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http://gnoobz.com/	[ "# Introduction\n\nThis was the last binary challenge released on the second day of the CTF, worth 300 points.\n\n```\\$ file lost.elf\nlost.elf: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.32, BuildID[sha1]=f9e196b2aad4d644223e22e424c8c192bdb41175, stripped\n\ngdb-peda\\$ checksec\nCANARY : disabled\nFORTIFY : disabled\nNX : ENABLED\nPIE : disabled\nRELRO : disabled\n```\n\nThis looks promising: no exotic architecture, no static linking, no stack canaries, no PIE. Should be a simple exploitation challenge, ready to pop a shell.\n\n# Reverse Engineering\n\n```int __cdecl main(int argc, const char argv, const char envp)\n{\nflagStream_ = fopen(\"flag.txt\", \"rb\");\nif ( !flagStream_ )\nexit(1);\nflagStream = flagStream_;\nfclose(flagStream);\ncreate_bpf_network_filter();\n// This initializes g_filter with the result from the bpf syscall.\n// Note: for my tests, I edited the binary to use the \"lo\" interface instead.\nlistenSock = create_raw_socket_bind_port_4352_udp(\"ens3\");\nlistenSock_ = listenSock;\n// SO_ATTACH_BPF.\nif ( listenSock < 0 \|\| (exitcode = setsockopt(listenSock, 1, 50, &g_filter, 4u)) != 0 )\n{\nexitcode = 1;\n}\nelse\n{\n// Receive somthing on port 4352 (only if it passes the packet filter code).\n{\nudpSocket = socket(2, 2, 17);\nrecipient.sa_family = 2;\n// port = 0x1337 = 4919\n(_WORD )&recipient.sa_data = 0x3713;\n// v20 contains the IP that sent the packet, so the binary\n// will send a reply with the flag to the same IP on port 4919.\n(_DWORD )&recipient.sa_data = v20;\n// < inline strlen nonsense code .. >\nsendto(\nudpSocket,\n&flag,\nlen,\n0,\n&recipient,\n0x10u);\n}\n}\nreturn exitcode;\n}\n```\n\nThe code for creating packet filter, binding the socket, etc. is fairly straight forward, so I will not dump the code here as well. So apparently, we do have to deal with some exotic architecture after all: BPF.\n\nI browsed the interwebs to find BPF (or in this case, eBPF) disassemblers, but none worked properly, so I started hand analyzing the code. It became pretty evident that I would not have enough time during the last few hours of the CTF to analyze the code (0x274 instructions of some weird architecture I do not know..), so I decided to move on to different challenges instead. After my flight home, equipped with stable non-hotel Internet, I started digging more into the architecture because I really wanted to solve this challenge. I am not sure if this is the inteded solution, but after a couple of hours I was able to solve it as well.\n\nKudos to StratumAuhuur for solving it so quickly during the CTF, deservedly getting first place!\n\n# Just In Time\n\nJust in time - get it? ...\n\nDuring my Googling-Frenzy on BPF and eBPF, I found several tools to debug and assemble these filters, but the disassemblers just would not work. Additionally, my kernel version does not support this, and during the CTF I did not have other VMs with newer kernels ready (lessons learned: prepare better). At last, I found something that eventually allowed me to solve the challenge: `/proc/sys/net/core/bpf_jit_enable` .\n\n```Values :\n0 - disable the JIT (default value)\n1 - enable the JIT\n2 - enable the JIT and ask the compiler to emit traces on kernel log.\n```\n\nIf we set this to 1, the kernel will emit x86_64 code for the network filter. Now we are getting somewhere! With option 2, not only will this generate sane code, it will also log it!\n\n```dmesg\n[ 3671.038259] flen=628 proglen=2479 pass=3 image=ffffffffa01035b2 from=lost.elf pid=90\n[ 3671.038300] JIT code: 00000000: 55 48 89 e5 48 81 ec 28 02 00 00 48 89 9d d8 fd\n[ 3671.038315] JIT code: 00000010: ff ff 4c 89 ad e0 fd ff ff 4c 89 b5 e8 fd ff ff\n[ 3671.038326] JIT code: 00000020: 4c 89 bd f0 fd ff ff 31 c0 4d 31 ed 48 89 85 f8\n...\n```\n\nGlorious!\n\n# Packet Filter Code\n\nWe know the address where the code is emitted to, so relocating it should not be difficult. Turns out, other than a few calls at the beginning to read out packet data, the whole code is a linear series of XORs and shifts and all that jazz, until it finally performs a comparison, deciding whether the packet should be allowed to pass through or not.\n\n```seg016:000000000000000A shl r14, 8\nseg016:000000000000000E and r14, 0FF0000h\nseg016:0000000000000015 mov rdi, rax\nseg016:0000000000000018 shl rdi, 18h\nseg016:000000000000001C or r14, rdi\nseg016:000000000000001F mov rdi, rax\nseg016:0000000000000022 shr rdi, 18h\nseg016:0000000000000026 and rdi, 0FFh\nseg016:000000000000002D or r14, rdi\nseg016:0000000000000030 shr rax, 8\nseg016:0000000000000034 and rax, 0FF00h\nseg016:000000000000003B or r14, rax\nseg016:000000000000003E mov r13, 0FFFFFFE0h\nseg016:0000000000000048 mov rdi, r14\nseg016:000000000000004B and rdi, r13\nseg016:000000000000004E shr rdi, 5\nseg016:0000000000000052 mov r15, r14\nseg016:0000000000000055 shl r15, 4\nseg016:0000000000000059 xor r15, rdi\nseg016:000000000000005C mov rax, rsi\nseg016:000000000000005F mov rdi, rax\nseg016:0000000000000062 shl rdi, 8\nseg016:0000000000000066 and rdi, 0FF0000h\nseg016:000000000000006D mov rsi, rax\nseg016:0000000000000070 shl rsi, 18h\nseg016:0000000000000074 or rdi, rsi\nseg016:0000000000000077 mov rsi, rax\nseg016:000000000000007A shr rsi, 18h\nseg016:000000000000007E and rsi, 0FFh\nseg016:0000000000000085 or rdi, rsi\nseg016:0000000000000088 shr rax, 8\nseg016:000000000000008C and rax, 0FF00h\n...\nseg016:00000000000008B2 mov edx, 46245443h ; CT\\$F\nseg016:00000000000008B7 cmp rsi, rd\n...\nseg016:00000000000008E5 mov edi, 42744968h ; hItB\nseg016:00000000000008EA cmp r14, rdi\n...\nseg016:0000000000000914 leave\nseg016:0000000000000915 retn\n```\n\nI parsed the dmesg output and extracted all the bytes for the emitted packet filter function. After analyzing the function prologue, I was able to replace all the function calls at the beginning with simple constants (these calls are used to extract the first 8 bytes of the user data in the UDP packet). To make the whole analysis easier, I simply converted this function into a function taking two integer arguments. The result is a function taking two arguments, returing 1 on success and 0 on failure.\n\n```int packet_filter(unsigned int a1, unsigned int a2) {\nv2 = (a1 >> 8) & 0xFF00 \| ((unsigned int)a1 >> 24) \| (a1 << 24) \| ((_DWORD)a1 << 8) & 0xFF0000;\nv3 = ((a2 >> 8) & 0xFF00 \| ((unsigned int)a2 >> 24) \| (a2 << 24) \| ((_DWORD)a2 << 8) & 0xFF0000)\n- ((v2 + (((unsigned __int64)((unsigned int)v2 & 0xFFFFFFE0) >> 5) ^ 16 * v2)) ^ 0x69859111);\nv4 = v2 - ((v3 + (((unsigned __int64)((unsigned int)v3 & 0xFFFFFFE0) >> 5) ^ 16 * v3)) ^ 0x6DCA2938);\nv5 = v3 - ((v4 + (((unsigned __int64)((unsigned int)v4 & 0xFFFFFFE0) >> 5) ^ 16 * v4)) ^ 0x6DCA2938);\nv6 = v4 - ((v5 + (((unsigned __int64)((unsigned int)v5 & 0xFFFFFFE0) >> 5) ^ 16 * v5)) ^ 0x2D169D9F);\nv7 = v5 - ((v6 + (((unsigned __int64)((unsigned int)v6 & 0xFFFFFFE0) >> 5) ^ 16 * v6)) ^ 0xFFFFFFFFF6F49763LL);\nv8 = v6 - ((v7 + (((unsigned __int64)((unsigned int)v7 & 0xFFFFFFE0) >> 5) ^ 16 * v7)) ^ 0xFFFFFFFFF7762178LL);\nv9 = v7 - ((v8 + (((unsigned __int64)((unsigned int)v8 & 0xFFFFFFE0) >> 5) ^ 16 * v8)) ^ 0xFFFFFFFFF7762178LL);\nv10 = v8 - ((v9 + (((unsigned __int64)((unsigned int)v9 & 0xFFFFFFE0) >> 5) ^ 16 * v9)) ^ 0xFFFFFFFFBA85A3F1LL);\nv11 = v9 - ((v10 + (((unsigned __int64)((unsigned int)v10 & 0xFFFFFFE0) >> 5) ^ 16 * v10)) ^ 0xFFFFFFFFF0A7AA2DLL);\nv12 = v10 - ((v11 + (((unsigned __int64)((unsigned int)v11 & 0xFFFFFFE0) >> 5) ^ 16 * v11)) ^ 0xFFFFFFFFF4EC4254LL);\nv13 = v11 - ((v12 + (((unsigned __int64)((unsigned int)v12 & 0xFFFFFFE0) >> 5) ^ 16 * v12)) ^ 0x52703074);\nv14 = v12 - ((v13 + (((unsigned __int64)((unsigned int)v13 & 0xFFFFFFE0) >> 5) ^ 16 * v13)) ^ 0xFFFFFFFFB438B6BBLL);\nv15 = v13 - ((v14 + (((unsigned __int64)((unsigned int)v14 & 0xFFFFFFE0) >> 5) ^ 16 * v14)) ^ 0x56B4C89B);\nv16 = v14 - ((v15 + (((unsigned __int64)((unsigned int)v15 & 0xFFFFFFE0) >> 5) ^ 16 * v15)) ^ 0x7E983A94);\nv17 = v15 - ((v16 + (((unsigned __int64)((unsigned int)v16 & 0xFFFFFFE0) >> 5) ^ 16 * v16)) ^ 0xFFFFFFFFDFDF36C6LL);\nv18 = v16 - ((v17 + (((unsigned __int64)((unsigned int)v17 & 0xFFFFFFE0) >> 5) ^ 16 * v17)) ^ 0x41A7BD0D);\nv19 = v17 - ((v18 + (((unsigned __int64)((unsigned int)v18 & 0xFFFFFFE0) >> 5) ^ 16 * v18)) ^ 0xFFFFFFFFE060C0DBLL);\nv20 = v18 - ((v19 + (((unsigned __int64)((unsigned int)v19 & 0xFFFFFFE0) >> 5) ^ 16 * v19)) ^ 0x7C0E5B70);\nv21 = v19 - ((v20 + (((unsigned __int64)((unsigned int)v20 & 0xFFFFFFE0) >> 5) ^ 16 * v20)) ^ 0x42294722);\nv22 = v20 - ((v21 + (((unsigned __int64)((unsigned int)v21 & 0xFFFFFFE0) >> 5) ^ 16 * v21)) ^ 0x3B5ACFD7);\nv23 = v21 - ((v22 + (((unsigned __int64)((unsigned int)v22 & 0xFFFFFFE0) >> 5) ^ 16 * v22)) ^ 0x3B5ACFD7);\nv24 = v22 - ((v23 + (((unsigned __int64)((unsigned int)v23 & 0xFFFFFFE0) >> 5) ^ 16 * v23)) ^ 0x5BA53B0);\nv25 = v23 - ((v24 + (((unsigned __int64)((unsigned int)v24 & 0xFFFFFFE0) >> 5) ^ 16 * v24)) ^ 0x3F9F67FE);\nv26 = v24 - ((v25 + (((unsigned __int64)((unsigned int)v25 & 0xFFFFFFE0) >> 5) ^ 16 * v25)) ^ 0xFFFFFFFFC8C9D629LL);\nv27 = v25 - ((v26 + (((unsigned __int64)((unsigned int)v26 & 0xFFFFFFE0) >> 5) ^ 16 * v26)) ^ 0xFFFFFFFFC8C9D629LL);\nv28 = v26 - ((v27 + (((unsigned __int64)((unsigned int)v27 & 0xFFFFFFE0) >> 5) ^ 16 * v27)) ^ 0x330748C);\nv29 = v27 - ((v28 + (((unsigned __int64)((unsigned int)v28 & 0xFFFFFFE0) >> 5) ^ 16 * v28)) ^ 0xFFFFFFFFC94B603ELL);\nv30 = v28 - ((v29 + (((unsigned __int64)((unsigned int)v29 & 0xFFFFFFE0) >> 5) ^ 16 * v29)) ^ 0xFFFFFFFFC27CE8F3LL);\nv31 = v29 - ((v30 + (((unsigned __int64)((unsigned int)v30 & 0xFFFFFFE0) >> 5) ^ 16 * v30)) ^ 0x2B13E685);\nv32 = v30 - ((v31 + (((unsigned __int64)((unsigned int)v31 & 0xFFFFFFE0) >> 5) ^ 16 * v31)) ^ 0xFFFFFFFF8CDC6CCCLL);\nv33 = v31 - ((v32 + (((unsigned __int64)((unsigned int)v32 & 0xFFFFFFE0) >> 5) ^ 16 * v32)) ^ 0x24456F3A);\nv34 = v32 - ((v33 + (((unsigned __int64)((unsigned int)v33 & 0xFFFFFFE0) >> 5) ^ 16 * v33)) ^ 0x4FEBEF45);\nv35 = v33 - ((v34 + (((unsigned __int64)((unsigned int)v34 & 0xFFFFFFE0) >> 5) ^ 16 * v34)) ^ 0x288A0761);\nv36 = v34 - ((v35 + (((unsigned __int64)((unsigned int)v35 & 0xFFFFFFE0) >> 5) ^ 16 * v35)) ^ 0xFFFFFFFF8A528DA8LL);\nv37 = v35 - ((v36 + (((unsigned __int64)((unsigned int)v36 & 0xFFFFFFE0) >> 5) ^ 16 * v36)) ^ 0xFFFFFFFFB1B4758CLL);\nv38 = v36 - ((v37 + (((unsigned __int64)((unsigned int)v37 & 0xFFFFFFE0) >> 5) ^ 16 * v37)) ^ 0x499F020F);\nv39 = v37 - ((v38 + (((unsigned __int64)((unsigned int)v38 & 0xFFFFFFE0) >> 5) ^ 16 * v38)) ^ 0xFFFFFFFFB235FFA1LL);\nv40 = v38 - ((v39 + (((unsigned __int64)((unsigned int)v39 & 0xFFFFFFE0) >> 5) ^ 16 * v39)) ^ 0x13FE85E8);\nv41 = v39 - ((v40 + (((unsigned __int64)((unsigned int)v40 & 0xFFFFFFE0) >> 5) ^ 16 * v40)) ^ 0x13FE85E8);\nv42 = v40 - ((v41 + (((unsigned __int64)((unsigned int)v41 & 0xFFFFFFE0) >> 5) ^ 16 * v41)) ^ 0xFFFFFFFFD70E0861LL);\nv43 = v41 - ((v42 + (((unsigned __int64)((unsigned int)v42 & 0xFFFFFFE0) >> 5) ^ 16 * v42)) ^ 0xD300E9D);\nv44 = v42 - ((v43 + (((unsigned __int64)((unsigned int)v43 & 0xFFFFFFE0) >> 5) ^ 16 * v43)) ^ 0x1174A6C4);\nv45 = v43 - ((v44 + (((unsigned __int64)((unsigned int)v44 & 0xFFFFFFE0) >> 5) ^ 16 * v44)) ^ 0x1174A6C4);\nv46 = v44 - ((v45 + (((unsigned __int64)((unsigned int)v45 & 0xFFFFFFE0) >> 5) ^ 16 * v45)) ^ 0xFFFFFFFFD0C11B2BLL);\nv47 = v45 - ((v46 + (((unsigned __int64)((unsigned int)v46 & 0xFFFFFFE0) >> 5) ^ 16 * v46)) ^ 0xFFFFFFFF9A9F14EFLL);\nv48 = v46 - ((v47 + (((unsigned __int64)((unsigned int)v47 & 0xFFFFFFE0) >> 5) ^ 16 * v47)) ^ 0xFFFFFFFF9B209F04LL);\nv49 = v47 - ((v48 + (((unsigned __int64)((unsigned int)v48 & 0xFFFFFFE0) >> 5) ^ 16 * v48)) ^ 0xFFFFFFFFFC679B36LL);\nv50 = v48 - ((v49 + (((unsigned __int64)((unsigned int)v49 & 0xFFFFFFE0) >> 5) ^ 16 * v49)) ^ 0x5E30217D);\nv51 = v49 - ((v50 + (((unsigned __int64)((unsigned int)v50 & 0xFFFFFFE0) >> 5) ^ 16 * v50)) ^ 0xFFFFFFFFFCE9254BLL);\nv52 = v50 - ((v51 + (((unsigned __int64)((unsigned int)v51 & 0xFFFFFFE0) >> 5) ^ 16 * v51)) ^ 0xFFFFFFFF9896BFE0LL);\nv53 = v51 - ((v52 + (((unsigned __int64)((unsigned int)v52 & 0xFFFFFFE0) >> 5) ^ 16 * v52)) ^ 0xFFFFFFFFF61AAE00LL);\nv54 = v52 - ((v53 + (((unsigned __int64)((unsigned int)v53 & 0xFFFFFFE0) >> 5) ^ 16 * v53)) ^ 0x57E33447);\nv55 = v53 - ((v54 + (((unsigned __int64)((unsigned int)v54 & 0xFFFFFFE0) >> 5) ^ 16 * v54)) ^ 0xFFFFFFFFFA5F4627LL);\nv56 = v54 - ((v55 + (((unsigned __int64)((unsigned int)v55 & 0xFFFFFFE0) >> 5) ^ 16 * v55)) ^ 0x2242B820);\nv57 = v55 - ((v56 + (((unsigned __int64)((unsigned int)v56 & 0xFFFFFFE0) >> 5) ^ 16 * v56)) ^ 0xFFFFFFFF8389B452LL);\nv58 = v56 - ((v57 + (((unsigned __int64)((unsigned int)v57 & 0xFFFFFFE0) >> 5) ^ 16 * v57)) ^ 0xFFFFFFFFE5523A99LL);\nv59 = v57 - ((v58 + (((unsigned __int64)((unsigned int)v58 & 0xFFFFFFE0) >> 5) ^ 16 * v58)) ^ 0xFFFFFFFFE5523A99LL);\nv60 = v58 - ((v59 + (((unsigned __int64)((unsigned int)v59 & 0xFFFFFFE0) >> 5) ^ 16 * v59)) ^ 0x1FB8D8FC);\nv61 = v59 - ((v60 + (((unsigned __int64)((unsigned int)v60 & 0xFFFFFFE0) >> 5) ^ 16 * v60)) ^ 0xFFFFFFFFE5D3C4AELL);\nv62 = v60 - ((v61 + (((unsigned __int64)((unsigned int)v61 & 0xFFFFFFE0) >> 5) ^ 16 * v61)) ^ 0xFFFFFFFFDF054D63LL);\nv63 = v61 - ((v62 + (((unsigned __int64)((unsigned int)v62 & 0xFFFFFFE0) >> 5) ^ 16 * v62)) ^ 0xFFFFFFFFDF054D63LL);\nv64 = v62 - ((v63 + (((unsigned __int64)((unsigned int)v63 & 0xFFFFFFE0) >> 5) ^ 16 * v63)) ^ 0xFFFFFFFFA964D13CLL);\nv65 = v63 - ((v64 + (((unsigned __int64)((unsigned int)v64 & 0xFFFFFFE0) >> 5) ^ 16 * v64)) ^ 0xFFFFFFFFE349E58ALL);\nv66 = 0xFFFFLL;\nif ( (v65 & 0xFFFFFFFF) != 0x46245443 )\nv66 = 0LL;\nif ( ((v64 - ((v65 + (((unsigned __int64)((unsigned int)v65 & 0xFFFFFFFF) >> 5) ^ 16 * v65)) ^ 0x6C7453B5)) & 0xFFFFFFFF) != 0x42744968 )\nv66 = 0LL;\nreturn v66;\n}\n```\n\nThis looks suspiciously like (X)TEA, but with non-standard constants. I could not figure out the reverse operations, so instead I opted for an automated solver. I had no high hopes for z3, but tried it regardless. As expected, it simply turned my laptop into a heating device for a couple of minutes, until I killed it.\n\n# KLEE\n\nI assumed that KLEE would also not fare too well, but surprisingly it coughed up a valid solution within approx. 3 seconds. I used the default docker image for klee and adjusted the C code:\n\n``` unsigned int a1, a2;\nklee_make_symbolic(&a1, sizeof(a1), \"a1\");\nklee_make_symbolic(&a2, sizeof(a2), \"a2\");\n\n// Code..\n\nv65 = v65 & 0xffffffff;\nv66 = v66 & 0xffffffff;\n\nif (v65 != 0x46245443)\nreturn 1;\n\nif (v66 != 0x42744968)\nreturn 1;\n\n// success.\nklee_assert(0);\nreturn 0;\n```\n```\\$ clang -emit-llvm -c -g x.c\n\\$ klee x.bc\nKLEE: output directory is \"/home/klee/x/klee-out-0\"\nUsing STP solver backend\nKLEE: WARNING: undefined reference to function: klee_assert\nKLEE: WARNING ONCE: calling external: klee_assert(0)\nKLEE: ERROR: /home/klee/x/x.c:157: failed external call: klee_assert\nKLEE: NOTE: now ignoring this error at this location\n\nKLEE: done: total instructions = 1312\nKLEE: done: completed paths = 3\nKLEE: done: generated tests = 3\n\n\\$ ktest-tool --write-ints klee-last/test000003.ktest\nktest file : 'klee-last/test000003.ktest'\nargs : ['x.bc']\nnum objects: 2\nobject 0: name: b'a1'\nobject 0: size: 4\nobject 0: data: -870510808\nobject 1: name: b'a2'\nobject 1: size: 4\nobject 1: data: 810027724\n```\n\nThese 2 integers fulfill the requirements:\n\n```>>> hex(-870510808&0xffffffff)\n'0xcc1d0f28'\n>>> hex(810027724&0xffffffff)\n'0x30480acc'\n```\n\nLets test this:\n\n```# Wait for the flag here..\n\\$ nc -u -l -p 4919\n\n# Send packet:\n\\$ echo -ne \"\\x28\\x0f\\x1d\\xcc\\xcc\\x0a\\x48\\x30\" \| nc -u localhost 4352\n\n# And on the nc we were listening:\nslowpoke{only_one_day_too_late_for_the_2000_euro...idiot}\n```\n\nWell, that is a bit harsh.. then again, I do not know what the real flag would be since the services have been turned off right after the CTF. ;) This would have given us enough points for second place.. but there is always next year.\n\nHuge thanks to the HITB CTF organizers and all the teams for an amazing event. The whole CTF infrastructure, challenges, and organization were impressive, there were no issues whatsoever, everything just worked (tm).\n\nCongratulations to StratumAuhuur, Hack.ERS, and KITCTF, well done!\n\n# References\n\n https://www.kernel.org/doc/Documentation/sysctl/net.txt\n\n## Sharif CTF 2016 - Misc 300 - impossible game\n\n10 February 2016 by sku\n\nWriteup for the misc challenge impossible-game\n\n## Sharif CTF 2016 - Web 250 - oldpersian\n\n06 February 2016 by sku\n\nWriteup for the web challenge oldpersian\n\n## HackIm CTF - Exploitation 200 - sandman\n\n03 February 2016 by sku\n\nWriteup for the pwnable challenge sandman\n\n## HackIm CTF - Exploitation 300 - cman\n\n03 February 2016 by sku\n\nWriteup for the pwnable challenge cman\n\n## HackIm CTF - Exploitation 100 - pinkfloyd\n\n31 January 2016 by sku\n\nWriteup for the pwnable challenge pinkfloyd\n\n## Insomnihack Teaser 2016 - rbaced1\n\n20 January 2016 by sku\n\nWriteup for the pwnable challenge rbaced1\n\n## Insomnihack Teaser 2016 - toasted\n\n20 January 2016 by sku\n\nWriteup for the pwnable challenge toasted\n\n## SECCON Online CTF 2015 - Exec dmesg\n\n06 December 2015 by sku\n\nWriteup for the forensics challenge Exec dmesg\n\n## SECCON Online CTF 2015 - Reverse-Engineering Android APK 2\n\n06 December 2015 by sku\n\nWriteup for the reverse engineering and web exploitation challenge APK 2." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57582027,"math_prob":0.9968877,"size":17284,"snap":"2021-43-2021-49","text_gpt3_token_len":6403,"char_repetition_ratio":0.3271412,"word_repetition_ratio":0.07396871,"special_character_ratio":0.47130293,"punctuation_ratio":0.15140967,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97260565,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T18:56:46Z\",\"WARC-Record-ID\":\"<urn:uuid:c004fa48-d43c-4ad7-9c87-9d2ac7433c92>\",\"Content-Length\":\"37799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec8ebba4-97c6-4514-823d-12d7e241b330>\",\"WARC-Concurrent-To\":\"<urn:uuid:5ec5dbff-197a-4197-863b-d5fdd18eb844>\",\"WARC-IP-Address\":\"78.47.69.36\",\"WARC-Target-URI\":\"http://gnoobz.com/\",\"WARC-Payload-Digest\":\"sha1:PGNKOQFQSNUAOY2IZNBIM3M5ZG3DJ4D6\",\"WARC-Block-Digest\":\"sha1:ERNB5KTEIYPWYNJ627EUW4WKS5LRH4LO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362287.26_warc_CC-MAIN-20211202175510-20211202205510-00556.warc.gz\"}"}
https://blog.myrawealth.com/insights/quantitative-investment-concepts	[ "", null, "You should review your portfolio at least every quarter unless you have a financial planner who monitors your investments. Looking at performance can be time-consuming and complex, so it’s important to find the most efficient way of doing so. You can save time by analyzing your portfolio with quantitative investment concepts.\n\nThis article will discuss the different concepts of quantitative investments. Included is information on distributions, diversifications, and different risk measurements. In particular, topics such as kurtosis, variance, covariance, and beta will be discussed.\n\nRelated Article \| The Finance Dictionary: Learn the jargon your Finance friends speak!", null, "## Distribution of Returns Types\n\n### Normal Distribution\n\nA normal distribution is represented by a bell-shaped curve. It has a single peak in the center consisting of mean, median, and mode.\n\n• A mean is the sum of observations divided by the number of observations.\n\n• The median is the midpoint of the values after they are ordered from smallest to largest.\n\n• The modes are the observation that has the greatest frequency.\n\nSymmetric distribution is where there is a 50% chance that an observation selected at random will fall to the right of the mean, and a 50% chance that it will fall to the left of the mean.\n\n### Skewness or Non-Symmetrical Distribution\n\nA non-symmetrical distribution can either be skewed to the right or skewed to the left. When the distribution is skewed to the right, it means that it is positively skewed. The mean is the highest of the three averages, followed by the median, then the mode.\n\nA distribution that is skewed to the left means that it is negatively skewed. In this case, the mode is the largest of the three averages, followed by the median, then the mean.\n\n### Kurtosis\n\nKurtosis measures whether the distribution is more peaked or less peaked than normal distributions. It measures the degrees of exceptional values that occur more frequently or less frequently. A leptokurtic means that the distribution is more peaked than a normal distribution.\n\nMore observations are clustered closely around the mean, which is good for investors that want to minimize volatility in their portfolios. Platykurtic is when the distribution is less peaked than a normal distribution. There is more observation with large deviations from the mean.\n\n### Lognormal Probability Distributions\n\nA lognormal distribution has a greater than 50% chance that an observation that is selected at random will fall to the left of the mean. The distribution is bound by zero, but in theory, the value of the portfolio could rise towards infinity. The distribution is skewed to the right and is often used to analyze the probability distribution of security and other asset prices. This distribution can be used as a representative of the expected future value of a client’s portfolio.\n\nRelated Article \| How To Interpret Your Financial Statements\n\n### Diversification\n\nDiversification is how to structure an investment portfolio to maximize risk and return. An investor should try to realize the amount of return that they are receiving per unit of risk. This can be done by structuring a portfolio that contains a low correlation to each other and having a longer investment time horizon. There is an inverse relationship between the number of stocks and the amount of unsystematic risk within the portfolio. As the number of stock increases, the level of unsystematic risk will decline.\n\n### Covariance\n\nCovariance measures the extent to which two variables move together, either positively or negatively. Covariance is the non-standardized version of the correlation coefficient. It ranges from negative infinity to positive infinity.\n\n### The Correlation Coefficient (R or ρ)\n\nA correlation coefficient measures the extent to which the returns on two securities are related. It measures the strength of the straight-line or linear relationship between the two variables. It has a range of +1 to -1.\n\n• Negative 1 means that the securities move in the opposite direction at the same rate.\n\n• Positive 1 means that the movement between the securities is moving at the same rate and in the same direction\n\n• A value of 0 means that the security movements are unrelated to one another.\n\nA correlation coefficient is a standardized version of a covariance. Combining securities that are not perfectly correlated will reduce the overall portfolio risk.\n\n### Coefficient of Determination (R-Squared)\n\nIt is calculated by squaring the correlation coefficient. This describes the percentage of variability of the dependent variables that are explained by the changes in the independent variable. If R squared is equal to 1 between the portfolio and the market, it means that the portfolio contains no unsystematic risk.\n\nIf R2 risk = 1 between a portfolio and the market, the portfolio contains no unsystematic risk.\n\n## Risk Measurements\n\n### Beta\n\nBeta is the relative measure of systematic risk, and it is useful for portfolios that are highly correlated with the market or benchmark portfolio. The market has a Beta of 1.\n\n• A beta of 1.25 will be considered to be 25% more volatile than the market.\n\n• A beta of less than 1 for a portfolio will be considered less risky than the market.\n\nAnother type of beta is the portfolio beta, which is a weighted average. Betas for individual securities are not considered stable over time. However, betas for portfolios are usually stable over time.\n\n### Statistical Risk Analysis\n\nStatistical risk analysis is the distribution of returns for different assets that are symmetrically distributed around the mean of the returns. The distribution will form to something that is referred to by statisticians as the normal distribution or bell-shaped curve.\n\nStandard deviation is the dispersion of outcomes that is around the mean that is used to measure the risk for normal distribution. The greater the standard deviation, the greater the risk or volatility. The total risk of systematic and unsystematic risk is measured by the standard deviation. The mean is the average return for the sample data. Variance is the standard of deviation squared. Standard deviation is calculated by taking the square root of the sum of the squared differences between the mean, and the individual's observations divided by the number of observations minus one. If the market has a standard deviation of 15%, then the stocks with a standard deviation greater than 15% are considered to be riskier than the market.\n\nConsequently, stocks with a standard deviation of less than 15% are considered less risky than the market. A z-statistic or z-score measures the number of standard deviations a data value is from the mean. The score is calculated by subtracting the mean from the given data value and dividing the result by the standard deviation.\n\nIf the data value is higher than the mean, the z-score will be positive. A z-score of 1.3 means that the data value is 1.3 standard deviation above the mean. If the data value is smaller than the mean, then the z-score will be negative. A z-score of -3.3 means that the data value is 3.3 standard deviations below the mean.\n\nRelated Article \| 4 Reasons Why Crypto May Not Be A Good Fit For You\n\n### Semivariance\n\nA semivariance considers the downside volatility of an investment and measures the variability of returns that are below the average or the expected return. When it is used in an analysis, the lower the semivariance of a security, the less likely that security can incur a substantial loss in value.\n\nCritics of variance and standard deviation state that investors are not concerned about the volatility above the average return. Instead, investors are concerned about the volatility below the average return. Thus, a portfolio manager with returns that have a large variance can be punished for having superior positive returns. Semivariance attempts to correct this flaw by only considering the returns and volatility below the expected or average return.\n\nRelated Article \| Top 10 Immigrant Money Questions" ]	[ null, "https://www.facebook.com/tr", null, "https://static.twentyoverten.com/5a942106a9d42e3adb339cce/HRnVm21MeMU/1587405897044.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93381476,"math_prob":0.9749525,"size":8160,"snap":"2022-27-2022-33","text_gpt3_token_len":1600,"char_repetition_ratio":0.15510054,"word_repetition_ratio":0.042424243,"special_character_ratio":0.18762255,"punctuation_ratio":0.08201241,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995181,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T00:07:14Z\",\"WARC-Record-ID\":\"<urn:uuid:07f97388-1399-44a1-87ff-9e9fc1c3f2e0>\",\"Content-Length\":\"83362\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62164897-501b-48a5-a5c9-3bb9d8a114c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:79f9f009-174a-4426-a67c-1377e8cdb291>\",\"WARC-IP-Address\":\"3.88.95.32\",\"WARC-Target-URI\":\"https://blog.myrawealth.com/insights/quantitative-investment-concepts\",\"WARC-Payload-Digest\":\"sha1:JRD6SC3RQPWE23VW3ACQZB5B5VSKNR4L\",\"WARC-Block-Digest\":\"sha1:5DOJY6ITT32CCH7JJBDJ3FTI5QEVBS4I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104506762.79_warc_CC-MAIN-20220704232527-20220705022527-00555.warc.gz\"}"}
https://www.splashlearn.com/math-vocabulary/algebra/term	[ "# Term in Math – Definition With Examples\n\nAlgebra is the part of mathematics that helps represent problems or situations in the form of mathematical expressions. For example, Tim’s grandma gave him many candy bars. He ate a few and now had 5 left. So, how many candy bars did Tim eat?\n\nLet us pose this question using algebra:\n\nWe know how many candy bars are left, so the number 5 would be called the constant. Constants are numbers that have a fixed numerical value. We do not know how many candy bars Tim’s grandma gave him, so we will call it x. The letter x represents the unknown quantity and is known as a variable. When we subtract the number of uneaten candy bars from the number of candy bars Tim’s grandma gave him, we will learn how many he has eaten already.\n\n## What Is an Algebraic Expression?\n\nAn algebraic expression consists of unknown variables, numbers, and arithmetic operators.\n\nx – 5 is a simple algebraic expression. When we combine constants and variables connected by mathematical operations such as +, -, x, and ÷, we get an algebraic expression.\n\nLet us look at a few more algebraic expressions.\n\nA coefficient is a number multiplied by the variable.The coefficient of 5 x + 10 is 5 and\n\n8 – 2 x is 2.\n\n## What Is a Term in Math?\n\nIn algebra, an algebraic expression is formed by a term or a group of terms together. Term in math is defined as the values on which mathematical operations occur in an algebraic expression. Let’s understand with an example of term.\n\nBoth 8x and 9 are terms of this algebraic expression.\n\n## What Are the Factors of a Term?\n\nThe factors of a term are the numbers or variables that are multiplied to form the term.\n\nFor example, the factors of the term 9xy are 9, x, and y.\n\n## Different Terms in Algebra\n\nThere are two kinds of terms in algebra: Like Terms and Unlike Terms.\n\nLike Terms: Like terms are terms whose variables and exponent power are the same. They can be simplified by combining them. The operations of addition and subtraction can be performed on them together.\n\nFor example, 5x + 8x is an algebraic expression with like terms.\n\nUnlike Terms: Unlike terms are those terms whose variables and their exponents are different from each other. They cannot be simplified by combining them. The operations of addition and subtraction cannot be performed on them together.\n\nFor example, 5x + 8y is an algebraic expression with unlike terms.\n\nPolynomial comprises two Greek words: the word “poly” means “many” and “nominal” means “terms”. So, we get the phrase “many terms”. Polynomials are classified into three different types based on the number of terms it consists of.\n\nThe three types of polynomials are:\n\n• Monomials\n• Binomials\n• Trinomials\n\nMonomial: It consists of only one term.\n\nHere are a few examples of monomials:\n\n• 4x\n• 12y\n• 5z\n\nBinomial: It is a polynomial that consists of exactly two terms.\n\nHere are a few examples of binomials:\n\n• 4x + 27\n• 12y -3\n• 5z+ 2x\n\nTrinomial: It is a polynomial that consists of exactly three terms.\n\nHere are a few examples of trinomials:\n\n• 4x + 27 – 3z\n• 12y -3 + 7z\n• 5z+ 2x + 8y\n\nExample 1: What are the terms, variables, and constants of the algebraic expression:\n\n9x 7y + 5?\n\n## Solved Examples\n\nSolution:\n\nIn the algebraic expression, 9x – 7y + 5 (Given)\n\nThe terms are 9x, -7y and 5\n\nThe variables are x and y\n\nVariables are numbers that can take various numerical values.\n\nThe constant is the number 5.\n\nConstants are numbers that have a fixed numerical value.\n\nExample 2: What are the factors of the algebraic expression 3abc?\n\nSolution:\n\nThe factors of a term are the numbers or variables that are multiplied to form the term.\n\nSo, the factors of 3abc are 3, a, b, and c.\n\nExample 3: Identify the like and unlike terms:\n\na) 4p 7q\n\nb) 12y – 5y\n\nSolution:\n\na) 4p – 7q\n\nThese are unlike terms. Their variables are different from each other.\n\nb) 12y – 5y These are like terms. Their variables are the same.\n\n## Practice Problems\n\n1\n\n### There were x number of sparrows sitting on the branches of a tree. Three flew away.Choose the correct algebraic expression to depict the statements given above.", null, "3 - x\n3 + x\nx - 3\nx + 3\nCorrectIncorrect\nCorrect answer is: x - 3\nThe number of sparrows sitting on branches of the tree is x (given). Three flew away, so the sparrows remaining are x – 3.\nTherefore, the algebraic expression for the given statement is x – 3.\n2\n\n### What are the terms of the algebraic expression 7x – 9?\n\nx and 9\n7 and 9\n7x and -9\n7x and 9\nCorrectIncorrect\nCorrect answer is: 7x and 9\nTerms are the values on which mathematical operations occur in an algebraic expression.\nThe terms of this algebraic expression are 7x and 9.\n3\n\n### What are the factors of the algebraic expression 11yz?\n\n11, y, and z\ny and z\n11 y and z\n11 and yz\nCorrectIncorrect\nCorrect answer is: 11, y, and z\nThe factors of a term are the numbers or variables that are multiplied to form the term.\nIn the given algebraic expression 11, y and z are multiplied to form 11yz.\n\nA Term in an algebraic expression can be:\n\n• A constant\n• A variable (with or without coefficients )\n• Both a constant and a variable\n\nThe terms add up to form an algebraic expression. So, they are known as the components of the expression.\n\nA polynomial comprises two Greek words: “poly” meaning “many” and “nominal” meaning “terms”. So, it makes up the phrase “many terms”.\n\nPolynomials are classified into three different types based on the number of terms they consist of. The three types of polynomials are:\n\nMonomial: It consists of only one term. For example, 3a.\n\nBinomial: It is a polynomial that consists of exactly two terms. For example, 3a + 7bTrinomial: It is a polynomial that consists of exactly three terms. For example, 3a + 7b – 5c" ]	[ null, "https://www.splashlearn.com/math-vocabulary/wp-content/uploads/2022/08/Practice-Problems-1-9.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92719895,"math_prob":0.9928266,"size":5605,"snap":"2022-40-2023-06","text_gpt3_token_len":1388,"char_repetition_ratio":0.18585967,"word_repetition_ratio":0.17179742,"special_character_ratio":0.23443355,"punctuation_ratio":0.121157326,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996393,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-30T02:41:56Z\",\"WARC-Record-ID\":\"<urn:uuid:677c9d5b-1924-453a-8c26-3648442de15c>\",\"Content-Length\":\"122493\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:85f41768-1ab0-4a08-a7a8-ed23f16f269f>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7338abc-5474-4b17-8c47-0ac33c0ae28e>\",\"WARC-IP-Address\":\"104.18.28.134\",\"WARC-Target-URI\":\"https://www.splashlearn.com/math-vocabulary/algebra/term\",\"WARC-Payload-Digest\":\"sha1:FHAWHLH6KZBF4BYMKTMI7WIQR43K43Z5\",\"WARC-Block-Digest\":\"sha1:U7EXH324I2VIUJHOWZFONNY3TFOSAUIL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499790.41_warc_CC-MAIN-20230130003215-20230130033215-00205.warc.gz\"}"}
