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 SG100.DAT      
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  1 to 24   2.164
  duplicate       number       number 
  spacings       observed     expected
        0          53.       67.668
        1         121.      135.335
        2         151.      135.335
        3          86.       90.224
        4          55.       45.112
        5          24.       18.045
  6 to INF         10.        8.282
 Chisquare with  6 d.o.f. =    11.20 p-value=  .917554
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  2 to 25   2.012
  duplicate       number       number 
  spacings       observed     expected
        0          65.       67.668
        1         129.      135.335
        2         143.      135.335
        3          91.       90.224
        4          50.       45.112
        5          16.       18.045
  6 to INF          6.        8.282
 Chisquare with  6 d.o.f. =     2.23 p-value=  .102885
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  3 to 26   1.998
  duplicate       number       number 
  spacings       observed     expected
        0          69.       67.668
        1         133.      135.335
        2         133.      135.335
        3          99.       90.224
        4          39.       45.112
        5          18.       18.045
  6 to INF          9.        8.282
 Chisquare with  6 d.o.f. =     1.85 p-value=  .067118
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  4 to 27   1.996
  duplicate       number       number 
  spacings       observed     expected
        0          63.       67.668
        1         131.      135.335
        2         146.      135.335
        3          93.       90.224
        4          48.       45.112
        5          12.       18.045
  6 to INF          7.        8.282
 Chisquare with  6 d.o.f. =     3.79 p-value=  .295589
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  5 to 28   1.912
  duplicate       number       number 
  spacings       observed     expected
        0          76.       67.668
        1         138.      135.335
        2         132.      135.335
        3          93.       90.224
        4          39.       45.112
        5          15.       18.045
  6 to INF          7.        8.282
 Chisquare with  6 d.o.f. =     2.79 p-value=  .164846
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  6 to 29   1.942
  duplicate       number       number 
  spacings       observed     expected
        0          70.       67.668
        1         142.      135.335
        2         139.      135.335
        3          86.       90.224
        4          36.       45.112
        5          18.       18.045
  6 to INF          9.        8.282
 Chisquare with  6 d.o.f. =     2.61 p-value=  .143850
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  7 to 30   1.954
  duplicate       number       number 
  spacings       observed     expected
        0          78.       67.668
        1         127.      135.335
        2         140.      135.335
        3          79.       90.224
        4          52.       45.112
        5          19.       18.045
  6 to INF          5.        8.282
 Chisquare with  6 d.o.f. =     6.05 p-value=  .582480
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  8 to 31   1.886
  duplicate       number       number 
  spacings       observed     expected
        0          83.       67.668
        1         138.      135.335
        2         138.      135.335
        3          68.       90.224
        4          50.       45.112
        5          15.       18.045
  6 to INF          8.        8.282
 Chisquare with  6 d.o.f. =    10.11 p-value=  .879742
  :::::::::::::::::::::::::::::::::::::::::
                   For a sample of size 500:     mean   
           SG100.DAT       using bits  9 to 32   2.028
  duplicate       number       number 
  spacings       observed     expected
        0          62.       67.668
        1         137.      135.335
        2         143.      135.335
        3          81.       90.224
        4          48.       45.112
        5          19.       18.045
  6 to INF         10.        8.282
 Chisquare with  6 d.o.f. =     2.46 p-value=  .127542
  :::::::::::::::::::::::::::::::::::::::::
   The 9 p-values were
        .917554   .102885   .067118   .295589   .164846
        .143850   .582480   .879742   .127542
  A KSTEST for the 9 p-values yields  .878397

