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 PRBIN For a sample of size 500: mean PRBIN using bits 1 to 24 509.070 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 2 to 25 507.570 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 3 to 26 504.624 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 4 to 27 499.262 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 5 to 28 489.948 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 6 to 29 473.746 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 7 to 30 446.928 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 8 to 31 404.514 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean PRBIN using bits 9 to 32 341.810 duplicate number number spacings observed expected 0 500. 67.668 1 0. 135.335 2 0. 135.335 3 0. 90.224 4 0. 45.112 5 0. 18.045 6 to INF 0. 8.282 Chisquare with 6 d.o.f. = 3194.53 p-value= 1.000000 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 A KSTEST for the 9 p-values yields 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE OVERLAPPING 5-PERMUTATION TEST :: :: This is the OPERM5 test. It looks at a sequence of one mill- :: :: ion 32-bit random integers. Each set of five consecutive :: :: integers can be in one of 120 states, for the 5! possible or- :: :: derings of five numbers. Thus the 5th, 6th, 7th,...numbers :: :: each provide a state. As many thousands of state transitions :: :: are observed, cumulative counts are made of the number of :: :: occurences of each state. Then the quadratic form in the :: :: weak inverse of the 120x120 covariance matrix yields a test :: :: equivalent to the likelihood ratio test that the 120 cell :: :: counts came from the specified (asymptotically) normal dis- :: :: tribution with the specified 120x120 covariance matrix (with :: :: rank 99). This version uses 1,000,000 integers, twice. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPERM5 test for file PRBIN For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=107.300; p-value= .732852 OPERM5 test for file PRBIN For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 87.984; p-value= .221690 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 PRBIN Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 40000 211.4******************* 29 0 5134.0******************* 30 0 23103.0******************* 31 0 11551.5******************* chisquare=****** for 3 d. of f.; p-value=1.000000 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 32x32 matrices. A random 32x :: :: 32 binary matrix is formed, each row a 32-bit random integer. :: :: The rank is determined. That rank can be from 0 to 32, ranks :: :: less than 29 are rare, and their counts are pooled with those :: :: for rank 29. Ranks are found for 40,000 such random matrices :: :: and a chisquare test is performed on counts for ranks 32,31, :: :: 30 and <=29. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for PRBIN Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 40000 211.4******************* 30 0 5134.0******************* 31 0 23103.0******************* 32 0 11551.5******************* chisquare=****** for 3 d. of f.; p-value=1.000000 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 6x8 matrices. From each of :: :: six random 32-bit integers from the generator under test, a :: :: specified byte is chosen, and the resulting six bytes form a :: :: 6x8 binary matrix whose rank is determined. That rank can be :: :: from 0 to 6, but ranks 0,1,2,3 are rare; their counts are :: :: pooled with those for rank 4. Ranks are found for 100,000 :: :: random matrices, and a chi-square test is performed on :: :: counts for ranks 6,5 and <=4. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary Rank Test for PRBIN Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 100000 944.310390790.00010390790.000 r =5 0 21743.9 21743.90010412530.000 r =6 0 77311.8 77311.80010489840.