OpenEuphoria

Re: 32-bit random numbers

• Posted by "Igor Kachan" <kinz at peterlink.ru> Jul 10, 2004
• 587 views

Hi Juergen,

You wrote:

----------
> From: Juergen Luethje <j.lue at gmx.de>
> To: EUforum at topica.com
> Subject: Re: 32-bit random numbers
> Sent: 9 jul 2004 y. 0:48
> 
> Ricardo Forno wrote:
> 
> > The resulting distributions are good, 
> > but there is a problem.
> > As rand(power(2, 30) -1) can generate 
> > only 2^30 - 1 different numbers, no matter
> > what you do with these numbers, you will
> > get possibly only 2^30 - 1
> > different numbers between 1 and 2^32,
> > and so you will never get one of the
> > remaining 2^32 - 2^30 numbers, or 3/4ths
> > of the total possibilities.
> 
> I see. 
> Thank you for this important information, Ricardo!
> 
> [snipped old text]


I see your    and can not sleep  
Dont     be     or  better    

Try please the next one RNG:

constant K = 1073741823 -- max EU integer
function rand_atom(integer N)
    return K * (rand(N) - 1) + rand(K) - 1
end function

A short explanation.

The (rand(N) - 1) part gives the random number
of each next standard flat range. 
Say, if N = 1, then next output result will be 
rand(K) - 1,  i.e. from the normal EU range.
If N = 2, then output result will be:

a.    rand(K) - 1
 
or

b.    K + rand(K) - 1     

The probability of a or b cases is 1/2, so,
the output result will has the flat range 0 .. K-1
or K + 0 .. K + K - 1,  i.e.   0 .. K-1 .. K .. K + K - 1,
i.e.  0 .. K + K - 1  with discrete flat distribution,
which has step = 1  -- all possible integer numbers.

Well, if N = 3, then (rand(N) - 1) will be 2 or 1 or 0
with probability 1/3 of each case.

So, the output result will have 
0 .. K-1 .. K .. K+K-1 .. K+K .. K+K+K - 1 range
with flat distribution and step = 1

And so on.

It seems to be correct, no?

The question is - what maximum N will give
still integer atoms on output - not EU integer type,
but inner EU integer atoms.

Regards,
Igor Kachan
kinz at peterlink.ru