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

     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::            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 SG100.DAT      
     For a sample of 1,000,000 consecutive 5-tuples,
 chisquare for 99 degrees of freedom=101.603; p-value= .591239
           OPERM5 test for file SG100.DAT      
     For a sample of 1,000,000 consecutive 5-tuples,
 chisquare for 99 degrees of freedom=105.512; p-value= .691597
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: 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 SG100.DAT      
         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       229     211.4  1.462156    1.462
        29      5116    5134.0   .063180    1.525
        30     23100   23103.0   .000402    1.526
        31     11555   11551.5   .001046    1.527
  chisquare= 1.527 for 3 d. of f.; p-value= .432196
--------------------------------------------------------------
     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: 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 SG100.DAT      
         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       200     211.4   .616651     .617
        30      5211    5134.0  1.154540    1.771
        31     22819   23103.0  3.492294    5.263
        32     11770   11551.5  4.132059    9.396
  chisquare= 9.396 for 3 d. of f.; p-value= .976619
--------------------------------------------------------------

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

     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     :: 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 SG100.DAT      
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  1 to  8
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          899       944.3       2.173       2.173
          r =5        21504     21743.9       2.647       4.820
          r =6        77597     77311.8       1.052       5.872
                        p=1-exp(-SUM/2)= .94693
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  2 to  9
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          916       944.3        .848        .848
          r =5        21545     21743.9       1.819       2.668
          r =6        77539     77311.8        .668       3.335
                        p=1-exp(-SUM/2)= .81131
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  3 to 10
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          969       944.3        .646        .646
          r =5        21433     21743.9       4.445       5.091
          r =6        77598     77311.8       1.059       6.151
                        p=1-exp(-SUM/2)= .95383
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  4 to 11
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          933       944.3        .135        .135
          r =5        21743     21743.9        .000        .135
          r =6        77324     77311.8        .002        .137
                        p=1-exp(-SUM/2)= .06631
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  5 to 12
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          988       944.3       2.022       2.022
          r =5        21677     21743.9        .206       2.228
          r =6        77335     77311.8        .007       2.235
                        p=1-exp(-SUM/2)= .67290
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  6 to 13
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          958       944.3        .199        .199
          r =5        21573     21743.9       1.343       1.542
          r =6        77469     77311.8        .320       1.862
                        p=1-exp(-SUM/2)= .60576
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  7 to 14
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          931       944.3        .187        .187
          r =5        21732     21743.9        .007        .194
          r =6        77337     77311.8        .008        .202
                        p=1-exp(-SUM/2)= .09610
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  8 to 15
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          948       944.3        .014        .014
          r =5        21946     21743.9       1.878       1.893
          r =6        77106     77311.8        .548       2.441
                        p=1-exp(-SUM/2)= .70488
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits  9 to 16
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          945       944.3        .001        .001
          r =5        21729     21743.9        .010        .011
          r =6        77326     77311.8        .003        .013
                        p=1-exp(-SUM/2)= .00664
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 10 to 17
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          899       944.3       2.173       2.173
          r =5        21656     21743.9        .355       2.529
          r =6        77445     77311.8        .229       2.758
                        p=1-exp(-SUM/2)= .74818
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 11 to 18
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          905       944.3       1.636       1.636
          r =5        21747     21743.9        .000       1.636
          r =6        77348     77311.8        .017       1.653
                        p=1-exp(-SUM/2)= .56244
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 12 to 19
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          941       944.3        .012        .012
          r =5        21834     21743.9        .373        .385
          r =6        77225     77311.8        .097        .482
                        p=1-exp(-SUM/2)= .21429
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 13 to 20
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          930       944.3        .217        .217
          r =5        21780     21743.9        .060        .277
          r =6        77290     77311.8        .006        .283
                        p=1-exp(-SUM/2)= .13180
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 14 to 21
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          915       944.3        .909        .909
          r =5        21836     21743.9        .390       1.299
          r =6        77249     77311.8        .051       1.350
                        p=1-exp(-SUM/2)= .49093
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 15 to 22
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          931       944.3        .187        .187
          r =5        21872     21743.9        .755        .942
          r =6        77197     77311.8        .170       1.113
                        p=1-exp(-SUM/2)= .42665
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 16 to 23
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          942       944.3        .006        .006
          r =5        21776     21743.9        .047        .053
          r =6        77282     77311.8        .011        .064
                        p=1-exp(-SUM/2)= .03173
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 17 to 24
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          993       944.3       2.511       2.511
          r =5        21686     21743.9        .154       2.666
          r =6        77321     77311.8        .001       2.667
                        p=1-exp(-SUM/2)= .73641
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 18 to 25
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          960       944.3        .261        .261
          r =5        21597     21743.9        .992       1.253
          r =6        77443     77311.8        .223       1.476
                        p=1-exp(-SUM/2)= .52195
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 19 to 26
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          952       944.3        .063        .063
          r =5        21893     21743.9       1.022       1.085
          r =6        77155     77311.8        .318       1.403
                        p=1-exp(-SUM/2)= .50421
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 20 to 27
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          959       944.3        .229        .229
          r =5        21661     21743.9        .316        .545
          r =6        77380     77311.8        .060        .605
                        p=1-exp(-SUM/2)= .26104
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 21 to 28
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          911       944.3       1.174       1.174
          r =5        21646     21743.9        .441       1.615
          r =6        77443     77311.8        .223       1.838
                        p=1-exp(-SUM/2)= .60104
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 22 to 29
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          941       944.3        .012        .012
          r =5        21645     21743.9        .450        .461
          r =6        77414     77311.8        .135        .596
                        p=1-exp(-SUM/2)= .25787
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 23 to 30
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          968       944.3        .595        .595
          r =5        21827     21743.9        .318        .912
          r =6        77205     77311.8        .148       1.060
                        p=1-exp(-SUM/2)= .41136
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 24 to 31
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          964       944.3        .411        .411
          r =5        21643     21743.9        .468        .879
          r =6        77393     77311.8        .085        .964
                        p=1-exp(-SUM/2)= .38258
        Rank of a 6x8 binary matrix,
     rows formed from eight bits of the RNG SG100.DAT      
     b-rank test for bits 25 to 32
                     OBSERVED   EXPECTED     (O-E)^2/E      SUM
          r<=4          954       944.3        .100        .100
          r =5        21674     21743.9        .225        .324
          r =6        77372     77311.8        .047        .371
                        p=1-exp(-SUM/2)= .16939
   TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
 These should be 25 uniform [0,1] random variables:
     .946926     .811309     .953829     .066306     .672905
     .605756     .096105     .704883     .006645     .748178
     .562440     .214295     .131803     .490927     .426648
     .031729     .736410     .521948     .504205     .261037
     .601044     .257873     .411363     .382583     .169391
   brank test summary for SG100.DAT      
       The KS test for those 25 supposed UNI's yields
                    KS p-value= .286560

