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Chapter 12: Problem 9

If \(G=(V, E)\) is a loop-free undirected graph, prove that \(G\) is a tree if there is a unique path between any two vertices of \(G\).

Short Answer

Expert verified
G is a tree. G is connected because there is a path between every pair of vertices. G is acyclic because the condition that there's a unique path between any two vertices in G excludes the possibility of cycles.

Step by step solution

01

Definition of a Tree

We know that a tree is an undirected graph that is connected and acyclic (i.e., it contains no cycles). So our goal is to show that if there's a unique path between any two vertices in G, then G is both connected and acyclic.
02

Show that G is Connected

To prove that G is connected, we just need to note the condition given in the problem statement - there is a unique path between any two vertices in G. By definition, a graph is connected if there is a path between every pair of vertices. So, this condition immediately gives us that G is connected.
03

Show that G is Acyclic

To show that G is acyclic, assume that there is a cycle in G. If there is a cycle, then we can find two vertices that have more than one path between them (one path along the cycle, and one path that goes directly from one vertex to the other), which contradicts the condition given in the problem statement that there is a unique path between any two vertices. Therefore G must be acyclic.
04

Final Conclusion

We have demonstrated that if there's a unique path between any two vertices in G, then G is both connected and acyclic. Hence, by the definition of a tree in graph theory, G is a tree.

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Most popular questions from this chapter

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If \(B_{1}, B_{2}, \ldots, B_{k}\) are the biconnected components of a loop-free connected undirected graph \(G\), how is \(\chi(G)\) related to \(\chi\left(B_{i}\right), 1 \leq i \leq k\) ?

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