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Dynamic Programming Reading Material: Chapter 7.

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Dynamic Programming Dynamic Programming algorithms address problems whose solution is recursive in nature, but has the following property: The direct implementation of the recursive solution results in identical recursive calls that are executed more than once. Dynamic programming implements such algorithms by evaluating the recurrence in a bottom-up manner, saving intermediate results that are later used in computing the desired solution

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Fibonacci Numbers What is the recursive algorithm that computes Fibonacci numbers? What is its time complexity? –Note that it can be shown that

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Computing the Binomial Coefficient Recursive Definition Actual Value

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Computing the Binomial Coefficient What is the direct recursive algorithm for computing the binomial coefficient? How much does it cost? –Note that

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Optimization Problems and Dynamic Programming Optimization problems with certain properties make another class of problems that can be solved more efficiently using dynamic programming. Development of a dynamic programming solution to an optimization problem involves four steps –Characterize the structure of an optimal solution Optimal substructures, where an optimal solution consists of sub- solutions that are optimal. Overlapping sub-problems where the space of sub-problems is small in the sense that the algorithm solves the same sub-problems over and over rather than generating new sub-problems. –Recursively define the value of an optimal solution. –Compute the value of an optimal solution in a bottom-up manner. –Construct an optimal solution from the computed optimal value.

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Longest Common Subsequence Problem Problem Definition: Given two strings A and B over alphabet , determine the length of the longest subsequence that is common in A and B. A subsequence of A=a 1 a 2 …a n is a string of the form a i1 a i2 …a ik where 1 i 1 <i 2 <…<i k n Example: Let = { x, y, z }, A = xyxyxxzy, B=yxyyzxy, and C= zzyyxyz –LCS(A,B)=yxyzyHence the length = –LCS(B,C)=Hence the length = –LCS(A,C)=Hence the length =

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Straight-Forward Solution Brute-force search –How many subsequences exist in a string of length n? –How much time needed to check a string whether it is a subsequence of another string of length m? –What is the time complexity of the brute-force search algorithm of finding the length of the longest common subsequence of two strings of sizes n and m?

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Dynamic Programming Solution Let L[i,j] denote the length of the longest common subsequence of a 1 a 2 …a i and b 1 b 2 …b j, which are substrings of A and B of lengths n and m, respectively. Then L[i,j] = when i = 0 or j = 0 L[i,j] = when i > 0, j > 0, a i =b j L[i,j] = when i > 0, j > 0, a i b j

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LCS Algorithm Algorithm LCS(A,B) Input: A and B strings of length n and m respectively Output: Length of longest common subsequence of A and B Initialize L[i,0] and L[0,j] to zero; for i ← 1 to n do for j ← 1 to m do if a i = b j then L[i,j] ← 1 + L[i-1,j-1] else L[i,j] ← max(L[i-1,j],L[i,j-1]) end if end for; return L[n,m];

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Example (Q7.5 pp. 220) Find the length of the longest common subsequence of A=xzyzzyx and B=zxyyzxz

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Example (Cont.) xzyzzyx 00000000 z0 x0 y0 y0 z0 x0 z0

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Complexity Analysis of LCS Algorithm What is the time and space complexity of the algorithm?

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Matrix Chain Multiplication Assume Matrices A, B, and C have dimensions 2 10, 10 2, and 2 10 respectively. The number of scalar multiplications using the standard Matrix multiplication algorithm for –(A B) C is –A (B C) is Problem Statement: Find the order of multiplying n matrices in which the number of scalar multiplications is minimum.

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Straight-Forward Solution Again, let us consider the brute-force method. We need to compute the number of different ways that we can parenthesize the product of n matrices. –e.g. how many different orderings do we have for the product of four matrices? –Let f(n) denote the number of ways to parenthesize the product M 1, M 2, …, M n. (M 1 M 2 …M k ) (M k+1 M k+2 …M n ) What is f(2), f(3) and f(1)?

