Cycle Detection in Undirected Graph in DSA C++ - Execution Trace

Concept Flow - Cycle Detection in Undirected Graph

Start at any node

↓

Mark node as visited

↓

For each neighbor

↓

If neighbor not visited

↓

Recurse DFS

↓

Cycle found

↓

Return result

Start DFS from any node, mark visited, explore neighbors. If a visited neighbor is not the parent, a cycle exists.

Execution Sample

DSA C++

void dfs(int node, int parent) {
  visited[node] = true;
  for (auto nbr : graph[node]) {
    if (!visited[nbr]) dfs(nbr, node);
    else if (nbr != parent) cycle = true;
  }
}

DFS visits nodes, marks them, and checks for back edges indicating cycles.

Execution Table

Step	Operation	Node Visited	Parent	Visited Set	Cycle Detected	Visual State
1	Start DFS	0	-1	[0]	False	0 -> [1,2] 1 -> [0] 2 -> [3] 3 -> [2,0]
2	Visit neighbor 1	1	0	[0,1]	False	0 -> [1,2] 1 -> [0] 2 -> [3] 3 -> [2,0]
3	Neighbor 1 has no unvisited neighbors	-	-	[0,1]	False	0 -> [1,2] 1 -> [0] 2 -> [3] 3 -> [2,0]
4	Back to node 0, visit neighbor 2	2	0	[0,1,2]	False	0 -> [1,2] 1 -> [0] 2 -> [3] 3 -> [2,0]
5	Visit neighbor 3 from 2	3	2	[0,1,2,3]	False	0 -> [1,2] 1 -> [0] 2 -> [3] 3 -> [2,0]
6	Neighbor 3 has neighbor 2 (parent), no cycle	-	-	[0,1,2,3]	False	0 -> [1,2] 1 -> [0] 2 -> [3] 3 -> [2,0]
7	At node 3, check neighbor 0 (visited, not parent)	0	3	[0,1,2,3]	True	Cycle detected via edge 3-0
8	Cycle found, stop DFS	-	-	[0,1,2,3]	True	Cycle detected
9	End	-	-	[0,1,2,3]	True	Final state with cycle

💡 Cycle detected when visiting a visited neighbor that is not the parent.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 4	After Step 5	After Step 7	Final
visited	[]	[0]	[0,1]	[0,1,2]	[0,1,2,3]	[0,1,2,3]	[0,1,2,3]
cycle	False	False	False	False	False	True	True
current_node	-	0	1	2	3	3	-
parent_node	-	-1	0	0	2	3	-

Key Moments - 3 Insights

Why do we check if the visited neighbor is not the parent to detect a cycle?

Because in an undirected graph, the edge back to the parent is normal traversal, not a cycle. The cycle is detected only if a visited neighbor is different from the parent (see execution_table step 7).

What happens if the graph is disconnected?

We run DFS from each unvisited node to cover all components. Cycle detection applies to each component separately.

Why do we mark nodes as visited before exploring neighbors?

Marking visited early prevents revisiting the same node and infinite loops. This is shown in execution_table step 1 where node 0 is marked visited before neighbors are checked.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, which node is currently visited and what is its parent?

A Node 3, parent 2

B Node 2, parent 0

C Node 1, parent 0

D Node 0, parent -1

💡 Hint

Check the 'Node Visited' and 'Parent' columns at step 5 in execution_table.

At which step does the cycle get detected according to the execution_table?

A Step 4

B Step 3

C Step 7

D Step 9

💡 Hint

Look at the 'Cycle Detected' column in execution_table to find when it changes to True.

If the graph had no back edge to a visited node other than the parent, what would be the final value of 'cycle'?

A True

B False

C Undefined

D Depends on starting node

💡 Hint

Refer to variable_tracker row for 'cycle' and see when it changes from False to True.

Concept Snapshot

Cycle Detection in Undirected Graph using DFS:
- Start DFS from any node, mark visited.
- For each neighbor:
  - If not visited, recurse DFS.
  - If visited and not parent, cycle found.
- Cycle means a back edge exists.
- Works for disconnected graphs by checking all nodes.

Full Transcript

Cycle detection in an undirected graph uses depth-first search (DFS). We start from any node, mark it visited, and explore neighbors. If we find a neighbor already visited and it is not the parent node, it means a cycle exists. We track visited nodes to avoid revisiting. The process repeats for all disconnected parts of the graph. The key is to distinguish between the parent edge and a back edge to detect cycles correctly.

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