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ENT TESTS:
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DIEHARD TESTS:
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| TEST 1 |
Entropy source: PIC18F2525 |
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 test1
For a sample of size 500: mean
test1
using bits 1 to 24 1.978
duplicate number number
spacings
observed expected
0 77. 67.668
1 132. 135.335
2 132.
135.335
3 84. 90.224
4 49. 45.112
5 17. 18.045
6 to
INF 9. 8.282
Chisquare with 6 d.o.f. = 2.34 p-value= .113930
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 2 to 25 1.942
duplicate
number number
spacings observed expected
0 73. 67.668
1 145. 135.335
2 126. 135.335
3 83. 90.224
4 47. 45.112
5 19. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. =
2.66 p-value= .149928
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 3 to 26 1.964
duplicate
number number
spacings observed expected
0 80. 67.668
1 120. 135.335
2 153. 135.335
3 75. 90.224
4 42. 45.112
5 20. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. =
9.64 p-value= .859468
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 4 to 27 1.996
duplicate
number number
spacings observed expected
0 74. 67.668
1 116. 135.335
2 142. 135.335
3 108. 90.224
4 33.
45.112
5 22. 18.045
6 to INF 5. 8.282
Chisquare with 6
d.o.f. = 12.60 p-value= .950243
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 5 to 28 1.868
duplicate
number number
spacings observed expected
0 70. 67.668
1 159. 135.335
2 124. 135.335
3 95. 90.224
4 31. 45.112
5 13. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. =
11.26 p-value= .919192
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 6 to 29 1.906
duplicate
number number
spacings observed expected
0 70. 67.668
1 133. 135.335
2 149. 135.335
3 84. 90.224
4 52. 45.112
5 11. 18.045
6 to INF 1. 8.282
Chisquare with 6 d.o.f. =
12.13 p-value= .940961
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 7 to 30 2.000
duplicate
number number
spacings observed expected
0 75. 67.668
1 129. 135.335
2 125. 135.335
3 102. 90.224
4 43.
45.112
5 16. 18.045
6 to INF 10. 8.282
Chisquare with 6
d.o.f. = 4.10 p-value= .337465
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 8 to 31 1.924
duplicate
number number
spacings observed expected
0 72. 67.668
1 145. 135.335
2 130. 135.335
3 81. 90.224
4 54. 45.112
5 12. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. =
6.53 p-value= .633049
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test1 using bits 9 to 32 1.986
duplicate
number number
spacings observed expected
0 73. 67.668
1 138. 135.335
2 136. 135.335
3 80. 90.224
4 39. 45.112
5 22. 18.045
6 to INF 12. 8.282
Chisquare with 6 d.o.f. =
5.00 p-value= .456022
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.113930 .149928 .859468 .950243 .919192
.940961 .337465
.633049 .456022
A KSTEST for the 9 p-values yields .687191
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
For a sample of 1,000,000
consecutive 5-tuples,
chisquare for 99 degrees of
freedom=131.403; p-value= .983682
OPERM5 test for file test1
For a sample of 1,000,000 consecutive 5-tuples,
chisquare
for 99 degrees of freedom=158.789; p-value= .999868
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
Rank test for 31x31 binary
matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 205 211.4 .194832
.195
29 5230 5134.0 1.794705 1.990
30 23003 23103.0
.433249 2.423
31 11562 11551.5 .009500 2.432
chisquare=
2.432 for 3 d. of f.; p-value= .571777
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
Rank test for 32x32 binary
matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 189 211.4 2.377126
2.377
30 5208 5134.0 1.066317 3.443
31 23080 23103.0
.022991 3.466
32 11523 11551.5 .070436 3.537
chisquare=
3.537 for 3 d. of f.; p-value= .713998
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG test1
b-rank test for
bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 921 944.3
.575 .575
r =5 21752 21743.9 .003 .578
r =6 77327 77311.8
.003 .581
p=1-exp(-SUM/2)= .25210
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 964 944.3 .411 .411
r =5 21675 21743.9 .218 .629
r =6
77361 77311.8 .031 .661
p=1-exp(-SUM/2)= .28128
Rank of a
6x8 binary matrix,
rows formed from eight bits of the RNG
test1
b-rank test for bits 3 to 10
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 939 944.3 .030 .030
r =5 21811 21743.9
.207 .237
r =6 77250 77311.8 .049 .286
p=1-exp(-SUM/2)=
.13335
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test1
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 949 944.3 .023 .023
r
=5 21646 21743.9 .441 .464
r =6 77405 77311.8 .112 .577
p=1-exp(-SUM/2)= .25043
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 956 944.3
.145 .145
r =5 21872 21743.9 .755 .900
r =6 77172 77311.8
.253 1.152
p=1-exp(-SUM/2)= .43797
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 919 944.3 .678 .678
r =5 21614 21743.9 .776 1.454
r
=6 77467 77311.8 .312 1.765
p=1-exp(-SUM/2)= .58635
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test1
b-rank test for bits 7 to 14
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 931 944.3 .187 .187
r =5 21705 21743.9
.070 .257
r =6 77364 77311.8 .035 .292
p=1-exp(-SUM/2)=
.13593
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test1
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 962 944.3 .332 .332
r
=5 21767 21743.9 .025 .356
r =6 77271 77311.8 .022 .378
p=1-exp(-SUM/2)= .17213
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 949 944.3
.023 .023
r =5 21574 21743.9 1.328 1.351
r =6 77477
77311.8 .353 1.704
p=1-exp(-SUM/2)= .57342
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 960 944.3 .261 .261
r =5 21504 21743.9 2.647
2.908
r =6 77536 77311.8 .650 3.558
p=1-exp(-SUM/2)=
.83119
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test1
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 943 944.3 .002 .002
r
=5 21634 21743.9 .555 .557
r =6 77423 77311.8 .160 .717
p=1-exp(-SUM/2)= .30134
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 920 944.3
.625 .625
r =5 21934 21743.9 1.662 2.287
r =6 77146
77311.8 .356 2.643
p=1-exp(-SUM/2)= .73326
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 897 944.3 2.369 2.369
r =5 21786 21743.9 .082
2.451
r =6 77317 77311.8 .000 2.451
p=1-exp(-SUM/2)=
.70642
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test1
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 926 944.3 .355 .355
r
=5 21612 21743.9 .800 1.155
r =6 77462 77311.8 .292 1.447
p=1-exp(-SUM/2)= .51485
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 910 944.3
1.246 1.246
r =5 21671 21743.9 .244 1.490
r =6 77419
77311.8 .149 1.639
p=1-exp(-SUM/2)= .55935
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 907 944.3 1.473 1.473
r =5 21663 21743.9 .301
1.774
r =6 77430 77311.8 .181 1.955
p=1-exp(-SUM/2)=
.62378
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test1
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 909 944.3 1.320 1.320
r =5 21730 21743.9 .009 1.329
r =6 77361 77311.8 .031 1.360
p=1-exp(-SUM/2)= .49335
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 940 944.3
.020 .020
r =5 21609 21743.9 .837 .857
r =6 77451 77311.8
.251 1.107
p=1-exp(-SUM/2)= .42510
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 967 944.3 .546 .546
r =5 21987 21743.9 2.718 3.264
r
=6 77046 77311.8 .914 4.177
p=1-exp(-SUM/2)= .87615
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test1
b-rank test for bits 20 to 27
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 997 944.3 2.941 2.941
r =5 21680
21743.9 .188 3.129
r =6 77323 77311.8 .002 3.130
p=1-exp(-SUM/2)= .79095
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 967 944.3
.546 .546
r =5 21879 21743.9 .839 1.385
r =6 77154 77311.8
.322 1.707
p=1-exp(-SUM/2)= .57411
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test1
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 943 944.3 .002 .002
r =5 21867 21743.9 .697 .699
r =6
77190 77311.8 .192 .891
p=1-exp(-SUM/2)= .35937
Rank of a
6x8 binary matrix,
rows formed from eight bits of the RNG
test1
b-rank test for bits 23 to 30
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 920 944.3 .625 .625
r =5 21772 21743.9
.036 .662
r =6 77308 77311.8 .000 .662
p=1-exp(-SUM/2)=
.28175
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test1
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r
=5 21650 21743.9 .406 .413
r =6 77403 77311.8 .108 .521
p=1-exp(-SUM/2)= .22925
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test1
b-rank test for bits
25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 958 944.3
.199 .199
r =5 21643 21743.9 .468 .667
r =6 77399 77311.8
.098 .765
p=1-exp(-SUM/2)= .31794
TEST SUMMARY, 25 tests
on 100,000 random 6x8 matrices
These should be 25 uniform
[0,1] random variables:
.252102 .281277 .133346 .250430
.437975
.586354 .135925 .172130 .573420 .831189
.301344
.733258 .706425 .514851 .559352
.623778 .493351 .425105
.876150 .790951
.574106 .359370 .281754 .229253 .317943
brank test summary for test1
The KS test for those 25
