DIEHARD test results for 64-bit Mersenne Twister (MT) version 1.7 output which
was 90 million bits.  MT was initialized with the original MT authors' default
value.  DIEHARD.EXE dated 1995-12-29 was used.

Read the "NOTE:" immediately following the line of low line characters, as well
as DIEHARD.TXT (a plain text file) for how to interpret these results.

I checked MT's performance by searching DIEHARD's output, looking for p-values
too close to 0.0 or 1.0, by searching for "0.0000" and also "0.9999" (no
quotes).  I found none.  The p-values also appear to have a uniform distibution
between 0 and 1.

Ron Charlton
2014-06-18

DIEHARD.EXE output starts after the next line:
_______________________________________________________________________________
       NOTE: Most of the tests in DIEHARD return a p-value, which               
       should be uniform on [0,1) if the input file contains truly              
       independent random bits.   Those p-values are obtained by                
       p=F(X), where F is the assumed distribution of the sample                
       random variable X---often normal. But that assumed F is just             
       an asymptotic approximation, for which the fit will be worst             
       in the tails. Thus you should not be surprised with                      
       occasional p-values near 0 or 1, such as .0012 or .9983.                 
       When a bit stream really FAILS BIG, you will get p's of 0 or             
       1 to six or more places.  By all means, do not, as a                     
       Statistician might, think that a p < .025 or p> .975 means               
       that the RNG has "failed the test at the .05 level".  Such               
       p's happen among the hundreds that DIEHARD produces, even                
       with good RNG's.  So keep in mind that " p happens".                     
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::            This is the BIRTHDAY SPACINGS TEST                 ::        
     :: Choose m birthdays in a year of n days.  List the spacings    ::        
     :: between the birthdays.  If j is the number of values that     ::        
     :: occur more than once in that list, then j is asymptotically   ::        
     :: Poisson distributed with mean m^3/(4n).  Experience shows n   ::        
     :: must be quite large, say n>=2^18, for comparing the results   ::        
     :: to the Poisson distribution with that mean.  This test uses   ::        
     :: n=2^24 and m=2^9,  so that the underlying distribution for j  ::        
     :: is taken to be Poisson with lambda=2^27/(2^26)=2.  A sample   ::        
     :: of 500 j's is taken, and a chi-square goodness of fit test    ::        
     :: provides a p value.  The first test uses bits 1-24 (counting  ::        
     :: from the left) from integers in the specified file.           ::        
     ::   Then the file is closed and reopened. Next, bits 2-25 are   ::        
     :: used to provide birthdays, then 3-26 and so on to bits 9-32.  ::        
     :: Each set of bits provides a p-value, and the nine p-values    ::        
     :: provide a sample for a KSTEST.                                ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
 BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA=  2.0000
           Results for t.out          
                   For a sample of size 500:     mean   
           t.out           using bits  1 to 24   1.934
  duplicate       number       number 
  spacings       observed     expected
        0          74.       67.668
        1         131.      135.335
        2         151.      135.335
        3          75.       90.224
        4          45.       45.112
        5          18.       18.045
  6 to INF          6.        8.282
 Chisquare with  6 d.o.f. =     5.74 p-value=  .547340
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  2 to 25   2.002
  duplicate       number       number 
  spacings       observed     expected
        0          77.       67.668
        1         124.      135.335
        2         137.      135.335
        3          84.       90.224
        4          51.       45.112
        5          16.       18.045
  6 to INF         11.        8.282
 Chisquare with  6 d.o.f. =     4.58 p-value=  .401128
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  3 to 26   1.954
  duplicate       number       number 
  spacings       observed     expected
        0          73.       67.668
        1         136.      135.335
        2         142.      135.335
        3          74.       90.224
        4          51.       45.112
        5          16.       18.045
  6 to INF          8.        8.282
 Chisquare with  6 d.o.f. =     4.68 p-value=  .414375
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  4 to 27   1.996
  duplicate       number       number 
  spacings       observed     expected
        0          79.       67.668
        1         133.      135.335
        2         123.      135.335
        3          88.       90.224
        4          42.       45.112
        5          28.       18.045
  6 to INF          7.        8.282
 Chisquare with  6 d.o.f. =     9.02 p-value=  .827691
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  5 to 28   2.014
  duplicate       number       number 
  spacings       observed     expected
        0          55.       67.668
        1         136.      135.335
        2         155.      135.335
        3          85.       90.224
        4          46.       45.112
        5          17.       18.045
  6 to INF          6.        8.282
 Chisquare with  6 d.o.f. =     6.24 p-value=  .603272
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  6 to 29   1.890
  duplicate       number       number 
  spacings       observed     expected
        0          69.       67.668
        1         143.      135.335
        2         145.      135.335
        3          88.       90.224
        4          34.       45.112
        5          15.       18.045
  6 to INF          6.        8.282
 Chisquare with  6 d.o.f. =     5.08 p-value=  .467009
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  7 to 30   1.962
  duplicate       number       number 
  spacings       observed     expected
        0          72.       67.668
        1         131.      135.335
        2         135.      135.335
        3         100.       90.224
        4          37.       45.112
        5          19.       18.045
  6 to INF          6.        8.282
 Chisquare with  6 d.o.f. =     3.61 p-value=  .271297
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  8 to 31   1.960
  duplicate       number       number 
  spacings       observed     expected
        0          64.       67.668
        1         149.      135.335
        2         130.      135.335
        3          91.       90.224
        4          43.       45.112
        5          14.       18.045
  6 to INF          9.        8.282
 Chisquare with  6 d.o.f. =     2.86 p-value=  .174199
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           t.out           using bits  9 to 32   1.996
  duplicate       number       number 
  spacings       observed     expected
        0          81.       67.668
        1         124.      135.335
        2         119.      135.335
        3         100.       90.224
        4          50.       45.112
        5          21.       18.045
  6 to INF          5.        8.282
 Chisquare with  6 d.o.f. =     8.92 p-value=  .821958
  :::::::::::::::::::::::::::::::::::::::::
   The 9 p-values were
        .547340   .401128   .414375   .827691   .603272
        .467009   .271297   .174199   .821958
  A KSTEST for the 9 p-values yields  .300906