http://calavi.crefap.org/daniel-arzani-vclek/oxidation-number-practice-825742	[ "Determine the oxidation number of the elements in each of the following compounds: a. H 2 CO 3 b. N 2 c. Zn(OH) 4 2-d. NO 2-e. LiH f. Fe 3 O 4 Hint; Identify the species being oxidized and reduced in each of the following reactions: a. Cr + + Sn 4+ Cr 3+ + Sn 2+ b. Practice Set 1: Oxidation Numbers and Redox Reactions 1. Also explore over 6 similar quizzes in this category. Oxidation numbers are very important and are used for 1) naming compounds, 2) balancing oxidation-reduction reactions, 3) calculations in electrochemistry and other areas of chemistry. This worksheet and quiz let you practice the following skills: Pure elements have an oxidation number of 0 2. Balance the following redox equation by the oxidation number methode...in acidic solution. Try this amazing Assigning Oxidation Numbers quiz which has been attempted 6569 times by avid quiz takers. The more-electronegative element in a binary compound is assigned the number equal to the charge it would have if it were an ion. The water molecule is neutral; therefore, the oxygen must have an oxidation number of to Rules: 1. AlBr3 + KMnO4 + H2SO4 = Al2(SO4)3 + K2SO4 + MnSO4 + Br2 + H2O Practice Problems: Redox Reactions. The oxidation number of any uncombined element is 0. The oxidation number of a monatomic ion equals the charge on the ion. The oxidation number of a monatomic ion equals the charge of the ion. The formula for water is . To become skilled at finding oxidation numbers you need lots of practice. Showing top 8 worksheets in the category - Oxidation Numbers. Some of the worksheets displayed are Work oxidation numbers name, Work 25, Oxidation number exercise, Work 1 determination of oxidation number or valence, Chapter 20 work redox, Work 25, , Redox practice work. Previous Oxidation Numbers. 4. Next Electron Transfer. To solve this question we need to calculate the oxidation number of oxygen in both molecules. The oxidation number of a free element is always 0. The oxidation number of Barium Identifying charges in given substances Identifying oxidation numbers in given compounds Skills Practiced. Oxidation Numbers: Rules 1) The oxidation number of the atoms in any free, uncombined element, is zero 2) The sum of the oxidation numbers of all atoms in a compound is zero 3) The sum of the oxidation numbers of all atoms in an ion is equal to the charge of the ion 4) The oxidation number of fluorine in all its compounds is –1 The oxidation number of hydrogen in a compound is +1 4. The oxidation state of hydrogen in most of its compounds is +1 unless it is combined with a metal, in which case it is -1. 2. 3. Since there are two of them, the hydrogen atoms contribute to a charge of +2. Discovery and Similarity Quiz: Discovery and Similarity Atomic Masses Quiz: Atomic Masses The Periodic Table Quiz: The … The oxidation number of fluorine in a compound is always -1. The usual oxidation number of hydrogen is +1. Oxygen has an oxidation number of -2 unless it is combined with F (when it is +2), or it is in a peroxide (such as H2O2 or Na2O2), when it is -1. If the compound is an ionic compound, the oxidation number for each element is the ion’s charge 3. The oxidation number of hydrogen is +1. The atoms in He and N 2, for example, have oxidation numbers of 0. For example, the oxidation number of Na + is +1; the oxidation number of N 3-is -3. Rule 0 The following rules are in the form of a hierarchy; that is, the first stated rule takes The oxidation number of fluorine in a compound is always -1. Determine the oxidation number of each element in the following compounds. To solve this question we need to calculate the oxidation number of N 3-is -3 the number. 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N 2, for example, have oxidation Numbers and Redox Reactions.... Monatomic ion equals the charge on the ion ’ s charge 3 solve this question we need calculate... Monatomic ion equals the charge of the ion for example, the hydrogen atoms contribute to a of. Charge of +2 let you practice the following compounds binary compound is assigned the number equal to the charge +2! Is an ionic compound, the hydrogen atoms contribute to a charge of ion... Are two of them, the oxidation number for each element in compound... A free element is always -1 Redox Reactions 1 6 similar quizzes in category... N 3-is -3 have oxidation Numbers similar quizzes in this category the category - Numbers... Is +1 4 assigned the number equal to the charge it would have it. For each element is the ion to the charge on the ion and N,! 1: oxidation Numbers a free element is always -1 of fluorine in a is. Each element in a compound is always 0 to become skilled at finding oxidation Numbers let you practice following. Compound, the oxidation number of fluorine in a compound is always.! ; the oxidation number of each element in a compound is +1 ; the number... Following compounds worksheets in the category - oxidation Numbers of 0 2 6 similar in... In He and N 2, for example, have oxidation Numbers need! Of them, the hydrogen atoms contribute to a charge of +2 in both molecules a monatomic equals... In the following skills: to become skilled at finding oxidation Numbers you need of! Charge of the ion ’ s charge 3 assigned the number equal to the charge of the ion s... This worksheet and quiz let you practice the following skills: to become skilled at finding oxidation Numbers lots practice! The number equal to the charge it would have if it were an.! N 3-is -3 +1 oxidation number practice the oxidation number of a monatomic ion equals the charge would. Following skills: to become skilled at finding oxidation Numbers it were an ion +1.... Reactions 1 an ionic compound, the oxidation number of N 3-is -3 in the following.. Element in a binary compound is +1 ; the oxidation number of oxygen both. Explore over 6 similar quizzes in this category since there are two of them the! To the charge of +2 element in the category - oxidation Numbers always 0 Numbers of 0....: to become skilled at finding oxidation Numbers you need lots of practice if the compound is assigned number! Are two of them, the hydrogen atoms contribute to a charge of +2 each element is ion. Need to calculate the oxidation number of Na + is +1 4 free element is oxidation number practice -1 Set 1 oxidation... Number for each element in a binary compound is an ionic compound the! Is always 0 atoms contribute to a charge of +2 need lots of practice oxidation of!\n\nChinese Tweed Heads, How To Defeat Crush In Spyro 2, Lego Island 2 Release Date, Matt Prater Longest Field Goal, Summarize In Bisaya, Lakeside Casino Winners, Daoist Traditions College Clinic, Lonely Life Clothing," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9066442,"math_prob":0.99149084,"size":7599,"snap":"2023-40-2023-50","text_gpt3_token_len":1773,"char_repetition_ratio":0.23844634,"word_repetition_ratio":0.26195732,"special_character_ratio":0.2359521,"punctuation_ratio":0.13650998,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9904397,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T18:48:17Z\",\"WARC-Record-ID\":\"<urn:uuid:b5f824a5-1708-45ae-9819-0e077cbf462c>\",\"Content-Length\":\"20258\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:92d87a8b-db07-43d1-8bb2-1a953bfd79ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:ccc6d674-d87f-43ad-89b6-ffb91963d388>\",\"WARC-IP-Address\":\"118.69.171.231\",\"WARC-Target-URI\":\"http://calavi.crefap.org/daniel-arzani-vclek/oxidation-number-practice-825742\",\"WARC-Payload-Digest\":\"sha1:L35I2KIEUNA2Z7SKXLXGXX3XTPPHHHRU\",\"WARC-Block-Digest\":\"sha1:XXECJGMF32GJCPJEHHSHOFMQVTY7DYKO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506029.42_warc_CC-MAIN-20230921174008-20230921204008-00689.warc.gz\"}"}
https://corevoila.in/tag/convert/	[ "# Convert Decimal to Binary (base -2)\n\nQuestion: Write a C# program to convert any decimal number (base-10 (0 to 9)) into binary number (base-2 (0 or 1))? Answer: Decimal Number – Decimal number is a base 10 number because it ranges from 0 to 9, there are total 10 digits between 0 to 9. Any combination of digits is decimal number such as 223, 585, 192, 0, 7 etc. Binary … Continue reading Convert Decimal to Binary (base -2)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8062558,"math_prob":0.9924345,"size":423,"snap":"2021-43-2021-49","text_gpt3_token_len":118,"char_repetition_ratio":0.18138425,"word_repetition_ratio":0.053333335,"special_character_ratio":0.31914893,"punctuation_ratio":0.11235955,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9883052,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T09:47:57Z\",\"WARC-Record-ID\":\"<urn:uuid:181c4fe0-a110-42f9-b264-671080dd8cbd>\",\"Content-Length\":\"47915\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3fd2c82-9a78-431e-b71f-8a94489f3e6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:514518d3-d88b-4c6f-9981-2d0dcc083e12>\",\"WARC-IP-Address\":\"116.202.192.24\",\"WARC-Target-URI\":\"https://corevoila.in/tag/convert/\",\"WARC-Payload-Digest\":\"sha1:6TAAYYBUDZN25EAVWLVWBNJY2EC3C3GC\",\"WARC-Block-Digest\":\"sha1:JRIDJQTIUN7HRGAMHUP4LMOQO3NM5MV2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585916.29_warc_CC-MAIN-20211024081003-20211024111003-00035.warc.gz\"}"}
https://jekel.me/2019/detect-number-of-line-segments-in-pwlf/	[ "# Attempting to detect the number of important line segments in data\n\n## March 12, 2019\n\nThere have been a number of people asking how to use pwlf when you don’t know how many line segments to use. I’ve been recommending the use of a regularization technique which penalizes the number of line segments (or model complexity). This is problematic in a few ways:\n\n• It’s an expensive three layer optimization problem (least squares fit, find break point locations, find number of line segments).\n• The result will be dependant upon the penalty parameter which is problem specific. Something like cross validation could be used to select the parameter, but again this can be expensive.\n\nA large portion of the expense results from searching for breakpoints on a continuous domain. Once the breakpoint locations are known, then the resulting piecewise continuous model is just a simple least squares fit. In many applications it is reasonable to assume some discrete set of possible breakpoint locations. For instance, it may be reasonable to assume that breakpoints can only occur at the points within the data. This would work well for a large amount of data that is evenly distributed in the domain. Alternatively, one could specify a large number of possible breakpoint locations over the domain. The 0.4.0 version of pwlf will include a function which returns the linear regression matrix, which will be used later on to perform Elastic Net fits to a complex function.\n\n# Least Squares Fits\n\nA least squares fit can be used to solve for the model parameters $$(\\mathbf{\\beta})$$ from a discrete list of possible breakpoint locations. Then the value of a model parameter $$(\|\\beta_3\|)$$ would tell whether a breakpoint was significant. The following example demonstrates this on a simple problem with two distinct line segments.\n\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport pwlf\n\nx = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,\n14, 15])\ny = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7,\n70.59, 84.47, 98.36, 112.25, 126.14,\n140.03])\n\n# initialize pwlf on data\nmy_pwlf = pwlf.PiecewiseLinFit(x, y)\n# perform a least squares fit with the breakpoints\n# occurring at the exact x locations\nssr = my_pwlf.fit_with_breaks(x.copy())\n\n# predict on the domain\nxhat = np.linspace(x.min(), x.max(), 100)\nyhat = my_pwlf.predict(xhat)\n\n\nThis results on the following fit which appears excellent.", null, "Let’s take a look at the model parameters.\n\nprint(my_pwlf.beta)\n\n[ 5.00000000e+00 2.00000000e+00 -8.72635297e-14\n4.06341627e-14 -1.99840144e-15 -6.66133815e-16\n1.19200000e+01 -3.00000000e-02 -1.24344979e-14\n1.29896094e-14 -1.00000000e-02 1.00000000e-02\n2.06085149e-14 -5.27564104e-14 7.40345285e-14]\n\n\nThe first parameter is the model offset and the second parameter is the slope of the first line. The remaining parameters are related to the slopes of subsequent lines. The small and near zero parameters imply that the slope didn’t change from the previous breakpoint. We can see that seventh parameter is significant, and it so happens that it corresponds to the seventh data point. This is the first data point on the second line!\n\nThere are a few obvious problems with this method. One problem is that the least squares problem becomes ill-posed in the case when you have more unknowns than data points. Additionally, this method is likely to result in a poor model that overfits the data. The following example follows the previous procedure to fit a continuous piecewise linear model to a noisy sine wave.\n\n# select random seed for reproducibility\nnp.random.seed(123)\n# generate sin wave data\nx = np.linspace(0, 10, num=100)\ny = np.sin(x * np.pi / 2)\nytrue = y.copy()\n# add noise to the data\ny = np.random.normal(0, 0.05, 100) + ytrue\n\n# initialize pwlf on data\nmy_pwlf = pwlf.PiecewiseLinFit(x, y)\n# perform a least squares fit with the breakpoints\n# occuring at the exact x locaitons\nssr = my_pwlf.fit_with_breaks(x.copy())\n\n# predict on the domain\nxhat = np.linspace(x.min(), x.max(), 1000)\nyhat = my_pwlf.predict(xhat)", null, "The result was a continuous piecewise linear function that has overfit the data. You’ll find that most of the $$(\\mathbf{\\beta})$$ parameters are active, with an average absolute value of 1.1.\n\n# Elastic Net for noisy data\n\nWe can use a linear model regularizer such as Elastic Net to prevent this overfitting. The least squares problem minimizes\n\n$$\\mathbf{\\hat{\\beta}} = {\\underset {\\mathbf{\\beta} }{\\operatorname {argmin} }}( \\\|\\mathbf{y}-\\mathbf{A}\\mathbf{\\beta} \\\|^{2} )$$\n\nwhich is prone to overfitting when there are a lot of possible $$(\\mathbf{\\beta})$$ parameters. The Elastic Net uses both L1 (LASSO) and L2 (Ridge regression) to penalize the model complexity. This prevents overfitting. The objective function of the Elastic Net is to minimize\n\n$$\\mathbf{\\hat{\\beta}} = {\\underset {\\mathbf{\\beta} }{\\operatorname {argmin} }}( \\\|\\mathbf{y}-\\mathbf{A}\\mathbf{\\beta} \\\|^{2} + \\lambda_2 \\\|\\mathbf{\\beta} \\\|^{2} + \\lambda_1 \\\|\\mathbf{\\beta} \\\|_{1})$$\n\nfor some $$(\\lambda_1, \\lambda_2)$$ penalty parameters. The Elastic Net solution stats each $$(\\mathbf{\\beta})$$ parameter at zero, and slowly activates a parameter one at a time through the iterations until convergence. This work well for continuous piecewise linear functions, where a zero parameter value corresponds to that breakpoint being inactive.\n\nScikit-learn has implemented the Elastic Net regularizer in ElasticNet. The important penalty parameters to select will be l1_ratio and alpha. However, we’ll use ElasticNetCV which will select these parameters based on cross validation results. The following example perform the Elastic Net fit to the noisy sine wave.\n\nfrom sklearn.linear_model import ElasticNetCV\nmy_pwlf_en = pwlf.PiecewiseLinFit(x, y)\n# copy the x data to use as break points\nbreaks = my_pwlf_en.x_data.copy()\n# new in 0.4.0; creates the linear regression matrix A\nA = my_pwlf_en.assemble_regression_matrix(breaks, my_pwlf_en.x_data)\n\n# set up the elastic net\nen_model = ElasticNetCV(cv=5,\nl1_ratio=[.1, .5, .7, .9,\n.95, .99, 1],\nfit_intercept=False,\nmax_iter=1000000, n_jobs=-1)\n# fit the model using the elastic net\nen_model.fit(A, my_pwlf_en.y_data)\n\n# predict from the elastic net parameters\nxhat = np.linspace(x.min(), x.max(), 1000)\nyhat_en = my_pwlf_en.predict(xhat, breaks=breaks,\nbeta=en_model.coef_)", null, "We can see the Elastic Net results are a much better fit to the sine wave than the least squares fit. In fact, the fit appears to resemble a result from support vector regression!\n\nPerhaps we can use the Elastic Net fit to identify important breakpoint locations. Again, the first two parameters are related to the first line segment, so we’ll always want to grab the first two parameters. However, after the first two parameters we’ll only grab parameters that have an absolute value larger than a threshold. The rational for this is that when $$(\\mathbf{\\beta})$$ parameters are small the breakpoint become inactive (effectively removing a column from the regression matrix).\n\ntrue_list = np.zeros(breaks.size, dtype=bool)\ntrue_list[:2] = True\nfor i in range(breaks.size):\nif np.abs(en_model.coef_[i]) > 2e-1:\ntrue_list[i] = True\nelse:\ntrue_list[i] = False\n# new list of important breakpoint locations\nnew_breaks = breaks[true_list]\nn_segments = new_breaks.size - 1.\n\n# get the y value from these new breakpoints form the EN model\nybreaks = my_pwlf_en.predict(new_breaks)\n\n# fit a least squares model from these new breakpoints\nssr = my_pwlf.fit_with_breaks(new_breaks)\nyhat = my_pwlf.predict(xhat)\n\n\nIf we filter the breakpoints with the above code, we’ll have 28 line segments and the following important breakpoints. Unfortunately, the new least squares fit at these new breakpoint locations still overfits the true function.", null, "# Conclusion\n\nA couple alternative strategies are presented to find the number of line segments instead of running an expensive three-layer optimization. If you do not have much noise in your data, you can try to perform a least squares fit where the breakpoints occur at the data points. Alternatively, if you have noise in your data you can use something like the Elastic Net regularizer to perform the fit. (I have a paper on approximating the noise in 1D data.) After the fits are performed, you can analyze the model parameters (or slopes) to identify unique line segments in your model. A small $$\\beta$$ parameter implies that the breakpoint is not active." ]	[ null, "https://jekel.me/assets/2019-03-12/simpleex.png", null, "https://jekel.me/assets/2019-03-12/sin_of.png", null, "https://jekel.me/assets/2019-03-12/sin.png", null, "https://jekel.me/assets/2019-03-12/sin_new.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8182654,"math_prob":0.9896476,"size":8059,"snap":"2019-13-2019-22","text_gpt3_token_len":2020,"char_repetition_ratio":0.13345748,"word_repetition_ratio":0.044319097,"special_character_ratio":0.26231542,"punctuation_ratio":0.14377406,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988109,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T14:17:26Z\",\"WARC-Record-ID\":\"<urn:uuid:5fa696ca-f5e0-4945-9283-ff7fa27f64ab>\",\"Content-Length\":\"26531\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea62c165-6249-4ef5-a483-ac138bb07e72>\",\"WARC-Concurrent-To\":\"<urn:uuid:5538266f-80c6-44e5-b0d7-f78d0cc0ecb2>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://jekel.me/2019/detect-number-of-line-segments-in-pwlf/\",\"WARC-Payload-Digest\":\"sha1:ON2TO5I5AQU6E5HHN2P7J4YU44SBI3XL\",\"WARC-Block-Digest\":\"sha1:HQWXM743OPMZY6YGTS4RKQKJZ4FV5A3F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203991.44_warc_CC-MAIN-20190325133117-20190325155117-00125.warc.gz\"}"}
https://clickhouse.com/docs/zh/sql-reference/aggregate-functions/reference/rankCorr	[ "# rankCorr\n\n## rankCorr\n\n``rankCorr(x, y)``\n\n• Returns a rank correlation coefficient of the ranks of x and y. The value of the correlation coefficient ranges from -1 to +1. If less than two arguments are passed, the function will return an exception. The value close to +1 denotes a high linear relationship, and with an increase of one random variable, the second random variable also increases. The value close to -1 denotes a high linear relationship, and with an increase of one random variable, the second random variable decreases. The value close or equal to 0 denotes no relationship between the two random variables.\n\n``SELECT rankCorr(number, number) FROM numbers(100);``\n\n``┌─rankCorr(number, number)─┐│ 1 │└──────────────────────────┘``\n\n``SELECT roundBankers(rankCorr(exp(number), sin(number)), 3) FROM numbers(100);``\n\n``┌─roundBankers(rankCorr(exp(number), sin(number)), 3)─┐│ -0.037 │└─────────────────────────────────────────────────────┘``" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6311482,"math_prob":0.99566156,"size":1037,"snap":"2023-40-2023-50","text_gpt3_token_len":289,"char_repetition_ratio":0.19941917,"word_repetition_ratio":0.18571429,"special_character_ratio":0.21697204,"punctuation_ratio":0.14772727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99818623,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T18:43:15Z\",\"WARC-Record-ID\":\"<urn:uuid:c1b4fc0d-b5ab-4c4c-8e49-7bc3ab78817b>\",\"Content-Length\":\"53593\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:84071c0c-ccfd-41e7-96b2-998cb0b63932>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d3d82c9-5a78-49ef-9463-96148bf643c9>\",\"WARC-IP-Address\":\"172.66.43.7\",\"WARC-Target-URI\":\"https://clickhouse.com/docs/zh/sql-reference/aggregate-functions/reference/rankCorr\",\"WARC-Payload-Digest\":\"sha1:ZGXNAW6WQZE6YUVRIBMBECMRFO7UEDUO\",\"WARC-Block-Digest\":\"sha1:F5CBP6DSCTED4U3C6WAJPMGBYDVXSDVB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100602.36_warc_CC-MAIN-20231206162528-20231206192528-00870.warc.gz\"}"}
http://arxiv-export-lb.library.cornell.edu/abs/2105.00247	[ "math.CA\n\n# Title: Extension of tetration to real and complex heights\n\nAuthors: Takeji Ueda\nAbstract: The continuous tetrational function ${^x}r=\\tau(r,x)$, the unique solution of equation $\\tau(r,x)=r^{\\tau(r,x-1)}$ and its differential equation $\\tau'(r,x) =q \\tau(r,x) \\tau'(r,x-1)$, is given explicitly as ${^x}r=\\exp_{r}^{\\lfloor x \\rfloor+1}[\\{x\\}]_q$, where $x$ is a real variable called height, $r$ is a real constant called base, $\\{x\\}=x-\\lfloor x \\rfloor$ is the sawtooth function, $\\lfloor x \\rfloor$ is the floor function of $x$, and $[\\{x\\}]_q=(q^{\\{x\\}}-1)/(q-1)$ is a q-analog of $\\{x\\}$ with $q=\\ln r$, respectively. Though ${^x}r$ is continuous at every point in the real $r-x$ plane, extensions to complex heights and bases have limited domains. The base $r$ can be extended to the complex plane if and only if $x\\in \\mathbb{Z}$. On the other hand, the height $x$ can be extended to the complex plane at $\\Re(x)\\notin \\mathbb{Z}$. Therefore $r$ and $x$ in ${^x}r$ cannot be complex values simultaneously. Tetrational laws are derived based on the explicit formula of ${^x}r$.\n Comments: 29pages, 5figures Subjects: Classical Analysis and ODEs (math.CA) Cite as: arXiv:2105.00247 [math.CA] (or arXiv:2105.00247v3 [math.CA] for this version)\n\n## Submission history\n\nFrom: Takeji Ueda [view email]\n[v1] Sat, 1 May 2021 13:37:41 GMT (3370kb,D)\n[v2] Tue, 4 May 2021 14:00:34 GMT (3370kb,D)\n[v3] Tue, 31 Aug 2021 08:09:18 GMT (3371kb,D)\n\nLink back to: arXiv, form interface, contact." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82929224,"math_prob":0.99960643,"size":1483,"snap":"2022-05-2022-21","text_gpt3_token_len":492,"char_repetition_ratio":0.1061528,"word_repetition_ratio":0.027777778,"special_character_ratio":0.34592044,"punctuation_ratio":0.14420062,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998729,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-16T13:26:56Z\",\"WARC-Record-ID\":\"<urn:uuid:74c0bbac-cecd-4349-b2ee-bbdc991623bc>\",\"Content-Length\":\"16164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8035d932-821b-4163-a7fe-44976b24b134>\",\"WARC-Concurrent-To\":\"<urn:uuid:7cd3d891-335d-47ab-8f1e-d5c5b9fe1c51>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"http://arxiv-export-lb.library.cornell.edu/abs/2105.00247\",\"WARC-Payload-Digest\":\"sha1:PH36T6MZ4BDYBE47SUF7SYQVOD4MGPDD\",\"WARC-Block-Digest\":\"sha1:TINRBHJVRDH2NPXYZEPUQMQOMPQMD4L5\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662510117.12_warc_CC-MAIN-20220516104933-20220516134933-00398.warc.gz\"}"}
https://www.mmusangeya.com/posts/day-10-least-common-multiple	[ "# Day 10: Least Common Multiple", null, "", null, "In mathematics, the least common multiple of a group of numbers is the smallest positive number that is divisible by all the numbers. For example, if our numbers are 12 and 15 then the lcm would be 60. It’s also known by other names such as lowest common multiple or smallest common multiple.\n\nFor my current project, it’s important that we calculate lcm. That’s another little hint as to what I’m working on😀. To find the lcm there are a few ways, the first I’ll talk about is the manual way. For this, you list all multiples of each number and then find the lowest common number. That’s your lcm\n\n## example\n\n12 => 12, 24, 36, 48, 60, 72, 84\n\n15 => 15, 30, 45, 60, 75, 90\n\nIn the example above the lowest common number is 60.\n\nThere is another way to calculate is to use the gcd.\n\n## Greatest Common Divisor\n\nIf we have two numbers a and b, then the gcd is the largest number that divides both a and b. For example, for 42 and 56 the gcd is 14.\n\nFor calculating gcd we will use the Euclidean algorithm. Here we do an integer division on the larger of the two numbers until one of them is 0. The remaining number will be our gcd.\n\n```.wp-block-code {\nborder: 0;\n}\n\n.wp-block-code > div {\noverflow: auto;\n}\n\n.shcb-language {\nborder: 0;\nclip: rect(1px, 1px, 1px, 1px);\n-webkit-clip-path: inset(50%);\nclip-path: inset(50%);\nheight: 1px;\nmargin: -1px;\noverflow: hidden;\nposition: absolute;\nwidth: 1px;\nword-wrap: normal;\nword-break: normal;\n}\n\n.hljs {\nbox-sizing: border-box;\n}\n\n.hljs.shcb-code-table {\ndisplay: table;\nwidth: 100%;\n}\n\n.hljs.shcb-code-table > .shcb-loc {\ncolor: inherit;\ndisplay: table-row;\nwidth: 100%;\n}\n\n.hljs.shcb-code-table .shcb-loc > span {\ndisplay: table-cell;\n}\n\n.wp-block-code code.hljs:not(.shcb-wrap-lines) {\nwhite-space: pre;\n}\n\n.wp-block-code code.hljs.shcb-wrap-lines {\nwhite-space: pre-wrap;\n}\n\n.hljs.shcb-line-numbers {\nborder-spacing: 0;\ncounter-reset: line;\n}\n\n.hljs.shcb-line-numbers > .shcb-loc {\ncounter-increment: line;\n}\n\n.hljs.shcb-line-numbers .shcb-loc > span {\n}\n\n.hljs.shcb-line-numbers .shcb-loc::before {\nborder-right: 1px solid #ddd;\ncontent: counter(line);\ndisplay: table-cell;\ntext-align: right;\n-webkit-user-select: none;\n-moz-user-select: none;\n-ms-user-select: none;\nuser-select: none;\nwhite-space: nowrap;\nwidth: 1%;\n}\n```use std::cmp;\n\npub fn gcd(a: i64, b: i64) -> i64 {\nif a == b {\nreturn a;\n}\nlet (mut a, mut b) = (a, b);\nloop {\nif cmp::min(a, b) == 0 {\nreturn cmp::max(a,b);\n}\n(a, b) = if a > b {\n(a % b, b)\n} else {\n(a, b % a)\n};\n}\n}\n```Code language: Rust (rust)```\n\n## Least Common Multiple\n\nNow we’re back to lcm. The formula is pretty simple.\n\n``````pub fn lcm (a: i64, b: i64) -> i64 {" ]	[ null, "data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27300%27%20height=%27200%27/%3e", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90171784,"math_prob":0.9297721,"size":2421,"snap":"2022-40-2023-06","text_gpt3_token_len":633,"char_repetition_ratio":0.11170873,"word_repetition_ratio":0.016877636,"special_character_ratio":0.28335398,"punctuation_ratio":0.14337568,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9803757,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T23:04:21Z\",\"WARC-Record-ID\":\"<urn:uuid:8943ebd9-a54d-40de-821a-bfa4ea2dec8b>\",\"Content-Length\":\"42660\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76830663-4da6-4c8c-b315-57cfad0d5e6a>\",\"WARC-Concurrent-To\":\"<urn:uuid:2fb2c7ea-eebc-4124-9110-393298ee61f8>\",\"WARC-IP-Address\":\"76.76.21.9\",\"WARC-Target-URI\":\"https://www.mmusangeya.com/posts/day-10-least-common-multiple\",\"WARC-Payload-Digest\":\"sha1:GFF52SPTOHCDLNVAZGYVSQIYKJWJLSOO\",\"WARC-Block-Digest\":\"sha1:NRHQY6TROWVVM7TSGWFPZSIGNNTO57AN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337889.44_warc_CC-MAIN-20221006222634-20221007012634-00785.warc.gz\"}"}
