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 no repeated edges or vertices (except the start/end vertex).

beginner

Which graph traversal methods can be used to detect cycles in an undirected graph?

Depth-First Search (DFS) and Breadth-First Search (BFS) can be used to detect cycles by checking if a visited vertex is encountered again through a different path.

intermediate

In cycle detection using DFS, why do we keep track of the parent vertex?

To avoid falsely detecting a cycle when revisiting the immediate parent vertex, since the edge back to the parent is not a cycle.

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.

beginner

Explain the role of the 'visited' array in cycle detection for undirected graphs.

The 'visited' array keeps track of which vertices have been explored to avoid repeated visits and to detect if a vertex is revisited through a different path, indicating a cycle.

What does detecting a cycle in an undirected graph mean?

A Counting the number of edges

B Finding the shortest path between two vertices

C Checking if the graph is connected

D Finding a path that starts and ends at the same vertex without repeating edges

A cycle is a path that starts and ends at the same vertex without repeating edges.

Which data structure is commonly used to keep track of visited vertices during DFS?

A Boolean array

B Stack

C Priority queue

D Queue

A boolean array is used to mark vertices as visited or not during DFS.

Why do we pass the parent vertex in DFS cycle detection?

A To avoid revisiting the parent and falsely detecting a cycle

B To find the shortest path

C To count the number of edges

D To sort the vertices

Passing the parent vertex helps avoid false cycle detection when revisiting the immediate parent.

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

A O(V^2)

B O(E^2)

C O(V + E)

D O(log V)

DFS visits each vertex and edge once, so the time complexity is O(V + E).

If a visited vertex is found that is not the parent during DFS, what does it indicate?

A Graph is disconnected

B A cycle exists

C No cycle

D Graph is directed

Finding a visited vertex other than the parent means a cycle is detected.

Describe the steps to detect a cycle in an undirected graph using DFS.

Think about how DFS explores neighbors and tracks parents.

Explain why cycle detection in undirected graphs requires tracking the parent vertex during traversal.

Consider what happens when you revisit the vertex you came from.

Practice

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

easy

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

B. If all nodes are visited exactly once, there is a cycle.

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 keeps track of the parent node to avoid false cycle detection.

2. Step 2: Detect cycle condition

   If during DFS, a visited node is found that is not the parent of the current node, it means a cycle exists.

3. Final Answer:

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

4. Quick Check:

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

Hint: Check visited nodes excluding parent during DFS [OK]

Common Mistakes:

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

2. Which of the following C# code snippets correctly initializes an adjacency list for an undirected graph with 3 nodes?

easy

A. var graph = new List<int>[3]; for(int i=0; i<3; i++) graph[i] = new List<int>();

B. var graph = new List<int>[3]; graph = new List<int>();

C. List<int> graph = new List<int>[3];

D. var graph = new List(); graph.Add(new List<int>());

Solution

1. Step 1: Understand adjacency list initialization

   We need an array of lists, each list initialized separately to hold neighbors.

2. Step 2: Check code correctness

   var graph = new List<int>[3]; for(int i=0; i<3; i++) graph[i] = new List<int>(); creates an array of List<int> and initializes each element with a new List<int>, which is correct.

3. Final Answer:

   var graph = new List<int>[3]; for(int i=0; i<3; i++) graph[i] = new List<int>(); -> Option A

4. Quick Check:

   Initialize array of lists with loop [OK]

Hint: Initialize each list in array separately [OK]

Common Mistakes:

• Not initializing each list element
• Using wrong data types for adjacency list
• Assigning list directly to array variable

3. Consider the following C# code snippet for cycle detection in an undirected graph:

bool DFS(int node, int parent, List<int>[] graph, bool[] visited) {
    visited[node] = true;
    foreach(var neighbor in graph[node]) {
        if(!visited[neighbor]) {
            if(DFS(neighbor, node, graph, visited)) return true;
        } else if(neighbor != parent) {
            return true;
        }
    }
    return false;
}

// Graph edges: 0-1, 1-2, 2-0
// Call: DFS(0, -1, graph, visited)

What will be the output of this DFS call?

medium

A. Compilation error

B. false

C. StackOverflowException

D. true

Solution

1. Step 1: Analyze graph structure and DFS call

   The graph has edges 0-1, 1-2, and 2-0 forming a triangle cycle. DFS starts at node 0 with parent -1.

2. Step 2: Trace DFS for cycle detection

   DFS visits 0, then 1, then 2. When 2 checks neighbors, it finds 0 already visited and not parent, so returns true indicating cycle.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Triangle graph cycle detected = true [OK]

Hint: Cycle if visited neighbor not parent found [OK]

Common Mistakes:

• Ignoring parent check causing false cycle
• Assuming no cycle in triangle graph
• Confusing return values in recursion

4. The following C# code attempts to detect a cycle in an undirected graph but has a bug:

bool DFS(int node, int parent, List<int>[] graph, bool[] visited) {
    visited[node] = true;
    foreach(var neighbor in graph[node]) {
        if(!visited[neighbor]) {
            if(DFS(neighbor, node, graph, visited)) return true;
        } else {
            return true;
        }
    }
    return false;
}

What is the bug in this code?

medium

A. It never marks nodes as visited.

B. It returns true even if the visited neighbor is the parent node.

C. It does not traverse all neighbors.

D. It causes infinite recursion.

Solution

1. Step 1: Review cycle detection condition

   The code returns true if a neighbor is visited, but does not check if neighbor is parent.

2. Step 2: Identify why this causes incorrect cycle detection

   In undirected graphs, the parent node is always visited, so returning true here falsely detects a cycle.

3. Final Answer:

   It returns true even if the visited neighbor is the parent node. -> Option B

4. Quick Check:

   Must exclude parent node in cycle check [OK]

Hint: Exclude parent node when checking visited neighbors [OK]

Common Mistakes:

• Missing parent check in visited condition
• Assuming any visited neighbor means cycle
• Not marking nodes visited before recursion

5. You have an undirected graph represented as adjacency list:

0: [1, 2]
1: [0, 3]
2: [0, 3]
3: [1, 2, 4]
4: [3]

Using DFS cycle detection, which node will first detect the cycle and why?

hard

A. Node 4, because it has only one neighbor.

B. Node 0, because it starts the DFS traversal.

C. Node 2, because it connects back to already visited node 0.

D. Node 1, because it has the smallest index among neighbors.

Solution

1. Step 1: Understand graph structure and DFS traversal

   Assuming neighbor lists are processed in listed order, DFS starts at node 0, visits 1 (first neighbor), then 3 (from 1), then 2 (from 3).

2. Step 2: Identify where cycle is detected

   Node 2 has neighbors 0 and 3; 0 is already visited and not parent (parent=3), so cycle is detected at node 2 first.

3. Final Answer:

   Node 2, because it connects back to already visited node 0. -> Option C

4. Quick Check:

   Cycle detected at node with visited neighbors not parent [OK]

Hint: Cycle detected where visited neighbors not parent found first [OK]

Common Mistakes:

• Assuming cycle detected at start node
• Ignoring multiple visited neighbors
• Confusing leaf nodes with cycle detection