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::            THE OVERLAPPING 5-PERMUTATION TEST                 ::        
     :: This is the OPERM5 test.  It looks at a sequence of one mill- ::        
     :: ion 32-bit random integers.  Each set of five consecutive     ::        
     :: integers can be in one of 120 states, for the 5! possible or- ::        
     :: derings of five numbers.  Thus the 5th, 6th, 7th,...numbers   ::        
     :: each provide a state. As many thousands of state transitions  ::        
     :: are observed,  cumulative counts are made of the number of    ::        
     :: occurences of each state.  Then the quadratic form in the     ::        
     :: weak inverse of the 120x120 covariance matrix yields a test   ::        
     :: equivalent to the likelihood ratio test that the 120 cell     ::        
     :: counts came from the specified (asymptotically) normal dis-   ::        
     :: tribution with the specified 120x120 covariance matrix (with  ::        
     :: rank 99).  This version uses 1,000,000 integers, twice.       ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
           OPERM5 test for file t.out          
     For a sample of 1,000,000 consecutive 5-tuples,
 chisquare for 99 degrees of freedom= 85.989; p-value= .178510
           OPERM5 test for file t.out          
     For a sample of 1,000,000 consecutive 5-tuples,
 chisquare for 99 degrees of freedom=109.708; p-value= .783032
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: 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 t.out          
         Rank test for 31x31 binary matrices:
        rows from leftmost 31 bits of each 32-bit integer
      rank   observed  expected (o-e)^2/e  sum
        28       191     211.4  1.971901    1.972
        29      5152    5134.0   .063037    2.035
        30     22904   23103.0  1.714911    3.750
        31     11753   11551.5  3.514031    7.264
  chisquare= 7.264 for 3 d. of f.; p-value= .939572
--------------------------------------------------------------
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: This is the BINARY RANK TEST for 32x32 matrices. A random 32x ::        
     :: 32 binary matrix is formed, each row a 32-bit random integer. ::        
     :: The rank is determined. That rank can be from 0 to 32, ranks  ::        
     :: less than 29 are rare, and their counts are pooled with those ::        
     :: for rank 29.  Ranks are found for 40,000 such random matrices ::        
     :: and a chisquare test is performed on counts for ranks  32,31, ::        
     :: 30 and <=29.                                                  ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
    Binary rank test for t.out          
         Rank test for 32x32 binary matrices:
        rows from leftmost 32 bits of each 32-bit integer
      rank   observed  expected (o-e)^2/e  sum
        29       220     211.4   .348364     .348
        30      5170    5134.0   .252290     .601
        31     23022   23103.0   .284317     .885
        32     11588   11551.5   .115177    1.000
  chisquare= 1.000 for 3 d. of f.; p-value= .357422
--------------------------------------------------------------

