Bipartite Graph Check in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the color array with -1 for all vertices.

DSA C

for (int i = 0; i < V; i++) {
    color[i] = [1];
}

Drag options to blanks, or click blank then click option'

A-1

B0

C1

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 instead of -1 for uncolored vertices.

Using the loop variable i as color.

We initialize all vertices with color -1 to indicate they are uncolored initially.

2fill in blank

medium

Complete the code to assign the first color (0) to the starting vertex in BFS.

DSA C

color[start] = [1];

Drag options to blanks, or click blank then click option'

A0

B-1

Cstart

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning -1 which means uncolored.

Assigning 1 instead of 0.

We assign color 0 to the starting vertex to begin the bipartite coloring.

3fill in blank

hard

Fix the error in the condition to check if adjacent vertex has the same color.

DSA C

if (color[v] == color[[1]]) {
    return false;
}

Drag options to blanks, or click blank then click option'

Astart

Bv

Cu

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing color[v] with color[v] (always true).

Using wrong variable like start or i.

We compare the color of adjacent vertex v with the current vertex u to detect conflict.

4fill in blank

hard

Fill both blanks to assign the opposite color to adjacent vertex and push it to the queue.

DSA C

color[v] = [1] - color[u];
queue[rear++] = [2];

Drag options to blanks, or click blank then click option'

A1

Bu

Cv

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning same color instead of opposite.

Pushing wrong vertex to queue.

We assign the opposite color by subtracting from 1 and enqueue the adjacent vertex v.

5fill in blank

hard

Fill all three blanks to complete the BFS loop for bipartite check.

DSA C

while (front != rear) {
    int u = queue[[1]++];
    for (int i = 0; i < adj[u].size; i++) {
        int v = adj[u].array[[2]];
        if (color[v] == -1) {
            color[v] = [3] - color[u];
            queue[rear++] = v;
        } else if (color[v] == color[u]) {
            return false;
        }
    }
}

Drag options to blanks, or click blank then click option'

Afront

Bi

C1

Drear

Attempts:

3 left

💡 Hint

Common Mistakes

Using rear instead of front for dequeue.

Using wrong index for adjacency array.

Assigning wrong color value.

We dequeue from front, iterate over neighbors using i, and assign opposite color 1 - color[u].

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