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\f48.raw For a sample of size 500: mean o:\tmp\f48.raw using bits 1 to 24 1.970 duplicate number number spacings observed expected 0 78. 67.668 1 127. 135.335 2 125. 135.335 3 98. 90.224 4 49. 45.112 5 20. 18.045 6 to INF 3. 8.282 Chisquare with 6 d.o.f. = 7.47 p-value= .720120 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 2 to 25 1.998 duplicate number number spacings observed expected 0 64. 67.668 1 133. 135.335 2 142. 135.335 3 97. 90.224 4 39. 45.112 5 17. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 1.97 p-value= .077958 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 3 to 26 2.072 duplicate number number spacings observed expected 0 57. 67.668 1 132. 135.335 2 138. 135.335 3 105. 90.224 4 41. 45.112 5 16. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 5.74 p-value= .546484 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 4 to 27 1.922 duplicate number number spacings observed expected 0 62. 67.668 1 154. 135.335 2 141. 135.335 3 74. 90.224 4 49. 45.112 5 14. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 8.07 p-value= .767240 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 5 to 28 2.028 duplicate number number spacings observed expected 0 64. 67.668 1 129. 135.335 2 140. 135.335 3 92. 90.224 4 53. 45.112 5 16. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 2.93 p-value= .182523 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 6 to 29 2.088 duplicate number number spacings observed expected 0 67. 67.668 1 127. 135.335 2 133. 135.335 3 89. 90.224 4 49. 45.112 5 24. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 3.77 p-value= .292171 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 7 to 30 2.072 duplicate number number spacings observed expected 0 71. 67.668 1 122. 135.335 2 145. 135.335 3 78. 90.224 4 48. 45.112 5 25. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 7.58 p-value= .729671 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 8 to 31 1.888 duplicate number number spacings observed expected 0 74. 67.668 1 149. 135.335 2 125. 135.335 3 89. 90.224 4 44. 45.112 5 14. 18.045 6 to INF 5. 8.282 Chisquare with 6 d.o.f. = 5.01 p-value= .457811 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48.raw using bits 9 to 32 1.966 duplicate number number spacings observed expected 0 61. 67.668 1 136. 135.335 2 156. 135.335 3 83. 90.224 4 44. 45.112 5 12. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 6.46 p-value= .625891 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .720120 .077958 .546484 .767240 .182523 .292171 .729671 .457811 .625891 A KSTEST for the 9 p-values yields .167374 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=101.609; p-value= .591407 OPERM5 test for file o:\tmp\f48.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=101.612; p-value= .591492 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48.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 28 192 211.4 1.783478 1.783 29 5073 5134.0 .725018 2.508 30 23020 23103.0 .298523 2.807 31 11715 11551.5 2.313484 5.121 chisquare= 5.121 for 3 d. of f.; p-value= .848639 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48.