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 random_neum.pcm
For a sample of size 500: mean
random_neum.pcm using bits 1 to 24 1.986
duplicate number number
spacings observed expected
0 69. 67.668
1 136. 135.335
2 124. 135.335
3 109. 90.224
4 38. 45.112
5 15. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 6.58 p-value= .638931
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 2 to 25 1.874
duplicate number number
spacings observed expected
0 72. 67.668
1 134. 135.335
2 157. 135.335
3 84. 90.224
4 36. 45.112
5 11. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 9.41 p-value= .848068
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 3 to 26 1.998
duplicate number number
spacings observed expected
0 58. 67.668
1 145. 135.335
2 136. 135.335
3 88. 90.224
4 55. 45.112
5 14. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 7.42 p-value= .716026
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 4 to 27 2.014
duplicate number number
spacings observed expected
0 66. 67.668
1 124. 135.335
2 144. 135.335
3 101. 90.224
4 46. 45.112
5 10. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 6.50 p-value= .630299
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 5 to 28 1.944
duplicate number number
spacings observed expected
0 75. 67.668
1 141. 135.335
2 132. 135.335
3 82. 90.224
4 39. 45.112
5 21. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. = 3.53 p-value= .260275
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 6 to 29 1.962
duplicate number number
spacings observed expected
0 66. 67.668
1 148. 135.335
2 126. 135.335
3 92. 90.224
4 42. 45.112
5 21. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. = 3.90 p-value= .310376
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 7 to 30 1.914
duplicate number number
spacings observed expected
0 81. 67.668
1 141. 135.335
2 115. 135.335
3 97. 90.224
4 43. 45.112
5 16. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 6.96 p-value= .675195
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 8 to 31 1.956
duplicate number number
spacings observed expected
0 78. 67.668
1 129. 135.335
2 137. 135.335
3 78. 90.224
4 54. 45.112
5 20. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 7.73 p-value= .741254
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
random_neum.pcm using bits 9 to 32 1.844
duplicate number number
spacings observed expected
0 77. 67.668
1 152. 135.335
2 136. 135.335
3 79. 90.224
4 35. 45.112
5 12. 18.045
6 to INF 9. 8.282
Chisquare with 6 d.o.f. = 9.09 p-value= .831544
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.638931 .848068 .716026 .630299 .260275
.310376 .675195 .741254 .831544
A KSTEST for the 9 p-values yields .829636
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 random_neum.pcm
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom= 85.543; p-value= .169563
OPERM5 test for file random_neum.pcm
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom= 74.066; p-value= .028815
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 random_neum.pcm
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 215 211.4 .060688 .061
29 5122 5134.0 .028096 .089
30 23219 23103.0 .581963 .671
31 11444 11551.5 1.000864 1.672
chisquare= 1.672 for 3 d. of f.; p-value= .454758
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 random_neum.pcm
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 232 211.4 2.003699 2.004
30 5136 5134.0 .000771 2.004
31 22948 23103.0 1.040535 3.045
32 11684 11551.5 1.519261 4.564
chisquare= 4.564 for 3 d. of f.; p-value= .809694
