Dynamic Programming

Dynamic Programming

Heejin Park

College of Information and Communications

Hanyang University

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Content

Introduction

Assembly-line scheduling

Matrix-chain multiplication

Elements of dynamic programming

Longest common subsequence

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Matrix-chain multiplication Given a chain of n matrices <A1, A2, ..., An>,

compute the product A1A2…An.

The number of ways to compute the product A1A2 …An.

= the number of ways to fully parenthesize the product. The number of ways to compute the product A1A2A3 is 2.

(A1 A2) A3

A1(A2 A3)

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The order of multiplications

The order of multiplications does not change the value of the product because matrix multiplication is associative.

For example, whether the left multiplication is done first or the right multiplication is done first does not matter.

( A1· A2) · A3 = A1· (A2 · A3)

However, the order of multiplication affects the number of scalar multiplications needed to compute the product.

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Multiplying two matrices A and B

We can multiply them if they are compatible: the number of columns of A must equal the number of rows of B.

If A is a p×q matrix and B is a q×r matrix,

the resulting matrix is a p×r matrix.

x =(A 2x3)

(B 3x2) (C 2x2)

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The number of scalar multiplications to multiply A and B. It is pqr because we compute pr elements and computing ea

ch element needs q scalar multiplications.

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The order of multiplications affects the number of scalar multiplications.

Computing A1A2A3 where A1: 10×100 A2: 100×5 A3 : 5×50

(A1 A2) A3 (A1 A2) = 10*100*5 = 5000, (10 ×5) A3 = 10*5*50 = 2500

=>5000 + 2500 = 7,500

A1 (A2 A3) (A2 A3) = 100*5*50 =25000, A1(100 × 50) = 10 *100*50 = 50000

=>25000 + 50000 = 75,000

Computing (A1 A2) A3 is 10 times faster.

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Matrix-chain multiplication problem

Given a chain A1, A2, ..., An of n matrices, where matrix Ai has dimension pi-1×pi, find the order of matrix multiplications minimizing the scalar multiplications to compute the product.

That is, to fully parenthesize the product of matrices minimizing scalar multiplications. For example, for the product A1 A2 A3 A4, a fully parenthe

sization is ((A1 A2) A3) A4.

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Solutions of the matrix-chain multiplication problem

Brute-force approach Enumerate all possible parenthesizations. Compute the number of scalar multiplications of each pa

renthesization. Select the parenthesization needing the least number of s

calar multiplications.

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The Brute-force approach is inefficient.

The number of enumerated parenthesizations is Ω(4n/n3/2).

The number of parenthesizations of a product of n matrices, denoted by P(n), is as follows.

1

1

)()(

1

)( n

k

knPkPnP

if n=1

if n≥2

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The product A1 A2 A3 A4 can be fully parenthesized in five distinct ways.

A1(A2(A3 A4)), A1((A2 A3) A4),

(A1 A2)(A3 A4),

(A1(A2 A3))A4, ((A1 A2) A3) A4.

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P(n): The number of alternative parenthesizations of a sequence of n matrices. The split between the two subproducts may occur betwee

n the kth and (k + 1)st matrices for any k = 1, 2, ..., n – 1.

1

1

)()(

1

)( n

k

knPkPnP

if n=1

if n≥2

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Dynamic programming

Optimal substructure m[i,j]: The minimum number of scalar multiplications for comp

uting Ai Ai+1 …Aj.

matrix Ai : pi-1 × pi

computing Ai..k Ak+1..j takes pi-1 pk pj scalar multiplications. s[i, j] stores the optimal k for tracing the optimal solution.

}],1[],[{min

0],[

1 jkijki

pppjkmkimjimif i=j

if i<j

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i j 1 2 3 4 5 6

1 0

2 0

3 0

4 0

5 0

6 0

i j 2 3 4 5 6

1

2

3

4

5

s

541

531

521

]5,5[]4,2[

]5,4[]3,2[

]5,3[]2,2[

min]5,2[

pppmm

pppmm

pppmm

m

i=2, j=5, i≤k<j

1

2

3

4

5

1

3

3

5

3

0 +2500+35·15·20=13000,

2625+1000+35·5·20=7125,

4375+0 +35·10·20 =11375

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3

3

3

m

15750 7875 9375 11875 15125

2625 4375 7125 10500

750 2500 5375

1000 3500

5000

3

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Running time O(n3) time in total

Θ(n2) subproblems O(n) time for each subproblem

Space consumption Θ(n2) space to store the m and s tables.

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Content

Introduction

Assembly-line scheduling



Longest common subsequence

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Elements of dynamic programming Optimal substructure Overlapping subproblems


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Element of dynamic programming

Subtleties Unweighted longest simple path problem

Does it have optimal substructure?

u w vp1 p2

p

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q r

s t

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Overlapping subproblems When a recursive algorithm revisits the same problem over an

d over again, that is the optimization problem has overlapping subproblems.

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1...4

1...1 2...4 1...2 3...4 1...3 4...4

2...2 3...4 2...3 4...4 1...1 2...2 3...3 4...4 1...1 2...3 1...2 3...3

3...3 4...4 2...2 3...3 2...2 3...3 1...1 2...2

Matrix chain multiplication: top-down vs. bottom-up

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Memoization Recursive solution but solve each subproblem only once. Fills the table in recursive way. In most case, it is slower than dynamic programming. It is useful when only a part of subproblems is solved.

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The running time of a dynamic-programming algorithm depends on the product of two factors.

The number of subproblems overall. How many choices each subproblem has.

Assembly line scheduling Θ(n) subproblems · 2 choices = Θ(n)

Matrix chain multiplication Θ(n2) subproblems · (n-1) choices = O(n3)

Dynamic Programming

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