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See the demo slide on the website.

See the demo slide on the website. Example 1: Whole Process. H G F E D C B A Divide H G F E D C B A Divide 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

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See the demo slide on the website.

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  1. See the demo slide on the website.

  2. Example 1: Whole Process H G F E D C B A Divide H G F ED C B A Divide H GF ED CB A Divide H G FE DCBA Merge G HE FC DA B Merge E F G HA B C D Merge A B C D E F G H

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

  4. 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 MergeA G H J L M O R S T

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

  6. 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).

  7. ((0.5)2+(0.5)2)0.5=(0.52)0.5<

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

  9. Sort the points according to the x-coordinates and get a list ListX={p1, …, pn}. Closest-Pair(p1, …, pn) { Compute separation line L such that half the points List1X={p1, p2, …, P0.5n} are on one side and the other half List2X={p0.5n+1, …pn} 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). }

  10. Example 1 • Input: points p1, p2,…,p8 in a plane. • p1=(1,4), p2=(2,5), p3=(4,2), p4=(7,2), p5=(10,3), p6=(13,4),p7=(14,4), p8=(15,3) 6 p6 p7 p2 4 p1 p8 p5 2 p4 p3 0 2 4 6 8 10 12 14 16

  11. Find a line L1 such that 4 points are on one side and the other 4 points are on the other side. 6 L1 p6 p7 p2 4 p1 p8 p5 2 p4 p3 0 2 4 6 8 10 12 14 16

  12. Consider the left four points p1,p2,p3,p4. Find a line L2 such that 2 points are on one side and the other 2 points are on the other side.δ1 =Closest-pair(Region 1) = dist(p1,p2) = . δ2 =Closest- pair (Region 2) = dist(p3,p4) = 3δ=min(δ1 ,δ2)= 6 L2 Reg 2 L1 p6 p7 p1 4 p2 3 p8 p5 2 p4 p3 Reg 1 0 2 4 6 8 10 12 14 16

  13. Delete the points in Region 1 and 2 further than δ= from L2 • Compare the distance dist(p1,p3) with δ. Here dist(p1,p3) = > δ, δ is not updated. • Closest-pair(Region 1 and 2) = dist(p1,p2) = . 6 L2 Reg 2 L1 p6 p7 p1 4 p8 p5 2 p4 p3 Reg 1 0 2 4 6 8 10 12 14 16

  14. For the four points p5,p6,p7,p8, we use a line L3 to divide them and use similar method to find δ1 =Closest-pair(Region 3) = dist(p5,p6) = δ2 =Closest- pair (Region 4) = dist(p7,p8) = δ= min(δ1 ,δ2)= 6 L3 L2 Reg 2 L1 Reg 3 Reg 4 p6 p7 p1 4 p2 p8 p5 2 p4 p3 Reg 1 0 2 4 6 8 10 12 14 16

  15. Delete the points in Region 1 and 2 further than δ= from L3 • Compare the distance dist(p6,p8) with δ. Here dist(p6,p8) = >δ, δ is not updated. • Compare the distance dist(p6,p7) with δ. Here dist(p6,p7) = 1 < δ, δ is updated. • Closest-pair(Region 3 and 4) = dist(p6,p7) = 1. 6 L3 L2 Reg 2 L1 Reg 3 Reg 4 p6 p7 1 p1 4 p2 p8 p5 2 p4 p3 Reg 1 0 2 4 6 8 10 12 14 16

  16. Consider Region 1 , 2 and Region 3, 4 as two large regions. δ1 =Closest-pair(Region 1 and 2) = dist(p1,p2) = δ2 =Closest- pair (Region 3 and 4) = dist(p6,p7) = 1 δ= min(δ1 ,δ2)= 1 6 L3 L2 Reg 2 L1 Reg 3 Reg 4 p6 p7 1 p1 4 p2 p8 p5 2 p4 p3 Reg 1 0 2 4 6 8 10 12 14 16

  17. Delete the points in Region 1, 2, 3 and 4 further than δ= 1 from L1 • Here only one point p4 is left, δ is not updated. • Closest-pair(Region 1, 2, 3 and 4) = dist(p6,p7) = 1. 6 L3 L2 Reg 2 L1 Reg 3 Reg 4 p6 p7 1 p1 4 p2 p8 p5 2 p3 p4 Reg 1 0 2 4 6 8 10 12 14 16

  18. 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.

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