection. If we are in single hypothesis testing scenario (one-step DISCO and propensity score methods), our proposed resampling technique is equivalent to permutation test (see Algorithm 1)\n\n555In all the algorithms here, without loss of generality we assume the computed statistics are P-values..\n\nIt is easy to see that Algorithm 1 controls FWER exactly, because all permutations have equal probability under the null. Under multiple hypotheses testing scenario, we modify Algorithm 1 a little (see Algorithm 2).\n\nWe claim that Algorithm 2 controls FWER under any hypotheses configuration.\n\n###### Theorem 3.2 ().\n\nAssume P-values are marginally Uniform(0,1). Under any covariance structure, Algorithm 2 controls FWER, so it controls FDR as well.\n\n###### Proof.\n\nThe proof is conceptually simple. Assume there are non-null and null hypotheses among the hypotheses. Then the probability of making at least one false rejection is the probability that the minimum of a subset of null P-values is less than the th percentile of the distribution of the minimum of total null P-values, which is less than . ∎\n\nIn practice, resampling can also be applied after hypotheses selection (see Algorithm 3).\n\nEmpirical results show that Algorithm 3 also controls FWER, thus FDR, when the selection rule satisfies certain properties. That is, when non-null hypotheses are more likely to be selected than null hypotheses. Intuitively, this controls FWER because the minimum of a subset of a total null P-values is stochastically larger than the minimum of all the null P-values.\n\nWe will compare this resampling technique to other multiplicity adjustment methods such as Holm’s procedure 666See https://en.wikipedia.org/wiki/Holm0-Bonferroni_method for Holm’s procedure., and the Benjamini-Yekutieli (BY) procedure proposed in (benjamini2001, ) in the next section.\n\n## 4. Simulations and analysis\n\nSince real data sets are huge and messy, we first use simulations to compare our methods introduced in the previous section. Heterogeneity in multi-dimensional distributions has two compositions: (a) heterogeneity in marginal distribution of one (or more) dimension(s); (b) heterogeneity in the covariance (interaction) structure among dimensions. In the following simulations, the data is generated with either (a) or (b), or both.\n\nFor randomized chi-square method, deciding the choices of the maximum number of columns to sample each time () and the number of times to sample () is tricky. When dimensions become large, we hope to increase to capture more complicated interaction structure. Yet the number in each cell of the table will decrease, which will diminish the power of chi-square test. Empirically, when the number of columns is not too large, and if the sum of each column of the table is greater than or equal to 5, we pick between and , where is the dimension of the data sets.\n\nWe do not do any feature selection to reduce dimension in our simulations, because in reality online marketing data has a great range of variety depending on the company, and each company has its own way of trimming data sets (either with general feature selection technique such as LASSO, or with domain knowledge). Thus, we do not apply Algorithm 3 in the previous section in this paper.\n\n### 4.1. Detection of heterogeneity in marginal distribution or interaction structure\n\nIn this first simulation, we consider detection of marginal distribution difference and interaction structure difference separately. We simulate four data sets, each with 10 columns (dimensions) and 100 rows. Three of them have iid entry with multinomial distribution with probability vector (0.25,0.25,0.25,0.25) (so the number of categories of each column is also four).\n\nIn Scenario 1, the last data set has different marginal distribution than the other three data sets: for weak signal, the probability vector is (0.3,0.25,0.25,0.2); for medium signal, it is (0.4,0.25,0.2,0.15); for strong signal, it is (0.5,0.2,0.2,0.1). For each heterogeneity we also vary the number of columns (from 1 to 10) that have heterogeneous marginal distribution to test the sensitivity of each method.\n\nIn Scenario 2, the last data set has different interaction structure among columns keeping the marginal distribution the same as the other three sets (the other three have independent columns): first sort each column, and then for strong signal, each column is rotated by 10 (turning an array to . Here\n\ndenotes concatenation) from the previous column, so the columns are positively correlated; for medium signal, after rotation 40% of the data in each column is permuted to mitigate the correlation; for weak signal, after rotation 80% of each column is permuted randomly. Like in Scenario 1, we also vary the number of columns (from 2 to 10) with heterogeneous interaction structure. We compare four methods: baseline t-test, propensity score method with resampling, DISCO method, and randomized chi-square test with resampling.", null, "Figure 1. All three plots are the results of detection of heterogeneity in marginal distribution, the top one for small difference, the middle one for medium difference, the bottom one for big difference", null, "Figure 2. All three plots are the results of detection of heterogeneity in interaction structure, the top one for weak correlation, the middle one for medium correlation, the bottom one for strong correlation\n\nFrom Figure 1 and 2 we see that baseline t-test has high power for detecting marginal heterogeneity, but not for detecting interaction heterogeneity. This agrees with our conjecture. Thus, in the following simulations, we will not test this method anymore. DISCO and propensity score methods have relatively higher power in detecting marginal difference, while randomized chi-square method has significant advantage in detecting interaction heterogeneity. Furthermore, while DISCO or propensity score methods can only tell whether the multi-dimensional distributions are the same, randomized chi-square test can also flag individual problematic columns when the distributions are imbalanced. The flagged columns may not be exhaustive due to randomness in column selection, but it definitely provides a starting point for the follow-up detail diagnosis procedure.\n\nWe also see from the power plots that interaction heterogeneity is in general harder to detect than marginal distribution heterogeneity since the former has convex power line (power only increase significantly when the number of heterogeneous columns is large) while the latter has concave power line.\n\n### 4.2. Varying dimension\n\nWe next vary the dimension of data sets from 10 to 50 while holding the number of rows to be 100. We compare the three methods, propensity methods, DISCO, and randomized chi-square test, on weak heterogeneity (weak marginal heterogeneity + weak interaction heterogeneity as defined in the previous simulation), medium heterogeneity (weak marginal heterogeneity + medium interaction heterogeneity), and strong heterogeneity (medium marginal heterogeneity + medium interaction heterogeneity). In each of the three signal level the number of heterogeneous columns is 1/5 of the dimensions.", null, "Figure 3. All three plots are the results of detection power of heterogeneity as dimension increases, the top one for weak heterogeneity, the middle one for medium heterogeneity, the bottom one for strong heterogeneity\n\nFrom Figure 3 we see randomized chi-square behaves almost the same, or a little better, when the difference in distribution is not too large, while the other two methods, especially DISCO, behave significantly better when the dimension of data sets is relatively large. In other words, randomized chi-square test has higher power in detecting many small effects.\n\nWe also notice that compared to the other two methods, increasing the dimension of the data has little positive effect in power for randomized chi-square test. The reason is that since the portion of heterogeneous columns is unchanged as dimension increases, the probability of picking the ”problematic columns” is also relatively unchanged as dimension increases. For the other two methods, the power has an increasing trend as dimension increases.\n\nWe use resamling procedure to get threshold of rejection in all the four methods, so here we also compare the power of resampling procedure, and one popular multiplicity adjustment procedure, the Holm’s procedure 777See https://en.wikipedia.org/wiki/Holm%E2%80%93Bonferroni_method for an introduction of Holm’s procedure. Both resampling and Holm’s procedure can control type I error under any P-value structure.", null, "Figure 4. Power comparison of resampling and Holm’s procedure for randomized chi-square method, the top one for weak heterogeneity, the middle one for medium heterogeneity, the bottom one for strong heterogeneity\n\nFigure 4 plots the power comparison of applying randomized chi-square test + resampling and applying randomized chi-square test + Holm’s procedure. We see that for any heterogeneity level, resampling has higher power than Holm’s procedure, which justifies the use of resampling procedure in multiple hypotheses testing problems.\n\n### 4.3. Simulated real-world scenario\n\nNext, we simulate a California-local real online marketing A/B testing scenario. Assume the company is an online retailer for fashion, skin care, and cosmetics. The data is collected from a 24-hour window from 12AM February 21st to 12AM February 22nd. There are 8 columns representing area of living, browser type, gender, age, employment status, income, accumulated number of visits (in the past year), and converted or not before respectively. Table 1 below summarizes the encoding details of these 8 features.\n\nThe number of categories for each column ranges from 2 to 5. The real interaction structure among these 8 columns can be complicated. We simplify this relationship and summarize it in Figure 5 below. Area of living and Browser type are independent variables, while the other 6 variables have complicated correlations. It is by no means accurate. For example conditioning on employment status, age and gender may not be independent in reality. We consider the scenario where the data traffic is not randomly split for a period of time. For example, almost all traffic is assigned to set A for two hours in the morning, which results in set A has more data than set B. To balance the number of data points in the two sets, the experimenter assigns most of the traffic in the evening to set B. A direct result is that employment status has different distributions in set A and B — unemployed people take up much higher proportion in the morning time when employed people and students are busy working than they do in the evening time.\n\nFigure 5 also shows a way to generate our data. All marginal distributions are multinomial. Area of living and Browser type are independent which can be easily generated. Then we generate the Employment status with certain probability vector. Conditioning on Employment status, we generate age (the edge with 1), gender (the edge with 2), and income (the edge with 3). Then conditioning on income, gender, age, we generate Accumulated number of visits (the edges with 4). Finally we generate Converted or not conditioning on income and Accumulated number of visits (edges with 5). A detailed data generation procedure is summarized in the Appendix. In particular, the probability vector for Employment status is (0.3,0.3,0.4) in set A (more people in the morning) and (0.4,0.4,0.3) in set B (more people in the evening), which results in difference in marginal distribution and correlation structure in other dimensions. We control the number of data points to be 800 in either data sets.\n\nThe data is simulated 100 times. The power for the propensity method, DISCO method, and randomized chi-square test is 0.93, 0.91, and 0.87 respectively. All three methods can detect the multi-dimensional distribution heterogeneity with high power. Again there is a trade-off: randomized chi-square method has comparably lower power but can point out the most imbalanced columns.\n\nTable 2 displays the number of time each column gets rejected in 200 simulations with (sampling columns 10 times per simulation) and (sampling maximum 3 columns each time). We see Area of living and Browser type have significantly fewer times of being rejected because they are balanced and independent. Employment status has the largest number of sampling times since it is directly affected by the timing of the day, with the rest 5 columns having milder imbalance depending on its correlation with Employment status.\n\n## 5. Anonymous real data\n\nWe also try our tests on some auto-personalization datasets provided by Adobe Digital Marketing Cloud. Auto-personalization is personalized recommendation to individual customers learned from past customer behavior in the database. Usually a random A/B splitting is needed to serve as baseline to evaluate any recommendation methodology. We obtain such a pair of randomly-split data sets, and test the propensity method, DISCO method, and randomized chi-square test on a data set from the online marketing section of a company. The data is split into set A and B, each of dimension , with 2 to 50 categories per column. For privacy concern, we do not get to know the encodings of the columns and the detailed data collecting procedure. However, propensity score with resampling, DISCO method, and randomized chi-square tests all reject the hypothesis that the two distributions are the same. Randomized chi-square also provides some combinations of columns that are the most imbalanced, one of which is columns 196, 57, 248, 260, 271, 342, 374, and 239. These 8 columns have 50 different combinations (for example if one column can take on 2 values, another 3 values, then the two columns together can take on at most values). Figure 6 shows a bar plot of the 15 combinations that have the most counts from both data sets, and a ratio of counts in set B to those in set A. We see the two categorical distributions do differ a great deal, with counts ratio being as high as 3. If we get more information on the these two sets we can do further analysis on these selected columns to identify the source of heterogeneity.", null, "Figure 6. comparison of a subset of columns from two categorical data sets, the red bar representing counts from set A, the green bar from set B, purple line denoting ratio of counts in A to counts in B\n\n## 6. Discussion\n\nIn summary, A/B testing has a wide range of application in online marketing and other areas, so making sure the test indeed provides valid results is crucial. This paper formulates the validity of A/B testing results as a hypothesis testing problem in multi-dimensional space. We propose three ways to test whether two or more data sets come from the same distribution and each has its own advantage. The propensity score and DISCO methods can both be generalized to continuous data, but randomized chi-square test is only applicable to categorical data. Of course, the simplest way is to categorize the continuous data. In fact, even when the data is one-dimensional normal with the same mean but difference variance, both propensity and DISCO methods have very low power, but categorize each data point to be the floor of and apply randomized chi-square test yields high power.\n\nBesides testing equivalence of distribution after the experiments have been carried out, we also hope to make sure the online experiments are correctly carried out from the beginning. This is a meaningful and promising area of research that involves multivariate analysis, multiple testing, feature selection, observational data and sequential analysis. There has been active research going on in some parts of this big area, including the papers we mentioned in Section 2. We also need to work closely with data engineers to effectively implement these methodology.\n\n## Appendix\n\n### .1. Real-world simulation data generation process\n\n• Generate Area of living Multinomial ((0.33,0.33,0.34)).\n\n• Generate Browser type Multinomial ((0.3,0.3,0.2,0.15,0.05)).\n\n• For data set A, generate Employment status Multinomial ((0.3,0.3,0.4)); for data set B, generate Employment status Multinomial ((0.4,0.3,0.3)).\n\n• gender Employment status Multinomial ((0.5,0.5)); gender Employment status Multinomial ((0.6,0.4)); gender Employment status Multinomial ((0.3,0.7)).\n\n• age Employment status Multinomial ((0.7,0.2,0.05,0.03,0.02)); age Employment status Multinomial ((0.1,0.2,0.3,0.3,0.1)); age Employment status Multinomial ((0.1,0.02,0.04,0.04,0.8)).\n\n• income Employment status Multinomial ((0.7,0.2,0.1,0)); income Employment status Multinomial ((0.2,0.3,0.3,0.2)); income Employment status Multinomial ((0.8,0.15,0.05,0)).\n\n• ANOV (Accumulated number of visits) (income & age ) Multinomial ((0.7,0.15,0.1,0.05)); ANOV (income & gender & age ) Multinomial ((0.5,0.2,0.2,0.1)); ANOV (income & gender & age ) Multinomial ((0.15,0.2,0.25,0.4)); ANOV other combinations of age, gender and income Multinomial ((0.25,0.25,0.25,0.25)).\n\n• Conv (converted or not) (income & ANOV ) Multinomial ((0.9,0.1)); Conv (income & ANOV ) Multinomial ((0.85,0.15)); Conv (income & ANOV ) Multinomial ((0.65,0.35)); Conv (income & ANOV ) Multinomial ((0.8,0.2)).\n\n## References\n\n• (1) Yoav Benjamini and Daniel Yekutieli. 2000 The control of the false discovery in multiple testing under dependency. The Annals of Statistics, 29, 4, 1165-88.\n• (2) Nikhil Bhat, Vivek F. Farias, Ciamac C. Moallemi, and Deeksha Sinha. 2017. Near optimal A-B testing.\n• (3) Diane Tang, et al. 2010. Overlapping experiment infrastructure: more, better, faster experimentation. Proceedings 16th Conference on Knowledge Discovery and Data Mining, 17-26.\n• (4) Brett Gordon, Florian Zettelmeyer, Neha Bhargava, and Dan Chapsky. 2016. A Comparison of approaches to advertising measurement: evidence from big field experiments at Facebook. White paper.\n• (5) Huan Gui, Ya Xu, Anmol Bhasin, and Jiawei Han. 2015. Network A/B testing: from sampling to estimation. Proceedings of the 24th international conference on World Wide Web, New York: ACM, 399-409.\n• (6) Ramesh Johari, Leo Pekelis, and David Walsh. 2015. Always valid inference: Brining sequential analysis to A/B testing, arXiv preprint: 1512.04922.\n• (7) Ron Kohavi, Roger Longbotham, Dan Sommerfield, and Randal Henne. 2009. Controlled experiments on the Web: Survey and practical guide. Data Mining and Knowledge Discovery, 18, 140-81.\n• (8) Ron Kohavi, Llew Mason, Rajesh Parekh, and Zijian Zheng. 2004. Lessons and challenges from mining retail e-commerce data. Machine Learn, 57, (1-2), 83-113.\n• (9) Maria L. Rizzo and Gábor J. Székely. 2010. DISCO analysis: A nonparametric extension of analysis of variance. The Annals of Applied Statistics, 4, 2, 1034-55.\n• (10) Pang-Ning Tan and Vipin Kumar. 2002. Discovery of web robot sessions based on their navigational patterns. Data Mining and Knowledge Discovery, 6, 1, 9-35.\n• (11) Jeffrey M. Wooldridge. 2007. Inverse probability weighted estimation for general missing data problems. Journal of Econometrics, 141, 1281-1301.\n• (12) Bo Zhang and Liwei Wang. 2017. Cold-start solution to A/B testing using adaptive sample size modification." ]
[ null, "https://deepai.org/static/images/logo.png", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null, "https://deepai.org/publication/None", null ]
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https://www.24houranswers.com/college-homework-library/Business/Economics/39890
[ "", null, "# Problem Set 3 Answer every question completely for a full grade. ...\n\n##", null, "Question\n\nProblem Set 3\nAnswer every question completely for a full grade. Please clearly show all of your work so I can appreciate how you got to a specific answer or equation. Any unexplained answer will only be judged on accuracy. This means that the correct answer will be granted full credit while the incorrect one will receive a zero. Therefore, I recommend that you show your work. Make sure to draw your graphs clearly indicating the title of each axis.\nThe graphs are estimates. The most important thing is that you show the equilibrium point both on the P axis and the Y axis and how graphs shift in case of policy implementation.\n1- Assume an economy with the following variables:\nYN = 12,500; t = 16% ; YA = 11,500; G = 5000 ; Ta = 975\na) Compute the amount of taxes at natural GDP.\nb) Compute the amount of taxes at real GDP\nc) Compute the actual deficit and express it as percent of actual GDP\nd) Compute the N.E.D (Natural Employment Deficit and express it as a percent of natural GDP.\n\n2- Assume an economy with the following variables:\nYN = 18,500; t = 23% ; YA = 16,650; G = 6,500 ; Ta = 700.\na) Compute the amount of taxes at actual GDP and express it as a percent of actual GDP.\nb) Compute the amount of taxes at natural GDP and express it as a percent of natural GDP.\nc) Policymakers decide to reduce the actual deficit to 6% of the natural GDP. What will be the size of the actual deficit under these conditions?\nd) Given no change in the tax rate t, compute by how much they should cut G to accomplish their goal in question c.\ne) Given no change in G, what should be the tax rate in order to accomplish the policy goal in question c.\nf) Suppose you are asked to make a policy recommendation that requires you to increase the tax rate (from 23%) and decrease government spending. What combination of tax rate and spending cut would you recommend to meet the policymakers’ goals in question c? Answers may vary.\n\n3- Consider the following economy with Ta = 875 and YA = 10000\na) If G = 3,950, Derive the equation of the actual budget deficit BDA as a function of t (tax rate).\nb) Find the tax rate at which the country will be running a budget balance.\nc) Find the values of the deficits for the following tax rates: 10%, 16%, 24%, 32%, 31%, 32% 36% and 40%, 44% and 50%.\nd) Graph your findings in c using any graphing tool.\n\n4- Consider the following economy with Ta = 975 and YA = 10,000.\na) If t is constant at 14%, derive the equation of the BDA as a function of G.\nb) Find the value of G for which the country will be running a budget balance.\nBudget Balance BD = 0.\nc) Find the values of the deficits for the following G spending levels: 10,000, 9000, 7,000, 5,000, 4,000, 3,000, 2,000, 1000 and 0\nd) Graph your findings in c using any graphing tool.\n\n##", null, "Solution Preview\n\nThese solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.\n\n1. a) Tax at natural GDP (T) = 0.16*12500 = 2000\nb) Tax at real GDP = 0.16*11500 = 1840\nc) 5000-1840=3160 Deficit as % of actual GDP = (3160/11500)*100 = 27.48%...\n\nBy purchasing this solution you'll be able to access the following files:\nSolution.docx.\n\n# 50% discount\n\nHours\nMinutes\nSeconds\n\\$25.00 \\$12.50\nfor this solution\n\nor FREE if you\nregister a new account!\n\nPayPal, G Pay, ApplePay, Amazon Pay, and all major credit cards accepted.\n\n### Find A Tutor\n\nView available Economics Tutors\n\nGet College Homework Help.\n\nAre you sure you don't want to upload any files?\n\nFast tutor response requires as much info as possible." ]
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https://math.stackexchange.com/questions/19838/idea-behind-defining-a-projective-system
[ "Idea behind defining a Projective System\n\nWhat is the idea behind defined a Projective system of Groups/Rings. In our class an example for the Projective system was given by Taking the Ring $\\mathbb{Z}/n\\mathbb{Z}$ over $\\mathbb{N}$. The order defined was:\n\n• $m \\leq n$ if $m \\mid n$.\n\nMoreover then we calculated the Projective Limit of this system and found it to be $\\hat{\\mathbb{Z}}$. I have not at all understood this concept fully so even while posing this question there might be errors. So people who can understand what i am talking about are welcome to pose answers. If one still has problems with the question please let me know i shall try to improve it or delete it.\n\n• The projective limit of the system you have described is $\\hat{\\mathbb{Z}}$ and not $\\mathbb{Z}_p$. In particular, there was no $p$ in the set-up, so where should it come from? – Alex B. Feb 1 '11 at 8:00\n• @Alex: Thanks Alex. As i told the question may contain errors and it did. – anonymous Feb 1 '11 at 8:51\n\nA useful heuristic when thinking about projective systems is that you want to specify a congruence condition modulo lots of numbers (or ideals) in a compatible way. For example, you can specify any congruence condition modulo 3 and any congruence condition modulo 5. No matter, what you specify, the two can be satisfied at the same time, since 3 and 5 are co-prime: this is the Chinese remainder theorem. But if you specify a congruence condition modulo 5 and modulo 25, then the two could turn out to be incompatible. E.g. there is no number that is 0 modulo 5, but 1 modulo 25. Why? Because a number that is 0 modulo 5, is divisible by 5, so it can only by congruent to 0,5,10,15, or 20 modulo 25. Similarly, once you have chosen a congruence class modulo 3 and modulo 5, the congruence class modulo 15 is uniquely determined, while the congruence class modulo 30 is not unique but heavily restricted.\n\nThe projective limit of $\\mathbb{Z}/n\\mathbb{Z}$ that you have described is the set of all legal simultaneous choices of congruence classes modulo all integers (with the obvious ring structure). The projective system $\\mathbb{Z}/m\\mathbb{Z}\\rightarrow \\mathbb{Z}/n\\mathbb{Z}$ for $n|m$ expresses all the above rules, that I have spelled out for 3 and 5.\n\nSimilarly, if you fix a prime $p$, then you can consider the projective system $\\mathbb{Z}/p^m\\mathbb{Z}\\rightarrow \\mathbb{Z}/p^n\\mathbb{Z}$ for $n\\leq m$. The projective limit is denoted by $\\mathbb{Z}_p$ and is the set of all possible legal simultaneous choices of congruence classes modulo higher and higher powers of $p$. Thus, the above example shows that $(0,5,\\ldots)$ could represent an element of the projective limit, while $(0,1,\\ldots)$ cannot.\n\nBasically, a projective system formalizes the idea of \"constructing elements by successive approximations\" in an algebraic setting.\n\nIf $G$ is a group (abelian, to fix ideas) and $G_n$ a sequence of smaller and smaller subgroups, the projective limit $\\Gamma=\\lim(G/G_n)$ is the set of elements that are defined by giving them up to an ambiguity that gets smaller and smaller.\n\nWhen $G$ is also a topological space and the $G_n$ are a local basis around the identity element, an element of $\\Gamma$ is morally an element given by assigning a Cauchy sequence converging to it.\n\nIncidentally, this is exactly the case for the $p$-adic integers ${\\Bbb Z}_p$ because it is possible to define a metric in $\\Bbb Z$ in such a way the ideal $p^n{\\Bbb Z}$ are open spheres." ]
[ null ]
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https://lists.racket-lang.org/users/archive/2011-March/044755.html
[ "# [racket] help with typed racket function?\n\n From: Eli Barzilay (eli at barzilay.org) Date: Sun Mar 13 20:42:33 EDT 2011 Previous message: [racket] help with typed racket function? Next message: [racket] help with typed racket function? Messages sorted by: [date] [thread] [subject] [author]\n\n```Two minutes ago, Danny Yoo wrote:\n> I'm trying to make a function that can discriminate typed boxes: I\n> tried the following in Typed Racket:\n>\n> (: ensure-number-box (Any -> (Boxof Number)))\n> (define (ensure-number-box x)\n> (if (and (box? x) (number? (unbox x)))\n> x\n> (error 'ensure-number-box \"~s\" x)))\n>\n> Unfortunately, I can't convince the type checker that x is a (Boxof\n> number). I'm not quite sure how to use the occurrence typing\n> properly to get this to type correctly.\n\nYou can't. Here's something I wrote for my class, and revised it just\nlast week, so it might have some bugs:\n\nThe lack of any subtype relations between (Boxof T) and (Boxof S)\nregardless of S and T can roughly be explained as follows.\n\nFirst, a box is a container that you can pull a value out of -- which\nmakes it similar to lists. In the case of lists, we have:\n\nif: S subtype-of T\nthen: (Listof S) subtype-of (Listof T)\n\nThis is true for all such containers that you can pull a value out of:\nif you expect to pull a T but is given a container of a subtype S, then\nthings are still fine. Such \"containers\" include functions that produce\na value -- for example:\n\nif: S subtype-of T\nthen: Q -> S subtype-of Q -> T\n\nHowever, functions also have the other side, where things are different\n-- instead of a side of some *produced* value, it's the side of the\n*consumed* value. There, we get the opposite rule:\n\nif: T subtype-of S\nthen: S -> Q subtype-of T -> Q\n\nTo see why this is right, use `Number' and `Integer' for S and T:\n\nif: Integer subtype-of Number\nthen: Number -> Q subtype-of Integer -> Q\n\nso -- if you expect a function that takes a number is a *subtype* of one\nthat takes an integer; in other words, every function that takes a\nnumber is also a function that takes an integer, but not the other way.\n\nNow, a (Boxof T) is a producer of T when you pull a value out of the\nbox, but it's also a consumer of T when you put such a value in it.\nThis means that -- using the above analogy -- the T is on both sides of\nthe arrow. This means that\n\nif: S subtype-of T *and* T subtype-of S\nthen: (Boxof S) subtype-of (Boxof T)\n\nwhich is actually:\n\nif: S is-the-same-type-as T\nthen: (Boxof S) is-the-same-type-as (Boxof T)\n\nA different way to look at this conclusion is to consider the function\ntype of (A -> A): when is it a subtype of some other (B -> B)? Only\nwhen A is a subtype of B and B is a subtype of A, which means that this\nhappens only when A and B are the same type.\n\n(Side note: this is related to the fact that in logic, `P => Q' is\nroughly equivalent to `not(P) or Q' -- so the left side, `P', is inside\na negation. It also explains why in ((S -> T) -> Q) the S obeys the\nfirst rule, as if it was on the right side -- because it's negated\ntwice.)\n\nThe following piece of code makes the analogy to function types more\nformally. Boxes behave as if their contents is on both sides of a\nfunction arrow -- on the right because they're readable, and on the left\nbecause they're writeable, which the conclusion that a (Boxof A) type is\na subtype of itself and no other (Boxof B).\n\n#lang pl\n\n;; a type for a \"read-only\" box\n(define-type (Boxof/R A) = (-> A))\n;; Boxof/R constructor\n(: box/r : (All (A) (A -> (Boxof/R A))))\n(define (box/r x) (lambda () x))\n;; we can see that (Boxof/R T1) is a subtype of (Boxof/R T2) if T1 is\n;; a subtype of T2 (this is not surprising, since these boxes are\n;; similar to any other container, like lists):\n(: foo1 : Integer -> (Boxof/R Integer))\n(define (foo1 b) (box/r b))\n(: bar1 : (Boxof/R Number) -> Number)\n(define (bar1 b) (b))\n(test (bar1 (foo1 123)) => 123)\n\n;; a type for a \"write-only\" box\n(define-type (Boxof/W A) = (A -> Void))\n;; Boxof/W constructor\n(: box/w : (All (A) (A -> (Boxof/W A))))\n(define (box/w x) (lambda (new) (set! x new)))\n;; in contrast to the above, (Boxof/W T1) is a subtype of (Boxof/W T2)\n;; if T2 is a subtype of T1, *not* the other way (and note how this is\n;; related to A being on the *left* side of the arrow in the `Boxof/W'\n;; type):\n(: foo2 : Number -> (Boxof/W Number))\n(define (foo2 b) (box/w b))\n(: bar2 : (Boxof/W Integer) Integer -> Void)\n(define (bar2 b new) (b new))\n(test (bar2 (foo2 123) 456))\n\n;; combining the above two into a type for a \"read/write\" box\n(define-type (Boxof/RW A) = (A -> A))\n;; Boxof/RW constructor\n(: box/rw : (All (A) (A -> (Boxof/RW A))))\n(define (box/rw x) (lambda (new) (let ([old x]) (set! x new) old)))\n;; this combines the above two: `A' appears on both sides of the\n;; arrow, so (Boxof/RW T1) is a subtype of (Boxof/RW T2) if T1 is a\n;; subtype of T2 (because there's an A on the right) *and* if T2 is a\n;; subtype of T1 (because there's another A on the left) -- and that\n;; can happen only when T1 and T2 are the same type. So this a type\n;; error:\n;; (: foo3 : Integer -> (Boxof/RW Integer))\n;; (define (foo3 b) (box/rw b))\n;; (: bar3 : (Boxof/RW Number) Number -> Number)\n;; (define (bar3 b new) (b new))\n;; (test (bar3 (foo3 123) 456) => 123)\n;; ** Expected (Number -> Number), but got (Integer -> Integer)\n;; And this a type error too:\n;; (: foo3 : Number -> (Boxof/RW Number))\n;; (define (foo3 b) (box/rw b))\n;; (: bar3 : (Boxof/RW Integer) Integer -> Integer)\n;; (define (bar3 b new) (b new))\n;; (test (bar3 (foo3 123) 456) => 123)\n;; ** Expected (Integer -> Integer), but got (Number -> Number)\n;; The two types must be the same for this to work:\n(: foo3 : Integer -> (Boxof/RW Integer))\n(define (foo3 b) (box/rw b))\n(: bar3 : (Boxof/RW Integer) Integer -> Integer)\n(define (bar3 b new) (b new))\n(test (bar3 (foo3 123) 456) => 123)\n\n--\n((lambda (x) (x x)) (lambda (x) (x x))) Eli Barzilay:\nhttp://barzilay.org/ Maze is Life!\n\n```\n\n Posted on the users mailing list. Previous message: [racket] help with typed racket function? Next message: [racket] help with typed racket function? Messages sorted by: [date] [thread] [subject] [author]" ]
