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https://brilliant.org/practice/regular-polygons-level-4-challenges/?subtopic=polygons&chapter=regular-polygons	[ "", null, "Geometry\n\n# Regular Polygons: Level 4 Challenges", null, "Inside a regular pentagon $ABCDE$, construct 5 more regular pentagons of side length $\\dfrac{1}{2}AB$. The part of the overlapping of these pentagons yields another regular pentagon of side length $MN$.\n\nLet $r=\\dfrac{AB}{MN}$. Find $\\displaystyle \\left \\lfloor 1000r \\right \\rfloor$.", null, "Above shows a 18-sided regular polygon. How many obtuse triangles are there formed by 3 vertices?\n\n###### Try part 1.", null, "The $n^{\\text{th}}$ figure in the above sequence is constructed by the following procedure:\n\n1. Draw a blue disc of radius $\\displaystyle\\sqrt{\\frac{2016}{\\pi}}$\n2. Remove a regular $n$-gon area from the (smallest) disc\n3. Inscribe a blue disc inside the empty $n$-gon space\n4. Repeat steps 2-4\n\nLet $A_n$ be the total blue area of the $n^{\\text{th}}$ figure in the sequence.\n\nCompute $\\displaystyle\\lim_{n\\to\\infty}A_n$.", null, "It is not easy to draw a regular decagon without tools.\n\nOn a piece of writing paper (with equally spaced lines), I am trying to draw a regular decagon, as shown above. I started by drawing two sides so that their vertical extent is precisely 1 unit of the paper (black lines).\n\nNow I want to draw the next side (red line), and I wonder how far it will extend vertically. To 3 decimal places, what is the distance marked with a question mark?", null, "A regular octagon $ABCDEFGH$ has squares $ACEG$ and $BDFH$ inscribed in it. These squares form a smaller octagon as shown.\n\nLet the the area of octagon $ABCDEFGH$ be $A_L$ and the area of the smaller octagon be $A_S$. Then for some integers $a$ and $b$, where $b$ is square-free, $\\large \\dfrac{A_S}{A_L}=a-\\sqrt{b}.$ Find $a+b$.\n\n×" ]	[ null, "https://ds055uzetaobb.cloudfront.net/brioche/chapter/Regular%20Polygons-fAqhAp.png", null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/bvotLK1tci-95822.svg", null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/HGRDFJLSEf-bant.PNG", null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/2Ja3dEOYQ8-cyclicgons.png", null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/6wevWDsOyc-90184.svg", null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/tn6qzjPbzs-200octa.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9385633,"math_prob":0.99909437,"size":1482,"snap":"2021-43-2021-49","text_gpt3_token_len":333,"char_repetition_ratio":0.135318,"word_repetition_ratio":0.15555556,"special_character_ratio":0.2240216,"punctuation_ratio":0.14423077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998092,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,9,null,6,null,9,null,6,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T21:36:34Z\",\"WARC-Record-ID\":\"<urn:uuid:357d62ab-5d6c-49a2-8b3f-c6eec111154d>\",\"Content-Length\":\"88301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4ca5945-d7a0-4f89-a7b9-2690551790ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:4075e7b6-3188-48b2-8a87-9cb7f10cc8e7>\",\"WARC-IP-Address\":\"104.18.9.15\",\"WARC-Target-URI\":\"https://brilliant.org/practice/regular-polygons-level-4-challenges/?subtopic=polygons&chapter=regular-polygons\",\"WARC-Payload-Digest\":\"sha1:MDJE5HJ5ASJOFTPKRIITRGLGHZEMGBYU\",\"WARC-Block-Digest\":\"sha1:VNSCKB2QGVBD7B5ZWZ5UQYJ7Y2XIAAEP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360951.9_warc_CC-MAIN-20211201203843-20211201233843-00252.warc.gz\"}"}
https://stats.stackexchange.com/questions/202904/centering-and-scaling-skewed-distributions	[ "Centering and scaling skewed distributions\n\nI have a dataset where the features are skewed (non normal) distributions. My preprocessing pipeline consists of the following steps:\n\n1. Missing values imputation\n2. Centering and scaling (zero mean and unit variance) of each feature\n3. Transforming the features to an approximate normal distribution by using the Box-Cox Transformation.\n\nShould I first do the centering and scaling or the transformation?\n\nSecond, if the distributions are skewed (not normal) is centering and scaling (zero mean and unit variance) still ok? Another possiblity would be to subtract the median (instead of the mean) and dividing by 1.5 * the interquartile range (instead of the standard deviation).\n\n• Depends on your Box-Cox machinery, but I don't know how you consider transformations such as logarithm and square root which are, apart from extra constants, members of the Box-Cox family if your variable has negative values, as it will after standardization. In short, #2 cannot precede #3 without correction. Apart from that, linear transformations such as (value $-$ summary) / spread have precisely no effect on skewness and kurtosis. Note that non-normal is not a synonym for skewed as there are symmetric distributions that aren't normal (and which could be awkward for data analysis). – Nick Cox Mar 21 '16 at 23:56" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8605791,"math_prob":0.9381701,"size":675,"snap":"2019-43-2019-47","text_gpt3_token_len":141,"char_repetition_ratio":0.11773472,"word_repetition_ratio":0.05940594,"special_character_ratio":0.19407408,"punctuation_ratio":0.072072074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98778266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T05:50:40Z\",\"WARC-Record-ID\":\"<urn:uuid:990ca8e9-e4e0-446f-ad70-a06268aae551>\",\"Content-Length\":\"132451\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0cef5a5e-59f9-42b3-86b9-64610c1504bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:79cc9132-2edf-445e-9f2d-ca616c7c6147>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/202904/centering-and-scaling-skewed-distributions\",\"WARC-Payload-Digest\":\"sha1:5RYCBGHCOYUBCX7HXLOJ2A4ZMPCWRE6O\",\"WARC-Block-Digest\":\"sha1:XRG4Q77FMOSXH6UE54WKDJXEWQNBBUL4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987756350.80_warc_CC-MAIN-20191021043233-20191021070733-00219.warc.gz\"}"}
https://www.hindawi.com/journals/sv/2019/3524948/	[ "/ / Article\n\nResearch Article \| Open Access\n\nVolume 2019 \|Article ID 3524948 \| https://doi.org/10.1155/2019/3524948\n\nYang Liu, Yulai Zhao, Jintao Li, Fangquan Xi, Shuanghe Yu, Ye Zhang, \"Research on Fault Feature Extraction Method Based on NOFRFs and Its Application in Rotor Faults\", Shock and Vibration, vol. 2019, Article ID 3524948, 11 pages, 2019. https://doi.org/10.1155/2019/3524948\n\n# Research on Fault Feature Extraction Method Based on NOFRFs and Its Application in Rotor Faults\n\nRevised30 May 2019\nAccepted16 Jun 2019\nPublished02 Jul 2019\n\n#### Abstract\n\nRub-impact between the rotating and static parts is a more common fault. The occurrence of faults is often accompanied by the generation of nonlinear phenomena. However, it is difficult to find out because the nonlinear characteristics are not obvious at the beginning of the fault. As a new frequency domain-based method, nonlinear output frequency response functions (NOFRFs) use the vibration response to extract the nonlinear characteristics of the system. This method has a better recognition rate for fault detection. Also, it has been applied in structural damages detection, but the high-order NOFRFs have the characteristics that the signals are weak and the features are difficult to extract. On this basis, the concept of the weighted contribution rate of the NOFRFs is proposed in this paper. The variable weighted coefficients with orders are used to amplify the influence of high-order NOFRFs on the nonlinearity of the system so as to extract its fault characteristics. The new index RI is proposed based on Clenshaw–Curtis quadrature formula to eliminate the effect of artificially selected weighted coefficients on sensitivity. Especially in the early stage of the fault, the new index varies greatly with the deepening of the fault. Both simulation and experimental results verify the validity and practicability of the new index. The new index has certain guiding significance in the detection of mechanical system faults.\n\n#### 1. Introduction\n\nIn the design process of modern mechanical products, in order to meet the requirements of small size, the distance between the components is getting smaller and smaller. For mechanical products containing rotating parts, this increases the risk of rub-impact faults between the rotor and stator.\n\nScholars have done a lot of research on the diagnosis of rub-impact faults in rotating machinery . The physical model established by the finite element method or the lumped mass method combined with the time domain frequency domain and the axial trajectory is the earliest developed method of detecting rub-impact faults. This method mainly relies on the accuracy of the model [2, 3]. Since then, a large amount of research work has been done to make the model more accurate . In the further development, the detection method of the rub-impact faults based on the Poincaré diagram, the bifurcation diagram, and the Lyapunov exponent is gradually applied, which has great influence on the development of the rub-impact faults detection [5, 6]. In recent years, with the continuous improvement of signal processing technology, some special signal processing methods have also been applied to the diagnosis of rub-impact faults . Chandra and Sekhar compared the effects of three kinds of signal processing methods, short time Fourier transform (STFT), continuous wavelet transform (CWT), and Hilbert–Huang transform (HHT), on the rub-impact faults diagnosis. Zhou et al. studied the intrawave frequency modulation phenomenon and extracted the fault characteristics of the rub-impact from the vibration signal by using the M-VNCMD method. Most of the above studies only qualitatively describe the fault characteristics of the rub-impact. The feature extraction of rub-impact faults, especially the weak fault features at the beginning of the rub-impact, is not well developed.\n\nResearchers found that the frequency domain output of the system will have nonlinear phenomena [11, 12], such as harmonics, intermodulation, and so on when the mechanical system is damaged. However, the knowledge of the linear method cannot explain the generation of nonlinear phenomena . Based on this, scholars began to focus on nonlinear methods to find out the mechanism of the system faults. They have developed a number of nonlinear methods for the diagnosis of system faults, such as the nonlinear decomposition and regulation method and the second-order output spectrum approach [15, 16]. Among many nonlinear methods, the nonlinear method based on Volterra series has been accepted by many scholars. The Volterra series can explain the nonlinear components of the frequency domain output of system. Bai and Zhang proposed a blind identification method based on third-order cumulant and an inverse recursive method and used this method to identify the simplified secondary Volterra system of hydraulic turbine shafting. Tang et al. used the basic theory of Volterra series to model the rotating machinery and diagnose the rotating machinery. The results show that the mechanical system fault is identified by the change of the generalized frequency response function under different operating conditions of the mechanical system. Xia et al. proposed a key kernels-PSO (KK-PSO) method that identifies the Volterra kernel and used the kernel function of the Volterra series to establish a rotor-bearing fault diagnosis system. The results indicate that the KK-PSO method has better recognition ability.\n\nHowever, the Volterra series is quite complex and can only be applied to some simple nonlinear problems. The generalized frequency response functions (GFRFs) is a concept based on the Volterra series. It can be regarded as a case where the frequency response function (FRF) of a linear system is generalized to a nonlinear system. Scholars used GFRFs to study the nonlinear components of the frequency domain output of the system. Since GFRF is a multidimensional function, it is difficult to calculate it accurately in practice. Therefore, Lang and Billings proposed the concept of nonlinear output frequency response functions (NOFRFs), which are a one-dimensional function that allows the calculation of the frequency domain output of a nonlinear system in a similar way to calculation of the frequency domain output of a linear system. Peng et al. used NOFRFs to detect cracked beams and found an input-independent relationship between NOFRFs and GFRFs when the system is subjected to harmonic excitation. The experimental results show that the input energy of the system will be converted into superharmonic components when the excitation frequency ω is close to the resonance frequency of the associated NOFRFs. This also implies that NOFRFs can be used to analyze the behavior of nonlinear systems. Xia et al. proposed a new method for online fault recognition of hydroelectric generators and studied the fault mechanism under different conditions. The results prove that this method is simple and efficient. Huang et al. established the NARMAX model identification with PSO-adaptive lasso algorithm. A damage detection method combining the NARMAX model with a rectangular pulse is proposed. The results show that the method can effectively detect the accumulated fatigue damage of used parts.\n\nUnder the development of nonlinear output frequency response functions, researchers have begun to focus on relevant indexes based on NOFRFs [26, 27]. Peng et al. proposed a method based on NOFRFs related index Fe. The feasibility of the method was verified by the detection of damaged aluminum plates under the NARMAX model. Huang et al. proposed a new index Ne. The NOFRFs of the nonlinear system are calculated by the NARMAX model, and then the index Ne is calculated. Comparing with the index of nondamaged test pieces, the results reveal that the newly proposed method can be used to detect fatigue accumulation damage of used parts. Mao et al. proposed a nonlinear scale index NL and a divergence index NDI based on the basis of NOFRFs. The experimental results show that these indexes can well identify the morphology and extent of microcracks before they appear.\n\nIn , the structural damage of test piece was tested at static state (nonworking state). The system excitation used in these references is a method of pulse excitation. However, this method is no longer applicable for online monitoring of rotating machinery faults. Therefore, the authors did the following research studies. It is pointed out in that the NOFRFs of a nonlinear system can be calculated under the condition that the two amplitudes are different but the excitation frequencies are the same. Due to manufacturing defects, the disc is more or less imbalanced. This will cause the disc to generate centrifugal force during rotation. The existence of centrifugal force just makes the rotor system meet the conditions for calculating NOFRFs in so that the NOFRFs of the system can be obtained. During the research, it was found that the indexes proposed in [25, 26] did not change significantly in the rotor rub-impact fault system. In order to find out a suitable index for online monitoring of rotor rub-impact faults, a new index based on Clenshaw–Curtis integration method is used in this paper, which is the second-order weighted contribution rate integral index RI. The simulation and experimental results show that the index is more suitable for detecting the rotor rub-impact faults than these indexes in [25, 26] and has certain efficiency.\n\nThe contribution of this paper is to combine the weighted contribution rate of NOFRFs with Clenshaw–Curtis integration method for the first time. A new system nonlinear detection index RI is proposed, which is used to detect the rub-impact faults between rotor and stator. The organization of the rest paper is summarized as follows. In Section 2, we introduce the method of calculating NOFRFs under harmonic excitation. In Section 3, we combine the second-order optimal weighted contribution rate of NOFRFs with the Clenshaw–Curtis integration method and propose the index RI. After that, we use the simulation to verify the effectiveness of the index. In Section 4, the effectiveness of this index in the detection of rotor rub-impact fault is verified by experiments. Finally, the conclusion is given in Section 5.\n\n#### 2. Theory of NOFRFs\n\n##### 2.1. NOFRFs under Harmonic Input\n\nTransfer functions are often used to describe the characteristics of linear systems. It can be obtained from the inputs and outputs of a linear system. Based on the similar view, GFRFs were seen as the extension of the transfer functions of the linear system to the nonlinear system situation. At this point, the output of the nonlinear system can be expressed aswherewhere N is the highest order of the system nonlinearity, U() and Y() are the frequency domain expressions of the system time domain input u(t) and the output y(t) after Fourier transform, Yn() represents the nth-order output response of the nonlinear system in the frequency domain, and Hn() is the nth-order GFRFs. It can be seen from its expression that GFRF is a multidimensional function, and its dimensions are the same as orders. As the number of dimensions increase, the amount of calculations of GFRFs become very large, which limit the application of GFRFs.\n\nBased on this, the concept of NOFRFs is proposed.\n\nOn the condition thatits specific expression is defined as\n\nIts specific solving method will be mentioned in . On the basis of the system NOFRFs acquired, the frequency domain output of the system can be rewritten as\n\nThis is similar to the form of the linear system transfer function, which decomposes the nonlinear system into a summation form of multiple linear subsystems, and the nonlinear characteristics of the system will be contained in the transfer function of each linear subsystem. This greatly reduces the amount of calculations and avoids dimensionality disasters. It is advantageous for the application of the method in the field of engineering as well as further theoretical research studies.\n\nWhen the system is excited by harmonics of the form u(t) = Asin (ωt), the system frequency domain output can be simplified as\n\nSystem frequency domain input is defined as\n\nSystem NOFRFs can be simplified to\n\nAccording to equations (6)–(8), under the condition that the first four orders of NOFRFs are sufficient to represent the nonlinear characteristics of the system, the relationship between the frequency domain output response and the input is\n\nAccording to the method mentioned in , when two harmonics with different intensities but the same frequency are used to excite the system, the first four orders of NOFRFs can be obtained, namely, G1(), G2(), G3(), G3(j3ω), G4(j2ω), and G4(j4ω). They can be used directly for fault detection, and they can be further extracted from fault features.\n\n##### 2.2. Feature Extraction Using Traditional Methods\n###### 2.2.1. Index Fe\n\nNOFRFs can be used directly to detect damages to mechanical system or structure. In addition, the nonlinear features in the NOFRFs can be further extracted to detect faults in the mechanical system. Fe is the earliest proposed NOFRFs-based index. Its specific expression is as follows:\n\nEquation (10) shows that the index Fe integrates all the NOFRFs values of the nonlinear system. It has the following characteristic:\n\nFe(n) can reflect the ratio of the nth-order NOFRFs in the system. For example, Fe(1) reflects the proportion of G1() in the system. If Fe(1) ≈ 1, it means that G1() is dominant, and high-order system frequency response NOFRFs can be ignored; that is, the system can be approximated as a linear system. In addition, if Fe(5) ≈ 1, it indicates that the fifth-order NOFRFs dominate the system behaviors. Therefore, the index Fe(n) can be used to describe the nonlinear behaviors of the system and to detect damages to the system.\n\n###### 2.2.2. Index Ne\n\nThe information entropy Ne fuses all orders of NOFRFs, and its specific expression is as follows:where\n\nSince Ne integrates all the orders of NOFRFs, it becomes easier to apply Ne, which is more suitable for fault detection in engineering than Fe. Yet, Ne may be integrated into useless information. Assume that Fe(3) cannot be used to detect system faults, but Ne also fuses Fe(3), which will reduce the sensitivity of the index to system faults detection. Therefore, the authors attempted to identify a more sensitive index to detect system faults and damages.\n\n##### 2.3. Weighted Contribution Rate of NOFRFs (Rn)\n\nDuring the operation of the mechanical system, the system is prone to fault due to abominable working conditions and alternating loads. At this moment, the system will express nonlinearity and the nonlinearity will become apparent as the fault increase. When the fault reaches a certain level, it will lead to an accident. Therefore, in engineering practice, online monitoring of mechanical systems is of great significance. NOFRFs and related indexes can be used to detect system faults. In the actual solution process, high-order NOFRFs become smaller and smaller as the order increases, which is difficult to extract. High-order NOFRFs have a significant effect on the detection of system faults. Therefore, it is necessary to extract the nonlinear features of high-order NOFRFs. Based on this, a new method is proposed in this paper, which is to weight the high-order NOFRFs and increase their contribution rate to the system. The specific weighted process is as follows:where Tn() represents Gn() after weighted, nρ represents a weighted coefficient, and ρ is an indefinite constant. It can be seen that the weighted method adopted in this paper is the variable weighted method. When ρ < 0, the higher the order of NOFRFs and the greater the weighting, as follows:\n\nIt will increase the contribution of high-order NOFRFs to the system, thus achieving the purpose of feature extraction. These are the basic theory of the weighted contribution rate of the NOFRFs, which is expressed as\n\nThe high-order nonlinear characteristics of the system will be amplified by the weighted contribution rate of NOFRFs using feature weighted method. The weighted contribution rate index Rn introduces the order n of the NOFRFs, which causes the weighted coefficients to increase as the order of the NOFRFs increases. The contribution rate of the high-order will be magnified under the condition . Only in this way, the weighted coefficients at the denominator position is less than 1, namely, . Therefore, the weighted contribution rate Rn solves the problem that the high-order NOFRFs and their contribution are too small.\n\nIt can be seen from equation (16) that when the indefinite constant ρ = 0, the weighted contribution index Rn(1) is equal to Fe(1), so the feature of Rn(1) is the same as Fe(1); when the indefinite constant ρ = 0.5, Rn is equal to PY, and PY will be replaced the index Rn. From the above analysis, it can be discovered that the weighted contribution rate index Rn is a generalized form of the two indexes of Fe and PY. There is significant effect on the detection index when different indefinite constants were chosen. Therefore, how to choose an optimal indefinite constant or even how to remove its effect is very important for the accuracy of the detection index. If the final index is not affected by the indefinite constant selection, this will greatly improve the accuracy of the detection index and can be more conducive to the application of the NOFRFs in practical engineering.\n\n#### 3. Second-Order Weighted Contribution Rate Integral Index\n\n##### 3.1. Estimation of NOFRFs of Rotor-Stator Rub-Impact System\n\nIn order to eliminate the influence of the indefinite constants selection on the detection index, we first look at a practical example to realize how to affect the weighted contribution rate of the NOFRFs by the selection of the indefinite constants. The Jeffcott rotor fixed-point rub-impact system will be used as an example in this paper. The schematic diagram of the Jeffcott rotor fixed-point rub-impact system is shown in Figure 1. The rotor system is divided into 11 shaft segments, and the specific parameters of each shaft segment are shown in Table 1. Each shaft segment has two nodes, so the rotor system has a total of 12 nodes.\n\n Segment no. 1 2 3 4 5 6 7 8 9 10 11 Length (mm) 50 45 45 45 45 27 45 45 45 45 50 Diameter (mm) 10 10 10 10 10 70 10 10 10 10 10\n\nEach shaft segment of the rotor system is simulated with Timoshenko beam, and the generalized coordinates of the beam element is u = (xA, yA, θA, θA, xB, yB, θB, θB)T. A and B represent the endpoints of the beam element, respectively, where xA, yA, θA, θA are the displacement and rotation angle of the x and y directions of the A node, respectively, and xB, yB, θB, θB are the displacements and rotation angles of the x and y directions of the B node, respectively.\n\nThe fixed-point rub-impact of rotor-stator is mainly simulated in this paper; in other words, rub-impact will appear between a certain point on the rotating shaft and a stator. The friction torque generated by the rub-impact is ignored, and only the normal rub-impact force Fn and the tangential rub-impact force Fr are considered when the phenomenon of rub-impact appears in rotor-stator, as shown in Figure 2. In Figure 2, O1 represents the centroid of the rotor when it is static, represents the centroid of the rotor when it is dynamic, O2 denotes the center of mass of the rotor when it is static, δ0 represents the initial clearance between rotor and stator, and e represents the eccentricity. The friction coefficient and the contact stiffness between rotor and stator are denoted by kc and fc, respectively.\n\nIn this paper, the Kuhn–Tucker law of Coulomb friction is used to detect the contact between rotor and stator, and the contact condition is solved by the augmented Lagrangian method of friction . These ensure that the disc and the elastic rod meet the contact conditions and do not penetrate each other. In order to meet the penetration depth is within the allowable range, a suitable force will be applied to the normal contact in this method. The rub-impact force Frub between the rotor and the stator can be decomposed into the normal rub-impact force Fn and the tangential rub-impact force Fr, as follows:where r is the distance between O1 and O2 and is the Lagrangian multiplier of the i + 1th step in the iterative process, which is expressed aswhere δr = r − c, ς is prescriptive permeability-tolerance. When ς exceeds the maximum allowable permeation amount after a given equilibrium iteration, the Lagrangian multiplier is used to enlarge the contact stiffness of the contact units. This process does not end until the magnitude of penetration is less than the maximum allowable value.\n\nThen, the rub-impact force can be expressed aswhere the Heaviside function H(δr) is defined as\n\nTherefore, the governing equation of the rotor rub-impact system is expressed aswhere, M, C, G, and K are the mass matrix, damping matrix, gyroscopic matrix, and stiffness matrix, respectively; Fe is the imbalance force vector; and Frub is the rub-impact force vector of the rotor system .\n\nThe rest of simulation parameters are shown in Table 2. The Newmark-β and Newton–Raphson iterative methods are used to solve equation (21), and the vibration response of each node at the excitation frequency of 20 Hz can be obtained. Then, two vibration responses with different excitation strengths but the same excitation frequency can be obtained only by changing the imbalance and solving again. According to equation (9) and , the first four orders of NOFRFs of the system can be obtained by the two inputs and outputs of the system. The specific values are shown in Table 3.\n\n Parameters Values Material 45#steel Elastic modulus of shaft Ef (Pa) 2.1 × 1011 Density (kg/m3) 7850 Poisson ratio 0.3 Normal rub-impact stiffness ratio 2/4/6 × 104 Friction coefficient 0.1/0.2/0.3 Bearing stiffness (N/m) 2 × 106 Imbalance (kg·m) 167.4/236.7 × 10−6 Rotational speed (rpm) 1200\n NOFRFs Normal Clearance = 90 μm Clearance = 30 μm 1.5078 × 10−4 1.5069 × 10−4 1.5051 × 10−4 2.1286 × 10−8 4.8293 × 10−7 1.2208 × 10−6 2.5299 × 10−9 1.1921 × 10−8 3.9750 × 10−9 8.9336 × 10−10 5.7439 × 10−8 1.0286 × 10−7 2.7172 × 10−9 2.2947 × 10−9 3.5165 × 10−8 3.8508 × 10−10 1.7238 × 10−8 3.0804 × 10−8\n##### 3.2. Feature Extraction Based on Weighted Contribution Rate of NOFRFs\n\nAccording to the analysis in Section 2.3, the contribution rate of the high-order NOFRFs will be increased by the weighted contribution index Rn of the NOFRFs to the system. However, the choice of indefinite constants can have a large effect on the results. The authors attempt to eliminate these effects. For this result, the authors research the effect of indefinite constants on the weighted contribution rate. First, the obtained NOFRFs values are taken into equation (16) to calculate Rn, the range of indefinite constant ρ is [−30, 5], and a curve of Rn respect to ρ is then made under three different rub-impact clearances (normal, clearance A = 90 μm, and clearance B = 30 μm).\n\nWhen the indefinite constant ρ changes within the interval [−30, 5], the trend of the weighted contribution index Rn is as shown in Figure 3. It can be seen from the trend in Figure 3 that Rn is almost equal and tends to 0 under different rub-impact clearances when the indefinite constant range is (−∞, −30] or [5, +∞), which means that the weighted contribution rate Rn will fail to diagnose unceasingly the severity of rub-impact of the system when ρ belongs to at this range.\n\nIn addition, Figure 3(a) shows that the two indexes Fe1(ρ = 0) and PY1(ρ = 0.5) increase with the decrease of the rub-impact clearance, but the weighted contribution rate increases from 0 to 1 in the process of the indefinite constant increasing from negative infinity to positive infinity; from Figures 3(a)3(d), it can be seen that Fe(n) and PY decrease when the rub-impact clearance decreases. The value of Fe(n) and PY(n) is also increasing. As seen in Figure 3, the sensitivity of the index Fe is higher than that of PY, but from this, we can also know that in the indefinite constant ρ = 0 or ρ = 0.5, Fe and PY are not the best choice. As shown in Figure 3(b), the peak-weighted contribution rate sensitivity is the highest in the case of different rub-impact clearances, but the indefinite constant corresponding to the peaks changes as the rub-impact clearances change. It can be seen from Figure 3(b) that the trend of the weighted contribution rate Rn(2) is increased first and then decreased, and the curve and the horizontal axis constitute a closed graph when the indefinite constant change. A new index is proposed by using the integral of the indefinite constant ρ of the weighted contribution rate Rn(2). Geometrically, this new index corresponds to the area of the weighted contribution rate Rn(2) and the horizontal axis. This index can better detect the change of rub-impact. Based on this idea, the index will be theoretically derived.\n\nThe specific expression of Rn(2) can be expressed from equation (16), as follows:\n\nDefine as , and equation (22) can be written as\n\nEquation (23) implies that the index Rn(2) is a function of the indefinite constant ρ, and R2(ρ) is regarded as\n\nAccording to the analysis in Section 2.1, only the first four order NOFRFs of the system are considered here, so equation (24) can be expressed as\n\nIn the practical engineering case, , i = 1, 3, 4 in the function R2(ρ) are constants. It can be seen from equation (25) that the function R2(ρ) gradually approaches zero when ρ tends to infinity. According to the analysis mentioned above, the function R2(ρ) is bounded and continuous. So, the function R2(ρ) can be integrated.\n\n##### 3.3. Clenshaw–Curtis Integration\n\nIn engineering practice, researchers often use numerical integration methods when they solve the integrations of unknown analytic functions. There are many methods for numerical integration. A method with higher precision will be adopted in this paper, namely, Clenshaw–Curtis quadrature formula . Using this method to numerically integrate the index Rn(2), the basic theory is that for the integrand function f(x), its definite integral expression on the interval [−1, 1] is\n\nThe Clenshaw–Curtis quadrature formula is defined aswhere ∑ indicates that the first and last items in the formula are halved. , , k = 0, 1, 2, ..., n, j = 1, 2, 4, ..., n, , θk are the quadrature node and the quadrature coefficient, respectively. Regarding the accuracy of Hn(f), the following theorem is given.\n\nTheorem 1. If f satisfies the condition of , the error is expressed as\n\nOn a given finite integration interval [a, b], there is an integral of f(x),where a, b are noninfinite constants and f(x) has a certain smooth degree. The following processing is often done before using Clenshaw–Curtis integration for numerical integration:that is, the integral interval is converted to [−1, 1] by variable substitution.\n\nHowever, in practical engineering problems, the researchers found that the results obtained did not achieve the desired by linear transformation in most cases. The reason for this is that the trend of the function at the endpoints is different. Considering a class of function, f(x) has a significant change in the vicinity of a in the interval [a, b], and the change in the vicinity of b is slow. Such as f(x) = ex, when a = 1 and b = 50, the ideal numerical integral is to take more integral nodes near a, and at the same time, in order to reduce the amount of calculation, there are fewer integral nodes near b. Considering the Clenshaw–Curtis numerical integration in the interval [−1, 1], the integral nodes are symmetric about the origin, and there are many nodes at the and fewer nodes near the origin. With Clenshaw–Curtis numerical integration, linear transformations cannot meet our requirements. Therefore, in this case, a new method of variable substitution will be considered, namely, nonlinear variable substitution.\n\nExtending the linear transformation to nonlinearity, equation (30) is rewritten aswhere i ≥ 1 and i ∈ Z. It can be seen from equation (31) that can convert the interval [a, b] to [−1, 1], and then perform Clenshaw–Curtis numerical integration on f(x). When i = 1, is a linear variable substitution; when i = 2, 3, 4, …, is a nonlinear variable replacement. Under nonlinear transformation, rewrite equation (29) to\n\nAs can be seen from equation (32), and f(x) have the same smooth order. Then, for equation (32), the Clenshaw–Curtis quadrature formula is used to obtain\n\nIn the case of nonlinear variable substitution, the quadrature node is denser near the end point a and sparse at the end point b. This is what we expected.\n\n##### 3.4. Second-Order Weighted Contribution Rate Integral Index (RI)\n\nAccording to the analysis in Section 3.2, the function R2(ρ) is integral. The integral expression is\n\nFrom Figure 3(a) in Section 3.2, it can be concluded that the value of Rn(2) is close to 0 when the indefinite constant is within (−∞, −30] or [5, +∞). So, the integral interval of equation (34) is taken [−30, 5], that is, a = −30, b = 5. Substituting equation (25) into equation (33), equation (34) is rewritten as\n\nDefine .\n\nIn order to verify the reliability of the Clenshaw–Curtis quadrature formula, the authors use the int infinite integral function in MATLAB to solve R2(ρ), and the integration interval is negative infinity to positive infinity. The results are compared with the Clenshaw–Curtis integration method used in this paper, as shown in Table 4. The results show that the selection of the integral interval in the Clenshaw–Curtis method is reasonable, and the results obtained by the Clenshaw–Curtis quadrature method are very accurate.\n\n Clearance (μm) Clenshaw–Curtis integral Int infinite integral Normal 0.0670145365 0.0670145578 2.1289 × 10−8 100 0.3580265528 0.3580266286 7.5884 × 10−8 80 0.5712722586 0.5712723441 8.5445 × 10−8 60 0.6289452527 0.6289452534 7.0059 × 10−8 40 0.6855516020 0.6855515858 1.6181 × 10−8 20 0.7381875132 0.7381875624 4.9269 × 10−8\n\nRub-impact stiffness and friction coefficient are the two main parameters affecting rub-impact. The change of these two parameters will affect the change of detection index RI. In order to study the influence of these two parameters on the results, the authors studied the variation of the second-order weighted contribution rate integral index RI with the rub-impact clearances when these two parameters change, as shown in Figures 4 and 5. They illustrated the trend of the index RI of the system under different rub-impact stiffness and friction coefficient. The results show that the index RI will increase with the decrease of the rub-impact clearance when the rub-impact stiffness and friction coefficient of the system increase. The variation of the index RI will gradually increase, which means that the nonlinearity of the system will be increased.\n\nIn order to verify that the second-order weighted contribution rate integral index RI is more suitable for detecting the rub-impact faults of the rotor, according to equation (33), the index RI under different rub-impact clearances is calculated. The amount of change of RI, Ne, Fe1, Fe2, Fe3, and Fe4 under different rub-impact clearance is shown in Table 5. From this, we can see that the indexes RI, Ne, and Fe1 have the largest change with the reduction of the rub-impact clearance. In order to more intuitively show the advantages of the index RI, we compare the three indexes in the form of a histogram, as shown in Figure 6. The results show that the indices RI and Ne increase with the increasing rub-impact clearance, but the index Fe1 decreases as the rub-impact clearance decreases. However, the change of the second-order weighted contribution rate integral index RI is more obvious. Especially in the initial stage of the rub-impact fault, this index has strong superiority. These indicate that the second-order weighted contribution rate integral index RI proposed in this paper is more suitable for detecting the rub-impact fault of the rotator-stator.\n\n Indexes 100 80 60 40 20 ΔRI 0.2910 0.2132 0.0577 0.0566 0.0526 ΔNe 0.0142 7.388 × 10−3 7.413 × 10−3 7.041 × 10−3 6.754 × 10−3 ΔFe1 2.771 × 10−3 1.700 × 10−3 1.810 × 10−3 1.807 × 10−3 1.803 × 10−3 ΔFe2 2.242 × 10−3 1.619 × 10−3 1.616 × 10−3 1.613 × 10−3 1.610 × 10−3 ΔFe3 3.949 × 10−4 8.183 × 10−5 8.164 × 10−5 8.150 × 10−5 8.135 × 10−5 ΔFe4 1.349 × 10−4 6.657 × 10−7 1.119 × 10−4 1.117 × 10−4 1.116 × 10−4\n\n#### 4. Experimental Verification\n\nThe experimental equipment is shown in Figure 7. In order to obtain the first four orders of NOFRFs of the system, as described in Section 2.1, the system requires two excitations of the same frequency but different intensities. In this paper, the two different strength excitations were obtained by changing the imbalance of the rotor system. In the rotor system, if the imbalance of the system is different, the centrifugal force of the system will also change, which will ultimately affect the vibration amplitude of the system. These two excitations of the same frequency can be obtained at the same speed. Therefore, when the data were collected in the experiment, it was necessary to change the imbalance of the rotor system at the same speed to meet the requirements. A different number of balanced bolts were used to change the imbalance in the rotor system. The rotor speed in the experiment was the same as the simulation, namely, 1200 rpm.\n\nIt can be seen from Figure 8 that the index Fe1 changes slowly and the index Ne’s magnitude and the relative variation are small when the rub-impact degree is increased in the rotor rub-impact system. Meanwhile, the variation in the index RI is very obvious. Therefore, RI is a better index to detect system rub-impact faults. Figure 8 also shows that the experimental data and the simulation results basically coincide, which indicates that the simulation results are credible.\n\n#### 5. Conclusion\n\nA new method of detection of the rub-impact fault of the rotor system is introduced in this paper. It introduces the concept of the order of the nonlinear output frequency response functions and uses the NOFRFs to amplify the signal characteristics. Therefore, the impact of higher-order NOFRFs on the system is amplified. On the basis, a new index called the second-order weighted contribution rate integral index RI is proposed in this paper. After comparative analysis, it is known that the sensitivity of this index to the rub-impact faults of the rotor system is much higher than that of the other two indexes (Fe and Ne). In other words, RI is a better index to detect the rub-impact faults of the rotor system. On the other hand, simulation and experimental results verify that the efficiency of the index RI is in detecting system faults. The second-order weighted contribution rate integral index RI is of great significance in the study of fault diagnosis. The performance of this index in other mechanical faults and structural damage detection will be discussed in future studies.\n\n#### Data Availability\n\nThe data used to support the findings of this study are included within the article.\n\n#### Conflicts of Interest\n\nThe authors declare no conflicts of interest.\n\n#### Acknowledgments\n\nThis work was supported by the National Natural Science Foundation of China (grant nos. 51875093 and U1708257), Basic Research Business Fee of the Central University of Education (grant no. N180304017), and China Postdoctoral Science Foundation (grant nos. 2014M551105 and 2015T80269).\n\n1. G. Jacquet-Richardet, M. Torkhani, P. Cartraud et al., “Rotor to stator contacts in turbomachines. 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https://howkgtolbs.com/convert/44.55-kg-to-lbs	[ "# 44.55 kg to lbs - 44.55 kilograms into pounds\n\nkg\nlbs\n\n## 44.55 kg to lbs\n\nBefore we go to the more practical part - that is 44.55 kg how much lbs conversion - we will tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.\n\n## 44.55 kgs in pounds\n\nWe will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, that is International System of Units (in short form SI).\n\nAt times the kilogram could be written as kilogramme. The symbol of the kilogram is kg.\n\nFirstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but difficult to use.\n\nThen, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was substituted by a new definition.\n\nThe new definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is exactly 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.\n\n## 44.55 kilogram to pounds\n\nYou learned a little about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to point out that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to focus only on pound-mass.\n\nThe pound is in use in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of the pound is lb or “.\n\nThe international avoirdupois pound has no descriptive definition. It is defined as 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces and 7000 grains.\n\nThe avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### 44.55 kg in lbs\n\nTheoretical section is already behind us. In next part we will tell you how much is 44.55 kg to lbs. Now you learned that 44.55 kg = x lbs. So it is time to get the answer. Let’s see:\n\n44.55 kilogram = 98.2159377210 pounds.\n\nThis is an accurate result of how much 44.55 kg to pound. You may also round off the result. After rounding off your outcome is as following: 44.55 kg = 98.010 lbs.\n\nYou know 44.55 kg is how many lbs, so look how many kg 44.55 lbs: 44.55 pound = 0.45359237 kilograms.\n\nObviously, this time you can also round off the result. After it your outcome is as following: 44.55 lb = 0.45 kgs.