sol3 - a subcircuit will be formed, violating rule 2, so at...

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AMS 301.3 Fall, 2009 Homework Set # 3 — Solution Notes #2, 2.1: (a). In order for K n to have an Euler cycle, every vertex must have even degree; thus, we must have n - 1 even, which means that n must be odd (and 3, to avoid the trivial case of K 1 ). (b). K 2 has an Euler trail but no Euler cycle. (c). K r,s has an Euler cycle if both r and s are even, as then the degree of every node is either r or s . For each of the graphs in problem #4(e,g) section 2.2, determine if there exists an Euler cycle; if so, give it, and if not, explain why not. #4(e): No Euler cycle, some nodes have odd degree such as b,c,h,d,e,g. (Also no Euler trail.) #4(g): No Euler cycle, some nodes (in fact all nodes) have odd degree; no Euler trail (since more than 2 odd-degree vertices). #4, 2.2: (d). One Hamilton path is: 4-5-6-1-2-3-7. There is no Hamilton circuit: Rule (1) at vertex 4, both edges (4,5) and (4,3) must be in a Hamilton circuit. Now consider node 7, if we use both edges (7,5) and (7,3)
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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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