NOTE: Most of the tests in DIEHARD return a p-value, which should be uniform on [0,1) if the input file contains truly independent random bits. Those p-values are obtained by p=F(X), where F is the assumed distribution of the sample random variable X---often normal. But that assumed F is just an asymptotic approximation, for which the fit will be worst in the tails. Thus you should not be surprised with occasional p-values near 0 or 1, such as .0012 or .9983. When a bit stream really FAILS BIG, you will get p's of 0 or 1 to six or more places. By all means, do not, as a Statistician might, think that a p < .025 or p> .975 means that the RNG has "failed the test at the .05 level". Such p's happen among the hundreds that DIEHARD produces, even with good RNG's. So keep in mind that " p happens". ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BIRTHDAY SPACINGS TEST :: :: Choose m birthdays in a year of n days. List the spacings :: :: between the birthdays. If j is the number of values that :: :: occur more than once in that list, then j is asymptotically :: :: Poisson distributed with mean m^3/(4n). Experience shows n :: :: must be quite large, say n>=2^18, for comparing the results :: :: to the Poisson distribution with that mean. This test uses :: :: n=2^24 and m=2^9, so that the underlying distribution for j :: :: is taken to be Poisson with lambda=2^27/(2^26)=2. A sample :: :: of 500 j's is taken, and a chi-square goodness of fit test :: :: provides a p value. The first test uses bits 1-24 (counting :: :: from the left) from integers in the specified file. :: :: Then the file is closed and reopened. Next, bits 2-25 are :: :: used to provide birthdays, then 3-26 and so on to bits 9-32. :: :: Each set of bits provides a p-value, and the nine p-values :: :: provide a sample for a KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA= 2.0000 Results for TRUERBIN For a sample of size 500: mean TRUERBIN using bits 1 to 24 431.950 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 2 to 25 419.490 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 3 to 26 406.782 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 4 to 27 395.550 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 5 to 28 387.364 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 6 to 29 381.880 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 7 to 30 377.850 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 8 to 31 377.836 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean TRUERBIN using bits 9 to 32 377.832 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 A KSTEST for the 9 p-values yields 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE OVERLAPPING 5-PERMUTATION TEST :: :: This is the OPERM5 test. It looks at a sequence of one mill- :: :: ion 32-bit random integers. Each set of five consecutive :: :: integers can be in one of 120 states, for the 5! possible or- :: :: derings of five numbers. Thus the 5th, 6th, 7th,...numbers :: :: each provide a state. As many thousands of state transitions :: :: are observed, cumulative counts are made of the number of :: :: occurences of each state. Then the quadratic form in the :: :: weak inverse of the 120x120 covariance matrix yields a test :: :: equivalent to the likelihood ratio test that the 120 cell :: :: counts came from the specified (asymptotically) normal dis- :: :: tribution with the specified 120x120 covariance matrix (with :: :: rank 99). This version uses 1,000,000 integers, twice. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPERM5 test for file TRUERBIN For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=*******; p-value=1.000000 OPERM5 test for file TRUERBIN For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=*******; p-value=1.000000 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 31x31 matrices. The leftmost :: :: 31 bits of 31 random integers from the test sequence are used :: :: to form a 31x31 binary matrix over the field {0,1}. The rank :: :: is determined. That rank can be from 0 to 31, but ranks< 28 :: :: are rare, and their counts are pooled with those for rank 28. :: :: Ranks are found for 40,000 such random matrices and a chisqua-:: :: re test is performed on counts for ranks 31,30,29 and <=28. