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Unformatted text preview: U.C. Berkeley — CS70: Discrete Mathematics and Probability Problem Set 7 Lecturers: Umesh Vazirani & Christos H. Papadimitriou Due October 6, 2006 at 4:00pm Problem Set 7 Solutions 1. Eulerian and Rudrata Cycles a) There are 5 Platonic solids, each characterized by its vertex degree d and the number of sides on each of its faces f . The following table shows the solids and whether they have an Eulerian cycle or a Rudrata cycle. Name d f Eulerian Rudrata Tetrahedron 3 3 N Y Cube 3 4 N Y Octahedron 4 3 Y Y b Dodecahedron 3 5 N Y Icosahedron 5 3 N Y b) This holds for n > 1. H 1 trivially has a Eulerian path. Recall that a graph has a Eulerian path if and only if all but at most two of its vertices have even degree. In the case of H n , all 2 n vertices have degree n . If n is even, this implies the existence of a Eulerian path. If n is odd, there are 2 n > 2 vertices with odd degree, indicating that no Eulerian path can exist. c) Proof by induction on n . For n = 1, H 1 , a single edge trivially has a Rudrata cycle. We use the recursive definition of H n as the union of two copies H n − 1 linked by a matching. Consider a Rudrata cycle C on H n − 1 visiting vertices v 1 ,v 2 , ··· v N ,v 1 . Consider now the two copies of H n − 1 making up H n : denote by v and v ′ a matching pair of vertices in these copies. Then, we can construct a Rudrata cycle of H n by visiting v 1 , ··· ,v N on the first copy of H n − 1 , using a matching edge to move to v ′ N on the second copy and then running C backwards v ′ N ,v ′ N − 1 , ··· ,v ′ 1 . At this point, it suffices to take the matching edge ( v ′ 1 ,v 1 ) to complete the Rudrata cycle....
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- Fall '08
- Graph Theory, Playing card, Shortest path problem, shortest path, Hn