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Unformatted text preview: a subcircuit will be formed, violating rule 2, so at least one (in fact exactly one) of theses 2 edges cannot be used. They are symmetric, so without loss of generality assume that we do not use (5,7). By rule 1 at node 7, we must use (7,1) and (7,3). By rule 3 at node 3, we cannot use edges (3,2) nor (3,6). By rule 1 at node 2, we must use edges (2,1) and (2,5) which forms a proper subcircuit, violating rule 2. (f). One Hamilton path is: d-e-c-b-a-g-f-h-i-n-k-j-l-m. No Hamilton circuit: Use Theorem 0 from class: Removing 2 nodes a,n leaves 3 connected components. (g). One Hamilton path is: m-g-a-f-e-k-l-d-j-i-c-b-h. No Hamilton circuit: Use exc 9, the graph is bipartite, with an odd number of nodes (or, the number of nodes on one side, 6, is not equal to the number of nodes on the other side, 7)....
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- Spring '08