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for TRUERBIN Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 40000 211.4******************* 29 0 5134.0******************* 30 0 23103.0******************* 31 0 11551.5******************* chisquare=****** for 3 d. of f.; p-value=1.000000 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 32x32 matrices. A random 32x :: :: 32 binary matrix is formed, each row a 32-bit random integer. :: :: The rank is determined. That rank can be from 0 to 32, ranks :: :: less than 29 are rare, and their counts are pooled with those :: :: for rank 29. Ranks are found for 40,000 such random matrices :: :: and a chisquare test is performed on counts for ranks 32,31, :: :: 30 and <=29. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for TRUERBIN Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 40000 211.4******************* 30 0 5134.0******************* 31 0 23103.0******************* 32 0 11551.5******************* chisquare=****** for 3 d. of f.; p-value=1.000000 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 6x8 matrices. From each of :: :: six random 32-bit integers from the generator under test, a :: :: specified byte is chosen, and the resulting six bytes form a :: :: 6x8 binary matrix whose rank is determined. That rank can be :: :: from 0 to 6, but ranks 0,1,2,3 are rare; their counts are :: :: pooled with those for rank 4. Ranks are found for 100,000 :: :: random matrices, and a chi-square test is performed on :: :: counts for ranks 6,5 and <=4. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary Rank Test for TRUERBIN Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 99886 944.310366880.00010366880.000 r =5 114 21743.9 21516.50010388400.000 r =6 0 77311.8 77311.80010465710.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 90012 944.3 8400979.000 8400979.000 r =5 9977 21743.9 6367.760 8407347.000 r =6 11 77311.8 77289.800 8484637.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 73815 944.3 5623353.000 5623353.000 r =5 26064 21743.9 858.322 5624211.000 r =6 121 77311.8 77069.990 5701281.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 62746 944.3 4044737.000 4044737.000 r =5 33455 21743.9 6307.509 4051044.000 r =6 3799 77311.8 69900.480 4120945.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 25146 944.3 620270.500 620270.500 r =5 54702 21743.9 49955.910 670226.400 r =6 20152 77311.8 42260.600 712487.100 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 10567 944.3 98058.040 98058.040 r =5 48778 21743.9 33611.380 131669.400 r =6 40655 77311.8 17380.540 149050.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1933 944.3 1035.184 1035.184 r =5 35837 21743.9 9134.307 10169.490 r =6 62230 77311.8 2942.123 13111.610 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 886 944.3 3.600 3.600 r =5 31027 21743.9 3963.224 3966.823 r =6 68087 77311.8 1100.699 5067.522 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 565 944.3 152.355 152.355 r =5 25506 21743.9 650.913 803.269 r =6 73929 77311.8 148.016 951.285 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 558 944.3 158.031 158.031 r =5 24771 21743.9 421.421 579.452 r =6 74671 77311.8 90.204 669.656 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 482 944.3 226.329 226.329 r =5 24449 21743.9 336.534 562.863 r =6 75069 77311.8 65.063 627.926 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 320 944.3 412.741 412.741 r =5 22511 21743.9 27.062 439.804 r =6 77169 77311.8 .264 440.067 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 260 944.3 495.889 495.889 r =5 13247 21743.9 3320.348 3816.237 r =6 86493 77311.8 1090.317 4906.553 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 734 944.3 46.835 46.835 r =5 21366 21743.9 6.568 53.403 r =6 77900 77311.8 4.475 57.878 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 839 944.3 11.742 11.742 r =5 16253 21743.9 1386.595 1398.338 r =6 82908 77311.8 405.079 1803.417 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 956 944.3 .145 .145 r =5 15889 21743.9 1576.528 1576.672 r =6 83155 77311.8 441.626 2018.299 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 424 944.3 286.681 286.681 r =5 8103 21743.9 8557.534 8844.215 r =6 91473 77311.8 2593.905 11438.120 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1069 944.3 16.467 16.467 r =5 42948 21743.9 20677.700 20694.170 r =6 55983 77311.8 5884.197 26578.360 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG TRUERBIN b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 19389 944.3 360273.700 360273.700 r =5 67417 21743.9 95936.420 456210.100 r =6 13194 