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: This is the BINARY RANK TEST for 6x8 matrices.  From each of  ::        
     :: six random 32-bit integers from the generator under test, a   ::        
     :: specified byte is chosen, and the resulting six bytes form a  ::        
     :: 6x8 binary matrix whose rank is determined.  That rank can be ::        
     :: from 0 to 6, but ranks 0,1,2,3 are rare; their counts are     ::        
     :: pooled with those for rank 4. Ranks are found for 100,000     ::        
     :: random matrices, and a chi-square test is performed on        ::        
     :: counts for ranks 6,5 and <=4.                                 ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
         Binary Rank Test for t.out          
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  1 to  8
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          922       944.3        .527        .527
          r =5        21771     21743.9        .034        .560
          r =6        77307     77311.8        .000        .561
                        p=1-exp(-SUM/2)= .24450
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  2 to  9
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          977       944.3       1.132       1.132
          r =5        21745     21743.9        .000       1.132
          r =6        77278     77311.8        .015       1.147
                        p=1-exp(-SUM/2)= .43648
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  3 to 10
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          959       944.3        .229        .229
          r =5        21609     21743.9        .837       1.066
          r =6        77432     77311.8        .187       1.253
                        p=1-exp(-SUM/2)= .46543
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  4 to 11
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          920       944.3        .625        .625
          r =5        21678     21743.9        .200        .825
          r =6        77402     77311.8        .105        .930
                        p=1-exp(-SUM/2)= .37197
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  5 to 12
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          924       944.3        .436        .436
          r =5        21790     21743.9        .098        .534
          r =6        77286     77311.8        .009        .543
                        p=1-exp(-SUM/2)= .23769
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  6 to 13
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4         1008       944.3       4.297       4.297
          r =5        21724     21743.9        .018       4.315
          r =6        77268     77311.8        .025       4.340
                        p=1-exp(-SUM/2)= .88582
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  7 to 14
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          906       944.3       1.554       1.554
          r =5        21838     21743.9        .407       1.961
          r =6        77256     77311.8        .040       2.001
                        p=1-exp(-SUM/2)= .63231
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  8 to 15
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          960       944.3        .261        .261
          r =5        21727     21743.9        .013        .274
          r =6        77313     77311.8        .000        .274
                        p=1-exp(-SUM/2)= .12809
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits  9 to 16
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          912       944.3       1.105       1.105
          r =5        21870     21743.9        .731       1.836
          r =6        77218     77311.8        .114       1.950
                        p=1-exp(-SUM/2)= .62281
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 10 to 17
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          947       944.3        .008        .008
          r =5        21882     21743.9        .877        .885
          r =6        77171     77311.8        .256       1.141
                        p=1-exp(-SUM/2)= .43483
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 11 to 18
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          949       944.3        .023        .023
          r =5        21754     21743.9        .005        .028
          r =6        77297     77311.8        .003        .031
                        p=1-exp(-SUM/2)= .01533
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 12 to 19
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          907       944.3       1.473       1.473
          r =5        21892     21743.9       1.009       2.482
          r =6        77201     77311.8        .159       2.641
                        p=1-exp(-SUM/2)= .73300
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 13 to 20
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          974       944.3        .934        .934
          r =5        21586     21743.9       1.147       2.081
          r =6        77440     77311.8        .213       2.293
                        p=1-exp(-SUM/2)= .68229
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 14 to 21
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          928       944.3        .281        .281
          r =5        21419     21743.9       4.855       5.136
          r =6        77653     77311.8       1.506       6.642
                        p=1-exp(-SUM/2)= .96388
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 15 to 22
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          987       944.3       1.931       1.931
          r =5        21618     21743.9        .729       2.660
          r =6        77395     77311.8        .090       2.749
                        p=1-exp(-SUM/2)= .74706
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 16 to 23
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4         1007       944.3       4.163       4.163
          r =5        21657     21743.9        .347       4.510
          r =6        77336     77311.8        .008       4.518
                        p=1-exp(-SUM/2)= .89554
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 17 to 24
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          937       944.3        .056        .056
          r =5        21851     21743.9        .528        .584
          r =6        77212     77311.8        .129        .713
                        p=1-exp(-SUM/2)= .29981
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 18 to 25
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          940       944.3        .020        .020
          r =5        21749     21743.9        .001        .021
          r =6        77311     77311.8        .000        .021
                        p=1-exp(-SUM/2)= .01034
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 19 to 26
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          954       944.3        .100        .100
          r =5        21706     21743.9        .066        .166
          r =6        77340     77311.8        .010        .176
                        p=1-exp(-SUM/2)= .08422
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 20 to 27
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          933       944.3        .135        .135
          r =5        21749     21743.9        .001        .136
          r =6        77318     77311.8        .000        .137
                        p=1-exp(-SUM/2)= .06618
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 21 to 28
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          957       944.3        .171        .171
          r =5        21659     21743.9        .331        .502
          r =6        77384     77311.8        .067        .570
                        p=1-exp(-SUM/2)= .24787
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 22 to 29
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          961       944.3        .295        .295
          r =5        21680     21743.9        .188        .483
          r =6        77359     77311.8        .029        .512
                        p=1-exp(-SUM/2)= .22582
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 23 to 30
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          977       944.3       1.132       1.132
          r =5        21659     21743.9        .331       1.464
          r =6        77364     77311.8        .035       1.499
                        p=1-exp(-SUM/2)= .52740
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 24 to 31
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          930       944.3        .217        .217
          r =5        21808     21743.9        .189        .406
          r =6        77262     77311.8        .032        .438
                        p=1-exp(-SUM/2)= .19653
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG t.out          
     b-rank test for bits 25 to 32
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          906       944.3       1.554       1.554
          r =5        21825     21743.9        .302       1.856
          r =6        77269     77311.8        .024       1.880
                        p=1-exp(-SUM/2)= .60931
   TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
 These should be 25 uniform [0,1] random variables:
     .244502     .436481     .465432     .371971     .237689
     .885816     .632309     .128091     .622812     .434828
     .015334     .732996     .682293     .963881     .747063
     .895539     .299812     .010344     .084220     .066181
     .247868     .225817     .527400     .196532     .609313
   brank test summary for t.out          
       The KS test for those 25 supposed UNI's yields
                    KS p-value= .579177

