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Unformatted text preview: The University of Sydney MATH2969/2069 Graph Theory Tutorial 2 (Week 9) 2008 1. Show that the graph on the left is Hamiltonian, but that the other two are not. Solution. To show that the graph is Hamiltonian, simply find a Hamiltonian cycle. (That is, a cycle which passes through every vertex exactly once.) Unfortunately, finding such a cycle may be tricky, even with so few vertices and edges. It may help to note that exactly two of the edges incident to the central vertex must be used (else this central vertex would not be used exactly once, contrary to requirement). Try assuming that, with the labelling shown, [ k, h ] and [ k, i ] are used, and [ k, f ], [ k, g ], [ k, j ] are not. Then, since f , g , j are to be used, edges [ e, f ], [ f, b ], [ a, g ], [ g, c ], [ a, j ], [ j, d ] must all be used. (See diagram above right.) With this start, it is not hard to complete a Hamiltonian cycle. An example is shown in the second diagram: a b c d e f g h i j k a b c d e f g h i j k The vertices of the second graph can be coloured, each black or white, in such a way that each edge joins a black vertex to a white vertex (eg., as shown in the diagram). Any path using all vertices would then have to alternate between black and white vertices, eg.: v 1 = black, v 2 = white, v 3 = black, v 4 = white, . . . , v 13 = black (7 blacks and 6 whites), but this last vertex could not be adjacent with the first to complete the cycle, both having the same colour. Hence no Hamiltonian cycle exists. 2 In the third graph, there is an outer pentagon (5 sided figure) of 5 vertices, an innermost pentagon of 5 vertices, and an inbetween decagon (10 sided figure) of 10 vertices. On the decagon, 5 of the vertices are of degree 2, which means that in any Hamil tonian cycle all 10 edges of this decagon must be used. (For a path to go through a vertex of degree 2, both incident edges must be used.) Since each vertex must be used only once, this means that none of the edges directed radially outwards from the centre of the graph as drawn can be used. That is, none of the edges connecting the outer pentagon, the decagon, and the inner pentagon can be used. So we cannot get from either pentagon to the decagon. Hence no Hamiltonian cycle can exist. 2. Is it possible for a simple connected graph containing a bridge to be Hamilto nian? Solution. A bridge is an edge whose removal would disconnect the graph. Clearly, any Hamiltonian cycle in a graph with a bridge would have to include the bridge. But a bridge is not part of any cycle (since if it were, its removal would not disconnect the graph). So any graph with a bridge is not Hamiltonian. 3. What is wrong with the following argument?...
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This document was uploaded on 03/20/2011.
 Spring '09
 Math, Graph Theory

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