77311.8 53175.490 509385.600 p=1-exp(-SUM/2)=1.00000 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 brank test summary for TRUERBIN The KS test for those 25 supposed UNI's yields KS p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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, 20 BITS PER WORD, N words This test uses N=2^21 and samples the bitstream 20 times. No. missing words should average 141909. with sigma=428. --------------------------------------------------------- tst no 1: 950192 missing words, 1888.51 sigmas from mean, p-value=1.00000 tst no 2: 950118 missing words, 1888.34 sigmas from mean, p-value=1.00000 tst no 3: 950121 missing words, 1888.34 sigmas from mean, p-value=1.00000 tst no 4: 950594 missing words, 1889.45 sigmas from mean, p-value=1.00000 tst no 5: 949833 missing words, 1887.67 sigmas from mean, p-value=1.00000 tst no 6: 951014 missing words, 1890.43 sigmas from mean, p-value=1.00000 tst no 7: 950696 missing words, 1889.69 sigmas from mean, p-value=1.00000 tst no 8: 950357 missing words, 1888.90 sigmas from mean, p-value=1.00000 tst no 9: 951193 missing words, 1890.85 sigmas from mean, p-value=1.00000 tst no 10: 950361 missing words, 1888.91 sigmas from mean, p-value=1.00000 tst no 11: 950304 missing words, 1888.77 sigmas from mean, p-value=1.00000 tst no 12: 950589 missing words, 1889.44 sigmas from mean, p-value=1.00000 tst no 13: 950487 missing words, 1889.20 sigmas from mean, p-value=1.00000 tst no 14: 951253 missing words, 1890.99 sigmas from mean, p-value=1.00000 tst no 15: 950100 missing words, 1888.30 sigmas from mean, p-value=1.00000 tst no 16: 950603 missing words, 1889.47 sigmas from mean, p-value=1.00000 tst no 17: 950417 missing words, 1889.04 sigmas from mean, p-value=1.00000 tst no 18: 949747 missing words, 1887.47 sigmas from mean, p-value=1.00000 tst no 19: 949771 missing words, 1887.53 sigmas from mean, p-value=1.00000 tst no 20: 950207 missing words, 1888.55 sigmas from mean, p-value=1.00000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The tests OPSO, OQSO and DNA :: :: 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. :: :: :: :: 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. :: :: :: :: 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. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPSO test for generator TRUERBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for TRUERBIN using bits 23 to 32 1013594******* 1.0000 OPSO for TRUERBIN using bits 22 to 31 994156******* 1.0000 OPSO for TRUERBIN using bits 21 to 30 979589******* 1.0000 OPSO for TRUERBIN using bits 20 to 29 980217******* 1.0000 OPSO for TRUERBIN using bits 19 to 28 980039******* 1.0000 OPSO for TRUERBIN using bits 18 to 27 982663******* 1.0000 OPSO for TRUERBIN using bits 17 to 26 987734******* 1.0000 OPSO for TRUERBIN using bits 16 to 25 987336******* 1.0000 OPSO for TRUERBIN using bits 15 to 24 989282******* 1.0000 OPSO for TRUERBIN using bits 14 to 23 985598******* 1.0000 OPSO for TRUERBIN using bits 13 to 22 975638******* 1.0000 OPSO for TRUERBIN using bits 12 to 21 964422******* 1.0000 OPSO for TRUERBIN using bits 11 to 20 952437******* 1.0000 OPSO for TRUERBIN using bits 10 to 19 981384******* 1.0000 OPSO for TRUERBIN using bits 9 to 18 1009827******* 1.0000 OPSO for TRUERBIN using bits 8 to 17 1026205******* 1.0000 OPSO for TRUERBIN using bits 7 to 16 1039575******* 1.0000 OPSO for TRUERBIN using bits 6 to 15 1044639******* 1.0000 OPSO for TRUERBIN using bits 5 to 14 1046619******* 1.0000 OPSO for TRUERBIN using bits 4 to 13 1047405******* 1.0000 OPSO for TRUERBIN using bits 3 to 12 1047790******* 1.0000 OPSO for TRUERBIN using bits 2 to 11 1048031******* 1.0000 OPSO for TRUERBIN using bits 1 to 10 1048208******* 1.0000 OQSO test for generator TRUERBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for TRUERBIN using bits 28 to 32 1041022******* 1.0000 OQSO for TRUERBIN using bits 27 to 31 1006903******* 1.0000 OQSO for TRUERBIN using bits 26 to 30 928408******* 1.0000 OQSO for TRUERBIN using bits 25 to 29 950562******* 1.0000 OQSO for TRUERBIN using bits 24 to 28 958891******* 1.0000 OQSO for TRUERBIN using bits 23 to 27 964199******* 1.0000 OQSO for TRUERBIN using bits 22 to 26 981131******* 1.0000 OQSO for TRUERBIN using bits 21 to 25 991352******* 1.0000 OQSO for TRUERBIN using bits 20 to 24 996691******* 1.0000 OQSO for TRUERBIN using bits 19 to 23 999580******* 1.0000 OQSO for