Bipartite Graph Check in DSA C - Execution Trace

Concept Flow - Bipartite Graph Check

Start with graph

↓

Pick uncolored node

↓

Assign color 0

↓

Use BFS/DFS to visit neighbors

↓

If neighbor uncolored -> assign opposite color

↓

If neighbor colored same as current -> Not Bipartite

↓

Repeat for all nodes

↓

If no conflicts -> Graph is Bipartite

We start with an uncolored graph, pick nodes one by one, color them and their neighbors with opposite colors using BFS or DFS. If a conflict arises, graph is not bipartite.

Execution Sample

DSA C

void bfsCheck(int start) {
  queue<int> q;
  color[start] = 0;
  q.push(start);
  while (!q.empty()) {
    int u = q.front(); q.pop();
    for (int v : adj[u]) {
      if (color[v] == -1) {
        color[v] = 1 - color[u];
        q.push(v);
      } else if (color[v] == color[u]) {
        bipartite = false;
      }
    }
  }
}

This BFS colors nodes starting from 'start'. If a neighbor has the same color, it marks graph as not bipartite.

Execution Table

Step	Operation	Node Visited	Neighbor	Color Assigned	Conflict Detected	Color Array State	Visual State
1	Start BFS at node 0	0	-	color[0]=0	No	[0:0,1:-1,2:-1,3:-1]	0(0) 1(-) 2(-) 3(-)
2	Visit neighbors of 0	0	1	color[1]=1	No	[0:0,1:1,2:-1,3:-1]	0(0) 1(1) 2(-) 3(-)
3	Visit neighbors of 0	0	3	color[3]=1	No	[0:0,1:1,2:-1,3:1]	0(0) 1(1) 2(-) 3(1)
4	Visit node 1	1	0	-	No	[0:0,1:1,2:-1,3:1]	0(0) 1(1) 2(-) 3(1)
5	Visit neighbors of 1	1	2	color[2]=0	No	[0:0,1:1,2:0,3:1]	0(0) 1(1) 2(0) 3(1)
6	Visit node 3	3	0	-	No	[0:0,1:1,2:0,3:1]	0(0) 1(1) 2(0) 3(1)
7	Visit neighbors of 3	3	2	-	No	[0:0,1:1,2:0,3:1]	0(0) 1(1) 2(0) 3(1)
8	Stop BFS	-	-	-	No	[0:0,1:1,2:0,3:1]	No conflict found, graph is bipartite

💡 No conflict detected; graph is bipartite.

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	After Step 7	Final
color	[-1,-1,-1,-1]	[0,-1,-1,-1]	[0,1,-1,1]	[0,1,0,1]	[0,1,0,1]	[0,1,0,1]
bipartite	true	true	true	true	true	true
queue	[]	[0]	[1,3]	[3,2]	[3,2]	[]

Key Moments - 3 Insights

Why do we assign opposite colors to neighbors?

Assigning opposite colors ensures no two connected nodes share the same color, which is the definition of bipartite. See execution_table steps 2 and 5 where neighbors get opposite colors.

What causes the conflict detection to mark graph as not bipartite?

When a neighbor already colored has the same color as the current node, it means the graph cannot be split into two sets properly. This happens at step 7 in execution_table.

Why do we start BFS from uncolored nodes?

Starting BFS from uncolored nodes ensures all disconnected parts of the graph are checked. The first step colors node 0 to start the process.

Visual Quiz - 3 Questions

Look at the execution_table, what color is assigned to node 3 at step 3?

A -1 (uncolored)

B 0

C 1

D Conflict detected

💡 Hint

Check the 'Color Assigned' and 'Color Array State' columns at step 3.

At which step does the conflict get detected?

A Step 7

B Step 4

C Step 5

D Step 2

💡 Hint

Look at the 'Conflict Detected' column in execution_table.

If node 2 was colored 0 instead of 1 at step 5, what would happen at step 7?

A Conflict detected earlier

B No conflict detected

C Queue would be empty

D Graph would be bipartite

💡 Hint

Refer to color assignments and conflict logic in execution_table steps 5 and 7.

Concept Snapshot

Bipartite Graph Check:
- Use BFS or DFS to color nodes with 0 or 1
- Assign opposite color to neighbors
- If neighbor has same color, graph is not bipartite
- Check all connected components
- Return true if no conflicts found

Full Transcript

To check if a graph is bipartite, we color nodes using two colors (0 and 1). We start from an uncolored node, assign color 0, then use BFS to assign opposite colors to neighbors. If at any point a neighbor has the same color as the current node, the graph is not bipartite. We repeat this for all nodes to cover disconnected parts. The execution table shows step-by-step how nodes are colored and when conflict is detected. The variable tracker shows color array changes and queue state. Key moments clarify why opposite colors are assigned and how conflicts arise. The quiz tests understanding of color assignments and conflict detection steps.

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