https://blueribbonwriters.com/your-question-a-sample-of-30-houses-that-were-sold-in-the-last-year-was-taken-the-value-of-the-house-y-was-estimated/	[ "# Your question: A sample of 30 houses that were sold in the last year was taken. The value of the house (Y) was estimated.\n\nYour question: A sample of 30 houses that were sold in the last year was taken. The value of the house (Y) was estimated. The independent variables included in the analysis were the number of rooms (X1), the size of the lot (X2), the number of bathrooms (X3), and a dummy variable (X4), which equals 0 if the house does not have a garage and equals 1 otherwise. The following results were obtained:Standard Error8,462.5778.02.22,229.3463.1Analysis of VarianceMS51,060.728,235.60a. Write out the estimated equation.b. Interpret the coefficient on the number of rooms (X1).c. Interpret the coefficient on the dummy variable (X4).d. What are the degrees of freedom for the sum of squares explained by the regression (SSR) and the sum of squares due to error (SSE)?e. Test whether or not there is a significant relationship between the value of a house and the independent variables. Use a .05 level of significance. Be sure to state the null and alternative hypotheses.f. Test the significance of 1 at the 5% level. Be sure to state the null and alternative hypotheses.g. Compute the coefficient of determination and interpret its meaning.h. Estimate the value of a house that has 9 rooms, a lot with an area of 7,500, 2 bathrooms, and 2 garages", null, "" ]	[ null, "https://blueribbonwriters.com/wp-content/uploads/2020/01/order-supreme-essay.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9297887,"math_prob":0.979581,"size":1242,"snap":"2021-43-2021-49","text_gpt3_token_len":303,"char_repetition_ratio":0.115508884,"word_repetition_ratio":0.05940594,"special_character_ratio":0.2584541,"punctuation_ratio":0.1598513,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99374783,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T02:06:02Z\",\"WARC-Record-ID\":\"<urn:uuid:9a735bb0-b6c4-4f70-9ae0-b58aef11ea66>\",\"Content-Length\":\"50249\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:39f42ff8-5a3e-45ac-a995-243853b539b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3aec3606-97aa-42ea-943d-14edef5f3ea2>\",\"WARC-IP-Address\":\"198.54.116.13\",\"WARC-Target-URI\":\"https://blueribbonwriters.com/your-question-a-sample-of-30-houses-that-were-sold-in-the-last-year-was-taken-the-value-of-the-house-y-was-estimated/\",\"WARC-Payload-Digest\":\"sha1:WP2F4BTMYLMRHZGMZNCU7DBFR6DRV5NI\",\"WARC-Block-Digest\":\"sha1:ZJ3DLXXEI4SIZA67EL4S6ODG4I336RDQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587608.86_warc_CC-MAIN-20211024235512-20211025025512-00028.warc.gz\"}"}
https://www.york.ac.uk/students/studying/manage/programmes/module-catalogue/module/MAT00079M/latest	[ "Accessibility statement\n\n# Partial Differential Equations II - MAT00079M\n\n« Back to module search\n\n• Department: Mathematics\n• Module co-ordinator: Dr. Konstantin Ilin\n• Credit value: 10 credits\n• Credit level: M\n• Academic year of delivery: 2022-23\n\n## Related modules\n\n• None\n\n### Prohibited combinations\n\nPre-requisite modules: students must have taken PDEs 1 - either MAT00040H or MAT00053M.\n\n## Module will run\n\nOccurrence Teaching cycle\nA Spring Term 2022-23\n\n## Module aims\n\n• To give an introduction to numerical methods for solving partial differential equations (PDEs) and to show how numerical algorithms can be implemented in practice.\n\n• To analyse the error of various numerical algorithms for solving PDEs and discuss practical problems that arise when we apply these algorithms.\n\n• To illustrate numerical methods by solving real problems from various areas of natural sciences such as physics, biology, fluid mechanics, etc.).\n\n## Module learning outcomes\n\n• know basic finite-difference methods for solving partial differential equations\n\n• be able to analyse the error for a particular numerical method and appreciate the efficiency in implementation of numerical algorithms\n\n• be able to obtain numerical solutions of simple PDEs with the help of MATLAB\n\n## Module content\n\nSyllabus\n\n• Finite-differences, truncation error, convergence and stability.\n\n• Explicit and implicit finite-difference schemes for parabolic PDEs. The alternating-direction method.\n\n• Finite-difference schemes for elliptic PDEs. Relaxation methods.\n\n• Finite-difference methods for hyperbolic PDEs: explicit and implicit schemes for wave equation; Lax-Wendroff scheme for hyperbolic systems.\n\n• Spectral methods: polynomial interpolation on Chebyshev points; Chebyshev differentiation matrices; boundary value problems. (These more advanced topics are not taught in the H-level variant of this module.)\n\n• Academic skills: the numerical techniques taught are used in many areas of applied mathematics.\n\n• Graduate skills: through lectures, examples, computer classes, students will develop their ability to assimilate, process and engage with new material quickly and efficiently. They develop problem solving-skills and learn how to apply techniques to unseen problems. Students on this module will learn to work more independently and assimilate advanced material at a greater rate than those on the H-level variant.\n\n## Assessment\n\nTask Length % of module mark\nEssay/coursework\nCoursework\nN/A 25\nOnline Exam\nPartial Differential Equations II\nN/A 75\n\nNone\n\n### Reassessment\n\nTask Length % of module mark\nOnline Exam\nPartial Differential Equations II\nN/A 100\n\n## Module feedback\n\nCurrent Department policy on feedback is available in the student handbook. Coursework and examinations will be marked and returned in accordance with this policy." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8346653,"math_prob":0.64448017,"size":2639,"snap":"2022-05-2022-21","text_gpt3_token_len":570,"char_repetition_ratio":0.12333966,"word_repetition_ratio":0.038356163,"special_character_ratio":0.1906025,"punctuation_ratio":0.10672854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95913744,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T12:08:54Z\",\"WARC-Record-ID\":\"<urn:uuid:7f2e99dc-6d70-49bc-8929-38491b478334>\",\"Content-Length\":\"26374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ead3641-842b-47b5-bc68-d3a59f64075d>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1e8be57-ef23-47e0-9c19-c896d9713218>\",\"WARC-IP-Address\":\"99.84.191.25\",\"WARC-Target-URI\":\"https://www.york.ac.uk/students/studying/manage/programmes/module-catalogue/module/MAT00079M/latest\",\"WARC-Payload-Digest\":\"sha1:4F7IYVIFXIRYVG7SLL4533MADB4TMXZS\",\"WARC-Block-Digest\":\"sha1:4B5CBJ3H4BOVIODC6RE5WSIBM6FWNYTW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517245.1_warc_CC-MAIN-20220517095022-20220517125022-00169.warc.gz\"}"}
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-10-analytic-geometry-10-2-the-parabola-10-2-assess-your-understanding-page-646/18	[ "## Precalculus (10th Edition)\n\nIf the major axis of the parabola is parallel to the y-axis then it is in the form of $k(y-a)=(x-b)^2$, where $(b,a)$ is the vertex of the parabola and $k$ is a constant (positive if open up, negative if open down). If the major axis of the parabola is parallel to the x-axis then it is in the form of $k(x-a)=(y-b)^2$, where $(a,b)$ is the vertex of the parabola and $k$ is a constant. (positive if open right, negative if open left) Here the major axis is parallel to the x-axis, $k=\\pm4$ in all cases, but here it is open right, hence $k=4$ and the vertex is in $(0,0)$. Hence the equation: $4(x-0)=(y-0)^2\\\\4x=y^2$ Thus the answer is a)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82662404,"math_prob":0.99995995,"size":670,"snap":"2021-43-2021-49","text_gpt3_token_len":210,"char_repetition_ratio":0.14714715,"word_repetition_ratio":0.26666668,"special_character_ratio":0.32388058,"punctuation_ratio":0.09638554,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T20:38:36Z\",\"WARC-Record-ID\":\"<urn:uuid:5dc91a6c-721b-44f6-a10b-f84dc4f3322e>\",\"Content-Length\":\"63798\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6c68d85c-435f-407f-9c58-1e68198938f6>\",\"WARC-Concurrent-To\":\"<urn:uuid:db365cb1-7a6d-4493-ae85-af7d1900c7ad>\",\"WARC-IP-Address\":\"34.193.95.253\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-10-analytic-geometry-10-2-the-parabola-10-2-assess-your-understanding-page-646/18\",\"WARC-Payload-Digest\":\"sha1:7SFAL4SGOSMY5P4AUFTFJVQFVYPQ342J\",\"WARC-Block-Digest\":\"sha1:ZEHMFA526VUQIY3TWJQNVRDDYPWD7RJQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359073.63_warc_CC-MAIN-20211130201935-20211130231935-00125.warc.gz\"}"}
https://discretize.simpeg.xyz/en/main/api/generated/discretize.base.BaseMesh.boundary_face_scalar_integral.html	[ "# discretize.base.BaseMesh.boundary_face_scalar_integral#\n\nproperty BaseMesh.boundary_face_scalar_integral#\n\nRepresent the operation of integrating a scalar function on the boundary.\n\nThis matrix represents the boundary surface integral of a scalar function multiplied with a finite volume test function on the mesh.\n\nReturns:\n(n_faces, n_boundary_faces) scipy.sparse.csr_matrix\n\nNotes\n\nThe integral we are representing on the boundary of the mesh is\n\n$\\int_{\\Omega} u\\vec{w} \\cdot \\hat{n} \\partial \\Omega$\n\nIn discrete form this is:\n\n$w^T * P @ u_b$\n\nwhere w is defined on all faces, and u_b is defined on boundary faces." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62188435,"math_prob":0.9769843,"size":623,"snap":"2023-40-2023-50","text_gpt3_token_len":150,"char_repetition_ratio":0.15347335,"word_repetition_ratio":0.0,"special_character_ratio":0.23113964,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935121,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T15:39:49Z\",\"WARC-Record-ID\":\"<urn:uuid:33cc305f-69b6-4ce2-b52b-bc7e1412a2fb>\",\"Content-Length\":\"221462\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97dc7ddd-d47e-4d6f-bbf8-1c1eb5ccf94b>\",\"WARC-Concurrent-To\":\"<urn:uuid:73c8ac2d-e0cc-46d9-a23b-9f15833951b5>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://discretize.simpeg.xyz/en/main/api/generated/discretize.base.BaseMesh.boundary_face_scalar_integral.html\",\"WARC-Payload-Digest\":\"sha1:XBENMCQYENXZYIZYNTDFABUDTAP4XCO3\",\"WARC-Block-Digest\":\"sha1:TYLJMF2BOC6MBUFAXRJU7J6TLLYJCBIU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510214.81_warc_CC-MAIN-20230926143354-20230926173354-00012.warc.gz\"}"}
https://homeworkdave.com/product/impulse-momentum-lab-4-2/	[ "Sale!\n\n# Impulse Momentum Lab 4\n\n\\$30.00\n\nCategory:\n\n## Description\n\nRate this product\n\nImpulse Momentum\nFigure 1: Overview of the setup for the Impulse Momentum Lab, make sure the range switch is in the 10N\nposition and you have zeroed the sensor at rest.\nLab Objectives\n• Data collection\n• Data Analysis\n• Integration\n• Propagation of Error\nLab Equipment\n• Vernier Motion Detector\n• Vernier Dynamics Cart\n• Vernier Dual Range Force Sensor\n1\n• Vernier Hoop Bumper\n• Flat Cart Track\n• Mass Balance\nOverview\nImpulse (J~ ) is a quantity that is closely related to momentum, today we will be looking only in one dimension\nand so Impulse is defined as the integral of the force applied to an object over time,\nJ~ =\nZ t2\nt1\nF~ (t)dt (1)\nImpulse is related to momentum (~p ) by,\nJ~ = ~p2 − ~p1 = 4~p (2)\nIn our lab we can collect force data over time with our force sensor, this allows us to plot the function of Force\nvs time and we can combine these equations into,\n~p2 − ~p1 =\nZ t2\nt1\nF~ (t)dt (3)\nWe can also measure position over time to calculate velocity with the motion sensors, and we can measure\nmass using the mass balances. With this in mind, we can rearrange the above equation to,\nmv2 − mv1 =\nZ t2\nt1\nF~ (t)dt (4)\nwhich we will use to test if the change in momentum is equal to the force applied.\nProcedure\nData Collection\nYou will be performing three variations of the same experiment and employing the same analysis techniques on\nall of them. The variations will be in how fast the cart is going when it encounters the hoop bumper on the force\nsensor. The three speeds will be: slow, slower and slowest. The hoop bumper should never fully compress, if it\ndoes then you will have to take another round of data.\n• Make sure that the force sensor is zeroed and that the range selector switch is in the 10N position.\n• Adjust the motion sensor so that it picks up the cart motion well for the whole range of motion on the track.\n• Take your data. Your data should be of sufficient quality to perform the analysis on in the next section, if it\nis not, you should repeat that trial.\nAnalysis\nIntegral Fit\nLogger Pro offers an integral fit tool that will measure the area under a curve numerically. Much like the\nother fits and statistic functions, you will need to specify the range of the graph that you want the function to\noperate on.\n2\nFigure 2: A view inside of the force sensor, the horizontal fiberglass piece is the strain gauge that will change\nresistance when it is flexed. Be sure to be careful when not to apply too much force on the hook as this could\nbreak the fiberglass strain mechanism.\nVelocity\nUse the statistics function to find the velocity before and after the collision, while the velocity may not be exactly\nconstant, avoid areas of large change. Make sure your group agrees on how to get the uncertainty in the mass of\nPropagation of Uncertainties\nPropagate your uncertainties for the difference in momentums, you should have uncertainties for your velocities\nand mass. There is a quick review of the propagation of uncertainty equations at the bottom of this lab.\nInclude your working and commented python code for the propagation of uncertainty. You should copy and\npaste the code into the answer sheet.\nWriting\nBased on the data that you took today, write and answer the questions in the following sections. Remember\nthat even though you will have the same data as your partner, the writing in these sections should be done\nindividually.\n3\nExperimental Method\n• In complete sentences, communicate the steps that you took when collecting and analyzing your data.\nPretend you are writing this so a fellow student that missed this lab could take and analyze the data\nusing only this section. For example, you do not need to tell them to press start in Logger Pro or open\nthe program, but you would want to tell them what sensors you used to collect data and setting to use\non the force sensor.\nResults\n• Report the results of your experiment in complete sentences using your calculated numbers, you should\nalso include you numbers in table. You should place your graphs with captions in this section as well.\nAt minimum, your table should have the following columns: Trial Number, Type of motion, Impulse\nand Change in momentum. Don’t forget to include units and significant figures.\nConclusion Write a sentence or two for each question asked below. Back up your conclusions with evidence.\nFor example, use equations, measurements, references to figures, etc when appropriate.\n• Based on your results listed above, does the change in momentum equal the impulse within your measured and calculated uncertainty (Include your numbers)?\n• Does your conclusion hold true for the different speed collisions?\n• If so, what might be an interesting extension to this experiment?\nGraph and Data Checklist You should have six graphs with complete captions and answered all of the questions highlighted by the gray boxes.\n4\nReview\nThe Figures and Captions\nThere are a few very important aspects to creating a proper figure and caption. If you follow these rules, not only\nwill you get points on your physics lab grades, you will impress your instructors and peers in the future.\nThe Caption\n• The caption should start with a label so you can reference the figure from other places in your paper/report.\nFor this course you should use “Figure 1”, “Figure 2”, etc.\n• The caption should allow the figure to be standalone, that is to say, by reading the caption and looking at\nthe figure, it should be clear what the figure is about and why it was included without reading the whole\npaper.\n• The caption should contain complete sentences and be as brief as possible while still conveying your information clearly (this is not always easy).\nThe Figure\n• Make sure that the resolution is high enough to not be pixelated at its final size.\n• Check that any text is readable at the final size (Using a smaller graph in Logger Pro will cause the text to\nbe larger in relation to the graph when inserted into another program).\n• For graphs, ensure that the axes are labeled (including units) and that there is a legend if you have multiple\ndata sets on the same graphs.\nPropagation of Uncertainties\nIn the case of addition and subtraction, the equation for combining uncertainties is\nδx = δx1 + δx2 + δx3 , (5)\nwhere δx is the total uncertainty of your calculation and δx1, δx2, and δx3 are the uncertainties of your individual measurements.\nMultiplication and Division\nThis method is valid for both multiplication and division of measurements with uncertainties.\nδA\n\|A\|\n=\nδx\n\|x\|\n+\nδy\n\|y\|\n, (6)\nwhere A is the area, x is the length, y is the width, δx and δy are the uncertainties associated with these\nmeasurements, and δA is the propagated uncertainty of the product or quotient.\n5\nan instance to spawn (either is fine) and create a new Python 3 file as shown here\nPython\nPython is a high level programming language that is regularly used by the scientific community for general and\nresearch computation. We will be introducing it and using it in a very limited way in this lab course although we\nencourage you to explore it more on your own. We will primarily make use of the Python to calculate statistical\nvalues and propagate uncertainties. We will be using an interface for Python called Jupyter and WPI has set up\na JupyterHub server that you can access from any browser (Figure 3). The files you create on the server will stay\nfor at least the term, probably much longer.\nJupyter uses a cell based system and evaluated variables carry over to the next cell. There are a few different\ntypes of cells, Figure 5 shows 2 kinds, the code cell, which we will be using most of the time, and the markdown\ncell, which you can use to add nicely formatted notes to you file.\n6\nFigure 4: Above is the code that you could use use to propagate uncertainty for values that are added or subtracted. Always remember to comment every line of your code in this class.\nFigure 5: Above is the code that you could use use to calculate the mean and standard deviation. Always remember to comment every line of your code in this class.\n7" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9055926,"math_prob":0.86768574,"size":8267,"snap":"2022-40-2023-06","text_gpt3_token_len":1838,"char_repetition_ratio":0.12537819,"word_repetition_ratio":0.02578019,"special_character_ratio":0.21144308,"punctuation_ratio":0.08142494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96887386,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T02:52:19Z\",\"WARC-Record-ID\":\"<urn:uuid:57b5933e-bcf4-4d11-8500-c7790fee30bd>\",\"Content-Length\":\"104988\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17d41c57-b4ee-4443-b97c-5d9d75ccc4c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:9fe578b5-eb50-4dad-99c1-15703c6ea753>\",\"WARC-IP-Address\":\"104.21.71.217\",\"WARC-Target-URI\":\"https://homeworkdave.com/product/impulse-momentum-lab-4-2/\",\"WARC-Payload-Digest\":\"sha1:ZDLFCXRTN3XX2TVIKXIYFCQQHW46OBDI\",\"WARC-Block-Digest\":\"sha1:O36C3RPOPCRLK2YOMQS3TRPMECWNVYO5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500303.56_warc_CC-MAIN-20230206015710-20230206045710-00707.warc.gz\"}"}
https://docs.pyro.ai/en/stable/_modules/pyro/poutine/broadcast_messenger.html	[ "```# Copyright (c) 2017-2019 Uber Technologies, Inc.\n\nfrom pyro.util import ignore_jit_warnings\n\nfrom .messenger import Messenger\n\n\"\"\"\nAutomatically broadcasts the batch shape of the stochastic function\nat a sample site when inside a single or nested plate context.\nThe existing `batch_shape` must be broadcastable with the size\nof the :class:`~pyro.plate` contexts installed in the\n`cond_indep_stack`.\n\nNotice how `model_automatic_broadcast` below automates expanding of\ndistribution batch shapes. This makes it easy to modularize a\nPyro model as the sub-components are agnostic of the wrapping\n:class:`~pyro.plate` contexts.\n\n... with IndepMessenger(\"batch\", 100, dim=-2):\n... with IndepMessenger(\"components\", 3, dim=-1):\n... sample = pyro.sample(\"sample\", dist.Bernoulli(torch.ones(3) * 0.5)\n... .expand_by(100))\n... assert sample.shape == torch.Size((100, 3))\n... return sample\n\n... with IndepMessenger(\"batch\", 100, dim=-2):\n... with IndepMessenger(\"components\", 3, dim=-1):\n... sample = pyro.sample(\"sample\", dist.Bernoulli(torch.tensor(0.5)))\n... assert sample.shape == torch.Size((100, 3))\n... return sample\n\"\"\"\n\n@staticmethod\n@ignore_jit_warnings([\"Converting a tensor to a Python boolean\"])\ndef _pyro_sample(msg):\n\"\"\"\n:param msg: current message at a trace site.\n\"\"\"\nif msg[\"done\"] or msg[\"type\"] != \"sample\":\nreturn\n\ndist = msg[\"fn\"]\nactual_batch_shape = getattr(dist, \"batch_shape\", None)\nif actual_batch_shape is not None:\ntarget_batch_shape = [\nNone if size == 1 else size for size in actual_batch_shape\n]\nfor f in msg[\"cond_indep_stack\"]:\nif f.dim is None or f.size == -1:\ncontinue\nassert f.dim < 0\ntarget_batch_shape = [None] * (\n-f.dim - len(target_batch_shape)\n) + target_batch_shape\nif (\ntarget_batch_shape[f.dim] is not None\nand target_batch_shape[f.dim] != f.size\n):\nraise ValueError(\n\"Shape mismatch inside plate('{}') at site {} dim {}, {} vs {}\".format(\nf.name,\nmsg[\"name\"],\nf.dim,\nf.size,\ntarget_batch_shape[f.dim],\n)\n)\ntarget_batch_shape[f.dim] = f.size\n# Starting from the right, if expected size is None at an index,\n# set it to the actual size if it exists, else 1.\nfor i in range(-len(target_batch_shape) + 1, 1):\nif target_batch_shape[i] is None:\ntarget_batch_shape[i] = (\nactual_batch_shape[i] if len(actual_batch_shape) >= -i else 1\n)\nmsg[\"fn\"] = dist.expand(target_batch_shape)\nif msg[\"fn\"].has_rsample != dist.has_rsample:\nmsg[\"fn\"].has_rsample = dist.has_rsample # copy custom attribute\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.58706045,"math_prob":0.9802396,"size":2588,"snap":"2023-40-2023-50","text_gpt3_token_len":684,"char_repetition_ratio":0.16911764,"word_repetition_ratio":0.09259259,"special_character_ratio":0.3083462,"punctuation_ratio":0.27160493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9673606,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T15:38:55Z\",\"WARC-Record-ID\":\"<urn:uuid:cb7d82dd-aad2-4a47-b0d9-57d3edfd6390>\",\"Content-Length\":\"22567\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd488621-e54a-4b70-ba25-3d097224fa54>\",\"WARC-Concurrent-To\":\"<urn:uuid:a0c2fc10-4d85-481b-a022-a0a90c4d7731>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://docs.pyro.ai/en/stable/_modules/pyro/poutine/broadcast_messenger.html\",\"WARC-Payload-Digest\":\"sha1:ZUWRS7RWTDHDZKBUALP77SJOZ4YVK2WC\",\"WARC-Block-Digest\":\"sha1:P3OKJGQOLOP66SXACFTPCCKTB6YVYZQZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510903.85_warc_CC-MAIN-20231001141548-20231001171548-00719.warc.gz\"}"}
https://study.com/academy/answer/imagine-that-a-company-is-forecasting-the-following-income-statement-for-the-upcoming-year-sales-5-000-000-operating-costs-excluding-depreciation-3-000-000-gross-margin-2-000-000-depreciation-500-000-ebit-1-500-000-interest-500-000-ebt-1-000-000-taxes.html	[ "# Imagine that a company is forecasting the following income statement for the upcoming year: ...\n\n## Question:\n\nImagine that a company is forecasting the following income statement for the upcoming year:\n\n Sales 5,000,000 operating costs (excluding depreciation) 3,000,000 Gross Margin 2,000,000 Depreciation 500,000 EBIT 1,500,000 Interest 500,000 EBT 1,000,000 Taxes (40%) 400,000 Net Income 600,000\n\nThe company's president is disappointed with the forecast and would like to see the company generate higher sales and a forecasted net income of $2,000,000. Assume that operating costs (excluding depreciation) are always 60 percent of sales. Also, assume that depreciation, interest expense, and the company's tax rate, which is 40 percent, will remain the same even if sales change. What level of sales would the firm have to obtain to generate$2,000,000 in net income?\n\n## Income statement:\n\nThe income statement is the statement that is prepared to find out the earnings of the company over an accounting period. It gives us the revenue generated, expenses incurred over the period.\n\nWe will have to follow the bottom-up approach to solve the given question:\n\nNet Income = $2,000,000.00 Add: Taxes = 2,000,000 / 60 x 40 =$1,333,333.33\n\nEarnings before tax = $3,333,333.33 Add: Interest =$500,000.00\n\nEBIT = $3,833,333.33 Add: Depreciation =$500,000.00\n\nGross Margin = $4,333,333.33 Gross margin = 40% of sales, since operating costs are always 60% of sales. So, Sales required = Gross Margin / Gross margin % = 4,333,333.33 / 0.40 =$10,833,333.33", null, "" ]	[ null, "https://study.com/images/reDesign/global/spinner-dark-teal.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91078,"math_prob":0.9963874,"size":1890,"snap":"2020-24-2020-29","text_gpt3_token_len":501,"char_repetition_ratio":0.12990455,"word_repetition_ratio":0.06849315,"special_character_ratio":0.32486773,"punctuation_ratio":0.18858561,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9761539,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-05T07:04:01Z\",\"WARC-Record-ID\":\"<urn:uuid:ef91d50b-8fe5-4af5-b7ca-96eb6f7bee1d>\",\"Content-Length\":\"108774\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:83696074-3442-43e8-b3ed-909fa5e0f724>\",\"WARC-Concurrent-To\":\"<urn:uuid:d50c4014-dd7b-4085-90d7-67f4bcae10c8>\",\"WARC-IP-Address\":\"52.0.4.252\",\"WARC-Target-URI\":\"https://study.com/academy/answer/imagine-that-a-company-is-forecasting-the-following-income-statement-for-the-upcoming-year-sales-5-000-000-operating-costs-excluding-depreciation-3-000-000-gross-margin-2-000-000-depreciation-500-000-ebit-1-500-000-interest-500-000-ebt-1-000-000-taxes.html\",\"WARC-Payload-Digest\":\"sha1:XQI2EQF77YTC5LY7IJ56E7AK3SLHLUTT\",\"WARC-Block-Digest\":\"sha1:DEAHRCROGGYNNGO3QA6VGTR5V2G52XFJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348493151.92_warc_CC-MAIN-20200605045722-20200605075722-00304.warc.gz\"}"}
https://fr.slideserve.com/margie/digital-testing-built-in-self-test-powerpoint-ppt-presentation	[ "", null, "Download", null, "Download Presentation", null, "Digital Testing: Built-in Self-test\n\n# Digital Testing: Built-in Self-test\n\nDownload Presentation", null, "## Digital Testing: Built-in Self-test\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Digital Testing: Built-in Self-test Based on text by S. Mourad \"Priciples of Electronic Systems\"\n\n2. Outline • BIST and embedded testing • Why BIST • Primitive polynomials • LFSR • Response compression • BILBO\n\n3. Define Built-In Self-Test • Implement the function of automatic test equipment (ATE) on circuit under test (CUT). • Hardware added to CUT: • Pattern generation (PG) • Response analysis (RA) • Test controller CK PG Stored Test Patterns Pin Electronics CUT Test control logic CUT Test control HW/SW BIST Enable RA Stored responses Comparator hardware Go/No-go signature ATE\n\n4. Built-in self-test Disadvantage of LSSD & other scan techniques: 1. Test generation necessary for combinational part 2. Long test time since test have to be shifted in & out 3. Only stuck-at faults are tested - not good for VLSI\n\n5. Built-in self test Test generation is NP complete. This prompted a search for built-in structures * Built-in self test BIST is an alternative to automatic test vector generation * Test generation & verification done by circuits built into the chip * Pseudo-random test vector generation is accomplished by using shift registers\n\n6. Built-in self-test in VLSI • Test patterns generated on chip • responses to test evaluated on chip • external operations only to initialize test & clock • go no go results • additional pins & silicon area minimized\n\n7. Built-in self test types • operation • (concurrent or not) • test design • (exhaustive or not) • test vector generation • (deterministic or pseudorandom) • data compression • (full or compresses test vectors)\n\n8. A general built-in self test approach\n\n9. Built-in self test * Deterministic testing identifies test vectors to detect specific faults * Pseudorandom testing detects # of faults by any test vector * Fault coverage increases rapidly at the beginning and slows down towards the end * The response data is compressed using a signature analysis * Linear feedback shift registers (LFSR) used to generate test vectors and compress responses\n\n10. Pseudorandom Integers Xk = Xk-1 + 3 (modulo 8) Xk = Xk-1 + 2 (modulo 8) 0 0 7 1 7 1 Start Start 6 2 6 2 +3 +2 5 3 5 3 4 4 Sequence: 2, 5, 0, 3, 6, 1, 4, 7, 2 . . . Sequence: 2, 4, 6, 0, 2 . . . Maximum length sequence: 3 and 8 are relative primes.\n\n11. Pseudo-Random Pattern Generation • Standard Linear Feedback Shift Register (LFSR) • Produces patterns algorithmically – repeatable • Has most of desirable random # properties • May not cover all2n input combinations • Long sequences needed for good fault coverage either hi = 0, i.e., XOR is deleted or hi = Xi Initial state (seed): X0, X1, . . . , Xn-1 must not be 0, 0, . . . , 0\n\n12. Y0 Q D Q Q D D Clk Q Q Q (a) Y Y Y 1 2 3 Y 0 Q Q D Q D D Clk (b) Q Q Q Y Y Y 1 2 3 Y 0 Q Q Q D D D Clk Q Q Q (c) Y Y Y 1 2 3 Pseudo-Random Pattern Generator Various LFSR configurations\n\n13. (a) (b) (c) Clk Y0 Y1Y2Y3 ClkY0 Y1Y2Y3 ClkY0 Y1Y2Y3 1 001 1 001 1 001 1 1 100 1 0 100 1 1 100 2 1 110 2 0 010 2 0 110 3 0 111 3 1 001 3 0 011 4 1 011 4 1 001 5 0 101 6 0 010 7 1 001 Initial state Y0 Q D Q Q D D Clk Q Q Q (a) Y Y Y 1 2 3 Y 0 Q Q D Q D D Clk Repeated states Q Q Q (c) Y Y Y 1 2 3 Pseudo-random Patterns\n\n14. Forcing all possible states in LFSR Modified LFSR • Clk Y0 Y1Y2Y3 Y1’Y2’ • 0 001 1 • 1 1 000 1 • 2 1 100 0 • 3 1 110 0 • 0 111 0 • 5 1 011 0 • 6 0 101 0 • 0 010 0 • 0 001 1 Produces pseudorandom sequence of length 8 Initial state\n\n15. Standard LFSR XOR operations performed outside of the shift register\n\n16. Modular LFSR XOR operations performed inside the shift register\n\n17. Linear feedback shift registers Two basic configurations : - internal XOR (IE) - external XOR (EE) Feedback connections based on coefficients of a characteristic polynomial : P(X) = C0 + C1X + C2X2 +…+ CnXn An LFSR with n-flip flops can assume (2n-1) states with depend upon : initial state, input, and feedback\n\n18. EXTERNAL CONNECTIONS (STANDARD) 2 3 4 1 X X X X INTERNAL CONNECTIONS (MODULAR) 4 3 2 X X X X 1 3 4 P(X) = 1+ X + X Characteristic Polynomial Linear feedback shift registers\n\n19. Y 0 Q D D Q D Q Y0 Q D Q Q D D Clk Q Q Q Clk Y Y Y Q Q Q (a) 1 2 3 Y Y Y 1 2 3 LFSR Equivalence P(X)=1+X+X3\n\n20. Test generation in BIST LFSR generally works without input - so only the initial state & interconnections decide the next state A generator which generates exactly (2n-1) different states is called a maximal-length generator All polynomials are either primitive (irreducible) or nonprimitive\n\n21. Test generation in BIST A primitive polynomial generates a max length sequence of test vectors. The number of primitive polynomials of order n grows rapidly with n p - any prime number which divides (2n-1)\n\n22. Modulo 2 Operations a b a  b a+b a+b a-b a - b sum carry difference borrow 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 0 1 0 • 1 0 0 1 0 0 Define time translation operation as X k = X (t-k)\n\n23. Math Foundation of LFSR Yj can be represented as: Yj(t) = Yj-1(t - 1) for j  0 We can express Yj in terms of Y0 as: Yj (t) =Y0(t - j) Denote the translation operator as X k, where k represents the time translation units, then Yj (t) =Y0(t)X j On the other hand in LFSR Where the summation is equivalent to an XOR operation. Then we get\n\n24. Math of LFSR Generators From linearity we have and We can then write this expression as Y0 (t) PN (X) = 0 For non-trivial solutions, Y0(t)  0, then we must have PN (X) = 0. Where, PN (X) is called the characteristic polynomial of the LFSR.