On 3/8/07, Karthik Singaram L <[EMAIL PROTECTED]> wrote: > > The same triangle reasoning holds good here too right? If we choose the end > points of the side that we are bisecting and the new point we are choosing > on the perpendicular bisector as the triangle the other vertex will be > inside the triange? >
I thought you were trying to build the answer incrementally, and found that logic flawed. Let me explain my reasoning. If we consider three points, we can say that the probability that the convex hull is a triangle will tend to 1 as the size of the plane tends to infinity. Now we're to introduce the fourth point. I was simply stating that there are several other positions which are outside the triangle where we can place this fourth point and still yield a triangle as the convex hull of these 4 points. My example was one such case, which yields infinitely many positions _outside_ the triangle. -- Regards, Rajiv Mathews --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
