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 PRNBIN For a sample of size 500: mean PRNBIN using bits 1 to 24 499.522 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 PRNBIN using bits 2 to 25 493.788 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 PRNBIN using bits 3 to 26 485.536 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 PRNBIN using bits 4 to 27 473.066 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 PRNBIN using bits 5 to 28 454.580 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 PRNBIN using bits 6 to 29 426.624 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 PRNBIN using bits 7 to 30 385.332 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 PRNBIN using bits 8 to 31 327.000 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 PRNBIN using bits 9 to 32 251.656 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 PRNBIN For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 84.524; p-value= .150097 OPERM5 test for file PRNBIN For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 93.447; p-value= .361347 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 PRNBIN 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 PRNBIN 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 PRNBIN Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN 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 PRNBIN 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 PRNBIN 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 PRNBIN 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 PRNBIN 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 PRNBIN 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 PRNBIN b-rank test for bits 7 to 14 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 PRNBIN b-rank test for bits 8 to 15 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 PRNBIN b-rank test for bits 9 to 16 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 PRNBIN b-rank test for bits 10 to 17 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 PRNBIN b-rank test for bits 11 to 18 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 PRNBIN b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 99999 944.310390580.00010390580.000 r =5 1 21743.9 21741.90010412320.000 r =6 0 77311.8 77311.80010489630.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 88522 944.3 8122253.000 8122253.000 r =5 11478 21743.9 4846.817 8127100.000 r =6 0 77311.8 77311.800 8204412.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 42536 944.3 1831904.000 1831904.000 r =5 51671 21743.9 41190.000 1873094.000 r =6 5793 77311.8 66159.880 1939254.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 14643 944.3 198723.000 198723.000 r =5 53562 21743.9 46559.790 245282.800 r =6 31795 77311.8 26797.710 272080.500 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 4196 944.3 11197.210 11197.210 r =5 37442 21743.9 11333.310 22530.520 r =6 58362 77311.8 4644.764 27175.290 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1441 944.3 261.262 261.262 r =5 24768 21743.9 420.586 681.847 r =6 73791 77311.8 160.339 842.186 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1131 944.3 36.912 36.912 r =5 22639 21743.9 36.847 73.760 r =6 76230 77311.8 15.137 88.897 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1012 944.3 4.853 4.853 r =5 22234 21743.9 11.047 15.900 r =6 76754 77311.8 4.025 19.925 p=1-exp(-SUM/2)= .99995 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 992 944.3 2.409 2.409 r =5 21961 21743.9 2.168 4.577 r =6 77047 77311.8 .907 5.484 p=1-exp(-SUM/2)= .93556 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 992 944.3 2.409 2.409 r =5 21954 21743.9 2.030 4.439 r =6 77054 77311.8 .860 5.299 p=1-exp(-SUM/2)= .92932 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 992 944.3 2.409 2.409 r =5 21688 21743.9 .144 2.553 r =6 77320 77311.8 .001 2.554 p=1-exp(-SUM/2)= .72112 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 963 944.3 .370 .370 r =5 21777 21743.9 .050 .421 r =6 77260 77311.8 .035 .455 p=1-exp(-SUM/2)= .20362 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 940 944.3 .020 .020 r =5 21734 21743.9 .005 .024 r =6 77326 77311.8 .003 .027 p=1-exp(-SUM/2)= .01326 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRNBIN b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 886 944.3 3.600 3.600 r =5 21776 21743.9 .047 3.647 r =6 77338 77311.8 .009 3.656 p=1-exp(-SUM/2)= .83925 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 .999953 .935558 .929318 .721120 .203623 .013264 .839248 brank test summary for PRNBIN 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: 