# Show that every graph $G$ has a path of length $\delta(G)$ [closed]

A path in a graph is a subgraph isomorphic to $P_n$, for some $n$. The length of a path is its number of edges. Prove that every graph $G$ has a path of length $\delta(G)$, where $\delta(G)$ is the minimum degree of $G$.

## closed as not constructive by Andrés E. Caicedo, Micah, Lord_Farin, Joe, Dennis GulkoApr 25 '13 at 16:41

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• Also, I think it's impolite to disregard suggestions on how to improve your question (such as those here). – Douglas S. Stones Apr 23 '13 at 14:12

## 2 Answers

I hope its a simple graph. Let $P$ be the maximal path in the graph $G$. On the contrary assume $|P|\le\delta(G)$.

Suppose $v\in V(P)$ be an endpoint in the path. As $v$ is adjacent to $\delta(G)$ vertices. All of its neighbours can't be in the path $P$. But then its contradicts the fact that its the maximal path.

Hence $|P|\ge\delta(G)+1$.

• I think you need to choose $v$ to be the end vertex of the path. – Easy Apr 23 '13 at 11:49
• Yeah, I edited it. – Grobber Apr 23 '13 at 11:57
• I wish people wouldn't answer questions, when the person asking the question has put in no effort beyond copying the question out of a problem set. – Gerry Myerson Apr 23 '13 at 13:31

Let $P$ be the longest path in $G$ from $v_0$ to $v_k$. As min degree is $\delta(G)$, this path contains every neighbor of $v_k$ are on this path and if it is not like this, then let us assume that there is a vertex $v_j$ neighbor of $v_k$ not in the path. then add this to the given path. and thus we get a path of longer length than $P$. A contradiction

• I wish people wouldn't answer questions, when the person asking the question has put in no effort beyond copying the question out of a problem set. – Gerry Myerson Apr 23 '13 at 13:31