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     :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::        
     ::                   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:  142407 missing words,    1.16 sigmas from mean, p-value= .87754
 tst no  2:  141541 missing words,    -.86 sigmas from mean, p-value= .19473
 tst no  3:  141911 missing words,     .00 sigmas from mean, p-value= .50156
 tst no  4:  141300 missing words,   -1.42 sigmas from mean, p-value= .07727
 tst no  5:  142023 missing words,     .27 sigmas from mean, p-value= .60472
 tst no  6:  141841 missing words,    -.16 sigmas from mean, p-value= .43658
 tst no  7:  142325 missing words,     .97 sigmas from mean, p-value= .83427
 tst no  8:  141689 missing words,    -.51 sigmas from mean, p-value= .30335
 tst no  9:  142087 missing words,     .42 sigmas from mean, p-value= .66097
 tst no 10:  142340 missing words,    1.01 sigmas from mean, p-value= .84285
 tst no 11:  141340 missing words,   -1.33 sigmas from mean, p-value= .09173
 tst no 12:  142234 missing words,     .76 sigmas from mean, p-value= .77595
 tst no 13:  141629 missing words,    -.65 sigmas from mean, p-value= .25624
 tst no 14:  141704 missing words,    -.48 sigmas from mean, p-value= .31571
 tst no 15:  141934 missing words,     .06 sigmas from mean, p-value= .52298
 tst no 16:  142176 missing words,     .62 sigmas from mean, p-value= .73338
 tst no 17:  142170 missing words,     .61 sigmas from mean, p-value= .72875
 tst no 18:  141332 missing words,   -1.35 sigmas from mean, p-value= .08868
 tst no 19:  142431 missing words,    1.22 sigmas from mean, p-value= .88855
 tst no 20:  141749 missing words,    -.37 sigmas from mean, p-value= .35398