901495 missing words, 1774.73 sigmas from mean, p-value=1.00000 tst no 2: 901635 missing words, 1775.06 sigmas from mean, p-value=1.00000 tst no 3: 901392 missing words, 1774.49 sigmas from mean, p-value=1.00000 tst no 4: 901438 missing words, 1774.60 sigmas from mean, p-value=1.00000 tst no 5: 901636 missing words, 1775.06 sigmas from mean, p-value=1.00000 tst no 6: 901441 missing words, 1774.61 sigmas from mean, p-value=1.00000 tst no 7: 901843 missing words, 1775.55 sigmas from mean, p-value=1.00000 tst no 8: 901466 missing words, 1774.67 sigmas from mean, p-value=1.00000 tst no 9: 900865 missing words, 1773.26 sigmas from mean, p-value=1.00000 tst no 10: 901795 missing words, 1775.43 sigmas from mean, p-value=1.00000 tst no 11: 901852 missing words, 1775.57 sigmas from mean, p-value=1.00000 tst no 12: 902010 missing words, 1775.94 sigmas from mean, p-value=1.00000 tst no 13: 901214 missing words, 1774.08 sigmas from mean, p-value=1.00000 tst no 14: 901908 missing words, 1775.70 sigmas from mean, p-value=1.00000 tst no 15: 900939 missing words, 1773.43 sigmas from mean, p-value=1.00000 tst no 16: 901651 missing words, 1775.10 sigmas from mean, p-value=1.00000 tst no 17: 902488 missing words, 1777.05 sigmas from mean, p-value=1.00000 tst no 18: 901880 missing words, 1775.63 sigmas from mean, p-value=1.00000 tst no 19: 900703 missing words, 1772.88 sigmas from mean, p-value=1.00000 tst no 20: 901796 missing words, 1775.44 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 PRNBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for PRNBIN using bits 23 to 32 142308 1.375 .9154 OPSO for PRNBIN using bits 22 to 31 142188 .961 .8317 OPSO for PRNBIN using bits 21 to 30 142432 1.802 .9643 OPSO for PRNBIN using bits 20 to 29 143607 5.854 1.0000 OPSO for PRNBIN using bits 19 to 28 150075 28.157 1.0000 OPSO for PRNBIN using bits 18 to 27 173316108.299 1.0000 OPSO for PRNBIN using bits 17 to 26 291020514.175 1.0000 OPSO for PRNBIN using bits 16 to 25 678929******* 1.0000 OPSO for PRNBIN using bits 15 to 24 920365******* 1.0000 OPSO for PRNBIN using bits 14 to 23 1007598******* 1.0000 OPSO for PRNBIN using bits 13 to 22 1036098******* 1.0000 OPSO for PRNBIN using bits 12 to 21 1044886******* 1.0000 OPSO for PRNBIN using bits 11 to 20 1047516******* 1.0000 OPSO for PRNBIN using bits 10 to 19 1048275******* 1.0000 OPSO for PRNBIN using bits 9 to 18 1048486******* 1.0000 OPSO for PRNBIN using bits 8 to 17 1048548******* 1.0000 OPSO for PRNBIN using bits 7 to 16 1048568******* 1.0000 OPSO for PRNBIN using bits 6 to 15 1048573******* 1.0000 OPSO for PRNBIN using bits 5 to 14 1048575******* 1.0000 OPSO for PRNBIN using bits 4 to 13 1048575******* 1.0000 OPSO for PRNBIN using bits 3 to 12 1048575******* 1.0000 OPSO for PRNBIN using bits 2 to 11 1048575******* 1.0000 OPSO for PRNBIN using bits 1 to 10 1048575******* 1.0000 OQSO test for generator PRNBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for PRNBIN using bits 28 to 32 141645 -.896 .1851 OQSO for PRNBIN using bits 27 to 31 141640 -.913 .1806 OQSO for PRNBIN using bits 26 to 30 141937 .094 .5374 OQSO for PRNBIN using bits 25 to 29 142055 .494 .6893 OQSO for PRNBIN using bits 24 to 28 141425 -1.642 .0503 OQSO for PRNBIN using bits 23 to 27 141655 -.862 .1943 OQSO for PRNBIN using bits 22 to 26 142189 .948 .8284 OQSO for PRNBIN using bits 21 to 25 143017 3.755 .9999 OQSO for PRNBIN using bits 20 to 24 145929 13.626 1.0000 OQSO for PRNBIN using bits 19 to 23 157791 53.836 1.0000 OQSO for PRNBIN using bits 18 to 22 201913203.402 1.0000 OQSO for PRNBIN using bits 17 to 21 393717853.585 1.0000 OQSO for PRNBIN using bits 16 to 20 863747******* 1.0000 OQSO for PRNBIN using bits 15 to 19 1023954******* 1.0000 OQSO for PRNBIN using bits 14 to 18 1045855******* 1.0000 OQSO for PRNBIN using bits 13 to 17 1048274******* 1.0000 OQSO for PRNBIN using bits 12 to 16 1048543******* 1.0000 OQSO for PRNBIN using bits 11 to 15 1048571******* 1.0000 OQSO for PRNBIN using bits 10 to 14 1048575******* 1.0000 OQSO for PRNBIN using bits 9 to 13 1048575******* 1.0000 OQSO for PRNBIN using bits 8 to 12 1048575******* 1.0000 OQSO for PRNBIN using bits 7 to 11 1048575******* 1.0000 OQSO for PRNBIN using bits 6 to 10 1048575******* 1.0000 OQSO for PRNBIN using bits 5 to 9 1048575******* 1.0000 OQSO for PRNBIN using bits 4 to 8 1048575******* 1.0000 OQSO for PRNBIN using bits 3 to 7 1048575******* 1.0000 OQSO for PRNBIN using bits 2 to 6 1048575******* 1.0000 OQSO for PRNBIN using bits 1 to 5 1048575******* 1.0000 DNA test for generator PRNBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for PRNBIN using