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 o:\tmp\i32.raw For a sample of size 500: mean o:\tmp\i32.raw using bits 1 to 24 2.014 duplicate number number spacings observed expected 0 74. 67.668 1 126. 135.335 2 128. 135.335 3 94. 90.224 4 57. 45.112 5 13. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 6.34 p-value= .614326 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 2 to 25 2.062 duplicate number number spacings observed expected 0 57. 67.668 1 136. 135.335 2 140. 135.335 3 97. 90.224 4 42. 45.112 5 17. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 3.52 p-value= .258962 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 3 to 26 2.020 duplicate number number spacings observed expected 0 77. 67.668 1 119. 135.335 2 130. 135.335 3 104. 90.224 4 38. 45.112 5 26. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 10.83 p-value= .906217 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 4 to 27 1.874 duplicate number number spacings observed expected 0 78. 67.668 1 146. 135.335 2 133. 135.335 3 81. 90.224 4 40. 45.112 5 13. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 5.45 p-value= .512861 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 5 to 28 1.988 duplicate number number spacings observed expected 0 59. 67.668 1 131. 135.335 2 153. 135.335 3 96. 90.224 4 40. 45.112 5 17. 18.045 6 to INF 4. 8.282 Chisquare with 6 d.o.f. = 6.78 p-value= .658146 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 6 to 29 1.972 duplicate number number spacings observed expected 0 67. 67.668 1 140. 135.335 2 129. 135.335 3 99. 90.224 4 42. 45.112 5 16. 18.045 6 to INF 7. 8.282 Chisquare with 6 d.o.f. = 1.96 p-value= .076876 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 7 to 30 1.988 duplicate number number spacings observed expected 0 65. 67.668 1 137. 135.335 2 141. 135.335 3 88. 90.224 4 49. 45.112 5 11. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 3.57 p-value= .264729 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 8 to 31 2.072 duplicate number number spacings observed expected 0 60. 67.668 1 141. 135.335 2 133. 135.335 3 79. 90.224 4 58. 45.112 5 20. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 6.50 p-value= .630299 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\i32.raw using bits 9 to 32 1.974 duplicate number number spacings observed expected 0 61. 67.668 1 149. 135.335 2 141. 135.335 3 83. 90.224 4 38. 45.112 5 18. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 4.33 p-value= .367871 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .614326 .258962 .906217 .512861 .658146 .076876 .264729 .630299 .367871 A KSTEST for the 9 p-values yields .107174 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 o:\tmp\i32.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 94.475; p-value= .390031 OPERM5 test for file o:\tmp\i32.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 61.110; p-value= .001007 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 o:\tmp\i32.raw Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 206 211.4 .138848 .139 29 5139 5134.0 .004850 .144 30 22974 23103.0 .720818 .865 31 11681 11551.5 1.451231 2.316 chisquare= 2.316 for 3 d. of f.; p-value= .554524 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 o:\tmp\i32.raw Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 203 211.4 .335179 .335 30 5211 5134.0 1.154540 1.490 31 