Checking If An Undirected Graph Is Bipartite

2021/01/05 - The previous C# code sample implemented the algorithm incorrectly. A corrected Java implementation has been added in its place.

If you are unfamiliar with graphs, check out some of my earlier posts on them.

Resources:

1. Video explanation
2. Brief video overview (clip in larger video)
3. Explanation

Takeaways:

• A bipartite graph (bigraph) is a graph where the vertices can be divided into two disjoint, independent, sets u and v. Every edge will connect a vertex from one set to the other (without self referencing edges - I.E edges going from a vertex in u to another vertex in u).
• One way to visualize a bipartite graph, is to colour all the vertices in a set the same colour. Set u could be red vertices, whereas v could be black. This would mean an edge would always consist of a red and black pair of vertices.
• This type of two-colouring is impossible in non-bipartite graphs. Think of a graph with three vertices arranged in a triangle. We cannot represent this graph as two independent sets, and we cannot two-colour it in such a way that will allow each edge to have different coloured endpoints.
• One way in which we can check if a graph is bipartite, is to run a depth-first search (DFS) over the vertices. Applying two colouring to the graph.
  • Start at a random vertex v and colour it colour1 (red, for example).
  • Colour all adjacent vertices u the opposite colour of v. For each adjacent u, also recursively call our DFS routine.
  • If a graph is bipartite, we can complete this two-colouring without a contradiction.
  • If the graph is not bipartite, then at some point a vertex will get both colours - and this contradiction means we cannot achieve a two-colouring of the graph.
• Time complexity is O(v + e) for an adjacency list. Space complexity is O(v). For an adjacency matrix, the time & space complexity would be O(v^2).

Undirected graph that can be two-coloured:

Undirected graph that cannot be two-coloured:

Below are implementations for checking if undirected graphs are bipartite. There is solutions for both undirected adjacency list & adjacency matrix representations of graphs:

As always, if you found any errors in this post please let me know!