Random number between any two numbers (a,b) can also be obtained as
follows.
let rand() return a random number between 0 and 1 ;
int random(a,b)
{
if(a==b)
return a ;
mid = a + (b-a)/2 ;
if( rand() )
return random(mid,b) ;
else
return random(a,mid) ;
}
I think probability of each number will be equal in this case.
On Sat, Sep 19, 2009 at 1:27 AM, Gene <gene.ress...@gmail.com> wrote:
>
> On Sep 7, 11:56 am, ankur aggarwal <ankur.mast....@gmail.com> wrote:
> > Given a random number generator that generates numbers in the range 1
> to
> > 5, how can u create a random number generator to generate numbers in the
> > range 1 to 7. (remember that the generated random numbers should follow a
> > uniform distribution in the corresponding range)
>
> If you generate 2 random numbers, there are 25 possible outcomes.
> This doesn't map nicely to the 7 desired outcomes. The standard way
> out of this mess is to ignore 4 outcomes leaving 21. Then map these
> to the range 1 to 7:
>
> int n;
> do n = (random1to5() - 1) * 5 + (random1to5() - 1); while n >= 7 * 3;
> return 1 + n % 7
>
> You can avoid throwing away random numbers with higher probabilty by
> generating more to begin with.
>
> For example, generating 3 provides 5 * 5 * 5 = 125 possible outcomes.
> Discard 6 cases to leave 119 = 17 * 7. So we have
>
> int n;
> do n = (random1to5() - 1) * 5 * 5 + (random1to5() - 1) * 5 +
> (random1to5() - 1); while n >= 7 * 17;
> return 1 + n % 7
>
> The probability of discarding a number is 4/25 in the first case, but
> only 6 / 125 in the second.
>
>
> >
>
--
Neeraj Kumar
B.Tech (Part III)
Department of Computer Sc. & Engineering
IT-BHU,Varanasi
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