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 41319 944.3 1726267.000 1726267.000 r =5 58681 21743.9 62746.300 1789014.000 r =6 0 77311.8 77311.800 1866325.000 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 12842 944.3 149904.700 149904.700 r =5 57773 21743.9 59699.320 209604.000 r =6 29385 77311.8 29710.580 239314.600 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 3617 944.3 7564.661 7564.661 r =5 38007 21743.9 12163.800 19728.460 r =6 58376 77311.8 4637.904 24366.360 p=1-exp(-SUM/2)=1.00000 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 942 944.3 .006 .006 r =5 21829 21743.9 .333 .339 r =6 77229 77311.8 .089 .427 p=1-exp(-SUM/2)= .19239 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 956 944.3 .145 .145 r =5 21840 21743.9 .425 .570 r =6 77204 77311.8 .150 .720 p=1-exp(-SUM/2)= .30232 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 947 944.3 .008 .008 r =5 21818 21743.9 .253 .260 r =6 77235 77311.8 .076 .337 p=1-exp(-SUM/2)= .15487 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 967 944.3 .546 .546 r =5 21550 21743.9 1.729 2.275 r =6 77483 77311.8 .379 2.654 p=1-exp(-SUM/2)= .73470 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 911 944.3 1.174 1.174 r =5 21982 21743.9 2.607 3.782 r =6 77107 77311.8 .543 4.324 p=1-exp(-SUM/2)= .88491 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 899 944.3 2.173 2.173 r =5 21777 21743.9 .050 2.224 r =6 77324 77311.8 .002 2.226 p=1-exp(-SUM/2)= .67136 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 941 944.3 .012 .012 r =5 21504 21743.9 2.647 2.658 r =6 77555 77311.8 .765 3.423 p=1-exp(-SUM/2)= .81944 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG PRBIN b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 931 944.3 .187 .187 r =5 21624 21743.9 .661 .849 r =6 77445 77311.8 .229 1.078 p=1-exp(-SUM/2)= .41667 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 .192390 .302317 .154871 .734703 .884915 .671356 .819438 .416665 brank test summary for PRBIN The KS test for those 25 supposed UNI's yields KS p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE BITSTREAM TEST :: :: The file under test is viewed as a stream of bits. Call them :: :: b1,b2,... . Consider an alphabet with two "letters", 0 and 1 :: :: and think of the stream of bits as a succession of 20-letter :: :: "words", overlapping. Thus the first word is b1b2...b20, the :: :: second is b2b3...b21, and so on. The bitstream test counts :: :: the number of missing 20-letter (20-bit) words in a string of :: :: 2^21 overlapping 20-letter words. There are 2^20 possible 20 :: :: letter words. For a truly random string of 2^21+19 bits, the :: :: number of missing words j should be (very close to) normally :: :: distributed with mean 141,909 and sigma 428. Thus :: :: (j-141909)/428 should be a standard normal variate (z score) :: :: that leads to a uniform [0,1) p value. The test is repeated :: :: twenty times. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: THE OVERLAPPING 20-tuples BITSTREAM TEST, 20 BITS PER WORD, N words This test uses N=2^21 and samples the bitstream 20 times. No. missing words should average 141909. with sigma=428. --------------------------------------------------------- tst no 1: 940677 missing words, 1866.28 sigmas from mean, p-value=1.00000 tst no 2: 940786 missing words, 1866.53 sigmas from mean, p-value=1.00000 tst no 3: 940894 missing words, 1866.79 sigmas from mean, p-value=1.00000 tst no 4: 941036 missing words, 1867.12 sigmas from mean, p-value=1.00000 tst no 5: 941131 missing words, 1867.34 sigmas from mean, p-value=1.00000 tst no 6: 940740 missing words, 1866.43 sigmas from mean, p-value=1.00000 tst no 7: 941084 missing words, 1867.23 sigmas from mean, p-value=1.00000 tst no 8: 940749 missing words, 1866.45 sigmas from mean, p-value=1.00000 tst no 9: 941160 missing words, 1867.41 sigmas from mean, p-value=1.00000 tst no 10: 940748 missing words, 1866.45 sigmas from mean, p-value=1.00000 tst no 11: 940913 missing words, 1866.83 sigmas from mean, p-value=1.00000 tst no 12: 940972 