[ null ]
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https://www.roseindia.net/java/javascript-array/javascript-array-merge.shtml
[ "Tutorials\n\n# JavaScript array merge\n\nIn this Tutorial we want to describe you a code that help you in understanding how to merge two different array into one array using Java Script.\n\nIn this Tutorial we want to describe you a code that help you in understanding how to merge two different array into one array using Java Script.\n\n# JavaScript array merge\n\nIn this Tutorial we want to describe you a code that help you in understanding how to merge two different array into one array using Java Script. We declare a array variable that is used to store array object. An array component  is accessed by an array expression that consists of an expression whose value is an array reference indicated by subscripts indexing expression enclosed by `[]`. An array with size length of x can be indexed by the integers from `0` to x`-1`.The document. write prints the elements in an array. The for loop execute the script till the array variable is less than array length.The document. write print the list of element in an array. In the same way we declare another array variable array1 that is used to used to store an array object and each array reference hold the list of element in array1 variable indicated by the indexes. The document.write print the list of element in array.\n\nOn clicking a button the function1( )is invoked that include\n\n1)array.concat(array1) - concatenated the two array and store in a variable c\n\nFinally the document. write print the concatenated array by merging two array.\n\nArraymerge.html\n\n ```     Array Declaration            function function1(){   var c=array.concat(array1);   document.write(\"\"+\"Array after merging is:    \"+\"\"+\"\n\");   document.write(\"====================\"+\"\n\");   for(var i=0;i\")   }    }    var array=new Array();   array=\"Girish\";   array=\"Santosh\";   array=\"Komal\";   array=\"Mahendra\";   document.write(\"\"+\"Element in the array are    \"+\"\"+ \"\");   document.write(\"=================\"+ \"\");   for(var i=0;i\")   }      var array1=new Array();   array1=\"Tewari\";   array1=\"Kashyap\";   array1=\"Chaoudry\";   array1=\"Singh\";   document.write(\"\"+\"Element in the array are    \"+\"\"+ \"\");   document.write(\"=================\"+ \"\");   for(var j=0;j\")   }   document.write(\"\");          Merge Arrays    ```\n\nOutput of the program", null, "", null, "" ]
[ null, "https://www.roseindia.net/tutorialfiles/1/502121.array-merge.gif", null, "https://www.roseindia.net/tutorialfiles/1/502122.arraymerge.gif", null ]
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https://fightinggg.github.io/shoka/PW1ZIX.html
[ "###name\nTransformation\n\n###description\nYuanfang is puzzled with the question below:\nThere are n integers, \\$a_1, a_2, …, a_n\\$. The initial values of them are 0. There are four kinds of operations.\nOperation 1: Add c to each number between ax and ay inclusive. In other words, do transformation \\$a_k=a_k+c\\$, k = x,x+1,…,y.\nOperation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation \\$a_k=a_k×c\\$, k = x,x+1,…,y.\nOperation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation \\$a_k=c\\$, k = x,x+1,…,y.\nOperation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of \\$a_x^p+a_{x+1}^p+…+a_y^p\\$.\nYuanfang has no idea of how to do it. So he wants to ask you to help him.\n\n###input\nThere are no more than 10 test cases.\nFor each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.\nEach the following m lines contains an operation. Operation 1 to 3 is in this format: “1 x y c” or “2 x y c” or “3 x y c”. Operation 4 is in this format: “4 x y p”. (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)\nThe input ends with 0 0.\n\n###output\nFor each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.\n\n###sample input\n5 5\n3 3 5 7\n1 2 4 4\n4 1 5 2\n2 2 5 8\n4 3 5 3\n0 0\n\n###sample output\n307\n7489\n\n###tutorial\n\n###cdoe" ]
[ null ]
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https://softmath.com/math-com-calculator/solving-a-triangle/algebra-and-trigonometry-a.html
[ "", null, "Search Engine visitors came to this page yesterday by using these algebra terms:\n\n• printable GED practice worksheets\n• 7th grade math problems decimals and fractions worksheets\n• sample problems-conics\n• mathpower answers for subtracting fractions\n• nonlinear differential equation in matlab\n• writing absolute values in two forms\n• integral calculator by parts\n• GCE A Level Mathematics Past Years' Exam Papers' Questions And Solutions Manual\n• lesson plan on fractions least to greatest\n• why use absolute values for radicals\n• Writing Linear Equations\n• can you factor equations on graphing calculators\n• math faction example\n• log on ti89\n• standard notation-math\n• Math Problem Solvers\n• ti 84 storing formulas\n• solve my algebra free\n• algebra aptitude algebra test preparation\n• convert the following mixed number into decimal equivalent\n• vertex form\n• ERB SAMPLE TEST\n• How to find the vertex on a graphing calculator\n• properties using integers worksheets\n• homogeneous equation only one root\n• colllege algebra book solutions\n• printable 4th grade fraction activities\n• online linear combination calculator\n• answers to math book McDougal Littell course 1\n• least common multiple and fifth grade lesson\n• how to find scalefactor\n• hardest math problems in the world\n• properties of logarithms 10-3 glencoe answers\n• online basic algebra 1 calculator\n• polynomial program in java\n• Elementary Algebra tutorial\n• how to write the coding using matlab to find the root of three equations\n• Elementary And Intermediate Algebra: 4TH EDITION\n• free printable maths worksheets for grade 6\n• mathmatical tricks\n• finding domain and range from a formula\n• printable worksheets ks3\n• interactive exponent games\n• college algebra with trigonometry seventh edition answers 7-5\n• Math tests PDF year 8 UK\n• matlab nonlinear system equations\n• basic algebra calculator\n• find the vertex calculate\n• adding dividing timesing and subtracting fractions\n• variable exponents\n• hardest math\n• interactive algebra explanations video\n• solve boolean algebra\n• free printable 6th grade math test\n• clep test: College Algebra\n• free college algebra software\n• how to use ladder method\n• solving linear systems pdf calculator ti 83\n• Fractions on the Ti-84\n• cost accounting concepts free online\n• how to do cube root on a scientific calculator\n• lesson plan for squares, square roots, cubes, cube roots\n• self checking decimal worksheet coloring\n• Polynoimial Factor Calculator\n• free cpm worksheet\n• algebraic scientific online calculator\n• summation in a Java program\n• ti 83 \"complex fractions\" solver variable\n• free math tutor probability\n• java how to determine decimal equivalent of a fraction\n• Answers to my math homework\n• sample paper for class viii\n• algebra 2 answers mcdougal littell\n• add equations with square roots\n• elementary math lessons - 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18.\n\nCalculator boolean product, generator for solving quadratic equations, how to store notes on a ti 89, equations with variables on each side printable worksheets, working with pizzazz worksheet, arithmatic sequences.\n\nScience progects, CONVERT 55 TO FRACTIONS, exponents + hands-on, java convert Number to BigDecimal.\n\nEquation multiple variable solver, pizzazz worksheets, math ratio poems.\n\nMixed number conversion, exponents worksheets, excel to solve math Slope, FRACTIONS FROM GREATEST TO LEAST, heaviside function ti-89.\n\nExercices des matric, Multiply, Add, Divide and subtract fractions machines, substitution method calculator, multiply quadratics using radicals, algebra program, fraction of expression and equation for a fifth grader.\n\nVertex form formula, converting fraction worksheets level 2, algebra simplification calculator, graphing a strict nonlinear inequality in two variables.\n\nFree ebooks for aptitude, ti 83 root, math problem solver with radicals.\n\nHow solve vertex form, trigonometry games online free ks3, nonlinear equations, ti-89, free online ged exam in hamilton indiana, \"grade 6 math\" +graphs +ontario.\n\nAlg 2 Answers, what is the square root of 128 simplified?, Help with solving elimination fraction equations.\n\nFree pre-algebra word problem worksheets, comparing worksheets triangles, Algebra - Expanding - Worksheet, ALEKS collge test answers, kumon - download.\n\n\"linear equations\" + \"used for\", easy way to find roots of a cubic, solving equations worksheet, graphing parabolas solver, taks worksheets for high school, algebra calculator roots.\n\nAlgebra tile for solving two-step equations powerpoint, example of math trivia, square root conversion.\n\nLcm factions, Math Permutation and Combination word problems, formulae for converting decimals back to fractions.\n\nRational expressions adding worksheet, factors worksheets ks2, how to convert decimal into roots, clock math poems, elementary intermediate algebra edition 2 McKeague, how to remove radical expression polynomials.\n\nAnalytic accounting howto, prentice hall algebra 2 book online, free online calculator for integers and albebraic expressions, trivias in geometry, \"equation factoring\" on a ti 89, vocabulary english 11 answers by McDougal littell.\n\nLowest common multiple + highest common factor, solving 6 order equations in excel, generalize patterns to identify the nth term lesson plans, lang algebra solutions.\n\nExtrapolation calculator online, tutorkansas.com, solve using the principles together.\n\nMaths hardest question, lesson on expanding trinomials?, intermediate algebra help, help with algebra one.\n\nAdding and subtracting matrix exercises, trigonometry degree chart, hardest easy geometry problem in the world, solving third degree inequalities, calculate lineal metre, simplifying square and cube roots, gcf lcf calculator.\n\nHow to complete the square in word equations, free worksheets comparing and ordering decimals, history worksheet printouts, math worksheet logarithm.\n\nHolt Mathematics answers, GCSE Maths Work Sheets, algebra factoring word problems borders, complex numbers solver, Pizzazz Worksheets - Solving Equations, physics practice worksheets.\n\nDistribution property worksheets, McGraw Hill, College algebra, 8th, help, writing variable expressions worksheets, pictograph worksheets.\n\nRoots and fractional exponents, free grade 8 math tutorials, greatest common factor of 65, Techniques of Integration substitution, quadratics factoring calculator, math worksheet with word problems figuring out percentages.\n\nSteps in a chemical change, Algebra Online Solver, online 2nd grade math tutorial, ap bio book online solomon cheat, cognitive tutor cheat codes.\n\nNotation+binary+octal+decimal, proportion worksheets for 5th grade, solve simultaneous equations on excell cramer's method, Multiplying variables with exponents worksheets.\n\nAlgebra Problem Solver, write a system of equations +ordered pairs, 3rd grade algebra, ading fractions, 2nd order ode matlab.\n\nSimplify radicals by factoring, online polynomial divider, algebra caculator java, where can i get triginometry problems worked out, factorising eqautions, \"Real Life Math\" and \"One Step Problems\".\n\nFree slope equation solvers, california middle school mathematics McDougal Littell \"verbal model\", mathematics+formulae and questions+trigonometry.\n\nSUBTRACT INTEGERS, cube root of a fraction, how to order fractions, decimals, and percentages from least to greatest.\n\nInteger operations worksheet, mcdougal littell algebra 2, chapter 5 answers, solving a polynomial equation homework solution.\n\nSIMPLIFYING EQUATIONS, ti-83 logarithm, T1-84 graphic calculators in sg shops, slope math games, algebra 2 McDougal Littell answers, how to program the quadratic in a ti 84.\n\nCalculas formulas, square roots with exponents, how to store answers on a ti 89, printable maths worksheets for kids ages 8-9, circles tangent function graph quadratic equation free ebook, ellipse calculator solvers, equations with variables third grade." ]
[ null, "https://softmath.com/r-solver/images/tutor.png", null ]
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https://kidsworksheetfun.com/multiplication-timed-test-4th-grade/
[ "", null, "# Multiplication Timed Test 4th Grade\n\nIf you are looking for a solid math fact test the set at the end that omits the times one and times zero problems are excellent for keeping multiplication facts razor sharp. Begin by reinforcing their times tables knowledge with basic multiplication equations or let them jump right into multi digit multiplication word problems and finding factors.\n\n### A multiplication math drill is a worksheet with all of the single digit problems for multiplication on one page.", null, "Multiplication timed test 4th grade. When you re done be sure to check out the unique spiral and bullseye multiplication worksheets to get a. You may choose from a vertical or horizontal format for. Math games by subject.\n\nA student should be able to work out the 60 problems correctly in 1 minute. Increase your mental math speed as you work through these times multiplication worksheets. Each student keeps a personal record of his her timed test accomplishments.\n\nThis is a worksheet for testing the students knowledge of the times tables. A student should be able to work out the 100 problems correctly in 5 minutes 60 problems in 3 minutes or 20 problems in 1 minute. You may select which multiplication times table to use.\n\nHelp your students kick their math skills up a notch with these fourth grade multiplication worksheets and printables. Multiplication worksheets 1 3 or 5 minute drill multiplication worksheets. From minute math to fill in the blank and even some worksheets about time itself these worksheets.\n\nFourth grade timed math drills division facts 0 12 set 2 this is a set of four timed division worksheets with one hundred problems on each page. Times table timed drill worksheets. Periodically in class we will be practicing our multiplication facts via a timed test.\n\nShare to google classroom math games by grade. And dad has a strategy for learning those multiplication facts that you don t want to miss. Multiply divide.\n\nStudents are given 3 minutes to complete up to 100 problems with multiplication facts through 12 12. Two minute timed test 4. This multiplication worksheet is appropriate for kindergarten 1st grade 2nd grade and 3rd grade.\n\nAddition subtraction. Timed multiplication practice 0 7 students will strive to accurately solve one hundred multiplication problems within five minutes or less. Two minute timed test 4.\n\nThis is another free math resource for teachers from the curriculum corner. Use this multiplication timed test set to help your students master their facts. We worked extensively with these two minute tests for placement exams that required demonstrating multiplication fact fluency prior to school admission.\n\nMath timed tests instructions choose a subject choose number of questions email your teacher the report get brownie points. These multiplication worksheets include timed math fact drills fill in multiplication tables multiple digit multiplication multiplication with decimals and much more. Includes multiple pages for each fact family.", null, "The 100 Vertical Questions Multiplying 1 To 12 By 1 To 11 A Printable Multiplication Worksheets Multiplication Worksheets Multiplication Facts Worksheets", null, "Multiplication Worksheets 6th Grade Multiplication In 2020 Math Fact Worksheets 4th Grade Multiplication Worksheets 4th Grade Math Worksheets", null, "Printable Multiplication Worksheets For 4th Grade Math Fact Worksheets Printable Multiplication Worksheets Multiplication Facts Worksheets", null, "Pin On Multiplication Facts", null, "Multiplication Printable Worksheets 4th Grade Paul S House Printable Multiplication Worksheets Math Pages Math Multiplication Worksheets", null, "Multiplication Sheets 4th Grade 4th Grade Math Worksheets Printable Math Worksheets Math Exercises", null, "Space Theme 4th Grade Math Practice Sheets Multiplication Facts 2 Digit Multiplication Practice Rounding Dividing Math Fact Worksheets Free Printable Math Worksheets Fun Math Worksheets", null, "Multiplication Facts To 144 No Zeros No Ones A Multiplication Worksh Math Fact Worksheets Printable Multiplication Worksheets Multiplication Facts Worksheets", null, "Multiplication Worksheets Two Minute Tests 80 And 100 Problem Spaceship Ma Math Multiplication Worksheets Free Multiplication Worksheets Division Worksheets", null, "Pin On Printable Worksheet Template", null, "Multiplication Sheet 4th Grade 4th Grade Math Worksheets Printable Multiplication Worksheets Free Printable Math Worksheets", null, "Multiplication Worksheets Math Fact Worksheets 4th Grade Multiplication Worksheets 4th Grade Math Worksheets", null, "", null, "Two Minute Multiplication Worksheet Education Com Math Drills 3rd Grade Math Worksheets Math Worksheets", null, "Multiplication Worksheets 0 12 Printable Pin On Multiplication Timed Test In 2020 Multiplication Worksheets Fact Fluency Worksheets Math Fact Worksheets", null, "100 Multiplication Worksheet In 2020 Math Fact Worksheets Multiplication Facts Worksheets Multiplication Worksheets", null, "Previous post 7th Grade Simplifying Rational Expressions Worksheet", null, "Next post 10 Single Digit Addition Problems" ]
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https://mathlearnit.com/what-is-75-191-as-a-decimal
[ "What is 75/191 as a decimal?\n\nSolution and how to convert 75 / 191 into a decimal\n\n75 / 191 = 0.393\n\nAs a refresher, we encourage you to check out our fractions section of the website. Also check out our introduction to decimal numbers article. Fractions can also be represented as decimal numbers which is a number with a whole part and a fractional part. In this section, we show you how to calculate the solution to such problems on your own.\n\n• Step 1 (only step): Divide 75 by 191 to get the number as a decimal. 75 / 191 = 0.393. That's all there is to it. After all, a fraction is another way of expressing a division.\nIt is often easier to use a calculator to solve these types of problems, but it is also possible to do it by hand. If you want to practice doing it by hand, you may want to double check your solution with a calculator afterward.\n\nWhen is this useful?\n\nSometimes it is more useful to convey a number as a decimal number rather than a fraction or a percentage. In those cases, this type of conversion is helpful." ]
[ null ]
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https://math.rutgers.edu/news-events/calendar-weekly-view/icalrepeat.detail/2022/10/03/11635/-/quantitative-problems-in-infinite-graph-ramsey-theory
[ "# Calendar\n\nDiscrete Math\n\n## Quantitative problems in infinite graph Ramsey theory\n\n#### Louis DeBiasio (Miami University)\n\nLocation:  Hill Center Room 705\nDate & time: Monday, 03 October 2022 at 2:00PM - 3:00PM\n\nAbstract: Two well-studied problems in Ramsey theory are (1) given a graph G on n vertices, what is the smallest integer N such that there is a monochromatic copy of G in every 2-coloring of a complete graph on N vertices, and (2) given a directed acyclic graph D on n vertices, what is the smallest integer N such that there is a copy of D in every tournament on N vertices.  Note that for both problems, the family of trees has turned out to be an interesting special case, each with a long history and a relatively recent resolution (for sufficiently large n).\n\nWe consider quantitative analogues of these problems in the infinite setting; that is, (1) given a countably infinite graph G what is the supremum of the set of real numbers r such that in every 2-coloring of the complete graph on the natural numbers there is a monochromatic copy of G whose vertex set has upper/lower density at least r, and (2) given a countably infinite directed acyclic graph D what is the supremum of the set of real numbers r such that in every tournament on the natural numbers there is a copy of D whose vertex set has upper/lower density at least r?  As it relates to these problems, I will discuss two very surprising results.\n\nBased on joint work with Alistair Benford, Jan Corsten, and Paul McKenney." ]
[ null ]
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https://ch.mathworks.com/help/textanalytics/ref/htmltree.findelement.html
[ "# findElement\n\nFind elements in HTML tree\n\n## Syntax\n\n``subtrees = findElement(tree,selector)``\n\n## Description\n\nexample\n\n````subtrees = findElement(tree,selector)` returns the elements in `tree` matching the CSS selector.```\n\n## Examples\n\ncollapse all\n\nRead HTML code from the URL `https://www.mathworks.com/help/textanalytics` using the `webread` function.\n\n```url = \"https://www.mathworks.com/help/textanalytics\"; code = webread(url);```\n\nParse the HTML code using `htmlTree`.\n\n`tree = htmlTree(code);`\n\nFind all the hyperlinks in the HTML tree using `findElement`. The hyperlinks are nodes with element name `\"A\"`.\n\n```selector = \"A\"; subtrees = findElement(tree,selector);```\n\nView the first few subtrees.\n\n`subtrees(1:10)`\n```ans = 10×1 htmlTree: <A class=\"skip_link sr-only\" href=\"#content_container\">Skip to content</A> <A href=\"https://www.mathworks.com?s_tid=gn_logo\" class=\"svg_link navbar-brand\"><IMG src=\"/images/responsive/global/pic-header-mathworks-logo.svg\" class=\"mw_logo\" alt=\"MathWorks\"/></A> <A href=\"https://www.mathworks.com/products.html?s_tid=gn_ps\">Products</A> <A href=\"https://www.mathworks.com/solutions.html?s_tid=gn_sol\">Solutions</A> <A href=\"https://www.mathworks.com/academia.html?s_tid=gn_acad\">Academia</A> <A href=\"https://www.mathworks.com/support.html?s_tid=gn_supp\">Support</A> <A href=\"https://www.mathworks.com/matlabcentral/?s_tid=gn_mlc\">Community</A> <A href=\"https://www.mathworks.com/company/events.html?s_tid=gn_ev\">Events</A> <A href=\"https://www.mathworks.com/products/get-matlab.html?s_tid=gn_getml\">Get MATLAB</A> <A href=\"https://www.mathworks.com?s_tid=gn_logo\" class=\"svg_link pull-left\"><IMG src=\"/images/responsive/global/pic-header-mathworks-logo.svg\" class=\"mw_logo\" alt=\"MathWorks\"/></A> ```\n\nExtract the text from the subtrees using `extractHTMLText`. The result contains the link text from each link on the page.\n\n```str = extractHTMLText(subtrees); str(1:10)```\n```ans = 10×1 string \"Skip to content\" \"\" \"Products\" \"Solutions\" \"Academia\" \"Support\" \"Community\" \"Events\" \"Get MATLAB\" \"\" ```\n\n## Input Arguments\n\ncollapse all\n\nHTML tree, specified as a scalar `htmlTree` object.\n\nCSS selector, specified as a string scalar or a character vector. For more information, see CSS Selectors.\n\n## Output Arguments\n\ncollapse all\n\nMatching HTML subtrees, returned as an `htmlTree` array.\n\ncollapse all\n\n### HTML Elements\n\nA typical HTML element contains the following components:\n\n• Element name – Name of the HTML tag. The element name corresponds to the `Name` property of the HTML tree.\n\n• Attributes – Additional information about the tag. HTML attributes have the form `name=\"value\"`, where `name` and `value` denote the attribute name and value respectively. The attributes appear inside the opening HTML tag. To get the attribute values from an HTML tree, use `getAttribute`.\n\n• Content – Element content. The content appears between opening and closing HTML tags. The content can be text data or nested HTML elements. To extract the text from an `htmlTree` object, use `extractHTMLText`. To get the nested HTML elements of an `htmlTree` object, use the `Children` property.\n\nFor example, the HTML element ```<a href=\"https://www.mathworks.com\">Home</a>``` comprises the following components:\n\nComponentValueDescription\nElement name`a`Element is a hyperlink\nAttributeAttribute name`href`Hyperlink reference\nAttribute value`\"https://www.mathworks.com\"`Hyperlink reference value\nContent`Home`Text to display\n\n### CSS Selectors\n\nCSS selectors specify patterns to match elements in a tree.\n\nThis table shows some examples showing how to extract different HTML elements from an HTML tree:\n\nFind all paragraph (`<p>`) elements.`\"p\"``findElement(tree,\"p\")`\nFind all paragraph (`<p>`) and list item (`<li>`) elements.`\"p,li\"``findElement(tree,\"p,li\")`\nFind all paragraph (`<p>`) elements that are inside table (`<table>`) elements.`\"table p\"``findElement(tree,\"table p\")`\nFind all hyperlink (`<a>`) elements with hyperlink reference attribute (`href`) values ending with `\".pdf\"`.`\"a[href\\$=\"\".pdf\"\"]\"``findElement(tree,\"a[href\\$=\"\".pdf\"\"]\")`\nFind all paragraph (`<p>`) elements that are the first child of their parent.`\"p:first-child\"``findElement(tr,\"p:first-child\")`\nFind all paragraph (`<p>`) elements that are the first paragraph element of their parent.`\"p:first-of-type\"``findElement(tr,\"p:first-of-type\")`\nFind all emphasis (`<em>`) elements where the parent is a paragraph (`<p>`) element.`\"p > em\"``findElement(tr,\"p > em\")`\nFind all paragraph (`<p>`) elements appearing immediately after a heading 1 (`<h1>`) element`\"h1 + p\"``findElement(tr,\"h1 + p\")`\nFind all empty elements.`\":empty\"``findElement(tr,\":empty\")`\nFind all nonempty label (`<label>`) elements.`\"label:not(:empty)\"``findElement(tr,\"label:not(:empty)\")`\n\nThe `findElement` function supports all of CSS level 3, except for the selectors `\":lang\"`, `\":checked\"`, `\":link\"`, `\":active\"`, `\":hover\"`, `\":focus\"`, `\":target\"`, `\":enabled\"`, and `\":disabled\"`." ]
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http://webiwip.com/quantitative-aptitude-questions-answers/trigonometry/70447
[ "# Trigonometry - Aptitude Questions and Answers\n\n If sinA= 8/17 and cosA=15/17, then cotA=? A) 17/8 B) 8/15 C) 17/15 D) 15/8 Correct Answer :15/8 Explanation :As given sinA = 8/17 = perpendicular/hypotenuse cosA= 15/17 = base/hypotenuse tanA = sinA/cosA = perpendicular/base = 8/15 cotA = 1/tanA = base/perpendicular = 15/8 Post/View Answer Post comment Cancel Thanks for your comment.! Write a comment(Click here) ...\n What is the value of cot 4π/3 ? A) 1/√3 B) -√3 C) -1 D) √3 Correct Answer :1/√3 Explanation :cot 4π/3 = cos 4π/3 / sin 4π/3 = -1/2 / -√3/2 = 1/√3 What is the value of sinx √( 1 /1+cosx + 1 /1-cosx ) ? A) √2 B) 2√2 C) √2 tanx D) 0 Correct Answer :√2 Explanation :sinx √( 1 /1+cosx + 1 /1-cosx ) =sinx √( 1-cosx + 1+cosx / (1+cosx ) (1-cosx) ) =sinx √( 2 / (1 - cos2x ) ) =sinx √ (2 / sin2x) =sinx √ 2 / sinx =√ 2 What is cos4 A - sin4 A / cos2 A - sin2 A equal to? A) cos2 A - sin2 A B) cos A - sin A C) 1 D) 2 Correct Answer :1 Explanation :cos4 A - sin4 A / cos2 A - sin2 A = (cos2 A)2 - (sin2 A)2 / cos2 A - sin2 A = (cos2 A + sin2 A) (cos2 A - sin2 A) / (cos2 A - sin2 A) = (cos2 A + sin2 A) (cos2 A - sin2 A) / (cos2 A - sin2 A) = (cos2 A + sin2 A) ​​​​​​​​​​​​​​= 1 If 7sin2 x +3cos2 x = 4, 0 < x < 90º ,then what is the value of tan x? A) √2 B) 1 C) √3/2 D) 1/√3 Correct Answer :1/√3 Explanation :7sin2 x +3cos2 x = 4 7sin2 x/cos2 x + 3cos2 x/cos2 x = 4/cos2 x 7tan2 x + 3 = 4sec2 x ​​​​​​​7tan2 x + 3 = 4(1+tan2 x) ​​​​​​​​​​​​​​3tan2 x = 1 ​​​​​​​​​​​​​​tan x = 1/√3 If x = a cosθ + b sinθ and y = a sinθ - b cosθ then what is x2 + y2 equal to? A) 2ab B) a + b C) a2 + b2 D) a2 - b2 Correct Answer :a2 + b2 Explanation :x2 + y2 = (a cosθ + b sinθ)2 + (a sinθ - b cosθ)2 = (a2 cos2θ + b2 sin2θ + 2*ab* cosθ*sinθ) + (a2 cos2θ + b2 sin2θ - 2*ab* cosθ*sinθ) = (a2 cos2θ + b2 sin2θ + a2 cos2θ + b2 sin2θ ) = a2(cos2θ + sin2θ) + b2 (cos2θ + sin2θ)​​​​​​​ = a2+ b2 2 tan 45° /( 1+ tan2 45°) = ? A) sin 90° B) tan 30° C) tan 90° D) sin 45° Correct Answer :sin 90° Explanation :2 tan 45° /( 1+ tan2 45°) = 2 / (1 + 12) (tan 45°=1) = 1 = sin 90°" ]
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https://www.keyword-suggest-tool.com/search/odds+ratio+example/
