@Prabhat
CustRand(n){
sum = n*(n+1)/2;
a = rand(sum);
for(i = 1; i <= n; i++){
h = i*(i+1)/2;
l = i*(i-1)/2;
if(a <= h && a > l)
return i;
}
}
Take an example of say 4
so sum = 10
and a = <any number b/w 1...10>
therefore, if 0 < a <= 1 return 1 (only 1 case possible)
if 1 < a <= 3 return 2 (only 2 cases are possible)
if 3 < a <= 6 return 3 (only 3 cases are possible)
if 6 < a <= 10 return 4 (only 4 cases are possible)
On Tue, Aug 30, 2011 at 3:17 AM, Don <[email protected]> wrote:
> Here is how to do it with a single call to rand and no looping.
>
> int custRand(int n)
> {
> int a = rand(n*n+n);
> int b = a / n;
> a %= n;
> return (b > a) ? n+a-b+1 : n-a+b;
> }
>
> On Aug 29, 10:48 am, Piyush Grover <[email protected]> wrote:
> > Given a function rand(n) which returns a random value between 1...n
> assuming
> > equal probability.
> > Write a function CustRand(n) using rand(n) which returns a value between
> > 1...n such that
> > the probability of occurrence of each number is proportional to its
> value.
> >
> > I have a solution but wondering if I can get better than this or some
> other
> > approaches:
> >
> > CustRand(n){
> >
> > sum = n*(n+1)/2;
> >
> > a = rand(sum);
> > for(i = 1; i <= n; i++){
> > h = i*(i+1)/2;
> > l = i*(i-1)/2;
> > if(a <= h && a > l)
> > return i;
> > }
> >
> > }
>
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