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Lecture 30

Lecture 30 . CSE 331 Nov 11, 2016. Mini project video due Mon. Homework 8. Solutions to Homework 7. At the END of the lecture. Counting Inversions. Input: n distinct numbers a 1 ,a 2 ,…,a n. Inversion: (i,j) with i < j s.t. a i > a j. Output: Number of inversions.

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Lecture 30

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  1. Lecture 30 CSE 331 Nov 11, 2016

  2. Mini project video due Mon

  3. Homework 8

  4. Solutions to Homework 7 At the END of the lecture

  5. Counting Inversions Input: n distinct numbers a1,a2,…,an Inversion: (i,j) with i < j s.t. ai > aj Output: Number of inversions

  6. Divide and Conquer Divide up the problem into at least two sub-problems Solve all sub-problems: Mergesort Recursively solve the sub-problems Solve some sub-problems: Multiplication Solve stronger sub-problems: Inversions “Patch up” the solutions to the sub-problems for the final solution

  7. Handling crossing inversions aL aR 1 100 2 4 Why should aL and aR be sorted? http://www.dovecoteidea.com/ Sort aL and aRrecursively!

  8. Mergesort-Count algorithm Input: a1, a2, …, an Output: Numbers in sorted order+ #inversion T(2) = c MergeSortCount( a, n ) T(n) = 2T(n/2) + cn If n = 1 return( 0 , a1) O(n log n) time If n = 2 return( a1 > a2, min(a1,a2); max(a1,a2)) aL = a1,…, an/2 aR = an/2+1,…, an O(n) (cL, aL) = MergeSortCount(aL, n/2) (cR, aR) = MergeSortCount(aR, n/2) Counts #crossing-inversions+ MERGE (c, a) = MERGE-COUNT(aL,aR) return (c+cL+cR,a)

  9. 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

  10. Group Talk time O(n2) time algorithm? 1-D problem in time O(n log n) ?

  11. Sorting to rescue in 2-D? Pick pairs of points closest in x co-ordinate Pick pairs of points closest in y co-ordinate Choose the better of the two

  12. A property of Euclidean distance yj d(pi,pj) = ( (xi-xj)2+(yi-yj)2)1/2 yi xj xi The distance is larger than the x or y-coord difference

  13. Rest of Today’s agenda Divide and Conquer based algorithm

  14. Dividing up P R Q First n/2 points according to the x-coord

  15. Recursively find closest pairs R Q δ = min (blue, green)

  16. An aside: maintain sorted lists Px and Pyare P sorted by x-coord and y-coord Qx, Qy, Rx, Rycan be computed from Px and Pyin O(n) time

  17. An easy case R > δ Q All “crossing” pairs have distance > δ δ = min (blue, green)

  18. Life is not so easy though R Q δ = min (blue, green)

  19. Rest of Today’s agenda Divide and Conquer based algorithm

  20. Euclid to the rescue (?) yj d(pi,pj) = ( (xi-xj)2+(yi-yj)2)1/2 yi xj xi The distance is larger than the x or y-coord difference

  21. Life is not so easy though δ δ >δ R Q > δ > δ δ = min (blue, green)

  22. All we have to do now δ δ R Q S Figure if a pair in S has distance <δ δ = min (blue, green)

  23. 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 Pxand R is the rest O(n) Compute Qx, Qy, Rxand 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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