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 random_neum.pcm
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 893 944.3 2.787 2.787
r =5 21871 21743.9 .743 3.530
r =6 77236 77311.8 .074 3.604
p=1-exp(-SUM/2)= .83506
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 931 944.3 .187 .187
r =5 21547 21743.9 1.783 1.970
r =6 77522 77311.8 .571 2.542
p=1-exp(-SUM/2)= .71943
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 3 to 10
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 887 944.3 3.477 3.477
r =5 21794 21743.9 .115 3.593
r =6 77319 77311.8 .001 3.593
p=1-exp(-SUM/2)= .83414
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 958 944.3 .199 .199
r =5 21469 21743.9 3.475 3.674
r =6 77573 77311.8 .882 4.557
p=1-exp(-SUM/2)= .89754
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 945 944.3 .001 .001
r =5 21485 21743.9 3.083 3.083
r =6 77570 77311.8 .862 3.945
p=1-exp(-SUM/2)= .86092
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 946 944.3 .003 .003
r =5 21623 21743.9 .672 .675
r =6 77431 77311.8 .184 .859
p=1-exp(-SUM/2)= .34918
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 936 944.3 .073 .073
r =5 21761 21743.9 .013 .086
r =6 77303 77311.8 .001 .087
p=1-exp(-SUM/2)= .04277
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 951 944.3 .048 .048
r =5 21865 21743.9 .674 .722
r =6 77184 77311.8 .211 .933
p=1-exp(-SUM/2)= .37288
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 1015 944.3 5.293 5.293
r =5 21626 21743.9 .639 5.932
r =6 77359 77311.8 .029 5.961
p=1-exp(-SUM/2)= .94924
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 942 944.3 .006 .006
r =5 21704 21743.9 .073 .079
r =6 77354 77311.8 .023 .102
p=1-exp(-SUM/2)= .04965
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 954 944.3 .100 .100
r =5 21822 21743.9 .281 .380
r =6 77224 77311.8 .100 .480
p=1-exp(-SUM/2)= .21331
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 904 944.3 1.720 1.720
r =5 21805 21743.9 .172 1.892
r =6 77291 77311.8 .006 1.897
p=1-exp(-SUM/2)= .61273
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 964 944.3 .411 .411
r =5 21785 21743.9 .078 .489
r =6 77251 77311.8 .048 .536
p=1-exp(-SUM/2)= .23526
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 932 944.3 .160 .160
r =5 22095 21743.9 5.669 5.829
r =6 76973 77311.8 1.485 7.314
p=1-exp(-SUM/2)= .97419
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 1002 944.3 3.526 3.526
r =5 21552 21743.9 1.694 5.219
r =6 77446 77311.8 .233 5.452
p=1-exp(-SUM/2)= .93452
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r =5 21518 21743.9 2.347 2.355
r =6 77535 77311.8 .644 2.999
p=1-exp(-SUM/2)= .77676
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 955 944.3 .121 .121
r =5 21766 21743.9 .022 .144
r =6 77279 77311.8 .014 .158
p=1-exp(-SUM/2)= .07577
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 950 944.3 .034 .034
r =5 21916 21743.9 1.362 1.397
r =6 77134 77311.8 .409 1.805
p=1-exp(-SUM/2)= .59454
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21750 21743.9 .002 .093
r =6 77315 77311.8 .000 .093
p=1-exp(-SUM/2)= .04565
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 977 944.3 1.132 1.132
r =5 21867 21743.9 .697 1.829
r =6 77156 77311.8 .314 2.143
p=1-exp(-SUM/2)= .65753
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21787 21743.9 .085 .127
r =6 77275 77311.8 .018 .145
p=1-exp(-SUM/2)= .06993
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 964 944.3 .411 .411
r =5 21695 21743.9 .110 .521
r =6 77341 77311.8 .011 .532
