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 SG100.DAT
For a sample of size 500: mean
SG100.DAT using bits 1 to 24 2.164
duplicate number number
spacings observed expected
0 53. 67.668
1 121. 135.335
2 151. 135.335
3 86. 90.224
4 55. 45.112
5 24. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. = 11.20 p-value= .917554
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 2 to 25 2.012
duplicate number number
spacings observed expected
0 65. 67.668
1 129. 135.335
2 143. 135.335
3 91. 90.224
4 50. 45.112
5 16. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 2.23 p-value= .102885
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 3 to 26 1.998
duplicate number number
spacings observed expected
0 69. 67.668
1 133. 135.335
2 133. 135.335
3 99. 90.224
4 39. 45.112
5 18. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 1.85 p-value= .067118
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 4 to 27 1.996
duplicate number number
spacings observed expected
0 63. 67.668
1 131. 135.335
2 146. 135.335
3 93. 90.224
4 48. 45.112
5 12. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 3.79 p-value= .295589
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 5 to 28 1.912
duplicate number number
spacings observed expected
0 76. 67.668
1 138. 135.335
2 132. 135.335
3 93. 90.224
4 39. 45.112
5 15. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 2.79 p-value= .164846
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 6 to 29 1.942
duplicate number number
spacings observed expected
0 70. 67.668
1 142. 135.335
2 139. 135.335
3 86. 90.224
4 36. 45.112
5 18. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 2.61 p-value= .143850
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 7 to 30 1.954
duplicate number number
spacings observed expected
0 78. 67.668
1 127. 135.335
2 140. 135.335
3 79. 90.224
4 52. 45.112
5 19. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. = 6.05 p-value= .582480
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 8 to 31 1.886
duplicate number number
spacings observed expected
0 83. 67.668
1 138. 135.335
2 138. 135.335
3 68. 90.224
4 50. 45.112
5 15. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. = 10.11 p-value= .879742
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
SG100.DAT using bits 9 to 32 2.028
duplicate number number
spacings observed expected
0 62. 67.668
1 137. 135.335
2 143. 135.335
3 81. 90.224
4 48. 45.112
5 19. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. = 2.46 p-value= .127542
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.917554 .102885 .067118 .295589 .164846
.143850 .582480 .879742 .127542
A KSTEST for the 9 p-values yields .878397
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 SG100.DAT
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=101.603; p-value= .591239
OPERM5 test for file SG100.DAT
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=105.512; p-value= .691597
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 SG100.DAT
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 229 211.4 1.462156 1.462
29 5116 5134.0 .063180 1.525
30 23100 23103.0 .000402 1.526
31 11555 11551.5 .001046 1.527
chisquare= 1.527 for 3 d. of f.; p-value= .432196
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 SG100.DAT
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 200 211.4 .616651 .617
30 5211 5134.0 1.154540 1.771
31 22819 23103.0 3.492294 5.263