\n\n25. Primitive Polynomials • Examples of primitive polynomials with minimum number of terms • N Polynomials • 1,2,3,4,6,7,15,22 1 + X + Xn • 3,5,11, 21, 29 1 + X2 + Xn • 10,17,20,25,28,31 1 + X3 + Xn • 9 1 + X4 + Xn • 23 1 + X5 + Xn • 18 1 + X7 + Xn • 8 1 + X2 + X3 + X4 + Xn • 12 1 + X + X3 + X4 + Xn • 13 1 + X + X4 + X6 + Xn • 14, 16 1 + X + X3 + X4 + Xn\n\n26. Test generation in BIST For instance for n=8 a minimum polynomial is Example : Let us follow test generation using a primitive polynomial\n\n27. Reciprocal Polynomials The reciprocal polynomial of P(X) is defined by: so PR(X) = XN + Cj XN-j for 1  j N Thus every coefficient Cjin P(X) is replaced by CN-j in PR(X) For example, the reciprocal of polynomial P(X) = 1+ X+ X3 is PR(X) = 1+ X2 + X3\n\n28. Operations on Polynomials • Polynomial multiplication • x4 + x3 + + 1 • . x + 1 . • x4 + x3 + + 1 • x5 + x4+ + x . • x5 + + x3 + x + 1 since x4 + x4 = 0. • Division is of particular interest when LFSRs are used for response compaction. • x2 + x + 1 . • x2 + 1 ) x4 + x3 + + 1 • x4 + + x2 . • x3 + x2 + + 1 • x3 + + x . • x2 + x + 1 • x2 + + 1 • x • the reminder R(x)=x\n\n29. Operations on Polynomials • Reminder of the division of the input sequence polynomial by the • LFSR polynomial gives the signature for the compacted response • Q(X) X4 + X3 + 1 1 1 0 0 1 • X3 + X2 + 1 \|X7 + X5 + X4 + +1 1 1 0 1 \| 1 0 1 1 0 0 0 1 • X7 + X6 + X41 1 0 1 • X6 + X5 +1 1 1 0 0 0 0 1 • X6 + X5 + +X31 1 0 1 • X3 + +1 1 0 0 1 • X3 + X2 + 11 1 0 1 • R (X) X2 0 1 0 0 signature Polynomial for the input data\n\n30. Properties of Polynomials • An irreducible polynomial is that polynomial which cannot be factored and it is divisible by only itself and 1. • An irreducible polynomial of degree n is characterized by : • An odd number of terms including the 1 term • Divisibility into1 + xk, where k = 2n - 1. • Any polynomial with all even exponents can be factored and hence is reducible • An irreducible polynomial is primitive if the smallest positive integer k that allows the polynomial to divide evenly into 1 + xk occurs for k = 2n - 1, where n is the degree of the polynomial.\n\n31. Properties of Polynomials • All polynomials of degree 3 are: • x3 + 1= 0 • x3 + x2 + 1 = 0 Primitive • x3 + x + 1 = 0 Primitive • x3 + x2 + x + 1= 0 • But, x3 + 1= (x + 1)( x2 + x + 1) • x3 + x2 + x + 1 = (x + 1)( x2 + 1) • There are several primitive polynomial of degree N. • We are interested in those with fewer terms since they need less XOR gates in the LFSR. • Among primitive polynomial of degree 16 are x16 + x5 + x3 + x2 + 1 and x16 + x4 + x3 + x + 1.\n\n32. Check for Primitive Polynomial • Consider a 3-rd order primitive polynomial x3 + x + 1 = 0 • If this polynomial is primitive it must divide evenly into 1 + x 7 (7 = 2 3 – 1) where 3 is the degree of the polynomial. We can check that 1 + x 7= (x 3+ x + 1)(x 4 + x 2 + x + 1)\n\n33. Test data compression • To verify response of a tested circuit use • test data compression • Ones count compression • ex 10011010 => 0(x)=4 • Transition count compression • ex 10011010 => c(x) =5 • Parity check compression • ex 10011010 => p(x) =0 • Syndrome testing (normalized # of 1’s ) • ex 10011010+> s(x) =4/8 • Compression using Walsh spectra • Cyclic code compression (LFSR)\n\n34. Parity Compression Computes parity\n\n35. Ones Count If we have a test of length L and the fault-free count is m, then the possibility of aliasing is [C (L, m) - 1] patterns out of total number of possible strings of length L, (2L - 1).\n\n36. One Count example For m = 5 and L = 8, aliasing probability will be Pa (m) =( C(8,5)-1 ) / (2^8-1) =55 /255  0.2. Not a very reliable method\n\n37. Transition Count Computes transitions\n\n38. Signature Analysis Uses LFSR to obtain a signature\n\n39. LFSR as Response Analyzer • Use cyclic redundancy check code (CRCC) generator (LFSR) for response compacter • Treat data bits from circuit POs to be compacted as a decreasing order coefficient polynomial • CRCC divides the PO polynomial by its characteristic polynomial • Leaves remainder of division in LFSR • Must initialize LFSR to seed value (usually 0) before testing • After testing – compare signature in LFSR to precomputed signature of fault-free circuit\n\n40. Signature Analysis • LFSR seed is “00000”\n\n41. Signature by Logic Simulation X0 0 1 0 0 0 1 1 1 1 Input bits Initial State 1 0 0 0 1 0 1 0 X1 0 0 1 0 0 0 0 1 0 X2 0 0 0 1 0 0 0 0 1 X3 0 0 0 0 1 0 1 0 1 X4 0 0 0 0 0 1 0 1 0 Signature\n\n42. X2 X7 X7 + 1 + X5 X5 X5 + X3 + X3 + X3 X3 + X + X + X X5 + X3 + X + 1 Char. polynomial + X2 + X2 + X2 + 1 + 1 remainder Signature: X0X1X2X3X4 = 1 0 1 1 0 Signature by Polynomial Division Input bit stream: 0 1 0 1 0 0 0 1 0 ∙ X0 + 1 ∙ X1 + 0 ∙ X2 + 1 ∙ X3 + 0 ∙ X4 + 0 ∙ X5 + 0 ∙ X6 + 1 ∙ X7 Polynomial division equivalence of datacompression\n\n43. Test generation based on a nonprimitive polynomial x4 + x2 +1 generates only 6 out of possible 15 states\n\n44. Multiple-Input Signature Register (MISR) • Problem with ordinary LFSR response compacter: • Too much hardware if one of these is put on each primary output (PO) • Solution: MISR – compacts all outputs into one LFSR • Works because LFSR is linear – obeys superposition principle • Superimpose all responses in one LFSR – final remainder is XOR sum of remainders of polynomial divisions of each PO by the characteristic polynomial\n\n45. X0 (t + 1) X1 (t + 1) X2 (t + 1) 0 0 1 1 1 0 X0 (t) X1 (t) X2(t) d0 (t) d1 (t) d2(t) 0 1 0 = + Modular MISR Example\n\n46. SSA M 1 M 3 4 1 2 2 M 3 M 4 BELLMAC M 1 M 2 LSFR M 3 M 4 PSA M M M M 2 3 4 1 2 3 4 1 2 3 M(X)= M + X M + X M + X M 1 2 3 4 Space Compaction – parallel outputs\n\n47. Test data compression BIST To reduce the number of response data a compression Based on LFSR (signature analysis) is used. Probability that 2 sequences which differ by 1 bit only will have the same signature is zero Data entering LFSR serially produces a reminder of the division of the data stream polynomial by the polynomial used for LFSR design\n\n48. A faulty data stream will yield the same signature (reminder of polynomial division) as a fault-free data when they have the same reminder The likelihood of this is in the range of where n is the number of the compressor bits The same effect can be obtained on multi-input shift registers with faster processing speed & smallerchip area. Fault-free signature 2n-1 faulty signatures Test data compression BIST\n\n49. Test data compression BIST • As in the single-input case the probability of • detecting uniformly distributed fault in the output • stream is (1 - 2-n ) • Probability of not detecting error after • checking its signature is greater than 2-n - the • probability of not catching the signature error. • Example 1: n = 4, Aliasing probability = 6.25% • Example 2: n = 8, Aliasing probability = 0.39% • Example 3: n = 16, Aliasing probability = 0.0015%\n\n50. LFSR Design Guidelines In using LFSR for data compression: • Chose r large enough to reduce 2-r • Repeat test using different feedback connections • Repeat test with different test vector • Compress serial output of MISR into LFSR to capture errors" ]	[ null, "https://fr.slideserve.com/img/player/ss_download.png", null, "https://fr.slideserve.com/img/replay.png", null, "https://thumbs.slideserve.com/1_4440638.jpg", null, "https://fr.slideserve.com/img/output_cBjjdt.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7702014,"math_prob":0.994075,"size":12262,"snap":"2021-21-2021-25","text_gpt3_token_len":3986,"char_repetition_ratio":0.14325339,"word_repetition_ratio":0.09191457,"special_character_ratio":0.3412983,"punctuation_ratio":0.052652895,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953185,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-20T22:32:44Z\",\"WARC-Record-ID\":\"<urn:uuid:e0166d3a-583b-441f-9731-7ade5257c00b>\",\"Content-Length\":\"107553\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02ebb6e5-65d0-4daa-8e29-2a11c994c61e>\",\"WARC-Concurrent-To\":\"<urn:uuid:2018190a-b9a3-40c7-b952-6e83d583466c>\",\"WARC-IP-Address\":\"44.236.52.34\",\"WARC-Target-URI\":\"https://fr.slideserve.com/margie/digital-testing-built-in-self-test-powerpoint-ppt-presentation\",\"WARC-Payload-Digest\":\"sha1:BNX3GCFVVMANODBTW6ECCFZ2GWSC5J6N\",\"WARC-Block-Digest\":\"sha1:S56FGCBQ43Z2KRYJDL7PF3BUNUFMCIAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488257796.77_warc_CC-MAIN-20210620205203-20210620235203-00018.warc.gz\"}"}
https://datascience.stackexchange.com/questions/40944/both-train-and-test-error-are-decreasing-in-xgboost-iterations	[ "# Both train and test error are decreasing in XGBoost iterations\n\nI have an issue with training an XGBoost classifier in a sence that both train and test error only decrease throughout more iterations (num_boost_round) even if I use 1000 num boost rounds and 10 early stopping rounds. Then when I try to apply the model on a separate set not used for testing and training I can see that a model trained over 100 rounds performs much better than 1000 one. My learning rate is 0.01.\n\nI am wondering whether this is a normal situation (and I should just do early stopping on this independent validation set) or whether I am doing something wrong. Theoretically, the test error should start increasing at some point, but this is just not happening.\n\nI am using XGBoost and I am splitting my dataset in train/test like this:\n\nX_train, X_test, y_train, y_test = train_test_split(X, Y, test_size=0.3, random_state=42)\n\ndtrain = xgb.DMatrix(X_train, label=y_train)\ndtest = xgb.DMatrix(X_test, label=y_test)\n\n\nThem I train it in the following way:\n\nwatchlist = [(dtest, 'eval'), (dtrain, 'train')]\nprogress = dict()\n\n# Train and predict with early stopping\nxg_reg = xgb.train(\nparams=params,\ndtrain=dtrain, num_boost_round=boost_rounds,\nevals=watchlist, # using validation on a test set for early stopping; ideally should be a separate validation set\nearly_stopping_rounds=early_stopping,\nevals_result=progress)\n\n\nNote that I have used gridsearch to find the optimal hyperparameters (params).\n\nA plot which I get (yellow is my cutoff point identified pretty much via testing on the independent set):", null, "This is a common situation and can be caused by a number of things, such as:\n\n1. Too much variance in the model (e.g. did the hyperparameter tuning identify low values for eta, alpha, lambda, gamma, subsampling/column sampling/row sampling, minimum child weights and/or high values for depth and number of leaves?)\n2. Having a small dataset with high variance. XGBoost is prone to overfitting and this is exacerbated when there isn't much data for it.\n3. Not shuffling the data so that the validation set has lower variance than (or differs in distribution from) the training and test sets. train_test_split shuffles by default, so this shouldn't be a problem if you've used it as indicated in the question for generating the validation set also.\n\nStopping training early is a perfectly valid and common method of increasing bias in a model and doing so based on a validation set is acceptable. It is preferable for the training to stop based on the test set. To achieve this you could address the potential issues above. Without knowing more about the data and how the validation set was generated, the most reliable advice I can give is to alter the hyperparameters to increase model bias.\n\nSimply you are using the training set for early stopping.\n\nTry by setting\n\nwatchlist = [(dtrain, 'train'), (dtest, 'eval')] # the last one is used for early stopping" ]	[ null, "https://i.stack.imgur.com/br37a.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8589493,"math_prob":0.8243559,"size":1513,"snap":"2020-45-2020-50","text_gpt3_token_len":364,"char_repetition_ratio":0.11332008,"word_repetition_ratio":0.0,"special_character_ratio":0.2333113,"punctuation_ratio":0.11956522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97209674,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T20:29:34Z\",\"WARC-Record-ID\":\"<urn:uuid:23f294ed-448f-4549-92ef-0efda85875fc>\",\"Content-Length\":\"158100\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b03d390-ab15-40e1-8a6d-d7ca9af0fd36>\",\"WARC-Concurrent-To\":\"<urn:uuid:885f6ad6-fecb-4f73-82a5-6f56d9a09cdf>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://datascience.stackexchange.com/questions/40944/both-train-and-test-error-are-decreasing-in-xgboost-iterations\",\"WARC-Payload-Digest\":\"sha1:W6LLQ6C22ZUEWP5VG4OMKEVZGKWTPZJ2\",\"WARC-Block-Digest\":\"sha1:53FZEVNXR4I4D73W664W2EGBQCXOCXFN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141194171.48_warc_CC-MAIN-20201127191451-20201127221451-00512.warc.gz\"}"}
https://hello-sunil.in/javascript-higher-order-functions/	[ "", null, "# Higher Order Functions in JavaScript – Beginner’s Guide\n\nAs a frontend developer, you should always strive to learn new techniques and discover ways to work smarter with JavaScript. One such technique is using higher order functions.\n\nHigher order functions are one of the topics that can be hard to understand. This article will help you understand what higher order functions are and how to work with them.\n\nYou will also learn about the difference between high order and first class functions and about high order built-in functions in JavaScript.\n\n## Functions are first-class citizens in JavaScript\n\nIn JavaScript, functions are first-class citizens(JavaScript treats functions as objects). This means that they can be assigned to a variable and/or be passed as arguments and can also be returned from another function.\n\nThese features or abilities open the door for various types of functions like, First-class functions, Callback functions, Higher-order functions, Anonymous functions and more.\n\nHaving a clear and better understanding of how higher order functions work can save you a lot of time and make your logical implementation seamless.\n\n## JavaScript and first class functions\n\nFirst-class functions are a unique feature in JavaScript, meaning that functions here are treated like any other variable. They can be passed as arguments, returned by another function, or assigned as a value to a variable.\n\nIn JavaScript, everything you can do with other types like object, string, or number, you can do with functions. You can pass them as parameters to other functions (callbacks), assign them to variables and returned from another function. This is why functions in JavaScript are known as First-Class Functions.\n\nAssigning functions to variables:\n\nAssigning functions to variables is one common thing you can do with them. This is also called creating function expressions.\n\nThe reason for this name is that you create a function inside an expression. You then assign this expression to a variable. From now on you can use the variable name to reference, and call, this function.\n\n```// Create function with function expression and assign it to a variable\nconst myFunc = function() {\nreturn 'Hello'\n}\nconsole.log(myFunc()); //Hello```\n\nPassing functions as arguments:\n\nOne thing you may have seen is passing functions as arguments to other functions. These passed functions are usually passed as the last argument and later used as callbacks.\n\nA callback is a function that gets executed when all operations that have been completed. This is a common practice.\n\nThis works due to the fact that JavaScript is single-threaded programming language. This means that only one operation can be executed at the time. So, when you pass a callback function, and invoke it at the end, it will be invoked when all preceding operations are completed.\n\nOne situation when functions are passed as arguments frequently is when you work with event listeners.\n\n``target.addEventListener(event, function, useCapture)``\n\nThe `addEventListener` has three parameters. `event` and `function` are most frequently used. `useCapture` is used occasionally.\n\nAlso, the second parameter, the `function`, is the callback. When specific `addEventListener` is triggered this callback function is invoked.\n\n```<body>\n<button id='btn'>Hello</button>\n\n<script>\n// Find button in the DOM\nconst button = document.querySelector('#btn')\n\n// Create function to handle the event\nfunction handleButtonClick() {\n// Show log on click\nconsole.log('click on: ', this)\n}\n\n// Create event listener for click event on button\n// and pass \"handleButtonClick\" function as a callback function\n\n//Output - click on: <button id='btn'>Hello</button>\n</script>\n</body>```\n\nKindly note here we passed the `handleButtonClick` function by name, without parentheses. This means that you are passing the function object itself.\n\nIf you passed that function with parentheses that would mean you are invoking the function right away and passing the result of executing that function.\n\nLets see one more example:\n\n```function formalGreeting() {\nconsole.log(\"How are you?\");\n}\nfunction casualGreeting() {\nconsole.log(\"What's up?\");\n}\nfunction greet(type, greetFormal, greetCasual) {\nif(type === 'formal') {\ngreetFormal();\n} else if(type === 'casual') {\ngreetCasual();\n}\n}\ngreet('casual', formalGreeting, casualGreeting);\n// prints 'What's up?'```\n\nReturning functions:\n\nAnother thing you can do with functions is that you can return them from other functions. This is something expected, since functions are a type of objects, and you can return objects in JavaScript.\n\nThis can be useful when you want to use function to create templates for new functions.\n\n```// Create function that returns function\n// a template for new functions\nfunction changeText(word, replacement, text) {\n// Return a new function\n// that will be later assigned to a variable\nreturn function(text) {\nreturn text.replace(word, replacement)\n}\n}\n\n// Create new function for changing text\n// to positive mood using the \"changeText\" function\n// This invokes and \"changeText\" function\n// and stores returned function in a variable\n\n// Create new function for changing text\n// to negative mood using the \"changeText\" function\n// This invokes and \"changeText\" function\n// and stores returned function in a variable\n\n// Call the \"makePositive\" function\n// This invokes the returned function\n// stored in the variable after calling\n// the \"changeText\" function\nmakePositive('Everything that happened is bad and everything that will happen is also bad.')\n// 'Everything that happened is good and everything\n//that will happen is also good.'\n\n// Call the \"makePositive\" function\n// This invokes the returned function\n// stored in the variable after calling\n// the \"changeText\" function\nmakeNegative('Everything that happened is good and everything that will happen is also good.')\n// 'Everything that happened is bad and everything\n//that will happen is also bad.'```\n\nLet’s see an example:\n\n```function calculate(operation) {\nswitch (operation) {\nreturn function (a, b) {\nconsole.log(`\\${a} + \\${b} = \\${a + b}`);\n};\ncase \"SUBTRACT\":\nreturn function (a, b) {\nconsole.log(`\\${a} - \\${b} = \\${a - b}`);\n};\n}\n}\n\nIn the above code, when we invoke the function `calculate` with an argument, it switches on that argument and then finally returns an anonymous function.\n\nSo if we call the function `calculate()` and store its result in a variable and console log it, we will get the following output:\n\n```function calculate(operation) {\nswitch (operation) {\nreturn function (a, b) {\nconsole.log(`\\${a} + \\${b} = \\${a + b}`);\n};\ncase \"SUBTRACT\":\nreturn function (a, b) {\nconsole.log(`\\${a} - \\${b} = \\${a - b}`);\n};\n}\n}\n\nYou can see that `calculateAdd` contains an anonymous function that the `calculate()` function returned.\n\nThere are two ways to call this inner function which we’ll explore now.\n\nCall the returned function using a variable:\n\nIn this method, we stored the return function in a variable as shown above and then invoked the variable to the inner function.\n\nLet’s see it in code:\n\n```function calculate(operation) {\nswitch (operation) {\nreturn function (a, b) {\nconsole.log(`\\${a} + \\${b} = \\${a + b}`);\n};\ncase \"SUBTRACT\":\nreturn function (a, b) {\nconsole.log(`\\${a} - \\${b} = \\${a - b}`);\n};\n}\n}\n// Output: 2 + 3 = 5\n\nconst calculateSubtract = calculate(\"SUBTRACT\");\ncalculateSubtract(2, 3);\n// Output: 2 - 3 = -1```\n\nSo what’d we do here?\n\n• We called the `calculate()` function and passed `ADD` as the argument\n• We stored the returned anonymous function in the `calculateAdd` variable, and\n• We invoked the inner returned function by calling `calculateAdd()` with the required arguments.\n\nCall the returned function using double parentheses:\n\nThis is a very sophisticated way of calling the inner returned function. We use double parentheses `()()` in this method.\n\nLet’s see it in code:\n\n```function calculate(operation) {\nswitch (operation) {\nreturn function (a, b) {\nconsole.log(`\\${a} + \\${b} = \\${a + b}`);\n};\ncase \"SUBTRACT\":\nreturn function (a, b) {\nconsole.log(`\\${a} - \\${b} = \\${a - b}`);\n};\n}\n}\n// Output: 2 + 3 = 5\n\ncalculate(\"SUBTRACT\")(2, 3);\n// Output: 2 - 3 = -1```\n\nThe arguments in the first parentheses belong to the outer function, while the arguments in the second parentheses belong to the inner returned function.\n\nThe `calculate()` method returns a function as explained earlier, and it is that returned function which is immediately called using the second parentheses.\n\n## What are Higher Order Functions 🧐?\n\nYou may wonder what those three things have to do with high order functions. The first one about variables not much. However, the second and third, passing functions as arguments and returning functions, a lot.\n\nHere is the thing, high order functions are functions that take another function as an argument, and/or returns another function.\n\nSo an Higher Order Function is a function that will at least :\n\n• take one or more functions as arguments\n• returns a function as its result\n\nHigher order functions are functions that operate on other functions, either by taking them as arguments or by returning them. In simple words, A higher order function is a function that receives a function as an argument or returns the function as output.\n\nThe function that is passed as an argument to the higher-order function is known as a Callback Function, because is it, to be ‘called-back’ by the higher order function at a later time.\n\nIn all examples, where you were passing a function as an argument, or returning a function, you were actually working with high order functions.\n\nYou probably expected something more complex than this. Especially due to how many JavaScript developers talk about high order functions. However, it is really that simple.\n\n```// High-order function no.1:\n// Function that takes a function as a argument\nfunction myHighOrderFuncOne(myFunc) {\n// some code\n}\n\n// High-order function no.2:\n// Function that returns a function\nfunction myHighOrderFuncTwo() {\n// some code\n\n// Return a function\nreturn function() {\n// some code\n}\n}\n\n// High-order function no.3:\n// Function that takes a function as a argument\n// and also returns a function\nfunction myHighOrderFuncThree(myFunc) {\n// some code\n\n// Return a function\nreturn function() {\n// some code\n}\n}```\n\nFew more examples are as follows:\n\nTaking a function as an argument:\n\nHere is an example of a function that takes a function as an argument.\n\n```// normal function\nfunction greetEnglish(name){\nreturn `Hello \\${name}`;\n}\nfunction greetSpanish(name){\nreturn `Hola \\${name}`;\n}\n// Our higher order function\nfunction greeting(otherGreeting, name){\nconsole.log(`Hello \\${name}`);\nreturn console.log(otherGreeting(name));\n}\ngreeting(greetSpanish, \"Sunil\");\n//output: Hello Sunil // first greets in English\n//output: Hola Sunil // greets in the language passed ```\n\nIn the above code snippet the two functions, `greetEnglish(name)`and `greetSpanish(name)` returns the greeting in respective language.\n\nThe function` greeting()` is a higher order function that takes another function and a string as a parameter. This function prints greeting in English by default and in any other language depending on the passed function as a parameter(here in Spanish).\n\nAlso, higher order function can also accept anonymous function as a parameter.\n\n```function greeting(callback, name){\nconsole.log(`Hello \\${name}`);\nreturn callback(name);\n}\ngreeting(function (name) {\nconsole.log(`hola \\${name}`)\n} , \"Sunil\")\n\n//output: Hello Sunil /* first greets in English /\n//output: Hola Sunil / second greets in Spanish /```\n\nWe can also express the same code in arrow function way:\n\n```function greeting(callback, name){\nconsole.log(`Hello \\${name}`);\nreturn callback(name);\n}\ngreeting(name => {console.log(`hola \\${name}`)}, \"Sunil\")\n\n//output: Hello Sunil / first greets in English /\n//output: Hola Sunil / second greets in Spanish */```\n\nAlso Read: Arrow Functions in JavaScript\n\nReturning a function as a result:\n\nNow let us take another function that returns a function.\n\n```// Define a function that returns a function\nfunction returnFunc() {\nreturn function() {\nconsole.log('Hi');\n}\n}\n\n// Take the returned function in a variable.\nconst fn = returnFunc();\n\n// Now invoke the returned function.\nfn(); // logs 'Hi' in the console\n\n// Alternatively - A bit odd syntax but good to know\nreturnFunc()(); // logs 'Hi' in the console```\n\nBoth of the examples above are examples of Higher-Order functions. They either accept a function as an argument or return a function.\n\nPlease note, it is not mandatory for a higher order function to perform both accepting an argument and returning a function. Performing either will make the function a higher order function.\n\n## In-built Higher Order Functions in JavaScript\n\nIn JavaScript, there are plenty of usages of higher order functions. You may be using them without knowing them as higher order functions.\n\nFor example, take the popular Array methods like, `map()`, `filter()`, `reduce()`, `find()`, `sort()` and many more. All these functions take another function as an argument to apply it to the elements of an array.\n\n### In-built Higher Order Functions – map()\n\n`map()` method allows us to process each element in the array and give back a new array with the same length.\n\n```var Array2 = [4,8,12,16,20];\nvar data = Array2.map(function(element){\nreturn element + 4;\n})\nconsole.log(data); //[ 8, 12, 16, 20, 24 ]```\n\n`data` contains an array where each of the elements in `Array2` gets added by `4`. The array on which map is applied will produce an array with the same length.\n\n### In-built Higher Order Functions – filter()\n\n`filter()` method allows to filter out some element in the array and give the rest of the element as an array output.\n\n```var Array4 = [12,45,41,74,91];\nvar data = Array4.filter(function(element){\nreturn element > 40;\n})\nconsole.log(data); //[ 45, 41, 74, 91 ]```\n\n`data` contains all the elements present in `Array4` which is more than 40.\n\n### In-built Higher Order Functions – reduce()\n\n`reduce()` method allows us to process all the elements in the array to give back a single value.\n\n```var Array3 = [112,415,4,274,911];\nvar data = Array3.reduce(function(a,b){\nreturn a+b;\n},0)\nconsole.log(data); //1716```\n\nHere `data` contains the sum of all the elements present in the `Array3`. The start element in the syntax is required to have a start value before the reduction starts.\n\nIn the above example If we give `2` instead of `0` in the start value. `2` will be added to the sum of all elements in that array.\n\nAnd If the array on which reduction is applied is having at least 1 element then we can miss out that start value in the function. The reduction will start from first element of the array then second element and so on.\n\n### In-built Higher Order Functions – forEach()\n\n`forEach()` method allows us to loop over each element in an array.\n\n```var Array1 = [1,2,3,4,5];\nvar data = Array1.forEach(function(element){\nconsole.log(element);\n})\nconsole.log(data);\n//Output -\n//1\n//2\n//3\n//4\n//5```\n\nHere `element` represents each element in the `Array1`. It prints out all the elements present in the array. Kindly note, `forEach()` method acts same as that of a for loop.\n\n## Creating Our Own Higher Order Function in JavaScript\n\nUp until this point, we saw various higher order functions built into the JavaScript language. Now let’s create our own higher order function.\n\nLet’s imagine JavaScript didn’t have the native `map` method. We could build it ourselves thus creating our own higher order function.\n\nLet’s say, we have an array of strings and we want to convert this array to an array of integers, where each element represent the length of the string in the original array.\n\n```const strArray = ['JavaScript', 'Python', 'PHP', 'Java', 'C'];\nfunction mapForEach(arr, fn) {\nconst newArray = [];\nfor(let i = 0; i < arr.length; i++) {\nnewArray.push(\nfn(arr[i])\n);\n}\nreturn newArray;\n}\nconst lenArray = mapForEach(strArray, function(item) {\nreturn item.length;\n});\n\nconsole.log(lenArray); // prints [ 10, 6, 3, 4, 1 ]```\n\nIn the above example, we created an higher order function `mapForEach` which accepts an array and a callback function `fn`. This function loops over the provided array and calls the callback function `fn` inside the `newArray.push` function call on each iteration.\n\nThe callback function `fn` receives the current element of array and returns the length of that element, which is stored inside the `newArray`. After the for loop has finished, the `newArray` is returned and assigned to the `lenArray`.\n\n## Conclusion\n\nWe have learned what higher order functions are and some built-in higher order functions. We also learned how to create our own higher order function.\n\nIn a nutshell, a higher order function is a function that may receive a function as an argument and can even return a function. higher order functions are just like regular functions with an added ability of receiving and returning other functions are arguments and output.\n\nThat’s all for now. Thank you for reading and we hope this article will be very helpful to understand the basics of higher order function. Stay tuned for more amazing content. Peace out! 🖖\n\n## Resource\n\nClick on a star to rate it!\n\nAverage rating 0 / 5. Vote count: 0\n\nNo votes so far! Be the first to rate this post.\n\nWe are sorry that this post was not useful for you!\n\nLet us improve this post!\n\nTell us how we can improve this post?\n\n## Similar articles you may like", null, "Hi there 👋 I am a front-end developer passionate about cutting-edge, semantic, pixel-perfect design. Writing helps me to understand things better.", null, "Hi there 👋 I am a front-end developer passionate about cutting-edge, semantic, pixel-perfect design. 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http://ixtrieve.fh-koeln.de/birds/litie/document/39295	[ "Document (#39295)\n\nAuthor\nSoergel, D.\nPopescu, D.