p=1-exp(-SUM/2)= .23353
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 895 944.3 2.574 2.574
r =5 21807 21743.9 .183 2.757
r =6 77298 77311.8 .002 2.760
p=1-exp(-SUM/2)= .74837
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 974 944.3 .934 .934
r =5 21714 21743.9 .041 .975
r =6 77312 77311.8 .000 .975
p=1-exp(-SUM/2)= .38589
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG random_neum.pcm
b-rank test for bits 25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 1006 944.3 4.031 4.031
r =5 21544 21743.9 1.838 5.869
r =6 77450 77311.8 .247 6.116
p=1-exp(-SUM/2)= .95302
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.835058 .719429 .834139 .897543 .860925
.349183 .042771 .372882 .949238 .049653
.213314 .612733 .235259 .974193 .934521
.776756 .075773 .594538 .045655 .657535
.069934 .233533 .748366 .385888 .953020
brank test summary for random_neum.pcm
The KS test for those 25 supposed UNI's yields
KS p-value= .697660
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 141444 missing words, -1.09 sigmas from mean, p-value= .13847
tst no 2: 141625 missing words, -.66 sigmas from mean, p-value= .25324
tst no 3: 141345 missing words, -1.32 sigmas from mean, p-value= .09366
tst no 4: 141240 missing words, -1.56 sigmas from mean, p-value= .05893
tst no 5: 141105 missing words, -1.88 sigmas from mean, p-value= .03010
tst no 6: 141679 missing words, -.54 sigmas from mean, p-value= .29524
tst no 7: 141876 missing words, -.08 sigmas from mean, p-value= .46897
tst no 8: 141350 missing words, -1.31 sigmas from mean, p-value= .09563
tst no 9: 141057 missing words, -1.99 sigmas from mean, p-value= .02322
tst no 10: 142087 missing words, .42 sigmas from mean, p-value= .66097
tst no 11: 141882 missing words, -.06 sigmas from mean, p-value= .47454
tst no 12: 141576 missing words, -.78 sigmas from mean, p-value= .21805
tst no 13: 142225 missing words, .74 sigmas from mean, p-value= .76961
tst no 14: 141718 missing words, -.45 sigmas from mean, p-value= .32743
tst no 15: 142743 missing words, 1.95 sigmas from mean, p-value= .97428
tst no 16: 141561 missing words, -.81 sigmas from mean, p-value= .20787
tst no 17: 141678 missing words, -.54 sigmas from mean, p-value= .29443
tst no 18: 142057 missing words, .35 sigmas from mean, p-value= .63496
tst no 19: 141849 missing words, -.14 sigmas from mean, p-value= .44395
tst no 20: 141433 missing words, -1.11 sigmas from mean, p-value= .13287
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 random_neum.pcm
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OPSO for random_neum.pcm using bits 23 to 32 142269 1.240 .8926
OPSO for random_neum.pcm using bits 22 to 31 142374 1.602 .9455
OPSO for random_neum.pcm using bits 21 to 30 141921 .040 .5161
OPSO for random_neum.pcm using bits 20 to 29 141764 -.501 .3081
OPSO for random_neum.pcm using bits 19 to 28 141502 -1.405 .0801
OPSO for random_neum.pcm using bits 18 to 27 141823 -.298 .3830
OPSO for random_neum.pcm using bits 17 to 26 141750 -.549 .2914
OPSO for random_neum.pcm using bits 16 to 25 141992 .285 .6122
OPSO for random_neum.pcm using bits 15 to 24 141622 -.991 .1609
OPSO for random_neum.pcm using bits 14 to 23 141982 .251 .5989
OPSO for random_neum.pcm using bits 13 to 22 142145 .813 .7918
OPSO for random_neum.pcm using bits 12 to 21 141962 .182 .5721
OPSO for random_neum.pcm using bits 11 to 20 142232 1.113 .8671
OPSO for random_neum.pcm using bits 10 to 19 141896 -.046 .4817
OPSO for random_neum.pcm using bits 9 to 18 142268 1.237 .8919
