# Hardware RNG

Creating truly random numbers is no easy task. Doing it in digital hardware is even harder. Generally the best you can do is generate pseudo random numbers. Luckily there are a few ways to accomplish this. This page doesn't explain any theory, or claim to be the best RNG, but it's a starting point for anyone who needs a quick hardware RNG solution.

## The Pseudo Random Number Generator

The pseudo random number generator used here is based upon a Cellular Automata. Take a look at the Wikipedia Entry for details. The CA rules and structure in this case were developed at the HP labs. Below are the details of the paper describing describing the CA.

FPGA Implementation of Neighborhood-of-Four Cellular Automata Random Number Generators

, 10th International Symposium on Field Programmable Gate Arrays, 2002, Monterey, California, USA

The CA is an 8x8 2D grid of cells, each connected to 4 neighbours. The state of each cell is updated every clock cycle with a value that is a function of the states of these neighbours.

## The VHDL

Below is the VHDL that describes the RNG. To initialise the RNG the reset
input has to be pulled high. Then the output, `dOut`

, contains the random numbers.

`RandomNumberGenerator.vhdl`

library ieee; use ieee.std_logic_1164.all; use ieee.numeric_std.all; entity RandomNumberGenerator is port ( clk : in std_logic; enable : in std_logic; reset : in std_logic; dOut : out std_logic_vector(63 downto 0)); end RandomNumberGenerator; architecture General of RandomNumberGenerator is -- Declare types subtype CELL_ROW is std_logic_vector(0 to 7); type CELL_SQUARE is array (0 to 7) of CELL_ROW; -- Declare cell array signal cells : CELL_SQUARE := ((others => '0'),(others => '0'), (others => '0'),(others => '0'), (others => '0'),(others => '0'), (others => '0'),(others => '0')); -- Declare function to calculate cell next state function getCellOutput( cell3 : std_logic; cell2 : std_logic; cell1 : std_logic; cell0 : std_logic) return std_logic is variable output : std_logic := '0'; variable input : std_logic_vector(3 downto 0) := (others => '0'); begin input := cell3 & cell2 & cell1 & cell0; -- CA 27225 -- h6A59 -- b0110 1010 0101 1001 case input is when b"0000" => output := '1'; when b"0001" => output := '0'; when b"0010" => output := '0'; when b"0011" => output := '1'; when b"0100" => output := '1'; when b"0101" => output := '0'; when b"0110" => output := '1'; when b"0111" => output := '0'; when b"1000" => output := '0'; when b"1001" => output := '1'; when b"1010" => output := '0'; when b"1011" => output := '1'; when b"1100" => output := '0'; when b"1101" => output := '1'; when b"1110" => output := '1'; when b"1111" => output := '0'; when others => output := '0'; end case; return output; end function; begin -- Convert the cell array into a linear output OutputConnect_row : for row in 0 to 7 generate begin OutputConnect_col : for col in 0 to 7 generate constant cellNum : natural := (row * 8) + col; begin dOut(cellNum) <= cells(row)(col); end generate; end generate; -- Connect the cell array Connect_row : for row in 0 to 7 generate constant cell0row : natural := (row-2) mod 8; -- 2n constant cell1row : natural := row; -- c constant cell2row : natural := (row-1) mod 8; -- n constant cell3row : natural := (row+2) mod 8; -- 2s begin Connect_col : for col in 0 to 7 generate constant cell0col : natural := (col-2) mod 8; -- 2w constant cell1col : natural := col; -- c constant cell2col : natural := (col+2) mod 8; -- 2e constant cell3col : natural := (col+1) mod 8; -- e begin CellUpdate : process(clk) variable cell0 : std_logic; variable cell1 : std_logic; variable cell2 : std_logic; variable cell3 : std_logic; begin if (clk'event and clk='1') then if (reset='1') then if (row=7 and col=7) then cells(row)(col) <= '1'; -- Reset cell 7,7 to 1 else cells(row)(col) <= '0'; -- Reset all other cells to '0' end if; elsif (enable='1') then cell3 := cells(cell3row)(cell3col); cell2 := cells(cell2row)(cell2col); cell1 := cells(cell1row)(cell1col); cell0 := cells(cell0row)(cell0col); cells(row)(col) <= getCellOutput(cell3, cell2, cell1, cell0); end if; end if; end process; end generate; end generate; end General;

## Testing the RNG

ModelSim can be used to test the RNG. Below is a testbench file for the RNG. This instantiates a single
`RandomNumberGenerator`

, performs a reset, and lets the RNG run.

