OpenEuphoria

Re: 32-bit random numbers

• Posted by "Juergen Luethje" <j.lue at gmx.de> Jul 06, 2004
• 607 views

Hi Igor, you wrote:

> Hi again Jurgen,
>
> Me wrote:

<snip>

>> Get one trick more:
>>
>> To get the 32 bit rnd flat distribution
>> you can do the following steps:
>>
>> a. Get rnd number in [0,1] range from the EU standard rand()
>> b. Get rnd number in needed range from [0,1] range.
>>
>> It may be done as:
>>
>> }}}
<eucode>
>> atom k
>> k = 1 / 1073741823 -- the biggest EU integer
>> atom rnd_0_1
>>
>> function rand_0_N(atom N)
>>   rnd_0_1 = k * (rand(1073741823)-1)
>>   return floor(N * rnd_0_1)
>> end function
>>
>> for i=1 to 1000 do
>>     ? 1 + rand_0_N(#FFFFFFFF) -- this distribution
>>                               -- is in [1,#FFFFFFFF] range
>> end for
>> </eucode>
{{{


Cool! Why didn't I think of that? 

> Yes, it seems to be:
>
> for i=1 to 1000 do
>     ? 1 + rand_0_N(#FFFFFFFF - 1)
>  -- this distribution
>  -- is in [1,#FFFFFFFF] range
> end for   ----    

It seems to me, that your *first* version was correct, wasn't it?
For testing, I used N=6:

atom k
k = 1 / 1073741823 -- the biggest EU integer
atom rnd_0_1

function rand_0_N(atom N)
  rnd_0_1 = k * (rand(1073741823)-1)
  return floor(N * rnd_0_1)
end function

atom N, x, min, max

N = 6
min = 1 + rand_0_N(N)
max = min
for i=1 to 1000 do
   x = 1 + rand_0_N(N)  -- this distribution is in [1,6] range
   if x < min then
      min = x
   elsif x > max then
      max = x
   end if
end for
printf(1, "range [%d,%d]", {min, max})

Thanks, Igor!

Regards,
   Juergen