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 29 189 211.4 2.377126 2.377 30 5225 5134.0 1.612606 3.990 31 23138 23103.0 .052881 4.043 32 11448 11551.5 .927783 4.970 chisquare= 4.970 for 3 d. of f.; p-value= .838915 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48.raw Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 948 944.3 .014 .014 r =5 21807 21743.9 .183 .198 r =6 77245 77311.8 .058 .255 p=1-exp(-SUM/2)= .11985 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 908 944.3 1.396 1.396 r =5 21791 21743.9 .102 1.498 r =6 77301 77311.8 .002 1.499 p=1-exp(-SUM/2)= .52741 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 907 944.3 1.473 1.473 r =5 21902 21743.9 1.150 2.623 r =6 77191 77311.8 .189 2.812 p=1-exp(-SUM/2)= .75485 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 933 944.3 .135 .135 r =5 21924 21743.9 1.492 1.627 r =6 77143 77311.8 .369 1.996 p=1-exp(-SUM/2)= .63130 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1006 944.3 4.031 4.031 r =5 21774 21743.9 .042 4.073 r =6 77220 77311.8 .109 4.182 p=1-exp(-SUM/2)= .87643 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 970 944.3 .699 .699 r =5 22019 21743.9 3.481 4.180 r =6 77011 77311.8 1.170 5.350 p=1-exp(-SUM/2)= .93110 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 955 944.3 .121 .121 r =5 21857 21743.9 .588 .709 r =6 77188 77311.8 .198 .908 p=1-exp(-SUM/2)= .36484 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 980 944.3 1.350 1.350 r =5 21753 21743.9 .004 1.353 r =6 77267 77311.8 .026 1.379 p=1-exp(-SUM/2)= .49826 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 985 944.3 1.754 1.754 r =5 21824 21743.9 .295 2.049 r =6 77191 77311.8 .189 2.238 p=1-exp(-SUM/2)= .67338 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 940 944.3 .020 .020 r =5 21857 21743.9 .588 .608 r =6 77203 77311.8 .153 .761 p=1-exp(-SUM/2)= .31648 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1001 944.3 3.404 3.404 r =5 21779 21743.9 .057 3.461 r =6 77220 77311.8 .109 3.570 p=1-exp(-SUM/2)= .83221 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 926 944.3 .355 .355 r =5 21678 21743.9 .200 .554 r =6 77396 77311.8 .092 .646 p=1-exp(-SUM/2)= .27607 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 926 944.3 .355 .355 r =5 21826 21743.9 .310 .665 r =6 77248 77311.8 .053 .717 p=1-exp(-SUM/2)= .30139 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 927 944.3 .317 .317 r =5 21770 21743.9 .031 .348 r =6 77303 77311.8 .001 .349 p=1-exp(-SUM/2)= .16026 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 953 944.3 .080 .080 r =5 21813 21743.9 .220 .300 r =6 77234 77311.8 .078 .378 p=1-exp(-SUM/2)= .17222 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 948 944.3 .014 .014 r =5 21769 21743.9 .029 .043 r =6 77283 77311.8 .011 .054 p=1-exp(-SUM/2)= .02673 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 910 944.3 1.246 1.246 r =5 21749 21743.9 .001 1.247 r =6 77341 77311.8 .011 1.258 p=1-exp(-SUM/2)= .46693 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 907 944.3 1.473 1.473 r =5 21699 21743.9 .093 1.566 r =6 77394 77311.8 .087 1.654 p=1-exp(-SUM/2)= .56254 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw 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 21899 21743.9 1.106 2.281 r =6 77190 77311.8 .192 2.473 p=1-exp(-SUM/2)= .70955 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 935 944.3 .092 .092 r =5 21513 21743.9 2.452 2.544 r =6 77552 77311.8 .746 3.290 p=1-exp(-SUM/2)= .80697 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 950 944.3 .034 .034 r =5 21470 21743.9 3.450 