@Sony. No. Consider the following table:
rand5() (2/3)*rand5()
_________________________
1 0
2 1
3 2
4 2
5 3
Thus, (2/3)*rand5() is not uniformly distributed, nor is it in the
range 0 to 2.
Dave
On Jul 31, 7:49 am, Sony Jose <sonyj...@gmail.com> wrote:
> 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- Hide quoted text -
>
> - Show quoted text -
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