Oh, by overlaps I meant it was landing on the same point; for example if you had two at 50,50 or something like that.
I could probably rewrite it in C++ and test it after I finish some project work (dont know enough java to modify it)
seededRandom is not random at all.

 Developer
 Posts: 2936
 Joined: Thu Jul 24, 2003 9:53 pm
 Contact:
Thanks! I've changed seededRandom to use your algorithm in the next beta. It's way better now.pixelfck wrote:...
Meanwhile, I did a quick extension to display randomness, based on your JavaScript code: https://www.dropbox.com/s/29buowdqmeuzt ... ssTest.xml
To use it:
1. Drop the file in Extensions
2. Start a new game
3.
4. Click on the "Randomness" menu item.
 pixelfck
 Militia Captain
 Posts: 571
 Joined: Tue Aug 11, 2009 8:47 pm
 Location: Travelling around in Europe
@Wolfy: thanks! (it is javascript, not Java; so the algorithmic operators should be the same as in c++).
Anyhow, I would be interested to see what algorithm you came up with that passes all the diehard tests, I think it is quite a feat.
@george
Nice, I'll put it to good use!
~Pixelfck
Anyhow, I would be interested to see what algorithm you came up with that passes all the diehard tests, I think it is quite a feat.
@george
Nice, I'll put it to good use!
~Pixelfck
 Wolfy
 Fleet Admiral
 Posts: 5363
 Joined: Tue Feb 05, 2008 1:10 am
 Location: Somewhere in the Frontier on a Hycrotan station, working on new ships.
Here's the results from diehard for yours (seed 123412341234  assuming order of operations was the same)
While the rate of outlier p values (p < 0.01 or p> 0.99) was high for the number of samples (also there were two p=1.000000 values [See DNA tests for bits 3230]), its probably sufficient levels of randomness for transcendence as well as being far faster than mine. (I tested; yours is ~34 orders of magnitude faster than mine)
While the rate of outlier p values (p < 0.01 or p> 0.99) was high for the number of samples (also there were two p=1.000000 values [See DNA tests for bits 3230]), its probably sufficient levels of randomness for transcendence as well as being far faster than mine. (I tested; yours is ~34 orders of magnitude faster than mine)
Code: Select all
NOTE: Most of the tests in DIEHARD return a pvalue, which
should be uniform on [0,1) if the input file contains truly
independent random bits. Those pvalues are obtained by
p=F(X), where F is the assumed distribution of the sample
random variable Xoften 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 pvalues 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 chisquare goodness of fit test ::
:: provides a p value. The first test uses bits 124 (counting ::
:: from the left) from integers in the specified file. ::
:: Then the file is closed and reopened. Next, bits 225 are ::
:: used to provide birthdays, then 326 and so on to bits 932. ::
:: Each set of bits provides a pvalue, and the nine pvalues ::
:: provide a sample for a KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA= 2.0000
Results for pxl_out.rbnf
For a sample of size 500: mean
pxl_out.rbnf using bits 1 to 24 1.998
duplicate number number
spacings observed expected
0 63. 67.668
1 130. 135.335
2 151. 135.335
3 91. 90.224
4 40. 45.112
5 19. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 3.61 pvalue= .270802
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 2 to 25 1.920
duplicate number number
spacings observed expected
0 71. 67.668
1 148. 135.335
2 133. 135.335
3 84. 90.224
4 39. 45.112
5 18. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 2.85 pvalue= .172013
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 3 to 26 2.036
duplicate number number
spacings observed expected
0 66. 67.668
1 127. 135.335
2 144. 135.335
3 86. 90.224
4 48. 45.112
5 22. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 2.56 pvalue= .137990
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 4 to 27 1.930
duplicate number number
spacings observed expected
0 63. 67.668
1 149. 135.335
2 142. 135.335
3 79. 90.224
4 44. 45.112
5 19. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 5.72 pvalue= .544466
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 5 to 28 2.004
duplicate number number
spacings observed expected
0 63. 67.668
1 140. 135.335
2 135. 135.335
3 90. 90.224
4 46. 45.112
5 19. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = .75 pvalue= .006666
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 6 to 29 2.014
duplicate number number
spacings observed expected
0 75. 67.668
1 134. 135.335
2 120. 135.335
3 99. 90.224
4 44. 45.112
5 17. 18.045
6 to INF 11. 8.282
Chisquare with 6 d.o.f. = 4.38 pvalue= .374489
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 7 to 30 1.988
duplicate number number
spacings observed expected
0 65. 67.668
1 138. 135.335
2 142. 135.335
3 86. 90.224
4 40. 45.112
5 21. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. = 1.76 pvalue= .059314
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 8 to 31 1.960
duplicate number number
spacings observed expected
0 77. 67.668
1 137. 135.335
2 125. 135.335
3 76. 90.224
4 65. 45.112
5 15. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. = 14.92 pvalue= .979123
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 9 to 32 2.056
duplicate number number
spacings observed expected
0 71. 67.668
1 134. 135.335
2 128. 135.335
3 82. 90.224
4 49. 45.112
5 22. 18.045
6 to INF 14. 8.282
Chisquare with 6 d.o.f. = 6.47 pvalue= .627831
:::::::::::::::::::::::::::::::::::::::::
The 9 pvalues were
.270802 .172013 .137990 .544466 .006666
.374489 .059314 .979123 .627831
A KSTEST for the 9 pvalues yields .846869
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE OVERLAPPING 5PERMUTATION TEST ::
:: This is the OPERM5 test. It looks at a sequence of one mill ::
:: ion 32bit 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 pxl_out.rbnf
For a sample of 1,000,000 consecutive 5tuples,
chisquare for 99 degrees of freedom= 81.249; pvalue= .097134
OPERM5 test for file pxl_out.rbnf
For a sample of 1,000,000 consecutive 5tuples,
chisquare for 99 degrees of freedom=102.731; pvalue= .621446
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32bit integer
rank observed expected (oe)^2/e sum
28 194 211.4 1.435011 1.435
29 5125 5134.0 .015813 1.451
30 23199 23103.0 .398519 1.849
31 11482 11551.5 .418442 2.268
chisquare= 2.268 for 3 d. of f.; pvalue= .547324

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 32x32 matrices. A random 32x ::
:: 32 binary matrix is formed, each row a 32bit 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 pxl_out.rbnf
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32bit integer
rank observed expected (oe)^2/e sum
29 193 211.4 1.604514 1.605
30 5222 5134.0 1.508021 3.113
31 23159 23103.0 .135513 3.248
32 11426 11551.5 1.364009 4.612
chisquare= 4.612 for 3 d. of f.; pvalue= .813364

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 6x8 matrices. From each of ::
:: six random 32bit 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 chisquare test is performed on ::
:: counts for ranks 6,5 and <=4. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Binary Rank Test for pxl_out.rbnf
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 1 to 8
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 949 944.3 .023 .023
r =5 21791 21743.9 .102 .125
r =6 77260 77311.8 .035 .160
p=1exp(SUM/2)= .07694
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 2 to 9
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 931 944.3 .187 .187
r =5 21739 21743.9 .001 .188
r =6 77330 77311.8 .004 .193
p=1exp(SUM/2)= .09187
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 3 to 10
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 913 944.3 1.038 1.038
r =5 21718 21743.9 .031 1.068
r =6 77369 77311.8 .042 1.111
p=1exp(SUM/2)= .42614
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 4 to 11
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 864 944.3 6.829 6.829
r =5 21623 21743.9 .672 7.501
r =6 77513 77311.8 .524 8.024
p=1exp(SUM/2)= .98191
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 5 to 12
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 942 944.3 .006 .006
r =5 21643 21743.9 .468 .474
r =6 77415 77311.8 .138 .612
p=1exp(SUM/2)= .26346
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 6 to 13
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 911 944.3 1.174 1.174
r =5 21802 21743.9 .155 1.330
r =6 77287 77311.8 .008 1.338
p=1exp(SUM/2)= .48767
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 7 to 14
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 954 944.3 .100 .100
r =5 21830 21743.9 .341 .441
r =6 77216 77311.8 .119 .559
p=1exp(SUM/2)= .24394
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 8 to 15
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21914 21743.9 1.331 2.179
r =6 77170 77311.8 .260 2.439
p=1exp(SUM/2)= .70462
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 9 to 16
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 971 944.3 .755 .755
r =5 21644 21743.9 .459 1.214
r =6 77385 77311.8 .069 1.283
p=1exp(SUM/2)= .47354
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 10 to 17
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 997 944.3 2.941 2.941
r =5 21458 21743.9 3.759 6.700
r =6 77545 77311.8 .703 7.404
p=1exp(SUM/2)= .97532
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 11 to 18
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21684 21743.9 .165 .207
r =6 77378 77311.8 .057 .264
p=1exp(SUM/2)= .12354
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 12 to 19
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 992 944.3 2.409 2.409
r =5 21916 21743.9 1.362 3.772
r =6 77092 77311.8 .625 4.396
p=1exp(SUM/2)= .88900
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 13 to 20
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21967 21743.9 2.289 2.381
r =6 77098 77311.8 .591 2.972
p=1exp(SUM/2)= .77372
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 14 to 21
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 957 944.3 .171 .171
r =5 21553 21743.9 1.676 1.847
r =6 77490 77311.8 .411 2.257
p=1exp(SUM/2)= .67656
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 15 to 22