OPSO for random_neum.pcm using bits 8 to 17 141832 -.267 .3949
OPSO for random_neum.pcm using bits 7 to 16 141969 .206 .5815
OPSO for random_neum.pcm using bits 6 to 15 141964 .189 .5748
OPSO for random_neum.pcm using bits 5 to 14 141652 -.887 .1874
OPSO for random_neum.pcm using bits 4 to 13 142241 1.144 .8736
OPSO for random_neum.pcm using bits 3 to 12 142088 .616 .7311
OPSO for random_neum.pcm using bits 2 to 11 142145 .813 .7918
OPSO for random_neum.pcm using bits 1 to 10 142167 .889 .8129
OQSO test for generator random_neum.pcm
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OQSO for random_neum.pcm using bits 28 to 32 142224 1.067 .8569
OQSO for random_neum.pcm using bits 27 to 31 142151 .819 .7937
OQSO for random_neum.pcm using bits 26 to 30 141935 .087 .5347
OQSO for random_neum.pcm using bits 25 to 29 142269 1.219 .8886
OQSO for random_neum.pcm using bits 24 to 28 142061 .514 .6964
OQSO for random_neum.pcm using bits 23 to 27 141857 -.177 .4296
OQSO for random_neum.pcm using bits 22 to 26 142558 2.199 .9861
OQSO for random_neum.pcm using bits 21 to 25 141876 -.113 .4550
OQSO for random_neum.pcm using bits 20 to 24 141769 -.476 .3171
OQSO for random_neum.pcm using bits 19 to 23 141741 -.571 .2841
OQSO for random_neum.pcm using bits 18 to 22 142076 .565 .7140
OQSO for random_neum.pcm using bits 17 to 21 141874 -.120 .4523
OQSO for random_neum.pcm using bits 16 to 20 142031 .412 .6600
OQSO for random_neum.pcm using bits 15 to 19 141843 -.225 .4111
OQSO for random_neum.pcm using bits 14 to 18 142022 .382 .6487
OQSO for random_neum.pcm using bits 13 to 17 141464 -1.510 .0656
OQSO for random_neum.pcm using bits 12 to 16 141636 -.927 .1771
OQSO for random_neum.pcm using bits 11 to 15 141808 -.343 .3656
OQSO for random_neum.pcm using bits 10 to 14 141768 -.479 .3159
OQSO for random_neum.pcm using bits 9 to 13 141774 -.459 .3232
OQSO for random_neum.pcm using bits 8 to 12 141605 -1.032 .1511
OQSO for random_neum.pcm using bits 7 to 11 142394 1.643 .9498
OQSO for random_neum.pcm using bits 6 to 10 142588 2.301 .9893
OQSO for random_neum.pcm using bits 5 to 9 142074 .558 .7116
OQSO for random_neum.pcm using bits 4 to 8 141937 .094 .5374
OQSO for random_neum.pcm using bits 3 to 7 141637 -.923 .1780
OQSO for random_neum.pcm using bits 2 to 6 142144 .795 .7868
OQSO for random_neum.pcm using bits 1 to 5 142345 1.477 .9301
DNA test for generator random_neum.pcm
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for random_neum.pcm using bits 31 to 32 141724 -.547 .2923
DNA for random_neum.pcm using bits 30 to 31 141780 -.381 .3514
DNA for random_neum.pcm using bits 29 to 30 142267 1.055 .8543
DNA for random_neum.pcm using bits 28 to 29 141728 -.535 .2964
DNA for random_neum.pcm using bits 27 to 28 142011 .300 .6179
DNA for random_neum.pcm using bits 26 to 27 142636 2.144 .9840
DNA for random_neum.pcm using bits 25 to 26 141608 -.889 .1870
DNA for random_neum.pcm using bits 24 to 25 141693 -.638 .2617
DNA for random_neum.pcm using bits 23 to 24 142405 1.462 .9282
DNA for random_neum.pcm using bits 22 to 23 141497 -1.216 .1119
DNA for random_neum.pcm using bits 21 to 22 141875 -.101 .4597
DNA for random_neum.pcm using bits 20 to 21 142076 .492 .6885
DNA for random_neum.pcm using bits 19 to 20 142087 .524 .6999
DNA for random_neum.pcm using bits 18 to 19 142472 1.660 .9515
DNA for random_neum.pcm using bits 17 to 18 141716 -.570 .2842
DNA for random_neum.pcm using bits 16 to 17 141707 -.597 .2753
DNA for random_neum.pcm using bits 15 to 16 142020 .326 .6280
DNA for random_neum.pcm using bits 14 to 15 141794 -.340 .3669
