Cycle Detection in Undirected Graph in DSA Javascript - Build from Scratch

Cycle Detection in Undirected Graph

📖 Scenario: Imagine you are working with a network of friends where each friendship is a connection between two people. You want to find out if there is any circle of friends where you can start from one person and come back to the same person by following friendships without repeating any connection.

🎯 Goal: You will build a program to detect if there is a cycle (a closed loop) in an undirected graph representing friendships.

📋 What You'll Learn

Create an adjacency list to represent the undirected graph

Use a helper variable to track visited nodes

Implement a depth-first search (DFS) function to detect cycles

Print true if a cycle exists, otherwise false

💡 Why This Matters

🌍 Real World

Cycle detection helps find loops in networks like social connections, computer networks, or road maps to avoid repeated paths or deadlocks.

💼 Career

Understanding cycle detection is important for software engineers working on graph algorithms, network analysis, and debugging complex data structures.

1

Create the graph as an adjacency list

Create a constant called graph as an object representing the undirected graph with these exact edges: 0 connected to 1 and 2, 1 connected to 0 and 2, 2 connected to 0, 1, and 3, 3 connected to 2.

DSA Javascript

# Your code here

Hint

Use an object where keys are node numbers and values are arrays of connected nodes.

2

Create a visited set to track visited nodes

Create a constant called visited and set it to a new empty Set() to keep track of visited nodes.

DSA Javascript

const graph = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};

# Your code here

Hint

Use new Set() to create an empty set for visited nodes.

3

Implement DFS function to detect cycle

Write a function called hasCycle that takes node and parent as parameters. Inside, add node to visited. Then, use a for loop with variable neighbor to iterate over graph[node]. If neighbor is not in visited, recursively call hasCycle(neighbor, node) and return true if it returns true. If neighbor is in visited and neighbor is not equal to parent, return true. After the loop, return false.

DSA Javascript

const graph = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};

const visited = new Set();

// Your code here

Hint

Use recursion and track the parent node to detect cycles in an undirected graph.

4

Check all nodes and print if cycle exists

Create a variable called cycleExists and set it to false. Use a for loop with variable node to iterate over Object.keys(graph). Inside the loop, convert node to a number. If visited does not have node, call hasCycle(node, -1) and if it returns true, set cycleExists to true and break the loop. Finally, print cycleExists.

DSA Javascript

const graph = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};

const visited = new Set();

function hasCycle(node, parent) {
  visited.add(node);
  for (const neighbor of graph[node]) {
    if (!visited.has(neighbor)) {
      if (hasCycle(neighbor, node)) {
        return true;
      }
    } else if (neighbor !== parent) {
      return true;
    }
  }
  return false;
}

// Your code here

Hint

Check each node to find cycles and print the result.

Practice

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

easy

A. If we find a node with no neighbors, a cycle exists.

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 always has a cycle.

D. If all nodes are visited exactly once, a cycle exists.

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 cycle exists.

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 node not parent [OK]

Hint: Check visited neighbors excluding parent to find cycles [OK]

Common Mistakes:

• Ignoring the parent node in cycle check
• Assuming more edges than nodes always means cycle
• Confusing cycle detection logic with directed graphs

2. Which of the following is the correct way to represent an undirected graph adjacency list in JavaScript?

easy

A. const graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]};

B. const graph = [[0,1], [1,2], [2,3]];

C. const graph = {0: 1, 1: 2, 2: 3};

D. const graph = [[1, 2], [0], [0], [1]];

Solution

1. Step 1: Understand adjacency list format

   An adjacency list maps each node to an array of its neighbors.

2. Step 2: Check each option

   const graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]}; correctly maps nodes to arrays of neighbors. const graph = [[0,1], [1,2], [2,3]]; is edge list, not adjacency list. const graph = {0: 1, 1: 2, 2: 3}; maps nodes to single values, not arrays. const graph = [[1, 2], [0], [0], [1]]; is an array of arrays but connections are not symmetric (node 1 missing neighbor 3).

3. Final Answer:

   const graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]}; -> Option A

4. Quick Check:

   Adjacency list = node keys with neighbor arrays [OK]

Hint: Adjacency list uses node keys with arrays of neighbors [OK]

Common Mistakes:

• Using edge list instead of adjacency list
• Mapping nodes to single values, not arrays
• Using arrays without keys for adjacency list

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

const graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]};

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

const visited = {};
const hasCycle = dfs(0, -1, visited);
console.log(hasCycle);

medium

A. true

B. undefined

C. false

D. TypeError

Solution

1. Step 1: Trace DFS calls and visited nodes

   Start at node 0, mark visited. Visit neighbors 1 and 2. Both lead to nodes without cycles.

2. Step 2: Check for cycle condition

   No neighbor is visited again except parent, so no cycle detected. DFS returns false.

3. Final Answer:

   false -> Option C

4. Quick Check:

   No back edge found = false [OK]

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

Common Mistakes:

• Confusing parent node with cycle neighbor
• Assuming any visited node means cycle
• Not initializing visited properly

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

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

medium

A. The function should return false inside the for loop.

B. The visited array is not initialized before DFS call.

C. The graph variable is not defined inside the function.

D. The condition 'neighbor === parent' should be 'neighbor !== parent' to detect cycles.

Solution

1. Step 1: Analyze cycle detection condition

   The code returns true if neighbor equals parent, which is incorrect because parent is allowed to be visited.

2. Step 2: Correct condition for cycle detection

   It should return true if neighbor is visited and neighbor is NOT the parent, indicating a cycle.

3. Final Answer:

   The condition 'neighbor === parent' should be 'neighbor !== parent' to detect cycles. -> Option D

4. Quick Check:

   Cycle if visited neighbor not parent [OK]

Hint: Cycle check: visited neighbor must not be parent [OK]

Common Mistakes:

• Returning true when neighbor is parent
• Not initializing visited before DFS
• Returning false too early inside loop

5. Given the undirected graph below, which edges form a cycle?

const graph = {
  0: [1, 2],
  1: [0, 2, 3],
  2: [0, 1, 4],
  3: [1],
  4: [2]
};

Select the correct set of edges that create a cycle.

hard

A. Edges (0-1), (1-2), (2-0)

B. Edges (1-3), (3-4), (4-2)

C. Edges (0-2), (2-4), (4-3)

D. Edges (0-1), (1-3), (3-0)

Solution

1. Step 1: Identify cycles by checking connected edges

   Edges (0-1), (1-2), and (2-0) connect nodes in a triangle, forming a cycle.

2. Step 2: Verify other edges

   Edges involving nodes 3 and 4 do not form cycles as they are connected linearly.

3. Final Answer:

   Edges (0-1), (1-2), (2-0) -> Option A

4. Quick Check:

   Triangle edges form cycle [OK]

Hint: Look for triangle or loop edges to find cycles [OK]

Common Mistakes:

• Assuming linear chains form cycles
• Ignoring the triangle formed by nodes 0,1,2
• Confusing edge directions in undirected graph