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::                   THE BITSTREAM TEST                          ::        
     :: The file under test is viewed as a stream of bits. Call them  ::        
     :: b1,b2,... .  Consider an alphabet with two "letters", 0 and 1 ::        
     :: and think of the stream of bits as a succession of 20-letter  ::        
     :: "words", overlapping.  Thus the first word is b1b2...b20, the ::        
     :: second is b2b3...b21, and so on.  The bitstream test counts   ::        
     :: the number of missing 20-letter (20-bit) words in a string of ::        
     :: 2^21 overlapping 20-letter words.  There are 2^20 possible 20 ::        
     :: letter words.  For a truly random string of 2^21+19 bits, the ::        
     :: number of missing words j should be (very close to) normally  ::        
     :: distributed with mean 141,909 and sigma 428.  Thus            ::        
     ::  (j-141909)/428 should be a standard normal variate (z score) ::        
     :: that leads to a uniform [0,1) p value.  The test is repeated  ::        
     :: twenty times.                                                 ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
 THE OVERLAPPING 20-tuples BITSTREAM  TEST, 20 BITS PER WORD, N words
   This test uses N=2^21 and samples the bitstream 20 times.
  No. missing words should average  141909. with sigma=428.
---------------------------------------------------------
 tst no  1:  142299 missing words,     .91 sigmas from mean, p-value= .81871
 tst no  2:  141617 missing words,    -.68 sigmas from mean, p-value= .24730
 tst no  3:  141759 missing words,    -.35 sigmas from mean, p-value= .36271
 tst no  4:  142285 missing words,     .88 sigmas from mean, p-value= .80996
 tst no  5:  141457 missing words,   -1.06 sigmas from mean, p-value= .14529
 tst no  6:  141822 missing words,    -.20 sigmas from mean, p-value= .41916
 tst no  7:  141818 missing words,    -.21 sigmas from mean, p-value= .41551
 tst no  8:  141969 missing words,     .14 sigmas from mean, p-value= .55544
 tst no  9:  141468 missing words,   -1.03 sigmas from mean, p-value= .15124
 tst no 10:  141732 missing words,    -.41 sigmas from mean, p-value= .33932
 tst no 11:  141927 missing words,     .04 sigmas from mean, p-value= .51647
 tst no 12:  141320 missing words,   -1.38 sigmas from mean, p-value= .08427
 tst no 13:  142642 missing words,    1.71 sigmas from mean, p-value= .95654
 tst no 14:  141752 missing words,    -.37 sigmas from mean, p-value= .35659
 tst no 15:  141670 missing words,    -.56 sigmas from mean, p-value= .28802
 tst no 16:  142077 missing words,     .39 sigmas from mean, p-value= .65238
 tst no 17:  141751 missing words,    -.37 sigmas from mean, p-value= .35572
 tst no 18:  142421 missing words,    1.20 sigmas from mean, p-value= .88405
 tst no 19:  141707 missing words,    -.47 sigmas from mean, p-value= .31820
 tst no 20:  141983 missing words,     .17 sigmas from mean, p-value= .56833

