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Dynamic Programming Mani Chandy mani@cs.caltech.edu The Pattern • Given a problem P, obtain a sequence of problems Q0, Q1, …., Qm, where: – You have a solution to Q0 – The solution to P can be obtained from the solution to Qm, – The solution to a problem Qj, j > 0, can be obtained from solutions to problems Qk, k < j, that appear earlier in the sequence. Dynamic Progamming Pattern Given problem P P Propose a partial ordering of problems You can compute the Q0 Qm solution to P from the solution to Qm You know how to compute solution to Q0 Creative Step Finding problems Qi from problem P More mechanical step: Determining the function that computes the solution Sk for problem Qk from the solutions Sj of problems Qj for j < k. Example: Matrix Multiplication 1XN N NXN NXN X 1 What is the cost of multiplying matrices of these sizes? Cost of multiplying 2 matrices pxq qxr p rows and q columns. Cost is 2pqr because resultant matrix has pr elements, and The cost of computing each element is 2q operations. Parenthesization 1XN N NXN NXN X 1 If we multiply these matrices first the cost is 2N3. Resulting matrix NXN Parenthesization 1XN N NXN X 1 Cost of multiplication is N2. Thus, total cost is proportional to N3 + N2 + N if we parenthesize the expression in this way. Different Ordering 1XN N NXN NXN X 1 Cost is proportional to N2 The Ordering Matters! 1XN N One ordering costs O(N3) NXN NXN X 1 The other ordering costs O(N2) 1XN N NXN NXN X 1 Generalization: Parenthesization ( A1 op ( A2 op A3 ) ) ( op …. op An ) Associative operation Cost of operation depends on parameters of the operands. Parenthesize to minimize total cost. Creative Step Come up with partial-ordering of problems Qi given problem P. Propose a partial ordering of problems Q0 Qm Creative Step: Solution Qi,j is: optimally parenthesize the expression Ai op ….. op Aj Relatively “mechanical” steps: 1. Find partial ordering of problems Qi,j 2. Find function f that computes solution Si,j from solutions of problems earlier in the ordering. Partial Ordering Structure Q1,4 Solution to given problem obtained from solution to this Q1,3 Q2,4 problem Q1,n. Depends on Q1,2 Q2,3 Q3,4 Q1,1 Q2,2 Q3,3 Q4,4 Solutions known The Recurrence Relation Let C[j,k] be the minimum cost of executing Aj op … op Ak. Base Case: ???? C[j,j] = 0 Induction Step: ???? C[j,k] for k > j is: min over all v of C[j,v]+C[v+1,k] + cost of the operation combining Proof: ??? [j…v] and [v+1 … k] For matrix multiplication Let j-th matrix have size: p(j-1) X pj Then the size of matrix obtained by combining [ j … v] is: ? p (j-1) X pv Then the size of matrix obtained by combining [ v+1 … k] is: ? pv X pk Cost of multiplying [j … v] and [v+1 … k] is p (j-1) X pv X pk Proof Structure What is the theorem that we are proving? We make an assertion about the meaning of a term. For instance, C[j,k] is the minimum cost of executing Aj op …. op Ak We are proving that this assertion is correct. Proof Structure Almost always, we use induction. Base case: establish that the value of C[j,j] is correct. Induction step: Assume that the value of C[j, j+u] is correct for all u where u is less than V, and prove that the value of C[j, j+V] is correct. Remember what we are proving: C[j,k] is the minimum cost of executing Aj op …. op Ak The Central Idea Bellman’s optimality principle Qa,z The discarded solutions for the smaller problem remain Qi,j Qu,v discarded because the optimal solution dominates them. Pick optimal Discard others All-Points Shortest Path Given a weighted directed graph. • The edge-weight W[j,k] represents the distance from vertex j to vertex k. • There are no cycles of negative weight. • For all j, k, compute D[j,k] where D[j,k] is the length of the shortest path from vertex j to vertex k. The Creative Step Come up with partial-ordering of problems Qi given problem P. There are different problem sets Qi some better than others. Creative Step Let F[j,k,m] be the length of the shortest path from vertex j to vertex k that has at most m hops. What is the partial-ordering of problems Q[j,k,m]? A recurrence relation F[j,k,m] = min over all r of F[j,r,m-1] + W[r,k] Base case: F[j,k,1] = ????? W[j,k] (assume W[j,j] = 0 for all j) Obtaining solution for given problem P D[j,k] = F[j,k,n-1] Proof of Correctness What are we proving? We are proving that the meaning we gave to F[j,k,m] is correct Base Case We show that F[j,k,1] is indeed the length of the shortest path from vertex j to vertex k that traverses at most one edge. Induction Step Assume that F[j,k,m] is the length of the shortest path from j to k that traverses at most m edges, for all m less than p, and prove that F[j,k,p] is the min length from j to k that traverses at most p edges Complexity? n4 Can you do better? Come up with partial-ordering of problems Qi given problem P. Let F[j,k,m] be the length of the shortest path from vertex j to vertex k that has at most m hops. 2m

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