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 bbs.32
For a sample of size 500: mean
bbs.32 using bits 1 to 24 2.046
duplicate number number
spacings observed expected
0 63. 67.668
1 131. 135.335
2 142. 135.335
3 93. 90.224
4 41. 45.112
5 19. 18.045
6 to INF 11. 8.282
Chisquare with 6 d.o.f. = 2.19 p-value= .098778
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 2 to 25 1.968
duplicate number number
spacings observed expected
0 75. 67.668
1 132. 135.335
2 134. 135.335
3 91. 90.224
4 40. 45.112
5 19. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 1.59 p-value= .046611
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 3 to 26 1.940
duplicate number number
spacings observed expected
0 76. 67.668
1 142. 135.335
2 121. 135.335
3 91. 90.224
4 46. 45.112
5 17. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 3.16 p-value= .210937
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 4 to 27 2.058
duplicate number number
spacings observed expected
0 64. 67.668
1 138. 135.335
2 131. 135.335
3 89. 90.224
4 47. 45.112
5 18. 18.045
6 to INF 13. 8.282
Chisquare with 6 d.o.f. = 3.17 p-value= .213268
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 5 to 28 1.916
duplicate number number
spacings observed expected
0 68. 67.668
1 154. 135.335
2 118. 135.335
3 96. 90.224
4 47. 45.112
5 13. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 8.87 p-value= .818933
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 6 to 29 1.958
duplicate number number
spacings observed expected
0 61. 67.668
1 157. 135.335
2 125. 135.335
3 89. 90.224
4 47. 45.112
5 13. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. = 6.43 p-value= .623214
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 7 to 30 2.094
duplicate number number
spacings observed expected
0 56. 67.668
1 123. 135.335
2 149. 135.335
3 93. 90.224
4 54. 45.112
5 21. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 9.05 p-value= .829229
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 8 to 31 2.030
duplicate number number
spacings observed expected
0 63. 67.668
1 118. 135.335
2 154. 135.335
3 96. 90.224
4 48. 45.112
5 17. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 7.95 p-value= .757884
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
bbs.32 using bits 9 to 32 1.986
duplicate number number
spacings observed expected
0 61. 67.668
1 137. 135.335
2 142. 135.335
3 94. 90.224
4 48. 45.112
5 12. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 4.00 p-value= .323627
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.098778 .046611 .210937 .213268 .818933
.623214 .829229 .757884 .323627
A KSTEST for the 9 p-values yields .327992
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 bbs.32
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=147.002; p-value= .998737
OPERM5 test for file bbs.32
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=187.354; p-value=1.000000
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 bbs.32
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 202 211.4 .419543 .420
29 5163 5134.0 .163694 .583
30 23139 23103.0 .055951 .639
31 11496 11551.5 .266888 .906
chisquare= .906 for 3 d. of f.; p-value= .346618
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 bbs.32
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 183 211.4 3.819843 3.820
30 5130 5134.0 .003132 3.823
31 23198 23103.0 .390256 4.213
32 11489 11551.5 .338423 4.552
chisquare= 4.552 for 3 d. of f.; p-value= .808715
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 bbs.32
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21662 21743.9 .308 .400
r =6 77403 77311.8 .108 .508
p=1-exp(-SUM/2)= .22418
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 949 944.3 .023 .023
r =5 21585 21743.9 1.161 1.185
r =6 77466 77311.8 .308 1.492
p=1-exp(-SUM/2)= .52577
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 3 to 10
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 939 944.3 .030 .030
r =5 21377 21743.9 6.191 6.221
r =6 77684 77311.8 1.792 8.013
p=1-exp(-SUM/2)= .98180