supposed UNI's yields
KS p-value= .770427
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 141679 missing words, -.54 sigmas from mean, p-value=
.29524
tst no 2: 142013 missing words, .24 sigmas from mean,
p-value= .59570
tst no 3: 141398 missing words, -1.19 sigmas
from mean, p-value= .11610
tst no 4: 141891 missing words,
-.04 sigmas from mean, p-value= .48292
tst no 5: 141621
missing words, -.67 sigmas from mean, p-value= .25026
tst no
6: 141309 missing words, -1.40 sigmas from mean, p-value= .08036
tst no 7: 141498 missing words, -.96 sigmas from mean, p-value=
.16826
tst no 8: 142029 missing words, .28 sigmas from mean,
p-value= .61011
tst no 9: 141884 missing words, -.06 sigmas
from mean, p-value= .47641
tst no 10: 142204 missing words,
.69 sigmas from mean, p-value= .75443
tst no 11: 142234
missing words, .76 sigmas from mean, p-value= .77595
tst no
12: 142438 missing words, 1.24 sigmas from mean, p-value= .89162
tst no 13: 142409 missing words, 1.17 sigmas from mean, p-value=
.87849
tst no 14: 141894 missing words, -.04 sigmas from
mean, p-value= .48572
tst no 15: 142045 missing words, .32
sigmas from mean, p-value= .62437
tst no 16: 142504 missing
words, 1.39 sigmas from mean, p-value= .91765
tst no 17:
142618 missing words, 1.66 sigmas from mean, p-value= .95112
tst no 18: 140938 missing words, -2.27 sigmas from mean,
p-value= .01162
tst no 19: 142041 missing words, .31 sigmas
from mean, p-value= .62082
tst no 20: 142630 missing words,
1.68 sigmas from mean, p-value= .95389
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
Output: No. missing words
(mw), equiv normal variate (z), p-value (p)
mw z p
OPSO
for test1 using bits 23 to 32 141858 -.177 .4298
OPSO for
test1 using bits 22 to 31 142527 2.130 .9834
OPSO for test1
using bits 21 to 30 142508 2.064 .9805
OPSO for test1 using
bits 20 to 29 142000 .313 .6227
OPSO for test1 using bits 19
to 28 141834 -.260 .3975
OPSO for test1 using bits 18 to 27
142388 1.651 .9506
OPSO for test1 using bits 17 to 26 142009
.344 .6345
OPSO for test1 using bits 16 to 25 142294 1.326
.9077
OPSO for test1 using bits 15 to 24 142453 1.875 .9696
OPSO for test1 using bits 14 to 23 141954 .154 .5612
OPSO for
test1 using bits 13 to 22 141926 .057 .5229
OPSO for test1
using bits 12 to 21 141783 -.436 .3316
OPSO for test1 using
bits 11 to 20 142430 1.795 .9637
OPSO for test1 using bits 10
to 19 142038 .444 .6714
OPSO for test1 using bits 9 to 18
141837 -.249 .4015
OPSO for test1 using bits 8 to 17 141892
-.060 .4762
OPSO for test1 using bits 7 to 16 141728 -.625
.2659
OPSO for test1 using bits 6 to 15 141812 -.336 .3686
OPSO for test1 using bits 5 to 14 142098 .651 .7423
OPSO for
test1 using bits 4 to 13 141864 -.156 .4379
OPSO for test1
using bits 3 to 12 141353 -1.918 .0275
OPSO for test1 using
bits 2 to 11 141573 -1.160 .1231
OPSO for test1 using bits 1
to 10 142226 1.092 .8626
OQSO test for generator test1
Output: No. missing words (mw), equiv normal variate (z),
p-value (p)
mw z p
OQSO for test1 using bits 28 to 32
141575 -1.133 .1285
OQSO for test1 using bits 27 to 31 141998
.301 .6181
OQSO for test1 using bits 26 to 30 142348 1.487
.9315
OQSO for test1 using bits 25 to 29 142109 .677 .7508
OQSO for test1 using bits 24 to 28 141587 -1.093 .1373
OQSO
for test1 using bits 23 to 27 141918 .029 .5117
OQSO for
test1 using bits 22 to 26 141951 .141 .5562
OQSO for test1
using bits 21 to 25 141818 -.310 .3784
OQSO for test1 using
bits 20 to 24 141722 -.635 .2627
OQSO for test1 using bits 19
to 23 142065 .528 .7011
OQSO for test1 using bits 18 to 22
142413 1.707 .9561
OQSO for test1 using bits 17 to 21 141944
.118 .5468
OQSO for test1 using bits 16 to 20 141507 -1.364
.0863
OQSO for test1 using bits 15 to 19 141431 -1.621 .0525
OQSO for test1 using bits 14 to 18 141972 .212 .5841
OQSO for
test1 using bits 13 to 17 141766 -.486 .3135
OQSO for test1
using bits 12 to 16 142554 2.185 .9856
OQSO for test1 using
bits 11 to 15 142229 1.084 .8607
OQSO for test1 using bits 10
to 14 141522 -1.313 .0946
OQSO for test1 using bits 9 to 13
141858 -.174 .4309
OQSO for test1 using bits 8 to 12 142087
.602 .7265
OQSO for test1 using bits 7 to 11 142044 .457
.6760
OQSO for test1 using bits 6 to 10 141988 .267 .6051
OQSO for test1 using bits 5 to 9 141717 -.652 .2572
OQSO for
test1 using bits 4 to 8 141941 .107 .5427
OQSO for test1
using bits 3 to 7 141920 .036 .5144
OQSO for test1 using bits
2 to 6 141308 -2.038 .0208
OQSO for test1 using bits 1 to 5
141267 -2.177 .0147
DNA test for generator test1
Output:
No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for test1 using bits 31 to 32 141750 -.470 .3192
DNA for test1 using bits 30 to 31 141995 .253 .5998
DNA for
test1 using bits 29 to 30 141748 -.476 .3171
DNA for test1
using bits 28 to 29 141565 -1.016 .1549
DNA for test1 using
bits 27 to 28 142418 1.501 .9333
DNA for test1 using bits 26
to 27 141920 .031 .5126
DNA for test1 using bits 25 to 26
141537 -1.098 .1360
DNA for test1 using bits 24 to 25 142262
1.040 .8509
DNA for test1 using bits 23 to 24 142189 .825
.7953
DNA for test1 using bits 22 to 23 141842 -.199 .4213
DNA for test1 using bits 21 to 22 141653 -.756 .2248
DNA for
test1 using bits 20 to 21 141521 -1.146 .1260
DNA for test1
using bits 19 to 20 141635 -.809 .2092
DNA for test1 using
bits 18 to 19 141569 -1.004 .1577
DNA for test1 using bits 17
to 18 141798 -.328 .3713
DNA for test1 using bits 16 to 17
141524 -1.137 .1278
DNA for test1 using bits 15 to 16 142248
.999 .8411
DNA for test1 using bits 14 to 15 142210 .887
.8124
DNA for test1 using bits 13 to 14 141775 -.396 .3460
DNA for test1 using bits 12 to 13 142154 .722 .7648
DNA for
test1 using bits 11 to 12 141563 -1.022 .1535
DNA for test1
using bits 10 to 11 141415 -1.458 .0724
DNA for test1 using
bits 9 to 10 141473 -1.287 .0990
DNA for test1 using bits 8
to 9 141827 -.243 .4041
DNA for test1 using bits 7 to 8
142167 .760 .7764
DNA for test1 using bits 6 to 7 141183
-2.143 .0161
DNA for test1 using bits 5 to 6 141995 .253
.5998
DNA for test1 using bits 4 to 5 142017 .318 .6246
DNA for test1 using bits 3 to 4 141877 -.095 .4620
DNA for
test1 using bits 2 to 3 141990 .238 .5940
DNA for test1 using
bits 1 to 2 141794 -.340 .3669
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNT-THE-1's TEST on a stream of bytes. ::
:: Consider the file under test as a stream of bytes (four
per ::
:: 32 bit integer). Each byte can contain from 0 to 8
1's, ::
:: with probabilities 1,8,28,56,70,56,28,8,1 over
256. Now let ::
:: the stream of bytes provide a string of
overlapping 5-letter ::
:: words, each "letter" taking
values A,B,C,D,E. The letters are ::
:: determined by the
number of 1's in a byte:: 0,1,or 2 yield A,::
:: 3 yields B,
4 yields C, 5 yields D and 6,7 or 8 yield E. Thus ::
:: we
have a monkey at a typewriter hitting five keys with vari- ::
:: ous probabilities (37,56,70,56,37 over 256). There are
5^5 ::
:: possible 5-letter words, and from a string of
256,000 (over- ::
:: lapping) 5-letter words, counts are
made on the frequencies ::
:: for each word. The quadratic
form in the weak inverse of ::
:: the covariance matrix of
the cell counts provides a chisquare ::
:: test:: Q5-Q4, the
difference of the naive Pearson sums of ::
::
(OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for test1
Chi-square with 5^5-5^4=2500 d.of
f. for sample size:2560000
chisquare equiv normal p-value
Results fo COUNT-THE-1's in successive bytes:
byte stream for
test1 2488.88 -.157 .437511
byte stream for test1 2531.95
.452 .674322
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 2658.39 2.240 .987453
bits 2
to 9 2532.28 .457 .676006
bits 3 to 10 2487.61 -.175 .430442
bits 4 to 11 2448.76 -.725 .234350
bits 5 to 12 2515.46 .219
.586534
bits 6 to 13 2574.13 1.048 .852776
bits 7 to 14
2536.15 .511 .695401
bits 8 to 15 2438.96 -.863 .193994
bits 9 to 16 2582.73 1.170 .878993
bits 10 to 17 2564.05 .906
.817471
bits 11 to 18 2492.92 -.100 .460105
bits 12 to 19
2540.93 .579 .718649
bits 13 to 20 2588.11 1.246 .893627
bits 14 to 21 2605.01 1.485 .931239
bits 15 to 22 2452.08
-.678 .248973
bits 16 to 23 2543.26 .612 .729671
bits 17
to 24 2639.70 1.976 .975906
bits 18 to 25 2460.11 -.564
.286328
bits 19 to 26 2548.82 .690 .755048
bits 20 to 27
2483.48 -.234 .407637
bits 21 to 28 2623.50 1.747 .959639
bits 22 to 29 2536.86 .521 .698897
bits 23 to 30 2555.09 .779
.782039
bits 24 to 31 2474.44 -.361 .358876
bits 25 to 32
2440.51 -.841 .200087
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
Of 12,000
tries, the average no. of successes
should be 3523 with
sigma=21.9
Successes: 3505 z-score: -.822 p-value: .205562
Successes: 3508 z-score: -.685 p-value: .246694
Successes:
3536 z-score: .594 p-value: .723613
Successes: 3533 z-score:
.457 p-value: .676028
Successes: 3544 z-score: .959 p-value:
.831196
Successes: 3538 z-score: .685 p-value: .753306
Successes: 3521 z-score: -.091 p-value: .463618
Successes:
3553 z-score: 1.370 p-value: .914635
Successes: 3551 z-score:
1.279 p-value: .899470
Successes: 3550 z-score: 1.233
p-value: .891189