missing words, 1866.97 sigmas from mean, p-value=1.00000 tst no 13: 940801 missing words, 1866.57 sigmas from mean, p-value=1.00000 tst no 14: 940716 missing words, 1866.37 sigmas from mean, p-value=1.00000 tst no 15: 940969 missing words, 1866.96 sigmas from mean, p-value=1.00000 tst no 16: 940954 missing words, 1866.93 sigmas from mean, p-value=1.00000 tst no 17: 940833 missing words, 1866.64 sigmas from mean, p-value=1.00000 tst no 18: 940544 missing words, 1865.97 sigmas from mean, p-value=1.00000 tst no 19: 940950 missing words, 1866.92 sigmas from mean, p-value=1.00000 tst no 20: 940831 missing words, 1866.64 sigmas from mean, p-value=1.00000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The tests OPSO, OQSO and DNA :: :: OPSO means Overlapping-Pairs-Sparse-Occupancy :: :: The OPSO test considers 2-letter words from an alphabet of :: :: 1024 letters. Each letter is determined by a specified ten :: :: bits from a 32-bit integer in the sequence to be tested. OPSO :: :: generates 2^21 (overlapping) 2-letter words (from 2^21+1 :: :: "keystrokes") and counts the number of missing words---that :: :: is 2-letter words which do not appear in the entire sequence. :: :: That count should be very close to normally distributed with :: :: mean 141,909, sigma 290. Thus (missingwrds-141909)/290 should :: :: be a standard normal variable. The OPSO test takes 32 bits at :: :: a time from the test file and uses a designated set of ten :: :: consecutive bits. It then restarts the file for the next de- :: :: signated 10 bits, and so on. :: :: :: :: OQSO means Overlapping-Quadruples-Sparse-Occupancy :: :: The test OQSO is similar, except that it considers 4-letter :: :: words from an alphabet of 32 letters, each letter determined :: :: by a designated string of 5 consecutive bits from the test :: :: file, elements of which are assumed 32-bit random integers. :: :: The mean number of missing words in a sequence of 2^21 four- :: :: letter words, (2^21+3 "keystrokes"), is again 141909, with :: :: sigma = 295. The mean is based on theory; sigma comes from :: :: extensive simulation. :: :: :: :: The DNA test considers an alphabet of 4 letters:: C,G,A,T,:: :: determined by two designated bits in the sequence of random :: :: integers being tested. It considers 10-letter words, so that :: :: as in OPSO and OQSO, there are 2^20 possible words, and the :: :: mean number of missing words from a string of 2^21 (over- :: :: lapping) 10-letter words (2^21+9 "keystrokes") is 141909. :: :: The standard deviation sigma=339 was determined as for OQSO :: :: by simulation. (Sigma for OPSO, 290, is the true value (to :: :: three places), not determined by simulation. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPSO test for generator PRBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for PRBIN using bits 23 to 32 137837-14.043 .0000 OPSO for PRBIN using bits 22 to 31 139976 -6.667 .0000 OPSO for PRBIN using bits 21 to 30 140834 -3.708 .0001 OPSO for PRBIN using bits 20 to 29 141322 -2.025 .0214 OPSO for PRBIN using bits 19 to 28 141404 -1.743 .0407 OPSO for PRBIN using bits 18 to 27 141843 -.229 .4095 OPSO for PRBIN using bits 17 to 26 786521******* 1.0000 OPSO for PRBIN using bits 16 to 25 983040******* 1.0000 OPSO for PRBIN using bits 15 to 24 1032192******* 1.0000 OPSO for PRBIN using bits 14 to 23 1044480******* 1.0000 OPSO for PRBIN using bits 13 to 22 1047552******* 1.0000 OPSO for PRBIN using bits 12 to 21 1048320******* 1.0000 OPSO for PRBIN using bits 11 to 20 1048512******* 1.0000 OPSO for PRBIN using bits 10 to 19 1048560******* 1.0000 OPSO for PRBIN using bits 9 to 18 1048572******* 1.0000 OPSO for PRBIN using bits 8 to 17 1048575******* 1.0000 OPSO for PRBIN using bits 7 to 16 1048575******* 1.0000 OPSO for PRBIN using bits 6 to 15 1048575******* 1.0000 OPSO for PRBIN using bits 5 to 14 1048575******* 1.0000 OPSO for PRBIN using bits 4 to 13 1048575******* 1.0000 OPSO for PRBIN using bits 3 to 12 