32 11770 11551.5 4.132059 9.396
chisquare= 9.396 for 3 d. of f.; p-value= .976619
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 SG100.DAT
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 899 944.3 2.173 2.173
r =5 21504 21743.9 2.647 4.820
r =6 77597 77311.8 1.052 5.872
p=1-exp(-SUM/2)= .94693
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21545 21743.9 1.819 2.668
r =6 77539 77311.8 .668 3.335
p=1-exp(-SUM/2)= .81131
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 3 to 10
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 969 944.3 .646 .646
r =5 21433 21743.9 4.445 5.091
r =6 77598 77311.8 1.059 6.151
p=1-exp(-SUM/2)= .95383
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 933 944.3 .135 .135
r =5 21743 21743.9 .000 .135
r =6 77324 77311.8 .002 .137
p=1-exp(-SUM/2)= .06631
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 988 944.3 2.022 2.022
r =5 21677 21743.9 .206 2.228
r =6 77335 77311.8 .007 2.235
p=1-exp(-SUM/2)= .67290
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 958 944.3 .199 .199
r =5 21573 21743.9 1.343 1.542
r =6 77469 77311.8 .320 1.862
p=1-exp(-SUM/2)= .60576
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 931 944.3 .187 .187
r =5 21732 21743.9 .007 .194
r =6 77337 77311.8 .008 .202
p=1-exp(-SUM/2)= .09610
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 948 944.3 .014 .014
r =5 21946 21743.9 1.878 1.893
r =6 77106 77311.8 .548 2.441
p=1-exp(-SUM/2)= .70488
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 945 944.3 .001 .001
r =5 21729 21743.9 .010 .011
r =6 77326 77311.8 .003 .013
p=1-exp(-SUM/2)= .00664
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 899 944.3 2.173 2.173
r =5 21656 21743.9 .355 2.529
r =6 77445 77311.8 .229 2.758
p=1-exp(-SUM/2)= .74818
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 905 944.3 1.636 1.636
r =5 21747 21743.9 .000 1.636
r =6 77348 77311.8 .017 1.653
p=1-exp(-SUM/2)= .56244
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 941 944.3 .012 .012
r =5 21834 21743.9 .373 .385
r =6 77225 77311.8 .097 .482
p=1-exp(-SUM/2)= .21429
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 930 944.3 .217 .217
r =5 21780 21743.9 .060 .277
r =6 77290 77311.8 .006 .283
p=1-exp(-SUM/2)= .13180
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 915 944.3 .909 .909
r =5 21836 21743.9 .390 1.299
r =6 77249 77311.8 .051 1.350
p=1-exp(-SUM/2)= .49093
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 931 944.3 .187 .187
r =5 21872 21743.9 .755 .942
r =6 77197 77311.8 .170 1.113
p=1-exp(-SUM/2)= .42665
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 942 944.3 .006 .006
r =5 21776 21743.9 .047 .053
r =6 77282 77311.8 .011 .064
p=1-exp(-SUM/2)= .03173
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 993 944.3 2.511 2.511
r =5 21686 21743.9 .154 2.666
r =6 77321 77311.8 .001 2.667
p=1-exp(-SUM/2)= .73641
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 960 944.3 .261 .261
r =5 21597 21743.9 .992 1.253
r =6 77443 77311.8 .223 1.476
p=1-exp(-SUM/2)= .52195
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 952 944.3 .063 .063
r =5 21893 21743.9 1.022 1.085
r =6 77155 77311.8 .318 1.403
p=1-exp(-SUM/2)= .50421
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 959 944.3 .229 .229
r =5 21661 21743.9 .316 .545
r =6 77380 77311.8 .060 .605
p=1-exp(-SUM/2)= .26104
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 911 944.3 1.174 1.174