square size avg. no. parked sample sigma
100. 3533.900 16.456
KSTEST for the above 10: p= .893812
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100
times:: choose n=8000 random points in a ::
:: square of
side 10000. Find d, the minimum distance between ::
:: the
(n^2-n)/2 pairs of points. If the points are truly inde- ::
:: pendent uniform, then d^2, the square of the minimum distance
::
:: should be (very close to) exponentially distributed
with mean ::
:: .995 . Thus 1-exp(-d^2/.995) should be
uniform on [0,1) and ::
:: a KSTEST on the resulting 100
values serves as a test of uni- ::
:: formity for random
points in the square. Test numbers=0 mod 5 ::
:: are printed
but the KSTEST is based on the full set of 100 ::
:: random
choices of 8000 points in the 10000x10000 square. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
This is the MINIMUM DISTANCE test
for random integers in
the file test1
Sample no. d^2 avg equiv uni
5 1.6352
.9600 .806678
10 2.7827 1.2902 .938986
15 .7697 1.0387
.538614
20 .4672 .9708 .374726
25 .2354 1.1740 .210667
30 .9094 1.2991 .599068
35 .9584 1.2461 .618352
40 1.2219
1.3726 .707140
45 1.0710 1.3186 .659162
50 .5160 1.3305
.404665
55 .2977 1.2463 .258556
60 3.4069 1.2907 .967420
65 .8146 1.2926 .559012
70 .3462 1.2310 .293870
75 4.5329
1.2606 .989493
80 .3873 1.2460 .322418
85 .6620 1.2551
.485879
90 .7637 1.2560 .535855
95 .5836 1.2410 .443751
100 .5891 1.2453 .446825
MINIMUM DISTANCE TEST for test1
Result of KS test on 20 transformed mindist^2's:
p-value=
.992516
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE 3DSPHERES TEST ::
:: Choose 4000 random points in
a cube of edge 1000. At each ::
:: point, center a sphere
large enough to reach the next closest ::
:: point. Then the
volume of the smallest such sphere is (very ::
:: close to)
exponentially distributed with mean 120pi/3. Thus ::
:: the
radius cubed is exponential with mean 30. (The mean is ::
::
obtained by extensive simulation). The 3DSPHERES test gener- ::
:: ates 4000 such spheres 20 times. Each min radius cubed
leads ::
:: to a uniform variable by means of
1-exp(-r^3/30.), then a ::
:: KSTEST is done on the 20
p-values. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file test1
sample no: 1 r^3=
1.378 p-value= .04489
sample no: 2 r^3= 2.102 p-value= .06765
sample no: 3 r^3= 5.058 p-value= .15517
sample no: 4 r^3=
38.238 p-value= .72046
sample no: 5 r^3= 54.851 p-value=
.83932
sample no: 6 r^3= 33.678 p-value= .67456
sample no:
7 r^3= 31.142 p-value= .64586
sample no: 8 r^3= 4.200
p-value= .13063
sample no: 9 r^3= 48.705 p-value= .80279
sample no: 10 r^3= 10.270 p-value= .28988
sample no: 11 r^3=
34.434 p-value= .68267
sample no: 12 r^3= 5.079 p-value=
.15575
sample no: 13 r^3= 39.034 p-value= .72777
sample
no: 14 r^3= 49.623 p-value= .80874
sample no: 15 r^3= 19.014
p-value= .46943
sample no: 16 r^3= 24.378 p-value= .55630
sample no: 17 r^3= 33.022 p-value= .66737
sample no: 18 r^3=
58.761 p-value= .85896
sample no: 19 r^3= 24.770 p-value=
.56206
sample no: 20 r^3= .370 p-value= .01227
A KS test
is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file test1 p-value= .504318
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the SQEEZE test ::
:: Random integers are
floated to get uniforms on [0,1). Start- ::
:: ing with
k=2^31=2147483647, the test finds j, the number of ::
::
iterations necessary to reduce k to 1, using the reduction ::
:: k=ceiling(k*U), with U provided by floating integers from
::
:: the file being tested. Such j's are found 100,000
times, ::
:: then counts for the number of times j was
<=6,7,...,47,>=48 ::
:: are used to provide a chi-square
test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESULTS OF SQUEEZE TEST FOR test1
Table of standardized
frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking
values <=6,7,8,...,47,>=48:
-.1 -.3 .3 -1.1 1.6 .3
.2 .5
-1.0 -2.2 .2 -.6
-1.6 .3 .1 -.7 1.4 .1
.3 .9 .7 -.5 1.5
-2.1
.0 2.1 .0 -1.3 .1 .5
-.9 .1 -.7 .3 .5 -.5
.0 .2 .9
-.1 .1 -1.0
-.1
Chi-square with 42 degrees of freedom:
34.423
z-score= -.827 p-value= .209137
______________________________________________________________
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The OVERLAPPING SUMS test ::
:: Integers are floated
to get a sequence U(1),U(2),... of uni- ::
:: form [0,1)
variables. Then overlapping sums, ::
::
S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. ::
:: The S's are virtually normal with a certain covariance mat-
::
:: rix. A linear transformation of the S's converts them
to a ::
:: sequence of independent standard normals, which
are converted ::
:: to uniform variables for a KSTEST. The
p-values from ten ::
:: KSTESTs are given still another
KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test no. 1 p-value .913771
Test no. 2 p-value .280257
Test no. 3 p-value .554307
Test no. 4 p-value .793627
Test
no. 5 p-value .673222
Test no. 6 p-value .235984
Test no.
7 p-value .105186
Test no. 8 p-value .114326
Test no. 9
p-value .852223
Test no. 10 p-value .265052
Results of the
OSUM test for test1
KSTEST on the above 10 p-values: .083267
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the RUNS test. It counts runs up, and runs down,
::
:: in a sequence of uniform [0,1) variables, obtained by
float- ::
:: ing the 32-bit integers in the specified file.
This example ::
:: shows how runs are counted:
.123,.357,.789,.425,.224,.416,.95::
:: contains an up-run of
length 3, a down-run of length 2 and an ::
:: up-run of (at
least) 2, depending on the next values. The ::
:: covariance
matrices for the runs-up and runs-down are well ::
:: known,
leading to chisquare tests for quadratic forms in the ::
::
weak inverses of the covariance matrices. Runs are counted ::
:: for sequences of length 10,000. This is done ten times.
Then ::
:: repeated. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The RUNS test for file test1
Up and down runs in a
sample of 10000
_________________________________________________
Run test
for test1 :
runs up; ks test for 10 p's: .412041
runs
down; ks test for 10 p's: .119478
Run test for test1 :
runs up; ks test for 10 p's: .022722
runs down; ks test for
10 p's: .873031
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test1
No. of wins: Observed
Expected
98730 98585.86
98730= No. of wins, z-score= .645
pvalue= .74043
Analysis of Throws-per-Game:
Chisq= 19.52
for 20 degrees of freedom, p= .51186
Throws Observed Expected
Chisq Sum
1 66591 66666.7 .086 .086
2 37609 37654.3 .055
.140
3 26715 26954.7 2.132 2.273
4 19153 19313.5 1.333
3.606
5 14010 13851.4 1.816 5.421
6 9919 9943.5 .061 5.482
7 7291 7145.0 2.982 8.464
8 5215 5139.1 1.122 9.586
9 3808
3699.9 3.160 12.746
10 2741 2666.3 2.093 14.839
11 1916
1923.3 .028 14.867
12 1350 1388.7 1.081 15.948
13 1024
1003.7 .410 16.358
14 744 726.1 .439 16.797
15 535 525.8
.160 16.957
16 379 381.2 .012 16.969
17 253 276.5 2.004
18.973
18 199 200.8 .017 18.990
19 144 146.0 .027 19.016
20 105 106.2 .014 19.030
21 299 287.1 .492 19.522
SUMMARY
FOR test1
p-value for no. of wins: .740433
p-value for
throws/game: .511855
| TEST 2 |
Entropy source: PIC24EP512GP202 |
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 test2
For a sample of size 500: mean
test2
using bits 1 to 24 2.060
duplicate number number
spacings
observed expected
0 53. 67.668
1 131. 135.335
2 158.
135.335
3 88. 90.224
4 45. 45.112
5 13. 18.045
6 to
INF 12. 8.282
Chisquare with 6 d.o.f. = 10.25 p-value=
.885436
:::::::::::::::::::::::::::::::::::::::::
For a
sample of size 500: mean
test2 using bits 2 to 25 2.034
duplicate number number
spacings observed expected
0 64.
67.668
1 128. 135.335
2 143. 135.335
3 94. 90.224
4
44. 45.112
5 19. 18.045
6 to INF 8. 8.282
Chisquare
with 6 d.o.f. = 1.28 p-value= .027040
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 3 to 26 1.910
duplicate
number number
spacings observed expected
0 66. 67.668
1 140. 135.335
2 153. 135.335
3 78. 90.224
4 45. 45.112
5 14. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. =
7.28 p-value= .704644
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 4 to 27 1.902
duplicate
number number
spacings observed expected
0 73. 67.668
1 144. 135.335
2 132. 135.335
3 95. 90.224
4 33. 45.112
5 15. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. =
5.09 p-value= .467061
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 5 to 28 2.022
duplicate
number number
spacings observed expected
0 58. 67.668
1 148. 135.335
2 131. 135.335
3 90. 90.224
4 47. 45.112
5 16. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. =
3.37 p-value= .239218
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 6 to 29 2.018
duplicate
number number
spacings observed expected
0 69. 67.668
1 116. 135.335
2 152. 135.335
3 97. 90.224
4 44. 45.112
5 12. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. =
7.76 p-value= .743657
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 7 to 30 2.020
duplicate
number number
spacings observed expected
0 63. 67.668
1 135. 135.335
2 124. 135.335
3 111. 90.224
4 45.