23122 23103.0 .015549 1.505 32 11464 11551.5 .663161 2.168 chisquare= 2.168 for 3 d. of f.; p-value= .532242 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 o:\tmp\i32.raw Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 959 944.3 .229 .229 r =5 21886 21743.9 .929 1.157 r =6 77155 77311.8 .318 1.475 p=1-exp(-SUM/2)= .52180 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 941 944.3 .012 .012 r =5 21920 21743.9 1.426 1.438 r =6 77139 77311.8 .386 1.824 p=1-exp(-SUM/2)= .59828 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 970 944.3 .699 .699 r =5 21760 21743.9 .012 .711 r =6 77270 77311.8 .023 .734 p=1-exp(-SUM/2)= .30716 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 946 944.3 .003 .003 r =5 21552 21743.9 1.694 1.697 r =6 77502 77311.8 .468 2.165 p=1-exp(-SUM/2)= .66118 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 933 944.3 .135 .135 r =5 21305 21743.9 8.859 8.994 r =6 77762 77311.8 2.622 11.616 p=1-exp(-SUM/2)= .99700 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21753 21743.9 .004 .285 r =6 77319 77311.8 .001 .286 p=1-exp(-SUM/2)= .13320 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 959 944.3 .229 .229 r =5 21450 21743.9 3.972 4.201 r =6 77591 77311.8 1.008 5.210 p=1-exp(-SUM/2)= .92608 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 957 944.3 .171 .171 r =5 21788 21743.9 .089 .260 r =6 77255 77311.8 .042 .302 p=1-exp(-SUM/2)= .14013 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 930 944.3 .217 .217 r =5 21862 21743.9 .641 .858 r =6 77208 77311.8 .139 .997 p=1-exp(-SUM/2)= .39268 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 948 944.3 .014 .014 r =5 21724 21743.9 .018 .033 r =6 77328 77311.8 .003 .036 p=1-exp(-SUM/2)= .01788 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 956 944.3 .145 .145 r =5 21676 21743.9 .212 .357 r =6 77368 77311.8 .041 .398 p=1-exp(-SUM/2)= .18037 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 962 944.3 .332 .332 r =5 21649 21743.9 .414 .746 r =6 77389 77311.8 .077 .823 p=1-exp(-SUM/2)= .33734 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 980 944.3 1.350 1.350 r =5 21741 21743.9 .000 1.350 r =6 77279 77311.8 .014 1.364 p=1-exp(-SUM/2)= .49436 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21863 21743.9 .652 .934 r =6 77209 77311.8 .137 1.070 p=1-exp(-SUM/2)= .41447 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 946 944.3 .003 .003 r =5 21633 21743.9 .566 .569 r =6 77421 77311.8 .154 .723 p=1-exp(-SUM/2)= .30334 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 943 944.3 .002 .002 r =5 22038 21743.9 3.978 3.980 r =6 77019 77311.8 1.109 5.089 p=1-exp(-SUM/2)= .92147 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 960 944.3 .261 .261 r =5 21906 21743.9 1.208 1.469 r =6 77134 77311.8 .409 1.878 p=1-exp(-SUM/2)= .60905 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 915 944.3 .909 .909 r =5 21627 21743.9 .628 1.538 r =6 77458 77311.8 .276 1.814 p=1-exp(-SUM/2)= .59630 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 955 944.3 .121 .121 r =5 21872 21743.9 .755 .876 r =6 77173 77311.8 .249 1.125 p=1-exp(-SUM/2)= .43024 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 944 944.3 .000 .000 r =5 21637 21743.9 .526 .526 r =6 77419 77311.8 .149 .674 p=1-exp(-SUM/2)= .28619 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 912 944.3 1.105 1.105 r =5 21936 21743.9 1.697 2.802 r =6 77152 77311.8 .330 3.132 p=1-exp(-SUM/2)= .79116 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 933 944.3 .135 .135 r =5 21572 21743.9 1.359 1.494 r =6 77495 77311.8 .434 1.928 p=1-exp(-SUM/2)= .61870 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 911 944.3 1.174 1.174 r =5 21783 21743.9 .070 1.245 r =6 77306 77311.8 .000 1.245 p=1-exp(-SUM/2)= .46343 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 929 944.3 .248 .248 r =5 22025 21743.9 3.634 3.882 r =6 77046 77311.8 .914 4.796 p=1-exp(-SUM/2)= .90909 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\i32.raw b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 955 944.3 .121 .121 r =5 22040 21743.9 4.032 4.153 r =6 77005 77311.8 1.218 5.371 p=1-exp(-SUM/2)= .93181 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .521805 .598277 .307157 .661180 .996997 .133195 .926080 .140129 .392684 .017885 .180374 .337341 .494364 .414466 .303338 .921473 .609050 .596295 .430244 .286193 .791159 .618700 .463434 .909091 .931810 brank test summary for o:\tmp\i32.raw The KS test for those 25 supposed UNI's yields KS p-value= .249569 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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: 142823 missing words, 2.13 sigmas from mean, p-value= .98361 tst no 2: 141503 missing words, -.95 sigmas from mean, p-value= .17122 tst no 3: 142345 missing words, 1.02 sigmas from mean, p-value= .84564 tst no 4: 141294 missing words, -1.44 sigmas from mean, p-value= .07526 tst no 5: 141858 missing words, -.12 sigmas from mean, p-value= .45227 tst no 6: 141802 missing words, -.25 sigmas from mean, p-value= .40100 tst no 7: 141242 missing words, -1.56 sigmas from mean, p-value= .05948 tst no 8: 142355 missing words, 1.04 sigmas from mean, p-value= .85113 tst no 9: 141771 missing words, -.32 sigmas from mean, p-value= .37327 tst no 10: 141857 missing words, -.12 sigmas from mean, p-value= .45135 tst no 11: 142640 missing words, 1.71 sigmas from mean, p-value= .95611 tst no 12: 142136 missing words, .53 sigmas from mean, p-value= .70181 tst no 13: 142416 missing words, 1.18 sigmas from mean, p-value= .88176 tst no 14: 142077 missing words, .39 sigmas from mean, p-value= .65238 tst no 15: 141962 missing words, .12 sigmas from mean, p-value= .54897 tst no 16: 141618 missing words, -.68 sigmas from mean, p-value= .24804 tst no 17: 141172 missing words, -1.72 sigmas from mean, p-value= .04247 tst no 18: 142220 missing words, .73 sigmas from mean, p-value= .76604 tst no 19: 141864 missing words, -.11 sigmas from mean, p-value= .45783 tst no 20: 141834 missing words, -.18 sigmas from mean, p-value= .43015 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 o:\tmp\i32.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for o:\tmp\i32.raw using bits 23 to 32 141824 -.294 .3843 OPSO for o:\tmp\i32.raw using bits 22 to 31 142024 .395 .6537 OPSO for o:\tmp\i32.raw using bits 21 to 30 141911 .006 .5023 OPSO for o:\tmp\i32.raw using bits 20 to 29 142313 1.392 .9180 OPSO for o:\tmp\i32.raw using bits 19 to 28 141499 -1.415 .0785 OPSO for o:\tmp\i32.raw using bits 18 to 27 141542 -1.267 .1026 OPSO for o:\tmp\i32.raw using bits 17 to 26 141647 -.905 .1828 OPSO for o:\tmp\i32.raw using bits 16 to 25 142421 1.764 .9612 OPSO for o:\tmp\i32.raw using bits 15 to 24 141648 -.901 .1838 OPSO for o:\tmp\i32.raw using bits 14 to 23 141881 -.098 .4611 OPSO for o:\tmp\i32.raw using