Question

Is it possible for wheel $W_n$ ($n \ge 3$) to be bipartite?

• A. No
• B. Yes
• C. Do not say
• D. None of these

Hint:

Bipartite graphs have no odd cycles. $W_n$ contains triangles for $n\ge3$.

Answer 1

The Correct Option is A

To determine if the wheel graph $W_n$ (where $n \ge 3$) is bipartite, we first need to understand what a wheel graph is and what it means for a graph to be bipartite.

Definition of a Wheel Graph:
A wheel graph $W_n$ is created by connecting a single central vertex to all vertices of a cycle graph $C_n$. Essentially, it consists of $n$ vertices in a cycle (which we call the outer cycle) and one additional central vertex that is connected to all vertices of the cycle.

Definition of a Bipartite Graph:
A graph is called bipartite if its vertex set can be divided into two disjoint sets such that no two graph vertices within the same set are adjacent. A common characteristic of bipartite graphs is that they contain no odd cycles.

Step-by-Step Analysis:

1. The cycle component of the wheel graph $W_n$ has $n$ vertices forming a closed loop. This is $C_n$.
2. If $n$ is odd, then $C_n$ itself is an odd cycle. A necessary condition for a graph to be bipartite is that it should not contain any odd cycles.
3. Since $C_n$ is an odd cycle when $n$ is odd, and it is part of the wheel graph, it directly implies the presence of an odd cycle in $W_n$.

Conclusion:
Since the wheel graph $W_n$ with an odd $n$ has an odd cycle (the cycle $C_n$), it is not bipartite. Even for the smallest wheel graph, $W_3$, which corresponds to connecting a central vertex to all vertices of a triangle (which is an odd cycle), it is not bipartite. Therefore, for any $W_n$, it is not possible to partition the vertices into two disjoint sets without having edges within the same set.

Thus, the correct answer is: No