Unformatted text preview: 4 4 F B R L Cube 4 B R W G Here is the solution indicated. Cube 3 R W G B Cube 2 G B R W Cube 1 W G B R (a) N;1 (b) Dl!1 (c) It is not possible. For the conditions in 4(a) and 4(b) to be hold simultaneously,the graph would have to have exactly three odd vertices. This is impossible by Corollary 9.2.6....
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- Summer '10
- Graph Theory, Tesseract, Cube Zero, Cube 2: Hypercube, Ernie Barbarash, edge disjoint subgraphs