DNA for random_neum.pcm using bits 13 to 14 141718 -.564 .2862
DNA for random_neum.pcm using bits 12 to 13 141552 -1.054 .1459
DNA for random_neum.pcm using bits 11 to 12 142094 .545 .7070
DNA for random_neum.pcm using bits 10 to 11 141572 -.995 .1599
DNA for random_neum.pcm using bits 9 to 10 142114 .604 .7270
DNA for random_neum.pcm using bits 8 to 9 141903 -.019 .4926
DNA for random_neum.pcm using bits 7 to 8 141470 -1.296 .0975
DNA for random_neum.pcm using bits 6 to 7 141910 .002 .5008
DNA for random_neum.pcm using bits 5 to 6 142046 .403 .6566
DNA for random_neum.pcm using bits 4 to 5 141616 -.865 .1934
DNA for random_neum.pcm using bits 3 to 4 142002 .273 .6077
DNA for random_neum.pcm using bits 2 to 3 141542 -1.084 .1393
DNA for random_neum.pcm using bits 1 to 2 142236 .964 .8324
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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 random_neum.pcm
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 random_neum.pcm 2469.79 -.427 .334625
byte stream for random_neum.pcm 2371.06 -1.823 .034114
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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 2482.32 -.250 .401277
bits 2 to 9 2641.16 1.996 .977048
bits 3 to 10 2428.74 -1.008 .156770
bits 4 to 11 2395.79 -1.474 .070270
bits 5 to 12 2556.75 .803 .788887
bits 6 to 13 2612.59 1.592 .944343
bits 7 to 14 2499.45 -.008 .496907
bits 8 to 15 2581.35 1.151 .875036
bits 9 to 16 2591.59 1.295 .902385
bits 10 to 17 2453.90 -.652 .257204
bits 11 to 18 2533.11 .468 .680192
bits 12 to 19 2353.43 -2.073 .019093
bits 13 to 20 2580.44 1.138 .872362
bits 14 to 21 2512.97 .183 .572750
bits 15 to 22 2467.03 -.466 .320526
bits 16 to 23 2579.50 1.124 .869558
bits 17 to 24 2399.28 -1.424 .077169
bits 18 to 25 2464.52 -.502 .307932
bits 19 to 26 2505.69 .081 .532087
bits 20 to 27 2485.04 -.212 .416239
bits 21 to 28 2540.26 .569 .715443
bits 22 to 29 2414.46 -1.210 .113187
bits 23 to 30 2610.10 1.557 .940276
bits 24 to 31 2533.09 .468 .680097
bits 25 to 32 2527.44 .388 .651019
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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 random_neum.pcm
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3536 z-score: .594 p-value: .723613
Successes: 3489 z-score: -1.553 p-value: .060270
Successes: 3521 z-score: -.091 p-value: .463618
Successes: 3563 z-score: 1.826 p-value: .966111
Successes: 3516 z-score: -.320 p-value: .374623
Successes: 3501 z-score: -1.005 p-value: .157553
Successes: 3559 z-score: 1.644 p-value: .949895
Successes: 3527 z-score: .183 p-value: .572463
Successes: 3502 z-score: -.959 p-value: .168804
Successes: 3539 z-score: .731 p-value: .767486
square size avg. no. parked sample sigma
100. 3525.300 23.253
KSTEST for the above 10: p= .062662
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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 random_neum.pcm
Sample no. d^2 avg equiv uni
5 .1170 .5952 .110978
10 .6266 .6687 .467268
15 3.3600 1.1726 .965846
20 .3280 1.2314 .280824
25 .8728 1.3343 .584043
30 .0542 1.2599 .053008
35 .8124 1.3687 .558013
40 .1722 1.2599 .158949
45 2.1677 1.2648 .886797
50 .1140 1.1815 .108248
55 .0724 1.1136 .070210
60 .4450 1.0911 .360635
65 .1412 1.0970 .132278
70 2.6878 1.1233 .932886
75 .2092 1.1579 .189621
80 1.3690 1.1651 .747370
85 2.0927 1.1417 .877938
90 .3649 1.1304 .306989
95 1.9834 1.2000 .863760
100 .7416 1.1887 .525436
MINIMUM DISTANCE TEST for random_neum.pcm
Result of KS test on 20 transformed mindist^2's:
