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Chapter 12: Problem 5

Let \(G=(V, E)\) be a loop-free undirected graph. If \(\operatorname{deg}(v) \geq 2\) for all \(v \in V\), prove that \(G\) contains a cycle.

Short Answer

Expert verified
If all vertices in a loop-free undirected graph have a degree of at least 2, it can be proven by contradiction that the graph contains a cycle. The contradiction arises from the fact that a sequence of vertices and edges, which is supposed to be maximal, comes back to an already visited vertex, thus forming a cycle.

Step by step solution

01

Assume the converse scenario

Assume the contrary, that there is no cycle in the graph \(G=(V,E)\). Now, suppose there is a maximal sequence (i.e. a sequence that cannot be made longer by appending vertices) of different vertices and edges (a path) \(v_1, e_1, v_2, e_2,...,v_n\). This sequence is maximal, meaning there are no other vertices adjacent to \(v_n\) that have not already been included in the sequence.
02

Inspect the adjacency of vertices in the sequence

From the assumption, since the degree of every vertex is at least 2, vertex \(v_n\) must have another adjacent vertex \(v_{m}\) not included in the sequence. But since the sequence is maximal, all the vertices adjacent to \(v_n\) must have appeared in the sequence, which is contradictory. This implies that \(v_{m}=v_k, k<n\). However, we assumed that there is no cycle in the graph, so all the vertices in the sequence should be different, which leads to another contradiction.
03

Conclude the proof

Since our original assumption (that the graph doesn't contain any cycles) leads to a contradiction, it must be incorrect. Therefore, the graph \(G\) must contain a cycle.

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Most popular questions from this chapter

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