`RandomNumberGenerator_TB.vhdl`

library ieee; use ieee.std_logic_1164.all; use ieee.numeric_std.all; entity RandomNumberGenerator_TB is end RandomNumberGenerator_TB; architecture General of RandomNumberGenerator_TB is signal clk : std_logic; signal enable : std_logic := '0'; signal reset : std_logic := '0'; signal dOut : std_logic_vector(63 downto 0); constant INT_DELAY : time := 1 ns; component RandomNumberGenerator is port ( clk : in std_logic; enable : in std_logic; reset : in std_logic; dOut : out std_logic_vector(63 downto 0)); end component; begin RNG : RandomNumberGenerator port map ( clk => clk, enable => enable, reset => reset, dOut => dOut); TestClk : process variable clkVar : std_logic := '0'; begin clk <= clkVar; clkVar := not clkVar; wait for INT_DELAY; end process; TB : process begin reset <= '1'; enable <= '0'; wait until clk='1'; reset <= '0'; enable <= '1'; wait; end process;

The do file below will perform the ModelSim test and also generate a file
containing the RNG output. The file `rng.lst`

can become very large if
you run the simulation for a long time. If you're going to run a very long
simulation, it is worth commenting out the `view`

and `add`

wave
commands

`RandomNumberGenerator_TB.do`

vlib work vcom -93 RandomNumberGenerator.vhdl vcom -93 RandomNumberGenerator_TB.vhdl vsim -t 100ps -lib work RandomNumberGenerator_TB view wave -undock add wave -height 25 -color red clk add wave -height 25 -color red reset add wave -height 25 -color pink enable add wave -height 25 -hex -color pink RNG/cells add wave -height 25 -color pink dOut add list -hex dOut configure list -delta none run 5 us write list rng.lst

## Checking for Randomness

The DIEHARD test suite can be used to test the generator's output for randomness. These aren't the easiest tools to use, but they seem to be the most recommended.

Download and decompress the C Source Code tar. I've had more success with
`source.tar.gz`

, rather than `die-c.tar.gz`

. You'll need to
compile the `asc2bin.c`

and the `diehard.c`

files. You'll need to
have the Fortran to c (f2c) libraries installed.

> gcc -o asc2bin asc2bin.c -lf2c -lm > gcc -o diehard diehard.c -lf2c -lm

Before you can put DIEHARD to use, you need to convert the output from
ModelSim into a different format. The one-liner below will do this. First, `awk`

is used to pull out the 2nd column of data (the random numbers) and ignore the
first 10 rows (whilst the RNG settles). Next, `tr`

removes the `EOL`

characters.
Finally, fold cuts everything into 80 character long lines. Everything is output
to `rng.data`

> awk '{if (NR>10) {print $2}}' rng.lst | tr -d '\n' | fold -w 80 > rng.data

Use the freshly compiled `asc2bin`

program to convert the text file,
`rng.data`

into a binary format understood by the Diehard program.

```
> ./asc2bin
```

Finally, run the Diehard program.

```
> ./diehard
```

## Randomness Results

The results below are for a 25ms simulation. DIEHARD was run using all the 15 tests available. Interpreting the result is even harder than designing a RNG! But, as far as I can tell if the p-values are evenly distributed, and neither 0 or 1, then the RNG is doing a good job. If anyone can provide further details on the reading/understanding of DIEHARD results, I'll be very happy to hear from them.

