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Re: non-Markov base ten random number generator based on Pi

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  • Subject: [mg52048] Re: non-Markov base ten random number generator based on Pi
  • From: Bill Rowe <readnewsciv at>
  • Date: Mon, 8 Nov 2004 03:13:41 -0500 (EST)
  • Sender: owner-wri-mathgroup at

On 11/7/04 at 1:03 AM, tftn at (Roger Bagula) wrote:

>This result  is an experimental random number generator. It uses the
>PSLQ Bailey Pi digits rational polynomial to generate digits in the 0
>to 9 range using a process that loses information , but generally
>behaves like the Pi digits. It is not Markov, in that there is no
>previous behavior involved in calculating the next random number.

>(* Bailey formula with digit drop base 80*)
>(* base 10 random number generator that isn't Markov *)
>(* use integer seed as the number of digits in to start calculation*)
>(* other PSLQ functions of transcendental numbers could be used to do 
>this same kind of random number*)
>(* Rowe Count*)
>d1=Flatten@{0,Length/@Split[Sort@c1], 0}

I gather from your various posts you are trying to use the plot of c1 as a visual indicator of randomness. If that is your goal, then there are far simpler plots that are more meaningful. One of the simplest and easiest to interpret would be a plot of the cumulative distribution function for the data set.

For example:

dataSize = 10000;
data = Table[Random[Integer,{0,9}],{dataSize}];

For a uniform distribution, the cumulative distribution function is x/(b-a) where a, b are the end points. That is, the cumulative distribution function for a uniform distribution should plot as a straight line from 0 to 1 over the interval (a,b).

Contrast this with

data = Table[Random[BinomialDistribution[5, .5]], {dataSize}];
The plot is clearly not a straight line indicating the data is not from a uniform distribution.

But to really demonstrate a pseudo random number generator adequately generates uniform deviates you will need more than such a simple plot.
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