Hi Dave,
i just checked;
int i;
int j;
for(i=1;i<6;i++) {
j = 2*(i/3);
printf("\n%d\n",j);
}
gives me output 0,0,2,2,2. and there was no 3;
and since we are anyway generating "random" numbers would this difference in
distribution really matter?
On Sat, Jul 31, 2010 at 6:30 PM, Dave <[email protected]> wrote:
> @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 <[email protected]> wrote:
> > what about
> >
> > int i = rand5() + (2/3)*rand5();
> >
> > won't this work?
> >
> >
> >
> >
> >
> > On Sat, Jul 31, 2010 at 5:46 PM, Dave <[email protected]> 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 <[email protected]> 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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Regards,
Sony
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