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

Cycle Detection in Directed Graph

📖 Scenario: Imagine you are building a task manager app. Each task can depend on other tasks to be completed first. This creates a directed graph where tasks point to their dependencies. You want to check if there is a cycle in these dependencies, which would make it impossible to complete all tasks.

🎯 Goal: Build a program to detect if a directed graph of tasks has a cycle. You will represent the graph as an adjacency list, then write code to detect cycles using depth-first search.

📋 What You'll Learn

Create a directed graph as an adjacency list using a JavaScript object

Add a helper variable to track visited nodes

Implement a depth-first search function to detect cycles

Print whether the graph contains a cycle or not

💡 Why This Matters

🌍 Real World

Cycle detection is important in task scheduling, dependency resolution, and deadlock detection in operating systems.

💼 Career

Understanding cycle detection helps in software development roles involving graph algorithms, backend systems, and project management tools.

1

Create the directed graph as an adjacency list

Create a JavaScript object called graph representing a directed graph with these edges: 1 -> 2, 2 -> 3, 3 -> 4, 4 -> 2, and 4 -> 5. Use numbers as keys and arrays of numbers as values.

DSA Javascript

# Your code here

Hint

Use an object where each key is a node number and the value is an array of nodes it points to.

2

Add helper variables to track visited nodes

Add two JavaScript objects called visited and recStack to track visited nodes and nodes in the current recursion stack. Initialize both as empty objects.

DSA Javascript

const graph = {
  1: [2],
  2: [3],
  3: [4],
  4: [2, 5],
  5: []
};

// Your code here

Hint

Use empty objects to mark nodes as visited or currently in the recursion stack.

3

Implement the depth-first search function to detect cycles

Write a function called isCyclicUtil that takes a node v and returns true if a cycle is detected starting from v. Use visited and recStack to track nodes. Mark v as visited and add it to recStack. For each neighbor of v in graph[v], if it is not visited, recursively call isCyclicUtil. If a neighbor is in recStack, return true. Remove v from recStack before returning false. Then write a function called isCyclic that checks all nodes in graph by calling isCyclicUtil and returns true if any call returns true.

DSA Javascript

const graph = {
  1: [2],
  2: [3],
  3: [4],
  4: [2, 5],
  5: []
};

const visited = {};
const recStack = {};

// Your code here

Hint

Use recursion to explore neighbors. Track nodes in the current path with recStack to detect cycles.

4

Print whether the graph contains a cycle

Write a console.log statement to print "Graph contains cycle" if isCyclic() returns true, otherwise print "Graph does not contain cycle".

DSA Javascript

const graph = {
  1: [2],
  2: [3],
  3: [4],
  4: [2, 5],
  5: []
};

const visited = {};
const recStack = {};

function isCyclicUtil(v) {
  if (!visited[v]) {
    visited[v] = true;
    recStack[v] = true;

    for (const neighbor of graph[v]) {
      if (!visited[neighbor] && isCyclicUtil(neighbor)) {
        return true;
      } else if (recStack[neighbor]) {
        return true;
      }
    }
  }
  recStack[v] = false;
  return false;
}

function isCyclic() {
  for (const node in graph) {
    if (isCyclicUtil(node)) {
      return true;
    }
  }
  return false;
}

// Your code here

Hint

Call isCyclic() and print the result with console.log.

Practice

1. What is the main purpose of cycle detection in a directed graph?

easy

A. To find if there is a path that starts and ends at the same node following the direction of edges

B. To count the total number of edges in the graph

C. To find the shortest path between two nodes

D. To check if the graph is connected

Solution

1. Step 1: Understand cycle in directed graph

   A cycle means a path that starts and ends at the same node following edge directions.

2. Step 2: Purpose of cycle detection

   Cycle detection checks if such a path exists in the graph.

3. Final Answer:

   To find if there is a path that starts and ends at the same node following the direction of edges -> Option A

4. Quick Check:

   Cycle detection = find directed loops [OK]

Hint: Cycle means a directed loop returning to start [OK]

Common Mistakes:

• Confusing cycle detection with counting edges
• Thinking cycle detection finds shortest paths
• Assuming cycle detection checks connectivity

2. Which of the following JavaScript code snippets correctly initializes a graph as an adjacency list for cycle detection?

easy

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

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

C. const graph = new Map([0, [1, 2], 1, [2], 2, [0]]);

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

Solution

1. Step 1: Check adjacency list syntax in JavaScript

   An adjacency list is an object with keys as nodes and values as arrays of neighbors.

