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# Hw3s - ENGRI 1101 Engineering Applications of OR Fall 2008...

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ENGRI 1101 Engineering Applications of OR Fall 2008 Homework 3 Homework 3 Solutions 1. (a) Each sequence corresponds to a node of the graph, so node 1=0000000101, node 2=1100111111, node 3=1100111110, node 4=0000001001, and node 5=0000111111. There is an edge between every pair of nodes, and the cost edge { i, j } is the number of bits that need to be changed to convert sequence i into sequence j (or vice versa). They are given in the following table: Node 1 2 3 4 5 1 6 7 2 4 2 1 6 2 3 7 3 4 4 5 The graph is depicted here: 5 2 4 3 1 6 7 2 4 1 6 2 7 3 4 (b) The minimum spanning tree calculated using Kruskal’s algorithm is presented below. Note that there are several edges of the same cost in the tree, so, depending on how you broke the ties between these edges, you may have ended up with a different tree. The cost of you minimum spanning tree should be the same, though. 1

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5 4 3 2 1 4 4 7 3 2 6 1 7 2 5 4 3 2 1 4 4 7 3 2 6 1 7 2 5 4 3 2 1 4 4 7 3 2 6 1 7 2 5 4 3 2 1 4 4 7 3 2 6 1 7 2 Edges were considered in the following order: • { 2 , 3 } — added • { 2 , 5 } — added
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Hw3s - ENGRI 1101 Engineering Applications of OR Fall 2008...

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