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Isomorphism of graphsThe simple graphs Example. Show that the graphs
Solution: The function f with Example. Show that the graph G and H are not isomorphic.
Solution: Both G and H have five vertices and six edges. However, H has a vertex of degree 1, namely e, whereas G has no vertices of degree 1. It follows that G and H are not isomorphic. The number of vertices, the number of edges, and the degrees of the vertices are all variants under isomorphism. If any of these quantities differ in two simple graphs, these graphs cannot be isomorphic. However, when these invariants are the same, it does not necessarily mean that the two graphs are isomorphic. Example. Determine whether the graphs G and H are isomorphic.
Solution: The graphs G and H have 8 vertices and 10 edges. They also both have 4 vertices of degree 2 and 4 of degree 3. Since these invariants all agree, it is still conceivable that these graphs are isomorphic. However, G and H are not isomorphic. To see this, note that since deg(a) = 2 in G, a must correspond to either t, u, x or y in H, since these are the vertices of degree 2 in H. However, each of these four vertices in H is adjacent to another vertex of degree 2 in H, which is not true for a in G. To show that a function f from the vertex set of a graph G to the vertex set of a graph H is an isomorphism, we need to show that f preserves edges. One helpful way to do this is to use adjacency matrices. In particular, to show that f is an isomorphism, we can show that the adjacency matrix of G is the same as the adjacency matrix of H, when rows and columns are labeled to correspond to the images under f of the vertices in G that are the labels of these rows and columns in the adjacency matrix of G.
Paths A path of length n from u to v, where n is a positive integer, in an undirected graph is a sequence of edges Example. In this simple graph a, d, c, f, e is a simple path of length 4, since {a, d}, {d, c}, {c, f} and {f, e} are all edges. However, d, e, c, a is not a path, since {e, c} is not an edge. Note that b, c, f, e, b is a circuit of length 4 since {b, c}, {c, f}, {f, e} and {e, b} are edges, and this path begins and ends at b. The path a, b, e, d, a, b, which is of length 5, is not simple since it contains the edge {a, b} twice. A path from a to b in a directed graph G is a sequence of one or more edges A path of length n from u to v in a directed multigraph is a sequence of edges
Date: 2015-01-02; view: 1810
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