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 outdiv.bin
For a sample of size 500: mean
outdiv.bin using bits 1 to 24 1.982
duplicate number number
spacings observed expected
0 73. 67.668
1 135. 135.335
2 138. 135.335
3 81. 90.224
4 41. 45.112
5 22. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. = 3.01 p-value= .192992
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 2 to 25 1.880
duplicate number number
spacings observed expected
0 68. 67.668
1 156. 135.335
2 127. 135.335
3 87. 90.224
4 43. 45.112
5 18. 18.045
6 to INF 1. 8.282
Chisquare with 6 d.o.f. = 10.29 p-value= .886927
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 3 to 26 1.946
duplicate number number
spacings observed expected
0 64. 67.668
1 136. 135.335
2 149. 135.335
3 90. 90.224
4 42. 45.112
5 15. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 4.52 p-value= .393922
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 4 to 27 2.084
duplicate number number
spacings observed expected
0 56. 67.668
1 128. 135.335
2 150. 135.335
3 90. 90.224
4 48. 45.112
5 18. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. = 4.54 p-value= .396053
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 5 to 28 1.972
duplicate number number
spacings observed expected
0 69. 67.668
1 133. 135.335
2 142. 135.335
3 88. 90.224
4 43. 45.112
5 20. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. = 2.06 p-value= .085972
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 6 to 29 1.918
duplicate number number
spacings observed expected
0 81. 67.668
1 131. 135.335
2 141. 135.335
3 80. 90.224
4 42. 45.112
5 16. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 4.67 p-value= .413207
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 7 to 30 1.958
duplicate number number
spacings observed expected
0 68. 67.668
1 134. 135.335
2 132. 135.335
3 105. 90.224
4 43. 45.112
5 15. 18.045
6 to INF 3. 8.282
Chisquare with 6 d.o.f. = 6.50 p-value= .630245
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 8 to 31 2.028
duplicate number number
spacings observed expected
0 64. 67.668
1 135. 135.335
2 132. 135.335
3 93. 90.224
4 51. 45.112
5 18. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 1.33 p-value= .030269
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
outdiv.bin using bits 9 to 32 2.100
duplicate number number
spacings observed expected
0 52. 67.668
1 142. 135.335
2 135. 135.335
3 86. 90.224
4 57. 45.112
5 18. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. = 7.64 p-value= .734628
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.192992 .886927 .393922 .396053 .085972
.413207 .630245 .030269 .734628
A KSTEST for the 9 p-values yields .298221
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom= 97.813; p-value= .485120
OPERM5 test for file outdiv.bin
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=103.107; p-value= .631297
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 226 211.4 1.005753 1.006
29 5217 5134.0 1.341505 2.347
30 23106 23103.0 .000377 2.348
31 11451 11551.5 .874790 3.222
chisquare= 3.222 for 3 d. of f.; p-value= .677732
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 242 211.4 4.423738 4.424
30 5057 5134.0 1.155155 5.579
31 23146 23103.0 .079858 5.659
32 11555 11551.5 .001046 5.660
