第二次作业

1 Divide and Conquer

A group of n Ghostbusters is battling n ghosts. Each Ghostbuster is armed with a proton pack, which shoots a stream at a ghost, eradicating it. A stream goes in a straight line and terminates when it hits the ghost. The Ghostbusters decide upon the following strategy. They will pair off with the ghosts, forming n Ghostbuster-ghost pairs, and then simultaneously each Ghostbuster will shoot a stream at his chosen ghost. As we all know, it is very dangerous to let streams cross, and so the Ghostbusters must choose pairings for which no streams will cross. Assume that the position of each Ghostbuster and each ghost is a fixed point in the plane and that no three positions are collinear.

i Argue that there exists a line passing through one Ghostbuster and one ghost such the number of Ghostbusters on one side of the line equals the number of ghosts on the same side. Describe how to find such a line in O(n log n) time.

ii Give an O(n2 log n)-time algorithm to pair Ghostbusters with ghosts in such a way that no streams cross.

Answer:

i -ii: We assume that all the points in the plane are not in same X or Y axis using mark to stand for ghostbuster or ghosts(mark 0 stand for ghostbuster and mark 1 stand for ghosts).(This can not affect the problem because when we get the same X or Y , we can add or minus a very little value to let all points’ X or Y are not same). First we get the deepest point (in other word, the minimum Y value). Also we compute the other points with this point ‘s angle in X axis and sort these other points with this angle. Then we iterate in this order, and get another mark which is different from the deepest point's mark also we line this two

point ,and compute the bottom mark 0 's number and mark 1's number, they are the same , then we find the pair which is the deepest point and the iterate point. So we use this line to divide all points into two parts and recursive do the same above operation until the points' number is only 2 then we only get the line to connect this two points.

Algorithm:

Get_Pair(points[1-2N])

if N>1

initial a array points[1-2N] stand for the points of ghostbusters and ghosts

find the deepest point min_Point in the array points

initial the pair_Point and the index with 0

get the other 2N-1 points with this min_Point 's angle in X axis and sort these angles

initial num1 with 0 stand for the number of the other 2N-1 points which's mark is same with min_Point

initial num2 with 0 stand for the number of the other 2N-1 points which's mark is not same with min_Point

for i=1 to 2N-1

if min_Point's mark is same with points[i]'s mark

num1++

else

num2++

endif

if num1+1==num2