1048575******* 1.0000 OPSO for PRBIN using bits 2 to 11 1048575******* 1.0000 OPSO for PRBIN using bits 1 to 10 1048575******* 1.0000 OQSO test for generator PRBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for PRBIN using bits 28 to 32 728903******* 1.0000 OQSO for PRBIN using bits 27 to 31 615378******* 1.0000 OQSO for PRBIN using bits 26 to 30 286600490.477 1.0000 OQSO for PRBIN using bits 25 to 29 128676-44.859 .0000 OQSO for PRBIN using bits 24 to 28 134869-23.866 .0000 OQSO for PRBIN using bits 23 to 27 139337 -8.720 .0000 OQSO for PRBIN using bits 22 to 26 141366 -1.842 .0328 OQSO for PRBIN using bits 21 to 25 140875 -3.506 .0002 OQSO for PRBIN using bits 20 to 24 141556 -1.198 .1155 OQSO for PRBIN using bits 19 to 23 141893 -.055 .4779 OQSO for PRBIN using bits 18 to 22 141991 .277 .6091 OQSO for PRBIN using bits 17 to 21 983040******* 1.0000 OQSO for PRBIN using bits 16 to 20 1044480******* 1.0000 OQSO for PRBIN using bits 15 to 19 1048320******* 1.0000 OQSO for PRBIN using bits 14 to 18 1048560******* 1.0000 OQSO for PRBIN using bits 13 to 17 1048575******* 1.0000 OQSO for PRBIN using bits 12 to 16 1048575******* 1.0000 OQSO for PRBIN using bits 11 to 15 1048575******* 1.0000 OQSO for PRBIN using bits 10 to 14 1048575******* 1.0000 OQSO for PRBIN using bits 9 to 13 1048575******* 1.0000 OQSO for PRBIN using bits 8 to 12 1048575******* 1.0000 OQSO for PRBIN using bits 7 to 11 1048575******* 1.0000 OQSO for PRBIN using bits 6 to 10 1048575******* 1.0000 OQSO for PRBIN using bits 5 to 9 1048575******* 1.0000 OQSO for PRBIN using bits 4 to 8 1048575******* 1.0000 OQSO for PRBIN using bits 3 to 7 1048575******* 1.0000 OQSO for PRBIN using bits 2 to 6 1048575******* 1.0000 OQSO for PRBIN using bits 1 to 5 1048575******* 1.0000 DNA test for generator PRBIN Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for PRBIN using bits 31 to 32 804113******* 1.0000 DNA for PRBIN using bits 30 to 31 613250******* 1.0000 DNA for PRBIN using bits 29 to 30 369234670.574 1.0000 DNA for PRBIN using bits 28 to 29 152965 32.613 1.0000 DNA for PRBIN using bits 27 to 28 135697-18.325 .0000 DNA for PRBIN using bits 26 to 27 144077 6.394 1.0000 DNA for PRBIN using bits 25 to 26 136155-16.974 .0000 DNA for PRBIN using bits 24 to 25 138672 -9.550 .0000 DNA for PRBIN using bits 23 to 24 139645 -6.679 .0000 DNA for PRBIN using bits 22 to 23 140628 -3.780 .0001 DNA for PRBIN using bits 21 to 22 141785 -.367 .3569 DNA for PRBIN using bits 20 to 21 141935 .076 .5302 DNA for PRBIN using bits 19 to 20 141349 -1.653 .0492 DNA for PRBIN using bits 18 to 19 141810 -.293 .3848 DNA for PRBIN using bits 17 to 18 1047552******* 1.0000 DNA for PRBIN using bits 16 to 17 1048575******* 1.0000 DNA for PRBIN using bits 15 to 16 1048575******* 1.0000 DNA for PRBIN using bits 14 to 15 1048575******* 1.0000 DNA for PRBIN using bits 13 to 14 1048575******* 1.0000 DNA for PRBIN using bits 12 to 13 1048575******* 1.0000 DNA for PRBIN using bits 11 to 12 1048575******* 1.0000 DNA for PRBIN using bits 10 to 11 1048575******* 1.0000 DNA for PRBIN using bits 9 to 10 1048575******* 1.0000 DNA for PRBIN using bits 8 to 9 1048575******* 1.0000 DNA for PRBIN using bits 7 to 8 1048575******* 1.0000 DNA for PRBIN using bits 6 to 7 1048575******* 1.0000 DNA for PRBIN using bits 5 to 6 1048575******* 1.0000 DNA for PRBIN using bits 4 to 5 1048575******* 1.0000 DNA for PRBIN using bits 3 to 4 1048575******* 1.0000 DNA for PRBIN using bits 2 to 3 1048575******* 1.0000 DNA for PRBIN using bits 1 to 2 1048575******* 1.0000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for PRBIN Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for PRBIN *********1778493.000 1.000000 byte stream for PRBIN *********1775986.000 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST for specific bytes. :: :: Consider the file under test as a stream of 32-bit integers. :: :: From each integer, a specific byte is chosen , say the left- :: :: most:: bits 1 to 8. Each