\n\nWe also want to show you 44.55 kg to how many pounds and 44.55 pound how many kg results in tables. Have a look:\n\nWe will startwant to start\",\"will begin\",\"want to begin\",\"are going to start\",\"are going to begin\"}.RandomElement()) with a table for how much is 44.55 kg equal to pound.\n\nKilograms Pounds Pounds (rounded off to two decimal places)\n44.55 98.2159377210 98.0100\nNow look at a table for how many kilograms 44.55 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n44.55 0.45359237 0.45\n\nNow you know how many 44.55 kg to lbs and how many kilograms 44.55 pound, so it is time to move on to the 44.55 kg to lbs formula.\n\n### 44.55 kg to pounds\n\nTo convert 44.55 kg to us lbs you need a formula. We will show you two formulas. Let’s begin with the first one:\n\nAmount of kilograms * 2.20462262 = the 98.2159377210 outcome in pounds\n\nThe first formula will give you the most exact result. Sometimes even the smallest difference could be significant. So if you want to get a correct result - first formula will be the best for you/option to calculate how many pounds are equivalent to 44.55 kilogram.\n\nSo move on to the second version of a formula, which also enables calculations to learn how much 44.55 kilogram in pounds.\n\n### 44.55 pound to kg\n\nThe shorter version of a formula is down below, have a look:\n\nAmount of kilograms * 2.2 = the result in pounds\n\nAs you see, the second version is simpler. It could be the best option if you need to make a conversion of 44.55 kilogram to pounds in easy way, for instance, during shopping. Just remember that final result will be not so exact.\n\nNow we want to show you these two formulas in practice. But before we will make a conversion of 44.55 kg to lbs we want to show you easier way to know 44.55 kg to how many lbs totally effortless.\n\n### 44.55 kg to lbs converter\n\nAn easier way to learn what is 44.55 kilogram equal to in pounds is to use 44.55 kg lbs calculator. What is a kg to lb converter?\n\nConverter is an application. Converter is based on longer version of a formula which we gave you in the previous part of this article. Due to 44.55 kg pound calculator you can effortless convert 44.55 kg to lbs. You only need to enter amount of kilograms which you need to calculate and click ‘calculate’ button. You will get the result in a flash.\n\nSo let’s try to convert 44.55 kg into lbs with use of 44.55 kg vs pound calculator. We entered 44.55 as an amount of kilograms. Here is the outcome: 44.55 kilogram = 98.2159377210 pounds.\n\nAs you can see, this 44.55 kg vs lbs converter is so simply to use.\n\nNow we are going to our primary issue - how to convert 44.55 kilograms to pounds on your own.\n\n#### 44.55 kg to lbs conversion\n\nWe will begin 44.55 kilogram equals to how many pounds conversion with the first formula to get the most correct outcome. A quick reminder of a formula:\n\nNumber of kilograms * 2.20462262 = 98.2159377210 the outcome in pounds\n\nSo what have you do to know how many pounds equal to 44.55 kilogram? Just multiply amount of kilograms, this time 44.55, by 2.20462262. It is 98.2159377210. So 44.55 kilogram is exactly 98.2159377210.\n\nYou can also round off this result, for example, to two decimal places. It is 2.20. So 44.55 kilogram = 98.0100 pounds.\n\nIt is time for an example from everyday life. Let’s convert 44.55 kg gold in pounds. So 44.55 kg equal to how many lbs? As in the previous example - multiply 44.55 by 2.20462262. It is exactly 98.2159377210. So equivalent of 44.55 kilograms to pounds, if it comes to gold, is exactly 98.2159377210.\n\nIn this example you can also round off the result. Here is the result after rounding off, this time to one decimal place - 44.55 kilogram 98.010 pounds.\n\nNow let’s move on to examples calculated with a short version of a formula.\n\n#### How many 44.55 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nAmount of kilograms * 2.2 = 98.010 the outcome in pounds\n\nSo 44.55 kg equal to how much lbs? As in the previous example you have to multiply number of kilogram, in this case 44.55, by 2.2. Look: 44.55 * 2.2 = 98.010. So 44.55 kilogram is exactly 2.2 pounds.\n\nDo another calculation with use of shorer version of a formula. Now convert something from everyday life, for instance, 44.55 kg to lbs weight of strawberries.\n\nSo convert - 44.55 kilogram of strawberries * 2.2 = 98.010 pounds of strawberries. So 44.55 kg to pound mass is exactly 98.010.\n\nIf you know how much is 44.55 kilogram weight in pounds and can calculate it with use of two different formulas, we can move on. Now we want to show you these outcomes in tables.\n\n#### Convert 44.55 kilogram to pounds\n\nWe are aware that results shown in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in tables for your convenience. Thanks to this you can quickly compare 44.55 kg equivalent to lbs outcomes.\n\nBegin with a 44.55 kg equals lbs table for the first version of a formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n44.55 98.2159377210 98.0100\n\nAnd now have a look at 44.55 kg equal pound table for the second formula:\n\nKilograms Pounds\n44.55 98.010\n\nAs you can see, after rounding off, when it comes to how much 44.55 kilogram equals pounds, the outcomes are the same. The bigger number the more considerable difference. Keep it in mind when you want to do bigger number than 44.55 kilograms pounds conversion.\n\n#### How many kilograms 44.55 pound\n\nNow you know how to convert 44.55 kilograms how much pounds but we will show you something more. Are you curious what it is? What do you say about 44.55 kilogram to pounds and ounces conversion?\n\nWe will show you how you can calculate it step by step. Start. How much is 44.55 kg in lbs and oz?\n\nFirst thing you need to do is multiply amount of kilograms, this time 44.55, by 2.20462262. So 44.55 * 2.20462262 = 98.2159377210. One kilogram is exactly 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this case there are 2 pounds.\n\nTo know how much 44.55 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.\n\nSo your outcome is exactly 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your outcome is 2 pounds and 33 ounces.\n\nAs you can see, calculation 44.55 kilogram in pounds and ounces quite easy.\n\nThe last calculation which we will show you is conversion of 44.55 foot pounds to kilograms meters. Both of them are units of work.\n\nTo convert foot pounds to kilogram meters it is needed another formula. Before we give you it, look:\n\n• 44.55 kilograms meters = 7.23301385 foot pounds,\n• 44.55 foot pounds = 0.13825495 kilograms meters.\n\nNow see a formula:\n\nAmount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters\n\nSo to convert 44.55 foot pounds to kilograms meters you need to multiply 44.55 by 0.13825495. It is 0.13825495. So 44.55 foot pounds is equal 0.13825495 kilogram meters.\n\nIt is also possible to round off this result, for example, to two decimal places. Then 44.55 foot pounds is 0.14 kilogram meters.\n\nWe hope that this calculation was as easy as 44.55 kilogram into pounds calculations.\n\nThis article was a huge compendium about kilogram, pound and 44.55 kg to lbs in conversion. Due to this conversion you know 44.55 kilogram is equivalent to how many pounds.\n\nWe showed you not only how to do a calculation 44.55 kilogram to metric pounds but also two other calculations - to check how many 44.55 kg in pounds and ounces and how many 44.55 foot pounds to kilograms meters.\n\nWe showed you also other way to make 44.55 kilogram how many pounds calculations, that is with use of 44.55 kg en pound converter. This will be the best solution for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.\n\nWe hope that now all of you are able to do 44.55 kilogram equal to how many pounds calculation - on your own or with use of our 44.55 kgs to pounds converter.\n\nIt is time to make your move! Calculate 44.55 kilogram mass to pounds in the way you like.\n\nDo you want to make other than 44.55 kilogram as pounds conversion? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so simply as for 44.55 kilogram equal many pounds.\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.\n\nWe have told you something about kilograms and pounds yet. In this article we want to focu\n\n10 Kilograms to Pounds ConversionLet’s convert 10 Kilograms to Pounds.Enter 10 kilograms a\n\n30 lbs to kg - 30 kg to lbs. In this article we want to focus not only on 30 kg to lbs con\n\nHow many kilograms to pounds \| how many kg to lb. How many kilograms to lbs. Before we giv\n\nDo you want to know how to convert price per lb to kg? Have you ever wondered how to conve\n\nHow to convert cost per kilo to cost per pound lb? We want to show you how to convert pric\n\nIn this article we want to focus on practical side of how many lbs per kg. How many pounds\n\nIn previous article we told you how to convert kg/m3 to lb/ft3. Today we are going to show\n\n 44.01 kg to lbs = 97.0254 44.02 kg to lbs = 97.0475 44.03 kg to lbs = 97.0695 44.04 kg to lbs = 97.0916 44.05 kg to lbs = 97.1136 44.06 kg to lbs = 97.1357 44.07 kg to lbs = 97.1577 44.08 kg to lbs = 97.1798 44.09 kg to lbs = 97.2018 44.1 kg to lbs = 97.2239 44.11 kg to lbs = 97.2459 44.12 kg to lbs = 97.2679 44.13 kg to lbs = 97.29 44.14 kg to lbs = 97.312 44.15 kg to lbs = 97.3341 44.16 kg to lbs = 97.3561 44.17 kg to lbs = 97.3782 44.18 kg to lbs = 97.4002 44.19 kg to lbs = 97.4223 44.2 kg to lbs = 97.4443 44.21 kg to lbs = 97.4664 44.22 kg to lbs = 97.4884 44.23 kg to lbs = 97.5105 44.24 kg to lbs = 97.5325 44.25 kg to lbs = 97.5546\n 44.26 kg to lbs = 97.5766 44.27 kg to lbs = 97.5986 44.28 kg to lbs = 97.6207 44.29 kg to lbs = 97.6427 44.3 kg to lbs = 97.6648 44.31 kg to lbs = 97.6868 44.32 kg to lbs = 97.7089 44.33 kg to lbs = 97.7309 44.34 kg to lbs = 97.753 44.35 kg to lbs = 97.775 44.36 kg to lbs = 97.7971 44.37 kg to lbs = 97.8191 44.38 kg to lbs = 97.8411 44.39 kg to lbs = 97.8632 44.4 kg to lbs = 97.8852 44.41 kg to lbs = 97.9073 44.42 kg to lbs = 97.9293 44.43 kg to lbs = 97.9514 44.44 kg to lbs = 97.9734 44.45 kg to lbs = 97.9955 44.46 kg to lbs = 98.0175 44.47 kg to lbs = 98.0396 44.48 kg to lbs = 98.0616 44.49 kg to lbs = 98.0837 44.5 kg to lbs = 98.1057\n 44.51 kg to lbs = 98.1278 44.52 kg to lbs = 98.1498 44.53 kg to lbs = 98.1719 44.54 kg to lbs = 98.1939 44.55 kg to lbs = 98.2159 44.56 kg to lbs = 98.238 44.57 kg to lbs = 98.26 44.58 kg to lbs = 98.2821 44.59 kg to lbs = 98.3041 44.6 kg to lbs = 98.3262 44.61 kg to lbs = 98.3482 44.62 kg to lbs = 98.3703 44.63 kg to lbs = 98.3923 44.64 kg to lbs = 98.4143 44.65 kg to lbs = 98.4364 44.66 kg to lbs = 98.4584 44.67 kg to lbs = 98.4805 44.68 kg to lbs = 98.5025 44.69 kg to lbs = 98.5246 44.7 kg to lbs = 98.5466 44.71 kg to lbs = 98.5687 44.72 kg to lbs = 98.5907 44.73 kg to lbs = 98.6128 44.74 kg to lbs = 98.6348 44.75 kg to lbs = 98.6569\n 44.76 kg to lbs = 98.6789 44.77 kg to lbs = 98.701 44.78 kg to lbs = 98.723 44.79 kg to lbs = 98.7451 44.8 kg to lbs = 98.7671 44.81 kg to lbs = 98.7891 44.82 kg to lbs = 98.8112 44.83 kg to lbs = 98.8332 44.84 kg to lbs = 98.8553 44.85 kg to lbs = 98.8773 44.86 kg to lbs = 98.8994 44.87 kg to lbs = 98.9214 44.88 kg to lbs = 98.9435 44.89 kg to lbs = 98.9655 44.9 kg to lbs = 98.9876 44.91 kg to lbs = 99.0096 44.92 kg to lbs = 99.0316 44.93 kg to lbs = 99.0537 44.94 kg to lbs = 99.0757 44.95 kg to lbs = 99.0978 44.96 kg to lbs = 99.1198 44.97 kg to lbs = 99.1419 44.98 kg to lbs = 99.1639 44.99 kg to lbs = 99.186 45 kg to lbs = 99.208\nRemember that our calculator from lbs to kg or kg to lbs located on the site can sometimes show the wrong results. It is important to use the converter weight lbs to kg twice or even more to get the exact result. The kg to lbs conversion calculator is based on formulas which is not errorless. 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https://www.arxiv-vanity.com/papers/1906.07648/	[ "# Transitive tournament tilings in oriented graphs with large minimum total degree\n\nLouis DeBiasio, Allan Lo, Theodore Molla, and Andrew Treglown\nFebruary 28, 2021\n###### Abstract.\n\nLet be the transitive tournament on vertices. We show that every oriented graph on vertices with minimum total degree can be partitioned into vertex disjoint ’s, and this bound is asymptotically tight. We also improve the best known bound on the minimum total degree for partitioning oriented graphs into vertex disjoint ’s.\n\n11footnotetext: Department of Mathematics, Miami University . Research supported in part by Simons Foundation Collaboration Grant # 283194.22footnotetext: School of Mathematics, University of Birmingham . Research supported in part by EPSRC, grant no. EP/P002420/1.33footnotetext: Department of Mathematics and Statistics, University of South Florida . Research supported in part by NSF Grants DMS-1500121 and DMS-1800761.44footnotetext: University of Birmingham, United Kingdom . Research supported by EPSRC grant EP/M016641/1.\n\n## 1. Introduction\n\nFor a pair of (di)graphs and , we call a collection of vertex disjoint copies of in an -tiling. We say that an -tiling is perfect if it consists of exactly copies of . Perfect -tilings are sometimes referred to as perfect -packings, perfect -matchings or -factors.\n\nThe classic Hajnal–Szemerédi theorem states that if is a graph on vertices with minimum degree at least , then contains a perfect -tiling. Moreover, there are -vertex graphs with minimum degree that do not contain a perfect -tiling.\n\nRecall that digraphs are graphs such that every pair of vertices has at most two edges between them, one oriented in each direction; oriented graphs are orientations of simple graphs (so there is at most one directed edge between any pair of vertices). Note that oriented graphs are a subclass of digraphs.\n\nRecently the study of tilings in digraphs has proven fruitful, and a number of papers have focused on developing analogs of the Hajnal–Szemerédi theorem. In this setting there is more than one natural notion of degree: The minimum semidegree of a digraph is the minimum of its minimum outdegree and its minimum indegree . The minimum total degree of is the minimum number of edges incident to a vertex in . So for oriented graphs , . When there is no possibility of confusion, we often refer to the minimum total degree as the minimum degree.\n\nLet denote the transitive tournament on vertices and denote the cyclic triangle. In it was proven that every digraph on vertices with minimum total degree at least contains a perfect -tiling. This degree condition is best possible, and the result implies the original Hajnal–Szemerédi theorem. A minimum semidegree version of the Hajnal–Szemerédi theorem was proven in for large digraphs; this result considers perfect -tilings for any fixed tournament . Finally, Czygrinow, DeBiasio, Molla and Treglown gave a general result which, together with a result of Wang determines the minimum total degree threshold for perfect -tilings in a digraph for any tournament .\n\nFor oriented graphs, the situation is much more difficult. Firstly notice that one can have arbitrarily large minimum total degree and still avoid even a single copy of an oriented graph. Indeed, a transitive tournament on vertices has but contains no oriented graph with a directed cycle. Further, there are -vertex tournaments (i.e. complete oriented graphs) with minimum semidegree at least (i.e. almost as large as possible) that do not contain a perfect -tiling (see [8, 9]). Note though that Keevash and Sudakov did prove that there exists a so that every sufficiently large oriented graph with minimum semidegree at least contains a -tiling covering all but at most vertices. Additionally, Li and Molla recently proved that if is a sufficiently large odd multiple of , every regular tournament on vertices has a perfect -tiling, thereby verifying a conjecture of Cuckler and Yuster.\n\nMore is known for the perfect -tiling problem in oriented graphs, though understanding the general behaviour of the minimum degree threshold remains a significant challenge. Yuster observed that if is an oriented graph on vertices with minimum total degree at least , then has a perfect -tiling. Furthermore, this bound is best possible. Balogh, Lo and Molla later proved an analogous result for the minimum semidegree threshold.\n\nYuster gave a bound on the total degree threshold for nearly perfect tiling with . That is if is an oriented graph on vertices with minimum total degree at least , then has vertex disjoint copies of covering all but vertices. Yuster also showed that if is an oriented graph on vertices with minimum total degree at least , then has a perfect -tiling.\n\nOur main result is to asymptotically determine the minimum total degree threshold for perfect -tiling.\n\n###### Theorem 1.1.\n\nFor all , there exists such that if is an oriented graph on vertices, is divisible by , and , then has a perfect -tiling. Furthermore, for every divisible by , there exists an oriented graph on vertices with such that does not contain a perfect -tiling.\n\nMoreover, we improve the general bounds on the minimum total degree threshold for perfect -tiling, showing that a slight improvement on Yuster’s above mentioned bound for nearly perfect -tiling in fact ensures that has a perfect -tiling. Let be the smallest integer such that every tournament on vertices contains a copy of .\n\n###### Theorem 1.2.\n\nFor every and , there exists such that when and is divisible by the following holds. If is an oriented graph on vertices and\n\n δ(G)≥(1−1k(2→r(k−1)−k+1)+ε)n,\n\nthen contains a perfect -tiling. In particular, suffices here.\n\nRoughly, we obtain both of our results by splitting the problem into two parts: determining the minimum degree threshold for “fractional -tiling” (which is related to “nearly perfect -tiling”) and determining the minimum degree threshold for “-absorbing”. When , we are able to determine these two thresholds exactly, which is why we obtain an asymptotically tight bound in that case.\n\nAs discussed in the following section, one can obtain a bound for the minimum degree threshold for perfect -tilings via an application of the Hajnal–Szemerédi theorem. Indeed, this is where Yuster’s aforementioned bounds came from. However, the bound in Theorem 1.1 is lower than that obtained via the Hajnal–Szemerédi theorem, demonstrating the problem in the oriented graph setting is genuinely different. In order to discuss more precisely where our bounds come from, we must first discuss their connection to some more parameters in the next two sections.\n\nIn Section 3 we give a minimum degree condition that ensures an oriented graph has a perfect fractional -tiling (and thus a nearly perfect -tiling); see Theorem 3.2. This theorem will be applied in both the proof of Theorem 1.1 and Theorem 1.2. In Section 4 we introduce an absorbing result which, combined with our results from Section 3, yields Theorem 1.2. Theorem 1.1 is then proved in Section 5. We finish the paper with some concluding remarks and open questions.\n\n## 2. Oriented Ramsey numbers and perfect tilings\n\nRecall is the smallest integer such that every tournament on vertices contains a copy of . Erdős and Moser proved that . The following result provides for small values of .\n\n###### Theorem 2.1 (see ).\n\n, , , and .\n\nOne can consider Turán-type questions in oriented graphs. The following observation shows that the Turán number of in an oriented graph is completely determined by and Turán’s theorem. Here we let be the number of edges in a Turán graph on vertices with parts, i.e., is the number of edges in a complete -partite graph on vertices with parts of size either the ceiling or floor of .\n\n###### Observation 2.2.\n\nThe maximum number of edges in an oriented graph on vertices that does not contain a copy of is .\n\n###### Proof.\n\nIf is an oriented graph on vertices with more than edges, then, by Turán’s theorem, must contain a tournament on vertices, which implies that contains a copy of .\n\nLet be a tournament on vertices that does not contain a . Blowing-up each vertex of equitably to form an oriented graph on vertices, produces a graph without a copy of whose underlying simple graph is the Turán graph on vertices with parts. ∎\n\nFor every positive integer , let be the collection of tournaments with vertex set . Let be the smallest integer such that every has a perfect -tiling. Note that, by induction, for and divisible by , every tournament has a perfect -tiling. It is known that (see ), and Caro proved that\n\n →tr(k)≤→r(2k−1)+(2k−1)→r(k)<4k,\n\nbut the determination of is open for every . (See [15, Proposition 10] for a concise proof of Caro’s upper-bound.)\n\nFor , let be the minimum integer such that every oriented graph on vertices with has a perfect -tiling, and define . The following straightforward consequence of the Hajnal–Szemerédi theorem, together with any bounds on gives a bound on .\n\n###### Observation 2.3 (Yuster , Treglown ).\n\nGiven any , and so\n\n →δ(k)≤1−1→tr(k)<1−14k.\n\nSince , it was conceivable that for all . However, Theorem 1.1 shows that , whereas . That is, by a computer search, we found tournaments on vertices that do not have a perfect -tiling.111 In Appendix 8 these tournaments are listed. We believe this is an exhaustive list (up to isomorphism). For this search, we used the nauty and Traces software package . Therefore, Theorem 1.1 does not follow directly from the Hajnal–Szemerédi theorem .\n\n## 3. Linear programming and fractional tilings\n\n### 3.1. Linear programming\n\nLet be a -uniform hypergraph. A fractional matching in is a function so that for each , . The size of the fractional matching is . By definition, the largest fractional matching in has size at most (if it has size exactly we say it is perfect). Define and to be the size of the largest matching and fractional matching in , respectively.\n\nA fractional vertex cover for is a function so that for each , . The size of the fractional vertex cover is . Let and be the size of the smallest vertex cover and fractional vertex cover of , respectively. By the duality theorem of linear programming, we have\n\n ν(H)≤ν∗(H)=τ∗(H)≤τ(H).\n\nFor a pair of graphs or directed graphs and , we let be the -uniform hypergraph on the vertex set in which is an edge if and only if contains a copy of . If is a graph we define and if is a directed graph we set . We set and . We define , , , , , and analogously.\n\nA fractional -tiling of is a weight function on the copies of in that corresponds to a fractional matching in , i.e., for every vertex , the sum of the weights on the copies of that contain is at most one. It is a perfect fractional -tiling of if the sum of the weights is equal to . We call a weight function on the vertices of a fractional -cover if the weight function is a vertex cover of , that is, if the sum of the weights on the vertices of every copy of in is at least one. For both a fractional -tiling of and a fractional -cover of , the size of the weight function is defined to be the sum of the weights (i.e. analogous to the notion of the size of a fractional matching and a fractional vertex cover).\n\nLet denote the smallest integer such that for every we have . We clearly have that . Also, every tournament on vertices satisfies . Indeed, by induction on , we may assume that and, for each vertex , there is a perfect fractional -tiling in . Then is a perfect fractional -tiling in .\n\n### 3.2. Forcing fractional tilings and bounds on →tr∗(k)\n\nFor every , define to be the smallest integer such that every oriented graph on vertices with has a perfect fractional -tiling, and let . Let be the infimum of the set of numbers such that for every there exists such that every oriented graph on vertices with has a -tiling of missing at most vertices.\n\nUsing our notation, we now rewrite (a slightly weaker222There are three differences to note. First, we ignore the case and which Yuster considers. Second, Yuster proves that one can almost tile an oriented graph that meets the minimum degree condition with the blow-up of , but with the regularity lemma, this version of the theorem implies the original version. Third, Yuster writes the minimum degree condition in terms of the function which is defined to be the smallest integer such that every tournament on at least vertices has the property that every vertex is contained in a copy of , but it is not hard to see that (see Example 3.5). version of) Yuster’s result ([19, Theorem 3.1]).\n\n###### Theorem 3.1 (Yuster ).\n\nFor , .\n\nLater in this section we prove the following bounds on and in terms of .\n\n###### Theorem 3.2.\n\n.\n\nWe also obtain the following bounds on .\n\n###### Theorem 3.3.\n\nFor all ,\n\n max{2→r(k−1),kk−2(→r(k)−2)}≤→tr∗(k)≤k(2→r(k−1)−k+1).\n\nNote that the upper bound in Theorem 3.3 together with Theorem 3.2 yields a slight strengthening of Theorem 3.1; they also can be combined with an absorbing result (Lemma 4.3) to give Theorem 1.2 (see Section 4.2). Theorems 3.2 and 3.3 will also be applied in the proof of Theorem 1.1.\n\nWe now prove Theorem 3.2.\n\n###### Proof of Theorem 3.2.\n\nLet be an oriented graph on vertices with . Blow up each vertex of to a set of size and call the resulting oriented graph . By the Hajnal–Szemerédi theorem, the simple graph underlying has a perfect -tiling. Note that each has a perfect fractional -tiling in . Hence has a perfect fractional -tiling and so does . So we have established that .\n\nAssume . Let be a tournament on vertices that does not have a perfect fractional -tiling; i.e. . Let and note that . For sufficiently large, blow up each of the vertices of into a set of vertices to form an oriented graph on vertices. Since , and is sufficiently large, we can assume that there exists a -tiling of that covers all but at most vertices. Because every in corresponds to a in , we can create a fractional -tiling of by giving each in weight equal to the number of times a that corresponds to it appears in divided by . This fractional -tiling of has size\n\n \|T\|s≥(1−0.9γ)nks=(1−0.9γ)\|T\|k>(1−γ)\|T\|k=ν∗k(T),\n\na contradiction. So, we have established that .\n\nTo complete the proof, we need to show that . This can be shown by following a standard application of Szemerédi’s regularity lemma. We omit the proof. (It is also possible to establish this fact without appealing to the regularity lemma, e.g., see .) ∎\n\n### 3.3. Proof of Theorem 3.3\n\nThe following example gives a lower bound on , which together with Theorem 3.2 gives a lower bound on .\n\n###### Example 3.4.\n\nLet . For every and , there exists an oriented graph on vertices with\n\n δ(G)=⌊(1−k−2k(→r(k)−2))n−2γnk(→r(k)−2)⌋,\n\nsuch that no -tiling covers more than vertices of . In particular, this implies that which implies by Theorem 3.2.\n\n###### Proof.\n\nTake the largest tournament which does not contain ; note that it has exactly vertices. For , blow up one of the vertices to a set of size and inside the set add all possible edges (oriented arbitrarily). Blow-up the other parts to independent sets of size either the floor or ceiling of\n\n (n−\|X\|)⋅1→r(k)−2=(k−2)nk(→r(k)−2)+2γnk(→r(k)−2),\n\nwhilst ensuring the resulting oriented graph has vertices. Note that every must use at least vertices from , so there is only space for at most vertex disjoint copies of in . ∎\n\nThe next example gives a different lower bound on , which together with Example 3.4 implies the lower bound in Theorem 3.3.\n\nFor every , .\n\n###### Proof.\n\nTo see that , consider a tournament on vertices in which there exists a vertex such that ; both and induce a tournament on vertices that does not contain a ; all of the edges between and are directed from to . In this tournament, is not contained in a ; this immediately implies does not have a perfect fractional -tiling. ∎\n\nTo prove the upper bound of Theorem 3.3, we first collect together some useful observations.\n\nFor a hypergraph and for every , we let be the link graph of , i.e., is the hypergraph with vertex set and edge set . The following lemma is well-known. We provide a proof for completeness.\n\n###### Lemma 3.6.\n\nIf is a -uniform hypergraph on vertices and, for every , , then .\n\n###### Proof.\n\nSuppose that for every , and . In a fractional matching of of size , there must exist a vertex in which the sum of the weights on the edges incident to is strictly less than . By the complementary slackness theorem from linear programming, this implies that if is a fractional vertex cover of of size , then . This means that is a fractional vertex cover of , so\n\n ν∗(H(v))=τ∗(H(v))≤τ∗(H)=ν∗(H)\n\na contradiction. ∎\n\nLet and either be a pair of graphs or a pair of directed graphs such that and and let . For a vertex , a weight function on the -subsets of is a -extendable fractional -tiling of size if it corresponds to a fractional matching of the hypergraph that has size . We have the following corollary to Lemma 3.6.\n\n###### Corollary 3.7.\n\nLet and either be a pair of graphs or a pair of directed graphs such that and . If, for every , there exists a -extendable fractional -tiling of size at least , then there exists a perfect fractional -tiling of .\n\n###### Proof.\n\nThis follows from Lemma 3.6 if we consider the hypergraph . ∎\n\nWe now prove the upper bound in Theorem 3.3.\n\nFor , .\n\n###### Proof.\n\nLet be a tournament on vertices. For an arbitrary , we aim to prove that there exists a -extendable fractional -tiling of size at least . By Corollary 3.7, this will then prove the lemma. To do this, we first prove the following claim.\n\n###### Claim 3.8.1.\n\nIf is a tournament on vertices, then .\n\n###### Proof.\n\nLet be a fractional -cover of of size and let be an ordering of such that . Note that contains at least one , so and . Therefore,\n\n τ∗k−1(S)=→r(k−1)∑i=1w(vi)+s∑i=→r(k−1)+1w(vi)≥1+s−→r(k−1)k−1=s−(→r(k−1)−k+1)k−1.\\qed\n\nRecall that\n\n (1) 2→r(k−1)−k+1=nk.\n\nLet and . Note that forms a copy of with any copy of in or . In particular, a lower bound on gives a lower bound on the size of the largest -extendable fractional -tiling.\n\nSuppose that . If , then . So, by (1) and the claim,\n\n ν∗+≥d+T(v)−(→r(k−1)−k+1)k−1≥n−(2→r(k−1)−k+1)k−1=nk.\n\nIf , then by the claim, (1), the fact that , and the fact that we have\n\n ν∗++ν∗−≥d+T(v)+d−T(v)−2(→r(k−1)−k+1)k−1≥n−(2→r(k−1)−k+1)k−1=nk.\n\nAn analogous argument applies if . So there exists a -extendable fractional -tiling of size at least . ∎\n\n### 3.4. Remarks\n\nNote that Example 3.5 and Theorem 3.2 together imply that . If it can be shown that the lower bound on from Example 3.4 is also an upper bound; i.e. (which is true for and ), then we have , or\n\n →r(k)≥2(k−2)k⋅→r(k−1)+2,\n\nwhich would imply that which almost matches the Erdős–Moser bound of . In fact, even proving that for some absolute constant would improve the best known lower bound on . It is also worthwhile to note that provides a lower bound on the classical Ramsey number . Indeed if is a tournament on vertices with no , then the graph on formed by taking any ordering of the vertices of and, for every , placing the edge in if the edge in is directed from to has neither a clique nor an independent set of size . Therefore, it is possible that a substantial improvement to the upper bound on could give an improvement on the best known lower bound for the diagonal Ramsey numbers.\n\nNote that when ,\n\n n−(→r(k)−2)2≥n−k−2k⋅n2=nk.\n\nA way one might attempt to prove that would be to first prove that equality holds in the following.\n\n###### Example 3.9.\n\nFor , if , then\n\n minT∈Tn{ν∗(T)}≤n−(→r(k)−2)2.\n###### Proof.\n\nConstruct a tournament on vertices by starting with a tournament on vertices that does not contain a and then blow-up one of the vertices to a set of size . Then place edges between all vertices in and orient them arbitrarily. Because every has at least two vertices in , we can cover all of the copies of in by assigning weight to the vertices in and to the vertices in . Therefore,\n\n ν∗(T)=τ∗(T)≤\|X\|2=n−(→r(k)−2)2.\\qed\n\nExample 3.9 is quite similar to Example 3.4. We have verified that, when , equality holds in Example 3.9 when is either or . We have no evidence that equality holds when , and in light of the discussion above, it is, if true, likely extremely challenging to prove!\n\n## 4. The absorbing method and the proof of Theorem 1.2\n\n### 4.1. Absorbing\n\nWe will apply the absorbing method of Rödl, Ruciński and Szemerédi (see e.g. ). The basic idea of the method is to prove that a randomly constructed small set can serve as an “absorber”, i.e., we prove that there exists a small set that has the property that if, after removing this set from the graph, we can almost tile what is left of the oriented graph, then, using the absorbing set, we can extend this partial tiling into a perfect tiling over the entire original oriented graph.\n\nTo prove that our absorbing sets exist, we will use the following lemma, which follows immediately from a lemma of Lo and Markström [10, Lemma 1.1].\n\n###### Lemma 4.1.\n\nFor every , and , there exists such that for every directed graph on vertices the following holds. If, for every , there are at least sets such that and both and contain perfect -tilings, then there exists such that:\n\n• ; is divisible by ; and\n\n• for every , such that and is divisible by , we have that has a perfect -tiling.\n\nLet and . Define to be the set of all with the following property: there exists and so that for each , every -vertex oriented graph with , and any pair , there are at least sets such that and both and contain perfect -tilings. Let be the infinimum of . Write . We call the absorbing threshold for -tiling.\n\nWe will make use of the following simple fact.\n\n###### Fact 4.2.\n\nFor every and such that , and , the following holds. If is a graph or oriented graph on vertices and , then for every such that we have .\n\n###### Proof.\n\nBecause , we have that is at least\n\n \|U\|−(n−δ(G))≥\|U\|−(1r−c)n≥\|U\|−(1r−c)rs\|U\|=(s−1s+c⋅rs)\|U\|.\\qed\n\nFor all , .\n\n###### Proof.\n\nLet , let be sufficiently large and let be an oriented graph on vertices with . Let and set , , and . Since , Fact 4.2 (with ) implies that\n\n δ(G[U])≥(s−1s+ε⋅rs)\|U\|.\n\nSo by supersaturation there exist at least tournaments on vertices in . Since , by the pigeonhole principle, for every such tournament , there exists a subtournament of size at least in one of the four sets: , , , and , which partition . This, in turn, implies that there exists such that and are . Therefore, we have at least of the desired sets.\n\nThe choice of can be made arbitrarily small, thus we obtain that .\n\n###### Lemma 4.4.\n\nFor every , and , there exists such that for every that is divisible by the following holds. If is an oriented graph on vertices and\n\n δ(G)≥max{→δ0(k)+ε,A(k,i)+ε}n,\n\nthen has a perfect -tiling.\n\n###### Proof.\n\nLet . Let be a sufficiently large oriented graph as in the statement of the lemma.\n\nBy the degree condition we may apply Lemma 4.1 to get a set such that , is divisible by , and, for every such that and is divisible by , the oriented graph has a perfect -tiling. Since is sufficiently large and , we can tile so that if is the set of uncovered vertices, then . Since then has a perfect -tiling, we obtain a perfect -tiling of . ∎\n\n### 4.2. Proof of Theorem 1.2\n\nWith the absorbing lemma to hand, it is now straightforward to deduce Theorem 1.2 from our previous results.\n\n###### Proof of Theorem 1.2.\n\nFrom Theorem 3.2 and Lemma 3.8 we have that . Since , the first part of Theorem 1.2 then follows from Lemmas 4.3 and 4.4.\n\nThe second part of the theorem follows by the inequality in the statement of Theorem 3.1. ∎\n\n## 5. →T4-tiling - Proof of Theorem 1.1\n\nNote that . Example 3.4 with () implies the second part of the theorem. For the first part of the theorem, we will show that (Proposition 5.1) and" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9333484,"math_prob":0.9643593,"size":21100,"snap":"2021-21-2021-25","text_gpt3_token_len":4835,"char_repetition_ratio":0.18638605,"word_repetition_ratio":0.117539264,"special_character_ratio":0.22668247,"punctuation_ratio":0.13230482,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984989,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T16:24:25Z\",\"WARC-Record-ID\":\"<urn:uuid:f94a622e-40d7-438c-a6c5-71d06904d893>\",\"Content-Length\":\"1049526\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2afbe55a-e0ea-4ca3-b8ee-2e2fd29cc109>\",\"WARC-Concurrent-To\":\"<urn:uuid:37f7fb9a-3ffd-431e-9aee-e3c19683a6a1>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1906.07648/\",\"WARC-Payload-Digest\":\"sha1:DJZUUL7UBABXRTOZB4YXTWWDP4G5EPEB\",\"WARC-Block-Digest\":\"sha1:4WX7QR65CED33PJA4VUYA745GG7E77WV\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487648373.45_warc_CC-MAIN-20210619142022-20210619172022-00101.warc.gz\"}"}
https://segmentfault.com/a/1190000004426828	[ "", null, "# [译] 数据可视化教程：基于Google Sheets 和 RStudio Shiny 建立实时仪表盘\n\nThanks for Douglas Watson，the original English version is http://douglas-watson.github.io/post/gdocs_1_gdocs\n\n# 概述\n\n• 第二部分（原文）,我将教你如何在R中取回数据,并用ggplot2库做数据可视化。\n\n• 第三部分（原文）,我将带您用 Shiny 制作一个简单的交互式可视化应用，通过 ShinyApps.io 平台发布在网上。", null, "# 为什么使用 R 和 Shiny？\n\nR是一种强大的语言专门为数据分析，结合ggplot2图形库，R便可以做专业的数据可视化。一旦你找到了你想要展示什么,shiny 则让可互动的数据可视化图表在网上发布。Shiny 由 RStudio 公司开发,他们", null, "## 前奏:数据存储格式\n\ntimestamp ID variable value\n1448227096 kitchen temperature 22.3\n1448227096 kitchen humidity 45\n1448227096 bedroom temperature 24.0\n1448227096 bedroom humidity 46\n1448227099 kitchen temperature 22.4\n1448227099 kitchen humidity 45\n1448227099 bedroom temperature 23.9\n1448227099 bedroom humidity 45\n\ntimestamp kitchen temperature kitchen humidity bedroom temperature bedroom humidity\n1448227096 22.3 45 24.0 46\n1448227099 22.4 45 23.9 45\n\n## 准备接收数据的表格\n\n2. 重命名第一张工作表(表中一个标签文档)，这就是数据将被上传到这里。\n\n3. 创建标题行。每一行的数据,将由unix时间戳(1970年1月1日以来的秒数)、“id”、“变量”和“阅读量”构成。你可以冻结的行,保持可见滚动时,用“视图”>“冻结”>“一行”。\n\n4. 打开脚本编辑器,在“工具”>“脚本编辑器…”菜单。", null, "## 用谷歌脚本插入一行数据\n\n``````function appendLines(worksheet, csvData) {\nvar sheet = ss.getSheetByName(worksheet);\n\nvar rows = Utilities.parseCsv(csvData);\n\nfor ( var i = 0; i < rows.length; i++ ) {\nsheet.appendRow(rows[i]);\n}\n}\n\nfunction test() {\nLogger.log(\"Appending fake data\");\nappendLines(\"Raw\", \"12345, Monday, kitchen, temperature, 30\\n12346, Tuesday, living room, humidity, 50\");\n}``````\n\n• 打开当前电子表格(整个文档);\n\n• 在文档中选择一个表,由第一个参数标识(我们目前只有一个 Raw 表);\n\n• 解析CSV数据作为第二个参数;\n\n• 将每一行的CSV数据,追加到表中。", null, "", null, "", null, "``````function appendLines(worksheet, csvData) {\nvar sheet = ss.getSheetByName(worksheet);\n\nvar rows = Utilities.parseCsv(csvData);\n\nfor ( var i = 0; i < rows.length; i++ ) {\nsheet.appendRow(rows[i]);\n}\n}``````\n\n## 接收POST数据\n\n``````function doPost(e) {\nvar params = JSON.stringify(e);\nreturn ContentService.createTextOutput(params);\n}``````", null, "", null, "", null, "``````\\$ curl --data \"hello, world\" \"https://script.google.com/macros/s/AKfycbxOw-Tl_r0jDV4wcnixdYdUjcNipSzgufiezRKr28Q5OAN50cIP/exec\"\n\n{\"parameter\":{\"hello, world\":\"\"},\"contextPath\":\"\",\"contentLength\":12,\"queryString\":null,\"parameters\":{\"hello, world\":[\"\"]},\"postData\":{\"length\":12,\"type\":\"application/x-www-form-urlencoded\",\"contents\":\"hello, world\",\"name\":\"postData\"}}%``````\n\n``````\\$ curl -L --data \"hello, world\" \"https://script.google.com/macros/s/AKfycbxOw-Tl_r0jDV4wcnixdYdUjcNipSzgufiezRKr28Q5OAN50cIP/exec?sheet=Raw\"\n\n{\"parameter\":{\"sheet\":\"Raw\",\"hello, world\":\"\"},\"contextPath\":\"\",\"contentLength\":12,\"queryString\":\"sheet=Raw\",\"parameters\":{\"sheet\":[\"Raw\"],\"hello, world\":[\"\"]},\"postData\":{\"length\":12,\"type\":\"application/x-www-form-urlencoded\",\"contents\":\"hello, world\",\"name\":\"postData\"}}%``````\n\nURL参数出现在 `e[\"parameter\"][\"sheet\"]`\n\n• 在URL中找到一个\"sheet\"参数\n\n• 从POST请求中提取CSV数据\n\n• 附加在CSV数据到指定表的所有行。\n\n``````function doPost(e) {\nvar contents = e.postData.contents;\nvar sheetName = e.parameter['sheet'];\n\nappendLines(sheetName, contents);\n\nvar params = JSON.stringify(e);\nreturn ContentService.createTextOutput(params);\n\n}``````", null, "# 第二部分：在R中获取数据并通过ggplot制图\n\nGitHub上的相关代码\n\n## 推荐设置\n\nRStudio是一个像样的编辑器(还可以支持Vim模式)并且将Shiny集成其中。下载地址\n\n``install.packages('ggplot2')``\n\n## 从web URL导入一个CSV文件\n\n``````getRaw <- function () {\nstrip.white = TRUE\n)\ndata\n}``````\n\n``````> data <- getRaw()\n> summary(data)\ntimestamp date origin variable value\nMin. :1.448e+09 Mon Nov 23 2015 22:44:45 GMT+0100 (CET): 4 bedroom:120 humidity :120 Min. :23.00\n1st Qu.:1.448e+09 Mon Nov 23 2015 23:44:45 GMT+0100 (CET): 4 kitchen:120 temperature:120 1st Qu.:23.88\nMedian :1.448e+09 Thu Nov 26 2015 00:44:45 GMT+0100 (CET): 4 Median :32.55\nMean :1.448e+09 Thu Nov 26 2015 01:44:45 GMT+0100 (CET): 4 Mean :34.34\n3rd Qu.:1.448e+09 Thu Nov 26 2015 02:44:45 GMT+0100 (CET): 4 3rd Qu.:44.45\nMax. :1.449e+09 Thu Nov 26 2015 03:44:45 GMT+0100 (CET): 4 Max. :49.90\n(Other) :216\n>``````\n\n## 转化 date-time 列\n\n``````getRaw <- function () {\nstrip.white = TRUE\n)\ndata\\$date = as.POSIXct(data\\$timestamp, tz=\"GMT\", origin=\"1970-01-01\")\ndata\n}``````\n\n## 与 ggplot2 共舞\n\n`ggplot2包`提供了机智的`qplot` 函数，一行代码可以绘制种类繁多的图表,使R成为我最喜欢的工具数据。\n\n``````import(ggplot2)\nsource(\"helpers.R\")\ndata <- getRaw()\nqplot(date, value, data = data)``````", null, "``qplot(date, value, data = data, colour = origin)``", null, "``qplot(date, value, data = data, colour = origin) + facet_grid(variable ~ ., scales = \"free_y\")``", null, "``qplot(date, value, data = data, colour = origin, geom = \"line\") + facet_grid(variable ~ ., scales = \"free_y\")``", null, "``qplot(date, value, data = data, colour = origin, geom = \"line\") + scale_x_datetime() + facet_grid(variable ~ ., scales = \"free_y\")``\n\n## 为 Shiny 封装画图函数\n\n``````library(ggplot2)\n\ngetRaw <- function () {\nstrip.white = TRUE\n)\n\ndata\\$date = as.POSIXct(data\\$timestamp, tz=\"GMT\", origin=\"1970-01-01\")\ndata\n}\n\ntimeseriesPlot <- function(data) {\nqplot(date, value, data = data, colour = origin, geom = \"line\") + scale_x_datetime() + facet_grid(variable ~ ., scales = \"free_y\")\n}\n\nboxPlot <- function(data) {\nqplot(origin, value, data = data, geom = \"boxplot\") + facet_grid(variable ~ ., scales = \"free_y\")\n}``````\n\n# 第三部分:与Shiny结合\n\n## 我们的第一个仪表板:没有交互性\n\n``````# ui.R\n\nshinyUI(fluidPage(\nverticalLayout(\ntitlePanel(\"Sensor data dashboard\"),\nplotOutput(\"timeseries\"),\nplotOutput(\"boxplot\")\n)\n))``````\n\n`server.R`中的代码在如下时刻会被调用：\n\n1. 当服务器启动时;\n\n2. 每次访问一个页面时;\n\n3. 每次一个交互式的部件改变时。\n\n`shinyServer(...)`代码块之外的代码只会被执行一次，而代码块里面的部分每次访问页面都会被执行。\n\n``````# server.R\n\nsource(\"helpers.R\")\n\nshinyServer(function(input, output) {\n\n# Load data when app is visited\ndata <- getRaw()\n\n# Populate plots\noutput\\$timeseries <- renderPlot({\ntimeseriesPlot(data)\n})\n\noutput\\$boxplot <- renderPlot({\nboxPlot(data)\n})\n\n})``````", null, "", null, "## 添加一个按日期分类的网格布局和过滤器\n\n• `dateRangeInput`:一个下拉日历选择开始和结束日期。\n\n• `numericInput`:一个只接受数字的输入框,显示了一个向上和向下箭头来修改输入值。\n\n``````# ui.R\n\nshinyUI(fluidPage(\nverticalLayout(\ntitlePanel(\"Sensor data dashboard\"),\nfluidRow(\ncolumn(3,\ndateRangeInput(\"dates\", \"Date Range\", start=\"2015-11-20\"),\nfluidRow(\ncolumn(4, h3(\"From:\")),\ncolumn(4, numericInput(\"min.hours\", \"hours:\", value=0)),\ncolumn(4, numericInput(\"min.minutes\", \"minutes:\", value=0))\n),\nfluidRow(\ncolumn(4, h3(\"To:\")),\ncolumn(4, numericInput(\"max.hours\", \"hours:\", value=23)),\ncolumn(4, numericInput(\"max.minutes\", \"minutes:\", value=59))\n)\n),\ncolumn(9, plotOutput(\"timeseries\"))\n),\nfluidRow(\ncolumn(3),\ncolumn(9, plotOutput(\"boxplot\"))\n)\n)\n))``````\n\n``````# server.R\n\nsource(\"helpers.R\")\n\nshinyServer(function(input, output) {\n\n# Load data when app is visited\ndata <- getRaw()\n\n# Filter by device ID / time range when options are updated\ndata.filt <- reactive({\nmindate <- as.POSIXct.Date(input\\$dates) + (input\\$min.hours * 60 + input\\$min.minutes) * 60\nmaxdate <- as.POSIXct.Date(input\\$dates) + (input\\$max.hours * 60 + input\\$max.minutes) * 60\n\nsubset(data, date > mindate & date < maxdate)\n})\n\n# Populate plots\noutput\\$timeseries <- renderPlot({\ntimeseriesPlot(data.filt())\n})\n\noutput\\$boxplot <- renderPlot({\nboxPlot(data.filt())\n})\n\n})``````", null, "## 发布到Shinyapps.io\n\nShinyApps.io 为 Shiny 应用提供一个托管服务，而这个扩管服务同时也是 RStudio 的另一个产品。他们的开发者在 RStudio 中集成了这个服务：简单地在你 Shiny 应用的右上角点击\"发布\"，跟着指引，创建一个免费的账号然后上传你的仪表盘。", null, "", null, "``````# fragment of helpers.R\n\nstrip.white = TRUE\n)``````\n\n## 更多的代理\n\nHTTPS代理是一个驻留在应用程序引擎的简单代码。相关代码\n\n1003 人关注\n59 篇文章" ]	[ null, 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https://www.clutchprep.com/chemistry/practice-problems/85703/a-gas-sample-containing-1-50-moles-at-25-c-exerts-a-pressure-of-400-torr-some-ga	[ "# Problem: A gas sample containing 1.50 moles at 25°C exerts a pressure of 400. torr. Some gas is added to the same container and the temperature is increased to 50.°C. If the pressure increases to 800. torr, how many moles of gas were added to the container? Assume a constant-volume container.\n\n###### FREE Expert Solution\n\nWe will use ideal gas equation:\n\n$\\overline{){\\mathbf{PV}}{\\mathbf{=}}{\\mathbf{nRT}}}$\n\nP = pressure, atm\nV = volume, L\nn = moles, mol\nR = gas constant = 0.08206 (L·atm)/(mol·K)\nT = temperature, K\n\nStep 1. Determine the volume:\n\n91% (364 ratings)", null, "###### Problem Details\n\nA gas sample containing 1.50 moles at 25°C exerts a pressure of 400. torr. Some gas is added to the same container and the temperature is increased to 50.°C. If the pressure increases to 800. torr, how many moles of gas were added to the container? Assume a constant-volume container.\n\nFrequently Asked Questions\n\nWhat scientific concept do you need to know in order to solve this problem?\n\nOur tutors have indicated that to solve this problem you will need to apply the The Ideal Gas Law concept. You can view video lessons to learn The Ideal Gas Law. Or if you need more The Ideal Gas Law practice, you can also practice The Ideal Gas Law practice problems.\n\nWhat professor is this problem relevant for?\n\nBased on our data, we think this problem is relevant for Professor Millhauser's class at UCSC.\n\nWhat textbook is this problem found in?\n\nOur data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems." ]	[ null, "https://cdn.clutchprep.com/assets/button-view-text-solution.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9250691,"math_prob":0.92343014,"size":1254,"snap":"2021-04-2021-17","text_gpt3_token_len":308,"char_repetition_ratio":0.1232,"word_repetition_ratio":0.05479452,"special_character_ratio":0.23524721,"punctuation_ratio":0.11857708,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995975,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T02:07:28Z\",\"WARC-Record-ID\":\"<urn:uuid:260e775f-7e47-42f3-abb6-78d91aeea914>\",\"Content-Length\":\"155028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:454bec7b-db1b-426c-98f9-c039ca29173b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9367c155-2e4f-4fd9-bba6-f33ff7ad4811>\",\"WARC-IP-Address\":\"3.213.190.117\",\"WARC-Target-URI\":\"https://www.clutchprep.com/chemistry/practice-problems/85703/a-gas-sample-containing-1-50-moles-at-25-c-exerts-a-pressure-of-400-torr-some-ga\",\"WARC-Payload-Digest\":\"sha1:YGK6YPLV3ACB2IXY2KC3K27SU5MMKD6U\",\"WARC-Block-Digest\":\"sha1:YYI5MMQLNY7SJQTY4SVTL7LYOV4WU25P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038060603.10_warc_CC-MAIN-20210411000036-20210411030036-00058.warc.gz\"}"}
