Algorithms and Data StructuresAlgorithms and Data Structures 10th Lecture: Dynamic Programming Yutaka Watanobe, Jie Huang, Yan Pei, Wenxi Chen, Qiangfu Zhao, Wanming Chu University

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Algorithms and Data Structures10th Lecture: Dynamic Programming

Yutaka Watanobe, Jie Huang, Yan Pei,Wenxi Chen, Qiangfu Zhao, Wanming Chu

University of Aizu

Last Updated: 2020/12/1

Y. Watanobe, J. Huang, Y. Pei, W. Chen, Q. Zhao, W. Chu University of Aizu

Algorithms and Data Structures

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Outline

Dynamic ProgrammingRecursive vs. Dynamic ProgrammingMatrix Chain MultiplicationLongest Common Subsequence



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Dynamic Programming

Dynamic programming solves a problem by combining thesolutions to subproblems. (The word ”programming” does notmean writing computer code)The development of a dynamic programming algorithm can bebroken into a sequence of four steps:

1 Characterize the structure of an optimal solution.2 Recursively define the value of an optimal solution.3 Compute the value of an optimal solution in a bottom-up fashion.4 Construct an optimal solution from computed information.



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Recursive vs. Dynamic Programming: FibonacciNumber (1)

Fibonacci Numbers: fn = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .A recursive formula for Fibonacci Numbers:

f (n) =

1 if n = 01 if n = 1f (n − 2) + f (n − 1)



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A recursive program to calculate the n-th fibonacci number.

fibonacci(n)

if n == 0 || n == 1

return 1

return fibonacci(n - 2) + fibonacci(n - 1)



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A function f (5) calls f (4) and f (3), but f (4) calls f (3) again.

f(5)

f(2)

f(1) f(0)

f(3)

f(2) f(1)

f(2)

f(1) f(0)

f(4)

f(1) f(0)

f(3)

f(1)



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A recursive program with memorization to calculate the n-thfibonacci number.

fibonacci(n)

if n == 0 || n == 1

return F[n] = 1

if F[n] is already defined

return F[n]

return F[n] = fibonacci(n - 2) + fibonacci(n - 1)



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The generation of Fibonacci Numbers with memorization.

f(5)

F[2]f(3)

f(2) F[1]

f(4)

f(1) f(0)

F[3]

F[1] =

F[2] =

F[3] =

F[4] =

F[5] =



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The generation of Fibonacci Numbers by Dynamic Programming(implementation by a loop).

fibonacci(n)

F[0] = 1

F[1] = 1

for i = 2 to n

F[i] = F[i - 2] + F[i - 1]



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Recursive vs. Dynamic Programming: BinomialCoefficient (1)

We want to compare two simple programs.Consider the binomial coefficient defined recursively as follows:for any natural numbers k ,n ∈ N,(

n0

)=

(nn

)= 1

(nk

)=

(n − 1

k

)+

(n − 1k − 1

)



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Pascal’s triangle

n = 0 1

n = 1 1 1

n = 2 1 2 1

n = 3 1 3 3 1

n = 4 1 4 6 4 1

n = 5 1 5 10 10 5 1

n = 6 1 6 15 20 15 6 1



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recursiveBinomial(n, k)

if k == 0 or k == n

return 1

else

u = recursiveBinomial(n-1, k)

v = recursiveBinomial(n-1, k-1)

return u + v



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The following dynamic programming reduces the running time ofa recursive function.

dynamicBinomial(n, k)

if dp[n, k] is already defined

return dp[n, k]

else

if k == 0 or k == n

dp[n, k] = 1

else

u = dynamicBiomial(n-1, k)

v = dynamicBiomial(n-1, k-1)

dp[n, k] = u + v

return dp[n, k]



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Matrix-chain Multiplication (1)

We can multiply two matrices A and B only if they are compatible.If A is an l × m matrix and B is an m × n matrix, the resultingmatrix C is an l × n matrix.The time to compute C is dominated by the number of scalarmultiplications l × m × n.

l

m

m

n

n

l

A B C



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We are given a sequence (chain) {A1,A2, ...,An} of n matrices tobe multiplied, and we wish to compute the product A1A2...An.We can evaluate the expression using the standard algorithm formultiplying pairs of matrices as a subroutine once we haveparenthesized it to resolve all ambiguities in how the matricesare multiplied together.A product of matrices is fully parenthesized if it is either a singlematrix or the product of two fully parenthesized matrix products,surrounded by parentheses.



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Matrix multiplication is associative, and so all parenthesizationyield the same product.For example, if the chain of matrices is {A1,A2,A3,A4}, theproduct A1A2A3A4 can be fully parenthesized in five distinctways:

(A1(A2(A3A4)))(A1((A2A3)A4))((A1A2)(A3A4))((A1(A2A3))A4)(((A1A2)A3)A4)



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Matrix-chain Multiplication: Example 1 (1)

Matrix-chain multiplication of matrices A1A2...A6 where Ai is api−1 × pi matrix.

p0

p1p2 p3 p4 p5 p6

A 1 A 2 A 3 A 4 A 5 A 6



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Possible parenthesization: From the left to the right.