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 921 944.3 .575 .575
r =5 21489 21743.9 2.988 3.563
r =6 77590 77311.8 1.001 4.564
p=1exp(SUM/2)= .89793
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 16 to 23
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 949 944.3 .023 .023
r =5 21596 21743.9 1.006 1.029
r =6 77455 77311.8 .265 1.295
p=1exp(SUM/2)= .47655
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 17 to 24
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21531 21743.9 2.085 2.127
r =6 77531 77311.8 .621 2.748
p=1exp(SUM/2)= .74692
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 18 to 25
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 884 944.3 3.851 3.851
r =5 21744 21743.9 .000 3.851
r =6 77372 77311.8 .047 3.898
p=1exp(SUM/2)= .85755
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 19 to 26
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 912 944.3 1.105 1.105
r =5 21678 21743.9 .200 1.305
r =6 77410 77311.8 .125 1.429
p=1exp(SUM/2)= .51065
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 20 to 27
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 930 944.3 .217 .217
r =5 21719 21743.9 .029 .245
r =6 77351 77311.8 .020 .265
p=1exp(SUM/2)= .12409
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 21 to 28
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 926 944.3 .355 .355
r =5 21768 21743.9 .027 .381
r =6 77306 77311.8 .000 .382
p=1exp(SUM/2)= .17380
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 22 to 29
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 947 944.3 .008 .008
r =5 21793 21743.9 .111 .119
r =6 77260 77311.8 .035 .153
p=1exp(SUM/2)= .07378
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 23 to 30
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 927 944.3 .317 .317
r =5 21900 21743.9 1.121 1.438
r =6 77173 77311.8 .249 1.687
p=1exp(SUM/2)= .56976
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 24 to 31
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 911 944.3 1.174 1.174
r =5 21782 21743.9 .067 1.241
r =6 77307 77311.8 .000 1.241
p=1exp(SUM/2)= .46244
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
brank test for bits 25 to 32
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 959 944.3 .229 .229
r =5 21822 21743.9 .281 .509
r =6 77219 77311.8 .111 .621
p=1exp(SUM/2)= .26682
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.076937 .091875 .426136 .981907 .263455
.487674 .243938 .704618 .473537 .975320
.123545 .888999 .773720 .676562 .897929
.476546 .746917 .857555 .510652 .124087
.173801 .073785 .569763 .462444 .266816
brank test summary for pxl_out.rbnf
The KS test for those 25 supposed UNI's yields
KS pvalue= .057610
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 20letter ::
:: "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 20letter (20bit) words in a string of ::
:: 2^21 overlapping 20letter 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 ::
:: (j141909)/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 20tuples 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: 142375 missing words, 1.09 sigmas from mean, pvalue= .86171
tst no 2: 141977 missing words, .16 sigmas from mean, pvalue= .56282
tst no 3: 142098 missing words, .44 sigmas from mean, pvalue= .67033
tst no 4: 142733 missing words, 1.92 sigmas from mean, pvalue= .97285
tst no 5: 142001 missing words, .21 sigmas from mean, pvalue= .58480
tst no 6: 142032 missing words, .29 sigmas from mean, pvalue= .61280
tst no 7: 142444 missing words, 1.25 sigmas from mean, pvalue= .89421
tst no 8: 141918 missing words, .02 sigmas from mean, pvalue= .50808
tst no 9: 141131 missing words, 1.82 sigmas from mean, pvalue= .03449
tst no 10: 141558 missing words, .82 sigmas from mean, pvalue= .20586
tst no 11: 141643 missing words, .62 sigmas from mean, pvalue= .26688
tst no 12: 141339 missing words, 1.33 sigmas from mean, pvalue= .09134
tst no 13: 142243 missing words, .78 sigmas from mean, pvalue= .78219
tst no 14: 141973 missing words, .15 sigmas from mean, pvalue= .55913
tst no 15: 142963 missing words, 2.46 sigmas from mean, pvalue= .99309
tst no 16: 142042 missing words, .31 sigmas from mean, pvalue= .62171
tst no 17: 141591 missing words, .74 sigmas from mean, pvalue= .22851
tst no 18: 142126 missing words, .51 sigmas from mean, pvalue= .69366
tst no 19: 141795 missing words, .27 sigmas from mean, pvalue= .39469
tst no 20: 141858 missing words, .12 sigmas from mean, pvalue= .45227
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The tests OPSO, OQSO and DNA ::
:: OPSO means OverlappingPairsSparseOccupancy ::
:: The OPSO test considers 2letter words from an alphabet of ::
:: 1024 letters. Each letter is determined by a specified ten ::
:: bits from a 32bit integer in the sequence to be tested. OPSO ::
:: generates 2^21 (overlapping) 2letter words (from 2^21+1 ::
:: "keystrokes") and counts the number of missing wordsthat ::
:: is 2letter 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 (missingwrds141909)/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 OverlappingQuadruplesSparseOccupancy ::
:: The test OQSO is similar, except that it considers 4letter ::
:: 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 32bit 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 10letter 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) 10letter 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 pxl_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
OPSO for pxl_out.rbnf using bits 23 to 32 142170 .899 .8156
OPSO for pxl_out.rbnf using bits 22 to 31 141836 .253 .4002
OPSO for pxl_out.rbnf using bits 21 to 30 141805 .360 .3595
OPSO for pxl_out.rbnf using bits 20 to 29 142079 .585 .7208
OPSO for pxl_out.rbnf using bits 19 to 28 142195 .985 .8377
OPSO for pxl_out.rbnf using bits 18 to 27 141534 1.294 .0978
OPSO for pxl_out.rbnf using bits 17 to 26 142044 .464 .6788
OPSO for pxl_out.rbnf using bits 16 to 25 142158 .857 .8044
OPSO for pxl_out.rbnf using bits 15 to 24 141648 .901 .1838
OPSO for pxl_out.rbnf using bits 14 to 23 141405 1.739 .0410
OPSO for pxl_out.rbnf using bits 13 to 22 141582 1.129 .1295
OPSO for pxl_out.rbnf using bits 12 to 21 142001 .316 .6240
OPSO for pxl_out.rbnf using bits 11 to 20 141936 .092 .5366
OPSO for pxl_out.rbnf using bits 10 to 19 142088 .616 .7311
OPSO for pxl_out.rbnf using bits 9 to 18 141839 .243 .4042
OPSO for pxl_out.rbnf using bits 8 to 17 141888 .074 .4707
OPSO for pxl_out.rbnf using bits 7 to 16 142163 .875 .8091
OPSO for pxl_out.rbnf using bits 6 to 15 141597 1.077 .1407
OPSO for pxl_out.rbnf using bits 5 to 14 141966 .195 .5775
OPSO for pxl_out.rbnf using bits 4 to 13 142254 1.189 .8827
OPSO for pxl_out.rbnf using bits 3 to 12 142377 1.613 .9466
OPSO for pxl_out.rbnf using bits 2 to 11 141916 .023 .5092
OPSO for pxl_out.rbnf using bits 1 to 10 141579 1.139 .1273
OQSO test for generator pxl_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
OQSO for pxl_out.rbnf using bits 28 to 32 142194 .965 .8327
OQSO for pxl_out.rbnf using bits 27 to 31 142041 .446 .6723
OQSO for pxl_out.rbnf using bits 26 to 30 141826 .282 .3888
OQSO for pxl_out.rbnf using bits 25 to 29 141467 1.499 .0669
OQSO for pxl_out.rbnf using bits 24 to 28 141725 .625 .2660
OQSO for pxl_out.rbnf using bits 23 to 27 142053 .487 .6869
OQSO for pxl_out.rbnf using bits 22 to 26 141978 .233 .5920
OQSO for pxl_out.rbnf using bits 21 to 25 142068 .538 .7047
OQSO for pxl_out.rbnf using bits 20 to 24 142175 .901 .8161
OQSO for pxl_out.rbnf using bits 19 to 23 141648 .886 .1878
OQSO for pxl_out.rbnf using bits 18 to 22 141993 .284 .6117
OQSO for pxl_out.rbnf using bits 17 to 21 142200 .985 .8378
OQSO for pxl_out.rbnf using bits 16 to 20 142035 .426 .6649
OQSO for pxl_out.rbnf using bits 15 to 19 141482 1.449 .0737
OQSO for pxl_out.rbnf using bits 14 to 18 141736 .588 .2784
OQSO for pxl_out.rbnf using bits 13 to 17 142196 .972 .8344
OQSO for pxl_out.rbnf using bits 12 to 16 141593 1.072 .1418
OQSO for pxl_out.rbnf using bits 11 to 15 141317 2.008 .0223
OQSO for pxl_out.rbnf using bits 10 to 14 141444 1.577 .0574
OQSO for pxl_out.rbnf using bits 9 to 13 142662 2.551 .9946
OQSO for pxl_out.rbnf using bits 8 to 12 142596 2.328 .9900
OQSO for pxl_out.rbnf using bits 7 to 11 142165 .867 .8069
OQSO for pxl_out.rbnf using bits 6 to 10 142339 1.457 .9274
OQSO for pxl_out.rbnf using bits 5 to 9 141519 1.323 .0929
OQSO for pxl_out.rbnf using bits 4 to 8 141804 .357 .3605
OQSO for pxl_out.rbnf using bits 3 to 7 141734 .594 .2761
OQSO for pxl_out.rbnf using bits 2 to 6 141803 .360 .3593
OQSO for pxl_out.rbnf using bits 1 to 5 141243 2.259 .0119
DNA test for generator pxl_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
DNA for pxl_out.rbnf using bits 31 to 32 145144 9.542 1.0000
DNA for pxl_out.rbnf using bits 30 to 31 143875 5.798 1.0000
DNA for pxl_out.rbnf using bits 29 to 30 142390 1.418 .9219
DNA for pxl_out.rbnf using bits 28 to 29 142277 1.085 .8609
DNA for pxl_out.rbnf using bits 27 to 28 142144 .692 .7556
DNA for pxl_out.rbnf using bits 26 to 27 142371 1.362 .9134
DNA for pxl_out.rbnf using bits 25 to 26 141998 .262 .6032
DNA for pxl_out.rbnf using bits 24 to 25 141948 .114 .5454
DNA for pxl_out.rbnf using bits 23 to 24 141560 1.030 .1514
DNA for pxl_out.rbnf using bits 22 to 23 141927 .052 .5208
DNA for pxl_out.rbnf using bits 21 to 22 142071 .477 .6833
DNA for pxl_out.rbnf using bits 20 to 21 141631 .821 .2058
DNA for pxl_out.rbnf using bits 19 to 20 142239 .972 .8346
DNA for pxl_out.rbnf using bits 18 to 19 142113 .601 .7260
DNA for pxl_out.rbnf using bits 17 to 18 142182 .804 .7894
DNA for pxl_out.rbnf using bits 16 to 17 142296 1.141 .8730
DNA for pxl_out.rbnf using bits 15 to 16 141620 .853 .1967
DNA for pxl_out.rbnf using bits 14 to 15 141755 .455 .3245
DNA for pxl_out.rbnf using bits 13 to 14 142224 .928 .8234
DNA for pxl_out.rbnf using bits 12 to 13 141150 2.240 .0125
DNA for pxl_out.rbnf using bits 11 to 12 142648 2.179 .9853
DNA for pxl_out.rbnf using bits 10 to 11 142009 .294 .6156
DNA for pxl_out.rbnf using bits 9 to 10 142114 .604 .7270
DNA for pxl_out.rbnf using bits 8 to 9 142300 1.152 .8754
DNA for pxl_out.rbnf using bits 7 to 8 142361 1.332 .9086
DNA for pxl_out.rbnf using bits 6 to 7 142144 .692 .7556
DNA for pxl_out.rbnf using bits 5 to 6 142193 .837 .7986