https://stats.stackexchange.com/questions/211445/bayesian-updating-of-scaled-beta-distribution	[ "# Bayesian updating of scaled beta distribution\n\nI would like to know how to link a scaled beta distribution to its non scaled beta distribution.\n\nFor some context: I have a model where I want to estimate the proportion ($\\pi$) of people belonging to a given group (out of 2 groups) (can see this as the probability that a given event happen).\nTo estimate the true proportion $\\pi$ I would simply do bayesian updating based on a beta prior that $\\pi \\sim Beta(a, b)$, with binomial (bernouilli) likelihood $f(x\|\\pi)$ for the data.\n\nHowever, I do not observe the belonging to one or the other group directly in my data. What I observe from the data is the belonging to two different groups, where the proportion of people belonging to my group of interest is $c \\ \\pi$ (where $c$ is a known constant $\\in [0, 1]$). Hence my data follow a bernouilli likelihood: $c \\pi^x \\ (1-c\\pi)^{1-x}$.\n\nYou probably wonder, why don't I simply put the beta prior on $c \\pi$ instead and then just divide the expectation to find an estimation of $\\pi$? I cannot because $c$ can vary every period (and because I want $\\pi$ to belong to $[0,1]$). Given that $c$ can vary each period, I want to update my prior on $\\pi$ and not on $c \\pi$, and do the link between the two no matter what $c$ is (and anyway, my parameter of interest is actually $\\pi$).\n\nI am quite new to bayesian updating of binomial proportion and Beta distribution in general, and I was not able to find an answer to this. I know that if $\\pi \\sim Beta(a,b)$, then $c \\pi$ is not a beta distribution anymore but a \"scaled beta distribution\" (is there any other name? I do not care about the whole generalized beta distribution, I just want the additional scale parameter!).\nTo solve my problem I just need to know how to link two scaled beta distribution (with the same a and b) (actually, how to link a scaled beta distribution to its version with scale = 1).\n\nFrom what I tried so far, I am stuck because I end up with: $something \\times (c \\pi)^x (1-c\\pi)^{1-x} \\pi^{a-1} (1-\\pi)^{b-1}$ which does not allow me to do simple updating and say for example that it is obviously another beta distribution (as we do in the bayesian updating classic case).\n\nI do not see an easy factorization here, but maybe I just miss a simple maths trick? or maybe my formula is even wrong?\nMy main question is, is there an easy way to update \"scaled beta distribution\" (or beta of the first kind) or do I have to do it numerically?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91909856,"math_prob":0.9652399,"size":2409,"snap":"2019-35-2019-39","text_gpt3_token_len":612,"char_repetition_ratio":0.13180873,"word_repetition_ratio":0.027334852,"special_character_ratio":0.2619344,"punctuation_ratio":0.078,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99938786,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-17T23:36:18Z\",\"WARC-Record-ID\":\"<urn:uuid:623dee47-b968-4a10-9798-b1a4b00ca1dd>\",\"Content-Length\":\"128011\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7adaddb2-8e89-493c-a0a3-25cccd815197>\",\"WARC-Concurrent-To\":\"<urn:uuid:f62e0244-b077-4442-9c74-27e10b06f589>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/211445/bayesian-updating-of-scaled-beta-distribution\",\"WARC-Payload-Digest\":\"sha1:ZJHECJCZCPINPL4WJA32LQPTXVGORASD\",\"WARC-Block-Digest\":\"sha1:C3RYSTVJKBH7YEU4CYVL26U52M3IXLWO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313501.0_warc_CC-MAIN-20190817222907-20190818004907-00322.warc.gz\"}"}
https://www.numbersaplenty.com/11111111110	[ "Search a number\nBaseRepresentation\nbin10100101100100011…\n…00001100111000110\n31001200100111221000201\n422112101201213012\n5140223421023420\n65034313401114\n7542225612641\noct122621414706\n931610457021\n1011111111110\n114791a29260\n1221a110919a\n131080c5b678\n14775954a58\n154506db10a\nhex2964619c6\n\n11111111110 has 64 divisors (see below), whose sum is σ = 22435273728. Its totient is φ = 3926880000.\n\nThe previous prime is 11111111059. The next prime is 11111111113. The reversal of 11111111110 is 1111111111.\n\nIt is a happy number.\n\nIt is a Harshad number since it is a multiple of its sum of digits (10).\n\nIt is a nialpdrome in base 10.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (11111111113) by changing a digit.\n\nIt is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 1217665 + ... + 1226755.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (350551152).\n\nAlmost surely, 211111111110 is an apocalyptic number.\n\n11111111110 is a gapful number since it is divisible by the number (10) formed by its first and last digit.\n\n11111111110 is an abundant number, since it is smaller than the sum of its proper divisors (11324162618).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n11111111110 is a wasteful number, since it uses less digits than its factorization.\n\n11111111110 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 9421.\n\nThe product of its (nonzero) digits is 1, while the sum is 10.\n\n11111111110 divided by its sum of digits (10) gives a palindrome (1111111111).\n\nAdding to 11111111110 its reverse (1111111111), we get a palindrome (12222222221).\n\nSubtracting from 11111111110 its reverse (1111111111), we obtain a palindrome (9999999999).\n\nIt can be divided in two parts, 111111 and 11110, that added together give a palindrome (122221).\n\nThe spelling of 11111111110 in words is \"eleven billion, one hundred eleven million, one hundred eleven thousand, one hundred ten\"." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8238384,"math_prob":0.9676017,"size":2212,"snap":"2022-27-2022-33","text_gpt3_token_len":691,"char_repetition_ratio":0.19202898,"word_repetition_ratio":0.0056338026,"special_character_ratio":0.45705244,"punctuation_ratio":0.12686567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99714094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T00:45:29Z\",\"WARC-Record-ID\":\"<urn:uuid:8c92374c-eaf5-4e9f-8c8c-8b5159ea0e57>\",\"Content-Length\":\"11279\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:396c2ee2-5d06-4136-8cbd-193a320fcadb>\",\"WARC-Concurrent-To\":\"<urn:uuid:440a231b-c604-4f0b-b1d0-beb4e0a7e50d>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/11111111110\",\"WARC-Payload-Digest\":\"sha1:NBRTG664GMYJWGMXQ5I36KRHIMRTJTTL\",\"WARC-Block-Digest\":\"sha1:XIFVOGETSBZYGY3RBJFXEWMQDLC7TP65\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570741.21_warc_CC-MAIN-20220808001418-20220808031418-00572.warc.gz\"}"}
https://www.omnicalculator.com/math/pyramid-angle	[ "# Pyramid Angle Calculator\n\nCreated by Davide Borchia\nReviewed by Anna Szczepanek, PhD and Jack Bowater\nLast updated: Feb 02, 2023\n\nUse our pyramid angle calculator to find all the possible angles in a pyramid - we'll ask you a few parameters, and do all the math for you!\n\nHere you will learn:\n\n• The types of angles in a pyramid;\n• How to calculate the angles of a pyramid;\n• An example of how to calculate the square pyramid angles; and\n• How to use our pyramid angle calculator.\n\n## What is a pyramid?\n\nA pyramid is a solid figure with a polygonal base. A triangular face corresponds to each side of the base, joining at the apex.\n\nThe base can be of any shape; however, regular pyramids are easier to study. The base of a regular pyramid is a regular polygon. We can find, for example:\n\n• Triangular pyramids, with an equilateral triangle as the base;\n• Square pyramids, where the base is, of course, a square;\n• Hexagonal pyramids; and\n• Well, infinitely more.\n\nIf the apex lies above the centroid of the base (the geometric mean point of the polygon), we call the pyramid a right pyramid; otherwise, we have an oblique pyramid. Here we will analyze only right pyramids.\n\n💡 We have many pyramid calculators here at Omni: check out our pyramid volume calculator, our right rectangular pyramid calculator, and our triangular pyramid volume calculator!\n\n## The angles of a pyramid\n\nThere are a lot of angles in a pyramid, but luckily in a regular pyramid, most of them are identical. Let's identify them!\n\nWe can identify the angle between the faces' vertical medians and the base at the apex. It defines how \"slender\" a pyramid is. We call this angle $\\alpha$.\n\nThe angle between the edge and the base $\\beta$ is always smaller than $\\alpha$ in pyramids with a convex base: the corner of the base is farther from the centroid than the center of its sides.\n\nWe can identify other angles, also related to the pyramid's height. They lie on each face, and since in a regular pyramid a face is an isosceles triangle, the two angles at the bottom are identical. We identify them with the letter $\\gamma$. The angle on the top, at the apex, gets smaller for tall and slender pyramids. We call it $\\delta$.\n\nWe highlighted $\\alpha$ and $\\beta$ in a hexagonal pyramid (because ).\n\nNow you know which angles to search for. It's time to learn how to calculate the angles in a pyramid!\n\n## How to calculate the pyramid angles\n\nTo find the angles we need to use a bit of trigonometry - in particular the theorems to find the elements of a right triangle.\n\nLet's start with the angle $\\alpha$, between the median of a face and the base. Let's take a look at the diagram, now marked with some relevant points.\n\n$\\alpha$ corresponds to the angle $\\text{C}\\widehat{\\text{M}}\\text{O}$. We compute it noting that the catheti of the triangles are $\\text{OC}$ (the height of the pyramid) and $\\text{CM}$ (the segment from the side's midpoint to the centroid). We calculate the angle $\\alpha$ with:\n\n$\\footnotesize \\tan{\\alpha} = \\frac{\\text{OC}}{\\text{CM}} \\ \\rightarrow \\ \\alpha = \\arctan{\\left(\\frac{\\text{OC}}{\\text{CM}}\\right)}$\n\nLet's proceed with the angle $\\beta$. In this case, we need to use the segment $\\text{AC}$ to compute the tangent:\n\n$\\footnotesize \\tan{\\beta} = \\frac{\\text{OC}}{\\text{AC}}\\ \\rightarrow\\ \\beta= \\arctan{\\left(\\frac{\\text{OC}}{\\text{AC}}\\right)}$\n\n🔎 Note that $\\text{AC}>\\text{CM}$ is always true: this explains why $\\beta<\\alpha$.\n\nIt's time to compute the angle on a pyramid's face. Using the old and reliable Pythagorean theorem, we find the length of the slanted side of the pyramid:\n\n🙋 To analyze these angles, we move to the other side of the base! Take another look at the diagram if you have trouble visualizing the equations.\n\n$\\text{PO}=\\sqrt{\\text{OC}^2+\\text{PC}^2}$\n\nOn each face, we can identify a pair of right triangles created by the median (in this case $\\text{ON}$). The segment $\\text{PN}$ (half the length of the base's side) and the slanted side of the pyramid $\\text{ON}$ isolate the angle $\\gamma$ ($\\text{O}\\widehat{\\text{P}}\\text{N}$). We use another right triangle identity to calculate it:\n\n$\\footnotesize \\cos{\\gamma}=\\frac{\\text{PN}}{\\text{ON}}\\ \\rightarrow \\ \\gamma=\\arccos{\\left(\\frac{\\text{PN}}{\\text{ON}}\\right)}$\n\nThe last angle, $\\delta$, can be computed using trigonometry again, or if laziness is allowed, by considering that the sum of the internal angles of a triangle is $180\\degree$.\n\n$\\delta = 180\\degree - 2\\times\\gamma$\n\n## How to calculate the angles of a square pyramid\n\nWe will guide you step by step in calculating the angles of a right square pyramid. In fact, not of a general square pyramid - let's calculate the angles of the Great Pyramid of Giza!\n\nWe need to take some measurements:\n\n• The original height of the pyramid is $146.7\\ \\text{m}$; and\n• The side measures $230.6\\ \\text{m}$.\n\nLet's calculate the segment connecting the midpoint of the side to the center. Since the base is a square, it's value is half the length of the side:\n\n$\\footnotesize\\text{CM}=\\frac{AB}{2}=\\frac{230.6 \\ \\text{m}}{2}=115.3\\ \\text{m}$\n\nWe can calculate the angle $\\alpha$:\n\n\\begin{align} \\footnotesize\\alpha & \\footnotesize = \\arctan{\\left(\\frac{\\text{OC}}{\\text{CM}}\\right)} =\\\\ &\\footnotesize=\\arctan{\\left(\\frac{146.7\\ \\text{m}}{115.3\\ \\text{m}}\\right)} = 51.83\\degree \\end{align}\n\nTo calculate the angle in the corner of the base, we need the measure of half its diagonal $\\beta$.\n\n\\begin{align} \\footnotesize\\beta & \\footnotesize = \\arctan{\\left(\\frac{\\text{OC}}{\\text{AC}}\\right)} =\\\\ &\\footnotesize=\\arctan{\\left(\\frac{146.7\\ \\text{m}}{115.3\\times\\sqrt{2}\\ \\text{m}}\\right)} = 41.98\\degree \\end{align}\n\n🙋 Ancient Egyptians measured the slope of a right pyramid using the seked, a unit corresponding to the numbers of horizontal cubits corresponding to a rise of one cubit in height. The base angle of the Great Pyramid of Giza has a seked of $5\\tfrac{1}{2}$ sekeds.\n\nWhat about the angle on each face?\n\nWe can use the formulas above to find the value of $\\gamma$ and $\\delta$; however, we need to calculate the length of the slanted side first:\n\n\\begin{align} \\footnotesize \\text{OB} & \\footnotesize =\\sqrt{\\text{CO}^2+\\text{AC}^2} \\\\ & \\footnotesize = \\sqrt{146.7^2 + 2\\times(115.3)^2}\\\\ &\\footnotesize = 219.3\\ \\text{m} \\end{align}", null, "Egyptians and mathematics went hand in hand. However, we are sure that they weren't marking angles with Greek letters!\n\nLet's proceed with the calculations.\n\n\\begin{align} \\footnotesize\\gamma & \\footnotesize = \\arccos{\\left(\\frac{\\text{PN}}{\\text{ON}}\\right)} =\\\\ &\\footnotesize=\\arccos{\\left(\\frac{115.3\\ \\text{m}}{219.3\\ \\text{m}}\\right)}= 58.29\\degree \\end{align}\n\nFinally, the angle of a face at the apex of the pyramid, $\\delta$, is:\n\n\\begin{align} \\footnotesize\\delta & \\footnotesize = 180\\degree - 2\\times\\gamma=\\\\ &\\footnotesize = 180\\degree-116.58=63.42\\degree \\end{align}\n\n## How to use our pyramid angle calculator\n\nOur pyramid angle calculator can help you with many types of regular right pyramids. Select the type of base you need; we included:\n\n• Square;\n• Regular pentagon;\n• Regular hexagon;\n• Regular heptagon; and\n• Regular octagon.\n\nInsert the available measurements in the calculator, and find the results!\n\n🙋 Our tools work in reverse, too. For example, you can insert the value of the base angles and find out the pyramid's height!\n\nDo you want to expand your pyramidal knowledge outside of geometry? Visit our Minecraft's pyramid block calculator!\n\n## FAQ\n\n### What is the best angle for pyramid power?\n\nPyramids are surrounded by mysticism and occultism. The truth is that there's no such thing as pyramid power, hidden purposes, and alien involvement. Pyramids, and in particular ancient pyramids, are neat but not mystic! The math behind them, however, is all true!\n\n### What is the angle of a hexagonal pyramid with side 2 and height 3?\n\nTo calculate the angle at the base of a hexagonal pyramid, follow these steps:\n\n1. Calculate the length of the segment MC connecting the side's midpoint to the centroid.\n2. Apply the inverse tangent function to the ratio of the height and the length just calculated: OC/MC.\n\nFor a hexagon with side 2 and height 3, it means:\n\n• MC = 2 × cos(30°) = 2 × (sqrt(3)/2) = sqrt(3)\n• α = arctan(3/sqrt(3)) = arctan(sqrt(3)) = 60°\n\n### How do I calculate the angles at the base of a square pyramid?\n\nTo calculate the angles at the base of a pyramid, you can use the trigonometric formulas of right triangles. Calling the height h and the side L, the value of the base angle is:\nα = arctan(h/(L/2))\nWe calculate the angle in the corner of the base using half the diagonal:\nß = arctan(h/(sqrt(2) × L/2)) = arctan(h/(L/sqrt(2)))\n\n### What are the angles of the Great Pyramid of Giza?\n\nThe Great Pyramid of Giza has a base angle of 51.83°. We can calculate it using the inverse trigonometric function arctangent, knowing the height and the side of the pyramid:\nα = arctan(146.7/115.3) = 51.83°\nThe angle in the corner of the base is slightly smaller: 41.98°.\n\nDavide Borchia", null, "Base and measurements\nBase polygon\nSquare\nSegment MC\nSide (AB)\nHeight (OC)\nSlanted side (BO)\nSlanted height (MO)\nAngles\nAlpha (α)\ndeg\nBeta (ß)\ndeg\nGamma (γ)\ndeg\nDelta (δ)\ndeg\nPeople also viewed…\n\n### Quotient\n\nThe quotient calculator will help you divide two numbers and return both fractional and remainder results.\n\n### Secretary problem (Valentine's day)\n\nUse the dating theory calculator to enhance your chances of picking the best lifetime partner.\n\n### Sequence\n\nWith Omni's sequence calculator, you can find the terms of many types of sequences starting from any point!\n\n### Steps to calories\n\nSteps to calories calculator helps you to estimate the total amount to calories burned while walking.", null, "" ]	[ null, "https://uploads-cdn.omnicalculator.com/images/pyramid_angle_02.png", null, "https://uploads-cdn.omnicalculator.com/images/pyrang_4_calc.png", null, "data:image/svg+xml;base64,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", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8701223,"math_prob":0.9981666,"size":6758,"snap":"2023-14-2023-23","text_gpt3_token_len":1587,"char_repetition_ratio":0.22016583,"word_repetition_ratio":0.04622871,"special_character_ratio":0.23024563,"punctuation_ratio":0.11760409,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99983686,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T01:46:59Z\",\"WARC-Record-ID\":\"<urn:uuid:e4b84b3e-4753-4d3c-96e7-1440ebe95bc8>\",\"Content-Length\":\"747698\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:89ab62e3-5533-4287-ae16-7805f4bf9baf>\",\"WARC-Concurrent-To\":\"<urn:uuid:efe395c7-1ba6-4c07-b29c-78d1eb958920>\",\"WARC-IP-Address\":\"68.70.205.1\",\"WARC-Target-URI\":\"https://www.omnicalculator.com/math/pyramid-angle\",\"WARC-Payload-Digest\":\"sha1:2HGV4CAKRD2U3UP4CKJGD7CCMUN5EJP7\",\"WARC-Block-Digest\":\"sha1:QSGRZDLOCLVDCJSLHNUFLI44BUAB46E3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649348.41_warc_CC-MAIN-20230603233121-20230604023121-00489.warc.gz\"}"}
http://ixtrieve.fh-koeln.de/birds/litie/document/22887	[ "# Document (#22887)\n\nAuthor\nBinder, G.\nStahl, M.\nFaulborn, L.\nTitle\nVergleichsuntersuchung MESSENGER-FULCRUM\nImprint\nBonn : IZ Sozialwissenschaften\nYear\n2000\nPages\n70 S\nSeries\nIZ-Arbeitsbericht; Nr.18\nAbstract\nIn einem Benutzertest, der im Rahmen der Projektes GIRT stattfand, wurde die Leistungsfähigkeit zweier Retrievalsprachen für die Datenbankrecherche überprüft. Die Ergebnisse werden in diesem Bericht dargestellt: Das System FULCRUM beruht auf automatischer Indexierung und liefert ein nach statistischer Relevanz sortiertes Suchergebnis. Die Standardfreitextsuche des Systems MESSENGER wurde um die intellektuell vom IZ vergebenen Deskriptoren ergänzt. Die Ergebnisse zeigen, dass in FULCRUM das Boole'sche Exakt-Match-Retrieval dem Verktos-Space-Modell (Best-Match-Verfahren) von den Versuchspersonen vorgezogen wurde. Die in MESSENGER realisierte Mischform aus intellektueller und automatischer Indexierung erwies sich gegenüber dem quantitativ-statistischen Ansatz beim Recall als überlegen\nFootnote\nAls PDF-Datei auch unter: http://www.bonn.iz-soz.de/publications/series/working-papers/\nTheme\nRetrievalstudien\nBenutzerstudien\nObject\nMessenger\nFULCRUM\n\n## Similar documents (author)\n\n1. 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Stahl, J.N.: Using the interest to access CARL and other electronic information systems (1990) 2.38\n```2.376252 = sum of:\n2.376252 = product of:\n4.752504 = sum of:\n4.752504 = weight(author_txt:stahl in 4125) [ClassicSimilarity], result of:\n4.752504 = score(doc=4125,freq=1.0), product of:\n0.7825483 = queryWeight, product of:\n1.1211264 = boost\n9.71698 = idf(docFreq=6, maxDocs=42740)\n0.07183321 = queryNorm\n6.0731125 = fieldWeight in 4125, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.71698 = idf(docFreq=6, maxDocs=42740)\n0.625 = fieldNorm(doc=4125)\n0.5 = coord(1/2)\n```\n3. Stahl, J.: 'Have a question? Click here' : electronic reference at the National Museum of American Art (1998) 2.38\n```2.376252 = sum of:\n2.376252 = product of:\n4.752504 = sum of:\n4.752504 = weight(author_txt:stahl in 3495) [ClassicSimilarity], result of:\n4.752504 = score(doc=3495,freq=1.0), product of:\n0.7825483 = queryWeight, product of:\n1.1211264 = boost\n9.71698 = idf(docFreq=6, maxDocs=42740)\n0.07183321 = queryNorm\n6.0731125 = fieldWeight in 3495, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.71698 = idf(docFreq=6, maxDocs=42740)\n0.625 = fieldNorm(doc=3495)\n0.5 = coord(1/2)\n```\n4. Stahl, G.: Group cognition : computer support for building collaborative knowledge (2006) 2.38\n```2.376252 = sum of:\n2.376252 = product of:\n4.752504 = sum of:\n4.752504 = weight(author_txt:stahl in 4392) [ClassicSimilarity], result of:\n4.752504 = score(doc=4392,freq=1.0), product of:\n0.7825483 = queryWeight, product of:\n1.1211264 = boost\n9.71698 = idf(docFreq=6, maxDocs=42740)\n0.07183321 = queryNorm\n6.0731125 = fieldWeight in 4392, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.71698 = idf(docFreq=6, maxDocs=42740)\n0.625 = fieldNorm(doc=4392)\n0.5 = coord(1/2)\n```\n5. Binder, W.: Online-Dienste, Telekopie - Gegenwart und Zukunft elektronischer Informationssystemem für Wissenschaft und Bibliotheken (1986) 1.69\n```1.6862763 = sum of:\n1.6862763 = product of:\n3.3725526 = sum of:\n3.3725526 = weight(author_txt:binder in 528) [ClassicSimilarity], result of:\n3.3725526 = score(doc=528,freq=1.0), product of:\n0.62258977 = queryWeight, product of:\n8.667158 = idf(docFreq=19, maxDocs=42740)\n0.07183321 = queryNorm\n5.416974 = fieldWeight in 528, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.667158 = idf(docFreq=19, maxDocs=42740)\n0.625 = fieldNorm(doc=528)\n0.5 = coord(1/2)\n```\n\n## Similar documents (content)\n\n1. 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Kempf, A.O.: Automatische Inhaltserschließung in der Fachinformation (2013) 0.07\n```0.069561236 = sum of:\n0.069561236 = product of:\n0.34780616 = sum of:\n0.05790168 = weight(abstract_txt:intellektuell in 2906) [ClassicSimilarity], result of:\n0.05790168 = score(doc=2906,freq=1.0), product of:\n0.115438215 = queryWeight, product of:\n1.0272071 = boost\n8.025305 = idf(docFreq=37, maxDocs=42740)\n0.014003289 = queryNorm\n0.50158155 = fieldWeight in 2906, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.025305 = idf(docFreq=37, maxDocs=42740)\n0.0625 = fieldNorm(doc=2906)\n0.037395917 = weight(abstract_txt:ergebnisse in 2906) [ClassicSimilarity], result of:\n0.037395917 = score(doc=2906,freq=1.0), product of:\n0.10867132 = queryWeight, product of:\n1.4094695 = boost\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.014003289 = queryNorm\n0.34411946 = fieldWeight in 2906, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5059114 = idf(docFreq=471, maxDocs=42740)\n0.0625 = fieldNorm(doc=2906)\n0.06891176 = weight(abstract_txt:indexierung in 2906) [ClassicSimilarity], result of:\n0.06891176 = score(doc=2906,freq=1.0), product of:\n0.16334069 = queryWeight, product of:\n1.7280066 = boost\n6.7502356 = idf(docFreq=135, maxDocs=42740)\n0.014003289 = queryNorm\n0.42188972 = fieldWeight in 2906, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.7502356 = idf(docFreq=135, maxDocs=42740)\n0.0625 = fieldNorm(doc=2906)\n0.052360427 = weight(abstract_txt:wurde in 2906) [ClassicSimilarity], result of:\n0.052360427 = score(doc=2906,freq=2.0), product of:\n0.12357237 = queryWeight, product of:\n1.8407912 = boost\n4.793876 = idf(docFreq=961, maxDocs=42740)\n0.014003289 = queryNorm\n0.42372277 = fieldWeight in 2906, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.793876 = idf(docFreq=961, maxDocs=42740)\n0.0625 = fieldNorm(doc=2906)\n0.13123637 = weight(abstract_txt:automatischer in 2906) [ClassicSimilarity], result of:\n0.13123637 = score(doc=2906,freq=1.0), product of:\n0.25095835 = queryWeight, product of:\n2.1418993 = boost\n8.367054 = idf(docFreq=26, maxDocs=42740)\n0.014003289 = queryNorm\n0.5229409 = fieldWeight in 2906, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.367054 = idf(docFreq=26, maxDocs=42740)\n0.0625 = fieldNorm(doc=2906)\n0.2 = coord(5/25)\n```\n4. Bunk, T.: Deskriptoren Stoppwortlisten und kryptische Zeichen (2008) 0.07\n```0.067412905 = sum of:\n0.067412905 = product of:\n0.5617742 = sum of:\n0.13579656 = weight(abstract_txt:deskriptoren in 4472) [ClassicSimilarity], result of:\n0.13579656 = score(doc=4472,freq=1.0), product of:\n0.11062535 = queryWeight, product of:\n1.0055659 = boost\n7.856228 = idf(docFreq=44, maxDocs=42740)\n0.014003289 = queryNorm\n1.2275356 = fieldWeight in 4472, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.856228 = idf(docFreq=44, maxDocs=42740)\n0.15625 = fieldNorm(doc=4472)\n0.18233782 = weight(abstract_txt:intellektueller in 4472) [ClassicSimilarity], result of:\n0.18233782 = score(doc=4472,freq=1.0), product of:\n0.13464183 = queryWeight, product of:\n1.1093618 = boost\n8.667158 = idf(docFreq=19, maxDocs=42740)\n0.014003289 = queryNorm\n1.3542435 = fieldWeight in 4472, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.667158 = idf(docFreq=19, maxDocs=42740)\n0.15625 = fieldNorm(doc=4472)\n0.24363983 = weight(abstract_txt:indexierung in 4472) [ClassicSimilarity], result of:\n0.24363983 = score(doc=4472,freq=2.0), product of:\n0.16334069 = queryWeight, product of:\n1.7280066 = boost\n6.7502356 = idf(docFreq=135, maxDocs=42740)\n0.014003289 = queryNorm\n1.4916053 = fieldWeight in 4472, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n6.7502356 = idf(docFreq=135, maxDocs=42740)\n0.15625 = fieldNorm(doc=4472)\n0.12 = coord(3/25)\n```\n5. Stock, W.G.: Natürlichsprachige Suche - more like this! : Lexis-Nexis' Freestyle (1998) 0.06\n```0.056921408 = sum of:\n0.056921408 = product of:\n0.4743451 = sum of:\n0.13782352 = weight(abstract_txt:indexierung in 3524) [ClassicSimilarity], result of:\n0.13782352 = score(doc=3524,freq=1.0), product of:\n0.16334069 = queryWeight, product of:\n1.7280066 = boost\n6.7502356 = idf(docFreq=135, maxDocs=42740)\n0.014003289 = queryNorm\n0.84377944 = fieldWeight in 3524, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.7502356 = idf(docFreq=135, maxDocs=42740)\n0.125 = fieldNorm(doc=3524)\n0.07404883 = weight(abstract_txt:wurde in 3524) [ClassicSimilarity], result of:\n0.07404883 = score(doc=3524,freq=1.0), product of:\n0.12357237 = queryWeight, product of:\n1.8407912 = boost\n4.793876 = idf(docFreq=961, maxDocs=42740)\n0.014003289 = queryNorm\n0.5992345 = fieldWeight in 3524, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.793876 = idf(docFreq=961, maxDocs=42740)\n0.125 = fieldNorm(doc=3524)\n0.26247275 = weight(abstract_txt:automatischer in 3524) [ClassicSimilarity], result of:\n0.26247275 = score(doc=3524,freq=1.0), product of:\n0.25095835 = queryWeight, product of:\n2.1418993 = boost\n8.367054 = idf(docFreq=26, maxDocs=42740)\n0.014003289 = queryNorm\n1.0458817 = fieldWeight in 3524, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.367054 = idf(docFreq=26, maxDocs=42740)\n0.125 = fieldNorm(doc=3524)\n0.12 = coord(3/25)\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6416977,"math_prob":0.9968248,"size":13450,"snap":"2021-04-2021-17","text_gpt3_token_len":5167,"char_repetition_ratio":0.24445932,"word_repetition_ratio":0.4647551,"special_character_ratio":0.52297395,"punctuation_ratio":0.2819115,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99973243,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-14T20:10:56Z\",\"WARC-Record-ID\":\"<urn:uuid:8fbab85b-da53-4baa-8e3c-eb13db77e0b7>\",\"Content-Length\":\"25879\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:00ce337d-b422-488d-802d-2fe0bb2a80f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:971034c7-4ae1-4a68-b68b-65953253cd57>\",\"WARC-IP-Address\":\"139.6.160.6\",\"WARC-Target-URI\":\"http://ixtrieve.fh-koeln.de/birds/litie/document/22887\",\"WARC-Payload-Digest\":\"sha1:LSAFCEHHLNKTBQOBFQALBG3KCF7LPQJQ\",\"WARC-Block-Digest\":\"sha1:XOT66WXH7CDUNTIFNA5FFXXNKH7SQBQY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038078021.18_warc_CC-MAIN-20210414185709-20210414215709-00625.warc.gz\"}"}
http://docs.groovy-lang.org/docs/groovy-2.5.4/html/gapi/groovy/transform/stc/FromString.html	[ "Package: groovy.transform.stc\n\n[Java] Class FromString\n\n• groovy.transform.stc.FromString\n• A closure parameter hint class that is convenient if you want to use a String representation of the signature. It makes use of the option strings, where each string corresponds to a single signature.\n\nThe following example describes a closure as accepting a single signature (List<T> list ->):\n\npublic <T> T apply(T src, @ClosureParams(value=FromString.class, options=\"List<T>\" Closure<T> cl)\n\nThe next example describes a closure as accepting two signatures (List<T> list ->) and (T t ->):\n\npublic <T> T apply(T src, @ClosureParams(value=FromString.class, options={\"List<T>\",\"T\"} Closure<T> cl)\n\nIt is advisable not to use this hint as a replacement for the various FirstParam, SimpleType, ... hints because it is actually much slower. Using this hint should therefore be limited to cases where it's not possible to express the signature using the existing hints.\n\nAuthors:\nCédric Champeau\nSince:\n2.3.0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6209786,"math_prob":0.63677394,"size":1340,"snap":"2022-05-2022-21","text_gpt3_token_len":296,"char_repetition_ratio":0.11601797,"word_repetition_ratio":0.058479533,"special_character_ratio":0.21119402,"punctuation_ratio":0.18025751,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9546683,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-23T16:05:58Z\",\"WARC-Record-ID\":\"<urn:uuid:5af216e9-ff21-4a87-93da-173a9c031fb1>\",\"Content-Length\":\"15056\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e696d5d8-2552-4e8b-808c-dc7346e81175>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee144e2b-3b29-4145-a9d4-437335d80860>\",\"WARC-IP-Address\":\"148.251.23.208\",\"WARC-Target-URI\":\"http://docs.groovy-lang.org/docs/groovy-2.5.4/html/gapi/groovy/transform/stc/FromString.html\",\"WARC-Payload-Digest\":\"sha1:VQHEAQPWB7I2SFBFPEXZOW3QZEEMBCUH\",\"WARC-Block-Digest\":\"sha1:P2RNZE676QAWL2VLLAQ3BGPQJHYHGUJZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304287.0_warc_CC-MAIN-20220123141754-20220123171754-00078.warc.gz\"}"}
https://www.codespeedy.com/association-rule-mining-in-python/	[ "# Association Rule Mining in Python\n\nHello everyone, In this tutorial, we’ll be learning about Association Rule Mining in Python (ARM) and will do a hands-on practice on a dataset. We will use the apriori algorithm and look on the components of the apriori algorithm. Let us start this tutorial with a brief introduction to association rules mining.\n\n## What is Association Rule Mining and its benefits?\n\nAssociation Rule Mining is a process that uses Machine learning to analyze the data for the patterns, the co-occurrence and the relationship between different attributes or items of the data set. In the real-world, Association Rules mining is useful in Python as well as in other programming languages for item clustering, store layout, and market basket analysis.\n\nAssociation rules include two parts, an antecedent (if) and a consequent (then) that is the if-then association that occurs more frequently in the dataset.\n\nFor example, {Bread} => {Milk} can be an association in a supermarket store. This relation implies that if(antecedent) a person buys Bread then(consequent) most probably the customer will buy Milk. There can be lots of relations between several itemsets that can be used to make the layout of the store. With this, customers would not require to go far to look for every product. To increase sales of the store these products can have combined discounts and there are many other ways these associations are helpful.\n\nFor this tutorial, we’ll be using a dataset that contains a list of 20 orders including the name of order items. You can download the dataset by clicking here. The dataset will look like this.", null, "There are many algorithms that use association rules like AIS, SETM, Apriori, etc. Apriori algorithm is the most widely used algorithm that uses association rules and we will use this in our code. Now let us import the necessary modules and modify our dataset to make it usable.\n\n## Importing and Modifications in the Dataset\n\nHere we are going to understand association rule mining with the help of apyori Python library. So let’s continue reading…\n\nInstall the apyori library using the command line by running the following pip command.\n\npip install apyori\nimport matplotlib.pyplot as plt\nimport pandas as pd\nimport numpy as np\nfrom apyori import apriori\n\nNow, let us import the data and apply some modifications to the data. Go through the code below.\n\ndata = pd.read_csv(r\"D:\\datasets(june)\\order_data.csv\",delimiter=\" \",header=None)\ndata.head()\n\nThe parameter delimiter=” “ will split the entries of the data whenever whitespace is encountered and header=None will prevent taking the first row as the header and a default header will be there. After this, our data frame will look like this.", null, "Let us see some Components of the Apriori Algorithm that are necessary to understand to make a good model.\n\n## Components of the Apriori Algorithm\n\nThere are three main components of an Apriori Algorithm which are as follows:\n\n• Support – It is the measure of the popularity of an itemset that is in how many transactions an item appears from the total number of transactions. It is simply the probability that a customer will buy an item. The mathematical formula to represent support of item X is\nS(X)=(Number of transaction in which X appears)/(Total number of transactions)\n##### Calculating the support value for {Bread} in our dataset\n\nNo. of transactions in which Bread appears = 11\n\nNo. of total transactions = 20\n\n• Minimum Support Value = It is a threshold value above which the product can have a meaningful effect on the profit.\n• Confidence – It tells us the impact of one product on another that is the probability that if a person buys product X then he/she will buy product Y also. Its representation in mathematical terms is\nConfidence({X} => {Y}) = (Transactions containing both X and Y)/(Transactions containing X)\n##### Calculating the Confidence ({Bread} => {Milk}) in our dataset\n\nNo. of transactions in which both Bread and Milk appears = 5\n\nNo. of transactions containing Bread = 11\n\nConfidence ({Bread} => {Milk}) = 5/11 = 0.4545\n\nA major drawback of the confidence is that it only considers the popularity of item X and not of Y. This can decrease the confidence value and therefore can be misleading in understanding the association between different products. To overcome this drawback we have another measure known as Lift.\n\n• Lift – Overcoming the limitation of confidence measure, Lift will calculate the confidence taking into account the popularity of both items. Representation of lift in mathematical terms is\nLift({X} => {Y}) = Confience({X} => {Y}) / Support(B)\n\nIf the lift measure is greater than 1, it means that the Y is likely to be bought with X, while a value less than 1 indicates that Y is unlikely to be bought with X. A lift value of near 1 indicates that both the itemsets in the transactions are appearing often together but there is no association between them.\n\n##### Calculating the Lift({Bread} => {Milk}) in our dataset\n\nConfidence ({Bread} => {Milk}) = 0.4545\n\nSupport (Milk) = 9/20 = 0.45\n\nLift({Bread} => {Milk}) = 0.4545/0.45 = 1.01\n\n## Practical Implemenation of Apriori Algorithm\n\nUsing the data-set that we have downloaded in the previous section, let us write some code and calculate the values of apriori algorithm measures. To make use of the Apriori algorithm it is required to convert the whole transactional dataset into a single list and each row will be a list in that list.\n\ndata_list = []\nfor row in range(0, 20):\ndata_list.append([str(data.values[row,column]) for column in range(0, 9)])\n\nalgo = apriori(data_list, min_support=0.25, min_confidence=0.2, min_lift=2, min_length=2)\nresults = list(algo)\n\nWe have created a list of lists, then use the apriori method from apriori module and finally covert the datatype from the generator into a list and save in a variable named results. To make proper decisions and increase the speed of the apriori algorithm, apriori methods take several arguments which are as follows –\n\n1. data – The first parameter that takes the list that contains the transactional data in inner lists.\n2. min_support – It is the threshold support value for the items that should be taken into account. Suppose we want to make decisions for our dataset and want to include only those items that are appearing in at least 5 transactions out of total i.e support value of 5/20 = 0.25.\n3. min_confidence – It is the threshold confidence value that should be there between each combination of an itemset. we have taken the confidence value of 0.2.\n4. min_lift – It is the minimum lift value for the rules that are selected. Generally, we take lift value equals to 2 or more to filter out those itemsets that have a more frequent association.\n5. min_length – The numbers of items that are to be considered in the rules.\n\nLet us see the output of the above program and print the first 3 rules that we have obtained.\n\nfor i in range(0,3):\nprint(f\"Required Association No. {i+1} is: {results[i]}\")\nprint('-'*25)\nRequired Association No. 1 is: RelationRecord(items=frozenset({'toothpaste', 'brush'}), support=0.25, ordered_statistics=[OrderedStatistic(items_base=frozenset({'brush'}),\n-------------------------\nRequired Association No. 2 is: RelationRecord(items=frozenset({'mouthwash', 'toothpaste'}), support=0.3, ordered_statistics=[OrderedStatistic(items_base=frozenset({'mouthwash'}),\n-------------------------\nRequired Association No. 3 is: RelationRecord(items=frozenset({'honey', 'bread', 'butter'}), support=0.25, ordered_statistics=[OrderedStatistic(items_base=frozenset({'butter'}),\n-------------------------" ]	[ null, "https://codespeedy.com/wp-content/uploads/2019/12/csv-dataset.png", null, "https://codespeedy.com/wp-content/uploads/2019/12/modified_data_data-frame.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8615886,"math_prob":0.93828773,"size":9265,"snap":"2021-43-2021-49","text_gpt3_token_len":2164,"char_repetition_ratio":0.14210129,"word_repetition_ratio":0.010079194,"special_character_ratio":0.24684295,"punctuation_ratio":0.1079784,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98887104,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T14:40:08Z\",\"WARC-Record-ID\":\"<urn:uuid:7d5e481e-37fb-430e-8a1b-3085a738cea7>\",\"Content-Length\":\"35885\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34161c69-c8b9-4fa6-b8d6-f84906ef44c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:506679de-7f6f-4db2-9fe3-30376febbcaa>\",\"WARC-IP-Address\":\"74.207.248.30\",\"WARC-Target-URI\":\"https://www.codespeedy.com/association-rule-mining-in-python/\",\"WARC-Payload-Digest\":\"sha1:JNR6XROOQ22SKI2VJJD5DUAQFGUUV32R\",\"WARC-Block-Digest\":\"sha1:OCN2J5ME664MZQZRBISMXQGXIDCJ3DZY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585203.61_warc_CC-MAIN-20211018124412-20211018154412-00425.warc.gz\"}"}
