1. UMass Lowell Computer Science 91.503 Analysis of Algorithms Prof. Karen Daniels Spring, 2009 Design Patterns for Optimization Problems Dynamic Programming Lecture 1 (Part 2)

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

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 Matrix Parenthesization

5. 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: 91.503 textbook Cormen, et al.

6. 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: 91.503 textbook Cormen, et al. common DP proof technique: “cut-and-paste” proof by contradiction

7. 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: 91.503 textbook Cormen, et al.

8. 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: 91.503 textbook Cormen, et al.

9. 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: 91.503 textbook Cormen, et al.

10. 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: 91.503 textbook Cormen, et al. MEMOIZED-MATRIX-CHAIN(p) 1 n length[p] - 1 2 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]

11. Dynamic Programming Longest Common Subsequence

12. 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: 91.503 textbook Cormen, et al.

13. 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.

14. 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: 91.503 textbook Cormen, et al.

15. 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 ))

16. 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.

17. 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?

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

19. 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: 91.503 textbook Cormen, et al.

20. Dynamic Programming …leading to a Greedy Algorithm… Activity Selection

21. 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: 91.404 textbook Cormen, et al.

22. Activity Selection 12 3 4 5 6 7 8 9 10 12 11 13 1415 16 1 2 3 4 8 7 6 5 Activity Time Duration Activity Number

23. 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: 91.404 textbook Cormen, et al. c[i,j]=size of maximum-size subset of mutually compatible activities in S ij.