r =5 21646 21743.9 .441 1.615
r =6 77443 77311.8 .223 1.838
p=1-exp(-SUM/2)= .60104
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 941 944.3 .012 .012
r =5 21645 21743.9 .450 .461
r =6 77414 77311.8 .135 .596
p=1-exp(-SUM/2)= .25787
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 968 944.3 .595 .595
r =5 21827 21743.9 .318 .912
r =6 77205 77311.8 .148 1.060
p=1-exp(-SUM/2)= .41136
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 964 944.3 .411 .411
r =5 21643 21743.9 .468 .879
r =6 77393 77311.8 .085 .964
p=1-exp(-SUM/2)= .38258
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG SG100.DAT
b-rank test for bits 25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 954 944.3 .100 .100
r =5 21674 21743.9 .225 .324
r =6 77372 77311.8 .047 .371
p=1-exp(-SUM/2)= .16939
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.946926 .811309 .953829 .066306 .672905
.605756 .096105 .704883 .006645 .748178
.562440 .214295 .131803 .490927 .426648
.031729 .736410 .521948 .504205 .261037
.601044 .257873 .411363 .382583 .169391
brank test summary for SG100.DAT
The KS test for those 25 supposed UNI's yields
KS p-value= .286560
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 142407 missing words, 1.16 sigmas from mean, p-value= .87754
tst no 2: 141541 missing words, -.86 sigmas from mean, p-value= .19473
tst no 3: 141911 missing words, .00 sigmas from mean, p-value= .50156
tst no 4: 141300 missing words, -1.42 sigmas from mean, p-value= .07727
tst no 5: 142023 missing words, .27 sigmas from mean, p-value= .60472
tst no 6: 141841 missing words, -.16 sigmas from mean, p-value= .43658
tst no 7: 142325 missing words, .97 sigmas from mean, p-value= .83427
tst no 8: 141689 missing words, -.51 sigmas from mean, p-value= .30335
tst no 9: 142087 missing words, .42 sigmas from mean, p-value= .66097
tst no 10: 142340 missing words, 1.01 sigmas from mean, p-value= .84285
tst no 11: 141340 missing words, -1.33 sigmas from mean, p-value= .09173
tst no 12: 142234 missing words, .76 sigmas from mean, p-value= .77595
tst no 13: 141629 missing words, -.65 sigmas from mean, p-value= .25624
tst no 14: 141704 missing words, -.48 sigmas from mean, p-value= .31571
tst no 15: 141934 missing words, .06 sigmas from mean, p-value= .52298
tst no 16: 142176 missing words, .62 sigmas from mean, p-value= .73338
tst no 17: 142170 missing words, .61 sigmas from mean, p-value= .72875
tst no 18: 141332 missing words, -1.35 sigmas from mean, p-value= .08868
tst no 19: 142431 missing words, 1.22 sigmas from mean, p-value= .88855
tst no 20: 141749 missing words, -.37 sigmas from mean, p-value= .35398
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 SG100.DAT
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OPSO for SG100.DAT using bits 23 to 32 142035 .433 .6676
OPSO for SG100.DAT using bits 22 to 31 142206 1.023 .8468
OPSO for SG100.DAT using bits 21 to 30 142083 .599 .7254
OPSO for SG100.DAT using bits 20 to 29 141822 -.301 .3817
OPSO for SG100.DAT using bits 19 to 28 142119 .723 .7652
OPSO for SG100.DAT using bits 18 to 27 141808 -.349 .3634
OPSO for SG100.DAT using bits 17 to 26 142236 1.126 .8700
OPSO for SG100.DAT using bits 16 to 25 142060 .520 .6983
OPSO for SG100.DAT using bits 15 to 24 141929 .068 .5270
OPSO for SG100.DAT using bits 14 to 23 141975 .226 .5896
OPSO for SG100.DAT using bits 13 to 22 142114 .706 .7598
OPSO for SG100.DAT using bits 12 to 21 141535 -1.291 .0984
