@Don...Nice solution.. But the return statement should be a+b-1

On Mon, Aug 29, 2011 at 10:33 PM, Don <[email protected]> wrote:

> If you draw the nxn grid and assign a value to each diagonal:
>
> (For n = 5)
>
>  ----------b
> |  12345
> |  2345
> |  345
> |  45
> |  5
> a
>
> You want the result to be the orthogonal distance from the diagonal.
> That is what the formula computes.
> Don
>
> On Aug 29, 11:28 am, Piyush Grover <[email protected]> wrote:
> > I understand what you are doing in the loop but return statement is not
> > clear to me. Can you explain.
> >
> > On Mon, Aug 29, 2011 at 9:48 PM, Don <[email protected]> wrote:
> > > int custRand(int n)
> > > {
> > >        int a,b;
> > >        do
> > >        {
> > >                a = rand(n);
> > >                b = rand(n);
> > >        } while(a < b);
> > >        return 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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-- 
S.Nishaanth,
Computer Science and engineering,
IIT Madras.

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