UMass Lowell Computer Science Analysis of Algorithms Prof. Karen Daniels Spring, 2002 Lecture 2 Tuesday, 2/5/02 Dynamic Programming

Algorithmic Paradigm Context

Dynamic Programming Example: Matrix Parenthesization

Example: Matrix Parenthesization Memoization ä Provide Dynamic Progamming efficiency ä But with top-down strategy ä Use recursion ä Fill in table “on demand” ä Example: ä RECURSIVE-MATRIX-CHAIN: MEMOIZED-MATRIX-CHAIN(p) 1 n length[p] for i 1 to n 3 do for j i to n 4 do m[i,j] 5 return LOOKUP-CHAIN(p,1,n) LOOKUP-CHAIN(p,1,n) 1 if m[i,j] < 2 then return m[i,j] 3 if i=j 4 then m[i,j] 0 5 else for k i to j-1 6 do q LOOKUP-CHAIN(p,i,k) + LOOKUP-CHAIN(p,k+1,j) + p i-1 p k p j 7if q < m[i,j] 8 then m[i,j] q 9 return m[i,j]

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, guanine, cytosine 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

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.

Example: LCS Step 1: Characterizing 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 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 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)

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 ) LCS(X, Y)

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

Example: LCS Step 3: Computing Length of an LCS c table ( represent b table)

Example: LCS Step 4: Constructing an LCS

Example: LCS Improving the Code ä Can eliminate b table  c[i,j] depends only on 3 other c table entries: ä c[i-1,j-1] c[i-1,j] c[i,j-1]  given value of c[i,j], can pick the one in O(1) time ä reconstruct LCS in O(m+n) time similar to PRINT-LCS  same  (mn) space, but  (mn) was needed anyway... ä Asymptotic space reduction ä leverage: need only 2 rows of c table at a time ä row being computed ä previous row ä can also do it with ~ space for 1 row of c table ä but does not preserve LCS reconstruction data