Data Structures and Algorithms
Homework Assignment 9
Due Monday, April 25, 2005
1. In class we stated the following result: Suppose Gis a connected, undirected graph with n
vertices. If the number of edges in Gis n−1,then Gis a free tree.
First note that the result can be proved by drawing the different cases if Ghas 1 or 2 vertices.
So we may as well assume that n≥3.
Since a free tree is a connected graph with no simple cycles, one way to prove this is to look
at the contrapositive: Suppose Gis a connected, undirected graph with n≥3 vertices. If G
contains a simple cycle, then the number of edges in Gis at least n.
In order to prove this, we take a connected graph Gcontaining a simple cycle Con kvertices.
As noted in the text, the smallest simple cycle has 3 vertices. So k≥3.Now we can use
induction on the number mof vertices in the complement of C, and k, the number of vertices
in C, is fixed.
(a) Prove the base case. That is, show that if Gis a connected graph with k≥3 vertices
and Gcontains a simple cycle on all kvertices, then Gcontains at least kedges. Hint:
Count the edges in a simple cycle.
(b) State the induction hypothesis.
(c) What do you need to show in the induction step? What are you assuming? What do
you need to prove?
(d) Extra Credit. Complete the induction step. Hint: in going from the case m=m0+ 1
vertices in the complement of Cto the case m=m0vertices in the complement of C,
use a DFS of G. The DFS should visit the vertices in Cfirst. Then the vertex that needs
to be deleted can be chosen to be a leaf in the depth-first spanning tree.
2. Problem 7.13, page 221.
3. Problem 7.17, page 222.
4. Problem 7.23, page 222.
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