OPSO for SG100.DAT using bits 11 to 20 141747 -.560 .2878
OPSO for SG100.DAT using bits 10 to 19 142167 .889 .8129
OPSO for SG100.DAT using bits 9 to 18 141718 -.660 .2547
OPSO for SG100.DAT using bits 8 to 17 142209 1.033 .8493
OPSO for SG100.DAT using bits 7 to 16 141934 .085 .5339
OPSO for SG100.DAT using bits 6 to 15 141859 -.174 .4311
OPSO for SG100.DAT using bits 5 to 14 141410 -1.722 .0426
OPSO for SG100.DAT using bits 4 to 13 142222 1.078 .8595
OPSO for SG100.DAT using bits 3 to 12 141659 -.863 .1940
OPSO for SG100.DAT using bits 2 to 11 142047 .475 .6825
OPSO for SG100.DAT using bits 1 to 10 142193 .978 .8360
OQSO test for generator SG100.DAT
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OQSO for SG100.DAT using bits 28 to 32 142465 1.884 .9702
OQSO for SG100.DAT using bits 27 to 31 141617 -.991 .1609
OQSO for SG100.DAT using bits 26 to 30 141894 -.052 .4793
OQSO for SG100.DAT using bits 25 to 29 141470 -1.489 .0682
OQSO for SG100.DAT using bits 24 to 28 141928 .063 .5252
OQSO for SG100.DAT using bits 23 to 27 142267 1.212 .8873
OQSO for SG100.DAT using bits 22 to 26 141749 -.543 .2934
OQSO for SG100.DAT using bits 21 to 25 141473 -1.479 .0696
OQSO for SG100.DAT using bits 20 to 24 142092 .619 .7321
OQSO for SG100.DAT using bits 19 to 23 142120 .714 .7624
OQSO for SG100.DAT using bits 18 to 22 142048 .470 .6808
OQSO for SG100.DAT using bits 17 to 21 141858 -.174 .4309
OQSO for SG100.DAT using bits 16 to 20 142111 .684 .7529
OQSO for SG100.DAT using bits 15 to 19 141522 -1.313 .0946
OQSO for SG100.DAT using bits 14 to 18 142184 .931 .8241
OQSO for SG100.DAT using bits 13 to 17 141730 -.608 .2716
OQSO for SG100.DAT using bits 12 to 16 141629 -.950 .1710
OQSO for SG100.DAT using bits 11 to 15 141991 .277 .6091
OQSO for SG100.DAT using bits 10 to 14 141792 -.398 .3454
OQSO for SG100.DAT using bits 9 to 13 141863 -.157 .4376
OQSO for SG100.DAT using bits 8 to 12 141447 -1.567 .0585
OQSO for SG100.DAT using bits 7 to 11 141956 .158 .5629
OQSO for SG100.DAT using bits 6 to 10 141929 .067 .5266
OQSO for SG100.DAT using bits 5 to 9 142342 1.467 .9288
OQSO for SG100.DAT using bits 4 to 8 142129 .745 .7718
OQSO for SG100.DAT using bits 3 to 7 142006 .328 .6284
OQSO for SG100.DAT using bits 2 to 6 141572 -1.143 .1264
OQSO for SG100.DAT using bits 1 to 5 141426 -1.638 .0507
DNA test for generator SG100.DAT
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for SG100.DAT using bits 31 to 32 141565 -1.016 .1549
DNA for SG100.DAT using bits 30 to 31 141623 -.845 .1992
DNA for SG100.DAT using bits 29 to 30 141319 -1.741 .0408
DNA for SG100.DAT using bits 28 to 29 141545 -1.075 .1413
DNA for SG100.DAT using bits 27 to 28 141534 -1.107 .1341
DNA for SG100.DAT using bits 26 to 27 141978 .203 .5803
DNA for SG100.DAT using bits 25 to 26 142095 .548 .7081
DNA for SG100.DAT using bits 24 to 25 141437 -1.393 .0818
DNA for SG100.DAT using bits 23 to 24 141847 -.184 .4271
DNA for SG100.DAT using bits 22 to 23 141959 .147 .5582
DNA for SG100.DAT using bits 21 to 22 141939 .088 .5349
DNA for SG100.DAT using bits 20 to 21 141604 -.901 .1839
DNA for SG100.DAT using bits 19 to 20 142019 .324 .6268
DNA for SG100.DAT using bits 18 to 19 141617 -.862 .1943
DNA for SG100.DAT using bits 17 to 18 142339 1.267 .8975
DNA for SG100.DAT using bits 16 to 17 142107 .583 .7201
DNA for SG100.DAT using bits 15 to 16 142363 1.338 .9096
DNA for SG100.DAT using bits 14 to 15 141648 -.771 .2204