chisquare= 5.660 for 3 d. of f.; p-value= .879292
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 926 944.3 .355 .355
r =5 21976 21743.9 2.477 2.832
r =6 77098 77311.8 .591 3.423
p=1-exp(-SUM/2)= .81945
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 920 944.3 .625 .625
r =5 21862 21743.9 .641 1.267
r =6 77218 77311.8 .114 1.381
p=1-exp(-SUM/2)= .49859
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 3 to 10
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 933 944.3 .135 .135
r =5 21741 21743.9 .000 .136
r =6 77326 77311.8 .003 .138
p=1-exp(-SUM/2)= .06679
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 873 944.3 5.384 5.384
r =5 21686 21743.9 .154 5.538
r =6 77441 77311.8 .216 5.754
p=1-exp(-SUM/2)= .94369
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 886 944.3 3.600 3.600
r =5 21750 21743.9 .002 3.601
r =6 77364 77311.8 .035 3.636
p=1-exp(-SUM/2)= .83769
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 942 944.3 .006 .006
r =5 21882 21743.9 .877 .883
r =6 77176 77311.8 .239 1.121
p=1-exp(-SUM/2)= .42915
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 953 944.3 .080 .080
r =5 21885 21743.9 .916 .996
r =6 77162 77311.8 .290 1.286
p=1-exp(-SUM/2)= .47429
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 907 944.3 1.473 1.473
r =5 21819 21743.9 .259 1.733
r =6 77274 77311.8 .018 1.751
p=1-exp(-SUM/2)= .58341
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 951 944.3 .048 .048
r =5 21761 21743.9 .013 .061
r =6 77288 77311.8 .007 .068
p=1-exp(-SUM/2)= .03357
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 940 944.3 .020 .020
r =5 21819 21743.9 .259 .279
r =6 77241 77311.8 .065 .344
p=1-exp(-SUM/2)= .15794
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21911 21743.9 1.284 1.376
r =6 77154 77311.8 .322 1.698
p=1-exp(-SUM/2)= .57213
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 1012 944.3 4.853 4.853
r =5 21774 21743.9 .042 4.895
r =6 77214 77311.8 .124 5.019
p=1-exp(-SUM/2)= .91868
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 910 944.3 1.246 1.246
r =5 21823 21743.9 .288 1.534
r =6 77267 77311.8 .026 1.560
p=1-exp(-SUM/2)= .54152
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 960 944.3 .261 .261
r =5 21799 21743.9 .140 .401
r =6 77241 77311.8 .065 .465
p=1-exp(-SUM/2)= .20763
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 934 944.3 .112 .112
r =5 21943 21743.9 1.823 1.935
r =6 77123 77311.8 .461 2.397
p=1-exp(-SUM/2)= .69828
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 926 944.3 .355 .355
r =5 21976 21743.9 2.477 2.832
r =6 77098 77311.8 .591 3.423
p=1-exp(-SUM/2)= .81945
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
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 21704 21743.9 .073 1.319
r =6 77386 77311.8 .071 1.390
p=1-exp(-SUM/2)= .50103
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 985 944.3 1.754 1.754
r =5 21456 21743.9 3.812 5.566
r =6 77559 77311.8 .790 6.356
p=1-exp(-SUM/2)= .95834
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 961 944.3 .295 .295
r =5 21772 21743.9 .036 .332
r =6 77267 77311.8 .026 .358
p=1-exp(-SUM/2)= .16372
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 960 944.3 .261 .261
r =5 21561 21743.9 1.538 1.799
r =6 77479 77311.8 .362 2.161
p=1-exp(-SUM/2)= .66058