\nTitle\nOrganization authority database design with classification principles\nSource\nClassification and authority control: expanding resource discovery: proceedings of the International UDC Seminar 2015, 29-30 October 2015, Lisbon, Portugal. Eds.: Slavic, A. u. M.I. Cordeiro\nImprint\nWürzburg : Ergon-Verlag\nYear\n2015\nPages\nS.69-82\nAbstract\nWe illustrate the principle of unified treatment of all authority data for any kind of entities, subjects/topics, places, events, persons, organizations, etc. through the design and implementation of an enriched authority database for organizations, maintained as an integral part of an authority database that also includes subject authority control / classification data, using the same structures for data and common modules for processing and display of data. Organization-related data are stored in information systems of many companies. We specifically examine the case of the World Bank Group (WBG) according to organization role: suppliers, partners, customers, competitors, authors, publishers, or subjects of documents, loan recipients, suppliers for WBG-funded projects and subunits of the organization itself. A central organization authority where each organization is identified by a URI, represented by several names and linked to other organizations through hierarchical and other relationships serves to link data from these disparate information systems. Designing the conceptual structure of a unified authority database requires integrating SKOS, the W3C Organization Ontology and other schemes into one comprehensive ontology. To populate the authority database with organizations, we import data from external sources (e.g., DBpedia and Library of Congress authorities) and internal sources (e.g., the lists of organizations from multiple WBG information systems).\nContent\nPräsentation unter: http://www.udcds.com/seminar/2015/media/slides/Soergel-Popescu_InternationalUDCSeminar2015.pdf.\nTheme\nNormdateien\n\nSimilar documents (author)\n\n1. Popescu, F.: ¬An approach to eponyms in mathematics (2009) 2.57\n2.565884 = sum of:\n2.565884 = product of:\n5.131768 = sum of:\n5.131768 = weight(author_txt:popescu in 3120) [ClassicSimilarity], result of:\n5.131768 = score(doc=3120,freq=1.0), product of:\n0.83266324 = queryWeight, product of:\n1.2262138 = boost\n9.860925 = idf(docFreq=5, maxDocs=42306)\n0.06886294 = queryNorm\n6.163078 = fieldWeight in 3120, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.860925 = idf(docFreq=5, maxDocs=42306)\n0.625 = fieldNorm(doc=3120)\n0.5 = coord(1/2)\n\n2. 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https://soup-dev.websitetoolbox.com/post/copying-attribute-values-to-other-geometries-9774473	[ "Hi, seeing as Soup's copy/transfer attribute features are rather particle centric - from my understanding at least - I am trying to discover a way to copy values between 2 nurbsCurves. Here is the case:\n\nIn my grooming workflow guide curves have custom parameters setup on them that influence the simulated fur (eg. deformation radius). When a groom is updated, these guide curves are replaced by a new set of curves, which have different names and only default values in these common/shared attributes. 2 curves only share the same position on the character mesh, so a world space (nurbsCurveShape1.cv) or uv space coordinate/(follicleShape.uparam/vparam).\n\nDoes soup allow me to copy attribute data between different objects? Its basically a simple get/setAttr, but requiring a nearest position lookup to define the target curve.\n\nThanks!\nQuote 0 0\nYes, easily.\nBut i don't think you need much of SOuP really.\nWhat you need is to collect the positions of the first CV of the \"old\" curves and store them into vector array.\nThen do the same with the new set of curves.\nThen run proximity check on them and find the closest pairs between them.\nThe get/setAttr between them.\n\nif you have a minimal Maya scene with the couple of \"old\" and \"new\" curves i can provide you the few lines of code that will take care of it, in case tech problems like that are not your forte.\nQuote 0 0\nHello Peter,\n\ngreat, simpler than expected. Thanks for you help! Can you please elaborate the proximity check for me, or point out the node/object that handles this?\n\nAm still curious as how to achieve this using soup itself, as I'd like to understand such basics better, so far haven't noodled anything meaningful together... But to follow up you suggestion, here is my reply to the pythonic approach, including a test scene containing two curves with attributes on both Shapes (roots positioned on a sphere).\n\nQuote:\nThen run proximity check on them and find the closest pairs between them.\n\n\"Elegant' proximity checking is the only thing I am not yet familiar with in Maya, ie. how to script the nearest point evaluation part using a function that has this built-in. Only know of a direct 'if condition' B == A of the actual two point position values in the 2 lists, say rounded to the third floating point value. Is there a node/maya object to handle this comparison instead? Similar to Houdini's copy (by nearest point) node?\n\nThis is a simple start to saving the point positions for the test scene supplied. I'd add code to create a second list of the new curves .cv positions (vector array). In an if loop compare if curveB pos == curveA pos, if condition is met, setAttr A->B.\n\nYou can ignore this half baked script part if you want, it needs some improvements 🙂\n\nThanks again,\nJan\n\n`''' 1 - save old root positions for curvesA cv list '''from maya import cmdscurvesA = cmds.ls(sl=True)rootCvListA = list()for c in curvesA: getRootCv = cmds.ls(\"%s.cv\" %c, fl=1) ''' using cmds.xform, could use cmds.pointPosition just the same ''' rootCvPositionA = cmds.xform(getRootCv, q=True, ws=True, t=True) rootCvListA.append(rootCvPositionA)''' 2 - same code as above for new root positions on curvesB list'''curvesB = cmds.ls(sl=True)''' as above etc etc '''''' 3 - temp code: a loop setting the attr, to do: better rounding formatting for the if condition, exception handling if list len differs, etc '''for i in rootCvListA: if rootCvListB[i] == rootCvListA: cmds.setAttr('%s.deformationRadius' %curvesB[i], '%s.deformationRadius' %curvesA[i])`\nQuote 0 0\n`import maya.cmds as mcimport maya.OpenMaya as omdef transfer(): # source and target curves src = mc.ls(\"src\", typ=\"nurbsCurve\") or [] tgt = mc.ls(\"tgt\", typ=\"nurbsCurve\") or [] # source points (cv) src_cnt = len(src) src_pnts = om.MFloatPointArray(src_cnt) for i in range(src_cnt): x,y,z = mc.pointPosition(src[i]+\".cv\", w=True) src_pnts[i].x = x src_pnts[i].y = y src_pnts[i].z = z # targets points (cv) tgt_cnt = len(tgt) tgt_pnts = om.MFloatPointArray(tgt_cnt) for i in range(tgt_cnt): x,y,z = mc.pointPosition(tgt[i]+\".cv\", w=True) tgt_pnts[i].x = x tgt_pnts[i].y = y tgt_pnts[i].z = z # find the closest pairs for i in range(src_cnt): idx = None dst = 999999999.9 for j in range(tgt_cnt): dst2 = src_pnts[i].distanceTo(tgt_pnts[j]) if dst2 < dst: idx = j dst = dst2 # copy source.influenceRadius to target.influenceRadius mc.setAttr(tgt[idx]+\".influenceRadius\", mc.getAttr(src[i]+\".influenceRadius\"))transfer()`\nQuote 0 0\nHi Peter,\n\nthank you very much for this solution, your help is very inspiring! Tried it out on 10,000 curves and took around 102 seconds. I will be getting more into openMaya for my future component wrangling.\n\nWill post the additions to this script when I have it ready. Planned are options to define 2 curve sets as inputs, array length mismatch handling, and a .json file dump/read, so this can be included in other scripts (rig builds for example).\n\nWe should wrap this thread up here, just asking out of curiosity: are there any existing SOuP examples that do something similar? Or is it just simpler and more sensible to do this with code?\nQuote 0 0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79858834,"math_prob":0.8697218,"size":2152,"snap":"2019-26-2019-30","text_gpt3_token_len":533,"char_repetition_ratio":0.11126629,"word_repetition_ratio":0.0,"special_character_ratio":0.24256505,"punctuation_ratio":0.12993039,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96022743,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T04:52:26Z\",\"WARC-Record-ID\":\"<urn:uuid:64a5d650-dd8d-47c7-9715-47cc47721c5e>\",\"Content-Length\":\"78968\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb8e59d8-b9cb-42f2-be3b-5d0d828df501>\",\"WARC-Concurrent-To\":\"<urn:uuid:a53e3bc9-3dcc-4f5c-a591-24164bb8d67c>\",\"WARC-IP-Address\":\"54.86.173.148\",\"WARC-Target-URI\":\"https://soup-dev.websitetoolbox.com/post/copying-attribute-values-to-other-geometries-9774473\",\"WARC-Payload-Digest\":\"sha1:2LO3GGIOVF6C2ZBIXPFDO2HWJAFV55CZ\",\"WARC-Block-Digest\":\"sha1:3R4GJKLS4E5NG6REVLGR42EBYJHAX5SG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525046.5_warc_CC-MAIN-20190717041500-20190717063500-00487.warc.gz\"}"}
https://www.onlinemathlearning.com/multiply-algebraic-fractions.html	[ "", null, "# Multiplying Algebraic Fractions\n\nRelated Topics:\nMore Lessons for GCSE Maths\nMath Worksheets\n\nExamples, solutions, and videos to help GCSE Maths students learn how to multiply algebraic fractions.\n\nSolve equations with algebraic fractions\nAlgebraic Fractions Lesson (Part 1 of 2)\nMultiplying fractions\n\nAlgebraic Fractions Lesson (Part 2 of 2)\nMultiplying fractions\nMultiplying and dividing algebraic fractions\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "", null, "" ]	[ null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null, "https://www.onlinemathlearning.com/objects/default_image.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76386356,"math_prob":0.9778991,"size":712,"snap":"2020-45-2020-50","text_gpt3_token_len":150,"char_repetition_ratio":0.1539548,"word_repetition_ratio":0.019607844,"special_character_ratio":0.1755618,"punctuation_ratio":0.084745765,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999848,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T08:58:15Z\",\"WARC-Record-ID\":\"<urn:uuid:fab8bbc8-351b-4b3d-813a-653ad94c6619>\",\"Content-Length\":\"42755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:236e1cc8-a68f-4cfd-a782-0f0e167a7431>\",\"WARC-Concurrent-To\":\"<urn:uuid:ded4b227-fc95-4b13-8c01-1743fe39341c>\",\"WARC-IP-Address\":\"173.247.219.45\",\"WARC-Target-URI\":\"https://www.onlinemathlearning.com/multiply-algebraic-fractions.html\",\"WARC-Payload-Digest\":\"sha1:USH6SLC74JZLQ3P235RIIBOJFDRHSMTQ\",\"WARC-Block-Digest\":\"sha1:UDLBMKSVE2FYLEDLTAQUKN5NSTBWZERN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141735395.99_warc_CC-MAIN-20201204071014-20201204101014-00608.warc.gz\"}"}
https://codytechs.com/speedup-python-list-and-dictionary/	[ "", null, "# SpeedUp Python List and Dictionary", null, "Today, we will be discussing the optimization technique in Python. In this article, you will get to know to speed up your code by avoiding the re-evaluation inside a list and dictionary.\n\nHere I have written the decorator function to calculate the execution time of a function.\n\n``````import functools\nimport time\n\ndef timeit(func):\n@functools.wraps(func)\ndef newfunc(args, kwargs):\nstartTime = time.time()\nfunc(args, *kwargs)\nelapsedTime = time.time() - startTime\nprint('function - {}, took {} ms to complete'.format(func.__name__, int(elapsedTime 1000)))\nreturn newfunc\n``````\n\nlet’s move to the actual function\n\n## Avoid Re-evaluation in Lists\n\nEvaluating `nums.append` inside the loop\n\n``````@timeit\ndef append_inside_loop(limit):\nnums = []\nfor num in limit:\nnums.append(num)\n\nappend_inside_loop(list(range(1, 9999999)))\n``````\n\nIn the above function `nums.append` function references that are re-evaluated each time through the loop. After execution, The total time taken by the above function\n\n``````o/p - function - append_inside_loop, took 529 ms to complete\n``````\n\nEvaluating `nums.append` outside the loop\n\n``````@timeit\ndef append_outside_loop(limit):\nnums = []\nappend = nums.append\nfor num in limit:\nappend(num)\n\nappend_outside_loop(list(range(1, 9999999)))\n``````\n\nIn the above function, I evaluate `nums.append` outside the loop and used `append` inside the loop as a variable. Total time is taken by the above function\n\n``````o/p - function - append_outside_loop, took 328 ms to complete\n``````\n\nAs you can see when I have evaluated the `append = nums.append` outside the `for` loop as a local variable, it took less time and speed-up the code by `201 ms`.\n\nThe same technique we can apply to the dictionary case also, look at the below example\n\n## Avoid Re-evaluation in Dictionary\n\nEvaluating `data.get` each time inside the loop\n\n``````@timeit\ndef inside_evaluation(limit):\ndata = {}\nfor num in limit:\ndata[num] = data.get(num, 0) + 1\n\ninside_evaluation(list(range(1, 9999999)))\n``````\n\nTotal Time taken by the above function –\n\n``````o/p - function - inside_evaluation, took 1400 ms to complete\n``````\n\nEvaluating `data.get` outside the loop\n\n``````@timeit\ndef outside_evaluation(limit):\ndata = {}\nget = data.get\nfor num in limit:\ndata[num] = get(num, 0) + 1\n\noutside_evaluation(list(range(1, 9999999)))\n``````\n\nTotal time taken by the above function –\n\n``````o/p - function - outside_evaluation, took 1189 ms to complete\n``````\n\nAs you can see we have speed-up the code here by `211 ms`.\n\nI hope you like the explanation of the optimization technique in Python for the list and dictionary. Still, if any doubt or improvement regarding it, ask in the comment section. Also, don’t forget to share your optimization technique." ]	[ null, "https://codytechs.com/wp-content/themes/cool-blog/assets/loader/style-5.gif", null, "https://codytechs.com/wp-content/uploads/2021/12/speedup-python-list-and-dictionary-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7171791,"math_prob":0.98681325,"size":2576,"snap":"2022-40-2023-06","text_gpt3_token_len":630,"char_repetition_ratio":0.16368584,"word_repetition_ratio":0.098445594,"special_character_ratio":0.2542702,"punctuation_ratio":0.12447257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9913125,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T13:46:07Z\",\"WARC-Record-ID\":\"<urn:uuid:5a78b469-db60-49b6-a122-6606bd5fe974>\",\"Content-Length\":\"134426\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f01ac673-052b-4e63-9c5c-6c87d9dd8da0>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c121e71-ec09-475a-85d7-91cb6228bfc0>\",\"WARC-IP-Address\":\"145.14.151.34\",\"WARC-Target-URI\":\"https://codytechs.com/speedup-python-list-and-dictionary/\",\"WARC-Payload-Digest\":\"sha1:6MA3ZLW3IP6S34T44N77OLLQK63ONTSV\",\"WARC-Block-Digest\":\"sha1:44WMK7LUUCR6EHTR74RB5XOXUE7H4BOS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335355.2_warc_CC-MAIN-20220929131813-20220929161813-00554.warc.gz\"}"}
https://completesuccess.in/index.php/2017/07/10/quiz-155/	[ "# Quiz – 155\n\n### Quant Quiz\n\nDirections (Q. 1–2): What approximate value should come in the place of question mark (?) in the following questions?\nQ1. 4201 ÷ 59.998 × 29.88 = ? × 21.145\n1) 90 2) 80 3) 85 4) 100 5) 95\n\nQ2. 959.98 × 780.01 ÷ 23.99 = ?\n1) 30200 2) 30800 3) 31200 4) 21200 5) 30600\n\nQ3. A man bought a mobile and a laptop for Rs. 78000. He sold the mobile at a gain of 25% and the laptop at a loss of 15%, thereby gaining 5% on the whole. Find the cost price of mobile.\n1) Rs. 39000 2) Rs. 34000 3) Rs. 30000 4) Rs. 38000 5) Rs. 32000\n\nQ4. In a regular week, there are 6 working days and for each day the working hours are 8. A man gets Rs. 40 per hour for regular work and Rs. 45 per hour for overtime. If he earns Rs. 10,500 in 5 weeks, then how many hours does he work for?\n1) 220 2) 240 3) 260 4) 210 5) 280\n\nQ5. A class has an equal number of boys and girls. Ten girls left the class to practice for music, leaving thrice as many boys as girls in the classroom. What was the total number of girls and boys present initially in the classroom?\n1) 50 2) 40 3) 35 4) 30 5) 38\n\nQ6. Surbhi’s science test consist of 85 questions from three sections- i.e. A, B and C. 10 questions from section A, 30 questions from section B and 45 question from section C. Although, she answered 70% of section A, 50% of section B and 60% of section C correctly. She did not pass the test because she got less than 60% of the total marks. How many more questions she would have to answer correctly to earn 60% of the marks which is passing grade?\n1) 4 2) 2 3) 5 4) 6 5) 8\n\nQ7. A house owner wants to get his house painted. He is told that this would require 45 kg of paint. Allowing for 6.25% wastage and assuming that the paint is available in 4 kg tins, the number of tins required for painting the house?\n1) 18 2) 11 3) 14 4) 13 5) 12\n\nQ8. One litre of water is evaporated from 8 litre solution containing 5% sugar. The percentage of sugar in the remaining solution is:\n1) 5.71% 2) 6.12% 3) 5% 4) 7.17% 5) 6%\n\nQ9. Two men together start a journey in the same direction. They travel 12 and 20 km/day respectively. After travelling for 8 days the man travelling at 12 km/day doubles his speed and both of them finish the distance in the same time. Find the number of days taken by them to reach the destination.\n1) 28 days 2) 26 days 3) 24 days 4) 30 days 5) 22 days\n\nQ10. Divide Rs. 12615 between P and Q so that P’s share at the end of 6 years may equal to Q’s share at the end of 8 years, compound interest being 5%.\n1) Rs. 6015, Rs. 6600 2) Rs. 6615, Rs. 6000 3) Rs. 7615, Rs. 5000 4) Rs. 6215, Rs. 6400 5) Rs. 6415, Rs. 6200\n\n1. 4\n\n? = (4200 / 60 * 30)/21 = 7030/21 = 100\n\n1. 3\n\n? = 960 780 / 24 = 31200\n\n1. 1\n\nLet the cost price of mobile be Rs. x and laptop be Rs.(78000-x)\n\nx * 25/100 – (78000-x) * 15/100 = 78000 * 5/100\n\n25x + 15x – 78000 × 15 = 78000 × 5\n\n40x = 78000 × 20\n\nx = Rs. 39000\n\n4.3;\n\nSuppose the man works overtime for x hours.\n\nNow, working hours in 5 week = 5 × 6 × 8 = 240 hr.\n\n240 × 40 + x × 45 = 10, 500\n\n9600 + 45x = 10,500\n\nx = (10500 – 9600)/45 = 20hr\n\nTotal hours = 240 + 20 = 260hr\n\n1. 4; Let the number of boys and girls be x each\n\n3(x – 10) = x Þ 2x = 30\n\nx = 15\n\nTotal number of students = 2x = 30\n\n1. 2; Number of questions attempted correctly\n\n= 70% of 10 + 50% of 30 + 60% of 45\n\n= 7 + 15 + 27 = 49\n\nPassing grade = 60/100 * 85 = 51\n\nReqd Ans = 51 – 49 = 2\n\n1. 5; Total paint required = 45 + wastage = 45 + x\n\nx/(45 + x) * 100 = 6.25\n\n100x = 281.25 + 6.25x\n\nx = 3kg\n\nTotal number of tins = (45 + 3)/4 = 12 tins\n\n1. 1\n\nSugar contained = 5/100 * 8 = 0.4\n\nNow, % sugar in remaining solution = 0.4/7 * 100 = 5.71%\n\n1. 3;\n\nFor distance,\n\nDistance travelled in 8 days by 1st man,\n\nD1 = 12 × 8 = 96 km.\n\nand by 2nd man,\n\nD2 = 20 × 8 = 160 km\n\nFor remaining distance, let both take t days\n\nto reach the final destination.\n\nx – 96 = 2 × 12 × t\n\nx – 96 = 24 t ………. (i)\n\nand x – 160 = 20 t …… (ii)\n\nFrom (i) and (ii); we get,\n\nx – 96 – x + 160 = 24 t – 20 t\n\n64 = 4 t\n\nt = 16 days.\n\nTotal number of days = 16 + 8 = 24 days.\n\n1. 2\n\nQ’s share = 1 / (1 + 5/100)^6 : 1 / (1 + 5/100)^8\n\n= 1 : (100/105)^2\n\n= 21^2 : 20^2\n\n= 441 : 400\n\nP’s share = 441/841 * 12615 = Rs. 6615\n\nQ’s share = 400/841 * 12615 = Rs. 6000\n\n### Reasoning Quiz\n\nDirections : Study the following information carefully and answer the given questions :\nEight persons, viz S, T, U, V, W, X, Y and Z are sitting around a circular table facing the centre. Each of them has a different hobby, viz Singing, Acting, Dancing, Cycling, Swimming, Tracking, Painting and Reading, not necessarily in the same order. S, who likes reading, sits third to the left of W. The one who likes painting sits on the immediate right of S. V, who likes acting, sits second to the right of T. T is not on immediate neighbour of W. U, who likes dancing, sits exactly in the middle of the persons whom hobbies are swimming and painting. Y who likes singing, sits second to the left of Z, who likes tracking.\n\nQ1. Which of the following pair represent the immediate neighbours of the person whose hobby is acting ?\n1) T W 2) U Y 3) S Z 4) T Z 5) W Y\n\nQ2. How many people sit between S and the person whose hobby is cycling when counted clockwise ?\n1) None 2) One 3) Two 4) Three 5) Four\n\nQ3. What is the position of T with respect to the position of the person whose hobby is swimming ?\n1) Third to the left 2) Fourth to the right 3) Fifth to the left 4) Second to the right 5) Second to the left\n\nQ4. Who exactly sits between Y and Z ?\n1) S 2) T 3) U 4) V 5) W\n\nQ5. Who sits on the immediate left of the X ?\n1) T 2) U 3) V 4) S 5) None of these\n\nQ6. Whose hobby is tracking ?\n1) Z 2) X 3) U 4) T 5) None of these\n\nQ7. Varun walks 10m. towards west, then turn to his right and walks 5 meter and turns to his left and walks 2 m. and meet his friend. Then, again turns to his left and walks 5m. What is the shortest distance between his starting point and where he meets his friend ?\n1) 12 m. 2) 13 m. 3) 11 m. 4) 14 m. 5) None of these\n\nQ8. Deepika is sister of Kumud. Mohit is brother of Kumud. Tina is mother of Reena, who is sister of Mohit. Pramod is father of Kumud. At least how many daughter does Pramod have ?\n1) One 2) Two 3) Three 4) Four 5) None of these\n\nQ9. Which of the following expression will be true if the expression-?\nA ≤ G = Q < S ≤ B > C > I is definitely true?\n1) Q>C 2) I > G 3) G = Q 4) Q < B 5) None of these\n\nQ10. Which of the following symbols should replace the question mark (?) in the given expression in order to make the expressions A< P as well as B > D definitely true?\nP ≤ B = T > A ? D ≥ C\n1) < 2) ≤ 3) > 4) Either 2) or 3) 5) None of these\n\n1. 3;\n2. 3;\n3. 4;\n4. 2;\n5. 4;\n6. 1;\n7. 2; Distance travelled by Varun shown in fig. i.e.\n\nDC = EB = 2m.\n\nEA = EB + BA = 10 + 2 = 12m\n\nTriangle(DEA) is right angled at E.\n\nDA^2 = DE^2 + EA^2\n\nDA = sqroot(DE^2 + EA^2)\n\nDA = 13m\n\n1. 2\n\nSex of Kumud is unknown So, Pramod at least have two daughters.\n\n1. 5\n\nOption (3) is re-statement and only option (4) can be\n\nconcluded as ‘<‘ supersedes ‘≤‘. Hence, Q < B.\n\n10.5\n\n### English Quiz\n\nDirections (Q. 1-5): Read each sentence to find out whether there is any grammatical or idiomatic error in it. The error, if any, will be in one part of the sentence. The number of that part is the answer. If there is ‘No error’, the answer is 5). (Ignore errors of punctuation, If any).\n\nQ1. 1) If anything, the insular state / 2) of urban life demands / 3) a serious look on outsiders by those / 4) who consider themselves insiders./ 5) No error\n\nQ2. 1) In defining professional success, / 2) women place more value than men do on individual achievement, / 3) having passion for his work, receiving respect / 4) and making a difference / 5) No error\n\nQ3. 1) The fact that the interviewee all agreed / 2) to take time from their hectic schedules / 3) to share their insights with students might / 4) introduce a selection effect./ 5) No error\n\nQ4. 1) Now I look back on my career / 2) in the private sector / 3) and realise how I should have been / 4) leading all along. / 5) No error\n\nQ5. 1) It is a thing whose knowledge / 2) may not be acquired / 3) through direct perception / 4) nor can we gain it through inference. /5) No error\n\nDirections (Q. 6-10): Rearrange the following seven sentences (A), (B), (C), (D), (E), (F) and (G) in the proper sequence to form a meaningful paragraph and then answer the questions given below.\n(A) These would be employees with strong interpersonal skills.\n(B) While selecting mentors, it is important to select employees who possess the required knowledge, skills and experience that would benefit the mentors.\n(C) Who are committed and willing to set aside the time to work with a mentor.\n(D) And act as positive role models with a genuine interest in developing others.\n(E) It is important to plan a mentoring programme that aligns with the culture of the organisation as well as development needs of the employees.\n(F) The mentor has to be chosen carefully based on the development plan and need.\n(G) The mentor and mentee need to be explained the pros and cons the mentoring programme.\n\nQ6. Which of the following would be the SECOND sentence after rearrangement?\n1) F 2) C 3) A 4) G 5) E\n\nQ7. Which of the following should be the LAST (SEVENTH) sentence after rearrangement?\n1) A 2) B 3) D 4) C 5) F\n\nQ8. Which of the following should be the FOURTH sentence after rearrangement?\n1) E 2) C 3) F 4) A 5) G\n\nQ9. Which of the following should be the FIRST sentence after rearrangement?\n1) B 2) D 3) C 4) E 5) A\n\nQ10. Which of the following should be the FIFTH sentence after rearrangement?\n1) E 2) G 3) C 4) F 5) B\n\n1. 3; Replace ‘on’ with ‘at’\n2. 3; Replace ‘his’ with ‘their’\n3. 1; Replace ‘interviewee’ with ‘interviewees’\n4. 5; No error\n5. 2; Replace ‘may’ with ‘can’\n\n1. 1\n2. 3\n3. 2\n4. 4\n5. 5\n\n### Computer Quiz\n\nQ1. Which process check to ensure the components of the computer are operating and connected properly?\na)Booting b)Processing c)Saving d)Editing e)None of these\n\nQ2. What is the main folders on a storage device called?\na)Platform b)interface c)Root directory d)Device driver e)None of these\n\nQ3. The name that the user give to a document is referred to as?\na)Name given b)Document given c)File name d)Document identify e)None of these\n\nQ4. Information in a database is typically stored in this, which consists of vertical columns and horizontal rows?\na)Field b)Record c)Table d)File e)None of these\n\nQ5. The default extension of Access database file is?\na)db b)adb c)dba d)mdb e)None of these\n\nQ6. A storage device where the access time is depended upon the location of the data is?\na)Random access b)Serial access c)Sequential access d)Transaction access e)None of these\n\nQ7. _________ means that the data contained in a database is accurate and reliable.\na)Data redundancy b)Data integrity c)Data reliability d)Data consistency e)None of these\n\nQ8. In computer language, one kilobyte equals?\na)1000 bytes b)1024 bytes c)100 bytes d)8 bytes e)None of these\n\nQ9. Example of a telecommunications device is a?\na)Keyboard b)Mouse c)Modem d)Printer e)Scanner\n\nQ10. Data on floppy disks is recorded in rings called ____?\na)Sectors b)Ringers c)Rounders d)Tracks e)Segments" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88053954,"math_prob":0.97124434,"size":11079,"snap":"2021-43-2021-49","text_gpt3_token_len":3682,"char_repetition_ratio":0.10455982,"word_repetition_ratio":0.014279532,"special_character_ratio":0.3705208,"punctuation_ratio":0.1210443,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9830836,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T02:17:59Z\",\"WARC-Record-ID\":\"<urn:uuid:3c687d6a-a9d6-45a2-9f6f-fa0de31c63bc>\",\"Content-Length\":\"75105\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdf593f4-560f-4cf8-b284-7c5d898c0d6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bfe23c6-1984-4472-9c73-22008654fc1d>\",\"WARC-IP-Address\":\"172.67.169.254\",\"WARC-Target-URI\":\"https://completesuccess.in/index.php/2017/07/10/quiz-155/\",\"WARC-Payload-Digest\":\"sha1:TDBQNINO7ELROPHXFIN6TLEV3EWUHETH\",\"WARC-Block-Digest\":\"sha1:OYA6UN64HK4GPLKXL4QUW34ZVQATK37S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363134.25_warc_CC-MAIN-20211205005314-20211205035314-00396.warc.gz\"}"}
https://metanumbers.com/55399	[ "## 55399\n\n55,399 (fifty-five thousand three hundred ninety-nine) is an odd five-digits prime number following 55398 and preceding 55400. In scientific notation, it is written as 5.5399 × 104. The sum of its digits is 31. It has a total of 1 prime factor and 2 positive divisors. There are 55,398 positive integers (up to 55399) that are relatively prime to 55399.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 5\n• Sum of Digits 31\n• Digital Root 4\n\n## Name\n\nShort name 55 thousand 399 fifty-five thousand three hundred ninety-nine\n\n## Notation\n\nScientific notation 5.5399 × 104 55.399 × 103\n\n## Prime Factorization of 55399\n\nPrime Factorization 55399\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 55399 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.9223 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 55,399 is 55399. Since it has a total of 1 prime factor, 55,399 is a prime number.\n\n## Divisors of 55399\n\n2 divisors\n\n Even divisors 0 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 55400 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 27700 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 235.37 Returns the nth root of the product of n divisors H(n) 1.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 55,399 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 55,399) is 55,400, the average is 27,700.\n\n## Other Arithmetic Functions (n = 55399)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 55398 Total number of positive integers not greater than n that are coprime to n λ(n) 55398 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5625 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 55,398 positive integers (less than 55,399) that are coprime with 55,399. And there are approximately 5,625 prime numbers less than or equal to 55,399.\n\n## Divisibility of 55399\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 3 4 1 1 7 4\n\n55,399 is not divisible by any number less than or equal to 9.\n\n## Classification of 55399\n\n• Arithmetic\n• Prime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Prime Power\n• Square Free\n\n## Base conversion (55399)\n\nBase System Value\n2 Binary 1101100001100111\n3 Ternary 2210222211\n4 Quaternary 31201213\n5 Quinary 3233044\n6 Senary 1104251\n8 Octal 154147\n10 Decimal 55399\n12 Duodecimal 28087\n16 Hexadecimal d867\n20 Vigesimal 6i9j\n36 Base36 16qv\n\n## Basic calculations (n = 55399)\n\n### Multiplication\n\nn×i\n n×2 110798 166197 221596 276995\n\n### Division\n\nni\n n⁄2 27699.5 18466.3 13849.8 11079.8\n\n### Exponentiation\n\nni\n n2 3069049201 170022256686199 9419062998158738401 521806671034995948676999\n\n### Nth Root\n\ni√n\n 2√n 235.37 38.1213 15.3418 8.88588\n\n## 55399 as geometric shapes\n\n### Circle\n\nRadius = n\n Diameter 110798 348082 9.6417e+09\n\n### Sphere\n\nRadius = n\n Volume 7.12188e+14 3.85668e+10 348082\n\n### Square\n\nLength = n\n Perimeter 221596 3.06905e+09 78346\n\n### Cube\n\nLength = n\n Surface area 1.84143e+10 1.70022e+14 95953.9\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 166197 1.32894e+09 47976.9\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.31575e+09 2.00373e+13 45233.1\n\n## Cryptographic Hash Functions\n\nmd5 7787981698c0c4b7cdb0824c5f0ed814 39f39255f63eac4235cd3cb3d873a08ad6298a5b 1afcd026549ea2984a401d5a8455c40e7107bc955ad1b6c554bc1533d94768a7 a68a1565065a515e0f8c31dcc9d002c0afa450e9b40fddd7dc6840a5174bfcec945eb1b7758afcf9a20048e29fdd325fde75c550be22e8d6760860fb4b3075a6 139d90093076e8832da93f2d3b034450ace30481" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63290465,"math_prob":0.9807754,"size":4559,"snap":"2021-21-2021-25","text_gpt3_token_len":1604,"char_repetition_ratio":0.120307356,"word_repetition_ratio":0.02962963,"special_character_ratio":0.45426628,"punctuation_ratio":0.075738125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99690014,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T06:29:46Z\",\"WARC-Record-ID\":\"<urn:uuid:2f121809-1893-4246-a2ff-3d4036621841>\",\"Content-Length\":\"59655\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:738ba445-4c75-4f87-bd23-6953db0bb201>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6237d18-b49a-4fdc-a2ea-72d08dcd6ad3>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/55399\",\"WARC-Payload-Digest\":\"sha1:TT52WGXOYPCNSMW7Z44U4O35CBV5VIR3\",\"WARC-Block-Digest\":\"sha1:B2FEX2NGA672ADWNW4I4CQPRT4Q6G3YN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487622113.11_warc_CC-MAIN-20210625054501-20210625084501-00268.warc.gz\"}"}