`rng.out`

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 rng.bin For a sample of size 500: mean rng.bin using bits 1 to 24 2.146 duplicate number number spacings observed expected 0 57. 67.668 1 113. 135.335 2 147. 135.335 3 104. 90.224 4 50. 45.112 5 22. 18.045 6 to INF 7. 8.282 Chisquare with 6 d.o.f. = 10.07 p-value= .878343 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 2 to 25 2.004 duplicate number number spacings observed expected 0 68. 67.668 1 123. 135.335 2 149. 135.335 3 93. 90.224 4 45. 45.112 5 14. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 3.51 p-value= .257040 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 3 to 26 1.900 duplicate number number spacings observed expected 0 73. 67.668 1 136. 135.335 2 142. 135.335 3 91. 90.224 4 39. 45.112 5 14. 18.045 6 to INF 5. 8.282 Chisquare with 6 d.o.f. = 3.79 p-value= .295399 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 4 to 27 1.880 duplicate number number spacings observed expected 0 68. 67.668 1 167. 135.335 2 119. 135.335 3 83. 90.224 4 37. 45.112 5 20. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 12.26 p-value= .943576 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 5 to 28 1.878 duplicate number number spacings observed expected 0 77. 67.668 1 155. 135.335 2 119. 135.335 3 82. 90.224 4 44. 45.112 5 15. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 7.42 p-value= .715952 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 6 to 29 2.016 duplicate number number spacings observed expected 0 63. 67.668 1 135. 135.335 2 136. 135.335 3 93. 90.224 4 50. 45.112 5 17. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 1.63 p-value= .049630 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 7 to 30 2.042 duplicate number number spacings observed expected 0 60. 67.668 1 127. 135.335 2 159. 135.335 3 79. 90.224 4 47. 45.112 5 20. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 7.22 p-value= .698746 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 8 to 31 2.032 duplicate number number spacings observed expected 0 66. 67.668 1 142. 135.335 2 126. 135.335 3 80. 90.224 4 56. 45.112 5 24. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 7.39 p-value= .714044 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean rng.bin using bits 9 to 32 1.982 duplicate number number spacings observed expected 0 71. 67.668 1 133. 135.335 2 137. 135.335 3 85. 90.224 4 49. 45.112 5 17. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = .93 p-value= .011958 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .878343 .257040 .295399 .943576 .715952 .049630 .698746 .714044 .011958 A KSTEST for the 9 p-values yields .383382 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 rng.bin For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 97.547; p-value= .477514 OPERM5 test for file rng.bin For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 99.109; p-value= .521976 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 rng.bin Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 218 211.4 .204914 .205 29 5164 5134.0 .175182 .380 30 23070 23103.0 .047271 .427 31 11548 11551.5 .001075 .428 chisquare= .428 for 3 d. of f.; p-value= .322071 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 rng.bin Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 207 211.4 .092324 .092 30 5171 5134.0 .266505 .359 31 22969 23103.0 .777757 1.137 32 11653 11551.5 .891423 2.028 chisquare= 2.028 for 3 d. of f.; p-value= .510598 