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wo nonadjacent vertices of $C,$ then we call $e$ a chord of $C.$ Order all of ay so that they form a proof for the following statment:

$G$ has degree at least $3,$ then $G$ contains a cycle with a chord.

Your Proof:

$p$ can be extended, which is a contradiction to the assumption that $p$ is a longest path.

Take a longest simple path $p=(u_0,u_1,$dots,$u_k)$ in $G.$

deg$(u_0)3$ and $u_1$ is a neighbor of $u_0.$

Since $p$ is longest, every neighbor of $u_0$ must be in $\{u_1,$dots,$u_k\}$ because otherwise

Then $C=(u_0,u_1,$dots,$u_j)$ is a cycle with a chord $e=(u_0,u_i).$

Therefore the vertex $u_0$ must have two neighbors $u_i$ and $u_j$ with $G2$i because

$We$ have found a cycle with a chord $inG,so$ the proof $is$ completed.