Cycle Detection in Undirected Graph in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the adjacency list for the graph.

DSA C

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

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

ANULL

B-1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing adjacency list elements to 0 or 1 instead of NULL.

Not initializing adjacency list before use.

The adjacency list should be initialized to NULL pointers before adding edges.

2fill in blank

medium

Complete the code to add an edge between two vertices u and v in the adjacency list.

DSA C

void addEdge(struct Node* adj[], int u, int v) {
    struct Node* newNode = malloc(sizeof(struct Node));
    newNode->vertex = v;
    newNode->next = adj[u];
    adj[u] = [1];
}

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

AnewNode

Bv

Cadj[u]

DnewNode->next

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning adj[u] to newNode->next instead of the other way.

Assigning wrong pointer to adj[u].

The new node should be assigned to adj[u] to update the adjacency list.

3fill in blank

hard

Fix the error in the DFS function to detect a cycle in the undirected graph.

DSA C

bool dfs(int v, bool visited[], int parent, struct Node* adj[]) {
    visited[v] = true;
    struct Node* temp = adj[v];
    while (temp != NULL) {
        int adjVertex = temp->vertex;
        if (!visited[adjVertex]) {
            if (dfs(adjVertex, visited, v, adj) == [1]) {
                return true;
            }
        } else if (adjVertex != parent) {
            return true;
        }
        temp = temp->next;
    }
    return false;
}

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Afalse

Btrue

Cvisited[adjVertex]

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false instead of true when cycle detected in recursion.

Confusing parent vertex check.

If the recursive DFS call returns true, it means a cycle is detected, so return true.

4fill in blank

hard

Fill both blanks to complete the cycle detection function that iterates over all vertices.

DSA C

bool isCycle(struct Node* adj[], int V) {
    bool visited[V];
    for (int i = 0; i < V; i++)
        visited[i] = [1];
    for (int i = 0; i < V; i++) {
        if (!visited[i]) {
            if (dfs(i, visited, [2], adj)) {
                return true;
            }
        }
    }
    return false;
}

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

Afalse

Btrue

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing visited[] to true instead of false.

Using 0 as parent for root vertex causing false cycle detection.

Visited array should be initialized to false; parent of root vertex is -1.

5fill in blank

hard

Fill all three blanks to complete the main function that builds the graph and checks for cycles.

DSA C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

struct Node {
    int vertex;
    struct Node* next;
};

int main() {
    int V = 5;
    struct Node* adj[V];
    for (int i = 0; i < V; i++)
        adj[i] = NULL;

    addEdge(adj, 0, 1);
    addEdge(adj, 1, 2);
    addEdge(adj, 2, 3);
    addEdge(adj, 3, 4);
    addEdge(adj, 4, [1]);

    if (isCycle(adj, V))
        printf("Cycle detected\n");
    else
        printf("No cycle detected\n");

    return [2];
}

void addEdge(struct Node* adj[], int u, int v) {
    struct Node* newNode = malloc(sizeof(struct Node));
    newNode->vertex = v;
    newNode->next = adj[u];
    adj[u] = newNode;

    newNode = malloc(sizeof(struct Node));
    newNode->vertex = u;
    newNode->next = adj[v];
    adj[v] = newNode;
}

bool isCycle(struct Node* adj[], int V);
bool dfs(int v, bool visited[], int parent, struct Node* adj[]);

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

A0

B1

C5

Attempts:

3 left

💡 Hint

Common Mistakes

Connecting vertex 4 to 0 instead of 1, changing the cycle structure.

Returning non-zero value from main.

The last edge connects vertex 4 to 1 to form a cycle; main returns 0 for success.

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