Bipartite Graph Check in DSA C# - Build from Scratch

Bipartite Graph Check

📖 Scenario: You are working with a social network where people are connected by friendships. You want to check if the network can be split into two groups so that no one in the same group is friends with each other. This is called checking if the network is a bipartite graph.

🎯 Goal: Build a program that represents the friendship network as a graph and checks if it is bipartite using colors to separate groups.

📋 What You'll Learn

Create a graph using an adjacency list representation with exact nodes and edges

Create a color array to track group assignments

Implement a function to check if the graph is bipartite using BFS

Print the result as 'True' or 'False'

💡 Why This Matters

🌍 Real World

Checking if a network can be divided into two groups without internal connections helps in social network analysis, scheduling problems, and resource allocation.

💼 Career

Understanding bipartite graphs is important for software engineers working on graph algorithms, network design, and optimization problems.

1

Create the graph adjacency list

Create a dictionary called graph representing the graph adjacency list with these exact entries: 0: new List {1, 3}, 1: new List {0, 2}, 2: new List {1, 3}, 3: new List {0, 2}.

DSA C#

# Your code here

Hint

Use a Dictionary<int, List<int>> to represent the graph. Each key is a node, and the value is a list of connected nodes.

2

Create the color array

Create an integer array called color with length 4 and initialize all elements to -1 to represent uncolored nodes.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        Dictionary> graph = new Dictionary>
        {
            {0, new List {1, 3}},
            {1, new List {0, 2}},
            {2, new List {1, 3}},
            {3, new List {0, 2}}
        };

        // Create the color array
        # Your code here
    }
}

Hint

Use int[] color = new int[4] and set all values to -1 to mark nodes as uncolored.

3

Implement the bipartite check function

Add a static method called IsBipartite that takes Dictionary<int, List<int>> graph and int[] color as parameters and returns a bool. Use a queue to perform BFS starting from node 0, assign colors 0 and 1 to nodes, and return false if a conflict is found; otherwise, return true. Call IsBipartite(graph, color) inside Main and store the result in a variable called result.

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        Dictionary> graph = new Dictionary>
        {
            {0, new List {1, 3}},
            {1, new List {0, 2}},
            {2, new List {1, 3}},
            {3, new List {0, 2}}
        };

        int[] color = new int[4] {-1, -1, -1, -1};

        // Call IsBipartite and store result
        # Your code here
    }

    // Implement IsBipartite method here
    # Your code here
}

Hint

Use BFS with a queue. Assign color 0 to the start node, then assign opposite colors to neighbors. If a neighbor has the same color as the current node, return false.

4

Print the bipartite check result

Print the value of the variable result using Console.WriteLine(result).

DSA C#

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        Dictionary> graph = new Dictionary>
        {
            {0, new List {1, 3}},
            {1, new List {0, 2}},
            {2, new List {1, 3}},
            {3, new List {0, 2}}
        };

        int[] color = new int[4] {-1, -1, -1, -1};

        bool result = IsBipartite(graph, color);

        // Print the result
        # Your code here
    }

    static bool IsBipartite(Dictionary> graph, int[] color)
    {
        Queue queue = new Queue();
        color[0] = 0;
        queue.Enqueue(0);

        while (queue.Count > 0)
        {
            int node = queue.Dequeue();
            foreach (int neighbor in graph[node])
            {
                if (color[neighbor] == -1)
                {
                    color[neighbor] = 1 - color[node];
                    queue.Enqueue(neighbor);
                }
                else if (color[neighbor] == color[node])
                {
                    return false;
                }
            }
        }
        return true;
    }
}

Hint

Use Console.WriteLine(result) to print the boolean result.

Practice

1. What does it mean for a graph to be bipartite?

easy

A. All vertices have the same degree

B. It has an equal number of vertices and edges

C. It contains at least one cycle

D. Its vertices can be divided into two groups with no edges inside the same group

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two sets with edges only between sets, not inside a set.

2. Step 2: Compare options with definition

   Its vertices can be divided into two groups with no edges inside the same group matches this definition exactly, while others describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two groups with no edges inside the same group -> Option D

4. Quick Check:

   Bipartite graph = two groups, no internal edges [OK]

Hint: Bipartite means two groups with no edges inside groups [OK]

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite means no cycles
• Assuming all vertices have same degree

2. Which C# data structure is best to represent a graph for bipartite check?

easy

A. List adjacency list

B. Dictionary adjacency list

C. int[,] adjacency matrix

D. HashSet<int> for vertices only

Solution

1. Step 1: Identify graph representation needs

   We need to store edges efficiently and access neighbors quickly for bipartite check.

