ics 353: design and analysis of algorithms

ICS 353: Design and Analysis of Algorithms

Dynamic Programming

King Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsInformation & Computer Science DepartmentInformation & Computer Science Department

Dynamic Programming092 ICS 353: Design and Analysis of Algorithms

Reading Assignment• M. Alsuwaiyel, Introduction to Algorithms:

Design Techniques and Analysis, World Scientific Publishing Co., Inc. 1999.• Chapter 7 Sections 1 - 4 and 6.

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Dynamic Programming• Dynamic Programming algorithms address

problems whose solution is recursive in nature, but has the following property: The direct implementation of the recursive solution results in identical recursive calls that are executed more than once.

• Dynamic programming implements such algorithms by evaluating the recurrence in a bottom-up manner, saving intermediate results that are later used in computing the desired solution

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Fibonacci Numbers•

• What is the recursive algorithm that computes Fibonacci numbers? What is its time complexity?• Note that it can be shown that

210

21

1

0

nfffff

nnn

251,2

nnf

4


Computing the Binomial Coefficient• Recursive Definition

• Actual Value

nk

kn

kn

nkk

kn

01

11

or01

!!!

knkn

kn

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Computing the Binomial Coefficient• What is the direct recursive algorithm for

computing the binomial coefficient? How much does it cost?• Note that

nnnn

nn n

2

!2/!2/!

2/

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Optimization Problems and Dynamic Programming• Optimization problems with certain properties make

another class of problems that can be solved more efficiently using dynamic programming.

• Development of a dynamic programming solution to an optimization problem involves four steps

• Characterize the structure of an optimal solution • Optimal substructures, where an optimal solution consists of sub-

solutions that are optimal.• Overlapping sub-problems where the space of sub-problems is small in

the sense that the algorithm solves the same sub-problems over and over rather than generating new sub-problems.

• Recursively define the value of an optimal solution.• Compute the value of an optimal solution in a bottom-up

manner.• Construct an optimal solution from the computed optimal value.

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Longest Common Subsequence Problem• Problem Definition: Given two strings A and B

over alphabet , determine the length of the longest subsequence that is common in A and B.

• A subsequence of A=a1a2…an is a string of the form ai1ai2…aik where 1i1<i2<…<ik n

• Example: Let = { x , y , z }, A = xyxyxxzy, B=yxyyzxy, and C= zzyyxyz

• LCS(A,B)=yxyzy Hence the length =• LCS(B,C)= Hence the length =• LCS(A,C)= Hence the length =

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Straight-Forward Solution• Brute-force search

• How many subsequences exist in a string of length n?

• How much time needed to check a string whether it is a subsequence of another string of length m?

• What is the time complexity of the brute-force search algorithm of finding the length of the longest common subsequence of two strings of sizes n and m?

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Dynamic Programming Solution

• Let L[i,j] denote the length of the longest common subsequence of a1a2…ai and b1b2…bj, which are substrings of A and B of lengths n and m, respectively. ThenL[i,j] = when i = 0 or j = 0L[i,j] = when i > 0, j > 0, ai=bj

L[i,j] = when i > 0, j > 0, aibj

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LCS AlgorithmAlgorithm LCS(A,B)Input: A and B strings of length n and m respectivelyOutput: Length of longest common subsequence of A and BInitialize L[i,0] and L[0,j] to zero;for i ← 1 to n do for j ← 1 to m do if ai = bj then

L[i,j] ← 1 + L[i-1,j-1] else L[i,j] ← max(L[i-1,j],L[i,j-1]) end if end for;end for;return L[n,m];

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Example (Q7.5 pp. 220)

• Find the length of the longest common subsequence of A=xzyzzyx and B=zxyyzxz

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Example (Cont.)x z y z z y x

0 0 0 0 0 0 0 0

z 0

x 0

y 0

y 0

z 0

x 0

z 0

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Complexity Analysis of LCS Algorithm• What is the time and space complexity of the

algorithm?

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Matrix Chain Multiplication• Assume Matrices A, B, and C have dimensions

210, 102, and 210 respectively. The number of scalar multiplications using the standard Matrix multiplication algorithm for• (A B) C is• A (B C) is

• Problem Statement: Find the order of multiplying n matrices in which the number of scalar multiplications is minimum.

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Straight-Forward Solution• Again, let us consider the brute-force method. We

need to compute the number of different ways that we can parenthesize the product of n matrices.• e.g. how many different orderings do we have for the

product of four matrices?• Let f(n) denote the number of ways to parenthesize the

product M1, M2, …, Mn.• (M1M2…Mk) (M k+1M k+2…Mn)

• What is f(2), f(3) and f(1)?