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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 SG100.DAT      
  Output: No. missing words (mw), equiv normal variate (z), p-value (p)
                                                           mw     z     p
    OPSO for SG100.DAT       using bits 23 to 32        142035   .433  .6676
    OPSO for SG100.DAT       using bits 22 to 31        142206  1.023  .8468
    OPSO for SG100.DAT       using bits 21 to 30        142083   .599  .7254
    OPSO for SG100.DAT       using bits 20 to 29        141822  -.301  .3817
    OPSO for SG100.DAT       using bits 19 to 28        142119   .723  .7652
    OPSO for SG100.DAT       using bits 18 to 27        141808  -.349  .3634
    OPSO for SG100.DAT       using bits 17 to 26        142236  1.126  .8700
    OPSO for SG100.DAT       using bits 16 to 25        142060   .520  .6983
    OPSO for SG100.DAT       using bits 15 to 24        141929   .068  .5270
    OPSO for SG100.DAT       using bits 14 to 23        141975   .226  .5896
    OPSO for SG100.DAT       using bits 13 to 22        142114   .706  .7598
    OPSO for SG100.DAT       using bits 12 to 21        141535 -1.291  .0984
    OPSO for SG100.DAT       using bits 11 to 20        141747  -.560  .2878
    OPSO for SG100.DAT       using bits 10 to 19        142167   .889  .8129
    OPSO for SG100.DAT       using bits  9 to 18        141718  -.660  .2547
    OPSO for SG100.DAT       using bits  8 to 17        142209  1.033  .8493
    OPSO for SG100.DAT       using bits  7 to 16        141934   .085  .5339
    OPSO for SG100.DAT       using bits  6 to 15        141859  -.174  .4311
    OPSO for SG100.DAT       using bits  5 to 14        141410 -1.722  .0426
    OPSO for SG100.DAT       using bits  4 to 13        142222  1.078  .8595
    OPSO for SG100.DAT       using bits  3 to 12        141659  -.863  .1940
    OPSO for SG100.DAT       using bits  2 to 11        142047   .475  .6825
    OPSO for SG100.DAT       using bits  1 to 10        142193   .978  .8360
 OQSO test for generator SG100.DAT      
  Output: No. missing words (mw), equiv normal variate (z), p-value (p)
                                                           mw     z     p
    OQSO for SG100.DAT       using bits 28 to 32        142465  1.884  .9702
    OQSO for SG100.DAT       using bits 27 to 31        141617  -.991  .1609
    OQSO for SG100.DAT       using bits 26 to 30        141894  -.052  .4793
    OQSO for SG100.DAT       using bits 25 to 29        141470 -1.489  .0682
    OQSO for SG100.DAT       using bits 24 to 28        141928   .063  .5252
    OQSO for SG100.DAT       using bits 23 to 27        142267  1.212  .8873
    OQSO for SG100.DAT       using bits 22 to 26        141749  -.543  .2934
    OQSO for SG100.DAT       using bits 21 to 25        141473 -1.479  .0696
    OQSO for SG100.DAT       using bits 20 to 24        142092   .619  .7321
    OQSO for SG100.DAT       using bits 19 to 23        142120   .714  .7624
    OQSO for SG100.DAT       using bits 18 to 22        142048   .470  .6808
    OQSO for SG100.DAT       using bits 17 to 21        141858  -.174  .4309
    OQSO for SG100.DAT       using bits 16 to 20        142111   .684  .7529
    OQSO for SG100.DAT       using bits 15 to 19        141522 -1.313  .0946
    OQSO for SG100.DAT       using bits 14 to 18        142184   .931  .8241
    OQSO for SG100.DAT       using bits 13 to 17        141730  -.608  .2716