DNA for pxl_out.rbnf using bits 4 to 5 141593 .933 .1754
DNA for pxl_out.rbnf using bits 3 to 4 141863 .137 .4456
DNA for pxl_out.rbnf using bits 2 to 3 141599 .915 .1800
DNA for pxl_out.rbnf using bits 1 to 2 141740 .499 .3087
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNTTHE1'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 5letter ::
:: 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 5letter words, and from a string of 256,000 (over ::
:: lapping) 5letter 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:: Q5Q4, the difference of the naive Pearson sums of ::
:: (OBSEXP)^2/EXP on counts for 5 and 4letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for pxl_out.rbnf
Chisquare with 5^55^4=2500 d.of f. for sample size:2560000
chisquare equiv normal pvalue
Results fo COUNTTHE1's in successive bytes:
byte stream for pxl_out.rbnf 2503.60 .051 .520298
byte stream for pxl_out.rbnf 2499.89 .002 .499391
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNTTHE1's TEST for specific bytes. ::
:: Consider the file under test as a stream of 32bit 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) 5letter 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 5letter words, and ::
:: from a string of 256,000 (overlapping) 5letter 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:: Q5Q4, the difference of ::
:: the naive Pearson sums of (OBSEXP)^2/EXP on counts for 5 ::
:: and 4letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Chisquare with 5^55^4=2500 d.of f. for sample size: 256000
chisquare equiv normal p value
Results for COUNTTHE1's in specified bytes:
bits 1 to 8 2649.09 2.108 .982505
bits 2 to 9 2464.47 .502 .307691
bits 3 to 10 2508.01 .113 .545073
bits 4 to 11 2459.82 .568 .284947
bits 5 to 12 2531.40 .444 .671514
bits 6 to 13 2533.45 .473 .681897
bits 7 to 14 2427.66 1.023 .153138
bits 8 to 15 2495.70 .061 .475736
bits 9 to 16 2525.30 .358 .639753
bits 10 to 17 2483.34 .236 .406858
bits 11 to 18 2493.98 .085 .466052
bits 12 to 19 2525.69 .363 .641815
bits 13 to 20 2491.37 .122 .451442
bits 14 to 21 2684.34 2.607 .995433
bits 15 to 22 2664.91 2.332 .990153
bits 16 to 23 2526.33 .372 .645175
bits 17 to 24 2549.44 .699 .757781
bits 18 to 25 2493.49 .092 .463333
bits 19 to 26 2633.46 1.887 .970447
bits 20 to 27 2556.29 .796 .787002
bits 21 to 28 2423.79 1.078 .140572
bits 22 to 29 2559.75 .845 .800959
bits 23 to 30 2579.56 1.125 .869725
bits 24 to 31 2425.69 1.051 .146651
bits 25 to 32 2431.76 .965 .167257
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THIS IS A PARKING LOT TEST ::
:: In a square of side 100, randomly "park" a cara 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 (k3523)/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 pxl_out.rbnf
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3558 zscore: 1.598 pvalue: .944998
Successes: 3522 zscore: .046 pvalue: .481790
Successes: 3533 zscore: .457 pvalue: .676028
Successes: 3560 zscore: 1.689 pvalue: .954438
Successes: 3548 zscore: 1.142 pvalue: .873180
Successes: 3477 zscore: 2.100 pvalue: .017844
Successes: 3525 zscore: .091 pvalue: .536382
Successes: 3544 zscore: .959 pvalue: .831196
Successes: 3504 zscore: .868 pvalue: .192812
Successes: 3528 zscore: .228 pvalue: .590298
square size avg. no. parked sample sigma
100. 3529.900 24.058
KSTEST for the above 10: p= .676456
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100 times:: choose n=8000 random points in a ::
:: square of side 10000. Find d, the minimum distance between ::
:: the (n^2n)/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 1exp(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 pxl_out.rbnf
Sample no. d^2 avg equiv uni
5 .4158 .5669 .341561
10 .6495 .7930 .479413
15 1.4765 .9208 .773248
20 .1684 .8054 .155709
25 .8864 .7717 .589707
30 .1966 .7591 .179272
35 .5151 .7458 .404086
40 .2314 .6891 .207473
45 1.5354 .7787 .786285
50 .3581 .7369 .302228
55 2.1589 .7722 .885796
60 .7537 .7463 .531158
65 .5487 .7249 .423875
70 .3815 .7405 .318455
75 .4195 .7450 .343980
80 .2645 .7724 .233401
85 1.5017 .8094 .778917
90 .3021 .8095 .261850
95 .5573 .7981 .428828
100 .4472 .8611 .361995
MINIMUM DISTANCE TEST for pxl_out.rbnf
Result of KS test on 20 transformed mindist^2's:
pvalue= .777127
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE 3DSPHERES TEST ::
:: Choose 4000 random points in a cube of edge 1000. At each ::
:: point, center a sphere large enough to reach the next closest ::
:: point. Then the volume of the smallest such sphere is (very ::
:: close to) exponentially distributed with mean 120pi/3. Thus ::
:: the radius cubed is exponential with mean 30. (The mean is ::
:: obtained by extensive simulation). The 3DSPHERES test gener ::
:: ates 4000 such spheres 20 times. Each min radius cubed leads ::
:: to a uniform variable by means of 1exp(r^3/30.), then a ::
:: KSTEST is done on the 20 pvalues. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file pxl_out.rbnf
sample no: 1 r^3= 62.595 pvalue= .87588
sample no: 2 r^3= 9.175 pvalue= .26349
sample no: 3 r^3= .822 pvalue= .02704
sample no: 4 r^3= 1.004 pvalue= .03291
sample no: 5 r^3= 6.329 pvalue= .19021
sample no: 6 r^3= 26.698 pvalue= .58931
sample no: 7 r^3= 31.373 pvalue= .64857
sample no: 8 r^3= 59.190 pvalue= .86096
sample no: 9 r^3= 6.742 pvalue= .20128
sample no: 10 r^3= 15.087 pvalue= .39522
sample no: 11 r^3= 4.257 pvalue= .13229
sample no: 12 r^3= 37.958 pvalue= .71783
sample no: 13 r^3= 18.202 pvalue= .45487
sample no: 14 r^3= 44.742 pvalue= .77494
sample no: 15 r^3= 66.820 pvalue= .89218
sample no: 16 r^3= 99.728 pvalue= .96400
sample no: 17 r^3= 17.679 pvalue= .44528
sample no: 18 r^3= 3.214 pvalue= .10161
sample no: 19 r^3= 58.317 pvalue= .85686
sample no: 20 r^3= 6.864 pvalue= .20450
A KS test is applied to those 20 pvalues.

3DSPHERES test for file pxl_out.rbnf pvalue= .161536
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the SQEEZE test ::
:: Random integers are floated to get uniforms on [0,1). Start ::
:: ing with k=2^31=2147483647, the test finds j, the number of ::
:: iterations necessary to reduce k to 1, using the reduction ::
:: k=ceiling(k*U), with U provided by floating integers from ::
:: the file being tested. Such j's are found 100,000 times, ::
:: then counts for the number of times j was <=6,7,...,47,>=48 ::
:: are used to provide a chisquare test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESULTS OF SQUEEZE TEST FOR pxl_out.rbnf
Table of standardized frequency counts
( (obsexp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
.1 .3 .1 .2 .3 .7
.1 .2 1.7 .8 .3 .6
.3 .7 .8 .4 2.5 .8
.7 .7 3.0 1.0 .2 .6
.2 .0 1.5 .5 1.5 .2
.1 1.4 .1 .4 1.6 .5
.7 1.5 .8 1.5 1.3 1.0
1.1
Chisquare with 42 degrees of freedom: 43.397
zscore= .152 pvalue= .588427
______________________________________________________________
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The OVERLAPPING SUMS test ::
:: Integers are floated to get a sequence U(1),U(2),... of uni ::
:: form [0,1) variables. Then overlapping sums, ::
:: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. ::
:: The S's are virtually normal with a certain covariance mat ::
:: rix. A linear transformation of the S's converts them to a ::
:: sequence of independent standard normals, which are converted ::
:: to uniform variables for a KSTEST. The pvalues from ten ::
:: KSTESTs are given still another KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test no. 1 pvalue .471938
Test no. 2 pvalue .661696
Test no. 3 pvalue .127312
Test no. 4 pvalue .138076
Test no. 5 pvalue .252853
Test no. 6 pvalue .966210
Test no. 7 pvalue .981441
Test no. 8 pvalue .961647
Test no. 9 pvalue .877290
Test no. 10 pvalue .974672
Results of the OSUM test for pxl_out.rbnf
KSTEST on the above 10 pvalues: .974109
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the RUNS test. It counts runs up, and runs down, ::
:: in a sequence of uniform [0,1) variables, obtained by float ::
:: ing the 32bit integers in the specified file. This example ::
:: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95::
:: contains an uprun of length 3, a downrun of length 2 and an ::
:: uprun of (at least) 2, depending on the next values. The ::
:: covariance matrices for the runsup and runsdown 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 pxl_out.rbnf
Up and down runs in a sample of 10000
_________________________________________________
Run test for pxl_out.rbnf :
runs up; ks test for 10 p's: .137403
runs down; ks test for 10 p's: .458294
Run test for pxl_out.rbnf :
runs up; ks test for 10 p's: .896885
runs down; ks test for 10 p's: .347245
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the CRAPS TEST. It plays 200,000 games of craps, finds::
:: the number of wins and the number of throws necessary to end ::
:: each game. The number of wins should be (very close to) a ::
:: normal with mean 200000p and variance 200000p(1p), 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 chisquare test is made on the no.ofthrows cell counts. ::
:: Each 32bit 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 pxl_out.rbnf
No. of wins: Observed Expected
98448 98585.86
98448= No. of wins, zscore= .617 pvalue= .26875
Analysis of ThrowsperGame:
Chisq= 25.52 for 20 degrees of freedom, p= .81772
Throws Observed Expected Chisq Sum
1 66332 66666.7 1.680 1.680
2 37909 37654.3 1.723 3.403
3 26940 26954.7 .008 3.411
4 19418 19313.5 .566 3.976
5 13744 13851.4 .833 4.810
6 9937 9943.5 .004 4.814
7 7026 7145.0 1.983 6.797
8 5303 5139.1 5.229 12.026
9 3782 3699.9 1.823 13.849
10 2708 2666.3 .652 14.501
11 1929 1923.3 .017 14.518
12 1352 1388.7 .972 15.490
13 1015 1003.7 .127 15.617
14 708 726.1 .453 16.070
15 529 525.8 .019 16.089
16 373 381.2 .174 16.263
17 244 276.5 3.829 20.092
18 179 200.8 2.373 22.465
19 157 146.0 .831 23.296
20 121 106.2 2.058 25.354
21 294 287.1 .165 25.519
SUMMARY FOR pxl_out.rbnf
pvalue for no. of wins: .268753
pvalue for throws/game: .817722
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Results of DIEHARD battery of tests sent to file pxl_out.txt
 Wolfy
 Fleet Admiral
 Posts: 5363
 Joined: Tue Feb 05, 2008 1:10 am
 Location: Somewhere in the Frontier on a Hycrotan station, working on new ships.
