@Sony. I took (2/3)*rand5() as if (2/3) was a double. Otherwise,
(2/3)*rand5() = 0*rand5() = 0, which is not what you calculated when
you wrote 2*(i/3) instead of (2/3)*i.
And yes, the fact that the distribution is not uniform does matter,
since we want the results to be uniform. Using your interpretation of
(2/3)*rand5(), to get i = 1, the first rand5() must be 1 and the
second rand5() must be 1 or 2. Thus, the probability of i = 1 is 1/5 *
2/5 = 2/25. To get i = 2, the first rand5() must be 2 and the second
rand5() must be 1 or 2. Thus, i = 2 also with probability 1/5 * 2/5 =
2/25. For i = 3, either the first rand5() must be 1 and the second
rand5() must be 3, 4, or 5, or the first rand5 must be 3 and the
second rand5() must be 1 or 2. Thus, i = 3 with probability 1/5 * 3/5
+ 1/5 * 2/5 = 1/5. Continuing in this way, we see that the
probabilities of the various results are 2/25, 2/25, 1/5, 1/5, 1/5,
3/25, 3/25. These are not uniform!
Dave
On Jul 31, 8:12 am, Sony Jose <sonyj...@gmail.com> wrote:
> 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 <dave_and_da...@juno.com> 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 <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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>
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>
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