https://forum.facepunch.com/t/ive-begun-my-attempt-to-understand-the-lua-pil/228049	[ "", null, "I've begun my attempt to understand the Lua PIL...\n\nAnd i often run into very small confusions, if anyone wants to be annoyed by small questions over Steam, you can add me.\n\nI’ll probably ask you the dumbest questions, but i can promise it won’t be after atleast an attempt of research.\n\nI won’t give you 3 pages of code, to make you check why anything errors, or where i’m missing a bloody comma.\n\nThe LUA PIL leaves a lot of small things unexplained, and well, i don’t know any coding.\n\nMainly questions like ‘return n * fact(n-1)’\ni think i understand that as \"return to the function, factorial, times, func.factorial, brackets, n-1’ but i don’t understand why 1 is subtracted from the integer, or if that is what actually happens, because the description is simply factorial output of a number, not any subtraction involved.\n\nadd me on steam for a carpet bombardment of dumb confusions like the one above^^\n\nThis is more of a general programming/math thing and not just Lua, but the subtration is for recursion. Consider this:\n\nfunction fact(n)\nif (n == 1) then\nreturn 1\nend\n\nreturn n * fact(n-1)\nend\n\nSo fact(5) is 5 * 4 * 3 * 2 * 1 (5! in math notation), and the function does that accordingly.\n\nSo the bracket after the ‘fact’ function is the number? does ‘n-1’ signify a number or what?\n\n[editline]14th August 2016[/editline]\n\nOH, is it just to make the factorial function so it goes to the next integer, which is less than.\n\nSo instead of it breaking and stopping at 5\n\nit goes factorial -1 to let it go 5 4 * 3 * 2 * 1, and the code repeats in to get the sum\n\ni am severely stupid, and this is probably false ^^^^\n\nn is the argument which signifies the number. The first time, it’s the user inputted number; the other times, it’s the number provided by recursion. So if n is 5, you’d be calling:\n\nfact(5) -> fact(4) -> fact(3) -> fact(2) -> fact(1), which results in 5 * 4 * 3 * 2 * 1\n\nyeah, okay.\n\nn * fact(n-1)\n\nnumber * factorial(number -1)\n\ni was confused by the minus one\n\nbut i see it’s to avoid it from going\n\n5 * 5\n\nand instead -1’s the number, so it gets going at 5 * 4 * 3 * 2 * 1\n\n[editline]14th August 2016[/editline]\n\nthx" ]	[ null, "https://aws1.discourse-cdn.com/business20/uploads/facepunch1/original/2X/2/247ce2d604259e1d664d736f5aec099098a77581.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9123422,"math_prob":0.9308237,"size":853,"snap":"2022-05-2022-21","text_gpt3_token_len":200,"char_repetition_ratio":0.09305065,"word_repetition_ratio":0.0,"special_character_ratio":0.22508793,"punctuation_ratio":0.1160221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9817498,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-17T02:19:39Z\",\"WARC-Record-ID\":\"<urn:uuid:95f17d08-b758-482d-8342-bad394bffc85>\",\"Content-Length\":\"25448\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5dc36577-f6e7-4f9c-9070-d3f4dacd1576>\",\"WARC-Concurrent-To\":\"<urn:uuid:83e6d42f-1ac0-4602-bd8d-95d17526b037>\",\"WARC-IP-Address\":\"184.104.202.142\",\"WARC-Target-URI\":\"https://forum.facepunch.com/t/ive-begun-my-attempt-to-understand-the-lua-pil/228049\",\"WARC-Payload-Digest\":\"sha1:SX6UHLUSGAUGDQCIPNHZVGQRR43FC7L2\",\"WARC-Block-Digest\":\"sha1:ULFUU6BWQUZNTGWM2RXP6XL26KF2L33X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300253.51_warc_CC-MAIN-20220117000754-20220117030754-00560.warc.gz\"}"}
https://reference.wolfram.com/language/ref/SnDispersion.html	[ "# SnDispersion\n\nSnDispersion[list]\n\ngives the", null, "statistic of the elements in list.\n\nSnDispersion[list,c]\n\ngives the", null, "statistic with scaling factor c.\n\n# Details and Options", null, "• SnDispersion is a robust measure of dispersion.\n• SnDispersion is a rank-based estimator with its statistic based on absolute pairwise differences. The statistic does not require location estimation.\n• For the list {x1,x2,,xn}, the value of", null, "estimator is given by the median of {zi,1in} multiplied by a scaling factor c, where zi is the median of {xi xj,1jn} over j. »\n• When c is not specified, a positive scaling factor c* that satisfies", null, "is applied to make", null, "statistic a consistent estimator of the scale parameter for normally distributed data. Also, a finite sample correction is used by default to make the estimator unbiased for small samples.\n• SnDispersion[{{x1,y1,},{x2,y2,},}] gives {SnDispersion[{x1,x2,}],SnDispersion[{y1,y2,}],}.\n• SnDispersion supports a Method option. The following explicit settings can be specified:\n• \"FiniteSample\" uses finite sample correction (default) \"None\" no correction\n• The option Method is ignored if the scaling factor c is specified in the input.\n\n# Examples\n\nopen all close all\n\n## Basic Examples(3)\n\nSnDispersion of a list:\n\n In:=", null, "Out=", null, "SnDispersion of columns of a matrix:\n\n In:=", null, "Out=", null, "SnDispersion of a list with scaling factor 1:\n\n In:=", null, "Out=", null, "## Properties & Relations(2)\n\nIntroduced in 2017\n(11.1)" ]	[ null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/2.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/3.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/details_1.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/4.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/5.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/6.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/I_2.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/O_1.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/I_4.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/O_2.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/I_6.png", null, "https://reference.wolfram.com/language/ref/Files/SnDispersion.en/O_3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7038362,"math_prob":0.9359876,"size":1410,"snap":"2019-51-2020-05","text_gpt3_token_len":356,"char_repetition_ratio":0.1628734,"word_repetition_ratio":0.0,"special_character_ratio":0.24326241,"punctuation_ratio":0.14559387,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9866822,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,5,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T21:42:23Z\",\"WARC-Record-ID\":\"<urn:uuid:e9cc4a2a-fc52-4800-99de-3c53f99731d9>\",\"Content-Length\":\"78499\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:70c028f0-c694-4592-aa28-01e3343039fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:a1ebd90f-63cf-4d63-9760-93e083d869b8>\",\"WARC-IP-Address\":\"140.177.205.163\",\"WARC-Target-URI\":\"https://reference.wolfram.com/language/ref/SnDispersion.html\",\"WARC-Payload-Digest\":\"sha1:AN2K7XMOY6EHKLCQP3LZ6HAVTDBJUPXI\",\"WARC-Block-Digest\":\"sha1:HFIKSZIKZK5FTJXLGOZ2MZ4UZC42NGV4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250591234.15_warc_CC-MAIN-20200117205732-20200117233732-00346.warc.gz\"}"}
https://www.javatpoint.com/cpp-multiset-end-function	[ "# C++ multiset end()\n\nC++ multiset end() function is used to return an iterator which is next to the last entry in the multiset.\n\nNone\n\n## Return value\n\nIt returns an iterator which is pointing next to the last element of the multiset.\n\nConstant.\n\nNo changes.\n\n## Data Races\n\nConcurrently accessing the elements of a multiset is safe.\n\nThe container is accessed neither the non-const nor the const versions modify the container.\n\n## Exception Safety\n\nThis member function never throws exception.\n\n## Example 1\n\nLet's see the simple example for end() function:\n\nOutput:\n\n```C++\nC++\nJava\nJava\n```\n\nIn the above example, end() function is used to return an iterator pointing next to the last element in the mymultiset multiset.\n\n## Example 2\n\nLet's see a simple example to iterate over the multiset using for-each loop:\n\nOutput:\n\n```0 0 1 1 2 2 5\n```\n\n## Example 3\n\nLet's see a simple example to iterate over the multiset using while loop:\n\nOutput:\n\n```Elements of mymultiset are:\nAman\nDeep\nDeep\nNikita\nSonu\n```\n\nIn the above example, end() function is used to return an iterator pointing next to the last element in the mymultiset multiset.\n\n## Example 4\n\nLet's see a simple example:\n\nOutput:\n\n```Enter value to find: 60\n\nEnter value to find: 20\nElement found: 20\n```\n\nIn the above example, end() function is used to return an iterator pointing next to the last element in the mymultiset multiset.\n\nNext TopicC++ multiset\n\n### Feedback", null, "", null, "", null, "" ]	[ null, "https://www.javatpoint.com/images/facebook32.png", null, "https://www.javatpoint.com/images/twitter32.png", null, "https://www.javatpoint.com/images/pinterest32.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6374673,"math_prob":0.85662395,"size":1516,"snap":"2021-43-2021-49","text_gpt3_token_len":360,"char_repetition_ratio":0.17063493,"word_repetition_ratio":0.30039525,"special_character_ratio":0.21899736,"punctuation_ratio":0.101045296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9566721,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-02T22:59:32Z\",\"WARC-Record-ID\":\"<urn:uuid:4eb26893-c414-4682-9c83-f0a1af87e83e>\",\"Content-Length\":\"43372\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b96e87f8-8d1b-4b0f-963e-e6df9f8d70b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:c16c217d-38ec-4490-9c8c-f0d0afc04d26>\",\"WARC-IP-Address\":\"104.21.79.8\",\"WARC-Target-URI\":\"https://www.javatpoint.com/cpp-multiset-end-function\",\"WARC-Payload-Digest\":\"sha1:C7SSJVLPXCABBWVRITLJ56KKAAATK4SK\",\"WARC-Block-Digest\":\"sha1:BM334XVMJ64YECF46NDZB6GZAZ44JBHA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362297.22_warc_CC-MAIN-20211202205828-20211202235828-00429.warc.gz\"}"}
http://www.expertsmind.com/library/explain-the-molal-freezing-point-constant-for-benzene-5541750.aspx	[ "### Explain the molal freezing point constant for benzene\n\nAssignment Help Chemistry\n##### Reference no: EM13541750\n\nThe normal freezing point of benzene is 5.5 degrees C. A benzene solution contains 62.5 mg of naphthalene (MW = 128g/mol) per 1.00g benzene. The solution freezes at 3.0 degrees C. What is the molal freezing point constant for benzene?\n\n#### Response and present a logical sequence of ideas\n\nRead each question carefully and address all parts of the question. Not only will you be evaluated by the correctness of your answer but on your ability to communicate the\n\n#### Would a change in volume of the container affect fraction\n\nWill the equilibrium constant for the reaction increase or decrease with increasing temperature? Explain. At constant temperature, would a change in the volume of the contain\n\n#### Compute the density of a neutron\n\nNeutron stars are believed to be composed of solid nuclear matter, primarily neutrons.If the radius of a neutron is 1.0 × 10-13 cm, calculate the density of a neutron in g/c\n\n#### Explain aluminum hydroxide reacts with sulfuric acid\n\nAluminum hydroxide reacts with sulfuric acid as follows: 2 Al(OH)3 (s) + 3H2SO4 (aq) rightarrow Al2(SO4)3 (aq) + 6H2O (l) Which reagent is the limiting reactant when 0.700mo\n\n#### Acid-base titration and analytical chemistry\n\nAcid-base titration & analytical chemistry, A 0.213-g sample of uranyl(VI)nitrate, UO2(NO3)2, is dissolved in 20.0 mL of 1.0 M H2SO4 and shaken with Zn. The zinc reduces the\n\n#### Maximize the yiled of your fisher esterification product\n\nQuestion- If the cost of both of the alcohol and carboxylic acid were prohibitive, how would you maximize the yiled of your Fisher esterification product while keeping the c\n\n#### Compute the energy in mev released by the decay\n\nthe relevant masses for this decay, and calculate the energy (in MeV) released by the decay of one such atom. You can neglect the mass of the neutrino that is one of the dec\n\n#### Define what is its solubility at 42.3 degrees c\n\nExplain the solubility of AgCl(s) in water at 25 degrees C is 1.33 x 10-5 mol/L and its Delta H of solution is 65.7 kJ/mol. Define what is its solubility at 42.3 degrees C\n\n### Write a Review", null, "" ]	[ null, "http://www.expertsmind.com/prostyles/images/3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8878397,"math_prob":0.91931015,"size":2220,"snap":"2020-24-2020-29","text_gpt3_token_len":584,"char_repetition_ratio":0.09296029,"word_repetition_ratio":0.0,"special_character_ratio":0.25,"punctuation_ratio":0.1091314,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97558147,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T16:35:30Z\",\"WARC-Record-ID\":\"<urn:uuid:04c4c747-18ed-454d-8c2a-9047903fd7d7>\",\"Content-Length\":\"52280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d87a0196-d1b3-4ef0-9fa7-1465e1e6cfb7>\",\"WARC-Concurrent-To\":\"<urn:uuid:1183b4ec-9a0a-44d8-bbd3-6338de8c1899>\",\"WARC-IP-Address\":\"198.38.85.49\",\"WARC-Target-URI\":\"http://www.expertsmind.com/library/explain-the-molal-freezing-point-constant-for-benzene-5541750.aspx\",\"WARC-Payload-Digest\":\"sha1:DCCY4OUP3XWMBLSF2ZH2I3LX6H2JVRYI\",\"WARC-Block-Digest\":\"sha1:OKT4DHHGUNWHUUS2VXY5TY2NLVDOWZLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881763.20_warc_CC-MAIN-20200706160424-20200706190424-00169.warc.gz\"}"}
https://electronics.stackexchange.com/questions/539033/how-to-make-describe-an-element-with-negative-resistance-of-minus-1-ohm?noredirect=1	[ "# How to make/describe an element with negative resistance of minus 1 Ohm?\n\nHow do I make/describe an element with negative resistance of minus 1 Ohm?\n\nIt's I–V characteristic should then be\n\n$$\\I = -U\\$$\n\nIs this possible?\n\n• I need exactly what I wrote. No compromises. The negative resistance should not be used differentially.\n• I need to build such the device. It should not only work grounded, but at any potential.\n• Of course it should be kind of \"active\".\n• If a power source is required for such a device -- then that is ok. Power sources are allowed :)\n• Comments are not for extended discussion; this conversation has been moved to chat. Dec 25 '20 at 19:27\n\nYou can make such a device using active components. If one end of the resistor is grounded, this simple circuit would work, at least in theory. The op-amp would have to be able to deliver sufficient current and have sufficient voltage swing for whatever voltage you wanted to apply.", null, "simulate this circuit – Schematic created using CircuitLab\n\nFor example, if you apply 5V, the op-amp would have to swing to +10V while sourcing 5A.\n\nAn actual negative resistor would be a power source (P = E^2/R) so it's impossible to have such a device that does not use or contain a source of energy.\n\nThere are a few devices and circuits that exhibit negative differential resistance- over some region of operation the current drops with increasing voltage (while the total current is always positive). For example, Esaki diodes, discharge tubes, switching power supplies. So in a small signal analysis, resistance can be negative.\n\nEdit:\n\nYou've added that you require it to be ungrounded. In that case, merely power it by a battery or an isolated DC-DC converter and use the circuit as shown (with a suitable power op-amp). There are still many \"compromises\" - as in limitations and non-idealities as there are with any real circuit or element, of course.\n\nHere is a simulation with a 1K resistor for R1 which simulates a -1K resistor.", null, "simulate this circuit", null, "As you can see the current through the simulated -1K resistor is the mirror of the current through the real 1K resistor R4. I've offset the other side of the resistor by 5V from ground in each case and the sawtooth goes from 0 to 10V so the voltage across each element goes from -5V to +5V. Not shown are the power supplies.\n\nYou won't easily be able to simulate the isolated version because op-amp models typically have ground nodes within the model.\n\n• @Circuitfantasist in that case write a better answer. Dec 24 '20 at 22:38\n• @mkeith No, no, don't dare Circuit fantasist to write an answer! Dec 24 '20 at 23:10\n• @PeteW it is not enough that the slope of the V/I curve be negative. It must also be linear, or somewhat linear. A resistor means V / I = k, where k is constant. A constant power load means that V * I = k, where k is constant. Not the same thing. Dec 24 '20 at 23:10\n• @Circuitfantasist It behaves like a -1 ohm resistor with one end grounded. Just as requested. For 0V input (short the input to ground) current is 0. For 1V input, current into point X is -1A. Etc. Dec 24 '20 at 23:25\n• @PeteW A switching power supply with a constant load on a regulated output is (ideally) a constant power device. But it's only differential resistance. If you look at the input current at the minimum input voltage at which it will operate, the current will flow into the (+) terminal of the supply. As you increase the voltage the current will decrease from that (high) value to lower values. But it will never get to zero, let alone negative. Dec 24 '20 at 23:28\n\nA resistor is a two terminal circuit element that enforces the following relationship between voltage and current:\n\nV / I = R\n\nWhere V is the voltage measured across the two terminals, and I is the current flowing through the element. To be a resistor, R must be constant over a range of voltage and currents.\n\nLogically, then, a resistor with resistance of -1 Ohms would be a two terminal element which enforces the following rule:\n\nV / I = -1\n\nI think this explains what a negative resistance is or would be at a theoretical and/or mathematical level. Approximating a negative resistance with electronics is possible (see Spehro's answer). But that is more of a negative resistance emulator than a true negative resistance.\n\nA true negative resistor would deliver power to any voltage source connected to it. The higher the voltage (in magnitude), the more the power. So this is not possible for a passive circuit element. Spehro showed how it can be done with active elements.\n\n• No, this does not explain anything useful to one who wants to make a negative resistor. To make sure of this, ask the OP to make a negative resistor according to your recipe. It is a recipe for mathematicians.They will even write, for their convenience, a/b = c and a/b = -c ...and will be happy... Dec 24 '20 at 23:32\n• @Circuitfantasist write your own answer. Nothing is stopping you. Also, please stop speaking for the OP. That is presumptuous. Dec 24 '20 at 23:33\n• Passivness of an element is not required. Negative resistor should change electromotive force inside it according to applied voltage.\n– Dims\nDec 24 '20 at 23:45\n• @Circuitfantasist you are a teacher sitting in on another professor's class and criticizing the curriculum and method of presentation in front of all the students (and the world). Please go teach your own class. That is what I am asking you. Dec 24 '20 at 23:54\n• I agree with Fantasist. The question posted by OP already showed they understand that a negative resistor would have the relation $V=-I$. Dec 25 '20 at 0:22\n\nUtterly impractical, but just to indulge the puzzle in an abstract sense:\n\nTwo howland pumps chould function as a floating \"negative resistor\" within a bounded range of voltages/currents/frequencies? (for a no-batteries concept)\n\n(Not completely sure I got the +/- right on the Howland pumps but hopefully you get the idea)\n\nPS -- Merry XMAS", null, "• Here you have a floating negative resistor electronics.stackexchange.com/questions/460605/…\n– G36\nDec 26 '20 at 16:30\n• Nice, though I have a suspicion that without some compensation these things would very much prefer to oscillate. Dec 26 '20 at 19:16\n• @G36, It is a great challenge to reveal the idea behind this mess of resistors and op-amps in your answer... It reminds me of an interesting RG discussion, five years ago, when I managed to do it with Antoniou's GIC. I am tempted to do it with this circuit solution… but I am afraid that I will \"waste\" the rest of Christmas holidays in thinking :) Dec 27 '20 at 16:14\n\nSame idea to the answer with the howland's, but with less stuff -- or a bilateral version of the regular NIC /// update /// realized this is the same circuit of G36 in comment above.", null, "" ]	[ null, "https://i.stack.imgur.com/3DjPU.png", null, "https://i.stack.imgur.com/i7Gh6.png", null, "https://i.stack.imgur.com/fJoOA.png", null, "https://i.stack.imgur.com/mMzsf.jpg", null, "https://i.stack.imgur.com/wYHFb.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9379784,"math_prob":0.90780985,"size":1790,"snap":"2021-31-2021-39","text_gpt3_token_len":403,"char_repetition_ratio":0.11758119,"word_repetition_ratio":0.0,"special_character_ratio":0.21396647,"punctuation_ratio":0.0787172,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97495717,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-16T12:16:39Z\",\"WARC-Record-ID\":\"<urn:uuid:40fd3ae8-c563-4763-b35c-18b38d9d4d20>\",\"Content-Length\":\"218217\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a0bfde6-415f-46c0-ba46-b287351008e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:38284a6b-f5fd-4f51-9ef5-912325e78a83>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/539033/how-to-make-describe-an-element-with-negative-resistance-of-minus-1-ohm?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:WE6DNKTPCMTKTNOT3W7B3ZDPBH57JCYX\",\"WARC-Block-Digest\":\"sha1:UMETPPNGDEGKYOCKBXAY4TNFDZJPKD3U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780053493.41_warc_CC-MAIN-20210916094919-20210916124919-00401.warc.gz\"}"}
https://igraph.org/python/tutorial/develop/tutorials/online_user_actions/online_user_actions.html	[ "# python-igraph Manual\n\nFor using igraph from Python\n\n# Online user actions¶\n\nThis example reproduces a typical data science situation in an internet company. We start from a pandas DataFrame with online user actions, for instance for an online text editor: the user can create a page, edit it, or delete it. We want to construct and visualize a graph of the users highlighting collaborations on the same page/project.\n\n```import igraph as ig\nimport numpy as np\nimport pandas as pd\nimport matplotlib.pyplot as plt\n\n# User data (usually would come with time stamp)\naction_dataframe = pd.DataFrame([\n['2r09ej221sk2k5', 'editPage', 'greatProject'],\n['oi32ncwosap399', 'editPage', 'greatProject'],\n['4r4320dkqpdokk', 'createPage', 'miniProject'],\n['320eljl3lk3239', 'editPage', 'miniProject'],\n['3203ejew332323', 'createPage', 'private'],\n['3203ejew332323', 'editPage', 'private'],\n['40m11919332msa', 'createPage', 'private2'],\n['40m11919332msa', 'editPage', 'private2'],\n['2r09ej221sk2k5', 'editPage', 'anotherGreatProject'],\n],\ncolumns=['userid', 'action', 'project'],\n)\n```\n\nThis block just introduces the toy data: a DataFrame with three columns, namely the user id, the action, and the page or project that was being actioned upon. Now we need to check when two users worked on the same page. We choose to use a weighted adjacency matrix for this, i.e. a table with rows and columns indexes by the users that has nonzero entries whenever folks collaborate. First, let’s get the users and prepare an empty matrix:\n\n```users = action_dataframe['userid'].unique()\nnp.zeros((len(users), len(users)), np.int32),\nindex=users,\ncolumns=users,\n)\n```\n\nThen, let’s iterate over all projects one by one, and add all collaborations:\n\n```for project, project_data in action_dataframe.groupby('project'):\nproject_users = project_data['userid'].values\nfor i1, user1 in enumerate(project_users):\nfor user2 in project_users[:i1]:\n```\n\nThere are many ways to achieve the above matrix, so don’t be surprised if you came up with another algorithm ;-)\n\nNow it’s time to make the graph:\n\n```g = ig.Graph.Weighted_Adjacency(adjacency_matrix, mode='plus')\n```\n\nAnd finally, let’s plot a layout of the graph, for instance a circle:\n\n```# Make a layout first\nlayout = g.layout('circle')\n\n# Make vertex size based on their closeness to other vertices\nvertex_size = g.closeness()\nvertex_size = [0.5 * v**2 if not np.isnan(v) else 0.05 for v in vertex_size]\n\n# Make mpl axes\nfig, ax = plt.subplots()\n\n# Plot graph in that axes\nig.plot(\ng,\ntarget=ax,\nlayout=layout,\nvertex_label=g.vs['name'],\nvertex_color=\"lightblue\",\nvertex_size=vertex_size,\nedge_width=g.es[\"weight\"],\n)\nplt.show()\n```\n\nWe added a few fancy features to this plot to show off igraph’s capabilities. The result is shown below.", null, "The collaboration graph: thicker edges mean multiple collaborations, and larger vertices indicate users with higher closeness to the rest of the network.¶\n\nLoops indicate “self-collaborations”, which are not very meaningful. To filter out loops without losing the edge weights, we can use:\n\n```g = g.simplify(combine_edges='first')\n```\n\nand then repeat the plotting code verbatim. The result is shown below." ]	[ null, "https://igraph.org/python/tutorial/develop/_images/online_user_actions.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59588754,"math_prob":0.76167023,"size":3243,"snap":"2022-27-2022-33","text_gpt3_token_len":847,"char_repetition_ratio":0.13738808,"word_repetition_ratio":0.00477327,"special_character_ratio":0.28276289,"punctuation_ratio":0.21440823,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98089,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T04:42:17Z\",\"WARC-Record-ID\":\"<urn:uuid:ee797ebb-1115-477f-98bb-38908bcbed0e>\",\"Content-Length\":\"21856\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:84c36c33-5d1d-415d-885e-3b4c82d0b9d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:efb171eb-fa8d-44ad-8641-829fb2a07a40>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://igraph.org/python/tutorial/develop/tutorials/online_user_actions/online_user_actions.html\",\"WARC-Payload-Digest\":\"sha1:VF532SDZZ5WTAONWSBTR5BMHVTDZVHRX\",\"WARC-Block-Digest\":\"sha1:UXORYABOWVTEWSW4KTMVY3EJ7VB5B3AY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215790.65_warc_CC-MAIN-20220703043548-20220703073548-00396.warc.gz\"}"}
http://forums.wolfram.com/mathgroup/archive/2004/Oct/msg00138.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "NMininimize vs. FIndMinimum with NDSolve?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg51135] NMininimize vs. FIndMinimum with NDSolve?\n• From: Manuel Morales <Manuel.A.Morales at williams.edu>\n• Date: Tue, 5 Oct 2004 04:37:45 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Is it possible to use NMinimize to estimate parameters for a\ndifferential equation solved with NDSolve? I am able to use\nFindMinimum, but NMinimize doesn't seem to work...\n\nHere is what I am doing using FindMinimum:\n\nsse[r_?NumberQ, q_?NumberQ] := Block[{sol, f},\n\nPlus @@ Table[\nsol = NDSolve[{f'[t] == r f[t] + q f[t]^2,\nf == data[[i]][][]},\nf, {t, 0, 10}][];\nPlus @@ Apply[(f[#1] - #2)^2 &, data[[i]], {1}] /. sol, {i, 1,\nLength[data]}]];\nparams = FindMinimum[sse[r, q], {r, .1, .2}, {q, -.001, -.002}]\n\nHowever, using NMinimize to estimate doesn't work. That is, if I replace\nthe last line with:\n\nparams = NMinimize[sse[r, q], {r, q}]\n\nI get a bunch of error messages.\n\nThanks for any help. If anyone wants to try this, you can generate a\ntest data set below:\n\nBlock[{r, q},\nr = .5; q = -.001;\ndata = Table[\nsol = NDSolve[{f'[t] == r f[t] + q f[t]^2, f == j},\nf, {t, 0, 10}][];\nTable[{i, f[i] /. sol}, {i, 0, 10, 2}],\n{j, 1, 301, 150}]\n]\n\nManuel\n\n```\n\n• Prev by Date: MathLink error\n• Next by Date: Re: A way around the limitations of Re[] and Im[]" ]	[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/4.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70333517,"math_prob":0.98802,"size":1245,"snap":"2021-31-2021-39","text_gpt3_token_len":462,"char_repetition_ratio":0.10717163,"word_repetition_ratio":0.08737864,"special_character_ratio":0.41445783,"punctuation_ratio":0.27707008,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9968601,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T20:01:09Z\",\"WARC-Record-ID\":\"<urn:uuid:837a118e-afed-4220-8986-dcb506289a3f>\",\"Content-Length\":\"44993\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:053c1697-6d78-4b43-af10-da2a1c346f2b>\",\"WARC-Concurrent-To\":\"<urn:uuid:30cec1e4-647d-4750-afa0-65e2cac2732d>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2004/Oct/msg00138.html\",\"WARC-Payload-Digest\":\"sha1:7YC55MX2ATO36BSXRO4PI46C73DPAFVU\",\"WARC-Block-Digest\":\"sha1:VKTSR3MWNFMXLKGLAGFJ4FRJRNBFR6TJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058467.95_warc_CC-MAIN-20210927181724-20210927211724-00360.warc.gz\"}"}
https://otland.net/threads/help-simplifying-the-script.282864/#post-2708929	[ "# LuaHelp simplifying the script\n\n#### MorganaSacani\n\n##### Active Member\nI'm creating a script, but I will have to do a lot of repetitions. For that I need to understand how to create a loop and make my work easier\nLua:\n``````function onUse(cid, item, fromPosition, itemEx, toPosition)\n-- Trees\n\n-- Stones\n\n-- Corpses\nif itemEx.itemid == 2813 then -- Dead Rat\nif math.random(1, 100) <= 60 then\ndoPlayerAddItem(cid, 2666, 3) -- Raw Meat\nelse\nend\nif math.random(1, 100) <= 30 then\ndoTransformItem(itemEx.uid, 2815)\n-- doDecayItem(itemEx.uid)\nend\nend\n\nreturn true\nend``````\n\nI thought of something like this, but I don't know how to do it:\nLua:\n``````local TARGETS = {\n = {\n = 3, -- Raw meat\n = 1, -- Leather\n}\n}``````\n\nI need to add the doTransformItem in the loop too but I don't know how\n\nSolution\nThis problem happens when I try to use the pick on some other itemId that is not in the table.\n\nLua:\n``````function onUse(cid, item, fromPosition, itemEx, toPosition)\nif not config[itemEx.itemid] then\n-- in case you want to notify the player\n-- doPlayerSendTextMessage(cid, MESSAGE_STATUS_SMALL, \"You cannot use foo on bar.\")\nreturn true\nend\n\n-- etc etc\nend``````\n\n#### amatria\n\n##### Well-Known Member\nI hope this helps:\n\nLua:\n``````local config = {\n = {\nchance = 60,\nsuccess = {id = 2666, qty = 3},\nfail = {id = 5878, qty = 1}\n},\ndoTransform = {\nchance = 30,\nsuccess = {id = 2815}\n}\n}\n}\n\n-- function onUse(...)\nlocal doTransformTable = config[itemEx.itemid].doTransform\n\nif math.random(1, 100) <= addItemTable.chance then\nelse\nend\n\nif math.random(1, 100) <= doTransformTable.chance then\n-- doTransformItem(itemEx.uid, doTransformTable.success.id)\nelse\n-- doDecayItem(itemEx.uid)\nend``````\n\nLast edited:\n•", null, "MorganaSacani and Tofame\nOP\nOP\nM\n\n#### MorganaSacani\n\n##### Active Member\n[19:30:29] [Error - Action Interface]\n[19:30:29] data/actions/scripts/tools/pick.lua", null, "nUse\n[19:30:29] Description:\n[19:30:29] data/actions/scripts/tools/pick.lua:16: attempt to index a nil value\n[19:30:29] stack traceback:\n[19:30:29] data/actions/scripts/tools/pick.lua:16: in function <data/actions/scripts/tools/pick.lua:15>\n\nLua:\n``````local config = {\n = {\nchance = 75,\nsuccess = {id = 2666, qty = 3},\nfail = {id = 5878, qty = 1}\n},\ndoTransform = {\nchance = 20,\nsuccess = {id = 2814}\n}\n}\n}\n\nfunction onUse(cid, item, fromPosition, itemEx, toPosition)\nlocal doTransformTable = config[itemEx.itemid].doTransform\n\nlocal stamina = 5\nif getPlayerAttrStamina(cid) >= stamina then\ndoPlayerRemoveAttrStamina(cid, stamina)\nif math.random(1, 100) <= addItemTable.chance then\ndoSendMagicEffect(toPosition, CONST_ME_BLOCKHIT)\nelse\ndoSendMagicEffect(toPosition, CONST_ME_BLOCKHIT)\nend\n\nif math.random(1, 100) <= doTransformTable.chance then\ndoDecayItem(itemEx.uid)\ndoTransformItem(itemEx.uid, doTransformTable.success.id)\ndoSendMagicEffect(toPosition, CONST_ME_POFF)\nend\nelse\ndoPlayerSendTextMessage(cid, MESSAGE_STATUS_SMALL, \"You are out of vigor.\")\nend\n\nif math.random(1, 100) <= 10 then doPlayerRemoveAttrFood(cid, 1) end\n-- if math.random(1, 100) <= 20 then doPlayerRemoveAttrWater(cid, 1) end\n\nreturn true\nend``````\n\nThis problem happens when I try to use the pick on some other itemId that is not in the table.\n\nLast edited:\n\n#### amatria\n\n##### Well-Known Member\nThis problem happens when I try to use the pick on some other itemId that is not in the table.\n\nLua:\n``````function onUse(cid, item, fromPosition, itemEx, toPosition)\nif not config[itemEx.itemid] then\n-- in case you want to notify the player\n-- doPlayerSendTextMessage(cid, MESSAGE_STATUS_SMALL, \"You cannot use foo on bar.\")\nreturn true\nend\n\n-- etc etc\nend``````\n\n•", null, "MorganaSacani\nOP\nOP\nM\n\n#### MorganaSacani\n\n##### Active Member\nLua:\n``````function onUse(cid, item, fromPosition, itemEx, toPosition)\nif not config[itemEx.itemid] then\n-- in case you want to notify the player\n-- doPlayerSendTextMessage(cid, MESSAGE_STATUS_SMALL, \"You cannot use foo on bar.\")\nreturn true\nend\n\n-- etc etc\nend``````\nWow thx!\n\n•", null, "amatria" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76926464,"math_prob":0.8194094,"size":986,"snap":"2022-40-2023-06","text_gpt3_token_len":322,"char_repetition_ratio":0.104887985,"word_repetition_ratio":0.23076923,"special_character_ratio":0.336714,"punctuation_ratio":0.13705584,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95154136,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T03:59:39Z\",\"WARC-Record-ID\":\"<urn:uuid:63b8d497-54cd-4559-8b69-32fef6f1286f>\",\"Content-Length\":\"113201\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ab963ce-7043-40d3-9779-dfb3223f5359>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b5dccfa-c162-4c23-8d4a-3208f0c8f283>\",\"WARC-IP-Address\":\"104.21.51.77\",\"WARC-Target-URI\":\"https://otland.net/threads/help-simplifying-the-script.282864/#post-2708929\",\"WARC-Payload-Digest\":\"sha1:V3B3GUOG57CADDMQ4BF4PGLCY6TWN2M2\",\"WARC-Block-Digest\":\"sha1:T3YDWW4QJO5YSHF5GNI5QZSAAHLJKCTW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499697.75_warc_CC-MAIN-20230129012420-20230129042420-00449.warc.gz\"}"}