DNA for SG100.DAT using bits 13 to 14 142102 .568 .7151
DNA for SG100.DAT using bits 12 to 13 141713 -.579 .2812
DNA for SG100.DAT using bits 11 to 12 142524 1.813 .9651
DNA for SG100.DAT using bits 10 to 11 141723 -.550 .2913
DNA for SG100.DAT using bits 9 to 10 142046 .403 .6566
DNA for SG100.DAT using bits 8 to 9 142373 1.368 .9143
DNA for SG100.DAT using bits 7 to 8 141894 -.045 .4820
DNA for SG100.DAT using bits 6 to 7 141222 -2.028 .0213
DNA for SG100.DAT using bits 5 to 6 141682 -.671 .2512
DNA for SG100.DAT using bits 4 to 5 142131 .654 .7434
DNA for SG100.DAT using bits 3 to 4 142307 1.173 .8796
DNA for SG100.DAT using bits 2 to 3 142155 .725 .7657
DNA for SG100.DAT using bits 1 to 2 142025 .341 .6335
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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 SG100.DAT
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 SG100.DAT 2489.05 -.155 .438457
byte stream for SG100.DAT 2559.93 .848 .801665
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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 2413.74 -1.220 .111244
bits 2 to 9 2548.37 .684 .753024
bits 3 to 10 2537.38 .529 .701486
bits 4 to 11 2428.28 -1.014 .155212
bits 5 to 12 2636.03 1.924 .972811
bits 6 to 13 2464.13 -.507 .305996
bits 7 to 14 2456.64 -.613 .269886
bits 8 to 15 2589.84 1.271 .898054
bits 9 to 16 2511.27 .159 .563300
bits 10 to 17 2430.62 -.981 .163254
bits 11 to 18 2527.38 .387 .650707
bits 12 to 19 2541.18 .582 .719840
bits 13 to 20 2461.69 -.542 .293971
bits 14 to 21 2330.63 -2.395 .008304
bits 15 to 22 2507.48 .106 .542138
bits 16 to 23 2522.38 .316 .624174
bits 17 to 24 2468.29 -.448 .326932
bits 18 to 25 2435.00 -.919 .178980
bits 19 to 26 2491.51 -.120 .452231
bits 20 to 27 2482.81 -.243 .403953
bits 21 to 28 2637.01 1.938 .973668
bits 22 to 29 2466.95 -.467 .320100
bits 23 to 30 2444.05 -.791 .214394
bits 24 to 31 2584.92 1.201 .885126
bits 25 to 32 2577.95 1.102 .864839
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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 SG100.DAT
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3557 z-score: 1.553 p-value: .939730
Successes: 3550 z-score: 1.233 p-value: .891189
Successes: 3506 z-score: -.776 p-value: .218799
Successes: 3502 z-score: -.959 p-value: .168804
Successes: 3556 z-score: 1.507 p-value: .934075
Successes: 3523 z-score: .000 p-value: .500000
Successes: 3517 z-score: -.274 p-value: .392053
Successes: 3539 z-score: .731 p-value: .767486
Successes: 3557 z-score: 1.553 p-value: .939730
Successes: 3549 z-score: 1.187 p-value: .882429
square size avg. no. parked sample sigma
100. 3535.600 20.587
KSTEST for the above 10: p= .945715
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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 SG100.DAT
Sample no. d^2 avg equiv uni
5 .3073 .4196 .265694
10 .6224 .6057 .465034
15 .0719 .7005 .069731
20 .0141 .7617 .014051
25 2.5423 .9465 .922313
30 1.0586 .8824 .654903
35 .8508 .8214 .574768
40 .2105 .7611 .190671
45 .7400 .7450 .524675
50 1.3736 .7899 .748560
55 .5317 .8405 .413977
60 .8544 .8232 .576305
65 .5630 .8249 .432100
70 .0735 .8211 .071233
75 4.2369 .8391 .985852
80 .5104 .8073 .401262
85 1.2002 .8116 .700678
90 .3804 .8260 .317741
95 1.0486 .8048 .651408
100 1.5839 .8236 .796456
MINIMUM DISTANCE TEST for SG100.DAT
Result of KS test on 20 transformed mindist^2's:
p-value= .752837
