comments for assignment 1
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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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
2 4 6 8 10 12 14 16
2
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
2 4 6 8 10 12 14 16
2
4
6
0
L1L2
Reg 1
Reg 2 L3
10 2
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
2 4 6 8 10 12 14 16
2
4
6
0
L1L2
Reg 1
Reg 2 L3Reg 3 Reg 4
2
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
2 4 6 8 10 12 14 16
2
4
6
0
L1L2
Reg 1
Reg 2 L3Reg 3 Reg 4
1
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
2
4
6
0
L1L2
Reg 1
Reg 2 L3Reg 3 Reg 4
1
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Closest pair of points
Question: How to handle the case, where two points can have the same x-coordinate?