3.485 r =6 77580 77311.8 .930 4.415 p=1-exp(-SUM/2)= .89002 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 949 944.3 .023 .023 r =5 21753 21743.9 .004 .027 r =6 77298 77311.8 .002 .030 p=1-exp(-SUM/2)= .01472 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 972 944.3 .812 .812 r =5 21804 21743.9 .166 .979 r =6 77224 77311.8 .100 1.078 p=1-exp(-SUM/2)= .41676 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 894 944.3 2.679 2.679 r =5 21744 21743.9 .000 2.679 r =6 77362 77311.8 .033 2.712 p=1-exp(-SUM/2)= .74232 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48.raw b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 942 944.3 .006 .006 r =5 21720 21743.9 .026 .032 r =6 77338 77311.8 .009 .041 p=1-exp(-SUM/2)= .02017 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .119849 .527407 .754849 .631301 .876433 .931102 .364839 .498259 .673381 .316480 .832206 .276067 .301394 .160258 .172223 .026733 .466928 .562544 .709546 .806970 .890024 .014717 .416759 .742317 .020171 brank test summary for o:\tmp\f48.raw The KS test for those 25 supposed UNI's yields KS p-value= .095690 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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: 140735 missing words, -2.74 sigmas from mean, p-value= .00304 tst no 2: 141446 missing words, -1.08 sigmas from mean, p-value= .13951 tst no 3: 142188 missing words, .65 sigmas from mean, p-value= .74251 tst no 4: 142282 missing words, .87 sigmas from mean, p-value= .80805 tst no 5: 141759 missing words, -.35 sigmas from mean, p-value= .36271 tst no 6: 142411 missing words, 1.17 sigmas from mean, p-value= .87943 tst no 7: 141082 missing words, -1.93 sigmas from mean, p-value= .02662 tst no 8: 141449 missing words, -1.08 sigmas from mean, p-value= .14107 tst no 9: 142451 missing words, 1.27 sigmas from mean, p-value= .89717 tst no 10: 142082 missing words, .40 sigmas from mean, p-value= .65669 tst no 11: 142389 missing words, 1.12 sigmas from mean, p-value= .86880 tst no 12: 142457 missing words, 1.28 sigmas from mean, p-value= .89966 tst no 13: 141302 missing words, -1.42 sigmas from mean, p-value= .07795 tst no 14: 141988 missing words, .18 sigmas from mean, p-value= .57292 tst no 15: 141608 missing words, -.70 sigmas from mean, p-value= .24070 tst no 16: 142068 missing words, .37 sigmas from mean, p-value= .64458 tst no 17: 141600 missing words, -.72 sigmas from mean, p-value= .23492 tst no 18: 143204 missing words, 3.02 sigmas from mean, p-value= .99876 tst no 19: 141751 missing words, -.37 sigmas from mean, p-value= .35572 tst no 20: 141865 missing words, -.10 sigmas from mean, p-value= .45876 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for o:\tmp\f48.raw using bits 23 to 32 142554 2.223 .9869 OPSO for o:\tmp\f48.raw using bits 22 to 31 142372 1.595 .9447 OPSO for o:\tmp\f48.raw using bits 21 to 30 141786 -.425 .3353 OPSO for o:\tmp\f48.raw using bits 20 to 29 142015 .364 .6422 OPSO for o:\tmp\f48.raw using bits 19 to 28 141620 -.998 .1592 OPSO for o:\tmp\f48.raw using bits 18 to 27 142180 .933 .8247 OPSO for o:\tmp\f48.raw using bits 17 to 26 142000 .313 .6227 OPSO for o:\tmp\f48.raw using bits 16 to 25 142069 .551 .7090 OPSO for o:\tmp\f48.raw using bits 15 to 24 141735 -.601 .2739 OPSO for o:\tmp\f48.raw using bits 14 to 23 141826 -.287 .3869 OPSO for o:\tmp\f48.raw using bits 13 to 22 141856 -.184 .4270 OPSO for