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 907 944.3 1.473 1.473
r =5 21674 21743.9 .225 1.698
r =6 77419 77311.8 .149 1.847
p=1-exp(-SUM/2)= .60283
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 979 944.3 1.275 1.275
r =5 21642 21743.9 .478 1.753
r =6 77379 77311.8 .058 1.811
p=1-exp(-SUM/2)= .59565
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 983 944.3 1.586 1.586
r =5 21618 21743.9 .729 2.315
r =6 77399 77311.8 .098 2.413
p=1-exp(-SUM/2)= .70079
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 922 944.3 .527 .527
r =5 21705 21743.9 .070 .596
r =6 77373 77311.8 .048 .645
p=1-exp(-SUM/2)= .27556
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 977 944.3 1.132 1.132
r =5 21588 21743.9 1.118 2.250
r =6 77435 77311.8 .196 2.446
p=1-exp(-SUM/2)= .70571
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 920 944.3 .625 .625
r =5 21831 21743.9 .349 .974
r =6 77249 77311.8 .051 1.025
p=1-exp(-SUM/2)= .40109
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 932 944.3 .160 .160
r =5 21629 21743.9 .607 .767
r =6 77439 77311.8 .209 .977
p=1-exp(-SUM/2)= .38635
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21551 21743.9 1.711 2.560
r =6 77533 77311.8 .633 3.192
p=1-exp(-SUM/2)= .79733
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 945 944.3 .001 .001
r =5 21599 21743.9 .966 .966
r =6 77456 77311.8 .269 1.235
p=1-exp(-SUM/2)= .46073
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r =5 21622 21743.9 .683 .691
r =6 77431 77311.8 .184 .875
p=1-exp(-SUM/2)= .35431
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 962 944.3 .332 .332
r =5 21542 21743.9 1.875 2.206
r =6 77496 77311.8 .439 2.645
p=1-exp(-SUM/2)= .73357
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21585 21743.9 1.161 1.253
r =6 77480 77311.8 .366 1.619
p=1-exp(-SUM/2)= .55486
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 925 944.3 .395 .395
r =5 21541 21743.9 1.893 2.288
r =6 77534 77311.8 .639 2.926
p=1-exp(-SUM/2)= .76851
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21639 21743.9 .506 1.354
r =6 77445 77311.8 .229 1.584
p=1-exp(-SUM/2)= .54701
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 980 944.3 1.350 1.350
r =5 21584 21743.9 1.176 2.525
r =6 77436 77311.8 .200 2.725
p=1-exp(-SUM/2)= .74397
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 920 944.3 .625 .625
r =5 21666 21743.9 .279 .904
r =6 77414 77311.8 .135 1.040
p=1-exp(-SUM/2)= .40535
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 964 944.3 .411 .411
r =5 21529 21743.9 2.124 2.535
r =6 77507 77311.8 .493 3.028
p=1-exp(-SUM/2)= .77994
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 976 944.3 1.064 1.064
r =5 21359 21743.9 6.813 7.877
r =6 77665 77311.8 1.614 9.491
p=1-exp(-SUM/2)= .99131
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 976 944.3 1.064 1.064
r =5 21665 21743.9 .286 1.350
r =6 77359 77311.8 .029 1.379
p=1-exp(-SUM/2)= .49822
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 945 944.3 .001 .001
r =5 21659 21743.9 .331 .332
r =6 77396 77311.8 .092 .424
p=1-exp(-SUM/2)= .19092
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 929 944.3 .248 .248
r =5 21739 21743.9 .001 .249
r =6 77332 77311.8 .005 .254
p=1-exp(-SUM/2)= .11941
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG bbs.32
b-rank test for bits 25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 930 944.3 .217 .217
r =5 21860 21743.9 .620 .836
r =6 77210 77311.8 .134 .971
p=1-exp(-SUM/2)= .38447
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.224182 .525771 .981799 .602832 .595654
.700795 .275561 .705708 .401094 .386354
.797332 .460727 .354312 .733570 .554863
.768511 .547007 .743974 .405349 .779935
.991309 .498220 .190917 .119407 .384472
brank test summary for bbs.32
The KS test for those 25 supposed UNI's yields
KS p-value= .745974
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 143655 missing words, 4.08 sigmas from mean, p-value= .99998
tst no 2: 143492 missing words, 3.70 sigmas from mean, p-value= .99989
tst no 3: 143648 missing words, 4.06 sigmas from mean, p-value= .99998