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 rng.bin Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 984 944.3 1.669 1.669 r =5 21889 21743.9 .968 2.637 r =6 77127 77311.8 .442 3.079 p=1-exp(-SUM/2)= .78551 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 984 944.3 1.669 1.669 r =5 21561 21743.9 1.538 3.207 r =6 77455 77311.8 .265 3.473 p=1-exp(-SUM/2)= .82383 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 921 944.3 .575 .575 r =5 21732 21743.9 .007 .581 r =6 77347 77311.8 .016 .598 p=1-exp(-SUM/2)= .25826 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 943 944.3 .002 .002 r =5 21725 21743.9 .016 .018 r =6 77332 77311.8 .005 .023 p=1-exp(-SUM/2)= .01168 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 956 944.3 .145 .145 r =5 21727 21743.9 .013 .158 r =6 77317 77311.8 .000 .158 p=1-exp(-SUM/2)= .07615 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 978 944.3 1.203 1.203 r =5 21646 21743.9 .441 1.643 r =6 77376 77311.8 .053 1.697 p=1-exp(-SUM/2)= .57188 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 947 944.3 .008 .008 r =5 21950 21743.9 1.954 1.961 r =6 77103 77311.8 .564 2.525 p=1-exp(-SUM/2)= .71708 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 990 944.3 2.212 2.212 r =5 21653 21743.9 .380 2.592 r =6 77357 77311.8 .026 2.618 p=1-exp(-SUM/2)= .72991 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 981 944.3 1.426 1.426 r =5 21640 21743.9 .496 1.923 r =6 77379 77311.8 .058 1.981 p=1-exp(-SUM/2)= .62863 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 977 944.3 1.132 1.132 r =5 21880 21743.9 .852 1.984 r =6 77143 77311.8 .369 2.353 p=1-exp(-SUM/2)= .69160 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 961 944.3 .295 .295 r =5 22096 21743.9 5.702 5.997 r =6 76943 77311.8 1.759 7.756 p=1-exp(-SUM/2)= .97931 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1005 944.3 3.902 3.902 r =5 21595 21743.9 1.020 4.921 r =6 77400 77311.8 .101 5.022 p=1-exp(-SUM/2)= .91881 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 918 944.3 .733 .733 r =5 21694 21743.9 .115 .847 r =6 77388 77311.8 .075 .922 p=1-exp(-SUM/2)= .36940 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 965 944.3 .454 .454 r =5 21673 21743.9 .231 .685 r =6 77362 77311.8 .033 .717 p=1-exp(-SUM/2)= .30145 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 967 944.3 .546 .546 r =5 21641 21743.9 .487 1.033 r =6 77392 77311.8 .083 1.116 p=1-exp(-SUM/2)= .42758 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 918 944.3 .733 .733 r =5 21856 21743.9 .578 1.310 r =6 77226 77311.8 .095 1.406 p=1-exp(-SUM/2)= .50483 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 948 944.3 .014 .014 r =5 21851 21743.9 .528 .542 r =6 77201 77311.8 .159 .701 p=1-exp(-SUM/2)= .29560 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21796 21743.9 .125 .406 r =6 77276 77311.8 .017 .423 p=1-exp(-SUM/2)= .19056 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 959 944.3 .229 .229 r =5 21826 21743.9 .310 .539 r =6 77215 77311.8 .121 .660 p=1-exp(-SUM/2)= .28108 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 947 944.3 .008 .008 r =5 21670 21743.9 .251 .259 r =6 77383 77311.8 .066 .324 p=1-exp(-SUM/2)= .14975 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 930 944.3 .217 .217 r =5 21767 21743.9 .025 .241 r =6 77303 77311.8 .001 .242 p=1-exp(-SUM/2)= .11402 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 909 944.3 1.320 1.320 r =5 21630 21743.9 .597 1.916 r =6 77461 