what about
int i = rand5() + (2/3)*rand5();
won't this work?
On Sat, Jul 31, 2010 at 5:46 PM, Dave <dave_and_da...@juno.com> wrote:
> Use the rejection method...
>
> int rand7()
> {
> int i;
> do
> i = 5 * rand5() + rand5() - 3;
> while( i > 23 );
> return i / 3;
> }
>
> The loop assigns i a value between 5*1+1-3 = 3 and 5*5+5-3 = 27 with
> uniform probability, and then rejects any value of i > 23. Thus, after
> the loop, i has a uniform distribution on the interval 3 to 23.
> Dividing by 3 gives the desired result.
>
> Under the assumptions of the problem, a value of i is rejected with
> probability 4/25, so the loop executes an average of 1/(1 - 4/25) =
> 25/21 times. Therefore, on average, it takes 50/21 rand5's to make a
> rand7.
>
> Dave
>
> On Jul 30, 8:33 am, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
> > given a rand5 function which generates numbers from 1 to 5 with equal
> > probability.. give an algorithm which uses rand5 function and generates
> > numbers from 1 to 7 with equal probability
> > --
> > With Regards,
> > Jalaj Jaiswal
> > +919026283397
> > B.TECH IT
> > IIIT ALLAHABAD
>
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--
Regards,
Sony
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