also heres the mk3b algorithm I used (super slow though)  the different versions were just slight refinements that lowered the number of (0.5p0.5) < 1%  it also seems that 123412341234 is a particularly rough seed:
statistics from the different versions:
mk1: normP < 1% = 5 ____ normP < 0.1% = 2
mk2: (didnt test, had a bug so just redid the part)
mk3: nomrP < 1% = 4 ____ normP < 0.1% = 3 (total number of outliers went down.. but the outliers got worse..)
mk3b: normP < 1% = 2 _____ normP < 0.1% = 0 (Reasonable results given the number of tests  also one was P=0.009.. so barely fell in that category at that..)
this one was slightly edited to run in C# so I could play around with doing some image stuff but I decided not to and just work on my class project instead >.>
Its extremely slow (~6070kb/s)  (this is the mk3b)
statistics from the different versions:
mk1: normP < 1% = 5 ____ normP < 0.1% = 2
mk2: (didnt test, had a bug so just redid the part)
mk3: nomrP < 1% = 4 ____ normP < 0.1% = 3 (total number of outliers went down.. but the outliers got worse..)
mk3b: normP < 1% = 2 _____ normP < 0.1% = 0 (Reasonable results given the number of tests  also one was P=0.009.. so barely fell in that category at that..)
this one was slightly edited to run in C# so I could play around with doing some image stuff but I decided not to and just work on my class project instead >.>
Its extremely slow (~6070kb/s)  (this is the mk3b)
Code: Select all
unchecked
{
rdnm = 0;
ulong[] p = new ulong[128];
for (int k = 0; k < 8; k++)
{
nseed = 784579452832475417 * seed + 599874909874320983;
ulong s = 0;
for (int j = 0; j < 128; j++)
{
p[j] = nseed;
nseed = (nseed * (ulong)j + nseed % 1200) * 499885209843298032;
nseed = cseed + nseed >> 5;
qseed = 423109432143 * cseed;
if (1 == (nseed & 0x01))
{
nseed = (fseed >> 5) * (cseed);
p[j] += qseed * (ulong)j + cseed;
p[j] = nseed * p[j] + 9839403928324 * ((qseed >> 5) % 117);
}
else
{
nseed += (cseed >> 10);
p[j] += qseed * (ulong)j  cseed;
p[j] += nseed * p[j]  9839403928324 * ((qseed >> 5) % 117);
}
if ((ulong)j < (cseed % 127)  j == 117)
{
cseed = p[j] * fseed * cseed;
cseed += nseed;
}
s += p[j];
nseed = qseed * 23498900909821;
qseed = 23439802347 * qseed + 74587924370984;
}
seed = nseed * qseed + cseed;
s = s * (nseed >> (char)(qseed % 3 + 1)) + (qseed + (cseed >> (char)(p[7] % 5 +1 )));
if (seed == lseed)
{
seed = 324897329807 * seed + qseed >> 4;
k;
}
else
{
lseed = seed;
rdnm = (rdnm << 8) + (s >> (64  8));
}
}
 Wolfy
 Fleet Admiral
 Posts: 5363
 Joined: Tue Feb 05, 2008 1:10 am
 Location: Somewhere in the Frontier on a Hycrotan station, working on new ships.
Ok, so for fun (since I already know the standard c rand() function is awful I didnt bother to test it) I decided to try out the simple .net rng...
Turns out that pixels knocks it flat (with around the same runtime!)  even with its own two 1.0's, the standard .net one is... just look at it! The number of 1.000000's is... This is without a doubt a complete fail.
Turns out that pixels knocks it flat (with around the same runtime!)  even with its own two 1.0's, the standard .net one is... just look at it! The number of 1.000000's is... This is without a doubt a complete fail.
Code: Select all
NOTE: Most of the tests in DIEHARD return a pvalue, which
should be uniform on [0,1) if the input file contains truly
independent random bits. Those pvalues are obtained by
p=F(X), where F is the assumed distribution of the sample
random variable Xoften 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 pvalues 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 chisquare goodness of fit test ::
:: provides a p value. The first test uses bits 124 (counting ::
:: from the left) from integers in the specified file. ::
:: Then the file is closed and reopened. Next, bits 225 are ::
:: used to provide birthdays, then 326 and so on to bits 932. ::
:: Each set of bits provides a pvalue, and the nine pvalues ::
:: provide a sample for a KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA= 2.0000
Results for net_out.rbnf
For a sample of size 500: mean
net_out.rbnf using bits 1 to 24 5.122
duplicate number number
spacings observed expected
0 4. 67.668
1 12. 135.335
2 36. 135.335
3 63. 90.224
4 98. 45.112
5 83. 18.045
6 to INF 204. 8.282
Chisquare with 6 d.o.f. = 5174.52 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 2 to 25 2.548
duplicate number number
spacings observed expected
0 34. 67.668
1 87. 135.335
2 155. 135.335
3 103. 90.224
4 72. 45.112
5 27. 18.045
6 to INF 22. 8.282
Chisquare with 6 d.o.f. = 81.87 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 3 to 26 2.570
duplicate number number
spacings observed expected
0 38. 67.668
1 86. 135.335
2 149. 135.335
3 99. 90.224
4 68. 45.112
5 35. 18.045
6 to INF 25. 8.282
Chisquare with 6 d.o.f. = 94.52 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 4 to 27 2.742
duplicate number number
spacings observed expected
0 37. 67.668
1 90. 135.335
2 99. 135.335
3 135. 90.224
4 70. 45.112
5 36. 18.045
6 to INF 33. 8.282
Chisquare with 6 d.o.f. = 166.43 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 5 to 28 2.600
duplicate number number
spacings observed expected
0 37. 67.668
1 101. 135.335
2 110. 135.335
3 125. 90.224
4 70. 45.112
5 28. 18.045
6 to INF 29. 8.282
Chisquare with 6 d.o.f. = 111.81 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 6 to 29 2.640
duplicate number number
spacings observed expected
0 32. 67.668
1 100. 135.335
2 112. 135.335
3 123. 90.224
4 72. 45.112
5 41. 18.045
6 to INF 20. 8.282
Chisquare with 6 d.o.f. = 105.77 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 7 to 30 2.540
duplicate number number
spacings observed expected
0 42. 67.668
1 98. 135.335
2 124. 135.335
3 106. 90.224
4 75. 45.112
5 34. 18.045
6 to INF 21. 8.282
Chisquare with 6 d.o.f. = 77.19 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 8 to 31 2.788
duplicate number number
spacings observed expected
0 30. 67.668
1 87. 135.335
2 116. 135.335
3 114. 90.224
4 74. 45.112
5 55. 18.045
6 to INF 24. 8.282
Chisquare with 6 d.o.f. = 171.27 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
net_out.rbnf using bits 9 to 32 3.028
duplicate number number
spacings observed expected
0 35. 67.668
1 66. 135.335
2 102. 135.335
3 106. 90.224
4 100. 45.112
5 46. 18.045
6 to INF 45. 8.282
Chisquare with 6 d.o.f. = 335.15 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
The 9 pvalues were
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 1.000000 1.000000
A KSTEST for the 9 pvalues yields 1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE OVERLAPPING 5PERMUTATION TEST ::
:: This is the OPERM5 test. It looks at a sequence of one mill ::
:: ion 32bit 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 net_out.rbnf
For a sample of 1,000,000 consecutive 5tuples,
chisquare for 99 degrees of freedom=143.711; pvalue= .997739
OPERM5 test for file net_out.rbnf
For a sample of 1,000,000 consecutive 5tuples,
chisquare for 99 degrees of freedom=107.195; pvalue= .730519
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 net_out.rbnf
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32bit integer
rank observed expected (oe)^2/e sum
28 9369 211.4*******************
29 30631 5134.0*******************
30 0 23103.0*******************
31 0 11551.5*******************
chisquare=****** for 3 d. of f.; pvalue=1.000000

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 32x32 matrices. A random 32x ::
:: 32 binary matrix is formed, each row a 32bit 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 net_out.rbnf
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32bit integer
rank observed expected (oe)^2/e sum
29 9169 211.4*******************
30 30831 5134.0*******************
31 0 23103.0*******************
32 0 11551.5*******************
chisquare=****** for 3 d. of f.; pvalue=1.000000

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 6x8 matrices. From each of ::
:: six random 32bit 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 chisquare test is performed on ::
:: counts for ranks 6,5 and <=4. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Binary Rank Test for net_out.rbnf
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 1 to 8
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3542 944.3 7146.065 7146.065
r =5 37959 21743.9 12092.100 19238.170
r =6 58499 77311.8 4577.847 23816.020
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 2 to 9
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21679 21743.9 .194 1.042
r =6 77405 77311.8 .112 1.154
p=1exp(SUM/2)= .43849
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 3 to 10
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 927 944.3 .317 .317
r =5 21719 21743.9 .029 .346
r =6 77354 77311.8 .023 .369
p=1exp(SUM/2)= .16829
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 4 to 11
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 922 944.3 .527 .527
r =5 21871 21743.9 .743 1.270
r =6 77207 77311.8 .142 1.412
p=1exp(SUM/2)= .50631
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 5 to 12
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 920 944.3 .625 .625
r =5 21712 21743.9 .047 .672
r =6 77368 77311.8 .041 .713
p=1exp(SUM/2)= .29989
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 6 to 13
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 978 944.3 1.203 1.203
r =5 21726 21743.9 .015 1.217
r =6 77296 77311.8 .003 1.221
p=1exp(SUM/2)= .45680