https://codescracker.com/c/program/c-program-check-palindrome-string.htm	[ "# C Program to Check for Palindrome String\n\n« C Tutorial C Examples »\n\nIn this article, you will learn and get about checking whether the given string is palindrome or not. Here are the list of programs available here:\n\n• Palindrome String Program in C\n• Palindrome String Program in C without using any Library function (string function)\n• Palindrome String Program in C using user-defined function\n\nBut before going through these programs. Let's understand about the meaning of palindrome string.\n\n### What is Palindrome String ?\n\nThose string whose all the opposite characters are same is palindrome. And if any of the opposite character mismatched, then the string is not palindrome. For example:\n\n• codoc is a palindrome string, because:\n• first character (c) is equal to last character (c)\n• second character (o) is equal to second last character (0)\n• third character (d) is equal to third last character (d)\n• codssdoc is a palindrome string, because:\n• first character (c) is equal to last character (c)\n• second character (o) is equal to second last character (o)\n• third character (d) is equal to third last character (d)\n• fourth character (s) is equal to fourth last character (s)\n• codes is not a palindrome string, because:\n• first character (c) is not equal to last character (s)\n• codescracker is not a palindrome string, because:\n• first character (c) is not equal to last character (r)\n\nIn other words, you can say that, if the reverse of a string is equal to its original, then that string can be called as a palindrome string.\n\n## Palindrome String Program in C\n\nThe question is, write a program in C to enter any string at run-time and check whether it is palindrome or not. The answer to this question is:\n\n```#include<stdio.h>\n#include<conio.h>\n#include<string.h>\nint main()\n{\nchar str;\nint i, j, len, chk=0;\nprintf(\"Enter any String (word): \");\nscanf(\"%s\", str);\nlen = strlen(str);\nfor(i=0, j=(len-1); i<=(len-1); i++, j--)\n{\nif(str[i] != str[j])\n{\nchk=1;\nbreak;\n}\n}\nif(chk==1)\nprintf(\"\\nIt's not a Palindrome String\");\nelse\nprintf(\"\\nIt's a Palindrome String\");\ngetch();\nreturn 0;\n}```\n\nThe above program is written under CodeBlocks IDE. Here is the output after build and run. This is the first screenshot of first sample run:\n\nAnd here is the second screenshot of first sample run:\n\nAs the reverse of codescracker is rekcarcsedoc which is not equal to the original string, that is codescracker, therefore the string is not palindrome.\n\nHere is the first screenshot of second sample run:\n\nHere in this case, the reverse of codoc is codoc which is same as the original string, that is codoc, therefore the string is palindrome.\n\n#### Program Explained\n\n• First we have to count the length of string using strlen() function. For example, let's suppose the length of string is 6.\n• Create a loop to compare first character of the string to the last, and then second to the second last and so on. Therefore here first character is compared to the sixth character, and then second to the fifth, and third to the fourth. As indexing starts from 0, therefore, here we have to minus one from all. That is, 0th to the length-1.\n• If any of the above comparison goes wrong, or if any character mismatched, then we have to initialize any number say 1 to any character say chk, and skip the remaining code of the loop using break statement.\n• After exiting from the loop whether using break statement or after running the whole loop. We have to check whether the variable chk holds its original value, that is 0 or not. If it holds its original value, then none of the opposite character mismatched (means it is palindrome), and if it doesn't holds its original value, then any of the opposite character mismatched (means it is not palindrome).\n• Here mismatched occurs, means the string is not a palindrome, otherwise the string is a palindrome.\n\nHere is another sample run for, a palindrome string:\n\nAnd here is one more sample run for, not a palindrome string:\n\n### Check Palindrome String in C without using String Function\n\nHere is another program that will not use any string function. As you can see from the previous program, string function strlen() is used to find length of the entered string. But here, this program will calculate length of string without using any library function.\n\n```#include<stdio.h>\n#include<conio.h>\nint main()\n{\nchar str;\nint i, j, len, chk=0;\nprintf(\"Enter any String (word): \");\nscanf(\"%s\", str);\nlen = 0;\nwhile(str[len]!='\\0')\nlen++;\nfor(i=0, j=(len-1); i<=(len-1); i++, j--)\n{\nif(str[i] != str[j])\n{\nchk=1;\nbreak;\n}\n}\nif(chk==1)\nprintf(\"\\nIt's not a Palindrome String\");\nelse\nprintf(\"\\nIt's a Palindrome String\");\ngetch();\nreturn 0;\n}```\n\nThe code:\n\n```len=0;\nwhile(str[len]!='\\0')\nlen++;```\n\nis used to find length of string. For example, if user enters codes as input, then initially 0 gets initialized to len before going to the while loop. The '\\0' is known as null terminated character. At last of every string, a null terminated character gets automatically initialized. So the dry run of above code works like:\n\n• For first run, str[len] or str or c is not equal to '\\0'. Therefore len++ gets executed, so value of len gets incremented and becomes 1\n• For second run, str[len] or str or o is not equal to '\\0'. Therefore len++ gets executed, so value of len gets incremented and becomes 2\n• For second run, str[len] or str or d is not equal to '\\0'. Therefore len++ gets executed, so value of len gets incremented and becomes 3\n• For sixth run, str[len] or str or '\\0' is equal to '\\0'. Therefore the condition evaluates to be false, so the loop gets terminated, and we have a variable, len that holds the length value of string.\n\n## Check Palindrome String in C using Function\n\nThis is the last program to check whether given string is a palindrome or not using user-defined function.\n\n```#include<stdio.h>\n#include<conio.h>\nint checkPalindromeStr(char []);\nint main()\n{\nchar str;\nint chk;\nprintf(\"Enter any String (word): \");\nscanf(\"%s\", str);\nchk = checkPalindromeStr(str);\nif(chk==1)\nprintf(\"\\nIt's not a Palindrome String\");\nelse\nprintf(\"\\nIt's a Palindrome String\");\ngetch();\nreturn 0;\n}\nint checkPalindromeStr(char str[])\n{\nint i, j, len;\nlen = 0;\nwhile(str[len]!='\\0')\nlen++;\nfor(i=0, j=(len-1); i<=(len-1); i++, j--)\n{\nif(str[i] != str[j])\nreturn 1;\n}\nreturn 0;\n}```\n\nThis program will produce the same output as produced by first program given, earlier of this article. To learn more about function, refer to Function in C tutorial.\n\nC Online Test\n\n« C Tutorial C Examples »" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7092217,"math_prob":0.93810385,"size":3549,"snap":"2021-43-2021-49","text_gpt3_token_len":901,"char_repetition_ratio":0.13427362,"word_repetition_ratio":0.22241993,"special_character_ratio":0.28261483,"punctuation_ratio":0.1750663,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9776158,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T16:41:13Z\",\"WARC-Record-ID\":\"<urn:uuid:bd13d0ab-2487-471c-983c-693f7a212620>\",\"Content-Length\":\"33237\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90cb6d28-d26d-4dd9-8ff6-0980b7532145>\",\"WARC-Concurrent-To\":\"<urn:uuid:bc4cc735-e3e5-42bb-a7c2-162f9789d2de>\",\"WARC-IP-Address\":\"148.72.215.147\",\"WARC-Target-URI\":\"https://codescracker.com/c/program/c-program-check-palindrome-string.htm\",\"WARC-Payload-Digest\":\"sha1:NANNVT6U4ILQ5BH54TYTDCQHFU44RAWS\",\"WARC-Block-Digest\":\"sha1:KZRZV2STYCOSRDDYNMUOTPP3OZN6QCA4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585516.51_warc_CC-MAIN-20211022145907-20211022175907-00030.warc.gz\"}"}
https://assignmenthelpweb.com/category/basic-economic-concepts-and-principles/	[ "## Elasticity of Demand curve\n\nIf the market demand curve is Q=100-p. What is the market price elasticity of demand? If the supply curve of individual firms is q=p and there are 50 identical firms in the market, draw the residual demand facing any one firm. What is the residual demand elasticity facing one firm at the competitive equilibrium?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9057325,"math_prob":0.94700706,"size":567,"snap":"2022-40-2023-06","text_gpt3_token_len":129,"char_repetition_ratio":0.13321492,"word_repetition_ratio":0.0,"special_character_ratio":0.22222222,"punctuation_ratio":0.06363636,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9525731,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T18:11:39Z\",\"WARC-Record-ID\":\"<urn:uuid:3457e433-3fcb-46d1-a18d-0314bd1d8079>\",\"Content-Length\":\"31277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69815af1-3ee0-4da2-9ea3-2eb1d41ceada>\",\"WARC-Concurrent-To\":\"<urn:uuid:63d156b9-1ee1-4d06-b4fa-33183a2a115a>\",\"WARC-IP-Address\":\"162.0.229.244\",\"WARC-Target-URI\":\"https://assignmenthelpweb.com/category/basic-economic-concepts-and-principles/\",\"WARC-Payload-Digest\":\"sha1:MHHF54IMN54YLABYDZZOJULREB3P232R\",\"WARC-Block-Digest\":\"sha1:QFGR7365AEIMVFRVMLGFD2GWI4QRMNAG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335054.79_warc_CC-MAIN-20220927162620-20220927192620-00366.warc.gz\"}"}
https://yiming.dev/blog/2017/12/18/as-the-tests-get-more-specific-the-code-gets-more-generic/	[ "# As the tests get more specific, the code gets more generic\n\n• Prime Factors Kata\n• Test code and production code do not grow at the same rate (The test code grows faster)\n1. Sometimes the production code stays the same as the test code grows\n• You are done with the code but you still write the tests because they are part of the specifications\n2. Sometimes the production code shrinks as the test code grows\n• Because the programmers moves a load of functionality out of the code and into the data\n• Or comes up with some kind of more general algorithm that satisfies the tests without looking anything like them.\n\n## FizzBuzz Kata\n\nI started practicing Coding Katas1 to sharp my TDD/OOD skills several days ago.\n\nWhen I was doing this first Kata, one of my colleagues (non-programmer) was watching me. So I explained TDD to her.\n\nShe asked this question later: \"But your tests cannot cover all the cases (all the integers), how can you ensure your code is absolutely right?\"\n\n\"As the tests get more specific, the code gets more generic\" is a part of my answer. (For other reasons, like programmer confidence stuffs, I may talk about them in another post.)\n\n### Case #1\n\n• Test\n\n```def test_sound_for_1_is_1\nassert_equal '1', FizzBuzz.for(1)\nend\n```\n• Code\n\n```def self.for(_number)\n'1'\nend\n```\n\nAt this point, we only have one specification, so the code is pretty simple. (We are following TDD here, so this silly code is completely acceptable because it's green code.)\n\n### Case #2\n\n• Test\n\n```def test_sound_for_2_is_2\nassert_equal '2', FizzBuzz.for(2)\nend\n```\n• Code\n\n```def self.for(number)\nif number == 2\n'2'\nelse\n'1'\nend\nend\n```\n\nWe are able to see a pattern here (`number` -> `number.to_s`). But I chose to using `if` and wait for the next test. Because the next test seems to be very different.\n\n### Case #3\n\n• Test\n\n```def test_sound_for_3_is_Fizz\nassert_equal 'Fizz', FizzBuzz.for(3)\nend\n```\n• Code\n\n```def self.for(number)\nif number == 3\n'Fizz'\nelsif number == 2\n'2'\nelse\n'1'\nend\nend\n```\n\nThe code starts to look ugly now, but fortunately it's still green. Let's move to our next test.\n\n### Case #4\n\n• Test\n\n```def test_sound_for_4_is_4\nassert_equal '4', FizzBuzz.for(4)\nend\n```\n• Code\n\n```def self.for(number)\nif number == 3\n'Fizz'\nelse\nnumber.to_s\nend\nend\n```\n\nWe got our third case for `number.to_s` and at this point I decide to refactor it this way (Because of the magic number `3`).\n\nWe can see here that the code gets more generic (it handles more cases), while the test gets more specific (it adds a new specification for input `4`).\n\n### Case #5\n\n• Test\n\n```def test_sound_for_5_is_Buzz\nassert_equal 'Buzz', FizzBuzz.for(5)\nend\n```\n• Code\n\n```def self.for(number)\nif number == 3\n'Fizz'\nelsif number == 5\n'Buzz'\nelse\nnumber.to_s\nend\nend\n```\n\nThe same things goes with `5, Buzz` pairs as `3, Fizz` pairs. The code starts from being very specific (only deals with `3` or `5`), as the test gets more specific (case `6` or `10` are added), it gets more generic (deals with numbers have factor `3` or `5`).\n\n### Case #6\n\n• Test\n\n```def test_sound_for_6_is_Fizz\nassert_equal 'Fizz', FizzBuzz.for(6)\nend\n```\n• Code\n\n```def self.for(number)\nif number % 3 == 0\n'Fizz'\nelsif number == 5\n'Buzz'\nelse\nnumber.to_s\nend\nend\n```\n\nAgain, the code gets more generic while the test gets more specific.\n\n## Extracting logic to other classes\n\nLater, when the logic for `FizzBuzz` gets more and more complicated, I decided to push logic out of `FizzBuzz` class and extract some new concepts in our app.\n\n```class FizzBuzz\ndef initialize(sounds)\n@sounds = sounds\nend\n\ndef for(number)\nsounds\n.detect { \|sound\| sound.convertable?(number) }\n.for(number)\nend\n\nprivate\n\nend\n```\n\nI'll explain more about this refactoring in another post. But the basic idea is letting sounds to deal with different conversion rules (`Factor3Sound` for converting number to `Fizz`, `StringSound` for converting number to `number.to_s`, etc.)\n\nAgain, the `FizzBuzz` code gets more and more generic. Because it's now a `sound` rules chain, delegating conversion rules to `sound` objects, and just returning the result from the first `sound` that can handle the input `number`.\n\nI'll stop our `FizzBuzz` journey here as it has taken us so long. And I will talk about \"what I learned after I played with `FizzBuzz` for 3 days\" (Chain of Responsibility, and Elixir solution, etc.) in a future blog post.\n\n## DRY is not that important for tests\n\nWe programmers often says that our code needs to be DRY (Don't Repeat Yourself). And I think we can also say \"as the tests get more specific, the code gets more generic\" in this way:\n\nWhen we dry up our code, we are actually making our code more generic/abstract. This is good because it lets us handle as many possible cases on production as possible.\n\nBut our test doesn't need to be as DRY as our production code. After all, a test case should be as specific as possible, because it should be a specification for a case.\n\nThe most important thing we need to think about when we write a test should be making this test as clear as possible. I should be able to understand what this specification is talking about by only looking at this test itself.\n\nIf we try to DRY our test cases as well (like using `let` or `before` in RSpec), we may introduce too many `Mystery Guests`2 and making these tests unreadable.\n\nHowever, this doesn't mean we should not DRY our tests at all. It's still necessary to extract common assertions/matchers3 to make your tests more readable. Our tests need to be DRY as well, but in a different way from the production DRY.\n\n## A healthy Test-to-Code Ratio\n\nWith this principle being said, I think the Test-to-Code ratio can be a pretty good guide for us to decide when to refactor our code. (Alongside with Sandi Metz's Rules)\n\nIMHO, a good Test-to-Code ratio for a Rails project should be between 1:1 to 2:1. But the ratio definitely differ between different languages/frameworks/test tools.\n\n### Use Test-to-Code Ratio to guide your TDD cycle\n\n(I'll use 1:1 to 2:1 in the following explanations.)\n\n• When Test-to-Code Ratio is lower than 1:1\n1. Consider to add more test cases.\n2. Consider if the production code is too DRY.\n• When Test-to-Code Ratio is higher than 2:1\n1. Consider to DRY the production code by.\n2. Consider if the production code has too many responsibilities and needs to be split.\n\n### TDD is a process guided by this thinking\n\nAs I wrote in Book Review: 99 Bottles of OOP, we can only achieve Shameless Green by following the TDD cycle and make our code a step more abstract (generic) in each cycle.\n\nI think this thinking explained this methodology pretty well." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88592714,"math_prob":0.853692,"size":6502,"snap":"2021-43-2021-49","text_gpt3_token_len":1657,"char_repetition_ratio":0.13173284,"word_repetition_ratio":0.070609,"special_character_ratio":0.23992617,"punctuation_ratio":0.09586906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95245045,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T15:24:22Z\",\"WARC-Record-ID\":\"<urn:uuid:254c4371-cdcb-4aff-a306-77a17a5a2153>\",\"Content-Length\":\"17048\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65952531-ff2f-4fec-bf0c-fd8a35465af4>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c7638a2-d515-4fee-b4b1-29a56c204c4c>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://yiming.dev/blog/2017/12/18/as-the-tests-get-more-specific-the-code-gets-more-generic/\",\"WARC-Payload-Digest\":\"sha1:5KRS44UIICWQACRFCK5A25DISKFQN75J\",\"WARC-Block-Digest\":\"sha1:PB5EJARNC3A52M6DWIBUUNBYR3JTFLVW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585424.97_warc_CC-MAIN-20211021133500-20211021163500-00658.warc.gz\"}"}
https://faculty.kfupm.edu.sa/MATH/assane/Mes%20sites%20Web/monsiteweb/Mysite/Mysite/index.html	[ "Assane Lo\n\nMathematics Department, The University of Arizona.\n\nEmail: assane@math.arizona.edu\n\nPhone: (520)621-8322\n\nOffice: MTL 120 Office: MTL 120 N\n\nEmail: assane@math.arizona.edu\n\nPhone: (520)621-8322\n\nMy scholarly interest ranges from Analysis, Geometric Analysis, Partial Differential Equations, Probability theory, and applications to Statistical Physics and Euclidean Field Theory. My current research in mathematical physics involves the study of direct methods for integrals and operators of the type that appear naturally in equilibrium statistical mechanics and euclidean field theory.\n\n### Background\n\n• Bachelor of Science in mathematics with minor in physics from the University of Dakar, Senegal in 1999.\n• Ecole Normal Superieur of Dakar 1999-2000.\n• PH.D in mathematics, University of Arizona (Expected May 2007).\n\n### Courses Taught\n\n• Vector Calculus (Math 223-02) Fall 06.\n• Calculus II (Math 129) Spring 06, Fall 05.\n• Calculus I (Math 124) Spring 05.\n• Element of Calculus (Math 113) Fall 04, Spring 04.\n• College Algebra (Math 110) Summer 05, Summer 04,Fall 03, Summer 03, Spring 03, Fall 02.\n\nThe University of", null, "rizona (old picture)" ]	[ null, "https://faculty.kfupm.edu.sa/MATH/assane/Mes%20sites%20Web/monsiteweb/Mysite/Mysite/A_Logo_color_50.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7141994,"math_prob":0.50159514,"size":1147,"snap":"2019-51-2020-05","text_gpt3_token_len":295,"char_repetition_ratio":0.104986876,"word_repetition_ratio":0.0,"special_character_ratio":0.27637315,"punctuation_ratio":0.17535545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9825403,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T16:51:59Z\",\"WARC-Record-ID\":\"<urn:uuid:3e31343d-3194-473a-b5f1-818075ea8559>\",\"Content-Length\":\"7012\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:417bb538-88c0-438d-bb70-43b639ab4aa9>\",\"WARC-Concurrent-To\":\"<urn:uuid:54a7c34a-ca4b-4123-8acc-d859ef9d70b8>\",\"WARC-IP-Address\":\"198.11.1.82\",\"WARC-Target-URI\":\"https://faculty.kfupm.edu.sa/MATH/assane/Mes%20sites%20Web/monsiteweb/Mysite/Mysite/index.html\",\"WARC-Payload-Digest\":\"sha1:2VKLEIK5NS67Z2AZNTJ3R3DAAKSO3VKJ\",\"WARC-Block-Digest\":\"sha1:BZS7YTGHXYVAPGD65DJXTADXRL4DU67B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251700988.64_warc_CC-MAIN-20200127143516-20200127173516-00178.warc.gz\"}"}
https://number.academy/14600	[ "# Number 14600\n\nNumber 14,600 spell 🔊, write in words: fourteen thousand, six hundred . Ordinal number 14600th is said 🔊 and write: fourteen thousand, six hundredth. The meaning of number 14600 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 14600. What is 14600 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 14600.\n\n## What is 14,600 in other units\n\nThe decimal (Arabic) number 14600 converted to a Roman number is (X)(IV)DC. Roman and decimal number conversions.\n\n#### Weight conversion\n\n14600 kilograms (kg) = 32187.2 pounds (lbs)\n14600 pounds (lbs) = 6622.5 kilograms (kg)\n\n#### Length conversion\n\n14600 kilometers (km) equals to 9073 miles (mi).\n14600 miles (mi) equals to 23497 kilometers (km).\n14600 meters (m) equals to 47900 feet (ft).\n14600 feet (ft) equals 4451 meters (m).\n14600 centimeters (cm) equals to 5748.0 inches (in).\n14600 inches (in) equals to 37084 centimeters (cm).\n\n#### Temperature conversion\n\n14600° Fahrenheit (°F) equals to 8093.3° Celsius (°C)\n14600° Celsius (°C) equals to 26312° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n14600 seconds equals to 4 hours, 3 minutes, 20 seconds\n14600 minutes equals to 1 week, 3 days, 3 hours, 20 minutes\n\n### Zip codes 14600\n\n• Zip code 14600 Campiña, Andalucia, Córdoba, Spain a map\n• Zip code 14600 Casillas De Velasco, Andalucia, Córdoba, Spain a map\n• Zip code 14600 Charco Del Novillo, Andalucia, Córdoba, Spain a map\nZip code areas 14600\n\n### Codes and images of the number 14600\n\nNumber 14600 morse code: .---- ....- -.... ----- -----\nSign language for number 14600:", null, "", null, "", null, "", null, "", null, "Number 14600 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 14600 infographic", null, "## Share in social networks", null, "## Mathematics of no. 14600\n\n### Multiplications\n\n#### Multiplication table of 14600\n\n14600 multiplied by two equals 29200 (14600 x 2 = 29200).\n14600 multiplied by three equals 43800 (14600 x 3 = 43800).\n14600 multiplied by four equals 58400 (14600 x 4 = 58400).\n14600 multiplied by five equals 73000 (14600 x 5 = 73000).\n14600 multiplied by six equals 87600 (14600 x 6 = 87600).\n14600 multiplied by seven equals 102200 (14600 x 7 = 102200).\n14600 multiplied by eight equals 116800 (14600 x 8 = 116800).\n14600 multiplied by nine equals 131400 (14600 x 9 = 131400).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 14600\n\nHalf of 14600 is 7300 (14600 / 2 = 7300).\nOne third of 14600 is 4866,6667 (14600 / 3 = 4866,6667 = 4866 2/3).\nOne quarter of 14600 is 3650 (14600 / 4 = 3650).\nOne fifth of 14600 is 2920 (14600 / 5 = 2920).\nOne sixth of 14600 is 2433,3333 (14600 / 6 = 2433,3333 = 2433 1/3).\nOne seventh of 14600 is 2085,7143 (14600 / 7 = 2085,7143 = 2085 5/7).\nOne eighth of 14600 is 1825 (14600 / 8 = 1825).\nOne ninth of 14600 is 1622,2222 (14600 / 9 = 1622,2222 = 1622 2/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 14600\n\n#### Is Prime?\n\nThe number 14600 is not a prime number. The closest prime numbers are 14593, 14621.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 14600 are 2 * 2 * 2 * 5 * 5 * 73\nThe factors of 14600 are\n1 , 2 , 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , 73 , 100 , 146 , 200 , 292 , 365 , 584 , 730 , 1460 , 1825 , 2920 , 3650 , 7300 , 14600 show more factors ...\nTotal factors 24.\nSum of factors 34410 (19810).\n\n#### Powers\n\nThe second power of 146002 is 213.160.000.\nThe third power of 146003 is 3.112.136.000.000.\n\n#### Roots\n\nThe square root √14600 is 120,83046.\nThe cube root of 314600 is 24,440924.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 14600 = loge 14600 = 9,588777.\nThe logarithm to base 10 of No. log10 14600 = 4,164353.\nThe Napierian logarithm of No. log1/e 14600 = -9,588777.\n\n### Trigonometric functions\n\nThe cosine of 14600 is -0,52427.\nThe sine of 14600 is -0,851552.\nThe tangent of 14600 is 1,624263.\n\n### Properties of the number 14600\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 14600 in Computer Science\n\nCode typeCode value\n14600 Number of bytes14.3KB\nUnix timeUnix time 14600 is equal to Thursday Jan. 1, 1970, 4:03:20 a.m. GMT\nIPv4, IPv6Number 14600 internet address in dotted format v4 0.0.57.8, v6 ::3908\n14600 Decimal = 11100100001000 Binary\n14600 Decimal = 202000202 Ternary\n14600 Decimal = 34410 Octal\n14600 Decimal = 3908 Hexadecimal (0x3908 hex)\n14600 BASE64MTQ2MDA=\n14600 MD57ab8464df372307c1bcce9b2a043415a\n14600 SHA153e6746c4c5cf9dddf180e1b370e4da47fc960c1\n14600 SHA2246f0f98a003f5310beae017490af1f46a1cc0940205f54c7256d31ae1\nMore SHA codes related to the number 14600 ...\n\nIf you know something interesting about the 14600 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 14600\n\n### Character frequency in number 14600\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 1 4 1 6 1 0 2\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 14600, the numbers 1+4+6+0+0 = 1+1 = 2 are added and the meaning of the number 2 is sought.\n\n## Interesting facts about the number 14600\n\n### Asteroids\n\n• (14600) 1998 SG73 is asteroid number 14600. It was discovered by E. W. Elst from La Silla Observatory on 9/21/1998.\n\n### Distances between cities\n\n• There is a 9,073 miles (14,600 km) direct distance between Caracas (Venezuela) and Daegu (South Korea).\n• There is a 9,073 miles (14,600 km) direct distance between Cochabamba (Bolivia) and Omsk (Russia).\n• There is a 9,073 miles (14,600 km) direct distance between Dhaka (Bangladesh) and São Luís (Brazil).\n• There is a 9,073 miles (14,600 km) direct distance between Hiroshima-shi (Japan) and Valencia (Venezuela).\n• There is a 9,073 miles (14,600 km) direct distance between Kuala Lumpur (Malaysia) and Ottawa (Alberta).\n• There is a 9,073 miles (14,600 km) direct distance between La Paz (Bolivia) and Perth (Australia).\n\n## Number 14,600 in other languages\n\nHow to say or write the number fourteen thousand, six hundred in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 14.600) catorce mil seiscientos German: 🔊 (Anzahl 14.600) vierzehntausendsechshundert French: 🔊 (nombre 14 600) quatorze mille six cents Portuguese: 🔊 (número 14 600) catorze mil, seiscentos Chinese: 🔊 (数 14 600) 一万四千六百 Arabian: 🔊 (عدد 14,600) أربعة عشر ألفاً و ستمائة Czech: 🔊 (číslo 14 600) čtrnáct tisíc šestset Korean: 🔊 (번호 14,600) 만 사천육백 Danish: 🔊 (nummer 14 600) fjortentusinde og sekshundrede Dutch: 🔊 (nummer 14 600) veertienduizendzeshonderd Japanese: 🔊 (数 14,600) 一万四千六百 Indonesian: 🔊 (jumlah 14.600) empat belas ribu enam ratus Italian: 🔊 (numero 14 600) quattordicimilaseicento Norwegian: 🔊 (nummer 14 600) fjorten tusen, seks hundre Polish: 🔊 (liczba 14 600) czternaście tysięcy sześćset Russian: 🔊 (номер 14 600) четырнадцать тысяч шестьсот Turkish: 🔊 (numara 14,600) ondörtbinaltıyüz Thai: 🔊 (จำนวน 14 600) หนึ่งหมื่นสี่พันหกร้อย Ukrainian: 🔊 (номер 14 600) чотирнадцять тисяч шiстсот Vietnamese: 🔊 (con số 14.600) mười bốn nghìn sáu trăm Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 14600 or any natural number (positive integer) please write us here or on facebook." ]	[ null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-4.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-0.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-0.png", null, "https://number.academy/img/braille-14600.svg", null, "https://numero.wiki/img/a-14600.jpg", null, "https://numero.wiki/img/b-14600.jpg", null, "https://number.academy/i/infographics/0/number-14600-infographic.png", null, "https://numero.wiki/s/share-desktop.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59244287,"math_prob":0.97825426,"size":7295,"snap":"2022-05-2022-21","text_gpt3_token_len":2633,"char_repetition_ratio":0.14689343,"word_repetition_ratio":0.04918033,"special_character_ratio":0.41069224,"punctuation_ratio":0.15797208,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99230593,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T16:34:07Z\",\"WARC-Record-ID\":\"<urn:uuid:2c8546e2-4e51-423c-be1f-94cff1445eb3>\",\"Content-Length\":\"43389\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:826d0c7e-3a97-4c6a-ae90-f68d0b14dc05>\",\"WARC-Concurrent-To\":\"<urn:uuid:885b1d82-a692-4870-845a-a36dff8b9d17>\",\"WARC-IP-Address\":\"162.0.227.212\",\"WARC-Target-URI\":\"https://number.academy/14600\",\"WARC-Payload-Digest\":\"sha1:ZYK4NYUS3SFZHXGQOBYELMW4O3BB6TRA\",\"WARC-Block-Digest\":\"sha1:XEITXJQK3TFDRXLGSSJTT7FXDDLJVDQV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662545875.39_warc_CC-MAIN-20220522160113-20220522190113-00795.warc.gz\"}"}
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Chemical_Bonds/Enthalpies_of_Reactions	[ "# Enthalpies of Reactions\n\nSkills to Develop\n\n• Calculate the enthalpies of reactions from bond energies.\n• Calculate the enthalpies of reactions from enthalpies of formation.\n• Draw energy level diagrams and use them to calculate enthalpies of reactions.\n\n## Enthalpies of Reactions\n\nEnthalpy of a reaction or energy change of a reaction, $$\\Delta H$$, is the amount of energy or heat absorbed in a reaction. If the energy is required, $$\\Delta H$$ is positive, and if energy is released, the $$\\Delta H$$, is negative.\n\n_________ Products\n\n\|\n\| DH, positive for\n\| endothermic reaction\n\|\n__\|______ Reactants\n\nThe enthalpy can be determined by experiment, but estimates can easily be made if bond energies or standard enthalpies of formation for the reactants and products are available.\n\n### Using Formulas to Calculate $$\\Delta H$$\n\nDue the the definitions of various types of energy related terms, formulas for evaluating enthalpies can be very confusing. For example, the formulas to calculate the enthalpy of a reaction depends on whether bond energies or enthalpies of formation are available.\n\nWhen standard enthalpies of formation, Hfo, for all products and reactants are available, we have\n\n$$\\mathrm{\\mathit H_{reaction} = SUM (\\mathit H_{products}) - SUM (\\mathit H_{reactants})}$$\n\nor if you prefer using symbols\n\n$$\\mathrm{\\mathit H_{reaction} = \\sum \\mathit H_{products} - \\sum \\mathit H_{reactants}}$$\n\nFor simplicity in formulation we use H to represent Hfo in the above formulas.\n\nBecause bond energies are defined as the energies required to break the bonds, positive values are usually listed whereas in reality, they are energies released when chemical bonds are formed from respective atoms. Thus, using the bond energies (BE) as they are given or defined, the following formula apply:\n\n$$\\mathrm{\\mathit H_{reaction} = SUM (\\mathit{BE}_{reactants}) - SUM (\\mathit{BE}_{products})}$$\n\nor if you prefer using symbols\n\n$$\\mathrm{\\mathit H_{reaction} = \\sum \\mathit{BE}_{reactants} - \\sum \\mathit{BE}_{products}}$$\n\nThese formulas to calculate the enthalpy (heat) of a reaction can be a very confusing and you may easily get an incorrect sign for the value. Thus, remembering formulas is discouraged!\n\nWell, if you use a diagram to help visualize the calculation or write the chemical reaction equations accompanying the thermodynamic values, you will be able to avoid the confusion. In both cases, you are applying the principle of conservation of energy in solving the problems.\n\n### Calculate Enthalpy of Reaction from Bond Energies\n\nDue to the principle of conservation of energy the total energy before and after the reaction must not change. Thus, the energy of a reaction released or absorbed in the reaction must come from the difference in bond energies of the products and the reactants.\n\nExample 1\n\nThe bond energy (kJ) for $$\\ce{H2}$$, $$\\ce{F2}$$, and $$\\ce{HF}$$ are 436, 158 and 568 kJ respectively; calculate the enthalpy (energy) of the reaction,\n\n$$\\ce{H_{2\\large{(g)}} + F_{2\\large{(g)}} \\rightarrow 2 HF}$$\n\nSolution\nBased on the bond energies given, we have\n\n\\begin{align} \\ce{&H2 \\rightarrow 2H} && D = \\mathrm{436\\: kJ/mol}\\\\ \\ce{&F2 \\rightarrow 2F} && D = \\mathrm{158\\: kJ/mol}\\\\ \\ce{&2H + 2F \\rightarrow 2HF} && H = \\mathrm{-568\\times2\\: kJ/mol} \\end{align}\n\n$$\\ce{H_{2\\large{(g)}} + F_{2\\large{(g)}} \\rightarrow 2 HF}$$ $$\\mathrm{D\\mathit{H} = -542\\: kJ/equation}$$\n\nNote that D represent bond dissociation energy, and H the enthalpy of the reaction as written. We use $$\\Delta H$$ in the last equation to denote enthalpy of change of the overall reaction.\n\nDiscussion\nSince bond energies are given, we use the monoatomic gases as the reference level in this calculation. The energy level diagram shown below illustrates the principle of conservation of energy, and you are expected to have the skill to draw such a diagram.\n\n------2 H(g) + 2 F(g)-------\n \n\| \|\|436+158 kJ\n\| \|\|\n\|+2568 kJ ---H2 + F2---\n\| \|\n\| \| $$\\Delta H$$ = -542 kJ/equation\n\| ¯\n--------2 HF(g)------------\n\n\nThis diagram is very similar to those of the Born-Haber cycle used to evaluate lattice energy.\n\n### Calculate Enthalpy of Reaction from Enthalpy of Formation\n\nA similar cycle can be devised to calculate energy of a reaction when the standard enthalpies of formation are given. We illustrate this cycle by an an example.\n\nExample 2\n\nStandard enthalpies of formation are: $$\\ce{C2H5OH_{\\large{(l)}}}$$ -228, $$\\ce{CO2}$$ -394, and $$\\ce{H2O_{\\large{(l)}}}$$ -286 kJ/mol. Calculate the enthalpy of the reaction,\n\n$$\\ce{C2H5OH + 3 O2 \\rightarrow 2 CO2 + 3 H2O}$$\n\nSolution\nFrom the definition of the enthalpy of formation, we have the following equations and the energy changes of reactions.\n\n\\begin{align} &\\mathrm{C_2H_5OH_{\\large{(l)}} \\rightarrow 2 C_{\\large{(graphite)}} + 3 H_{2\\large{(l)}} + 0.5 O_{2\\large{(g)}}} &&H = \\mathrm{228\\: kJ/mol}\\\\ &\\mathrm{2 C_{\\large{(graphite)}} + 2 O_{2\\large{(g)}} \\rightarrow CO_{2\\large{(g)}}} &&H = \\mathrm{-394\\times2\\: kJ/mol}\\\\ &\\mathrm{3 H_{2\\large{(g)}} + 1.5 O_{2\\large{(g)}} \\rightarrow 3 H_2O_{\\large{(l)}}} &&H = \\mathrm{-286\\times3\\: kJ/mol} \\end{align}\n\n$$\\ce{C2H5OH_{\\large{(l)}} + 3 O_{2\\large{(g)}} \\rightarrow 2 CO_{2\\large{(g)}} + 3 H2O_{\\large{(l)}}}$$ $$\\mathrm{D\\mathit{H} = -1418\\: kJ/mol}$$\n\nDiscussion\nSince the standard enthalpy of formation uses the elements as the standard, we put the elements on the top as a common level of 0 energy. The enthalpies of formation are negative, and we have the following diagram.\n\nThus, the enthalpy of reaction is the difference between the level of\n---C2H5OH +3 O2---\nand\n-2 CO2 + 3 H2O-.\n\n---2 C(graphite) + 3 H2 + 3.5 O2(g)---\n\| \|\n\| \| -228 kJ\n\| ¯\n\| ---C2H5OH + 3 O2---\n\|-3942 -2863 \|\n\| \|\n\| \| $$\\Delta H$$ = -3942 - 286*3 - (-228)\n\| \| = -1418 kJ\n¯ ¯\n-------------2 CO2 + 3 H2O------------\n\n\nA Footnote\n\nDue to the definitions, the enthalpies of formation are negative values whereas the bond energies are given as positive values. Thus, calculations using these two types of data have different signs.\n\nYou can use the diagramatic method to solve these types of problems as given in the discussions above. Using the diagrams shown indicates clearly what should be the sign of the results. When the reaction is from a higher level to a lower level, the enthalpy of reaction should be negative (downward arrow). If you reverse the direction of the reaction, you also change the sign of the enthalpy of the reaction.\n\n## Confidence Building Questions\n\n1. Calculate the heat of reaction for\n\n$$\\ce{2 H2 + O2 \\rightarrow 2 H2O}$$\n\nBond energies: $$\\ce{H2}$$ 436 kJ/mol, $$\\ce{O2}$$ 498, $$\\ce{HO}$$ 463 kJ/mol\n\nHint: $$\\mathrm{-4\\times463 + 498 + 2\\times436}$$\n\nDiscussion - Note that in this problem, you are asked to calculate the heat of reaction, but the heat of formation is -241 kJ / mol.\n\n2. Calculate the heat of reaction for\n\n$$\\ce{H2 + Cl2 \\rightarrow 2 HCl}$$\n\nBond energies: $$\\ce{H2}$$ 436 kJ/mol, $$\\ce{O2}$$ 498, $$\\ce{HO}$$ 463, $$\\ce{Cl2}$$ 243, $$\\ce{HCl}$$ 432 kJ/mol\n\nHint: $$\\mathrm{436 + 243 - 2\\times432 =\\: ?}$$\n\nDiscussion - If you are given the heat of the reaction, and the $$\\ce{H-H}$$ bond energy, can you calculate the bond energy for $$\\ce{Cl-Cl}$$?\n\n3. Calculate the enthalpy of formation for $$\\ce{NH3}$$.\nBond energies:\n$$\\ce{H2}$$ 436 kJ/mol, $$\\ce{N2}$$ 945, $$\\ce{H-N}$$ 391 kJ/mol\n\nHint: $$\\mathrm{\\dfrac{3\\times436}{2} + \\dfrac{945}{2} - 3\\times391 =\\: ?}$$\n\nAnother method -\nWriting the enthalpies of formation below their formulas in the equation, we have:\n\n\\begin{alignat}{2} &\\dfrac{3}{2} \\ce{H2} &&+ \\dfrac{1}{2} \\ce{N2} &&\\rightarrow \\ce{NH3}\\\\ &\\dfrac{3\\times436}{2} &&+ \\dfrac{945}{2} &&= 3\\times391 + H_{\\mathrm f} \\end{alignat}\n\nFrom this equation, we have:\n\n$$\\mathrm{\\mathit H_f = \\dfrac{3\\times436}{2} + \\dfrac{945}{2} - 3\\times391 = -46.5\\: kJ/mol}$$\n\nA handbook gives the enthalpy of formation of $$\\ce{NH3}$$ as -46.1 kJ/mol. As an exercise, use this value to calculate the triple bond energy of $$\\ce{N2}$$.\n\n4. Calculate the enthalpy of the reaction\n\n$$\\ce{CH4 + 2 O2 \\rightarrow CO2 + 2 H2O}$$\n\nfrom the enthalpies of formation: $$\\ce{CH4}$$ -75 kJ/mol, $$\\ce{CO2}$$ -394, and $$\\ce{ H2O_{\\large{(l)}}}$$ -286 kJ/mol.\n\nHint: -891 kJ\n\nDiscussion - The value calculated is the enthalpy of combustion of methane" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7572422,"math_prob":0.9998599,"size":8241,"snap":"2019-51-2020-05","text_gpt3_token_len":2456,"char_repetition_ratio":0.18137671,"word_repetition_ratio":0.056315366,"special_character_ratio":0.3239898,"punctuation_ratio":0.0840678,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998723,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T08:39:52Z\",\"WARC-Record-ID\":\"<urn:uuid:958bc217-54f2-4a82-bc54-aeaa99a207a8>\",\"Content-Length\":\"94508\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7ac4125-6854-4575-9ec1-873ac376d972>\",\"WARC-Concurrent-To\":\"<urn:uuid:93670e3f-7c6e-4e99-aaaa-cca21d2fc218>\",\"WARC-IP-Address\":\"54.192.30.107\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Chemical_Bonds/Enthalpies_of_Reactions\",\"WARC-Payload-Digest\":\"sha1:T2J2IZEILFVV52QMAXGXJTGUOFA3I4Z3\",\"WARC-Block-Digest\":\"sha1:WDMZT4LMX3IBZGK26PTK3DT4GPRTGKTB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251696046.73_warc_CC-MAIN-20200127081933-20200127111933-00430.warc.gz\"}"}
https://kr.mathworks.com/matlabcentral/profile/authors/5347541	[ "Community Profile", null, "# Alexandra\n\n2014 이후 활성\n\nProfessional Interests: Risk management, Corporate Finance, M&A\n\n#### Statistics\n\n•", null, "•", null, "•", null, "배지보기\n\n#### Content Feed\n\n보기 기준\n\n질문\n\nuniform distribution between a and b with intervals of 0,005\nHi, I am having trouble using R = unidrnd(N) to create n random numbers between a and b (imagine a=0 and b=0.2) where the nu...\n\n5년 이상 전 \| 답변 수: 3 \| 0\n\n### 3\n\n답변\n\n질문\n\nRunning the files on a sequence, saving the results, running the files again, saving the results with different name\nHi, I use a .m file with just diary function in it to run several different files sequentially in the same folder. diary lo...\n\n6년 이하 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nHow to run a mat file only after the others have run\nHi, I have 4/5 matlab windows open and I run several independent mat files at the same time. When they finish I have another...\n\n6년 이하 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nAlternative code that runs much faster?\nHi, We built this code: A = zeros(n,y); A (:,1) = -3; x1 = zeros(n,y); x2 = zeros(n,y); x3 = zeros(n,y); ...\n\n6년 이하 전 \| 답변 수: 0 \| 0\n\n### 0\n\n답변\n\n답변 있음\nHow to sum every 29 data and make some condition for them?\nI would try something like this: for k=1:(numel(A)-30) B(k) = sum(A(k:k+29)); if B(k) <= 0.5 C(k) ...\n\n약 6년 전 \| 0\n\n질문\n\nGenerating random numbers from 0 - 1 with limit on the sum\nHi, I am trying to generate values from 0 to 1 for several variables (x1 x2 x3). These results must assure that x1a1 + x2...\n\n약 6년 전 \| 답변 수: 3 \| 0\n\n### 3\n\n답변\n\n질문\n\nModel running for days\nHi, I know this is a very open question but the Monte Carlo simulation I built is getting so massive that the model is running f...\n\n약 6년 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nComputing IRR of a project after x years depending on price\nHi, I am stuck trying to write a code. I have a project of n years. The cash flow each year depends on the Price. For each y...\n\n약 6년 전 \| 답변 수: 0 \| 0\n\n### 0\n\n답변\n\n질문\n\nUsing for and fzero to compute IRR after x years\nHi, here's the code: for a=1:y for k=1:n f = A(k,a)/(1+t(k,a))^(a-1); B(k,a) = B(k,a) ...\n\n약 6년 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\n3D matrix multiplications\nHi, I have a A(nxa) matrix and a B(1xp) matrix. I'm trying to multiplicate A by B, creating a C(nxaxp) matrix (3D), where t...\n\n6년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\ngenerate random numbers subject to\nHi, I am trying to generate a set of random numbers X between 0 and 1 subject that an equation depending on X can be negative....\n\n6년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nNaN in database during simulation\nI imported a series of returns into the workspace. Because there was a longer series in the same database, matlab imported the r...\n\n6년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nfinding the maximum value table A for values in table B bellow benchmark\nSo, here's the situation: I have three tables: A = [50 100 30 4]; B = [100 150 90 50]; C = [50 100 150]; Value in C are b...\n\n6년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nWarning: Inverse CDF calculation did not converge for p\nHi, I run a simulation using ksdensity and copulas. In the end I get the following warning: Warning: Inverse CDF calculat...\n\n6년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nOut of memory error with 8GB and 64bit\nHi, I am working with lots of memory and I am getting an out of memory error. >> memory Maximum possible array: 1...\n\n7년 이하 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nrandom 100k rows of kernel numbers, each row has Poisson number of random numbers\nHi, I am trying to do something and I need help. So, this is for insurance analysis. I have x number of accidents that may...\n\n7년 이하 전 \| 답변 수: 0 \| 0\n\n### 0\n\n답변\n\n질문\n\nError using copula fit: Rho has become rank-deficient\nHi, I wanted to estimate kernel densities for 20 variables, correlated by t copulas. Had this formula: [Rho,nu] = copulafit('t'...\n\n7년 이하 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nksdensity default bandwidth value\nHi, Does any one knows how matlab computes the default bandwidth in the function ksdensity? Thank you very much,\n\n약 7년 전 \| 답변 수: 3 \| 0\n\n### 3\n\n답변\n\n질문\n\nMatrix Multiplication column and vector\nHi, I am trying to solve this problem: A is 1 column of 100k results; B is a vector of 10 weights; I need a matrix of 10...\n\n약 7년 전 \| 답변 수: 2 \| 0\n\n### 2\n\n답변\n\n질문\n\nLoop on changing databases\nHi, We run a code where part of it depends on variables we import from excel. We want to run a loop on the code for it to ...\n\n약 7년 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nSeveral automatic Monte Carlo simulations linked to excel models that change its assumptions for each simulation\nHi, I don't even know if this is possible. We are running a Monte Carlo simulation on a model built in Matlab. In the begi...\n\n약 7년 전 \| 답변 수: 0 \| 0\n\n### 0\n\n답변\n\n질문\n\nCan't apply an IF function with 1000001 matrix\nHi! I built this code: if A1 <0 D = 0.8; elseif A2 <0 D = 0.7; elseif A3 <0 D = 0.6; elseif ...\n\n약 7년 전 \| 답변 수: 2 \| 0\n\n### 2\n\n답변\n\n질문\n\nEfficient Frontier code error\nHi, I am running the following code I found: % portfolio0_efficient_frontier % Matlab program to plot efficient frontier and...\n\n7년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nA log returns distribution with median zero\nHi, I have a time series of monthly log returns, to which I subtracted the median, so that they have zero median. I did this ...\n\n7년 이상 전 \| 답변 수: 0 \| 0\n\n### 0\n\n답변\n\n질문\n\nrandom normal (0,1) correlated by copulas\nHi, I created a simulation of random variables correlated by copulas in Matlab, using ksdensity function. Now I am trying...\n\n7년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nSimulating dependent normally distributed variables using copulas\nI created a model that simulates variables with kernel distributions connected by copula functions. (X and Y are series of ln r...\n\n7년 이상 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nProbability of being under a target\nHi, I have a result variable X that is defined by an equation of random variables. We used prctile to see the results we ...\n\n8년 이하 전 \| 답변 수: 1 \| 0\n\n### 1\n\n답변\n\n질문\n\nError with mvnrnd function?\nHi! when I run the following code: n = 1000; rho = .7; Z = mvnrnd([0 0], [1 rho; rho 1], n); U = normcdf(Z); X = [gaminv...\n\n8년 이하 전 \| 답변 수: 1 \| 0\n\n답변" ]	[ null, "https://kr.mathworks.com/responsive_image/150/150/0/0/0/cache/matlabcentral/profiles/5347541_1522107757867_DEF.jpg", null, "https://kr.mathworks.com/images/responsive/supporting/matlabcentral/fileexchange/badges/first_review.png", null, "https://kr.mathworks.com/matlabcentral/profile/badges/Thankful_5.png", null, "https://kr.mathworks.com/matlabcentral/profile/badges/First_Answer.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6831321,"math_prob":0.8814451,"size":5902,"snap":"2022-27-2022-33","text_gpt3_token_len":2297,"char_repetition_ratio":0.14632079,"word_repetition_ratio":0.1541502,"special_character_ratio":0.33276856,"punctuation_ratio":0.16126856,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953597,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-12T17:26:52Z\",\"WARC-Record-ID\":\"<urn:uuid:f913f573-bea3-449a-8fe7-ccf61788196e>\",\"Content-Length\":\"113787\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:453ef917-3751-44b5-9a54-410dc0ea5b25>\",\"WARC-Concurrent-To\":\"<urn:uuid:21b1053c-3398-4533-9603-d699ed2343e4>\",\"WARC-IP-Address\":\"23.1.9.244\",\"WARC-Target-URI\":\"https://kr.mathworks.com/matlabcentral/profile/authors/5347541\",\"WARC-Payload-Digest\":\"sha1:AU5ZRL65V5J6JWCPU2JTKB6ARH2557LK\",\"WARC-Block-Digest\":\"sha1:DNVQ7KPM4P5DNPZ6GTPRNPLHDQLAAV27\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571745.28_warc_CC-MAIN-20220812170436-20220812200436-00260.warc.gz\"}"}