TRUERBIN using bits 18 to 22 1002324******* 1.0000 OQSO for TRUERBIN using bits 17 to 21 1003578******* 1.0000 OQSO for TRUERBIN using bits 16 to 20 993426******* 1.0000 OQSO for TRUERBIN using bits 15 to 19 982282******* 1.0000 OQSO for TRUERBIN using bits 14 to 18 959326******* 1.0000 OQSO for TRUERBIN using bits 13 to 17 919022******* 1.0000 OQSO for TRUERBIN using bits 12 to 16 930972******* 1.0000 OQSO for TRUERBIN using bits 11 to 15 955326******* 1.0000 OQSO for TRUERBIN using bits 10 to 14 1008911******* 1.0000 OQSO for TRUERBIN using bits 9 to 13 1038293******* 1.0000 OQSO for TRUERBIN using bits 8 to 12 1046069******* 1.0000 OQSO for TRUERBIN using bits 7 to 11 1047843******* 1.0000 OQSO for TRUERBIN using bits 6 to 10 1048296******* 1.0000 OQSO for TRUERBIN using bits 5 to 9 1048431******* 1.0000 OQSO for TRUERBIN using bits 4 to 8 1048444******* 1.0000 OQSO for TRUERBIN using bits 3 to 7 1048444******* 1.0000 OQSO for TRUERBIN using bits 2 to 6 1048453******* 1.0000 OQSO for TRUERBIN using bits 1 to 5 1048513******* 1.0000 DNA test for generator TRUERBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for TRUERBIN using bits 31 to 32 1048537******* 1.0000 DNA for TRUERBIN using bits 30 to 31 1043876******* 1.0000 DNA for TRUERBIN using bits 29 to 30 825404******* 1.0000 DNA for TRUERBIN using bits 28 to 29 820060******* 1.0000 DNA for TRUERBIN using bits 27 to 28 926639******* 1.0000 DNA for TRUERBIN using bits 26 to 27 871999******* 1.0000 DNA for TRUERBIN using bits 25 to 26 856020******* 1.0000 DNA for TRUERBIN using bits 24 to 25 965890******* 1.0000 DNA for TRUERBIN using bits 23 to 24 1003026******* 1.0000 DNA for TRUERBIN using bits 22 to 23 1016047******* 1.0000 DNA for TRUERBIN using bits 21 to 22 1018693******* 1.0000 DNA for TRUERBIN using bits 20 to 21 1022023******* 1.0000 DNA for TRUERBIN using bits 19 to 20 1023803******* 1.0000 DNA for TRUERBIN using bits 18 to 19 1023666******* 1.0000 DNA for TRUERBIN using bits 17 to 18 1026432******* 1.0000 DNA for TRUERBIN using bits 16 to 17 960965******* 1.0000 DNA for TRUERBIN using bits 15 to 16 937073******* 1.0000 DNA for TRUERBIN using bits 14 to 15 939180******* 1.0000 DNA for TRUERBIN using bits 13 to 14 944705******* 1.0000 DNA for TRUERBIN using bits 12 to 13 960386******* 1.0000 DNA for TRUERBIN using bits 11 to 12 1004972******* 1.0000 DNA for TRUERBIN using bits 10 to 11 1043661******* 1.0000 DNA for TRUERBIN using bits 9 to 10 1048190******* 1.0000 DNA for TRUERBIN using bits 8 to 9 1048509******* 1.0000 DNA for TRUERBIN using bits 7 to 8 1048530******* 1.0000 DNA for TRUERBIN using bits 6 to 7 1048537******* 1.0000 DNA for TRUERBIN using bits 5 to 6 1048521******* 1.0000 DNA for TRUERBIN using bits 4 to 5 1048529******* 1.0000 DNA for TRUERBIN using bits 3 to 4 1048537******* 1.0000 DNA for TRUERBIN using bits 2 to 3 1048545******* 1.0000 DNA for TRUERBIN using bits 1 to 2 1048565******* 1.0000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for TRUERBIN Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for TRUERBIN *********1698628.000 1.000000 byte stream for TRUERBIN *********1734969.000 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST for specific bytes. :: :: Consider the file under test as a stream of 32-bit integers. :: :: From each integer, a specific byte is chosen , say the left- :: :: most:: bits 1 to 8. Each byte can contain from 0 to 8 1's, :: :: with probabilitie 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the specified bytes from successive integers 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 that byte:: 0,1,or 2 ---> A, 3 ---> B, 4 ---> C, 5 ---> D,:: :: and 6,7 or 8 ---> E. Thus we have a monkey at a typewriter :: :: hitting five keys with with various probabilities:: 37,56,70,:: :: 56,37 over 256. There are 5^5 possible 5-letter words, and :: :: from a string of 256,000 (overlapping) 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 (OBS-EXP)^2/EXP on counts for 5- :: :: and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Chi-square with 5^5-5^4=2500 d.of f. for sample size: 256000 chisquare equiv normal p value Results for COUNT-THE-1's in specified bytes: bits 1 to 8******************** 1.000000 bits 2 to 9******************** 1.000000 bits 3 to 10******************** 1.000000 bits 4 to 11******************** 1.000000 bits 5 to 12******************** 1.000000 bits 6 to 13******************** 1.000000 bits 7 to 