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 data.txt For a sample of size 500: mean data.txt using bits 1 to 24 1.964 duplicate number number spacings observed expected 0 70. 67.668 1 132. 135.335 2 142. 135.335 3 84. 90.224 4 52. 45.112 5 14. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 3.51 p-value= .256985 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 2 to 25 2.052 duplicate number number spacings observed expected 0 65. 67.668 1 128. 135.335 2 144. 135.335 3 87. 90.224 4 49. 45.112 5 14. 18.045 6 to INF 13. 8.282 Chisquare with 6 d.o.f. = 5.10 p-value= .469251 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 3 to 26 1.976 duplicate number number spacings observed expected 0 68. 67.668 1 132. 135.335 2 137. 135.335 3 97. 90.224 4 44. 45.112 5 18. 18.045 6 to INF 4. 8.282 Chisquare with 6 d.o.f. = 2.85 p-value= .173130 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 4 to 27 2.064 duplicate number number spacings observed expected 0 60. 67.668 1 129. 135.335 2 134. 135.335 3 102. 90.224 4 54. 45.112 5 16. 18.045 6 to INF 5. 8.282 Chisquare with 6 d.o.f. = 6.00 p-value= .576710 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 5 to 28 2.042 duplicate number number spacings observed expected 0 70. 67.668 1 126. 135.335 2 142. 135.335 3 85. 90.224 4 45. 45.112 5 22. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 2.58 p-value= .140439 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 6 to 29 2.066 duplicate number number spacings observed expected 0 62. 67.668 1 136. 135.335 2 127. 135.335 3 100. 90.224 4 42. 45.112 5 25. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 4.96 p-value= .450511 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 7 to 30 1.926 duplicate number number spacings observed expected 0 68. 67.668 1 152. 135.335 2 127. 135.335 3 82. 90.224 4 47. 45.112 5 21. 18.045 6 to INF 3. 8.282 Chisquare with 6 d.o.f. = 7.25 p-value= .701493 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 8 to 31 2.024 duplicate number number spacings observed expected 0 72. 67.668 1 134. 135.335 2 126. 135.335 3 90. 90.224 4 51. 45.112 5 19. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 1.76 p-value= .059907 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean data.txt using bits 9 to 32 2.062 duplicate number number spacings observed expected 0 68. 67.668 1 132. 135.335 2 130. 135.335 3 86. 90.224 4 51. 45.112 5 22. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 3.02 p-value= .193612 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .256985 .469251 .173130 .576710 .140439 .450511 .701493 .059907 .193612 A KSTEST for the 9 p-values yields .860231 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 data.txt For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=110.329; p-value= .794921 OPERM5 test for file data.txt For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=102.959; p-value= .627430 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 data.txt Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 201 211.4 .513367 .513 29 5114 5134.0 .077992 .591 30 23269 23103.0 1.192070 1.783 31 11416 11551.5 1.589995 3.373 chisquare= 3.373 for 3 d. of f.; p-value= .695575 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 data.txt Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 204 211.4 .260276 .260 30 5074 5134.0 .701446 .962 31 23113 23103.0 .004288 .966 32 11609 11551.5 .285975 1.252 chisquare= 1.252 for 3 d. of f.; p-value= .391022 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 data.txt Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 921 944.3 .575 .575 r =5 21638 21743.9 .516 1.091 r =6 77441 77311.8 .216 1.307 p=1-exp(-SUM/2)= .47969 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 969 944.3 .646 .646 r =5 21553 21743.9 1.676 2.322 r =6 77478 77311.8 .357 2.679 p=1-exp(-SUM/2)= .73806 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 968 944.3 .595 .595 r =5 21529 21743.9 2.124 