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

Depth-First Search (DFS) is commonly used to detect cycles by checking if a visited vertex is encountered again and is not the parent of the current vertex.

intermediate

In cycle detection using DFS, why do we check if the visited vertex is not the parent?

Because in an undirected graph, the edge to the parent is always visited, so revisiting the parent does not mean a cycle. We look for visited vertices other than the parent to confirm 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.

advanced

How does the Union-Find (Disjoint Set) data structure help in cycle detection?

Union-Find keeps track of connected components. When adding an edge, if both vertices belong to the same set, a cycle is detected. Otherwise, the sets are united.

What indicates a cycle during DFS traversal in an undirected graph?

A Visiting a vertex that is already visited and is not the parent

B Visiting a vertex for the first time

C Visiting the parent vertex again

D No edges left to explore

A cycle is detected if a visited vertex is found again during DFS and it is not the parent of the current vertex.

Which of the following is NOT true about cycle detection in undirected graphs?

A DFS can be used to detect cycles

B A cycle can be detected by checking only the degree of vertices

C Union-Find can detect cycles efficiently

D A cycle must include at least one vertex repeated twice

Checking only the degree of vertices is not sufficient to detect cycles in undirected graphs.

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

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

B To mark the starting vertex

C To count the number of edges

D To store the depth of the vertex

The parent parameter helps ignore the edge leading back to the parent vertex, which is not a cycle.

What is the worst-case time complexity of cycle detection using Union-Find?

A O(V + E)

B O(E log V)

C O(V^2)

D O(E α(V)) where α is the inverse Ackermann function

Union-Find operations run in almost constant time, O(α(V)), making total complexity O(E α(V)) for E edges.

If a graph has no cycles, what kind of graph is it called?

A Disconnected graph

B Complete graph

C Tree

D Directed graph

A connected graph with no cycles is called a tree.

Explain how DFS can be used to detect a cycle in an undirected graph.

Think about revisiting vertices during traversal.

Describe how the Union-Find data structure detects cycles when edges are added.

Focus on connected components and set membership.

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