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++

std::vector<int> color(V, [1]);

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

A-1

B0

C1

DV

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 as initial color which conflicts with actual colors.

Initializing with the number of vertices instead of a color marker.

We use -1 to mark all vertices as 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'

A-1

B1

Cstart

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning -1 which means uncolored.

Assigning the vertex index as color.

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 the adjacent vertex has the same color.

DSA C++

if (color[adj] == [1]) return false;

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

A0

B-1

Ccolor[u]

Dadj

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with -1 which means uncolored.

Comparing with 0 or adj vertex index instead of current vertex color.

If the adjacent vertex has the same color as the current vertex, the graph is not bipartite.

4fill in blank

hard

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

DSA C++

color[adj] = [1];
q.push([2]);

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

A1 - color[u]

Bcolor[u]

Cadj

Du

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning same color as current vertex.

Pushing the current vertex instead of adjacent vertex.

Assign opposite color to adjacent vertex and push it into the queue for BFS.

5fill in blank

hard

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

DSA C++

while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int adj : graph[[1]]) {
        if (color[adj] == -1) {
            color[adj] = [2];
            q.push([3]);
        } else if (color[adj] == color[u]) {
            return false;
        }
    }
}

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

Au

B1 - color[u]

Cadj

Dcolor[u]

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable for adjacency list.

Assigning wrong color or pushing wrong vertex.

Iterate over neighbors of u, assign opposite color to uncolored adj, and push adj to queue.

Practice

1. What does it mean if a graph is bipartite?

easy

A. It contains at least one cycle of odd length

B. It has an equal number of vertices and edges

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

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph means vertices can be split into two sets where no edge connects vertices within the same 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 C

4. Quick Check:

   Bipartite = 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 graphs must have cycles
• Assuming all vertices have same degree

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

easy

A. vector graph(n);

B. vector<int> graph(n);

C. int graph[n][n];

D. map graph;

Solution

1. Step 1: Identify adjacency list structure

   An adjacency list for a graph with n vertices is best represented as vector of vectors: vector graph(n);

2. Step 2: Check other options

   vector<int> graph(n); is a vector of ints, not suitable for adjacency lists. int graph[n][n]; is a 2D array but less flexible. map graph; is a map of int to int, not adjacency list.

3. Final Answer:

   vector graph(n); -> Option A

4. Quick Check:

   Adjacency list = vector of vectors [OK]

Hint: Use vector of vectors for adjacency list in C++ [OK]

Common Mistakes:

• Using vector<int> instead of vector
• Using fixed size arrays limiting flexibility
• Using map which is not adjacency list

3. What is the output of the following code snippet that checks if a graph is bipartite?

int n = 4;
vector graph = {{1,3}, {0,2}, {1,3}, {0,2}};
vector<int> color(n, -1);

function dfs = [&](int node, int c) {
    color[node] = c;
    for (int neigh : graph[node]) {
        if (color[neigh] == -1) {
            if (!dfs(neigh, 1 - c)) return false;
        } else if (color[neigh] == c) {
            return false;
        }
    }
    return true;
};

bool result = true;
for (int i = 0; i < n; i++) {
    if (color[i] == -1) {
        if (!dfs(i, 0)) {
            result = false;
            break;
        }
    }
}
cout << (result ? "YES" : "NO") << endl;

medium

A. Runtime Error

B. NO

C. Compilation Error

D. YES

Solution

1. Step 1: Analyze graph structure

   The graph edges connect vertices 0-1, 0-3, 1-2, 2-3 forming a square (even cycle), which is bipartite.

2. Step 2: Trace DFS coloring

   DFS colors nodes alternatingly. No adjacent nodes share the same color, so dfs returns true for all nodes.

3. Final Answer:

   YES -> Option D

4. Quick Check:

   Even cycle graph is bipartite = YES [OK]

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

Common Mistakes:

• Assuming all cycles are non-bipartite
• Confusing color assignment logic
• Ignoring disconnected components

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

bool dfs(int node, int c, vector& graph, vector<int>& color) {
    color[node] = c;
    for (int neigh : graph[node]) {
        if (color[neigh] == -1) {
            dfs(neigh, 1 - c, graph, color);
        } else if (color[neigh] == c) {
            return false;
        }
    }
    return true;
}

medium

A. Missing return value check for recursive dfs call

B. Wrong color assignment to nodes

C. Using vector<int> instead of vector<bool>

D. Incorrect graph representation

Solution

1. Step 1: Check recursive call handling

   The recursive dfs call result is not checked. If it returns false, the function should also return false immediately.

2. Step 2: Explain fix

   Adding 'if (!dfs(...)) return false;' ensures early exit on conflict detection.

3. Final Answer:

   Missing return value check for recursive dfs call -> Option A

4. Quick Check:

   Always check recursive call results in DFS [OK]

Hint: Return false immediately if recursive DFS fails [OK]

Common Mistakes:

• Ignoring recursive call return values
• Assigning wrong colors
• Using wrong data structures for graph

5. Given a graph with 5 vertices and edges: (0-1), (1-2), (2-3), (3-4), (4-1), which statement is true about its bipartite property?

hard

A. The graph is bipartite because it has no odd cycles

B. The graph is not bipartite because it contains an odd cycle

C. The graph is bipartite because all vertices have degree 2

D. The graph is not bipartite because it is disconnected

Solution

1. Step 1: Identify cycles in the graph

   The edges form a cycle: 1-2-3-4-1 which has 4 edges (even length). However, the full cycle including vertices 0-1-4-3-2-1 forms a cycle of length 5 (odd cycle): 0-1-4-3-2-1 (0-1,1-4,4-3,3-2,2-1).

2. Step 2: Check bipartite condition

   Since the graph contains an odd cycle of length 5, it is not bipartite.

3. Step 3: Re-examine options

   The graph is bipartite because it has no odd cycles is false. The graph is not bipartite because it contains an odd cycle is true. The graph is bipartite because all vertices have degree 2 is incorrect reasoning. The graph is not bipartite because it is disconnected is false because the graph is connected.

4. Final Answer:

   The graph is not bipartite because it contains an odd cycle -> Option B

5. Quick Check:

   Odd cycle means not bipartite [OK]

Hint: Check cycle lengths: odd cycle means not bipartite [OK]

Common Mistakes:

• Miscounting cycle length
• Confusing connectivity with bipartite property
• Assuming degree affects bipartite