Correctness theorem if there is an intersection the

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Unformatted text preview: to store dynamic set S. Total running time: O(n lg n). Correctness Theorem: If there is an intersection, the algorithm finds it. Correctness Theorem: If there is an intersection, the algorithm finds it. Proof: Let X be the leftmost intersection point. Assume for simplicity that • only two segments s1, s2 pass through X, and • no two points have the same x-coordinate. At some point before we reach X, s1 and s2 become consecutive in the order of S. Either initially consecutive when s1 or s2 inserted, or became consecutive when another deleted. Closest Pair Closest Pair • Find a closest pair among p1…pn !Rd • Easy to do in O(dn2) time – For all pi pj, compute ||pi – pj|| and choose the minimum • We will aim for better time, as long as d is “small” • For now, focus on d=2 October 30, 2003 Lecture 17: Closest Pair 2 Divide & Conquer Divide and conquer • Divide: – Compute the median of xcoordinates – Split the points into PL and PR, each of size n/2 • Conquer: compute the closest pairs for PL and PR • Combine the results (the h a r d p a r t) Divide & Conquer Combine 2k • Let k=min(k1,k2) • Observe: – Need to check only pairs which cross the dividing line – Only interested in pairs within distance < k ...
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