[ "# Odds ratio example\n\nOdds ratio example keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website\n\n## Keyword Suggestions\n\nOdds ratio example\n\nOdds ratio example problems\n\nOdds ratio example physicians heart attack\n\n( Please select at least 2 keywords )\n\n#### Most Searched Keywords\n\n## Websites Listing\n\nWe found at least 10 Websites Listing below when search with odds ratio example on Search Engine\n\nRelative Risk and Odds Ratios Examples\n\nRelative Risk and Odds Ratios: Examples Calculating Relative Risk Calculating Relative Risk Imagine that the incidence of gun violence is compared in two cities, one with relaxed gun laws (A), the other with strict gun laws (B). In the city with relaxed gun laws, there were 50 shootings in a\n\nhttps://atrium.lib.uoguelph.ca/xmlui/bitstream/handle/10214/1873/B_Relative_Risk_and_Odds_Ratios_Examples.pdf?sequence=8\n\nA beginner’s guide to interpreting odds ratios, confidence ...\n\nhttps://www.students4bestevidence.net/a-beginners-guide-to-interpreting-odds-ratios-confidence-intervals-and-p-values-the-nuts-and-bolts-20-minute-tutorial/\n\nOdds ratio - Wikipedia\n\nThe sample odds ratio n 11 n 00 / n 10 n 01 is easy to calculate, and for moderate and large samples performs well as an estimator of the population odds ratio. When one or more of the cells in the contingency table can have a small value, the sample odds ratio can be biased and exhibit high variance .\n\nhttps://en.m.wikipedia.org/wiki/Odds_ratio\n\nHow do I interpret odds ratios in logistic regression ...\n\n e b = e [log(odds male /odds female)] = odds male /odds female = OR . which means the the exponentiated value of the coefficient b results in the odds ratio for gender. In our particular example, e 1.694596 = 5.44 which implies that the odds of being admitted for males is 5.44 times that of females.\n\nhttps://stats.idre.ucla.edu/stata/faq/how-do-i-interpret-odds-ratios-in-logistic-regression/\n\nExplaining Odds Ratios - ncbi.nlm.nih.gov\n\nOdds ratios and logistic regression. When a logistic regression is calculated, the regression coefficient (b1) is the estimated increase in the log odds of the outcome per unit increase in the value of the exposure. In other words, the exponential function of the regression coefficient (e b1) is the odds ratio associated with a one-unit increase in the exposure.\n\nhttps://www.ncbi.nlm.nih.gov/pmc/articles/PMC2938757/\n\nHow do I interpret odds ratios in logistic regression ...\n\nodds(failure) = q/p = .2/.8 = .25. This looks a little strange but it is really saying that the odds of failure are 1 to 4. The odds of success and the odds of failure are just reciprocals of one another, i.e., 1/4 = .25 and 1/.25 = 4. Next, we will add another variable to the equation so that we can compute and odds ratio. Another example\n\nhttps://stats.idre.ucla.edu/spss/faq/how-do-i-interpret-odds-ratios-in-logistic-regression/\n\nOdds Ratio Calculator - calculate odds ratio. Confidence ...\n\nOdds Ratio Calculator. Use this odds ratio calculator to easily calculate the ratio of odds, confidence intervals and p-values for the odds ratio (OR) between an exposed and control group. One and two-sided confidence intervals are reported, as well as Z-scores.\n\nhttps://www.gigacalculator.com/calculators/odds-ratio-calculator.php\n\nThe Difference Between Relative Risk and Odds Ratios - The ...\n\nThe basic difference is that the odds ratio is a ratio of two odds (yep, it’s that obvious) whereas the relative risk is a ratio of two probabilities. (The relative risk is also called the risk ratio). Let’s look at an example. Relative Risk/Risk Ratio. Suppose you have a school that wants to test out a new tutoring program.\n\nhttps://www.theanalysisfactor.com/the-difference-between-relative-risk-and-odds-ratios/\n\nHow to Calculate an Adjusted Odds Ratio | Sciencing\n\nAn odds ratio is the statistical measure of association between an exposure and an outcome. In other words, the odds ratio is the statistical chance than an outcome will occur under a specific condition: in the case of our example, the odds ratio represents the chance that taking one of two asthma medications may still lead to a hospital visit." ]
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https://thcscience.wiki/colab/term/terpene/?rdp_we_resource=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FChemical_formula
[ "# Chemical formula\n\n Aluminium sulfate has the chemical formula Al2(SO4)3. The form of aluminium sulfate hexadecahydrate is Al2(SO4)3·16H2O.\n Structural formula for butane. Examples of other chemical formulas for butane are the empirical formula C2H5, the molecular formula C4H10 and the condensed (or semi-structural) formula CH3CH2CH2CH3.\n\nA chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, commas and plus (+) and minus (−) signs. These are limited to a single typographic line of symbols, which may include subscripts and superscripts. A chemical formula is not a chemical name, and it contains no words. Although a chemical formula may imply certain simple chemical structures, it is not the same as a full chemical structural formula. Chemical formulas can fully specify the structure of only the simplest of molecules and chemical substances, and are generally more limited in power than are chemical names and structural formulas.\n\nThe simplest types of chemical formulas are called empirical formulas, which use letters and numbers indicating the numerical proportions of atoms of each type. Molecular formulas indicate the simple numbers of each type of atom in a molecule, with no information on structure. For example, the empirical formula for glucose is CH2O (twice as many hydrogen atoms as carbon and oxygen), while its molecular formula is C6H12O6 (12 hydrogen atoms, six carbon and oxygen atoms).\n\nSometimes a chemical formula is complicated by being written as a condensed formula (or condensed molecular formula, occasionally called a \"semi-structural formula\"), which conveys additional information about the particular ways in which the atoms are chemically bonded together, either in covalent bonds, ionic bonds, or various combinations of these types. This is possible if the relevant bonding is easy to show in one dimension. An example is the condensed molecular/chemical formula for ethanol, which is CH3-CH2-OH or CH3CH2OH. However, even a condensed chemical formula is necessarily limited in its ability to show complex bonding relationships between atoms, especially atoms that have bonds to four or more different substituents.\n\nSince a chemical formula must be expressed as a single line of chemical element symbols, it often cannot be as informative as a true structural formula, which is a graphical representation of the spatial relationship between atoms in chemical compounds (see for example the figure for butane structural and chemical formulas, at right). For reasons of structural complexity, a single condensed chemical formula (or semi-structural formula) may correspond to different molecules, known as isomers. For example glucose shares its molecular formula C6H12O6 with a number of other sugars, including fructose, galactose and mannose. Linear equivalent chemical names exist that can and do specify uniquely any complex structural formula (see chemical nomenclature), but such names must use many terms (words), rather than the simple element symbols, numbers, and simple typographical symbols that define a chemical formula.\n\nChemical formulas may be used in chemical equations to describe chemical reactions and other chemical transformations, such as the dissolving of ionic compounds into solution. While, as noted, chemical formulas do not have the full power of structural formulas to show chemical relationships between atoms, they are sufficient to keep track of numbers of atoms and numbers of electrical charges in chemical reactions, thus balancing chemical equations so that these equations can be used in chemical problems involving conservation of atoms, and conservation of electric charge.\n\n## Overview\n\nA chemical formula identifies each constituent element by its chemical symbol and indicates the proportionate number of atoms of each element. In empirical formulas, these proportions begin with a key element and then assign numbers of atoms of the other elements in the compound, by ratios to the key element. For molecular compounds, these ratio numbers can all be expressed as whole numbers. For example, the empirical formula of ethanol may be written C2H6O because the molecules of ethanol all contain two carbon atoms, six hydrogen atoms, and one oxygen atom. Some types of ionic compounds, however, cannot be written with entirely whole-number empirical formulas. An example is boron carbide, whose formula of CBn is a variable non-whole number ratio with n ranging from over 4 to more than 6.5.\n\nWhen the chemical compound of the formula consists of simple molecules, chemical formulas often employ ways to suggest the structure of the molecule. These types of formulas are variously known as molecular formulas and condensed formulas. A molecular formula enumerates the number of atoms to reflect those in the molecule, so that the molecular formula for glucose is C6H12O6 rather than the glucose empirical formula, which is CH2O. However, except for very simple substances, molecular chemical formulas lack needed structural information, and are ambiguous.\n\nFor simple molecules, a condensed (or semi-structural) formula is a type of chemical formula that may fully imply a correct structural formula. For example, ethanol may be represented by the condensed chemical formula CH3CH2OH, and dimethyl ether by the condensed formula CH3OCH3. These two molecules have the same empirical and molecular formulas (C2H6O), but may be differentiated by the condensed formulas shown, which are sufficient to represent the full structure of these simple organic compounds.\n\nCondensed chemical formulas may also be used to represent ionic compounds that do not exist as discrete molecules, but nonetheless do contain covalently bound clusters within them. These polyatomic ions are groups of atoms that are covalently bound together and have an overall ionic charge, such as the sulfate [SO\n4\n]2−\nion. Each polyatomic ion in a compound is written individually in order to illustrate the separate groupings. For example, the compound dichlorine hexoxide has an empirical formula ClO\n3\n, and molecular formula Cl\n2\nO\n6\n, but in liquid or solid forms, this compound is more correctly shown by an ionic condensed formula [ClO\n2\n]+\n[ClO\n4\n]\n, which illustrates that this compound consists of [ClO\n2\n]+\nions and [ClO\n4\n]\nions. In such cases, the condensed formula only need be complex enough to show at least one of each ionic species.\n\nChemical formulas as described here are distinct from the far more complex chemical systematic names that are used in various systems of chemical nomenclature. For example, one systematic name for glucose is (2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal. This name, interpreted by the rules behind it, fully specifies glucose's structural formula, but the name is not a chemical formula as usually understood, and uses terms and words not used in chemical formulas. Such names, unlike basic formulas, may be able to represent full structural formulas without graphs.\n\n## Empirical formula\n\nIn chemistry, the empirical formula of a chemical is a simple expression of the relative number of each type of atom or ratio of the elements in the compound. Empirical formulas are the standard for ionic compounds, such as CaCl\n2\n, and for macromolecules, such as SiO\n2\n. An empirical formula makes no reference to isomerism, structure, or absolute number of atoms. The term empirical refers to the process of elemental analysis, a technique of analytical chemistry used to determine the relative percent composition of a pure chemical substance by element.\n\nFor example, hexane has a molecular formula of C\n6\nH\n14\n, or structurally CH\n3\nCH\n2\nCH\n2\nCH\n2\nCH\n2\nCH\n3\n, implying that it has a chain structure of 6 carbon atoms, and 14 hydrogen atoms. However, the empirical formula for hexane is C\n3\nH\n7\n. Likewise the empirical formula for hydrogen peroxide, H\n2\nO\n2\n, is simply HO expressing the 1:1 ratio of component elements. Formaldehyde and acetic acid have the same empirical formula, CH\n2\nO\n. This is the actual chemical formula for formaldehyde, but acetic acid has double the number of atoms.\n\n## Molecular formula\n\nMolecular formulas indicate the simple numbers of each type of atom in a molecule of a molecular substance. They are the same as empirical formulas for molecules that only have one atom of a particular type, but otherwise may have larger numbers. An example of the difference is the empirical formula for glucose, which is CH2O (ratio 1:2:1), while its molecular formula is C6H12O6 (number of atoms 6:12:6). For water, both formulas are H2O. A molecular formula provides more information about a molecule than its empirical formula, but is more difficult to establish.\n\nA molecular formula shows the number of elements in a molecule, and determines whether it is a binary compound, ternary compound, quaternary compound, or has even more elements.\n\n## Condensed formula", null, "n-Butane structural formula\nMolecular formula: C4H10\nCondensed or semi-structural formula: CH3CH2CH2CH3\n\nThe connectivity of a molecule often has a strong influence on its physical and chemical properties and behavior. Two molecules composed of the same numbers of the same types of atoms (i.e. a pair of isomers) might have completely different chemical and/or physical properties if the atoms are connected differently or in different positions. In such cases, a structural formula is useful, as it illustrates which atoms are bonded to which other ones. From the connectivity, it is often possible to deduce the approximate shape of the molecule.\n\nA condensed chemical formula may represent the types and spatial arrangement of bonds in a simple chemical substance, though it does not necessarily specify isomers or complex structures. For example, ethane consists of two carbon atoms single-bonded to each other, with each carbon atom having three hydrogen atoms bonded to it. Its chemical formula can be rendered as CH3CH3. In ethylene there is a double bond between the carbon atoms (and thus each carbon only has two hydrogens), therefore the chemical formula may be written: CH2CH2, and the fact that there is a double bond between the carbons is implicit because carbon has a valence of four. However, a more explicit method is to write H2C=CH2 or less commonly H2C::CH2. The two lines (or two pairs of dots) indicate that a double bond connects the atoms on either side of them.\n\nA triple bond may be expressed with three lines (HC≡CH) or three pairs of dots (HC:::CH), and if there may be ambiguity, a single line or pair of dots may be used to indicate a single bond.\n\nMolecules with multiple functional groups that are the same may be expressed by enclosing the repeated group in round brackets. For example, isobutane may be written (CH3)3CH. This condensed structural formula implies a different connectivity from other molecules that can be formed using the same atoms in the same proportions (isomers). The formula (CH3)3CH implies a central carbon atom connected to one hydrogen atom and three CH3 groups. The same number of atoms of each element (10 hydrogens and 4 carbons, or C4H10) may be used to make a straight chain molecule, n-butane: CH3CH2CH2CH3.\n\n## Law of composition\n\nIn any given chemical compound, the elements always combine in the same proportion with each other. This is the law of constant composition.\n\nThe law of constant composition says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the same proportion or ratio. For example, any water molecule is always made up of two hydrogen atoms and one oxygen atom in a 2:1 ratio. If we look at the relative masses of oxygen and hydrogen in a water molecule, we see that 94% of the mass of a water molecule is accounted for by oxygen and the remaining 6% is the mass of hydrogen. This mass proportion will be the same for any water molecule.\n\n### Chemical names in answer to limitations of chemical formulas\n\nThe alkene called but-2-ene has two isomers, which the chemical formula CH3CH=CHCH3 does not identify. The relative position of the two methyl groups must be indicated by additional notation denoting whether the methyl groups are on the same side of the double bond (cis or Z) or on the opposite sides from each other (trans or E). Such extra symbols violate the rules for chemical formulas, and begin to enter the territory of more complex naming systems.[citation needed]\n\nAs noted above, in order to represent the full structural formulas of many complex organic and inorganic compounds, chemical nomenclature may be needed which goes well beyond the available resources used above in simple condensed formulas. See IUPAC nomenclature of organic chemistry and IUPAC nomenclature of inorganic chemistry 2005 for examples. In addition, linear naming systems such as International Chemical Identifier (InChI) allow a computer to construct a structural formula, and simplified molecular-input line-entry system (SMILES) allows a more human-readable ASCII input. However, all these nomenclature systems go beyond the standards of chemical formulas, and technically are chemical naming systems, not formula systems.[citation needed]\n\n### Polymers in condensed formulas\n\nFor polymers in condensed chemical formulas, parentheses are placed around the repeating unit. For example, a hydrocarbon molecule that is described as CH3(CH2)50CH3, is a molecule with fifty repeating units. If the number of repeating units is unknown or variable, the letter n may be used to indicate this formula: CH3(CH2)nCH3.\n\n### Ions in condensed formulas\n\nFor ions, the charge on a particular atom may be denoted with a right-hand superscript. For example, Na+, or Cu2+. The total charge on a charged molecule or a polyatomic ion may also be shown in this way. For example: H3O+ or SO42−. Note that + and - are used in place of +1 and -1, respectively.\n\nFor more complex ions, brackets [ ] are often used to enclose the ionic formula, as in [B12H12]2−, which is found in compounds such as Cs2[B12H12]. Parentheses ( ) can be nested inside brackets to indicate a repeating unit, as in [Co(NH3)6]3+Cl3. Here, (NH3)6 indicates that the ion contains six NH3 groups bonded to cobalt, and [ ] encloses the entire formula of the ion with charge +3.[further explanation needed]\n\nThis is strictly optional; a chemical formula is valid with or without ionization information, and Hexamminecobalt(III) chloride may be written as [Co(NH3)6]3+Cl3 or [Co(NH3)6]Cl3. Brackets, like parentheses, behave in chemistry as they do in mathematics, grouping terms together – they are not specifically employed only for ionization states. In the latter case here, the parentheses indicate 6 groups all of the same shape, bonded to another group of size 1 (the cobalt atom), and then the entire bundle, as a group, is bonded to 3 chlorine atoms. In the former case, it is clearer that the bond connecting the chlorines is ionic, rather than covalent.\n\n## Isotopes\n\nAlthough isotopes are more relevant to nuclear chemistry or stable isotope chemistry than to conventional chemistry, different isotopes may be indicated with a prefixed superscript in a chemical formula. For example, the phosphate ion containing radioactive phosphorus-32 is [32PO4]3−. Also a study involving stable isotope ratios might include the molecule 18O16O.\n\nA left-hand subscript is sometimes used redundantly to indicate the atomic number. For example, 8O2 for dioxygen, and 16\n8\nO\n2\nfor the most abundant isotopic species of dioxygen. This is convenient when writing equations for nuclear reactions, in order to show the balance of charge more clearly.\n\n## Trapped atoms\n\nThe @ symbol (at sign) indicates an atom or molecule trapped inside a cage but not chemically bound to it. For example, a buckminsterfullerene (C60) with an atom (M) would simply be represented as MC60 regardless of whether M was inside the fullerene without chemical bonding or outside, bound to one of the carbon atoms. Using the @ symbol, this would be denoted M@C60 if M was inside the carbon network. A non-fullerene example is [As@Ni12As20]3−, an ion in which one As atom is trapped in a cage formed by the other 32 atoms.\n\nThis notation was proposed in 1991 with the discovery of fullerene cages (endohedral fullerenes), which can trap atoms such as La to form, for example, La@C60 or La@C82. The choice of the symbol has been explained by the authors as being concise, readily printed and transmitted electronically (the at sign is included in ASCII, which most modern character encoding schemes are based on), and the visual aspects suggesting the structure of an endohedral fullerene.\n\n## Non-stoichiometric chemical formulas\n\nChemical formulas most often use integers for each element. However, there is a class of compounds, called non-stoichiometric compounds, that cannot be represented by small integers. Such a formula might be written using decimal fractions, as in Fe0.95O, or it might include a variable part represented by a letter, as in Fe1–xO, where x is normally much less than 1.\n\n## General forms for organic compounds\n\nA chemical formula used for a series of compounds that differ from each other by a constant unit is called a general formula. It generates a homologous series of chemical formulas. For example, alcohols may be represented by the formula CnH(2n + 1)OH (n ≥ 1), giving the homologs methanol, ethanol, propanol for n=1–3.\n\n## Hill system\n\nThe Hill system (or Hill notation) is a system of writing empirical chemical formulas, molecular chemical formulas and components of a condensed formula such that the number of carbon atoms in a molecule is indicated first, the number of hydrogen atoms next, and then the number of all other chemical elements subsequently, in alphabetical order of the chemical symbols. When the formula contains no carbon, all the elements, including hydrogen, are listed alphabetically.\n\nBy sorting formulas according to the number of atoms of each element present in the formula according to these rules, with differences in earlier elements or numbers being treated as more significant than differences in any later element or number—like sorting text strings into lexicographical order—it is possible to collate chemical formulas into what is known as Hill system order.\n\nThe Hill system was first published by Edwin A. Hill of the United States Patent and Trademark Office in 1900. It is the most commonly used system in chemical databases and printed indexes to sort lists of compounds.\n\nA list of formulas in Hill system order is arranged alphabetically, as above, with single-letter elements coming before two-letter symbols when the symbols begin with the same letter (so \"B\" comes before \"Be\", which comes before \"Br\").\n\nThe following example formulae are written using the Hill system, and listed in Hill order:\n\n• BrI\n• CCl4\n• CH3I\n• C2H5Br\n• H2O4S" ]
[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/9832d7ae154c8aa976748d951691b600c631232b", null ]
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https://swag.outpostbbs.net/POINTERS/0013.PAS.html
[ "``````{\n³ I'm trying to understand the rudiments of linked lists\n\n³ 4) What are common uses for linked lists? Is any one particular form\n³ (oneway, circular etc ) preferred or used over any other form?\n\nOne use is to maintain queues. New people, requests, or jobs come in at\nthe end of the line (or break in with priority), but once the head of\nthe line has been serviced, there is no need to maintain its location in\nthe queue. I wrote the following last semester:\n---------------------------------------------------------------\nPurpose:\nMaintains a queue of jobs and priorities of those jobs in a linked list.\nThe user will be prompted for job number and priority and can list the\nqueue, remove a job from the front of the queue (as if it ran), and stop\nthe program. A count of jobs outstanding at the end will be displayed. }\n\ntype\nPriRange = 0 .. 9;\nJobPnt = ^JobNode;\nJobnode = RECORD\nNumb : integer;\nPriority : PriRange;\nLink : JobPnt\nEND;\n\nprocedure addrec(var Start : JobPnt; comprec : Jobnode);\nvar\ncurr,\nnext,\nthis : JobPnt;\nfound : boolean;\nbegin\nnew(this);\nthis^.Numb := comprec.Numb;\nthis^.Priority := comprec.Priority;\nif Start = NIL then\nbegin\nStart := this; {Points to node just built}\nStart^.Link := NIL; {Is end of list}\nend\nelse {Chain exists, find a place to insert it}\nif comprec.Priority > Start^.Priority then\nbegin\nthis^.Link := Start; {Prep for a new beg of chain}\nStart := this\nend {Condition for insert at beg of chain}\nelse\nbegin {Begin loop to insert after beg of chain}\nfound := false; {To initialize}\ncurr := start;\nwhile not found do\nbegin\nnext := curr^.link;\nif (next = NIL) or (comprec.Priority > next^.Priority) then\nfound := true;\nif not found then\ncurr:= next {another iteration needed}\nend;\n{Have found this^ goes after curr^ and before next^}\nthis^.Link := next; {Chain to end (even if NIL)}\ncurr^.Link := this; {Insertion complete}\nend;\nend;\n\nprocedure remove(Var Start : JobPnt);\nvar\nhold : JobPnt;\nbegin\nif Start = NIL then\nWriteln('Cannot remove from empty queue', chr(7))\nelse\nbegin\nhold := Start^.Link; {Save 1st node of new chain}\ndispose(Start); {Delete org from chain}\nStart := hold; {Reset to new next job}\nend;\nend;\n\nprocedure list(Start : JobPnt); {List all jobs in queue. \"var\" omitted}\nbegin\nif Start = NIL then\nWriteln('No jobs in queue')\nelse\nbegin\nWriteln('Job No Priority');\nWriteln;\nwhile Start <> NIL do\nbegin\nWriteln(' ',Start^.Numb : 3, ' ', Start^.Priority);\nStart:=Start^.Link\nend;\nWriteln;\nWriteln('End of List');\nend;\nend;\n\n{Main Procedure starts here}\nvar\ncntr : integer;\nbuild : JobNode;\nwork,\nStart : JobPnt;\nAchar : char;\n\nbegin\nStart := NIL; {Empty at first}\ncntr := 0;\nREPEAT\nWrite('Enter (S)top, (R)emove, (L)ist, or A jobnumb priority to');\nWriteln(' add to queue');\nRead(Achar);\n\nCASE Achar of\n'A', 'a' :\nbegin\nRead(build.Numb);\nREPEAT\nReadln(build.Priority);\nif (build.Priority < 0) or (build.priority > 9) then\nWrite(chr(7), 'Priority between 0 and 9, try again ');\nUNTIL (build.Priority >= 0) and (build.Priority <= 9);\naddrec(Start, build);\nend;\n\n'R', 'r' :\nbegin\nReadln;\nremove(Start);\nend;\n\n'L', 'l' :\nbegin\nReadln;\nlist(Start);\nend;\n\n'S', 's' : Readln; {Will wait until out of CASE loop}\n\nelse\nbegin\nReadln;\nWriteln('Invalid option',chr(7))\nend;\nend;\n\nUNTIL (Achar = 's') or (Achar = 'S');\nwork := start;\nwhile work <> NIL do\nbegin\ncntr := cntr + 1;\nwork := work^.link\nend;\nWriteln('Number of jobs remaining in queue: ', cntr);\nend.\n\n``````" ]
[ null ]
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https://www.gkstudies.in/2018/09/computer-database-reletive-question.html
[ "Computer Database Reletive Question & Answer-2 - GK Study\n\n# GK Study\n\n## Breaking\n\n### Computer General knowledge for all competitive exam like ibps,ssc,mppsc,vyapam,RRB. Don't forget to follow our website to get notifications on your email\n\nQ.1 It is better to use files than a DBMS when there are\n(A) Stringent real-time requirements.\n(B) Multiple users wish to access the data.\n(C) Complex relationships among data.\n(D) All of the above.\nAns: B\nQ.2 The conceptual model is\n(A) dependent on hardware.\n(B) dependent on software.\n(C) dependent on both hardware and software .\n(D) independent of both hardware and software.\nAns: D\nQ.3 What is a relationship called when it is maintained between two entities?\n(A) Unary (B) Binary\n(C) Ternary (D) Quaternary\nAns: B\nQ.4 Which of the following operation is used if we are interested in only certain columns of a\ntable?\n(A) PROJECTION (B) SELECTION\n(C) UNION (D) JOIN\nAns: A\nQ.5 Which of the following is a valid SQL type?\n(A) CHARACTER (B) NUMERIC\n(C) FLOAT (D) All of the above\nAns: D\nQ.6 The RDBMS terminology for a row is\n(A) tuple. (B) relation.\n(C) attribute. D) degree.\nAns: A\nQ.7 Which of the following operations need the participating relations to be union compatible?\n(A) UNION (B) INTERSECTION\n(C) DIFFERENCEqq (D) All of the above\nAns: D\nQ.8 The full form of DDL is\n(A Dynamic Data Language (B) Detailed Data Language\n(C) Data Definition Language (D) Data Derivation Language\nAns: C\nQ.9 Which of the following is an advantage of view?\n(A) Data security (B) Derived columns\n(C) Hiding of complex queries (D) All of the above\nAns: D\nQ.10 Which of the following is a legal expression in SQL?\n(A) SELECT NULL FROM EMPLOYEE;\n(B) SELECT NAME FROM EMPLOYEE;\n(C) SELECT NAME FROM EMPLOYEE WHERE SALARY = NULL;\n(D) None of the above\nAns: B\nQ.11 The users who use easy-to-use menu are called\n(A) Sophisticated end users. (B) Naïve users.\n(C) Stand-alone users. (D) Casual end users.\nAns: B\nQ.12 Which database level is closest to the users?\n(A) External (B) Internal\n(C) Physical (D) Conceptual\nAns: A\nQ.13 Which are the two ways in which entities can participate in a relationship?\n(A) Passive and active (B) Total and partial\n(C) Simple and Complex (D) All of the above\nAns: B\nQ.14 The result of the UNION operation between R1 and R2 is a relation that includes\n(A) all the tuples of R1\n(B) all the tuples of R2\n(C) all the tuples of R1 and R2\n(D) all the tuples of R1 and R2 which have common columns\nAns: D\nQ.15 Which of the following is a comparison operator in SQL?\n(A) = (B) LIKE\n(C) BETWEEN (D) All of the above\nAns: D\nQ.16 A set of possible data values is called\n(A) attribute. (B) degree.\n(C) tuple. (D) domain.\nAns: D\nQ.17 Which of the operations constitute a basic set of operations for manipulating relational\ndata?\n(A) Predicate calculus (B) Relational calculus\n(C) Relational algebra (D) None of the above\nAns:C\nQ.18 Which of the following is another name for weak entity?\n(A) Child (B) Owner\n(C) Dominant (D) All of the above\nAns: A\nQ.19 Which of the following database object does not physically exist?\n(A) base table (B) index\n(C) view\n(D) none of the above\nAns: C\nQ.20 NULL is\n(A) the same as 0 for integer\n(B) the same as blank for character\n(C) the same as 0 for integer and blank for character\n(D) not a value\nAns: D\nQ.21 Which of the following is record based logical model?\n(A) Network Model (B) Object oriented model\n(C) E-R Model (D) None of these\nAns: A\nQ.22 A data dictionary is a special file that contains:\n(A) The name of all fields in all files.\n(B) The width of all fields in all files.\n(C) The data type of all fields in all files.\n(D) All of the above.\nAns: D\nQ.23 A file manipulation command that extracts some of the records from a file is called\n(A) SELECT (B) PROJECT\n(C) JOIN (D) PRODUCT\nAns: A\nQ.24 The physical location of a record is determined by a mathematical formula that trAnsforms\na file key into a record location is :\n(A) B-Tree File (B) Hashed File\n(C) Indexed File (D) Sequential file.\nAns: B\nQ.25 Using Relational Algebra the query that finds customers, who have a balance of over\n1000 is\n(A) PCustomer_name(s balance >1000(Deposit))\n(B) s Customer_name(Pbalance >1000(Deposit))\n(C) PCustomer_name(s balance >1000(Borrow))\n(D) s Customer_name(Pbalance >1000(Borrow))\nAns: A\nQ.26 A primary key is combined with a foreign key creates\n(A) Parent-Child relation ship between the tables that connect them.\n(B) Many to many relationship between the tables that connect them.\n(C) Network model between the tables that connect them.\n(D) None of the above.\nAns: A\nQ.27 In E-R Diagram derived attribute are represented by\n(A) Ellipse (B) Dashed ellipse\n(C) Rectangle (D) Triangle\nAns B\nQ.28 Cross Product is a:\n(A) Unary Operator (B) Ternary Operator\n(C) Binary Operator (D) Not an operator\nAns: C\nQ.29 An instance of relational schema R (A, B, C) has distinct values of A including NULL\nvalues. Which one of the following is true?\n(A) A is a candidate key (B) A is not a candidate key\n(C) A is a primary Key (D) Both (A) and (C)\nAns: B\nQ.30 Consider the join of a relation R with relation S. If R has m tuples and S has n tuples,\nthen the maximum size of join is:\n(A) mn (B) m+n\n(C) (m+n)/2 (D) 2(m+n)\nAns: A\nQ.31 The natural join is equal to :\n(A) Cartesian Product\n(B) Combination of Union and Cartesian product\n(C) Combination of selection and Cartesian product\n(D) Combination of projection and Cartesian product\nAns: D\nQ.32 Which one of the following is not true for a view:\n(A) View is derived from other tables.\n(B) View is a virtual table.\n(C) A view definition is permanently stored as part of the database.\n(D) View never contains derived columns.\nAns: C\nQ.33 A primary key if combined with a foreign key creates\n(A) Parent-Child relationship between the tables that connect them.\n(B) Many to many relationship between the tables that connect them.\n(C) Network model between the tables that connect them.\n(D) None of the above.\nAns: A\nQ.34 In E-R Diagram relationship type is represented by\n(A) Ellipse (B) Dashed ellipse\n(C) Rectangle (D) Diamond\nAns: D\nQ.35 Hierarchical model is also called\n(A) Tree structure (B) Plex Structure\n(C) Normalize Structure (D) Table Structure\nAns: A\nQ.36 To delete a particular column in a relation the command used is:\n(A) UPDATE (B) DROP\n(C) ALTER (D) DELETE\nAns: C\nQ.37 The ______ operator is used to compare a value to a list of literals values that have been\nspecified.\n(A) BETWEEN (B) ANY\n(C) IN (D) ALL\nAns: A\nQ.38 A logical schema\nA) is the entire database\nB) is a standard way of organizing information into a accessible part\nC) describe how data is actually stored on disk\nD) none of these\nAns: D\nQ.39 A B-tree of order m has maximum of _____________ children\n(A) m (B) m+1\n(C) m-1 (D) m/2\nAns: A\nQ.40 _____________ function divides one numeric expression by another and returns the\nremainder.\n(A) POWER (B) MOD\n(C) ROUND (D) REMAINDER\nAns: B\nQ.41 A data manipulation command the combines the records from one or more tables is called\n(A) SELECT (B) PROJECT\n(C) JOIN (D) PRODUCT\nAns: C\nQ.42 In E-R diagram generalization is represented by\n(A) Ellipse (B) Dashed ellipse\n(C) Rectangle (D) Triangle\nAns: D\nQ.43 _________ is a virtual table that draws its data from the result of an SQL\nSELECT statement.\n(A) View (B) Synonym\n(C) Sequence (D) TrAnsaction\nAns: A\nQ.44 The method of access which uses key trAnsformation is known as\n(A) Direct (B) Hash\n(C) Random (D) Sequential\nAns: B\nQ.45 A table joined with itself is called\n(A) Join (B) Self Join\n(C) Outer Join (D) Equi Join\nAns: B\nQ.46 _________ data type can store unstructured data\n(A) RAW (B) CHAR\n(C) NUMERIC (D) VARCHAR\nAns: A" ]
