## Related questions with answers

Let k be a fixed positive integer and let G = (V, E) be a loop-free undirected graph, where $\operatorname{deg}(\mathrm{v}) \geq \mathrm{k}$ for all $\mathrm{v} \in \mathrm{V}.$ Prove that G contains a path of length k.

Solution

VerifiedProof by induction.

For $k = 1$ the claim is trivial.

Suppose that the claim holds for some $k \in \mathbb{N}$.

Let now $G$ be any graph such that $\deg v \ge k+1$ for every $v \in V$. We then specially have that $\deg v \ge k$ for every $v \in V$, so we can apply the above assumption on $G$.

By assumption there exists a path of length $k$ in $G$ which passes through the vertices $v_0$, ..., $v_k$. Now since $\deg v_k \ge k+1$, it follows that there exists a vertex $v_{k+1} \neq v_i$ for all $i \le k$ which is adjacent to $v_k$.

Now the path passing through $v_0$, ..., $v_{k+1}$ is a path of length $k+1$ in $G$, which proves that the claim holds for $k+1$.

Now by the induction principle, the claim holds for every $k \in \mathbb{N}$.

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