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35 Pages

### DynamicProgramming

Course: RXC 064000, Fall 2009
School: Dallas
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Word Count: 1562

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Programming Dynamic Dynamic Programming 1 Outline and Reading Matrix ChainProduct (5.3.1) The General Technique (5.3.2) 01 Knapsack Problem (5.3.3) Dynamic Programming 2 Matrix ChainProducts Dynamic Programming is a general algorithm design paradigm. Review: Matrix Multiplication. Rather than give the general structure, let us first give a motivating example: Matrix ChainProducts f B e j C = A*B A...

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Programming Dynamic Dynamic Programming 1 Outline and Reading Matrix ChainProduct (5.3.1) The General Technique (5.3.2) 01 Knapsack Problem (5.3.3) Dynamic Programming 2 Matrix ChainProducts Dynamic Programming is a general algorithm design paradigm. Review: Matrix Multiplication. Rather than give the general structure, let us first give a motivating example: Matrix ChainProducts f B e j C = A*B A is d e and B is e f O(d e f ) time e -1 k =0 e A i C i,j f 3 C[i, j ] = A[i, k ] * B[k , j ] d d Dynamic Programming Matrix ChainProducts Matrix ChainProduct: Dynamic Programming 4 Enumeration Approach Matrix ChainProduct Alg.: Try all possible ways to parenthesize Dynamic Programming 5 Greedy Approach Idea #1: repeatedly select the product that uses the fewest operations. Counterexample: The greedy approach is not giving us the optimal value. Dynamic Programming 6 A is 101 11 B is 11 9 C is 9 100 D is 100 99 Greedy idea #1 gives A*((B*C)*D)), which takes 109989+9900+108900=228789 ops (A*B)*(C*D) takes 9999+89991+89100=189090 ops "Recursive" Approach Define subproblems: Subproblem optimality: The optimal solution can be defined in terms of optimal subproblems Find the best parenthesization of Ai*Ai+1*...*Aj. Let Ni,j denote the number of operations done by this subproblem. The optimal solution for the whole problem is N0,n1. There has to be a final multiplication (root of the expression tree) for the optimal solution. Say, the final multiplication is at index i: (A0*...*Ai)*(Ai+1*...*An1). Then the optimal solution N0,n1 is the sum of two optimal subproblems, N0,i and Ni+1,n1 plus the time for the last multiplication. If the global optimum did not have these optimal subproblems, we could define an even better "optimal" solution. Dynamic Programming 7 Characterizing Equation The global optimal has to be defined in terms of optimal subproblems, depending on where the final multiplication is at. Let us consider all possible places for that final multiplication: N i , j = min{N i ,k + N k +1, j + d i d k +1d j +1} i k < j Dynamic Programming 8 Subproblem Overlap Algorithm RecursiveMatrixChain(S, 0, n-1): Input: sequence S of n matrices to be multiplied Output: number of operations in an optimal parenthesization of S if i=j then return 0 k i j Dynamic Programming 9 Subproblem Overlap 1..4 1..1 2..2 3..4 2..4 2..3 4..4 1..2 1..1 2..2 3..4 3..3 4..4 1..3 4..4 ... 3..3 4..4 2..2 3..3 Dynamic Programming 10 Dynamic Programming Algorithm Visualization The bottomup N i , j = min{N i ,k + N k +1, j + d i d k +1d j +1} i k < j construction fills in the N array by diagonals j ... n1 N 0 1 2 i Ni,j gets values from 0 previous entries in ith 1 row and jth column ... answer Filling in each entry in the i N table takes O(n) time. Total run time: O(n3) j Getting actual parenthesization can be done by remembering "k" n1 for each N entry Dynamic Programming 11 Dynamic Programming Algorithm Visualization A0: 30 X 35; A1: 35 X15; A2: 15X5; A3: 5X10; A4: 10X20; A5: 20 X 25 N i , j = min{N i ,k + N k +1, j + d i d k +1d j +1} i k < j N1, 4 = min{ N1,1 + N 2, 4 + d1d 2 d 5 = 0 + 2500 + 35 *15 * 20 = 13000, N1, 2 + N 3, 4 + d1d 3 d 5 = 2625 + 1000 + 35 * 5 * 20 = 7125, N1,3 + N 4, 4 + d1d 4 d 5 = 4375 + 0 + 35 *10 * 20 = 11375 } = 7125 Dynamic Programming 12 Dynamic Programming Algorithm Visualization (A0*(A1*A2))*((A3*A4)*A5) Dynamic Programming 13 Dynamic Programming Algorithm Since Algorithm matrixChain(S): subproblems Input: sequence S of n matrices to be overlap, we