2.719 r =6 77503 77311.8 .473 3.192 p=1-exp(-SUM/2)= .79724 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 922 944.3 .527 .527 r =5 21743 21743.9 .000 .527 r =6 77335 77311.8 .007 .534 p=1-exp(-SUM/2)= .23420 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 997 944.3 2.941 2.941 r =5 21687 21743.9 .149 3.090 r =6 77316 77311.8 .000 3.090 p=1-exp(-SUM/2)= .78670 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 939 944.3 .030 .030 r =5 21669 21743.9 .258 .288 r =6 77392 77311.8 .083 .371 p=1-exp(-SUM/2)= .16929 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 897 944.3 2.369 2.369 r =5 21736 21743.9 .003 2.372 r =6 77367 77311.8 .039 2.412 p=1-exp(-SUM/2)= .70056 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 964 944.3 .411 .411 r =5 21771 21743.9 .034 .445 r =6 77265 77311.8 .028 .473 p=1-exp(-SUM/2)= .21063 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 980 944.3 1.350 1.350 r =5 21770 21743.9 .031 1.381 r =6 77250 77311.8 .049 1.430 p=1-exp(-SUM/2)= .51088 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 955 944.3 .121 .121 r =5 21829 21743.9 .333 .454 r =6 77216 77311.8 .119 .573 p=1-exp(-SUM/2)= .24911 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 997 944.3 2.941 2.941 r =5 21670 21743.9 .251 3.192 r =6 77333 77311.8 .006 3.198 p=1-exp(-SUM/2)= .79790 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 964 944.3 .411 .411 r =5 21847 21743.9 .489 .900 r =6 77189 77311.8 .195 1.095 p=1-exp(-SUM/2)= .42156 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 981 944.3 1.426 1.426 r =5 21890 21743.9 .982 2.408 r =6 77129 77311.8 .432 2.840 p=1-exp(-SUM/2)= .75830 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 986 944.3 1.841 1.841 r =5 21709 21743.9 .056 1.897 r =6 77305 77311.8 .001 1.898 p=1-exp(-SUM/2)= .61286 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 938 944.3 .042 .042 r =5 21711 21743.9 .050 .092 r =6 77351 77311.8 .020 .112 p=1-exp(-SUM/2)= .05432 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 997 944.3 2.941 2.941 r =5 21623 21743.9 .672 3.613 r =6 77380 77311.8 .060 3.673 p=1-exp(-SUM/2)= .84065 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 902 944.3 1.895 1.895 r =5 21755 21743.9 .006 1.901 r =6 77343 77311.8 .013 1.913 p=1-exp(-SUM/2)= .61580 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21568 21743.9 1.423 1.704 r =6 77504 77311.8 .478 2.182 p=1-exp(-SUM/2)= .66415 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 911 944.3 1.174 1.174 r =5 21680 21743.9 .188 1.362 r =6 77409 77311.8 .122 1.484 p=1-exp(-SUM/2)= .52393 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 955 944.3 .121 .121 r =5 21584 21743.9 1.176 1.297 r =6 77461 77311.8 .288 1.585 p=1-exp(-SUM/2)= .54729 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 940 944.3 .020 .020 r =5 21688 21743.9 .144 .163 r =6 77372 77311.8 .047 .210 p=1-exp(-SUM/2)= .09975 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 995 944.3 2.722 2.722 r =5 21678 21743.9 .200 2.922 r =6 77327 77311.8 .003 2.925 p=1-exp(-SUM/2)= .76831 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 956 944.3 .145 .145 r =5 21683 21743.9 .171 .316 r =6 77361 77311.8 .031 .347 p=1-exp(-SUM/2)= .15920 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 996 944.3 2.830 2.830 r =5 21912 21743.9 1.300 4.130 r =6 77092 77311.8 .625 4.755 p=1-exp(-SUM/2)= .90721 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG data.txt b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 960 944.3 .261 .261 r =5 21947 21743.9 1.897 2.158 r =6 77093 77311.8 .619 2.777 p=1-exp(-SUM/2)= .75059 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .479686 .738061 .797244 .234204 .786698 .169292 .700556 .210629 .510882 .249110 .797895 .421562 .758303 .612864 .054319 .840653 .615802 .664148 .523927 .547290 .099753 .768307 .159202 .907213 .750588 brank test summary for data.txt The KS test for those 25 supposed UNI's yields KS p-value= .579773 