Re: ISI INTERVIEW PREP

Posted by deepak on Jul 02, 2012; 6:47pm
URL: http://discussion-forum.276.s1.nabble.com/ISI-INTERVIEW-PREP-tp7578819p7578980.html

This is how I thought about the problem.
Number of points that can be placed such that minimum distance of any 2 points is 2^.5 = (Area of the binding square)/(Area of a circle of diameter (2)^.5)
The binding square, I take to have side = 4(which is the given square) + 2 * (2^.5)/2 . The second term I get because, assuming a circle to be centered at one of the vertices, the binding square will have to envelope this. Since I need to extend on both directions by the radius, I get the above term.
If I compute the area of the square as (4+2^.5) ^2 and the area of each circle as pi *(1/2^.5)^2, I get n = 13.2.
Since you can place maximum 13 points such that each pair is atleast (2)^.5 apart, there is no way we can place 17 points such that they have a pairwise distance greater than root2.
Does this look sensible at all?