don't multiplied use recursion. Instead, we Output: number of operations in an optimal construct optimal parenthesization of subproblems S "bottomup." Ni,i's are easy, so start with them Then do problems of "length" 2,3,... subproblems, and so on. Running time: O(n3) Dynamic Programming 14 The General Dynamic Programming Technique Applies to a problem that at first seems to require a lot of time (possibly exponential), provided we have: Simple subproblems: the subproblems can be defined in terms of a few variables, such as j, k, l, m, and so on. Subproblem optimality: the global optimum value can be defined in terms of optimal subproblems Subproblem overlap: the subproblems are not independent, but instead they overlap (hence, should be constructed bottomup). Dynamic Programming 15 The 0/1 Knapsack Problem Given: A set S of n items, with each item i having Goal: Choose items with maximum total benefit but with weight at most W. If we are not allowed to take fractional amounts, then this is the 0/1 knapsack problem. wi a positive weight bi a positive benefit In this case, we let T denote the set of items we take Objective: maximize Constraint: b iT i w iT i W 16 Dynamic Programming Example Given: A set S of n items, with each item i having Goal: Choose with items maximum total benefit but with weight at most W. "knapsack" Items: 1 2 3 4 5 bi a positive "benefit" wi a positive "weight" box of width 9 in Weight: Benefit: 4 in \$20 2 in \$3 2 in \$6 6 in \$25 2 in \$80 Solution: Dynamic Programming item 5 (\$80, 2 in) item 3 (\$6, 2in) item 1 (\$20, 4in) 17 A 0/1 Knapsack Algorithm, First Attempt Sk: Set of items numbered 1 to k. Define B[k] = best selection from Sk. Problem: does not have subproblem optimality: Consider set S={(3,2),(5,4),(8,5),(4,3),(10,9)} of (benefit, weight) pairs and total weight W = 20 Dynamic Programming 18 A 0/1 Knapsack Algorithm, Second Attempt Sk: Set of items numbered 1 to k. Define B[k,w] to be the best selection from Sk with weight at most w Good news: this does have subproblem optimality. B[k - 1, w] if wk > w B[k , w] = else max{B[k - 1, w], B[k - 1, w - wk ] + bk } I.e., the best subset of Sk with weight at most w is either the best subset of Sk1 with weight at most w or the best subset of Sk1 with weight at most w- k plus item k w Dynamic Programming 19 0/1 Knapsack Algorithm Consider set S={(1,1),(2,2),(4,3),(2,2),(5,5)} of (benefit, weight) pairs and total weight W = 10 Dynamic Programming 20 0/1 Knapsack Algorithm Trace back to find the items picked Dynamic Programming 21 0/1 Knapsack Algorithm Each diagonal arrow corresponds to adding one item into the bag Pick items 2,3,5 {(2,2),(4,3),(5,5)} are what you will take away Dynamic Programming 22 0/1 Knapsack Algorithm B[k - 1, w] if wk > w B[k , w] = else max{B[k - 1, w], B[k - 1, w - wk ] + bk } Algorithm 01Knapsack(S, W): Recall the definition of Input: set S of n items with benefit B[k,w] Since B[k,w] is defined in terms of B[k- 1,*], we can use two arrays of instead of a matrix Running time: O(nW). Not a polynomialtime algorithm since W may be large This is a pseudopolynomial time algorithm k 1 n Dynamic Programming 23 Longest Common Subsequence Given two strings, find a longest subsequence that they share substring vs. subsequence of a string Consider ababc and abdcb alignment 1 ababc. abd.cb Substring: the characters in a substring of S must occur contiguously in S Substring: the characters can be interspersed with gaps. the longest common subsequence is ab..c with length 3 alignment 2 aba.bc abdcb. the longest common subsequence is ab..b with length 3 Dynamic Programming 24 Longest Common Subsequence Let's give a score M an alignment in this way, Dynamic Programming 25 2 1 2 3 j Longest Common Subsequence Subproblem optimality Consider two sequences Let the optimal alignment be x1x2x3...xn1xn y1y2y3...yn1yn There are three possible cases for the last pair (xn,yn): Dynamic Programming 26 2 1 2 3 j Longest Common Subsequence Mi,j = MAX {Mi-1, j-1 + S (ai,bj) (match/mismatch) Mi,j-1 + 0 (gap in sequence #1) Mi-1,j + 0 (gap in sequence #2) } Mi,j is the score for optimal alignment between strings a[...

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