bits 13 to 22 142184 .947 .8282 OPSO for o:\tmp\i32.raw using bits 12 to 21 141753 -.539 .2949 OPSO for o:\tmp\i32.raw using bits 11 to 20 141794 -.398 .3454 OPSO for o:\tmp\i32.raw using bits 10 to 19 141649 -.898 .1847 OPSO for o:\tmp\i32.raw using bits 9 to 18 142185 .951 .8291 OPSO for o:\tmp\i32.raw using bits 8 to 17 141903 -.022 .4913 OPSO for o:\tmp\i32.raw using bits 7 to 16 141715 -.670 .2514 OPSO for o:\tmp\i32.raw using bits 6 to 15 141750 -.549 .2914 OPSO for o:\tmp\i32.raw using bits 5 to 14 141717 -.663 .2536 OPSO for o:\tmp\i32.raw using bits 4 to 13 142257 1.199 .8847 OPSO for o:\tmp\i32.raw using bits 3 to 12 142005 .330 .6293 OPSO for o:\tmp\i32.raw using bits 2 to 11 141799 -.380 .3518 OPSO for o:\tmp\i32.raw using bits 1 to 10 141937 .095 .5380 OQSO test for generator o:\tmp\i32.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for o:\tmp\i32.raw using bits 28 to 32 142208 1.012 .8443 OQSO for o:\tmp\i32.raw using bits 27 to 31 142016 .362 .6412 OQSO for o:\tmp\i32.raw using bits 26 to 30 142077 .568 .7151 OQSO for o:\tmp\i32.raw using bits 25 to 29 142026 .395 .6538 OQSO for o:\tmp\i32.raw using bits 24 to 28 141624 -.967 .1667 OQSO for o:\tmp\i32.raw using bits 23 to 27 142106 .667 .7475 OQSO for o:\tmp\i32.raw using bits 22 to 26 141678 -.784 .2165 OQSO for o:\tmp\i32.raw using bits 21 to 25 142161 .853 .8032 OQSO for o:\tmp\i32.raw using bits 20 to 24 141813 -.327 .3720 OQSO for o:\tmp\i32.raw using bits 19 to 23 142482 1.941 .9739 OQSO for o:\tmp\i32.raw using bits 18 to 22 141979 .236 .5934 OQSO for o:\tmp\i32.raw using bits 17 to 21 142079 .575 .7174 OQSO for o:\tmp\i32.raw using bits 16 to 20 141832 -.262 .3966 OQSO for o:\tmp\i32.raw using bits 15 to 19 141876 -.113 .4550 OQSO for o:\tmp\i32.raw using bits 14 to 18 141791 -.401 .3442 OQSO for o:\tmp\i32.raw using bits 13 to 17 141819 -.306 .3797 OQSO for o:\tmp\i32.raw using bits 12 to 16 142426 1.751 .9601 OQSO for o:\tmp\i32.raw using bits 11 to 15 142511 2.040 .9793 OQSO for o:\tmp\i32.raw using bits 10 to 14 141541 -1.249 .1059 OQSO for o:\tmp\i32.raw using bits 9 to 13 142076 .565 .7140 OQSO for o:\tmp\i32.raw using bits 8 to 12 141577 -1.127 .1300 OQSO for o:\tmp\i32.raw using bits 7 to 11 142098 .640 .7388 OQSO for o:\tmp\i32.raw using bits 6 to 10 141272 -2.160 .0154 OQSO for o:\tmp\i32.raw using bits 5 to 9 142137 .772 .7799 OQSO for o:\tmp\i32.raw using bits 4 to 8 141963 .182 .5722 OQSO for o:\tmp\i32.raw using bits 3 to 7 142078 .572 .7163 OQSO for o:\tmp\i32.raw using bits 2 to 6 142354 1.507 .9341 OQSO for o:\tmp\i32.raw using bits 1 to 5 142075 .562 .7128 DNA test for generator o:\tmp\i32.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for o:\tmp\i32.raw using bits 31 to 32 142294 1.135 .8718 DNA for o:\tmp\i32.raw using bits 30 to 31 142044 .397 .6544 DNA for o:\tmp\i32.raw using bits 29 to 30 142206 .875 .8093 DNA for o:\tmp\i32.raw using bits 28 to 29 141910 .002 .5008 DNA for o:\tmp\i32.raw using bits 27 to 28 141304 -1.786 .0371 DNA for o:\tmp\i32.raw using bits 26 to 27 142179 .795 .7868 DNA for o:\tmp\i32.raw using bits 25 to 26 141634 -.812 .2083 DNA for o:\tmp\i32.raw using bits 24 to 25 142352 1.306 .9042 DNA for o:\tmp\i32.raw using bits 23 to 24 142008 .291 .6145 DNA for o:\tmp\i32.raw using bits 22 to 23 141753 -.461 .3223 DNA for o:\tmp\i32.raw using bits 21 to 22 141541 -1.087 .1386 DNA for o:\tmp\i32.raw using bits 20 to 