tst no 4: 143430 missing words, 3.55 sigmas from mean, p-value= .99981
tst no 5: 144164 missing words, 5.27 sigmas from mean, p-value=1.00000
tst no 6: 143382 missing words, 3.44 sigmas from mean, p-value= .99971
tst no 7: 143469 missing words, 3.64 sigmas from mean, p-value= .99987
tst no 8: 143690 missing words, 4.16 sigmas from mean, p-value= .99998
tst no 9: 143449 missing words, 3.60 sigmas from mean, p-value= .99984
tst no 10: 144108 missing words, 5.14 sigmas from mean, p-value=1.00000
tst no 11: 143443 missing words, 3.58 sigmas from mean, p-value= .99983
tst no 12: 143521 missing words, 3.77 sigmas from mean, p-value= .99992
tst no 13: 143785 missing words, 4.38 sigmas from mean, p-value= .99999
tst no 14: 143482 missing words, 3.67 sigmas from mean, p-value= .99988
tst no 15: 144113 missing words, 5.15 sigmas from mean, p-value=1.00000
tst no 16: 143448 missing words, 3.60 sigmas from mean, p-value= .99984
tst no 17: 143471 missing words, 3.65 sigmas from mean, p-value= .99987
tst no 18: 143875 missing words, 4.59 sigmas from mean, p-value=1.00000
tst no 19: 143541 missing words, 3.81 sigmas from mean, p-value= .99993
tst no 20: 144047 missing words, 4.99 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 bbs.32
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OPSO for bbs.32 using bits 23 to 32 556044******* 1.0000
OPSO for bbs.32 using bits 22 to 31 556021******* 1.0000
OPSO for bbs.32 using bits 21 to 30 556191******* 1.0000
OPSO for bbs.32 using bits 20 to 29 556130******* 1.0000
OPSO for bbs.32 using bits 19 to 28 556033******* 1.0000
OPSO for bbs.32 using bits 18 to 27 556680******* 1.0000
OPSO for bbs.32 using bits 17 to 26 556939******* 1.0000
OPSO for bbs.32 using bits 16 to 25 555897******* 1.0000
OPSO for bbs.32 using bits 15 to 24 556034******* 1.0000
OPSO for bbs.32 using bits 14 to 23 556063******* 1.0000
OPSO for bbs.32 using bits 13 to 22 556181******* 1.0000
OPSO for bbs.32 using bits 12 to 21 556088******* 1.0000
OPSO for bbs.32 using bits 11 to 20 555978******* 1.0000
OPSO for bbs.32 using bits 10 to 19 556623******* 1.0000
OPSO for bbs.32 using bits 9 to 18 557106******* 1.0000
OPSO for bbs.32 using bits 8 to 17 556027******* 1.0000
OPSO for bbs.32 using bits 7 to 16 556000******* 1.0000
OPSO for bbs.32 using bits 6 to 15 556117******* 1.0000
OPSO for bbs.32 using bits 5 to 14 556231******* 1.0000
OPSO for bbs.32 using bits 4 to 13 556157******* 1.0000
OPSO for bbs.32 using bits 3 to 12 555983******* 1.0000
OPSO for bbs.32 using bits 2 to 11 556656******* 1.0000
OPSO for bbs.32 using bits 1 to 10 557002******* 1.0000
OQSO test for generator bbs.32
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OQSO for bbs.32 using bits 28 to 32 554451******* 1.0000
OQSO for bbs.32 using bits 27 to 31 554657******* 1.0000
OQSO for bbs.32 using bits 26 to 30 554543******* 1.0000
OQSO for bbs.32 using bits 25 to 29 554300******* 1.0000
OQSO for bbs.32 using bits 24 to 28 554419******* 1.0000
OQSO for bbs.32 using bits 23 to 27 554326******* 1.0000
OQSO for bbs.32 using bits 22 to 26 554390******* 1.0000
OQSO for bbs.32 using bits 21 to 25 554629******* 1.0000
OQSO for bbs.32 using bits 20 to 24 554508******* 1.0000
OQSO for bbs.32 using bits 19 to 23 554570******* 1.0000
OQSO for bbs.32 using bits 18 to 22 554574******* 1.0000
OQSO for bbs.32 using bits 17 to 21 554179******* 1.0000
OQSO for bbs.32 using bits 16 to 20 554394******* 1.0000
OQSO for bbs.32 using bits 15 to 19 554367******* 1.0000
OQSO for bbs.32 using bits 14 to 18 554465******* 1.0000
OQSO for bbs.32 using bits 13 to 17 554580******* 1.0000
OQSO for bbs.32 using bits 12 to 16 554406******* 1.0000
OQSO for bbs.32 using bits 11 to 15 554556******* 1.0000
OQSO for bbs.32 using bits 10 to 14 554551******* 1.0000
OQSO for bbs.32 using bits 9 to 13 554235******* 1.0000
OQSO for bbs.32 using bits 8 to 12 554477******* 1.0000
OQSO for bbs.32 using bits 7 to 11 554273******* 1.0000
OQSO for bbs.32 using bits 6 to 10 554595******* 1.0000
OQSO for bbs.32 using bits 5 to 9 554624******* 1.0000
