Question

Devise an algorithm based on breadth-first search for finding the connected components of a graph.

Short answer

Using the breadth-first search and algorithm helps in finding the connected components of graph G.

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 that contains all vertices of G.

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

The breadth-first search method.

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.

Procedure for finding the connected components.

Select the first vertex. And add all vertices connected by an edge to the list L and M then remove the edges from the graph. L will keep track of all vertices checked in the algorithm, while M keeps track of all vertices in the last step of the algorithm. It then adds all vertices connected by an edge to a vertex in M. If no vertex was added to the L, then it needs to choose a new vertex to add to the first list and repeat the procedure for the vertex, then it is moving on to a different connected component and thus it increase the connected components C by 1.

Algorithm.

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

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

\(\begin{array}{c}{\rm{while }}{\bf{L}} \ne {\bf{M}}\\{\rm{if }}{\bf{M = }}\phi ;\forall j = 1{\rm{ to }}n.\\{\rm{if }}{{\bf{v}}_{\bf{j}}} \notin {\bf{L}}{\rm{ then L = L}} \cup \left\{ {{v_j}} \right\}\\{\rm{M = M}} \cup \left\{ {{v_j}} \right\}\\N = \phi \end{array}\)

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 \({{\bf{v}}_{\bf{j}}} \in L\).

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.

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

Therefore, this is the required algorithm.