45.112
5 18. 18.045
6 to INF 4. 8.282
Chisquare with 6
d.o.f. = 8.27 p-value= .781067
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 8 to 31 2.078
duplicate
number number
spacings observed expected
0 70. 67.668
1 121. 135.335
2 132. 135.335
3 98. 90.224
4 50. 45.112
5 18. 18.045
6 to INF 11. 8.282
Chisquare with 6 d.o.f. =
3.77 p-value= .292669
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test2 using bits 9 to 32 2.060
duplicate
number number
spacings observed expected
0 63. 67.668
1 128. 135.335
2 137. 135.335
3 98. 90.224
4 45. 45.112
5 21. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. =
1.90 p-value= .071702
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.885436 .027040 .704644 .467061 .239218
.743657 .781067
.292669 .071702
A KSTEST for the 9 p-values yields .116131
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test2
For a sample of 1,000,000
consecutive 5-tuples,
chisquare for 99 degrees of
freedom=127.105; p-value= .970001
OPERM5 test for file test2
For a sample of 1,000,000 consecutive 5-tuples,
chisquare
for 99 degrees of freedom=100.421; p-value= .558774
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test2
Rank test for 31x31 binary
matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 214 211.4 .031533
.032
29 5144 5134.0 .019438 .051
30 23014 23103.0 .343216
.394
31 11628 11551.5 .506298 .900
chisquare= .900 for 3
d. of f.; p-value= .346016
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test2
Rank test for 32x32 binary
matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 202 211.4 .419543
.420
30 5160 5134.0 .131567 .551
31 23204 23103.0 .441134
.992
32 11434 11551.5 1.195685 2.188
chisquare= 2.188 for
3 d. of f.; p-value= .535219
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test2
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG test2
b-rank test for
bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 950 944.3
.034 .034
r =5 21815 21743.9 .232 .267
r =6 77235 77311.8
.076 .343
p=1-exp(-SUM/2)= .15767
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 936 944.3 .073 .073
r =5 21693 21743.9 .119 .192
r =6
77371 77311.8 .045 .237
p=1-exp(-SUM/2)= .11195
Rank of a
6x8 binary matrix,
rows formed from eight bits of the RNG
test2
b-rank test for bits 3 to 10
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 966 944.3 .499 .499
r =5 21805 21743.9
.172 .670
r =6 77229 77311.8 .089 .759
p=1-exp(-SUM/2)=
.31579
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test2
b-rank test for bits 4 to 11
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 test2
b-rank test for bits
5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 940 944.3
.020 .020
r =5 21806 21743.9 .177 .197
r =6 77254 77311.8
.043 .240
p=1-exp(-SUM/2)= .11315
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 941 944.3 .012 .012
r =5 21726 21743.9 .015 .026
r =6
77333 77311.8 .006 .032
p=1-exp(-SUM/2)= .01592
Rank of a
6x8 binary matrix,
rows formed from eight bits of the RNG
test2
b-rank test for bits 7 to 14
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 922 944.3 .527 .527
r =5 21673 21743.9
.231 .758
r =6 77405 77311.8 .112 .870
p=1-exp(-SUM/2)=
.35280
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test2
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 943 944.3 .002 .002
r
=5 21613 21743.9 .788 .790
r =6 77444 77311.8 .226 1.016
p=1-exp(-SUM/2)= .39826
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test2
b-rank test for bits
9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 955 944.3
.121 .121
r =5 21462 21743.9 3.655 3.776
r =6 77583
77311.8 .951 4.727
p=1-exp(-SUM/2)= .90592
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 932 944.3 .160 .160
r =5 21362 21743.9 6.708
6.868
r =6 77706 77311.8 2.010 8.878
p=1-exp(-SUM/2)=
.98819
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test2
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 885 944.3 3.724 3.724
r =5 21582 21743.9 1.205 4.930
r =6 77533 77311.8 .633 5.562
p=1-exp(-SUM/2)= .93804
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test2
b-rank test for bits
12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 918 944.3
.733 .733
r =5 21734 21743.9 .005 .737
r =6 77348 77311.8
.017 .754
p=1-exp(-SUM/2)= .31409
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 902 944.3 1.895 1.895
r =5 21542 21743.9 1.875 3.770
r =6 77556 77311.8 .771 4.541
p=1-exp(-SUM/2)= .89674
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test2
b-rank test for bits 14 to 21
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 883 944.3 3.979 3.979
r =5 21598
21743.9 .979 4.958
r =6 77519 77311.8 .555 5.514
p=1-exp(-SUM/2)= .93651
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test2
b-rank test for bits
15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 980 944.3
1.350 1.350
r =5 21560 21743.9 1.555 2.905
r =6 77460
77311.8 .284 3.189
p=1-exp(-SUM/2)= .79699
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 960 944.3 .261 .261
r =5 21603 21743.9 .913 1.174
r =6 77437 77311.8 .203 1.377
p=1-exp(-SUM/2)= .49761
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test2
b-rank test for bits 17 to 24
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 921 944.3 .575 .575
r =5 21743 21743.9
.000 .575
r =6 77336 77311.8 .008 .583
p=1-exp(-SUM/2)=
.25270
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test2
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 963 944.3 .370 .370
r
=5 21280 21743.9 9.897 10.267
r =6 77757 77311.8 2.564 12.831
p=1-exp(-SUM/2)= .99836
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test2
b-rank test for bits
19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 911 944.3
1.174 1.174
r =5 21730 21743.9 .009 1.183
r =6 77359
77311.8 .029 1.212
p=1-exp(-SUM/2)= .45449
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 958 944.3 .199 .199
r =5 21855 21743.9 .568 .766
r =6 77187 77311.8 .201 .968
p=1-exp(-SUM/2)= .38364
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test2
b-rank test for bits 21 to 28
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 962 944.3 .332 .332
r =5 21804 21743.9
.166 .498
r =6 77234 77311.8 .078 .576
p=1-exp(-SUM/2)=
.25029
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test2
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 885 944.3 3.724 3.724
r =5 21629 21743.9 .607 4.331
r =6 77486 77311.8 .392 4.724
p=1-exp(-SUM/2)= .90576
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test2
b-rank test for bits
23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 958 944.3
.199 .199
r =5 21784 21743.9 .074 .273
r =6 77258 77311.8
.037 .310
p=1-exp(-SUM/2)= .14364
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test2
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 959 944.3 .229 .229
r =5 21560 21743.9 1.555 1.784
r
=6 77481 77311.8 .370 2.154
p=1-exp(-SUM/2)= .65946
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test2
b-rank test for bits 25 to 32
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 962 944.3 .332 .332
r =5 21583 21743.9
1.191 1.522
r =6 77455 77311.8 .265 1.788
p=1-exp(-SUM/2)=
.59090
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.157675
.111949 .315791 .605756 .113153
.015916 .352803 .398262
.905921 .988190
.938036 .314088 .896738 .936511 .796989
.497610 .252703 .998364 .454494 .383642
.250289 .905755
.143636 .659457 .590897
brank test summary for test2
The
KS test for those 25 supposed UNI's yields
KS p-value=
.550253
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 141999 missing words, .21 sigmas from mean, p-value=
.58298
tst no 2: 141428 missing words, -1.12 sigmas from
mean, p-value= .13038
tst no 3: 142119 missing words, .49
sigmas from mean, p-value= .68789
tst no 4: 141780 missing
words, -.30 sigmas from mean, p-value= .38126
tst no 5:
141111 missing words, -1.87 sigmas from mean, p-value= .03107
tst no 6: 142333 missing words, .99 sigmas from mean, p-value=
.83889
tst no 7: 141956 missing words, .11 sigmas from mean,
p-value= .54342
tst no 8: 141808 missing words, -.24 sigmas
from mean, p-value= .40643
tst no 9: 141713 missing words,
-.46 sigmas from mean, p-value= .32322
tst no 10: 142234
missing words, .76 sigmas from mean, p-value= .77595
tst no
11: 141440 missing words, -1.10 sigmas from mean, p-value=
.13642
tst no 12: 141590 missing words, -.75 sigmas from
mean, p-value= .22781
tst no 13: 141436 missing words, -1.11
sigmas from mean, p-value= .13438
tst no 14: 140896 missing
words, -2.37 sigmas from mean, p-value= .00895
tst no 15:
140879 missing words, -2.41 sigmas from mean, p-value= .00804
tst no 16: 142698 missing words, 1.84 sigmas from mean, p-value=
.96731
tst no 17: 142288 missing words, .88 sigmas from mean,
p-value= .81185
tst no 18: 141412 missing words, -1.16 sigmas
from mean, p-value= .12262
tst no 19: 142813 missing words,
2.11 sigmas from mean, p-value= .98263
tst no 20: 141881
missing words, -.07 sigmas from mean, p-value= .47361
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test2
Output: No. missing words
(mw), equiv normal variate (z), p-value (p)
mw z p
OPSO
for test2 using bits 23 to 32 142050 .485 .6862
OPSO for
test2 using bits 22 to 31 141966 .195 .5775
OPSO for test2
using bits 21 to 30 142635 2.502 .9938
OPSO for test2 using
bits 20 to 29 141944 .120 .5476
OPSO for test2 using bits 19
to 28 141874 -.122 .4515
OPSO for test2 using bits 18 to 27
142123 .737 .7694
OPSO for test2 using bits 17 to 26 142439
1.826 .9661
OPSO for test2 using bits 16 to 25 142134 .775
.7808
OPSO for test2 using bits 15 to 24 141554 -1.225 .1102
OPSO for test2 using bits 14 to 23 141775 -.463 .3216
OPSO
for test2 using bits 13 to 22 141661 -.856 .1959
OPSO for
test2 using bits 12 to 21 142576 2.299 .9892
OPSO for test2
using bits 11 to 20 142016 .368 .6435
OPSO for test2 using
bits 10 to 19 142126 .747 .7725
OPSO for test2 using bits 9
to 18 141954 .154 .5612
OPSO for test2 using bits 8 to 17
142382 1.630 .9484
OPSO for test2 using bits 7 to 16 141896
-.046 .4817
OPSO for test2 using bits 6 to 15 142236 1.126
.8700
OPSO for test2 using bits 5 to 14 142380 1.623 .9477
OPSO for test2 using bits 4 to 13 141916 .023 .5092
OPSO for
test2 using bits 3 to 12 141484 -1.467 .0712
OPSO for test2
using bits 2 to 11 141729 -.622 .2670
OPSO for test2 using
bits 1 to 10 141511 -1.374 .0848
OQSO test for generator
test2
Output: No. missing words (mw), equiv normal variate
(z), p-value (p)
mw z p
OQSO for test2 using bits 28 to 32
141854 -.188 .4256
OQSO for test2 using bits 27 to 31 141881
-.096 .4617
OQSO for test2 using bits 26 to 30 142218 1.046
.8523
OQSO for test2 using bits 25 to 29 141878 -.106 .4577
OQSO for test2 using bits 24 to 28 141656 -.859 .1952
OQSO
