Here's another way, using a rejection technique on the bits of the
mantissa of p. Each iteration of the do-while loop exposes another
high-order bit of p, and the do-while loop iterates as long as the
random bits produced by f match the high order bit sequence of p. This
most likely will use fewer evaluations of f() than Don's approach.
int g(double p)
{
int i;
do
{
i = p + p;
p += p - i;
} while( i == f() );
return 1 - i;
}
Dave
On Sep 12, 10:19 am, Don <[email protected]> wrote:
> For particular values of p we might be able to do better, but for
> unknown values of p, I can't think of anything better than this:
>
> int g(double p)
> {
> int n = 0;
> for(int i = 0; i < 30; ++i)
> n += n+f();
> return n > (int)(p*1073741824.0);
>
> }
>
> On Sep 12, 9:55 am, JITESH KUMAR <[email protected]> wrote:
>
>
>
> > Hi
> > You are given a function f() that returns either 0 or 1 with equal
> > probability.
> > Write a function g() using f() that return 0 with probability p (where 0<p<1
> > )
>
> > --
> > *Regards
> > Jitesh Kumar*- Hide quoted text -
>
> - Show quoted text -
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