Okk...I got my mistake...
Thank you all for clearing my doubts..

On Sun, Feb 27, 2011 at 10:53 AM, sankalp srivastava <
richi.sankalp1...@gmail.com> wrote:

> The points must satisfy the equation
>
> (x-x1)(x-x2)+(y-y1)(y-y2)=0
>
> Circle centered at origin
>
> x2+y2=Some radius .With N points on the circle , we find out the
> radius
>
> In order to find if the two points are antipodal , we check the first
> equation putting the two points and checking for any other point on
> the circle if it satisfies the equation . Test for antinodality .
>
> This will do in O(1) .
> Given N points and we have to find if two points are antinodal
>
> On Feb 26, 11:15 am, Mohan Mangal <mohan.mangal...@gmail.com> wrote:
> > Hi Vinay,
> >
> > Here the condition is Point lies on same circle..
> > hope you got it.
> >
> >
> >
> > On Sat, Feb 26, 2011 at 10:58 AM, vinay reddy <gvina...@gmail.com>
> wrote:
> > > Hi Dave,
> > > I don't think ur logic will cover all cases like   (1,1)(-3,-3),
>  (1,1)
> > > (2,2)  a line connecting these points passes through origin,
> > > i think the solution is, we need to compute the slope of the point at
> index
> > > i with origin and build a binary tree with theses slopes.
> > > but worst cases of this algo is N*N , if we try balancing the tree
> while
> > > inserting I guess it can be done in NlogN
> > > Thanks
> > > Vinay
> >
> > > On Fri, Feb 25, 2011 at 9:20 AM, Gene <gene.ress...@gmail.com> wrote:
> >
> > >> Dave's solution is best if numerical error is possible.
> > >> If the points are precise, you can also do it in linear time.  Just
> hash
> > >> the points on abs(y/x).
> >
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> > --
> > Regards,
> > Mohan Mangal
> > Software Engineer, Bangalore
> > Mob- 80952-03670
>
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