for test2 using bits 23 to 27 141917 .026 .5104
OQSO for
test2 using bits 22 to 26 142291 1.294 .9021
OQSO for test2
using bits 21 to 25 141838 -.242 .4045
OQSO for test2 using
bits 20 to 24 142064 .524 .7000
OQSO for test2 using bits 19
to 23 141142 -2.601 .0046
OQSO for test2 using bits 18 to 22
142020 .375 .6462
OQSO for test2 using bits 17 to 21 141955
.155 .5615
OQSO for test2 using bits 16 to 20 141792 -.398
.3454
OQSO for test2 using bits 15 to 19 142450 1.833 .9666
OQSO for test2 using bits 14 to 18 141819 -.306 .3797
OQSO
for test2 using bits 13 to 17 142042 .450 .6735
OQSO for
test2 using bits 12 to 16 142070 .545 .7070
OQSO for test2
using bits 11 to 15 141454 -1.543 .0614
OQSO for test2 using
bits 10 to 14 142284 1.270 .8980
OQSO for test2 using bits 9
to 13 141981 .243 .5960
OQSO for test2 using bits 8 to 12
142114 .694 .7561
OQSO for test2 using bits 7 to 11 141648
-.886 .1878
OQSO for test2 using bits 6 to 10 141772 -.466
.3208
OQSO for test2 using bits 5 to 9 141672 -.805 .2106
OQSO for test2 using bits 4 to 8 141966 .192 .5762
OQSO for
test2 using bits 3 to 7 141685 -.760 .2235
OQSO for test2
using bits 2 to 6 141864 -.154 .4389
OQSO for test2 using
bits 1 to 5 142258 1.182 .8814
DNA test for generator test2
Output: No. missing words (mw), equiv normal variate (z),
p-value (p)
mw z p
DNA for test2 using bits 31 to 32
142173 .778 .7817
DNA for test2 using bits 30 to 31 141741
-.497 .3098
DNA for test2 using bits 29 to 30 141914 .014
.5055
DNA for test2 using bits 28 to 29 142116 .610 .7290
DNA for test2 using bits 27 to 28 141427 -1.423 .0774
DNA for
test2 using bits 26 to 27 141958 .144 .5571
DNA for test2
using bits 25 to 26 141477 -1.275 .1011
DNA for test2 using
bits 24 to 25 142005 .282 .6111
DNA for test2 using bits 23
to 24 141989 .235 .5929
DNA for test2 using bits 22 to 23
141542 -1.084 .1393
DNA for test2 using bits 21 to 22 142529
1.828 .9662
DNA for test2 using bits 20 to 21 142445 1.580
.9430
DNA for test2 using bits 19 to 20 142053 .424 .6641
DNA for test2 using bits 18 to 19 141439 -1.387 .0827
DNA for
test2 using bits 17 to 18 142300 1.152 .8754
DNA for test2
using bits 16 to 17 141606 -.895 .1855
DNA for test2 using
bits 15 to 16 142192 .834 .7978
DNA for test2 using bits 14
to 15 141618 -.859 .1951
DNA for test2 using bits 13 to 14
141624 -.842 .2000
DNA for test2 using bits 12 to 13 142709
2.359 .9908
DNA for test2 using bits 11 to 12 142231 .949
.8287
DNA for test2 using bits 10 to 11 141980 .208 .5826
DNA for test2 using bits 9 to 10 142249 1.002 .8418
DNA for
test2 using bits 8 to 9 142080 .503 .6927
DNA for test2 using
bits 7 to 8 142446 1.583 .9433
DNA for test2 using bits 6 to
7 141838 -.210 .4167
DNA for test2 using bits 5 to 6 142292
1.129 .8705
DNA for test2 using bits 4 to 5 142160 .739 .7702
DNA for test2 using bits 3 to 4 141934 .073 .5290
DNA for
test2 using bits 2 to 3 141701 -.615 .2694
DNA for test2
using bits 1 to 2 142273 1.073 .8583
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNT-THE-1's TEST on a stream of bytes. ::
:: Consider the file under test as a stream of bytes (four
per ::
:: 32 bit integer). Each byte can contain from 0 to 8
1's, ::
:: with probabilities 1,8,28,56,70,56,28,8,1 over
256. Now let ::
:: the stream of bytes provide a string of
overlapping 5-letter ::
:: words, each "letter" taking
values A,B,C,D,E. The letters are ::
:: determined by the
number of 1's in a byte:: 0,1,or 2 yield A,::
:: 3 yields B,
4 yields C, 5 yields D and 6,7 or 8 yield E. Thus ::
:: we
have a monkey at a typewriter hitting five keys with vari- ::
:: ous probabilities (37,56,70,56,37 over 256). There are
5^5 ::
:: possible 5-letter words, and from a string of
256,000 (over- ::
:: lapping) 5-letter words, counts are
made on the frequencies ::
:: for each word. The quadratic
form in the weak inverse of ::
:: the covariance matrix of
the cell counts provides a chisquare ::
:: test:: Q5-Q4, the
difference of the naive Pearson sums of ::
::
(OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for test2
Chi-square with 5^5-5^4=2500 d.of
f. for sample size:2560000
chisquare equiv normal p-value
Results fo COUNT-THE-1's in successive bytes:
byte stream for
test2 2534.65 .490 .687927
byte stream for test2 2530.42 .430
.666480
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 2379.32 -1.707 .043937
bits 2
to 9 2557.94 .819 .793737
bits 3 to 10 2599.84 1.412 .921016
bits 4 to 11 2542.56 .602 .726360
bits 5 to 12 2476.99 -.325
.372445
bits 6 to 13 2383.61 -1.646 .049888
bits 7 to 14
2586.83 1.228 .890274
bits 8 to 15 2504.97 .070 .528040
bits 9 to 16 2540.97 .579 .718854
bits 10 to 17 2412.05
-1.244 .106789
bits 11 to 18 2538.03 .538 .704663
bits 12
to 19 2502.92 .041 .516495
bits 13 to 20 2556.32 .796 .787107
bits 14 to 21 2483.68 -.231 .408739
bits 15 to 22 2522.06
.312 .622476
bits 16 to 23 2543.70 .618 .731696
bits 17 to
24 2546.72 .661 .745625
bits 18 to 25 2535.45 .501 .691938
bits 19 to 26 2560.10 .850 .802313
bits 20 to 27 2427.84
-1.021 .153734
bits 21 to 28 2591.24 1.290 .901520
bits 22
to 29 2524.49 .346 .635455
bits 23 to 30 2418.28 -1.156
.123909
bits 24 to 31 2630.67 1.848 .967694
bits 25 to 32
2474.78 -.357 .360670
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test2
Of 12,000
tries, the average no. of successes
should be 3523 with
sigma=21.9
Successes: 3523 z-score: .000 p-value: .500000
Successes: 3530 z-score: .320 p-value: .625377
Successes:
3506 z-score: -.776 p-value: .218799
Successes: 3534 z-score:
.502 p-value: .692266
Successes: 3531 z-score: .365 p-value:
.642555
Successes: 3522 z-score: -.046 p-value: .481790
Successes: 3535 z-score: .548 p-value: .708135
Successes:
3537 z-score: .639 p-value: .738676
Successes: 3551 z-score:
1.279 p-value: .899470
Successes: 3544 z-score: .959 p-value:
.831196
square size avg. no. parked sample sigma
100.
3531.300 11.811
KSTEST for the above 10: p= .857477
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100
times:: choose n=8000 random points in a ::
:: square of
side 10000. Find d, the minimum distance between ::
:: the
(n^2-n)/2 pairs of points. If the points are truly inde- ::
:: pendent uniform, then d^2, the square of the minimum distance
::
:: should be (very close to) exponentially distributed
with mean ::
:: .995 . Thus 1-exp(-d^2/.995) should be
uniform on [0,1) and ::
:: a KSTEST on the resulting 100
values serves as a test of uni- ::
:: formity for random
points in the square. Test numbers=0 mod 5 ::
:: are printed
but the KSTEST is based on the full set of 100 ::
:: random
choices of 8000 points in the 10000x10000 square. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
This is the MINIMUM DISTANCE test
for random integers in
the file test2
Sample no. d^2 avg equiv uni
5 1.5700
.8965 .793580
10 .8236 .8487 .562978
15 .2478 .7923
.220484
20 .6932 .9070 .501790
25 .3200 1.0408 .275001
30 .9178 1.0026 .602434
35 .5345 1.0161 .415618
40 .0951
.9276 .091141
45 .3578 .9107 .302048
50 .0422 .9004
.041495
55 .0527 .8899 .051624
60 .0524 .8654 .051309
65 1.2588 .8663 .717803
70 .0685 .8710 .066492
75 1.1848
.8597 .696010
80 .4625 .8643 .371770
85 .5492 .8344
.424163
90 .5521 .8427 .425877
95 .5565 .8500 .428383
100 .3326 .8418 .284108
MINIMUM DISTANCE TEST for test2
Result of KS test on 20 transformed mindist^2's:
p-value=
.621339
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE 3DSPHERES TEST ::
:: Choose 4000 random points in
a cube of edge 1000. At each ::
:: point, center a sphere
large enough to reach the next closest ::
:: point. Then the
volume of the smallest such sphere is (very ::
:: close to)
exponentially distributed with mean 120pi/3. Thus ::
:: the
radius cubed is exponential with mean 30. (The mean is ::
::
obtained by extensive simulation). The 3DSPHERES test gener- ::
:: ates 4000 such spheres 20 times. Each min radius cubed
leads ::
:: to a uniform variable by means of
1-exp(-r^3/30.), then a ::
:: KSTEST is done on the 20
p-values. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file test2
sample no: 1 r^3=
1.829 p-value= .05914
sample no: 2 r^3= 19.172 p-value=
.47222
sample no: 3 r^3= 85.899 p-value= .94292
sample no:
4 r^3= 25.478 p-value= .57227
sample no: 5 r^3= 45.964
p-value= .78393
sample no: 6 r^3= 10.062 p-value= .28496
sample no: 7 r^3= 42.853 p-value= .76031
sample no: 8 r^3=
15.238 p-value= .39827
sample no: 9 r^3= 7.405 p-value=
.21874
sample no: 10 r^3= 14.219 p-value= .37746
sample
no: 11 r^3= 54.344 p-value= .83658
sample no: 12 r^3= 9.248
p-value= .26529
sample no: 13 r^3= 4.479 p-value= .13869
sample no: 14 r^3= 38.643 p-value= .72421
sample no: 15 r^3=
10.574 p-value= .29705
sample no: 16 r^3= 26.240 p-value=
.58300
sample no: 17 r^3= 15.308 p-value= .39967
sample
no: 18 r^3= 14.764 p-value= .38868
sample no: 19 r^3= 47.634
p-value= .79563
sample no: 20 r^3= 17.227 p-value= .43686
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file test2 p-value= .296908
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the SQEEZE test ::
:: Random integers are
floated to get uniforms on [0,1). Start- ::
:: ing with
k=2^31=2147483647, the test finds j, the number of ::
::
iterations necessary to reduce k to 1, using the reduction ::
:: k=ceiling(k*U), with U provided by floating integers from
::
:: the file being tested. Such j's are found 100,000
times, ::
:: then counts for the number of times j was
<=6,7,...,47,>=48 ::
:: are used to provide a chi-square
test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESULTS OF SQUEEZE TEST FOR test2
Table of standardized
frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking
values <=6,7,8,...,47,>=48:
.6 -2.0 1.8 .6 -2.5 1.8
-.8 .0
-1.3 -1.0 1.2 .6
.0 .4 -.2 -.5 -1.4 -.4
-.9 1.4 .1 .1 -.2
3.1
-.5 -.2 -.1 .8 .4 .7
-.9 -.7 -.7 -.1 .1 -.3
-2.6
-.1 1.3 -.7 -1.3 -1.0
-1.1
Chi-square with 42 degrees of
freedom: 53.598
z-score= 1.265 p-value= .891928
______________________________________________________________
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The OVERLAPPING SUMS test ::
:: Integers are floated
to get a sequence U(1),U(2),... of uni- ::
:: form [0,1)
variables. Then overlapping sums, ::
::
S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. ::
:: The S's are virtually normal with a certain covariance mat-
::
:: rix. A linear transformation of the S's converts them
to a ::
:: sequence of independent standard normals, which
are converted ::
:: to uniform variables for a KSTEST. The
p-values from ten ::
:: KSTESTs are given still another
KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test no. 1 p-value .594600
Test no. 2 p-value .180036
Test no. 3 p-value .002648
Test no. 4 p-value .588924
Test
no. 5 p-value .590185
Test no. 6 p-value .116015
Test no.