77311.8 .288 2.204 p=1-exp(-SUM/2)= .66783 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 910 944.3 1.246 1.246 r =5 21684 21743.9 .165 1.411 r =6 77406 77311.8 .115 1.526 p=1-exp(-SUM/2)= .53368 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 938 944.3 .042 .042 r =5 21470 21743.9 3.450 3.492 r =6 77592 77311.8 1.015 4.508 p=1-exp(-SUM/2)= .89501 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG rng.bin b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 949 944.3 .023 .023 r =5 21572 21743.9 1.359 1.382 r =6 77479 77311.8 .362 1.744 p=1-exp(-SUM/2)= .58187 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .785508 .823833 .258259 .011680 .076153 .571875 .717078 .729908 .628630 .691601 .979310 .918810 .369401 .301445 .427582 .504831 .295599 .190558 .281075 .149745 .114025 .667834 .533678 .895009 .581875 brank test summary for rng.bin The KS test for those 25 supposed UNI's yields KS p-value= .000284 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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, 2^21 words. This test samples the bitstream 20 times. No. missing words should average 141909. with sigma=428. ----------------------------------- --------------- tst no 1: 143298 missing words, 3.24 sigmas from mean, p-value= .99941 tst no 2: 143119 missing words, 2.83 sigmas from mean, p-value= .99765 tst no 3: 143320 missing words, 3.30 sigmas from mean, p-value= .99951 tst no 4: 143079 missing words, 2.73 sigmas from mean, p-value= .99686 tst no 5: 143379 missing words, 3.43 sigmas from mean, p-value= .99970 tst no 6: 144148 missing words, 5.23 sigmas from mean, p-value=1.00000 tst no 7: 143034 missing words, 2.63 sigmas from mean, p-value= .99570 tst no 8: 142911 missing words, 2.34 sigmas from mean, p-value= .99037 tst no 9: 142724 missing words, 1.90 sigmas from mean, p-value= .97151 tst no 10: 143635 missing words, 4.03 sigmas from mean, p-value= .99997 tst no 11: 143687 missing words, 4.15 sigmas from mean, p-value= .99998 tst no 12: 143914 missing words, 4.68 sigmas from mean, p-value=1.00000 tst no 13: 143386 missing words, 3.45 sigmas from mean, p-value= .99972 tst no 14: 143748 missing words, 4.30 sigmas from mean, p-value= .99999 tst no 15: 143845 missing words, 4.52 sigmas from mean, p-value=1.00000 tst no 16: 143915 missing words, 4.69 sigmas from mean, p-value=1.00000 tst no 17: 143531 missing words, 3.79 sigmas from mean, p-value= .99992 tst no 18: 144044 missing words, 4.99 sigmas from mean, p-value=1.00000 tst no 19: 144056 missing words, 5.02 sigmas from mean, p-value=1.00000 tst no 20: 143526 missing words, 3.78 sigmas from mean, p-value= .99992 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 rng.bin Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for rng.bin using bits 23 to 32 142830 3.175 .9993 OPSO for rng.bin using bits 22 to 31 142271 1.247 .8938 OPSO for rng.bin using bits 21 to 30 142220 1.071 .8580 OPSO for rng.bin using bits 20 to 29 141856 -.184 .4270 OPSO for rng.bin using bits 19 to 28 141673 -.815 .2076 OPSO for rng.bin using bits 18 to 27 141381 -1.822 .0342 OPSO for rng.bin using bits 17 to 26 142049 .482 .6850 OPSO for rng.bin using bits 16 to 25 141967 .199 .5788 OPSO for rng.bin using bits 15 to 24 141905 -.015 .4940 OPSO for rng.bin using bits 14 to 23 141809 -.346 .3647 OPSO for rng.bin using bits 13 to 22 141957 .164 .5653 OPSO for rng.bin using bits 12 to 21 142148 .823 .7947 OPSO for rng.bin using bits 11 to 20 141825 -.291 .3856 OPSO