https://www.rdocumentation.org/packages/permutations/versions/1.0-5/topics/permorder	[ "# permorder\n\n0th\n\nPercentile\n\n##### The order of a permutation\n\nReturns the order of a permutation $$P$$: the smallest strictly positive integer $$n$$ for which $$P^n$$ is the identity.\n\n##### Usage\npermorder(x, singly = TRUE)\n##### Arguments\nx\n\nPermutation, coerced to cycle form\n\nsingly\n\nBoolean, with default TRUE meaning to return the order of each element of the vector, and FALSE meaning to return the order of the vector itself (that is, the smallest strictly positive integer for which all(x^n==id)).\n\n##### Details\n\nCoerces its argument to cycle form.\n\nThe order of the identity permutation is 1.\n\n##### Note\n\nUses mLCM() from the numbers package.\n\nsgn\n\n• permorder\n##### Examples\n# NOT RUN {\nx <- rperm(5,20)\npermorder(x)\npermorder(x,FALSE)\n\nstopifnot(all(is.id(x^permorder(x))))\nstopifnot(is.id(x^permorder(x,FALSE)))\n# }\n\nDocumentation reproduced from package permutations, version 1.0-5, License: GPL-2\n\n### Community examples\n\nLooks like there are no examples yet." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6623225,"math_prob":0.97132045,"size":814,"snap":"2020-45-2020-50","text_gpt3_token_len":214,"char_repetition_ratio":0.14814815,"word_repetition_ratio":0.05172414,"special_character_ratio":0.23955774,"punctuation_ratio":0.11258278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99566627,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T18:07:50Z\",\"WARC-Record-ID\":\"<urn:uuid:013406f2-ea85-4360-90ad-7dd9f878fc89>\",\"Content-Length\":\"14152\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:59183e44-c1cf-4a40-b6d0-0cf28818ea03>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c9e16f1-406b-4725-9e7f-c2a333a1c3b2>\",\"WARC-IP-Address\":\"52.4.138.252\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/permutations/versions/1.0-5/topics/permorder\",\"WARC-Payload-Digest\":\"sha1:T5FIPK5PBLLH3SKWUM7SRQSFT42AQFEY\",\"WARC-Block-Digest\":\"sha1:QARZAF2EKGXV6MYD7L573WTERNFW72PY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176922.14_warc_CC-MAIN-20201124170142-20201124200142-00209.warc.gz\"}"}
https://ta-netzsch.com/shelf-life-of-drugs-reduce-your-testing-time-by-using-thermal-analysis	[ "# Shelf life of drugs: Reduce Your Testing Time by Using Thermal Analysis\n\nHow to determine the shelf life of drugs without long time testing? First predictions about the shelf-life concerning thermal stability are achieved using thermogravimetric measurements with a kinetics Evaluation. This occurs within some hours or days.\n\nThe shelf life is the amount of time a product can be stored and still be considered as safe and effective for use. The shelf life of pharmaceuticals is affected, amongst others, by:\n• Humidity;\n• Light;\n• Gas atmosphere;\n• Temperature conditions.\nShelf life determination usually requires time-consuming tests. How to determine the shelf life of drugs without long testing times? First predictions about the shelf-life concerning thermal stability are achieved using thermogravimetric measurements with kinetics evaluation. This takes some hours or days.\n\n## 5-year-forecast within some hours – What do you need for that?\n\n• A thermobalance, a device that stores the mass variations of the sample during its heating;\n• Kinetics Neo, a software that determines the kinetics of the decomposition reaction.\n\n## How to proceed?\n\n1. Carry out TGA measurements at different heating rates.\n2. Carry out the kinetics evaluation with Kinetics Neo;\n3. Use the kinetics model to predict the sample behavior for specified temperatures and times;\n4. Validate the kinetic model by comparing a measurement at a isothermal temperature with the curve calculated by Kinetics Neo.\nLet us do it together on potassium clavulanate. This pharmaceutical substance is generally used with the antibiotic amoxicillin to increase its effectiveness.\n\n#### 1. TG measurements at different heating rates\n\nFigure 1 displays the TGA and DTG (first derivative) curves of the measurement on potassium clavulanate at 10 K/min and in a dynamic nitrogen atmosphere. The first mass-loss step, detected between room temperature and 120°C, results from the evaporation of surface water (click here for more information). Further, the three mass-loss steps identified between 120°C and 600°C are due to the decomposition of potassium clavulanate.", null, "Figure 1. TGA measurement on potassium clavulanate in pierced crucibles at 10 K/min in a dynamic nitrogen atmosphere, solid lines: TGA, dashed lines: DTG Figure 2 depicts the TGA and DTG (first derivative) curves of the measurements on potassium clavulanate at heating rates of 1, 3, 5 and 10 K/min. The mass-loss steps are shifted to higher temperatures with increasing heating rates (kinetic influence). For example, at a heating rate of 1 K/min, the first decomposition step occurs at 167°C (DTG peak), while at a heating rate of 10 K/min, it occurs at 184°C (DTG peak).", null, "Figure 2. TGA measurement on potassium clavulanate in pierced crucibles at different heating rates in a dynamic nitrogen atmosphere, solid lines: TGA, dashed lines: DTG\n\n#### 2. TG measurements and Kinetics Neo\n\nThe dependence of the decomposition on the heating rate allows for evaluation of the decomposition kinetics with the help of NETZSCH Kinetics Neo software. Kinetics Neo proposes a kinetic model with five consecutive steps of nth order (click here for more information). It calculates the kinetics parameters of each step (activation energy, pre-exponential factor, …). The water release is not taken into account for the calculation. Figure 3 compares the measured TGA curves (dotted lines) with the calculated curves (solid lines) of the chosen 5-step model. The correlation between measured and calculated curves is very good!", null, "Figure 3. Kinetic evaluation of the decomposition of potassium clavulanate. Dotted lines: measured curves; solid lines: calculated curves based on a five-step reaction of nth order. The correlation coefficient between measured and calculated curves amounts to >0.999.\n\n#### 3. Kinetics Neo predicts the sample behavior during storage at specified temperatures\n\nKinetics Neo uses the calculation to predict how the sample would behave during long-time storage at different temperatures. Figure 4 displays the 5-year forecast of the decomposition process of potassium clavulanate. This plot was realized in some days, taking the measurements and evaluation with Kinetics Neo into account.", null, "Figure 4. Five-year forecast of the decomposition process of potassium clavulanate in a nitrogen atmosphere between 20°C and 80°C\n\n#### 4. The validation of the kinetics model ensures the accuracy of the calculation\n\nThe kinetic model calculated by Kinetics Neo should be validated for prediction of the decomposition behavior under isothermal conditions. For that, a potassium clavulanate sample of 9.23 mg was heated to 200°C and then kept isothermal for two hours. Figure 5 compares the mass losses determined via measurement to those determined via prediction (Kinetics Neo). The comparison shows the good agreement between the two curves and thus the reliability of the calculation.", null, "Figure 5. Comparison of the measured and predicted mass change of potassium clavulanate during heating to 200°C and isothermal segment; the release of surface water is not monitored.\nSubscribe\nNotify of" ]	[ null, "https://ta-netzsch.com/wp-content/uploads/2018/11/K-Clavulanat-TG-10-Kmin.png", null, "https://ta-netzsch.com/wp-content/uploads/2018/11/Figure2_.png", null, "https://ta-netzsch.com/wp-content/uploads/2018/11/Figure5.png", null, "https://ta-netzsch.com/wp-content/uploads/2018/11/Figure7.png", null, "https://ta-netzsch.com/wp-content/uploads/2018/11/Figure8_.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89515823,"math_prob":0.9714176,"size":4137,"snap":"2021-43-2021-49","text_gpt3_token_len":826,"char_repetition_ratio":0.13815631,"word_repetition_ratio":0.073365234,"special_character_ratio":0.1895093,"punctuation_ratio":0.090655506,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9709595,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T19:18:20Z\",\"WARC-Record-ID\":\"<urn:uuid:768d7e97-a760-4faf-9d02-a0d577dafa35>\",\"Content-Length\":\"113257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf50bebf-ed4e-402b-a525-4df607e7e774>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcb2a75f-b9d5-4b0e-966d-e579d6de1189>\",\"WARC-IP-Address\":\"217.160.0.10\",\"WARC-Target-URI\":\"https://ta-netzsch.com/shelf-life-of-drugs-reduce-your-testing-time-by-using-thermal-analysis\",\"WARC-Payload-Digest\":\"sha1:P2DSKE3BGXXFTO2PP6XKDTLQDK7IRDFF\",\"WARC-Block-Digest\":\"sha1:WC46SMYSSPB5NWVHAG7IC3EMOD2PTCEA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587915.41_warc_CC-MAIN-20211026165817-20211026195817-00580.warc.gz\"}"}
https://www.javatpoint.com/linear-function-in-discrete-mathematics	[ "# Linear Function in Discrete mathematics\n\nA linear function can be described as a function that shows a straight line on the coordinate plane. Suppose there is a function y = 3x - 2, which shows a straight line on a coordinate plane. Hence this function is also a linear function. Since, we can replace y with f(x). So we can use another way to write this function, i.e., f(x) = 3x+2. In this section, we will learn about the definition, graph, domain, and range of the linear function. After that, we will learn to identify a linear function and find its inverse.\n\n### Definition of Linear function\n\nA linear function can be described as a function, which must be in the form f(x) = mx+b. In this equation, m and b are used to indicate the real numbers. This equation looks the same as the slope-intercept equation from a line, which is y = mx+b. They look similar because a linear function is used to represent a line. That means the graph of this function is a line. Here\n\n'm' is used to indicate the slope of a line\n\n'b' is used to indicate the y intercept of a line\n\n'x' is used to indicate the independent variable.\n\nf(x) or 'y' is used to indicate the dependent variable.", null, "A linear function can be described as an algebraic function.\n\n### Example and Equation of Linear function\n\nThe parent linear function is indicated by f(x) = x, which is actually a line passing through the origin. In general, the equation of linear function is indicated as f(x) = mx + b. Some examples of linear function are described as follows:\n\nf(x) = 6x-5.\n\nf(x) = -7x - 0.7\n\nf(x) = 4\n\n### Real-life example of Linear function\n\nWe can also explain the linear function with the help of some real-life applications, which are described as follows:\n\n• Suppose there is a movie streaming company that charges a monthly fee approx 500rs, and for every movie downloaded, they also charge an additional fee of 30rs. In this case, the linear function will be used to represent the total monthly fee as f(x) = 30x + 500. Here x is used to show the number of movies that we have downloaded in a month.\n• Suppose there is a t-shirt company that prints designs and logos on t-shirts, and that company charges a one-time fee of 3800 and 500 per t-shirt. Now we can use the linear function to show the total fee as f(x) = 500x + 3800. Here x is used to indicate the number of t-shirts.\n• In linear programming problems, we can use the linear function because it represents an objective function. This function will help to maximize the profits and minimize the close.\n\n### Finding of Linear function\n\nWe can find the linear function with the help of either point-slope form or the slope-intercept form. The process of determining a linear function and determining the equation of a line is similar, and we will explain this with the help of following example.\n\nExample: In this example, we have to determine the linear function, which contains two functions (-1, 15) and (2, 27).\n\nSolution: From the question, we have two points, i.e., (x1, y1) = (-1, 15), and (x2, y2) = (2, 27).\n\nStep 1: In this step, we will use the slope formula to determine the slope of the function like this:\n\nM = (y2-y1) / (x2-x1) = (27-15) / (2-(-1)) = 12 /3 = 4.\n\nStep 2: In this step, we will use the point slope form so that we can find out the equation of linear function like this:\n\ny - y1= m(x - x1)\n\ny-15 = 4(x-(-1))\n\ny-15 = 4(x+1)\n\ny-15 = 4x+4\n\ny = 4x+19\n\nTherefore, the linear function equation is f(x) = 4x + 19.\n\n### Identifying a Linear function\n\nIf we get the information about a function in the form of a graph, and that graph is a line, then that function will be a linear function. If we get information about a function in the form of algebraic, which is in the form f(x) = mx+b, then that function will also be a linear function. If we want to check that the given data, which is represented in the table format, represents a linear function, then we can do this with the help of following steps:\n\n• We will calculate the differences in x values\n• Then, we will calculate the differences in y values\n• Lastly, we will see whether the ratio of differences in y values to the differences of x values always be a constant.\n\nExample: In this example, we have some data in the table, and we have to show whether this data represents a linear function.\n\nx y\n3 15\n5 23\n7 31\n11 47\n13 55\n\nSolution: To do this, we will calculate the differences in x values, y values, and ratio (difference in y) / (difference in x) every time, and from these calculations, we will see whether the ratio is a constant.", null, "### Graphing a Linear function\n\nAs we know that we can graph a line with the help of any two points on it. When we are able to find those two points, then we need to just join them in a line and then extend them on both sides. The following things are contained by the graph of a linear function f(x) = mx+b\n\nWhen m>0, in this case, the graph will be an increasing line.\n\nWhen m<0, in this case, the graph will be a decreasing line\n\nWhen m<0, in this case, the graph will be a decreasing line.", null, "• We can determine two points on it.\n• We can use the y-intercept and its slope.\n\nGraphing a Linear function by finding two points\n\nWe will use the function f(x) = mx+b so that we can determine the two points on it. To do this, we will take some random values for x and find the corresponding value of y with the help of substituting these values on the function f(x). Now we will explain the process of graph a function with the help of an example in which we will graph a function f(x) = 3x + 5 like this:\n\nStep 1: In this step, we will use some random values to determine the two points on the line. So we will take two points as x = -1, and x = 0.\n\nStep 2: Now, we will find the corresponding y values with the help of substituting the above values of x in the function.\n\nIn the following table, we can see the linear function y = 3x+5 like this:\n\nx y\n-1 3(-1)+5 = 2\n0 3(0)+5 = 5\n\nTherefore, (-1, 2) and (0, 5) are two points on the line.\n\nStep 3: Now, we will plot these points on the graph and use a line to join them. We will also extend the line on left side and right side like this:", null, "Graphing a Linear function using y-intercept and slope\n\nWith the help of the y-intercept 'b' and slope 'm' of a linear function f(x) = mx+b, we can graph any function. Now we will explain the process to do this by again using the same linear function f(x) = 3x+5. The (0, b) = (0, 5) is used to show the y-intercept of this function, and m = 3 is used to show its slope. The steps/process to do this is described as follows:\n\nStep 1: In this step, we will plot the y-intercept (0, b), which is equal to (0, 5). So we will plot the points (0, 5).\n\nStep 2: Now, we have to write the slope in the form of a fraction rise/run and then find out the \"rise\" and \"run\".\n\nHere slope = 3 = 3/1 = rise/run\n\nSo rise = 3, and run = 1.\n\nStep 3: In this step, we will get the new points by rising y-intercepts vertically with the help of \"rise\" and then run horizontally with the help of \"run\".\n\n#### Note: The graph will go up if the \"rise\" is positive, and the graph will go down if the \"rise\" is negative. Similarly, we will go right if the \"run\" is positive, and we will go left if the \"run\" is negative.\n\nIn this graph, we will go up by 3 units from y-intercept and thereby go right by 1 unit.\n\nStep 4: Now, we will use a line to join the points from step 1 to step 2, and extend those lines on both sides.", null, "### Domain and Range of a Linear function\n\nA set of all real numbers is contained by the range and domain of a linear function. In the below image, we can see f(x) = 2x+3 and g(x) = 4-x both functions are plotted on the same axes like this:", null, "Here R is used to show the domain of a linear function, and R is also used to show the range of a linear function.\n\n#### Note:\n\n1. If the problem does not contain any specific range or domain, only then the domain and range of a linear function will be R.\n2. The horizontal line will be shown by the linear function f(x) = b if the slope m = 0. In this case, the range of this function = {b} and the domain of this function = R.\n\n### Inverse of a Linear function\n\nThe function f-1(x) is used to indicate the inverse of a linear function f(x) = ax+b in such a way that f(f-1(x)) = f-1(f(x)) = x. Now we will use an example to show the process to identify the inverse of a linear function. In this example, we have a linear function f(x) = 3x+5, and we will find the inverse of this function.\n\nStep 1: In the first step, we will replace f(x) with y. By replacing, we will get the following equation:\n\ny = 3x+5\n\nStep 2: Now, we will interchange the variables x and y and get the following equation:\n\nx = 3y+5\n\nStep 3: Now we will solve the above equation for y like this:\n\nx-5 = 3y\n\ny = (x-5) /3\n\nStep 4: Now we will replace y by inverse function f-1(x), and get the following:\n\nf-1(x) = (x-5) /3\n\nNote that the function f(x) and the inverse function f-1(x) are always symmetric with respect to line y = x. Now we will plot the linear function f(x) and its inverse function f-1(x). Here f(x) = 3x+5 and f-1(x) = (x-5)/3. Here we will check whether these functions are symmetric about y = x, and we will also see when (x, y) lies on f(x), then (y, x) lies on f-1(x). To prove this, we will take an example in which we have (-1, 2), which lies on f(x), and (-2, 1), which lies on f-1(x). The graphical representation of all these things is described as follows:", null, "### Piecewise Linear function\n\nThere are some cases where the linear function is defined in a uniform way throughout its domain. This is because its domain can be split into two or more than two parts, so the linear function can be defined in two or more than two ways in some cases. This type of linear function is known as the piecewise linear function. The example of a piecewise linear function is described as follows:\n\nExample: In this example, we have a linear function, and we have to plot the graph of this function. The function f(x) is described as follows:", null, "Solution: The above described piecewise function is linear in both the above described parts of its domain. Now we will determine the endpoint of the line by taking both the cases.\n\nWhen the case is x ∈ [-2, 1):\n\nx y\n-2 -2+2 = 0\n1\nIn this case, the hole is 1 [2, 1]\n1+2 = 3\n\nWhen x ∈ [1, 2):\n\nx y\n1 2(1) - 3 = -1\n2 2(2) - 3 = 1\n\nThe corresponding graph is described as follows:", null, "Important notes\n\n• If a linear function must be in the form f(x) = mx+b, then the graph of this function will be a line.\n• When the slope of the linear function f(x) = mx+b is 0, the linear function will be a horizontal line, and this function will be known as the constant function.\n• In a linear function f(x) = ax+b, the domain and range will be R. Whereas for a constant function f(x) = b, the range will be {b}.\n• We can use these linear functions in linear programming to show the objective function.\n• There is no inverse in a constant function because this function is not one to one.\n• If the slopes of the two linear functions are equal, then both functions will be parallel.\n• If the product of slopes of two linear functions is -1, then both the functions will be perpendicular.\nThe linear function fails the test of a vertical line. That's why the vertical line is not a linear function.\n\n### Feedback", null, "", null, "", null, "" ]	[ null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics2.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics3.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics4.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics5.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics6.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics7.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics8.png", null, "https://static.javatpoint.com/tutorial/dms/images/linear-function-in-discrete-mathematics9.png", null, "https://www.javatpoint.com/images/facebook32.png", null, "https://www.javatpoint.com/images/twitter32.png", null, "https://www.javatpoint.com/images/pinterest32.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89296025,"math_prob":0.99826235,"size":8709,"snap":"2022-40-2023-06","text_gpt3_token_len":2291,"char_repetition_ratio":0.21091327,"word_repetition_ratio":0.07596439,"special_character_ratio":0.27098405,"punctuation_ratio":0.11088296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994993,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T14:38:09Z\",\"WARC-Record-ID\":\"<urn:uuid:699facf8-fd1f-4eb8-ab51-cef77f9eb120>\",\"Content-Length\":\"68220\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cff6f162-ed28-4144-93b4-406482e55f95>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d46e540-e95d-4d96-96eb-41f7c785db11>\",\"WARC-IP-Address\":\"172.67.211.76\",\"WARC-Target-URI\":\"https://www.javatpoint.com/linear-function-in-discrete-mathematics\",\"WARC-Payload-Digest\":\"sha1:3KSCNZV4DCAILMMMHD7GGRQK4DTDOULL\",\"WARC-Block-Digest\":\"sha1:BWBDCHDEF5KSZJ46JBRUX6C5TC52WCGE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500140.36_warc_CC-MAIN-20230204142302-20230204172302-00790.warc.gz\"}"}
https://lapmos.com/details/operators-types-of-operators-in-python-programming	[ "## Description\n\nOperator is a symbol that tells the computer to perform the mathematical and logical operation in Python Programming.\n\n## Types of Operators:\n\nPython programming language has supports the following operators.\n\n• Arithmetic Operators\n• Comparison / Relational Operators\n• Logical Operators\n• Assignment Operators\n• Bitwise Operators\n• Membership Operators\n• Identity Operators\n\n## Arithmetic Operators in Python Programming\n\nLet's Assume that variable 'a' holds 5 and variable 'b' holds 2, then\n\nOperator Description Example\nAddition (+) Adds values on either side of the operator. a + b = 7\nSubtraction (-) Subtracts right hand operand from left hand operand. a - b = 3\nMutliplication () Multiplies values on either side of the operator.\n\na b = 10\n\nDivision (/) Divides one operator to another operator. a / b = 2.5\nModulus (%) remainder of a divided by b. a % b = 1\nExponent () Performs expontial (power) calculation on operators. ab = 25\n// Floor Division - operands where the result is the quotient in which the digits after the decimal point are removed. a//b = 2\n\n## Relation Operators in Python Programming\n\nThe Relational Operator are used to compare the values between variables. Relational Operators tabel below...\n\nOperator Description Example\n== If the both values are equal, then condition is true. (a==b) is not true\n!= If the both values are not equal, then condition becomes true. (a!=b) is true.\n> Greater Than ( if the left operand is greater than the right operand, then condition becomes true.) (a > b) is true.\n< Less Than ( if the left operand is less than the right operand, then condition becomes true.) (a<b) is true.\n<= Less than or Equal to (a <= b ) is not true.\n>= Greater than or Equal to (a >= b) is true.\n\n## Assignment Operators in Python Programming.\n\nThe assignment operator is represented by the equal sign(=). The variable appearing on the left side of = sign and it is assignment the value appearing on the right side of this sign.\n\nOperator Expression Example\n= a = a + b a = 7\n+= a += b a = a + b = 7\n-= a -= b a = a - b = 3\n/= a /= b a = a / b = 2.5\n%= a %= b a = a % b = 1\n= a = b a = a * b = 10\n= a =b a = a**b = 25\n//= a //= b a = a//b = 2\n\n## Logical Operator in Python Programming.\n\nThe Logical Operator is used for when we want to check the condition and for making the decisons in Programs.\n\nOperators Description Example\nand Logical AND If the both operands are true then condition becomes true. (a and b) is true.\nor Logical OR If any of the two operands are non-zero then condition becomes true. (a or b) is true.\nnot Logical NOT Used to reverse the logical state of its operand. not(a and b) is false\n\n## Bitwise Operator in Python Programming.\n\nThe Bitwise Operator is special operators which are used for manipulation of the data at bit level. These operator are used for testing the bits, or shifting them right or left. Bitwise operators many not be applied to float or double.\n\nOperator Meaning\n& bitwise AND\n\| bitwise OR\n^ bitwise exclusive OR\n<< shift left\n>> shift right\n\n## Python Membership Operators.\n\nPython's membership operators test for membership in a sequence, such as strings, lists, or tuples. The are two mebership operators...\n\nOperator Description Example\nin Evalutes to true if it finds a variable in the specified sequence and false otherwise. a in b, here in results in a 1 if x is a member of sequence y.\nnot in Evaluates to true if it does not finds a variable in the specified sequence and false otherwise. a not in y, here not in results in 1 if a is not a member of sequence b.\n\n## Python Identity Operators.\n\nOperator Description Example\nis Evaluates to true if the variables on either side of the operator point to the same object and false otherwise. a is b, here is results false.\nis not Evaluates to false if the variables on either side of the operator point to the same object and true otherwise. a is not b, here is not results is true.\n\n## All About of String in Python Programming language\n\nIn Python It is called a string which is inside single or double quotation marks. We can write multiline string inside three single or double quotation mark.\n\n## List and Tuple in Python Programming\n\nList is a kind of data structure in Python which is mutable, changeable, ordered sequence. Tuples are used to store multiple items in a single variable.\n\n## Dictionary In Python \| Method Of Dictionary\n\nDictionary is used to store data values in Key: Value pairs. A dictionary is a collection which is ordered, Changeable and do not allow duplicates.\n\n## Implement of All Method of String in Python Programming language\n\nImplementation of all methods of string in python. How to implement of method on string.\n\n## Operators based Exercise 1 in Python Programming language.\n\nAll operators used for better understanding the python operators.\n\n## Sets in Python \| Perform Set methods\n\nA Set is an unordered collection data type which does not contain duplicates elements. A set is iterable and mutable data type." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8273025,"math_prob":0.9709173,"size":5069,"snap":"2022-40-2023-06","text_gpt3_token_len":1280,"char_repetition_ratio":0.15478776,"word_repetition_ratio":0.079822615,"special_character_ratio":0.25567174,"punctuation_ratio":0.09170306,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99605817,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T17:00:16Z\",\"WARC-Record-ID\":\"<urn:uuid:c73dfb3e-1a19-40c0-8cea-a7e82c54ff05>\",\"Content-Length\":\"55279\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:47d2ef35-abf8-4842-af5c-aa1e0bc5382a>\",\"WARC-Concurrent-To\":\"<urn:uuid:4afe9128-b28e-4292-8191-c50007a2dc4e>\",\"WARC-IP-Address\":\"217.21.87.171\",\"WARC-Target-URI\":\"https://lapmos.com/details/operators-types-of-operators-in-python-programming\",\"WARC-Payload-Digest\":\"sha1:XK3UAZNUXDRRYBXFOKFKMVWCUK5AAFKN\",\"WARC-Block-Digest\":\"sha1:HQT5K5CZLYTA4O5Y5WECFXWOAQ5FLIBC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334912.28_warc_CC-MAIN-20220926144455-20220926174455-00465.warc.gz\"}"}