o:\tmp\f48.raw using bits 12 to 21 141969 .206 .5815 OPSO for o:\tmp\f48.raw using bits 11 to 20 141989 .275 .6082 OPSO for o:\tmp\f48.raw using bits 10 to 19 142406 1.713 .9566 OPSO for o:\tmp\f48.raw using bits 9 to 18 142392 1.664 .9520 OPSO for o:\tmp\f48.raw using bits 8 to 17 141314 -2.053 .0200 OPSO for o:\tmp\f48.raw using bits 7 to 16 141568 -1.177 .1196 OPSO for o:\tmp\f48.raw using bits 6 to 15 141650 -.894 .1856 OPSO for o:\tmp\f48.raw using bits 5 to 14 141816 -.322 .3738 OPSO for o:\tmp\f48.raw using bits 4 to 13 141869 -.139 .4447 OPSO for o:\tmp\f48.raw using bits 3 to 12 141819 -.311 .3777 OPSO for o:\tmp\f48.raw using bits 2 to 11 141723 -.643 .2603 OPSO for o:\tmp\f48.raw using bits 1 to 10 141894 -.053 .4789 OQSO test for generator o:\tmp\f48.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for o:\tmp\f48.raw using bits 28 to 32 141588 -1.089 .1380 OQSO for o:\tmp\f48.raw using bits 27 to 31 142127 .738 .7697 OQSO for o:\tmp\f48.raw using bits 26 to 30 141859 -.171 .4323 OQSO for o:\tmp\f48.raw using bits 25 to 29 142298 1.318 .9062 OQSO for o:\tmp\f48.raw using bits 24 to 28 142099 .643 .7399 OQSO for o:\tmp\f48.raw using bits 23 to 27 141945 .121 .5481 OQSO for o:\tmp\f48.raw using bits 22 to 26 141948 .131 .5521 OQSO for o:\tmp\f48.raw using bits 21 to 25 142412 1.704 .9558 OQSO for o:\tmp\f48.raw using bits 20 to 24 141986 .260 .6025 OQSO for o:\tmp\f48.raw using bits 19 to 23 142212 1.026 .8476 OQSO for o:\tmp\f48.raw using bits 18 to 22 141992 .280 .6104 OQSO for o:\tmp\f48.raw using bits 17 to 21 142427 1.755 .9604 OQSO for o:\tmp\f48.raw using bits 16 to 20 141415 -1.676 .0469 OQSO for o:\tmp\f48.raw using bits 15 to 19 141651 -.876 .1906 OQSO for o:\tmp\f48.raw using bits 14 to 18 142113 .690 .7550 OQSO for o:\tmp\f48.raw using bits 13 to 17 142384 1.609 .9462 OQSO for o:\tmp\f48.raw using bits 12 to 16 141896 -.045 .4820 OQSO for o:\tmp\f48.raw using bits 11 to 15 142105 .663 .7464 OQSO for o:\tmp\f48.raw using bits 10 to 14 141828 -.276 .3914 OQSO for o:\tmp\f48.raw using bits 9 to 13 142102 .653 .7432 OQSO for o:\tmp\f48.raw using bits 8 to 12 141535 -1.269 .1022 OQSO for o:\tmp\f48.raw using bits 7 to 11 142115 .697 .7572 OQSO for o:\tmp\f48.raw using bits 6 to 10 141784 -.425 .3355 OQSO for o:\tmp\f48.raw using bits 5 to 9 142207 1.009 .8435 OQSO for o:\tmp\f48.raw using bits 4 to 8 142256 1.175 .8800 OQSO for o:\tmp\f48.raw using bits 3 to 7 141612 -1.008 .1568 OQSO for o:\tmp\f48.raw using bits 2 to 6 141932 .077 .5306 OQSO for o:\tmp\f48.raw using bits 1 to 5 141696 -.723 .2348 DNA test for generator o:\tmp\f48.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for o:\tmp\f48.raw using bits 31 to 32 142149 .707 .7602 DNA for o:\tmp\f48.raw using bits 30 to 31 141903 -.019 .4926 DNA for o:\tmp\f48.raw using bits 29 to 30 142880 2.863 .9979 DNA for o:\tmp\f48.raw using bits 28 to 29 142003 .276 .6088 DNA for o:\tmp\f48.raw using bits 27 to 28 141641 -.792 .2143 DNA for o:\tmp\f48.raw using bits 26 to 27 141622 -.848 .1983 DNA for o:\tmp\f48.raw using bits 25 to 26 141410 -1.473 .0704 DNA for o:\tmp\f48.raw using bits 24 to 25 141486 -1.249 .1059 DNA for o:\tmp\f48.raw using bits 23 to 24 141631 -.821 .2058 DNA for o:\tmp\f48.raw using bits 22 to 23 141585 -.957 .1694 DNA for o:\tmp\f48.raw using bits 21 to 22 142058 .439 .6695 DNA for o:\tmp\f48.raw using bits 20 to 21 141684 -.665 .2531 DNA