for rng.bin using bits 10 to 19 141988 .271 .6069 OPSO for rng.bin using bits 9 to 18 141498 -1.418 .0780 OPSO for rng.bin using bits 8 to 17 141777 -.456 .3241 OPSO for rng.bin using bits 7 to 16 141691 -.753 .2258 OPSO for rng.bin using bits 6 to 15 141738 -.591 .2773 OPSO for rng.bin using bits 5 to 14 142330 1.451 .9266 OPSO for rng.bin using bits 4 to 13 141562 -1.198 .1155 OPSO for rng.bin using bits 3 to 12 141901 -.029 .4885 OPSO for rng.bin using bits 2 to 11 142004 .326 .6280 OPSO for rng.bin using bits 1 to 10 141692 -.749 .2268 OQSO test for generator rng.bin Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for rng.bin using bits 28 to 32 142601 2.345 .9905 OQSO for rng.bin using bits 27 to 31 142473 1.911 .9720 OQSO for rng.bin using bits 26 to 30 143111 4.073 1.0000 OQSO for rng.bin using bits 25 to 29 142602 2.348 .9906 OQSO for rng.bin using bits 24 to 28 142275 1.240 .8924 OQSO for rng.bin using bits 23 to 27 141864 -.154 .4389 OQSO for rng.bin using bits 22 to 26 141916 .023 .5090 OQSO for rng.bin using bits 21 to 25 142088 .606 .7276 OQSO for rng.bin using bits 20 to 24 142328 1.419 .9221 OQSO for rng.bin using bits 19 to 23 142199 .982 .8369 OQSO for rng.bin using bits 18 to 22 143060 3.901 1.0000 OQSO for rng.bin using bits 17 to 21 142957 3.551 .9998 OQSO for rng.bin using bits 16 to 20 141592 -1.076 .1410 OQSO for rng.bin using bits 15 to 19 142025 .392 .6525 OQSO for rng.bin using bits 14 to 18 142443 1.809 .9648 OQSO for rng.bin using bits 13 to 17 142157 .840 .7994 OQSO for rng.bin using bits 12 to 16 142478 1.928 .9731 OQSO for rng.bin using bits 11 to 15 142632 2.450 .9929 OQSO for rng.bin using bits 10 to 14 142575 2.257 .9880 OQSO for rng.bin using bits 9 to 13 142405 1.680 .9535 OQSO for rng.bin using bits 8 to 12 141835 -.252 .4005 OQSO for rng.bin using bits 7 to 11 141296 -2.079 .0188 OQSO for rng.bin using bits 6 to 10 142536 2.124 .9832 OQSO for rng.bin using bits 5 to 9 141565 -1.167 .1216 OQSO for rng.bin using bits 4 to 8 142735 2.799 .9974 OQSO for rng.bin using bits 3 to 7 142315 1.375 .9155 OQSO for rng.bin using bits 2 to 6 142764 2.897 .9981 OQSO for rng.bin using bits 1 to 5 143207 4.399 1.0000 DNA test for generator rng.bin Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for rng.bin using bits 31 to 32 141215 -2.048 .0203 DNA for rng.bin using bits 30 to 31 142079 .501 .6916 DNA for rng.bin using bits 29 to 30 142072 .480 .6843 DNA for rng.bin using bits 28 to 29 141875 -.101 .4597 DNA for rng.bin using bits 27 to 28 142070 .474 .6822 DNA for rng.bin using bits 26 to 27 141981 .211 .5837 DNA for rng.bin using bits 25 to 26 142215 .902 .8164 DNA for rng.bin using bits 24 to 25 141825 -.249 .4018 DNA for rng.bin using bits 23 to 24 141439 -1.387 .0827 DNA for rng.bin using bits 22 to 23 141777 -.390 .3481 DNA for rng.bin using bits 21 to 22 141487 -1.246 .1064 DNA for rng.bin using bits 20 to 21 141858 -.151 .4398 DNA for rng.bin using bits 19 to 20 141741 -.497 .3098 DNA for rng.bin using bits 18 to 19 142294 1.135 .8718 DNA for rng.bin using bits 17 to 18 141473 -1.287 .0990 DNA for rng.bin using bits 16 to 17 141416 -1.455 .0728 DNA for rng.bin using bits 15 to 16 141942 .096 .5384 DNA for rng.bin using bits 14 to 15 141987 .229 .5906 DNA for rng.bin using bits 13 to 14 141888 -.063 .4749 DNA for rng.bin using bits 12 to 13 141714 -.576 .2822 DNA for rng.bin using bits 11 to 12 142271 1.067 .8570 DNA for rng.bin using bits 10 to 11 141540 -1.089 .1380 DNA for rng.bin