bits 31 to 32 142201 .860 .8052 DNA for PRNBIN using bits 30 to 31 141928 .055 .5220 DNA for PRNBIN using bits 29 to 30 141900 -.028 .4890 DNA for PRNBIN using bits 28 to 29 141783 -.373 .3547 DNA for PRNBIN using bits 27 to 28 141748 -.476 .3171 DNA for PRNBIN using bits 26 to 27 141993 .247 .5975 DNA for PRNBIN using bits 25 to 26 142812 2.663 .9961 DNA for PRNBIN using bits 24 to 25 142101 .565 .7141 DNA for PRNBIN using bits 23 to 24 142022 .332 .6302 DNA for PRNBIN using bits 22 to 23 142274 1.076 .8590 DNA for PRNBIN using bits 21 to 22 143624 5.058 1.0000 DNA for PRNBIN using bits 20 to 21 151372 27.913 1.0000 DNA for PRNBIN using bits 19 to 20 176294101.430 1.0000 DNA for PRNBIN using bits 18 to 19 266997368.990 1.0000 DNA for PRNBIN using bits 17 to 18 577029******* 1.0000 DNA for PRNBIN using bits 16 to 17 1009197******* 1.0000 DNA for PRNBIN using bits 15 to 16 1047946******* 1.0000 DNA for PRNBIN using bits 14 to 15 1048565******* 1.0000 DNA for PRNBIN using bits 13 to 14 1048575******* 1.0000 DNA for PRNBIN using bits 12 to 13 1048575******* 1.0000 DNA for PRNBIN using bits 11 to 12 1048575******* 1.0000 DNA for PRNBIN using bits 10 to 11 1048575******* 1.0000 DNA for PRNBIN using bits 9 to 10 1048575******* 1.0000 DNA for PRNBIN using bits 8 to 9 1048575******* 1.0000 DNA for PRNBIN using bits 7 to 8 1048575******* 1.0000 DNA for PRNBIN using bits 6 to 7 1048575******* 1.0000 DNA for PRNBIN using bits 5 to 6 1048575******* 1.0000 DNA for PRNBIN using bits 4 to 5 1048575******* 1.0000 DNA for PRNBIN using bits 3 to 4 1048575******* 1.0000 DNA for PRNBIN using bits 2 to 3 1048575******* 1.0000 DNA for PRNBIN using bits 1 to 2 1048575******* 1.0000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for PRNBIN Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for PRNBIN *********1713035.000 1.000000 byte stream for PRNBIN *********1709880.000 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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******************** 1.000000 bits 9 to 16******************** 1.000000 bits 10 to 17******************** 1.000000 bits 11 to 18******************** 1.000000 bits 12 to 19******************** 1.000000 bits 13 to 20*********7779339.000 1.000000 bits 14 to 21********* 849214.000 1.000000 bits 15 to 22********* 67312.310 1.000000 bits 16 to 23369744.00 5193.614 1.000000 bits 17 to 24 34423.55 451.467 1.000000 bits 18 to 25 7946.24 77.021 1.000000 bits 19 to 26 3595.80 15.497 1.000000 bits 20 to 27 2831.52 4.688 .999999 bits 21 to 28 2690.37 2.692 .996451 bits 22 to 29 2639.75 1.976 .975943 bits 23 to 30 2454.03 -.650 .257817 bits 24 to 31 2456.00 -.622 .266881 bits 25 to 32 2688.54 2.666 .996166 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 PRNBIN Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 square size avg. no. parked sample sigma 100. 1.000 .000 KSTEST for the above 10: p= 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file PRNBIN Sample no. d^2 avg equiv uni 5 .0000 .0000 .000000 10 .0000 .0000 .000000 15 .0000 .0000 .000000 20 .0000 .0000 .000000 25 .0000 .0000 .000000 30 .0000 .0000 .000000 35 .0000 .0000 .000000 40 .0000 .0000 .000000 45 .0000 .0000 .000000 50 .0000 .0000 .000000 55 .0000 .0000 .000000 60 .0000 .0000 .000000 65 .0000 .0000 .000000 70 .0000 .0000 .000000 75 .0000 .0000 .000000 80 .0000 .0000 .000000 85 .0000 .0000 .000000 90 .0000 .0000 .000000 95 .0000 .0000 .000000 100 .0000 .0000 .000000 MINIMUM DISTANCE TEST for PRNBIN Result of KS test on 20 transformed mindist^2's: p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file PRNBIN sample no: 1 r^3= .000 p-value= .00000 sample no: 2 r^3= .000 p-value= .00000 sample no: 3 r^3= .000 p-value= .00000 sample no: 4 r^3= .000 p-value= .00000 sample no: 5 r^3= .000 p-value= .00000 sample no: 6 r^3= .000 p-value= .00000 sample no: 7 r^3= .000 p-value= .00000 sample no: 8 r^3= .000 p-value= .00000 sample no: 9 r^3= .000 p-value= .00000 sample no: 10 r^3= .000 p-value= .00000 sample no: 11 r^3= .000 p-value= .00000 sample no: 12 r^3= .000 p-value= .00000 sample no: 13 r^3= .000 p-value= .00000 sample no: 14 r^3= .000 p-value= .00000 sample no: 15 r^3= .000 p-value= .00000 sample no: 16 r^3= .000 p-value= .00000 sample no: 17 r^3= .000 p-value= .00000 sample no: 18 r^3= .000 p-value= .00000 sample no: 19 r^3= .000 p-value= .00000 sample no: 20 r^3= .000 p-value= .00000 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file PRNBIN p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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. :: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::