UMass Lowell Computer Science Analysis of Algorithms Prof. Karen Daniels Spring, 2009 Design Patterns for Optimization Problems Dynamic Programming Lecture 1 (Part 2)

Algorithmic Paradigm Context Subproblem solution order Make choice, then solve subproblem(s) Solve subproblem(s), then make choice

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: textbook Cormen, et al.

Dynamic Programming Matrix Parenthesization

Example: Matrix Parenthesization Definitions ä Given “chain” of n matrices: ä Given “chain” of n matrices: ä Compute product A 1 A 2 … A n efficiently ä Minimize “cost” = number of scalar multiplications ä Multiplication order matters! source: textbook Cormen, et al.

Example: Matrix Parenthesization Step 1: Characterizing an Optimal Solution Observation: Any parenthesization of A i A i+1 … A j must split it between A k and A k+1 for some k. THM: Optimal Matrix Parenthesization: If an optimal parenthesization of A i A i+1 … A j splits at k, then parenthesization of prefix A i A i+1 … A k must be an optimal parenthesization. Why? If existed less costly way to parenthesize prefix, then substituting that parenthesization would yield less costly way to parenthesize A i A i+1 … A j, contradicting optimality of that parenthesization. If existed less costly way to parenthesize prefix, then substituting that parenthesization would yield less costly way to parenthesize A i A i+1 … A j, contradicting optimality of that parenthesization. source: textbook Cormen, et al. common DP proof technique: “cut-and-paste” proof by contradiction

Example: Matrix Parenthesization Step 2: A Recursive Solution ä Recursive definition of minimum parenthesization cost: m[i,j]= min{m[i,k] + m[k+1,j] + p i-1 p k p j } if i < j 0 if i = j How many distinct subproblems? i <= k < j each matrix A i has dimensions p i-1 x p i source: textbook Cormen, et al.

Example: Matrix Parenthesization Step 3: Computing Optimal Costs 0 2,625 2,500 1,000 s: value of k that achieves optimal cost in computing m[i, j] source: textbook Cormen, et al.

Example: Matrix Parenthesization Step 4: Constructing an Optimal Solution PRINT-OPTIMAL-PARENS(s, i, j) if i = j then print “A” i else print “(“ PRINT-OPTIMAL-PARENS(s, i, s[i, j]) PRINT-OPTIMAL-PARENS(s, i, s[i, j]) PRINT-OPTIMAL-PARENS(s, s[i, j]+1, j) PRINT-OPTIMAL-PARENS(s, s[i, j]+1, j) print “)“ print “)“ source: textbook Cormen, et al.

Example: Matrix Parenthesization Memoization ä Provide Dynamic Programming efficiency ä But with top-down strategy ä Use recursion ä Fill in table “on demand” ä Example: ä RECURSIVE-MATRIX-CHAIN: source: textbook Cormen, et al. 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,i,j) 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]

Dynamic Programming Longest Common Subsequence

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 source: textbook Cormen, et al.

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: textbook Cormen, et al.

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 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) source: textbook Cormen, et al.

Example: LCS Step 2: A Recursive Solution ä Implications of Theorem 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 1 (X, Y) = LCS(X m-1, Y n-1 ) + x m LCS 2 (X, Y) = max(LCS(X m-1, Y), LCS(X, Y n-1 ))

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: textbook Cormen, et al.

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

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

Example: LCS Improve 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 source: textbook Cormen, et al.

Dynamic Programming …leading to a Greedy Algorithm… Activity Selection

Activity Selection Optimization Problem ä Problem Instance: ä Set S = {1,2,...,n} of n activities ä Each activity i has: ä start time: s i ä finish time: f i ä Activities i, j are compatible iff non-overlapping: ä Objective: ä select a maximum-sized set of mutually compatible activities source: textbook Cormen, et al.

Activity Selection Activity Time Duration Activity Number

Activity Selection Solution to S ij including a k produces 2 subproblems: 1) S ik (start after a i finishes; finish before a k starts) 2) S kj (start after a k finishes; finish before a j starts) source: textbook Cormen, et al. c[i,j]=size of maximum-size subset of mutually compatible activities in S ij.