2. Step 2: Evaluate options

   Dictionary allows flexible vertex keys and fast neighbor lookup, better than fixed-size arrays or sets.

3. Final Answer:

   Dictionary adjacency list -> Option B

4. Quick Check:

   Dictionary adjacency list = best for flexible graphs [OK]

Hint: Use Dictionary for flexible graph edges [OK]

Common Mistakes:

• Using HashSet only stores vertices, no edges
• Adjacency matrix wastes space for sparse graphs
• List lacks vertex key mapping

3. What is the output of this C# code snippet checking bipartite for a graph with edges: 0-1, 1-2, 2-3, 3-0?

var graph = new Dictionary {
  {0, new List<int>{1,3}},
  {1, new List<int>{0,2}},
  {2, new List<int>{1,3}},
  {3, new List<int>{2,0}}
};
bool IsBipartite(Dictionary g) {
  var color = new Dictionary();
  foreach(var node in g.Keys) {
    if(!color.ContainsKey(node)) {
      var queue = new Queue<int>();
      queue.Enqueue(node);
      color[node] = 0;
      while(queue.Count > 0) {
        var u = queue.Dequeue();
        foreach(var v in g[u]) {
          if(!color.ContainsKey(v)) {
            color[v] = 1 - color[u];
            queue.Enqueue(v);
          } else if(color[v] == color[u]) {
            return false;
          }
        }
      }
    }
  }
  return true;
}
Console.WriteLine(IsBipartite(graph));

medium

A. True

B. False

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Analyze the graph structure

   The graph forms a cycle of length 4: 0-1-2-3-0.

2. Step 2: Check bipartite condition on even cycle

   Even cycles are bipartite, but here the cycle is length 4 (even), so it should be bipartite. However, the code returns false if adjacent nodes have same color.

3. Step 3: Trace coloring

   Coloring alternates 0 and 1 around the cycle, no conflict occurs, so function returns true.

4. Final Answer:

   True -> Option A

5. Quick Check:

   Even cycle graph is bipartite = True [OK]

Hint: Even cycles are bipartite; odd cycles are not [OK]

Common Mistakes:

• Assuming all cycles are not bipartite
• Confusing color assignment logic
• Ignoring queue BFS traversal

4. Identify the bug in this bipartite check code snippet:

bool CheckBipartite(Dictionary graph) {
  var color = new Dictionary();
  foreach(var node in graph.Keys) {
    if(!color.ContainsKey(node)) {
      color[node] = 0;
      var stack = new Stack<int>();
      stack.Push(node);
      while(stack.Count > 0) {
        var u = stack.Pop();
        foreach(var v in graph[u]) {
          if(!color.ContainsKey(v)) {
            color[v] = color[u];
            stack.Push(v);
          } else if(color[v] == color[u]) {
            return false;
          }
        }
      }
    }
  }
  return true;
}

medium

A. Colors assigned to neighbors are same as current node, should be opposite

B. Stack should be replaced with queue for BFS

C. Missing initialization of color dictionary

D. No return statement for true case

Solution

1. Step 1: Review color assignment logic

   Neighbors must have opposite colors to current node for bipartite check.

2. Step 2: Identify incorrect assignment

   Code assigns neighbor color same as current node, causing false negatives.

3. Final Answer:

   Colors assigned to neighbors are same as current node, should be opposite -> Option A

4. Quick Check:

   Neighbor colors must differ [OK]

Hint: Neighbors must have opposite colors, not same [OK]

Common Mistakes:

• Assigning same color to neighbors
• Confusing stack vs queue (both work but BFS preferred)
• Forgetting to initialize colors

5. Given a graph with vertices 0 to 5 and edges: 0-1, 1-2, 2-3, 3-4, 4-5, 5-0, and 1-4, is the graph bipartite? Explain why.

hard

A. Yes, because all cycles are even length

B. No, because it has a self-loop

C. No, because it contains an odd cycle including edges 1-2-3-4-1

D. Yes, because it has no cycles

Solution

1. Step 1: Identify cycles in the graph

   The graph has a large cycle 0-1-2-3-4-5-0 (length 6, even), and with edge 1-4, cycles 1-2-3-4-1 (length 4, even) and 0-1-4-5-0 (length 4, even).

2. Step 2: Check for odd-length cycles

   No odd-length cycles exist. A graph is bipartite if and only if it has no odd-length cycles.

3. Final Answer:

   No, because it contains an odd cycle including edges 1-2-3-4-1 -> Option C

4. Quick Check:

   Odd cycle means not bipartite [OK]

Hint: Check for odd cycles to decide bipartite [OK]

Common Mistakes:

• Miscounting cycle lengths
• Assuming any cycle means not bipartite
• Ignoring extra edges creating odd cycles