Question

Devise an algorithm for constructing the spanning forest of a graph based on breadth-first searching.

Short answer

By using the breadth-first search I can construct the spanning forest.

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. And a tree with n vertices has \({\bf{n - 1}}\) edges.

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

\(\begin{array}{c}{\bf{T = G}}\\{\bf{V = \{ }}\,{{\bf{v}}_{\bf{1}}}{\bf{,}}{{\bf{v}}_{\bf{2}}}{\bf{,}}......{\bf{,}}{{\bf{v}}_{\bf{n}}}{\bf{\} }}\\{\bf{L = \{ }}\,{{\bf{v}}_{\bf{1}}}{\bf{\} }}\\{\bf{M = \{ }}\,{{\bf{v}}_{\bf{1}}}{\bf{\} }}\end{array}\)

\(\begin{array}{c}{\bf{while}}\,{\bf{L}} \ne {\bf{M}}\\{\bf{if}}\,{\bf{M = }}\phi \,{\bf{then}}\\{\bf{for}}\,{\bf{j = 1}}\,\,{\bf{ton}}\\{\bf{if}}\,{{\bf{v}}_{\bf{j}}} \notin {\bf{L}}\,{\bf{then}}\end{array}\)

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

For every vertex v in M

\({\bf{for}}\,{\bf{j = 1}}\,\,{\bf{ton}}\)

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

\({\bf{if}}\,{{\bf{v}}_{\bf{j}}} \in {\bf{L}}\,{\bf{then}}\)

Remove the edge \({\bf{(}}{{\bf{v}}_{\bf{i}}}{\bf{,}}{{\bf{v}}_{\bf{j}}}{\bf{)}}\) from T

Remove the edge \({\bf{(}}{{\bf{v}}_{\bf{i}}}{\bf{,}}{{\bf{v}}_{\bf{j}}}{\bf{)}}\) from G

Else

Remove the edge from the graph G

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

Therefore, this is the require algorithm.