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## Lecture 13

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**Dynamic Programming: Steps**• The seven steps in the development of a dynamic programming algorithm are as follows: • Establish a recursive property that gives the solution to an instance of the problem. • Develop a recursive algorithm as per recursive property • See if same instance of the problem is being solved again an again in recursive calls • Develop a memoized recursive algorithm • See the pattern in storing the data in the memory • Convert the memoized recursive algorithm into iterative algorithm • Optimize the iterative algorithm by using the storage as required (storage optimization)**Exercise 1**• Initialise a two dimensional array A[1..n, 1..n] using following definition with an efficient algorithm. • A[i, j] = i + j if i = j • A[i, j] = Min(A[i, j-1], A[i+1, j])+ (i+j) if i < j • A[i, j] = Max(A[i-1, j], A[i, j+1])+ (i *j) if i > j • Where i is for row index and j is for column index.**Quiz**Develop a recursive algorithm for the following definition, where P[] is a one-dimensional array having indices from i-1 to j,and m is a two-dimensional array of size ixj**Solution for the Quiz**• RMC(p, i, j) • { • if i = j • return 0 • m[i, j] = • for(k = i; k <= j-1; k++) • { q = RMC(p, i, k) + RMC(p, k+1, j) + pi-1pkpj • if(q < m[i, j]) • m[i, j] = q • } • return m[i, j]; • }**Quiz**Convert the algorithm of the previous Quiz into an iterative algorithm**Quiz**Develop a recursive algorithm for the following definition, 0 if i = 0 or j = 0 C(i, j) = c(i-1, j-1) + 1 if i, j > 0 and xi = yj max (c(i, j-1), c(i-1, j)) if i, j > 0 and xi yj X = <x1, x2, x3,…..,xm> Y = <y1, y2, y3,…..,yn> i varies from m to 0 and j varies from n to 0**Quiz**• Develop a memoized recursive algorithm for the following definition 0 if i = 0 or j = 0 C(i, j) = c(i-1, j-1) + 1 if i, j > 0 and xi = yj max (c(i, j-1), c(i-1, j)) if i, j > 0 and xi yj X = <x1, x2, x3,…..,xm> Y = <y1, y2, y3,…..,yn> i varies from m to 0 and j varies from n to 0**Quiz**Convert the algorithm of the previous Quiz into an iterative algorithm**Third Practice Problem – Quiz**Develop a recursive and memoized recursive algorithm for the following recursive definition P is an array of j elements A[i, i] = Pi A[i, i-1] and A[j+1, j] are defined to be 0**Home Exercise**• Convert the iterative algorithm of the quizzes into one which consume memory as per requirement of the problem**Matrix Multiplications**• We can multiply two matrices A and B if Columns of A = Rows of B I.e. order of A = p * q and order of B = q * r void Matrix_Multiply(A, B) { If (columns of A != Rows of B) error else {for(I = 1; I <= n; I++) for(j = 1; j <= N; j++) {C[ I ][ j ] = 0; for(k = 1; k <= N; k++) C[ I ][ j ] += A[ I ][k ] * B[ k ][ j ] } } } No. of Scalar Multiplications = p * q * r**Matrix Chain Multiplication**• Problem arises when we want to multiply a chain of matrices. For example A = 50 * 10 B = 10 * 40 C = 40 * 30 D = 30 * 5 • We want to compute A * B * C * D. • In which sequence should we call matrix multiply? • Matrix multiplication is associative.**Matrix Chain Multiplication**A = 50 * 10, B = 10 * 40, C = 40 * 30,D = 30 * 5**Matrix Chain Multiplication Problem**• Given a chain <A1, A2, ….,An> of n matrices. • Multiply all the matrices with minimum cost • First Compute the order in which matrices should be multiplied • Or in other words • We have to find an optimum sequence or an optimum parenthesization of matrix chain multiplication. • Then by using that order multiply the matrices.**Matrix Chain Multiplication Problem Bottom Up Approach with**Memoization Matrix-Chain-Multiply(A [ ] , s[ ], i, j) // A is an array of matrices // s is an array of order, s[i,j] represents parenthesization position for // matrix i to 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 Ai } The problem is to compute s[i,j] for a given chain of matrices from i to j.