7 p-value .814074
Test no. 8 p-value .912421
Test no. 9
p-value .861501
Test no. 10 p-value .542335
Results of the
OSUM test for test2
KSTEST on the above 10 p-values: .364078
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the RUNS test. It counts runs up, and runs down,
::
:: in a sequence of uniform [0,1) variables, obtained by
float- ::
:: ing the 32-bit integers in the specified file.
This example ::
:: shows how runs are counted:
.123,.357,.789,.425,.224,.416,.95::
:: contains an up-run of
length 3, a down-run of length 2 and an ::
:: up-run of (at
least) 2, depending on the next values. The ::
:: covariance
matrices for the runs-up and runs-down are well ::
:: known,
leading to chisquare tests for quadratic forms in the ::
::
weak inverses of the covariance matrices. Runs are counted ::
:: for sequences of length 10,000. This is done ten times.
Then ::
:: repeated. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The RUNS test for file test2
Up and down runs in a
sample of 10000
_________________________________________________
Run test
for test2 :
runs up; ks test for 10 p's: .734043
runs
down; ks test for 10 p's: .801918
Run test for test2 :
runs up; ks test for 10 p's: .434815
runs down; ks test for
10 p's: .394612
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the CRAPS TEST. It plays 200,000 games of craps,
finds::
:: the number of wins and the number of throws
necessary to end ::
:: each game. The number of wins should
be (very close to) a ::
:: normal with mean 200000p and
variance 200000p(1-p), with ::
:: p=244/495. Throws
necessary to complete the game can vary ::
:: from 1 to
infinity, but counts for all>21 are lumped with 21. ::
:: A
chi-square test is made on the no.-of-throws cell counts. ::
:: Each 32-bit integer from the test file provides the value for
::
:: the throw of a die, by floating to [0,1), multiplying
by 6 ::
:: and taking 1 plus the integer part of the result.
::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Results of craps test for test2
No. of wins: Observed
Expected
98776 98585.86
98776= No. of wins, z-score= .850
pvalue= .80245
Analysis of Throws-per-Game:
Chisq= 17.71
for 20 degrees of freedom, p= .39370
Throws Observed Expected
Chisq Sum
1 66689 66666.7 .007 .007
2 37904 37654.3 1.656
1.663
3 26687 26954.7 2.659 4.322
4 19433 19313.5 .740
5.062
5 14023 13851.4 2.125 7.188
6 9755 9943.5 3.575
10.763
7 7235 7145.0 1.133 11.896
8 5039 5139.1 1.949
13.844
9 3683 3699.9 .077 13.921
10 2696 2666.3 .331
14.252
11 1886 1923.3 .724 14.977
12 1384 1388.7 .016
14.993
13 966 1003.7 1.417 16.410
14 717 726.1 .115 16.525
15 513 525.8 .313 16.838
16 386 381.2 .062 16.900
17 278
276.5 .008 16.908
18 196 200.8 .116 17.024
19 141 146.0
.170 17.194
20 99 106.2 .490 17.684
21 290 287.1 .029
17.713
SUMMARY FOR test2
p-value for no. of wins: .802454
p-value for throws/game: .393697
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
Results of DIEHARD battery of tests sent to file t2.txt
| TEST 3 |
Entropy source: SRAM CY62177EV30 |
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 test3
For a sample of size 500: mean
test3
using bits 1 to 24 2.004
duplicate number number
spacings
observed expected
0 72. 67.668
1 135. 135.335
2 122.
135.335
3 103. 90.224
4 41. 45.112
5 14. 18.045
6 to
INF 13. 8.282
Chisquare with 6 d.o.f. = 7.37 p-value= .712091
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 2 to 25 1.964
duplicate
number number
spacings observed expected
0 64. 67.668
1 147. 135.335
2 126. 135.335
3 93. 90.224
4 52. 45.112
5 14. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. =
6.11 p-value= .588548
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 3 to 26 2.008
duplicate
number number
spacings observed expected
0 64. 67.668
1 135. 135.335
2 141. 135.335
3 92. 90.224
4 43. 45.112
5 15. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. =
1.44 p-value= .036668
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 4 to 27 1.874
duplicate
number number
spacings observed expected
0 85. 67.668
1 120. 135.335
2 144. 135.335
3 94. 90.224
4 44. 45.112
5 8. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. =
13.81 p-value= .968160
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 5 to 28 2.088
duplicate
number number
spacings observed expected
0 52. 67.668
1 133. 135.335
2 153. 135.335
3 90. 90.224
4 39. 45.112
5 22. 18.045
6 to INF 11. 8.282
Chisquare with 6 d.o.f. =
8.56 p-value= .800209
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 6 to 29 2.040
duplicate
number number
spacings observed expected
0 71. 67.668
1 113. 135.335
2 150. 135.335
3 89. 90.224
4 53. 45.112
5 17. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. =
7.09 p-value= .687767
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 7 to 30 2.036
duplicate
number number
spacings observed expected
0 65. 67.668
1 131. 135.335
2 147. 135.335
3 80. 90.224
4 47. 45.112
5 20. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. =
3.06 p-value= .198122
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 8 to 31 2.116
duplicate
number number
spacings observed expected
0 65. 67.668
1 120. 135.335
2 142. 135.335
3 86. 90.224
4 53. 45.112
5 24. 18.045
6 to INF 10. 8.282
Chisquare with 6 d.o.f. =
6.07 p-value= .584606
:::::::::::::::::::::::::::::::::::::::::
For a sample of
size 500: mean
test3 using bits 9 to 32 2.004
duplicate
number number
spacings observed expected
0 58. 67.668
1 144. 135.335
2 148. 135.335
3 74. 90.224
4 45. 45.112
5 26. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. =
10.85 p-value= .906756
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.712091 .588548 .036668 .968160 .800209
.687767 .198122
.584606 .906756
A KSTEST for the 9 p-values yields .613718
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test3
For a sample of 1,000,000
consecutive 5-tuples,
chisquare for 99 degrees of freedom=
87.274; p-value= .205777
OPERM5 test for file test3
For a
sample of 1,000,000 consecutive 5-tuples,
chisquare for 99
degrees of freedom=129.770; p-value= .979327
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test3
Rank test for 31x31 binary
matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 212 211.4 .001602
.002
29 5142 5134.0 .012434 .014
30 23125 23103.0 .020860
.035
31 11521 11551.5 .080659 .116
chisquare= .116 for 3
d. of f.; p-value= .377046
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test3
Rank test for 32x32 binary
matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 220 211.4 .348364
.348
30 5223 5134.0 1.542493 1.891
31 23167 23103.0
.177033 2.068
32 11390 11551.5 2.258588 4.326
chisquare=
4.326 for 3 d. of f.; p-value= .790462
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test3
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for
bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 960 944.3
.261 .261
r =5 21568 21743.9 1.423 1.684
r =6 77472
77311.8 .332 2.016
p=1-exp(-SUM/2)= .63503
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 944 944.3 .000 .000
r =5 21597 21743.9 .992 .993
r =6 77459 77311.8 .280 1.273
p=1-exp(-SUM/2)= .47080
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test3
b-rank test for bits 3 to 10
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 983 944.3 1.586 1.586
r =5 21716
21743.9 .036 1.622
r =6 77301 77311.8 .002 1.623
p=1-exp(-SUM/2)= .55586
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3
.008 .008
r =5 21954 21743.9 2.030 2.038
r =6 77099
77311.8 .586 2.624
p=1-exp(-SUM/2)= .73066
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 933 944.3 .135 .135
r =5 21801 21743.9 .150 .285
r =6 77266 77311.8 .027 .312
p=1-exp(-SUM/2)= .14458
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test3
b-rank test for bits 6 to 13
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 925 944.3 .395 .395
r =5 21879 21743.9
.839 1.234
r =6 77196 77311.8 .173 1.407
p=1-exp(-SUM/2)=
.50524
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test3
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 945 944.3 .001 .001
r
=5 21766 21743.9 .022 .023
r =6 77289 77311.8 .007 .030
p=1-exp(-SUM/2)= .01474
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 953 944.3
.080 .080
r =5 21761 21743.9 .013 .094
r =6 77286 77311.8
.009 .102
p=1-exp(-SUM/2)= .04981
Rank of a 6x8 binary
matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 933 944.3 .135 .135
r =5 21894 21743.9 1.036 1.171
r
=6 77173 77311.8 .249 1.421
p=1-exp(-SUM/2)= .50851
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test3
b-rank test for bits 10 to 17
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 990 944.3 2.212 2.212
r =5 21812
21743.9 .213 2.425
r =6 77198 77311.8 .168 2.592
p=1-exp(-SUM/2)= .72642
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 993 944.3
2.511 2.511
r =5 21751 21743.9 .002 2.514
r =6 77256
77311.8 .040 2.554
p=1-exp(-SUM/2)= .72113
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 934 944.3 .112 .112
r =5 21732 21743.9 .007 .119
r =6 77334 77311.8 .006 .125
p=1-exp(-SUM/2)= .06071
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test3
b-rank test for bits 13 to 20
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 941 944.3 .012 .012
r =5 21582 21743.9
1.205 1.217
r =6 77477 77311.8 .353 1.570
p=1-exp(-SUM/2)=
.54388
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test3
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 968 944.3 .595 .595
r
=5 21804 21743.9 .166 .761
r =6 77228 77311.8 .091 .852
p=1-exp(-SUM/2)= .34679
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 993 944.3
2.511 2.511
r =5 21862 21743.9 .641 3.153
r =6 77145
77311.8 .360 3.513
p=1-exp(-SUM/2)= .82733