21 142237 .967 .8331 DNA for o:\tmp\i32.raw using bits 19 to 20 142230 .946 .8279 DNA for o:\tmp\i32.raw using bits 18 to 19 142036 .374 .6457 DNA for o:\tmp\i32.raw using bits 17 to 18 142352 1.306 .9042 DNA for o:\tmp\i32.raw using bits 16 to 17 142038 .380 .6479 DNA for o:\tmp\i32.raw using bits 15 to 16 141367 -1.600 .0548 DNA for o:\tmp\i32.raw using bits 14 to 15 141904 -.016 .4937 DNA for o:\tmp\i32.raw using bits 13 to 14 141742 -.494 .3108 DNA for o:\tmp\i32.raw using bits 12 to 13 141987 .229 .5906 DNA for o:\tmp\i32.raw using bits 11 to 12 141611 -.880 .1894 DNA for o:\tmp\i32.raw using bits 10 to 11 141862 -.140 .4445 DNA for o:\tmp\i32.raw using bits 9 to 10 141940 .090 .5360 DNA for o:\tmp\i32.raw using bits 8 to 9 141714 -.576 .2822 DNA for o:\tmp\i32.raw using bits 7 to 8 142549 1.887 .9704 DNA for o:\tmp\i32.raw using bits 6 to 7 142788 2.592 .9952 DNA for o:\tmp\i32.raw using bits 5 to 6 142054 .427 .6652 DNA for o:\tmp\i32.raw using bits 4 to 5 142381 1.391 .9179 DNA for o:\tmp\i32.raw using bits 3 to 4 141695 -.632 .2636 DNA for o:\tmp\i32.raw using bits 2 to 3 142132 .657 .7444 DNA for o:\tmp\i32.raw using bits 1 to 2 141375 -1.576 .0575 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for o:\tmp\i32.raw Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for o:\tmp\i32.raw 2426.83 -1.035 .150390 byte stream for o:\tmp\i32.raw 2563.91 .904 .816947 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 2545.41 .642 .739634 bits 2 to 9 2483.13 -.239 .405705 bits 3 to 10 2594.13 1.331 .908448 bits 4 to 11 2486.54 -.190 .424527 bits 5 to 12 2477.41 -.319 .374693 bits 6 to 13 2289.88 -2.972 .001481 bits 7 to 14 2677.38 2.509 .993939 bits 8 to 15 2456.92 -.609 .271185 bits 9 to 16 2566.53 .941 .826615 bits 10 to 17 2452.31 -.674 .250006 bits 11 to 18 2556.54 .800 .788045 bits 12 to 19 2529.26 .414 .660468 bits 13 to 20 2428.21 -1.015 .154991 bits 14 to 21 2551.09 .722 .764996 bits 15 to 22 2594.89 1.342 .910201 bits 16 to 23 2571.00 1.004 .842339 bits 17 to 24 2529.58 .418 .662156 bits 18 to 25 2424.96 -1.061 .144279 bits 19 to 26 2438.97 -.863 .194039 bits 20 to 27 2459.28 -.576 .282336 bits 21 to 28 2572.95 1.032 .848893 bits 22 to 29 2378.58 -1.717 .042978 bits 23 to 30 2532.75 .463 .678387 bits 24 to 31 2501.44 .020 .508128 bits 25 to 32 2396.71 -1.461 .072035 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 o:\tmp\i32.raw Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3503 z-score: -.913 p-value: .180558 Successes: 3528 z-score: .228 p-value: .590298 Successes: 3478 z-score: -2.055 p-value: .019949 Successes: 3517 z-score: -.274 p-value: .392053 Successes: 3541 z-score: .822 p-value: .794438 Successes: 3533 z-score: .457 p-value: .676028 Successes: 3520 z-score: -.137 p-value: .445521 Successes: 3563 z-score: 1.826 p-value: .966111 Successes: 3480 z-score: -1.963 p-value: .024796 Successes: 3559 z-score: 1.644 p-value: .949895 square size avg. no. parked sample sigma 100. 