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Comments for Assignment 1Q1 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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See the demo slide on the website.

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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

Merge A B C D E F G H

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

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Example 2: Whole Process

A L G O R I T H M SDivide A L G O R | J T H M SDivide A L | G O R | J T | H M SDivide A | L | G | O R | J | T | H | M SDivide A | L | G | O | R | J | T | H | M | SMerge A | L | G | O R | J | T | H | M SMerge A L | G O R | J T | H M SMerge A G L O R | H J M S TMerge A G H J L M O R S T

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R1: g, q, p, r, m1, 2, 3, 4, 5

R2: q, p, r, g, m2, 3, 4, 1, 5

We can assume that the first rank is 1, 2, ,…n.

example

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

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Closest-Pair(p1, …, pn) {xg: array sorted based on x in increasing orderyg: array sorted based on y in increasing orderCompute separation line L such that half the points are on one side and half on th

e 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 .}

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The closest Pair (see the program )• Combine (The last step)

(0,0)

x

y

10 20 30 40 50 60 70 80

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)

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The closest Pair (see the program )• Combine (The last step)

(0,0)

x

y

10 20 30 40 50 60 70 80

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)*

*

*

1st:blue. 2nd:green,

3rd: orange

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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)

p2

p3

p1

p4

p7

p8

p6

p5

2 4 6 8 10 12 14 16

2

4

6

0

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Find a line L1 such that 4 points are on one side and the other 4 points are on the other side.

p2

p3

p1

p4

p7

p8

p6

p5

2 4 6 8 10 12 14 16

2

4

6

0

L1

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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)=

2

p2

p3

p1

p4

p7

p8

p6

p5

2 4 6 8 10 12 14 16

2

4

6

0

L1L2

Reg 1

Reg 2

3

2

2

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• 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) = .

p3

p1

p4

p7

p8

p6

p5

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4

6

0

L1L2

Reg 1

Reg 2

2

2

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• 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)=

2

p2

p3

p1

p4

p7

p8

p6

p5

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2

4

6

0

L1L2

Reg 1

Reg 2 L3

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10

2

2

Reg 3 Reg 4

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• 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.

2

p2

p3

p1

p4

p7

p8

p6

p5

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6

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L1L2

Reg 1

Reg 2 L3Reg 3 Reg 4

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5

1

5

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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

2

p2

p3

p1

p4

p7

p8

p6

p5

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4

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L1L2

Reg 1

Reg 2 L3Reg 3 Reg 4

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2

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• 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.

2

p2

p3

p1

p4

p7

p8

p6

p5

2 4 6 8 10 12 14 16

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4

6

0

L1L2

Reg 1

Reg 2 L3Reg 3 Reg 4

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Closest pair of points

Question: How to handle the case, where two points can have the same x-coordinate?