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::             The tests OPSO, OQSO and DNA                      ::        
     ::         OPSO means Overlapping-Pairs-Sparse-Occupancy         ::        
     :: The OPSO test considers 2-letter words from an alphabet of    ::        
     :: 1024 letters.  Each letter is determined by a specified ten   ::        
     :: bits from a 32-bit integer in the sequence to be tested. OPSO ::        
     :: generates  2^21 (overlapping) 2-letter words  (from 2^21+1    ::        
     :: "keystrokes")  and counts the number of missing words---that  ::        
     :: is 2-letter words which do not appear in the entire sequence. ::        
     :: That count should be very close to normally distributed with  ::        
     :: mean 141,909, sigma 290. Thus (missingwrds-141909)/290 should ::        
     :: be a standard normal variable. The OPSO test takes 32 bits at ::        
     :: a time from the test file and uses a designated set of ten    ::        
     :: consecutive bits. It then restarts the file for the next de-  ::        
     :: signated 10 bits, and so on.                                  ::        
     ::                                                               ::        
     ::     OQSO means Overlapping-Quadruples-Sparse-Occupancy        ::        
     ::   The test OQSO is similar, except that it considers 4-letter ::        
     :: words from an alphabet of 32 letters, each letter determined  ::        
     :: by a designated string of 5 consecutive bits from the test    ::        
     :: file, elements of which are assumed 32-bit random integers.   ::        
     :: The mean number of missing words in a sequence of 2^21 four-  ::        
     :: letter words,  (2^21+3 "keystrokes"), is again 141909, with   ::        
     :: sigma = 295.  The mean is based on theory; sigma comes from   ::        
     :: extensive simulation.                                         ::        
     ::                                                               ::        
     ::    The DNA test considers an alphabet of 4 letters::  C,G,A,T,::        
     :: determined by two designated bits in the sequence of random   ::        
     :: integers being tested.  It considers 10-letter words, so that ::        
     :: as in OPSO and OQSO, there are 2^20 possible words, and the   ::        
     :: mean number of missing words from a string of 2^21  (over-    ::        
     :: lapping)  10-letter  words (2^21+9 "keystrokes") is 141909.   ::        
     :: The standard deviation sigma=339 was determined as for OQSO   ::        
     :: by simulation.  (Sigma for OPSO, 290, is the true value (to   ::        
     :: three places), not determined by simulation.                  ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
 OPSO test for generator t.out          
  Output: No. missing words (mw), equiv normal variate (z), p-value (p)
                                                           mw     z     p
    OPSO for t.out           using bits 23 to 32        142165   .882  .8110
    OPSO for t.out           using bits 22 to 31        142313  1.392  .9180
    OPSO for t.out           using bits 21 to 30        142009   .344  .6345
    OPSO for t.out           using bits 20 to 29        141867  -.146  .4420
    OPSO for t.out           using bits 19 to 28        141899  -.036  .4858
    OPSO for t.out           using bits 18 to 27        142232  1.113  .8671
    OPSO for t.out           using bits 17 to 26        142470  1.933  .9734
    OPSO for t.out           using bits 16 to 25        142296  1.333  .9088
    OPSO for t.out           using bits 15 to 24        141814  -.329  .3712
    OPSO for t.out           using bits 14 to 23        142097   .647  .7412
    OPSO for t.out           using bits 13 to 22        142015   .364  .6422
    OPSO for t.out           using bits 12 to 21        141706  -.701  .2416
    OPSO for t.out           using bits 11 to 20        142440  1.830  .9664
    OPSO for t.out           using bits 10 to 19        142116   .713  .7620
    OPSO for t.out           using bits  9 to 18        142015   .364  .6422
    OPSO for t.out           using bits  8 to 17        141962   .182  .5721
    OPSO for t.out           using bits  7 to 16        141695  -.739  .2299
    OPSO for t.out           using bits  6 to 15        142395  1.675  .9530
    OPSO for t.out           using bits  5 to 14        141641  -.925  .1774
    OPSO for t.out           using bits  4 to 13        141964   .189  .5748
    OPSO for t.out           using bits  3 to 12        142644  2.533  .9944
    OPSO for t.out           using bits  2 to 11        142164   .878  .8101
    OPSO for t.out           using bits  1 to 10        142315  1.399  .9191
 OQSO test for generator t.out          
  Output: No. missing words (mw), equiv normal variate (z), p-value (p)
                                                           mw     z     p
    OQSO for t.out           using bits 28 to 32        142125   .731  .7676
    OQSO for t.out           using bits 27 to 31        142017   .365  .6424
    OQSO for t.out           using bits 26 to 30        142203   .995  .8403
    OQSO for t.out           using bits 25 to 29        141630  -.947  .1719
    OQSO for t.out           using bits 24 to 28        141772  -.466  .3208
    OQSO for t.out           using bits 23 to 27        142059   .507  .6940
    OQSO for t.out           using bits 22 to 26        141584 -1.103  .1351
    OQSO for t.out           using bits 21 to 25        142465  1.884  .9702
    OQSO for t.out           using bits 20 to 24        142000   .307  .6207
    OQSO for t.out           using bits 19 to 23        141788  -.411  .3404
    OQSO for t.out           using bits 18 to 22        142001   .311  .6220
    OQSO for t.out           using bits 17 to 21        142080   .579  .7186
    OQSO for t.out           using bits 16 to 20        141956   .158  .5629
    OQSO for t.out           using bits 15 to 19        141618  -.988  .1617
    OQSO for t.out           using bits 14 to 18        142238  1.114  .8674
    OQSO for t.out           using bits 13 to 17        141576 -1.130  .1293
    OQSO for t.out           using bits 12 to 16        141506 -1.367  .0858
    OQSO for t.out           using bits 11 to 15        142000   .307  .6207
    OQSO for t.out           using bits 10 to 14        141605 -1.032  .1511
    OQSO for t.out           using bits  9 to 13        142108   .673  .7497
    OQSO for t.out           using bits  8 to 12        141570 -1.150  .1250
    OQSO for t.out           using bits  7 to 11        141782  -.432  .3330
    OQSO for t.out           using bits  6 to 10        142100   .646  .7410
    OQSO for t.out           using bits  5 to  9        141588 -1.089  .1380
    OQSO for t.out           using bits  4 to  8        141683  -.767  .2215
    OQSO for t.out           using bits  3 to  7        142034   .423  .6637
    OQSO for t.out           using bits  2 to  6        142348  1.487  .9315
    OQSO for t.out           using bits  1 to  5        141587 -1.093  .1373
  DNA test for generator t.out          
  Output: No. missing words (mw), equiv normal variate (z), p-value (p)
                                                           mw     z     p
     DNA for t.out           using bits 31 to 32        142039   .383  .6490
     DNA for t.out           using bits 30 to 31        141916   .020  .5079
     DNA for t.out           using bits 29 to 30        141793  -.343  .3657
     DNA for t.out           using bits 28 to 29        141957   .141  .5559
     DNA for t.out           using bits 27 to 28        141789  -.355  .3613
     DNA for t.out           using bits 26 to 27        142801  2.630  .9957
     DNA for t.out           using bits 25 to 26        141894  -.045  .4820
     DNA for t.out           using bits 24 to 25        142160   .739  .7702
     DNA for t.out           using bits 23 to 24        141708  -.594  .2763
     DNA for t.out           using bits 22 to 23        142427  1.527  .9366
     DNA for t.out           using bits 21 to 22        142210   .887  .8124
     DNA for t.out           using bits 20 to 21        142081   .506  .6937
     DNA for t.out           using bits 19 to 20        142503  1.751  .9600
     DNA for t.out           using bits 18 to 19        141152 -2.234  .0127
     DNA for t.out           using bits 17 to 18        141622  -.848  .1983
     DNA for t.out           using bits 16 to 17        141601  -.910  .1815
     DNA for t.out           using bits 15 to 16        141874  -.104  .4585
     DNA for t.out           using bits 14 to 15        141919   .029  .5114
     DNA for t.out           using bits 13 to 14        141376 -1.573  .0578
     DNA for t.out           using bits 12 to 13        142278  1.088  .8616
     DNA for t.out           using bits 11 to 12        142208   .881  .8109
     DNA for t.out           using bits 10 to 11        141464 -1.314  .0945
     DNA for t.out           using bits  9 to 10        141368 -1.597  .0552
     DNA for t.out           using bits  8 to  9        141609  -.886  .1878
     DNA for t.out           using bits  7 to  8        142217   .908  .8180
     DNA for t.out           using bits  6 to  7        141789  -.355  .3613
     DNA for t.out           using bits  5 to  6        141589  -.945  .1723
     DNA for t.out           using bits  4 to  5        141743  -.491  .3118
     DNA for t.out           using bits  3 to  4        142478  1.677  .9533
     DNA for t.out           using bits  2 to  3        141976   .197  .5780
     DNA for t.out           using bits  1 to  2        141978   .203  .5803