3522.200 27.669 KSTEST for the above 10: p= .372058 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file o:\tmp\i32.raw Sample no. d^2 avg equiv uni 5 2.9371 1.2763 .947759 10 .4516 .8738 .364850 15 .5758 .7306 .439383 20 .5513 .7433 .425394 25 .0497 .6630 .048708 30 1.0874 .8099 .664753 35 4.4673 .9871 .988776 40 1.5316 1.0197 .785469 45 1.3086 .9976 .731569 50 2.4756 .9633 .916926 55 .0293 .9742 .029041 60 .0162 .9061 .016156 65 1.1299 .8795 .678748 70 1.2469 .8913 .714407 75 .8915 .9159 .591777 80 .3829 .9180 .319459 85 .0951 1.0149 .091160 90 .1103 .9828 .104929 95 .0103 .9773 .010291 100 1.3122 .9975 .732553 MINIMUM DISTANCE TEST for o:\tmp\i32.raw Result of KS test on 20 transformed mindist^2's: p-value= .566228 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file o:\tmp\i32.raw sample no: 1 r^3= 24.253 p-value= .55445 sample no: 2 r^3= 58.488 p-value= .85767 sample no: 3 r^3= 9.832 p-value= .27944 sample no: 4 r^3= 33.970 p-value= .67772 sample no: 5 r^3= 2.999 p-value= .09515 sample no: 6 r^3= 7.646 p-value= .22497 sample no: 7 r^3= 47.624 p-value= .79556 sample no: 8 r^3= 5.811 p-value= .17610 sample no: 9 r^3= 40.461 p-value= .74042 sample no: 10 r^3= 20.406 p-value= .49349 sample no: 11 r^3= 12.784 p-value= .34698 sample no: 12 r^3= 101.512 p-value= .96608 sample no: 13 r^3= 6.206 p-value= .18688 sample no: 14 r^3= 8.516 p-value= .24714 sample no: 15 r^3= 66.166 p-value= .88981 sample no: 16 r^3= 10.749 p-value= .30114 sample no: 17 r^3= 14.427 p-value= .38177 sample no: 18 r^3= 94.393 p-value= .95700 sample no: 19 r^3= 54.110 p-value= .83530 sample no: 20 r^3= 26.765 p-value= .59023 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file o:\tmp\i32.raw p-value= .145648 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR o:\tmp\i32.raw Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: -.8 .5 .1 1.2 .4 .2 -.1 1.0 .6 .9 -1.5 -.2 -.7 -.3 .5 -1.7 2.6 1.5 -1.6 .2 -1.8 .4 .8 -2.0 .0 1.3 -.8 -1.0 1.6 .4 .8 .5 1.2 2.0 1.3 .2 -1.9 .5 -2.0 1.5 -1.3 .0 -1.1 Chi-square with 42 degrees of freedom: 57.126 z-score= 1.650 p-value= .940272 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .804984 Test no. 2 p-value .520573 Test no. 3 p-value .163327 Test no. 4 p-value .294671 Test no. 5 p-value .608291 Test no. 6 p-value .298100 Test no. 7 p-value .749557 Test no. 8 p-value .418009 Test no. 9 p-value .285274 Test no. 10 p-value .574868 Results of the OSUM test for o:\tmp\i32.raw KSTEST on the above 10 p-values: .427061 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file o:\tmp\i32.raw Up and down runs in a sample of 10000 _________________________________________________ Run test for o:\tmp\i32.raw : runs up; ks test for 10 p's: .573029 runs down; ks test for 10 p's: .616733 Run test for o:\tmp\i32.raw : runs up; ks test for 10 p's: .797989 runs down; ks test for 10 p's: .358424 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for o:\tmp\i32.raw No. of wins: Observed Expected 98643 98585.86 98643= No. of wins, z-score= .256 pvalue= .60086 Analysis of Throws-per-Game: Chisq= 18.83 for 20 degrees of freedom, p= .46679 Throws Observed Expected Chisq Sum 1 66588 66666.7 .093 .093 2 37813 37654.3 .669 .762 3 26830 26954.7 .577 1.339 4 19403 19313.5 .415 1.754 5 13832 13851.4 .027 1.781 6 9865 9943.5 .620 2.401 7 7165 7145.0 .056 2.457 8 5053 5139.1 1.442 3.899 9 3697 3699.9 .002 3.901 10 2646 2666.3 .155 4.056 11 1922 1923.3 .001 4.056 12 1458 1388.7 3.454 7.511 13 1052 1003.7 2.323 9.833 14 691 726.1 1.701 11.534 15 537 525.8 .237 11.771 16 431 381.2 6.520 18.291 17 270 276.5 .155 18.445 18 197 200.8 .073 18.518 19 152 146.0 .248 18.766 20 108 106.2 .030 18.796 21 290 287.1 .029 18.825 SUMMARY FOR o:\tmp\i32.raw p-value for no. of wins: .600857 p-value for throws/game: .466789 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file o:\tmp\i32.txt