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Catalan Numbers

•

• Cn=f(n+1)

• Using Stirling’s Formula, it can be shown that f(n) is approximately

1221)(

nn

nnf

5.144

n

n

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Cost of Brute Force Method

• How many possibilities do we have for parenthesizing n matrices?

• How much does it cost to find the number of scalar multiplications for one parenthesized expression?

• Therefore, the total cost is

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The Recursive Solution• Since the number of columns of each matrix Mi is equal

to the number of rows of Mi+1, we only need to specify the number of rows of all the matrices, plus the number of columns of the last matrix, r1, r2, …, rn+1 respectively.

• Let the cost of multiplying the chain Mi…Mj (denoted by Mi,j) be C[i,j]

• If k is an index between i+1 and j, what is the cost of multiplying Mi,j considering multiplying Mi,k-1 with Mk,j?

• Therefore, C[1,n]=

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The Dynamic Programming AlgorithmC[1,1] C[1,2] C[1,3] C[1,4] C[1,5] C[1,6]

C[2,2] C[2,3] C[2,4] C[2,5] C[2,6]

C[3,3] C[3,4] C[3,5] C[3,6]

C[4,4] C[4,5] C[4,6]

C[5,5] C[5,6]

C[6,6]

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Example (Q7.11 pp. 221-222)• Given as input 2 , 3 , 6 , 4 , 2 , 7 compute the

minimum number of scalar multiplications:

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Example (Q7.11 pp. 221-222)0

M1

0M2

0M3

0M4

0M5

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MatChain AlgorithmAlgorithm MatChainInput: r[1..n+1] of +ve integers corresponding to the dimensions

of a chain of matricesOutput: Least number of scalar multiplications required to

multiply the n matrices for i := 1 to n do C[i,i] := 0; // diagonal d0

for d := 1 to n-1 do // for diagonals d1 to dn-1

for i := 1 to n-d do j := i+d; C[i,j] := ; for k := i+1 to j do C[i,j] := min{C[i,j],C[i,k-1]+C[k,j]+r[i]r[k]r[j+1]; end for; end for; return C[1,n];

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Time and Space Complexity of MatChain Algorithm• Time Complexity

• Space Complexity

)(6

12121

)(1

1

3

1

1

2

1

11

1

11

1

1

11

1

111

1

1

n

nnncnncnddnc

dndcdcdc

cc

n

d

n

d

dn

i

n

d

dn

i

n

d

di

ik

dn

i

n

d

j

ik

dn

i

n

d

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The Knapsack Problem• Let U = {u1, u2, …, un} be a set of n items to

be packed in a knapsack of size C.• Let sj and vj be the size and value of the jth

item, where sj, vj , 1 j n.• The objective is to fill the knapsack with some

items from U whose total size does not exceed C and whose total value is maximum.• Assume that the size of each item does not exceed

C.

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The Knapsack Problem Formulation• Given n +ve integers in U, we want to find a

subset SU s.t.

is maximized subject to the constraint

Su

ii

v

CsSu

ii

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Inductive Solution• Let V[i,j] denote the value obtained by filling a

knapsack of size j with items taken from the first i items {u1, u2, …, ui} in an optimal way:• The range of i is • The range of j is• The objective is to find V[ , ]

• V[i,0] = V[0,j] =

• V[i,j] = V[i-1,j] if = max {V[i-1,j], V[ , ]+vi} if

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Example (pp. 223 Question 7.22)• There are five items of sizes 3, 5, 7, 8, and 9 with values 4,

6, 7, 9, and 10 respectively. The size of the knapsack is 22.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

0 0 0

0 0 0

0 0 0

0 0 0

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Algorithm KnapsackAlgorithm KnapsackInput: A set of items U = {u1,u2,…,un} with sizes s1,s2,…,sn and

values v1,v2,…,vn, respectively and knapsack capacity C.

Output: the maximum value of subject to for i := 0 to n do V[i,0] := 0; for j := 0 to C do V[0,j] := 0; for i := 1 to n do for j := 1 to C do V[i,j] := V[i-1,j]; if si j then

V[i,j] := max{V[i,j], V[i-1,j-si]+vi}

end for; end for; return V[n,C];

Su

ii

v CsSu

ii

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Time and Space Complexity of the Knapsack Algorithm

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ics 353: design and analysis of algorithms

Documents

dynamic programming

design techniques

recursive solution results

desired solution

space of subproblems

new subproblems

class of problems

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