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::     This is the COUNT-THE-1's TEST on a stream of bytes.      ::        
     :: Consider the file under test as a stream of bytes (four per   ::        
     :: 32 bit integer).  Each byte can contain from 0 to 8 1's,      ::        
     :: with probabilities 1,8,28,56,70,56,28,8,1 over 256.  Now let  ::        
     :: the stream of bytes provide a string of overlapping  5-letter ::        
     :: words, each "letter" taking values A,B,C,D,E. The letters are ::        
     :: determined by the number of 1's in a byte::  0,1,or 2 yield A,::        
     :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus ::        
     :: we have a monkey at a typewriter hitting five keys with vari- ::        
     :: ous probabilities (37,56,70,56,37 over 256).  There are 5^5   ::        
     :: possible 5-letter words, and from a string of 256,000 (over-  ::        
     :: lapping) 5-letter words, counts are made on the frequencies   ::        
     :: for each word.   The quadratic form in the weak inverse of    ::        
     :: the covariance matrix of the cell counts provides a chisquare ::        
     :: test::  Q5-Q4, the difference of the naive Pearson sums of    ::        
     :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts.    ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
   Test results for t.out          
 Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000
                               chisquare  equiv normal  p-value
  Results fo COUNT-THE-1's in successive bytes:
 byte stream for t.out            2546.78       .662      .745859
 byte stream for t.out            2497.17      -.040      .484052

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::     This is the COUNT-THE-1's TEST for specific bytes.        ::        
     :: Consider the file under test as a stream of 32-bit integers.  ::        
     :: From each integer, a specific byte is chosen , say the left-  ::        
     :: most::  bits 1 to 8. Each byte can contain from 0 to 8 1's,   ::        
     :: with probabilitie 1,8,28,56,70,56,28,8,1 over 256.  Now let   ::        
     :: the specified bytes from successive integers provide a string ::        
     :: of (overlapping) 5-letter words, each "letter" taking values  ::        
     :: A,B,C,D,E. The letters are determined  by the number of 1's,  ::        
     :: in that byte::  0,1,or 2 ---> A, 3 ---> B, 4 ---> C, 5 ---> D,::        
     :: and  6,7 or 8 ---> E.  Thus we have a monkey at a typewriter  ::        
     :: hitting five keys with with various probabilities::  37,56,70,::        
     :: 56,37 over 256. There are 5^5 possible 5-letter words, and    ::        
     :: from a string of 256,000 (overlapping) 5-letter words, counts ::        
     :: are made on the frequencies for each word. The quadratic form ::        
     :: in the weak inverse of the covariance matrix of the cell      ::        
     :: counts provides a chisquare test::  Q5-Q4, the difference of  ::        
     :: the naive Pearson  sums of (OBS-EXP)^2/EXP on counts for 5-   ::        
     :: and 4-letter cell counts.                                     ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
 Chi-square with 5^5-5^4=2500 d.of f. for sample size: 256000
                      chisquare  equiv normal  p value
  Results for COUNT-THE-1's in specified bytes:
           bits  1 to  8  2647.48      2.086      .981496
           bits  2 to  9  2568.53       .969      .833753
           bits  3 to 10  2493.99      -.085      .466131
           bits  4 to 11  2417.85     -1.162      .122660
           bits  5 to 12  2538.82       .549      .708518
           bits  6 to 13  2551.62       .730      .767323
           bits  7 to 14  2526.01       .368      .643523
           bits  8 to 15  2568.30       .966      .832944
           bits  9 to 16  2509.82       .139      .555224
           bits 10 to 17  2346.08     -2.177      .014751
           bits 11 to 18  2373.24     -1.793      .036512
           bits 12 to 19  2478.16      -.309      .378734
           bits 13 to 20  2564.33       .910      .818531
           bits 14 to 21  2529.71       .420      .662840
           bits 15 to 22  2521.78       .308      .620953
           bits 16 to 23  2470.05      -.424      .335938
           bits 17 to 24  2526.81       .379      .647712
           bits 18 to 25  2460.22      -.563      .286860
           bits 19 to 26  2571.04      1.005      .842475
           bits 20 to 27  2459.63      -.571      .284044
           bits 21 to 28  2467.98      -.453      .325340
           bits 22 to 29  2530.62       .433      .667516
           bits 23 to 30  2523.79       .336      .631729
           bits 24 to 31  2397.51     -1.449      .073613
           bits 25 to 32  2452.86      -.667      .252479

