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MIT6_042JF10_rec08_sol

# MIT6_042JF10_rec08_sol - 6.042/18.062J Mathematics for...

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6.042/18.062J Mathematics for Computer Science October 5, 2010 Tom Leighton and Marten van Dijk Notes for Recitation 8 1 Build-up error Recall a graph is connected iff there is a path between every pair of its vertices. False Claim. If every vertex in a graph has positive degree, then the graph is connected. 1. Prove that this Claim is indeed false by providing a counterexample. Solution. There are many counterexamples; here is one: 2. Since the Claim is false, there must be a logical mistake in the following bogus proof. Pinpoint the first logical mistake (unjustified step) in the proof. Proof. We prove the Claim above by induction. Let P ( n ) be the proposition that if every vertex in an n -vertex graph has positive degree, then the graph is connected. Base cases : ( n 2). In a graph with 1 vertex, that vertex cannot have positive degree, so P (1) holds vacuously. P (2) holds because there is only one graph with two vertices of positive degree, namely, the graph with an edge between the vertices, and this graph is connected. Inductive step : We must show that P ( n ) implies P ( n + 1) for all n 2. Consider an n -vertex graph in which every vertex has positive degree. By the assumption P ( n ), this graph is connected; that is, there is a path between every pair of vertices. Now we add one more vertex x to obtain an ( n + 1)-vertex graph:

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2 Recitation 8 z y X n - vertex graph All that remains is to check that there is a path from x to every other vertex z . Since x has positive degree, there is an edge from x to some other vertex, y . Thus, we can obtain a path from x to z by going from x to y and then following the path from y to z . This proves P ( n + 1). By the principle of induction, P ( n ) is true for all n 0, which proves the Claim. Solution. This one is tricky: the proof is actually a good proof of something else. The first error in the proof is only in the final statement of the inductive step: “This proves P ( n + 1)”. The issue is that to prove P ( n + 1), every ( n + 1)-vertex positive-degree graph must be shown to be connected. But the proof doesn’t show this. Instead, it shows that every ( n + 1)-vertex positive-degree graph that can be built up by adding a vertex of positive degree to an n -vertex connected graph , is connected. The problem is that not every ( n + 1)-vertex positive-degree graph can be built up in this way. The counterexample above illustrates this: there is no way to build that 4-vertex positive-degree graph from a 3-vertex positive-degree graph.
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MIT6_042JF10_rec08_sol - 6.042/18.062J Mathematics for...

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