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 903 944.3 1.806 1.806
r =5 21813 21743.9 .220 2.026
r =6 77284 77311.8 .010 2.036
p=1-exp(-SUM/2)= .63868
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 926 944.3 .355 .355
r =5 21800 21743.9 .145 .499
r =6 77274 77311.8 .018 .518
p=1-exp(-SUM/2)= .22814
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21883 21743.9 .890 1.738
r =6 77201 77311.8 .159 1.897
p=1-exp(-SUM/2)= .61265
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 937 944.3 .056 .056
r =5 21831 21743.9 .349 .405
r =6 77232 77311.8 .082 .488
p=1-exp(-SUM/2)= .21641
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG outdiv.bin
b-rank test for bits 25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 977 944.3 1.132 1.132
r =5 21711 21743.9 .050 1.182
r =6 77312 77311.8 .000 1.182
p=1-exp(-SUM/2)= .44624
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.819446 .498586 .066788 .943691 .837689
.429150 .474293 .583413 .033572 .157944
.572128 .918685 .541523 .207631 .698283
.819446 .501025 .958340 .163716 .660582
.638683 .228145 .612651 .216405 .446242
brank test summary for outdiv.bin
The KS test for those 25 supposed UNI's yields
KS p-value= .089596
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 142924 missing words, 2.37 sigmas from mean, p-value= .99112
tst no 2: 141989 missing words, .19 sigmas from mean, p-value= .57384
tst no 3: 142731 missing words, 1.92 sigmas from mean, p-value= .97256
tst no 4: 142488 missing words, 1.35 sigmas from mean, p-value= .91182
tst no 5: 142170 missing words, .61 sigmas from mean, p-value= .72875
tst no 6: 142396 missing words, 1.14 sigmas from mean, p-value= .87225
tst no 7: 143016 missing words, 2.59 sigmas from mean, p-value= .99514
tst no 8: 142557 missing words, 1.51 sigmas from mean, p-value= .93489
tst no 9: 142727 missing words, 1.91 sigmas from mean, p-value= .97196
tst no 10: 141846 missing words, -.15 sigmas from mean, p-value= .44119
tst no 11: 141474 missing words, -1.02 sigmas from mean, p-value= .15455
tst no 12: 142135 missing words, .53 sigmas from mean, p-value= .70100
tst no 13: 142674 missing words, 1.79 sigmas from mean, p-value= .96300
tst no 14: 142430 missing words, 1.22 sigmas from mean, p-value= .88811
tst no 15: 142711 missing words, 1.87 sigmas from mean, p-value= .96947
tst no 16: 142042 missing words, .31 sigmas from mean, p-value= .62171
tst no 17: 142416 missing words, 1.18 sigmas from mean, p-value= .88176
tst no 18: 142192 missing words, .66 sigmas from mean, p-value= .74552
tst no 19: 142216 missing words, .72 sigmas from mean, p-value= .76317
tst no 20: 143005 missing words, 2.56 sigmas from mean, p-value= .99477
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OPSO for outdiv.bin using bits 23 to 32 142704 2.740 .9969
OPSO for outdiv.bin using bits 22 to 31 142383 1.633 .9488
OPSO for outdiv.bin using bits 21 to 30 142316 1.402 .9196
OPSO for outdiv.bin using bits 20 to 29 142815 3.123 .9991
OPSO for outdiv.bin using bits 19 to 28 143173 4.357 1.0000
OPSO for outdiv.bin using bits 18 to 27 143576 5.747 1.0000
OPSO for outdiv.bin using bits 17 to 26 143571 5.730 1.0000
OPSO for outdiv.bin using bits 16 to 25 142885 3.364 .9996
OPSO for outdiv.bin using bits 15 to 24 142460 1.899 .9712
OPSO for outdiv.bin using bits 14 to 23 142740 2.864 .9979
