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 <[email protected]> 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 <[email protected]> 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 <[email protected]> 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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