1. 1 Announcement Change of Venue for Tutorial Session #T01 and #TA1 from Week 6 to Week 13 (except in Week 8): Section Day Time T01 & TA1 Friday 15:00 - 15:50 New Venue: Week 6: 9 Oct - Room 1406, Academic 2 Week 7: 16 Oct - Room 1406, Academic 2 Week 8: 23 Oct - Room 2308, Academic 2 Week 9: 30 Oct - Room 1406, Academic 2 Week 10: 6 Nov - Room 1406, Academic 2 Week 11: 13 Nov - Room 1406, Academic 2 Week 12: 20 Nov - Room 1406, Academic 2 Week 13: 27 Nov - Room 1406, Academic 2

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4. 4 See the demo slide on the website.

5. 5 Example 1: Whole Process H G F E D C B A Divide H G F E D C B A Merge G H E F C D A B Merge E F G H A B C D Merge A B C D E F G H

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7. 7 (T(n/2)=2T(n/4)+n/2)

8. 8 Example 2: Whole Process A L G O R I T H M S Divide A L G O R | J T H M S Divide A L | G O R | J T | H M S Divide A | L | G | O R | J | T | H | M S Divide A | L | G | O | R | J | T | H | M | S Merge A | L | G | O R | J | T | H | M S Merge A L | G O R | J T | H M S Merge A G L O R | H J M S T Merge A G H J L M O R S T

9. 9 R1: g, q, p, r, m  1, 2, 3, 4, 5 R2: q, p, r, g, m  2, 3, 4, 1, 5 We can assume that the first rank is 1, 2,,…n. example

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17. 17 Merge and Count Process: Another example. 1, 2, 8, 10, 11, 12; 3, 4, 5, 6, 7, 9; 4 4 4 4 4 3 # of inversions 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Time for merge : O(n).

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28. 28 ((0.5  ) 2 +(0.5  ) 2 ) 0.5 =(0.5  2 ) 0.5 < 

29. 29 Can be done in O(n) time if we sort all the points first.

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31. Sort the points according to the x-coordinates and get a list ListX={p 1, …, p n }. Closest-Pair(p 1, …, p n ) { Compute separation line L such that half the points List1X={p 1, p 2, …, P 0.5n } are on one side and the other half List2X={p 0.5n+1, …p n } on the other side. (  1, List1Y) = Closest-Pair(List1X) ; (  2, LIst2Y)= Closest-Pair(List2X) (  = min((  1,  2)) Merge: merge List1Y and list2Y to form ListY. (O(n) time) Delete all points further than  from separation line L (O(n) time) Create array new-yg for remaining points sorted by y according to ListY (O(n) time). for (i=0; i<= size of new-yg; i++) for (j=1; j<=11; j++) (the nested loop takes O(n) time) if (d(new-yg[i], new-yg[i+j]<  ) then  =d(new-yg[i], new-yg[j]; return ( , ListY). }

32. 32 Example 1 Input: points p 1, p 2,…,p 8 in a plane. p 1 =(1,4), p 2 =(2,5), p 3 =(4,2), p 4 =(7,2), p 5 =(10,3), p 6 =(13,4),p 7 =(14,4), p 8 =(15,3) p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0

33. 33 Find a line L 1 such that 4 points are on one side and the other 4 points are on the other side. p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1

34. 34 Consider the left four points p 1,p 2,p 3,p 4. Find a line L 2 such that 2 points are on one side and the other 2 points are on the other side. δ 1 =Closest-pair (Region 1) = dist(p 1,p 2 ) =. δ 2 =Closest- pair (Region 2) = dist(p 3,p 4 ) = 3 δ=min(δ 1,δ 2 )= p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1 L2L2 Reg 1 Reg 2 3

35. 35 Delete the points in Region 1 and 2 further than δ= from L 2 Compare the distance dist(p 1,p 3 ) with δ. Here dist(p 1,p 3 ) = > δ, δ is not updated. Closest-pair (Region 1 and 2) = dist(p 1,p 2 ) =. p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1 L2L2 Reg 1 Reg 2

36. 36 For the four points p 5,p 6,p 7,p 8, we use a line L 3 to divide them and use similar method to find δ 1 = Closest-pair (Region 3) = dist(p 5,p 6 ) = δ 2 =Closest- pair (Region 4) = dist(p 7,p 8 ) = δ= min(δ 1,δ 2 )= p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1 L2L2 Reg 1 Reg 2 L3L3 Reg 3Reg 4

37. 37 Delete the points in Region 1 and 2 further than δ= from L 3 Compare the distance dist(p 6,p 8 ) with δ. Here dist(p 6,p 8 ) = >δ, δ is not updated. Compare the distance dist(p 6,p 7 ) with δ. Here dist(p 6,p 7 ) = 1 < δ, δ is updated. Closest-pair (Region 3 and 4) = dist(p 6,p 7 ) = 1. p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1 L2L2 Reg 1 Reg 2 L3L3 Reg 3Reg 4 1

38. 38 Consider Region 1, 2 and Region 3, 4 as two large regions. δ 1 =Closest-pair (Region 1 and 2) = dist(p 1,p 2 ) = δ 2 =Closest- pair (Region 3 and 4) = dist(p 6,p 7 ) = 1 δ= min(δ 1,δ 2 )= 1 p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1 L2L2 Reg 1 Reg 2 L3L3 Reg 3Reg 4 1

39. 39 Delete the points in Region 1, 2, 3 and 4 further than δ= 1 from L 1 Here only one point p4 is left, δ is not updated. Closest-pair (Region 1, 2, 3 and 4) = dist(p 6,p 7 ) = 1. p2p2 p3p3 p1p1 p4p4 p7p7 p8p8 p6p6 p5p5 246810121416 2 4 6 0 L1L1 L2L2 Reg 1 Reg 2 L3L3 Reg 3Reg 4 1

40. 40 Closest pair of points Question: How to handle the case, where two points can have the same x-coordinate? For those points with the same x-coordinates, sort them based on y-coordinates.