## Related questions with answers

Draw all nonisomorphic simple graphs having three vertices.

Solution

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

Two simple graphs $G_1=(V_1,E_1)$ and $G_2=(V_2,E_2)$ are $\textbf{isomorphic}$ if there exists a one-to-one and onto function $f:V_1\rightarrow V_2$ such that $a$ and $b$ are adjacent in $G_1$ if and only if $f(a)$ and $f(b)$ are adjacent in $G_2$.

The graph requires $n=3$ vertices.

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

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

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

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

$\textbf{3 edges}$: 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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