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Unformatted text preview: (b) Show that a graph with 98 vertices cannot have 49 vertices of degree 4 and 49 of degree 3. (c) Show that if G is a graph with 100 vertices, in which u and v have degree 3 and the other 98 vertices all have degree 4, then there must be a path from u to v in G . (d) Show that if G is a 4-regular graph then G cannot have a bridge. 4. [6 marks] Answer ONE of the following. (a) Show that in a graph with p vertices, p 2, there must be two vertices which have the same degree. ( Hint: if all vertex degrees are dierent, what must the smallest and largest degrees be? Show that this cannot happen. ) OR (b) Show that if G is a connected graph and its longest path length is k , then any two paths P and Q in G of length k must have at least one vertex in common. ( Hint: if P and Q have no vertex in common, nd a longer path, which is a contradiction. )...
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This note was uploaded on 11/26/2011 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
- Spring '09