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 7 to 14
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 22000 21743.9 3.016 3.865
r =6 77084 77311.8 .671 4.536
p=1exp(SUM/2)= .89647
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 8 to 15
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21497 21743.9 2.804 3.652
r =6 77587 77311.8 .980 4.631
p=1exp(SUM/2)= .90130
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 9 to 16
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 940 944.3 .020 .020
r =5 21664 21743.9 .294 .313
r =6 77396 77311.8 .092 .405
p=1exp(SUM/2)= .18327
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 10 to 17
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3583 944.3 7373.421 7373.421
r =5 38417 21743.9 12784.840 20158.260
r =6 58000 77311.8 4823.918 24982.180
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 11 to 18
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3576 944.3 7334.353 7334.353
r =5 38318 21743.9 12633.460 19967.820
r =6 58106 77311.8 4771.107 24738.920
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 12 to 19
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3526 944.3 7058.307 7058.307
r =5 38414 21743.9 12780.240 19838.540
r =6 58060 77311.8 4793.990 24632.540
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 13 to 20
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3639 944.3 7689.708 7689.708
r =5 38527 21743.9 12954.090 20643.800
r =6 57834 77311.8 4907.205 25551.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 14 to 21
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3553 944.3 7206.714 7206.714
r =5 38410 21743.9 12774.110 19980.820
r =6 58037 77311.8 4805.451 24786.270
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 15 to 22
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3563 944.3 7262.071 7262.071
r =5 38131 21743.9 12349.990 19612.060
r =6 58306 77311.8 4672.257 24284.320
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 16 to 23
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3651 944.3 7758.348 7758.348
r =5 37995 21743.9 12145.850 19904.200
r =6 58354 77311.8 4648.687 24552.890
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 17 to 24
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3675 944.3 7896.543 7896.543
r =5 37866 21743.9 11953.790 19850.340
r =6 58459 77311.8 4597.335 24447.670
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 18 to 25
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 907 944.3 1.473 1.473
r =5 21766 21743.9 .022 1.496
r =6 77327 77311.8 .003 1.499
p=1exp(SUM/2)= .52737
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 19 to 26
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 921 944.3 .575 .575
r =5 21711 21743.9 .050 .625
r =6 77368 77311.8 .041 .666
p=1exp(SUM/2)= .28309
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 20 to 27
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 960 944.3 .261 .261
r =5 21914 21743.9 1.331 1.592
r =6 77126 77311.8 .447 2.038
p=1exp(SUM/2)= .63908
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 21 to 28
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 952 944.3 .063 .063
r =5 21729 21743.9 .010 .073
r =6 77319 77311.8 .001 .074
p=1exp(SUM/2)= .03615
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 22 to 29
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 963 944.3 .370 .370
r =5 21961 21743.9 2.168 2.538
r =6 77076 77311.8 .719 3.257
p=1exp(SUM/2)= .80379
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 23 to 30
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 921 944.3 .575 .575
r =5 21780 21743.9 .060 .635
r =6 77299 77311.8 .002 .637
p=1exp(SUM/2)= .27277
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 24 to 31
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 904 944.3 1.720 1.720
r =5 21686 21743.9 .154 1.874
r =6 77410 77311.8 .125 1.999
p=1exp(SUM/2)= .63192
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG net_out.rbnf
brank test for bits 25 to 32
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 956 944.3 .145 .145
r =5 21684 21743.9 .165 .310
r =6 77360 77311.8 .030 .340
p=1exp(SUM/2)= .15633
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
1.000000 .438492 .168286 .506309 .299889
.456800 .896470 .901299 .183267 1.000000
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 .527374 .283087 .639080
.036154 .803785 .272771 .631917 .156332
brank test summary for net_out.rbnf
The KS test for those 25 supposed UNI's yields
KS pvalue=1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 20letter ::
:: "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 20letter (20bit) words in a string of ::
:: 2^21 overlapping 20letter 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 ::
:: (j141909)/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 20tuples 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: 169573 missing words, 64.63 sigmas from mean, pvalue=1.00000
tst no 2: 169451 missing words, 64.35 sigmas from mean, pvalue=1.00000
tst no 3: 169912 missing words, 65.43 sigmas from mean, pvalue=1.00000
tst no 4: 169295 missing words, 63.99 sigmas from mean, pvalue=1.00000
tst no 5: 169775 missing words, 65.11 sigmas from mean, pvalue=1.00000
tst no 6: 170435 missing words, 66.65 sigmas from mean, pvalue=1.00000
tst no 7: 168946 missing words, 63.17 sigmas from mean, pvalue=1.00000
tst no 8: 170116 missing words, 65.90 sigmas from mean, pvalue=1.00000
tst no 9: 169784 missing words, 65.13 sigmas from mean, pvalue=1.00000
tst no 10: 169100 missing words, 63.53 sigmas from mean, pvalue=1.00000
tst no 11: 169270 missing words, 63.93 sigmas from mean, pvalue=1.00000
tst no 12: 169535 missing words, 64.55 sigmas from mean, pvalue=1.00000
tst no 13: 170433 missing words, 66.64 sigmas from mean, pvalue=1.00000
tst no 14: 170230 missing words, 66.17 sigmas from mean, pvalue=1.00000
tst no 15: 168796 missing words, 62.82 sigmas from mean, pvalue=1.00000
tst no 16: 169419 missing words, 64.27 sigmas from mean, pvalue=1.00000
tst no 17: 169751 missing words, 65.05 sigmas from mean, pvalue=1.00000
tst no 18: 169532 missing words, 64.54 sigmas from mean, pvalue=1.00000
tst no 19: 168830 missing words, 62.90 sigmas from mean, pvalue=1.00000
tst no 20: 169772 missing words, 65.10 sigmas from mean, pvalue=1.00000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The tests OPSO, OQSO and DNA ::
:: OPSO means OverlappingPairsSparseOccupancy ::
:: The OPSO test considers 2letter words from an alphabet of ::
:: 1024 letters. Each letter is determined by a specified ten ::
:: bits from a 32bit integer in the sequence to be tested. OPSO ::
:: generates 2^21 (overlapping) 2letter words (from 2^21+1 ::
:: "keystrokes") and counts the number of missing wordsthat ::
:: is 2letter 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 (missingwrds141909)/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 OverlappingQuadruplesSparseOccupancy ::
:: The test OQSO is similar, except that it considers 4letter ::
:: 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 32bit 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 10letter 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) 10letter 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 net_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
OPSO for net_out.rbnf using bits 23 to 32 142349 1.516 .9353
OPSO for net_out.rbnf using bits 22 to 31 142016 .368 .6435
OPSO for net_out.rbnf using bits 21 to 30 141868 .143 .4433
OPSO for net_out.rbnf using bits 20 to 29 141893 .056 .4775
OPSO for net_out.rbnf using bits 19 to 28 142364 1.568 .9415
OPSO for net_out.rbnf using bits 18 to 27 141751 .546 .2925
OPSO for net_out.rbnf using bits 17 to 26 786515******* 1.0000
OPSO for net_out.rbnf using bits 16 to 25 786511******* 1.0000
OPSO for net_out.rbnf using bits 15 to 24 786504******* 1.0000
OPSO for net_out.rbnf using bits 14 to 23 786503******* 1.0000
OPSO for net_out.rbnf using bits 13 to 22 786514******* 1.0000
OPSO for net_out.rbnf using bits 12 to 21 786527******* 1.0000
OPSO for net_out.rbnf using bits 11 to 20 786518******* 1.0000
OPSO for net_out.rbnf using bits 10 to 19 786511******* 1.0000
OPSO for net_out.rbnf using bits 9 to 18 786524******* 1.0000
OPSO for net_out.rbnf using bits 8 to 17 786517******* 1.0000
OPSO for net_out.rbnf using bits 7 to 16 141683 .780 .2176
OPSO for net_out.rbnf using bits 6 to 15 142121 .730 .7673
OPSO for net_out.rbnf using bits 5 to 14 142451 1.868 .9691
OPSO for net_out.rbnf using bits 4 to 13 141813 .332 .3699
OPSO for net_out.rbnf using bits 3 to 12 142192 .975 .8352
OPSO for net_out.rbnf using bits 2 to 11 141768 .487 .3130
OPSO for net_out.rbnf using bits 1 to 10 786525******* 1.0000
OQSO test for generator net_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
OQSO for net_out.rbnf using bits 28 to 32 141761 .503 .3075
OQSO for net_out.rbnf using bits 27 to 31 141736 .588 .2784
OQSO for net_out.rbnf using bits 26 to 30 141866 .147 .4416
OQSO for net_out.rbnf using bits 25 to 29 142024 .389 .6513
OQSO for net_out.rbnf using bits 24 to 28 142091 .616 .7310
OQSO for net_out.rbnf using bits 23 to 27 142143 .792 .7859
OQSO for net_out.rbnf using bits 22 to 26 141707 .686 .2464
OQSO for net_out.rbnf using bits 21 to 25 142085 .595 .7242
OQSO for net_out.rbnf using bits 20 to 24 141259 2.205 .0137
OQSO for net_out.rbnf using bits 19 to 23 141907 .008 .4969
OQSO for net_out.rbnf using bits 18 to 22 142032 .416 .6612