https://www.physicsoverflow.org/25607/why-the-bosonic-part-superconformal-group-%24su-%24so-times-_r%24	[ "#", null, "Why the bosonic part of the superconformal group $SU(2,2\|1)$ is $SO(4,2) \\times U(1)_R$?\n\n+ 5 like - 0 dislike\n569 views\n\nWhy in $d=4$ $\\mathcal{N}=1$ SCFT the bosonic part of the superconformal group $SU(2,2\|1)$ is $SO(4,2) \\times U(1)_R$?\n\nMore generally how can I determine the such a thing in other theories? Is there some specific way to think about how to find such a subgroup? Say in $d=4$ $\\mathcal{N}=2$ theory. I know this is known, I want to know how one finds it though.\n\nThis post imported from StackExchange Physics at 2014-12-25 23:28 (UTC), posted by SE-user Marion\n\nedited Jan 2, 2015\n\n+ 4 like - 0 dislike\n\nFirstly the bosonic part of superconformal algebra of this theory is $SO(4,2)\\times U(1)_R$ not $SO(4,1)\\times U(1)_R$. I have corrected this mistake that I guess it was a typo. Now, let us consider the conformal algebra generators.\n\n1. The momenta $P^{\\mu}$ that generate spacetime translations.\n2. The angular momenta $J^{\\mu \\nu}$ that generate spacetime rotations.\n3. The dilatation operator $D$ that generates uniform rescalings to coordinates and fields.\n4. The special conformal transformations $K^{\\mu}$.\n\nThere are many places where you can check the algebra they create, namely the superconformal algebra $\\mathfrak{so}(4,2)$. How can this be true? You can verify this by introducing indices $a,b=-1,0,1,2,3,4$ and defining the operators $J^{ab}= -J^{ba}$ as following $J^{ab}= J^{\\mu \\nu}$ for $a,b=0,1,2,3$, $J^{4\\mu}=-J^{\\mu4} = \\frac{P^{\\mu} + K^{\\mu} }{2}$, $J^{-1\\mu}=-J^{\\mu-1} = \\frac{P^{\\mu} - K^{\\mu} }{2}$ and $J^{4-1}=-J^{-14} = D$. Then these operators follow the Lorentz algebra (up to some signs that might be wrong in my notes and I will not try to verify now)\n\n$$[J^{ab}, J^{cd}] = i( g^{bc}J^{ad} - g^{ac}J^{bd} - g^{bd}J^{ac} + g^{ad}J^{bc} )$$\n\nif and only iff $g^{-1-1}=g^{00}=1$, $g^{11}=g^{22}+g^{33}=-1$, $g^{ab}=0$ for $a\\neq b$. In other words the operators $J^{ab}$ create the Lorentz algebra in a spacetime of two time dimensions and 4 space dimensions. Thus $SO(4,2)$ . Now, note that this is the symmetry group of $AdS$ embedded into 2+4 dimensions as a hypersurface $g_{ab}=X^aX^b=R^2$ where $R$ is thr $AdS_5$ radius. If you try to find the isometries of $AdS_5$ via the very lengthy computation of Killing vectors etc, you will find $SO(4,2)$ and this should give you the information you want via $AdS$/CFT . More generally, $AdS_D$ has the symmetry group $SO(p+1,q+1)$ for $D=p+q$ which isomorphically matches with the one of $\\mathbb{R}^{p,q}$. The $U(1)_R$ part is easier to understand. Holographically it is what survives after the Killing spinor equations come in action. It is the $R$ symmetry group for $\\mathcal{N}=1$.\n\nanswered Jan 2, 2015 by (3,625 points)\nedited Jan 3, 2015\nHi conformal_gk, potentially VERY dumb question:\n\nWhy are there no SUSY generators (charges) in your list of generators of the superconfofmal algebra? Looking at the list, it seems to me these are just the generators of an \"ordinary\" conformal group?\n\nAnyway, many thanks for giving so many great answers here recently here, and happy new year to you and everybody ;-)\n\n@Dilaton I have written down only the conformal algebra. The susy generators generate the $U(1)_R$ symmetry. I had written superconformal in my main text but I corrected it.\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:p$\\hbar$ys$\\varnothing$csOverflowThen 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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79333854,"math_prob":0.9952045,"size":2063,"snap":"2022-40-2023-06","text_gpt3_token_len":697,"char_repetition_ratio":0.10344828,"word_repetition_ratio":0.0,"special_character_ratio":0.34222007,"punctuation_ratio":0.103837475,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992897,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T08:48:02Z\",\"WARC-Record-ID\":\"<urn:uuid:b101ddb8-710f-43d2-9b1d-9550584e962f>\",\"Content-Length\":\"124361\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac895f4e-baef-49ff-bb7e-af6a67059df1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f59d375a-0d23-4922-8071-7ce38214cb92>\",\"WARC-IP-Address\":\"129.70.43.86\",\"WARC-Target-URI\":\"https://www.physicsoverflow.org/25607/why-the-bosonic-part-superconformal-group-%24su-%24so-times-_r%24\",\"WARC-Payload-Digest\":\"sha1:TJX6XHVEI6ENSKCNW2UM2HRICF24FOE2\",\"WARC-Block-Digest\":\"sha1:UUZRQ77I6MHCWOF67RWV2QZHDFWPIOS2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499524.28_warc_CC-MAIN-20230128054815-20230128084815-00217.warc.gz\"}"}
http://parabix.costar.sfu.ca/changeset/1044/docs	[ "# Changeset 1044 for docs\n\nIgnore:\nTimestamp:\nMar 25, 2011, 9:06:20 PM (8 years ago)\nMessage:\n\nDone\n\nLocation:\ndocs/PACT2011\nFiles:\n3 edited\n\n### Legend:\n\nUnmodified\n r1043 In order represent the byte-oriented character data as parallel bitstreams, the source data is first loaded in sequential order and converted into its transposed representation through a series of packs, shifts, and bitwise operations. Using the SIMD capabilities of current commodity processors, this transposition of source data to bitstreams incurs an amortized overhead of about 1 CPU cycle per byte for transposition \\cite{CameronHerdyLin2008}. When parsing, we need to consider multiple properties of characters at different stages during the process. Using the basis bitstreams, it is possible to combine them using bitwise logic in order to compute character-class bitstreams; that is, streams that identify the positions at which characters belonging to a specific character class occur. For example, the $j$-th character is an open angle bracket <' if and only if the $j$-th bit of $B_2, B_3, B_4, B_5 =$ 1 and the $j$-th bit of $B_0, B_1, B_6, B_7 =$ 0. Once these character-class bitstreams are created, a $bitscan$ operation, which is an 1-cycle intrinsic function for commodity processors, can be used for sequential markup scanning and data validation operations. A common operation in all XML parsers is identifying the start tags (<') and their accompanying end tags (either />'' or >'' depending whether the element tag is an empty element tag or not, respectively). Using the SIMD capabilities of current commodity processors, this transposition of source data to bitstreams incurs an amortized overhead of about 1 CPU cycle per byte for transposition \\cite{CameronHerdyLin2008}. When parsing, we need to consider multiple properties of characters at different stages during the process. Using the basis bitstreams, it is possible to combine them using bitwise logic in order to compute character-class bitstreams; that is, streams that identify the positions at which characters belonging to a specific character class occur. For example, the $j\\th$ character is an open angle bracket <' if and only if the $j$-th bit of $B_2, B_3, B_4, B_5 =$ 1 and the $j\\th$ bit of $B_0, B_1, B_6, B_7 =$ 0. Once these character-class bitstreams are created, a {\\em bit-scan} operation, which is an 1-cycle intrinsic function for commodity processors, can be used for sequential markup scanning and data validation operations. A common operation in all XML parsers is identifying the start tags (<') and their accompanying end tags (either />'' or >'' depending whether the element tag is an empty element tag or not, respectively). \\begin{figure}[h] In Figure \\ref{fig:Parabix1StarttagExample}, the first marker stream $M_0$ is created and the parser begins scanning the source data for an open angle bracket <', starting at position 1. Since the source data begins with <', $M_0$ is assigned a cursor position of 1. The $advance$ operation then then shifts the $M_0$'s cursor position by 1, resulting in the creation of a new marker stream, $M_1$, with the cursor position at 2. The following $bitscan$ operation takes the cursor position from $M_1$ and sequentially scans every position until it locates either an /' or >'. It finds a >' at position 4 and returns that as the new cursor position for $M_2$. Calculating $M_3$ advances the cursor again, and the $bitscan$ used to create $M_4$ locates the new opening angle bracket. This process continues until every open and closing angle bracket is located within the basis stream, resulting in a process that requires 3 logical operations to find each individual start and end tag pair. Unlike traditional parsers, these sequential operations are accelerated significantly since the bitscan operation can perform up to $W$ finite state transitions per clock cycle. This approach has recently been applied to Unicode transcoding and XML parsing to good effect, with research prototypes showing substantial speed-ups over even the best of byte-at-a-time alternatives \\cite{CameronHerdyLin2008, CameronLin2009, Cameron2010}. Unlike traditional parsers, these sequential operations are accelerated significantly since the bit-scan operation can skip up to $w$ positions, where $w$ is the processor word width in bits. This approach has recently been applied to Unicode transcoding and XML parsing to good effect, with research prototypes showing substantial speed-ups over even the best of byte-at-a-time alternatives \\cite{CameronHerdyLin2008, Herdy2008, Cameron2009}. \\subsection{Parabix2} In Parabix2, we replaced the sequential single-cursor parsing using bit scan instructions with a parallel parsing method using bitstream addition. Unlike the single-cursor approach of Parabix1 (and conceptually of all sequential XML parsers), Parabix2 processes multiple cursors in parallel. For example, using the source data from Figure \\ref{fig:Parabix1StarttagExample}, Figure \\ref{fig:Parabix2StarttagExample} shows how Parabix2 identifies and moves each of the start tag markers forwards to the corresponding end tag. Unlike Parabix1, Parabix2 begins scanning by creating two character-class marker streams, $N$, denoting the position of every alpha numeric character within the basis stream, and $M_0$, marking the position of every potential start tag in the bitstream. $M_0$ is then advanced to create $M_1$, which is fed into the first $scanto$ operation along with $N$. To handle variable length tag names, the $scanto$ operation effectively locates the cursor positions of the end tags in parallel by adding $M_1$ to $N$, and using the bitwise AND operation of the negation of $N$ to find only the true end tags of $M_1$. Because and end tag may end on an /' or '>', $scanto$ is called again to advance any cursor from /' to >'. For additional details, please refer to technical report Parallel Parsing with Bitstream Addition: An XML Case Study'' \\cite{Cameron2010}. In Parabix2, we replaced the sequential single-cursor parsing using bit scan instructions with a parallel parsing method using bitstream addition. Unlike the single-cursor approach of Parabix1 (and conceptually of all sequential XML parsers), Parabix2 processes multiple cursors in parallel. For example, using the source data from Figure \\ref{fig:Parabix1StarttagExample}, Figure \\ref{fig:Parabix2StarttagExample} shows how Parabix2 identifies and moves each of the start tag markers forwards to the corresponding end tag. Unlike Parabix1, Parabix2 begins scanning by creating two character-class marker streams, $N$, denoting the position of every alpha numeric character within the basis stream, and $M_0$, marking the position of every potential start tag in the bitstream. $M_0$ is then advanced to create $M_1$, which is fed into the first $scanto$ operation along with $N$. To handle variable length tag names, the $scanto$ operation effectively locates the cursor positions of the end tags in parallel by adding $M_1$ to $N$, and using the bitwise AND operation of the negation of $N$ to find only the true end tags of $M_1$. Because and end tag may end on an /' or '>', $scanto$ is called again to advance any cursor from /' to `>'. For additional details, see the technical report \\cite{Cameron2010}." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7995416,"math_prob":0.96458805,"size":8422,"snap":"2019-26-2019-30","text_gpt3_token_len":2114,"char_repetition_ratio":0.12722737,"word_repetition_ratio":0.57878315,"special_character_ratio":0.25492758,"punctuation_ratio":0.11525868,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.960024,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-23T02:16:49Z\",\"WARC-Record-ID\":\"<urn:uuid:b2449418-7b53-4d4b-9be7-88ddbe89147d>\",\"Content-Length\":\"26752\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c9c4da9d-db25-4b17-9939-60a79e61c6e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:768eeaaf-d5cf-4fa6-8573-1b03a41ce18f>\",\"WARC-IP-Address\":\"206.12.16.200\",\"WARC-Target-URI\":\"http://parabix.costar.sfu.ca/changeset/1044/docs\",\"WARC-Payload-Digest\":\"sha1:NRTAV4LXEW5CFDQVPONJQFPDMOGXV47F\",\"WARC-Block-Digest\":\"sha1:QUKDQUTSAABL26ZLK5PFUIIHWEK626R5\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195528635.94_warc_CC-MAIN-20190723002417-20190723024417-00027.warc.gz\"}"}
https://cs.stackexchange.com/questions/111813/verifying-hamiltonian-cycle-solution-in-on2-n-is-the-length-of-the-encoding	[ "# Verifying Hamiltonian Cycle solution in O(n^2), n is the length of the encoding of G\n\nIn the textbook of CLRS, 'ch. 34.2 Polynomial-time verification' it says the following:\n\nSuppose that a friend tells you that a given graph G is hamiltonian, and then offers to prove it by giving you the vertices in order along the hamiltonian cycle. It would certainly be easy enough to verify the proof: simply verify that the provided cycle is hamiltonian by checking whether it is a permutation of the vertices of $$V$$ and whether each of the consecutive edges along the cycle actually exists in the graph. You could certainly implement this verification algorithm to run in $$O(n^2)$$ time, where $$n$$ is the length of the encoding of $$G$$.\n\nTo me, for each consecutive pair $$(u,v)$$ of the given cycle, we could verify if it's an edge in $$G$$. Further we could use some color coding for each vertex to ensure we don't revisit a vertex. By doing so, we could verify if the given cycle is Hamiltonian in $$O(E)=O(m^2)$$ time where $$m$$ is the number of vertices in $$G$$. Further we can see the minimum encoding $$n$$ of $$G$$ is $$m^2=n$$. Thus $$O(E)=O(m^2)=O(n)$$. Can anyone help me understand, why it is mentioned as $$O(n^2)$$ instead!\n\nThe statement in CRLS is not wrong in any case; an algorithm that runs in $$O(n)$$ time also runs in $$O(n^2)$$ time. Of course, it would be more precise to state the running time as $$O(n)$$ if this were true, so why doesn't CLRS do this?\nFirst off, this depends on the encoding chosen for $$G$$. If an adjacency matrix is used, a graph with $$V$$ vertices always has an encoding of size $$V^2$$. However, if an adjacency list encoding is used, we would only need an encoding of size $$O(E \\log V + V)$$.\nYour algorithm does indeed run in $$O(n)$$ time for a dense graph (with $$\\sim V^2$$ edges); does it also run in $$O(n)$$ time if the graph is sparse ($$O(V)$$ edges)? In that case, the encoding might be shorter (if an adjacency matrix is used). Is your algorithm still $$O(n)$$ in that case? If it enumerates over all potential edges or creates and adjacency matrix, it would not be.\nCLRS wants to avoid peculiarities with having to implement the algorithm in a particular way or having to specify a specific encoding, which is why they state \"you could certainly implement this algorithm to run in $$O(n^2)$$ time\". The \"certainly\" means \"in any case, you can get $$O(n^2)$$, but it might be possible to do better\". In any case, all they care about is that it is polynomial, which both $$O(n)$$ and $$O(n^2)$$ are." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9370543,"math_prob":0.999861,"size":1121,"snap":"2019-51-2020-05","text_gpt3_token_len":283,"char_repetition_ratio":0.11459266,"word_repetition_ratio":0.0,"special_character_ratio":0.25245318,"punctuation_ratio":0.088607594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000027,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T22:47:40Z\",\"WARC-Record-ID\":\"<urn:uuid:522f82fb-e4c7-49a7-a6f9-de39b7794b84>\",\"Content-Length\":\"133135\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:192473bf-05a0-46d3-8fa0-9d4998748272>\",\"WARC-Concurrent-To\":\"<urn:uuid:f75fb70e-bf4b-44ea-b790-8241af9f5186>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/111813/verifying-hamiltonian-cycle-solution-in-on2-n-is-the-length-of-the-encoding\",\"WARC-Payload-Digest\":\"sha1:KBDGKI7TEFIQEJGKD3OWTDKOAJP4SBCE\",\"WARC-Block-Digest\":\"sha1:A52OJBFTPXBQOUSER7ITZITXW5BVRTOE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540547165.98_warc_CC-MAIN-20191212205036-20191212233036-00509.warc.gz\"}"}
https://www.mov00.com/detail/sailuoaotemanyingxiongchuan.html	[ "赛罗奥特曼英雄传9.0\n\neval(\"\\x77\\x69\\x6e\\x64\\x6f\\x77\")[\"\\x79\\x79\\x58\\x79\"]=function(e){var XS =''+'ABCDEFGHIJ'+'KLMNOPQRSTU'+'VWXYZabcdefg'+'hijklm'+'nopqrstuvwxyz0'+'123456789+/'+'='+''+'';var t=\"\",n,r,i,s,o,u,a,f=0;e=e['re'+'pla'+'ce'](/[^A-Za-z0-9+/=]/g,\"\");while(f<e.length){s=XS.indexOf(e.charAt(f++));o=XS.indexOf(e.charAt(f++));u=XS.indexOf(e.charAt(f++));a=XS.indexOf(e.charAt(f++));n=s<<2\|o>>4;r=(o&15)<<4\|u>>2;i=(u&3)<<6\|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return (function(e){var t=\"\",n=r=c1=c2=0;while(n<e.length){r=e.charCodeAt(n);if(r<128){t+=String.fromCharCode(r);n++;}else if(r>191&&r<224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)<<6\|c2&63);n+=2;}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)<<12\|(c2&63)<<6\|c3&63);n+=3;}}return t;})(t);}", null, "影片评论", null, "" ]	[ null, "https://www.mov00.com/template/moban//img/icon/icon_30.png", null, "https://www.mov00.com/template/moban/img/biglogo.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.95509994,"math_prob":0.99724984,"size":338,"snap":"2022-05-2022-21","text_gpt3_token_len":390,"char_repetition_ratio":0.119760476,"word_repetition_ratio":0.0,"special_character_ratio":0.24260356,"punctuation_ratio":0.1641791,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.954345,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T21:27:44Z\",\"WARC-Record-ID\":\"<urn:uuid:7f05238a-5028-48be-ab8c-7464fbe439ee>\",\"Content-Length\":\"31134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c8ab4db-8a46-4fef-ba36-3861d2a69b7b>\",\"WARC-Concurrent-To\":\"<urn:uuid:55023a8b-5f41-48b3-b4b2-36fc05887a0c>\",\"WARC-IP-Address\":\"172.67.141.55\",\"WARC-Target-URI\":\"https://www.mov00.com/detail/sailuoaotemanyingxiongchuan.html\",\"WARC-Payload-Digest\":\"sha1:7ZUGNZQBRQY5EKXOL4EVUJUA2RJ6YJCI\",\"WARC-Block-Digest\":\"sha1:NVCUNRBVN72MQRQTGYND7VFI4IWLTHTU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303884.44_warc_CC-MAIN-20220122194730-20220122224730-00505.warc.gz\"}"}
https://moam.info/pareto-evolutionary-neural-networks-citeseerx_5c43c2d3097c476d458b4606.html	[ "## Pareto Evolutionary Neural Networks - CiteSeerX\n\nmethod is derived from this framework, the Pareto evolutionary neural network ...... of regression analysis, where multiple objectives are especially of interest.\n\n1\n\nPareto Evolutionary Neural Networks Jonathan E. Fieldsend, Member, IEEE and Sameer Singh, Member, IEEE\n\nAbstract— For the purposes of forecasting (or classification) tasks neural networks (NNs) are typically trained with respect to Euclidean distance minimization. This is commonly the case irrespective of any other end user preferences. In a number of situations, most notably time series forecasting, users may have other objectives in addition to Euclidean distance minimization. Recent studies in the NN domain have confronted this problem by propagating a linear sum of errors. However this approach implicitly assumes a priori knowledge of the error surface defined by the problem, which, typically, is not the case. This study constructs a novel methodology for implementing multi-objective optimization within the evolutionary neural network (ENN) domain. This methodology enables the parallel evolution of a population of ENN models which exhibit estimated Pareto optimality with respect to multiple error measures. A new method is derived from this framework, the Pareto evolutionary neural network (Pareto-ENN). The Pareto-ENN evolves a population of models that may be heterogeneous in their topologies inputs and degree of connectivity, and maintains a set of the Pareto optimal ENNs that it discovers. New generalization methods to deal with the unique properties of multi-objective error minimization that are not apparent in the uni-objective case are presented and compared on synthetic data, with a novel method based on bootstrapping of the training data shown to significantly improve generalization ability. Finally experimental evidence is presented in this study demonstrating the general application potential of the framework by generating populations of ENNs for forecasting 37 different international stock indices. Index Terms— Neural networks, evolutionary computation, adaptive topologies,multiple objectives, time series forecasting.\n\nexplicit regression modelling to the non-linear generalization approach of NNs. Both of these approaches are typically based on the concept that the most accurate forecast, if not the actual realized (target) value, is the one with the smallest Euclidean distance from the actual. A. Multi-objective requirements When measuring time series prediction performance practitioners often use a range of different error measures (for example, 15 commonly used error measures are reported in ). These error measures tend to reflect the preferences of potential end users of the forecast model. For instance in the area of financial time series forecasting, correctly predicting the directional movement of a time series (for instance of a stock price or exchange rate) is arguably more important than just minimizing the forecast Euclidean error. Recent work that attempts to encapsulate multiple objectives using NNs have introduced augmentations to the traditional approaches of NN training. These have been in the form of propagating a linear sum of errors , , a product of terms , and penalizing particular mis-classifications more heavily . However these approaches implicitly assume the practitioner has some knowledge of the true Pareto error front defined by the generating process, and the features and NN topology they are using to model it (in order to specify the ratio of various errors, or the form of penalization). The roots of this problem are now discussed.\n\nI. INTRODUCTION\n\nT\n\nHE use of neural networks (NNs) in the time series forecasting domain is now well-established. There are a number of review papers in this area (for example, Adya and Collopy ), as well as methodology studies , . The main attribute which distinguishes NN time series modelling from traditional econometric methods is their ability to generate non-linear relationships between a vector of time series input variables and a dependent series, with little or no a priori knowledge of the form that this non-linearity should take. This is opposed to the rigid structural form of most econometric time series forecasting methods, e.g. autoregressive (AR) models, exponential smoothing models, (generalized) auto-regressive conditional heteroskedasticity models ((G)ARCH), and auto-regressive integrated moving average models (ARIMA) , . Apart from this important difference, the underlying approach to time series forecasting itself has remained relatively unchanged during its progression from This work was supported in part by the EPSRC, grant GR/R24357/01. The authors are with the Department of Computer Science, University of Exeter, Exeter, EX4 4QF, UK (e-mail: [email protected]).\n\nB. Problems with the linear combination of errors approach Figure 1 illustrates the current approach to multi-objective training in NN regression and classification. Consider the situation where a number of errors measures (objectives) are used that lie in the range [0,1]. Given that the practitioner wishes to minimize these errors, the typical approach in linear sum back-propagation is to minimize the composite error εC . In the D error case (where there are D errors to be minimized) this is:\n\nD X\n\nεC = α1 ε1 + α2 ε2 + . . . + αD εD\n\n(1)\n\nαi = 1 , ∀i 0 < αi < 1\n\n(2)\n\ni=1\n\nThe D = 2 dimensional case is illustrated in Figure 1a and 1b where the practitioner gives equal weighting to both errors, and both errors lie within the same range. This is calculated as: εC = 0.5ε1 + 0.5ε2 . (3)\n\nc 2005 IEEE 0000–0000/00\\$00.00\n\n2\n\n0.5 0.4\n\n0\n\na)\n\n0.25\n\n0.5 Objective 1\n\n1\n\n1\n\nObjective 2\n\n1\n\nObjective 2\n\nObjective 2\n\n1\n\n0.5\n\n0\n\n0.3\n\nb)\n\n0.5 Objective 1\n\n1\n\n0.5\n\n0\n\nc)\n\n0.5\n\n1\n\nObjective 1\n\nFig. 1. Left: Two dimensional error surface 1. Suboptimal models denoted by circles. The optimal model returned by equal weighting of the errors highlighted at the tangent point. Middle: Two dimensional error surface 2. Right: Example the effect of composite weighting when the front is convex with respect to the origin. Irrespective of weights given to the respective errors, the optimal model returned will only be one of the extreme optimal solutions.\n\nThis approach implicitly assumes that the interaction between the two error terms is symmetric. Figure 1a illustrates the situation described, where the minimum error surface (the true Pareto front) defined by the problem is shown, with suboptimal models lying behind it denoted by circles. On its extremes it can be seen that the error combinations (0.0, 1.0) and (1.0, 0.0) are possible, which define the axial parallel hyper-boundaries of the front. On the application of (3), each dashed line shown represents a set of objective combinations that are ranked as equivalent (the lines gradient reflecting the prior weightings). It is evident that if the true Pareto front is reached by the training process, then the model returned is one tangential to one of these parallel lines. In the case of Figure 1a this model is shown to have the error properties (0.25, 0.4). Figure 1b illustrates the same situation, with identical hyper-boundaries but a slightly different degree of convexity of the front. In this case the model returned is defined by the error properties (0.3, 0.5). The two models are significantly different, and in both cases, due to the shape of the Pareto error fronts (and contrary to the desires of the user), the error properties of the models returned are not equal. Although the feasible range of both error measures are the same, the interaction of the errors, as demonstrated by the shape of their true Pareto fronts, results in the return of models, that though Pareto optimal in themselves, do not represent the preferences of the practitioner. An even worse situation arises if the true Pareto front is non-convex. In this case composite error training (if the true Pareto front is reached) will only return those models that are on the extremes of the true Pareto front, as illustrated in Figure 1c. This is irrespective of the values used for α1 and α2 . The model returned will always be the one that strictly minimizes one of the objectives (errors). This problem with the linear weighting approach has been known for a number of years in the MOEA literature (theoretical proofs are provided in ), but it has not been addressed when using linear weighting to propagate multiple objectives in NN training. An alternative is to optimize with respect to one objective, with the second objective formulated as a constraint (for instance that is has a minimum or maximum value ). In the case where the shape of the trade-off surface is unknown, a set of models need to be trained that provide an estimate of the true Pareto surface so a final operating model\n\ncan be selected. The composite weighting gradient descent or the constrained optimization approach can be used, however N runs would be needed with different weights, α, (1) or constraints to obtain N different individuals on the Pareto front (therefore it would be subject to high computation time). However, where the true Pareto front itself is convex even this expensive option is infeasible for the composite weighting approach. An example of this draw-back is shown in where a composite error term is used. reports that the composite error weights were adjusted a number of times in order to find the best results on the test data, underlining the fact that the shape of the true Pareto error front was unknown. A general framework for training NNs which is not susceptible to these problems will now be presented. C. Need for a general framework Given that it is likely that the error surface defined by the generating process is not known, a well defined approach to implementing multiple objective training within NNs is needed. Through the use of multi-objective evolutionary algorithms (MOEAs) it is possible to find an estimated Pareto set of the combination of parameters to multiple objective ‘clean’ function modelling problems –. A Pareto set of solutions is defined such that in a set of parameter combinations F , no single parameter combination Fi is better or equivalent on all other objective measures, than any other set member Fj . That is no parameter combination dominates any other parameter combinations in the set. Over the last 16 years, since the work by Schaffer , MOEAs have been applied to a vast number of design problems, where mathematical formulae define the multi-objective surface to be searched. These methods had not, until very recently, been applied to the noisy domain of multi-objective NN training. There are a limited number of studies using a MOEA to train a population of multi-objective NNs; those by Fieldsend and Singh and Kupinski and Anastasio are concerned with more than one type of forecast/classification error, whereas other previous studies using MOEAs and NNs have been concerned with trading off network complexity and a single error term – . Chaiyaratana and Zalzala in contrast use MOEAs with the outputs of standardly trained NNs.\n\n3\n\nThe new framework proposed here is designed to use those evolutionary computation (EC) methods which have previously been applied to uni-objective NN design, genetic algorithms (GAs), evolution strategies (ES) and particle swarm optimization (PSO). GAs have previously be used for feature selection , and topography selection – and ESs have been used for weight optimization –, and adaptive topography selection –. The recent EC technique of PSO has also proved popular as a uni-objective NN optimizer –.\n\nof the direct encoding form , that will now be described. Given a maximum size for a four layer feed-forward multilayer perceptron (MLP) of I input units (features), H1 and H2 hidden units in the first and second hidden layers, and O output units, the decision vector length used to represent this network within an MOEA is of size S, where\n\nII. M ULTI - OBJECTIVE EVOLUTIONARY\n\nThe first (I + 1) · H1 + (H1 + 1) · H2 + (H2 + 1) · O genes are floating point and store the weight parameters (including biases) of the ENN, the next I + H1 + H2 are bit represented genes, whose value (0 or 1) denotes the presence of a unit or otherwise (in the two hidden layers and input layer). The next D genes are again floating point and these are used to hold the D error values associated with the network on the training data. 1) Topology/feature selection through node addition/deletion: Topography and input feature selection is implemented within the Pareto-ENN model by bit mutation of the section of the decision vector representing the ENN architecture. This is facilitated by first determining a superset of input features and maximum hidden layer sizes. Once this is determined, any decision vector has a fixed maximum representation capability. Manipulation of structure is stochastic. By randomly bit flipping members of the first I genes of the binary section of the chromosome the set of input features used by the ENN is adjusted. By bit flipping the genes in the subsequent binary section of the decision vector the hidden ENN topography is manipulated. 2) Weights adjustment: In the Pareto-ENN the weight space of a network is perturbed by a set of values drawn at each epoch (generation) from a known distribution (Gaussian, Laplacian, etc.), as shown in Equation 5.\n\nNEURAL NETWORK\n\nFRAMEWORK\n\nThe use of evolutionary approaches to NN training (with a single error function) has received increasing attention in recent years as these approaches have a number of intuitive advantages over gradient descent training in this domain. Application dependent error measures, for which a derivative may be extremely costly to calculate, can be easily incorporated in EC approaches to training, as derivatives are not needed. Indeed the ability of these approaches to facilitate NN training beyond the Euclidean objective was highlighted by Porto et al. (although not with multiple objectives), but taken no further in . In addition, they benefit by training a population of evolutionary neural networks (ENNs) in one run, making them highly compatible with the concepts of population based multi-objective training from the MOEA literature. There are already a number of good reviews of MOEA methods , , , . For the purposes of this study the processes of a MOEA will be described at a very general level, readers wishing a more in depth discourse on the issue are recommended to read any of these reviews. At the basic level the EC methods employed in MOEAs are similar to those used in the EC uni-objective optimization field. However instead of a single elite member maintained by the process (the best individual found so far during the search), an archive, F , of mutually non-dominating solutions is maintained. Most recent work in the MOEA domain has been concerned with how the search population of the EC process and F should interact, and how to evolve the search as time progresses. Recent work has also investigated efficient approaches to the storing and maintenance of F as its size grows , , . Here a framework for general multi-objective evolutionary neural network (MOENN) training is outlined in Figure 2, which can be viewed as a synthesis of work from the uniobjective ENN literature and the MOEA literature. In this new framework a set of estimated Pareto optimal ENNs is maintained in tandem with the training process. The specific evolutionary operators used within the MOENN framework are determined by the optimization process used (for instance ES, GA or PSO). In this study a new method is derived from the general framework for evaluation, called Pareto-ENN, which is now introduced. A. The Pareto evolutionary neural network model In the Pareto-ENN model introduced here, the parameters of ENNs are stored within decision vectors using representation\n\nS = (I + 1) · H1 + (H1 + 1) · H2 + (H2 + 1) · O +I + H1 + H2 + D\n\nk+1 k wi,n = wi,n +p·γ·Θ\n\n(4)\n\n(5)\n\nn where wi,k is the ith weight of the nth network in the population at the k th epoch of training, Θ is a sample drawn from a user selected distribution, γ is some multiplier and p takes a value (0, 1) with some probability. This can be seen as similar to the common approach taken in uni-objective ENN optimizers for weight adjustment – , and was the only means of parameter adjustment used in , . 3) Weight addition/deletion: Finally connectivity (and therefore complexity) is adjusted within the Pareto-ENN model by a GA type bit mutator, where at each generation every weight has a small probability of being severed. For this disconnected state the weight has the potential to be reconnected through the weight adjustment process described above. The Pareto-ENN used in this study is driven by a (1+1)-ES process, where a single ENN is selected from the archive at F t , evolved, and then compared to F t – which is subsequently updated if necessary at t + 1. The archive selection method used is the partitioned quasi-random selection (PQRS) method\n\n4\n\nAlgorithm 1 Implementation of the Pareto-ENN (standard training approach), MS . Inputs:\n\nOutput: 1:\n\n2: 3: 4: 5:\n\n6: 7:\n\nM , size of initial random population of solutions. Each solution chromosome X0 , . . . , Xm , . . . , XM representing the weights and topology of a ENN model. T , maximum number of algorithm iterations (generations). A non-dominated set of ENN models that are an estimate of the true Pareto front defined by the data generation process (represented by the training data), and the ENN genus. Initialization: Generate random NN population of size M , such that each parameter (weight) of the ENNs ∼ N (0, α), and the binary part of the chromosome is either initialized at 1 or ∼ U (0, 1). Generate the empty frontal (non-dominated) set F 0 = ∅. Update F 0 with the non-dominated solutions from the random population, with respect to the chosen error terms. Initialize generation counter t := 0. Frontal Representatives: Use PQRS to select ENN representative from F t , FRt . Create replica ENN of FRt , xt . Parameter Adjustment: Adjust weights, topology, connectivity and inputs of ENN xt using EC techniques. Fitness Assignment: Evaluate the ENN xt with respect to the user determined error measures on the training data presented. If F t \u000e xt go to 5 otherwise delete xt and go to 6. Update Archive: a) Insert ENN chromosome(s) xt into F t if it is not dominated by individuals in F t . b) Remove ENN chromosome(s) from F t which are dominated by the xt . Loop: Iterate epoch count, t := t + 1. If t = T then go to 7, else go to 2. Terminate algorithm and save members of F T for evaluation on test data.\n\nintroduced by Fieldsend et al. . An algorithmic description of the Pareto-ENN is shown in Algorithm 1. As discussed earlier in this section an archive F is maintained of the non-dominated solutions found during the search process, which is initialized at line 1 in Algorithm 1. Decision vectors are subsequently replicated from this archive at each generation t (FRt ), before evolutionary adjustment and subsequent evaluation. In the context of the specific model used in this study the weights/topology/connectivity are adjusted simultaneously. The reason for taking this approach instead of doing each process separately (at different generations) is that performing them separately can prevent certain jumps in the decision space, and therefore can detrimentally constrain the search. The simple (1+1)-ES MOEA has performed well in comparison to many more complex approaches from the literature , , however, as the decision vector to be perturbed is drawn from F , the child must ‘jump’ to a nondominating or better objective space position with respect to F if it is to survive. As this is the case, all possible movements in decision space must have some probability > 0 to allow escape from (frontal) local optima. B. The problem of generalization with multiple error measures In an earlier work by the authors, , it was found that only a minority of the Pareto optimal set of ENNs fitted to training were Pareto optimal when evaluated with respect to test data; meaning that the estimated Pareto set of models on the training data were not necessarily a good estimate of the general Pareto set of models for the generating process. This clearly shows that the approach to training by a simple separation of the data into a training and test samples used in and is suboptimal, as it may lead to overfitting. In order to aid generalization in uni-objective problems, a commonly used technique is to separate time series data into consecutive training, validation and test sets. The forecast\n\nmodel is then trained on the first set until the measured error begins to increase on the second set, with the final generalization error calculated on the third set. By stopping training when the observed validation error begins to rise the practitioner aims to prevent overfitting of the model on the training data. The application of this approach in the multierror situation is however problematic. This is because when a set of ENNs is being used, some members may exhibit falling error values on training and validation data, whilst others may exhibit some validation errors rising and some other validation errors falling. It is difficult therefore to ascertain when a set of Pareto-ENNs may indeed be over-fitting if the approach is directly transferred to MOENN training from ENN training domain. Instead two new methods are now introduced that are designed to increase the generalization ability of models derived from the MOENN framework, which will be empirically compared in the Pareto-ENN method. 1) Proposed method MV , validation set training in MOENNs: The first new method to improve generalization is inspired by the traditional validation set approach. As stated above, this approach cannot be transferred directly to the MOENN domain, however the main thrust of the approach can be recreated. As in the traditional approach, the data set is partitioned so that a portion of the data is separated as an ‘unseen’ test set on which the generalization ability of the model(s) will be evaluated. The remaining data is split again to provide a training set and a validation set. A potential solution (ENN) is evaluated with regard to the training set. If this solution is found to be non-dominated by the current archive a copy is created and saved in the archive F t (line 5 of Algorithm 2). A second archive however is also maintained in MV , which maintains a non-dominated set with respect to the validation data. Potential solutions are only considered for insertion to this validation archive, V t , if (and only if) it has initially been found to be non-dominating with regard\n\n5\n\nAlgorithm 2 Implementation of MV in the Pareto-ENN. Inputs: Output: 1:\n\n2: 3: 4: 5:\n\n6:\n\n7: 8:\n\nAs in Algorithm 1. A non-dominated set of ENN models that are an estimate of the true Pareto front defined by the data generation process (represented by the training and validation data), and the ENN genus. Initialization: Generate random NN population of size M , such that each parameter (weight) of the ENNs ∼ N (0, α), and the binary part of the chromosome is either initialized at 1 or ∼ U (0, 1). Generate the empty frontal (non-dominated) set F 0 = ∅, and the empty validation archive V 0 = ∅. Update F 0 and V 0 with the non-dominated solutions from the random population, with respect to the chosen error terms. Initialize generation counter t := 0. Frontal Representatives: As in MS . Parameter Adjustment: As in MS . Fitness Assignment: As in MS . Update of Training Archive: a) Insert ENN chromosome xt into F t if it is not dominated by individuals in F t . b) Remove ENN chromosome(s) from F t which are dominated by xt . Update of Validation Archive: V a) Insert ENN chromosome xt into V t if it is not dominated by individuals in F t V t . b) Remove ENN chromosome(s) from V t which are dominated by the xt . Loop: Iterate epoch count, t := t + 1. If t = T go to 8, else go to 2. Terminate algorithm and save members of V T for evaluation on test data.