Cycle Detection in Undirected Graph in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a cycle in an undirected graph?

A cycle is a path that starts and ends at the same vertex with at least one edge, and no edges or vertices are repeated except the starting and ending vertex.

beginner

Which data structure is commonly used to detect cycles in an undirected graph using DFS?

A recursive Depth-First Search (DFS) with a parent tracking variable is used to detect cycles by checking if a visited vertex is encountered again that is not the parent.

intermediate

Why do we need to track the parent node during DFS for cycle detection in undirected graphs?

Because in undirected graphs, edges are bidirectional, so visiting the parent node again does not mean a cycle. Tracking the parent helps ignore this edge and detect real cycles.

intermediate

What is the time complexity of cycle detection in an undirected graph using DFS?

The time complexity is O(V + E), where V is the number of vertices and E is the number of edges, because each vertex and edge is visited once.

advanced

Explain the difference between cycle detection in directed and undirected graphs.

In undirected graphs, cycles are detected by checking visited nodes excluding the parent. In directed graphs, cycles are detected by tracking recursion stack to find back edges.

What does it mean if during DFS in an undirected graph, you visit a vertex that is already visited and is not the parent?

A Graph is disconnected

B No cycle, just a back edge to parent

C A cycle is detected

D Graph is a tree

Visiting a previously visited vertex that is not the parent means a cycle exists.

Which of these is NOT required for cycle detection in an undirected graph using DFS?

A Recursion stack tracking

B Parent tracking

C Visited array

D Graph adjacency list

Recursion stack tracking is used in directed graphs, not undirected.

What is the initial state of the visited array before DFS starts?

A All true

B Depends on graph size

C Random values

D All false

All vertices are initially unvisited, so visited array is all false.

If an undirected graph has no cycles, what is it called?

A Complete graph

B Tree

C Disconnected graph

D Directed graph

A connected undirected graph with no cycles is called a tree.

What is the role of the parent parameter in the DFS function for cycle detection?

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

B To mark the starting vertex

C To store the depth of the vertex

D To count the number of cycles

Parent helps ignore the edge leading back to the immediate previous vertex.

Describe the step-by-step process of detecting a cycle in an undirected graph using DFS.

Think about how DFS explores neighbors and how parent helps avoid false positives.

Explain why cycle detection in undirected graphs requires different handling than in directed graphs.

Focus on edge direction and how it affects cycle detection logic.

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