CSE 421 Closest Pair of Points, Master Theorem, Integer Multiplication Shayan Oveis Gharan 1
Finding the Closest Pair of Points
Closest Pair of Points (non geometric) Given n points and arbitrary distances between them, find the closest pair. (E.g., think of distance as airfare definitely not Euclidean distance!) ( and all the rest of the ( n ) edges ) 2 Must look at all n choose 2 pairwise distances, else any one you didn t check might be the shortest. i.e., you have to read the whole input
Closest Pair of Points (1-dimension) Given n points on the real line, find the closest pair, e.g., given 11, 2, 4, 19, 4.8, 7, 8.2, 16, 11.5, 13, 1 find the closest pair 1 2 4 4.8 7 8.2 11 11.5 13 16 19 Fact: Closest pair is adjacent in ordered list So, first sort, then scan adjacent pairs. Time O(n log n) to sort, if needed, Plus O(n) to scan adjacent pairs Key point: do not need to calc distances between all pairs: exploit geometry + ordering
Closest Pair of Points (2-dimensions) Given n points in the plane, find a pair with smallest Euclidean distance between them. Fundamental geometric primitive. Graphics, computer vision, geographic information systems, molecular modeling, air traffic control. Special case of nearest neighbor, Euclidean MST, Voronoi. Brute force: Check all pairs of points p and q with (n 2 ) time. Assumption: No two points have same x coordinate.
A Divide and Conquer Alg Divide: draw vertical line L with n/2 points on each side. Conquer: find closest pair on each side, recursively. Combine to find closest pair overall Return best solutions seems like (n 2 )? L 8 21 12
Key Observation Suppose δ is the minimum distance of all pairs in left/right of L. δ = min 12,21 = 12. Key Observation: suffices to consider points within of line L. Almost the one-d problem again: Sort points in 2 -strip by their y coordinate. Only check pts within 11 in sorted list! 7 6 L 4 5 21 12 3 2 1 =12
Almost 1D Problem Partition each side of L into δ 2 δ 2 squares Claim: No two points lie in the same δ 2 δ 2 box. Pf: Such points would be within 39 j δ 2 2 + δ 2 2 = δ 1 2 0.7δ < δ 31 Let s i have the i th smallest y-coordinate among points in the 2δ-width-strip. Claim: If i j > 11, then the distance i 29 27 30 28 29 ½ ½ between s i and s j is > δ. Pf: only 11 boxes within of y(s i ). 26 25
Closest Pair (2Dim Algorithm) Closest-Pair(p 1,, p n ) { if(n <=??) return?? Compute separation line L such that half the points are on one side and half on the other side. 1 = Closest-Pair(left half) 2 = Closest-Pair(right half) = min( 1, 2 ) Delete all points further than from separation line L Sort remaining points p[1] p[m] by y-coordinate. for i = 1..m i for k = 1 11 if i+k <= m = min(, distance(p[i], p[i+k])); } return.
Closest Pair Analysis I Let D(n) be the number of pairwise distance calculations in the Closest-Pair Algorithm when run on n 1 points 1 if n = 1 D n ቐ 2D n D n = O(nlog n) + 11 n o. w. 2 BUT, that s only the number of distance calculations What if we counted running time? 1 if n = 1 T n ቐ 2T n + O(n log n) 2 o. w. D n = O(nlog2 n)
Can we do better? (Analysis II) Yes!! Don t sort by y-coordinates each time. Sort by x at top level only. This is enough to divide into two equal subproblems in O(n) Each recursive call returns and list of all points sorted by y Sort points by y-coordinate by merging two pre-sorted lists. 1 if n = 1 T n ቐ 2T n + O n 2 o. w. D n = O(n log n)
Recurrences Above: Where they come from, how to find them Next: how to solve them
Master Theorem Suppose T n = a T n b + cnk for all n > b. Then, If a > b k then T n = Θ n log ba If a < b k then T n = Θ n k If a = b k then T n = Θ n k log n Works even if it is n b instead of n b. We also need a 1, b > 1, k 0 and T n = O(1) for n b.
Master Theorem Suppose T n = a T n + b cnk for all n > b. Then, If a > b k then T n = Θ n log ba If a < b k then T n = Θ n k If a = b k then T n = Θ n k log n Example: For mergesort algorithm we have T n = 2T n + O n. 2 So, k = 1, a = b k and T n = Θ(n log n)
Proving Master Theorem Problem size n T(n) = at(n/b) + cn k a # probs 1 cost cn k n/b a c a n k /b k n/b 2 b 1 a 2 a d c a 2 n k /b 2k =c n k (a/b k ) 2 c n k (a/b k ) d d=log b n T n = a i c n k i=0 b i
A Useful Identity Theorem: 1 + x + x 2 + + x d = xd+1 1 x 1 Pf: Let S = 1 + x + x 2 + + x d Then, xs = x + x 2 + + x d+1 So, xs S = x d+1 1 i.e., S x 1 = x d+1 1 Therefore, S = xd+1 1 x 1
Solve: T n = at n b + cnk, a > b k b k log b n log b n T n = a i c n k i=0 b i = cn k log b n a i=0 b k = cn k a b k i log b n+1 1 a b k 1 a b k a 1 alogb n x k+1 1 = b log n k b n k = n k c b k log b n b k a log b n 2c a log b n = O(n log b a ) for x = a b k using x 1 x 1 = (b log b a ) log b n = (b log b n ) log b a = n log b a
Solve: T n = at n b + cnk, a = b k log b n T n = a i c n k i=0 b i = cn k log b n a i=0 b k = cn k log b n i
Master Theorem Suppose T n = a T n b + cnk for all n > b. Then, If a > b k then T n = Θ n log ba If a < b k then T n = Θ n k If a = b k then T n = Θ n k log n Works even if it is n b instead of n b. We also need a 1, b > 1, k 0 and T n = O(1) for n b.