<p> CS 331-01 Design and Analysis of Algorithms Homework #4</p><p>1. Given A=[20x2], B=[2x15], C=[15x40], and D=[40x4]. Compute the C(i, j)'s defined in the class for all i < j and trace your computation to get the multiplication order for this instance of chained matrix multiplication.</p><p>2. Let A be the adjacency matrix of a directed graph G=(V, E), where A(i, j)=1 if <i, j>E and A(i, j)=0 otherwise. Define the transitive closure, A+, of A to be a matrix with the property A+(i, j)=1 iff G has a directed path, containing at least one edge, from vertex i to vertex j. A+(i, j)=0 otherwise. The reflexive transitive closure, A*, is a matrix with the property A*(i, j)=1 iff G has a path, containing zero or more edges, from i to j. A*(i, j)=0 otherwise. 1 2</p><p>4 3</p><p>(a) Obtain A, A+ and A* for the above directed graph. (b) Let Ak(i, j)=1 iff there is a path with zero or more edges from i to j going through no vertex of index greater than k. Define A0. (c) Obtain a recurrence definition between Ak and Ak-1 using the logical operators or and and rather than min and +. (d) Write an algorithm, using the recurrence of (c), to find A*. What is the time complexity of your algorithm?</p><p>3. Two character strings may have many common sub-strings. For example, photograph and tomography have several common sub-strings of length one (i.e., single letters), and common sub-strings ph, to, and ograph (as well as all the sub-strings of ograph). The maximum common sub-string length is 6. Let X = x1x2...xn and Y = y1y2...ym be two character strings. The Maximum Common Sub-string (MCS) Problem is to find the maximum common sub-string length for X and Y. </p><p>Let s[i, j] = the length of the longest common sub-string ending at xi and yj, define the functional equation for s[i, j] = s[i-1, j-1]+1 if xi = yj 0 otherwise Give an algorithm to implement this functional equation. What is the time complexity of your algorithm? </p><p>4. Given the cost adjacent matrix for a directed graph as below, find a tour of minimum cost starting from vertex 1 using the functional equation given in class. Show steps. 1 2 3 4 5 1 0 8 13 18 20 2 3 0 7 8 10 3 4 11 0 10 7 4 6 6 7 0 11 5 10 6 2 1 0</p>