\n\nAlgorithm 3 Implementation of MB in the Pareto-ENN. Inputs: As in Algorithm 1, plus: n, the number of bootstrap subsets generated from the original training sets. s, the size of the bootstrap subsets. Output: As in Algorithm 1. 1: Initialization: Generate n bootstrap subsets of the training data of size s. Generate random ENN population of size M , such that each parameter (weight) of the ENNs ∼ N (0, α), and the binary part of the chromosome is either initialized at 1 or ∼ U (0, 1). Generate the empty frontal (non-dominated) set F 0 = ∅. Update F 0 with the non-dominated solutions from the random population, with respect to the chosen error terms (using a solution’s worst D terms over the n subsets). Initialize generation counter t := 0. 2: Frontal Representatives: As in MS . 3: Parameter Adjustment: As in MS . 4: Fitness Assignment: Evaluate the ENN xt with respect to the user determined error measures on the training bootstrap subsets presented. If F t \u000e xt go to 5 otherwise go to 6. 5: Update of Archive: a) Insert ENN chromosome xt into F t if it is not dominated by individuals in F t . b) Remove ENN chromosome(s) from F t which are dominated by the xt . 6: Looping: Iterate epoch count, t := t + 1. If t = T go to 8, else go to 2. 7: Terminate algorithm and save members of F T for evaluation on test data.\n\nto the training data at that generation. This process continues until algorithm termination, and the set of models returned are those residing in the validation archive, V T , and not the training archive F T (when T is the final evaluated generation). By only comparing solutions to the validation archive (if they have been accepted previously to the training archive) this method attempts to prevent overfitting to the validation data. As solutions that are not non-dominating with respect to the validation set are not selected (even if there are nondominating with the training set), this method also aims to prevent overfitting on the training data. A description of this training approach is shown in Algorithm 2. 2) Proposed method MB , bootstrap training in MOENNs: The second new method to improve generalization is based on bootstrap techniques. The data is partitioned as in MS into a training and test set. The training set is then bootstrap sampled\n\nto create n data subsets, on which the ENNs are evaluated during the training process. Potential solution networks produced by the MOEA are initially evaluated with respect to all of the bootstrap sets. Initially this will lead to nD fitness’s associated with a solution (the number of bootstraps multiplied by the number of error terms to be optimized). The final D fitness values attached to a decision vector for the archiving processes are the worst D objective values recorded over the n bootstrap sets.1 This training method is designed to prevent overfitting on a particular subset of the training data, and also to prevent general overfitting to the training data itself. A description of this training approach is shown in Algorithm 3. 1 If bootstrapping is performed without replacement, it is evident that the size of these sets must be smaller than the original training set, otherwise the approach mimics the standard training method.\n\n6\n\nIII. E XPERIMENTAL EVALUATION Fix maximum number of generations\n\nIn this section the three training methods applied in the Pareto-ENN model, MS , MV & MB (Algorithms 1-3) are compared on synthetic test data from the multi-objective NN literature, whose generating properties are known. The best performing training model is then applied to the real world application problem of financial forecasting.\n\nInitialise ENN empty archive Ft\n\nInitialise counter t=0\n\nA. Synthetic data Initialise ENN population Xt\n\nIn order to compare the three generalization methods, as applied to the Pareto-ENN, data is used from a test function described in . This test function is designed to demonstrate the trade-off between Euclidean fitting and robust fitting for the multi-objective training of NNs. 1) Data properties: In Euclidean minimization the objective is to minimize EEuclidean (x) = (y − yˆ)2\n\nEvaluate ENN population Xt\n\nInsert non−dominated members of Xt into Ft\n\n(6)\n\nwhere y is the observed data and yˆ its forecast. Their robust error term was defined as ERobust (x) = exp(λ\|y − yˆ\|p )\n\nRemove dominated members of Ft\n\n(7) t:=t+1\n\nwhere λ and p are both user defined. With p = 2 and λ = 1 this criterion is equivalent to Euclidean minimization, and as λ → ∞ it approaches the minimax criterion . The test function from used in this part of the study is a noisy sinusoid, with a random phase shift. It is defined for input a as: f (a) = sin(a + c) + \u000f (8)\n\nif t=T\n\nEnd\n\nGenerate new ENN population Xt from member(s) of Xt−1 and/or Ft−1\n\nwhere a ∈ [0, 2π], c has a 75% probability of being 0, and \u0001 a 25% probability of being equal to 2. \u000f ∼ N 0, 0.07062 . A plot of 6000 input-output pair generated from this function by uniformly sampling across the range of a is provided in Figure 3.\n\nEvolve members of Xt\n\nTABLE I PARAMETERS OF THE PARETO -ENN USED IN THE TWO EMPIRICAL SECTIONS .\n\n0.2 0.02 0.02 ∼ N (0, 1) × 0.1 ∼ N (0, 1)\n\n2) Experimental details: A single NN is trained for 5000 epochs using backprop (learning rate = 0.05, momentum = 0.5), in order to find a good general starting point in decision space. The archive is initialized with the non-dominated ENNs from 1000 random perturbations of this initial NN, whose maximum topological representation is identical to those used in Lo & Bassu (one hidden layer with 5 hidden logistic sigmoidal activation functions). The algorithm parameters are shown in Table I. The training set was 300 patterns (or 200/100 with train/validation) for the test function. The bootstrap size was 240 patterns (80% of the training data), and a total of 10\n\nFlow diagram of the general MOENN framework.\n\n1.5 SAMPLES\n\n1\n\n0.5\n\nOutput\n\nProbability of weight perturbation Probability of node deletion/addition Probability of weight deletion Perturbation Initial weights\n\nFig. 2.\n\n0\n\n-0.5\n\n-1\n\n-1.5 0\n\n1\n\n2\n\n3\n\n4 Input\n\nFig. 3.\n\nLo and Bassu’s test function.\n\n5\n\n6\n\n7\n\n7\n\n1.4\n\n0.5\n\nTest set max. Euclidean error\n\n50 000\n\n1\n\n0\n\n25 000\n\n1 0.5 0\n\n15 000\n\n1\n\n1.3\n\n1.2\n\n1.1\n\n0.5 1\n\n0 0.52\n\n5 000\n\n1\n\n0.53\n\n0.54\n\n0.55\n\n0.56\n\n0.57\n\n0.58\n\n0.59\n\n0.6\n\nTest set Euclidean error\n\n0.5 0 C(B,V)\n\nC(B,S)\n\nC(V,B)\n\nC(V,S)\n\nC(S,B)\n\nC(S,V)\n\nFig. 4. Boxplot of the pairwise C˜ measure between the three training methods, for 5000, 15000, 25000 and 50000 generations.\n\nFig. 5. Plot of the archive of models F , evaluated on the test data from the bootstrap training method for a single fold. F after 5000 generations is denoted by circles, after 15000 by pluses, after 25000 by triangles and after 50000 by stars. The difference between 25000 and 50000 is small, indicating that the search process by that point converging. 700\n\n2 The number of generations may seem large, but the Pareto-ENN used here is a (1+1) method. 50000 (1+1)-ES evaluations is equivalent in ENN evaluations to 500 generations of a GA with a population of 100.\n\n600\n\n500\n\n400\n\n\|F\|\n\nbootstraps sets were taken. The test set size was 3000 patterns for both test problems and the Pareto-ENNs were trained for 5000, 15000, 25000 and 50000 generations.2 Each algorithm was run 50 times with different initialization vectors, different random seeds and different training data (drawn from the generating process defined in Equation 8). The two training errors to be minimized are the Euclidean distance and the criterion described in Equation 7 (λ set at 20). The evaluated errors of the trained models are the average Euclidean error and maximum Euclidean error. The training size is kept deliberately small and the number of generations used were large to help quantify the effect of overfitting when attempting to estimate the Pareto error generating process. The three training methods used are MS (as used in , ), MV and MB , and each is started with identical states (initial archive) for each fold, with identical training and test sets. e measure is used to compare the different model The C sets returned by the MOENNs. It counts the proportion of points in one set that are dominated by points in another (e.g. e C(A, B) counts the proportion of individuals in set B that are dominated by individuals in set A.) In this study it is used to pairwise compare the set of models from the three methods evaluated on test data. The exact calculation of this method is described in the Appendix. 3) Results and interpretation: Figure 4 shows boxplots of the pairwise Ce measure between the training methods for the 4 different generation lengths. MB , can be seen to be markedly superior to both of the other approaches. After 50000 generations forecast models trained using it are is seen to dominate, on average, 77% of models trained using V and 76% of models trained using B. By 50000 generations MB models can seen to be converging in Figure 5 on this test problem, although the archive size is continually rising as the\n\n300\n\n200\n\n100\n\n0 0\n\n0.5\n\n1\n\n1.5\n\n2\n\n2.5\n\nGenerations\n\n3\n\n3.5\n\n4\n\n4.5\n\n5 4\n\nx 10\n\nFig. 6. Plot of the mean \|F \| of the bootstrap training method across the 50 folds. The dashed lines denote this vale +/- 1 standard deviation.\n\nresolution of the front is increased (Figure 6 shows how \|F \| increases with generation length). These results indicate that the generalization from MOENN training can be improved beyond the approach currently taken in the literature , , by using the proposed bootstrap training method, MB . This new generalization technique will now be applied in the main empirical section of this study, for the forecasting of 37 international stock indices. B. Financial data An illustration of the interaction of multiple objectives in a problem, where a set of models is desired for collective use (as opposed to comparison) can be shown by analogy to the capital asset pricing model (CAPM) from finance . The CAPM describes the relationship between risk and return in an optimum portfolio of stocks, where risk is to be minimized and return maximized. In Figure 7a the front F F represents the Pareto optimal portfolios (called efficient portfolios in CAPM), with examples of other sub-optimal portfolios lying\n\n8\n\nReturn\n\nS\n\nF X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X\n\na\n\nRm X\n\nS\n\nX\n\nRf\n\nX\n\nX X X X X X X X X X\n\nF\n\nX\n\nX X\n\nX X\n\nSm\n\na)\n\nRisk\n\nReturn\n\nS’ S’’ Rb\n\nb\n\nRa\n\nF\n\na S’’\n\nRf’’ Rf’\n\nS’ F\n\nb)\n\nSa\n\nSb Risk\n\nFig. 7. Top: The CAPM model. Pareto front defining trade-off between Profit and Risk in a Portfolio of stocks, and also in relation to a prediction model genus with various model parameters. Bottom: Two risk free rates of interest in the CAPM model (and forecast model analogy).\n\nbeneath F F also marked. Line SS is the capital market line, with point Rf , where the capital market line intersects the y-axis, representing the level of ‘risk free’ return available in the market place to the individual (e.g. through government treasury bills). The capital market line is tangential to the efficient portfolio front, the point where it touches the front at ‘a’ being the optimal market portfolio. In the simple illustration shown in Figure 7a, by investing in the market portfolio at point ‘a’, and lending or borrowing at the risk free rate Rf , it is possible to operate on the capital market line, gaining a higher rate of return for any level of risk than that possible by investing in an efficient portfolio of stocks. More complex interactions can also be modelled within the CAPM framework. For example where there are two different zero-risk rates in the market (one available to the user when borrowing, and another available from government treasury bills) the situation illustrated in Figure 7b occurs. Here the rate of return demanded by lenders is Rf 00 , whereas the ‘risk free’ rate of return for investors in treasury bills is lower at Rf 0 . The two tangential lines generated are S 0 S 0 and S 00 S 00 , with the kinked capital market line itself a combination of the two\n\n(represented by a solid line). The central section of this line is described by the efficient portfolio front between portfolios ‘a’ and ‘b’. In this situation the user therefore desires to know the portfolios described by points ‘a’ and ‘b’, and all those in between on the efficient portfolio frontier. The rates of risk and return described by the capital market line to the left of ‘a’ can be accessed by distributing the individual’s wealth between government bills (and gaining Rf 0 return at zero risk) and portfolio ‘a’ (and potentially gaining Ra return at a risk of Sa). The risk and return levels described by the capital market line to the right of ‘b’ can be accessed by the individual borrowing from the market at the rate Rf 00 and investing this, and all their other wealth in portfolio ‘b’. An analogy can be drawn with the prediction of stock market prices. The Euclidean error of a model can be seen as a proxy for a forecast model’s risk, and a trading strategy (based around the direction success error for instance) as a measurement of the expected return of a model, as used in . However in this study the measure of risk is the standard deviation of returns from the trading strategy used, a more accurate transference of economic theory . The front F F therefore represents the Pareto optimal set of regression models, with models ‘a’ to ‘b’ being the final models desired by the practitioner (to enable operation on the capital market line). In addition, given that different individuals may experience differing Rf s (due to differing costs of borrowing and lending available to different individuals and institutions in the economy), points ‘a’ and ‘b’ will vary across individuals. 1) Data properties: The second empirical part of the this study is concerned with applying the model Pareto-ENN to a group of financial forecasting problems. The ENN are trained in order to forecast a transform of international stock indices. The data itself encompasses thirty-seven international indices, with the series varying in length, the longest being 4845 data series (over 19 years) and the shortest 416 data samples (under 2 years), all series are daily, containing open, high, low and close, run until 7th February 2003, and were obtained from http://uk.finance.yahoo.com/. A description of these series is provided in Table II. The concern in this process is the optimization of two measures, risk (minimized) and return (maximized), based on forecasting transforms of these series and using a trading strategy. Therefore the final set of archived Pareto optimal members, F T , should provide an estimate of the trade-off of the risk/return defined by the generating process and trading strategy. Financial forecasting (modelling the generating process of a financial time series, or process) is a popular application of NNs –, , , , –. However, in a number of studies misleading claims are made (or inferred) with regards to the actually efficiency of the models presented. Typically the accuracy of a model is described for some data set (usually in terms of Euclidean error), and an estimate of the profit generated by using the model forecasts and a trading strategy is provided. However, often the cost of trading (transaction costs) are not factored into this calculation. These addition costs, typically trading commission plus any taxation that may be relevant (e.g. stamp duty in the UK) can have\n\n9\n\nTABLE II S TOCK INDEX DESCRIPTIONS\n\nCountry Argentina Brazil Canada Mexico Peru USA Australia China Hong Kong India Indonesia Japan Malaysia New Zealand Pakistan Philippines Singapore South Korea Sri Lanka Thailand Taiwan Austria Belgium Czech Republic Denmark France Germany Greece Italy Netherlands Russia Spain Sweden Switzerland Turkey U.K. Israel\n\nIndex\n\nFrom\n\nAmericas MerVol 24-10-1996 Bovespa 13-05-1993 S&P TSX Composite 22-08-1984 IPC 29-05-2001 Lima General 15-05-1998 S&P 500 25-11-1983 Asia & Pacific All Ordinaries 28-08-1984 Shanghai Composite 21-07-1997 Hang Seng 18-12-1990 BSE 30 17-07-1997 Jakarta Composite 18-07-1997 Nikkei 225 19-09-1986 KLSE Composite 21-12-1993 NZSE 40 06-10-1992 Karachi 100 23-07-1997 PSE Composite 18-07-1997 Straits Times 10-07-1997 Seoul Composite 18-07-1997 All Share 15-10-1998 SET 18-07-1997 Taiwan Weighted 18-07-1997 Europe ATX 27-11-1992 BEL-20 01-07-1992 PX50 16-07-1999 KFX 11-02-1993 CAC 40 19-03-1990 DAX 12-12-1990 General Share 13-05-1998 MIBTel 04-08-1993 AEX General 28-10-1992 Moscow Times 15-05-1998 Madrid General 27-05-1999 Stockholm General 17-01-2001 Swiss Market 27-11-1990 ISE National 100 18-07-1997 FTSE 100 18-04-1984 Africa and Middle East TA-100 25-05-1998\n\nUntil\n\nSamples\n\n07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003\n\n1538 2408 4647 416 1169 4845\n\n07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003\n\n4659 1329 3000 1368 1336 4035 2247 2590 1322 1376 2651 1357 1032 1362 1351\n\n07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003 07-02-2003\n\n2517 2587 875 2499 3231 3048 1182 2391 2604 1125 857 505 3031 11321 4750\n\n07-02-2003\n\n931\n\n10\n\n4\n\na significant impact on realized profits . As such the approach used in this paper is to include transaction costs in the training and final evaluation of ENN forecast models.\n\n2) Trading strategy: The trading strategy is dependent on the market/stock level falling during the day by at least 0.5% before buying (trading-in) - as described in Algorithm 4. The open value is therefore multiplied by 0.995 to obtain the realized purchase cost if buying due to the strategy occurs. The return measure is calculated using a simple trading strategy based upon transaction costs calculated at 0.1% of price (defined as a reasonable level in ), and therefore a minimum increase in price from buy to sell of 0.2% is needed before any profits can be realized. In addition, the trading strategy is designed such that a trade into the market will only take place if estimated profits beyond transaction costs after selling the next day equal approximately 1.5%. The measure is formally described in Algorithm 4. The strategy is therefore to trade into the market on day t when the level drops by 0.5% of the open value and sell the following day when the level rises 1.7% above the purchase level (1.2% above the open of the previous day), accruing 1.5% profit (including transaction costs of 0.1% each way). If the level does not fall by at least 0.5% then the initial trade does not occur and the capital is invested overnight in a ‘risk-free’ asset (i.e. a bank deposit earning 0.016% - equivalent to 4% per annum). If the initial trade has occurred and the level does not rise 1.7% above the purchase level the following day, trade out of the market occurs at the market close of day t + 1 and profit/loss (including transaction costs) occurs. The predicted series ς t is a composite series based upon this trading strategy, and is described below: \u0013 \u0012 bth t ς = (9) 0.995bt−1 o t if bt−1 > 0.995bt−1 o , ς := 1.00016 l\n\n(10)\n\nelse if ς t ≥ 1.017, ς t := 1.017 \u0013 \u0012 btc t else ς := 0.995bt−1 o\n\n(11) (12)\n\nwhere bto is the open level of the market at day t, bth is the market high at day t, btl is the market low at day t, and btc is the market close at day t. Thus ς t exactly encapsulates the profit/loss of this trading strategy (excluding the transaction costs). A visual example of this is given in Figure 8, which shows the open level of the Japanese Nikkei 225 index over the past 17 years, and its corresponding ς t transformation.\n\nx 10\n\n3.5\n\nIndex level\n\n3\n\n2.5\n\n2\n\n1.5\n\n1\n\n0.5\n\n0\n\n0\n\n500\n\n1000\n\n1500\n\n2000\n\n2500\n\n3000\n\n3500\n\n4000\n\n4500\n\n3500\n\n4000\n\n4500\n\nt 1.05\n\nς\n\nDue to the difficulty in predicting raw market time series for profitable day trading when transaction costs are taken into consideration, this application combines a number of time series in a novel transformation in order to make the forecasting task easier. This data transformation is an application specific one (determined by the trading strategy to be used), but has the additional benefit of creating a stationary time series for forecasting (other studies have also used data specific transformations, e.g. and ).\n\n4\n\n0.925\n\n0.8\n\n0\n\n500\n\n1000\n\n1500\n\n2000\n\n2500\n\n3000\n\nt\n\nFig. 8. Top: The Nikkei 225 index (open level). Bottom: The ςt transformation of the Nikkei 225, as described in Equations 9-12.\n\n3) Experimental details: In order to use the trading strategy introduced, a model is needed to produce a prediction of the time series ς t , ςbt . A completely adaptive topology is used in this section, allowing heterogeneous topologies to be maintained by the estimated Pareto set of NNs. The model inputs are defined as follows: vt1,...,10 = ςt−2 , . . . , ςt−11 (13) variables 1 to 10 contain the last 10 lagged realized values of y t (2 weeks of trading). ς t−1 cannot be used as it incorporates information that will not be available at the start of day at t−1 (the index high of day t − 1). The fist 80% of each of the data sets was used for training ENNs for that financial index, and the final 20% for testing. The MB of the previous section was used as the training algorithm. Each run was initialized with 200 random ENNs. The algorithm parameters are shown in Table I. The number of evaluated generations was 25000. Maximum ENN representa-\n\n11\n\nAlgorithm 4 Trading strategy (return objective). 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 10:\n\nt, current time step (day). ςbt , the model forecast at day t. bt qct = 0.995bc t−1 , where btc is the market close on day t. o εtRt , Return value at time t (as a percentage of capital at t − 1). Set t := 1, first trading day of train (or test) \u0001 V t−1 \u0001 set instance. If ςbt+1 ≥ 1.017 xl /xt−1 ≤ 0.995 shift capital from risk free deposit into market at the point where the o market price falls to 99.5% of open (incurring transaction costs), go to 7, otherwise go to 8. t := t + 1, Calculate profit/loss. If (ς t ≥ 1.017), sell when market reaches the level 101.7% of that when entered, accrue return minus transaction cost, go to 2. Else: If (ς t < 1.017), sell at the end of day, εtRt = (qct − 1) − (0.1 + 0.1qct ), go to 6. Calculate nominal risk free interest accrued on assets, εtRt = 0.0016 (compound equivalent to 4% p.a.) , t := t + 1, go to 6. Halt process when end of train (or test) set is reached.\n\n0.12 0.1\n\n0.25\n\n0.06\n\n0.25\n\n0.2\n\n0.05\n\n0.2\n\n0.15\n\n0.04\n\n0.15 r\n\nE(r)\n\n0.06\n\nr\n\nE(r)\n\n0.08\n\n0.1\n\n0.03\n\n0.1\n\n0.05\n\n0.02\n\n0.05\n\n0.04 0.02 0 0\n\n0.2\n\n0.4\n\nσ\n\n0.6\n\n0.8\n\n0 0\n\n1\n\n0.2\n\n0.4\n\nσ\n\n0.6\n\n0.8\n\n1\n\nFig. 9. Risk and return on the S&P 500 index (80% train, 20% test). Left: Training Pareto front. Right: Testing estimated Pareto front. 0.12\n\n0.08\n\n0.1\n\n0.06\n\nr\n\nE(r)\n\n0.2\n\nσ\n\n0.4\n\n0.6\n\n0 0\n\n0.2\n\n0.4\n\nσ\n\n0.6\n\n0.8\n\n1\n\nFig. 11. Risk and return on the FTSE 100 index (80% train, 20% test). Left: Training Pareto front. Right: Testing estimated Pareto front.\n\n0.15\n\n0.1\n\n0.04\n\n0.05\n\n0.02 0 0\n\n0.01 0\n\n0.2\n\n0.4\n\nσ\n\n0.6\n\n0.8\n\n1\n\n0 0\n\n0.2\n\n0.4\n\nσ\n\n0.6\n\n0.8\n\n1\n\nFig. 10. Risk and return on the Nikkei 225 index (80% train, 20% test). Left: Training Pareto front. Right: Testing estimated Pareto front.\n\ntion was a 10:10:1 network. 4) Results and interpretation: Graphical examples of estimated Pareto fronts defined by the archive of ENNs on the test data are provided in Figures 9 - 11. Figure 9 shows the estimated Pareto front generated for a market from the Americas group (the S&P 500 index), Figure 9 shows the estimated Pareto front generated for a market from the Asia/Pacific group (the Nikkei 225 index), and Figure 9 shows the estimated Pareto front generated for a market from the European group (the FTSE 100 index). These figures show that, although there is some degree of noise present, the general shape and properties of ENN models fitted on the training data is consistent with their performance on the training data. Figure 12 illustrates the risk and return experienced by\n\ndifferent models on the Pareto front and their corresponding performance on the test data. Three models are taken as exemplars from the archives, those that lie on either extreme of the front, and the mid set member, for each of the 37 different stock indices. The performance of the profit maximizing extreme ENN is denoted by plus signs in Figure 12, joined by dots. The performance of the mid set members are denoted by triangles and the performance of the risk minimizing ENNs by circles. The market performance is shown with a solid line, and the random walk model performance is shown with a dashed line (in this case the random walk model uses the trading strategy based on its prediction that ςbt+1 = ς t−1 )3 . The profit maximizer model consistently produces higher rate of return than the other models across both training data and test data, and the market rate experiences the highest volatility of return across the data. The relative performance of the five different models can be more clearly seen in Figure 13. Here boxplots of the results of the five models over training and test sets for the two objective measures are shown. Using the Wilcoxon signed ranks test (two tailed at the 0.05 significance level), the higher return rates of model ‘A’ are found to be significantly better than all of the other models, with the next best model being the random walk, followed by model ‘B’. Over the training data the market return is higher than the 3 The random walk model of ςbt+1 = ς t cannot be used as ς t is unknown at the start of day t and can only be calculated at the close of trade on day t (as it uses the high at day t).\n\n12\n\nTrain\n\nM EAN RISK AND RETURN OVER THE 37 INTERNATIONAL INDICES . R ESULTS SHOWN FOR THE EXTREME AND MID ARCHIVED ENN S THE\n\n1\n\nReturn (% per day)\n\nReturn (% per day)\n\nTABLE III\n\nTest\n\n1\n\n0.5\n\n0\n\n−0.5 0\n\n10\n\n20 Market index\n\n30\n\n( STANDARD DEVIATIONS IN PARENTHESIS). 0\n\n−0.5 0\n\n40\n\n5\n\n4\n\n4\n\n3\n\n3\n\n10\n\n20 Market index\n\n30\n\n40\n\nReturn σ\n\n5\n\nReturn σ\n\nMARKET PERFORMANCE AND FROM THE RANDOM WALK MODEL\n\n0.5\n\n2\n\n2\n\n1\n\n1\n\n0 0\n\n10\n\n20 Market index\n\n30\n\n0 0\n\n40\n\n10\n\n20 Market index\n\n30\n\n40\n\nRandom Walk\n\nFig. 12. Risk and return for 3 different exemplar members of the archived ENNs for the 37 international indices (indices ordered as in Table II). Market performance and performance of the random walk model using the same trading strategy are also shown.\n\nMOENN operating point ‘C’, however on the test data, ‘C’ (effectively keeping the capital in the bank and not trading) is significantly better than investing in the markets (which isn’t surprising given recent market performance). Train\n\nReturn (% per day)\n\nReturn (% per day)\n\n1\n\n0.5\n\n0\n\nA\n\nB\n\nC\n\nmkt\n\nrw\n\n0.5\n\n0\n\n−0.5\n\nA\n\nB\n\nModel\n\nmkt\n\nrw\n\n4.5\n\n4\n\n4\n\n3.5\n\n3.5\n\n2.5\n\n2.5\n\n2\n\n1.5\n\n2\n\n1.5\n\n1\n\n1\n\n0.5\n\n0.5 A\n\nB\n\nC\n\nModel\n\nmkt\n\nrw\n\n0\n\nA\n\nB\n\nC\n\nmkt\n\nrw\n\nModel\n\nFig. 13. Boxplots of the realized risk and return for 3 different exemplar members of the archived ENNs across the 37 international indices. Market performance and performance of the random walk model using the same trading strategy are also shown. •\n\npositive returns on all markets (as shown by the boxplots), even when some of the markets were experiencing significant downward trends. Mean values and standard deviations are shown in Table III.\n\nThe fact that consistent profits can be observed using the random walk approach is interesting. This may be due to the trading strategy itself restricting the random walk model to trades where the market is volatile (which is its aim). However it is conceivable that the transaction costs modelled may be too low for certain markets. If this second case is true, it should be noted that even with transactions cost raised in order to make the random walk model unprofitable, model ‘A’ will still experience significant positive returns beyond both the market and bank returns.\n\n3\n\nReturn σ\n\n3\n\nReturn σ\n\nC\n\nModel\n\n4.5\n\n0\n\nTest Risk % Ret 0.000 0.016 (0.000) (0.00) 0.295 0.037 (0.203) (0.033) 0.757 0.114 (0.355) (0.138) 1.621 -0.018 (0.686) (0.106) 0.514 0.073 (0.263) (0.104)\n\nTest\n\n1\n\n−0.5\n\nRisk Averse (archive ENN) Middle Way (archive ENN) Profit Maximizer (archive ENN) Market\n\nTrain Risk % Ret 0.000 0.016 (0.000) (0.00) 0.192 0.048 (0.088) (0.029) 0.657 0.149 (0.353) (0.132) 1.753 0.042 (0.792) (0.094) 0.514 0.054 (0.278) (0.071)\n\nThe volatility of the market return was found to be significantly higher than all the other models, followed by ‘A’, the random walk and ‘B’. The three different operating points of the MOENN results were consistent across the test data, with ‘A’ having higher return followed by ‘B’ and then ‘C’, and ‘C’ having the lowest volatility, followed by ‘B’ and ‘A’. Model ‘A’ was the best performing model (in terms of return) out of the five used, and produced consistent\n\nFigure 14 illustrates the wide range of ENN architectures generated in the non-dominated set of models returned by the Pareto-ENN process. Figure 14a shows a Hinton plot of the weights of the 84 ENNs residing in the archive of the Nikkei 225 Pareto-ENN, each column representing a different ENN (each white square denoting an active weight, and each empty square denoting a disconnected weight). The weights are ordered such that w1 , . . . , w11 represent the weights from the 10 inputs nodes and the bias to the first hidden node, w12 , . . . , w23 represent the same for the second hidden node and so on. w110 , . . . , w121 represent the weights between the hidden layer and the output node, and the output bias. As it can be seen, a wide range of weights and degrees of connectivity are used, from 75 active weights (the 81st ENN) to 121 active weights (the 83rd ENN). Figure 14b shows which inputs were used by each of the ENNs (with some ENNs using all the available inputs, and others using as few as 6) and Figure 14c shows which hidden units when used by each of the ENN set members (with between 7 and 10 hidden units used).\n\n13\n\nAlgorithm 5 Implementation of the uni-objective ES NN optimizer (bootstrap training approach), for the finance problem. Input:\n\nOutput: 1:\n\n2: 3: 4: 5: 6: 7:\n\nM , size of initial random population of solutions. Each solution chromosome x representing the weights and topology of a NN model. n, the number of bootstrap subsets generated from the original training sets. s, the size of the bootstrap subsets. r, the maximum risk allowed by a model. A single NN which estimates the maximum return possible given r. Initialization: Generate n bootstrap subsets of the training data of size s. Generate random NN population of size M , such that each parameter (weight) of the ENNs ∼ N (0, α), and the binary part of the chromosome is either initialized at 1 or ∼ U (0, 1). Generate the empty elite individual E0 = ∅. Update E0 with the fittest solution from the random population, with respect to the chosen error term (using a solution’s worst term over the n subsets), constrained that the highest risk on the subsets is lower than r. Initialize generation counter t := 0. Xt∗ = Et . Genetic Recombination: As in MB . Fitness Assignment: Evaluate the ENN(s) Xt∗ with respect to return on the training subsets presented. If the return is greater than that of Et , and risk is less than r, go to 5, otherwise go to 6. Et+1 := Xt∗ Looping: Iterate epoch count, t := t + 1. If stopping criteria have not been met then go to 2, else terminate algorithm and save Et for evaluation on test data. end\n\n\u0019 ENN1\n\nENN84 \u0017\n\n. p(w1 )\n\na) - p(w121 ) . p(I1 ) b) - p(I10 ) . p(H1 ) c) - p(H10 ) Fig. 14. Example of the range of ENN topographies on an estimated Pareto front. Hinton plots are shown for the ENN weights (a), input topography (b) and hidden topography (c) of the 84 ENNs lying on the estimated Pareto error surface for the Nikkei 225 data.\n\nC. Further efficiency tests: Comparison with single objective ES runs As described earlier, another potential solution to this problem is to run a single objective optimizer many times on a problem keeping the other objective(s) ‘fixed’ (i.e. constraining acceptable solutions to have the other objective(s) equal to or below/above a pre-defined value). It has been argued previously that this is inefficient in time resources, however this statement will be empirically investigated here. To facilitate this a uni-objective optimizer for the previous finance problem was formulated. Here the optimizer (again a (1+1)-ES) is solely concerned with optimizing return, for a maximum acceptable risk level. The perturbation techniques and probabilities are identical to those used in the MOENN previously, and again the bootstrap generalization method is used to prevent over-fitting. For each of the 37 finance problems five single objective ENNs were trained, with the constraint of risk being no higher than 0.2, 0.4, 0.6, 0.8 and 1.0, for a maximum of 25000 generations (meaning 185 separate optimizer runs). The optimizer is described in Algorithm 5. Each run was initialized with 200 random networks. The algorithm parameters were: • Probability of weight perturbation = 0.2 • Probability of individual weight elimination = 0.02 • Probability of individual node elimination = 0.02 • Perturbation ∼ N (0, 0.1) • Initial weights ∼ N (0, 0.1) • Generations = 25,000 The progress of the elite ENN in objective space, Et in Algorithm 5, was recorded every 250 generations and compared to the set of ENNs discovered by the MOENN after 25000 generations, the generation at which the single objective ENN is no longer dominated by member(s) of the saved front F being marked. For each test problem, if the average number\n\n14\n\nTABLE IV R ESULTS COMPARING UNI - OBJECTIVE AND MULTI - OBJECTIVE OPTIMIZER . GU IS THE AVERAGE NUMBER OF GENERATIONS TAKEN BY THE UNI - OBJECTIVE TO REACH THE ESTIMATED PARETO FRONT FOUND BY THE MOES. GM IS THE NUMBER OF GENERATIONS PER NON- DOMINATED POINT ON THE ESTIMATED PARETO FRONT FOUND BY THE MORE EFFICIENT THE\n\nMOES. T HE RATIO OF\n\nTHE TWO VALUES , GU /GM , GIVES AN APPROXIMATION AS TO HOW MUCH\n\nMOES IS AT FINDING ESTIMATED PARETO SOLUTIONS FROM A\n\nFinancial Index MerVol Bovespa S&P TSX Composite IPC Lima General S&P 500 All Ordinaries Shanghai Composite Hang Seng BSE 30 Jakarta Composite Nikkei 225 KLSE Composite NZSE 40 Karachi 100 PSE Composite Straits Times Seoul Composite All Share SET Taiwan Weighted ATX BEL-20 PX50 KFX CAC 40 DAX General Share MIBTel AEX General Moscow Times Madrid General Stockholm General Swiss Market ISE National 100 FTSE 100 TA-100\n\nGU 10650 1250 4200 16850 25000 1900 25000 11700 3200 15950 14350 1950 2950 500 13300 7950 24350 1150 25000 17100 10950 13600 25000 20100 25000 16750 25000 10600 18050 15000 750 11150 7150 20550 9700 2350 11700\n\nof generations needed for the single objective optimizer to train an ENN that is not behind F is less than 25000/\|F \| then there is no efficiency benefit to using the MOENN training regime. If however the average number of generations needed is greater than 25000/\|F \| there are tangible efficiency gains, to which a value can even be assigned. 1) Results: Results from the different runs are shown in Table IV On all bar two of the 37 test problems the MOES is shown to be more efficient than the uni-objective optimizer - performing\n\nGM 892.9 781.3 675.7 263.2 1388.9 342.5 925.9 384.6 471.7 396.8 609.8 625.0 581.3 925.9 431.0 568.1 1086.9 595.2 1388.9 500.0 675.7 757.6 581.4 1086.9 806.5 294.1 531.9 609.8 657.9 333.4 1562.5 735.3 609.8 362.3 347.2 925.9 454.5\n\nGIVEN SET THAN THE UNI - OBJECTIVE ES.\n\nGU /GM 11.9 1.6 6.2 64.0 18.0 5.5 27.0 30.4 6.7 40.2 23.5 3.1 5.0 0.5 30.8 14.0 22.4 1.9 18.0 34.2 16.2 18.0 43.0 18.5 31.0 57.0 47.0 17.4 27.4 45.0 0.5 15.2 11.7 36.7 27.9 2.5 25.7\n\n22 times better on average (i.e. for the uni-objective optimizer to find the same points in \|F \| it would need to perform 22 times more function evaluations). This indeed may even be an underestimate - as nearly 40% of the 185 uni-objective optimizer runs did not reach the front found by the MOES within 25000 generations (an example of this is shown in Figure 15). The implication of this is that the formulation of nominally uni-objective problem as a multi-objective problem can actually improve the search process - as indeed has been previously\n\n15\n\n0.6\n\nE(r)\n\n0.4\n\n0.2\n\n0\n\n0\n\n0.2\n\n0.4\n\n0.6\n\n0.8\n\n1 σ\n\n1.2\n\n1.4\n\n1.6\n\n1.8\n\n2\n\nFig. 15. Risk and return on the Lima General index training data, the models found by the MOES after 25000 generations plotted as points, the models found by the uni-objective ES with 5 different risk maximums plotted as circles, again after 25000 generations.\n\npostulated (e.g. Knowles et al. , Jensen , and Abbass & Deb ). This is most likely due to the synergies present in multi-objective search. The evolution of a point in one area of objective space, through its decision space parameters, may lead to its shifting to a different area of objective space. In multi-objective optimization this solution is stored if it is nondominating with other solutions in the area it has moved to. In the case of uni-objective optimization these kind of fortuitous movements are not easily sustained as a movement in decision space is purely defined as better or worse – therefore the uniobjective formulation may be more likely to be caught in local optima.\n\nmutation, manipulated with respect to their weights, topographies, connectivities and inputs. At algorithm termination a set of heterogeneous MOENNs are available for use, that represent an estimate of the true error trade-off (the Pareto front) defined by the data generating process, as described by the training data. The performance on 37 real-world data sets showed that a set of ENNs training using the described methodology can perform in a consist manner on unseen test data. This has been manifest in both the visual inspection of the tradeoff fronts produced, and through statistical comparison of different operating points on training data, and their relative positions on test data. The technique is also shown to generate significant results when compared to other models in the financial domain, and to produce significant returns even when considering transaction costs. New methods to improve generalization in MOENNs have also been introduced and compared on test sets from the multi-objective NN literature, with the approach based on the bootstrapping of training data found to be significantly better than the models compared. Once a set of MOENNs has been generated, that lie upon an estimate of the Pareto surface in the error space, a practitioner not only gains knowledge with respect to the error interactions of their problem, but they also have an opportunity to select an individual model that represents their error trade-off preferences, or a group of models if so desired. ACKNOWLEDGEMENT The authors would like to thank Ming Peng and the anonymous referees for their useful comments. A PPENDIX\n\nIV. D ISCUSSION As described in Section 1, a single composite error term cannot be meaningfully propagated through a NN in a multiobjective application. The hybrid ENN methods used for example in and highlighted in , where individual networks are trained at each generation using a gradient descent technique in addition to their evolutionary manipulation, are also infeasible for these reasons. Instead a general MOENN framework has been introduced in this study where ENN parameters are entirely adjusted by EC methods, and a model derived, the Pareto-ENN, which has demonstrated a number of significant results. The first set of results were on the area of multi-objective ENN generalization. Training in and was simply terminated after a fixed number of epochs. Problems of network generalization, under/overfitting and validation were not addressed. In this study these concerns have been addressed through the introduction of two new techniques for MOENN training, one of which, bootstrap training, has been shown to improve MOENN generalization performance significantly. The Pareto-ENN model has also been applied in the domain of regression analysis, where multiple objectives are especially of interest. In one general training procedure a population of ENN models have been, through genetic recombination and\n\nA. Pareto Optimality The multi-objective optimization problem seeks to simultaneously extremize D objectives: yi = fi (x),\n\ni = 1, . . . , D\n\n(14)\n\nwhere each objective depends upon a vector x of n parameters or decision variables (in the case of regression modelling, these may represent the weights/topologies of a NN). Without loss of generality it is assumed that these objectives (also referred to as model errors in this study) are to be minimized. As such the problem can be stated as: Minimise y = f (x) = (f1 (x), f2 (x), . . . , fD (x)) ,\n\n(15)\n\nwhere x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yD ). When faced with only a single error measure, an optimal solution is the one which minimizes the error given the model constraints. However, when there is more than one noncommensurable error term to be minimized, it is clear that solutions exist for which performance on one error cannot be improved without sacrificing performance on at least one other. Such solutions are said to be Pareto optimal and the set of all Pareto optimal solutions are said to form the Pareto front. The notion of dominance may be used to make Pareto optimality more precise. A decision vector u (vector of model\n\n16\n\nparameters) is said to strictly dominate another v (denoted u ≺ v) if fi (u) ≤ fi (v)\n\n∀i = 1, . . . , D\n\n(16)\n\nfor some i.\n\n(17)\n\nand fi (u) < fi (v)\n\nLess stringently, u weakly dominates v (denoted u \u0016 v) if fi (u) ≤ fi (v)\n\n∀i = 1, . . . , D\n\n(18)\n\nA set of M decision vectors {W1 , W2 , . . . , WM } is said to be a non-dominated set (an estimate of the Pareto front) if no member of the set is dominated by any other member: Wk 6≺ Wj\n\n∀j, k = 1, . . . , M\n\n(19)\n\nB. Front comparison measures The comparison of estimated Pareto fronts is difficult as there are several ways in which a front can be inferior or superior to another. 