4 x 2 x 3 = 24

4 x 3 x 1 = 12

4 x 1 x 2 = 8

4 x 2 x 2 = 16

4 x 2 x 3 = 24

24 + 12 + 8 + 16 + 24 = 84



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Possible parenthesization: From the right to the left.

2 x 2 x 3 = 12

1 x 2 x 3 = 6

3 x 1 x 3 = 9

2 x 3 x 3 = 18

4 x 2 x 3 = 24

12 + 6 + 9 + 18 + 24 = 69



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Possible parenthesization: the optimal solution.

2 x 3 x 1 = 6

1 x 2 x 2 = 4

4 x 2 x 1 = 8

1 x 2 x 3 = 6

4 x 1 x 3 = 12

6 + 4 + 8 + 6 + 12 = 36



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Matrix-chain Multiplication: Example 2

Consider the problem of a chaining {A1,A2,A3}. Suppose thatthe dimensions of the matrices are 10× 100, 100× 5, and 5× 50respectively.

((A1A2)A3)cost of (A1A2) = A12 : 10 × 100 × 5 = 5000cost of (A12A3) : 10 × 5 × 50 = 2500cost of ((A1A2)A3) : 5000 + 2500 = 7500

(A1(A2A3))cost of (A2A3) = A23 : 100 × 5 × 50 = 25000cost of (A1A23) : 10 × 100 × 50 = 50000cost of (A1(A2A3)) : 25000 + 50000 = 75000



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The Structure of an Optimal Parenthesization (1)

Let us adopt the notation Ai..j , where i ≤ j , for the matrix thatresults from evaluating the product AiAi+1..Aj .If the problem is nontrivial, i.e., i < j , then any parenthesizationof the product AiAi+1...Aj must split the product between Ak andAk+1 for some integer k in the range i ≤ k < j .(AiAi+1...Ak )(Ak+1....Aj)



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The Structure of an Optimal Parenthesization (2)

For some value of k , we first compute the matrices Ai..k andAk+1..j and then multiply them together to produce the finalproduct Ai..j .The cost of this parenthesization is thus the cost of computingthe matrix Ai..k , plus the cost of computing Ak+1..j , plus the costof multiplying them together.



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A Recursive Solution

Let m[i , j] be the minimum number of scalar multiplicationsneeded to compute the matrix Ai..j ; for the full problem, the costof a cheapest way to compute A1..n would thus be m[1,n].We can define m[i , j] recursively as follows.

m[i , j] ={

0 if i = jmini≤k

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A Recursive Solution: Implementation

matrixChainOrder(p)

n = p.length-1

for i = 1 to n

m[i, i] = 0

for l = 2 to n

for i = 1 to n-l+1

j = i+l-1

m[i, j] = INF

for k = i to j-1

q = m[i,k] + m[k+1, j] + p[i-1]*p[k]*p[j]

m[i,j] = min(m[i,j], q)



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Computing the Optimal Costs (1)

4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

Initializationm[i , i] = 0



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4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

60

m[1,2] = min1≤k

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4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

60 15

m[2,3] = min2≤k

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4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

60 15 30

m[3,4] = min3≤k

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4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

60 15 30

27

m[1,3] = min1≤k

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4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

60 15 30

27 33

m[2,4] = min2≤k

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4 3 5 1 6

p0 p1 p2 p3 p4

1

2

3

4 1

2

3

4

j i

m[i,j]

A1 A2 A3 A4

0 0 0 0

60 15 30

27 33

51

m[1,4] = min1≤k

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Computing the Optimal Costs: Example 2 (1)

15,125

11,875 10,500

9,375 7,125

7,875 4,375 2,500 3,500

5,375

15,750 2,625 750 1,000

0 0 0 0 0 0

5,0001

2

3

4

5

6 1

2

3

4

5

6

A1 A2 A3 A4 A5 A6

30×35 35×15 15×5 5×10 10×20 20×25

ij

m

p 30 35 15 5 10 20 25



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Computing the Optimal Costs: Example 2 (2)

m[2,5] =

min

m[2,2] + m[3,5] + p1p2p5 = 0 + 2500 + 35 × 15 × 20 = 13000m[2,3] + m[4,5] + p1p3p5 = 2625 + 1000 + 35 × 5 × 20 = 7125m[2,4] + m[5,5] + p1p4p5 = 4375 + 0 + 35 × 10 × 20 = 11375= 7125



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Matrix-chain Multiplication by DP: Complexity

A simple inspection of the nested loop structure ofmatrixChainOrder(p) yields a running time of O(n3) for thealgorithm.The loops are nested three deep, and each loop index (l , i , andk ) takes on at most n − 1 values.