using bits 9 to 10 141367 -1.600 .0548 DNA for rng.bin using bits 8 to 9 141810 -.293 .3848 DNA for rng.bin using bits 7 to 8 142004 .279 .6100 DNA for rng.bin using bits 6 to 7 141379 -1.564 .0589 DNA for rng.bin using bits 5 to 6 142357 1.321 .9067 DNA for rng.bin using bits 4 to 5 141929 .058 .5231 DNA for rng.bin using bits 3 to 4 142306 1.170 .8790 DNA for rng.bin using bits 2 to 3 141718 -.564 .2862 DNA for rng.bin using bits 1 to 2 142148 .704 .7593 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for rng.bin Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for rng.bin 2466.94 -.467 .320078 byte stream for rng.bin 2451.27 -.689 .245355 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 2727.18 3.213 .999343 bits 2 to 9 2440.25 -.845 .199063 bits 3 to 10 2431.63 -.967 .166802 bits 4 to 11 2614.24 1.616 .946916 bits 5 to 12 2494.23 -.082 .467507 bits 6 to 13 2476.92 -.326 .372037 bits 7 to 14 2491.96 -.114 .454719 bits 8 to 15 2438.49 -.870 .192196 bits 9 to 16 2544.50 .629 .735423 bits 10 to 17 2597.59 1.380 .916223 bits 11 to 18 2426.74 -1.036 .150082 bits 12 to 19 2341.97 -2.235 .012715 bits 13 to 20 2511.80 .167 .566266 bits 14 to 21 2539.23 .555 .710472 bits 15 to 22 2664.43 2.325 .989975 bits 16 to 23 2539.10 .553 .709873 bits 17 to 24 2619.61 1.691 .954628 bits 18 to 25 2401.88 -1.388 .082631 bits 19 to 26 2508.30 .117 .546708 bits 20 to 27 2651.73 2.146 .984054 bits 21 to 28 2573.41 1.038 .850392 bits 22 to 29 2434.89 -.921 .178573 bits 23 to 30 2443.39 -.801 .211692 bits 24 to 31 2521.79 .308 .621023 bits 25 to 32 2604.64 1.480 .930547 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 rng.bin Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3545 z-score: 1.005 p-value: .842447 Successes: 3545 z-score: 1.005 p-value: .842447 Successes: 3530 z-score: .320 p-value: .625377 Successes: 3539 z-score: .731 p-value: .767486 Successes: 3542 z-score: .868 p-value: .807188 Successes: 3531 z-score: .365 p-value: .642555 Successes: 3539 z-score: .731 p-value: .767486 Successes: 3496 z-score: -1.233 p-value: .108811 Successes: 3518 z-score: -.228 p-value: .409702 Successes: 3520 z-score: -.137 p-value: .445521 square size avg. no. parked sample sigma 100. 3530.500 14.702 KSTEST for the above 10: p= .823155 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file rng.bin Sample no. d^2 avg equiv uni 5 .1194 .3292 .113094 10 1.0545 .5702 .653467 15 2.6312 .8902 .928955 20 2.6907 .9965 .933080 25 .2643 1.0946 .233292 30 2.4415 1.1991 .914035 35 .1677 1.4590 .155138 40 .3458 1.3471 .293607 45 .2835 1.2585 .247907 50 .0340 1.1628 .033610 55 .0860 1.1680 .082846 60 2.6036 1.2162 .926955 65 1.0217 1.2002 .641851 70 .8516 1.1808 .575097 75 .0720 1.1774 .069766 80 1.3485 1.1477 .742123 85 .2336 1.1157 .209233 90 .0144 1.0993 .014364 95 .8251 1.1296 .563615 100 3.4086 1.1380 .967474 MINIMUM DISTANCE TEST for rng.bin Result of KS test on 20 transformed mindist^2's: p-value= .427126 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file rng.bin sample no: 1 r^3= 5.077 p-value= .15569 sample no: 2 r^3= 11.540 p-value= .31932 sample no: 3 r^3= 85.572 p-value= .94229 sample no: 4 r^3= 29.439 p-value= .62518 sample no: 5 r^3= 34.526 p-value= .68364 sample no: 6 r^3= 65.367 p-value= .88683 sample no: 7 r^3= 10.023 p-value= .28401 sample no: 8 r^3= 24.019 p-value= .55095 sample no: 9 r^3= 58.162 p-value= .85612 sample no: 10 r^3= .810 p-value= .02665 sample no: 11 r^3= 13.453 p-value= .36137 sample no: 12 r^3= 