[ null ]
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https://forum.linqpad.net/profile/gmazel
[ "# gmazel\n\n//Am I missing a reference in LinqPad? //I get an error in LinqPad on line 9 below from Listing 2-14. Generate Logo commands to draw a //Serpinski triangle.linq: //\"CS0103 The name 'serpinskyTriangle' does not exist in the current context\" * string serpinskiTriangle = \"A\"; * Func transformA = x => x.Replace(\"A\",\"B-A-B\"); * Func markBs = x => x.Replace(\"B\",\"[B]\"); * Func transformB = x => x.Replace(\"[B]\",\"A+B+A\"); * int length = 6; * Enumerable.Range(1,length) * .ToList() * .ForEach (k => serpinskiTriangle = * transformB(transformA(markBs(serpinskyTriangle)))); * * serpinskiTriangle * .Replace(\"A\", \"forward 5\" + Environment.NewLine) * .Replace(\"B\", \"forward 5\" + Environment.NewLine) * .Replace(\"+\", \"left 60\" + Environment.NewLine) * .Replace(\"-\", \"right 60\" + Environment.NewLine) * .Dump(\"LOGO Commands for drawing Serpinsky Triangle\"); // Thanks! // Gregg\n\n## Activity\n\n• Not much happening here, yet." ]
[ null ]
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https://stats.stackexchange.com/questions/424558/why-is-it-hard-for-a-neural-network-to-learn-the-identity-function
[ "# Why is it hard for a neural network to learn the identity function?\n\nI wanted to see if a neural network could learn the identity function using the MNIST handwritten dataset.\n\nHere is the full code\n\nimport keras\nfrom keras.datasets import mnist\nfrom keras.models import Sequential\nfrom keras.layers import Dense\nfrom keras.optimizers import RMSprop\n\nbatch_size = 128\nepochs = 20\n(x_train, y_train), (x_test, y_test) = mnist.load_data()\nx_train = x_train.reshape(60000, 784)\nx_test = x_test.reshape(10000, 784)\n\nmodel = Sequential()\nmodel.summary()\nmodel.compile(loss='mean_squared_error',\noptimizer=RMSprop(),\nmetrics=['mean_absolute_percentage_error'])\n\nhistory = model.fit(x_train, x_train,\nbatch_size=batch_size,\nepochs=epochs,\nverbose=1,\nvalidation_data=(x_test, x_test))\nscore = model.evaluate(x_test, x_test, verbose=0)\nprint('Test loss:', score)\nprint('Test MAPE:', score)\n\n\nand the output\n\n**4 dense layers**\nEpoch 20/20\n60000/60000 [==============================] - 50s 840us/step - loss: 456.7581 - mean_absolute_percentage_error: 351097677.7045 - val_loss: 523.7151 - val_mean_absolute_percentage_error: 504905991.0656\nTest loss: 523.7150838867187\nTest MAPE: 504905988.5056\n\n\nWhat I can't quite get my head around is why training cannot find the perfect solution to the problem and why it takes so long to even get close to it? Even with one dense layer the exact solution cannot be found:\n\n**1 dense layer**\nEpoch 20/20\n60000/60000 [==============================] - 16s 268us/step - loss: 180.6187 - mean_absolute_percentage_error: 209296481.2373 - val_loss: 167.9543 - val_mean_absolute_percentage_error: 192590419.9936\nTest loss: 167.954341796875\nTest MAPE: 192590420.1984\n\n\nConceptually I can see that there is a solution space (not just the exact identity function) as it is likely there are some pixels that have the same value as each other in all the images which could be swapped in the training set at no loss (0's around the edge for example). With the knowledge that this is a local minimum can I learn anything from this to guide me out rather than playing with hyperparameters until I find something better?\n\n• Maybe I don't understand your code, but the output should be a linear layer not relu ( for regression) – seanv507 Sep 1 '19 at 21:21\n• IE the hidden layers just add an offset to push the relus into linear region, then the linear output removes offset/rescales. – seanv507 Sep 1 '19 at 21:23\n• Changing to linear activation for the last layer did not make an identifiable difference. In this case the last layer output should be all positive. – Shaped Sundew Sep 5 '19 at 19:45\n• Scaling the input to between 0 and 1, as was suggested in a deleted comment, did make a big difference in training time - and I get that now - thx Sycorax. – Shaped Sundew Sep 5 '19 at 19:47\n\nFor a single example, the network takes a 784-element vector as its input. So rephrasing the problem in OP's post, they wish to learn the function\n\n$$f(x) = Ix$$\n\nwhere $$I$$ is the $$784\\times 784$$ identity matrix.\n\n# Perfect fit is impossible with this model\n\nThe 1-layer network probably has an easier time because instead of attempting to \"line up\" four weight matrices through four nonlinearities, it only has to line up one, i.e. it is easier to find an approximation in $$W_1, b_1$$ for\n\n$$Ix = g(W_1 x+b_1).$$\n\nBut even the simple expression $$Ix = g(W_1 x+b_1)$$ should be an obvious warning that attempting to find a perfect fit is a fool's errand, because it is trying to approximate a linear function with a nonlinear function. In particular, because of how ReLUs are defined, any $$x<0$$ is set to 0, so this model will never achieve 0 error when any elements of $$x$$ are negative.\n\n## The UAT is an approximation theorem\n\nIndeed, for any choice of nonlinear activation $$g$$, I can find an $$x$$ for which the error is positive. So then the interesting question becomes \"Can we fit a model so that the error is at most $$\\epsilon$$ for $$x$$ in some interval $$\\mathcal{I}$$?\" And this statement of the problem is more-or-less compatible with the caveats of the UAT. It also points us in a more profitable direction: instead of seeking 0 error, we wish to find minimal error when the inputs are in some interval.\n\nIn other words, theorems about neural networks don't guarantee that you can achieve 0 error, they guarantee that you can bound error for inputs in some interval (subject to some terms and conditions).\n\n## The UAT doesn't comment on whether it's easy to train any particular network.\n\nActually finding the weights and biases which achieve the minimum error is a very challenging problem. In particular, we don't have much reason to believe that the choice of initialization, optimizer, learning rate and number of epochs, etc. in this code snippet are best for this task.\n\n# This optimization problem is hard\n\nA four-layer network with ReLU activations $$g(x)=\\max\\{0, x\\}$$ is given by\n\n$$h(x)=g(W_4g(W_3g(W_2g(W_1x+b_1)+b_2)+b_3)+b_4).$$\n\nSo what you seek in your question is solutions $$W_i, b_i$$ such that $$Ix = g(W_4g(W_3g(W_2g(W_1x+b_1)+b_2)+b_3)+b_4)$$ for all $$x$$, where $$W_i, b_i$$ are have appropriate shape.\n\nThis doesn't look particularly friendly to try and solve. Indeed, in light of my remarks about the UAT, we will have to restate this to bound the error and focus on an interval of inputs.\n\nEven if we restate the problem in this way, it is still challenging from the perspective of gradient descent because of the dying ReLU phenomenon, the weaknesses of gradient descent, and the poor conditioning of the optimization task due to the scale of the inputs.\n\n# Tuning a neural network is the greater part of using neural networks.\n\nIf you don't want to spend a lot of time changing hyper-paremters, then you should use a different model." ]
[ null ]
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http://quantwisetrading.com/article_ml.html
[ "", null, "## Applied Machine Learning – Bayesian Modeling in Ninja Trader Strategies\n\nby Marcio Costa\n\n##### Introduction\n\nOver time, the way traders approach to markets has become more complex, more fragmented, and more abstract. Today, there are multiple trading venues, hundreds of order types with modifiers, and unlimited algorithmic smart order routing systems, all with the express purpose of competing to get the best fill at the lowest total cost. But there is still, conceptually, one major progress that all market participants understand as the tool that brought more significant value to quantitative approach. Machine Learning is the fastest way to prove the evidence or the effectiveness of alphas during the strategy building process. When analyzing data, we can several methods to look for alphas: visual, back testing, Monte Carlo, and other simulations, but machine learning can show us the best statistical evidence over time. This article brings some light on the Bayesian model and a real application for Ninja Trader Platform. We will not explore machine learning definitions, and we assume traders are comfortable with basic ML, math, and statistical concepts.\n\n##### Bayesian Modeling\n\nThe Naive Bayes model can be utilized for twofold (binary) arrangement, for instance anticipating if the weather is warm or cold, we can track several patterns like rain, cloud coverage, season,…etc or for multiclass grouping. The classification process is the key factor for the ML statistics to be applied. Also, we have to make sure the variables we are using to our dataset are eligible for normal distribution. This variable feature is called stationary. Thus, stationarity implies that a time series does not have a trend or seasonal effects1. This way, I named alphas (https://www.investopedia.com/terms/a/alpha.asp) as the results extracted from indicators or even the market raw data. In this text, we will test alphas as indicators results, statistically called priors, to measure if some values individually or combined have the possibility to predict price movement. In this article I disclose how to make a credulous Bayes order framework when the indicator esteems are numeric, utilizing the C# language with any special code libraries.\n\n##### The Code\n\nThis coding process has been simplified because it can be used for many applications and situations so you can explorer more options if we leave like an open concept. All code runs on real time during OnBarUpdate() entry on Ninja Trader script. It is not recommend using it in high frequency or tick-by-tick due the nature of the operation and the number of resources used by the system. First step is to create and load a dataset. In this case we are using arrays, the fastest way to access from PC memory. The reference data that is used to determine the class of an unknown data item is hard-coded into an array-of-arrays style matrix. This structure allows several numbers of priors (explained in the next subtitle) and this way we can combine more probabilities. In order to create a more visual and practical, we have this table below to show some random numbers:\n\n Prior Bar 1 Bar X Bar Y - prediction Δ EMA 65 69 71 Δ SMA 480 515 452 Δ RSI 65 72 69 Price action (return in ticks) 3 2.5 ????\n\nThe matrix (arrays) will be loaded with values from indicators for several bars. From these arrays we will:\n\n``` ```\n1 compute class counts\n2 compute / display means\n3 compute / display variances\n4 set up item to predict\n5 compute / display conditional probabilities\n6 compute / display unconditional probabilities\n7 compute / display evidence terms\n8 compute / display prediction probabilities\n```\n```\n\nThe prediction probabilities will indicate the best chances for the price prediction for the next bar.\n\nLet’s move to real code and first thing is declare the variables:\n\n``` ```\ndouble[][] data;\n```\n```\n``` ```\nprotected override void OnStateChange()\n{\nif (State == State.SetDefaults)\n{\ndata = new double[];\n}\n{\nLista = new List();\n}\n}\n```\n```\n\nBesides the main data array, we have to create a new list. This redundancy allows us to work with functions on both datasets. Although it looks like a double storage, many accesses are provided to lists and other to arrays\n\n``` ```\nprivate\tList Lista;\n```\n```\n``` ```\nprivate class TestList\n{\npublic double lprior1;\npublic double lprior2;\npublic double lprior3;\npublic double lside;\n\npublic TestList(double myprior1, double myprior2, double myprior3, double myside)\n{\nlprior1 = myprior1;\nlprior2 = myprior2;\nlprior3 = myprior3;\nlside = myside;\n}\n}\n```\n```\n\nFor all machine learning applications, we have to follow the same workflow, that involves train in-sample data and test with out-of-sample data. The dataset is split into training and test data. The training data contained the historical data, so you must allow NT work with State.Historical. In the OnBarUpdate() class, we add the input values:\n\n``` ```\nprior1 = EMA(8) - EMA(8);\nprior2 = SMA(8) - SMA(8);\nprior3 = RSI(13) - RSI(13);\n```\n```\n\nAnd then we complete the array input instruction with the code below. Note for the “trnum” is our reference bar number and we do not use the CurrentBar() component from NT. Our data[] array is limited in 1000 registers, for this study, we had to limit the historical bar’s number.\n\n``` ```\nif(Closes> Open)\n{\ndata[trnum] = new double[] { prior1, prior2, prior3, 1 }; // 1 = Up Bar\nLista.Add( new TestList(prior1, prior2, prior3, 1));\n}\nif(Closes< Open)\n{\ndata[trnum] = new double[] { prior1, prior2, prior3, 0 }; // 0 = Down Bar\nLista.Add( new TestList(prior1, prior2, prior3, 0));\n}\n```\n```\n##### Preparing the Prediction\n\nThere are four steps to prepare a naive Bayes prediction for numeric data. You must compute the counts of each class to predict, the means of each predictor, the variances of each predictor, and set up the item to predict. The class counts are computed like so:\n\n``` ```\nint N = Lista.Count();\nint[] classCts = new int; // up bar, down bar\nfor (int i = 0; i < N; ++i)\n{\nint c = (int)data[i]; // class is at \n++classCts[c];\n}\n```\n```\n\nIn these instructions above we define the 2 class counts, for up bar and down bar. Next step we compute the means for each predictor variable, for each class.\n\n``` ```\ndouble[][] means = new double[];\nfor (int c = 0; c < 2; ++c)\nmeans[c] = new double;\n\nfor (int i = 0; i < N; ++i)\n{\nint c = (int)data[i];\nfor (int j = 0; j < 3; ++j) // EMA, SMA, RSI\nmeans[c][j] += data[i][j];\n}\n```\n```\n\nThe values are stored in an array-of-arrays matrix named means where the first index indicates which class (0 = down, 1 = up) and the second index indicates which predictor (0 = EMA, 1 = SMA, 2 = RSI). After the sums of the predictors are computed, they're converted to means by dividing by the number of items in each class:\n\n``` ```\nfor (int c = 0; c < 2; ++c)\n{\nfor (int j = 0; j < 3; ++j)\nmeans[c][j] /= classCts[c];\n}\n```\n```\n\nThe demo sets up storage for the variances of each predictor using the same indexing scheme that's used for the means:\n\n``` ```\ndouble[][] variances = new double[];\nfor (int c = 0; c < 2; ++c)\nvariances[c] = new double;\n\nfor (int i = 0; i < N; ++i)\n{\nint c = (int)data[i];\nfor (int j = 0; j < 3; ++j)\n{\ndouble x = data[i][j];\ndouble u = means[c][j];\nvariances[c][j] += (x - u) * (x - u);\n}\n}\n```\n```\n\nThen the sums are divided by one less than each class count to get the sample variances:\n\n``` ```\nfor (int c = 0; c < 2; ++c)\n{\nfor (int j = 0; j < 3; ++j)\nvariances[c][j] /= classCts[c] - 1;\n}\n```\n```\n##### Making the Prediction\n\nTo make a prediction, you use the item to predict, the means, and the variances to compute conditional probabilities. So, the formula for our basic Gaussian distribution formula and the code to solve it.", null, "Storage for the conditional probabilities utilizes the same indexing scheme that is used for means and variances, where the first index indicate the class and the second index indicates the predictor:\n\n``` ```\nstatic double ProbDensFunc(double u, double v, double x)\n{\ndouble left = 1.0 / Math.Sqrt(2 * Math.PI * v);\ndouble right = Math.Exp( -(x - u) * (x - u) / (2 * v) );\nreturn left * right;\n}\n\nstatic double ProbDensFuncStdDev(double u, double v, double x)\n{\ndouble left = ( x - u) > 0 ? (x - u) : (x - u) *-1;\ndouble right = left / v;\nreturn right;\n}\n```\n```\n\nThe probability density function defines the shape of the Gaussian bell-shaped curve. Note that these qualities truly aren't probabilities since they can be more prominent than 1.0, however they're typically called probabilities in any case. Unconditional probabilities of the classes to predict are ordinary probabilities. For example, the P(C = 0) = 0.5000 because four of the eight data items are class 0. The unconditional probabilities are computed and displayed by this code:\n\n``` ```\ndouble[] evidenceTerms = new double;\nfor (int c = 0; c < 2; ++c)\n{\nevidenceTerms[c] = classProbs[c];\nfor (int j = 0; j < 3; ++j)\nevidenceTerms[c] *= condProbs[c][j];\n}\n```\n```\n\nThe last step in making a prediction is to convert the evidence terms to probabilities. You compute the sum of all evidence terms then divide each term by the sum:\n\n``` ```\ndouble sumEvidence = 0.0;\nfor (int c = 0; c < 2; ++c)\nsumOfEvidences += evidenceTerms[c];\n\ndouble[] predictProbs = new double;\nfor (int c = 0; c < 2; ++c)\npredictProbs[c] = evidenceTerms[c] / sumEvidence;\n```\n```\n##### Predication probabilities\n\nThe result returned by the sub is a number between 0 to 1, and this can be the class of 0 (probability for short) or 1 (for long). These classes can be integrated to a strategy and can also work with a second data layer processing results from this virtual trading layer. The virtual trading layer results can be analyzed and added to the original dataset.\n\nIn a ML environment, the concept of Bayesian modeling is getting more popularity due to the individual element for each prior, very applicable for financial analysis and the capacity to sum up with other priors.\n\n##### Alphas and priors\n\nWe can discuss a lot about the alphas and priors and that is the main reason we can apply ML. In order to generate alpha (or edge) we have analyze several set ups using market indicators, datasets, strategies and other tools. The predictor model is used to seek evidence between these factors. Most strategies rely in one or more indictors and during the backtest process we can measure trading efficiency and behavior. The evolution shown with the ML applications are relevant because have dynamic datasets, so it reduces the marginal error from bad quality data.\n\n##### Conclusion and future research\n\nIt is a constant struggle to find a way to implement new concepts like Machine Learning and its variants to the current platforms. Ninja Trader has a very strategy builder interface with NT Script sometimes the use of external libraries become quite tough. The way we explore in this article shows definite and easy path to proof results from ML stand point. This work will extend over new ways to improve the operation speed, resources efficiency and language optimization.\n\nFeel free to keep in touch at marcio@quantwisetrading.com\n\nReference\n\n1 Jansen, Stefan. Machine Learning for Algorithmic Trading: Predictive models to extract signals from market and alternative data for systematic trading strategies with Python, 2nd Edition (p. 433). Packt Publishing. Kindle Edition" ]
[ null, "http://quantwisetrading.com/images/ie8-panel/warning_bar_0000_us.jpg", null, "http://quantwisetrading.com/images/image001.png", null ]
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https://rallycoding.com/problems/200/comments
[ "19 days ago\n\nconst solve = (intArray, k) => { const val = (k + 1) * -1; return intArray.splice(val); };\n\na month ago\n\nconst solve = (intArray, k) => { k = ++k return intArray.splice(-k); };\n\n6 months ago\n\nconst solve = (intArray, k) => { return intArray[(intArray.length -1) - k]; };\n\n6 months ago\n\nvalue give in constant time return intArray[intArray.length-k-1];\n\n6 months ago\n\nconst solve = (intArray, k) => intArray.reverse()[k];\n\n7 months ago\n\nconst solve = (intArray, k) => { intArray.reverse() return intArray[k] };\n\n8 months ago\n\nO(1)\n\nconst solve = (intArray, k) => {\nreturn intArray[intArray.length-k-1];\n};\n\n8 months ago\n\nconst solve = (intArray, k) => { return intArray[intArray.length - k - 1] }\n\n8 months ago\n\nconst solve = (intArray, k) => {\n\nlet counter = 0;\n\nfor(let i = (intArray.length-1); i >= 0; i--){\n\nif(counter === k){\nreturn intArray[i];\n}\ncounter++;\n\n}\n\n};\n\n8 months ago\n\nYou can avoid a runtime complexity of O(n) or a linear runtime complexity with a for loop .If you already know the length of the array, you can do intArray.length-k go go backwards for 3 steps.As arrays index start from 0, you should substract the result of the previous calculation by 1.And the runtime complexity of your will become constant or O(1)\n\nreturn intArray[intArray.length-k-1];\n\n9 months ago\n\nreturn intArray.reverse()[k];\n\n10 months ago\n\nif there are no rules const solve = (intArray, k) => {\n\nreturn intArray.reverse()[k]; };\n\n10 months ago\n\nconst solve = (intArray, k) => {\nreturn intArray.slice(intArray.length, intArray.length-k)\n\n};\n\n4 months ago\n\nSorry it passed the test but it should'nt because of zero\n\n10 months ago\n\nconst solve = (intArray, k) => {\nreturn intArray.slice(intArray.length, intArray.length-1)\n\n};\n\n2 years ago\n\nbelow code also gives correct results. when i click submit it says no u dont qualify you fail. not sure what evaluation criteria is. but anyhow my results are also correct with below code\n\nconst solve = (intArray, k) => { let intArray1 = intArray.toString().replace(/[,-]/g, \"\"); return intArray1.charAt(intArray1.length-k-1); };\n\n2 years ago\n\nconst solve = (intArray, k) => { return intArray.find((el, i) => k === Math.abs( i - (intArray.length-1))); };\n\n2 years ago\n\nconst solve = (intArray, k) => { return intArray.reverse()[k]; };\n\na year ago\n\nActurally, you may omit { return } (curly braces and \"return\" key word.\n\n2 years ago\n\nDon't get too complicated!\n\nconst solve = (intArray, k) => {\nreturn intArray[intArray.length - k - 1];\n};\n\na year ago\n\nThis was exactly my solution and 'something went terribly wrong'." ]
[ null ]
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https://www.shaalaa.com/question-bank-solutions/concept-application-conversion-through-rectangular-field-length-90-m-breadth-60-m-two-roads-are-constructed-which-are-parallel-sides-cut-each-other-right-angles-through-centre-fields-if-width-each-road-3-m-find_17047
[ "Share\n\n# Through a Rectangular Field of Length 90 M and Breadth 60 M, Two Roads Are Constructed Which Are Parallel to the Sides and Cut Each Other at Right Angles Through the Centre of the Fields. If the Width of Each Road is 3 M, Find - CBSE Class 7 - Mathematics\n\nConceptConcept of Application of Conversion\n\n#### Question\n\nThrough a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find\n\n1)the area covered by the roads.\n\n2) the cost of constructing the roads at the rate of Rs 110 per m2.\n\n#### Solution", null, "Length (l) of field = 90 m\n\nBreadth (b) of field = 60 m\n\nArea of field = 90 × 60 = 5400 m2\n\nLength of road PQRS = 90 m\n\nLength of road ABCD = 60 m\n\nWidth of each road = 3 m\n\nArea of the roads = ar (PQRS) + ar (ABCD) − ar (KLMN)\n\n= (90 × 3) + (60 × 3) − (3 × 3)\n\n= 270 + 180 − 9 = 441 m2\n\nCost for constructing 1 m2 road = Rs 110\n\nCost for constructing 441 m2 road = 110 × 441 = Rs 48510\n\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nNCERT Solution for Mathematics for Class 7 (2018 to Current)\nChapter 11: Perimeter and Area\nEx. 11.40 | Q: 7 | Page no. 227\nSolution Through a Rectangular Field of Length 90 M and Breadth 60 M, Two Roads Are Constructed Which Are Parallel to the Sides and Cut Each Other at Right Angles Through the Centre of the Fields. If the Width of Each Road is 3 M, Find Concept: Concept of Application of Conversion.\nS" ]
[ null, "https://www.shaalaa.com/images/_4:82a5b6ba5ba242838a80e6ef9e0f77b4.png", null ]
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https://mathspace.co/textbooks/syllabuses/Syllabus-441/topics/Topic-8224/subtopics/Subtopic-107853/?activeTab=interactive
[ "", null, "Simplify expressions using multiple exponent laws with variable bases\n\nInteractive practice questions\n\nSimplify the expression, giving your answer in simplest exponential form: $\\left(3^5\\right)^2$(35)2.\n\nEasy\nLess than a minute\n\nSimplify the following expression: $\\left(x^3\\right)^5$(x3)5.\n\n$\\left(-5\\right)^{13}$(5)13 simplifies to which of the following?\n\n$\\left(-x^3\\right)^4$(x3)4 simplifies to which of the following:\n\nOutcomes\n\n9D.NA1.04\n\nExtend the multiplication rule to derive and understand the power of a power rule, and apply it to simplify expressions involving one and two variables with positive exponents" ]
[ null, "https://mathspace-production-static.mathspace.co/permalink/badges/v3/indices.svg", null ]
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https://forum.allaboutcircuits.com/threads/question-about-unbalanced-load-amperage-on-a-delta-system.153770/
[ "#### PGB1\n\nJoined Jan 15, 2013\n119\nGood Morning Everyone!\n\nThe question below is something that I certainly used to know the answer to, but the answer has apparently fallen out of my brain.\nI think it belongs in this section because it is education related. (I wasn't sure if it belonged here or in Physics)\n\nIt relates to the amperage load when connecting two heating elements on a 3-Phase delta supply.\nThe only way I can describe my question is by the examples below. I'll start with a Wye system to hopefully make my question clear. (Yeah, sure....)\n\nFor simplicity, the amp draw on the sample loads will be 20 each, even though the voltage is different on each example. There are no real parts involved, just hypothetical.\n\n----------------------------------\nWYE SYSTEM WE KNOW THE ANSWER-\nPicturing 2 purely resistance loads (for example heating elements) of 20 amps per item on a 120/208 volt Wye system as:\n\nElement One is wired across L-1 & L-2\n\nElement Two is wired across L-2 & L-3\n\nReading the amperage between L-1 & the element, we will have 20 amps.\n\nReading the amperage between L-3 & the element, we will have 20 amps.\n\nReading the amperage between L-2 & where both elements are connected, we will have 40 amps.\n\n-------------------------------------------------------------------------------------\nBUT ON A DELTA SYSTEM = ?\nPicturing 2 purely resistance loads (for example heating elements) of 20 amps per item on a 125/240 volt Delta system, where\n\nL-1 & L-3 to ground are each 125 volts\nL-2 to ground is 208 volts\n\nL-1 to L-2 reads 240 volts\nL-1 to L-3 reads 240 volts\nL-2 to L-3 reads 240 volts\n\nElement One is across L-1 and L-2\nElement Two is across L-3 and L-2\n\nReading the amperage between L-1 and the element, we will read 20 amps.\n\nReading the amperage between L-3 and the element, we will read 20 amps.\n\nTHE QUESTION-\nBut, reading the amperage between L-2 and where both elements are tied together, will we have 20 amps, 40 amps or\n20 amps times the square root of 3?\n\nThank You All for putting up with this \"I should remember this\" question. I appreciate your help!\nPaul\n\n#### WBahn\n\nJoined Mar 31, 2012\n26,398\nReading the amperage between L-2 & where both elements are connected, we will have 40 amps.\nWhy do you believe that this is the case?\n\nDo the math -- and don't forget that the line voltages are not in phase.\n\n•", null, "PGB1\n\n#### PGB1\n\nJoined Jan 15, 2013\n119\nThank You WBahn for taking time to reply.\n\nI apologize, but today is one of the all to frequent cloudy brain days. (Age is sometimes quite inconvenient.) I also apologize for using L-1, L-2, L-3. I should have named Phase A, B, C to be more conventional.\n\nQUOTE FROM Y FIRST POST-\nReading the amperage between L-2 & where both elements are connected, we will have 40 amps.\nWhy do you believe that this is the case?\n\nI intended to type that I believe that we would have 20 amps at L-2 where both elements (loads) tie together. At least in the WYE system...\n------------------------------------------------\nI probably went into confusion mode because if this were a single phase supply and I had 20 amp load between L1 & Neutral and 20 amp load between L2 and Neutral, the neutral current would be 20.\nAnd if it were a 3-Phase system with 20 amp loads from each of the three phases to neutral, the neutral current would read zero. (If I remember properly the math...)\n\nI (hope I still) remember the formula for unbalanced loads in three phase: (I being amperage L being the phase)\nIL = Square Root Of 1Asq + IBsq + ICsq - (IA x IB) - (IB x IC) - (IC x IA)\n\nBut, since we are not using a neutral in my example, what would the math be for the L-2 amperage?\nWould it be the same for Delta and Wye, since Delta has L-2 at higher voltage to ground?\nWould we look at the loads as two single phase loads since there are only 2 loads and find L-2 current is equal to either one of the (balanced) loads?\n\nThanks Again for refreshing my brain. Please pardon if these are ignorant questions. Perhaps tomorrow, or next week, the answer to my question will be crystal clear as if I never forgot how to reason them out in my mind.\n\nI appreciate your help & the education!\nPaul\n\nLast edited:\n\n#### WBahn\n\nJoined Mar 31, 2012\n26,398\nThank You WBahn for taking time to reply.\n\nI apologize, but today is one of the all to frequent cloudy brain days. (Age is sometimes quite inconvenient.) I also apologize for using L-1, L-2, L-3. I should have named Phase A, B, C to be more conventional.\n\nQUOTE FROM Y FIRST POST-\nReading the amperage between L-2 & where both elements are connected, we will have 40 amps.\nWhy do you believe that this is the case?\n\nI intended to type that I believe that we would have 20 amps at L-2 where both elements (loads) tie together. At least in the WYE system...\n------------------------------------------------\nI probably went into confusion mode because if this were a single phase supply and I had 20 amp load between L1 & Neutral and 20 amp load between L2 and Neutral, the neutral current would be 20.\nWhy do you thing that this would be the case?\n\nTake a single phase supply and put two equal loads in series between L1 and L2 such that they draw 20 A. How much current is flowing in the neutral? None! It's not even connected. What is the voltage at the junction of the two loads? Zero volts, by symmetry. So now connect the neutral to that junction. How much current will flow in it? Zero, because there's no voltage there.\n\n•", null, "PGB1\n\n#### PGB1\n\nJoined Jan 15, 2013\n119\nThank You Again for helping me out. After reading your above reply, it all is coming back to me (slowly, but surely).\n\nTo take this further, if I connect two 20 amp loads to a 3 phase delta supply (rough sketch attached) as-\nLoad One is across Phases A & B, Load Two is across Phases B & C (Each is line-To-Line. None go to neutral.)\n\nI realize that an ammeter connected between Load One and the transformer Phase A tap will read 20 amps and a meter connected between Load Two and transformer Phase C tap will read 20 amps. But, since it is an unbalanced load on the transformer, will a meter connected between both loads and transformer Phase B tap read 20 or 40? From the math I remember, I believe it will read 20 amps.\n\nDid I get this correct or goof it up again?\n\n#### WBahn\n\nJoined Mar 31, 2012\n26,398\nThank You Again for helping me out. After reading your above reply, it all is coming back to me (slowly, but surely).\n\nTo take this further, if I connect two 20 amp loads to a 3 phase delta supply (rough sketch attached) as-\nLoad One is across Phases A & B, Load Two is across Phases B & C (Each is line-To-Line. None go to neutral.)\n\nI realize that an ammeter connected between Load One and the transformer Phase A tap will read 20 amps and a meter connected between Load Two and transformer Phase C tap will read 20 amps. But, since it is an unbalanced load on the transformer, will a meter connected between both loads and transformer Phase B tap read 20 or 40? From the math I remember, I believe it will read 20 amps.\n\nDid I get this correct or goof it up again?\nThe two 20 A readings have a phase associated with them. Your meter is only showing the amplitude (probably scaled as RMS) with no information about the phase.\n\nWhen two such signals are connected and add together, the result can be anywhere from A-B to A+B, depending on the phase relationship.\n\nYou need to do the math -- taking the phase into account.\n\n•", null, "PGB1\n\n#### PGB1\n\nJoined Jan 15, 2013\n119\nThanks Once Again!\nI found & studied my formula books, did the math for both Delta and Wye with the unbalanced loads. The results were quite interesting.\n\nBy coincidence, I was working in a Delta system and had the opportunity to do a quick test of an un-balanced load on a low voltage (240 ph-ph) 3-phase branch circuit. (Balanced loads are easy for me because that is most of my experience. Unbalanced was what caused my original post.)\nThe loads were pure resistance loads, so I assumed my Phase Voltage would be just about the same as Line Voltage.\n\nAfter doing the math, I tested.\nMy amperages at Phase A & Phase C were just about as-expected. Surprisingly, my amperage reading at Phase B was also just about as expected. (A & C were each 36.67 amps with 8800 watt loads at 240 vac- 6.55 Ohm). B was 63.94 amps. (Amps Line-Line times 1.73) I account for the small difference between math & as-tested to be voltage drop with the loads running, meter variances, etc. It was less than 2 amps on each.\n\nIt would be fun to calculate an unbalanced induction load, such as a motor or lighting transformer, and test how my math worked out.\n\nThank You Again for helping. You are a great teacher!\nPaul" ]