for o:\tmp\f48.raw using bits 19 to 20 142166 .757 .7755 DNA for o:\tmp\f48.raw using bits 18 to 19 142238 .970 .8339 DNA for o:\tmp\f48.raw using bits 17 to 18 142219 .913 .8195 DNA for o:\tmp\f48.raw using bits 16 to 17 141993 .247 .5975 DNA for o:\tmp\f48.raw using bits 15 to 16 141989 .235 .5929 DNA for o:\tmp\f48.raw using bits 14 to 15 141912 .008 .5031 DNA for o:\tmp\f48.raw using bits 13 to 14 141642 -.789 .2152 DNA for o:\tmp\f48.raw using bits 12 to 13 141803 -.314 .3769 DNA for o:\tmp\f48.raw using bits 11 to 12 142278 1.088 .8616 DNA for o:\tmp\f48.raw using bits 10 to 11 141443 -1.376 .0845 DNA for o:\tmp\f48.raw using bits 9 to 10 142427 1.527 .9366 DNA for o:\tmp\f48.raw using bits 8 to 9 141784 -.370 .3558 DNA for o:\tmp\f48.raw using bits 7 to 8 142240 .975 .8353 DNA for o:\tmp\f48.raw using bits 6 to 7 141509 -1.181 .1188 DNA for o:\tmp\f48.raw using bits 5 to 6 141615 -.868 .1926 DNA for o:\tmp\f48.raw using bits 4 to 5 141494 -1.225 .1103 DNA for o:\tmp\f48.raw using bits 3 to 4 141918 .026 .5102 DNA for o:\tmp\f48.raw using bits 2 to 3 141792 -.346 .3646 DNA for o:\tmp\f48.raw using bits 1 to 2 142222 .922 .8218 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for o:\tmp\f48.raw Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for o:\tmp\f48.raw 2471.22 -.407 .342020 byte stream for o:\tmp\f48.raw 2337.97 -2.291 .010968 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 2495.87 -.058 .476700 bits 2 to 9 2448.87 -.723 .234797 bits 3 to 10 2580.55 1.139 .872669 bits 4 to 11 2618.72 1.679 .953422 bits 5 to 12 2579.92 1.130 .870819 bits 6 to 13 2484.16 -.224 .411351 bits 7 to 14 2581.03 1.146 .874103 bits 8 to 15 2458.35 -.589 .277918 bits 9 to 16 2469.23 -.435 .331701 bits 10 to 17 2497.74 -.032 .487249 bits 11 to 18 2487.75 -.173 .431218 bits 12 to 19 2464.37 -.504 .307157 bits 13 to 20 2477.76 -.314 .376579 bits 14 to 21 2443.61 -.797 .212596 bits 15 to 22 2521.95 .310 .621893 bits 16 to 23 2454.42 -.645 .259588 bits 17 to 24 2579.21 1.120 .868695 bits 18 to 25 2469.14 -.436 .331263 bits 19 to 26 2591.23 1.290 .901505 bits 20 to 27 2528.00 .396 .653939 bits 21 to 28 2413.30 -1.226 .110063 bits 22 to 29 2565.16 .921 .821598 bits 23 to 30 2483.94 -.227 .410156 bits 24 to 31 2454.86 -.638 .261627 bits 25 to 32 2576.20 1.078 .859406 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48.raw Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3515 z-score: -.365 p-value: .357445 Successes: 3556 z-score: 1.507 p-value: .934075 Successes: 3502 z-score: -.959 p-value: .168804 Successes: 3510 z-score: -.594 p-value: .276387 Successes: 3512 z-score: -.502 p-value: .307734 Successes: 3507 z-score: -.731 p-value: .232514 Successes: 3569 z-score: 2.100 p-value: .982156 Successes: 3515 z-score: -.365 p-value: .357445 Successes: 3524 z-score: .046 p-value: .518210 Successes: 3503 z-score: -.913 p-value: .180558 square size avg. no. parked sample sigma 100. 