13.447 p-value= .36124 sample no: 13 r^3= 52.152 p-value= .82420 sample no: 14 r^3= 34.210 p-value= .68028 sample no: 15 r^3= 16.248 p-value= .41818 sample no: 16 r^3= 67.559 p-value= .89481 sample no: 17 r^3= 21.954 p-value= .51896 sample no: 18 r^3= 14.856 p-value= .39055 sample no: 19 r^3= 34.521 p-value= .68358 sample no: 20 r^3= 19.331 p-value= .47500 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file rng.bin p-value= .366727 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR rng.bin Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: -1.5 -.3 -1.6 .2 .4 1.2 -.8 1.9 .3 1.0 -1.5 .2 -.6 .0 -2.2 -.3 .3 -.9 1.5 1.7 -1.7 -.8 .8 .6 .5 .3 1.8 -.5 1.1 -1.0 -.3 .4 -2.3 .9 -.2 1.1 1.4 .8 .5 -.1 .9 1.0 -1.1 Chi-square with 42 degrees of freedom: 48.814 z-score= .743 p-value= .782086 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .887849 Test no. 2 p-value .136395 Test no. 3 p-value .103668 Test no. 4 p-value .165459 Test no. 5 p-value .594303 Test no. 6 p-value .537640 Test no. 7 p-value .914682 Test no. 8 p-value .514679 Test no. 9 p-value .498268 Test no. 10 p-value .981655 Results of the OSUM test for rng.bin KSTEST on the above 10 p-values: .265369 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file rng.bin Up and down runs in a sample of 10000 _________________________________________________ Run test for rng.bin : runs up; ks test for 10 p's: .421407 runs down; ks test for 10 p's: .236630 Run test for rng.bin : runs up; ks test for 10 p's: .555940 runs down; ks test for 10 p's: .984019 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for rng.bin No. of wins: Observed Expected 98780 98585.86 98780= No. of wins, z-score= .868 pvalue= .80739 Analysis of Throws-per-Game: Chisq= 28.28 for 20 degrees of freedom, p= .89702 Throws Observed Expected Chisq Sum 1 66835 66666.7 .425 .425 2 37309 37654.3 3.167 3.592 3 26684 26954.7 2.719 6.311 4 19499 19313.5 1.782 8.094 5 13889 13851.4 .102 8.196 6 10057 9943.5 1.295 9.490 7 7188 7145.0 .258 9.749 8 5229 5139.1 1.574 11.322 9 3717 3699.9 .079 11.402 10 2583 2666.3 2.602 14.004 11 1953 1923.3 .458 14.462 12 1392 1388.7 .008 14.469 13 972 1003.7 1.002 15.471 14 749 726.1 .720 16.191 15 546 525.8 .773 16.964 16 351 381.2 2.385 19.349 17 293 276.5 .980 20.329 18 185 200.8 1.248 21.577 19 171 146.0 4.287 25.863 20 121 106.2 2.058 27.921 21 277 287.1 .356 28.278 SUMMARY FOR rng.bin p-value for no. of wins: .807388 p-value for throws/game: .897016 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file rng.out

All in all, probably not a bad pseudo random number generator at all...

## Xilinx ISE Implementation

It seems that the ISE synthesis tools have an issue with the mod operator used to calculate the cell neighbour connections. There is a simple fix to this problem. Rather than allowing any possible negative numbers to be formed, adjust the row and column calculations. The extract below shows the required changes.

`RandomNumberGenerator.vhdl`

-- Connect the cell array Connect_row : for row in 0 to 7 generate constant cell0row : integer := (row+6) mod 8; -- 2n constant cell1row : integer := row; -- c constant cell2row : integer := (row+7) mod 8; -- n constant cell3row : integer := (row+2) mod 8; -- 2s begin Connect_col : for col in 0 to 7 generate constant cell0col : integer := (col+6) mod 8; -- 2w constant cell1col : integer := col; -- c constant cell2col : integer := (col+2) mod 8; -- 2e constant cell3col : integer := (col+1) mod 8; -- e begin