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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: 142257 missing words, .81 sigmas from mean, p-value= .79170 tst no 2: 141843 missing words, -.15 sigmas from mean, p-value= .43842 tst no 3: 142125 missing words, .50 sigmas from mean, p-value= .69284 tst no 4: 141300 missing words, -1.42 sigmas from mean, p-value= .07727 tst no 5: 142120 missing words, .49 sigmas from mean, p-value= .68872 tst no 6: 142012 missing words, .24 sigmas from mean, p-value= .59479 tst no 7: 142062 missing words, .36 sigmas from mean, p-value= .63935 tst no 8: 140718 missing words, -2.78 sigmas from mean, p-value= .00269 tst no 9: 141439 missing words, -1.10 sigmas from mean, p-value= .13591 tst no 10: 141423 missing words, -1.14 sigmas from mean, p-value= .12792 tst no 11: 142110 missing words, .47 sigmas from mean, p-value= .68041 tst no 12: 141940 missing words, .07 sigmas from mean, p-value= .52857 tst no 13: 141879 missing words, -.07 sigmas from mean, p-value= .47175 tst no 14: 142012 missing words, .24 sigmas from mean, p-value= .59479 tst no 15: 141451 missing words, -1.07 sigmas from mean, p-value= .14212 tst no 16: 141744 missing words, -.39 sigmas from mean, p-value= .34964 tst no 17: 140999 missing words, -2.13 sigmas from mean, p-value= .01671 tst no 18: 141282 missing words, -1.47 sigmas from mean, p-value= .07136 tst no 19: 141771 missing words, -.32 sigmas from mean, p-value= .37327 tst no 20: 142183 missing words, .64 sigmas from mean, p-value= .73873 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 data.txt Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for data.txt using bits 23 to 32 142207 1.026 .8477 OPSO for data.txt using bits 22 to 31 142170 .899 .8156 OPSO for data.txt using bits 21 to 30 142143 .806 .7898 OPSO for data.txt using bits 20 to 29 141822 -.301 .3817 OPSO for data.txt using bits 19 to 28 142064 .533 .7031 OPSO for data.txt using bits 18 to 27 141824 -.294 .3843 OPSO for data.txt using bits 17 to 26 142152 .837 .7986 OPSO for data.txt using bits 16 to 25 141822 -.301 .3817 OPSO for data.txt using bits 15 to 24 142103 .668 .7479 OPSO for data.txt using bits 14 to 23 141963 .185 .5734 OPSO for data.txt using bits 13 to 22 141561 -1.201 .1149 OPSO for data.txt using bits 12 to 21 141972 .216 .5855 OPSO for data.txt using bits 11 to 20 141988 .271 .6069 OPSO for data.txt using bits 10 to 19 142138 .789 .7848 OPSO for data.txt using bits 9 to 18 141622 -.991 .1609 OPSO for data.txt using bits 8 to 17 141519 -1.346 .0892 OPSO for data.txt using bits 7 to 16 142303 1.357 .9127 OPSO for data.txt using bits 6 to 15 141543 -1.263 .1033 OPSO for data.txt using bits 5 to 14 141776 -.460 .3228 OPSO for data.txt using bits 4 to 13 141631 -.960 .1686 OPSO for data.txt using bits 3 to 12 141450 -1.584 .0566 OPSO for data.txt using bits 2 to 11 141686 -.770 .2206 OPSO for data.txt using bits 1 to 10 141277 -2.180 .0146 OQSO test for generator data.txt Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for data.txt using bits 28 to 32 142207 1.009 .8435 OQSO for data.txt using bits 27 to 31 141682 -.771 .2205 OQSO for data.txt using bits 26 to 30 141975 .223 .5881 OQSO for data.txt using bits 25 to 29 142047 .467 .6796 OQSO for data.txt using bits 24 to 28 141884 -.086 .4658 OQSO for data.txt using bits 23 to 27 141433 -1.615 .0532 OQSO for data.txt using bits 22 to 26 142298 1.318 .9062 OQSO for data.txt using bits 21 to 25 141604 -1.035 .1503 OQSO for data.txt using bits 20 to 24 142581 2.277 .9886 OQSO for data.txt using bits 19 to 23 141796 -.384 .3504 OQSO for data.txt using bits 18 to 22 141650 -.879 .1897 OQSO for data.txt using bits 17 to 21 141889 -.069 .4725 OQSO for data.txt using bits 16 to 20 141904 -.018 .4928 OQSO for data.txt using bits 15 to 19 142149 .812 .7917 OQSO for data.txt using bits 14 to 18 141716 -.655 .2561 OQSO for data.txt using bits 13 to 17 141818 -.310 .3784 OQSO for data.txt using bits 12 to 16 141980 .240 .5947 