p-value= .824783
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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 random_neum.pcm
sample no: 1 r^3= 25.427 p-value= .57154
sample no: 2 r^3= 4.851 p-value= .14930
sample no: 3 r^3= 6.465 p-value= .19387
sample no: 4 r^3= 107.266 p-value= .97200
sample no: 5 r^3= 15.842 p-value= .41026
sample no: 6 r^3= 14.712 p-value= .38761
sample no: 7 r^3= 11.740 p-value= .32386
sample no: 8 r^3= 16.846 p-value= .42967
sample no: 9 r^3= 76.981 p-value= .92316
sample no: 10 r^3= 19.757 p-value= .48241
sample no: 11 r^3= 14.147 p-value= .37598
sample no: 12 r^3= 17.966 p-value= .45056
sample no: 13 r^3= 20.122 p-value= .48867
sample no: 14 r^3= 79.401 p-value= .92911
sample no: 15 r^3= 5.389 p-value= .16442
sample no: 16 r^3= 5.308 p-value= .16216
sample no: 17 r^3= 20.841 p-value= .50077
sample no: 18 r^3= 82.404 p-value= .93587
sample no: 19 r^3= 12.190 p-value= .33392
sample no: 20 r^3= 68.509 p-value= .89809
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file random_neum.pcm p-value= .649917
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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 random_neum.pcm
Table of standardized frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
-.8 .5 -.8 -.1 .1 .3
.8 .2 -.2 -.1 -.7 .7
1.0 -.8 .1 -.4 .6 .5
.0 -1.1 1.0 -.2 -.7 .3
.1 .6 -1.2 .5 -.1 .8
-1.0 -.3 -1.7 .9 -.5 -.3
-.7 .2 -1.2 .4 -.6 .0
1.8
Chi-square with 42 degrees of freedom: 21.916
z-score= -2.191 p-value= .004567
______________________________________________________________
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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 .109101
Test no. 2 p-value .646255
Test no. 3 p-value .320501
Test no. 4 p-value .407480
Test no. 5 p-value .414983
Test no. 6 p-value .280204
Test no. 7 p-value .445494
Test no. 8 p-value .026272
Test no. 9 p-value .009500
Test no. 10 p-value .491763
Results of the OSUM test for random_neum.pcm
KSTEST on the above 10 p-values: .950267
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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 random_neum.pcm
Up and down runs in a sample of 10000
_________________________________________________
Run test for random_neum.pcm:
runs up; ks test for 10 p's: .959990
runs down; ks test for 10 p's: .553549
Run test for random_neum.pcm:
runs up; ks test for 10 p's: .995082
runs down; ks test for 10 p's: .716674
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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 random_neum.pcm
No. of wins: Observed Expected
98478 98585.86
98478= No. of wins, z-score= -.482 pvalue= .31476
Analysis of Throws-per-Game:
Chisq= 15.76 for 20 degrees of freedom, p= .26884
Throws Observed Expected Chisq Sum
1 66873 66666.7 .639 .639
2 37495 37654.3 .674 1.313
3 27075 26954.7 .537 1.849
4 19243 19313.5 .257 2.106
5 13991 13851.4 1.407 3.513
6 9824 9943.5 1.437 4.950
7 7173 7145.0 .110 5.060
8 5088 5139.1 .508 5.567
9 3568 3699.9 4.700 10.267
10 2674 2666.3 .022 10.289
11 1894 1923.3 .447 10.736
12 1412 1388.7 .390 11.126
13 1005 1003.7 .002 11.128
14 724 726.1 .006 11.134
15 556 525.8 1.730 12.864
16 400 381.2 .932 13.796
17 257 276.5 1.381 15.177
18 211 200.8 .515 15.692
19 147 146.0 .007 15.699
20 107 106.2 .006 15.705
21 283 287.1 .059 15.764
SUMMARY FOR random_neum.pcm
p-value for no. of wins: .314757
p-value for throws/game: .268841
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Results of DIEHARD battery of tests sent to file random_neum.txt