1. UMass Lowell Computer Science 91.404 Analysis of Algorithms Prof. Karen Daniels Fall, 2005 Design Patterns for Optimization Problems Dynamic Programming

2. Algorithmic Paradigm Context Divide & Conquer Dynamic Programming View problem as collection of subproblems “Recursive” nature Independent subproblems Number of subproblems depends on partitioning factors typically small Preprocessing Characteristic running time typically log function of n depends on number and difficulty of subproblems Primarily for optimization problems Optimal substructure: optimal solution to problem contains within it optimal solutions to subproblems Overlapping subproblems

3. Dynamic Programming Approach to Optimization Problems 1. Characterize structure of an optimal solution. 2. Recursively define value of an optimal solution. 3. Compute value of an optimal solution in bottom-up fashion. 4. Construct an optimal solution from computed information. source: 91.503 textbook Cormen, et al.

4. Dynamic Programming Longest Common Subsequence

5. Example: Longest Common Subsequence (LCS): Motivation ä Strand of DNA: string over finite set {A,C,G,T} ä each element of set is a base: adenine, cytosine, guanine or thymine ä Compare DNA similarities ä S 1 = ACCGGTCGAGTGCGCGGAAGCCGGCCGAA ä S 2 = GTCGTTCGGAATGCCGTTGCTCTGTAAA ä One measure of similarity: ä find the longest string S 3 containing bases that also appear (not necessarily consecutively) in S 1 and S 2 ä S 3 = GTCGTCGGAAGCCGGCCGAA source: 91.503 textbook Cormen, et al.

6. Example: LCS Definitions ä Sequence is a subsequence of if (strictly increasing indices of X) such that ä example: is subsequence of with index sequence ä Z is common subsequence of X and Y if Z is subsequence of both X and Y ä example: ä common subsequence but not longest ä common subsequence. Longest? Longest Common Subsequence Problem: Given 2 sequences X, Y, find maximum-length common subsequence Z. source: 91.503 textbook Cormen, et al.

7. Example: LCS Step 1: Characterize an LCS THM 15.1: Optimal LCS Substructure Given sequences: For any LCSof X and Y: 1 if thenand Z k-1 is an LCS of X m-1 and Y n-1 2 if thenZ is an LCS of X m-1 and Y 3 if thenZ is an LCS of X and Y n-1 PROOF: based on producing contradictions 1 a) Suppose. Appending to Z contradicts longest nature of Z. b) To establish longest nature of Z k-1, suppose common subsequence W of X m-1 and Y n-1 has length > k-1. Appending to W yields common subsequence of X, Y of length > k = contradiction. b) To establish longest nature of Z k-1, suppose common subsequence W of X m-1 and Y n-1 has length > k-1. Appending to W yields common subsequence of X, Y of length > k = contradiction. 2 Common subsequence W of X m-1 and Y of length > k would also be common subsequence of X m, Y, contradicting longest nature of Z. 3 Similar to proof of (2) source: 91.503 textbook Cormen, et al.

8. Example: LCS Step 2: A Recursive Solution ä Implications of Thm 15.1: ? yes no Find LCS(X m-1, Y n-1 ) Find LCS(X m-1, Y) Find LCS(X, Y n-1 ) LCS(X, Y) = LCS(X m-1, Y n-1 ) + x m LCS(X, Y) = max(LCS(X m-1, Y), LCS(X, Y n-1 ))

9. Example: LCS Step 2: A Recursive Solution (continued) ä Overlapping subproblem structure: ä Recurrence for length of optimal solution: Conditions of problem can exclude some subproblems! c[i,j]= c[i-1,j-1]+1 if i,j > 0 and x i =y j max(c[i,j-1], c[i-1,j])if i,j > 0 and x i =y j 0 if i=0 or j=0  (mn) distinct subproblems source: 91.503 textbook Cormen, et al.

10. Example: LCS Step 3: Compute Length of an LCS source: 91.503 textbook Cormen, et al. c table c table (represent b table) (represent b table) B CB A B C B A 0 1 2 3 4 What is the asymptotic worst- case time complexity?

11. Example: LCS Step 4: Construct an LCS source: 91.503 textbook Cormen, et al.