OQSO for bbs.32 using bits 4 to 8 554468******* 1.0000
OQSO for bbs.32 using bits 3 to 7 554684******* 1.0000
OQSO for bbs.32 using bits 2 to 6 554602******* 1.0000
OQSO for bbs.32 using bits 1 to 5 554299******* 1.0000
DNA test for generator bbs.32
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for bbs.32 using bits 31 to 32 553983******* 1.0000
DNA for bbs.32 using bits 30 to 31 554454******* 1.0000
DNA for bbs.32 using bits 29 to 30 554507******* 1.0000
DNA for bbs.32 using bits 28 to 29 554595******* 1.0000
DNA for bbs.32 using bits 27 to 28 553789******* 1.0000
DNA for bbs.32 using bits 26 to 27 554648******* 1.0000
DNA for bbs.32 using bits 25 to 26 554889******* 1.0000
DNA for bbs.32 using bits 24 to 25 554417******* 1.0000
DNA for bbs.32 using bits 23 to 24 553857******* 1.0000
DNA for bbs.32 using bits 22 to 23 554470******* 1.0000
DNA for bbs.32 using bits 21 to 22 554379******* 1.0000
DNA for bbs.32 using bits 20 to 21 554478******* 1.0000
DNA for bbs.32 using bits 19 to 20 553947******* 1.0000
DNA for bbs.32 using bits 18 to 19 554751******* 1.0000
DNA for bbs.32 using bits 17 to 18 554720******* 1.0000
DNA for bbs.32 using bits 16 to 17 554601******* 1.0000
DNA for bbs.32 using bits 15 to 16 553971******* 1.0000
DNA for bbs.32 using bits 14 to 15 554489******* 1.0000
DNA for bbs.32 using bits 13 to 14 554487******* 1.0000
DNA for bbs.32 using bits 12 to 13 554490******* 1.0000
DNA for bbs.32 using bits 11 to 12 553876******* 1.0000
DNA for bbs.32 using bits 10 to 11 554751******* 1.0000
DNA for bbs.32 using bits 9 to 10 554617******* 1.0000
DNA for bbs.32 using bits 8 to 9 554448******* 1.0000
DNA for bbs.32 using bits 7 to 8 554003******* 1.0000
DNA for bbs.32 using bits 6 to 7 554534******* 1.0000
DNA for bbs.32 using bits 5 to 6 554456******* 1.0000
DNA for bbs.32 using bits 4 to 5 554431******* 1.0000
DNA for bbs.32 using bits 3 to 4 553940******* 1.0000
DNA for bbs.32 using bits 2 to 3 554672******* 1.0000
DNA for bbs.32 using bits 1 to 2 554748******* 1.0000
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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 bbs.32
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 bbs.32 10844.84 118.014 1.000000
byte stream for bbs.32 11197.53 123.002 1.000000
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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 2502.96 .042 .516674
bits 2 to 9 2431.26 -.972 .165510
bits 3 to 10 2638.48 1.958 .974911
bits 4 to 11 2446.33 -.759 .223936
bits 5 to 12 2619.88 1.695 .954994
bits 6 to 13 2487.37 -.179 .429143
bits 7 to 14 2495.13 -.069 .472543
bits 8 to 15 2493.78 -.088 .464977
bits 9 to 16 2477.33 -.321 .374258
bits 10 to 17 2584.70 1.198 .884515
bits 11 to 18 2638.29 1.956 .974750
bits 12 to 19 2493.99 -.085 .466143
bits 13 to 20 2539.12 .553 .709964
bits 14 to 21 2433.14 -.946 .172180
bits 15 to 22 2488.89 -.157 .437574
bits 16 to 23 2612.95 1.597 .944906
bits 17 to 24 2509.74 .138 .554775
bits 18 to 25 2529.68 .420 .662638
bits 19 to 26 2452.17 -.676 .249398
bits 20 to 27 2404.22 -1.355 .087774
bits 21 to 28 2411.41 -1.253 .105123
bits 22 to 29 2480.67 -.273 .392258
bits 23 to 30 2506.54 .093 .536866
bits 24 to 31 2519.82 .280 .610355
bits 25 to 32 2507.45 .105 .541979
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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 bbs.32
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3506 z-score: -.776 p-value: .218799
Successes: 3541 z-score: .822 p-value: .794438
Successes: 3535 z-score: .548 p-value: .708135
Successes: 3480 z-score: -1.963 p-value: .024796
Successes: 3547 z-score: 1.096 p-value: .863437
Successes: 3476 z-score: -2.146 p-value: .015932
Successes: 3548 z-score: 1.142 p-value: .873180
Successes: 3540 z-score: .776 p-value: .781201
Successes: 3551 z-score: 1.279 p-value: .899470
Successes: 3524 z-score: .046 p-value: .518210
square size avg. no. parked sample sigma
100. 3524.800 26.551
KSTEST for the above 10: p= .781492