OQSO for data.txt using bits 11 to 15 141720 -.642 .2605 OQSO for data.txt using bits 10 to 14 142025 .392 .6525 OQSO for data.txt using bits 9 to 13 142126 .734 .7687 OQSO for data.txt using bits 8 to 12 142153 .826 .7956 OQSO for data.txt using bits 7 to 11 142256 1.175 .8800 OQSO for data.txt using bits 6 to 10 141621 -.977 .1642 OQSO for data.txt using bits 5 to 9 141484 -1.442 .0747 OQSO for data.txt using bits 4 to 8 142585 2.290 .9890 OQSO for data.txt using bits 3 to 7 141328 -1.971 .0244 OQSO for data.txt using bits 2 to 6 141962 .179 .5709 OQSO for data.txt using bits 1 to 5 142121 .718 .7635 DNA test for generator data.txt Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for data.txt using bits 31 to 32 143030 3.306 .9995 DNA for data.txt using bits 30 to 31 141920 .031 .5126 DNA for data.txt using bits 29 to 30 142008 .291 .6145 DNA for data.txt using bits 28 to 29 143039 3.332 .9996 DNA for data.txt using bits 27 to 28 142153 .719 .7639 DNA for data.txt using bits 26 to 27 142078 .498 .6906 DNA for data.txt using bits 25 to 26 141710 -.588 .2783 DNA for data.txt using bits 24 to 25 141915 .017 .5067 DNA for data.txt using bits 23 to 24 141280 -1.856 .0317 DNA for data.txt using bits 22 to 23 141705 -.603 .2733 DNA for data.txt using bits 21 to 22 142168 .763 .7773 DNA for data.txt using bits 20 to 21 141719 -.561 .2872 DNA for data.txt using bits 19 to 20 142099 .560 .7121 DNA for data.txt using bits 18 to 19 141676 -.688 .2456 DNA for data.txt using bits 17 to 18 142188 .822 .7945 DNA for data.txt using bits 16 to 17 142641 2.158 .9845 DNA for data.txt using bits 15 to 16 141174 -2.169 .0150 DNA for data.txt using bits 14 to 15 141854 -.163 .4352 DNA for data.txt using bits 13 to 14 141844 -.193 .4236 DNA for data.txt using bits 12 to 13 141723 -.550 .2913 DNA for data.txt using bits 11 to 12 141656 -.747 .2274 DNA for data.txt using bits 10 to 11 141829 -.237 .4063 DNA for data.txt using bits 9 to 10 141702 -.612 .2704 DNA for data.txt using bits 8 to 9 142276 1.082 .8603 DNA for data.txt using bits 7 to 8 141781 -.379 .3525 DNA for data.txt using bits 6 to 7 141499 -1.210 .1131 DNA for data.txt using bits 5 to 6 141404 -1.491 .0680 DNA for data.txt using bits 4 to 5 142029 .353 .6380 DNA for data.txt using bits 3 to 4 141734 -.517 .3025 DNA for data.txt using bits 2 to 3 141530 -1.119 .1316 DNA for data.txt using bits 1 to 2 142032 .362 .6413 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for data.txt Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for data.txt 2559.30 .839 .799142 byte stream for data.txt 2494.73 -.074 .470310 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 2487.47 -.177 .429682 bits 2 to 9 2430.02 -.990 .161161 bits 3 to 10 2366.24 -1.892 .029267 bits 4 to 11 2544.83 .634 .736937 bits 5 to 12 2478.97 -.297 .383084 bits 6 to 13 2606.53 1.506 .934030 bits 7 to 14 2396.89 -1.458 .072395 bits 8 to 15 2438.68 -.867 .192918 bits 9 to 16 2563.73 .901 .816274 bits 10 to 17 2418.92 -1.147 .125774 bits 11 to 18 2461.71 -.542 .294071 bits 12 to 19 2583.36 1.179 .880776 bits 13 to 20 2378.72 -1.715 .043164 bits 14 to 21 2518.29 .259 .602069 bits 15 to 22 2532.51 .460 .677142 bits 16 to 23 2568.23 .965 .832721 bits 17 to 24 2519.51 .276 .608707 bits 18 to 25 2579.38 1.123 .869186 bits 19 to 26 2568.26 .965 .832807 bits 20 to 27 2537.42 .529 .701667 bits 21 to 28 2458.66 -.585 .279403 bits 22 to 29 2643.15 2.024 .978540 bits 23 to 30 2491.25 -.124 .450743 bits 24 to 31 2532.04 .453 .674758 bits 25 to 32 2495.48 -.064 .474490 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 data.txt Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3539 z-score: .731 p-value: .767486 Successes: 3565 z-score: 1.918 p-value: .972432 Successes: 3497 z-score: -1.187 p-value: .117571 Successes: 3523 z-score: .000 p-value: .500000 Successes: 3525 z-score: .091 p-value: .536382 Successes: 3535 z-score: .548 p-value: .708135 Successes: 3532 z-score: .411 p-value: .659449 Successes: 3517 z-score: -.274 p-value: .392053 Successes: 3499 z-score: -1.096 p-value: .136563 Successes: 3533 z-score: .457 p-value: .676028 square size avg. no. parked sample sigma 100. 