Bipartite Graph Check in DSA C - Build from Scratch

Bipartite Graph Check

📖 Scenario: You are working on a social network app where users can be grouped into two sets such that no two users within the same set are friends. This means the friendship graph is bipartite. You want to check if the friendship connections form a bipartite graph.

🎯 Goal: Build a program in C that checks if a given undirected graph is bipartite using a coloring method.

📋 What You'll Learn

Use an adjacency list to represent the graph

Use an array to store colors of nodes

Implement a BFS or DFS to assign colors and check bipartiteness

Print 1 if the graph is bipartite, else print 0

💡 Why This Matters

🌍 Real World

Bipartite graphs are used in matching problems like job assignments, social networks, and scheduling.

💼 Career

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

1

Create the graph adjacency list

Create an integer variable n and set it to 4. Create an adjacency list array graph of size 4, where each element is an integer array representing connected nodes. Initialize graph[0] with {1, 3}, graph[1] with {0, 2}, graph[2] with {1, 3}, and graph[3] with {0, 2}. Also create an integer array graph_sizes with values {2, 2, 2, 2} representing the number of neighbors for each node.

DSA C

# Your code here

Hint

Use a 2D array for the adjacency list and an array for sizes of each node's neighbors.

2

Create the color array and initialize it

Create an integer array color of size n and initialize all elements to -1 to represent uncolored nodes.

DSA C

int n = 4;
int graph[4][2] = {{1, 3}, {0, 2}, {1, 3}, {0, 2}};
int graph_sizes[4] = {2, 2, 2, 2};
// Your code here

Hint

Use -1 to mark nodes as uncolored initially.

3

Implement BFS to check bipartite

Write a function int bfs_check(int start) that uses a queue to assign colors 0 and 1 to nodes starting from start. Initialize color[start] to 0. While the queue is not empty, dequeue a node u and for each neighbor v in graph[u], if color[v] is -1, assign color[v] to 1 - color[u] and enqueue v. If color[v] equals color[u], return 0 (not bipartite). If BFS finishes without conflict, return 1.

DSA C

int n = 4;
int graph[4][2] = {{1, 3}, {0, 2}, {1, 3}, {0, 2}};
int graph_sizes[4] = {2, 2, 2, 2};
int color[4] = {-1, -1, -1, -1};

// Your code here

Hint

Use a simple array as a queue with front and rear indices. Assign colors alternately and check for conflicts.

4

Check all nodes and print result

In main(), iterate over all nodes from 0 to n-1. If color[i] is -1, call bfs_check(i). If any call returns 0, print 0 and return 0. If all calls return 1, print 1.

DSA C

int n = 4;
int graph[4][2] = {{1, 3}, {0, 2}, {1, 3}, {0, 2}};
int graph_sizes[4] = {2, 2, 2, 2};
int color[4] = {-1, -1, -1, -1};

int bfs_check(int start) {
    int queue[100];
    int front = 0, rear = 0;
    color[start] = 0;
    queue[rear++] = start;

    while (front < rear) {
        int u = queue[front++];
        for (int i = 0; i < graph_sizes[u]; i++) {
            int v = graph[u][i];
            if (color[v] == -1) {
                color[v] = 1 - color[u];
                queue[rear++] = v;
            } else if (color[v] == color[u]) {
                return 0;
            }
        }
    }
    return 1;
}

int main() {
    for (int i = 0; i < n; i++) {
        if (color[i] == -1) {
            // Your code here
        }
    }
    return 0;
}

Hint

Check each node if uncolored, run bfs_check. If any returns 0, print 0 and exit. Otherwise print 1.

Practice

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

easy

A. It contains at least one cycle

B. Its vertices can be divided into two sets with no edges inside the same set

C. It has an equal number of vertices and edges

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two groups so that no edge connects vertices within the same group.

2. Step 2: Compare options with definition

   Only Its vertices can be divided into two sets with no edges inside the same set matches this definition exactly. Other options describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two sets with no edges inside the same set -> Option B

4. Quick Check:

   Bipartite graph = two sets, no same-set edges [OK]

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

Common Mistakes:

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

2. Which of the following C code snippets correctly initializes a graph adjacency list for bipartite check?

easy

A. int graph[5][5] = {{0}};

B. int graph[5]; for(int i=0; i<5; i++) graph[i] = NULL;

C. int graph[][] = {0};

D. int* graph[5]; for(int i=0; i<5; i++) graph[i] = malloc(sizeof(int)*5); memset(graph[i], 0, sizeof(int)*5);

Solution

1. Step 1: Understand adjacency list initialization in C

   Adjacency list uses array of pointers to arrays or linked lists. int* graph[5]; for(int i=0; i<5; i++) graph[i] = malloc(sizeof(int)*5); memset(graph[i], 0, sizeof(int)*5); correctly allocates memory and initializes it.