OQSO for net_out.rbnf using bits 17 to 21 983040******* 1.0000
OQSO for net_out.rbnf using bits 16 to 20 983040******* 1.0000
OQSO for net_out.rbnf using bits 15 to 19 983040******* 1.0000
OQSO for net_out.rbnf using bits 14 to 18 983040******* 1.0000
OQSO for net_out.rbnf using bits 13 to 17 983040******* 1.0000
OQSO for net_out.rbnf using bits 12 to 16 142272 1.229 .8905
OQSO for net_out.rbnf using bits 11 to 15 141695 .727 .2338
OQSO for net_out.rbnf using bits 10 to 14 141582 1.110 .1336
OQSO for net_out.rbnf using bits 9 to 13 141567 1.160 .1229
OQSO for net_out.rbnf using bits 8 to 12 142117 .704 .7593
OQSO for net_out.rbnf using bits 7 to 11 142030 .409 .6588
OQSO for net_out.rbnf using bits 6 to 10 141754 .527 .2993
OQSO for net_out.rbnf using bits 5 to 9 142027 .399 .6550
OQSO for net_out.rbnf using bits 4 to 8 141863 .157 .4376
OQSO for net_out.rbnf using bits 3 to 7 141750 .540 .2946
OQSO for net_out.rbnf using bits 2 to 6 141959 .168 .5669
OQSO for net_out.rbnf using bits 1 to 5 983040******* 1.0000
DNA test for generator net_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
DNA for net_out.rbnf using bits 31 to 32 141962 .155 .5617
DNA for net_out.rbnf using bits 30 to 31 141435 1.399 .0809
DNA for net_out.rbnf using bits 29 to 30 142229 .943 .8272
DNA for net_out.rbnf using bits 28 to 29 141108 2.364 .0090
DNA for net_out.rbnf using bits 27 to 28 141704 .606 .2724
DNA for net_out.rbnf using bits 26 to 27 141578 .977 .1642
DNA for net_out.rbnf using bits 25 to 26 142114 .604 .7270
DNA for net_out.rbnf using bits 24 to 25 141327 1.718 .0429
DNA for net_out.rbnf using bits 23 to 24 141896 .039 .4843
DNA for net_out.rbnf using bits 22 to 23 141885 .072 .4714
DNA for net_out.rbnf using bits 21 to 22 142166 .757 .7755
DNA for net_out.rbnf using bits 20 to 21 142333 1.250 .8943
DNA for net_out.rbnf using bits 19 to 20 141897 .036 .4855
DNA for net_out.rbnf using bits 18 to 19 141705 .603 .2733
DNA for net_out.rbnf using bits 17 to 18 1047552******* 1.0000
DNA for net_out.rbnf using bits 16 to 17 1047552******* 1.0000
DNA for net_out.rbnf using bits 15 to 16 142103 .571 .7161
DNA for net_out.rbnf using bits 14 to 15 141882 .081 .4679
DNA for net_out.rbnf using bits 13 to 14 142237 .967 .8331
DNA for net_out.rbnf using bits 12 to 13 141532 1.113 .1328
DNA for net_out.rbnf using bits 11 to 12 142020 .326 .6280
DNA for net_out.rbnf using bits 10 to 11 141657 .744 .2283
DNA for net_out.rbnf using bits 9 to 10 141828 .240 .4052
DNA for net_out.rbnf using bits 8 to 9 142480 1.683 .9539
DNA for net_out.rbnf using bits 7 to 8 141927 .052 .5208
DNA for net_out.rbnf using bits 6 to 7 141893 .048 .4808
DNA for net_out.rbnf using bits 5 to 6 141460 1.325 .0925
DNA for net_out.rbnf using bits 4 to 5 141657 .744 .2283
DNA for net_out.rbnf using bits 3 to 4 142061 .447 .6727
DNA for net_out.rbnf using bits 2 to 3 141776 .393 .3470
DNA for net_out.rbnf using bits 1 to 2 1047552******* 1.0000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNTTHE1's TEST on a stream of bytes. ::
:: Consider the file under test as a stream of bytes (four per ::
:: 32 bit integer). Each byte can contain from 0 to 8 1's, ::
:: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let ::
:: the stream of bytes provide a string of overlapping 5letter ::
:: words, each "letter" taking values A,B,C,D,E. The letters are ::
:: determined by the number of 1's in a byte:: 0,1,or 2 yield A,::
:: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus ::
:: we have a monkey at a typewriter hitting five keys with vari ::
:: ous probabilities (37,56,70,56,37 over 256). There are 5^5 ::
:: possible 5letter words, and from a string of 256,000 (over ::
:: lapping) 5letter words, counts are made on the frequencies ::
:: for each word. The quadratic form in the weak inverse of ::
:: the covariance matrix of the cell counts provides a chisquare ::
:: test:: Q5Q4, the difference of the naive Pearson sums of ::
:: (OBSEXP)^2/EXP on counts for 5 and 4letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for net_out.rbnf
Chisquare with 5^55^4=2500 d.of f. for sample size:2560000
chisquare equiv normal pvalue
Results fo COUNTTHE1's in successive bytes:
byte stream for net_out.rbnf 99146.78 1366.792 1.000000
byte stream for net_out.rbnf 100118.30 1380.531 1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNTTHE1's TEST for specific bytes. ::
:: Consider the file under test as a stream of 32bit 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) 5letter 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 5letter words, and ::
:: from a string of 256,000 (overlapping) 5letter 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:: Q5Q4, the difference of ::
:: the naive Pearson sums of (OBSEXP)^2/EXP on counts for 5 ::
:: and 4letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Chisquare with 5^55^4=2500 d.of f. for sample size: 256000
chisquare equiv normal p value
Results for COUNTTHE1's in specified bytes:
bits 1 to 8 50405.77 677.490 1.000000
bits 2 to 9 2552.36 .740 .770498
bits 3 to 10 2601.35 1.433 .924113
bits 4 to 11 2552.19 .738 .769753
bits 5 to 12 2537.58 .531 .702440
bits 6 to 13 2502.22 .031 .512521
bits 7 to 14 2391.28 1.538 .062074
bits 8 to 15 2491.19 .125 .450430
bits 9 to 16 2512.03 .170 .567536
bits 10 to 17 51356.86 690.940 1.000000
bits 11 to 18 51619.03 694.648 1.000000
bits 12 to 19 50927.70 684.871 1.000000
bits 13 to 20 51476.92 692.638 1.000000
bits 14 to 21 51668.66 695.350 1.000000
bits 15 to 22 51569.03 693.941 1.000000
bits 16 to 23 50675.28 681.301 1.000000
bits 17 to 24 51223.92 689.060 1.000000
bits 18 to 25 2480.85 .271 .393260
bits 19 to 26 2503.63 .051 .520446
bits 20 to 27 2514.93 .211 .583635
bits 21 to 28 2498.62 .020 .492213
bits 22 to 29 2583.72 1.184 .881778
bits 23 to 30 2373.18 1.794 .036445
bits 24 to 31 2749.17 3.524 .999787
bits 25 to 32 2550.96 .721 .764437
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THIS IS A PARKING LOT TEST ::
:: In a square of side 100, randomly "park" a cara 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 (k3523)/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 net_out.rbnf
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 1173 zscore:******* pvalue: .000000
Successes: 1175 zscore:******* pvalue: .000000
Successes: 1176 zscore:******* pvalue: .000000
Successes: 1193 zscore:******* pvalue: .000000
Successes: 1170 zscore:******* pvalue: .000000
Successes: 1183 zscore:******* pvalue: .000000
Successes: 1168 zscore:******* pvalue: .000000
Successes: 1179 zscore:******* pvalue: .000000
Successes: 1188 zscore:******* pvalue: .000000
Successes: 1177 zscore:******* pvalue: .000000
square size avg. no. parked sample sigma
100. 1178.200 7.440
KSTEST for the above 10: p= 1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100 times:: choose n=8000 random points in a ::
:: square of side 10000. Find d, the minimum distance between ::
:: the (n^2n)/2 pairs of points. If the points are truly inde ::
:: pendent uniform, then d^2, the square of the minimum distance ::
:: should be (very close to) exponentially distributed with mean ::
:: .995 . Thus 1exp(d^2/.995) should be uniform on [0,1) and ::
:: a KSTEST on the resulting 100 values serves as a test of uni ::
:: formity for random points in the square. Test numbers=0 mod 5 ::
:: are printed but the KSTEST is based on the full set of 100 ::
:: random choices of 8000 points in the 10000x10000 square. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
This is the MINIMUM DISTANCE test
for random integers in the file net_out.rbnf
Sample no. d^2 avg equiv uni
5 .2258 .1921 .203040
10 .6656 .2594 .487744
15 .4633 .2489 .372287
20 .0424 .2266 .041749
25 .2217 .2325 .199727
30 .0218 .2448 .021711
35 .0516 .2395 .050492
40 .0798 .2489 .077066
45 .0629 .2331 .061242
50 .2636 .2418 .232726
55 .0162 .2333 .016121
60 .2499 .2399 .222073
65 .0768 .2305 .074252
70 .0000 .2198 .000001
75 .8635 .2422 .580162
80 .0902 .2379 .086628
85 .4389 .2418 .356683
90 .7755 .2504 .541302
95 .5807 .2512 .442114
100 .0246 .2486 .024457
MINIMUM DISTANCE TEST for net_out.rbnf
Result of KS test on 20 transformed mindist^2's:
pvalue=1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 1exp(r^3/30.), then a ::
:: KSTEST is done on the 20 pvalues. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file net_out.rbnf
sample no: 1 r^3= 1.004 pvalue= .03291
sample no: 2 r^3= 1.228 pvalue= .04010
sample no: 3 r^3= 7.467 pvalue= .22033
sample no: 4 r^3= 5.840 pvalue= .17688
sample no: 5 r^3= .698 pvalue= .02299
sample no: 6 r^3= 1.947 pvalue= .06285
sample no: 7 r^3= 5.116 pvalue= .15680
sample no: 8 r^3= 11.201 pvalue= .31158
sample no: 9 r^3= .130 pvalue= .00433
sample no: 10 r^3= 8.450 pvalue= .24547
sample no: 11 r^3= .772 pvalue= .02542
sample no: 12 r^3= .814 pvalue= .02678
sample no: 13 r^3= 5.034 pvalue= .15448
sample no: 14 r^3= 2.730 pvalue= .08698
sample no: 15 r^3= .497 pvalue= .01642
sample no: 16 r^3= 2.151 pvalue= .06918
sample no: 17 r^3= 2.345 pvalue= .07520
sample no: 18 r^3= 2.304 pvalue= .07391
sample no: 19 r^3= 14.069 pvalue= .37435
sample no: 20 r^3= .918 pvalue= .03015
A KS test is applied to those 20 pvalues.

3DSPHERES test for file net_out.rbnf pvalue=1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 chisquare test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
 Wolfy
 Fleet Admiral
 Posts: 5363
 Joined: Tue Feb 05, 2008 1:10 am
 Location: Somewhere in the Frontier on a Hycrotan station, working on new ships.