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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 bbs.32
Sample no. d^2 avg equiv uni
5 1.5626 .8398 .792047
10 1.5134 1.1995 .781510
15 .7072 1.0606 .508735
20 .0317 .8967 .031319
25 .9343 .9578 .608993
30 .2187 .8499 .197353
35 1.1373 .8655 .681157
40 .2205 .8226 .198787
45 1.4047 .8169 .756292
50 .3830 .7987 .319490
55 .1387 .7652 .130142
60 1.1169 .7551 .674540
65 .3054 .7424 .264329
70 2.8579 .7551 .943431
75 2.7509 .7713 .937006
80 2.8622 .8273 .943674
85 .4990 .8110 .394369
90 3.6190 .8614 .973672
95 .1502 .8747 .140152
100 .9247 .8628 .605176
MINIMUM DISTANCE TEST for bbs.32
Result of KS test on 20 transformed mindist^2's:
p-value= .495479
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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 bbs.32
sample no: 1 r^3= 6.376 p-value= .19147
sample no: 2 r^3= 16.483 p-value= .42272
sample no: 3 r^3= 57.796 p-value= .85435
sample no: 4 r^3= 15.579 p-value= .40506
sample no: 5 r^3= 214.059 p-value= .99920
sample no: 6 r^3= 23.327 p-value= .54047
sample no: 7 r^3= 2.732 p-value= .08703
sample no: 8 r^3= 34.392 p-value= .68222
sample no: 9 r^3= 19.009 p-value= .46935
sample no: 10 r^3= 71.366 p-value= .90735
sample no: 11 r^3= 20.297 p-value= .49165
sample no: 12 r^3= 22.199 p-value= .52287
sample no: 13 r^3= 83.751 p-value= .93868
sample no: 14 r^3= 22.177 p-value= .52252
sample no: 15 r^3= 51.553 p-value= .82065
sample no: 16 r^3= 45.434 p-value= .78008
sample no: 17 r^3= 11.800 p-value= .32520
sample no: 18 r^3= 61.038 p-value= .86927
sample no: 19 r^3= 28.374 p-value= .61163
sample no: 20 r^3= 14.798 p-value= .38936
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file bbs.32 p-value= .801257
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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 bbs.32
Table of standardized frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
-1.5 .5 -2.8 -2.4 -4.3 .1
.5 -.5 1.1 -2.1 3.0 -1.1
-3.2 -1.5 -1.1 -1.4 -2.1 -.2
1.4 1.6 -1.8 4.8 3.0 1.7
1.8 -1.9 .2 -.2 .5 .0
-1.6 .4 2.1 1.4 -2.6 .1
2.1 5.5 3.0 -1.8 .9 -1.0
-1.1
Chi-square with 42 degrees of freedom:186.535
z-score= 15.770 p-value=1.000000
______________________________________________________________
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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 .208261
Test no. 2 p-value .389122
Test no. 3 p-value .823191
Test no. 4 p-value .427107
Test no. 5 p-value .164340
Test no. 6 p-value .077852
Test no. 7 p-value .715234
Test no. 8 p-value .820594
Test no. 9 p-value .269488
Test no. 10 p-value .856772
Results of the OSUM test for bbs.32
KSTEST on the above 10 p-values: .108177
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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 bbs.32
Up and down runs in a sample of 10000
_________________________________________________
Run test for bbs.32 :
runs up; ks test for 10 p's: .894457
runs down; ks test for 10 p's: .668952
Run test for bbs.32 :
runs up; ks test for 10 p's: .394742
runs down; ks test for 10 p's: .231693
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 bbs.32
No. of wins: Observed Expected
98508 98585.86
98508= No. of wins, z-score= -.348 pvalue= .36383
Analysis of Throws-per-Game:
Chisq= 13.37 for 20 degrees of freedom, p= .13905
Throws Observed Expected Chisq Sum
1 66442 66666.7 .757 .757
2 37808 37654.3 .627 1.384
3 26863 26954.7 .312 1.697
4 19321 19313.5 .003 1.699
5 13927 13851.4 .412 2.112
6 10140 9943.5 3.881 5.993
7 7050 7145.0 1.264 7.257
8 5109 5139.1 .176 7.433
9 3768 3699.9 1.255 8.688
10 2630 2666.3 .494 9.182
11 1905 1923.3 .175 9.357
12 1399 1388.7 .076 9.432
13 1012 1003.7 .068 9.501
14 743 726.1 .391 9.892
15 512 525.8 .364 10.256
16 352 381.2 2.229 12.486
17 283 276.5 .151 12.637
18 205 200.8 .087 12.723
19 147 146.0 .007 12.730
20 98 106.2 .635 13.366
21 286 287.1 .004 13.370
SUMMARY FOR bbs.32
p-value for no. of wins: .363833
p-value for throws/game: .139054
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Results of DIEHARD battery of tests sent to file bbs.out2