t05s_y2008

# t05s_y2008 - The University of Sydney MATH2969/2069 Graph...

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Unformatted text preview: The University of Sydney MATH2969/2069 Graph Theory Tutorial 5 (Week 12) Solutions 2008 1. ( i ) Let G be the disconnected planar graph shown. Draw its dual G , and the dual of the dual ( G ) . ( ii ) Show that if G is a disconnected planar graph, then G is connected. Deduce that ( G ) is not isomorphic to G . G Solution. ( i ) G ( G ) ( ii ) A connected graph is one in which there exists a path between any pair of vertices. Let G be a disconnected planar graph, and consider its dual G . In G , there is clearly a path between any two vertices which correspond to faces within one of the components of G . Let v be the vertex in G corresponding to the infinite face of G . Then v is adjacent to at least one vertex (corresponding to a face) in each component of G . So if u and w are two vertices of G which correspond to faces in different components of G , then there is a path from u to w , via v . Hence G is connected. Since G is connected, its dual ( G ) is also connected. But G is discon- nected, and so ( G ) is not isomorphic to G . 2. A certain polyhedron has faces which are triangles and pentagons, with each triangle surrounded by pentagons and each pentagon surrounded by triangles. If every vertex has the same degree, p say, show that 1 e = 1 p- 7 30 . Deduce that p = 4, and that there are 20 triangles and 12 pentagons. Can you construct such a polyhedron? State the dual result. Solution. Let T be the number of triangles, and P the number of pentagons. Then the number of faces, f = T + P . Since each triangle is bounded by 3 edges, and each pentagon by 5, we have 3 T + 5 P = 2 e . But each edge adjoins exactly one triangle and exactly one pentagon, 2 so 3 T = 5 P . Hence, T = 2 e 6 = e 3 , P = 2 e 10 = e 5 , f = P + T = e 3 + e 5 = 8 e 15 . Also, assuming every vertex has degree p , we have pv = 2 e , or v = 2 e/p . Now substitute into Eulers formula: 2 e p- e + 8 e 15 = 2 . Divide by 2 e and simplify: 1 e = 1 p- 7 30 Clearly, 1 e must be a positive number, so we must have 1 p- 7 30 0, or p 30 / 7. But p must be an integer 3. So the only possible values for p are 3 or 4. If p = 3, e = 10, but then v = 2 e/p = 20 / 3, which is not an integer. (Actually, it is easy enough to see that it is not possible to put together triangles and pentagons in the way specified, such that the degree of each vertex is three. Try drawing it and see!) If p = 4, e = 60 and T = 20 and P = 12, as required. Such a polyhedron may be constructed in two halves: We then stitch together along the edges which correspond in the label- ing. a b c d e f g h i j f g h i j a b c d e It is certainly possible to draw the full graph, with a little more effort: Note that the outside region is one of the pentagons....
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t05s_y2008 - The University of Sydney MATH2969/2069 Graph...

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