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Longest Common Subsequence (LCS)

Given two sequences X and Y , a sequence Z is a commonsubsequence of X and Y if Z is a subsequence of both X and Y .For example, if X = {A,B,C,B,D,A,B} andY = {B,D,C,A,B,A}, the sequence {B,C,A} is a commonsubsequence of both X and Y .The sequence {B,C,A} is not a longest common subsequence(LCS) of X and Y , since it has length 3 and the sequence{B,C,B,A}, which is also common to both X and Y , has length4.The sequence {B,C,B,A} is an LCS of X and Y , since there isno common subsequence of length 5 or greater.



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LCS: Recursive Solution (1)

In the longest common subsequence problem, we are given twosequences X = {x1, x2, ..., xm} and Y = {y1, y2, ..., yn} and wishto find a maximum-length common subsequence of X and Y .To be precise, given a sequence X = {x1, x2, ..., xm}, we definethe i th prefix of X , for i = 0,1, ...,m, as Xi = {x1, x2, ..., xi}. X0 isthe empty sequence.There are either one or two subproblems to examine whenfinding an LCS of X = {x1, x2, ..., xm} and Y = {y1, y2, ..., yn}.if xm = yn, we must find an LCS of Xm−1 and Yn−1. Appendingxm = yn to this LCS yields an LCS of X and Y .if xm ̸= yn, then we must solve two subproblems: finding an LCSof Xm−1 and Y and finding an LCS of X and Yn−1. Whichever ofthese two LCS’s is longer is an LCS of X and Y .



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To find an LCS of X and Y , we may need to find the LCS’s of Xand Yn−1 and of Xm−1 and Y .But each of these subproblems has the subsubproblem of findingthe LCS of Xm−1 and Yn−1.Many other subproblems share subsubproblems.



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Let us define c[i , j] to be the length of an LCS of the sequencesXi and Yj .The optimal substructure of the LCS problem gives the recursiveformula:

c[i , j] =

0 if i = 0 or j = 0c[i − 1, j − 1] + 1 if i , j > 0 and xi = yjmax(c[i , j − 1], c[i − 1, j]) if i , j > 0 and xi ̸= yj



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Computing the Length of an LCS (1)

B

0

0 0 0 0 0 0 0

0

0

0

0

0

0

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 41/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0

0

0

0

0

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 42/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0

0

0

0

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 43/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0 1 1 2 2 2 2

0

0

0

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 44/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0 1 1 2 2 2 2

0 1 1 2 2 3 3

0

0

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 45/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0 1 1 2 2 2 2

0 1 1 2 2 3 3

0 1 2 2 2 3 3

0

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 46/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0 1 1 2 2 2 2

0 1 1 2 2 3 3

0 1 2 2 2 3 3

0 1 2 2 3 3 4

0

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 47/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0 1 1 2 2 2 2

0 1 1 2 2 3 3

0 1 2 2 2 3 3

0 1 2 2 3 3 4

0 1 2 2 3 4 4

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



. . . . . . 48/1


B

0

0 0 0 0 0 0 0

0 0 0 0 1 1 1

0 1 1 1 1 2 2

0 1 1 2 2 2 2

0 1 1 2 2 3 3

0 1 2 2 2 3 3

0 1 2 2 3 3 4

0 1 2 2 3 4 4

D C A B A

A

B

C

B

D

A

B

1 2 3 4 5 6

0

1

2

3

4

5

6

7

i

j



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Dynamic Programming to Compute the Length of anLCS (1)

Since there are only O(mn) distinct subproblems, we can usedynamic programming to compute the solutions bottom up.

LCSLength(X, Y)

m = X.length

n = Y.length

for i = 1 to m

c[i,0] = 0

for j = 0 to n

c[0,j] = 0



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Dynamic Programming to Compute the Length of anLCS (2)

for i = 1 to m

for j = 1 to n

if X[i] == Y[j]

c[i,j] = c[i-1, j-1] + 1

else if c[i-1,j] >= c[i, j-1]

c[i,j] = c[i-1,j]

else

c[i,j] = c[i,j-1]

Note that there may be more than one solution if we have twosub-problems with the optimal solution.



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LCS by DP: Complexity

The running time of the procedure is O(mn), since each tableentry takes O(1) time to compute.



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Reference

1 Introduction to Algorithms (third edition), Thomas H.Cormen,Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. TheMIT Press, 2012.



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Algorithms and Data StructuresAlgorithms and Data Structures 10th Lecture: Dynamic Programming Yutaka Watanobe, Jie Huang, Yan Pei, Wenxi Chen, Qiangfu Zhao, Wanming Chu University