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CS 3345 – Spring 2012 Homework 5 Solutions Q1. We will find the longest distance travelled away from every vertex and report back the largest number. The following algorithm will accomplish this and the analysis follows. Algorithm: findDiameter(T) Input: a tree T Output: the length of the longest path between two vertices Pseudocode: diameter 0 for all v ∈ T do T1 BFS(T, v) for all w such that w ≠ v do if distance(v, w) > diameter then diameter distance(v, w) return diameter This algorithm runs in O(n(n + m)) time. For each of the n vertices in T, breadth-first search will run in O(n + m) time. Since T is a tree, it follows that m = n – 1. Thus, the algorithm will run in O(n(n + n – 1)) = O(n2) time. Q2. If we think of the stations as vertices and the links as edges, we are basically just performing a modified breath-first search to find the vertices that can be reached by at most 4 edges from the vertex. If we are given a graph in which every vertex is at most 3 edges away from any other vertex, we would be performing n complete BFS traversals, which gives a worst case run time of O(n(n + m) time. Q3. This strategy does not always work. Consider the following graph as a counter-example: 1 a 20 b 10 start 10 30 40 10 goal 15 c Using the above algorithm, the vertices are visited in the order start, a, c, goal, for a total path length of 30. However, the shortest path from start to goal is through the vertex c; that is, start c goal for a total path length of 25. Q4. 3 b d 7 -5 c starta 12 Q5. By contradiction: Let T be a minimum spanning tree of a graph G (with distinctly weighted edges) generated by Kruskal’s algorithm. Assume that T’ is another minimum spanning tree for G. Now, suppose e is the first edge of T which does not belong to T’. We denote by C the cycle in T’ + e that contains e and f to be an edge of C which is not in T’. Thus, by Kruskal’s algorithm, it follows that the weight of f must be greater than the weight of e. If weight(f) < weight(e), then Kruskal’s algorithm would have called for tree of weight less than T’, and from this it follows that T’ is not a minimum spanning tree for G. Hence, this is a contradiction. 2