homework1solns-spring08

homework1solns-spring08 - CS500 Homework #1, Spring 2008 1....

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CS500 Homework #1, Spring 2008 Solution. The graph on the left is colored so that each edge connected a light vertex to a dark one. Formally, it is bipartite . Any Hamiltonian cycle would have to have an equal number of light and dark vertices, but there are an odd number of vertices, and one more dark one than light ones. More generally, no bipartite graph with an odd number of vertices can be Hamiltonian. For the Petersen graph, we use the fact that it is highly symmetric. In particular, all the paths of four vertices are equivalent to each other. So, we can avoid most of the search tree by choosing one of these paths and trying to complete it. Let’s take four vertices going clockwise around the outside of the graph. The fifth vertex in this cycle has to be on the Hamiltonian cycle, so it has to connect to the first or last vertex in the path of four. Then there is no way to visit all five of the inner vertices. Solution. Euler’s argument, that every time we enter a vertex we have to leave it along a different edge, shows that every vertex must have an equal number of incoming and outgoing edges. Solution. There are many ways to solve this problem. Here’s one. Suppose the original graph G has n vertices. If the oracle says “yes” to the question of whether G is Hamiltonian, remove one of G ’s edges, and ask her about the resulting graph G 0 . If she says “no”, remove a different edge from G instead, until she says “yes” to G 0 . Then remove an edge from
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This note was uploaded on 04/29/2011 for the course CS 500 taught by Professor Moore,c during the Spring '08 term at New Mexico.

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homework1solns-spring08 - CS500 Homework #1, Spring 2008 1....

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