    OQSO for SG100.DAT       using bits 12 to 16        141629  -.950  .1710
    OQSO for SG100.DAT       using bits 11 to 15        141991   .277  .6091
    OQSO for SG100.DAT       using bits 10 to 14        141792  -.398  .3454
    OQSO for SG100.DAT       using bits  9 to 13        141863  -.157  .4376
    OQSO for SG100.DAT       using bits  8 to 12        141447 -1.567  .0585
    OQSO for SG100.DAT       using bits  7 to 11        141956   .158  .5629
    OQSO for SG100.DAT       using bits  6 to 10        141929   .067  .5266
    OQSO for SG100.DAT       using bits  5 to  9        142342  1.467  .9288
    OQSO for SG100.DAT       using bits  4 to  8        142129   .745  .7718
    OQSO for SG100.DAT       using bits  3 to  7        142006   .328  .6284
    OQSO for SG100.DAT       using bits  2 to  6        141572 -1.143  .1264
    OQSO for SG100.DAT       using bits  1 to  5        141426 -1.638  .0507
  DNA test for generator SG100.DAT      
  Output: No. missing words (mw), equiv normal variate (z), p-value (p)
                                                           mw     z     p
     DNA for SG100.DAT       using bits 31 to 32        141565 -1.016  .1549
     DNA for SG100.DAT       using bits 30 to 31        141623  -.845  .1992
     DNA for SG100.DAT       using bits 29 to 30        141319 -1.741  .0408
     DNA for SG100.DAT       using bits 28 to 29        141545 -1.075  .1413
     DNA for SG100.DAT       using bits 27 to 28        141534 -1.107  .1341
     DNA for SG100.DAT       using bits 26 to 27        141978   .203  .5803
     DNA for SG100.DAT       using bits 25 to 26        142095   .548  .7081
     DNA for SG100.DAT       using bits 24 to 25        141437 -1.393  .0818
     DNA for SG100.DAT       using bits 23 to 24        141847  -.184  .4271
     DNA for SG100.DAT       using bits 22 to 23        141959   .147  .5582
     DNA for SG100.DAT       using bits 21 to 22        141939   .088  .5349
     DNA for SG100.DAT       using bits 20 to 21        141604  -.901  .1839
     DNA for SG100.DAT       using bits 19 to 20        142019   .324  .6268
     DNA for SG100.DAT       using bits 18 to 19        141617  -.862  .1943
     DNA for SG100.DAT       using bits 17 to 18        142339  1.267  .8975
     DNA for SG100.DAT       using bits 16 to 17        142107   .583  .7201
     DNA for SG100.DAT       using bits 15 to 16        142363  1.338  .9096
     DNA for SG100.DAT       using bits 14 to 15        141648  -.771  .2204
     DNA for SG100.DAT       using bits 13 to 14        142102   .568  .7151
     DNA for SG100.DAT       using bits 12 to 13        141713  -.579  .2812
     DNA for SG100.DAT       using bits 11 to 12        142524  1.813  .9651
     DNA for SG100.DAT       using bits 10 to 11        141723  -.550  .2913
     DNA for SG100.DAT       using bits  9 to 10        142046   .403  .6566
     DNA for SG100.DAT       using bits  8 to  9        142373  1.368  .9143
     DNA for SG100.DAT       using bits  7 to  8        141894  -.045  .4820
     DNA for SG100.DAT       using bits  6 to  7        141222 -2.028  .0213
     DNA for SG100.DAT       using bits  5 to  6        141682  -.671  .2512
     DNA for SG100.DAT       using bits  4 to  5        142131   .654  .7434
     DNA for SG100.DAT       using bits  3 to  4        142307  1.173  .8796
     DNA for SG100.DAT       using bits  2 to  3        142155   .725  .7657
     DNA for SG100.DAT       using bits  1 to  2        142025   .341  .6335

