Question

Use Exercise 60 to construct an algorithm for determining whether a directed graph contains a circuit.

Short answer

By using the algorithm get that a directed graph contains a circuit.

Step-by-step solution

Compare with the definition.

A spanning tree of a simple graph G is a subgraph of G that is a tree and which contains all vertices of G.

A tree is an undirected graph that is connected and does not contain any single circuit n vertices has \({\bf{n - 1}}\) edges of a tree.

A rooted spanning tree is a directed graph is a rooted tree that contains the edge of the directed graph and every vertex of the directed graph should be the endpoint of an edge in the rooted tree.

Method for breadth-first search.

In breadth-first search first, choose a root. Add all edges incident to the root. Then each of the vertices at level 1, add all edges incident with ta vertex not included in the tree yet. And repeat until all vertices were added to the tree.

Algorithm.

Let G be undirected graph having vertices\({\bf{V = \{ }}{{\bf{v}}_{\bf{1}}}{\bf{,}}{{\bf{v}}_{\bf{2}}}{\bf{,}}......{\bf{,}}{{\bf{v}}_{\bf{n}}}{\bf{\} }}\)

\(\begin{array}{c}{\bf{T = G}}\\{\bf{V = \{ }}{{\bf{v}}_{\bf{1}}}{\bf{,}}{{\bf{v}}_{\bf{2}}}{\bf{,}}......{\bf{,}}{{\bf{v}}_{\bf{n}}}{\bf{\} }}\\{\bf{L = \{ }}{{\bf{v}}_{\bf{1}}}{\bf{\} }}\\{\bf{M = \{ }}{{\bf{v}}_{\bf{1}}}{\bf{\} }}\\{\bf{circuit = No}}\\{\bf{whileL}} \ne {\bf{Vandcircuit = No}}\end{array}\)

For every vertex v in M

\({\bf{forj = 1ton}}\)

If \({\bf{(}}{{\bf{v}}_{\bf{i}}}{\bf{,}}{{\bf{v}}_{\bf{j}}}{\bf{)}}\)is an edge in G then \({\bf{if}}{{\bf{v}}_{\bf{j}}} \in L{\bf{then}}\)

\({\bf{circuit = yes}}\)

Else

Remove the edge from the graph G.

\(\begin{array}{c}{\bf{L = L}} \cup {\bf{\{ }}{{\bf{v}}_{\bf{j}}}{\bf{\} }}\\{\bf{N = N}} \cup {\bf{\{ }}{{\bf{v}}_{\bf{j}}}{\bf{\} }}\\{\bf{M = N}}\\{\bf{returncircuit}}\end{array}\)

Therefore, by using the algorithm get that a directed graph contains a circuit.