OpenEuphoria

Re: 32-bit random numbers

• Posted by "Juergen Luethje" <j.lue at gmx.de> Jul 11, 2004
• 587 views

Hi Igor, you wrote:

>> From: Juergen Luethje
>> 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    

Don't worry! I'm sleeping very good, and I hope that you do so, too. 

> Try please the next one RNG:
>
> }}}
<eucode>
>
> constant K = 1073741823 -- max EU integer
> function rand_atom(integer N)
>     return K * (rand(N) - 1) + rand(K) - 1
> end function
>
> </eucode>
{{{

>
> 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?

I don't know. It has already been said, that it's not beneficial to
combine successive results of rand(). I'm afraid it's all a bit over my
head. Things concerning randomness are rather difficult IMHO. Anyway,
thanks Igor!

> 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

Regards,
   Juergen