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https://mizugadro.mydns.jp/t/index.php?title=Superfunctions&oldid=41698
[ "# Superfunctions\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\n«Superfunctions» is book about evaluation of abelfunctions, superfunctions, and the non-integer iterates of holomorphic functions. In particular, results for tetration, arctetration and iterates of exponential are presented.\n\nThe book is available at https://www.morebooks.de/store/gb/book/superfunctions/isbn/978-620-2-67286-3\n\nThe Book is loaded also as\n\nThere is also Russian version \"Суперфункции\" (y. 2014) \n\n## Covers\n\nAt the front cover, the complex map of natural tetration is shown.\n\nThe following topics are indicated:\nNon-integer iterates of holomorphic functions.\nTetration and other superfunctions.\nFormulas, algorithms, tables, graphics and complex maps.\n\nThe back cover suggests short abstract of the Book and few notes about the Author.\n\nAssume some given holomorphic function $$T$$. The superfunction is holomorphic solution $$F$$ of equation $$T(F(z))=F(z+1)$$\n\nThe Abel function (or abelfunction) is the inverse of superfunction, $$G=F^{-1}$$\n\nThe abelfunction is solution of the Abel equation $$G(T(z))=G(z)+1$$\n\nAdditional conditions are considered to provide uniqueness of the solution $$F$$ (and uniqueness of $$G$$).\n\nAs the superfunction $$F$$ and the abelfunction $$G=F^{-1}$$ are established, the $$n$$th iterate of transfer function $$T$$ can be expressed as follows: $$T^n(z)=F(n+G(z))$$\n\nThis expression allows to evaluate the non-integer iterates. The number n of iterate can be real or even complex. In particular, for integer $$n$$, the iterates have the common meaning: $$T^{-1}$$ is inverse function of $$T$$,\n$$T^0(z)=z$$,\n$$T^1(z)=T(z)$$,\n$$T^2(z)=T(T(z))$$,\n$$T^3(z)=T(T(T(z)))$$,\nand so on. The group property holds: $$T^m(T^n(z))=T^{m+n}(z)$$\n\nThe book is about evaluation of the superfunction $$F$$, the abelfunction $$G$$ and the non-integer iterates of various transfer functions $$T$$.\n\nThe special notation is used in through the book; the number of iterate is indicated as superscript. For example, In these notations, $$\\sin^2(z)=\\sin(\\sin(z))$$, but never $$\\sin(z)^2$$.\nThis notation is borrowed from the Quantum mechanics, where $$P^2(\\psi)=P(P(\\psi))$$, but never $$P(\\psi)^2$$.\n\nTools for evaluation of superfunctions, abelfunctions and non-integer iterates of holomorphic functions are collected. For a given transfer function T, the superfunction is solution F of the transfer equation F(z+1)=T(F(z)) . The abelfuction is inverse of F. In particular, the superfunctions of factorial, exponent, sin; the holomorphic extensions of the logistic sequence and of the Ackermann functions are suggested. From ackermanns, the tetration (mainly to the base b>1) and pentation (to base e) are presented. The efficient algorithm for the evaluation of superfunctions and abelfunctions are described. The graphics and complex maps are plotted. The possible applications are discussed. Superfunctions significantly extend the set of functions that can be used in scientific research and technical design. Generators of figures are loaded to the site TORI, http://mizugadro.mydns.jp for the free downloading. With these generators, the Readers can reproduce (and modify) the figures from the Book. The Book is intended to be applied and popular. I try to avoid the complicated formulas, but some basic knowledge of the complex arithmetics, Cauchi integral and the principles of the asymptotical analysis should help at the reading.\n\nDmitrii Kouznetsov\n\nGraduated from the Physics Department of the Moscow State University (1980). Work: USSR, Mexico, USA, Japan.\nCentury 20: Proven the quantum stability of the optical soliton, suggested the low bound of the quantum noise of nonlinear amplifier, indicated the limit of the single mode approximation in the quantum optics.\nCentury 21: Theorem about boundary behaviour of modes of Dirichlet laplacian, Theory of ridged atomic mirrors, Formalism of superfunctions, TORI axioms.\n\n## Summary\n\nThe summary suggests main notations used in the Book:\n\n$$T$$$$~ ~ ~ ~ ~$$ Transfer function\n\n$$T\\big(F(z)\\big)=F(z\\!+\\!1)$$ $$~ ~ ~$$ Transfer equation, superfunction\n\n$$G\\big(T(z)\\big)=G(z)+1$$ $$~ ~ ~$$ Abel equation, abelfunction\n\n$$F\\big(G(z)\\big)=z$$ $$~ ~ ~ ~ ~$$ Identity function\n\n$$T^n(z)=F\\big(n+G(z)\\big)$$ $$~ ~ ~$$ $$n$$th iterate\n\n$$\\displaystyle F(z)=\\frac{1}{2\\pi \\mathrm i} \\oint \\frac{F(t) \\, \\mathrm d t}{t-z}$$ $$~ ~ ~$$ Cauchi integral\n\n$$\\mathrm{tet}_b(z\\!+\\!1)=b^{\\mathrm{tet}_b(z)}$$ $$~ ~ ~$$ tetration to base $$b$$\n\n$$\\mathrm{tet}_b(0)=1$$ $$~, ~ ~$$ $$\\mathrm{tet}_b\\big(\\mathrm{ate}_b(z)\\big)=z$$\n\n$$\\mathrm{ate}_b(b^z)=\\mathrm{ate}_b(z)+1$$ $$~ ~$$ arctetration to base $$b$$\n\n$$\\exp_b^{~n}(z)=\\mathrm{tet}_b\\big(n+\\mathrm{ate}_b(z)\\big)$$ $$~ ~$$ $$n$$th iterate of function $$~$$ $$z\\!\\mapsto\\! b^z$$\n\n$$\\displaystyle \\mathrm{Tania}^{\\prime}(z)=\\frac{\\mathrm{Tania}(z)}{\\mathrm{Tania}(z)\\!+\\!1}$$ $$~ ~$$ Tania function,$$~$$ $$\\mathrm{Tania}(0)\\!=\\!1$$\n\n$$\\displaystyle \\mathrm{Doya}(z)=\\mathrm{Tania}\\big(1\\!+\\!\\mathrm{ArcTania}(z)\\big)$$ $$~ ~$$ Doya function\n\n$$\\displaystyle \\mathrm{Shoka}(z)=z+\\ln(\\mathrm e^{-z}\\!+\\!\\mathrm e \\!-\\! 1)$$ $$~$$ Shoka function\n\n$$\\displaystyle \\mathrm{Keller}(z)=\\mathrm{Shoka}\\big(1\\!+\\!\\mathrm{ArcShoka}(z)\\big)$$ $$~ ~$$ Keller function\n\n$$\\displaystyle \\mathrm{tra}(z)=z+\\exp(z)$$ $$~ ~ ~$$ Trappmann function\n\n$$\\displaystyle \\mathrm{zex}(z)=z\\,\\exp(z)$$ $$~ ~ ~ ~$$ Zex function\n\n$$\\displaystyle \\mathrm{Nem}_q(z)=z+z^3+qz^4$$ $$~ ~ ~ ~$$ Nemtsov function\n\n## Pictures\n\nThe pictures in the book are supplied with generators. The colleagues may download them, execute and reproduce the pics in *.eps and *.pdf format.\n\nAnd, of course, modify them, adopt for the new similar problems.\n\nThe plotter of complex map conto.cin is original. It may have some bug(s), but still it seems to be best plotter for the complex maps, at least for year 2022. Any map from the Book can be generated in \"real time\".\n\nThe hand-drawn pics show connection of the topic of the book with the common sense of reality. They can be used also separately.\n\nMost of results, presented in the book, are published in scientific journals; the links (without numbers) are supplied at the bottom.\n\nAfter the appearance of the first version of the Book (2014), certain advances are observed about evaluation of tetration of complex argument; the new algorithm is suggested . These results are not included in the book.\n\n## Refutation\n\nSeveral transfer functions are considered in the book. I was searching for a transfer function such that for it, no suprefuction can be constructed. I failed. Perhaps, such a function does not exist. This can be considered as a conjecture.\n\nIf some colleague can prove this conjecture,\nor refute it, id est, to construct such a function, for which the superfuntion cannot be constricted with methods, described in this Book –\nthen, please, publish your result and let me know (dmitriiKouznetsov(at)gmail.com), in order that I can cite your publication here.\n\n## Misprints\n\nplaceprintedshould be\np.236, line 1 In the movel . In the novel\np.236, line 3 16.10 or 16.11, (16.10) or (16.11),\nhttp://www.ils.uec.ac.jp/~dima/BOOK/443.pdf (its preliminary version)\n2. https://www.lap-publishing.com/catalog/details//store/gb/book/978-3-659-56202-0/Суперфункции Дмитрий Кузнецов. Суперфункции. ISBN-13: 978-3-659-56202-0 ISBN-10: 3659562025 EAN: 9783659562020\nhttp://www.ils.uec.ac.jp/~dima/BOOK/202.pdf\n3. http://journal.kkms.org/index.php/kjm/article/view/428 William Paulsen. Finding the natural solution to f(f(x))=exp(x). Korean J. Math. Vol 24, No 1 (2016) pp.81-106.\n4. https://link.springer.com/article/10.1007/s10444-017-9524-1 William Paulsen, Samuel Cowgill. Solving F(z + 1) = b F(z) in the complex plane. Advances in Computational Mathematics, December 2017, Volume 43, Issue 6, pp 1261–1282\n5. https://search.proquest.com/openview/cb7af40083915e275005ffca4bfd4685/1?pq-origsite=gscholar&cbl=18750&diss=y Cowgill, Samuel. Exploring Tetration in the Complex Plane. Arkansas State University, ProQuest Dissertations Publishing, 2017. 10263680.\n6. https://link.springer.com/article/10.1007/s10444-018-9615-7 William Paulsen. Tetration for complex bases. Advances in Computational Mathematics, 2018.06.02.\n\nThe book combines the main results from the following publications:\n\nhttp://www.ams.org/mcom/2009-78-267/S0025-5718-09-02188-7/home.html\nD.Kouznetsov. Analytic solution of F(z+1)=exp(F(z)) in complex z-plane. Mathematics of Computation 78 (2009), 1647-1670.\n\nhttp://www.jointmathematicsmeetings.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/S0025-5718-10-02342-2.pdf\nhttp://www.ams.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/home.html\nhttp://eretrandre.org/rb/files/Kouznetsov2009_215.pdf\nhttp://www.ils.uec.ac.jp/~dima/PAPERS/2010q2.pdf\nD.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756.\n\nD.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12.\n\nD.Kouznetsov. Tetration as special function. Vladikavkaz Mathematical Journal, 2010, v.12, issue 2, p.31-45.\n\nD.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12\n\nhttp://mizugadro.mydns.jp/t/index.php/Place_of_science_in_the_human_knowledge D.Kouznetsov. Place of science and physics in human knowledge. English translation from http://ufn.ru/tribune/trib120111 Russian Physics:Uspekhi, v.191, Tribune, p.1-9 (2010)\n\nhttp://www.ils.uec.ac.jp/~dima/PAPERS/2010logistie.pdf http://mizugadro.mydns.jp/PAPERS/2010logistie.pdf D.Kouznetsov. Continual generalisation of the Logistic sequence. Moscow State University Physics Bulletin, 3 (2010) No.2, стр.23-30.\n\nhttp://www.ams.org/journals/mcom/0000-000-00/S0025-5718-2012-02590-7/S0025-5718-2012-02590-7.pdf\nhttp://www.ils.uec.ac.jp/~dima/PAPERS/2012e1eMcom2590.pdf\nH.Trappmann, D.Kouznetsov. Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of Computation, 2012, 81, February 8. p.2207-2227.\n\nhttp://www.ils.uec.ac.jp/~dima/PAPERS/2012or.pdf\nDmitrii Kouznetsov. Superfunctions for optical amplifiers. Optical Review, July 2013, Volume 20, Issue 4, pp 321-326.\n\nhttp://www.scirp.org/journal/PaperInformation.aspx?PaperID=36560" ]
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https://ethereum.stackexchange.com/questions/55568/ethereum-go-source-hextocompact-function
[ "# Ethereum Go Source: hexToCompact function\n\nI have a question regarding the following code snippet. Specifically the following function:\n\n``````func hexToCompact(hex []byte) []byte {\nterminator := byte(0)\nif hasTerm(hex) {\nterminator = 1\nhex = hex[:len(hex)-1]\n}\nbuf := make([]byte, len(hex)/2+1)\nbuf = terminator << 5 // the flag byte\nif len(hex)&1 == 1 {\nbuf |= 1 << 4 // odd flag\nbuf |= hex // first nibble is contained in the first byte\nhex = hex[1:]\n}\ndecodeNibbles(hex, buf[1:])\nreturn buf\n}\n``````\n\nThis is a implementation of Appendix-C Hex-Prefix encoding from the Yellow Paper and when I did a search on this in the repo. I found a function within slasher.go calls hexToCompact.\n\n``````func (h *hasher) hashChildren(original node, db *Database) (node, node, error) {\nvar err error\n\nswitch n := original.(type) {\ncase *shortNode:\n// Hash the short node's child, caching the newly hashed subtree\ncollapsed, cached := n.copy(), n.copy()\ncollapsed.Key = hexToCompact(n.Key)\ncached.Key = common.CopyBytes(n.Key)\n\nif _, ok := n.Val.(valueNode); !ok {\ncollapsed.Val, cached.Val, err = h.hash(n.Val, db, false)\nif err != nil {\nreturn original, original, err\n}\n}\nreturn collapsed, cached, nil\n\ncase *fullNode:\n// Hash the full node's children, caching the newly hashed subtrees\ncollapsed, cached := n.copy(), n.copy()\n\nfor i := 0; i < 16; i++ {\nif n.Children[i] != nil {\ncollapsed.Children[i], cached.Children[i], err = h.hash(n.Children[i], db, false)\nif err != nil {\nreturn original, original, err\n}\n}\n}\ncached.Children = n.Children\nreturn collapsed, cached, nil\n\ndefault:\n// Value and hash nodes don't have children so they're left as were\nreturn n, original, nil\n}\n}\n``````\n\nMy questions are\n\n1. What is the need for Hex-Prefix compactness as specified in Yellow Paper?\n\n2. What is slasher.go is trying to do here by compacting a shortnode and why just shortnode?." ]
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https://toph.co/p/easy-sequence-1
[ "# Practice on Toph\n\nParticipate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.\n\n# Easy Sequence 1\n\nBy ridowan007 · Limits 1s, 512 MB\n\nLet’s define an easy sequence F. Where\n\n$F(1) = 1$\n\n$F(2X) = F(X)^2 + 1$\n\n$F(2X+1) = F(X) \\times F(X+1) + 2$\n\nHere X is a positive integer. The first few elements of the sequence are 1, 2, 4, 5, 10, 17, 22, 26, 52, 101.\n\nNow given a value of $F(X)$, can you compute X?\n\n## Input\n\nThe first line of the input is a positive integer T (T ≤ 100), the number of test cases. Then T lines follow.\n\nIn each line there will be a value of $F(X)$ (1 ≤ $F(X)$ ≤ 109). It is guaranteed that for every $F(X)$ in the input, there exists a valid X.\n\n## Output\n\nFor each case print a line in “Case I: X” format where I is the test case number and X is the corresponding value for the given $F(X)$ in the input.\n\n## Sample\n\nInputOutput\n10\n1\n2\n4\n5\n10\n17\n22\n26\n52\n101\nCase 1: 1\nCase 2: 2\nCase 3: 3\nCase 4: 4\nCase 5: 5\nCase 6: 6\nCase 7: 7\nCase 8: 8\nCase 9: 9\nCase 10: 10\n\n•", null, "DP\n\n### Statistics\n\n86% Solution Ratio\n\nmizan_rEarliest, Jun '16\n\ndrifter.pustFastest, 0.0s\n\ntamzid_secLightest, 0 B\n\nWhiteCiferShortest, 178B\n\n### Submit", null, "" ]
[ null, "https://static.toph.co/assets/icons/pricetags-outline.svg", null, "https://static.toph.co/assets/images/loading.svg", null ]
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https://www.cprogramming.com/tips/tip/xor-swap-swaping-the-values-of-2-variables
[ "# XOR swap: Swaping the values of 2 variables.\n\nThis tip submitted by John Rambo on 2013-02-03 17:45:25. It has been viewed 8437 times.\nRating of 7.3 with 27 votes\n\nI am not going to be explaining how XOR works, so I suggest you google \"C bitwise operators\" and learn about them.\n\nIn a tutorial I found, it states this: \\\" ^ The bitwise-exclusive-OR operator compares each bit of its first operand to the corresponding bit of its second operand. If one bit is 0 and the other bit is 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.\\\"\n\nHere's the code:\n```/* This is a XOR swap. It works\nby using the XOR bitwise operator\nto compare variable values. */\n\n#include\n\nint main(int argc, char *argv[])\n{\nint i = 6; //0110 in binary.\nint y = 12; //1100 in binary.\n\nprintf(\"i = %d, and y = %d\\n\", i, y);\n\ni ^= y; // i = |1010| |1010|\ny ^= i; // y = |0110|, because we're comparing y to new i. |1100|\ni ^= y; // i = |1100|\n\nprintf(\"i = %d, and y = %d\\n\", i, y);\n\nreturn 0;\n}```\n\nMore tips" ]
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https://www.bartleby.com/questions-and-answers/the-graph-of-a-linear-equation-in-two-variables-is-a-line-in-a-twodimensional-coordinate-system.-the/6fde6a2b-1608-4bc4-9222-50c35b44e34a
[ "# The graph of a linear equation in two variables is a line in a two-dimensional coordinate system. The graph of a linear equation in three variables is a __________ in a three-dimensional coordinate system.\n\nQuestion\n2 views\n\nThe graph of a linear equation in two variables is a line in a two-dimensional coordinate system. The graph of a linear equation in three variables is a __________ in a three-dimensional coordinate system.\n\ncheck_circle\n\nStep 1\n\nAs the graph of a linear equation in two variables is a line in two-dimensional coordinate system.\n\nFor example:\n\nAn equat...\n\n### Want to see the full answer?\n\nSee Solution\n\n#### Want to see this answer and more?\n\nSolutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*\n\nSee Solution\n*Response times may vary by subject and question.\nTagged in\n\n### Algebra", null, "" ]
[ null, "https://www.bartleby.com/static/logo-full-footer.svg", null ]
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https://www.24houranswers.com/college-homework-library/Mathematics/Pre-Calculus/8773
[ "", null, "# Pre-Calculus Problem\n\nSubject Mathematics Pre-Calculus\n\n## Question\n\nFind the polynomial with real coefficients of the smallest possible degree for which i and 1 + i are zeros and in which the coefficient of the highest power is 1.\n\n## Solution Preview\n\nThis material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.\n\nSince i and 1+i are roots of this problem then -i and 1- i are also roots\n...\n\nThis is only a preview of the solution. Please use the purchase button to see the entire solution\n\n## Related Homework Solutions\n\nPrecalculus Questions", null, "\\$5.00\nLim\nDerivative\nFraction\nL'Hospital\nRule\nSquare\nRoot\n0/0\nDifferentiate\nArctan\nSin\nLn\nData Analysis Questions", null, "\\$15.00\nMathematics\nData Analysis\nPredictions\nModels\nPlots\nPolynomials\nCurves\nFitting Data\nDiscussion\nScatterplot\nSlope-Intercept Form\nLinear Regression\nPoints\nAxes\nPre-Calculus Problem", null, "\\$3.00\nPre-Calclus\nBreak Even Analysis\nPersonal Income Tax\nMonthly Fixed Costs\nBreak-Even Point\nProfit\nPre-Calculus Problem", null, "\\$3.00\nPre-Calculus\nDecision Analysis\nCost\nCost of Manufacturing\nPrice\nProfit\nFixed Cost\nMaximising the Profit", null, "\\$15.00\nMathematics\nFunctions\nVariables\nDiscussion\nPeople\nNumbers\nDomain\nRange\nSystems\nRelations\nCar Registration\nApplication of Systems of Equations", null, "\\$13.00\nMathematics\nSystem of Equations\nPerimeter\nArea\nVolume\nLinearity\nApplication Problems\nInformation\nStatements\nSolution Techniques\nVariables\nDiscussion\nLive Chats" ]
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http://drjj101.tripod.com/calculusproject/id1.html
[ "", null, "", null, "", null, "", null, "", null, "", null, "AP Calculus BC Free Response Questions Problem #1", null, "", null, "", null, "", null, "HOME", null, "Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6", null, "Let f and g be the functions given by f(x) = e^x and g(x) = ln x.\n\n(a) Find the area of the region encloed by the graphs of f and g between x = 1/2 and x = 1.\n\n(b) Find the volume of the solid generated when the region eclosed by the graphs of f and g between x = 1/2 and x = 1 is revolved about the line y = 4.\n\n(c) Let h be the function given h(x) = f(x) - g(x). Find the absolute minimum value of h(x) on the closed interval 1/2 less than or equal to x less than or equal to 1, and find the absolute maximum value of h(x) on the closed interval 1/2 less than or equal to x less than or equal to 1. Show the analysis that leads to your answer.", null, "", null, "", null, "", null, "", null, "This problem wasn't very hard, I just couldn't remember how to find volume around a line, so that made things a little more difficult on the day of the test.", null, "", null, "" ]
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https://mathoverflow.net/questions/182444/power-series-with-funny-behavior-at-the-boundary
[ "# Power series with funny behavior at the boundary\n\nConsider a power series $$\\sum_{n=0}^{\\infty}a_nz^n$$ where $a_n$ and $z$ are complex numbers. There is radius $R$ of convergence. Let us assume that is a positive real number. It is well known that for $|z|<R$ the series converges absolutely; for $|z|>R$ it does not converge.\n\nOn the other hand, when $|z|=R$, the series can have very different behaviors. This has been discussed in many posts, e.g. Behaviour of power series on their circle of convergence , Seeking a Geometric Proof of a Generalized Alternating Series' Convergence .\n\nI am looking for some relatively easy explicit examples of power series with funny behavior at the boundary. The only I know is $$\\sum_n\\frac{1}{n}z^n$$ or small variations, e.g. replacing $z$ with $z^k$.\n\nthanks\n\n• This is hard to answer without knowing what you mean by \"funny\". One remark: given $f(z) = \\sum_n z^n/n$, which has its only singularity on the unit circle at $z=1$, you can form linear combinations $\\sum_k c_k f(e^{i\\theta_k}z)$ to get singuarities at any finite number of boundary points, or even an infinite number if $\\sum |c_k|$ converges. You could also consider $\\sum_m z^{2^m}$, which has a singularity at every $2^j$th root of unity. – Greg Martin Oct 3 '14 at 7:43\n• Funny means just that the set $S$ of singular points is not the all circle and maybe it is dense or I do not know. I would like just explicit examples. How do you show that $\\sum_m z^{2^m}$ converges outside the $2^j$ roots of unity??? This would be the sort of example I am looking for. – Giulio Oct 3 '14 at 7:50\n• It looks like $\\sum z^{2^k}$ does not converge at any root of unity. At a $2^j$th root it diverges to infinity. At other roots the partial sums are periodic (try $e^{2\\pi i/3}.$) I think that would converge every place else. Suppose $\\sum c_k$ diverges with the $c_i$ positive and decreasing to $0.$ Then $\\sum z^{2^k}c_k$ would diverge to infinity at a $2^j$th root of unity but should converge everywhere else. – Aaron Meyerowitz Oct 3 '14 at 8:48\n• what I am thrilled by is how to prove convergence everywhere else. Say at a point whose phase is irrational. – Giulio Oct 3 '14 at 9:00\n• I wonder if a method as in the post \"seeking...\" I linked in the question could work. – Giulio Oct 3 '14 at 9:04\n\nThe series $$f(z) = \\sum_{n=1}^\\infty \\dfrac{z^{2^n}}{n}$$ converges almost everywhere on the unit circle by Carleson's theorem (it is the Fourier series of an $L^2$ function). However, it diverges on a dense set, including all the $2^k$'th roots of unity: in fact at each of those points the real parts of the partial sums $S_k = \\sum_{n=1}^k z^{2^n}/n$ are unbounded above.\n\nNow for each positive integer $N$, the set $U_N$ of points $z$ such that $\\text{Re}(S_k) > N$ for some $k$ is open and dense in the unit circle. The intersection $G$ of the $U_N$ is a dense $G_\\delta$ by the Baire category theorem, and the series diverges at every point of $G$.\n\nPutting these facts together, we find that the set of points of the unit circle where the series diverges is negligible in the sense of Lebesgue measure but the set where it converges is negligible in the sense of Baire category. I'd call that funny...\n\n• What are you talking about? It doesn't diverge at every non-root of $2^j$, and neither does $\\sum_n \\frac{z^n}{n}$ (which is also an $L^2$ function). – Robert Israel Oct 5 '14 at 6:28\n• Sorry, it is taking me a while to understand. Let me try to rephrase your argument to see if I get it. – Giulio Oct 5 '14 at 15:07\n• Convergence. The sequence of S_k is a Fourier series and it is convergent sequence for the L2 norm on the circle. Because of Carleson's theorem it converges point-wise almost everywhere for the Lebesgue measure. – Giulio Oct 5 '14 at 17:14\n• Divergence. The series diverges at every $2^j$ root of unity. Let $U_N$ be the subset of the circle where $|S_k|>N$ for some k (or also $Re(S_k)$). Since $S_k$ are continuos functions the $U_N$ are open. Each $U_N$ contains the $2^j$ roots of unity, so they are dense. Call G the intersection of all $U_N$. The series can not converge on G because it is not Cauchy. By Baire's theorem G is dense in the circle; moreover it is a $G_{\\delta}$, so, for instance, it is not countable. – Giulio Oct 5 '14 at 18:17\n• Does this make sense?? – Giulio Oct 5 '14 at 18:17\n\nHere is a funny result. $\\newcommand{\\ve}{\\varepsilon}$ Consider the random series\n\n$$\\sum_{n\\geq 1}\\frac{\\ve_n}{n} z^n,$$\n\nwhere the signs $\\ve_n=\\pm 1$ are chosen randomly and independently with ${\\rm Prob}\\;(\\ve_n=\\pm 1)=\\frac{1}{2}$, $\\forall n$.\n\nThen, almost surely, the above series cannot be continued across the circle of convergence $|z|=1$; see Chapter 4 of Kahane's book on random series of functions.\n\nOn the other hand, Kolmogorov's three-series theorem implies that for any fixed $z$ with $|z|=1$ the above series converges for almost any choice of random signs $\\ve_n$.\n\nPut these two facts together to deduce that, with probability $1$, the above series cannot be extended across $|z|=1$, yet it converges for any $z$ of the form $z=\\exp( 2\\pi \\sqrt{-1} r)$, $r\\in \\mathbb{Q}$." ]
[ null ]
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https://projecteuclid.org/euclid.jgsp/1495764128
[ "Journal of Geometry and Symmetry in Physics\n\nForce Free Möbius Motions of the Circle\n\nAbstract\n\nLet $\\mathcal{M}$ be the Lie group of Möbius transformations of the circle. Suppose that the circle has initially a homogeneous distribution of mass and that the particles are allowed to move only in such a way that two configurations differ in an element of $\\mathcal{M}$. We describe all force free Möbius motions, that is, those curves in $\\mathcal{M}$ which are critical points of the kinetic energy. The main tool is a Riemannian metric on $\\mathcal{M}$ which turns out to be not complete (in particular not invariant, as happens with non-rigid motions) given by the kinetic energy.\n\nArticle information\n\nSource\nJ. Geom. Symmetry Phys., Volume 27 (2012), 59-65.\n\nDates\nFirst available in Project Euclid: 26 May 2017\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.jgsp/1495764128\n\nDigital Object Identifier\ndoi:10.7546/jgsp-27-2012-59-65\n\nMathematical Reviews number (MathSciNet)\nMR3026387\n\nZentralblatt MATH identifier\n1267.53016\n\nCitation\n\nEmmanuele, Daniela; Salvai, Marcos. Force Free Möbius Motions of the Circle. J. Geom. Symmetry Phys. 27 (2012), 59--65. doi:10.7546/jgsp-27-2012-59-65. https://projecteuclid.org/euclid.jgsp/1495764128" ]
[ null ]
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https://answers.everydaycalculation.com/simplify-fraction/9000-195
[ "Solutions by everydaycalculation.com\n\n## Reduce 9000/195 to lowest terms\n\nThe simplest form of 9000/195 is 600/13.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 9000 and 195 is 15\n2. Divide both the numerator and denominator by the GCD\n9000 ÷ 15/195 ÷ 15\n3. Reduced fraction: 600/13\nTherefore, 9000/195 simplified to lowest terms is 600/13.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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https://delteria.app/the-base-of-a-triangle-understanding-its-importance/
