Added the program to check Bipartite Graph

[shuvojitss]

Pull Request for PyVerse 💡

Requesting to submit a pull request to the PyVerse repository.

---

Issue Title

Please enter the title of the issue related to your pull request.
Bipartite graph

• I have provided the issue title.

---

Info about the Related Issue

What's the goal of the project?
Given an adjacency list / matrix representing a graph with V vertices indexed from 0, the task is to determine whether the graph is bipartite or not. You can use queue data structure to check the graph.

• I have described the aim of the project.

---

Name

Please mention your name.
Shuvojit Samanta

• I have provided my name.

---

GitHub ID

Please mention your GitHub ID.
shuvojitss

• I have provided my GitHub ID.

---

Email ID

Please mention your email ID for further communication.
shuvojitsamanta15@gmail.com

• I have provided my email ID.

---

Identify Yourself

Mention in which program you are contributing (e.g., WoB, GSSOC, SSOC, SWOC).
GSSOC

• I have mentioned my participant role.

---

Closes

Enter the issue number that will be closed through this PR.
Closes: #1078

• I have provided the issue number.

---

Describe the Add-ons or Changes You've Made

Give a clear description of what you have added or modified.

Algorithm Steps to Check if a Graph is Bipartite:

1. Initialize Colors: Create an array color of size V (number of vertices) and initialize all elements to -1, indicating that no vertices have been colored.

2. BFS for Each Component: For each vertex start from 0 to V-1:

   • If color[start] is not -1, continue to the next vertex (this vertex is already colored).

3. Start BFS:

   • Initialize a queue and push the start vertex into it.
   • Color start with color 0.

4. Process the Queue:

   • While the queue is not empty, dequeue a vertex and check its neighbors:
     • Color uncolored neighbors with the opposite color.
     • If a neighbor has the same color as the current vertex, return false (the graph is not bipartite).

5. Completion: If all vertices are processed without conflicts, return true (the graph is bipartite).

• I have described my changes.

---

Type of Change

Select the type of change:

• Bug fix (non-breaking change which fixes an issue)
• New feature (non-breaking change which adds functionality)
• Code style update (formatting, local variables)
• Breaking change (fix or feature that would cause existing functionality to not work as expected)
• This change requires a documentation update

---

How Has This Been Tested?

Describe how your changes have been tested.
I have tested it in all python editors and its working. I am providing a sample input output to describe it.

Sample Input:

Enter the number of vertices: 5
Enter the adjacency matrix:
0 1 1 0 0
1 0 0 1 0
1 0 0 1 0
0 1 1 0 1
0 0 0 1 0

Sample Output:

The graph is bipartite.

Explainataion of Sample:

Diagrammatic representation of the above input:

   0 ---- 1
   |      |
   2 ---- 1 --- 4                            

To determine if the graph is bipartite, we can attempt to color the vertices using two colors. The following coloring can be used:

• Color Vertex 0 with color 0.
• Color Vertex 1 and Vertex 2 with color 1 (neighbors of Vertex 0).
• Color Vertex 3 with color 0 (neighbor of Vertices 1 and 2).
• Color Vertex 4 with color 1 (neighbor of Vertex 3).

Since we can assign colors such that no two adjacent vertices share the same color, the graph is confirmed to be bipartite.

• I have described my testing process.

---

Checklist

Please confirm the following:

• My code follows the guidelines of this project.
• I have performed a self-review of my own code.
• I have commented my code, particularly wherever it was hard to understand.
• I have made corresponding changes to the documentation.
• My changes generate no new warnings.
• I have added things that prove my fix is effective or that my feature works.
• Any dependent changes have been merged and published in downstream modules.

[github-actions]

👋 Thank you for opening this pull request! We're excited to review your contribution. Please give us a moment, and we'll get back to you shortly!

Feel free to join our community on Discord to discuss more!