Cycle Detection in Undirected Graph in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

vector<vector<int>> graph[[1]];

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Aedges

Bn

Cnodes

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using edges count instead of nodes count for adjacency list size.

Using an unrelated variable like visited.

The adjacency list is declared with size n, which represents the number of nodes in the graph.

2fill in blank

medium

Complete the code to mark a node as visited during DFS.

DSA C++

visited[[1]] = true;

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Anode

Bparent

Cgraph

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Marking the parent node instead of the current node.

Using loop variable i incorrectly.

We mark the current node as visited to avoid revisiting it.

3fill in blank

hard

Fix the error in the DFS condition to detect a cycle.

DSA C++

if (visited[[1]] && neighbour != parent) return true;

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Aneighbour

Bparent

Cnode

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Checking visited status of the current node instead of neighbour.

Confusing parent with neighbour in the condition.

The condition checks if the neighbour is visited and is not the parent, indicating a cycle.

4fill in blank

hard

Fill both blanks to complete the DFS recursive call and check for cycle.

DSA C++

if (!visited[[1]] && dfs([2], node, visited, graph)) return true;

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Aneighbour

Bnode

Cparent

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using the current node instead of neighbour in the recursive call.

Passing wrong parent parameter.

The DFS is called recursively on the neighbour node, passing the current node as parent.

5fill in blank

hard

Fill all three blanks to complete the main function that detects cycle in the graph.

DSA C++

for (int [1] = 0; [2] < n; [3]++) {
    if (!visited[i] && dfs(i, -1, visited, graph)) {
        return true;
    }
}

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

Ai

Bj

Ck

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in loop parts causing syntax errors.

Using variables not declared or inconsistent.

The loop variable i iterates over all nodes to start DFS if unvisited.

Practice

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

easy

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

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

C. If the graph has more edges than nodes, it must have a cycle.

D. If the graph is disconnected, it contains 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 B

4. Quick Check:

   Cycle detection in undirected graph = check visited and parent [OK]

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

Common Mistakes:

• Ignoring the parent node in cycle check
• Assuming disconnected graph always has cycles
• Confusing cycle detection with number of edges

2. Which of the following C++ function signatures is correct for a DFS-based cycle detection helper in an undirected graph?

easy

A. bool dfs(int node, int parent, vector& graph, vector<bool>& visited);

B. void dfs(int node, vector<int>& graph, vector<bool>& visited);

C. bool dfs(int node, vector<int>& graph, vector<int>& visited, int parent);

D. int dfs(int node, int parent, vector<int>& graph, vector<bool>& visited);

Solution

1. Step 1: Analyze function parameters for DFS cycle detection

   We need current node, parent node, adjacency list (graph), and visited array.

2. Step 2: Check return type and parameter types

   Return type should be bool to indicate cycle presence; graph is vector of vectors; visited is bool vector.

3. Final Answer:

   bool dfs(int node, int parent, vector& graph, vector<bool>& visited); -> Option A

4. Quick Check:

   Correct DFS signature includes parent and returns bool [OK]

Hint: DFS cycle helper returns bool and takes parent node [OK]

Common Mistakes:

• Using wrong graph data type (vector<int> instead of vector)
• Missing parent parameter
• Wrong return type for cycle detection

3. What will be the output of the following code snippet detecting cycle in an undirected graph?

int main() {
  vector graph = {{1,2}, {0,2}, {0,1,3}, {2}};
  vector<bool> visited(4, false);
  bool hasCycle = false;
  function dfs = [&](int node, int parent) {
    visited[node] = true;
    for (int adj : graph[node]) {
      if (!visited[adj]) {
        if (dfs(adj, node)) return true;
      } else if (adj != parent) {
        return true;
      }
    }
    return false;
  };
  for (int i = 0; i < 4; i++) {
    if (!visited[i]) {
      if (dfs(i, -1)) {
        hasCycle = true;
        break;
      }
    }
  }
  cout << (hasCycle ? "Cycle Found" : "No Cycle") << endl;
  return 0;
}

medium

A. Runtime Error

B. No Cycle

C. Compilation Error

D. Cycle Found

Solution

1. Step 1: Analyze graph structure

   Graph edges: 0-1, 0-2, 1-2, 2-3. Nodes 0,1,2 form a triangle (cycle).

2. Step 2: Trace DFS cycle detection

   DFS visits nodes and finds back edge between 0 and 2 or 1 and 2, detecting cycle.

3. Final Answer:

   Cycle Found -> Option D

4. Quick Check:

   Triangle in graph means cycle detected [OK]

Hint: Triangle or more edges among nodes means cycle [OK]

Common Mistakes:

• Ignoring back edge condition
• Assuming no cycle if graph is connected
• Confusing parent node with visited node

4. Identify the bug in the following cycle detection code for an undirected graph:

bool dfs(int node, int parent, vector& graph, vector<bool>& visited) {
  visited[node] = true;
  for (int adj : graph[node]) {
    if (!visited[adj]) {
      if (dfs(adj, node, graph, visited)) return true;
    } else {
      return true;
    }
  }
  return false;
}

medium

A. It returns true even if adjacent node is the parent, causing false cycle detection.

B. It does not mark nodes as visited.

C. It uses wrong data types for graph and visited.

D. It never returns false.

Solution

1. Step 1: Review cycle detection condition

   The else block returns true if adjacent node is visited, but does not check if it is the parent.

2. Step 2: Identify impact of missing parent check

   Returning true for parent node causes false positive cycle detection.

3. Final Answer:

   It returns true even if adjacent node is the parent, causing false cycle detection. -> Option A

4. Quick Check:

   Check parent node before returning true for visited adjacency [OK]

Hint: Always exclude parent node when checking visited adjacency [OK]

Common Mistakes:

• Not checking if visited node is parent
• Returning true too early
• Incorrect recursion parameters

5. Given a graph with 5 nodes and edges: 0-1, 1-2, 2-3, 3-4, 4-1, which modification to the DFS cycle detection algorithm will correctly identify the cycle and also return the nodes involved in the cycle?

hard

A. Remove the parent check and mark all visited nodes as cycle nodes.

B. Use BFS instead of DFS to detect cycles and print nodes.

C. Modify DFS to store the path stack and return the cycle path when a back edge is found.

D. Count edges and nodes; if edges >= nodes, print all nodes as cycle.

Solution

1. Step 1: Understand cycle detection with node tracking

   DFS can track the current path using a stack or vector to record nodes visited in the current recursion.

2. Step 2: Modify DFS to return cycle path

   When a back edge is found (visited node not parent), extract nodes from the path stack to get cycle nodes.

3. Final Answer:

   Modify DFS to store the path stack and return the cycle path when a back edge is found. -> Option C

4. Quick Check:

   Track path in DFS to identify cycle nodes [OK]

Hint: Use path stack in DFS to retrieve cycle nodes [OK]

Common Mistakes:

• Using BFS which doesn't track path easily
• Ignoring parent check causing false cycles
• Using edge count alone to identify cycle nodes