You are given a graph containing n vertices and m edges and given that the graph doesn’t contain cycle of odd length. Time Complexity of the best known algorithm to find out whether the graph is bipartite or not isa)O(m+n)b)O(1)c)O(mn)d)O(n2)Correct answer is option 'B'. Can you explain this answer?

Question

You are given a graph containing n vertices and m edges and given that the graph doesn&#8217;t contain cycle of odd length. Time Complexity of the best known algorithm to find out whether the graph is bipartite or not isa)O(m+n)b)O(1)c)O(mn)d)O(n2)Correct answer is option 'B'. Can you explain this answer?

Answer

Explanation:

A graph is said to be bipartite if its vertices can be divided into two disjoint sets such that every edge of the graph connects a vertex from one set to a vertex from the other set. In other words, the graph can be colored with two colors such that no two adjacent vertices have the same color.

The best known algorithm to determine whether a graph is bipartite or not is the Breadth-First Search (BFS) algorithm. The BFS algorithm explores the graph level by level, starting from a given vertex. During the traversal, it assigns colors to the vertices in a way that no two adjacent vertices have the same color. If at any point, an edge is found between two vertices of the same color, it means the graph is not bipartite. Otherwise, if the traversal completes without any such contradiction, the graph is bipartite.

Time Complexity Analysis:

The time complexity of the BFS algorithm is determined by the number of vertices (n) and the number of edges (m) in the graph. Here's a breakdown of the time complexity:

1. Traversing through all the vertices: O(n)
2. Exploring the edges of each vertex: O(m)

Since the algorithm must visit every vertex and explore every edge once, the overall time complexity of the BFS algorithm is O(n + m).

However, in the given scenario where the graph doesn't contain cycles of odd length, we can make an observation:

- If there are no odd cycles in the graph, it implies that the graph is bipartite.
- The presence of an odd cycle would lead to a contradiction in the coloring process, as two adjacent vertices in the cycle would have the same color.

Therefore, we can conclude that if the graph doesn't contain cycles of odd length, it is bipartite. In this case, we don't need to run the BFS algorithm, and we can determine the bipartiteness of the graph in constant time, i.e., O(1).

Hence, the correct answer is option 'B' - O(1).

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