[ "# The Base of a Triangle: Understanding its Importance\n\nThe base of a triangle plays a crucial role in the determination of the triangle’s area and other geometric properties. In geometry, a triangle is a polygon with three sides, three vertices, and three angles. It is one of the most basic shapes that exist in mathematics, but its importance cannot be overlooked in many fields such as architecture, engineering, and physics.\n\nIn this article, we aim to explore the significance of the base of a triangle and understand its importance in various geometric contexts.\n\nDefinition of the Base of a Triangle\n\nIn a triangle, the base is defined as the side of the triangle that is opposite to the apex or vertex of the triangle. The apex or vertex is the point at which two sides of a triangle meet. To determine the height or altitude of a triangle, we need to drop a perpendicular line from the apex of the triangle to its base. Therefore, the base and height or altitude of a triangle are essential parameters in the determination of the triangle’s area.\n\nImportance of the Base of a Triangle\n\nThe base of a triangle plays a pivotal role in the determination of the triangle’s area, perimeter, and other geometric properties. Let us take a look at some of the ways the base influences these measures.\n\nArea of a Triangle\n\nThe area of a triangle is given by the formula:\n\nArea = 1/2 * base * height\n\nwhere the height is the length of the perpendicular line dropped from the apex of the triangle to its base. It is evident from the formula that the base and height are two critical parameters in determining the area of a triangle. For example, if we have two triangles with the same height but different bases, the triangle with the longer base will have a larger area compared to the triangle with the shorter base.\n\nPerimeter of a Triangle\n\nThe perimeter of a triangle is defined as the sum of the lengths of its sides. The base of a triangle is one of its sides, and therefore, it contributes to the total perimeter of the triangle. In a right-angled triangle, the hypotenuse is the longest side, and the other two sides are the base and height. The perimeter of a right-angled triangle is given by the formula:\n\nPerimeter = base + height + hypotenuse\n\nIt is clear from the formula that the base of a right-angled triangle contributes to the total perimeter of the triangle.\n\nCentroid of a Triangle\n\nThe centroid of a triangle is defined as the point where the three medians of a triangle intersect. A median of a triangle is a line segment drawn from one vertex of the triangle to the midpoint of the opposite side. The centroid divides each median into two segments in the ratio 2:1. It is an essential parameter in determining the stability of structures such as bridges and buildings. The centroid of a triangle is located at two-thirds the distance from the vertex to the midpoint of the opposite side along the median. The base of a triangle is one of its sides, and therefore, it contributes to the determination of the centroid of the triangle.\n\nInscribed Circle of a Triangle\n\nThe inscribed circle of a triangle is defined as the largest circle that can be inscribed within a triangle. It has a center that is equidistant from the three sides of the triangle. The radius of the inscribed circle depends on the area of the triangle, which, in turn, depends on the base and height of the triangle. Therefore, the base of a triangle plays a crucial role in determining the radius of the inscribed circle of the triangle.\n\nConclusion\n\nIn conclusion, the base of a triangle is a crucial parameter in determining its area, perimeter, centroid, and inscribed circle. We have explored some of the ways the base influences these geometric properties and have reiterated the importance of understanding the base’s significance in various contexts. Whether you’re an architect, engineer, or physicist, understanding the importance of the base of a triangle can help you make informed decisions and calculations." ]
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https://nyuscholars.nyu.edu/en/publications/classroom-examples-of-robustness-problems-in-geometric-computatio-2
[ "# Classroom examples of robustness problems in geometric computations\n\nLutz Kettner, Kurt Mehlhorn, Sylvain Pion, Stefan Schirra, Chee Yap\n\nResearch output: Contribution to journalArticlepeer-review\n\n## Abstract\n\nThe algorithms of computational geometry are designed for a machine model with exact real arithmetic. Substituting floating-point arithmetic for the assumed real arithmetic may cause implementations to fail. Although this is well known, there are no concrete examples with a comprehensive documentation of what can go wrong and why. In this paper, we provide a case study of what can go wrong and why. For our study, we have chosen two simple algorithms which are often taught, an algorithm for computing convex hulls in the plane and an algorithm for computing Delaunay triangulations in space. We give examples that make the algorithms fail in many different ways. We also show how to construct such examples systematically and discuss the geometry of the floating-point implementation of the orientation predicate. We hope that our work will be useful for teaching computational geometry.\n\nOriginal language English (US) 61-78 18 Computational Geometry: Theory and Applications 40 1 https://doi.org/10.1016/j.comgeo.2007.06.003 Published - May 2008\n\n## Keywords\n\n• Floating-point geometry\n• Implementation\n• Numerical robustness problems\n\n## ASJC Scopus subject areas\n\n• Computer Science Applications\n• Geometry and Topology\n• Control and Optimization\n• Computational Theory and Mathematics\n• Computational Mathematics\n\n## Fingerprint\n\nDive into the research topics of 'Classroom examples of robustness problems in geometric computations'. Together they form a unique fingerprint." ]
[ null ]
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https://iamdurga.github.io/2022/10/21/mathematics-behind-naivebayes/
[ "Here, in this blog, we are going to discuss the mathematics behind Naivebayes, taking one example problem.\n\n# Introduction to NaiveBayes\n\nNaiveBayes is a simple classification algorithm in the machine learning sense. It works on the basis of probability. It calculates the poster’s probability using some prior knowledge. It is assumed that each feature used is independent of the other. It is used for multi-class classification problems. When the probability of a particular class is high, then it confirms that features belong in that class. Let’s see one example to be clear how Naive Bayes worked Mathematically.\n\n[Q.N] Predict class level of the tuple: X = (age = youth, income = medium, student = yes, credit_rating = fair) using Bayesian classification.\n\nAge income Student Credit_rating Bayes_computer\nyouth high no fair no\nyouth high no excellent no\nmiddle_age high no fair yes\nsenior medium no fair yes\nsenior low yes fair yes\nsenior low yes excellent yes\nyouth medium no fair no\nyouth low yes fair yes\nsenior medium yes fair yes\nyouth medium yes excellent yes\nmiddle_age medium no excellent yes\nmiddle_age high yes fair yes\nsenior medium no excellent no\n\nSolution:\n\nStep I\n\nPrior probability of each class can be computed based on the training tuples\n\nP(Buy-computer => yes) = Total no of yes/Total class = 9/14 = 0.643\n\nP(Buy-computer => no) = Total no of no/Total class = 5/14 = 0.357\n\nStep II\n\nComputation of conditional probability,\n\nP(Age => youth/Buy computer => yes) = Total no of youth out of yes/Total yes = 2/9 = 0.222\n\nP(Age => youth/Buy-computer => no) = Total no of youth out of no/Total no = 3/5 = 0.6\n\nP(Income => medium/Buy computer => yes) = Total no of medium out of yes/Total yes = 4/9 = 0.444\n\nP(income => medium/Buy computer => no) = Total no of medium out of no/Total no = 2/5 = 0.4\n\nP(student => yes/Buy computer => yes) = Total no of yes out of yes/Total yes = 6/9 = 0.667\n\nP(Student => yes/Buy computer => no) = Total no of yes out of no/Total no = 1/5 = 0.2\n\nP(Credit rating => fair/Buy computer => yes) = Total no of fair out of yes/Total-yes = 6/9 = 0.667\n\nP(Credit-rating => fair}{Buy computer => no) = Total no of yes out of noTtotal no = 2/5 = 0.4\n\nStep III\n\nUsing above probability let’s calculate conditional probability\n\nP(features/Buy computer yes) = P(Age => yes/Buy computer => yes) * P(Income => medium}{Buy computer => yes) * P(Student => yes/Buy computer => yes) * P(Credit rating => fair/Buy computer => yes) = 0.222×0.444×0.667×0.667 = 0.044\n\nP(features/Buy computer no) = P(Age => yes/Buy computer => no) * P(Income => medium/Buy computer => no) * P(Student => yes/Buy computer => no) * P(Credit rating => fair/Buy computer => no) = 0.60.40.2*0.4 = 0.019\n\nStep IV\n\nComputation of Posterior Probability\n\nConclusion:\n\nSince," ]
[ null ]
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https://wiki.openjdk.java.net/exportword?pageId=11829253
[ "Date: Sat, 23 Oct 2021 02:35:07 +0000 (UTC) Message-ID: <2102456419.11956.1634956507464@aojmv0003.oracle.com> Subject: Exported From Confluence MIME-Version: 1.0 Content-Type: multipart/related; boundary=\"----=_Part_11955_1275614187.1634956507463\" ------=_Part_11955_1275614187.1634956507463 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Content-Location: file:///C:/exported.html RangeCheckElimination\n\n# RangeCheckElimination\n\n#### Iteration = range splitting\n\n=20\n\nFor loop-invariant arrays, range checks can usually be eliminated.\n\n= =20\n\n=20\n\nThis is carried out by means of iteration range splitting. A middle rang= e of loop index values is computed before the loop is entered. (It is often= the whole sequence of index values, but need not be if loop peeling or unr= olling is happening also.) The loop is cloned three times, with the three c= lones running in succession. The middle loop handles the middle range, and = the pre-loop (resp. post-loop) handles any index values before (resp. after= ) the middle range. The middle range is chosen so that the middle loop (mai= n loop) is as large as possible, but is constrained to values which can be = predicted not to cause array range checks to fail.\n\n=20\n\nHere is a simple case:\n\n=20\nSimple Loop\n=20\n```for (int index =3D Start; index < Limit; index++) {\nArray[index] =3D 0;\n}\n```\n=20\n=20\n\nThe loop is split like this:\n\n=20\nSimple Loop, Optimized\n=20\n```int MidStart =3D Math.max(Start, 0);\nint MidLimit =3D Math.min(Limit, Array.length);\nint index =3D Start;\nfor (; index < MidStart; index++) { // PRE-LOOP\nArray[index] =3D 0; // RANGE CHECK\n}\nfor (; index < MidLimit; index++) { // MAIN LOOP\nArray[index] =3D 0; // NO RANGE CHECK\n}\nfor (; index < Limit; index++) { // POST-LOOP\nArray[index] =3D 0; // RANGE CHECK\n}\n```\n=20\n=20\n\n#### When the optimi= zation applies\n\n=20\n\nThis typically happens when:\n\n=20\n=20\n• the array is a loop invariant,\n• =20\n• on a hot loop,\n• =20\n• whose index variable has a constant stride,\n• =20\n• and where the array is indexed by linear functions of the index variabl= e.\n• =20\n=20\nTypical Loop\n=20\n```for (int index =3D Start; index < Limit; index +=3D STRIDE) {\n... Array1[index * SCALE1 + Offset1] ...\n... Array2[index * SCALE2 + Offset2] ...\n}\n```\n=20\n=20\n\nCapitalized names (`Limit`, `Array1`) must be loop= -invariant, while all-caps names (`STRIDE`, `SCALE1`)= must be compile-time constants. Any number of arrays can be dealt with thi= s way, both for read and write. Other non-matching arrays will not disturb = the optimization, but they will be range-checked continually. The loop need= not be a literal Java `for` statement; the JIT can deal with a = wide range of loop idioms.\n\n=20\n\nThe values `MinStart` and `MinLimit` are computed = similarly as above, except that there are more min/max terms, and there may= be divisions if `SCALE` values are not unity.\n\n=20\n\nThis pattern is quite general, although it doesn't help with every imagi= nable loop. Pointer-chasing loops, and loops with non-constant strides, don= 't participate in this optimization. Key rules of thumb:\n\n=20\n=20\n• =20\n• Use at most simple linear expressions to index each array.\n• =20\n=20\n\nA constant value can be a `final static` variable.\n\n=20\n\nA loop-invariant value can be an object field, as long as the compiler c= an prove the field isn't changed by the loop. This will fail if the loop bo= dy is not totally inlined, and the object field cannot be proven non-escapi= ng.\n\n=20\n\n#### Source code\n\n=20\n=20\n• See logic related to `PhaseIdealLoop::add_constraint` in loopTra= nsform.cpp.\n• =20\n• The optimization is controlled by the flag RangeCheckElimination.\n• = =20\n------=_Part_11955_1275614187.1634956507463--" ]
[ null ]
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https://www.fda.gov/food/laboratory-methods-food/bam-appendix-2-most-probable-number-serial-dilutions
[ "", null, "An official website of the United States government\n\n# BAM Appendix 2: Most Probable Number from Serial Dilutions\n\nBacteriological Analytical Manual (BAM) Main Page\n\nAuthor: Robert Blodgett (retired)\n\nFor additional Information, contact: Guodong Zhang, Martine Ferguson or CFSAN Biostatistics.\n\nRevision History:\n\n• August 2023: Dilution selector EXCEL tool added\n• October 2020: MPN calculator app added\n• April 2015: Contact for this Appendix was updated\n• July 2003: Added Table 5. for 10 tubes at 10 ml inocula and link to spreadsheet\n\n## Background\n\nSerial dilution tests measure the concentration of a target microbe in a sample with an estimate called the most probable number (MPN). The MPN is particularly useful for low concentrations of organisms (<100/g), especially in milk and water, and for those foods whose particulate matter may interfere with accurate colony counts. The following background observations are adapted and extended from the article on MPN by James T. Peeler and Foster D. McClure in the Bacteriological Analytical Manual (BAM), 7th edition.\n\nOnly viable organisms are enumerated by the MPN determination. If, in the microbiologist's experience, the bacteria in the prepared sample in question can be found attached in chains that are not separated by the preparation and dilution, the MPN should be judged as an estimate of growth units (GUs) or colony-forming units (CFUs) instead of individual bacteria. For simplicity, however, this appendix will speak of these GUs or CFUs as individual bacteria. If a confirmation test involves selecting colonies to test, then a statistical adjustment not discussed in this appendix should be used (see Blodgett 2005a.)\n\nThe following assumptions are necessary to support the MPN method. The bacteria are distributed randomly within the sample. The bacteria are separate, not clustered together, and they do not repel each other. Every tube (or plate, etc.) whose inoculum contains even one viable organism will produce detectable growth or change. The individual tubes of the sample are independent.\n\nThe essence of the MPN method is to dilute the sample to such a degree that inocula in the tubes will sometimes but not always contain viable organisms. The \"outcome\", i.e., the number of tubes and the number of tubes with growth at each dilution, will imply an estimate of the original, undiluted concentration of bacteria in the sample. In order to obtain estimates over a broad range of possible concentrations, microbiologists use serial dilutions incubating tubes at several dilutions.\n\nThe MPN is the number which makes the observed outcome most probable. It is the solution for λ, concentration, in the following equation", null, "where exp(x) means ex, and\n\nK denotes the number of dilutions,\ngj denotes the number of positive (or growth) tubes in the jth dilution,\nmj denotes the amount of the original sample put in each tube in the jth dilution,\ntj denotes the number of tubes in the jth dilution.\n\nIn general, this equation can be solved by iteration.\n\nMcCrady (1915) published the first accurate estimation of the number of viable bacteria by the MPN method. Halvorson and Ziegler (1933), Eisenhart and Wilson (1943), and Cochran (1950) published articles on the statistical foundations of the MPN. Woodward (1957) recommended that MPN tables should omit those combinations of positive tubes (high for low concentrations and low for high concentrations) that are so improbable that they raise concerns about laboratory error or contamination. De Man (1983) published a confidence interval method that was modified to make the tables for this appendix.\n\n## Confidence Intervals\n\nThe 95 percent confidence intervals in the tables have the following meaning:\n\nBefore the tubes are inoculated, the chance is at least 95 percent that the confidence interval associated with the eventual result will enclose the actual concentration.\n\nIt is possible to construct many different sets of intervals that satisfy this criterion. This manual uses a modification of the method of de Man (1983). De Man calculated his confidence limits iteratively from the smallest concentrations upward. Because this manual emphasizes pathogens, the intervals have been shifted slightly upward by iterating from the largest concentrations downward.\n\nThe confidence intervals of the spreadsheet and the tables associated with this appendix may be different.  The MPN Excel spreadsheet uses a normal approximation to the log (MPN) to calculate its confidence intervals.  This approximation is similar to a normal approximation discussed in Haldane (1939).  This approximation is less computationally intense so more appropriate for a spreadsheet than de Man's confidence intervals.\n\n## Precision, Bias, and Extreme Outcomes\n\nThe MPNs and confidence limits have been expressed to 2 significant digits. For example, the entry \"400\" has been rounded from a number between 395 and 405.\n\nNumerous articles have noted a bias toward over-estimation of microbial concentrations by the MPN. Garthright (1993) has shown, however, that there is no appreciable bias when the concentrations are expressed as logarithms, the customary units used for regressions and for combining outcomes. Therefore, these MPNs have not been adjusted for bias.\n\nThe outcome with all positive tubes in each dilution gives no upper bound on the concentration. The tables in this appendix list the MPN for this outcome as greater than the highest MPN for an outcome with at least one negative tube. Similarly, the outcome with all negative tubes is listed as less than the lowest MPN for an outcome with at least one positive tube.\n\n## Cautionary Notes\n\n### Improbable Outcomes\n\nSeveral potential problems may cause improbable outcomes. For example, there may be interference at low dilutions or selecting too few colonies at low dilutions for a confirmation test may overlook the target microbe. If the problem is believed limited to the low dilutions, then using only the high dilutions with positive tubes might be more reliable. If the cause of the problem is unknown, then the estimate may be unreliable.\n\nWhen excluding improbable outcomes, de Man's (1983) preferred degree of improbability was adopted. The outcomes included are among the 99.985 percent most likely outcomes if their own MPNs were the actual bacterial concentrations. Therefore, among 10 different outcomes, all will be found in these tables at least 99 percent of the time.\n\nInconclusive Tubes\n\nIn special cases where tubes cannot be judged either positive or negative (e.g., plates overgrown by competing microflora at low dilutions), these tubes should be excluded from the outcome. The resulting outcome may have different numbers of tubes than any of the tables in this appendix. Its MPN can be solved by computer algorithms or estimated by Thomas's Rule below. Haldane's method can find the confidence limits as described below Thomas's rule.\n\n## Using Tables\n\n### Selecting Three Dilutions for Table Reference\n\nAn MPN can be computed for any positive number of tubes at any positive number of dilutions, but often serial dilutions use three or more dilutions and a decimal series (Each dilution has one tenth as much of the original sample as the previous dilution.) The tables in this appendix require reducing an outcome to three of its decimal dilutions. This procedure for selecting three dilutions was developed for the designs (numbers of tubes per dilution and ratio of dilutions) in these tables. They all have decimal dilutions and a fairly small number of tubes per dilution. For other designs, other procedures may be needed. When the MPN model holds, the three decimal dilutions are chosen to give a good approximation to the MPN of the entire outcome. Otherwise, the reduction may remove interference (possible from another species of microbe or a toxic substance) that can be diluted out. The remainder of this section tells how to select the three dilutions.\n\nFor serial dilution experiments with more than three dilutions, attached is a dilution selector tool which selects the three dilutions which give a good approximation to the MPN.\n\nFirst, remove the highest dilution (smallest sample volume) if it and the next lower dilution have all negative tubes. As long as this condition holds and at least four dilutions remain, continue removing these dilutions.\n\nNext, if only three dilutions remain, use them as illustrated in example A. In each example there are five tubes in each dilution. In example A, removing the two highest dilutions (0.001 and 0.01 grams) leaves three dilutions.\n\nIf more than three dilutions remain, then find the highest dilution with all positive tubes. There are three cases. In the first case, the highest dilution with all positive tubes is within the three highest remaining dilutions. Then use the three highest remaining dilutions. In example B, the first step removes the highest dilution (0.001 grams.) Since the highest dilution with all positive tubes (1 gram) is within the three highest remaining dilutions, (1, 0.1, and 0.01 grams,) use them. In example C, the highest dilution with all positive tubes (0.01 g) is within the three highest remaining dilutions (0.1, 0.01, and 0.001.)\n\nIn the second case, the highest dilution with all positive tubes is not within the three highest remaining dilutions. Then select the next two higher dilutions than the highest dilution with all positive tubes. Assign the sum of the positive tubes of any still higher dilutions to the third higher dilution. In example D, the highest dilution with all positive tubes has 1 gram. Select the two dilutions immediately higher which have 0.1 and 0.01 grams. There is only one higher dilution whose positive tubes are assigned to form the third dilution with 0.001 grams.\n\nIn the third case, there is no dilution with all positive tubes. Then select the two lowest dilutions. Assign the sum of the positive tubes of any higher dilutions to the third dilution. In example E no dilution has all positive tubes. The two lowest dilutions have 10 and 1 grams. The sum of the positive in the dilutions with 0.1, 0.01 and 0.001 grams is assigned to form the third dilution with 0.1 grams.\n\nIf the three dilutions selected are not in the tables, then something in the serial dilution probably was unusual. This is a warning that the outcome is sufficiently improbable that the basic assumptions of the MPN may be questionable. If possible, redoing the test may be the most reliable procedure. If an MPN value is still desired, use the three highest remaining dilutions. In example F, the three highest dilutions are used. If these dilutions are not in the tables, then use the highest dilution with any positive tubes. The section entitled 'MPN for a single dilution with any positive tubes' shows how to calculate the MPN.\n\n Examples 10 g 1 g .1 g .01 g .001 g A 4 1 0 0 0 410xx B 5 5 1 0 0 x510x C 4 5 4 5 1 xx451 D 4 5 4 3 1 xx431 E 4 3 0 1 1 432xx F 4 3 3 2 1 xx321\n\n### Conversion of Table Units\n\nThe tables below apply to inocula of 0.1, 0.01, and 0.001 g. When different inocula are selected for table reference, multiply the MPN/g and confidence limits by whatever multiplier makes the inocula match the table inocula. For example, if the inocula were 0.01, 0.001, and 0.0001 with three tubes per dilution, multiplying by 10 would make these inocula match the table inocula. If the outcome were (3, 1, 0), multiply the Table 1 MPN/g estimate, 43/g, by 10 to arrive at 430/g.\n\n### Bounds and approximations for a design without a table\n\nThe MPN for a serial dilutions not addressed by any table (e.g., resulting from accidental loss of some tubes) may be computed by iteration or bounded as follows.", null, "Where W and Q are two disjoint sets of dilutions that together contain all the dilutions. The lower bound allows low dilutions with all positive tubes to be deleted from the bound. Blodgett (2005b) introduces these and other bounds.\n\nThe following gives an estimate of the MPN. First, select the lowest dilution that doesn't have all positive tubes. Second, select the highest dilution with at least one positive tube. Finally, select all the dilutions between them. Use only the selected dilutions in the following formula of Thomas (1942):\n\nMPN/g = (∑ gj) / (∑ tjmj ∑ (tj-gj)mj) (½)\n\nwhere the summation is over the selected dilutions and\n\n∑ gj denotes the number of positive tubes in the selected dilutions,\n\n∑ tjmj denotes the grams of sample in all tubes in the selected dilutions,\n\n∑ (tj-gj)mj denotes the grams of sample in all negative tubes in the selected dilutions.\n\nThe following examples will illustrate the application of Thomas's formula. We assume that the dilutions are 1.0, 0.1, 0.01, 0.001, and 0.0001 g.\n\nExample (1). For outcome (5/5, 10/10, 4/10, 2/10, 0/5) use only (–,–, 4/10, 2/10,–); so ∑ tjmj = 10*0.01 + 10*0.001 = 0.11. Where * means multiplication. There are 6 negative tubes at 0.01 and 8 negative tubes at 0.001, so ∑ (tj-gj)mj = 6*0.01 + 8*0.001 = 0.068. There are 6 positive tubes, so\n\nMPN/g = 6/(0.068 * 0.11)(½) = 6/0.086 = 70/g\n\nExample (2). For outcome (5/5, 10/10, 10/10, 0/10, 0/5) use only(–, –, 10/10, 0/10,–), so by Thomas's formula,\n\nMPN/g = 10/(0.01 * 0.11)(½) = 10/.0332=300/g\n\nThese two approximated MPNs compare well with the MPNs for (10, 4, 2) and (10,10,0) (i.e., 70/g and 240/g, respectively).\n\nApproximate confidence limits for any dilution test outcome can be calculated by first estimating the standard error of log10(MPN) by the method of Haldane. We describe the method for 3 dilutions, but it can be shortened to 2 or extended to any positive number.\n\nLet m1, m2, m3 denote the inoculation amounts at the largest to the smallest amounts (e.g., m1 = 0.1 g, m2 = 0.01 g, m3 = 0.001 g in these tables).\n\nLet g1, g2, g3 denote the numbers of positive tubes at the corresponding dilutions. For legibility, we denote yx by \"y**x\" and \"y times x\" by \"y*x\".\n\nNow we compute\n\nT1 = exp(-mpn*m1), T2 = exp(-mpn*m2), etc.\n\nThen we compute\n\nB = [g1*m1*m1*T1/((T1 - 1)**2] + ... + [g3*m3*m3*T3/((T3 - 1)**2)].\n\nFinally, we compute\n\nStandard Error of Log10(mpn) = 1/(2.303*mpn*(B**0.5))\n\nNow the 95 percent confidence intervals, for example, are found at\n\nLog10(mpn) ± 1.96*(Standard Error).\n\n### MPN for a Single Dilution with any Positive Tubes\n\nIf just one dilution has any positive tubes, then a simpler expression gives its MPN.\n\nMPN/g = (1/m)*2.303*log10((∑ tjmj)/(∑ (tj-gj)mj))\n\nWhere m denotes the amount of sample in each tube in the dilution with a positive tube.\n\n### Special requirements and tables included\n\nThe attached spreadsheet should be able to handle most specialized designs. Garthright and Blodgett (2003) discusses this spreadsheet. Requests for special computations and different designs will be honored as resources permit. Designs may be requested with more or less than 3 dilutions, uneven numbers of tubes, different confidence levels, etc. (Telephone or write the Division of Mathematics, FDA/CFSAN, 5100 Paint Branch Parkway, HFS-205 Rm 2D-011, College Park, MD 20740) The most-published designs, three 10-fold dilutions with 3, 5, 8, or 10 tubes at each dilution, are presented here.\n\nClick on the link https://mpncalc.galaxytrakr.org/ to automatically calculate the MPN point estimate and confidence intervals of varying levels (80%, 90%, 95% or 99%). The confidence interval method implemented in this application uses the large sample theory approach of Jarvis et al. (2010), with a slight modification for all-positive or all-negative cases Ferguson and Ihrie (2018) and differs from the BAM Tables confidence intervals which use the de Man method. The web application also computes the rarity index, Jarvis et al. (2010) and Blodgett RJ (2002), an assessment of how likely the observed experimental outcome is." ]
[ null, "https://www.fda.gov/themes/custom/preview/assets/images/US_Flag.png", null, "https://www.fda.gov/media/83858/download", null, "https://www.fda.gov/media/83865/download", null ]
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https://www.guptamechanical.com/2022/08/solution-dominant-element-codechef.html
[ "## GUPTA MECHANICAL\n\nIN THIS WEBSITE I CAN TELL ALL ABOUT TECH. TIPS AND TRICKS APP REVIEWS AND UNBOXINGS ALSO TECH. NEWS .............\n\n# [Solution] Dominant Element CodeChef Solution\n\nProblem\n\nYou are given an array $A$ of length $N$. An element $X$ is said to be dominant if the frequency of $X$ in $A$ is strictly greater than the frequency of any other element in the $A$.\n\nFor example, if $A = [2, 1, 4, 4, 4]$ then $4$ is a dominant element since its frequency is higher than the frequency of any other element in $A$.\n\nFind if there exists any dominant element in $A$.\n\n### Input Format\n\n• The first line of input contains a single integer $T$ — the number of test cases. Then the test cases follow.\n• The first line of each test case contains an integer $N$ — the size of the array $A$.\n• The second line of each test case contains $N$ space-separated integers $A_1, A_2, \\ldots, A_N$ denoting the array $A$.\n\nSolution Click Below:-  👉\n👇👇👇👇👇\n\n### Output Format\n\nFor each test case, output YES if there exists any dominant element in $A$. Otherwise, output NO.\n\nYou may print each character of YES and NO in uppercase or lowercase (for example, yesyEsYes will be considered identical).\n\n### Explanation:\n\nTest case $1$: $2$ is the dominant element.\n\nTest case $2$: There does not exist any dominant element.\n\nTest case $3$: $3$ is the dominant element." ]
[ null ]
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https://completesuccess.in/index.php/2018/05/22/circle-set-73/
[ "# Circle Set 73\n\nDirections (1-5): Study the following information carefully and answer the questions given below:\nA, B, C, D, E, F, G and H are sitting around a circular table facing the centre. H is third to the right of C and second to the left of E. B is not an immediate neighbour of H or C. F is second to the right of D and is an immediate neighbour of C. G is not the neighbour of E.\n\nExplanation\n\nQ1. Who among the following is second to the right of C?\n(a) H\n(b) G\n(c) F\n(d) E\n(e) None of these\n\nQ2. Who among the following is an immediate neighbour of H and E(both)?\n(a) A\n(b) B\n(c) C\n(d) G\n(e) None of these\n\nQ3. In which of the following pairs the second person is sitting on the immediate right of the first person?\n(a) A, H\n(b) C, D\n(c) G, H\n(d) E, H\n(e) F, C\n\nQ4. Who among the following is second to the left of B?\n(a) C\n(b) H\n(c) F\n(d) A\n(e) None of these\n\nQ5. Who among the following is opposite D?\n(a) A\n(b) G\n(c) H\n(d) E\n(e) None of these" ]
[ null ]
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https://www.hindawi.com/journals/algebra/2013/272104/