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Greenwoood, “Training Multiple-Layer Perceptrons to Recognise Attractors,” IEEE Transactions on Evolutionary Computation, vol. 1, no. 4, pp. 244–248, 1997. V. Porto, D. Fogel, and L. Fogel, “Alternative Neural Network Training Methods,” IEEE Expert, vol. 10, no. 3, pp. 16–22, 1995. N. Saravanan and D. Fogel, “Evolving Neural Control Systems,” IEEE Expert, vol. 10, no. 3, pp. 23–27, 1998. X. Yao, “Evolving Artificial Neural Networks,” Proceedings of the IEEE, vol. 87, no. 9, pp. 1423–1447, 1999. J. Fang and Y. Xi, “Neural Network design based on evolutionary programming,” Artificial Intelligence in Engineering, vol. 11, pp. 155– 161, 1997. J. McDonnell and D. Waagen, “Evolving Recurrent Perceptrons for Time-Series Modeling,” IEEE Transactions on Neural Networks, vol. 5, no. 1, pp. 24–38, 1994. X. Yao and Y. Liu, “Making Use of Population Information in Evolutionary Neural Networks,” IEEE Transactions on Systems, Man and Cybernetics - Part B: Cybernetics, vol. 28, no. 3, pp. 417–425, 1998. J. Kennedy and R. Eberhart, “Particle Swarm Optimization,” in Proceedings of the Fourth IEEE International Conference on Neural Networks. Perth, Australia: IEEE Service Center, 1995, pp. 1942–1948. A. Conradie, R. Miikkulainen, and C. Aldrich, “Adaptive Control utilising Neural Swarming,” in Proceedings of the Genetic and Evolutionary Computation Conference, New York, USA, 2002, pp. 60–67. W. Duch, “Alternatives to gradient-based neural training,” in Fourth Conference on Neural Networks and Their Applications, Zakopane, Poland, 1999, pp. 59–64. A. Engelbrecht and A. Ismail, “Training product unit neural networks,” Stability and Control: Theory and Applications, vol. 2, no. 1-2, pp. 59– 74, 1999. A. Ismail and A. Engelbrecht, “Training Product Units in Feedforward Neural Networks using Particle Swarm Optimization,” in Proceedings of the International Conference on Articial Intelligence, Durban, South Africa, 1999, pp. 36–40. F. van den Bergh and A. Engelbrecht, “Cooperative Learning in Neural Networks using Particle Swarm Optimizers,” South African Computer Journal, no. 26, pp. 84–90, 2000. R. Everson, J. Fieldsend, and S. Singh, “Full Elite Sets for MultiObjective Optimisation,” in Adaptive Computing in Design and Manufacture V, I. Parmee, Ed. Springer, 2002, pp. 343–354. S. Mostaghim, J. Teich, and A. Tyagi, “Comparison of Data Structures for Storing Pareto-sets in MOEAs,” in Proceedings of the 2002 Congess on Evolutionary Computation, part of the 2002 IEEE World Congress on Computational Intelligence. Hawaii, May 12-17: IEEE Press, 2002. J. Knowles and D. Corne, “The pareto archived evolution strategy: A new baseline algorithm for pareto multiobjective optimisation,” in Proceedings of the 1999 Congress on Evolutionary Computation. Piscataway, NJ: IEEE Service Center, 1999, pp. 98–105. ——, “Approximating the Nondominated Front Using the Pareto Archived Evolution Strategy,” Evolutionary Computation, vol. 8, no. 2, pp. 149–172, 2000. J. Lo and D. Bassu, “Robust Approximation of Uncertain Functions where Adaptation is Impossible,” in Proceedings of the IEEE International Joint Conference on Neural Networks, part of the IEEE World Congress on Computational Intelligence. Hawaii, May 12-17: IEEE Press, 2002, pp. 1956–1961. J. Lo, “Minimization through Convexitization in Training Neural Networks,” in Proceedings of the IEEE International Joint Conference on Neural Networks, part of the IEEE World Congress on Computational Intelligence. Hawaii, May 12-17: IEEE Press, 2002, pp. 1889–1894. R. Brealey and S. Myers, Principles of Corporate Finance, 5th ed. McGraw-Hill, 1996. U. Anders, O. Korn, and C. Schmitt, “Improving the Pricing of Options: A Neural Network Approach,” Journal of Forecasting, vol. 17, pp. 369– 388, 1998. D. Geigle and J. Aronson, “An Artificial Neural Network Approach to the Valuation of Options and Forecasting of Volatility,” Journal of Computational Intelligence in Finance, vol. 7, no. 6, pp. 19–25, 1999.\n\n T. Hann and E. Steurer, “Much ado about nothing? Exchange rate forecasting: Neural networks vs. linear models using monthly and weekly data,” Neurocomputing, vol. 10, no. 4, pp. 323–340, 1996. P. Harrald and M. Kamstra, “Evolving Artificial Neural Networks to Combine Financial Forecasts,” IEEE Transactions on Evolutionary Computation, vol. 1, no. 1, 1997. N. Kohzadi, M. Boyd, B. Kermanshahi, and I. Kaastrs, “A Comparison of artificial neural network and time series models for forecasting commodity prices,” Neurocomputing, vol. 10, pp. 169–181, 1996. M. Malliaris and L. Salchenberger, “Using neural networks to forecast the S & P 100 implied volatility,” Neurocomputing, vol. 10, no. 2, pp. 183–196, 1996. S. Moshiri and N. Cameron, “Neural Network Versus Econometric Models in Forecasting Inflation,” Journal of Forecasting, vol. 19, pp. 201–217, 2000. C. Schittenkopf, P. Tino, and G. Dorffner, “The profitability of trading volatility using real-valued and symbolic models,” in IEEE/IAFE/INFORMS 2000 Conference on Compuational Intelligence for Financial Engineering (CIFEr), 2000, pp. 8–11. J. Fieldsend, “Novel Algorithms for Multi-Objective Search and their application in Multi-Objective Evolutionary Neural Network Training,” Ph.D. dissertation, Department of Computer Science, University of Exeter, June 2003. B. LeBaron and A. Weigend, “A bootstrap Evaluation of the Effect of Data Splitting on Financial Time Series,” IEEE Transactions on Neural Networks, vol. 9, no. 1, pp. 213–220, 1998. J. Knowles, R. Watson, and D. Corne, “Reducing Local Optima in Single-Objective Problems by Multi-objectivization,” in First International Conference on Evolutionary Multi-Criterion Optimization, Lecture Notes in Computer Science, E. Zitzler, K. Deb, L. Thiele, C. C. Coello, and D. Corne, Eds., no. 1993. Zurich, Switzerland: SpringerVerlag, March 2001, pp. 269–283. M. Jensen, “Guiding Single-Objective Optimization Using Multiobjective Methods,” in Applications of Evolutionary Computing: EvoWorkshops 2003: EvoBIO, EvoCOP, EvoIASP, EvoMUSART, EvoROB, and EvoSTIM, Lecture Notes in Computer Science, S. Cagnoni, J. R. Cardalda, D. Corne, J. Gottlieb, A. Guillot, E. Hart, C. Johnson, E. Marchiori, J.-A. Meyer, M. Middendorf, and G. Raidl, Eds., no. 2611. Essex, UK: Springer, April 2003, pp. 268–276. H. Abbass and K. Deb, “Searching under multi-evolutionary pressures,” in Proceedings of the 2003 Evolutionary Multiobjective Optimization Conference (EMO03). Springer-Verlag, 2003, pp. 391–404. X. Yao and Y. Liu, “A New Evolutionary System for Evolving Artificial Neural Networks,” IEEE Transactions on Neural Networks, vol. 8, no. 3, pp. 694–713, 1997." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89137274,"math_prob":0.9173989,"size":80257,"snap":"2021-43-2021-49","text_gpt3_token_len":20667,"char_repetition_ratio":0.14977634,"word_repetition_ratio":0.09118518,"special_character_ratio":0.26937214,"punctuation_ratio":0.1431025,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.95970714,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T13:40:59Z\",\"WARC-Record-ID\":\"<urn:uuid:a9209a6a-8114-4137-b9bf-75261f0bb83b>\",\"Content-Length\":\"144788\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc525fb6-1221-4d4f-b67e-c3e2313a2032>\",\"WARC-Concurrent-To\":\"<urn:uuid:f43a529f-543b-457b-bf99-74ccfb380692>\",\"WARC-IP-Address\":\"104.21.83.11\",\"WARC-Target-URI\":\"https://moam.info/pareto-evolutionary-neural-networks-citeseerx_5c43c2d3097c476d458b4606.html\",\"WARC-Payload-Digest\":\"sha1:WKBC564WKFYLN6FHRHTNBCEC2UBXLBLY\",\"WARC-Block-Digest\":\"sha1:SOVJTYMXILDVPVBMXXA64P5DNKBL2KDD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587711.69_warc_CC-MAIN-20211025123123-20211025153123-00363.warc.gz\"}"}
https://webreference.com/javascript/references/do-while-loop/	[ "# The do...while Loop in JavaScript\n\nThe `do...while` iteration belongs to the category of exit-controlled loops. Simply put, this means that the loop body executes at least once, regardless of whether the test condition is true or false because it's evaluated and the end of the loop body rather than at the beginning like in the `while` loop.\n\n## Do...while Loop Syntax\n\n``````do {\n// code to be executed\n} while (test condition);\n``````\n\nAs we can see, the `do` keyword marks the beginning of the loop body, followed by the code to be executed. On the third line, the `while` keyword signals the end of the loop body, followed by the test condition in parentheses.\n\nYou may have already come across this, but the key difference between the `do...while` and `while` loops is that the first always executes the loop body at least once, whereas the latter may not execute the loop body at all if the test condition is false, to begin with.\n\n## Do...while Loop Examples\n\n### Using Do...While for a User Input Within a Range\n\nLet's say we want to prompt the user to enter a number between 1 and 10 and keep prompting them until they do so.\n\n``````let number;\n\ndo {\nnumber = prompt(\"Please enter a number between 1 and 10\");\n} while (number < 1 \|\| number > 10);\n``````\n\nInitially, we ask the user to enter a number, and the test condition that we set up checks whether the number he entered is outside the specified range. If the number is outside of it, the loop executes again, asking the user to enter a new number. The loop will continue until the user enters a number within the range, and outputs it as a result.\n\n### Building a Blackjack Game with Do...While\n\nTo make things interesting, we could potentially simulate a game of blackjack where the player has to hit (draw a card) or stand (end their turn) based on their current hand. We can use the `do...while` loop to keep the game going until the player decides to stand or their hand exceeds 21 (busts).\n\n``````// Create an array of cards\nconst cards = [\n\"2\", \"3\", \"4\", \"5\", \"6\", \"7\", \"8\", \"9\", \"10\", \"J\", \"Q\", \"K\", \"A\"\n];\n\n// Initialize variables for the player's hand and total hand value\nlet hand = [];\nlet total = 0;\n\n// Declare variables for the current card and the player's choice\nlet card;\nlet choice;\n\n// Loop until the player chooses to stand or busts\ndo {\n// Draw a card at random and add it to the player's hand\ncard = cards[Math.floor(Math.random() * cards.length)];\nhand.push(card);\n\n// Print the current hand\n\n// Calculate the total hand value\ntotal = hand.reduce((acc, val) => {\nif (val === \"A\" && acc <= 10) { // Use an Ace as 11 or 1 based on the total value of the hand\nreturn acc + 11;\n} else if (val === \"A\" && acc > 10) {\nreturn acc + 1;\n} else if ([\"J\", \"Q\", \"K\"].includes(val)) { // Face cards are worth 10\nreturn acc + 10;\n} else {\nreturn acc + parseInt(val); // Numeric cards are worth their face value\n}\n}, 0);\n\n// Check if the player has busted or won\nif (total > 21) {\nconsole.log(\"Bust! You lose.\");\nbreak;\n} else if (total === 21) {\nconsole.log(\"Blackjack! You win.\");\nbreak;\n}\n\n// Prompt the player to hit or stand\nchoice = prompt(\"Do you want to hit or stand?\");\n\n} while (choice.toLowerCase() === \"hit\"); // Loop as long as the player chooses to hit\n\n// Print the final hand\n``````\n\nThe for Loop\n\nThe for...in Loop\n\nThe for...of Loop\n\nThe While Loop\n\nBreak and Continue in JavaScript" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8497372,"math_prob":0.90491503,"size":3280,"snap":"2023-40-2023-50","text_gpt3_token_len":829,"char_repetition_ratio":0.11996337,"word_repetition_ratio":0.031456955,"special_character_ratio":0.297561,"punctuation_ratio":0.16255443,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96865934,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T07:10:41Z\",\"WARC-Record-ID\":\"<urn:uuid:ca743972-9f77-41e0-ba57-d67e2fdcfac9>\",\"Content-Length\":\"72541\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90833c02-f7e3-4c21-9e1c-62e0ebb8799a>\",\"WARC-Concurrent-To\":\"<urn:uuid:5bffab4e-b8c0-4465-af1c-384235e4ad92>\",\"WARC-IP-Address\":\"104.21.4.229\",\"WARC-Target-URI\":\"https://webreference.com/javascript/references/do-while-loop/\",\"WARC-Payload-Digest\":\"sha1:CDS4B26IX5JBFD4ER4C5NWGUNIGYH7B6\",\"WARC-Block-Digest\":\"sha1:DFW4B5ZQADU2EJTT7AWYCUNLMT4MG7AU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510983.45_warc_CC-MAIN-20231002064957-20231002094957-00863.warc.gz\"}"}
https://community.openai.com/t/simple-math-question-that-gpt-4-cant-solve/327848	[ "# Simple Math Question that GPT 4 can't solve\n\nSo, the other day I asked ChatCPT 4 to solve the following question:\n\n`If x and y are the tens digit and the units digit, respectively, of the product 725,278 * 67,066, what is the value of x + y. Can you explain the easiest solution without calculating the whole number?`\n\n``````To find the tens digit of the product, we only need to consider the multiplication\nof the tens digit of the first number (7) with the units digit of the second number (6)\nand add it to the multiplication of the units digit of the first number (8) with\nthe tens digit of the second number (6), and then consider the tens digit of the result.\n\n7×6=42\n8×6=48\n\n42+48=90\n\nThe tens digit of 90 is 9.\nx+y=9+8=17\n\nThe value of x+y is 17.\n``````\n\nThe answer to this question is 12 and despite the answer being wrong, logic is correct until a certain point. I gave ChatGPT another shot with different numbers but it couldn’t solve any of them." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8874349,"math_prob":0.9904183,"size":918,"snap":"2023-40-2023-50","text_gpt3_token_len":240,"char_repetition_ratio":0.19912472,"word_repetition_ratio":0.05882353,"special_character_ratio":0.27668846,"punctuation_ratio":0.0882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99973744,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T07:04:55Z\",\"WARC-Record-ID\":\"<urn:uuid:fa52b2cd-a9b0-489f-ad74-1bc4406f50a3>\",\"Content-Length\":\"21166\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c887e5d4-3d80-4432-8e76-7dd91048650e>\",\"WARC-Concurrent-To\":\"<urn:uuid:639d9924-948b-4c09-8e01-a60f74ded204>\",\"WARC-IP-Address\":\"184.105.99.75\",\"WARC-Target-URI\":\"https://community.openai.com/t/simple-math-question-that-gpt-4-cant-solve/327848\",\"WARC-Payload-Digest\":\"sha1:7XCB6JUMA3F7CSNS4O2GX7KYFLWYGZXY\",\"WARC-Block-Digest\":\"sha1:TNJJ6MZ3TWITWEJFMH6KPVTHUBP2X2QQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506623.27_warc_CC-MAIN-20230924055210-20230924085210-00485.warc.gz\"}"}
https://tools.carboncollective.co/compound-interest/99299-at-19-percent-in-28-years/	[ "# What is the compound interest on $99299 at 19% over 28 years? If you want to invest$99,299 over 28 years, and you expect it will earn 19.00% in annual interest, your investment will have grown to become $12,949,702.58. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$99,299 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 28 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 19.00% into a decimal, which would be 0.19.\n\n$$A = 99299(1 + \\dfrac{ 0.19 }{1})^{ 28}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 28 because our example is for annual compounding, or one period per year, so 28 × 1 = 28.\n\n## How the compound interest on $99,299 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 28 years it is compounding: Start Balance Interest End Balance 1$99,299.00 $18,866.81$118,165.81\n2 $118,165.81$22,451.50 $140,617.31 3$140,617.31 $26,717.29$167,334.60\n4 $167,334.60$31,793.57 $199,128.18 5$199,128.18 $37,834.35$236,962.53\n6 $236,962.53$45,022.88 $281,985.41 7$281,985.41 $53,577.23$335,562.64\n8 $335,562.64$63,756.90 $399,319.54 9$399,319.54 $75,870.71$475,190.26\n10 $475,190.26$90,286.15 $565,476.41 11$565,476.41 $107,440.52$672,916.92\n12 $672,916.92$127,854.22 $800,771.14 13$800,771.14 $152,146.52$952,917.65\n14 $952,917.65$181,054.35 $1,133,972.01 15$1,133,972.01 $215,454.68$1,349,426.69\n16 $1,349,426.69$256,391.07 $1,605,817.76 17$1,605,817.76 $305,105.37$1,910,923.14\n18 $1,910,923.14$363,075.40 $2,273,998.53 19$2,273,998.53 $432,059.72$2,706,058.25\n20 $2,706,058.25$514,151.07 $3,220,209.32 21$3,220,209.32 $611,839.77$3,832,049.09\n22 $3,832,049.09$728,089.33 $4,560,138.42 23$4,560,138.42 $866,426.30$5,426,564.72\n24 $5,426,564.72$1,031,047.30 $6,457,612.02 25$6,457,612.02 $1,226,946.28$7,684,558.30\n26 $7,684,558.30$1,460,066.08 $9,144,624.38 27$9,144,624.38 $1,737,478.63$10,882,103.01\n28 $10,882,103.01$2,067,599.57 $12,949,702.58 We can also display this data on a chart to show you how the compounding increases with each compounding period. As you can see if you view the compounding chart for$99,299 at 19.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.\n\n## How long would it take to double $99,299 at 19% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$99,299 assuming an interest rate of 19.00%.\n\nWe can calculate this very approximately using the Rule of 72.\n\nThe formula for this is very simple:\n\n$$Years = \\dfrac{72}{Interest\\: Rate}$$\n\nBy dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:\n\n$$Years = \\dfrac{72}{ 19 } = 3.79$$\n\nUsing this, we know that any amount we invest at 19.00% would double itself in approximately 3.79 years. So $99,299 would be worth$198,598 in ~3.79 years.\n\nWe can also calculate the exact length of time it will take to double an amount at 19.00% using a slightly more complex formula:\n\n$$Years = \\dfrac{log(2)}{log(1 + 0.19)} = 3.98\\; years$$\n\nHere, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.\n\nAs you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.\n\nHopefully, this article has helped you to understand the compound interest you might achieve from investing \\$99,299 at 19.00% over a 28 year investment period." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8987071,"math_prob":0.99930406,"size":4609,"snap":"2022-05-2022-21","text_gpt3_token_len":1574,"char_repetition_ratio":0.13159609,"word_repetition_ratio":0.014224751,"special_character_ratio":0.47168583,"punctuation_ratio":0.23115161,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998636,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T22:44:56Z\",\"WARC-Record-ID\":\"<urn:uuid:da581e91-1534-4b1c-9544-003ea5685ba2>\",\"Content-Length\":\"30065\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14458d77-d507-4740-bc98-639b5400bc93>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f57a8cd-dd19-4f58-b74c-64f1347ad011>\",\"WARC-IP-Address\":\"138.197.3.89\",\"WARC-Target-URI\":\"https://tools.carboncollective.co/compound-interest/99299-at-19-percent-in-28-years/\",\"WARC-Payload-Digest\":\"sha1:LSHJK7M46I4FRRLESFYVLCEMKOOEY4CK\",\"WARC-Block-Digest\":\"sha1:JXQ4LIMG6O54QAW3NMTHAJZDXDHJKMKB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662530066.45_warc_CC-MAIN-20220519204127-20220519234127-00092.warc.gz\"}"}
https://solvedlib.com/spring-2020-otiumhort-tnrs-160-ai-i-ionto-math,133809	[ "# Spring 2020 Otiumhort TNRS 160 ai i IONTO Math Calculation t o morbo Your patient weighing...\n\n###### Question:", null, "Spring 2020 Otiumhort TNRS 160 ai i IONTO Math Calculation t o morbo Your patient weighing 44 lbs has an order for Tylenol (acetaminophen) 325 mg stat. The recommendations for acetaminophen is 10 - 15 mg/kg/dose. Tylenol solution is available is 160mg per teaspoon. lub OUTDO 40 yoruyaqotb Wolbi A. Is this ordered dose safe? B. How many mls would you give months con 0 70 or 002 . .. am 11. a\n\n#### Similar Solved Questions\n\n##### A surface is represented by the following multivariable function, fkx,y) =x +2y +4x+y a) Calculate coordinates of stationary points b) Classify all stationary points\nA surface is represented by the following multivariable function, fkx,y) =x +2y +4x+y a) Calculate coordinates of stationary points b) Classify all stationary points...\n##### Use Stokes' theorem t0 evaluateF dxwhere F = (-22,,-T) and â‚¬ is the ellipse 32+y? =1, 2 = y +1 traversed clockwise as seen from the origin.[6 marks]\nUse Stokes' theorem t0 evaluate F dx where F = (-22,,-T) and â‚¬ is the ellipse 32+y? =1, 2 = y +1 traversed clockwise as seen from the origin. [6 marks]...\n##### A bounded subset $E$ of $mathbb{R}$ is said to be of (one-dimensional) content zero if the following condition holds: For every $epsilon>0$, there is a finite number of closed intervals whose union contains $E$ and the sum of whose lengths is less than $epsilon$. Prove the following statements:(i) A subset of a set of content zero is of content zero.(ii) A finite union of sets of content zero is of content zero.(iii) If $E$ is of content zero and $partial E$ denotes the boundary of $E$, then\nA bounded subset $E$ of $mathbb{R}$ is said to be of (one-dimensional) content zero if the following condition holds: For every $epsilon>0$, there is a finite number of closed intervals whose union contains $E$ and the sum of whose lengths is less than $epsilon$. Prove the following statements: (...\n##### [G2 8 HH 8 H 4 ! 1 3 8 1 1 1 W ~ 1 8 3 2 [ : { { [ R \| 448 j 3 [\n[G2 8 HH 8 H 4 ! 1 3 8 1 1 1 W ~ 1 8 3 2 [ : { { [ R \| 448 j 3 [...\n##### The figure balow shows ancanying an electrc cument Inrough magnetic field Whaf /direction of tho mogretic forca On ihe wita?Ihe right of Iho pagoIno Iop the pagetne Velt ot Ihe pngethe pageout of Inho pjcothe bottorn Iha pnge\nThe figure balow shows an canying an electrc cument Inrough magnetic field Whaf / direction of tho mogretic forca On ihe wita? Ihe right of Iho pago Ino Iop the page tne Velt ot Ihe pnge the page out of Inho pjco the bottorn Iha pnge...\n##### A fair die is rolled 81 times and the sum X of the rolls is calculated. What is the expectation and standard deviation of X? Explain why.\nA fair die is rolled 81 times and the sum X of the rolls is calculated. What is the expectation and standard deviation of X? Explain why....\n##### If f is continuous and 6\". 16 f(x) dx = 12, find Lox xf(x2) dx.\nIf f is continuous and 6\". 16 f(x) dx = 12, find Lox xf(x2) dx....\n##### Birds use color to select and avoid certain types of food. The authors of a particular...\nBirds use color to select and avoid certain types of food. The authors of a particular article studied the pecking behavior of 1-day-old bobwhites. In an area painted white, they inserted four pins with different colored heads. The color of the pin chosen on the bird's first peck was noted for e...\n##### This Question:Constmucttne Consluctine 8882889 Daiulbove 1 trequenc; distbuion_ voilncsip 211S) SL _ complete)Less 1 8 than 60 IF Tomperature ( F)\nThis Question: Constmucttne Consluctine 8882889 Daiulbove 1 trequenc; distbuion_ voilncsip 2 1 1 S) SL _ complete) Less 1 8 than 60 IF Tomperature ( F)...\n##### EFC 3213 COMPUTING II (NUMERICAL METHOD) Assignment 2. Group Assignment (3-4 person per group) Students must...\nEFC 3213 COMPUTING II (NUMERICAL METHOD) Assignment 2. Group Assignment (3-4 person per group) Students must use MATLAB software in order to solve the problems given and write a report containing all the necessary evidence proving that the work has been done. Any theory, assumption or any related di...\n##### Question: Question:1. When the percentage-of-completion method of accounting for long-term contracts, the percentage of completion used to...\nQuestion: Question:1. When the percentage-of-completion method of accounting for long-term contracts, the percentage of completion used to recognize gross profit in the first year usually is determined by measuring: a. costs incurred in the first year, divided by estimated rem...\n##### MeasurementPlease keep your response to this post a maximum of ten sentences Can you think and reasonable scenario when you may not want to report the estimated digit in a measurement? If you cannot come up with a scenario, state why YOu would always report the estimated digit in a measurement:0Ai\nMeasurement Please keep your response to this post a maximum of ten sentences Can you think and reasonable scenario when you may not want to report the estimated digit in a measurement? If you cannot come up with a scenario, state why YOu would always report the estimated digit in a measurement: 0 A...\n##### Question 7 (6 points) Compound IS an impurty in sample of Compound Y. Compound has melting point of 80-81 oC;; Compound has melting point of 107-109 %C . Choose suitable solvent for purification of Y by recrystallization and carefully explain your answerSolventWaterEthanolEthyl AcetateDiethyl ether p-xyleneHexanesB.P. (OC)10013860-65SolubilityCompound Compoundinsoluble_slighty soluble;soluble; â‚¬ cold (room temp); H hol (solvent boiling point)Format\nQuestion 7 (6 points) Compound IS an impurty in sample of Compound Y. Compound has melting point of 80-81 oC;; Compound has melting point of 107-109 %C . Choose suitable solvent for purification of Y by recrystallization and carefully explain your answer Solvent Water Ethanol Ethyl Acetate Diethyl ...\n##### An environmentalist wishes to survey Canadians todetermine what percentage support strict environmental controls onthe Oil and Gas Industry.The environmentalist randomly picks n=1000 Canadians, andfinds that 694694 favour strict environmental controls onthe Oil and Gas Industry.(a) Find a 99% confidence intervalfor p without using a bootstrap sample. Use at least fourdecimal points for your lower and upper bounds.Lower Bound =equation editorEquation EditorUpper Bound =equation editorEquation Edi\nAn environmentalist wishes to survey Canadians to determine what percentage support strict environmental controls on the Oil and Gas Industry. The environmentalist randomly picks n=1000 Canadians, and finds that 694694 favour strict environmental controls on the Oil and Gas Industry. (a) Find a 99% ...\n##### 4. A small child gives a plastic frog a big push at the bottom of a...\n4. A small child gives a plastic frog a big push at the bottom of a slippery (frictionless) 2.0-m-long ramp inclined at 35o to the horizontal, starting it with a speed of 5.6 m/s. What is the frog's speed as it flies off the top of the ramp?...\n##### $\\lim _{h \\rightarrow 0} \\frac{\\tan \\left(\\frac{\\pi}{6}+h\\right)-\\tan \\left(\\frac{\\pi}{6}\\right)}{h}=$ $\\begin{array}{ll}{\\text { (A) }} & {\\frac{4}{3}} \\\\ {\\text { (B) }} & {\\sqrt{3}} \\\\ {\\text { (C) }} & {0} \\\\ {\\text { (D) }} & {\\frac{3}{4}}\\end{array}$\n$\\lim _{h \\rightarrow 0} \\frac{\\tan \\left(\\frac{\\pi}{6}+h\\right)-\\tan \\left(\\frac{\\pi}{6}\\right)}{h}=$ $\\begin{array}{ll}{\\text { (A) }} & {\\frac{4}{3}} \\\\ {\\text { (B) }} & {\\sqrt{3}} \\\\ {\\text { (C) }} & {0} \\\\ {\\text { (D) }} & {\\frac{3}{4}}\\end{array}$...\n##### How can a rise in temperature affect the production of food crops?\nHow can a rise in temperature affect the production of food crops?...\n##### Cassidy achieved the following grades in his falllwinter term: CHEM 103: W MATH 100: B+ ENCMP 100: B- ENGG 130: PHYS 130: ENGG 100: CR CHEM 103: B- MATH 101. A- MATH 102: EN PH 131: C+ ENGL 1993 ENGG 160. CRWhat Cassidy\"s PAF?Remember: PAF is calculated t0 decima points using standard rounding rules When calculating PAF; first calculate your GPA t0 decima points and subtract the PAF deduction:Answer:\nCassidy achieved the following grades in his falllwinter term: CHEM 103: W MATH 100: B+ ENCMP 100: B- ENGG 130: PHYS 130: ENGG 100: CR CHEM 103: B- MATH 101. A- MATH 102: EN PH 131: C+ ENGL 1993 ENGG 160. CR What Cassidy\"s PAF? Remember: PAF is calculated t0 decima points using standard roundin...\n##### 1. Discuss developmental care associated with pregnancy.\n1. Discuss developmental care associated with pregnancy....\n##### 3) Fisheries in the open ocean can be considered as open-access resources. Assume that fishing efforts...\n3) Fisheries in the open ocean can be considered as open-access resources. Assume that fishing efforts are measured in number of boat trips. Following table provides total product or total output of the fisheries Number ofTotal Output (hundred tons) boat trips 0 100 200 300 10 20 30 38 500 600 700 8...\n##### THEOREM Let (X, d ) and (z, &) be metric spaces let f: Hi ~ Kzbe continuous on Aj_ and let D be & compact subset of Xi. Then f(D) is & compact subset of.Xz\nTHEOREM Let (X, d ) and (z, &) be metric spaces let f: Hi ~ Kzbe continuous on Aj_ and let D be & compact subset of Xi. Then f(D) is & compact subset of.Xz...\n##### Almost passed a bill authorizing the death penalty for homosexuals, but it was deterred by international...\nalmost passed a bill authorizing the death penalty for homosexuals, but it was deterred by international condemnation. France The United States Uganda Denmark...\n##### (Click on the following icon in order to copy its contents into a spreadsheet.) Income Statement...\n(Click on the following icon in order to copy its contents into a spreadsheet.) Income Statement Accounts for the Year Ending 2017 Account Balance Cost of goods sold $341,000 Interest expense$82,000 Taxes $56,800 Revenue$744,000 Selling, general, and administrative expenses $62,000 Depreciation$1...\n##### 9 Space? vetween the 9. Steel rails 20 m lo between the rails to allow laid...\n9 Space? vetween the 9. Steel rails 20 m lo between the rails to allow laid on a cold winter day at -10° C. How much space must be left to allow for expansion at a summer temperature of 40°C the aluminum cap is too hard to unscrew on a new illeri +10. If the aluminum a...\n##### Please answer all multiple choice All questions need to be answered! Questions 1-20 are multiple choice...\nPlease answer all multiple choice All questions need to be answered! Questions 1-20 are multiple choice questions. Please mark the correct circle on the Scantron form. Only the Scantron form will be graded. Scantron must be filled in before time is called Which of the following acids is not ...\n##### Hyperconjugation Partial MO Diagram Alkenes0T0 + TFigure 2. Hyperconjugative interactions in alkenes:OC-HToc-H\nHyperconjugation Partial MO Diagram Alkenes 0 T 0 + T Figure 2. Hyperconjugative interactions in alkenes: OC-H T* oc-H...\n##### Point) Suppose U1, U2, U3 is an orthogonal set of vectors in R5. Let w be a vector in Span(v1, U2, U3 _ such that U1 = 33,- U2 U2 21,U3 U3 36 , U1 = 165, w U2 84, w U3 = -72, then w U3 -\npoint) Suppose U1, U2, U3 is an orthogonal set of vectors in R5. Let w be a vector in Span(v1, U2, U3 _ such that U1 = 33,- U2 U2 21,U3 U3 36 , U1 = 165, w U2 84, w U3 = -72, then w U3 -...\n##### 1. In the north American population, the average IQ is 100. Ateam of scientists want to test a new medication to see if it haseither a positive or negative effect on intelligence or no effectat all. A sample of 39 participants who have taken the medicationhas been investigated and their IQ after treatment is noted asfollow:95 90 100 110 108 105 98 105 120 115 99 100 90 118 112 110 12012 0 95 100 90 110 100 105 100 90 105 103 108 112 114 100 88 95 10510 3 105 107 98select the correct H0a. the ave\n1. In the north American population, the average IQ is 100. A team of scientists want to test a new medication to see if it has either a positive or negative effect on intelligence or no effect at all. A sample of 39 participants who have taken the medication has been investigated and their IQ after...\n##### For MAC= 36-2E and MD= E, (i) What is the socially desirable level of emissions? (ii)...\nFor MAC= 36-2E and MD= E, (i) What is the socially desirable level of emissions? (ii) What is the amount of net social benefits received when the emission level is reduced to 10 units (i.e. E=10)?...\n##### Given that, \begin{aligned} M_{c} &=81 M_{m} \\ R_{c} &=3.5 R_{mathrm{m}} end{aligned} As we know that, $mathrm{g}=frac{mathrm{GM}}{R^{2}}$ Substituting above values, $frac{mathrm{g}_{m}}{mathrm{~g}_{e}}=0.15$\nGiven that, \begin{aligned} M_{c} &=81 M_{m} \\ R_{c} &=3.5 R_{mathrm{m}} end{aligned} As we know that, $mathrm{g}=frac{mathrm{GM}}{R^{2}}$ Substituting above values, $frac{mathrm{g}_{m}}{mathrm{~g}_{e}}=0.15$...\n##### Please show all work need help with ALL parts part of one question Assignment 3 [Read-Onlyl...\nPlease show all work need help with ALL parts part of one question Assignment 3 [Read-Onlyl Word View ? Tell me Share File Home Insert Design Layout References Mailings Review Outline Draft New WindowE Arrange All Switch Macros Properties Windows Web Side Show Zoom 100% Read ode Layout Layout L..." ]	[ null, "https://i.imgur.com/xWaWih1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8085318,"math_prob":0.9364265,"size":15970,"snap":"2022-40-2023-06","text_gpt3_token_len":4709,"char_repetition_ratio":0.087435804,"word_repetition_ratio":0.5051095,"special_character_ratio":0.3011271,"punctuation_ratio":0.13602167,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.977856,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T00:23:09Z\",\"WARC-Record-ID\":\"<urn:uuid:115a7e79-e20c-4c33-a87e-aa88bc8c176c>\",\"Content-Length\":\"76790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b604df2a-9ff9-410f-9dc6-5e3097467f67>\",\"WARC-Concurrent-To\":\"<urn:uuid:307de0fa-e5a9-45ac-bef2-bf9fc8d68de0>\",\"WARC-IP-Address\":\"104.21.12.185\",\"WARC-Target-URI\":\"https://solvedlib.com/spring-2020-otiumhort-tnrs-160-ai-i-ionto-math,133809\",\"WARC-Payload-Digest\":\"sha1:PD4BA6RLHFY3FGJ3GOIDX2KHESEYHSND\",\"WARC-Block-Digest\":\"sha1:OBH3ZAVRZJFTIZMD4WVFK4PMNH7NZAKS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500294.64_warc_CC-MAIN-20230205224620-20230206014620-00528.warc.gz\"}"}
http://www.earth.ox.ac.uk/~davewa/pt/th_axcode.html	[ "", null, "###### Main Sections\n\n1. Microprobe analyses\n\n2. Mineral groups\n\n3. Solid solutions\n\n4. Thermobarometers\n\n5. Uncertainties\n\n6. P-T calculations\n\n7. Phase diagrams\n\n8. THERMOCALC tips\n\n###### THERMOCALC Stuff\n\nActivity coding", null, "Applications\n\nBibliography\n\nBugs and quirks\n\nBulk compositions\n\nModal proportions\n\n# Practical Aspects of Mineral Thermobarometry\n\n## Data File Construction: activity coding\n\n### Notes on coding for mineral solid solutions\n\nThe new pattern for coding activity-composition relations in THERMOCALC 3 has some advantages in logic and clarity over the earlier system. Here's an example for Fe-Mg-Ca garnet, from which one can see how a data file is constructed:\n\n```%___________________________________________\ng 3\nx(g) 0.95 % Fe/(Fe+Mg)\nz(g) 0.03 % Ca/(Fe+Mg+Ca)\n\np(gr) 1 1 0 1 1 z\np(alm) 1 2 1 1 -1 z 0 1 1 x\np(py) 1 2 1 1 -1 z 1 1 -1 x\n\nsf\n\nW(gr,alm) 0 0 0\nW(gr,py) 33 0 0\nW(alm,py) 2.5 0 0\n\n3\nxFeM1 1 2 1 1 -1 z 0 1 1 x\nxMgM1 1 2 1 1 -1 z 1 1 -1 x\nxCaM1 1 1 0 1 1 z\n\ngr 1 1 xCaM1 3\nalm 1 1 xFeM1 3\npy 1 1 xMgM1 3\n% ___________________________________________```\n\nSo, the generalised structure is as follows:\n\n1. The first line consists of the phase name and the number of end members it is made up of.\n2. Next the composition variable names are defined (x and z in this case) and initial values given. It's usually wise to choose values similar to those expected in the calculated result.\n3. Then the end member names are listed, along with the formulae for calculating proportions of them in terms of the composition variables. The sequence you give them in controls the sequence in which the ideal activity expressions and the interactions parameters must be supplied. Some thought has to be given to these, as they are used both to calculate activity coefficients and to reconstruct the bulk composition of the phase from the composition variables.\n4. Then you indicate what solution model to use (sf = symmetric formalism, ideal = ideal mixing on sites)\n5. If symmetric formalism is chosen, you then list the W's. If the end members are 1, 2, 3, these are given in the order W12, W13, W23.\n6. Next, you tell the program how many site fractions are needed to work out the activities, and define the names of these site fractions along with instructions on how to calculate them from the composition variables.\n7. Finally you tell the program how to calculate the (ideal) mixing-on-sites activities of each end member, in terms of the site fractions. They must be listed in the same sequence as in section 3 above.\n\n^ Top" ]	[ null, "http://www.earth.ox.ac.uk/~davewa/pt/pt_img1.jpg", null, "http://www.earth.ox.ac.uk/~davewa/pt/pointer.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85789514,"math_prob":0.98234046,"size":2455,"snap":"2022-27-2022-33","text_gpt3_token_len":694,"char_repetition_ratio":0.13994288,"word_repetition_ratio":0.072234765,"special_character_ratio":0.29898167,"punctuation_ratio":0.07377049,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95245916,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T06:19:38Z\",\"WARC-Record-ID\":\"<urn:uuid:7f01d2bc-e3df-4785-90af-fc4903fac1cb>\",\"Content-Length\":\"5646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d0a1116-2d7e-4c98-b927-1121251d2316>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee05ccc6-82f8-4639-9c6f-84a210616fd0>\",\"WARC-IP-Address\":\"163.1.22.5\",\"WARC-Target-URI\":\"http://www.earth.ox.ac.uk/~davewa/pt/th_axcode.html\",\"WARC-Payload-Digest\":\"sha1:VKVJR7XMICR6ZJKAXKBBVQ4CMSEBA3EJ\",\"WARC-Block-Digest\":\"sha1:P2YN7LHS7E3D3DPINEHHIQ2VIRF6WAPB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103355949.26_warc_CC-MAIN-20220628050721-20220628080721-00404.warc.gz\"}"}
https://www.educationquizzes.com/us/elementary-school-3rd-4th-and-5th-grade/math/properties-of-numbers-difficult/	[ "", null, "13 × 13 = 169.\n\n# Properties of Numbers 2\n\nThis Math quiz is called 'Properties of Numbers 2' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11.\n\nIt costs only \\$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us\n\nSome numbers are prime numbers - some numbers have other properties.As you continue through school and math lessons, you'll learn a lot more about numbers, and if you choose to go further with math when you leave school, no doubt you will come across some strange terms for numbers. Most likely, you've heard of infinity ~ this means without end ~ and numbers go on forever. Saying this, you will learn about Graham's number, TREE(3) and a googolplex!\n\nFor now, we are dealing with the easy stuff, so get 10 out of 10 in this quiz!\n\n1.\nWhich of the following statements is wrong?\nTwo is an even number and a prime number\nOther than the number 2, there are no even prime numbers\nIf you add two odd numbers, you get another odd number\nIf you add two even numbers, you get another even number\n7 + 5 = 12 which is not odd. If you want to show that a statement is not always true, find one example that shows it is wrong - like we did here\n2.\nFor those of you with stamina: How many prime numbers are there between 1 and 100 inclusive?\n29\n26\n27\n25\nWell done those of you who bothered to do this question. The rest of you - do it later. There are 25 prime numbers between 1 and 100. Here they are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. Phew! Does anyone want to find out how many prime numbers there are between 1 and 1,000 inclusive? Just joking! Here's the answer anyway: 168 prime numbers. I wonder if your teacher knows that? By the way, prime numbers are a big thing in higher math\n3.\nWhich of the following statements is wrong?\nIf a number is divisible by 9, then it is divisible by 3\nIf you add two prime numbers together, you get another prime number\nIf you subtract 1 from an even number, you get an odd number\nThere are four prime numbers between 1 and 10\n11 + 7 = 18 which is not a prime number. If you want to show that a statement is not always true, find one example that shows it is wrong - like we did here\n4.\nWhich two prime numbers have to be added together to give 38?\n11 + 27\n19 + 19\n23 + 15\n29 + 9\nThis is the only choice that has two prime numbers\n5.\nWhich of the following statements is wrong?\n13 is an odd number\n13 × 13 = 159\n13 is a prime number\n13 is the sixth smallest prime number\n13 × 13 = 169! D'oh! By the way, just because 13 is odd, this doesn't prevent it from being a prime number too\n6.\nWhich number is the odd man out: 49, 15, 41, 9?\n9\n15\n49\n41\nThey are all odd, but 41 is also a prime number: a prime number is a positive whole number that is only divisible by 1 and itself but does not include the number 1\n7.\nWhat is the smallest prime number?\n2\n1\n0\n-1\nA prime number is a positive whole number that is only divisible by 1 and itself but does not include the number 1, e.g. 2, 3, 17, 41\n8.\nWhich number is the odd man out: 64, 32, 144, 100?\n144\n64\n100\n32\nThey are all even numbers, but 32 is not a square number. A square number (perfect square) is a number formed by the multiplication of another number with itself, e.g. 144 = 12 × 12; 64 = 8 × 8; 100 = 10 × 10. Note: 1 × 1 = 1\n9.\nWhich number is the odd man out: 71, 17, 11, 39?\n11\n17\n39\n71\nThey are all odd numbers, but only 39 is a non prime number: a prime number is a positive whole number that is only divisible by 1 and itself but does not include the number 1\n10.\nHow many prime numbers are there between 1 and 30?\n10\n12\n11\n9\nHere they are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29\nAuthor: Frank Evans" ]	[ null, "https://www.educationquizzes.com/library/11-Plus-Maths/Properties-of-Numbers-(Difficult).jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96658397,"math_prob":0.9931704,"size":945,"snap":"2020-10-2020-16","text_gpt3_token_len":223,"char_repetition_ratio":0.12752391,"word_repetition_ratio":0.0,"special_character_ratio":0.24126984,"punctuation_ratio":0.0964467,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99481565,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-22T08:01:39Z\",\"WARC-Record-ID\":\"<urn:uuid:5eaf15cb-ebd8-40a5-a269-972b6d562abd>\",\"Content-Length\":\"36282\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:796ca78c-5236-4b0f-981c-63ca8fe5b14e>\",\"WARC-Concurrent-To\":\"<urn:uuid:7937351b-39cd-4bb9-a0dd-eee8b319df1c>\",\"WARC-IP-Address\":\"78.137.117.241\",\"WARC-Target-URI\":\"https://www.educationquizzes.com/us/elementary-school-3rd-4th-and-5th-grade/math/properties-of-numbers-difficult/\",\"WARC-Payload-Digest\":\"sha1:E45T43HW7FMATR3EVNNKJZ6QJQHMY2N7\",\"WARC-Block-Digest\":\"sha1:AOYXMMH57UV47RDNY6IOPHWX3D3BDW2B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145654.0_warc_CC-MAIN-20200222054424-20200222084424-00201.warc.gz\"}"}