14******************** 1.000000 bits 8 to 15*********5221132.000 1.000000 bits 9 to 16********* 940554.100 1.000000 bits 10 to 17********* 343562.600 1.000000 bits 11 to 18********* 264956.200 1.000000 bits 12 to 19********* 225216.900 1.000000 bits 13 to 20********* 185480.700 1.000000 bits 14 to 21********* 158024.800 1.000000 bits 15 to 22********* 131707.900 1.000000 bits 16 to 23********* 154073.900 1.000000 bits 17 to 24********* 165081.200 1.000000 bits 18 to 25********* 149311.500 1.000000 bits 19 to 26********* 174810.400 1.000000 bits 20 to 27********* 139874.800 1.000000 bits 21 to 28********* 168589.300 1.000000 bits 22 to 29********* 206564.000 1.000000 bits 23 to 30********* 186918.200 1.000000 bits 24 to 31********* 166791.300 1.000000 bits 25 to 32********* 206615.400 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THIS IS A PARKING LOT TEST :: :: In a square of side 100, randomly "park" a car---a circle of :: :: radius 1. Then try to park a 2nd, a 3rd, and so on, each :: :: time parking "by ear". That is, if an attempt to park a car :: :: causes a crash with one already parked, try again at a new :: :: random location. (To avoid path problems, consider parking :: :: helicopters rather than cars.) Each attempt leads to either :: :: a crash or a success, the latter followed by an increment to :: :: the list of cars already parked. If we plot n: the number of :: :: attempts, versus k:: the number successfully parked, we get a:: :: curve that should be similar to those provided by a perfect :: :: random number generator. Theory for the behavior of such a :: :: random curve seems beyond reach, and as graphics displays are :: :: not available for this battery of tests, a simple characteriz :: :: ation of the random experiment is used: k, the number of cars :: :: successfully parked after n=12,000 attempts. Simulation shows :: :: that k should average 3523 with sigma 21.9 and is very close :: :: to normally distributed. Thus (k-3523)/21.9 should be a st- :: :: andard normal variable, which, converted to a uniform varia- :: :: ble, provides input to a KSTEST based on a sample of 10. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: CDPARK: result of ten tests on file TRUERBIN Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 2 z-score:******* p-value: .000000 Successes: 5 z-score:******* p-value: .000000 Successes: 4 z-score:******* p-value: .000000 Successes: 2 z-score:******* p-value: .000000 Successes: 3 z-score:******* p-value: .000000 Successes: 3 z-score:******* p-value: .000000 Successes: 3 z-score:******* p-value: .000000 Successes: 3 z-score:******* p-value: .000000 Successes: 4 z-score:******* p-value: .000000 Successes: 3 z-score:******* p-value: .000000 square size avg. no. parked sample sigma 100. 3.200 .872 KSTEST for the above 10: p= 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file TRUERBIN Sample no. d^2 avg equiv uni 5 .0000 .0000 .000000 10 .0000 .0000 .000000 15 .0000 .0000 .000000 20 .0000 .0000 .000000 25 .0000 .0000 .000000 30 .0000 .0000 .000000 35 .0000 .0000 .000000 40 .0000 .0000 .000000 45 .0000 .0000 .000000 50 .0000 .0000 .000000 55 .0000 .0000 .000000 60 .0000 .0000 .000000 65 .0000 .0000 .000000 70 .0000 .0000 .000000 75 .0000 .0000 .000000 80 .0000 .0000 .000000 85 .0000 .0000 .000000 90 .0000 .0000 .000000 95 .0000 .0000 .000000 100 .0000 .0000 .000000 MINIMUM DISTANCE TEST for TRUERBIN Result of KS test on 20 transformed mindist^2's: p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file TRUERBIN sample no: 1 r^3= .000 p-value= .00000 sample no: 2 r^3= .000 p-value= .00000 sample no: 3 r^3= .000 p-value= .00000 sample no: 4 r^3= .000 p-value= .00000 sample no: 5 r^3= .000 p-value= .00000 sample no: 6 r^3= .000 p-value= .00000 sample no: 7 r^3= .000 p-value= .00000 sample no: 8 r^3= .000 p-value= .00000 sample no: 9 r^3= .000 p-value= .00000 sample no: 10 r^3= .000 p-value= .00000 sample no: 11 r^3= .000 p-value= .00000 sample no: 12 r^3= .000 p-value= .00000 sample no: 13 r^3= .000 p-value= .00000 sample no: 14 r^3= .000 p-value= .00000 sample no: 15 r^3= .000 p-value= .00000 sample no: 16 r^3= .000 p-value= .00000 sample no: 17 r^3= .000 p-value= .00000 sample no: 18 r^3= .000 p-value= .00000 sample no: 19 r^3= .000 p-value= .00000 sample no: 20 r^3= .000 p-value= .00000 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file TRUERBIN p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::