Xref: news2.mv.net comp.os.msdos.djgpp:8237
From: elric AT wheel DOT dcn DOT davis DOT ca DOT us (Jeffrey Taylor)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: random() and rand() question
Date: 3 Sep 1996 21:56:41 GMT
Organization: Davis Community Network - Davis, California, USA
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Message-ID: <50i9ip$qv5@mark.ucdavis.edu>
References: <DwvF3A DOT H2r AT interactive DOT net> <3225ABDD DOT 7AF48E03 AT alcyone DOT com> <507p1f$36f AT epx DOT cis DOT umn DOT edu>
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To: djgpp AT delorie DOT com
DJ-Gateway: from newsgroup comp.os.msdos.djgpp

Jason Hoffoss (hoffo002 AT gold DOT tc DOT umn DOT edu) wrote:
<snip>
: will work good, and give you good numbers.  If you want like a number
: from 1 to 100, however, doing random() % 100 + 1 isn't going to give
: you great results.  Why?  Lets look at the natural range of random().
: It's a number from 0 to 4294967296.  As you can see, with a %100, your
: chances of getting a number from 0-95 is a little higher than your
: chances of a number from 96-99.  Other mod values will give you
: different distributions.  What I would recommend instead is something
: like so, which keeps the probability flat:
: 
: int myrand(void)
: {
:    int n;
: 
:    do
:    {
:       n = random() & 127;
:    } while (n >= 100);
: 
:    return n;
: }
: 
: This gets quite a bit more complex when you want to generalize it for
: any range, but this should point you in the right direction.  If you
: can, limiting your ranges to powers of 2 is the best solution.
: 
:     -Jason
: 

Here is another solution.  IF meory serves, it comes from RC4 author.

// returns a random integer in range 0 <= X < sides
unsigned range(unsigned sides)
{
   unsigned const div = ((unsigned)RAND_MAX+1) / sides;
   unsigned result;

   while ((result = rand() / div) == sides)
      ;     // retry

   return result;
}

-- 
============================================
Without my guitar, I am a poet without arms.
                        - Michael Bloomfield
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