Cycle Detection in Undirected Graph in DSA Typescript - 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 nodes visited, check neighbors. If a visited neighbor is not the parent, cycle exists.

Execution Sample

DSA Typescript

const graph = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};
// Detect cycle using DFS

Detects cycle in an undirected graph using DFS and parent tracking.

Execution Table

Step	Operation	Current Node	Visited Set	Parent	Neighbor Checked	Cycle Detected	Visual State
1	Start DFS	0	{}	null	None	No	Nodes: 0 Edges: 0
2	Visit node 0	0	{0}	null	1	No	0 -> visited
3	Check neighbor 1 of 0	0	{0}	null	1	No	0 connected to 1
4	Neighbor 1 not visited, DFS to 1	1	{0}	0	None	No	Traverse 0->1
5	Visit node 1	1	{0,1}	0	0	No	1 -> visited
6	Check neighbor 0 of 1	1	{0,1}	0	0	No	Neighbor is parent, ignore
7	Check neighbor 2 of 1	1	{0,1}	0	2	No	Check 2 from 1
8	Neighbor 2 not visited, DFS to 2	2	{0,1}	1	None	No	Traverse 1->2
9	Visit node 2	2	{0,1,2}	1	0	No	2 -> visited
10	Check neighbor 0 of 2	2	{0,1,2}	1	0	Yes	0 visited and not parent, cycle found
11	Cycle detected, stop DFS	2	{0,1,2}	1	0	Yes	Cycle confirmed
12	Return True	N/A	{0,1,2}	N/A	N/A	Yes	Cycle exists in graph

💡 Cycle detected at step 10 when neighbor 0 of node 2 is visited and not parent

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 9	Final
Current Node	null	0	1	2	N/A
Visited Set	{}	{0}	{0,1}	{0,1,2}	{0,1,2}
Parent	null	null	0	1	N/A
Cycle Detected	No	No	No	No	Yes

Key Moments - 3 Insights

Why do we ignore the neighbor if it is the parent node during DFS?

Because the parent node is where we came from, so visiting it again does not mean a cycle. See step 6 where neighbor 0 is parent of 1 and no cycle is detected.

How do we know a cycle exists when we find a visited neighbor that is not the parent?

If a neighbor is visited and not the parent, it means we found a back edge forming a loop. This happens at step 10 where neighbor 0 of node 2 is visited but not parent, so cycle is detected.

What happens if the graph has no edges or is empty?

DFS will visit nodes but never find a visited neighbor that is not parent, so no cycle is detected and DFS ends normally.

Visual Quiz - 3 Questions

Look at the execution table, what is the visited set after visiting node 1?

A {0}

B {0,1}

C {1}

D {}

💡 Hint

Check step 5 in the execution table under Visited Set column

At which step does the cycle get detected in the graph?

A Step 10

B Step 8

C Step 6

D Step 12

💡 Hint

Look at Cycle Detected column in execution table rows

If the graph had no edges, how would the cycle detection result change?

A Algorithm would crash

B Cycle detected immediately

C No cycle detected

D Cycle detected at last step

💡 Hint

Refer to key moment about empty graph and no edges

Concept Snapshot

Cycle Detection in Undirected Graph using DFS:
- Start DFS from any node
- Mark node visited
- For each neighbor:
  - If not visited, DFS recursively
  - If visited and not parent, cycle found
- Return true if cycle detected, else false

Full Transcript

Cycle detection in an undirected graph uses depth-first search (DFS). We start from any node and mark it visited. For each neighbor, if it is not visited, we recursively DFS into it. If a neighbor is visited and is not the parent node, it means a cycle exists. We track the parent to avoid false positives from the edge we came from. The process stops when a cycle is found or all nodes are visited without finding a cycle. This method works because a back edge to a visited node other than the parent indicates a loop in the graph.

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 the graph has more edges than nodes, it always has a cycle.