Now, if you want to see something especially abysmal, these are the results from randu! (again for seed 123412341234)
http://en.wikipedia.org/wiki/RANDU
http://en.wikipedia.org/wiki/RANDU
Code: Select all
NOTE: Most of the tests in DIEHARD return a pvalue, which
should be uniform on [0,1) if the input file contains truly
independent random bits. Those pvalues are obtained by
p=F(X), where F is the assumed distribution of the sample
random variable Xoften 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 pvalues 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 chisquare goodness of fit test ::
:: provides a p value. The first test uses bits 124 (counting ::
:: from the left) from integers in the specified file. ::
:: Then the file is closed and reopened. Next, bits 225 are ::
:: used to provide birthdays, then 326 and so on to bits 932. ::
:: Each set of bits provides a pvalue, and the nine pvalues ::
:: provide a sample for a KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA= 2.0000
Results for randu_out.rbnf
For a sample of size 500: mean
randu_out.rbnf using bits 1 to 24 506.750
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 2 to 25 503.878
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 3 to 26 498.626
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 4 to 27 489.878
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 5 to 28 473.880
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 6 to 29 473.880
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 7 to 30 426.332
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 8 to 31 425.394
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
randu_out.rbnf using bits 9 to 32 425.394
duplicate number number
spacings observed expected
0 500. 67.668
1 0. 135.335
2 0. 135.335
3 0. 90.224
4 0. 45.112
5 0. 18.045
6 to INF 0. 8.282
Chisquare with 6 d.o.f. = 3194.53 pvalue= 1.000000
:::::::::::::::::::::::::::::::::::::::::
The 9 pvalues were
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 1.000000 1.000000
A KSTEST for the 9 pvalues yields 1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE OVERLAPPING 5PERMUTATION TEST ::
:: This is the OPERM5 test. It looks at a sequence of one mill ::
:: ion 32bit 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 randu_out.rbnf
For a sample of 1,000,000 consecutive 5tuples,
chisquare for 99 degrees of freedom=*******; pvalue=1.000000
OPERM5 test for file randu_out.rbnf
For a sample of 1,000,000 consecutive 5tuples,
chisquare for 99 degrees of freedom=*******; pvalue=1.000000
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 31x31 matrices. The leftmost ::
:: 31 bits of 31 random integers from the test sequence are used ::
:: to form a 31x31 binary matrix over the field {0,1}. The rank ::
:: is determined. That rank can be from 0 to 31, but ranks< 28 ::
:: are rare, and their counts are pooled with those for rank 28. ::
:: Ranks are found for 40,000 such random matrices and a chisqua::
:: re test is performed on counts for ranks 31,30,29 and <=28. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Binary rank test for randu_out.rbnf
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32bit integer
rank observed expected (oe)^2/e sum
28 40000 211.4*******************
29 0 5134.0*******************
30 0 23103.0*******************
31 0 11551.5*******************
chisquare=****** for 3 d. of f.; pvalue=1.000000

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 32x32 matrices. A random 32x ::
:: 32 binary matrix is formed, each row a 32bit 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 randu_out.rbnf
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32bit integer
rank observed expected (oe)^2/e sum
29 40000 211.4*******************
30 0 5134.0*******************
31 0 23103.0*******************
32 0 11551.5*******************
chisquare=****** for 3 d. of f.; pvalue=1.000000

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the BINARY RANK TEST for 6x8 matrices. From each of ::
:: six random 32bit 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 chisquare test is performed on ::
:: counts for ranks 6,5 and <=4. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Binary Rank Test for randu_out.rbnf
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 1 to 8
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 2 to 9
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 3 to 10
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 4 to 11
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 5 to 12
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 6 to 13
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 7 to 14
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 8 to 15
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 9 to 16
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 10 to 17
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 11 to 18
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 12 to 19
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 13 to 20
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 100000 944.310390790.00010390790.000
r =5 0 21743.9 21743.90010412530.000
r =6 0 77311.8 77311.80010489840.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 14 to 21
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 52545 944.3 2819685.000 2819685.000
r =5 47455 21743.9 30402.120 2850087.000
r =6 0 77311.8 77311.800 2927399.000
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 15 to 22
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 20486 944.3 404402.700 404402.700
r =5 56223 21743.9 54673.190 459075.900
r =6 23291 77311.8 37746.460 496822.400
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 16 to 23
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 6421 944.3 31763.410 31763.410
r =5 44090 21743.9 22964.980 54728.390
r =6 49489 77311.8 10012.810 64741.200
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 17 to 24
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 2247 944.3 1797.121 1797.121
r =5 25869 21743.9 782.585 2579.706
r =6 71884 77311.8 381.068 2960.774
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 18 to 25
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 2489 944.3 2526.836 2526.836
r =5 28278 21743.9 1963.514 4490.350
r =6 69233 77311.8 844.206 5334.556
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 19 to 26
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 2147 944.3 1531.804 1531.804
r =5 27937 21743.9 1763.919 3295.723
r =6 69916 77311.8 707.498 4003.221
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 20 to 27
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 1562 944.3 404.057 404.057
r =5 20314 21743.9 94.032 498.089
r =6 78124 77311.8 8.532 506.621
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 21 to 28
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 782 944.3 27.895 27.895
r =5 19527 21743.9 226.024 253.920
r =6 79691 77311.8 73.217 327.137
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 22 to 29
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 1563 944.3 405.367 405.367
r =5 39059 21743.9 13788.360 14193.720
r =6 59378 77311.8 4160.055 18353.780
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 23 to 30
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 3126 944.3 5040.563 5040.563
r =5 28126 21743.9 1873.224 6913.787
r =6 68748 77311.8 948.610 7862.396
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 24 to 31
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 0 944.3 944.301 944.301
r =5 25000 21743.9 487.594 1431.895
r =6 75000 77311.8 69.128 1501.023
p=1exp(SUM/2)=1.00000
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG randu_out.rbnf
brank test for bits 25 to 32
OBSERVED EXPECTED (OE)^2/E SUM
r<=4 0 944.3 944.301 944.301
r =5 50000 21743.9 36718.670 37662.970
r =6 50000 77311.8 9648.393 47311.370
p=1exp(SUM/2)=1.00000
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 1.000000 1.000000 1.000000
1.000000 1.000000 1.000000 1.000000 1.000000
brank test summary for randu_out.rbnf
The KS test for those 25 supposed UNI's yields
KS pvalue=1.000000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 20letter ::
:: "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 20letter (20bit) words in a string of ::
:: 2^21 overlapping 20letter 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 ::
:: (j141909)/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 20tuples 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: 977524 missing words, 1952.37 sigmas from mean, pvalue=1.00000
tst no 2: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 3: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 4: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 5: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 6: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 7: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 8: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 9: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 10: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 11: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 12: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 13: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 14: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 15: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 16: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 17: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 18: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 19: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
tst no 20: 977541 missing words, 1952.41 sigmas from mean, pvalue=1.00000
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The tests OPSO, OQSO and DNA ::
:: OPSO means OverlappingPairsSparseOccupancy ::
:: The OPSO test considers 2letter words from an alphabet of ::
:: 1024 letters. Each letter is determined by a specified ten ::
:: bits from a 32bit integer in the sequence to be tested. OPSO ::
:: generates 2^21 (overlapping) 2letter words (from 2^21+1 ::
:: "keystrokes") and counts the number of missing wordsthat ::
:: is 2letter 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 (missingwrds141909)/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 OverlappingQuadruplesSparseOccupancy ::
:: The test OQSO is similar, except that it considers 4letter ::
:: 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 32bit 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 10letter 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) 10letter 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 randu_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
OPSO for randu_out.rbnf using bits 23 to 32 1048448******* 1.0000
OPSO for randu_out.rbnf using bits 22 to 31 1048320******* 1.0000
OPSO for randu_out.rbnf using bits 21 to 30 1048064******* 1.0000
OPSO for randu_out.rbnf using bits 20 to 29 1047552******* 1.0000
OPSO for randu_out.rbnf using bits 19 to 28 1046528******* 1.0000
OPSO for randu_out.rbnf using bits 18 to 27 1045504******* 1.0000
OPSO for randu_out.rbnf using bits 17 to 26 1045504******* 1.0000
OPSO for randu_out.rbnf using bits 16 to 25 1047040******* 1.0000
OPSO for randu_out.rbnf using bits 15 to 24 1047808******* 1.0000
OPSO for randu_out.rbnf using bits 14 to 23 1048192******* 1.0000
OPSO for randu_out.rbnf using bits 13 to 22 1048384******* 1.0000
OPSO for randu_out.rbnf using bits 12 to 21 1048480******* 1.0000
OPSO for randu_out.rbnf using bits 11 to 20 1048528******* 1.0000
OPSO for randu_out.rbnf using bits 10 to 19 1048552******* 1.0000
OPSO for randu_out.rbnf using bits 9 to 18 1048564******* 1.0000
OPSO for randu_out.rbnf using bits 8 to 17 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 7 to 16 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 6 to 15 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 5 to 14 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 4 to 13 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 3 to 12 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 2 to 11 1048572******* 1.0000
OPSO for randu_out.rbnf using bits 1 to 10 1048572******* 1.0000
OQSO test for generator randu_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
OQSO for randu_out.rbnf using bits 28 to 32 1048572******* 1.0000
OQSO for randu_out.rbnf using bits 27 to 31 1048568******* 1.0000
OQSO for randu_out.rbnf using bits 26 to 30 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 25 to 29 1048544******* 1.0000
OQSO for randu_out.rbnf using bits 24 to 28 1048512******* 1.0000
OQSO for randu_out.rbnf using bits 23 to 27 1048448******* 1.0000
OQSO for randu_out.rbnf using bits 22 to 26 1048320******* 1.0000
OQSO for randu_out.rbnf using bits 21 to 25 1048096******* 1.0000
OQSO for randu_out.rbnf using bits 20 to 24 1047840******* 1.0000
OQSO for randu_out.rbnf using bits 19 to 23 1047712******* 1.0000
OQSO for randu_out.rbnf using bits 18 to 22 1047712******* 1.0000
OQSO for randu_out.rbnf using bits 17 to 21 1047712******* 1.0000
OQSO for randu_out.rbnf using bits 16 to 20 1048144******* 1.0000