2. Step 2: Check other options for correctness

   int graph[5][5] = {{0}}; is a fixed 2D array but only zero-initialized; B tries to assign NULL to int array elements (invalid); C has incomplete array declaration.

3. Final Answer:

   int* graph[5]; for(int i=0; i<5; i++) graph[i] = malloc(sizeof(int)*5); memset(graph[i], 0, sizeof(int)*5); -> Option D

4. Quick Check:

   Proper adjacency list init = malloc + memset [OK]

Hint: Use malloc and memset to init adjacency lists in C [OK]

Common Mistakes:

• Trying to assign NULL to int array elements
• Using incomplete array declarations
• Not allocating memory for adjacency lists

3. Given the following graph edges: 0-1, 1-2, 2-3, 3-0, what will the bipartite check function return?

medium

A. Runtime error due to cycle

B. false (graph is not bipartite)

C. true (graph is bipartite)

D. Indeterminate result

Solution

1. Step 1: Analyze the graph structure

   The edges form a cycle of length 4: 0-1-2-3-0. Even length cycles can be bipartite, so this graph is bipartite.

2. Step 2: Check bipartite condition on cycle

   Since the cycle length is even, the graph can be colored with two colors without conflict.

3. Final Answer:

   true (graph is bipartite) -> Option C

4. Quick Check:

   Even cycle = bipartite [OK]

Hint: Even length cycles are bipartite, odd length are not [OK]

Common Mistakes:

• Assuming all cycles are not bipartite
• Confusing cycle length parity
• Thinking runtime error occurs due to cycles

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

bool isBipartite(int graph[][3], int V) {
  int color[V];
  for(int i=0; i

medium

A. Function does not check disconnected components

B. Color array is not initialized to -1

C. Queue is not initialized properly

D. Incorrect color assignment logic

Solution

1. Step 1: Understand the code logic

   The code colors nodes starting from node 0 and uses BFS to assign colors to neighbors.

2. Step 2: Identify missing checks

   The code only starts BFS from node 0 and does not check other disconnected components, which may cause incorrect results if graph is disconnected.

3. Step 3: Verify other options

   Queue is properly initialized; color array is initialized to -1; color assignment logic is correct.

4. Final Answer:

   Function does not check disconnected components -> Option A

5. Quick Check:

   Check all components for bipartite [OK]

Hint: Always check all graph components in bipartite check [OK]

Common Mistakes:

• Ignoring disconnected graph parts
• Assuming single BFS covers entire graph
• Misunderstanding color assignment

5. You have a graph with 6 vertices and edges: 0-1, 0-3, 1-2, 2-3, 4-5. Which statement is true about its bipartite nature?

hard

A. The graph is bipartite because all components are bipartite

B. The graph is not bipartite because of the cycle 0-1-2-3-0

C. The graph is bipartite only if vertex 4 is connected to vertex 2

D. The graph is not bipartite because vertices 4 and 5 form an odd cycle

Solution

1. Step 1: Analyze graph components

   Vertices 0-1-2-3 form a cycle 0-1-2-3-0 of length 4 (even), which is bipartite. Vertices 4-5 form a separate edge, also bipartite.

2. Step 2: Check bipartite condition for entire graph

   Since all components are bipartite, the entire graph is bipartite.

3. Step 3: Evaluate other options

   The graph is not bipartite because of the cycle 0-1-2-3-0 is wrong because even cycles are bipartite. The graph is bipartite only if vertex 4 is connected to vertex 2 is irrelevant. The graph is not bipartite because vertices 4 and 5 form an odd cycle is wrong because 4-5 is just an edge, no cycle.

4. Final Answer:

   The graph is bipartite because all components are bipartite -> Option A

5. Quick Check:

   All components bipartite = graph bipartite [OK]

Hint: Check each component's bipartite status separately [OK]

Common Mistakes:

• Assuming any cycle means not bipartite
• Ignoring disconnected components
• Confusing edges with cycles