3521.300 21.650 KSTEST for the above 10: p= .728703 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file o:\tmp\f48.raw Sample no. d^2 avg equiv uni 5 2.3512 .6535 .905860 10 .1665 .5038 .154084 15 .2461 .4861 .219104 20 3.3976 .7351 .967114 25 .0441 .6708 .043390 30 1.5218 .8530 .783347 35 .5291 .8975 .412433 40 .6048 .8841 .455479 45 .6915 .8856 .500930 50 3.1368 .9240 .957257 55 3.1522 .9614 .957914 60 .4480 .9769 .362514 65 2.4458 .9608 .914404 70 .6497 .9542 .479498 75 .0435 .9283 .042804 80 .0586 .8989 .057210 85 2.3689 .8971 .907522 90 .4128 .8954 .339600 95 1.8775 .9033 .848469 100 .1496 .8913 .139593 MINIMUM DISTANCE TEST for o:\tmp\f48.raw Result of KS test on 20 transformed mindist^2's: p-value= .764059 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file o:\tmp\f48.raw sample no: 1 r^3= 11.662 p-value= .32209 sample no: 2 r^3= 22.264 p-value= .52391 sample no: 3 r^3= 4.343 p-value= .13478 sample no: 4 r^3= 17.207 p-value= .43649 sample no: 5 r^3= 26.931 p-value= .59250 sample no: 6 r^3= 15.827 p-value= .40995 sample no: 7 r^3= 35.645 p-value= .69522 sample no: 8 r^3= 48.816 p-value= .80352 sample no: 9 r^3= 60.367 p-value= .86631 sample no: 10 r^3= 13.549 p-value= .36342 sample no: 11 r^3= .950 p-value= .03118 sample no: 12 r^3= 21.660 p-value= .51423 sample no: 13 r^3= 44.801 p-value= .77538 sample no: 14 r^3= 20.318 p-value= .49199 sample no: 15 r^3= 112.944 p-value= .97683 sample no: 16 r^3= 10.584 p-value= .29729 sample no: 17 r^3= 9.382 p-value= .26856 sample no: 18 r^3= .649 p-value= .02139 sample no: 19 r^3= 15.991 p-value= .41318 sample no: 20 r^3= .827 p-value= .02720 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file o:\tmp\f48.raw p-value= .450922 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR o:\tmp\f48.raw Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: -.1 -1.2 -.6 -.4 1.0 -1.7 -.7 -1.3 -.1 -1.1 -1.3 .0 1.0 -.6 .5 -.3 -.2 .6 -1.0 -.5 1.1 .4 .9 .5 -.7 .0 .3 -.2 -.1 1.4 2.0 1.0 -1.2 .3 -.6 .1 1.2 -.1 .1 -.1 .1 .0 -1.1 Chi-square with 42 degrees of freedom: 28.830 z-score= -1.437 p-value= .060840 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .378168 Test no. 2 p-value .742759 Test no. 3 p-value .819471 Test no. 4 p-value .119108 Test no. 5 p-value .771191 Test no. 6 p-value .683488 Test no. 7 p-value .262310 Test no. 8 p-value .352634 Test no. 9 p-value .037435 Test no. 10 p-value .660939 Results of the OSUM test for o:\tmp\f48.raw KSTEST on the above 10 p-values: .113254 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file o:\tmp\f48.raw Up and down runs in a sample of 10000 _________________________________________________ Run test for o:\tmp\f48.raw : runs up; ks test for 10 p's: .776669 runs down; ks test for 10 p's: .477862 Run test for o:\tmp\f48.raw : runs up; ks test for 10 p's: .382740 runs down; ks test for 10 p's: .193532 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for o:\tmp\f48.raw No. of wins: Observed Expected 98337 98585.86 98337= No. of wins, z-score=-1.113 pvalue= .13284 Analysis of Throws-per-Game: Chisq= 29.93 for 20 degrees of freedom, p= .92907 Throws Observed Expected Chisq Sum 1 66354 66666.7 1.466 1.466 2 38055 37654.3 4.264 5.730 3 26941 26954.7 .007 5.737 4 19396 19313.5 .353 6.090 5 13994 13851.4 1.468 7.557 6 9647 9943.5 8.844 16.401 7 7162 7145.0 .040 16.441 8 5081 5139.1 .656 17.098 9 3594 3699.9 3.029 20.127 10 2664 2666.3 .002 20.129 11 1996 1923.3 2.746 22.874 12 1383 1388.7 .024 22.898 13 1034 1003.7 .914 23.812 14 748 726.1 .658 24.470 15 522 525.8 .028 24.498 16 378 381.2 .026 24.524 17 306 276.5 3.139 27.663 18 198 200.8 .040 27.702 19 134 146.0 .984 28.686 20 107 106.2 .006 28.692 21 306 287.1 1.242 29.934 SUMMARY FOR o:\tmp\f48.raw p-value for no. of wins: .132845 p-value for throws/game: .929073 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file o:\tmp\f48.txt