Cycle Detection in Undirected Graph in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of cycle detection using DFS

What is the output of the following C code that detects a cycle in an undirected graph using DFS?

DSA C

#include <stdio.h>
#include <stdbool.h>
#define V 4

bool dfs(int v, int parent, bool visited[], int graph[V][V]) {
    visited[v] = true;
    for (int i = 0; i < V; i++) {
        if (graph[v][i]) {
            if (!visited[i]) {
                if (dfs(i, v, visited, graph))
                    return true;
            } else if (i != parent) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    int graph[V][V] = {
        {0, 1, 0, 0},
        {1, 0, 1, 1},
        {0, 1, 0, 1},
        {0, 1, 1, 0}
    };
    bool visited[V] = {false};
    bool hasCycle = false;
    for (int i = 0; i < V; i++) {
        if (!visited[i]) {
            if (dfs(i, -1, visited, graph)) {
                hasCycle = true;
                break;
            }
        }
    }
    if (hasCycle) {
        printf("Cycle detected\n");
    } else {
        printf("No cycle detected\n");
    }
    return 0;
}

A Cycle detected

B Compilation error

C No cycle detected

D Runtime error

💡 Hint

Trace the DFS calls and check if any back edge is found that is not the parent.

The graph has edges forming a cycle: nodes 1, 2, and 3 form a cycle. The DFS detects this cycle and prints "Cycle detected".

❓ Predict Output

intermediate

Output when graph has no cycle

What will be printed by this C code detecting cycles in an undirected graph without any cycles?

DSA C

#include <stdio.h>
#include <stdbool.h>
#define V 3

bool dfs(int v, int parent, bool visited[], int graph[V][V]) {
    visited[v] = true;
    for (int i = 0; i < V; i++) {
        if (graph[v][i]) {
            if (!visited[i]) {
                if (dfs(i, v, visited, graph))
                    return true;
            } else if (i != parent) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    int graph[V][V] = {
        {0, 1, 0},
        {1, 0, 1},
        {0, 1, 0}
    };
    bool visited[V] = {false};
    bool hasCycle = false;
    for (int i = 0; i < V; i++) {
        if (!visited[i]) {
            if (dfs(i, -1, visited, graph)) {
                hasCycle = true;
                break;
            }
        }
    }
    if (hasCycle) {
        printf("Cycle detected\n");
    } else {
        printf("No cycle detected\n");
    }
    return 0;
}

A Cycle detected

B No cycle detected

C Segmentation fault

D Compilation error

💡 Hint

Check if any back edge exists that is not the parent.

The graph is a simple chain with no cycles. DFS will not find any back edge, so it prints "No cycle detected".

🔧 Debug

advanced

Identify the bug causing incorrect cycle detection

This code is intended to detect cycles in an undirected graph using DFS. What is the bug causing it to always print "No cycle detected" even when a cycle exists?

DSA C

#include <stdio.h>
#include <stdbool.h>
#define V 3

bool dfs(int v, int parent, bool visited[], int graph[V][V]) {
    visited[v] = true;
    for (int i = 0; i < V; i++) {
        if (graph[v][i]) {
            if (!visited[i]) {
                dfs(i, v, visited, graph);
            } else if (i != parent) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    int graph[V][V] = {
        {0, 1, 1},
        {1, 0, 1},
        {1, 1, 0}
    };
    bool visited[V] = {false};
    bool hasCycle = false;
    for (int i = 0; i < V; i++) {
        if (!visited[i]) {
            if (dfs(i, -1, visited, graph)) {
                hasCycle = true;
                break;
            }
        }
    }
    if (hasCycle) {
        printf("Cycle detected\n");
    } else {
        printf("No cycle detected\n");
    }
    return 0;
}

A The recursive DFS call result is not returned, so true is never propagated up

B The visited array is not initialized properly

C The graph adjacency matrix is incorrect

D The parent parameter is not used in the DFS function

💡 Hint

Check if the recursive DFS calls return values are used properly.

The DFS function calls itself recursively but does not return the result of the recursive call. So even if a cycle is found deeper, it is not propagated back, causing the function to always return false.

🧠 Conceptual

advanced

Why is parent tracking necessary in cycle detection DFS?

In cycle detection for an undirected graph using DFS, why do we pass the parent node as a parameter and check if a visited neighbor is not the parent?

A To store the path of the DFS traversal

B To mark the parent node as visited

C To avoid counting the edge back to the parent as a cycle

D To prevent visiting the parent node again in BFS

💡 Hint

Think about why an edge to the immediate previous node should not be considered a cycle.

In an undirected graph, each edge is bidirectional. Without ignoring the parent, the DFS would always find the edge back to the parent as a cycle. So we exclude the parent to detect only real cycles.

🚀 Application

expert

Number of connected components with cycles

Given an undirected graph represented by adjacency matrix below, how many connected components contain at least one cycle?

DSA C

#define V 7
int graph[V][V] = {
    {0,1,0,0,0,0,0},
    {1,0,1,0,0,0,0},
    {0,1,0,0,0,0,0},
    {0,0,0,0,1,1,0},
    {0,0,0,1,0,1,0},
    {0,0,0,1,1,0,0},
    {0,0,0,0,0,0,0}
};

A 0

B 3

C 2

D 1

💡 Hint

Identify connected components and check each for cycles using DFS.

The graph has three connected components: nodes {0,1,2}, nodes {3,4,5}, and node {6} alone. The first component is a chain (no cycle), the second component forms a triangle (cycle), and the third is a single isolated node (no cycle). So only one component has a cycle.

