@Dave: Very nice.
Don
On Sep 12, 10:51 pm, Dave <dave_and_da...@juno.com> wrote:
> 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 <dondod...@gmail.com> 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 <jkhas...@gmail.com> 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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