C. If all nodes are visited exactly once, there is 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 during DFS, 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 A

4. Quick Check:

   Cycle detection = visit visited node not parent [OK]

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

Common Mistakes:

• Ignoring the parent node check causes false positives.
• Assuming more edges than nodes always means cycle.
• Confusing disconnected graph with cycle presence.

2. Which of the following TypeScript function signatures correctly represents a DFS function for cycle detection in an undirected graph?

easy

A. function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean

B. function dfs(node: string, graph: number[][]): void

C. function dfs(visited: boolean[], graph: number[][]): number

D. function dfs(node: number, graph: number[][]): boolean

Solution

1. Step 1: Identify required parameters for DFS cycle detection

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

2. Step 2: Match function signature with parameters and return type

   function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean has all parameters and returns boolean indicating cycle presence.

3. Final Answer:

   function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean -> Option A

4. Quick Check:

   DFS needs node, parent, visited, graph, returns boolean [OK]

Hint: DFS cycle detection needs parent node parameter [OK]

Common Mistakes:

• Missing parent parameter causes incorrect cycle detection.
• Wrong return type (void instead of boolean).
• Using wrong node type (string instead of number).

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

const graph = [[1,2],[0,2],[0,1,3],[2]];
const visited = new Array(4).fill(false);
function dfs(node: number, parent: number): boolean {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      if (dfs(neighbor, node)) return true;
    } else if (neighbor !== parent) {
      return true;
    }
  }
  return false;
}
console.log(dfs(0, -1));

medium

A. undefined

B. true

C. false

D. Runtime Error

Solution

1. Step 1: Analyze graph structure and DFS traversal

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

2. Step 2: Trace DFS calls from node 0

   DFS visits 0, then 1, then 2. At 2, neighbor 0 is visited and not parent, cycle detected, returns true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Triangle cycle detected = true [OK]

Hint: Triangle in graph means cycle detected true [OK]

Common Mistakes:

• Ignoring parent check causes false cycle detection.
• Assuming no cycle in triangle graph.
• Confusing return values of DFS function.

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

function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      dfs(neighbor, node, visited, graph);
    } else if (neighbor !== parent) {
      return true;
    }
  }
  return false;
}

medium

A. Visited array is not updated correctly.

B. Parent parameter is not used in the function.

C. Missing return statement for recursive dfs call result.

D. Graph adjacency list is not passed correctly.

Solution

1. Step 1: Check recursive DFS call handling

   The recursive call dfs(neighbor, node, visited, graph) is made but its boolean result is ignored.

2. Step 2: Identify missing return for cycle detection

   If recursive call returns true (cycle found), it should be returned immediately to propagate detection.

3. Final Answer:

   Missing return statement for recursive dfs call result. -> Option C

4. Quick Check:

   Return recursive DFS result to detect cycle [OK]

Hint: Always return recursive DFS result for cycle detection [OK]

Common Mistakes:

• Ignoring return value of recursive DFS calls.
• Not updating visited array before recursion.
• Misusing parent parameter in cycle check.

5. Given an undirected graph represented as an adjacency list, how can you modify the cycle detection algorithm to also return the nodes forming the cycle?

hard

A. Mark nodes visited twice to identify cycle nodes.

B. Use BFS instead of DFS to find cycles and record nodes.

C. Count edges and nodes; if edges > nodes, return all nodes as cycle.

D. Track the path stack during DFS and return the cycle path when detected.

Solution

1. Step 1: Understand cycle detection with path tracking

   Maintain a stack or list of nodes visited in current DFS path to track traversal.

2. Step 2: Return cycle nodes when back edge found

   When a visited node (not parent) is found, extract nodes from stack forming the cycle path.

3. Final Answer:

   Track the path stack during DFS and return the cycle path when detected. -> Option D

4. Quick Check:

   Use DFS path stack to get cycle nodes [OK]

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

Common Mistakes:

• Using BFS which is less natural for cycle path retrieval.
• Assuming edge count alone identifies cycle nodes.
• Marking nodes visited twice without path context.