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Catalan Numbers C n =f(n+1) Using Stirling’s Formula, it can be shown that f(n) is approximately

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Cost of Brute Force Method How many possibilities do we have for parenthesizing n matrices? How much does it cost to find the number of scalar multiplications for one parenthesized expression? Therefore, the total cost is

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The Recursive Solution –Since the number of columns of each matrix M i is equal to the number of rows of M i+1, we only need to specify the number of rows of all the matrices, plus the number of columns of the last matrix, r 1, r 2, …, r n+1 respectively. –Let the cost of multiplying the chain M i …M j (denoted by M i,j ) be C[i,j] –If k is an index between i+1 and j, what is the cost of multiplying M i,j considering multiplying M i,k-1 with M k,j ? –Therefore, C[1,n]=

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The Dynamic Programming Algorithm C[1,1]C[1,2]C[1,3]C[1,4]C[1,5]C[1,6] C[2,2]C[2,3]C[2,4]C[2,5]C[2,6] C[3,3]C[3,4]C[3,5]C[3,6] C[4,4]C[4,5]C[4,6] C[5,5]C[5,6] C[6,6]

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Example (Q7.11 pp. 221-222) Given as input 2, 3, 6, 4, 2, 7 compute the minimum number of scalar multiplications:

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MatChain Algorithm Algorithm MatChain Input: r[1..n+1] of +ve integers corresponding to the dimensions of a chain of matrices Output: Least number of scalar multiplications required to multiply the n matrices for i := 1 to n do C[i,i] := 0; // diagonal d 0 for d := 1 to n-1 do // for diagonals d 1 to d n-1 for i := 1 to n-d do j := i+d; C[i,j] := ; for k := i+1 to j do C[i,j] := min{C[i,j],C[i,k-1]+C[k,j]+r[i]r[k]r[j+1]; end for; return C[1,n];

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Time and Space Complexity of MatChain Algorithm Time Complexity Space Complexity

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The Knapsack Problem Let U = {u 1, u 2, …, u n } be a set of n items to be packed in a knapsack of size C . Let s j and v j be the size and value of the j th item, where s j, v j , 1 j n. The objective is to fill the knapsack with some items from U whose total size does not exceed C and whose total value is maximum. –Assume that the size of each item does not exceed C.

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The Knapsack Problem Formulation Given n +ve integers in U, we want to find a subset S U s.t. is maximized subject to the constraint

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Inductive Solution Let V[i,j] denote the value obtained by filling a knapsack of size j with items taken from the first i items {u 1, u 2, …, u i } in an optimal way: –The range of i is –The range of j is –The objective is to find V[, ] V[i,0] =V[0,j] = V[i,j] = V[i-1,j] if = max {V[i-1,j], V[, ]+v i } if

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Example (pp. 223 Question 7.22) There are five items of sizes 3, 5, 7, 8, and 9 with values 4, 6, 7, 9, and 10 respectively. The size of the knapsack is 22. 00000000000000000000000 00044444444444444444444 000 000 000 000

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Algorithm Knapsack Input: A set of items U = {u 1,u 2,…,u n } with sizes s 1,s 2,…,s n and values v 1,v 2,…,v n, respectively and knapsack capacity C. Output: the maximum value of subject to for i := 0 to n do V[i,0] := 0; for j := 0 to C do V[0,j] := 0; for i := 1 to n do for j := 1 to C do V[i,j] := V[i-1,j]; if s i j then V[i,j] := max{V[i,j], V[i-1,j-s i ]+v i } end for; return V[n,C];

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Time and Space Complexity of the Knapsack Algorithm

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All-Pairs Shortest Paths Problem: For every vertex u, v V, calculate (u, v). Possible Solution 1: Cost of Possible Solution 1:

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Dynamic Programming Solution Define a k-path from u to v, where u,v {1, 2, …, n} to be any path whose intermediate vertices all have indices less than or equal to k. –What is a 0-path? –What is a 1-path? –… –What is an n-path?

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Floyd’s Algorithm Algorithm Floyd Input: An n n matrix length[1..n, 1..n] such that length[i,j] is the weight of the edge (i,j) in a directed graph G = ({1,2,…,n}, E) Output: A matrix D with D[i,j] = [i,j] 1 D = length; //copy the input matrix length into D 2 for k = 1 to n do 3 for i = 1 to n do 4 for j = 1 to n do 5 D[i,j] = min{D[i,j], D[i,k] + D[k,j]}

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Example 1 3 4 2 4 8 11 1 5 1215 5 11 2

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0-p1234 1 2 3 4 1-p1234 1 2 3 4 2-p1234 1 2 3 4 3-p1234 1 2 3 4 Example (Cont.)

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Time and Space Complexity Time Complexity: Space Complexity:

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0-p1234 10125 2150811 3 402 415 0 1-p1234 10125 2150811 3 402 41560 2-p1234 1012523 2150811 319402 41560 3-p1234 10957 2150810 319402 41560 Example (Cont.)

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4-p1234 10957 2110810 33402 41560 3-p1234 10957 2150810 319402 41560 Example (Cont.)

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