byte can contain from 0 to 8 1's, :: :: with probabilitie 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the specified bytes from successive integers provide a string :: :: of (overlapping) 5-letter words, each "letter" taking values :: :: A,B,C,D,E. The letters are determined by the number of 1's, :: :: in that byte:: 0,1,or 2 ---> A, 3 ---> B, 4 ---> C, 5 ---> D,:: :: and 6,7 or 8 ---> E. Thus we have a monkey at a typewriter :: :: hitting five keys with with various probabilities:: 37,56,70,:: :: 56,37 over 256. There are 5^5 possible 5-letter words, and :: :: from a string of 256,000 (overlapping) 5-letter words, counts :: :: are made on the frequencies for each word. The quadratic form :: :: in the weak inverse of the covariance matrix of the cell :: :: counts provides a chisquare test:: Q5-Q4, the difference of :: :: the naive Pearson sums of (OBS-EXP)^2/EXP on counts for 5- :: :: and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Chi-square with 5^5-5^4=2500 d.of f. for sample size: 256000 chisquare equiv normal p value Results for COUNT-THE-1's in specified bytes: bits 1 to 8******************** 1.000000 bits 2 to 9******************** 1.000000 bits 3 to 10******************** 1.000000 bits 4 to 11******************** 1.000000 bits 5 to 12******************** 1.000000 bits 6 to 13******************** 1.000000 bits 7 to 14******************** 1.000000 bits 8 to 15******************** 1.000000 bits 9 to 16******************** 1.000000 bits 10 to 17******************** 1.000000 bits 11 to 18******************** 1.000000 bits 12 to 19******************** 1.000000 bits 13 to 20******************** 1.000000 bits 14 to 21*********1518919.000 1.000000 bits 15 to 22********* 121121.400 1.000000 bits 16 to 23668675.30 9421.141 1.000000 bits 17 to 24 51397.13 691.510 1.000000 bits 18 to 25 2402.82 -1.374 .084675 bits 19 to 26 2429.01 -1.004 .157696 bits 20 to 27 2478.17 -.309 .378790 bits 21 to 28 2566.01 .934 .824730 bits 22 to 29 2523.82 .337 .631891 bits 23 to 30 2522.84 .323 .626678 bits 24 to 31 2452.90 -.666 .252657 bits 25 to 32 2522.71 .321 .625966 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 PRBIN Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 Successes: 1 z-score:******* p-value: .000000 square size avg. no. parked sample sigma 100. 1.000 .000 KSTEST for the above 10: p= 1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file PRBIN Sample no. d^2 avg equiv uni 5 .0000 .0000 .000000 10 .0000 .0000 .000000 15 .0000 .0000 .000000 20 .0000 .0000 .000000 25 .0000 .0000 .000000 30 .0000 .0000 .000000 35 .0000 .0000 .000000 40 .0000 .0000 .000000 45 .0000 .0000 .000000 50 .0000 .0000 .000000 55 .0000 .0000 .000000 60 .0000 .0000 .000000 65 .0000 .0000 .000000 70 .0000 .0000 .000000 75 .0000 .0000 .000000 80 .0000 .0000 .000000 85 .0000 .0000 .000000 90 .0000 .0000 .000000 95 .0000 .0000 .000000 100 .0000 .0000 .000000 MINIMUM DISTANCE TEST for PRBIN Result of KS test on 20 transformed mindist^2's: p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file PRBIN sample no: 1 r^3= .000 p-value= .00000 sample no: 2 r^3= .000 p-value= .00000 sample no: 3 r^3= .000 p-value= .00000 sample no: 4 r^3= .000 p-value= .00000 sample no: 5 r^3= .000 p-value= .00000 sample no: 6 r^3= .000 p-value= .00000 sample no: 7 r^3= .000 p-value= .00000 sample no: 8 r^3= .000 p-value= .00000 sample no: 9 r^3= .000 p-value= .00000 sample no: 10 r^3= .000 p-value= .00000 sample no: 11 r^3= .000 p-value= .00000 sample no: 12 r^3= .000 p-value= .00000 sample no: 13 r^3= .000 p-value= .00000 sample no: 14 r^3= .000 p-value= .00000 sample no: 15 r^3= .000 p-value= .00000 sample no: 16 r^3= .000 p-value= .00000 sample no: 17 r^3= .000 p-value= .00000 sample no: 18 r^3= .000 p-value= .00000 sample no: 19 r^3= .000 p-value= .00000 sample no: 20 r^3= .000 p-value= .00000 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file PRBIN p-value=1.000000 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::