📄 kiss.out
字号:
r=6 77048 77311.8 0.900 4.619 chi-square = 4.619 with df = 2; p-value = 0.099 -------------------------------------------------------------- bits 17 to 24 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 910 944.3 1.246 1.246 r=5 21817 21743.9 0.246 1.492 r=6 77273 77311.8 0.019 1.511 chi-square = 1.511 with df = 2; p-value = 0.470 -------------------------------------------------------------- bits 18 to 25 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 964 944.3 0.411 0.411 r=5 21683 21743.9 0.171 0.582 r=6 77353 77311.8 0.022 0.604 chi-square = 0.604 with df = 2; p-value = 0.740 -------------------------------------------------------------- bits 19 to 26 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 923 944.3 0.480 0.480 r=5 21521 21743.9 2.285 2.765 r=6 77556 77311.8 0.771 3.537 chi-square = 3.537 with df = 2; p-value = 0.171 -------------------------------------------------------------- bits 20 to 27 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 960 944.3 0.261 0.261 r=5 21616 21743.9 0.752 1.013 r=6 77424 77311.8 0.163 1.176 chi-square = 1.176 with df = 2; p-value = 0.555 -------------------------------------------------------------- bits 21 to 28 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 947 944.3 0.008 0.008 r=5 21665 21743.9 0.286 0.294 r=6 77388 77311.8 0.075 0.369 chi-square = 0.369 with df = 2; p-value = 0.831 -------------------------------------------------------------- bits 22 to 29 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 921 944.3 0.575 0.575 r=5 21678 21743.9 0.200 0.775 r=6 77401 77311.8 0.103 0.878 chi-square = 0.878 with df = 2; p-value = 0.645 -------------------------------------------------------------- bits 23 to 30 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 906 944.3 1.553 1.553 r=5 21807 21743.9 0.183 1.737 r=6 77287 77311.8 0.008 1.744 chi-square = 1.744 with df = 2; p-value = 0.418 -------------------------------------------------------------- bits 24 to 31 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 875 944.3 5.086 5.086 r=5 21445 21743.9 4.109 9.195 r=6 77680 77311.8 1.754 10.948 chi-square = 10.948 with df = 2; p-value = 0.004 -------------------------------------------------------------- bits 25 to 32 RANK OBSERVED EXPECTED (O-E)^2/E SUM r<=4 905 944.3 1.636 1.636 r=5 21645 21743.9 0.450 2.085 r=6 77450 77311.8 0.247 2.332 chi-square = 2.332 with df = 2; p-value = 0.312 -------------------------------------------------------------- TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variates: 0.101382 0.014654 0.387393 0.081280 0.455949 0.736504 0.564019 0.657683 0.488121 0.956049 0.976818 0.111338 0.228562 0.618236 0.551158 0.099323 0.469750 0.739521 0.170608 0.555386 0.831470 0.644824 0.418013 0.004194 0.311537 The KS test for those 25 supposed UNI's yields KS p-value = 0.489549 |-------------------------------------------------------------| | THE BITSTREAM TEST | |The file under test is viewed as a stream of bits. Call them | |b1,b2,... . Consider an alphabet with two "letters", 0 and 1| |and think of the stream of bits as a succession of 20-letter | |"words", overlapping. Thus the first word is b1b2...b20, the| |second is b2b3...b21, and so on. The bitstream test counts | |the number of missing 20-letter (20-bit) words in a string of| |2^21 overlapping 20-letter words. There are 2^20 possible 20| |letter words. For a truly random string of 2^21+19 bits, the| |number of missing words j should be (very close to) normally | |distributed with mean 141,909 and sigma 428. Thus | | (j-141909)/428 should be a standard normal variate (z score)| |that leads to a uniform [0,1) p