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 993 944.3 2.511 2.511
r =5 21759 21743.9 .010
2.522
r =6 77248 77311.8 .053 2.575
p=1-exp(-SUM/2)=
.72398
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test3
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r
=5 21843 21743.9 .452 .459
r =6 77210 77311.8 .134 .593
p=1-exp(-SUM/2)= .25674
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 899 944.3
2.173 2.173
r =5 21770 21743.9 .031 2.205
r =6 77331
77311.8 .005 2.209
p=1-exp(-SUM/2)= .66868
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 958 944.3 .199 .199
r =5 21619 21743.9 .717 .916
r =6 77423 77311.8 .160 1.076
p=1-exp(-SUM/2)= .41611
Rank
of a 6x8 binary matrix,
rows formed from eight bits of the
RNG test3
b-rank test for bits 20 to 27
OBSERVED EXPECTED
(O-E)^2/E SUM
r<=4 968 944.3 .595 .595
r =5 21906 21743.9
1.208 1.803
r =6 77126 77311.8 .447 2.250
p=1-exp(-SUM/2)=
.67531
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test3
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 1017 944.3 5.597 5.597
r =5 21820 21743.9 .266 5.863
r =6 77163 77311.8 .286 6.150
p=1-exp(-SUM/2)= .95380
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 921 944.3
.575 .575
r =5 21981 21743.9 2.585 3.160
r =6 77098
77311.8 .591 3.752
p=1-exp(-SUM/2)= .84677
Rank of a 6x8
binary matrix,
rows formed from eight bits of the RNG test3
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E
SUM
r<=4 971 944.3 .755 .755
r =5 21906 21743.9 1.208
1.963
r =6 77123 77311.8 .461 2.424
p=1-exp(-SUM/2)=
.70246
Rank of a 6x8 binary matrix,
rows formed from eight
bits of the RNG test3
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 990 944.3 2.212 2.212
r =5 21718 21743.9 .031 2.242
r =6 77292 77311.8 .005 2.247
p=1-exp(-SUM/2)= .67494
Rank of a 6x8 binary matrix,
rows
formed from eight bits of the RNG test3
b-rank test for bits
25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 982 944.3
1.505 1.505
r =5 21735 21743.9 .004 1.509
r =6 77283
77311.8 .011 1.519
p=1-exp(-SUM/2)= .53219
TEST SUMMARY,
25 tests on 100,000 random 6x8 matrices
These should be 25
uniform [0,1] random variables:
.635032 .470804 .555861
.730658 .144584
.505243 .014743 .049812 .508505 .726425
.721134 .060709 .543879 .346790 .827334
.723984 .256742
.668681 .416114 .675307
.953801 .846770 .702457 .674938
.532193
brank test summary for test3
The KS test for
those 25 supposed UNI's yields
KS p-value= .675361
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 142206 missing words, .69 sigmas from mean, p-value=
.75589
tst no 2: 142252 missing words, .80 sigmas from mean,
p-value= .78833
tst no 3: 142380 missing words, 1.10 sigmas
from mean, p-value= .86427
tst no 4: 141531 missing words,
-.88 sigmas from mean, p-value= .18836
tst no 5: 142649
missing words, 1.73 sigmas from mean, p-value= .95802
tst no
6: 141419 missing words, -1.15 sigmas from mean, p-value= .12597
tst no 7: 141590 missing words, -.75 sigmas from mean, p-value=
.22781
tst no 8: 141460 missing words, -1.05 sigmas from
mean, p-value= .14690
tst no 9: 142507 missing words, 1.40
sigmas from mean, p-value= .91871
tst no 10: 141807 missing
words, -.24 sigmas from mean, p-value= .40552
tst no 11:
142606 missing words, 1.63 sigmas from mean, p-value= .94821
tst no 12: 141377 missing words, -1.24 sigmas from mean,
p-value= .10679
tst no 13: 141467 missing words, -1.03 sigmas
from mean, p-value= .15069
tst no 14: 141668 missing words,
-.56 sigmas from mean, p-value= .28643
tst no 15: 142192
missing words, .66 sigmas from mean, p-value= .74552
tst no
16: 141802 missing words, -.25 sigmas from mean, p-value= .40100
tst no 17: 141616 missing words, -.69 sigmas from mean, p-value=
.24656
tst no 18: 141854 missing words, -.13 sigmas from
mean, p-value= .44857
tst no 19: 141877 missing words, -.08
sigmas from mean, p-value= .46990
tst no 20: 141788 missing
words, -.28 sigmas from mean, p-value= .38841
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test3
Output: No. missing words
(mw), equiv normal variate (z), p-value (p)
mw z p
OPSO
for test3 using bits 23 to 32 142206 1.023 .8468
OPSO for
test3 using bits 22 to 31 141519 -1.346 .0892
OPSO for test3
using bits 21 to 30 141949 .137 .5544
OPSO for test3 using
bits 20 to 29 141922 .044 .5174
OPSO for test3 using bits 19
to 28 141464 -1.536 .0623
OPSO for test3 using bits 18 to 27
142144 .809 .7908
OPSO for test3 using bits 17 to 26 141522
-1.336 .0908
OPSO for test3 using bits 16 to 25 141932 .078
.5312
OPSO for test3 using bits 15 to 24 141703 -.711 .2384
OPSO for test3 using bits 14 to 23 141884 -.087 .4652
OPSO
for test3 using bits 13 to 22 141530 -1.308 .0954
OPSO for
test3 using bits 12 to 21 141804 -.363 .3582
OPSO for test3
using bits 11 to 20 142154 .844 .8006
OPSO for test3 using
bits 10 to 19 142280 1.278 .8994
OPSO for test3 using bits 9
to 18 142218 1.064 .8564
OPSO for test3 using bits 8 to 17
142180 .933 .8247
OPSO for test3 using bits 7 to 16 141751
-.546 .2925
OPSO for test3 using bits 6 to 15 142307 1.371
.9149
OPSO for test3 using bits 5 to 14 141341 -1.960 .0250
OPSO for test3 using bits 4 to 13 141759 -.518 .3021
OPSO for
test3 using bits 3 to 12 141027 -3.043 .0012
OPSO for test3
using bits 2 to 11 142086 .609 .7288
OPSO for test3 using
bits 1 to 10 142533 2.151 .9842
OQSO test for generator test3
Output: No. missing words (mw), equiv normal variate (z),
p-value (p)
mw z p
OQSO for test3 using bits 28 to 32
142197 .975 .8353
OQSO for test3 using bits 27 to 31 141743
-.564 .2864
OQSO for test3 using bits 26 to 30 142250 1.155
.8759
OQSO for test3 using bits 25 to 29 142244 1.134 .8717
OQSO for test3 using bits 24 to 28 142062 .518 .6976
OQSO for
test3 using bits 23 to 27 141757 -.516 .3028
OQSO for test3
using bits 22 to 26 141630 -.947 .1719
OQSO for test3 using
bits 21 to 25 142249 1.151 .8752
OQSO for test3 using bits 20
to 24 142088 .606 .7276
OQSO for test3 using bits 19 to 23
141796 -.384 .3504
OQSO for test3 using bits 18 to 22 141863
-.157 .4376
OQSO for test3 using bits 17 to 21 141586 -1.096
.1365
OQSO for test3 using bits 16 to 20 141748 -.547 .2922
OQSO for test3 using bits 15 to 19 141597 -1.059 .1449
OQSO
for test3 using bits 14 to 18 141863 -.157 .4376
OQSO for
test3 using bits 13 to 17 142098 .640 .7388
OQSO for test3
using bits 12 to 16 142177 .907 .8179
OQSO for test3 using
bits 11 to 15 142430 1.765 .9612
OQSO for test3 using bits 10
to 14 141829 -.272 .3927
OQSO for test3 using bits 9 to 13
142503 2.012 .9779
OQSO for test3 using bits 8 to 12 141835
-.252 .4005
OQSO for test3 using bits 7 to 11 142167 .873
.8088
OQSO for test3 using bits 6 to 10 142123 .724 .7656
OQSO for test3 using bits 5 to 9 141524 -1.306 .0957
OQSO for
test3 using bits 4 to 8 141884 -.086 .4658
OQSO for test3
using bits 3 to 7 142066 .531 .7023
OQSO for test3 using bits
2 to 6 141889 -.069 .4725
OQSO for test3 using bits 1 to 5
141986 .260 .6025
DNA test for generator test3
Output:
No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for test3 using bits 31 to 32 142364 1.341 .9101
DNA for test3 using bits 30 to 31 141706 -.600 .2743
DNA for
test3 using bits 29 to 30 141569 -1.004 .1577
DNA for test3
using bits 28 to 29 142207 .878 .8101
DNA for test3 using
bits 27 to 28 141640 -.794 .2135
DNA for test3 using bits 26
to 27 142015 .312 .6224
DNA for test3 using bits 25 to 26
141386 -1.544 .0613
DNA for test3 using bits 24 to 25 141545
-1.075 .1413
DNA for test3 using bits 23 to 24 141884 -.075
.4702
DNA for test3 using bits 22 to 23 142734 2.433 .9925
DNA for test3 using bits 21 to 22 141782 -.376 .3536
DNA for
test3 using bits 20 to 21 142252 1.011 .8440
DNA for test3
using bits 19 to 20 142329 1.238 .8921
DNA for test3 using
bits 18 to 19 141493 -1.228 .1097
DNA for test3 using bits 17
to 18 141764 -.429 .3341
DNA for test3 using bits 16 to 17
142318 1.206 .8860
DNA for test3 using bits 15 to 16 142065
.459 .6770
DNA for test3 using bits 14 to 15 141855 -.160
.4363
DNA for test3 using bits 13 to 14 142042 .391 .6522
DNA for test3 using bits 12 to 13 142169 .766 .7782
DNA for
test3 using bits 11 to 12 141929 .058 .5231
DNA for test3
using bits 10 to 11 141374 -1.579 .0572
DNA for test3 using
bits 9 to 10 141955 .135 .5536
DNA for test3 using bits 8 to
9 141356 -1.632 .0513
DNA for test3 using bits 7 to 8 141824
-.252 .4006
DNA for test3 using bits 6 to 7 142310 1.182
.8814
DNA for test3 using bits 5 to 6 141692 -.641 .2607
DNA for test3 using bits 4 to 5 141942 .096 .5384
DNA for
test3 using bits 3 to 4 142680 2.273 .9885
DNA for test3
using bits 2 to 3 142048 .409 .6588
DNA for test3 using bits
1 to 2 141487 -1.246 .1064
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNT-THE-1's TEST on a stream of bytes. ::
:: Consider the file under test as a stream of bytes (four
per ::
:: 32 bit integer). Each byte can contain from 0 to 8
1's, ::
:: with probabilities 1,8,28,56,70,56,28,8,1 over
256. Now let ::
:: the stream of bytes provide a string of
overlapping 5-letter ::
:: words, each "letter" taking
values A,B,C,D,E. The letters are ::
:: determined by the
number of 1's in a byte:: 0,1,or 2 yield A,::
:: 3 yields B,
4 yields C, 5 yields D and 6,7 or 8 yield E. Thus ::
:: we
have a monkey at a typewriter hitting five keys with vari- ::
:: ous probabilities (37,56,70,56,37 over 256). There are
5^5 ::
:: possible 5-letter words, and from a string of
256,000 (over- ::
:: lapping) 5-letter words, counts are
made on the frequencies ::
:: for each word. The quadratic
form in the weak inverse of ::
:: the covariance matrix of
the cell counts provides a chisquare ::
:: test:: Q5-Q4, the
difference of the naive Pearson sums of ::
::
(OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for test3
Chi-square with 5^5-5^4=2500 d.of