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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 SG100.DAT
sample no: 1 r^3= 41.137 p-value= .74621
sample no: 2 r^3= 7.369 p-value= .21780
sample no: 3 r^3= 24.243 p-value= .55430
sample no: 4 r^3= 35.577 p-value= .69453
sample no: 5 r^3= 24.337 p-value= .55569
sample no: 6 r^3= 2.268 p-value= .07281
sample no: 7 r^3= 3.260 p-value= .10298
sample no: 8 r^3= 47.719 p-value= .79620
sample no: 9 r^3= 17.620 p-value= .44420
sample no: 10 r^3= .868 p-value= .02851
sample no: 11 r^3= 23.506 p-value= .54320
sample no: 12 r^3= 23.901 p-value= .54918
sample no: 13 r^3= 48.215 p-value= .79954
sample no: 14 r^3= 119.179 p-value= .98118
sample no: 15 r^3= 31.909 p-value= .65480
sample no: 16 r^3= 29.470 p-value= .62557
sample no: 17 r^3= 34.265 p-value= .68088
sample no: 18 r^3= 11.700 p-value= .32294
sample no: 19 r^3= 14.170 p-value= .37645
sample no: 20 r^3= 73.255 p-value= .91300
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file SG100.DAT p-value= .243269
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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 SG100.DAT
Table of standardized frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
.6 .1 -1.3 1.1 1.4 1.1
.2 1.4 .2 -.2 .9 -.1
-.4 .7 1.6 1.7 -1.9 -.8
-.7 -.4 .3 -.3 -.1 -1.5
.4 .9 -.8 -.4 -.2 .4
-.9 -.1 -.1 -.9 -.6 -1.0
.5 .2 .5 .4 -.6 -1.0
.8
Chi-square with 42 degrees of freedom: 30.647
z-score= -1.239 p-value= .097122
______________________________________________________________
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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 .330733
Test no. 2 p-value .232319
Test no. 3 p-value .893748
Test no. 4 p-value .057761
Test no. 5 p-value .662586
Test no. 6 p-value .447470
Test no. 7 p-value .916826
Test no. 8 p-value .584608
Test no. 9 p-value .919413
Test no. 10 p-value .258387
Results of the OSUM test for SG100.DAT
KSTEST on the above 10 p-values: .107964
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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 SG100.DAT
Up and down runs in a sample of 10000
_________________________________________________
Run test for SG100.DAT :
runs up; ks test for 10 p's: .508297
runs down; ks test for 10 p's: .373533
Run test for SG100.DAT :
runs up; ks test for 10 p's: .188019
runs down; ks test for 10 p's: .230541
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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 SG100.DAT
No. of wins: Observed Expected
98211 98585.86
98211= No. of wins, z-score=-1.677 pvalue= .04681
Analysis of Throws-per-Game:
Chisq= 22.20 for 20 degrees of freedom, p= .67012
Throws Observed Expected Chisq Sum
1 66367 66666.7 1.347 1.347
2 37696 37654.3 .046 1.393
3 27053 26954.7 .358 1.751
4 19610 19313.5 4.553 6.305
5 13787 13851.4 .300 6.604
6 9847 9943.5 .937 7.541
7 7198 7145.0 .393 7.934
8 5151 5139.1 .028 7.962
9 3723 3699.9 .145 8.107
10 2677 2666.3 .043 8.150
11 1880 1923.3 .976 9.126
12 1338 1388.7 1.854 10.980
13 997 1003.7 .045 11.024
14 772 726.1 2.896 13.921
15 530 525.8 .033 13.954
16 391 381.2 .255 14.208
17 256 276.5 1.526 15.734
18 198 200.8 .040 15.774
19 160 146.0 1.346 17.119
20 83 106.2 5.074 22.193
21 286 287.1 .004 22.198
SUMMARY FOR SG100.DAT
p-value for no. of wins: .046811
p-value for throws/game: .670116
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Results of DIEHARD battery of tests sent to file SG100.TXT