**Formulation of the problem**• Fully parenthesize the product A1 * A2….An in a way that minimizes the number of scalar multiplications.**Matrix Chain Multiplication Problem**• Ai..j is a notation for the matrix that results from evaluating the product AiAi+1…Aj. • An optimal parenthesization of the product A1A2….An splits the product between Ak and Ak+1 for some integer k in the range 1<=k<n • For some value of k we first • Compute the matrices A1..k and Ak+1..n and then • Multiply them together to produce the final product A1..n.**Matrix Chain Multiplication Problem**• The cost of this optimal parenthesization is thus the • Cost of computing the matrix A1..k, plus • The cost of computing Ak+1..n, plus • The cost of multiplying them together.**Matrix Chain Multiplication Problem**• Parenthesization of the “prefix” sub chain A1A2…Ak within this optimal parenthesization of A1A2…An must be • An optimal parenthesization of A1A2…Ak. • Why? • A similar observation holds for the parenthesization of the sub chain Ak+1Ak+2…An • Thus, an optimal solution to an instance of the matrix-chain multiplication problem contains within it optimal solutions to sub problem instances.**Matrix Chain Multiplication Problem**A Recursive Solution • Define the value of an optimal solution recursively in terms of the optimal solutions to subproblems. • For this problem, we pick as our subproblems • the problems of determining the minimum cost of a parenthesization of AiAi+1….Aj, for 1 <= i <= j <=n**A Recursive Solution**• Let m[i, j] be the minimum number of scalar multiplications needed to compute the matrix Ai..j; • The cost of cheapest way to compute A1..n would thus be m[1, n]**A Recursive Solution**• We can define m[i, j] recursively as follows: • If i = j, the chain consists of just one matrix Ai..i=Ai, so • No scalar multiplications are necessary to compute the product. Thus • m[i, i] = 0 for i = 1, 2, …..,n. • To compute m[i, j] when i < j, we take advantage of the structure of an optimal solution from step 1. • Assume that optimal parenthesization splits the product Ai, Ai+1,….Aj between Ak and Ak+1, where i<=k < j.**A Recursive Solution**• Now m[i, j] is equal to the minimum cost for • Computing the subproducts A1..k, and • Computing the subproducts Ak+1..j, plus • The cost of multiplying these two matrices together. • For i = 1, 2, ….., n, matrix Ai has dimensions Pi-1*Pi • Since computing the matrix product A1..kAk+1..j takes pi-1pkpj scalar multiplications, we obtain • m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj • But we don’t know the value of k for an optimum solution. • There are only j-1 possible values for k, namely k=i,i+1, …j-1.**A Recursive Solution**• Since, the optimal parenthesization must use one of these values for k, we need only check them all to find the best. • Thus, our recursive definition for the minimum cost of parenthesizing the product Ai, Ai+1, … Aj becomes**Matrix Chain Multiplication Problem**• Based on this recurrence, we can easily write the recursive algorithm to compute the minimum cost m[1..n] for multiplying A1, A2, ….,An. Recursive-Matrix-Chain(p, i, j) { • if i = j • return 0 • MinCost = • for(k = i; k <= j-1; k++) { q = Recursive-Matrix-Chain(p, i, k) + Recursive-Matrix- Chain(p, k+1, j) + pi-1pkpj • if(q < MinCost) • MinCost = q } • return MinCost; }**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.**Matrix Chain Multiplication ProblemTop 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); }**Matrix Chain Multiplication ProblemTop 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) + pi-1pkpj if(q < m[i, j]) { m[i, j] = q; s[i, j] = k; } } } }**Matrix Chain Multiplication ProblemBottom 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 Ai,Ai+1, … Aj 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] + pi-1pkpj.