f. for sample size:2560000
chisquare equiv normal p-value
Results fo COUNT-THE-1's in successive bytes:
byte stream for
test3 2456.45 -.616 .268965
byte stream for test3 2533.89
.479 .684111
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 2477.71 -.315 .376278
bits 2
to 9 2525.62 .362 .641422
bits 3 to 10 2485.85 -.200 .420702
bits 4 to 11 2549.86 .705 .759624
bits 5 to 12 2534.42 .487
.686809
bits 6 to 13 2496.46 -.050 .480063
bits 7 to 14
2585.96 1.216 .887950
bits 8 to 15 2358.91 -1.995 .023007
bits 9 to 16 2512.44 .176 .569849
bits 10 to 17 2426.54
-1.039 .149435
bits 11 to 18 2554.88 .776 .781162
bits 12
to 19 2396.97 -1.457 .072555
bits 13 to 20 2655.00 2.192
.985809
bits 14 to 21 2523.04 .326 .627704
bits 15 to 22
2578.38 1.109 .866181
bits 16 to 23 2586.25 1.220 .888725
bits 17 to 24 2549.18 .696 .756649
bits 18 to 25 2552.58 .744
.771425
bits 19 to 26 2558.83 .832 .797289
bits 20 to 27
2617.87 1.667 .952241
bits 21 to 28 2578.85 1.115 .867609
bits 22 to 29 2512.09 .171 .567892
bits 23 to 30 2511.25 .159
.563178
bits 24 to 31 2504.01 .057 .522637
bits 25 to 32
2484.70 -.216 .414342
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 test3
Of 12,000
tries, the average no. of successes
should be 3523 with
sigma=21.9
Successes: 3510 z-score: -.594 p-value: .276387
Successes: 3482 z-score: -1.872 p-value: .030593
Successes:
3505 z-score: -.822 p-value: .205562
Successes: 3547 z-score:
1.096 p-value: .863437
Successes: 3525 z-score: .091 p-value:
.536382
Successes: 3523 z-score: .000 p-value: .500000
Successes: 3501 z-score: -1.005 p-value: .157553
Successes:
3529 z-score: .274 p-value: .607947
Successes: 3522 z-score:
-.046 p-value: .481790
Successes: 3496 z-score: -1.233
p-value: .108811
square size avg. no. parked sample sigma
100. 3514.000 17.866
KSTEST for the above 10: p= .679832
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100
times:: choose n=8000 random points in a ::
:: square of
side 10000. Find d, the minimum distance between ::
:: the
(n^2-n)/2 pairs of points. If the points are truly inde- ::
:: pendent uniform, then d^2, the square of the minimum distance
::
:: should be (very close to) exponentially distributed
with mean ::
:: .995 . Thus 1-exp(-d^2/.995) should be
uniform on [0,1) and ::
:: a KSTEST on the resulting 100
values serves as a test of uni- ::
:: formity for random
points in the square. Test numbers=0 mod 5 ::
:: are printed
but the KSTEST is based on the full set of 100 ::
:: random
choices of 8000 points in the 10000x10000 square. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
This is the MINIMUM DISTANCE test
for random integers in
the file test3
Sample no. d^2 avg equiv uni
5 .0632
.3650 .061586
10 .3836 .4171 .319888
15 .5267 .5140
.411016
20 1.1033 .6271 .670070
25 .0213 .6262 .021185
30 1.1779 .6922 .693902
35 .6951 .8446 .502730
40 .3488
.8748 .295703
45 2.8859 1.0273 .944998
50 1.7469 1.0468
.827215
55 .0148 1.0692 .014813
60 .0643 1.0348 .062551
65 2.3781 1.0475 .908371
70 3.7897 1.0440 .977824
75 .3471
1.0123 .294499
80 2.8361 1.0028 .942174
85 1.7446 .9758
.826807
90 2.1385 .9824 .883432
95 .0485 .9420 .047623
100 .6420 .9372 .475462
MINIMUM DISTANCE TEST for test3
Result of KS test on 20 transformed mindist^2's:
p-value=
.819181
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE 3DSPHERES TEST ::
:: Choose 4000 random points in
a cube of edge 1000. At each ::
:: point, center a sphere
large enough to reach the next closest ::
:: point. Then the
volume of the smallest such sphere is (very ::
:: close to)
exponentially distributed with mean 120pi/3. Thus ::
:: the
radius cubed is exponential with mean 30. (The mean is ::
::
obtained by extensive simulation). The 3DSPHERES test gener- ::
:: ates 4000 such spheres 20 times. Each min radius cubed
leads ::
:: to a uniform variable by means of
1-exp(-r^3/30.), then a ::
:: KSTEST is done on the 20
p-values. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file test3
sample no: 1 r^3=
42.722 p-value= .75926
sample no: 2 r^3= 11.005 p-value=
.30708
sample no: 3 r^3= 27.425 p-value= .59915
sample no:
4 r^3= 1.494 p-value= .04859
sample no: 5 r^3= 139.928
p-value= .99057
sample no: 6 r^3= 38.329 p-value= .72131
sample no: 7 r^3= 8.875 p-value= .25609
sample no: 8 r^3=
2.849 p-value= .09060
sample no: 9 r^3= .667 p-value= .02198
sample no: 10 r^3= 15.606 p-value= .40561
sample no: 11 r^3=
44.121 p-value= .77023
sample no: 12 r^3= 68.041 p-value=
.89649
sample no: 13 r^3= 46.759 p-value= .78958
sample
no: 14 r^3= 100.402 p-value= .96480
sample no: 15 r^3= 15.220
p-value= .39790
sample no: 16 r^3= 55.755 p-value= .84409
sample no: 17 r^3= 11.275 p-value= .31328
sample no: 18 r^3=
73.755 p-value= .91444
sample no: 19 r^3= 126.098 p-value=
.98505
sample no: 20 r^3= 44.547 p-value= .77348
A KS test
is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file test3 p-value= .878239
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the SQEEZE test ::
:: Random integers are
floated to get uniforms on [0,1). Start- ::
:: ing with
k=2^31=2147483647, the test finds j, the number of ::
::
iterations necessary to reduce k to 1, using the reduction ::
:: k=ceiling(k*U), with U provided by floating integers from
::
:: the file being tested. Such j's are found 100,000
times, ::
:: then counts for the number of times j was
<=6,7,...,47,>=48 ::
:: are used to provide a chi-square
test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESULTS OF SQUEEZE TEST FOR test3
Table of standardized
frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking
values <=6,7,8,...,47,>=48:
-.8 -.7 -1.1 .2 -.9 1.3
.8 -.6
-1.4 .0 .3 -.9
-.7 .9 -.8 .0 1.1 -1.4
-.7 -.5 -.2 1.9 1.3
-.5
.1 1.9 -.3 -1.6 1.8 1.4
-.6 .1 -.4 -.5 .1 -1.6
.5
.2 .1 1.0 -1.3 .0
-1.1
Chi-square with 42 degrees of
freedom: 38.989
z-score= -.329 p-value= .395903
______________________________________________________________
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The OVERLAPPING SUMS test ::
:: Integers are floated
to get a sequence U(1),U(2),... of uni- ::
:: form [0,1)
variables. Then overlapping sums, ::
::
S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. ::
:: The S's are virtually normal with a certain covariance mat-
::
:: rix. A linear transformation of the S's converts them
to a ::
:: sequence of independent standard normals, which
are converted ::
:: to uniform variables for a KSTEST. The
p-values from ten ::
:: KSTESTs are given still another
KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test no. 1 p-value .516390
Test no. 2 p-value .248179
Test no. 3 p-value .230872
Test no. 4 p-value .227177
Test
no. 5 p-value .599579
Test no. 6 p-value .710847
Test no.
7 p-value .571424
Test no. 8 p-value .285338
Test no. 9
p-value .079077
Test no. 10 p-value .636207
Results of the
OSUM test for test3
KSTEST on the above 10 p-values: .660794
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the RUNS test. It counts runs up, and runs down,
::
:: in a sequence of uniform [0,1) variables, obtained by
float- ::
:: ing the 32-bit integers in the specified file.
This example ::
:: shows how runs are counted:
.123,.357,.789,.425,.224,.416,.95::
:: contains an up-run of
length 3, a down-run of length 2 and an ::
:: up-run of (at
least) 2, depending on the next values. The ::
:: covariance
matrices for the runs-up and runs-down are well ::
:: known,
leading to chisquare tests for quadratic forms in the ::
::
weak inverses of the covariance matrices. Runs are counted ::
:: for sequences of length 10,000. This is done ten times.
Then ::
:: repeated. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The RUNS test for file test3
Up and down runs in a
sample of 10000
_________________________________________________
Run test
for test3 :
runs up; ks test for 10 p's: .872171
runs
down; ks test for 10 p's: .650265
Run test for test3 :
runs up; ks test for 10 p's: .540054
runs down; ks test for
10 p's: .733981
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the CRAPS TEST. It plays 200,000 games of craps,
finds::
:: the number of wins and the number of throws
necessary to end ::
:: each game. The number of wins should
be (very close to) a ::
:: normal with mean 200000p and
variance 200000p(1-p), with ::
:: p=244/495. Throws
necessary to complete the game can vary ::
:: from 1 to
infinity, but counts for all>21 are lumped with 21. ::
:: A
chi-square test is made on the no.-of-throws cell counts. ::
:: Each 32-bit integer from the test file provides the value for
::
:: the throw of a die, by floating to [0,1), multiplying
by 6 ::
:: and taking 1 plus the integer part of the result.
::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Results of craps test for test3
No. of wins: Observed
Expected
98573 98585.86
98573= No. of wins, z-score= -.058
pvalue= .47707
Analysis of Throws-per-Game:
Chisq= 21.02
for 20 degrees of freedom, p= .60427
Throws Observed Expected
Chisq Sum
1 66634 66666.7 .016 .016
2 37722 37654.3 .122
.138
3 27250 26954.7 3.234 3.372
4 19063 19313.5 3.248
6.620
5 13772 13851.4 .455 7.075
6 9898 9943.5 .209 7.284
7 7090 7145.0 .424 7.708
8 5133 5139.1 .007 7.715
9 3740
3699.9 .435 8.150
10 2694 2666.3 .288 8.438
11 1966 1923.3
.947 9.385
12 1389 1388.7 .000 9.385
13 1024 1003.7 .410
9.795
14 791 726.1 5.793 15.588
15 489 525.8 2.580 18.169
16 365 381.2 .684 18.853
17 258 276.5 1.243 20.096
18 191
200.8 .481 20.577
19 138 146.0 .437 21.014
20 107 106.2
.006 21.020
21 286 287.1 .004 21.024
SUMMARY FOR test3
p-value for no. of wins: .477068
p-value for throws/game:
.604271
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
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