Mincost q 8 return mincost matrix chain

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MinCost = q } 8. return MinCost; }
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Matrix Chain Multiplication Problem A recursive algorithm may encounter each sub problem many times in different branches of its recursion tree. Instead of computing the same solution again and again, we can use the memoization strategy.
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Matrix Chain Multiplication Problem Top Down Recursive Approach with Memoization Memoized-Matrix-Chain(p) { n = length[p] – 1; for(i= 1; i <= n; i++) for(j = i; j <= n; j++) m[i, j] = return LookUp-Chain(p, 1, n); }
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Matrix Chain Multiplication Problem Top Down Recursive Approach with Memoization LookUp-Chain(p, i, j) { if( m[i, j] < ) return m[i, j] if(i == j) m[i, j] = 0; else { for(k = i; k <= j-1; k++) { q = LookUp-Chain(p, I, k) + LookUp-Chain(p, k+1, j) + p i- 1 p k p j if(q < m[i, j]) { m[i, j] = q; s[i, j] = k; } } } }
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Matrix Chain Multiplication Problem Bottom Up Heuristic Approach To keep track of how to construct an optimal solution, let us define s[i, j] to be a value of k at which we can split the product A i ,A i+1 , … A j to obtain an optimal parenthesization.i.e. s[i, j] equals a value k such that m[i, j] = m[i, k] + m[k+1, j] + p i-1 p k p j .
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Matrix Chain Multiplication Problem Bottom Up Approach with Memoization Matrix-Chain-Order(p) { n = length(p) –1 for(i = 1; i <= n; i++) m[i, i] = 0; for(l = 2; l <= n; l ++) for(i = l; i<=n-l+1; i++) { j = i + l – 1; m[i, j] = for(k = i; k<= j-1; k++) { q = m[i, k] + m[k+1, j] + p i-1 p k p j if (q < m[i, j]) { m[i, j] = q; s[i, j] = k; } } } return m and s; }
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Matrix Chain Multiplication Problem Bottom Up Approach with Memoization Matrix-Chain-Multiply(A , s, i, j) { if (j > i) { X = Matrix-Chain-Multiply(A, s, i, s[i, j]); Y = Matrix-Chain-Multiply(A, s, s[i, j]+1, j); return Matrix-Multiply(X, Y); } else return A i }
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Longest Common Subsequence (LCS) Problem Given two sequences of symbols, X and Y, determine the longest subsequence of symbols that appears in both X and Y. Example: X = <A, B, C, B, D, A, B> Y = <B, D, C, A, B, A> LCS of X, Y? <B, C, B, A> <B, D, A, B>
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1. Enumerate all subsequences of X. 2. Each subsequence of X corresponds to a subset of indices {1, 2, ….,m} of X. 3. There are 2 m subsequences of X, so this approach requires exponential time. Longest Common Subsequence (LCS) Problem
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Longest Common Subsequence (LCS) 7 1 2 3 4 5 6 i 0 j 0 1 2 3 4 5 6 y j B D C A B A x i A B C B D A B 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 2 2 0 1 1 2 2 2 2 0 1 1 2 2 3 3 0 1 2 2 2 3 3 0 1 2 2 3 3 4 0 1 2 2 3 4 4
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LCS – Algorithm: Constructing an LCS Table b[1..m, 1..n] can be used to construct an LCS. Initial invocation is Print_LCS(b, X, length[X], length[Y]). Print_LCS(b[], X[], i, j) { if(i = = 0 or j = = 0) return 0; if(b[i, j] = = “ // common elements { Print_LCS(b, X, I-1, j-1); print xi; } else if (b[I, j] = = “ ”) Print_LCS(b, X, I-1, j); else Print_LCS(b, X, I, j-1); } Devise a memoized version of LCS_Length();
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Longest Common Subsequence (LCS) Problem Dynamic Programming Approach I.
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