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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 SG100.DAT      
 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 SG100.DAT        2489.05      -.155      .438457
 byte stream for SG100.DAT        2559.93       .848      .801665

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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  2413.74     -1.220      .111244
           bits  2 to  9  2548.37       .684      .753024
           bits  3 to 10  2537.38       .529      .701486
           bits  4 to 11  2428.28     -1.014      .155212
           bits  5 to 12  2636.03      1.924      .972811
           bits  6 to 13  2464.13      -.507      .305996
           bits  7 to 14  2456.64      -.613      .269886
           bits  8 to 15  2589.84      1.271      .898054
           bits  9 to 16  2511.27       .159      .563300
           bits 10 to 17  2430.62      -.981      .163254
           bits 11 to 18  2527.38       .387      .650707
           bits 12 to 19  2541.18       .582      .719840
           bits 13 to 20  2461.69      -.542      .293971
           bits 14 to 21  2330.63     -2.395      .008304
           bits 15 to 22  2507.48       .106      .542138
           bits 16 to 23  2522.38       .316      .624174
           bits 17 to 24  2468.29      -.448      .326932
           bits 18 to 25  2435.00      -.919      .178980
           bits 19 to 26  2491.51      -.120      .452231
           bits 20 to 27  2482.81      -.243      .403953
           bits 21 to 28  2637.01      1.938      .973668
           bits 22 to 29  2466.95      -.467      .320100
           bits 23 to 30  2444.05      -.791      .214394
           bits 24 to 31  2584.92      1.201      .885126
           bits 25 to 32  2577.95      1.102      .864839

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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 SG100.DAT      
            Of 12,000 tries, the average no. of successes
                 should be 3523 with sigma=21.9
            Successes: 3557    z-score:  1.553 p-value: .939730
            Successes: 3550    z-score:  1.233 p-value: .891189
            Successes: 3506    z-score:  -.776 p-value: .218799
            Successes: 3502    z-score:  -.959 p-value: .168804
            Successes: 3556    z-score:  1.507 p-value: .934075
            Successes: 3523    z-score:   .000 p-value: .500000
            Successes: 3517    z-score:  -.274 p-value: .392053
            Successes: 3539    z-score:   .731 p-value: .767486
            Successes: 3557    z-score:  1.553 p-value: .939730
            Successes: 3549    z-score:  1.187 p-value: .882429
 
           square size   avg. no.  parked   sample sigma
             100.            3535.600       20.587
            KSTEST for the above 10: p=  .945715

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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 SG100.DAT      
     Sample no.    d^2     avg     equiv uni            
           5     .3073    .4196     .265694
          10     .6224    .6057     .465034
          15     .0719    .7005     .069731
          20     .0141    .7617     .014051
          25    2.5423    .9465     .922313
          30    1.0586    .8824     .654903
          35     .8508    .8214     .574768
          40     .2105    .7611     .190671
          45     .7400    .7450     .524675
          50    1.3736    .7899     .748560
          55     .5317    .8405     .413977
          60     .8544    .8232     .576305
          65     .5630    .8249     .432100
          70     .0735    .8211     .071233
          75    4.2369    .8391     .985852
          80     .5104    .8073     .401262
          85    1.2002    .8116     .700678
          90     .3804    .8260     .317741
          95    1.0486    .8048     .651408
         100    1.5839    .8236     .796456
     MINIMUM DISTANCE TEST for SG100.DAT      
          Result of KS test on 20 transformed mindist^2's:
                                  p-value= .752837