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::               THIS IS A PARKING LOT TEST                      ::        
     :: In a square of side 100, randomly "park" a car---a circle of  ::        
     :: radius 1.   Then try to park a 2nd, a 3rd, and so on, each    ::        
     :: time parking "by ear".  That is, if an attempt to park a car  ::        
     :: causes a crash with one already parked, try again at a new    ::        
     :: random location. (To avoid path problems, consider parking    ::        
     :: helicopters rather than cars.)   Each attempt leads to either ::        
     :: a crash or a success, the latter followed by an increment to  ::        
     :: the list of cars already parked. If we plot n:  the number of ::        
     :: attempts, versus k::  the number successfully parked, we get a::        
     :: curve that should be similar to those provided by a perfect   ::        
     :: random number generator.  Theory for the behavior of such a   ::        
     :: random curve seems beyond reach, and as graphics displays are ::        
     :: not available for this battery of tests, a simple characteriz ::        
     :: ation of the random experiment is used: k, the number of cars ::        
     :: successfully parked after n=12,000 attempts. Simulation shows ::        
     :: that k should average 3523 with sigma 21.9 and is very close  ::        
     :: to normally distributed.  Thus (k-3523)/21.9 should be a st-  ::        
     :: andard normal variable, which, converted to a uniform varia-  ::        
     :: ble, provides input to a KSTEST based on a sample of 10.      ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
           CDPARK: result of ten tests on file t.out          
            Of 12,000 tries, the average no. of successes
                 should be 3523 with sigma=21.9
            Successes: 3562    z-score:  1.781 p-value: .962529
            Successes: 3540    z-score:   .776 p-value: .781201
            Successes: 3560    z-score:  1.689 p-value: .954438
            Successes: 3575    z-score:  2.374 p-value: .991212
            Successes: 3501    z-score: -1.005 p-value: .157553
            Successes: 3501    z-score: -1.005 p-value: .157553
            Successes: 3488    z-score: -1.598 p-value: .055002
            Successes: 3550    z-score:  1.233 p-value: .891189
            Successes: 3523    z-score:   .000 p-value: .500000
            Successes: 3542    z-score:   .868 p-value: .807188
 
           square size   avg. no.  parked   sample sigma
             100.            3534.200       28.142
            KSTEST for the above 10: p=  .935121

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::               THE MINIMUM DISTANCE TEST                       ::        
     :: It does this 100 times::   choose n=8000 random points in a   ::        
     :: square of side 10000.  Find d, the minimum distance between   ::        
     :: the (n^2-n)/2 pairs of points.  If the points are truly inde- ::        
     :: pendent uniform, then d^2, the square of the minimum distance ::        
     :: should be (very close to) exponentially distributed with mean ::        
     :: .995 .  Thus 1-exp(-d^2/.995) should be uniform on [0,1) and  ::        
     :: a KSTEST on the resulting 100 values serves as a test of uni- ::        
     :: formity for random points in the square. Test numbers=0 mod 5 ::        
     :: are printed but the KSTEST is based on the full set of 100    ::        
     :: random choices of 8000 points in the 10000x10000 square.      ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
               This is the MINIMUM DISTANCE test
              for random integers in the file t.out          
     Sample no.    d^2     avg     equiv uni            
           5     .0707    .9643     .068577
          10     .7511    .6119     .529923
          15    2.5338    .8167     .921646
          20     .3494    .8375     .296103
          25    1.0591    .7867     .655083
          30    3.4795    .8736     .969712
          35     .4952    .9051     .392085
          40    2.2841    .9539     .899296
          45    1.2260    .9627     .708344
          50    1.1236    .9358     .676721
          55     .0576    .9547     .056233
          60     .0924    .9129     .088682
          65     .2947    .9051     .256313
          70     .1465    .8684     .136905
          75    2.8941    .9078     .945453
          80     .9679    .9303     .621946
          85     .5146    .9175     .403801
          90     .5801    .9069     .441804
          95     .5782    .9197     .440728
         100    3.0909    .9323     .955241
     MINIMUM DISTANCE TEST for t.out          
          Result of KS test on 20 transformed mindist^2's:
                                  p-value= .415298

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::              THE 3DSPHERES TEST                               ::        
     :: Choose  4000 random points in a cube of edge 1000.  At each   ::        
     :: point, center a sphere large enough to reach the next closest ::        
     :: point. Then the volume of the smallest such sphere is (very   ::        
     :: close to) exponentially distributed with mean 120pi/3.  Thus  ::        
     :: the radius cubed is exponential with mean 30. (The mean is    ::        
     :: obtained by extensive simulation).  The 3DSPHERES test gener- ::        
     :: ates 4000 such spheres 20 times.  Each min radius cubed leads ::        
     :: to a uniform variable by means of 1-exp(-r^3/30.), then a     ::        
     ::  KSTEST is done on the 20 p-values.                           ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
               The 3DSPHERES test for file t.out          
 sample no:  1     r^3=  70.114     p-value= .90340
 sample no:  2     r^3=   9.634     p-value= .27468
 sample no:  3     r^3=  10.010     p-value= .28370
 sample no:  4     r^3=   5.902     p-value= .17859
 sample no:  5     r^3=  18.834     p-value= .46624
 sample no:  6     r^3=   7.989     p-value= .23378
 sample no:  7     r^3=  74.742     p-value= .91721
 sample no:  8     r^3=  20.972     p-value= .50295
 sample no:  9     r^3=   3.125     p-value= .09891
 sample no: 10     r^3=  40.976     p-value= .74484
 sample no: 11     r^3=  11.057     p-value= .30827
 sample no: 12     r^3=   9.272     p-value= .26587
 sample no: 13     r^3=  12.001     p-value= .32971
 sample no: 14     r^3=   3.836     p-value= .12003
 sample no: 15     r^3=  27.954     p-value= .60616
 sample no: 16     r^3=  19.994     p-value= .48649
 sample no: 17     r^3=  13.063     p-value= .35301
 sample no: 18     r^3=  27.773     p-value= .60378
 sample no: 19     r^3=  48.002     p-value= .79811
 sample no: 20     r^3=  74.229     p-value= .91578
  A KS test is applied to those 20 p-values.
---------------------------------------------------------
       3DSPHERES test for file t.out                p-value= .351941
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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::      This is the SQEEZE test                                  ::        
     ::  Random integers are floated to get uniforms on [0,1). Start- ::        
     ::  ing with k=2^31=2147483647, the test finds j, the number of  ::        
     ::  iterations necessary to reduce k to 1, using the reduction   ::        
     ::  k=ceiling(k*U), with U provided by floating integers from    ::        
     ::  the file being tested.  Such j's are found 100,000 times,    ::        
     ::  then counts for the number of times j was <=6,7,...,47,>=48  ::        
     ::  are used to provide a chi-square test for cell frequencies.  ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
            RESULTS OF SQUEEZE TEST FOR t.out          
         Table of standardized frequency counts
     ( (obs-exp)/sqrt(exp) )^2
        for j taking values <=6,7,8,...,47,>=48:
      .6     1.3     -.6      .3     -.2     -.2
    -1.2    -2.4     1.5    -1.5     -.8     1.9
     -.9      .6      .0     2.1     -.3      .7
    -1.3      .0      .2     -.6     -.6      .4
      .7     -.5    -1.3     -.9     1.5      .3
     -.8     -.2      .9      .0      .8      .8
     1.0      .5      .9     1.0     -.6    -1.0
    -1.1
           Chi-square with 42 degrees of freedom: 41.162
              z-score=  -.091  p-value= .492294
______________________________________________________________