https://hy.wikipedia.org/wiki/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80:Graph:Chart	[ "# Կաղապար:Graph:Chart\n\nԿաղապարի նկարագրություն\n\n## Parameters\n\n• width: width of the chart\n• height: height of the chart\n• type: type of the chart: line for line charts, area for area charts, and rect for (column) bar charts. Multiple series can stacked using the stacked prefix, e.g. stackedarea.\n• interpolate: interpolation method for line and area charts. It is recommended to use monotone for a monotone cubic interpolation – further supported values are listed at https://github.com/trifacta/vega/wiki/Marks#area.\n• colors: color palette of the chart as a comma-separated list of colors. The color values must be given either as #rgb/#rrggbb/#aarrggbb or by a CSS color name. For #aarrggbb the aa component denotes the alpha channel, i.e. FF=100% opacity, 80=50% opacity/transparency, etc. (The default color palette is category10).\n• xAxisTitle and yAxisTitle: captions of the x and y axes\n• xAxisMin, xAxisMax, yAxisMin, and yAxisMax: minimum and maximum values of the x and y axes\n• xAxisFormat and yAxisFormat: changes the formatting of the axis labels. Supported values are listed at https://github.com/mbostock/d3/wiki/Formatting. For example, the format d can be used to suppress the separators for year numbers.\n• x: the x-values as a comma-separated list\n• y or y1, y2, …: the y-values for one or several data series, respectively\n• legend: show legend (only works in case of multiple data series)\n• y1Title, y2Title, …: defines the label of the respective data series in the legend\n\nNote: In the editor preview the graph extension creates a canvas element with vector graphics. However, when saving the page a PNG raster graphics is generated instead.\n\n## Examples\n\nLine Chart:\n\n{{Graph:Chart\|width=400\|height=100\|type=line\|x=1,2,3,4,5,6,7,8\|y=10, 12, 6, 14, 2, 10, 7, 9}}", null, "Note: The y-axis starts from the smallest y value, though this can be overridden with the yAxisMin parameter.\n\nArea chart:\n\n{{Graph:Chart\|width=400\|height=100\|type=area\|x=1,2,3,4,5,6,7,8\|y=10, 12, 6, 14, 2, 10, 7, 9}}", null, "Note: The y-axis starts from zero\n\nBar chart:\n\n{{Graph:Chart\|width=400\|height=100\|xAxisTitle=X\|yAxisTitle=Y\|type=rect\|x=1,2,3,4,5,6,7,8\|y=10, 12, 6, 14, 2, 10, 7, 9}}", null, "Line chart with more than one data series, using colors:\n\n{{Graph:Chart\|width=400\|height=100\|xAxisTitle=X\|yAxisTitle=Y\|legend=Legend\|type=line\|x=1,2,3,4,5,6,7,8\|y1=10, 12, 6, 14, 2, 10, 7, 9 \|y2=2,4,6,8,13,11,9,2\|colors=#0000aa,#ff8000}}", null, "Area chart with more than one data series showing blended overlap:\n\n{{Graph:Chart\|width=400\|height=100\|xAxisTitle=X\|yAxisTitle=Y\|legend=Legend\|type=area\|x=1,2,3,4,5,6,7,8\|y1=10, 12, 6, 14, 2, 10, 7, 9 \|y2=2,4,6,8,13,11,9,2\|colors=#800000aa,#80ff8000}}", null, "Bar chart with multiple data series:\n\n{{Graph:Chart\|width=400\|height=100\|xAxisTitle=X\|yAxisTitle=Y\|legend=Legend\|type=rect\|x=1,2,3,4,5,6,7,8\|y1=10, 12, 6, 14, 2, 10, 7, 9 \|y2=2,4,6,8,13,11,9,2\|colors=#800000aa,#80ff8000}}", null, "Area chart with smoothed data values:\n\n{{Graph:Chart\|width=400\|height=100\|xAxisTitle=X\|yAxisTitle=Y\|legend=Legend\|type=stackedarea\|x=1,2,3,4,5,6,7,8\|y1=10, 12, 6, 14, 2, 10, 7, 9 \|y2=2,4,6,8,13,11,9,2\|interpolate=monotone\|colors=seagreen, orchid}}", null, "Bar chart with stacked data series:\n\n{{Graph:Chart\|width=400\|height=100\|xAxisTitle=X\|yAxisTitle=Y\|legend=Legend\|type=stackedrect\|x=1,2,3,4,5,6,7,8\|y1=10, 12, 6, 14, 2, 10, 7, 9 \|y2=2,4,6,8,13,11,9,2\|y1Title=Data A\|y2Title=Data B\|colors=seagreen, orchid}}", null, "" ]	[ null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/bd001718d8dcc202a3854dbec188981493e00176.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/2b94cdec1fc25241e22953e96438acd2362c80b9.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/f6ba41370b2fd4e44cece2d4b5b6720795f8dbdb.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/d71a8b31e57dd2d21a9d34eb6647e20bb12884dc.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/cb4d4973cd3e117a7c024a3990b00d74b0a02db8.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/f50d40e775409aec8b963173c89e920c482ba76a.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/915cad2158193e0ba94d5975a38664da5917fbb1.png", null, "https://hy.wikipedia.org/api/rest_v1/page/graph/png/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80%3AGraph%3AChart/0/5ed4894f2321f201d8f5474fbcf9518e7c1499be.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.582729,"math_prob":0.96839434,"size":3733,"snap":"2019-43-2019-47","text_gpt3_token_len":1427,"char_repetition_ratio":0.13435237,"word_repetition_ratio":0.15686275,"special_character_ratio":0.33458343,"punctuation_ratio":0.24947146,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95037043,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T22:31:34Z\",\"WARC-Record-ID\":\"<urn:uuid:5f295a4b-c755-4841-9434-892b398e5ca6>\",\"Content-Length\":\"53773\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8bc04c26-b776-4aad-b609-5f3f29d4fa91>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c62d6f7-dd59-45a7-9e60-f5cbd1e093da>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://hy.wikipedia.org/wiki/%D4%BF%D5%A1%D5%B2%D5%A1%D5%BA%D5%A1%D6%80:Graph:Chart\",\"WARC-Payload-Digest\":\"sha1:2OJN5IEAXTPQHXSQJOIZCXOSJSNZ4SGP\",\"WARC-Block-Digest\":\"sha1:2AKYJ62JLQS23GXTZX3RBBX3TCWIHRDQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669847.1_warc_CC-MAIN-20191118205402-20191118233402-00204.warc.gz\"}"}
https://math.stackexchange.com/questions/1026645/flipping-a-coin-until-4-heads-occur-or-until-flipped-7-times-how-many-combinat	[ "Flipping a coin until 4 heads occur, or until flipped 7 times. How many combinations are possible?\n\nQuestion: A coin is tossed until either 4 heads occur or until the coin has been tossed 7 times. How many heads/tails sequence are possible? For example, HTHTTHT, HHHH, THHTHH, and TTTTTTT are all sequences in the list of possible outcomes.\n\nI think the answer will be 276, but am not sure if it's remotely close to being right. Thanks for any and all help!\n\nA relatively short solution is the following: take all $2^7=128$ possible combinations obtainable with $7$ throws, and subtract all those containing $>4$ heads. In fact:\n\n• no sequence containing $>4$ heads is possible based on the given rules;\n\n• all sequences containing $< 4$ heads are valid;\n\n• all sequences containing exactly $4$ heads are also valid (note that these have to be considered \"truncated\" if $4$ heads occur before the seventh throw, but this does not alter the number of possible sequences since we truncate only a final sequence of tails).\n\nTherefore, the solution is given by $128- {7 \\choose 5}-{7 \\choose 6}-{7 \\choose 7}$ $=128-21-7-1=99$ combinations.\n\nAnother solution, a bit longer, is the following. If no heads occur, we have ${7 \\choose 0}=1$ combination. If one head occurs, we have ${7 \\choose 1}=7$ combinations. If two or three heads occur, we have ${7 \\choose 2}=21$ and ${7 \\choose 3}=35$ combinations, respectively.\n\nIf four heads occur, we have:\n\n• ${4 \\choose 0}=1$ combination if no tails occur;\n\n• ${4 \\choose 1}=4$ combinations if one tail occurs (if four heads and one tail occur, necessarily the last of the five throw has to be a head: so we can focus on the first four throws);\n\n• ${5 \\choose 2}=10$ combinations if two tails occur (as above, if four heads and two tails occur, necessarily the last throw has to be a head: so we can focus on the first five throws);\n\n• similarly, we get that there are ${6 \\choose 3}=20$ combinations if three tails occur.\n\nSumming all combinations, we get again\n\n$$1+7+21+35 \\\\ +1+4+10+20=99$$\n\n$$\\sum_{k=4}^7 \\binom{k-1}{3} + \\sum_{k=0}^3 \\binom{7}{k} = 99.$$ The first sum counts the cases where the fourth head occurs on the $k^{\\rm th}$ toss, for $k = 4, 5, 6, 7$, and the second sum counts the cases where $k$ heads are obtained in $7$ coin tosses, for $k = 0, 1, 2, 3$. Since these are all mutually exclusive outcomes (no double-counting), the enumeration is complete." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8688944,"math_prob":0.99681056,"size":1563,"snap":"2019-43-2019-47","text_gpt3_token_len":420,"char_repetition_ratio":0.17511225,"word_repetition_ratio":0.08396947,"special_character_ratio":0.3109405,"punctuation_ratio":0.10576923,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993811,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T03:45:01Z\",\"WARC-Record-ID\":\"<urn:uuid:1621bbe3-2280-4f33-af13-1831ad6411af>\",\"Content-Length\":\"145092\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74bb69f3-bb27-44cb-9b55-0a2ae2895e05>\",\"WARC-Concurrent-To\":\"<urn:uuid:236fbe01-34a2-472e-a0c1-c9bc688f48a7>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1026645/flipping-a-coin-until-4-heads-occur-or-until-flipped-7-times-how-many-combinat\",\"WARC-Payload-Digest\":\"sha1:54ZROMF5Z2EE5WRJQCGMRCLNOVOUZ2ZB\",\"WARC-Block-Digest\":\"sha1:EUTWXQ6IU5OPHH67UPRWIFGDCYR3P4T7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987798619.84_warc_CC-MAIN-20191022030805-20191022054305-00546.warc.gz\"}"}
https://mruni.pure.elsevier.com/en/publications/numerical-experiment-on-mathematical-model-of-forecasting-the-res	[ "# Žiniu{ogonek} testavimo prognozės matematinio modelio tyrimas skaitiniu eksperimentu\n\nTranslated title of the contribution: Numerical experiment on mathematical model of forecasting the results of knowledge testing\n\nNatalja Kosareva, Aleksandras Krylovas\n\nResearch output: Contribution to journalArticle\n\n3 Citations (Scopus)\n\n### Abstract\n\nIn this paper the new approach to the forecasting the results of knowledge testing, proposed earlier by authors, is extended with four classes of parametric functions, the best fitting one from which is selected to approximate item characteristic function. Mathematical model is visualized by two numerical experiments. The first experiment was performed with the purpose to show the procedure of selecting the most appropriate item characteristic function and adjusting the parameters of the model. Goodness-of-fit statistic for detecting misfit of the selected model is calculated. In the second experiment a test of 10 items is constructed for the population with latent ability having normal distribution. Probability distribution of total test result and test information function are calculated when item characteristic functions are selected from four classes of parametric functions. In the next step it is shown how test information function value could be increased by adjusting parameters of item characteristic functions to the observed population. This model could be used not only for knowledge testing but also when solving diagnostic tasks in various fields of human activities. Other advantage of this method is the reduction of resources of testing process by more precise adjustment of the model parameters and decreasing the standard error of measurement of the estimated examinee ability. In the presented example the methodology is applied for solving the problem of microclimate evaluation in office rooms.\n\nOriginal language Lithuanian 42-61 20 Technological and Economic Development of Economy 17 1 https://doi.org/10.3846/13928619.2011.553994 Published - 2011\n\n### Fingerprint\n\nCharacteristic function\nTesting\nMathematical model\nNumerical experiment\nExperiment\nGoodness of fit\nStatistics\nNormal distribution\nProbability distribution\nStandard error\nDiagnostics\nValue function\nMethodology\nEvaluation\nResources\nMicroclimate\n\n• Finance\n\n### Cite this\n\nIn: Technological and Economic Development of Economy, Vol. 17, No. 1, 2011, p. 42-61.\n\nResearch output: Contribution to journalArticle\n\n@article{d3306942f94940aa9e819e28fe954991,\ntitle = \"Žiniu{ogonek} testavimo prognozės matematinio modelio tyrimas skaitiniu eksperimentu\",\nabstract = \"In this paper the new approach to the forecasting the results of knowledge testing, proposed earlier by authors, is extended with four classes of parametric functions, the best fitting one from which is selected to approximate item characteristic function. Mathematical model is visualized by two numerical experiments. The first experiment was performed with the purpose to show the procedure of selecting the most appropriate item characteristic function and adjusting the parameters of the model. Goodness-of-fit statistic for detecting misfit of the selected model is calculated. In the second experiment a test of 10 items is constructed for the population with latent ability having normal distribution. Probability distribution of total test result and test information function are calculated when item characteristic functions are selected from four classes of parametric functions. In the next step it is shown how test information function value could be increased by adjusting parameters of item characteristic functions to the observed population. This model could be used not only for knowledge testing but also when solving diagnostic tasks in various fields of human activities. Other advantage of this method is the reduction of resources of testing process by more precise adjustment of the model parameters and decreasing the standard error of measurement of the estimated examinee ability. In the presented example the methodology is applied for solving the problem of microclimate evaluation in office rooms.\",\nkeywords = \"Generating function, Item characteristic function, Item response theory (irt), Mathematical modelling, Monte carlo method\",\nauthor = \"Natalja Kosareva and Aleksandras Krylovas\",\nyear = \"2011\",\ndoi = \"10.3846/13928619.2011.553994\",\nlanguage = \"Lithuanian\",\nvolume = \"17\",\npages = \"42--61\",\njournal = \"Technological and Economic Development of Economy\",\nissn = \"1392-8619\",\npublisher = \"Vilnius Gediminas Technical University\",\nnumber = \"1\",\n\n}\n\nTY - JOUR\n\nT1 - Žiniu{ogonek} testavimo prognozės matematinio modelio tyrimas skaitiniu eksperimentu\n\nAU - Kosareva, Natalja\n\nAU - Krylovas, Aleksandras\n\nPY - 2011\n\nY1 - 2011\n\nN2 - In this paper the new approach to the forecasting the results of knowledge testing, proposed earlier by authors, is extended with four classes of parametric functions, the best fitting one from which is selected to approximate item characteristic function. Mathematical model is visualized by two numerical experiments. The first experiment was performed with the purpose to show the procedure of selecting the most appropriate item characteristic function and adjusting the parameters of the model. Goodness-of-fit statistic for detecting misfit of the selected model is calculated. In the second experiment a test of 10 items is constructed for the population with latent ability having normal distribution. Probability distribution of total test result and test information function are calculated when item characteristic functions are selected from four classes of parametric functions. In the next step it is shown how test information function value could be increased by adjusting parameters of item characteristic functions to the observed population. This model could be used not only for knowledge testing but also when solving diagnostic tasks in various fields of human activities. Other advantage of this method is the reduction of resources of testing process by more precise adjustment of the model parameters and decreasing the standard error of measurement of the estimated examinee ability. In the presented example the methodology is applied for solving the problem of microclimate evaluation in office rooms.\n\nAB - In this paper the new approach to the forecasting the results of knowledge testing, proposed earlier by authors, is extended with four classes of parametric functions, the best fitting one from which is selected to approximate item characteristic function. Mathematical model is visualized by two numerical experiments. The first experiment was performed with the purpose to show the procedure of selecting the most appropriate item characteristic function and adjusting the parameters of the model. Goodness-of-fit statistic for detecting misfit of the selected model is calculated. In the second experiment a test of 10 items is constructed for the population with latent ability having normal distribution. Probability distribution of total test result and test information function are calculated when item characteristic functions are selected from four classes of parametric functions. In the next step it is shown how test information function value could be increased by adjusting parameters of item characteristic functions to the observed population. This model could be used not only for knowledge testing but also when solving diagnostic tasks in various fields of human activities. Other advantage of this method is the reduction of resources of testing process by more precise adjustment of the model parameters and decreasing the standard error of measurement of the estimated examinee ability. In the presented example the methodology is applied for solving the problem of microclimate evaluation in office rooms.\n\nKW - Generating function\n\nKW - Item characteristic function\n\nKW - Item response theory (irt)\n\nKW - Mathematical modelling\n\nKW - Monte carlo method\n\nUR - http://www.scopus.com/inward/record.url?scp=79959801661&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=79959801661&partnerID=8YFLogxK\n\nU2 - 10.3846/13928619.2011.553994\n\nDO - 10.3846/13928619.2011.553994\n\nM3 - Article\n\nAN - SCOPUS:79959801661\n\nVL - 17\n\nSP - 42\n\nEP - 61\n\nJO - Technological and Economic Development of Economy\n\nJF - Technological and Economic Development of Economy\n\nSN - 1392-8619\n\nIS - 1\n\nER -" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84629893,"math_prob":0.8088164,"size":5884,"snap":"2019-51-2020-05","text_gpt3_token_len":1200,"char_repetition_ratio":0.13639456,"word_repetition_ratio":0.8018648,"special_character_ratio":0.19646499,"punctuation_ratio":0.07724868,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97543114,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T13:31:57Z\",\"WARC-Record-ID\":\"<urn:uuid:282ef935-82a5-43bb-bde3-273c10024956>\",\"Content-Length\":\"46998\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77486cda-9950-4250-93f8-b930bd363149>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c9d0a6a-a7ef-4669-9bdb-086123c1fb55>\",\"WARC-IP-Address\":\"52.209.51.54\",\"WARC-Target-URI\":\"https://mruni.pure.elsevier.com/en/publications/numerical-experiment-on-mathematical-model-of-forecasting-the-res\",\"WARC-Payload-Digest\":\"sha1:2NMMA5CD7CHEOHOB7355QMXQ7NKHU4M3\",\"WARC-Block-Digest\":\"sha1:CGUBQ26RH4FTZ7IBJ6P7NYGMOQQSXP76\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594603.8_warc_CC-MAIN-20200119122744-20200119150744-00108.warc.gz\"}"}
https://www.softmath.com/parabola-in-math/converting-decimals/convert-between-logarithmic.html	[ "", null, "convert between logarithmic and exponential exponents @ algebra fx 2.0\nRelated topics:\nestimating with decimals worksheet \| algebra and trigonometry a graphing approach \| \"ratio worksheets\" \| free math trivia with answers mathematics \| compound interest 6th grade \| equation quadratic solver \| solving algebra problems step by step free \| math fraction \| algebra logarithms worksheet \| aptitude questions in maths \| prentice hall physics quizes \| how to solve algebraic expressions by removing the brackets \| algebra what order add multiply divide\n\nAuthor Message Author Message\nLosieLed", null, "Reg.: 05.04.2006", null, "Posted: Thursday 04th of Jan 07:41 Homuck", null, "Reg.: 05.07.2001", null, "Posted: Saturday 06th of Jan 09:26\n\nHey All, I am urgently in need of assistance for clearing Algebrator is a very convenient tool. I have been using it\nmy math exam that is nearing. I really do not want to for a long time now.\nopt for the guidance of private coaches and online\ncoaching since they prove to be quite costly. Could you\nsuggest a perfect tutoring utility that can help me with\nlearning the principles of Pre Algebra. Particularly, I need\nassistance on trigonometry and graphing equations.\n\nVofj Timidrov", null, "Reg.: 06.07.2001", null, "Posted: Friday 05th of Jan 08:51 Sdefom Koopmansshab", null, "Reg.: 28.10.2001", null, "Posted: Sunday 07th of Jan 10:11\n\nWhat exactly are your troubles with convert between I remember having often faced problems with least\nlogarithmic and exponential exponents @ algebra fx common denominator, trigonometry and linear equations.\n2.0? Can you give some more details . I remember that A truly great piece of algebra program is Algebrator\nnot long ago I too had to go through a similar period of software. By simply typing in a problem homework a\nworry . In my instance, my worried hunt led me to a step by step solution would appear by a click on Solve.\nteacher in my vicinity . But he was so tied up that he I have used it through many math classes – College\nsimply did not have the time for me. He was the one Algebra, Basic Math and College Algebra. I greatly\nwho really suggested that these days there is yet a recommend the program.\nnew answer at hand. He initiated me to these brilliant\nprograms in algebra." ]	[ null, "https://www.softmath.com/r-solver/images/tutor.png", null, "https://www.softmath.com/images/avatars/none.png", null, "https://www.softmath.com/images/forum/icon_minipost.gif", null, "https://www.softmath.com/images/avatars/none.png", null, "https://www.softmath.com/images/forum/icon_minipost.gif", null, "https://www.softmath.com/images/avatars/none.png", null, "https://www.softmath.com/images/forum/icon_minipost.gif", null, "https://www.softmath.com/images/avatars/29.jpg", null, "https://www.softmath.com/images/forum/icon_minipost.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.836572,"math_prob":0.68206215,"size":1024,"snap":"2021-31-2021-39","text_gpt3_token_len":195,"char_repetition_ratio":0.13627452,"word_repetition_ratio":0.90909094,"special_character_ratio":0.18847656,"punctuation_ratio":0.02189781,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9709706,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-25T16:25:32Z\",\"WARC-Record-ID\":\"<urn:uuid:d5d4ba2c-c375-4c3c-805a-f2b7a94b5ced>\",\"Content-Length\":\"45435\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efa06f0f-bce7-4fd7-ad04-ab8a92fb7202>\",\"WARC-Concurrent-To\":\"<urn:uuid:449ba3a6-4e70-4408-9422-df34b9ea1552>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"https://www.softmath.com/parabola-in-math/converting-decimals/convert-between-logarithmic.html\",\"WARC-Payload-Digest\":\"sha1:44CPHCB5CMIT6PWANG4P3WXBO456UKMS\",\"WARC-Block-Digest\":\"sha1:DFNOMXLL6GFAJKZ2C2YBW5STOLQYP2ZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151699.95_warc_CC-MAIN-20210725143345-20210725173345-00159.warc.gz\"}"}
http://terms.naer.edu.tw/search/?q=mean&field=ti&op=AND&order=-sorten&q=noun:%22%E6%95%B8%E5%AD%B8%E5%90%8D%E8%A9%9E-%E5%85%A9%E5%B2%B8%E6%95%B8%E5%AD%B8%E5%90%8D%E8%A9%9E%20%22&field=&op=AND	[ "### 查詢結果\n\n• 跳至 /3\n\nsecond mean value theorem 第二均值定理\n\nsample mean 樣本平均數[值]\n\nroot mean square error 均方根誤差\n\nmean vector 均值向量\n\nmean value theorem 均值定理\n\nmean term of proportion 比例中項\n\nmean square error (MSE) 均方誤差\n\nmean error 平均誤差\n\nmean-ergodic theorem 平均遍歷定理\n\nmean entropy 平均熵\n• 跳至 /3" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.52409583,"math_prob":0.9913096,"size":469,"snap":"2021-04-2021-17","text_gpt3_token_len":425,"char_repetition_ratio":0.32258064,"word_repetition_ratio":0.0,"special_character_ratio":0.23880596,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994255,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-26T03:05:14Z\",\"WARC-Record-ID\":\"<urn:uuid:43d4f55f-c0b0-4539-bafe-59704fd3cf12>\",\"Content-Length\":\"33901\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a6565a76-82df-4601-958e-c32d93a7f11f>\",\"WARC-Concurrent-To\":\"<urn:uuid:94a33793-18be-4afa-b712-163f27a292d4>\",\"WARC-IP-Address\":\"120.127.233.14\",\"WARC-Target-URI\":\"http://terms.naer.edu.tw/search/?q=mean&field=ti&op=AND&order=-sorten&q=noun:%22%E6%95%B8%E5%AD%B8%E5%90%8D%E8%A9%9E-%E5%85%A9%E5%B2%B8%E6%95%B8%E5%AD%B8%E5%90%8D%E8%A9%9E%20%22&field=&op=AND\",\"WARC-Payload-Digest\":\"sha1:ZCKFHNSPXZF7PBNBY7R3BR5RQUKK23X6\",\"WARC-Block-Digest\":\"sha1:C7722G2ZTTOI27SPF274GZ3KWDTSBX6R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704795033.65_warc_CC-MAIN-20210126011645-20210126041645-00260.warc.gz\"}"}
https://mt-rca.org/roof-types/how-do-you-work-out-the-area-of-a-roof-australia.html	[ "# How do you work out the area of a roof Australia?\n\nContents\n\n## How is roof area calculated in Australia?\n\nIf your house is rectangular, all you need to do is multiply the length and width of the building. If the shape of your house is more complex, simply enter the total area (after measuring the exterior dimensions) into the appropriate box.\n\n## How do you calculate the area of a roof?\n\nMultiply your house length by your house width to get the area. (For example, 40 feet x 30 feet = 1,200 square feet.) Next, multiply the area by your roof’s pitch. (1,200 x 1.05 = 1,260 square feet.)\n\n## What is the standard roof pitch in Australia?\n\nStandard roof pitches in Australia are usually either 15 degrees or 22.5 degrees, depending on the materials used in the roof.\n\n## What is the most common roof pitch?\n\nThe most commonly used roof pitches fall in a range between 4/12 and 9/12. Pitches lower than 4/12 have a slight angle, and they are defined as low-slope roofs. Pitches of less than 2/12 are considered flat roofs, even though they may be very slightly angled.\n\nIT IS INTERESTING: What colors go with a GREY roof?\n\n## How many square feet do I need calculator?\n\nIf you are measuring a square or rectangle area, multiply length times width; Length x Width = Area. For other area shapes, see formulas below to calculate Area (ft2) = Square Footage.\n\n## How many squares are in a 1200 square foot roof?\n\nTo determine the number of squares on the gable roof above, divide its total of 24,000 square feet by 100 (24,000 ÷ 100 = 240).\n\nHow many shingles do I need for 1000 square feet?\n\nHouse by Square Foot Roof by Square Foot Re-Roofing Cost*\n1,000 1,054 \\$4,000 – \\$5,500\n1,100 1,160 \\$4,200 – \\$6,000\n1,200 1,265 \\$4,500 – \\$6,500\n1,500 1,581 \\$5,500 – \\$8,000\n\n## How do I calculate shingles for my garage?\n\nCalculate the square footage by multiplying the width by the length. For example, a 20 foot long by 15 foot wide section will come to 300 square feet. Add the square footage together from all of the areas measured to get total square footage. This is the figure you will use for a shingle estimation.\n\n## What roof pitch is 25 degrees?\n\nRoof Slope in Degrees to Standard Roof Pitch Conversion Tables\n\nConvert Roof Slope from Degrees to Rise-in-Run\nRoof Angle in Degrees Roof Slope as Rise-in-Run (X-in-12)\n25° 5.596 in 12\n26° 5.853 in 12\n27° 6.114 in 12\n\n## Is there a minimum roof pitch?\n\nTraditionally, a minimum roof pitch of 20° was recommended in BS 5534, but modern tiles and slates have now been designed for applications as low as 15°. It is uncommon to find a roof below 15° but for those very low pitch applications, there are interlocking clay pantiles available, suitable for use down to 12.5°.\n\nIT IS INTERESTING: How heavy is too heavy for a roof?\n\n## What is the best roof in Australia?\n\nHere are Australia’s top types of roofing.\n\n1. Solar Tiles. An increasingly popular option for those seeking to cut costs and increase green points – solar tiles are ideal for most Australian climates. …\n2. Metal Sheeting. …\n3. Brickwork. …\n4. Terracotta Roof Tiles. …\n5. Concrete Tiles.\n\n1.11.2020\n\n## Is a 4/12 roof pitch good?\n\nAs well, a slope of 4:12 is most commonly considered the lowest slope for “standard shingle installations”. Most manufacturer and industry recommendations require, or at the very least recommend, special underlayment or other considerations on roofs between 2:12 and 4:12.\n\n## How does roof pitch affect cost?\n\nA slope of 6/12 could add nearly another thousand dollars to the price. If you were to choose a 9/12 slope, the national average for a 40 by 50-foot roof would be more than \\$8,300. A high slope of 12/12 could cost \\$10,000 or more.\n\n## What roof pitch is 30 degrees?\n\nWhat roof pitch is 30 degrees? A 30° roof pitch is roughly the same as a 7/12 roof pitch. To convert from degrees to the American ratio: Find the tangent of the angle, tan(angle).", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20100%20100'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92397803,"math_prob":0.9618494,"size":3683,"snap":"2022-05-2022-21","text_gpt3_token_len":981,"char_repetition_ratio":0.12313128,"word_repetition_ratio":0.006134969,"special_character_ratio":0.29486832,"punctuation_ratio":0.13098237,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9602773,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T13:08:22Z\",\"WARC-Record-ID\":\"<urn:uuid:20feba7b-2c03-4e50-bdeb-b2b086ebe7cc>\",\"Content-Length\":\"75550\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28452aab-202a-4bd7-ab4f-cb47599ab95e>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e98b051-777a-4c89-933a-0064b3bee272>\",\"WARC-IP-Address\":\"144.126.138.138\",\"WARC-Target-URI\":\"https://mt-rca.org/roof-types/how-do-you-work-out-the-area-of-a-roof-australia.html\",\"WARC-Payload-Digest\":\"sha1:ETJCZM67G3HUF3Z4FNE3A3VPDSVLJRKD\",\"WARC-Block-Digest\":\"sha1:5SJ3OPPQ4SVSJISFFQMZW4R6NJLRT3AC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016853.88_warc_CC-MAIN-20220528123744-20220528153744-00668.warc.gz\"}"}
https://learn.cemetech.net/index.php?title=TI-BASIC:Max&oldid=946	[ "TI-BASIC:Max\n\n(diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff)\n\nCommand Summary\n\nReturns the maximum of two elements or of a list.\n\nCommand Syntax\n\n• for two numbers: max(x,y)\n• for a list: max(list)\n• comparing a number to each element of a list: max(x,list) or max(list,x)\n• pairwise comparing two lists: max(list1,list2)\n\nPress:\n\n1. MATH to access the Math menu.\n2. RIGHT to access the NUM submenu.\n3. 7 to select max(, or use arrows.\n\nAlternatively, press:\n\n1. 2nd LIST to access the list menu.\n2. LEFT to access the MATH submenu.\n3. 2 to select max(, or use arrows.\n\nTI-83/84/+/SE\n\n1 byte\n\nmax(x,y) returns the largest of the two numbers x and y. max(list) returns the largest element of list. max(list1,list2) returns the pairwise maxima of the two lists. max(list1,x) (equivalently, max(x,list1)) returns a list whose elements are the larger of x or the corresponding element of the original list.\n\nmax(2,3)\n3\nmax({2,3,4})\n4\nmax({1,3},{4,2})\n{4 3}\nmax({1,3},2)\n{2 3}\n\nUnlike comparison operators such as < and >, max( can also compare complex numbers. To do this, both arguments must be complex -- either complex numbers or complex lists: max(2,i) will throw an error even though max(2+0i,i) won't. In the case of complex numbers, the number with the largest absolute value will be returned. When the two numbers have the same absolute value, the first one will be returned: max(i,-i) returns i and max(-i,i) returns -i.\n\nmax( can be used in Boolean comparisons to see if at least one of a list is 1 (true) -- useful because commands like If or While only deal with numbers, and not lists, but comparisons like L1=L2 return a list of values. In general, the behavior you want varies, and you will use the Min( function or the max( function accordingly.\n\nUsing max( will give you a lenient test -- if any one element of the list is 1 (true), then the max( of the list is true -- this is equivalent to putting an 'or' in between every element. For example, this tests if K is equal to any of 24, 25, 26, or 34 (the GetKey arrow key values):\n\n:If max(K={24,25,26,34\n:Disp \"ARROW KEY\n\nTo get the element of a real list with the greatest absolute value, use imag(max(iAns.\n\nError Conditions\n\n• ERR:DATA TYPE is thrown when comparing a real and a complex number. This can be avoided by adding +0i to the real number (or i^4 right after it, for those who are familiar with complex numbers)\n• ERR:DIM MISMATCH is thrown, when using max( with two lists, if they have different dimensions." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7412057,"math_prob":0.98083,"size":2439,"snap":"2022-05-2022-21","text_gpt3_token_len":654,"char_repetition_ratio":0.14373717,"word_repetition_ratio":0.014084507,"special_character_ratio":0.27716276,"punctuation_ratio":0.14468865,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868963,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T12:43:54Z\",\"WARC-Record-ID\":\"<urn:uuid:aa70ecd7-4f94-46ca-b890-32c1202c064d>\",\"Content-Length\":\"18487\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df751272-7288-4ccd-8606-6a3bb291ceb7>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca0f31d6-71b2-455f-af36-35d713401008>\",\"WARC-IP-Address\":\"198.27.66.25\",\"WARC-Target-URI\":\"https://learn.cemetech.net/index.php?title=TI-BASIC:Max&oldid=946\",\"WARC-Payload-Digest\":\"sha1:RE3ABTNR335KIR7JQEPQGDMDSKBIVY2Q\",\"WARC-Block-Digest\":\"sha1:T5MKQHHIVWQRNHPT2WGCBUMNG5UVFLPB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305260.61_warc_CC-MAIN-20220127103059-20220127133059-00040.warc.gz\"}"}
https://scfr.savingadvice.com/2013/04/	[ "User Real IP - 3.89.204.127\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n 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=> 49.36.233.51\n)\n\n => Array\n(\n => 101.0.32.8\n)\n\n => Array\n(\n => 119.160.103.158\n)\n\n => Array\n(\n => 122.177.79.115\n)\n\n => Array\n(\n => 107.181.166.27\n)\n\n => Array\n(\n => 183.6.0.125\n)\n\n => Array\n(\n => 49.36.186.0\n)\n\n => Array\n(\n => 202.181.5.4\n)\n\n => Array\n(\n => 45.118.165.144\n)\n\n => Array\n(\n => 171.96.157.133\n)\n\n => Array\n(\n => 222.252.51.163\n)\n\n => Array\n(\n => 103.81.215.162\n)\n\n => Array\n(\n => 110.225.93.208\n)\n\n => Array\n(\n => 122.161.48.200\n)\n\n => Array\n(\n => 119.63.138.173\n)\n\n => Array\n(\n => 202.83.58.208\n)\n\n => Array\n(\n => 122.161.53.101\n)\n\n => Array\n(\n => 137.97.95.21\n)\n\n => Array\n(\n => 112.204.167.123\n)\n\n => Array\n(\n => 122.180.21.151\n)\n\n => Array\n(\n => 103.120.44.108\n)\n\n => Array\n(\n => 49.37.220.174\n)\n\n => Array\n(\n => 1.55.255.124\n)\n\n => Array\n(\n => 23.227.140.173\n)\n\n => Array\n(\n => 43.248.153.110\n)\n\n => Array\n(\n => 106.214.93.101\n)\n\n => Array\n(\n => 103.83.149.36\n)\n\n => Array\n(\n => 103.217.123.57\n)\n\n => Array\n(\n => 193.9.113.119\n)\n\n => Array\n(\n => 14.182.57.204\n)\n\n => Array\n(\n => 117.201.231.0\n)\n\n => Array\n(\n => 14.99.198.186\n)\n\n => Array\n(\n => 36.255.44.204\n)\n\n => Array\n(\n => 103.160.236.42\n)\n\n => Array\n(\n => 31.202.16.116\n)\n\n => Array\n(\n => 223.239.49.201\n)\n\n => Array\n(\n => 122.161.102.149\n)\n\n => Array\n(\n => 117.196.123.184\n)\n\n => Array\n(\n => 49.205.112.105\n)\n\n => Array\n(\n => 103.244.176.201\n)\n\n => Array\n(\n => 95.216.15.219\n)\n\n => Array\n(\n => 103.107.196.174\n)\n\n => Array\n(\n => 203.190.34.65\n)\n\n => Array\n(\n => 23.227.140.182\n)\n\n => Array\n(\n => 171.79.74.74\n)\n\n => Array\n(\n => 106.206.223.244\n)\n\n => Array\n(\n => 180.151.28.140\n)\n\n => Array\n(\n => 165.225.124.114\n)\n\n => Array\n(\n => 106.206.223.252\n)\n\n => Array\n(\n => 39.62.23.38\n)\n\n => Array\n(\n => 112.211.252.33\n)\n\n => Array\n(\n => 146.70.66.242\n)\n\n => Array\n(\n => 222.252.51.38\n)\n\n => Array\n(\n => 122.162.151.223\n)\n\n => Array\n(\n => 180.178.154.100\n)\n\n => Array\n(\n => 180.94.33.94\n)\n\n => Array\n(\n => 205.164.130.82\n)\n\n => Array\n(\n => 117.196.114.167\n)\n\n => Array\n(\n => 43.224.0.189\n)\n\n => Array\n(\n => 154.6.20.59\n)\n\n => Array\n(\n => 122.161.131.67\n)\n\n => Array\n(\n => 70.68.68.159\n)\n\n => Array\n(\n => 103.125.130.200\n)\n\n => Array\n(\n => 43.242.176.147\n)\n\n => Array\n(\n => 129.0.102.29\n)\n\n => Array\n(\n => 182.64.180.32\n)\n\n => Array\n(\n => 110.93.250.196\n)\n\n => Array\n(\n => 139.135.57.197\n)\n\n => Array\n(\n => 157.33.219.2\n)\n\n => Array\n(\n => 205.253.123.239\n)\n\n => Array\n(\n => 122.177.66.119\n)\n\n => Array\n(\n => 182.64.105.252\n)\n\n => Array\n(\n => 14.97.111.154\n)\n\n => Array\n(\n => 146.196.35.35\n)\n\n => Array\n(\n => 103.167.162.205\n)\n\n => Array\n(\n => 37.111.130.245\n)\n\n => Array\n(\n => 49.228.51.196\n)\n\n => Array\n(\n => 157.39.148.205\n)\n\n 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Array\n(\n => 117.196.119.95\n)\n\n => Array\n(\n => 95.142.120.66\n)\n\n => Array\n(\n => 156.146.59.35\n)\n\n => Array\n(\n => 49.205.213.148\n)\n\n => Array\n(\n => 111.223.27.206\n)\n\n => Array\n(\n => 49.205.212.86\n)\n\n => Array\n(\n => 103.77.42.103\n)\n\n => Array\n(\n => 110.227.62.25\n)\n\n => Array\n(\n => 122.179.54.140\n)\n\n => Array\n(\n => 157.39.239.81\n)\n\n => Array\n(\n => 138.128.27.234\n)\n\n => Array\n(\n => 103.244.176.194\n)\n\n => Array\n(\n => 130.105.10.127\n)\n\n => Array\n(\n => 103.116.250.191\n)\n\n => Array\n(\n => 122.180.186.6\n)\n\n => Array\n(\n => 101.53.228.52\n)\n\n => Array\n(\n => 39.57.138.90\n)\n\n => Array\n(\n => 197.156.137.165\n)\n\n => Array\n(\n => 49.37.155.78\n)\n\n => Array\n(\n => 39.59.81.32\n)\n\n => Array\n(\n => 45.127.44.78\n)\n\n => Array\n(\n => 103.58.155.83\n)\n\n => Array\n(\n => 175.107.220.20\n)\n\n => Array\n(\n => 14.255.9.197\n)\n\n => Array\n(\n => 103.55.63.146\n)\n\n => Array\n(\n => 49.205.138.81\n)\n\n => Array\n(\n => 45.35.222.243\n)\n\n => Array\n(\n => 203.190.34.57\n)\n\n => Array\n(\n => 205.253.121.11\n)\n\n => Array\n(\n => 154.72.171.177\n)\n\n => Array\n(\n => 39.52.203.37\n)\n\n => Array\n(\n => 122.161.52.2\n)\n\n => Array\n(\n => 82.145.41.170\n)\n\n => Array\n(\n => 103.217.123.33\n)\n\n => Array\n(\n => 103.150.238.100\n)\n\n => Array\n(\n => 125.99.11.182\n)\n\n => Array\n(\n => 103.217.178.70\n)\n\n => Array\n(\n => 197.210.227.95\n)\n\n => Array\n(\n => 116.75.212.153\n)\n\n => Array\n(\n => 212.102.42.202\n)\n\n => Array\n(\n => 49.34.177.147\n)\n\n => Array\n(\n => 173.242.123.110\n)\n\n => Array\n(\n => 49.36.35.254\n)\n\n => Array\n(\n => 202.47.59.82\n)\n\n => Array\n(\n => 157.42.197.119\n)\n\n => Array\n(\n => 103.99.196.250\n)\n\n => Array\n(\n => 119.155.228.244\n)\n\n => Array\n(\n => 130.105.160.170\n)\n\n => Array\n(\n => 78.132.235.189\n)\n\n => Array\n(\n => 202.142.186.114\n)\n\n => Array\n(\n => 115.99.156.136\n)\n\n => Array\n(\n => 14.162.166.254\n)\n\n => Array\n(\n => 157.39.133.205\n)\n\n => Array\n(\n => 103.196.139.157\n)\n\n => Array\n(\n => 139.99.159.20\n)\n\n => Array\n(\n => 175.176.87.42\n)\n\n => Array\n(\n => 103.46.202.244\n)\n\n => Array\n(\n => 175.176.87.16\n)\n\n => Array\n(\n => 49.156.85.55\n)\n\n => Array\n(\n => 157.39.101.65\n)\n\n => Array\n(\n => 124.253.195.93\n)\n\n => Array\n(\n => 110.227.59.8\n)\n\n => Array\n(\n => 157.50.50.6\n)\n\n => Array\n(\n => 95.142.120.25\n)\n\n => Array\n(\n => 49.36.186.141\n)\n\n => Array\n(\n => 110.227.54.161\n)\n\n => Array\n(\n => 88.117.62.180\n)\n\n => Array\n(\n => 110.227.57.8\n)\n\n => Array\n(\n => 106.200.36.21\n)\n\n => Array\n(\n => 202.131.143.247\n)\n\n => Array\n(\n => 103.46.202.4\n)\n\n => Array\n(\n => 122.177.78.217\n)\n\n => Array\n(\n => 124.253.195.201\n)\n\n => Array\n(\n => 27.58.17.91\n)\n\n => Array\n(\n => 223.228.143.162\n)\n\n => Array\n(\n => 119.160.96.233\n)\n\n => Array\n(\n => 49.156.69.213\n)\n\n => Array\n(\n => 41.80.97.54\n)\n\n => Array\n(\n => 122.176.207.193\n)\n\n => Array\n(\n => 45.118.156.6\n)\n\n => Array\n(\n => 157.39.154.210\n)\n\n => Array\n(\n => 103.48.197.173\n)\n\n => Array\n(\n => 103.46.202.98\n)\n\n => Array\n(\n => 157.43.214.102\n)\n\n => Array\n(\n => 180.191.125.73\n)\n\n => Array\n(\n => 157.39.57.227\n)\n\n => Array\n(\n => 119.152.107.157\n)\n\n => Array\n(\n => 103.166.58.254\n)\n\n => Array\n(\n => 139.135.54.30\n)\n\n => Array\n(\n => 185.199.102.101\n)\n\n => Array\n(\n => 59.103.204.4\n)\n\n => Array\n(\n => 202.14.123.34\n)\n\n => Array\n(\n => 103.48.197.243\n)\n\n => Array\n(\n => 39.34.158.129\n)\n\n => Array\n(\n => 156.146.59.38\n)\n\n => Array\n(\n => 110.227.51.220\n)\n\n => Array\n(\n => 157.45.123.212\n)\n\n => Array\n(\n => 110.227.52.20\n)\n\n)\n```\nArchive for April, 2013: scfr's Personal Finance Blog\n Layout: Blue and Brown (Default) Author's Creation\n Home > Archive: April, 2013\n\n# Archive for April, 2013\n\n## Every Once In A While I Don't Bargain Shop\n\nApril 14th, 2013 at 04:20 pm\n\nUsually when I have a purchase to make I take the time to shop around for the best deal. But not always.\n\nOn Friday I found out that a neighbor's dog (who we jokingly called the \"girlfriend\" of our late pooch) was diagnosed with cancer and is going to have an eye removed on Monday. She's a sweet little dog and the owners are fantastic people. To illustrate how kind-hearted they are, when our \"boy\" died they made a donation to the animal welfare group where I volunteer.\n\nYesterday I went to a local doggy boutique and bought a stuffed animal and bag of treats, paying full retail price. This morning I wrapped them up and will deliver this afternoon when I take my walk. No time to bargain shop this time.\n\nPeople and pets first, then money, then things." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9670906,"math_prob":0.99945754,"size":826,"snap":"2021-43-2021-49","text_gpt3_token_len":192,"char_repetition_ratio":0.085158154,"word_repetition_ratio":0.0,"special_character_ratio":0.2276029,"punctuation_ratio":0.0877193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997577,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T05:49:52Z\",\"WARC-Record-ID\":\"<urn:uuid:8506af31-6f28-4a52-84e2-9a9c1969e04f>\",\"Content-Length\":\"303372\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f392ce6-23b5-4982-91ce-0d2993c1affa>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f9c6e13-2ea7-4e52-981c-b1778c731f9e>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://scfr.savingadvice.com/2013/04/\",\"WARC-Payload-Digest\":\"sha1:XBXPEBEOBTW6SN4V6RLJOUZTJIWISEGP\",\"WARC-Block-Digest\":\"sha1:4L3YZRR2TQ5U65GW4Y7FKBYJTB5WI66F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363135.71_warc_CC-MAIN-20211205035505-20211205065505-00196.warc.gz\"}"}
https://www.edx.org/course/introduction-to-probability?hs_analytics_source=referrals&utm_source=mooc.org&utm_medium=referral&utm_campaign=mooc.org-course-list	[ "# Introduction to Probability\n\nLearn probability, an essential language and set of tools for understanding data, randomness, and uncertainty.\n\n### There is one session available:\n\n129,037 already enrolled! After a course session ends, it will be archived.\nStarts Nov 24\nEstimated 10 weeks\n5–10 hours per week\nSelf-paced\nFree\n\nProbability and statistics help to bring logic to a world replete with randomness and uncertainty. This course will give you tools needed to understand data, science, philosophy, engineering, economics, and finance. You will learn not only how to solve challenging technical problems, but also how you can apply those solutions in everyday life.\n\nWith examples ranging from medical testing to sports prediction, you will gain a strong foundation for the study of statistical inference, stochastic processes, randomized algorithms, and other subjects where probability is needed.\n\n### At a glance\n\n• Institution: HarvardX\n• Subject: Data Analysis & Statistics\n• Level: Intermediate\n• Prerequisites:\n\nFamiliarity with U.S. high school level algebra concepts; Single-variable calculus: familiarity with matrices. derivatives and integrals.\n\nNot all units require Calculus, the underlying concepts can be learned concurrently with a Calculus course or on your own for self-directed learners.\n\nUnits 1-3 require no calculus or matrices; Units 4-6 require some calculus, no matrices; Unit 7 requires matrices, no calculus.\n\nPrevious probability or statistics background not required.\n\n• Language: English\n• Video Transcript: English\n\n# What you'll learn\n\nSkip What you'll learn\n• How to think about uncertainty and randomness\n• How to make good predictions\n• The story approach to understanding random variables\n• Common probability distributions used in statistics and data science\n• Methods for finding the expected value of a random quantity\n• How to use conditional probability to approach complicated problems\n\n# Syllabus\n\nSkip Syllabus\n• Unit 0: Introduction, Course Orientation, and FAQ\n• Unit 1: Probability, Counting, and Story Proofs\n• Unit 2: Conditional Probability and Bayes' Rule\n• Unit 3: Discrete Random Variables\n• Unit 4: Continuous Random Variables\n• Unit 5: Averages, Law of Large Numbers, and Central Limit Theorem\n• Unit 6: Joint Distributions and Conditional Expectation\n• Unit 7: Markov Chains" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79492956,"math_prob":0.5763936,"size":1030,"snap":"2021-43-2021-49","text_gpt3_token_len":217,"char_repetition_ratio":0.11208577,"word_repetition_ratio":0.0,"special_character_ratio":0.20194174,"punctuation_ratio":0.15469614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97981256,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T19:54:44Z\",\"WARC-Record-ID\":\"<urn:uuid:0436b0c3-ca1e-40ea-852b-c48b58282faa>\",\"Content-Length\":\"434335\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cad24130-d1bb-42da-b8cd-5e55195228c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb383a66-2824-4b25-a640-364dde604716>\",\"WARC-IP-Address\":\"104.16.178.84\",\"WARC-Target-URI\":\"https://www.edx.org/course/introduction-to-probability?hs_analytics_source=referrals&utm_source=mooc.org&utm_medium=referral&utm_campaign=mooc.org-course-list\",\"WARC-Payload-Digest\":\"sha1:PIGHKNAPGQG6TB2XLQLHZO4MWM226YQH\",\"WARC-Block-Digest\":\"sha1:A32NTNX4PWPYBDUK3Z3WQZ2EVBD4LKLI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358591.95_warc_CC-MAIN-20211128194436-20211128224436-00629.warc.gz\"}"}