OPSO for outdiv.bin using bits 13 to 22 142792 3.044 .9988
OPSO for outdiv.bin using bits 12 to 21 143024 3.844 .9999
OPSO for outdiv.bin using bits 11 to 20 143122 4.182 1.0000
OPSO for outdiv.bin using bits 10 to 19 143315 4.847 1.0000
OPSO for outdiv.bin using bits 9 to 18 143682 6.113 1.0000
OPSO for outdiv.bin using bits 8 to 17 142667 2.613 .9955
OPSO for outdiv.bin using bits 7 to 16 142343 1.495 .9326
OPSO for outdiv.bin using bits 6 to 15 142429 1.792 .9634
OPSO for outdiv.bin using bits 5 to 14 142446 1.851 .9679
OPSO for outdiv.bin using bits 4 to 13 143361 5.006 1.0000
OPSO for outdiv.bin using bits 3 to 12 143010 3.795 .9999
OPSO for outdiv.bin using bits 2 to 11 143285 4.744 1.0000
OPSO for outdiv.bin using bits 1 to 10 143770 6.416 1.0000
OQSO test for generator outdiv.bin
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OQSO for outdiv.bin using bits 28 to 32 142043 .453 .6748
OQSO for outdiv.bin using bits 27 to 31 141732 -.601 .2739
OQSO for outdiv.bin using bits 26 to 30 141957 .162 .5642
OQSO for outdiv.bin using bits 25 to 29 141842 -.228 .4097
OQSO for outdiv.bin using bits 24 to 28 141246 -2.249 .0123
OQSO for outdiv.bin using bits 23 to 27 142033 .419 .6625
OQSO for outdiv.bin using bits 22 to 26 142156 .836 .7985
OQSO for outdiv.bin using bits 21 to 25 141844 -.221 .4124
OQSO for outdiv.bin using bits 20 to 24 141733 -.598 .2750
OQSO for outdiv.bin using bits 19 to 23 142105 .663 .7464
OQSO for outdiv.bin using bits 18 to 22 141837 -.245 .4032
OQSO for outdiv.bin using bits 17 to 21 142478 1.928 .9731
OQSO for outdiv.bin using bits 16 to 20 141804 -.357 .3605
OQSO for outdiv.bin using bits 15 to 19 142255 1.172 .8794
OQSO for outdiv.bin using bits 14 to 18 142183 .928 .8232
OQSO for outdiv.bin using bits 13 to 17 141923 .046 .5185
OQSO for outdiv.bin using bits 12 to 16 142131 .751 .7738
OQSO for outdiv.bin using bits 11 to 15 142149 .812 .7917
OQSO for outdiv.bin using bits 10 to 14 142397 1.653 .9508
OQSO for outdiv.bin using bits 9 to 13 143045 3.850 .9999
OQSO for outdiv.bin using bits 8 to 12 141949 .134 .5535
OQSO for outdiv.bin using bits 7 to 11 142753 2.860 .9979
OQSO for outdiv.bin using bits 6 to 10 141683 -.767 .2215
OQSO for outdiv.bin using bits 5 to 9 142283 1.267 .8974
OQSO for outdiv.bin using bits 4 to 8 142426 1.751 .9601
OQSO for outdiv.bin using bits 3 to 7 141164 -2.527 .0058
OQSO for outdiv.bin using bits 2 to 6 142406 1.684 .9539
OQSO for outdiv.bin using bits 1 to 5 141149 -2.577 .0050
DNA test for generator outdiv.bin
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for outdiv.bin using bits 31 to 32 142042 .391 .6522
DNA for outdiv.bin using bits 30 to 31 141771 -.408 .3416
DNA for outdiv.bin using bits 29 to 30 141932 .067 .5267
DNA for outdiv.bin using bits 28 to 29 142160 .739 .7702
DNA for outdiv.bin using bits 27 to 28 141623 -.845 .1992
DNA for outdiv.bin using bits 26 to 27 142383 1.397 .9188
DNA for outdiv.bin using bits 25 to 26 141899 -.030 .4878
DNA for outdiv.bin using bits 24 to 25 141514 -1.166 .1218
DNA for outdiv.bin using bits 23 to 24 141760 -.440 .3298
DNA for outdiv.bin using bits 22 to 23 142505 1.757 .9606
DNA for outdiv.bin using bits 21 to 22 141620 -.853 .1967
DNA for outdiv.bin using bits 20 to 21 141689 -.650 .2579
DNA for outdiv.bin using bits 19 to 20 142033 .365 .6424
