A graph $G=(V(G), E(G))$ is said to be Bipartite if and only if there exists a partition $V(G)=A\cup B$ and $A\cap B =\emptyset.$ Hence all edges share a vertex from both set $A$ and $B,$ and there are no edges formed between two vertices in the set $A,$ and there are not edges formed between the two vertices in $B.$
$$\textbf{(OR)}$$
A bipartite graph, also called a bigraph, is a set of graph vertices decomposed into two disjoint sets such that no two graph vertices within the same set are adjacent.
- $G$ is bipartite if and only if all cycles in $G$ are of even length.
- Every tree is bipartite.
- A graph is bipartite if and only if it is $2$-colorable $,\text{(i.e. its chromatic number is less than or equal to}\: 2).$
The first graph is known as the Peterson graph.
It is not a bipartite graph.
The Petersen graph has chromatic number $3,$ meaning that its vertices can be colored with three colors — but not with two — such that no edge connects vertices of the same color. It has a list colouring with $3$ colours, by Brooks' theorem for list colourings.
The Petersen graph has chromatic index $4;$ coloring the edges requires four colors. As a connected bridgeless cubic graph with chromatic index four, the Petersen graph is a snark.
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The second graph is a bipartite graph because it is $2$-colorable.
The third graph is the wheel graph.
In a wheel graph, the hub has degree $n-1$, and other nodes have degree $3.$ Wheel graphs are $3$-connected. $W_{4} = K_{4}$, where $K_{4}$ is the complete graph of order four. The chromatic number of $W_{n}$ is
$\chi(W_{n}) = \left\{\begin{matrix} 3 & \text{for n odd} \\ 4 & \text{for n even} \end{matrix}\right.$
It is not a bipartite graph.
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So, the correct answer is $(B).$