2. Step 2: Validate each option

   const graph = {0: [1, 2], 1: [2], 2: [0]}; uses correct object syntax with arrays as values. const graph = [[0,1], [1,2], [2,0]]; is an edge list, not adjacency list. const graph = new Map([0, [1, 2], 1, [2], 2, [0]]); uses Map incorrectly (missing array of pairs). const graph = {0 => [1, 2], 1 => [2], 2 => [0]}; uses invalid arrow syntax in object keys.

3. Final Answer:

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

4. Quick Check:

   Object with arrays = adjacency list [OK]

Hint: Adjacency list uses object keys with array values [OK]

Common Mistakes:

• Using edge list instead of adjacency list
• Incorrect Map initialization syntax
• Using invalid object key syntax

3. Consider the following JavaScript code for cycle detection in a directed graph using DFS:

const graph = {0: [1], 1: [2], 2: [0]};
const visited = new Set();
const recStack = new Set();

function dfs(node) {
  if (recStack.has(node)) return true;
  if (visited.has(node)) return false;
  visited.add(node);
  recStack.add(node);
  for (const neighbor of graph[node] || []) {
    if (dfs(neighbor)) return true;
  }
  recStack.delete(node);
  return false;
}

console.log(dfs(0));

What will be printed when this code runs?

medium

A. false

B. true

C. undefined

D. TypeError

Solution

1. Step 1: Trace DFS calls starting from node 0

   Start at 0: visited={0}, recStack={0}. Visit neighbor 1: visited={0,1}, recStack={0,1}. Visit neighbor 2: visited={0,1,2}, recStack={0,1,2}. Visit neighbor 0 again: recStack has 0, so return true (cycle found).

2. Step 2: Return true up the call stack

   Since cycle detected, dfs returns true and console.log prints true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Cycle detected = true [OK]

Hint: Cycle if node revisited in recursion stack [OK]

Common Mistakes:

• Ignoring recursion stack and returning false
• Confusing visited set with recursion stack
• Expecting false because all nodes visited

4. The following code attempts to detect a cycle in a directed graph but does not work correctly. What is the main error?

const graph = {0: [1], 1: [2], 2: [0]};
const visited = new Set();

function dfs(node) {
  if (visited.has(node)) return false;
  visited.add(node);
  for (const neighbor of graph[node] || []) {
    if (dfs(neighbor)) return true;
  }
  return false;
}

console.log(dfs(0));

medium

A. Visited set is not cleared after recursion

B. Function does not return any value

C. Graph adjacency list is incorrectly defined

D. Missing recursion stack to detect back edges

Solution

1. Step 1: Analyze cycle detection logic

   The code uses only a visited set but no recursion stack to track nodes in current path.

2. Step 2: Understand why recursion stack is needed

   Without recursion stack, the code cannot detect back edges that form cycles, so it misses cycles.

3. Final Answer:

   Missing recursion stack to detect back edges -> Option D

4. Quick Check:

   Recursion stack needed for cycle detection [OK]

Hint: Use recursion stack to detect cycles in DFS [OK]

Common Mistakes:

• Assuming visited set alone detects cycles
• Not tracking nodes in current recursion path
• Confusing visited set clearing with cycle detection

5. You have a directed graph represented as an adjacency list:

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

Which of the following modifications will break the cycle detection algorithm that uses DFS with a recursion stack?

hard

A. Starting DFS from node 0 instead of node 4

B. Adding a visited set to track visited nodes

C. Removing the recursion stack and using only a visited set

D. Using an iterative BFS instead of DFS

Solution

1. Step 1: Understand the role of recursion stack in cycle detection

   The recursion stack tracks nodes in the current path to detect cycles (back edges).

2. Step 2: Effect of removing recursion stack

   Without recursion stack, the algorithm cannot detect cycles correctly, even if visited set is used.

3. Step 3: Other options explained

   Adding visited set helps. Starting DFS from different nodes does not break detection if all nodes are visited. Using BFS does not detect cycles in directed graphs reliably.

4. Final Answer:

   Removing the recursion stack and using only a visited set -> Option C

5. Quick Check:

   Recursion stack needed for cycle detection [OK]

Hint: Recursion stack is essential for detecting cycles in DFS [OK]

Common Mistakes:

• Thinking visited set alone detects cycles
• Believing BFS can replace DFS for cycle detection
• Assuming start node affects cycle detection correctness