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::             The  OVERLAPPING SUMS test                        ::        
     :: Integers are floated to get a sequence U(1),U(2),... of uni-  ::        
     :: form [0,1) variables.  Then overlapping sums,                 ::        
     ::   S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed.    ::        
     :: The S's are virtually normal with a certain covariance mat-   ::        
     :: rix.  A linear transformation of the S's converts them to a   ::        
     :: sequence of independent standard normals, which are converted ::        
     :: to uniform variables for a KSTEST. The  p-values from ten     ::        
     :: KSTESTs are given still another KSTEST.                       ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
                Test no.  1      p-value  .667345
                Test no.  2      p-value  .178672
                Test no.  3      p-value  .647551
                Test no.  4      p-value  .824356
                Test no.  5      p-value  .220831
                Test no.  6      p-value  .101479
                Test no.  7      p-value  .241668
                Test no.  8      p-value  .236141
                Test no.  9      p-value  .515305
                Test no. 10      p-value  .410520
   Results of the OSUM test for t.out          
        KSTEST on the above 10 p-values:  .592965

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::     This is the RUNS test.  It counts runs up, and runs down, ::        
     :: in a sequence of uniform [0,1) variables, obtained by float-  ::        
     :: ing the 32-bit integers in the specified file. This example   ::        
     :: shows how runs are counted:  .123,.357,.789,.425,.224,.416,.95::        
     :: contains an up-run of length 3, a down-run of length 2 and an ::        
     :: up-run of (at least) 2, depending on the next values.  The    ::        
     :: covariance matrices for the runs-up and runs-down are well    ::        
     :: known, leading to chisquare tests for quadratic forms in the  ::        
     :: weak inverses of the covariance matrices.  Runs are counted   ::        
     :: for sequences of length 10,000.  This is done ten times. Then ::        
     :: repeated.                                                     ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
           The RUNS test for file t.out          
     Up and down runs in a sample of 10000
_________________________________________________ 
                 Run test for t.out          :
       runs up; ks test for 10 p's: .802361
     runs down; ks test for 10 p's: .495120
                 Run test for t.out          :
       runs up; ks test for 10 p's: .641606
     runs down; ks test for 10 p's: .838058

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: This is the CRAPS TEST. It plays 200,000 games of craps, finds::        
     :: the number of wins and the number of throws necessary to end  ::        
     :: each game.  The number of wins should be (very close to) a    ::        
     :: normal with mean 200000p and variance 200000p(1-p), with      ::        
     :: p=244/495.  Throws necessary to complete the game can vary    ::        
     :: from 1 to infinity, but counts for all>21 are lumped with 21. ::        
     :: A chi-square test is made on the no.-of-throws cell counts.   ::        
     :: Each 32-bit integer from the test file provides the value for ::        
     :: the throw of a die, by floating to [0,1), multiplying by 6    ::        
     :: and taking 1 plus the integer part of the result.             ::        
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
                Results of craps test for t.out          
  No. of wins:  Observed Expected
                                98530    98585.86
                  98530= No. of wins, z-score= -.250 pvalue= .40136
   Analysis of Throws-per-Game:
 Chisq=  29.24 for 20 degrees of freedom, p=  .91692
               Throws Observed Expected  Chisq     Sum
                  1    66373    66666.7   1.294    1.294
                  2    37837    37654.3    .886    2.180
                  3    26920    26954.7    .045    2.225
                  4    19190    19313.5    .789    3.014
                  5    13901    13851.4    .177    3.191
                  6    10106     9943.5   2.654    5.846
                  7     7162     7145.0    .040    5.886
                  8     5245     5139.1   2.183    8.069
                  9     3745     3699.9    .551    8.620
                 10     2628     2666.3    .550    9.170
                 11     1938     1923.3    .112    9.282
                 12     1309     1388.7   4.579   13.861
                 13     1022     1003.7    .333   14.194
                 14      712      726.1    .275   14.469
                 15      462      525.8   7.750   22.219
                 16      403      381.2   1.253   23.471
                 17      295      276.5   1.232   24.703
                 18      201      200.8    .000   24.704
                 19      168      146.0   3.320   28.024
                 20      112      106.2    .315   28.339
                 21      271      287.1    .905   29.243
            SUMMARY  FOR t.out          
                p-value for no. of wins: .401357
                p-value for throws/game: .916923

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 Results of DIEHARD battery of tests sent to file t.out.txt      