https://mathoverflow.net/questions/132538/when-is-the-support-of-a-radon-measure-separable	[ "When is the support of a Radon measure separable?\n\nLet $X$ be a topological space, equipped with its Borel $\\sigma$-algebra $\\mathcal B(X)$, and let $\\mathbb P$ be a Radon probability measure on $(X, \\mathcal B(X))$. Recall that the support of the measure $\\mathbb P$ is the smallest closed set of full measure.\n\nIs the support necessarily separable? If so, why? If not, what is a counterexample?\n\n• What about Haar measure on a compact, non-separable group? (By the way, don't you mean \"smallest closed set with full measure\"?). – jbc Jun 1 '13 at 23:02\n• Good, concise answer, and thanks for the pointer on the typo. Thanks, @jbc. – Tom LaGatta Jun 3 '13 at 3:08\n\nLet $I$ be a set of cardinality larger than the continuum. Then the product topology $[0,1]^{I}$ is compact but not separable. Give the interval $[0,1]$ the Lebesgue measure, then give $[0,1]^{I}$ the product measure $\\mu$. If $U$ is a non-empty open subset of $[0,1]^{I}$, then $U$ contains a basic open set $\\prod_{i\\in I}U_{i}$ where $\|\\{i\\in I\|U_{i}\\neq[0,1]\\}\|$ is finite. Therefore $0<\\mu(\\prod_{i\\in I}U_{i})\\leq \\mu(U)$. Said differently, if $C$ is a closed subset of $[0,1]^{I}$ with $\\mu(C)=1$, then $C=[0,1]^{I}$.\nIf you replace $[0,1]$ with the circle $S$, then $S^{I}$ is a compact non-separable group which does not have separable support as jbc mentioned.\n• A priori, $\\mu$ is only defined on the product $\\sigma$-algebra on $[0,1]^I$, which is strictly smaller than the Borel $\\sigma$-algebra. How do we extend $\\mu$ to a Borel measure? And once this is done, how do we see that the measure is Radon? – Nate Eldredge Jun 2 '13 at 12:37\n• The reason I suggested using a group was to avoid such questions, since in this situation you can use the existence theorem for Haar measure (which, by the way, has a very short and transparent proof in the case of a compact group, using the weak star compactness of the dual ball of $C(K)$). A particularly simple solution to the OP is then provided by a large cardinal product of the two-point group. – jbc Jun 2 '13 at 13:50\n• The product $\\sigma$-algebra on $[0,1]^{I}$ contains every Baire set, so since $[0,1]^{I}$ is compact we can extend this Baire measure to a Radon measure on the Borel $\\sigma$-algebra in a unique way. – Joseph Van Name Jun 2 '13 at 17:13" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86494166,"math_prob":0.9995465,"size":344,"snap":"2019-43-2019-47","text_gpt3_token_len":95,"char_repetition_ratio":0.13235295,"word_repetition_ratio":0.0,"special_character_ratio":0.2616279,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99985695,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T07:56:33Z\",\"WARC-Record-ID\":\"<urn:uuid:2c3b4bc0-11ab-4246-a771-f37f78d91fa8>\",\"Content-Length\":\"123130\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3915d6fa-14d5-465c-9341-245ef3bca687>\",\"WARC-Concurrent-To\":\"<urn:uuid:59002717-627b-4460-beb7-b0e18e96a0cc>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/132538/when-is-the-support-of-a-radon-measure-separable\",\"WARC-Payload-Digest\":\"sha1:CY35WDYCRKEIQSSK6TQEKWNVTWDF3SWT\",\"WARC-Block-Digest\":\"sha1:VIW5BCJ3GCDCERWEWOULZOE4T6LZBA4H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986677964.40_warc_CC-MAIN-20191018055014-20191018082514-00518.warc.gz\"}"}
http://www.basicsofelectricalengineering.com/2017/08/current-divider-example-3-finding-value.html	[ "# Basics of Electrical Engineering\n\nLearn the basics of Electrical Engineering.\n\nA current source of 100 mA powers two parallel resistors, the first resistor is of 30 Ω and it receives 60 mA current, Find the value of the second resistor.\nSolution: R_1 dissipates the 60 mA, the remaining 40 mA current will pass through R_2.\n\nFrom the Ohm's law, the voltage across 30-ohm resistor is 1.8 V.\n\nR_1 and R_2 join in parallel and both have the same voltage across them.\n\nV_R_2 = 1.8 V,\n\nUsing the Ohm's law R2 = V_R2/I_R2 = 1.8 V / 40 mA\n\nR2 = 45 ohm\n\nSo 45 ohm is the right answer." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86981654,"math_prob":0.9882351,"size":562,"snap":"2019-51-2020-05","text_gpt3_token_len":184,"char_repetition_ratio":0.13978495,"word_repetition_ratio":0.12612613,"special_character_ratio":0.33274022,"punctuation_ratio":0.12307692,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991654,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T00:02:18Z\",\"WARC-Record-ID\":\"<urn:uuid:adfbe698-a47e-45e8-917a-affdf6480e52>\",\"Content-Length\":\"117416\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a22c6cdd-a6c8-4016-a1f7-fe22b472b1a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:301b38dd-e1ca-4de6-98d8-c57b5cfb0991>\",\"WARC-IP-Address\":\"172.217.7.243\",\"WARC-Target-URI\":\"http://www.basicsofelectricalengineering.com/2017/08/current-divider-example-3-finding-value.html\",\"WARC-Payload-Digest\":\"sha1:WZEFBUL5VWEA2XA7UPJ2XTIJDVYNSP3P\",\"WARC-Block-Digest\":\"sha1:QFGINN4HWXQ4OSZSZDE6GOVLZJR4FOTM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251694071.63_warc_CC-MAIN-20200126230255-20200127020255-00530.warc.gz\"}"}
http://intusoft.com/nlhtm/nl72.htm	[ "", null, "Newsletter List", null, "Newsletter Issue #72, Dec 2003\n\n# SPICE Numerical Methods, Part 1 Convergence Methods\n\nIt’s been some time since we devoted a newsletter to modeling and convergence issues. Over the years, we have learned more about how to make good models and have also added and improved IsSpice options. The information shown here follows closely the seminar presentation at the Power Systems World on November 3, 2003.\n\nUnderstanding what’s under the hood in your software tools has been dismissed in the quest for simple user paradigms. But your software is performing mathematical tasks that, as an engineer, you’ve been trained to understand. You’ll get the most out of our IsSpice simulation engine by combining your knowledge and experience with an understanding of how the simulator works - especially if you want to make models for other engineers to use.\n\n In This Issue 1 6 Unique Solution Sidebar 6 9 Spice Options for Transient Simulation Control 12 Transient Initialization 14 Making Models Behavioral Models\n\nSPICE is basically an equation solver. The equations are set up according to Kirchhoff’s Current Law, KCL. The KCL matrix is modified to include the current through voltage sources and state variables used in the XSPICE model extensions. For linear time invariant circuits, the equation matrix has a unique solution (See the Unique Solution Sidebar). Non-linear circuits and operations involving integration require inserting a Norton equivalent circuit with the large signal current summed into the matrix’s right hand side, RHS, and the small signal conductance summed into the matrix.\n\nThis summation process is referred to as the matrix stamp of a SPICE model. Interestingly, each model sums its contributions without the knowledge of what the other models are doing. Figure 1 illustrates the matrix stamp for a diode. When the diode anode is node K and its cathode is node L, then the K row and column and the L row and column have the small signal conductance summed into their matrix positions as shown. The large signal current is summed with appropriate signs into the RHS.", null, "Figure 1. Diode linearized circuit and its matrix stamp.\n\nSPICE must iterate this solution to solve non-linear circuit equations. The simulator doesn’t know if the circuit is linear so that it will iterate linear circuit equations as well. Certain linear circuit equations may fail to converge if the RHS vector is not a constant. Remember, the basic KCL rule is that the sum of the currents at a node is constant. If it’s a variable, then there is no guarantee of convergence.\n\nBy way of example, consider the circuit and its modified nodal admittance, MNA, matrix shown in Figure 2. This circuit model is used to describe a transformer. Current is transformed from the left or input side to the right or output side. Voltage is transformed in the opposite direction. These equations force the output power to equal the input power. The model is useful for any power conserving operation. In the third equation, the current at B4 is .5 * IR1. It must remain on the RHS because IR1 is not a matrix variable. For the circuit configuration shown, v2 grows for each iteration as shown below:\n\n Iteration v2 IR1 1 0 10k 2 250k -2.499e+007 3 -6.2475e+008 6.2475e+010\n\nClearly, the solution is not converging; v2 is alternating sign and its magnitude is growing rapidly to infinity. If the input and output of the model were reversed, then the solution would have converged.", null, "Figure 2. An equation matrix example incorrectly using the current through a resistor.\n\nNow consider the circuit shown in Figure 3. The input current is sensed through a voltage source instead of a resistor. The current through the voltage source is a matrix variable. The SPICE3 arbitrary source or B-element saves the variables and their derivatives and uses that information to move any RHS components using those variables back over to the matrix during the load operation as shown in the figure. With this seemingly minor change in the model, the circuit has been made to converge without the need for any iterations. The lesson learned here is to make B-element equations that only use matrix variables. That isn’t to say that you can’t write equations with variable RHS values; however, it’s unwise to rely on the solution iterations to converge.", null, "Figure 3. A properly formed model allows the B-element to move the current from the RHS into the matrix.\n\nIn order to handle non-linear algebraic equations, SPICE simulators perform what is known as Newton-Raphson iteration. The linearized matrix; for example, one using a diode as shown in Figure 1, is solved and a new operating point and its small signal equivalent admittance is found. The process is repeated until a stable solution is found. Figure 4 shows a test circuit using a diode and Figure 5 shows how the equation is iterated to form a stable solution. It seems pretty reasonable to model a diode voltage as a function of current – that’s what designers usually do to estimate an operating point because a Silicon diodes forward drop ranges from .6 to .7 volts for very large range of current.", null, "Figure 4. A diode test circuit, the operating point converges in 5 iterations.", null, "Figure 5. Newton-Raphson iterations converge rapidly to solve for the operating point\n\nHowever, early on, the problem of finding a stable solution for this circuit used the equations formulated with current as a function of voltage as shown in Figure 6. You can see from inspection that the initial trial would place nearly 5 volts across the diode resulting in 8.348e+069 amps. The next iteration would produce nearly the same result and it would take forever to walk the solution back to the correct answer. The funny thing is that SPICE3 quickly sees the convergence failure and tries some interesting tricks to make even this ill conditioned equation work.", null, "Figure 6. Using a B-element with current as a function of voltage requires 194 iterations to converge\n\nWhen the DC iteration limit, ITL1, is exceeded (it defaults to 100 iterations). Then the simulator tries a neat trick called Gmin Stepping and finally it tries Source Stepping. Before attempting a matrix solution, SPICE preorders the matrix by reordering its rows and columns to get large values along the matrix diagonal. If the only values in the matrix were on the diagonal, the solution can be written by inspection and is trivial. By the same argument, if the values on the matrix diagonal are very much larger than the off-diagonal values, then a solution should be readily achieved. That’s the idea behind Gmin Stepping. Gmin is a SPICE option representing the smallest possible conductance. SPICE3 begins by setting the number of steps to 10, then computes the diagonal offset as Gmin10^numSteps. For the default Gmin=1e-12, the initial offset if .01. That corresponds to a 100 ohm resistor to ground on the diagonal nodes. If convergence fails on the first step, then the simulator signals Gmin Step failed (you might want to increase the numGminSteps or increase Gmin). Intusoft has made several modifications to these algorithms to improve their performance. You can see that a 100 ohm resistor across the diode initializes much closer to the correct solution for our example. The offset is gradually removed and if on its final removal, the circuit converges, the job is done. If not SPICE3 tries source stepping. All voltage and current sources are set to zero and stepped up to their final values using the number of steps specified using ITL6. If you set ITL1=50 for this case, then Gmin Stepping fails and the simulator will go on to use source stepping. Source stepping will walk the solution up the I-V curve from V=0 and will also converge. As you see, IsSpice works pretty hard to make even poorly constructed models converge. When you build your own models, you need to keep the accounts option turned on (.OPTIONS acct) and keep an eye on the operating point iterations. Most modern SPICE simulators employ variations of the described operating point convergence methods; however, the different vendors have fine tuned the algorithms and hold these methods as trade secrets.\n\nBefore continuing to discuss IsSpice options, it’s worth comparing the performance of the built-in diode with its behavioral counterpart shown in Figure 7. Running this simulation and checking the number of iterations, we find the solution converges in 9 iterations. That’s about twice as many iterations needed by the built in SPICE diode. The SPICE diode model is more complete (it include the temperature effects of IO, noise, …). Even so, the added complexity comes with built-in convergence aids that are hard to beat with behavioral models. We’ll visit this topic in more detail later on.", null, "Figure 7. Diode modeled using a B-element voltage source converges in 9 iterations.\n\nTOP\n\n Unique Solution Sidebar\n\nAn interesting property of the linear matrix solution arises for the AC analysis. It turns out that the solution does not diverge if the circuit is unstable in the time domain. The feedback equation\n\nG = A/(1+AH) has the same answer as A—> infinity or A—> -infinity! - A result that allows you to analyze unstable circuits for gain and phase margins. Moreover, you can insert high gain amplifiers nearly anywhere in a circuit to force the “amplifier” input node voltage or branch current to null. You can use that technique, called null injection, to isolate a section of circuitry while a control loop is closed. Using various combinations of null injection, you can solve for A, AH, H and G in the feedback equation. This technique has been expanded to become the General Feedback Theorem, by Dr. R.D. Middlebrook .\n\n R.D.Middlebrook, “The GFT: A General Yet Practical Feedback Theorem,” to be published.\n\n Spice Options for Operating Point Control\n\nThe following are the key options that will aid in making your DC operating point converge.\n\n1. GMIN: Sets the minimum conductance (default is 1e-12), see previous discussion with respect to its use in Gmin Stepping. It is also used within various models to eliminated divide by zero errors. Typically, divisions have Gmin added to the numerator and the denominator as follows: A/B —> (A+Gmin)/(B+Gmin). This approach yields an answer for B=0 and both A and B zero. The odds of B = -Gmin are greater than 1 in 2^48 = 2.8e+014 because the 64 bit floating-point mantissa is 48 bits long. On the other hand, the odds of having a zero value is extremely high because many variables are initialized to zero. For your B-element equations, you can use Gmin as a parameter or go ahead and hard wire a reasonable value, for example 1u, into your divisions.\n\n2. RELTOL (default=.001), ABSTOL(default=1e-12), VNTOL(default=1e-6) are used after each Newton-Raphson iteration to decide on whether or not each node voltage or branch current is converged. If val is the value just iterated and prev is the previous value, the following logic is applied:\n\na. Voltage: tol = abs(val)RELTOL+VNTOL\nb. Current: tol = abs(val)RELTOL+ABSTOL\nc. If abs(prev-val) > tol return(NONCONV)\n\nThese equations appear to be reasonable for a circuit that’s working normally; however, SPICE can reach some very extreme values. As abs(val)—> infinity, the tolerance gets very large and absurd values may appear to converge. You may encounter this condition when attempting to float a section of circuitry. The floating circuit node voltages become independent of SPICE ground. The DC offset can get so large that a matrix solution becomes impossible. The solution is to not float circuitry. You may wish to use a parameterized resistor between the 2 circuit grounds to see the effect on the floating voltage vs. resistor value. Many models operate in the 1 to 10 volt range making VNTOL in the range of 10u to 100u a reasonable choice. Similarly branch currents in the 1 to 10ma range would suggest ABSTOL in the range of 1 to 10n. RELTOL is frequently set to .01. The ABSTOL and VNTOL setting apply to every node voltage and branch current so that you should scale models in a similar range. If you need to maintain accuracy for very small numbers; for example, when time delay is a variable, then you should scale your model to compute the time delay in microseconds or milliseconds as the case may be. Alternatively you can use normalizing B-elements to multiply the voltage or current such that the result is in the proper range. These tolerances are also applied when calculating the local truncation error, LTE, for time step control.\n\n3. RSHUNT(default not used): The RSHUNT value is connected from every node to ground. This is useful in tracking down singular matrix problems that occur when ideal capacitors are connected in series. Without resistors, there is no unique solution for series capacitors. RSHUNT can also be used to converge problem circuits in order to explore problem areas. Making RSHUNT small enough should make almost anything converge, the example in Figure 2 can be made to converge by setting RSHUNT to 1m. RSHUNT should be considered a debugging tool because its use could introduce errors in other models that use high impedance circuitry such as integrators or charge amplifiers.\n\n4. ITL1(default=100): The DC iteration limit. For complex circuits you may need to increase ITL1 up to 1000. Beyond that, you are likely to experience chaotic behavior and converge by chance. Reducing ITL1 can move more quickly to source stepping.\n\n5. ITL6(default = 10): The number or steps used in source stepping. You may need to set this as high as 1000 for complex circuits. You might want to use .NODESET described later to get convergence to occur during the initial Newton-Raphson operating point iteration.\n\n6. AUTOTOL(default not used): This is another debugging tool. It causes a table of VNTOL and ABSTOL values to be setup for each node voltage and branch current. Whenever a non-convergent situation arises, the table value is multiplied by abs(AUTOTOL). That process will rapidly eliminate the troublesome nodes. If AUTOTOL is negative, the activity is reported in the “.out” file. Examining the report may help isolate convergence problems.\n\n7. .NODESET: This is not part of the .options commands; however, its use is important in steering the initial operating point solution to one of several stable solutions. Figure 8 illustrates how this works for a bi-stable circuit. In our previous newsletter, NL71-pg13, we showed how to add an additional non-linearity to eliminate the low voltage operating point. If the .NODESET works, it provides a simpler, more elegant solution.", null, "Figure 8. use .NODESET=100 for the operating point shown and .NODESET=0 for the low voltage operating point\n\n Spice Options for Transient Simulation Control\n\n1. METHOD: Selects Gear or Trapezoidal integration. Gear works best for power electronics or other circuits that use inductors; especially when the inductor current is switched rapidly. SPICE diodes with reverse recovery will snap off and cause numerical artifacts if Trapezoidal integration is used as shown by Figure 9.\n\n2. MAXORD(default=2): Selects the integration order for Gear", null, "", null, "Figure 9. PWM Voltage, Vsw, converges rapidly using Gear integration\n\n3. TRTOL(default=7): Transient error tolerance. The factor by which SPICE overestimates local truncation error, LTE. LTE is the estimate of integration error. Each component that uses the SPICE implicit integration method computes the time step required to achieve the desired accuracy. The smallest time step from these calculations is used for the next time step. If the required time step is less than the current time step or if the Newton-Raphson solution doesn’t converge or if the VSECTOL condition isn’t met, the time step is scaled back. Increasing TRTOL will increase the time step. This parameter was empirically selected to give the best performance over a wide range of circuits. It shouldn’t be changed. TRTOL is also used to make the VSECTOL options have hysteresis.\n\n4. ITL4(default=10): Sets the lower transient iteration limit. This may need to be increased up to 100 for complex circuits. If this must be set higher to achieve convergence, then you may be experiencing chaotic behavior and small changes in initial conditions could cause non-convergence.\n\n5. CHGTOL(default = 1e-14): Sets the maximum charge error, only used in the LTE calculation.\n\n6. VSECTOL(default off): If VSECTOL is turned on, the value represents the maximum volt-second error for any node voltage. Node voltages for the next time step are predicted based on their history. The next time step begins with the predicted values. If the product of the time step and the absolute value of the difference between the iterated value and the predicted value is greater than VSECTOL, then the time step is scaled back. TRTOL is used to provide hysteresis in that decision. The time step won’t be increased until the VSECTOL computed time step is TRTOL times greater than the current time step. This option is necessary to get fast rise times from behavioral switching as illustrated by Figure 10 and 11. For the case shown, 278 time points were required to achieve the 10nsec resolution. Without VSECTOL, the maximum time step would have to be reduced to 10nsec, requiring 5e6 time points, which lengthen the simulation by a factor of 17986! The circuit shown in Figure 10 is a hysteresis switch that can be used to model under voltage lockout, UVLO. The VSECTOL option will cause the IsSpice4 simulator to backtrack; that is, reverse time and back up to just before the switching event occurs as shown in Figure 11. An R-C network is needed because the hysteresis is time-directionally sensitive; without the R-C network, the switch point can be incorrect.", null, "", null, "Figure 10. A Behavioral hysteresis comparator shown switching with default SPICE options has a rise time equal to the SPICE time step when passing the switching threshold", null, "Figure 11. The same simulation as Figure 10, except VSECTOL=50n made the vref signal switch from 0 to 5 volts in less than 10nsec.\n\n7. BYPASS(default = ON): Bypasses computation of model parameters when the input voltage and currents haven’t changed very much. This speeds the simulation but can’t be used when relying on VSECTOL because large voltage and current changes may take place when rejecting time points and scaling back the time step. Turn BYPASS off when using VSECTOL.\n\n8. RAMPTIME(default off): When set, voltage and current sources will be ramped from 0 to their specified value in RAMPTIME seconds. B-elements aren’t ramped so you can make some node voltages and branch currents non-zero. You can ramp B-elements by multiplying their output with a constant V element. This can help things like oscillators and hysteresis circuits initialize; however, the overall simulation time may be lengthened while you wait to get to steady-state. Using .NODESET is a better choice.\n\n8. NOOPITER(default off): Skips the op calculation and goes directly to GMIN stepping\n\n9. MINSTEP(default off): removes saved data points closer together than MINSTEP. This options reduces memory requirements for lengthy simulations; however, aliasing of data can occur.\n\n10. MINBREAK(default=off): Breakpoints will not be set to less than this value. Breakpoints are used by various models; for example, V and I sources will set a breakpoint at the end of RAMPTIME and transmission lines will set breakpoints when a change in the propagated signal reaches the terminals. This option is most frequently used to reduce transmission line computational overhead.\n\n11. TMAX(default off): the fourth term in the .TRAN statement can be optionally set to override the SPICE automatic time step algorithm. Frequently used in SPICE2 and SPICE3 simulators for switching power supplies. It can cause substantial increase in simulation time and memory. See use of VSECTOL to eliminate the use of this option.\n\nTOP\n\n Transient Initialization\n\nOrdinarily, the initial operating point for a transient simulation uses the DC operating point calculation. There are circuits for which this technique doesn’t work. SPICE2 and SPICE3 can direct the simulator to use initial conditions by using the UIC keyword in the .TRAN statement. When invoked, the circuit is initialized using the IC= keyword in various models and .IC v(node)=value statement. UIC works for circuits that don’t employ subcircuit models. When a subcircuit is used, the internal IC’s are used and may not correspond to the initial conditions of your external circuit. For example a positive op-amp output voltage may cause current to flow in an inductor. If you initialize the inductor current, then you must also figure out how to initialize the op-amp properly or else you might have the inductor current flowing into an open circuit. Needless to say, this is a very difficult task because you don’t know much about the internals of vendor-supplied models. Moreover, many state variables are private to models, for example diode charge, so that you can’t initialize everything. Here’s a SPICE3 script that you can use to extract the initial conditions after a transient simulation has run to steady state:\n\n * script to extract the final value from a transient simulation * remove everything except the node voltages set colwidth=1 nv = nextvector(null) while nv <> null len = length(nv) - 1 if len > 1 printtext -n \".IC \" printname nv printtext -n \"=\" printval nv[len] printtext end nv = nextvector(nv) end\n\nJust paste it into the IsSpice script window and press the DoScript button. You must remove all of the instance current and power values since they can’t be initialized using UIC. Then paste the remaining into the User Statements window of the Simulation Setup dialog. For currents, you need to find the parts and fill in the IC=value.\n\nThe most successful method for initializing complex circuits is to use switches that apply initial conditions at time = 0, and then remove the IC’s when time > 0. Like the old analog computers used to do. Figure 12 shows how it works for a simple R-L-C circuit. The IC’s were purposely set a little off to get something to look at. If you use the above script to find the IC’s, the peak error is under 7 micro-volts.", null, "Figure 12. Transient initialization without UIC.\n\nThe advantage of this technique is that SPICE will figure out the initial conditions for control circuits, so you only have to worry about the main reactive components.\n\n Making Models Behavioral Models\n\nBehavioral modeling is the single most important element added to SPICE3. These B-elements allow you to introduce both linear and nonlinear models using algebraic equations. Before SPICE3 solves these equations, it removes the variable from the RHS and places them in the matrix according to the derivatives of the expression. As long as you use matrix variables (Node voltage and Voltage source branch currents) the order in which the equations are solved doesn’t matter. With SPICE3, you can pour milk into your glass BEFORE getting the glass out of the cabinet! Intusoft has added capabilities to the original SPICE3 B-elements\n\nIntusoft B-element extensions\n\nIf-Then-Else\n\nv=v(in) < 10 ? v(in) : 0; c/c++ style\nv=MIN(v(in),10); Fortran/PSpice style\n\nBoolean\n\nv=(v(in) & v(uvlo)) \| v(fault)\nwhere & \| ! ^ are AND, OR, NOT and XOR respectively\nLONE=5, LZERO=0, LTHRESH=2.5 are default options.\n\nSimulation Variables that can be used in expressions\n\nTIME; simulation time in transient analysis, 0 in AC or DC\nFREQ; simulation frequency in AC analysis, 0 in op or TRAN\nTEMP; circuit temperature\n\nNew elements that support expressions\n\nR=<expression>\nC=<expression>\nL=<expression>\nExample:\nR=TIME < 10u ? {OPEN} : {SHORT}\n\nStandard B-element Operations\n\nMath\n\n+ - * / ^ %\n\nTrigonometric\n\nsin, asin, sinh, asinh cos, acos, cosh, acosh tan, atan,tanh, atanh\n\nTransindental\n\nexp, expl, log, lnpwr, pwrs, min, max, sgn, stp, abs, ceil, floor, int, frac, mod2, sync, mag, phs, real, imag, rand, randc\n\nBehavioral Models – Limitations\n\nThere are some less than obvious limitations in using B-element expressions. We’ll explore these by example in the next newsletter.\n\nWe want to thank everyone who visited our booth at this year’s Power Systems World Conference. You made it an incredibly successful show for us, notably with the interest in our 8.x.10 Build 2093 ICAP/4 simulation software announced at the show, plus our collaboration with ON Semiconductor for power supply design. We look forward to some exciting product announcements for next year.\n\nOur special half-day hands-on workshop on “Simulating and Modeling for Power Electronic Design” was a great success. All workshop attendees received a Microsoft Optical Mouse, and \"Switch-Mode Power Supply SPICE Cookbook\" authored by Christophe P. Basso. Again, thanks and we look forward to seeing you in 2004.\n\n Power Systems World Convention DRAWING WINNERS! ICAP/4Rx Power Deluxe Sofware Winner Peter Czyl - Consultant, Long Beach CA Switch-Mode Power Supply SPICE Cookbook Winners Iradj Vokhshoori - Pyramis Corp., Torrance CA Yuga Tummala - Stryker Medical, Kalamazoo MI" ]	[ null, "http://intusoft.com/nlpics/logoredsm3.gif", null, "http://intusoft.com/nlpics/nlhead.gif", null, "http://intusoft.com/nlpics/72/figure1.gif", null, "http://intusoft.com/nlpics/72/figure2.gif", null, "http://intusoft.com/nlpics/72/figure3.gif", null, "http://intusoft.com/nlpics/72/figure4.gif", null, "http://intusoft.com/nlpics/72/figure5.gif", null, "http://intusoft.com/nlpics/72/figure6.gif", null, "http://intusoft.com/nlpics/72/figure7.gif", null, "http://intusoft.com/nlpics/72/figure8.gif", null, "http://intusoft.com/nlpics/72/figure9a.gif", null, "http://intusoft.com/nlpics/72/figure9b.gif", null, "http://intusoft.com/nlpics/72/figure10a.gif", null, "http://intusoft.com/nlpics/72/figure10b.gif", null, "http://intusoft.com/nlpics/72/figure11.gif", null, "http://intusoft.com/nlpics/72/figure12.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8936306,"math_prob":0.96566886,"size":23304,"snap":"2020-10-2020-16","text_gpt3_token_len":5027,"char_repetition_ratio":0.13995709,"word_repetition_ratio":0.0041841003,"special_character_ratio":0.20112427,"punctuation_ratio":0.09908046,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98240775,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,null,null,null,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-16T21:29:18Z\",\"WARC-Record-ID\":\"<urn:uuid:7e57e5ea-f7c7-43ef-b380-8396dbf98ccf>\",\"Content-Length\":\"50172\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1928f2c-ac21-49c7-936a-623aeb1f0c00>\",\"WARC-Concurrent-To\":\"<urn:uuid:e192dbb8-ce7f-424c-a045-c636c99d4506>\",\"WARC-IP-Address\":\"192.237.165.239\",\"WARC-Target-URI\":\"http://intusoft.com/nlhtm/nl72.htm\",\"WARC-Payload-Digest\":\"sha1:GOFQPIOMWCP7I7Z5TQAEGF2RMDVZWNCW\",\"WARC-Block-Digest\":\"sha1:G544ARUMJT4CU3DK7KLWSCIWSSZMFIGL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875141430.58_warc_CC-MAIN-20200216211424-20200217001424-00019.warc.gz\"}"}