value. The test is repeated | |twenty times. | |-------------------------------------------------------------| THE OVERLAPPING 20-TUPLES BITSTREAM TEST for kiss.32 (20 bits/word, 2097152 words 20 bitstreams. No. missing words should average 141909.33 with sigma=428.00.) ---------------------------------------------------------------- BITSTREAM test results for kiss.32. Bitstream No. missing words z-score p-value 1 142021 0.26 0.397080 2 141511 -0.93 0.823990 3 141928 0.04 0.482603 4 142357 1.05 0.147790 5 141199 -1.66 0.951508 6 141894 -0.04 0.514286 7 142279 0.86 0.193872 8 141488 -0.98 0.837544 9 141864 -0.11 0.542174 10 141879 -0.07 0.528247 11 142351 1.03 0.151050 12 141600 -0.72 0.765078 13 142388 1.12 0.131701 14 142755 1.98 0.024085 15 141585 -0.76 0.775709 16 142181 0.63 0.262798 17 141703 -0.48 0.685125 18 141505 -0.94 0.827593 19 141867 -0.10 0.539392 20 141233 -1.58 0.942971 ---------------------------------------------------------------- |-------------------------------------------------------------| | OPSO means Overlapping-Pairs-Sparse-Occupancy | |The OPSO test considers 2-letter words from an alphabet of | |1024 letters. Each letter is determined by a specified ten | |bits from a 32-bit integer in the sequence to be tested. OPSO| |generates 2^21 (overlapping) 2-letter words (from 2^21+1 | |"keystrokes") and counts the number of missing words---that | |is 2-letter words which do not appear in the entire sequence.| |That count should be very close to normally distributed with | |mean 141,909, sigma 290. Thus (missingwrds-141909)/290 should| |be a standard normal variable. The OPSO test takes 32 bits at| |a time from the test file and uses a designated set of ten | |consecutive bits. It then restarts the file for the next de- | |signated 10 bits, and so on. | |------------------------------------------------------------ | OPSO test for file kiss.32 Bits used No. missing words z-score p-value 23 to 32 142347 1.5092 0.065623 22 to 31 142167 0.8885 0.187131 21 to 30 142216 1.0575 0.145146 20 to 29 141554 -1.2253 0.889764 19 to 28 141714 -0.6736 0.749702 18 to 27 141849 -0.2080 0.582399 17 to 26 141945 0.1230 0.451053 16 to 25 141561 -1.2011 0.885151 15 to 24 141540 -1.2736 0.898589 14 to 23 142495 2.0196 0.021715 13 to 22 142211 1.0402 0.149114 12 to 21 141891 -0.0632 0.525199 11 to 20 142050 0.4851 0.313814 10 to 19 142234 1.1196 0.131452 9 to 18 142198 0.9954 0.159767 8 to 17 142386 1.6437 0.050120 7 to 16 141407 -1.7322 0.958379 6 to 15 141815 -0.3253 0.627514 5 to 14 141349 -1.9322 0.973331 4 to 13 142062 0.5264 0.299288 3 to 12 141779 -0.4494 0.673433 2 to 11 142258 1.2023 0.114622 1 to 10 141886 -0.0804 0.532060 ----------------------------------------------------------------- |------------------------------------------------------------ | | OQSO means Overlapping-Quadruples-Sparse-Occupancy | | The test OQSO is similar, except that it considers 4-letter| |words from an alphabet of 32 letters, each letter determined | |by a designated string of 5 consecutive bits from the test | |file, elements of which are assumed 32-bit random integers. | |The mean number of missing words in a sequence of 2^21 four- | |letter words, (2^21+3 "keystrokes"), is again 141909, with | |sigma = 295. The mean is based on theory; sigma comes from | |extensive simulation. | |------------------------------------------------------------ | OQSO test for file kiss.32 Bits used No. missing words z-score p-value 28 to 32 142130 0.7480 0.227220 27 to 31 141728 -0.6147 0.730616 26 to 30 142201 0.9887 0.161402 25 to 29 141944 0.1175 0.453222 24 to 28 142036 0.4294 0.333820 23 to 27 142191 0.9548 0.169836 22 