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Question

Draw all nonisomorphic simple graphs having three vertices.

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Simple graph\textbf{Simple graph}: Undirected edges, multiple/parallel edges not allowed, no loops allowed.

Two simple graphs G1=(V1,E1)G_1=(V_1,E_1) and G2=(V2,E2)G_2=(V_2,E_2) are isomorphic\textbf{isomorphic} if there exists a one-to-one and onto function f:V1V2f:V_1\rightarrow V_2 such that aa and bb are adjacent in G1G_1 if and only if f(a)f(a) and f(b)f(b) are adjacent in G2G_2.

The graph requires n=3n=3 vertices.

Since the graph is a simple graph, the graph contains at most n(n1)2=3(31)2=62=3\frac{n(n-1)}{2}=\frac{3(3-1)}{2}=\frac{6}{2}=3 edges.

0 edges\textbf{0 edges}: 1 nonisomorphic graph

1 edge\textbf{1 edge}: 1 nonisomorphic graph with a single edge between two of the three vertices.

2 edges\textbf{2 edges}: 1 nonisomorphic graph which is a simple path of length 2

3 edges\textbf{3 edges}: 1 nonisomorphic graph which is a simple cycle of length 3

Thus there are 4 nonisomorphic simple graphs with 3 vertices in total.

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