3526.500 18.744 KSTEST for the above 10: p= .216668 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file data.txt Sample no. d^2 avg equiv uni 5 .1067 .6313 .101717 10 .2075 .9050 .188264 15 .0256 .7512 .025447 20 2.3914 .8820 .909592 25 .3354 .8628 .286123 30 1.0764 .9570 .661026 35 1.3006 1.0329 .729395 40 2.8556 1.1036 .943298 45 .2084 1.0283 .188981 50 1.2425 .9882 .713131 55 .5131 1.0040 .402927 60 1.8918 1.0064 .850630 65 3.4570 1.0197 .969017 70 .3003 1.0513 .260531 75 2.1931 1.0814 .889653 80 2.5171 1.0716 .920324 85 1.4425 1.0582 .765368 90 .1767 1.0340 .162704 95 .5380 1.0306 .417649 100 .4978 1.0096 .393664 MINIMUM DISTANCE TEST for data.txt Result of KS test on 20 transformed mindist^2's: p-value= .025444 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file data.txt sample no: 1 r^3= 67.193 p-value= .89352 sample no: 2 r^3= 6.174 p-value= .18599 sample no: 3 r^3= 5.602 p-value= .17034 sample no: 4 r^3= 27.928 p-value= .60581 sample no: 5 r^3= 16.265 p-value= .41852 sample no: 6 r^3= 79.862 p-value= .93020 sample no: 7 r^3= 3.500 p-value= .11011 sample no: 8 r^3= 8.302 p-value= .24175 sample no: 9 r^3= 32.219 p-value= .65835 sample no: 10 r^3= 2.164 p-value= .06960 sample no: 11 r^3= 11.041 p-value= .30790 sample no: 12 r^3= 30.049 p-value= .63272 sample no: 13 r^3= .481 p-value= .01590 sample no: 14 r^3= 20.672 p-value= .49795 sample no: 15 r^3= 20.035 p-value= .48719 sample no: 16 r^3= 19.143 p-value= .47170 sample no: 17 r^3= 55.204 p-value= .84121 sample no: 18 r^3= 30.251 p-value= .63519 sample no: 19 r^3= 50.961 p-value= .81708 sample no: 20 r^3= 1.043 p-value= .03416 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file data.txt p-value= .240003 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR data.txt Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: -.8 -.3 .8 -.7 2.0 -1.7 .6 -2.2 -1.5 .5 1.2 -1.4 .0 1.0 .5 -.3 .2 .1 .3 .3 .1 -.2 -.9 -.3 -.1 -.3 1.4 -.1 -.1 -1.1 -.2 .5 1.5 1.0 -.1 .4 .0 -.1 -1.2 -.1 .1 1.0 -.1 Chi-square with 42 degrees of freedom: 32.090 z-score= -1.081 p-value= .134167 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .915488 Test no. 2 p-value .856182 Test no. 3 p-value .903568 Test no. 4 p-value .386940 Test no. 5 p-value .555228 Test no. 6 p-value .952627 Test no. 7 p-value .917217 Test no. 8 p-value .921442 Test no. 9 p-value .395145 Test no. 10 p-value .170341 Results of the OSUM test for data.txt KSTEST on the above 10 p-values: .981562 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file data.txt Up and down runs in a sample of 10000 _________________________________________________ Run test for data.txt : runs up; ks test for 10 p's: .434626 runs down; ks test for 10 p's: .328503 Run test for data.txt : runs up; ks test for 10 p's: .953026 runs down; ks test for 10 p's: .027564 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for data.txt No. of wins: Observed Expected 98541 98585.86 98541= No. of wins, z-score= -.201 pvalue= .42049 Analysis of Throws-per-Game: Chisq= 9.65 for 20 degrees of freedom, p= .02598 Throws Observed Expected Chisq Sum 1 66443 66666.7 .750 .750 2 37899 37654.3 1.590 2.340 3 26940 26954.7 .008 2.348 4 19224 19313.5 .414 2.763 5 13819 13851.4 .076 2.839 6 10021 9943.5 .603 3.442 7 7185 7145.0 .224 3.666 8 5098 5139.1 .328 3.994 9 3691 3699.9 .021 4.015 10 2735 2666.3 1.770 5.785 11 1912 1923.3 .067 5.852 12 1374 1388.7 .156 6.009 13 1012 1003.7 .068 6.077 14 707 726.1 .505 6.582 15 544 525.8 .627 7.209 16 398 381.2 .745 7.954 17 272 276.5 .075 8.028 18 199 200.8 .017 8.045 19 153 146.0 .337 8.382 20 96 106.2 .982 9.365 21 278 287.1 .289 9.654 SUMMARY FOR data.txt p-value for no. of wins: .420491 p-value for throws/game: .025980 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file output.txt