**Matrix Chain Multiplication ProblemBottom 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] + pi-1pkpj • if (q < m[i, j]) • { m[i, j] = q; • s[i, j] = k; • } • } • } • return m and s; }**Matrix Chain Multiplication ProblemBottom 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 Ai }**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>**Longest Common Subsequence (LCS) ProblemBrute Force Approach**• Enumerate all subsequences of X. • Each subsequence of X corresponds to a subset of indices {1, 2, ….,m} of X. • There are 2m subsequences of X, so this approach requires exponential time.**j**0 1 2 3 4 5 6 0 0 0 0 0 0 0 i B D C A B A 0 0 0 0 1 1 1 0 yj 0 1 1 1 1 2 2 A 1 xi 0 1 1 2 2 2 2 B 2 0 1 1 2 2 3 3 C 3 0 1 2 2 2 3 3 4 B 0 1 2 2 3 3 4 D 5 0 1 2 2 3 4 4 A 6 B 7 Longest Common Subsequence (LCS)**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();**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach • Characterize the LCS problem • The LCS problem has an optimal substructure property. • Can we formulate optimal substructure? • ith preifx of a sequence X is Xi, e.g., • X4 of sequence X = <A, B, C, B, D, A, B> is <A, B, C, B>**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach Optimal Substructure of an LCS: Let X = <x1, x2, x3,…..,xm> and Y = <y1, y2, y3,…..,yn> and let Z = <z1, z2, z3,….zk> be any LCS of X and Y • If xm = yn then zk = xm = yn and zk-1 is an LCS of Xm-1 and Yn-1. • If xm yn then zk xm implies that Z is an LCS of Xm-1 and Y. • If xm yn then zk yn implies that Z is an LCS of X and Yn-1**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach Step 1: Characterizing an LCS • LCS problem has an optimal substructure property. I.e. • There are either one or two subproblems to examine when finding an LCS of X = <x1, x2,…..,xm>, Y = <y1,y2,…..,yn> • if xm = yn • find an LCS of Xm-1 and Yn-1 appending xm and yn (xm = yn) to this LCS yields an LCS of X and Y. • If xm yn • solve two subproblems: • Finding an LCS of Xm-1 and Y • Finding an LCS of X and Yn-1 • Longer of I and II is an LCS of X and Y. • This shows overlapping subproblmes property in LCS.**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach Step II:Define a recursive solution • Establish a recurrence for the cost of an optimal solution. • Let c[i, j] to be the length of the longest subsequences (LCS) of Xi and Yj. 0 if i = 0 or j = 0 c[i, j] = c[i-1, j-1] + 1 if i, j > 0 and xi = yj max(c[i, j-1], c[i-1, j]) if i, j > 0 and xi yj • Write a recursive algorithm to compute the length of an LCS of two sequences?**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach Step III:Devise an algorithm to compute the length of an LCS and then construct the LCS. • Development of an algorithm • Input two sequences X and Y and a two dimensional table c[0..m, 0..n] • According to the definition for(i = 1; I <= m; I++) { for(j = 1; j <= n; j++) { if (X[i] = = Y[j]) c[i, j] = c[i-1, j-1] + 1 else if (c[i-1, j] >= c[i, j-1]) c[i, j] = c[i-1, j]; else c[i, j] = c[i, j-1]; } } /* It means to compute c[i, j] We need to know c[i-1, j-1], c[i, j-1] and c[i-1, j] */**j**0 1 2 3 4 5 6 0 0 0 0 0 0 0 i B D C A B A 0 0 0 0 1 1 1 0 yj 0 1 1 1 1 2 2 A 1 xi 0 1 1 2 2 2 2 B 2 0 1 1 2 2 3 3 C 3 0 1 2 2 2 3 3 4 B 0 1 2 2 3 3 4 D 5 0 1 2 2 3 4 4 A 6 B 7 Longest Common Subsequence (LCS) ProblemDynamic Programming Approach Example**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach Example • LCS – Algorithm: Computing the length LCS-Length(X, Y) { m = length[X]; n = length[Y]; for(i = 1; i <= m; i++) c[i, 0] = 0; for(j = 0; j <= n; j++) c[0, j] = 0; for(i = 1; i <= m; I++) { for(j = 1; j <= n; j++) { if(X[i] = = Y[i] { c[i, j] = c[i-1, j-1] + 1; b[I, j] = “” } else if(c[i-1, j] >= c[i, j-1]) { c[i, j] = c[i-1, j]; b[I, j] = “” } else { c[i, j] = c[i, j-1]; b[I, j] = “” } } } return c and b; } Complexity: O(m*n)**Longest Common Subsequence (LCS) ProblemDynamic Programming**Approach Example • LCS – Algorithm: Constructing an LCS • Table b[1..m, 1..n] returned by LCS-length() 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] = = “” { 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();