hw1a - . Since and are disjoint, there is no shared horse...

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Michael Sipser Introduction To The Theory Of Computation 2 nd Edition Chapter 0 (0.3) a. b. c. d. e. f. (0.5) (0.6) 1 6 2 7 3 6 4 7 5 6 6 7 8 9 10 1 10 10 10 10 10 2 7 8 9 10 6 3 7 7 8 8 9 4 9 8 7 6 10 5 6 6 6 6 6 a. b. c. d. e. (0.8) (0.10) If , dividing by is dividing by , which is a no-no. 1 2 3 4
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(0.11) Take . The induction hypothesis (and the base step, and common sense, for that matter) tells us that for a set of horse, all horses in the set are the same color. To prove for , we pull out one horse, and the second horse comprises set . Then we put it back and pull out the second horse, so the first is set
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Unformatted text preview: . Since and are disjoint, there is no shared horse forcing them to be the same color, so the proof falters. (0.12) Suppose there is a graph with nodes, . If no 2 nodes have the same degree, each node must have a different degree between (no connections to other nodes) and (connections to all other nodes), inclusive. Since there can’t be both an unconnected node and a node connected to every other node, at least 2 of the nodes must therefore have the same degree....
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This homework help was uploaded on 04/10/2008 for the course COSI 30a taught by Professor Di lillo during the Spring '08 term at Brandeis.

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hw1a - . Since and are disjoint, there is no shared horse...

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