1. Basic idea: Use smaller instances of the problem to find the solution of larger instances

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1 Chapter 8. Dynamic Programming CmSc Intro to Algorithms. Basic idea: Use smaller instances of the problem to find the solution of larger instances Example : Fibonacci numbers F = F = F n = F n- + F n- Recursion is exponential goes through all nodes in a tree of height n Dynamic solution: You need only the two previous values in order to compute the next number in the sequence F F Do n times: F F + F F F F F Output F Example : Computing a binomial coefficient (a + b) n = C(n,)a n + C(n,)a n- b + +C(n,k) a n-k b k + C(n,n) b n e.g (a + b) = C(,)a + C(,)a b +C(,) a b + C(,) a b + C(,) ab + C(,) b = a + a b + a b a b + a b + b Also: C(n,k): the number of combinations of k elements out of n elements: e.g. C(,) = {,}, {,}, {,} Formula: C(n,k) = n(n-)(n-) (n-k+) / k! = n! / ( k! (n-k)!) C(n,) = C(n,n) = Multiplication is a heavy operation Dynamic programming solution: Based on the relation C(n,k) = C(n-,k-) + C(n-,k) for n > k > ; C(n,) = C(n,n) =

2 k- k n- n k k k n- n- C(n-,k-) C(n-,k). n n C(n.k) n NOTE: recursive solution is possible but not efficient: Coeff(n,k) If k = or k = n return Else C(n,k) = c(n-,k) + c(n-,k-) Return C(n,k) Example: C(,) C(,) C(,) / \ C(,) C(,) C(,) C(,) / \ / \ / \ C(,) C(,) C(,) C(,) C(,) C(,) / \ / \ / \ C(,) C(,) C(,) C(,) C(,) C(,) Compare with Fibonacci recursion tree exponential complexity

3 Example : The Manhattan Tourist Problem Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions () in the Manhattan grid Source Sink Goal: Find the longest path in a weighted grid. Input: A weighted grid G with two distinct vertices, one labeled source and the other labeled sink Output: A longest path in G from source to sink Each edge has a weight reflects the usefulness of going along that edge

4 source The greedy algorithm is not optimal: sink source promising start, but leads to bad choices! 8 sink

5 The greedy algorithm is given by a simple recursive program MT(n,m) x MT(n-,m)+ length of the edge from (n-,m) to (n,m) y MT(n,m-)+ length of the edge from (n,m-) to (n,m) return max{x,y} Dynamic programming approach: Calculate optimal path score for each vertex in the graph : Each vertex s score is the maximum of the prior vertices score plus the weight of the respective edge in between source j i S, = S, =

6 source i S, = - S, = 8 S, = 8 We continue in this way until the sink is reached Computing the score for a point (i,j) by the recurrence relation: s i, j = max s i-, j + weight of the edge between (i-, j) and (i, j) s i, j- + weight of the edge between (i, j-) and (i, j) The running time is n x m for a n by m grid (n = # of rows, m = # of columns) 6

7 Manhattan is not a perfect grid A A A B What about the diagonals? The score at point B is given by: S B = max( S A + weight of edge (A,B), S A + weight of edge (A,B), S A + weight of edge (A,B) ) In the general case, when a point x has several predecessors, computing the score for point x is given by the recurrence relation: S x = max y (S y + weight of edge (y,x) where y is a predecessor of x) Predecessors (x) set of vertices that have edges leading to x The running time for a graph G(V, E) is O(E) since each edge is evaluated once (V is the set of all vertices and E is the set of all edges) Traveling in the Grid The only hitch is that one must decide on the order in which visit the vertices By the time the vertex x is analyzed, the values sy for all its predecessors y should be computed otherwise we are in trouble. We need to traverse the vertices in some order The order is determined by the topological sort of the vertexes in the graph 7

8 . Characteristics. The problem can be divided into stages with a decision required at each stage.. Each stage has a number of states associated with it.. The decision at one stage transforms one state into a state in the next stage.. The optimal decision for state j does not depend on the optimal decision for stage j+. There exists a recursive relationship that identifies the optimal decision for stage j, given that stage j- has already been solved. 6. The initial stage must be solvable by itself.. Some other problems that can be solved with dynamic programming Computing the transitive closure of a directed graph - Warshall s algorithm All-pairs shortest paths problem Floyd s algorithm Optimal binary search trees The Knapsack problem The Coin Change problem DNA sequence alignments 8

9/29/09 Comp /Comp Fall

9/29/09 Comp /Comp Fall 9/29/9 Comp 9-9/Comp 79-9 Fall 29 1 So far we ve tried: A greedy algorithm that does not work for all inputs (it is incorrect) An exhaustive search algorithm that is correct, but can take a long time A