OQSO for randu_out.rbnf using bits 15 to 19 1048360******* 1.0000
OQSO for randu_out.rbnf using bits 14 to 18 1048468******* 1.0000
OQSO for randu_out.rbnf using bits 13 to 17 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 12 to 16 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 11 to 15 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 10 to 14 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 9 to 13 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 8 to 12 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 7 to 11 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 6 to 10 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 5 to 9 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 4 to 8 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 3 to 7 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 2 to 6 1048560******* 1.0000
OQSO for randu_out.rbnf using bits 1 to 5 1048560******* 1.0000
DNA test for generator randu_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), pvalue (p)
mw z p
DNA for randu_out.rbnf using bits 31 to 32 1048575******* 1.0000
DNA for randu_out.rbnf using bits 30 to 31 1048574******* 1.0000
DNA for randu_out.rbnf using bits 29 to 30 1048574******* 1.0000
DNA for randu_out.rbnf using bits 28 to 29 1048572******* 1.0000
DNA for randu_out.rbnf using bits 27 to 28 1048568******* 1.0000
DNA for randu_out.rbnf using bits 26 to 27 1048560******* 1.0000
DNA for randu_out.rbnf using bits 25 to 26 1048544******* 1.0000
DNA for randu_out.rbnf using bits 24 to 25 1048512******* 1.0000
DNA for randu_out.rbnf using bits 23 to 24 1048448******* 1.0000
DNA for randu_out.rbnf using bits 22 to 23 1048320******* 1.0000
DNA for randu_out.rbnf using bits 21 to 22 1048064******* 1.0000
DNA for randu_out.rbnf using bits 20 to 21 1047552******* 1.0000
DNA for randu_out.rbnf using bits 19 to 20 1046528******* 1.0000
DNA for randu_out.rbnf using bits 18 to 19 1044480******* 1.0000
DNA for randu_out.rbnf using bits 17 to 18 1040384******* 1.0000
DNA for randu_out.rbnf using bits 16 to 17 1047580******* 1.0000
DNA for randu_out.rbnf using bits 15 to 16 1047580******* 1.0000
DNA for randu_out.rbnf using bits 14 to 15 1047580******* 1.0000
DNA for randu_out.rbnf using bits 13 to 14 1047580******* 1.0000
DNA for randu_out.rbnf using bits 12 to 13 1047580******* 1.0000
DNA for randu_out.rbnf using bits 11 to 12 1047580******* 1.0000
DNA for randu_out.rbnf using bits 10 to 11 1047580******* 1.0000
DNA for randu_out.rbnf using bits 9 to 10 1047580******* 1.0000
DNA for randu_out.rbnf using bits 8 to 9 1047580******* 1.0000
DNA for randu_out.rbnf using bits 7 to 8 1047580******* 1.0000
DNA for randu_out.rbnf using bits 6 to 7 1047580******* 1.0000
DNA for randu_out.rbnf using bits 5 to 6 1047580******* 1.0000
DNA for randu_out.rbnf using bits 4 to 5 1047580******* 1.0000
DNA for randu_out.rbnf using bits 3 to 4 1047580******* 1.0000
DNA for randu_out.rbnf using bits 2 to 3 1047580******* 1.0000
DNA for randu_out.rbnf using bits 1 to 2 1047580******* 1.0000
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:: This is the COUNTTHE1'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 5letter ::
:: 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 5letter words, and from a string of 256,000 (over ::
:: lapping) 5letter 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:: Q5Q4, the difference of the naive Pearson sums of ::
:: (OBSEXP)^2/EXP on counts for 5 and 4letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for randu_out.rbnf
Chisquare with 5^55^4=2500 d.of f. for sample size:2560000
chisquare equiv normal pvalue
Results fo COUNTTHE1's in successive bytes:
byte stream for randu_out.rbnf ********* 579173.600 1.000000
byte stream for randu_out.rbnf ********* 579286.100 1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNTTHE1's TEST for specific bytes. ::
:: Consider the file under test as a stream of 32bit 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) 5letter 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 5letter words, and ::
:: from a string of 256,000 (overlapping) 5letter 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:: Q5Q4, the difference of ::
:: the naive Pearson sums of (OBSEXP)^2/EXP on counts for 5 ::
:: and 4letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Chisquare with 5^55^4=2500 d.of f. for sample size: 256000
chisquare equiv normal p value
Results for COUNTTHE1's in specified bytes:
bits 1 to 8*********2022919.000 1.000000
bits 2 to 9*********2024444.000 1.000000
bits 3 to 10*********2022877.000 1.000000
bits 4 to 11*********2024529.000 1.000000
bits 5 to 12*********2022702.000 1.000000
bits 6 to 13*********2024642.000 1.000000
bits 7 to 14*********2022632.000 1.000000
bits 8 to 15*********2024575.000 1.000000
bits 9 to 16*********2022714.000 1.000000
bits 10 to 17*********2024586.000 1.000000
bits 11 to 18*********2022687.000 1.000000
bits 12 to 19*********2024626.000 1.000000
bits 13 to 20********* 500841.800 1.000000
bits 14 to 21********* 75988.340 1.000000
bits 15 to 22975377.00 13758.560 1.000000
bits 16 to 23378613.60 5319.049 1.000000
bits 17 to 24232668.00 3255.066 1.000000
bits 18 to 25293615.60 4116.996 1.000000
bits 19 to 26428343.20 6022.332 1.000000
bits 20 to 27738129.10 10403.370 1.000000
bits 21 to 28********* 18892.250 1.000000
bits 22 to 29********* 37849.360 1.000000
bits 23 to 30********* 65135.280 1.000000
bits 24 to 31********* 101677.500 1.000000
bits 25 to 32********* 216174.100 1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THIS IS A PARKING LOT TEST ::
:: In a square of side 100, randomly "park" a cara 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 (k3523)/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 randu_out.rbnf
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
Successes: 1 zscore:******* pvalue: .000000
square size avg. no. parked sample sigma
100. 1.000 .000
KSTEST for the above 10: p= 1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100 times:: choose n=8000 random points in a ::
:: square of side 10000. Find d, the minimum distance between ::
:: the (n^2n)/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 1exp(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 randu_out.rbnf
Sample no. d^2 avg equiv uni
5 .0000 .0000 .000000
10 .0000 .0000 .000000
15 .0000 .0000 .000000
20 .0000 .0000 .000000
25 .0000 .0000 .000000
30 .0000 .0000 .000000
35 .0000 .0000 .000000
40 .0000 .0000 .000000
45 .0000 .0000 .000000
50 .0000 .0000 .000000
55 .0000 .0000 .000000
60 .0000 .0000 .000000
65 .0000 .0000 .000000
70 .0000 .0000 .000000
75 .0000 .0000 .000000
80 .0000 .0000 .000000
85 .0000 .0000 .000000
90 .0000 .0000 .000000
95 .0000 .0000 .000000
100 .0000 .0000 .000000
MINIMUM DISTANCE TEST for randu_out.rbnf
Result of KS test on 20 transformed mindist^2's:
pvalue=1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE 3DSPHERES TEST ::
:: Choose 4000 random points in a cube of edge 1000. At each ::
:: point, center a sphere large enough to reach the next closest ::
:: point. Then the volume of the smallest such sphere is (very ::
:: close to) exponentially distributed with mean 120pi/3. Thus ::
:: the radius cubed is exponential with mean 30. (The mean is ::
:: obtained by extensive simulation). The 3DSPHERES test gener ::
:: ates 4000 such spheres 20 times. Each min radius cubed leads ::
:: to a uniform variable by means of 1exp(r^3/30.), then a ::
:: KSTEST is done on the 20 pvalues. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file randu_out.rbnf
sample no: 1 r^3= .000 pvalue= .00000
sample no: 2 r^3= .000 pvalue= .00000
sample no: 3 r^3= .000 pvalue= .00000
sample no: 4 r^3= .000 pvalue= .00000
sample no: 5 r^3= .000 pvalue= .00000
sample no: 6 r^3= .000 pvalue= .00000
sample no: 7 r^3= .000 pvalue= .00000
sample no: 8 r^3= .000 pvalue= .00000
sample no: 9 r^3= .000 pvalue= .00000
sample no: 10 r^3= .000 pvalue= .00000
sample no: 11 r^3= .000 pvalue= .00000
sample no: 12 r^3= .000 pvalue= .00000
sample no: 13 r^3= .000 pvalue= .00000
sample no: 14 r^3= .000 pvalue= .00000
sample no: 15 r^3= .000 pvalue= .00000
sample no: 16 r^3= .000 pvalue= .00000
sample no: 17 r^3= .000 pvalue= .00000
sample no: 18 r^3= .000 pvalue= .00000
sample no: 19 r^3= .000 pvalue= .00000
sample no: 20 r^3= .000 pvalue= .00000
A KS test is applied to those 20 pvalues.

3DSPHERES test for file randu_out.rbnf pvalue=1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the SQEEZE test ::
:: Random integers are floated to get uniforms on [0,1). Start ::
:: ing with k=2^31=2147483647, the test finds j, the number of ::
:: iterations necessary to reduce k to 1, using the reduction ::
:: k=ceiling(k*U), with U provided by floating integers from ::
:: the file being tested. Such j's are found 100,000 times, ::
:: then counts for the number of times j was <=6,7,...,47,>=48 ::
:: are used to provide a chisquare test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESULTS OF SQUEEZE TEST FOR randu_out.rbnf
Table of standardized frequency counts
( (obsexp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
1.5 2.4 4.2 6.8 10.5 15.4
21.5 28.7 36.9 45.8 54.9 63.9
72.1 79.3 84.9 88.7 90.6 90.5
88.6 85.0 80.0 74.0 67.2 60.1
52.9 37.1 1539.5 334.4 281.4 210.0
201.9 177.2 91.7 122.6 49.4 140.7
91.6 123.1 2.4 1.8 1.3 1.0
382.4
Chisquare with 42 degrees of freedom:*******
zscore=******* pvalue=1.000000
______________________________________________________________
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The OVERLAPPING SUMS test ::
:: Integers are floated to get a sequence U(1),U(2),... of uni ::
:: form [0,1) variables. Then overlapping sums, ::
:: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. ::
:: The S's are virtually normal with a certain covariance mat ::
:: rix. A linear transformation of the S's converts them to a ::
:: sequence of independent standard normals, which are converted ::
:: to uniform variables for a KSTEST. The pvalues from ten ::
:: KSTESTs are given still another KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test no. 1 pvalue 1.000000
Test no. 2 pvalue 1.000000
Test no. 3 pvalue 1.000000
Test no. 4 pvalue 1.000000
Test no. 5 pvalue 1.000000
Test no. 6 pvalue 1.000000
Test no. 7 pvalue 1.000000
Test no. 8 pvalue 1.000000
Test no. 9 pvalue 1.000000
Test no. 10 pvalue 1.000000
Results of the OSUM test for randu_out.rbnf
KSTEST on the above 10 pvalues: 1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the RUNS test. It counts runs up, and runs down, ::
:: in a sequence of uniform [0,1) variables, obtained by float ::
:: ing the 32bit integers in the specified file. This example ::
:: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95::
:: contains an uprun of length 3, a downrun of length 2 and an ::
:: uprun of (at least) 2, depending on the next values. The ::
:: covariance matrices for the runsup and runsdown 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 randu_out.rbnf
Up and down runs in a sample of 10000
_________________________________________________
Run test for randu_out.rbnf :
runs up; ks test for 10 p's:1.000000
runs down; ks test for 10 p's:1.000000
Run test for randu_out.rbnf :
runs up; ks test for 10 p's:1.000000
runs down; ks test for 10 p's:1.000000
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the CRAPS TEST. It plays 200,000 games of craps, finds::
:: the number of wins and the number of throws necessary to end ::
:: each game. The number of wins should be (very close to) a ::
:: normal with mean 200000p and variance 200000p(1p), 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 chisquare test is made on the no.ofthrows cell counts. ::
:: Each 32bit 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 randu_out.rbnf
No. of wins: Observed Expected
140155 98585.86
140155= No. of wins, zscore=****** pvalue=1.00000
Analysis of ThrowsperGame:
 pixelfck
 Militia Captain
 Posts: 571
 Joined: Tue Aug 11, 2009 8:47 pm
 Location: Travelling around in Europe
Nice to see those results. I'm a bit surprised by the failure of the .net rng. I would think that by now a standard rng would provide better quality output.
I'm most happy to see that my seededRandom is not that bad, although it seems that it is not particularly good either
It seems you put a lot more work in your, I'll take a better look at it later today.
Cheers,
Pixelfck
I'm most happy to see that my seededRandom is not that bad, although it seems that it is not particularly good either
It seems you put a lot more work in your, I'll take a better look at it later today.
Cheers,
Pixelfck
 Wolfy
 Fleet Admiral
 Posts: 5363
 Joined: Tue Feb 05, 2008 1:10 am
 Location: Somewhere in the Frontier on a Hycrotan station, working on new ships.
For its speed, its great  Generally failing one test with a simple fast algorithm isnt bad  I was working on getting mine to produce a distribution of pvalues that were appropriate (ie, a combination that should only happen 10% of the time should only happen ~10% of the time, etc), but at the cost of it plodding along slowly.