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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 SG100.DAT      
 sample no:  1     r^3=  41.137     p-value= .74621
 sample no:  2     r^3=   7.369     p-value= .21780
 sample no:  3     r^3=  24.243     p-value= .55430
 sample no:  4     r^3=  35.577     p-value= .69453
 sample no:  5     r^3=  24.337     p-value= .55569
 sample no:  6     r^3=   2.268     p-value= .07281
 sample no:  7     r^3=   3.260     p-value= .10298
 sample no:  8     r^3=  47.719     p-value= .79620
 sample no:  9     r^3=  17.620     p-value= .44420
 sample no: 10     r^3=    .868     p-value= .02851
 sample no: 11     r^3=  23.506     p-value= .54320
 sample no: 12     r^3=  23.901     p-value= .54918
 sample no: 13     r^3=  48.215     p-value= .79954
 sample no: 14     r^3= 119.179     p-value= .98118
 sample no: 15     r^3=  31.909     p-value= .65480
 sample no: 16     r^3=  29.470     p-value= .62557
 sample no: 17     r^3=  34.265     p-value= .68088
 sample no: 18     r^3=  11.700     p-value= .32294
 sample no: 19     r^3=  14.170     p-value= .37645
 sample no: 20     r^3=  73.255     p-value= .91300
  A KS test is applied to those 20 p-values.
---------------------------------------------------------
       3DSPHERES test for file SG100.DAT            p-value= .243269
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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 SG100.DAT      
         Table of standardized frequency counts
     ( (obs-exp)/sqrt(exp) )^2
        for j taking values <=6,7,8,...,47,>=48:
      .6      .1    -1.3     1.1     1.4     1.1
      .2     1.4      .2     -.2      .9     -.1
     -.4      .7     1.6     1.7    -1.9     -.8
     -.7     -.4      .3     -.3     -.1    -1.5
      .4      .9     -.8     -.4     -.2      .4
     -.9     -.1     -.1     -.9     -.6    -1.0
      .5      .2      .5      .4     -.6    -1.0
      .8
           Chi-square with 42 degrees of freedom: 30.647
              z-score= -1.239  p-value= .097122
______________________________________________________________

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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  .330733
                Test no.  2      p-value  .232319
                Test no.  3      p-value  .893748
                Test no.  4      p-value  .057761
                Test no.  5      p-value  .662586
                Test no.  6      p-value  .447470
                Test no.  7      p-value  .916826
                Test no.  8      p-value  .584608
                Test no.  9      p-value  .919413
                Test no. 10      p-value  .258387
   Results of the OSUM test for SG100.DAT      
        KSTEST on the above 10 p-values:  .107964

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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 SG100.DAT      
     Up and down runs in a sample of 10000
_________________________________________________ 
                 Run test for SG100.DAT      :
       runs up; ks test for 10 p's: .508297
     runs down; ks test for 10 p's: .373533
                 Run test for SG100.DAT      :
       runs up; ks test for 10 p's: .188019
     runs down; ks test for 10 p's: .230541

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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 SG100.DAT      
  No. of wins:  Observed Expected
                                98211    98585.86
                  98211= No. of wins, z-score=-1.677 pvalue= .04681
   Analysis of Throws-per-Game:
 Chisq=  22.20 for 20 degrees of freedom, p=  .67012
               Throws Observed Expected  Chisq     Sum
                  1    66367    66666.7   1.347    1.347
                  2    37696    37654.3    .046    1.393
                  3    27053    26954.7    .358    1.751
                  4    19610    19313.5   4.553    6.305
                  5    13787    13851.4    .300    6.604
                  6     9847     9943.5    .937    7.541
                  7     7198     7145.0    .393    7.934
                  8     5151     5139.1    .028    7.962
                  9     3723     3699.9    .145    8.107
                 10     2677     2666.3    .043    8.150
                 11     1880     1923.3    .976    9.126
                 12     1338     1388.7   1.854   10.980
                 13      997     1003.7    .045   11.024
                 14      772      726.1   2.896   13.921
                 15      530      525.8    .033   13.954
                 16      391      381.2    .255   14.208
                 17      256      276.5   1.526   15.734
                 18      198      200.8    .040   15.774
                 19      160      146.0   1.346   17.119
                 20       83      106.2   5.074   22.193
                 21      286      287.1    .004   22.198
            SUMMARY  FOR SG100.DAT      
                p-value for no. of wins: .046811
                p-value for throws/game: .670116

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 Results of DIEHARD battery of tests sent to file SG100.TXT