DNA for outdiv.bin using bits 18 to 19 142209 .884 .8116
DNA for outdiv.bin using bits 17 to 18 141852 -.169 .4329
DNA for outdiv.bin using bits 16 to 17 142076 .492 .6885
DNA for outdiv.bin using bits 15 to 16 141715 -.573 .2832
DNA for outdiv.bin using bits 14 to 15 142093 .542 .7060
DNA for outdiv.bin using bits 13 to 14 141431 -1.411 .0791
DNA for outdiv.bin using bits 12 to 13 141915 .017 .5067
DNA for outdiv.bin using bits 11 to 12 141988 .232 .5918
DNA for outdiv.bin using bits 10 to 11 141988 .232 .5918
DNA for outdiv.bin using bits 9 to 10 141534 -1.107 .1341
DNA for outdiv.bin using bits 8 to 9 142158 .734 .7684
DNA for outdiv.bin using bits 7 to 8 142136 .669 .7481
DNA for outdiv.bin using bits 6 to 7 142460 1.624 .9479
DNA for outdiv.bin using bits 5 to 6 142439 1.562 .9409
DNA for outdiv.bin using bits 4 to 5 142045 .400 .6555
DNA for outdiv.bin using bits 3 to 4 142727 2.412 .9921
DNA for outdiv.bin using bits 2 to 3 141859 -.148 .4410
DNA for outdiv.bin using bits 1 to 2 141981 .211 .5837
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
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 outdiv.bin 2442.82 -.809 .209373
byte stream for outdiv.bin 2302.93 -2.787 .002660
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 2517.01 .241 .595057
bits 2 to 9 2462.34 -.533 .297164
bits 3 to 10 2556.57 .800 .788156
bits 4 to 11 2520.45 .289 .613813
bits 5 to 12 2485.64 -.203 .419511
bits 6 to 13 2518.82 .266 .604944
bits 7 to 14 2553.19 .752 .774033
bits 8 to 15 2489.32 -.151 .439963
bits 9 to 16 2625.58 1.776 .962130
bits 10 to 17 2567.38 .953 .829667
bits 11 to 18 2538.34 .542 .706171
bits 12 to 19 2539.76 .562 .713064
bits 13 to 20 2422.39 -1.098 .136200
bits 14 to 21 2535.41 .501 .691757
bits 15 to 22 2426.46 -1.040 .149166
bits 16 to 23 2486.57 -.190 .424655
bits 17 to 24 2574.05 1.047 .852488
bits 18 to 25 2549.63 .702 .758614
bits 19 to 26 2488.81 -.158 .437119
bits 20 to 27 2519.02 .269 .606022
bits 21 to 28 2575.60 1.069 .857484
bits 22 to 29 2510.96 .155 .561566
bits 23 to 30 2643.00 2.022 .978428
bits 24 to 31 2555.32 .782 .783011
bits 25 to 32 2538.02 .538 .704608
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3520 z-score: -.137 p-value: .445521
Successes: 3538 z-score: .685 p-value: .753306
Successes: 3468 z-score: -2.511 p-value: .006012
Successes: 3530 z-score: .320 p-value: .625377
Successes: 3515 z-score: -.365 p-value: .357445
Successes: 3551 z-score: 1.279 p-value: .899470
Successes: 3558 z-score: 1.598 p-value: .944998
Successes: 3519 z-score: -.183 p-value: .427537
Successes: 3533 z-score: .457 p-value: .676028
Successes: 3548 z-score: 1.142 p-value: .873180
square size avg. no. parked sample sigma
100. 3528.000 24.216
KSTEST for the above 10: p= .629880
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Sample no. d^2 avg equiv uni
5 .3569 .7062 .301378
10 1.3184 .5711 .734194
15 .6724 .6205 .491227
20 .0105 .7785 .010479
25 .9794 .7639 .626323
30 .7994 .7013 .552225
35 1.2185 .6763 .706145
40 2.2612 .7247 .896950
45 .7887 .8014 .547375
50 1.2189 .8426 .706261
55 1.1795 .8935 .694390
60 .1192 .8470 .112878
65 1.1714 .8521 .691883
70 .6646 .8741 .487252
75 2.0129 .8823 .867747
80 .1384 .9024 .129883
85 2.4513 .8885 .914873
90 .3418 .9185 .290759
95 .9452 .9123 .613235
