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Lecture 31 CSE 331 Nov 13, 2017
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Mini project video due TODAY
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Two changes in HWs
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Closest pairs of points
Input: n 2-D points P = {p1,…,pn}; pi=(xi,yi) d(pi,pj) = ( (xi-xj)2+(yi-yj)2)1/2 Output: Points p and q that are closest
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Dividing up P Q R First n/2 points according to the x-coord
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Recursively find closest pairs
Q R δ = min (blue, green)
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An aside: maintain sorted lists
Px and Py are P sorted by x-coord and y-coord Qx, Qy, Rx, Ry can be computed from Px and Py in O(n) time
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An easy case R > δ Q All “crossing” pairs have distance > δ
δ = min (blue, green)
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Life is not so easy though
Q R δ = min (blue, green)
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Euclid to the rescue (?) d(pi,pj) = ( (xi-xj)2+(yi-yj)2)1/2 yj yi xi
The distance is larger than the x or y-coord difference
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Life is not so easy though
δ Q R > δ > δ > δ δ = min (blue, green)
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All we have to do now δ R Q S
Figure if a pair in S has distance < δ S δ = min (blue, green)
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Assume can be done in O(n)
The algorithm so far… O(n log n) + T(n) Input: n 2-D points P = {p1,…,pn}; pi=(xi,yi) Sort P to get Px and Py O(n log n) T(< 4) = c Closest-Pair (Px, Py) T(n) = 2T(n/2) + cn If n < 4 then find closest point by brute-force Q is first half of Px and R is the rest O(n) Compute Qx, Qy, Rx and Ry O(n) (q0,q1) = Closest-Pair (Qx, Qy) O(n log n) overall (r0,r1) = Closest-Pair (Rx, Ry) O(n) δ = min ( d(q0,q1), d(r0,r1) ) O(n) S = points (x,y) in P s.t. |x – x*| < δ return Closest-in-box (S, (q0,q1), (r0,r1)) Assume can be done in O(n)
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Rest of today’s agenda Implement Closest-in-box in O(n) time
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Three general techniques
High level view of CSE 331 Problem Statement Problem Definition Three general techniques Algorithm “Implementation” Data Structures Analysis Correctness+Runtime Analysis
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Greedy Algorithms Natural algorithms
Reduced exponential running time to polynomial
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Divide and Conquer Recursive algorithmic paradigm
Reduced large polynomial time to smaller polynomial time
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A new algorithmic technique
Dynamic Programming
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Dynamic programming vs. Divide & Conquer
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Same same because Both design recursive algorithms
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Different because Dynamic programming is smarter about solving recursive sub-problems
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End of Semester blues Can only do one thing at any day: what is the optimal schedule to obtain maximum value? Write up a term paper (30) (3) (2) (5) (10) Party! Exam study 331 HW Project Monday Tuesday Wednesday Thursday Friday
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Previous Greedy algorithm
Order by end time and pick jobs greedily Greedy value = 5+2+3= 10 Write up a term paper (10) OPT = 30 Party! (2) Exam study (5) 331 HW (3) Project (30) Monday Tuesday Wednesday Thursday Friday
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Today’s agenda Formal definition of the problem
Start designing a recursive algorithm for the problem
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