[ "/ / Article\n\nResearch Article | Open Access\n\nVolume 2013 |Article ID 272104 | 9 pages | https://doi.org/10.1155/2013/272104\n\n# On the Jacobson Radical of an -Semiring\n\nAccepted09 Jul 2013\nPublished07 Oct 2013\n\n#### Abstract\n\nThe notion of -ary semimodules is introduced so that the Jacobson radical of an -semiring is studied and some well-known results concerning the Jacobson radical of a ring (a semiring or a ternary semiring) are generalized to an -semiring.\n\n#### 1. Introduction\n\nThe concept of semigroups was generalized to that of ternary semigroups , that of -ary semigroups , and even to that of -semigroups . Similarly, it was natural to generalize the notion of rings to that of ternary semirings, that of -ary semirings, and even that of -semirings.\n\nIndeed, there were some research articles on semirings, (see, for example, ), specially on the radical of a semiring; see . Semigroups over semirings were studied in and semimodules over semirings were studied in . The notion of semirings can be generalized to ternary semirings and -semirings , even to -semirings . The radicals of ternary semirings and of -semirings were studied in [20, 21], respectively. The concept of -semirings was introduced and accordingly some simple properties were discussed in , where the concept of radicals was not mentioned.\n\nThe notion of the Jacobson radicals was first introduced by Jacobson in the ring theory in 1945. Jacobson defined the radical of , which we call the Jacobson radical, to be the join of all quasi-regular right ideals and verified that the radical is a two-sided ideal and can also be defined to be the join of the left quasi-regular ideals.\n\nThe concept of the Jacobson radical of a semiring has been introduced internally by Bourne , where it was proved that the right and left Jacobson radicals coincide; thus one could say the Jacobson radical briefly. These and some other results were generalizations of well-known results of Jacobson .\n\nIn 1958, by associating a suitable ring with the semiring, Bourne and Zassenhaus defined the semiradical of the semiring . In it was proved that the concepts of the Jacobson radical and the semiradical coincide.\n\nIizuka considered the Jacobson radical of a semiring from the point of view of the representation theory without reducing it to the ring theory. The external notion of the radical was proved to be related to internal one; at the same time, it was shown that the radical defined in coincides with the Jacobson radical and with the semiradical of the semiring.\n\nIn the present paper, we investigate -semirings by means of -ary semimodules so that we can define externally the Jacobson radical of an -semiring, and then we establish the radical properties of the Jacobson radical of an -semiring. Some necessary notions such as irreducible -ary semimodules over an -semiring are adequately defined. All results in this paper generalize the corresponding ones concerning the radical of a ring , of a semiring , or of a ternary semiring .\n\n#### 2. Preliminaries\n\nWe used following convention as followed by : The sequence is denoted by . Thus the following expression is represented as In the case when , then (2) is expressed as If , then (2) can be written as .\n\nRecall that an -ary semigroup is defined as a nonempty set with an -ary associative operation ; that is, for all and all . Whence we may denote by briefly. Generally, we have the notation for each positive and all . Thus for positive integer , is well defined if and only if ; see [7, Lemma 1.1]. An -ary semigroup is called cancellative if for all .\n\nThe next definition is a generalization of the concept of ternary semirings in and similar to the notion of the -semirings in .\n\nDefinition 1. A nonempty set together with an -ary operation , called addition, and an -ary operation , called multiplication, is said to be an -semiring if the following conditions are satisfied. (1) is an -ary semigroup and is an -ary semigroup.(2) is distributive with respect to operation ; that is, for every , , , (3) is commutative; that is, for every permutation of and all , (4)  There is an element , called the zero of , satisfying the following two properties: (4A) is an -identity; that is, for every ;(4B) is a -zero; that is, for all , whenever there exists such that .\n\nIt is clear that the zero of an -semiring is necessarily unique.\n\nDefinition 2. An -semiring is called additively cancellative if the -ary semigroup is cancellative and multiplicatively cancellative if the -ary semigroup is cancellative.\n\nRecall that for an -ary semigroup , a nonempty subset of is called a subsemigroup of if whenever all . For , we call an -ideal of if whenever . is called an ideal of if and only if it is an -ideal of . See, for example, [7, Definition 1.6].\n\nDefinition 3. A nonempty subset of an -semiring is called an -ary subsemiring of if is a subsemigroup of as well as a subsemigroup of and an (i-)ideal of if is a subsemigroup of as well as an (i-)ideal of (where ). An -ideal is also called a right ideal and an -ideal is also called a left ideal. An ideal of is called a -ideal if and imply that . An ideal of is called an -ideal if and imply that .\n\nLet be an ideal of . Then the -closure of , denoted by , is defined by . Similarly, the -closure of , denoted by , is defined by for some and some . One can show that is a -ideal and is an -ideal. Furthermore, it is shown that an ideal of is a -ideal if and only if and that is an -ideal if and only if .\n\nDefinition 4. An equivalence relation on an -semiring is said to be a congruence relation or simply a congruence of if the following conditions are satisfied:(1) for all , (2) for all .\n\nLet be a proper ideal of an -semiring . Then the congruence on , denoted by , and defined by setting if and only if for some , is called the Bourne congruence on defined by the ideal . We denote the Bourne congruence class of an element by and denote the set of all such congruence classes of by . If the Bourne congruence is proper, that is, , then we can define two operations, -ary addition and -ary multiplication on by and for all . Then is an -semiring and is called the Bourne factor -semiring.\n\nSimilarly, the congruence on , denoted by , and defined by setting if and only if for some and some , is called the Iizuka congruence on defined by the ideal . We denote the Iizuka congruence class of an element by and denote the set of all such congruence classes of by . If the Iizuka congruence is proper, that is, , then we can define two operations, -ary addition and -ary multiplication on by and for all . Then is an -semiring and we call it the Iizuka factor -semiring.\n\nThe next definition is a generalization of [20, Definition 2.13].\n\nDefinition 5. A commutative -ary semigroup with an identity (operation to be called addition) is called a right -ary semimodule over an -semiring or simply an -ary -semimodule if there exists a mapping (images to be denoted by or briefly by for all and ) satisfying the following conditions: (1) for all and all ; (2) for all and all ; (3) for all , , and ; (4) for all ; (5) whenever , , and for some .\n\nDefinition 6. A nonempty subset of a right -ary semimodule over an -semiring is called an -ary subsemimodule of if (i) and (ii) for all and .\nAn -ary subsemimodule of is called an -ary -subsemimodule if and imply that . An -ary subsemimodule of is called an -ary -subsemimodule if and imply that .\n\nFor example, an -semiring can be regarded as a right -ary -semimodule naturally. Then if is a -ideal (an -ideal) of the -semiring , then is also an -ary -(-)subsemimodule of this right -ary -semimodule .\n\nDefinition 7. A right -ary -semimodule is said to be cancellative if is a cancellative -ary semigroup.\n\nDefinition 8. An equivalence relation on right -ary -semimodule is said to be a congruence relation or simply a congruence of if the following conditions are satisfied:(1) for all , (2) for all and all . We say that a congruence of admits the cancellation law (of addition) if (3) and imply .\n\nLet be an -ary subsemimodule of an -ary right semimodule over an -semiring . Then the congruence on , denoted by , and defined by setting is called the Bourne congruence on defined by the -ary subsemimodule . We denote the Bourne congruence class of an element by and denote the set of all such congruence classes of by . Define two operations, -ary addition and -ary scalar multiplication on , by and for all and all . With these two operations, is an -ary right semimodule over and we call it the Bourne factor -ary semimodule.\n\nSimilarly, we can define the Iizuka congruence   and the Iizuka factor -ary semimodule  . It is easy to show that is cancellative.\n\nIn what follows, we always assume that the -ary right semimodule is cancellative.\n\n#### 3. Primitive -Semirings\n\nDefinition 9. Let be an -semiring with zero . Then the zeroid of , denoted by , is defined as\n\nClearly, the zero element of belongs to . Furthermore, we have the following.\n\nLemma 10. The zeroid of an -semiring is the smallest -ideal of .\n\nProof. It is easily verified that is an ideal of . To show is an -ideal of , we suppose , where and . For each there exist such that , so we have that is,\nIt follows that\nHence we obtain which shows that , so that is an -ideal of .\nAt last, suppose that is an arbitrary -ideal of . We aim to show . For this, let . Then there exist such that , so . It follows that since is an -ideal and . Thus .\n\nDefinition 11. Let be a right -ary -semimodule. The annihilator of in , denoted by or , is defined as the subset\n\nLemma 12. is an -ideal of .\n\nProof. It is obvious that is an ideal of . To show that it is an -ideal, suppose , where and . Then for all , that is, which deduces that since for each . Thus we have\nBy cancellation law of , . Hence , as required.\n\nDefinition 13. A right -ary -semimodule is said to be faithful if .\n\nOne of difficulties when studying the radical of an -semiring is how to give an appropriate definition of the irreducibility of a right -ary -semimodule. The next definition is a generalization of [20, Definition 3.9].\n\nDefinition 14. A right -ary -semimodule is said to be irreducible if for every arbitrary fixed pair with for some and for any , there exist with such that\n\nRemark 15. Since is cancellative, it is easily seen that a right -ary -semimodule is irreducible if and only if for every arbitrary fixed pair with for all and for any , there exist with such that equality (2) holds.\n\nLemma 16. Let be an -ideal of an -semiring . If is an irreducible right -ary -semimodule, then is an irreducible right -ary -semimodule.\n\nProof. Let be an irreducible right -ary -semimodule. Then we can define an -ary action on by for all and for all , and this makes into an irreducible right -ary -semimodule.\n\nThe converse of Lemma 16 is not necessarily true. But in particular we have the following lemma.\n\nLemma 17. If is a right -ary -semimodule then is a right -ary -semimodule, where is the Bourne factor semiring. Moreover, if is an irreducible right -ary -semimodule, then is also an irreducible right -ary -semimodule.\n\nProof. Suppose is a right -ary -semimodule. We define an -ary action on as follows: where , for all and for all . We now show that this definition is well-defined. If for each , , then , that is, there exist such that . It follows that , which implies that since . Thus , as required. It is easy to see that the above definition makes into a right -ary -semimodule.\nMoreover, if is an irreducible right -ary -semimodule then it is routine to verify that is also an irreducible right -ary -semimodule by (2).\n\nLemma 18. Let be a right -ary -semimodule. Then .\n\nProof. Let , where . Then for any whenever and for some . It follows that where for some . This shows that and so that . Consequently, .\n\nLemma 19. Any right -ary -semimodule is a faithful -semimodule.\n\nProof. Let be a right -ary -semimodule. Then in view of Lemma 17, is an -semimodule. On the one hand, by Lemma 12, is an -ideal of . On the other hand, by Lemma 10, is the smallest -ideal of . Thus . According to Lemma 18, . So , which means that is a faithful -semimodule.\n\nLemma 20. If is an -ideal of an -semiring then where is the Bourne factor semiring.\n\nProof. Suppose . Then we have for some . Thus we have which implies that for some . Hence . This shows that since is an -ideal of . Consequently, . Thus .\n\nDefinition 21. An -semiring is said to be primitive if it has a faithful irreducible cancellative -ary -semimodule. An ideal is said to be primitive if the Bourne factor semiring is primitive.\n\nEvidently, an -semiring is primitive if and only if is a primitive ideal of . The following theorem characterizes primitive ideals of an -semiring.\n\nTheorem 22. An -ideal of -semiring is primitive if and only if for some irreducible right -ary -semimodule .\n\nProof. Let be an -ideal of such that for some irreducible right -ary -semimodule . Then by Lemmas 17 and 19   is a faithful irreducible -ary -semimodule. This shows that is primitive and hence is a primitive -ideal of .\nConversely, let be a primitive -ideal of . Then is a primitive -semiring. So there exists a faithful irreducible -ary -semimodule . Now by Lemma 16   is an irreducible -ary -semimodule. It remains to show that . Now for all and , whenever for some whenever for some since is a faithful -ary -semimodule , by Lemma 20  . Thus as desired.\n\n#### 4. Jacobson Radical of an -Semiring\n\nLet us begin this section by defining the semi-irreducibility of a right -ary -semimodule.\n\nDefinition 23. A right -ary -semimodule is said to be semi-irreducible if ; that is, for some and some , and does not contain any -ary -subsemimodule other than and .\n\nLemma 24. Let be a subset of an -semiring and a right -ary -semimodule with for some . In the case where , we assume further that is a left ideal of . Then the following statements are true:(1)If is semi-irreducible and , then if and only if ; (2)If is irreducible and , then if and only if for all and all .\n\nProof. Suppose that is a semi-irreducible right -ary semimodule over an -semiring , and that is a subset of such that for some . In the case where , we further assume that is a left ideal of .\nAssume that is semi-irreducible. Let be such that\nSet\nIt is clear that , and it is easy to show that is a subsemimodule of . Let and . Then and . Thus ; that is, . This shows that is a -subsemimodule of . Since , . Since is semi-irreducible, and therefore .\nThe converse part is obvious.\nAssume that is irreducible. Let be such that . Set , for . Since , we have for some and . Since is irreducible, according to Definition 14, there exist with such that\nHence\nSince is cancellative and , at least one of the following equalities does not hold:\nSo we conclude that if for all and all , then .\nThe converse part follows easily.\n\nLemma 25. Let be a right -ary -semimodule and . Then is semi-irreducible if and only if for every nonzero , .\n\nProof. Assume that is a semi-irreducible right -ary -semimodule and . Let be such that . Then by Lemma 24  . Since is an -ary -subsemimodule of , .\nConversely, suppose that for any nonzero , . Let be an -ary -subsemimodule of . Then there exists such that . So by hypothesis, . Hence for any , there exist such that . Since , we have . Since is an -ary -subsemimodule, implies that . This shows that . Now if then for all . Hence . So we have , a contradiction. Therefore, . Thus is semi-irreducible.\n\nCorollary 26. If a right -ary -semimodule is irreducible, then it is semi-irreducible and .\n\nProof. Assume that is an irreducible right -ary -semimodule. Then and, consequently, there exists a nonzero . In view of (2) with and for , we obtain that for any there exist with such that so that\nIt follows that . Thus . By Lemma 25, is semi-irreducible.\nFurthermore, , which implies that . Since is an -ary -subsemimodule of , as required.\n\nNow we can define the Jacobson radical of an -semiring in an external way.\n\nDefinition 27. Let be an -semiring and be the set of all irreducible right -ary -semimodules. Then is called the Jacobson radical of . If is empty then itself is considered as ; that is, , and in this case, we say that is a radical -semiring. An -semiring is said to be Jacobson semisimple or J-semisimple if .\n\nBy Lemma 12, is an -ideal of . Note that the intersection of any family of -ideals is again an -ideal. Consequently, we obtain the following.\n\nLemma 28. is an -ideal of .\n\nLemma 29. If is a right -ary -semimodule then is a right -ary -semimodule, where is the Bourne factor semiring. Moreover, if is an irreducible right -ary -semimodule, then is also an irreducible right -ary -semimodule.\n\nProof. This lemma can be proved by the same method as in proving Lemma 17.\n\nTheorem 30. If is an -semiring, then the Bourne factor semiring is Jacobson semisimple.\n\nProof. By and , we denote the set of all irreducible right -ary -semimodules and the set of all irreducible right -ary -semimodules, respectively. Then according to Lemmas 28, 16, and, 29, we obtain that . For any and any , we have , which means that for any , whenever and for some . Thus whenever and for some , so for all . That is, . Hence . We have shown that . By Definition 27, is Jacobson semisimple.\n\nThe next theorem is a direct corollary of Theorem 22, giving an internal characterization of the Jacobson radical of an -semiring.\n\nTheorem 31. is the intersection of all primitive -ideals of .\n\nDefinition 32. Let be an -ideal of an -semiring for some . Then is said to be strongly seminilpotent if there exists a positive integer such that , where , times, . is said to be strongly nilpotent if there exists a positive integer such that .\n\nTheorem 33. If is a strongly semi-nilpotent left ideal of , then .\n\nProof. Suppose on the contrary that where is an -semiring and is the set of all irreducible right -ary -semimodules. Then there exists an such that . Thus there exists such that\nSince is strongly semi-nilpotent, there exists a positive integer such that . By Lemmas 10 and 12, . It follows that , which implies that If (29) holds for all positive integers ’s, then in particular it is true for and in this case we have , a contradiction to (28). If (29) does not hold for all , then there exist and positive such that Thus and there exists such that . It follows that\nso we have\nBy Lemma 24, we obtain , again a contradiction. This completes the proof.\n\nThe next result is a direct corollary of Theorem 33.\n\nCorollary 34. If an -semiring is Jacobson semisimple then does not contain any non-zero strongly semi-nilpotent left ideal and hence does not contain any nontrivial strongly nilpotent left ideal.\n\nLemma 35. If is a (semi-)irreducible right -ary -semimodule and is an arbitrary -subsemimodule (and ), then is (semi-)irreducible, and for any the following statement is true: the equality holds for all if and only if the same equality holds for all . Furthermore, .\n\nProof. Assume is an irreducible right -ary -semimodule. Then from (2), it follows that is irreducible. If is a semi-irreducible and , then is semi-irreducible by Definition 23 since any subsemimodule of is clearly a subsemimodule of .\nLet be such that the equality holds for all . Since is semi-irreducible, for any and any , there exist positive , and such that and\nThus we have the following two equalities:\nIt follows that\nObserving that , since is a submodule, we have and for all by the assumption. Hence by cancellation law, (35) deduces that . The converse implication is clear.\nFurthermore, letting for some , we get that the equality holds for all if and only if the same equality holds for all . Thus .\n\nLemma 36. Let be an ideal of an -semiring . (1)If is an (semi-)irreducible right -ary -semimodule (and ), then is an (semi-)irreducible right -ary -semimodule. (2)If is an irreducible right -ary -semimodule, then there exists an irreducible right -ary -semimodule , which can be regarded as an -subsemimodule of .\n\nProof. Let be an irreducible -semimodule and be such that for some . Without loss of generality, we suppose that . From (2) we deduce that . By Lemma 24, for some . Since is irreducible, by (19) there exist with such that that is, which means that is an irreducible -semimodule by (19) again since for all .\nAssume that is a semi-irreducible -semimodule and . According to Lemma 24, for any there exist such that . By Lemma 25, , so for any there exist positive integers and" ]
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http://www.matrix67.com/blog/archives/3507
[ "# 惊人的答案:平均要取多少个(0,1)中的随机数才能让和超过1\n\n数学常数最令人着迷的就是,它们常常出现在一些看似与之毫不相干的场合中。 随便取一个 0 到 1 之间的数,再加上另一个 0 到 1 之间的随机数,然后再加上一个 0 到 1 之间的随机数⋯⋯直到和超过 1 为止。一个有趣的问题:平均需要加多少次,才能让和超过 1 呢?答案是 e 次。", null, "为了证明这一点,让我们先来看一个更简单的问题:任取两个 0 到 1 之间的实数,它们的和小于 1 的概率有多大?容易想到,满足 x+y<1 的点 (x, y) 占据了正方形 (0, 1)×(0, 1) 的一半面积,因此这两个实数之和小于 1 的概率就是 1/2 。类似地,三个数之和小于 1 的概率则是 1/6 ,它是平面 x+y+z=1 在单位立方体中截得的一个三棱锥。这个 1/6 可以利用截面与底面的相似比关系,通过简单的积分求得:        ∫(0..1) (x^2)*1/2 dx = 1/6", null, "可以想到,四个 0 到 1 之间的随机数之和小于 1 的概率就等于四维立方体一角的“体积”,它的“底面”是一个体积为 1/6 的三维体,在第四维上对其进行积分便可得到其“体积”\n\n∫(0..1) (x^3)*1/6 dx = 1/24\n\n依此类推, n 个随机数之和不超过 1 的概率就是 1/n! ,反过来 n 个数之和大于 1 的概率就是 1 – 1/n! ,因此加到第 n 个数才刚好超过 1 的概率就是\n\n(1 – 1/n!) – (1 – 1/(n-1)!) = (n-1)/n!\n\n因此,要想让和超过 1 ,需要累加的期望次数为\n\n∑(n=2..∞) n * (n-1)/n! = ∑(n=1..∞) n/n! = e\n\n## 59 条评论\n\n•", null, "question1663\n\nBRILLIANT!\n\n•", null, "voices\n\n佩服!\n\n•", null, "wecing\n\n我靠,这个答案太悬疑了……\n\n•", null, "Phil\n\n您真的用 macbook 了?\n\n•", null, "掌柜的马甲\n\ne又是这它啊…\n\n•", null, "Kane\n\n突然发现源地址 http://www.mostlymaths.net 竟然被墙…\n\n•", null, "Eagle_Fantasy\n\n妙哉 无处不在的e\n\n•", null, "白左\n\n看见了mac疑似物\n\n•", null, "白左\n\n啊呀,所以说这是为什么呢,为什么宇宙中有一个数是大量规律的不变常数呢?\n一个可能的答案是,为了减轻模拟我们这个宇宙的超级计算机的负荷,所有的积分常数和偏移常数都是一样的\n\n超意识A:喂,给这个基准常数取一个数吧,整数?\n(超意识B在单向加密软件中输入了他的名字,很快,一个长度为无限的无理数MD5i码就生成好了,前三位是2.71。超意识B随手将它复制进变量池最后一个空缺的位置,回车)\n\n下一秒,宇宙诞生\n\n•", null, "crush\n\n不错!\n\n•", null, "严酷的魔王\n\n不错不错~\n\n•", null, "shenpeng\n\n刚刚用这个模型去求e的值,两次400000次重复试验试验,结果分别是2.71525和2.7154275,收敛好慢啊……\n\n•", null, "青年\n\n数学果然很美~\n\n•", null, "DarkRaven\n\n另有一个经典题目:\n一串物品,每一个都有一个喜好度,一个一个给你看,你只能选择一次.\n最佳方案是先看前n/e个,选出最大的,然后之后看到一个比这个最大值大的就选.\n额,怎么证明呢?\n\n•", null, "Melkor\n\n算法导论里讲到了这个问题,随机算法\n\n•", null, "DarkRaven\n\n@白左:\n无论在什么宇宙,只要数学系统相同,结果就是e,而数学是人类的创造!\n\n•", null, "sexla\n\nlaiguangguangle,实在看不懂\n\n•", null, "biohu\n\n很神奇…\n\n•", null, "剑影若兰\n\n严格来说应该是满足[0,1]均匀分布的随机数吧。。。\n\n•", null, "cbkid\n\n博主也用mac了?linux没人了吗?\n\n•", null, "ncmooc\n\n这个答案令人深思。\n惊现mac疑似物。\n\n•", null, "猜的是3。。。还挺近的。。。一开始想随机取一个期望是0.5,所以大于两个的话基本就行。。。没想到这么复杂:D\n\n•", null, "snaily\n\n15L的问题也很有趣,求证!\n\n•", null, "snaily\n\n15L的问题有证明不?\n\n•", null, "花荣\n\n再引申一下,这些随机数加在一起,和是多少?\n1.359?\n\n•", null, "呼吸\n\n看似毫不相干 我也这么想 幸会幸会\n\n•", null, "Ming\n\n一定要添加一条:\n随机数是均匀分布在[0,1]上,否则结论不成立\n\n我很喜欢你对于’n维立方体一角’体积的证明,很简单明了,得到1/n!的结果比我的笨方法好多了.\n但是最后一步有点多余了,因为E(n|S(n)>1)=E(n|S(n)<1)=E(n|S(n)=1).\n\n•", null, "wuzhengkai\n\n归纳的很精妙!\n\n•", null, "onlytest\n\n这个是什么软件…\n\n•", null, "Boleyn Su\n\n首先,将题目转化为平均要取多少个1到n-1中的随机数才能让和超过n。显然当n趋于无穷大时,这两个问题是等效的。\n然后,解:\n设f(x)表示平均取多少次才能大于x.\n显然 f(x)=0 x0来说,设之前最后一次取得了i,那么i在1到n-1之间,且取到任意一个数的概率为1/(n-1).\n所以 f(x)=sum{(1+f(x-i))/(n-1);1<=i<=n-1}\n=1+sum{f(x-i);1<=i<=n-1}/(n-1)\n=1+sum{f(t);0<=t<=x-1}/(n-1)\n令 F(x)=sum{f(i);0<=ioo) f(n)=(1+1/(n-1))^(n-1)*(1+1/(n-1))\n=e*1\n=e\n\n•", null, "Boleyn Su\n\n无语 XXXX为被和谐内容 这个HTML隔绝方式 也太奇特了吧\n首先,将题目转化为平均要取多少个1到n-1中的随机数才能让和超过n。显然当n趋于无穷大时,这两个问题是等效的。\n然后,解:\n设f(x)表示平均取多少次才能大于x.\n显然 f(x)=0 x<0\nf(x)=1 x=0\n对于x>0来说,设之前最后一次取得了i,那么i在1到n-1之间,且取到任意一个数的概率为1/(n-1).\n所以 f(x)=sum{(1+f(x-i))/(n-1);1<=i<=n-1}\n=1+sum{f(x-i);1<=i<=n-1}/(n-1)\n=1+sum{f(t);0<=t<=x-1}/(n-1)\n令 F(x)=sum{f(i);0<=i<=x}\n则 f(x)=1+F(x-1)/(n-1) ……\n所以 f(x-1)=1+F(x-2)/(n-1) ……\n式-式得 f(x)-f(x-1)=(F(x-1)-F(x-2))/(n-1)\n所以 (f(x)-f(x-1))*(n-1)=f(x-1)\n所以 f(x)=f(x-1)*n/(n-1)\n=(n/(n-1))^x*f(0)\n=(n/(n-1))^x\n所以 f(n)=(n/(n-1))^n\nlim(n->+oo) f(n)=(1+1/(n-1))^(n-1)*(1+1/(n-1))\n=e*1\n=e\n\n•", null, "math\n\n加到第 n 个数才刚好超过 1 的概率就是\n\n(1 – 1/n!) – (1 – 1/(n-1)!) = (n-1)/n!\n这步怎么感觉不对啊,减是啥意思\n\n•", null, "加力球\n\n对于任意离散的随机变量N:\nP{N = n} = P{N <= n} – P{N <= n – 1} 明白了吧\n\n•", null, "sugarzh\n\n关于15L的问题,魔方俱乐部里面有人讨论过这个\n\n结论在主贴,证明在14L\n\n•", null, "sugarzh\n\n关于15L的问题,中国魔方俱乐部有人讨论过\n\n结论在14L有证明\n\n•", null, "sugarzh\n\n– – 不小心风怒了。。。第一次卡了,刷新也刷不出来\n\n•", null, "Nice Guy\n\n从0到1取一个数小于1的概率也是1吗?!\n\n•", null, "Nice Guy\n\nn维空间与n-1维空间的运算这样不对吧\n\n•", null, "aLex\n\n15L本来是选秘书的题吧,大学概率书上有,很经典那,选老婆是不是也适用呀哈哈\n\n•", null, "paramecium\n\n昨天刚学了*中心极限定理*手痒想试试其威力的工科男解法:\n\n对每一个随机数字xi服从均匀分布U(0,1),均值E(xi)=1/2,方差D(xi)=1/12,令X=∑(i=1..n)xi,由中心极限定理,Zn = (X-∑E(xi)) / √∑D(xi)服从标准正态分布N(0,1),P{X>1} = P{Zn>(1-n/2)/√(n/12)} = Φ((1-n/2)/√(n/12))。令(1-n/2)/√(n/12) (7+√33)/2=6.372,即平均*最少*要取6.372个随机数,X=Σxi>1,置信度超过99.87%(=99.74%+0.13%)。\n\n•", null, "paramecium\n\n发现少了字,再发一遍。\n\n对每一个随机数字xi服从均匀分布U(0,1),均值E(xi)=1/2,方差D(xi)=1/12,令X=∑(i=1..n)xi,由中心极限定理,Zn = (X-∑E(xi)) / √∑D(xi)服从标准正态分布N(0,1),P{X>1} = P{Zn>(1-n/2)/√(n/12)} = Φ((1-n/2)/√(n/12))。令(1-n/2)/√(n/12)[小于号]-3σ=-3,解得n[大于号](7+√33)/2=6.372,即平均*最少*要取6.372个随机数,X=Σxi>1,置信度超过99.87%(=99.74%+0.13%)。\n\n“尖括号”里面的字被吃掉了,强大的html过滤…\n\n•", null, "paramecium\n\n还有,31L的问题与原问题并不等价,最多只能说两个问题的答案刚好相等。M牛的问题是(0,1)范围内的实数集(不可数集),而{1/n,2/n..1}是(0,1)范围内的有理数集的一个子集(可数集),不可数集和可数集明显是不能一一对应的,31L的解法把所有的无理数都忽略掉了(比如√2/2)。\n\n•", null, "yh\n\n15L那个问题若干年前的mm群光棍节模拟赛里有。。\n\n•", null, "ZWJ\n\n如果n个数地和大于一常数K,那n的期望和K是什么函数关系呢?好像是分段的。用体积法可以证明K<1时,n=e^K。1<K<2时,n=e^K-(K-1)*e^(K-1)。后面的维数太高,几何直观出不来了。跪求高人啊!\n\n•", null, "yang_bigarm\n\n这个解法太高明了,我以前做过这个问题,先求均匀分布之和的表达式f[x](这一步就巨复杂了),然后再算f[x]从0到1积分(又是一堆巨麻烦的积分),可以得到相同的答案。\n\n•", null, "mmaa01\n\n我曾经考虑过一个问题:\n一个数4,可以分解为4个相同的数之和吗?\n一个数5,可以分解为5个相同的数之和吗?\n一个数6,可以分解为6个相同的数之和吗?\n结果发现都是1.\n同样又想\n一个数4,可以分解为4个相同的数之和吗?\n一个数5,可以分解为5个相同的数之和吗?\n一个数6,可以分解为6个相同的数之和吗?\n答案是可以。但是每个分解都不一样。分别是4的4次根、5的5次根、6的6次根。大小不一样。\n自然联想到哪个数的分解是最大的呢,结果是3的分解最大。\n那么一个实数n可不可以考虑它是n(个)相等的数之乘积呢。就是n的1/n次方。\n发现在实数域里面,通过求导可以得到,最大的那个分解的数是e。\n用上面的语言描述就是e可以分解为e(个)相同的数之积。这个数是所有实数中按这个规律分解得到的最大的一个数1.4446678610097661336583391085964。(用xp自带的计算器算了一下,里面没有e,自己倒腾了几下不知道对不对)\n\n•", null, "orbea jersey\n\n这个理念不错 我很喜欢!\n\n•", null, "AirLee\n\n另一个问题:\n定义运算Ranint(1,x),x∈N,x>1,是从1~x中随机取一个整数,设Xn+1=Ranint(1,Xn),X0=m时,若Xn=1,则期望E(n)=Hm≈ln(m),Hm为调和级数.\n拓展:1、运算改为Ran(0,x),x∈R:从(0,x)中随机取一个实数,仿上迭代过程,当Xn<1时结束,E(n)=?\n2、运算改为Ran(0,kx),k>0.k>e时,Xn有增加的趋势,k<e,Xn有减少的趋势,猜想E(Xn/X0)≈(k/e)^n,n趋向于正无穷.\n\n两个拓展我都没搞出来,我觉得跟这个问题有渊源,求大牛解答.\n\n•", null, "gml777\n\n不错,不过看了一下博主都是 写概率方面的,其实数学除了概率还有很多很多\n\n•", null, "cervelo\n\n首先,将题目转化为平均要取多少个1到n-1中的随机数才能让和超过n。显然当n趋于无穷大时,这两个问题是等效的。\n\n•", null, "lipan\n\n31楼的答案跳跃的有点儿快。\n\n•", null, "qirenrui\n\n这个真神了……有没有本质些的解释\n\n•", null, "ruiyang\n\n厉害!!对于四维以上的,如四维 ∫(0..1) (x^3)*1/6 dx = 1/24,积分表达式不太理解。。。\n\n•", null, "加力球\n\n二维中面积比等于相似比平方,三维中体积比等于相似比立方,依此类推三维中面积比等于相似比立方, 可以类似推广到N维中的”体积”比等于相似比N次方\nN维中的”锥体”是底面为一个N-1维图形, 并且在”高度”x处截得图行与底面相似且相似比为 (1-x) (假设高为1) 所以体积(也就是概率)\nPn = ∫(0,1)(1-x)^(n-1)P(n-1)dx, 得递推公式Pn = P(n-1) / n, 又有初值P1 = 1所以Pn = 1/n!" ]
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https://ambitiousbaba.com/rrb-po-prelims-reasoning-ability-quiz-6/
[ "# RRB PO Prelims Reasoning Ability Quiz – 6\n\n## RRB PO Prelims Reasoning Ability Quiz\n\nReasoning Ability plays a very crucial role in every competitive examination. With consistent practice, candidates can ace this section in examination. In this article, we bring to you RRB PO Reasoning Ability Quiz to boost your preparation. This RRB PO Reasoning Ability Quiz contains various types of questions ranging from easy to difficult level. This RRB PO Reasoning Ability Quiz is absolutely FREE. Candidates will be provided with a detailed explanation of each question in this RRB PO Reasoning Ability Quiz. Question given in this RRB PO Reasoning Ability Quiz are based on the latest or updated exam pattern. In order to be able to answer questions quickly and efficiently in upcoming exams, aspirants must practice this RRB PO Reasoning Ability Quiz.\n\nDirections (1-5): Study the given information carefully to answer the given questions:\nTen persons are sitting in two parallel rows containing five persons each, in such a way that there is an equal distance between adjacent persons. In row —1, J, K, L, M, and N are seated (not necessarily in the same order) and all of them are facing south. In row-2, V, W, X, Y and Z are seated (not necessarily in the same order) and all of them are facing north. Therefore in the given seating arrangement each member seated in row faces another member of the other row.\nZ sits third to the right of W. V sits second to the left of Z. The person facing V sits to the immediate right of K. Only one person sits between K and M. J is not an immediate neighbour of K. Only two persons sit between J and L. Neither K nor J faces Y.\n\n1. Who amongst the following is facing N?\n(a) Y\n(b) V\n(c) X\n(d) W\n(e) Z\n\nAns. b", null, "2. Which of the following statements is true regarding M?\n(a) M faces one of the immediate neighbours of X.\n(b) K is one of the immediate neighbours of M.\n(c) None of the given statements is true.\n(d) L sits to the immediate right of M.\n(e) Only one person sits between M and N.\n\nAns. d", null, "3. Who amongst the following is facing X?\n(a) K\n(b) L\n(c) M\n(d) J\n(e) N\n\nAns. a", null, "4. What is the position of Z with respect of Y?\n(a) Third to the right\n(b) second to the right\n(c) Immediate left\n(d) Immediate right\n(e) Second to the left\n\nAns. c", null, "5. Four of the following five are alike in a certain way based on the given arrangement and hence form a group. Which of them does not belong to that group?\n(a) M\n(b) J\n(c) Y\n(d) W\n(e) N\n\nAns. e", null, "Directions (6-10): In these questions, relationship between different elements is shown in the statements. The statements are followed by conclusions. Study the conclusions based on the given statements and select the appropriate answer. Give answer-\n6. Statements: S ≤ L ≤ I = P ≥ E > R; L > Q\nConclusions: I. P ≥ S II. I > R\n(a) If only conclusion II is true\n(b) If only conclusion I is true\n(c) If both conclusions I and II are true.\n(d) If either conclusion I or II is true\n(e) If neither conclusion I nor II is true.\n\nAns. c\nExp. I. P ≥ S (True) II. I > R (True)\n\n7. Statements: G > R ≥ E = A ≤ T ≤ S; D ≤ A ≤ J\nConclusions: I. T ≥ D II. R > S\n(a) If only conclusion II is true\n(b) If only conclusion I is true\n(c) If both conclusions I and II are true.\n(d) If either conclusion I or II is true\n(e) If neither conclusion I nor II is true.\n\nAns. b\nExp. I. T ≥ D (True) II. R > S (False)\n\n8. Statements: A ≥ B > C ≤ D ≤ E < F\nConclusions: I. A ≥ E II. C < F\n(a) If only conclusion II is true\n(b) If only conclusion I is true\n(c) If both conclusions I and II are true.\n(d) If either conclusion I or II is true\n(e) If neither conclusion I nor II is true.\n\nAns. a\nExp. I. A ≥ E (False) II. C < F (True)\n\n9. Statements: G > R ≥ E = A ≤ T ≤ S; D ≤ A ≤ J\nConclusions: I. J > G II. J = G\n(a) If only conclusion II is true\n(b) If only conclusion I is true\n(c) If both conclusions I and II are true.\n(d) If either conclusion I or II is true\n(e) If neither conclusion I nor II is true.\n\nAns. e\nExp. I. J > G (False) II. J = G (False)\n\n10. Statements: S ≤ L ≤ I = P ≥ E > R; L > Q\nConclusions: I. L < R II. E ≥ Q\n(a) If only conclusion II is true\n(b) If only conclusion I is true\n(c) If both conclusions I and II are true.\n(d) If either conclusion I or II is true\n(e) If neither conclusion I nor II is true.\n\nAns. e\nExp. I. L < R (False) II. E ≥ Q (False)", null, "### Preparation Kit PDF\n\n#### Most important PDF’s for Bank, SSC, Railway and Other Government Exam : Download PDF Now\n\nAATMA-NIRBHAR Series- Static GK/Awareness Practice Ebook PDF Get PDF here\nThe Banking Awareness 500 MCQs E-book| Bilingual (Hindi + English) Get PDF here\nAATMA-NIRBHAR Series- Banking Awareness Practice Ebook PDF Get PDF here\nComputer Awareness Capsule 2.O Get PDF here\nAATMA-NIRBHAR Series Quantitative Aptitude Topic-Wise PDF 2020 Get PDF here\nMemory Based Puzzle E-book | 2016-19 Exams Covered Get PDF here\nCaselet Data Interpretation 200 Questions Get PDF here\nPuzzle & Seating Arrangement E-Book for BANK PO MAINS (Vol-1) Get PDF here\n\n3" ]
