Let G = ( V , E ) G = (V, E) G = ( V , E ) Be A Connected Graph Where Every Vertex Has Degree Either 1 1 1 Or 2 2 2 . Then G G G Must Be Either A Single Path Or A Single Cycle.

Let $G = (V, E)$ be a connected graph where every vertex has degree either $1$ or $2$. Then $G$ must be either a single path or a single cycle.

In graph theory, a connected graph is a graph in which there is a path between every pair of vertices. A graph with every vertex having a degree of either $1$ or $2$ is a specific type of connected graph. In this article, we will explore the properties of such graphs and prove that they must be either a single path or a single cycle.

Theorem Statement

The theorem states that if $G = (V, E)$ is a connected graph where every vertex has degree either $1$ or $2$, then $G$ must be either a single path or a single cycle.

Proof

To prove this theorem, we will use a combination of mathematical induction and graph theory concepts.

Base Case

Let's start with the base case. Suppose we have a graph with only one vertex. In this case, the graph is a single vertex, which is both a single path and a single cycle. Therefore, the base case is true.

Inductive Step

Now, let's assume that the theorem is true for all graphs with $n$ vertices, where $n \geq 1$. We need to show that the theorem is true for all graphs with $n+1$ vertices.

Let $G = (V, E)$ be a connected graph with $n+1$ vertices, where every vertex has degree either $1$ or $2$. We will show that $G$ must be either a single path or a single cycle.

Case 1: $G$ has a vertex with degree $1$

Let $v$ be a vertex in $G$ with degree $1$. Since $G$ is connected, there must be a vertex $u$ adjacent to $v$. Let $G'$ be the graph obtained by removing $v$ from $G$. Then $G'$ is a connected graph with $n$ vertices, where every vertex has degree either $1$ or $2$. By the inductive hypothesis, $G'$ must be either a single path or a single cycle.

Now, let's consider the following cases:

• If $G'$ is a single path, then $G$ is also a single path, since we can add $v$ to the end of the path.
• If $G'$ is a single cycle, then $G$ is also a single cycle, since we can add $v$ to the cycle.

Case 2: $G$ has a vertex with degree $2$

Let $v$ be a vertex in $G$ with degree $2$. Since $G$ is connected, there must be two vertices $u$ and $w$ adjacent to $v$. Let $G'$ be the graph obtained by removing $v$ from $G$. Then $G'$ is a connected graph with $n-1$ vertices, where every vertex has degree either $1$ or $2$. By the inductive hypothesis, $G'$ must be either a single path or a single cycle.

Now, let's consider the following cases:

• If $G'$ is a single path, then $G$ is also a single path, since we can add $v$ to the path.
• If $G'$ is a single cycle, then $G$ is also a single cycle, since we can add $v$ to the cycle.

Conclusion

In both cases, we have shown that $G$ must be either a single path or a single cycle. Therefore, the theorem is true for all connected graphs with every vertex having degree either $1$ or $2$.

The theorem has several interesting implications. For example, it shows that a connected graph with every vertex having degree either $1$ or $2$ cannot have any vertices with degree greater than $2$. This is because if a vertex had degree greater than $2$, it would not be possible to add any more vertices to the graph without violating the condition that every vertex has degree either $1$ or $2$.

The theorem also has implications for the structure of connected graphs. For example, it shows that a connected graph with every vertex having degree either $1$ or $2$ must be either a single path or a single cycle. This is because if a graph had more than one cycle, it would not be possible to add any more vertices to the graph without violating the condition that every vertex has degree either $1$ or $2$.

In conclusion, the theorem states that if $G = (V, E)$ is a connected graph where every vertex has degree either $1$ or $2$, then $G$ must be either a single path or a single cycle. We have provided a proof of this theorem using mathematical induction and graph theory concepts. The theorem has several interesting implications for the structure of connected graphs and the properties of vertices in these graphs.

• [1] Graph Theory by Reinhard Diestel
• [2] Introduction to Graph Theory by Douglas B. West

This is my attempt of a proof, but I am sure it is not perfect. I would appreciate any feedback or suggestions on how to improve the proof.
Q&A: Let $G = (V, E)$ be a connected graph where every vertex has degree either $1$ or $2$. Then $G$ must be either a single path or a single cycle.

Q: What is the significance of the theorem?

A: The theorem has several interesting implications for the structure of connected graphs. It shows that a connected graph with every vertex having degree either $1$ or $2$ cannot have any vertices with degree greater than $2$. This is because if a vertex had degree greater than $2$, it would not be possible to add any more vertices to the graph without violating the condition that every vertex has degree either $1$ or $2$.

Q: What are the possible structures of a connected graph with every vertex having degree either $1$ or $2$?

A: According to the theorem, a connected graph with every vertex having degree either $1$ or $2$ must be either a single path or a single cycle. This is because if a graph had more than one cycle, it would not be possible to add any more vertices to the graph without violating the condition that every vertex has degree either $1$ or $2$.

Q: Can a connected graph with every vertex having degree either $1$ or $2$ have multiple connected components?

A: No, a connected graph with every vertex having degree either $1$ or $2$ cannot have multiple connected components. This is because if a graph had multiple connected components, it would not be possible to add any more vertices to the graph without violating the condition that every vertex has degree either $1$ or $2$.

Q: Can a connected graph with every vertex having degree either $1$ or $2$ have vertices with degree $0$?

A: No, a connected graph with every vertex having degree either $1$ or $2$ cannot have vertices with degree $0$. This is because if a graph had vertices with degree $0$, it would not be possible to add any more vertices to the graph without violating the condition that every vertex has degree either $1$ or $2$.

Q: Can a connected graph with every vertex having degree either $1$ or $2$ be a tree?

A: No, a connected graph with every vertex having degree either $1$ or $2$ cannot be a tree. This is because a tree is a connected graph with no cycles, but a connected graph with every vertex having degree either $1$ or $2$ must be either a single path or a single cycle.

Q: Can a connected graph with every vertex having degree either $1$ or $2$ be a complete graph?

A: No, a connected graph with every vertex having degree either $1$ or $2$ cannot be a complete graph. This is because a complete graph is a graph in which every vertex is connected to every other vertex, but a connected graph with every vertex having degree either $1$ or $2$ must be either a single path or a single cycle.

Q: Can a connected graph with every vertex having degree either $1$ or $2$ be a bipartite graph?

A: Yes, a connected graph with every vertex having degree either $1$ or $2$ can be a bipartite graph. This is because a bipartite graph is a graph in which the vertices can be divided into two disjoint sets such that every edge connects a vertex in one set to a vertex in the other set, and a connected graph with every vertex having degree either $1$ or $2$ can be divided into two disjoint sets in this way.

In conclusion, the theorem states that if $G = (V, E)$ is a connected graph where every vertex has degree either $1$ or $2$, then $G$ must be either a single path or a single cycle. We have provided a proof of this theorem using mathematical induction and graph theory concepts. We have also answered several frequently asked questions about the theorem and its implications for the structure of connected graphs.