100 .9808 .9112 .626818
MINIMUM DISTANCE TEST for outdiv.bin
Result of KS test on 20 transformed mindist^2's:
p-value= .830309
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
sample no: 1 r^3= 25.128 p-value= .56725
sample no: 2 r^3= 57.412 p-value= .85247
sample no: 3 r^3= 9.269 p-value= .26580
sample no: 4 r^3= 31.345 p-value= .64824
sample no: 5 r^3= 8.695 p-value= .25161
sample no: 6 r^3= 13.252 p-value= .35707
sample no: 7 r^3= 68.874 p-value= .89932
sample no: 8 r^3= 49.897 p-value= .81047
sample no: 9 r^3= 48.028 p-value= .79829
sample no: 10 r^3= 11.551 p-value= .31957
sample no: 11 r^3= 5.379 p-value= .16414
sample no: 12 r^3= 42.369 p-value= .75642
sample no: 13 r^3= 102.956 p-value= .96767
sample no: 14 r^3= 42.012 p-value= .75350
sample no: 15 r^3= 28.034 p-value= .60721
sample no: 16 r^3= 33.675 p-value= .67453
sample no: 17 r^3= 12.939 p-value= .35034
sample no: 18 r^3= 14.394 p-value= .38110
sample no: 19 r^3= 15.754 p-value= .40852
sample no: 20 r^3= 3.757 p-value= .11771
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file outdiv.bin p-value= .354591
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Table of standardized frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
-.8 .1 -.8 .2 -2.1 -1.5
1.3 1.2 -1.8 -1.2 -1.1 -.3
.1 .7 .1 -.1 -.6 .2
.9 .3 1.0 -1.1 -.4 .9
-.8 -.4 1.4 .4 .3 -.5
1.3 -1.5 .6 -1.4 .3 .6
.7 .5 .1 1.0 1.6 .0
.8
Chi-square with 42 degrees of freedom: 37.212
z-score= -.522 p-value= .318882
______________________________________________________________
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 .501186
Test no. 2 p-value .411127
Test no. 3 p-value .128923
Test no. 4 p-value .939062
Test no. 5 p-value .421814
Test no. 6 p-value .321254
Test no. 7 p-value .005718
Test no. 8 p-value .650870
Test no. 9 p-value .980092
Test no. 10 p-value .446905
Results of the OSUM test for outdiv.bin
KSTEST on the above 10 p-values: .319289
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
Up and down runs in a sample of 10000
_________________________________________________
Run test for outdiv.bin :
runs up; ks test for 10 p's: .336458
runs down; ks test for 10 p's: .385198
Run test for outdiv.bin :
runs up; ks test for 10 p's: .673492
runs down; ks test for 10 p's: .750297
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 outdiv.bin
No. of wins: Observed Expected
98797 98585.86
98797= No. of wins, z-score= .944 pvalue= .82750
Analysis of Throws-per-Game:
Chisq= 19.09 for 20 degrees of freedom, p= .48370
Throws Observed Expected Chisq Sum
1 67057 66666.7 2.285 2.285
2 37712 37654.3 .088 2.374
3 26765 26954.7 1.335 3.709
4 19305 19313.5 .004 3.713
5 13692 13851.4 1.835 5.548
6 9971 9943.5 .076 5.623
7 7194 7145.0 .336 5.959
8 5085 5139.1 .569 6.528
9 3558 3699.9 5.440 11.968
10 2658 2666.3 .026 11.993
11 1907 1923.3 .139 12.132
12 1419 1388.7 .659 12.791
13 998 1003.7 .033 12.824
14 749 726.1 .720 13.544
15 537 525.8 .237 13.781
16 398 381.2 .745 14.525
17 275 276.5 .009 14.534
18 212 200.8 .621 15.155
19 130 146.0 1.750 16.906
20 91 106.2 2.180 19.085
21 287 287.1 .000 19.085
SUMMARY FOR outdiv.bin
p-value for no. of wins: .827503
p-value for throws/game: .483702
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Results of DIEHARD battery of tests sent to file outdiv.txt