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https://dev.to/shahor/comment/7co4
[ "re: AoC Day 6: Chronal Coordinates VIEW POST\n\nThe worst for this day was understanding the instructions.\nI just couldn't for the life of me understand what was being asked, until a friend helped me and then I was on my way.\n\nFull disclosure: I was feeling pretty dumb, and almost gave up. I'm glad I didn't 🙏\n\n``````import Fs from \"fs\"\nimport Path from \"path\"\n\n.toString()\n.trim()\n.split(\"\\n\")\n\ninterface Coordinates {\nx: number\ny: number\n}\ntype Distance = number\ntype Boundaries = [Coordinates, Coordinates]\ntype ID = string\ntype Pixel = null | ID\n\nfunction getIdFromCoordinates(coords: Coordinates): ID {\nreturn `\\${coords.x}\\${coords.y}`\n}\n\nfunction convertToCoordinates(line: string): Coordinates {\nconst [x, y] = line.split(\", \")\n\nreturn { x: parseInt(x, 10), y: parseInt(y, 10) }\n}\n\nfunction computeManhattanDistance(a: Coordinates, b: Coordinates): Distance {\nreturn Math.abs(a.x - b.x) + Math.abs(a.y - b.y)\n}\n\nfunction computeBoardBoundaries(coords: Array<Coordinates>): Boundaries {\nreturn coords.reduce(\n(boundaries: Boundaries, coordinate: Coordinates, index: number) => {\nlet [topLeftBoundary, bottomRightBoundary] = boundaries\n\nif (index === 0) {\nreturn [{ ...coordinate }, { ...coordinate }] as Boundaries\n}\n\nif (coordinate.x < topLeftBoundary.x) {\ntopLeftBoundary.x = coordinate.x\n}\nif (coordinate.y < topLeftBoundary.y) {\ntopLeftBoundary.y = coordinate.y\n}\n\nif (coordinate.x > bottomRightBoundary.x) {\nbottomRightBoundary.x = coordinate.x\n}\nif (coordinate.y > bottomRightBoundary.y) {\nbottomRightBoundary.y = coordinate.y\n}\n\nreturn [topLeftBoundary, bottomRightBoundary] as Boundaries\n},\n[\n// first element is top left boundary\n{ x: 0, y: 0 },\n// last element is bottom right boundary\n{ x: 0, y: 0 },\n] as Boundaries\n)\n}\n\nfunction doesBeaconHaveFiniteArea(\nbeacon: Coordinates,\nbeacons: Array<Coordinates>\n): Boolean {\nlet hasTopLeft = false\nlet hasTopRight = false\nlet hasBottomLeft = false\nlet hasBottomRight = false\n\nfor (let i = 0; i < beacons.length; i++) {\nconst comparedBeacon = beacons[i]\n\nif (\nhasTopRight === false &&\ncomparedBeacon.x >= beacon.x &&\ncomparedBeacon.y > beacon.y\n) {\nhasTopRight = true\n}\n\nif (\nhasBottomRight === false &&\ncomparedBeacon.x >= beacon.x &&\ncomparedBeacon.y < beacon.y\n) {\nhasBottomRight = true\n}\n\nif (\nhasBottomLeft === false &&\ncomparedBeacon.x <= beacon.x &&\ncomparedBeacon.y < beacon.y\n) {\nhasBottomLeft = true\n}\n\nif (\nhasTopLeft === false &&\ncomparedBeacon.x <= beacon.x &&\ncomparedBeacon.y > beacon.y\n) {\nhasTopLeft = true\n}\n}\n\nreturn hasTopLeft && hasTopRight && hasBottomLeft && hasBottomRight\n}\n\nconst coordinates = input.map(convertToCoordinates)\nconst beaconsWithFiniteArea: Array<Coordinates> = coordinates.reduce(\n(beacons, beacon) => {\nconst hasFiniteArea = doesBeaconHaveFiniteArea(\nbeacon,\ncoordinates.filter(ref => ref.x !== beacon.x || ref.y !== beacon.y)\n)\n\nif (hasFiniteArea) {\nreturn [...beacons, beacon]\n}\n\nreturn beacons\n},\n[] as Array<Coordinates>\n)\n\nconst boundaries: Boundaries = computeBoardBoundaries(coordinates)\nconsole.log(JSON.stringify(boundaries))\n\nfunction computePixelForCoords(\ncoordinates: Coordinates,\ncoordinatesList: Array<Coordinates>\n): Pixel {\nconst distances = coordinatesList\n.map(beacon => {\nreturn {\nid: getIdFromCoordinates(beacon),\ndistance: computeManhattanDistance(coordinates, beacon),\n}\n})\n.sort((a, b) => a.distance - b.distance)\n\nif (distances.distance === distances.distance) {\nreturn null\n}\n\nreturn distances.id\n}\n\nconst canvas: Array<Array<Pixel>> = []\nfor (let x = boundaries.x; x <= boundaries.x; x++) {\nfor (let y = boundaries.y; y <= boundaries.y; y++) {\nif (canvas[x] === undefined) {\ncanvas[x] = []\n}\n\ncanvas[x][y] = computePixelForCoords({ x, y }, coordinates)\n}\n}\n\nconst sorted = beaconsWithFiniteArea\n.map(beacon => {\nlet count = 0\nlet id = getIdFromCoordinates(beacon)\n\nfor (let x = boundaries.x; x <= boundaries.x; x++) {\nfor (let y = boundaries.y; y <= boundaries.y; y++) {\nif (canvas[x][y] === id) {\ncount++\n}\n}\n}\nreturn count\n})\n.sort((a, b) => b - a)\n\n// 01\nconsole.log(sorted)\n``````\n\nPart2:\n\n``````import Fs from \"fs\"\nimport Path from \"path\"\n\n.toString()\n.trim()\n.split(\"\\n\")\n\ninterface Coordinates {\nx: number\ny: number\n}\ntype Distance = number\ntype Boundaries = [Coordinates, Coordinates]\n\nfunction convertToCoordinates(line: string): Coordinates {\nconst [x, y] = line.split(\", \")\n\nreturn { x: parseInt(x, 10), y: parseInt(y, 10) }\n}\n\nfunction computeManhattanDistance(a: Coordinates, b: Coordinates): Distance {\nreturn Math.abs(a.x - b.x) + Math.abs(a.y - b.y)\n}\n\nfunction computeBoardBoundaries(coords: Array<Coordinates>): Boundaries {\nreturn coords.reduce(\n(boundaries: Boundaries, coordinate: Coordinates, index: number) => {\nlet [topLeftBoundary, bottomRightBoundary] = boundaries\n\nif (index === 0) {\nreturn [{ ...coordinate }, { ...coordinate }] as Boundaries\n}\n\nif (coordinate.x < topLeftBoundary.x) {\ntopLeftBoundary.x = coordinate.x\n}\nif (coordinate.y < topLeftBoundary.y) {\ntopLeftBoundary.y = coordinate.y\n}\n\nif (coordinate.x > bottomRightBoundary.x) {\nbottomRightBoundary.x = coordinate.x\n}\nif (coordinate.y > bottomRightBoundary.y) {\nbottomRightBoundary.y = coordinate.y\n}\n\nreturn [topLeftBoundary, bottomRightBoundary] as Boundaries\n},\n[\n// first element is top left boundary\n{ x: 0, y: 0 },\n// last element is bottom right boundary\n{ x: 0, y: 0 },\n] as Boundaries\n)\n}\n\nconst coordinates = input.map(convertToCoordinates)\nconst boundaries: Boundaries = computeBoardBoundaries(coordinates)\nconst MAX_DISTANCE = 10000\n\nlet closestSum: number = -Infinity\nlet position: Coordinates = { x: Infinity, y: Infinity }\nlet numberOfPixelsForRegion: number = 0\nfor (let x = boundaries.x; x <= boundaries.x; x++) {\nfor (let y = boundaries.y; y <= boundaries.y; y++) {\nconst beacon = { x, y }\nconst sum = coordinates.reduce((total, point) => {\n}, 0)\n\nif (sum < MAX_DISTANCE) {\nnumberOfPixelsForRegion++\n\nif (sum > closestSum) {\nclosestSum = sum\nposition = { x, y }\n}\n}\n}\n}\n\nconsole.log(\n`Chosen position \\${JSON.stringify(\nposition\n)} with sum \\${closestSum} and region size: \\${numberOfPixelsForRegion}`\n)\n\n``````\ncode of conduct - report abuse", null, "", null, "" ]
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https://morecpq.com/index.php/2020/04/21/cpq-line-editor-component-option-quantities-are-calculated-between-before-calculate-and-on-calculate/
[ "# CPQ – Line Editor – Component Option Quantities are Calculated Between Before Calculate and On Calculate\n\nYou have a Price Rule on before calculate that updates SBQQ__ListPrice__c using a formula which is something like: SBQQ__OriginalPrice__c * (SBQQ__Quantity__c *0.01). A formula that changes the list price based on quantity. You add a bundle product containing an option that is of type “Component” and save the configuration. The bundle product has a zero price. Component type options have their quantity calculated based on the parent’s quantity. It brings you out to the Line Editor. Things look good. You change the quantity to 50,000. Things don’t look good. Why didn’t the Price Rule update SBQQ__ListPrice__c of the child to the new price??\n\nComponent Option quantities get calculated between before calculate and on calculate. Your price rule doesn’t know about the new option quantity until on calculate.\n\nSimilarly, the Original Quantity (SBQQ__BundledQuantity__c) also gets calculated between before calculate and on calculate.\n\nThe optimal solution here would be to move the rule from before calculate to on calculate and done!\n\n## 3 thoughts on “CPQ – Line Editor – Component Option Quantities are Calculated Between Before Calculate and On Calculate”\n\n1.", null, "Pramod Dhokane says:\n\nTry apply immediately on product option records to update change in quantity within product configuration. I guess then price rule will get chance to see updated values\n\n2.", null, "Gilles Muys says:\n\nThere is nothing weird about it. That’s how the calculation sequence works: the final quantity for product options of type Component are indeed calculated between the BeforeCalculate and the Calculate events. If you absolutely need your rule to execute on BeforeCalculate, you can use this pseudo formula to determine the quantity of the component in question:\n\nQuantity of Parent Bundle x Original Quantity\n\nOtherwise, just move your rule to execute on Calculate.\n\n1.", null, "D P says:\n\nThanks for the feedback, Gilles. We would have used that formula, however, we found that Original Quantity is also calculated between before calculate and on calculate. In this case, we moved the event to on calculate.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]
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https://johnfsymons.com/7th-grade-math-puzzle-worksheets-free/
[ "As you know, a lot of practice increases the speed of the student’s action in the 7th grade fsa math exam. Free math worksheets for sixth, seventh, eighth, and ninth grade (w/ answer keys) the following printable math worksheets for 6th, 7th, 8th, and 9th grade include a complete answer key.\n\n### To find the answer to a funny riddle, solve the math problems.", null, "7th grade math puzzle worksheets free. Students will use all basic operator skills you can think of. Worksheets for learning multiplication and division. New available free printable 7th grade math worksheets we hope you find what you are looking for here.\n\nFree math puzzles worksheets pdf printable, math puzzles worksheets to practice and improve different math skills, addition, subtraction, ratios, fractions, division, multiplication, for kindergarten, 1st,. Each free math printables is a tool mathematics teachers and parents can print out for use in supplementing their course or for extra homework practice for parents who need to keep their kids busy after school. Top 10 math activities for 7th graders free printable 7th grade math worksheets.\n\nBring learning to life with worksheets, games, lesson plans, and more from education.com. Free math puzzle worksheets for sports, games for preschool, kindergarden, 1st grade, 2nd grade, 3rd grade, 4th grade and 5th grade. Ease into key concepts with our printable 7th grade math worksheets that are equipped with boundless learning to extend your understanding of ratios and proportions, order of operations, rational numbers, and help you in solving expressions and linear equations, in describing geometrical figures, calculating the area, volume and surface area, finding the pairs of angles, and getting an insight.\n\nIt involves recognizing patterns, building sequences, and coming up with logical solutions to problems. Brain teasers for 4th grade math worksheets free. Free math worksheets for grade 7.\n\nWater will first flow into the pipe to […] 7th grade math worksheets fractions worksheets math fractions maths dividing fractions equivalent fractions multiplication free printable worksheets free printables. Below are free videos, games, and worksheets that provide children with a variety of puzzling scenarios. Another fun box math puzzle.\n\nFree printable 7th grade math worksheets, you are visiting at the correct website. Help your 12 year old master math with these free printable 7th grade math worksheets. Skills include division, multiplication, subtraction, addition, place value, roman numerals, and many more.\n\nEach 7th grade math vocabulary crossword puzzle comes with a corresponding printable solution. Type keywords and hit enter. Free printable math practice worksheets for kids in early childhood, preschool, kindergarten, first grade on writing number, counting, addition, subtraction, shapes, ordering number, fill in missin.\n\nShading odds and primes puzzle. One of the ways to increase the speed of the 7th grade student’s action in the 7th grade fsa math test is to get acquainted with the type of question and the method of solving it. The drill may look technical somehow but the teacher must put them through.\n\nGrade 7 math brain teasers worksheets. Fraction problems will no longer intimidate your children! These worksheets are printable pdf.\n\nFill in the missing sum puzzle 0 to 9. Free math puzzles worksheets printable math champions math puzzle worksheets for kids in 1st to 6th grades math puzzle worksheets. This goes beyond assembling the pieces of a jigsaw puzzle.\n\nTeachers should also check out 7th grade math lesson plans. With a little practice, even the toughest problems can be solved. Bring learning to life with worksheets, games, lesson plans, and more from education.com.\n\nFind all the primes and circle them. Name, order sides, and edging shapes puzzle. Sums of numbers box puzzle." ]
[ null, "https://i0.wp.com/i.pinimg.com/originals/50/8a/9f/508a9fd106c8c859a00f170954099809.jpg", null ]
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https://www.geeksforgeeks.org/gate-gate-cs-2014-set-2-question-11/
[ "Skip to content\nRelated Articles\nGATE | GATE-CS-2014-(Set-2) | Question 11\n• Difficulty Level : Hard\n• Last Updated : 18 Apr, 2014\n\nThe security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p = _____________.\n(A) 11.85\n\nAnswer: (A)\n\nExplanation:\n\n```Total ways to pick 4 computers = 10*9*8*7\n\nTotal ways that at least three computers are fine =\nTotal ways that all 4 are fine + Total ways any 3 are fine\n\nTotal ways that all 4 are fine = 4*3*2*1\n\nTotal ways three are fine = 1st is Not working and other 3 working +\n2nd is Not working and other 3 working +\n3rd is Not working and other 3 working +\n4th is Not working and other 3 working +\n= 6*4*3*2 + 4*6*3*2 + 4*3*6*2 + 4*3*2*6\n= 6*4*3*2*4\n\nThe probability = Total ways that at least three computers are fine /\nTotal ways to pick 4 computers\n= (4*3*2*1 + 6*4*3*2*4) / (10*9*8*7)\n= (4*3*2*25) / (10*9*8*7)\n= 11.9% ```\n\nAttention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.\n\nMy Personal Notes arrow_drop_up" ]
[ null ]
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http://www.thescienceforum.com/physics/15390-electromagnetic-induction-faradays-law.html
[ "1. Hi, i am a secondary student and i have some doubts on Electromagnetic induction.\n\nIt is clearly stated in Faraday's law that 'The induced electromotive force or EMF in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.'\n\nNow is there a way to explain in terms of electron or particles as to why this phenomenal?\n\nSecondly, as i searched on wikipedia, the explanation for lenz's law is that ' If the magnetic field associated with this current were in the same direction as the change in magnetic field that created it, these two magnetic fields would combine to give a net magnetic field which would in turn induce a current with twice the magnitude. This process would continue creating infinite current from just moving a magnet; a violation of the law of conservation of energy'\n\nI am unable to understand this explanation, can anyone please help me explain these 2 laws in greater detail?", null, "", null, "2.\n\n3. Rest assured, induction is proven, and has been for years.\n\nI was looking for the animation on inducing current, but I can't find it anywhere. Basically you induce current by moving a wire through a magnetic field.\n\nIt's hard to explain, but flux is like a liquid moving through some object. At any given point you have a certain amount flowing through the cross-section at that point. That amount is called the flux. Due to many factors, magnetic fields are not all shaped exactly the same, so when you move a wire through them, the amount of \"field lines\" you pass through changes. The more field lines you pass through, the higher the induced current is. There is one small caveat. Instead of current in the direction of the field lines, you get current in the opposite direction. This is the \"equal but opposite reaction.\"\n\nLenz's law is simple. \"If you are inducing a current in an object, the current will be in the direction opposing the motion that created it.\"\n\nThe bit that you're asking about is an explanation, poorly worded. What they should have shown is the math. Basically, induction cannot work if the induced current is in the direction of the motion. This is because if it were in the direction of motion, it would create a larger field in the same direction, which would create a larger current, which would create a larger field, until eventually it hit the mechanical limit. This breaks the conservation of energy law because you are creating energy.", null, "", null, "4. oh i get it now. Thank you very much. Basically, it must oppose the motion so energy is not created. But is the energy lost in anyway?", null, "", null, "5.", null, "Originally Posted by tikai\noh i get it now. Thank you very much. Basically, it must oppose the motion so energy is not created. But is the energy lost in anyway?\nBasically the electromotive force (voltage) around a closed loop is equal to the tiime rate of change of the magnetic flux through the area enclosed by the loop.\n\nA proper explanation of this requires the machinery of vector calculus. You will see it done when you study Maxwell's equations after you have the necessary mathematics.\n\nThis is the principle by which electric generators work. Energy is never created, but mechanical energy (say from steam expanding in a turbine) is used to move a coil through a magnetic field and that converts mechanical energy into electrical energy.", null, "", null, "6. Yes, some energy is lost. That is the nature of real systems.\n\nIt's really hard to explain and understand without understanding 3D calculus in vector form. I'd suggest learning how to do that, then picking up Principles of Electrodynamics by Schwartz. It's a really good book, fairly cheap, and it goes in great detail.", null, "", null, "7. thank you very much for all your reply. I have gained at least the basic idea on why how electromagnetic induction works. I have yet to learn any higher level mathematics and i will do so once my o'level examinations ends.", null, "", null, "Bookmarks\n Posting Permissions\n You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement" ]
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https://reference.wolframcloud.com/language/ref/AngleBisector.html
[ "# AngleBisector\n\nAngleBisector[{q1,p,q2}]\n\ngives the bisector of the interior angle at p formed by the triangle with vertex points p, q1 and q2.\n\nAngleBisector[{q1,p,q2},\"type\"]\n\ngives the angle bisector of the specified type.\n\n# Details", null, "• AngleBisector gives an InfiniteLine object.\n• The qi and p in AngleBisector[{q1,p,q2}] can be lists of coordinates or explicit Point objects.\n• AngleBisector[p{q1,q2}] is equivalent to AngleBisector[{q1,p,q2}].\n• AngleBisector gives the line that divides the angle into two equal angles and that passes through the point p.\n•", null, "• The following bisector type specifications \"type\" can be given:\n• \"Interior\" bisector of the interior angle of the triangle at p \"Exterior\" bisector of the exterior angle of the triangle at p\n• AngleBisector only works in 2D.\n• AngleBisector can be used with symbolic points in GeometricScene.\n\n# Examples\n\nopen allclose all\n\n## Basic Examples(1)\n\nCalculate an angle bisector:\n\n## Scope(2)\n\nFind an interior angle bisector:\n\nFind an exterior angle bisector:\n\n## Properties & Relations(4)\n\nAngleBisector finds the interior angle bisector by default:\n\nThe angle bisector divides the angle into two equal angles:\n\nThe exterior angle bisector divides the exterior angle into two equal angles:\n\nTriangleConstruct[{a,b,c},\"AngleBisector\"] is equivalent to AngleBisector[{a,b,c}]:\n\n## Possible Issues(2)\n\nThe three points must be distinct:", null, "AngleBisector only works in 2D:" ]
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https://math.libretexts.org/Bookshelves/PreAlgebra/Book%3A_Prealgebra_(OpenStax)/04%3A_Fractions/4.10%3A_Add_and_Subtract_Mixed_Numbers_(Part_1)
[ "$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 4.10: Add and Subtract Mixed Numbers (Part 1)\n\n•", null, "• Contributed by OpenStax\n• Mathematics at OpenStax CNX\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\nSkills to Develop\n\n• Model addition of mixed numbers with a common denominator\n• Add mixed numbers with a common denominator\n• Model subtraction of mixed numbers\n• Subtract mixed numbers with a common denominator\n• Add and subtract mixed numbers with different denominators\n\nbe prepared!\n\nBefore you get started, take this readiness quiz.\n\n1. Draw figure to model $$\\dfrac{7}{3}$$. If you missed this problem, review Example 4.1.6.\n2. Change $$\\dfrac{11}{4}$$ to a mixed number. If you missed this problem, review Example 4.1.9.\n3. Change $$3 \\dfrac{1}{2}$$ to an improper fraction. If you missed this problem, review Example 4.1.11.\n\n## Model Addition of Mixed Numbers with a Common Denominator\n\nSo far, we’ve added and subtracted proper and improper fractions, but not mixed numbers. Let’s begin by thinking about addition of mixed numbers using money.\n\nIf Ron has $$1$$ dollar and $$1$$ quarter, he has $$1 \\dfrac{1}{4}$$ dollars. If Don has $$2$$ dollars and $$1$$ quarter, he has $$2 \\dfrac{1}{4}$$ dollars. What if Ron and Don put their money together? They would have $$3$$ dollars and $$2$$ quarters. They add the dollars and add the quarters. This makes $$3 \\dfrac{2}{4}$$ dollars. Because two quarters is half a dollar, they would have $$3$$ and a half dollars, or $$3 \\dfrac{1}{2}$$ dollars.\n\n$\\begin{split} & 1 \\dfrac{1}{4} \\\\ + & 2 \\dfrac{1}{4} \\\\ \\hline \\\\ & 3 \\dfrac{2}{4} = 3 \\dfrac{1}{2} \\end{split} \\nonumber$\n\n$1 \\dfrac{1}{4} + 2 \\dfrac{1}{4} \\nonumber$\n\nWe can use fraction circles to model this same example:\n\n Start with $$1 \\dfrac{1}{4}$$. one whole and one $$\\dfrac{1}{4}$$ pieces", null, "$$1 \\dfrac{1}{4}$$ Add $$2 \\dfrac{1}{4}$$ more. two wholes and one $$\\dfrac{1}{4}$$ pieces", null, "$$\\begin{split}+ & 2 \\dfrac{1}{4} \\\\ & \\hline \\end{split}$$ The sum is: three wholes and two $$\\dfrac{1}{4}$$'s", null, "$$3 \\dfrac{2}{4} = 3 \\dfrac{1}{2}$$\n\nExample $$\\PageIndex{1}$$: model\n\nModel $$2 \\dfrac{1}{3} + 1 \\dfrac{2}{3}$$ and give the sum.\n\nSolution\n\nWe will use fraction circles, whole circles for the whole numbers and $$\\dfrac{1}{3}$$ pieces for the fractions.\n\n two wholes and one $$\\dfrac{1}{3}$$", null, "$$2 \\dfrac{1}{3}$$ plus one whole and two $$\\dfrac{1}{3}$$s", null, "$$\\begin{split}+ & 1 \\dfrac{2}{3} \\\\ & \\hline \\end{split}$$ sum is three wholes and three $$\\dfrac{1}{3}$$s", null, "$$3 \\dfrac{3}{3} = 4$$\n\nThis is the same as $$4$$ wholes. So, $$2 \\dfrac{1}{3} + 1 \\dfrac{2}{3} = 4$$.\n\nExercise $$\\PageIndex{1}$$\n\nUse a model to add the following. Draw a picture to illustrate your model. $$1 \\dfrac{2}{5} + 3 \\dfrac{3}{5}$$\n\n$$5$$", null, "Exercise $$\\PageIndex{2}$$\n\nUse a model to add the following. Draw a picture to illustrate your model. $$2 \\dfrac{1}{6} + 2 \\dfrac{5}{6}$$\n\n$$5$$", null, "Example $$\\PageIndex{2}$$: model\n\nModel $$1 \\dfrac{3}{5} + 2 \\dfrac{3}{5}$$ and give the sum as a mixed number.\n\nSolution\n\nWe will use fraction circles, whole circles for the whole numbers and $$\\dfrac{1}{5}$$ pieces for the fractions.\n\n one whole and three $$\\dfrac{1}{5}$$s", null, "$$1 \\dfrac{3}{5}$$ plus two wholes and three $$\\dfrac{1}{5}$$s", null, "$$\\begin{split}+ & 2 \\dfrac{3}{5} \\\\ & \\hline \\end{split}$$ sum is three wholes and six $$\\dfrac{1}{5}$$s", null, "$$3 \\dfrac{6}{5} = 4 \\dfrac{1}{5}$$\n\nAdding the whole circles and fifth pieces, we got a sum of $$3 \\dfrac{6}{5}$$. We can see that $$\\dfrac{6}{5}$$ is equivalent to $$1 \\dfrac{1}{5}$$, so we add that to the $$3$$ to get $$4 \\dfrac{1}{5}$$.\n\nExercise $$\\PageIndex{3}$$\n\nModel, and give the sum as a mixed number. Draw a picture to illustrate your model. $$2 \\dfrac{5}{6} + 1 \\dfrac{5}{6}$$\n\n$$4\\dfrac{2}{3}$$", null, "Exercise $$\\PageIndex{4}$$\n\nModel, and give the sum as a mixed number. Draw a picture to illustrate your model. $$1 \\dfrac{5}{8} + 1 \\dfrac{7}{8}$$\n\n$$3\\dfrac{1}{2}$$", null, "Modeling with fraction circles helps illustrate the process for adding mixed numbers: We add the whole numbers and add the fractions, and then we simplify the result, if possible.\n\nHOW TO: ADD MIXED NUMBERS WITH A COMMON DENOMINATOR\n\nStep 1. Add the whole numbers.\n\nStep 3. Simplify, if possible.\n\nExample $$\\PageIndex{3}$$: add\n\nAdd: $$3 \\dfrac{4}{9} + 2 \\dfrac{2}{9}$$.\n\nSolution\n\n Add the whole numbers. $$\\begin{split} & \\textcolor{red}{3} \\dfrac{4}{9} \\\\ + & \\textcolor{red}{2} \\dfrac{2}{9} \\\\ \\hline \\\\ & \\textcolor{red}{5} \\end{split}$$ Add the fractions. $$\\begin{split} & 3 \\textcolor{red}{\\dfrac{4}{9}} \\\\ + & 2 \\textcolor{red}{\\dfrac{2}{9}} \\\\ \\hline \\\\ & 5 \\textcolor{red}{\\dfrac{6}{9}} \\end{split}$$ Simplify the fraction. $$\\begin{split} & 3 \\dfrac{4}{9} \\\\ + & 2 \\dfrac{2}{9} \\\\ \\hline \\\\ & \\textcolor{red}{5 \\dfrac{6}{9}} = 5 \\dfrac{2}{3} \\end{split}$$\n\nExercise $$\\PageIndex{5}$$\n\nFind the sum: $$4 \\dfrac{4}{7} + 1 \\dfrac{2}{7}$$.\n\n$$5\\dfrac{6}{7}$$\n\nExercise $$\\PageIndex{6}$$\n\nFind the sum: $$2 \\dfrac{3}{11} + 5 \\dfrac{6}{11}$$.\n\n$$7\\dfrac{9}{11}$$\n\nIn Example $$\\PageIndex{3}$$, the sum of the fractions was a proper fraction. Now we will work through an example where the sum is an improper fraction.\n\nExample $$\\PageIndex{4}$$: add\n\nFind the sum: $$9 \\dfrac{5}{9} + 5 \\dfrac{7}{9}$$.\n\nSolution\n\n Add the whole numbers and then add the fractions. $$\\begin{split} & 9 \\dfrac{5}{9} \\\\ + & 5 \\dfrac{7}{9} \\\\ \\hline \\\\ & 14 \\dfrac{12}{9} \\end{split}$$ Rewrite $$\\dfrac{12}{9}$$ as an improper fraction. $$14 + 1 \\dfrac{3}{9}$$ Add. $$15 \\dfrac{3}{9}$$ Simplify. $$15 \\dfrac{1}{3}$$\n\nExercise $$\\PageIndex{7}$$\n\nFind the sum: $$8 \\dfrac{7}{8} + 7 \\dfrac{5}{8}$$.\n\n$$16\\dfrac{1}{2}$$\n\nExercise $$\\PageIndex{8}$$\n\nFind the sum: $$6 \\dfrac{7}{9} + 8 \\dfrac{5}{9}$$.\n\n$$15\\dfrac{1}{3}$$\n\nAn alternate method for adding mixed numbers is to convert the mixed numbers to improper fractions and then add the improper fractions. This method is usually written horizontally.\n\nExample $$\\PageIndex{5}$$: add\n\nAdd by converting the mixed numbers to improper fractions: $$3 \\dfrac{7}{8} + 4 \\dfrac{3}{8}$$.\n\nSolution\n\n Convert to improper fractions. $$\\dfrac{31}{8} + \\dfrac{35}{8}$$ Add the fractions. $$\\dfrac{31 + 35}{8}$$ Simplify the numerator. $$\\dfrac{66}{8}$$ Rewrite as a mixed number. $$8 \\dfrac{2}{8}$$ Simplify the fraction. $$8 \\dfrac{1}{4}$$\n\nSince the problem was given in mixed number form, we will write the sum as a mixed number.\n\nExercise $$\\PageIndex{9}$$\n\nFind the sum by converting the mixed numbers to improper fractions: $$5 \\dfrac{5}{9} + 3 \\dfrac{7}{9}$$\n\n$$9\\dfrac{1}{3}$$\n\nExercise $$\\PageIndex{10}$$\n\nFind the sum by converting the mixed numbers to improper fractions: $$3 \\dfrac{7}{10} + 2 \\dfrac{9}{10}$$\n\n$$6\\dfrac{3}{5}$$\n\nTable $$\\PageIndex{1}$$ compares the two methods of addition, using the expression $$3 \\dfrac{2}{5} + 6 \\dfrac{4}{5}$$ as an example. Which way do you prefer?\n\nTable $$\\PageIndex{1}$$\nMixed Numbers Improper Fractions\n$$\\begin{split} & 3 \\dfrac{2}{5} \\\\ + & 6 \\dfrac{4}{5} \\\\ \\hline \\\\ & 9 \\dfrac{6}{5} \\end{split}$$ $$3 \\dfrac{2}{5} + 6 \\dfrac{4}{5}$$\n$$9 + \\dfrac{6}{5}$$ $$\\dfrac{17}{5} + \\dfrac{34}{5}$$\n$$9 + 1 \\dfrac{1}{5}$$ $$\\dfrac{51}{5}$$\n$$10 \\dfrac{1}{5}$$ $$10 \\dfrac{1}{5}$$\n\n## Model Subtraction of Mixed Numbers\n\nLet’s think of pizzas again to model subtraction of mixed numbers with a common denominator. Suppose you just baked a whole pizza and want to give your brother half of the pizza. What do you have to do to the pizza to give him half? You have to cut it into at least two pieces. Then you can give him half.\n\nWe will use fraction circles (pizzas!) to help us visualize the process. Start with one whole.", null, "Figure $$\\PageIndex{1}$$\n\nAlgebraically, you would write:", null, "Example $$\\PageIndex{6}$$: subtract\n\nUse a model to subtract: $$1 − \\dfrac{1}{3}$$.\n\nSolution", null, "Exercise $$\\PageIndex{11}$$\n\nUse a model to subtract: $$1 − \\dfrac{1}{4}$$.\n\n$$\\dfrac{3}{4}$$\n\nExercise $$\\PageIndex{12}$$\n\nUse a model to subtract: $$1 − \\dfrac{1}{5}$$.\n\n$$\\dfrac{4}{5}$$\n\nWhat if we start with more than one whole? Let’s find out.\n\nExample$$\\PageIndex{7}$$: subtract\n\nUse a model to subtract: $$2 − \\dfrac{3}{4}$$.\n\nSolution", null, "Exercise $$\\PageIndex{13}$$\n\nUse a model to subtract: $$2 − \\dfrac{1}{5}$$.\n\n$$\\dfrac{9}{5}$$\n\nExercise $$\\PageIndex{14}$$\n\nUse a model to subtract: $$2 − \\dfrac{1}{3}$$.\n\n$$\\dfrac{5}{3}$$\n\nIn the next example, we’ll subtract more than one whole.\n\nExample $$\\PageIndex{8}$$: subtract\n\nUse a model to subtract: $$2 − 1 \\dfrac{2}{5}$$.\n\nSolution", null, "Exercise $$\\PageIndex{15}$$\n\nUse a model to subtract: $$2 − 1 \\dfrac{1}{3}$$.\n\n$$\\dfrac{2}{3}$$\n\nExercise $$\\PageIndex{16}$$\n\nUse a model to subtract: $$2 − 1 \\dfrac{1}{4}$$.\n\n$$\\dfrac{3}{4}$$\n\nWhat if you start with a mixed number and need to subtract a fraction? Think about this situation: You need to put three quarters in a parking meter, but you have only a $$\\1$$ bill and one quarter. What could you do? You could change the dollar bill into $$4$$ quarters. The value of $$4$$ quarters is the same as one dollar bill, but the $$4$$ quarters are more useful for the parking meter. Now, instead of having a $$\\1$$ bill and one quarter, you have $$5$$ quarters and can put $$3$$ quarters in the meter.\n\nThis models what happens when we subtract a fraction from a mixed number. We subtracted three quarters from one dollar and one quarter.\n\nWe can also model this using fraction circles, much like we did for addition of mixed numbers.\n\nExample $$\\PageIndex{9}$$: subtract\n\nUse a model to subtract: $$1 \\dfrac{1}{4} − \\dfrac{3}{4}$$\n\nSolution\n\n Rewrite vertically. Start with one whole and one fourth.", null, "$$\\begin{split} & \\textcolor{red}{1 \\dfrac{1}{4}} \\\\ - & \\dfrac{3}{4} \\\\ \\hline \\end{split}$$ Since the fractions have denominator 4, cut the whole into 4 pieces. You now have $$\\dfrac{4}{4}$$ and $$\\dfrac{1}{4}$$ which is $$\\dfrac{5}{4}$$.", null, "$$\\begin{split} & \\textcolor{red}{\\dfrac{5}{4}} \\\\ - & \\dfrac{3}{4} \\\\ \\hline \\end{split}$$ Take away $$\\dfrac{3}{4}$$. There is $$\\dfrac{1}{2}$$ left.", null, "$$\\begin{split} & \\dfrac{5}{4}\\\\\\textcolor{red}{- & \\dfrac}3}{4}}\\\\ \\hline \\\\ & \\dfrac{2}{4} = \\dfrac{1}{2} \\end{split}$$\n\nExercise $$\\PageIndex{17}$$\n\nUse a model to subtract. Draw a picture to illustrate your model. $$1 \\dfrac{1}{3} − \\dfrac{2}{3}$$", null, "Exercise $$\\PageIndex{18}$$\nUse a model to subtract. Draw a picture to illustrate your model. $$1 \\dfrac{1}{5} − \\dfrac{4}{5}$$", null, "" ]
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