1. 1 Comments for Assignment 1 Q1 is basic part and is also part of outcome 1. Q2 is for outcome 1. Q3 is for outcome 2. To pass outcome 1, you need to get >=40 marks in total for Q1+Q2. To pass outcome 2, you need to give the algorithm for Q3. You have MANY chances to pass the three outcomes. You pass outcome 1, 2 or 3 for the whole course, if you pass each outcome ONCE. The chances are Assignment 1-4 plus mid-term in Week 7. I want every student to pass the three outcomes before the final exam. Do not worry about this.

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

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31. 31 Closest-Pair(p 1, …, p n ) { xg: array sorted based on x in increasing order yg: array sorted based on y in increasing order Compute separation line L such that half the points are on one side and half on the other side. Create sorted xg1 (left half), xg2(right half), yg1(left half) and yg2 (right half). O(n)  1 = Closest-Pair(xg1,yg1) (  2 = Closest-Pair(xg2,yg2) (  = min((  1,  2)) Merge: Delete all points further than  from separation line L (O(n) time) Create array new-yg for remaining points sorted by y-coordinatein (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 . }

32. 32 The closest Pair ( see the program ) Combine (The last step) (0,0) x y 1020304050607080 10 20 30 40 50 60 70 80 (0,5) (5,50) (17,40) (10,4) (12,60) (69,55) (74,43) (65,14) (60,1) (78,2) (31,0) (48,0) (37,8) (30,13) (43,9) (42,19) (38,20) (50,25) (41,30) (39,30) (36,40)(46,40) (40,50) (49,55) (34,60) (44,65) (33,70) (47,75)

33. 33 The closest Pair ( see the program ) Combine (The last step) (0,0) x y 1020304050607080 10 20 30 40 50 60 70 80 (0,5) (5,50) (17,40) (10,4) (12,60) (69,55) (74,43) (65,14) (60,1) (78,2) (31,0) (48,0) (37,8) (30,13) (43,9) (42,19) (38,20) (50,25) (41,30) (39,30) (36,40)(46,40) (40,50) (49,55) (34,60) (44,65) (33,70) (47,75) * * * 1 st :blue. 2 nd :green, 3 rd : orange

34. 34 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

35. 35 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

36. 36 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

37. 37 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

38. 38 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

39. 39 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

40. 40 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

41. 41 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

42. 42 Closest pair of points Question: How to handle the case, where two points can have the same x-coordinate?