to 26 142269 1.2192 0.111380 21 to 25 142139 0.7785 0.218125 20 to 24 141985 0.2565 0.398779 19 to 23 141656 -0.8587 0.804760 18 to 22 142345 1.4768 0.069858 17 to 21 141705 -0.6926 0.755733 16 to 20 142035 0.4260 0.335054 15 to 19 141747 -0.5503 0.708933 14 to 18 141801 -0.3672 0.643273 13 to 17 142143 0.7921 0.214151 12 to 16 141632 -0.9401 0.826417 11 to 15 142222 1.0599 0.144595 10 to 14 141931 0.0735 0.470721 9 to 13 141655 -0.8621 0.805694 8 to 12 142335 1.4429 0.074517 7 to 11 142287 1.2802 0.100231 6 to 10 141418 -1.6655 0.952096 5 to 9 141776 -0.4520 0.674353 4 to 8 142705 2.6972 0.003496 3 to 7 142123 0.7243 0.234439 2 to 6 141943 0.1141 0.454565 1 to 5 141793 -0.3943 0.653335 ----------------------------------------------------------------- |------------------------------------------------------------ | | The DNA test considers an alphabet of 4 letters: C,G,A,T,| |determined by two designated bits in the sequence of random | |integers being tested. It considers 10-letter words, so that| |as in OPSO and OQSO, there are 2^20 possible words, and the | |mean number of missing words from a string of 2^21 (over- | |lapping) 10-letter words (2^21+9 "keystrokes") is 141909. | |The standard deviation sigma=339 was determined as for OQSO | |by simulation. (Sigma for OPSO, 290, is the true value (to | |three places), not determined by simulation. | |------------------------------------------------------------ | DNA test for file kiss.32 Bits used No. missing words z-score p-value 31 to 32 141360 -1.6204 0.947431 30 to 31 141725 -0.5437 0.706692 29 to 30 142363 1.3383 0.090406 28 to 29 142059 0.4415 0.329424 27 to 28 141591 -0.9390 0.826141 26 to 27 141979 0.2055 0.418584 25 to 26 141941 0.0934 0.462784 24 to 25 142561 1.9223 0.027282 23 to 24 142089 0.5300 0.298056 22 to 23 141811 -0.2901 0.614114 21 to 22 141635 -0.8092 0.790809 20 to 21 141844 -0.1927 0.576408 19 to 20 141838 -0.2104 0.583327 18 to 19 142043 0.3943 0.346677 17 to 18 141965 0.1642 0.434780 16 to 17 142265 1.0492 0.147049 15 to 16 142192 0.8338 0.202187 14 to 15 141683 -0.6676 0.747818 13 to 14 142092 0.5388 0.294995 12 to 13 142582 1.9843 0.023612 11 to 12 142434 1.5477 0.060847 10 to 11 141887 -0.0659 0.526259 9 to 10 141593 -0.9331 0.824623 8 to 9 141763 -0.4317 0.667003 7 to 8 141691 -0.6440 0.740226 6 to 7 141992 0.2439 0.403668 5 to 6 142153 0.7188 0.236135 4 to 5 141575 -0.9862 0.837988 3 to 4 142069 0.4710 0.318819 2 to 3 142127 0.6421 0.260406 1 to 2 142371 1.3619 0.086621 ----------------------------------------------------------------- |-------------------------------------------------------------| | This is the COUNT-THE-1''s TEST on a stream of bytes. | |Consider the file under test as a stream of bytes (four per | |32 bit integer). Each byte can contain from 0 to 8 1''s, | |with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let | |the stream of bytes provide a string of overlapping 5-letter| |words, each "letter" taking values A,B,C,D,E. The letters are| |determined by the number of 1''s in a byte: 0,1,or 2 yield A,| |3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus| |we have a monkey at a typewriter hitting five keys with vari-| |ous probabilities (37,56,70,56,37 over 256). There are 5^5 | |possible 5-letter words, and from a string of 256,000 (over- | |lapping) 5-letter words, counts are made on the frequencies | |for each word. The quadratic form in the weak inverse of | |the covariance matrix of the cell counts provides a chisquare| |test: Q5-Q4, the difference of the naive Pearson sums of |⌨️ 快捷键说明
复制代码
Ctrl + C
搜索代码
Ctrl + F
全屏模式
F11
切换主题
Ctrl + Shift + D
显示快捷键
?
增大字号
Ctrl + =
减小字号
Ctrl + -