2.1 DynamicProgramming

8/13/2019 2.1 DynamicProgramming

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Assembly Line SchedulingProblem


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Step 2: A recursive solution


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Step 3: Computing the fastest times


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PRINT-STATIONS (l, n)


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Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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Matrix Chain MultiplicationProblem


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Step 1: The structure of an optimal parenthesization

let us adopt the notation Aij , where i j, for the matrix that resultsfrom evaluating the product Ai Ai+1 A j.

Ifthe problem is nontrivial, i.e., i < j, then any parenthesization of the

product Ai Ai+1 A j must split the product between A k and A k+1 for someinteger k in the range i k < j. That is, for some value of k, we firstcompute the matrices Aik and A k+1j and then multiply them together to

produce the final product A ij.

The cost of this parenthesization is thus the cost of computing thematrix Aik , plus the cost of computing Ak+1j , plus the cost ofmultiplying them together.


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Step 2: A recursive solution

C i h Th M G Hill C i I P i i i d f d i di l


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Step 3: Computing the optimal costs


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Cop right The McGra Hill Companies Inc Permission req ired for reprod ction or displa


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Copyright The McGraw Hill Companies Inc Permission required for reproduction or display


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Copyright The McGraw Hill Companies Inc Permission required for reproduction or display


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Copyright The McGraw-Hill Companies Inc Permission required for reproduction or display


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Longest Common SubsequenceProblem



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Copyright The McGraw Hill Companies, Inc. Permission required for reproduction or display.



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Copy g t e cG aw Co pa es, c. e ss o equ ed o ep oduct o o d sp ay.

0 if i=0, or j=0c[i,j] = c[i-1, j-1]+1 if i, j>0 and xi= y j,

max{ c[i-1, j], c[i, j-1]} if i, j>0 and xi y j,



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py g p , q p p y



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py g p q p p y



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Optimal Binary Search TreeProblem



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All Pair Shortest Path

The structure of a shortest path


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The structure of a shortest path

For the all-pairs shortest-paths problem on a graph G = (V, E), all subpaths of a shortest path should also be shortest paths.

Suppose that the graph is represented by an adjacency matrix W =(wij ). Consider a shortest path p from vertex i to vertex j, and

suppose that p contains at most m edges.

Assuming that there are no negative-weight cycles, m is finite. If i = j, then p has weight 0 and no edges.

If vertices i and j are distinct, then we decompose path p into ,where path p now contains at most m - 1 edges.

p is a shortest path from i to k, and so (i, j) = (i, k) + w kj.


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A recursive solution to the all-pairs shortest-paths problem

The latter equality follows since w jj = 0 for all j


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Computing the shortest-path weights bottom up


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0/1 Knapsack Problem


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Given a knapsack with maximum capacity W , and a set S consisting of n items.

Each item i has some weight wi and benefit value bi (all wi , b i and W are integer values)

Problem: How to pack the knapsack to achieve maximum

total value of packed items?

0-1 Knapsack problem


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W = 20

w i b i

109

85

54

43

32

Items

This is a knapsackMax weight: W = 20

0-1 Knapsack problem:


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T i

i

T i

i W wb subject tomax

Problem, in other words, is to find

The problem is called a 0 -1 problem, because each itemmust be entirely accepted or rejected.Just another version of this problem is the Fractional

Knapsack Problem , where we can take fractions of items.


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Defining a Subproblem

If items are labeled 1..n , then a subproblem would be to find anoptimal solution for S k = {items labeled 1, 2, .. k}

Recursive Formula for subproblems

else }],1[],,1[max{ if ],1[

],[k k

k

bwwk Bwk B

wwwk Bwk B

It means, that the best subset of S k that has total weight w is oneof the two:1) the best subset of S k-1 that has total weight w, or 2) the best subset of S k-1 that has total weight w-wk plus the item k


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The best subset of S k that has the total weight w, either contains item k or not.

First case: wk >w . Item k cant be part of thesolution, since if it was, the total weight would be> w , which is unacceptable

Second case: wk


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0-1 Knapsack Algorithmfor w = 0 to W

B[0,w] = 0for i = 0 to n

B[i,0] = 0

for w = 0 to Wif w i B[i-1,w]B[i,w] = b

i + B[i-1,w- w

i]

elseB[i,w] = B[i-1,w]

else B[i,w] = B[i-1,w] // w i > w


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Example:

n = 4 (Number of elements)W = 5 (max weight)Elements (weight, benefit):(2,3), (3,4), (4,5), (5,6)


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Example (2)

for w = 0 to WB[0,w] = 0

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

4


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Example (3)

for i = 0 to nB[i,0] = 0

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0

4


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Example (4)

if w i B[i-1,w]

B[i,w] = b i + B[i-1,w- w i]else

B[i,w] = B[i-1,w]else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=1

b i=3w i=2w=1 w-w i =-1

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0


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Example (5)

if w i B[i-1,w] B[i,w] = b i + B[i-1,w- w i]


else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=1

b i=3w i=2w=2 w-w i =0

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3


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Example (6)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=1

b i=3w i=2w=3 w-w i=1

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3


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Example (7)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=1

b i=3w i=2w=4 w-w i=2

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3


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Example (8)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=1

b i=3w i=2w=5 w-w i=2

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3

3


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Example (9)

if w i B[i-1,w]B[i,w] = b i + B[i-1,w- w i]


else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=2

b i=4w i=3w=1 w-w i=-2

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3

3

0

I


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Example (10)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=2

b i=4w i=3w=2 w-w i=-1

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3

3

0

3

I


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Example (11)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=2

b i=4w i=3w=3 w-w i=0

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3

3

0

3

4

I


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Example (12)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=2

b i=4w i=3w=4 w-w i=1

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3

3

0

3

4

4


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It


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Example (14)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W01

2

3

4

5

i 0 1 2 3

0 0 0 0i=3

b i=5w i=4w=1..3

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0

3

3

3

3

00

3

4

4

7

0

3

4

It


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Example (15)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W

01

2

3

4

5

i 0 1 2 3

0 0 0 0i=3

b i=5w i=4w=4w- w i=0

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0 00

3

4

4

7

0

3

4

5

3

3

3

3

It


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Example (15)


else B[i,w] = B[i-1,w]

else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W

01

2

3

4

5

i 0 1 2 3

0 0 0 0i=3

b i=5w i=4w=5w- w i=1

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0 00

3

4

4

7

0

3

4

5

7

3

3

3

3

Items:


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Example (16)



else B[i,w] = B[i-1,w] // w i > w

00

0

0

0

0

W

01

2

3

4

5

i 0 1 2 3

0 0 0 0i=4

b i=5w i=4w=1..4

Items:1: (2,3)2: (3,4)3: (4,5)4: (5,6)

4

0 00

3

4

4

7

0

3

4

5

7

0

3

4

5

3

3

3

3


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2.1 DynamicProgramming

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