Practice

1. What is the main idea behind detecting a cycle in an undirected graph using DFS?

easy

A. If the graph is disconnected, it contains a cycle.

B. If all nodes are visited exactly once, the graph has a cycle.

C. If we visit a node that is already visited and is not the parent, a cycle exists.

D. If the graph has more edges than nodes, it always has a cycle.

Solution

1. Step 1: Understand DFS traversal in undirected graph

   DFS visits nodes and tracks the parent node to avoid false cycle detection.

2. Step 2: Detect cycle condition

   If a visited node is found that is not the parent, it means a back edge exists, indicating a cycle.

3. Final Answer:

   If we visit a node that is already visited and is not the parent, a cycle exists. -> Option C

4. Quick Check:

   Cycle detection in undirected graph = check visited node not parent [OK]

Hint: Check visited neighbors excluding parent to find cycles [OK]

Common Mistakes:

• Ignoring the parent node in cycle check
• Assuming disconnected graph always has cycles
• Confusing edge count with cycle presence

2. Which of the following is the correct function signature in C for DFS used in cycle detection in an undirected graph?

easy

A. bool dfs(int node, int parent, int visited[], int graph[][100], int n)

B. int dfs(int node, int visited[], int graph[][100])

C. void dfs(int node, int graph[][100], int visited[])

D. int dfs(int node, int parent, int graph[][100])

Solution

1. Step 1: Identify parameters needed for DFS cycle detection

   We need current node, parent node, visited array, adjacency matrix, and number of nodes.

2. Step 2: Match function signature with parameters

   bool dfs(int node, int parent, int visited[], int graph[][100], int n) includes all required parameters with correct types and order.

3. Final Answer:

   bool dfs(int node, int parent, int visited[], int graph[][100], int n) -> Option A

4. Quick Check:

   DFS signature includes node, parent, visited, graph, size [OK]

Hint: Include parent parameter to avoid false cycle detection [OK]

Common Mistakes:

• Omitting parent parameter
• Wrong return type for DFS
• Missing graph size parameter

3. Consider the following adjacency matrix for an undirected graph with 4 nodes:

int graph[4][4] = {
  {0, 1, 0, 0},
  {1, 0, 1, 1},
  {0, 1, 0, 0},
  {0, 1, 0, 0}
};

If we run cycle detection using DFS starting from node 0, what will be the output?

medium

A. Cycle detected

B. No cycle detected

C. Runtime error due to invalid indexing

D. Infinite loop

Solution

1. Step 1: Analyze graph edges from adjacency matrix

   Edges: 0-1, 1-2, 1-3. Node 1 connects to nodes 0, 2, 3.

2. Step 2: Check for cycle using DFS

   Starting at 0, DFS visits 1, then 2 and 3. Node 1 connects back to 0 (parent), no cycle there. No back edge to visited non-parent node is found, so no cycle.

3. Final Answer:

   No cycle detected -> Option B

4. Quick Check:

   Graph edges form a tree structure = No cycle [OK]

Hint: Trace DFS and check visited nodes excluding parent [OK]

Common Mistakes:

• Assuming any connected graph has cycle
• Ignoring parent node in cycle check
• Confusing adjacency matrix indexing

4. The following C code snippet is intended to detect a cycle in an undirected graph using DFS. What is the bug?

bool dfs(int node, int parent, int visited[], int graph[][100], int n) {
  visited[node] = 1;
  for (int i = 0; i < n; i++) {
    if (graph[node][i] == 1) {
      if (!visited[i]) {
        if (dfs(i, node, visited, graph, n))
          return true;
      } else if (i != parent) {
        return true;
      }
    }
  }
  return false;
}

medium

A. The visited array is not initialized before calling dfs.

B. The parent parameter is not passed correctly in recursive calls.

C. The graph adjacency matrix is accessed out of bounds.

D. The function should return void, not bool.

Solution

1. Step 1: Check initialization of visited array

   The code snippet does not show visited array initialization, which is required to mark nodes unvisited before DFS.

2. Step 2: Verify other parameters and logic

   Parent is passed correctly, adjacency matrix access is within bounds, and return type bool is appropriate for cycle detection.

3. Final Answer:

   The visited array is not initialized before calling dfs. -> Option A

4. Quick Check:

   Initialize visited[] before DFS to avoid incorrect results [OK]

Hint: Always initialize visited array to zero before DFS [OK]

Common Mistakes:

• Forgetting to initialize visited array
• Passing wrong parent in recursion
• Misusing return type for DFS

5. You have an undirected graph represented by adjacency lists:

0: 1, 2
1: 0, 3
2: 0, 3
3: 1, 2

Using cycle detection with DFS, which nodes form the cycle and why?

hard

A. Nodes 0, 1, 3 form a cycle because 0 connects to 1 and 1 connects to 3.

B. Nodes 0, 2, 3 form a cycle because 0 connects to 2 and 2 connects to 3 which connects back to 0.

C. Nodes 1, 2, 3 form a cycle because 1 connects to 3 and 2 connects to 3.

D. Nodes 0, 1, 2, 3 form a cycle because they form a closed loop through all edges.

Solution

1. Step 1: Analyze adjacency lists for cycles

   Edges: 0-1, 0-2, 1-3, 2-3. These edges connect nodes forming a square shape.

2. Step 2: Identify cycle involving all nodes

   Starting at 0, path 0->1->3->2->0 forms a closed loop including all four nodes, indicating a cycle involving nodes 0,1,2,3.

3. Final Answer:

   Nodes 0, 1, 2, 3 form a cycle because they form a closed loop through all edges. -> Option D

4. Quick Check:

   Cycle includes all nodes forming closed loop [OK]

Hint: Look for closed loops connecting all nodes [OK]

Common Mistakes:

• Identifying partial cycles ignoring full loop
• Confusing connected nodes with cycle nodes
• Missing edges that complete the cycle