Articulation Points in Graph in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an articulation point in a graph?

An articulation point is a vertex in a graph that, if removed along with its edges, increases the number of connected components. It is also called a cut vertex.

intermediate

Which algorithm is commonly used to find articulation points in a graph?

Tarjan's algorithm using DFS (Depth First Search) with discovery and low time arrays is commonly used to find articulation points efficiently.

intermediate

In Tarjan's algorithm, what does the 'low' value represent for a vertex?

The 'low' value of a vertex is the earliest discovery time reachable from that vertex or its descendants by following back edges.

intermediate

Why is the root of DFS tree an articulation point only if it has more than one child?

Because if the root has more than one child, removing it disconnects those children subtrees, increasing connected components. If it has one or zero children, removal does not disconnect the graph.

intermediate

What is the time complexity of finding articulation points using Tarjan's algorithm?

The time complexity is O(V + E), where V is the number of vertices and E is the number of edges, because it uses a single DFS traversal.

What happens if you remove an articulation point from a connected graph?

A The graph becomes a tree

B The graph remains connected

C The graph becomes disconnected or increases connected components

D Nothing changes

Removing an articulation point increases the number of connected components, disconnecting the graph.

In Tarjan's algorithm, which data structure is primarily used to explore the graph?

A Queue

B Depth First Search (DFS) recursion

C Breadth First Search (BFS)

D Stack

Tarjan's algorithm uses DFS recursion to explore vertices and compute discovery and low values.

What does a 'back edge' in DFS indicate?

A An edge to a child vertex

B An edge that forms a cycle with the root only

C An edge to a disconnected vertex

D An edge to a previously visited ancestor vertex

A back edge connects a vertex to an ancestor in the DFS tree, indicating a cycle.

Which condition identifies a vertex u as an articulation point when u is not the root?

A If low[v] >= disc[u] for any child v of u

B If disc[u] > low[v] for any child v of u

C If u has no children

D If u is connected to all vertices

If for any child v of u, low[v] >= disc[u], u is an articulation point because v or its descendants cannot reach an ancestor of u.

What is the initial value of discovery time for all vertices before DFS starts?

A -1 or undefined

B Infinity

C 0

D 1

Discovery times are initialized as -1 or undefined to mark unvisited vertices before DFS.

Explain how Tarjan's algorithm finds articulation points in a graph using DFS.

Think about how discovery and low values help detect if a vertex connects different parts of the graph.

Describe why the root of the DFS tree is treated differently when identifying articulation points.

Consider how removing the root affects connectivity compared to other vertices.

Practice

1. What is an articulation point (or cut vertex) in a graph?

easy

A. A vertex that, when removed, increases the number of connected components

B. A vertex with the highest degree in the graph

C. A vertex connected to all other vertices

D. A vertex that has no edges connected to it

Solution

1. Step 1: Understand the definition of articulation point

   An articulation point is a vertex that, if removed, breaks the graph into more pieces (connected components).

2. Step 2: Compare options with definition

   Only A vertex that, when removed, increases the number of connected components matches this definition exactly.

3. Final Answer:

   A vertex that, when removed, increases the number of connected components -> Option A

4. Quick Check:

   Articulation point = vertex increasing components when removed [OK]

Hint: Articulation point breaks graph connectivity when removed [OK]

Common Mistakes:

• Confusing articulation point with highest degree vertex
• Thinking articulation point connects to all vertices
• Assuming isolated vertex is articulation point

2. Which of the following is the correct TypeScript function signature to find articulation points in an undirected graph represented as adjacency list?

easy

A. function findArticulationPoints(graph: number): number[]

B. function findArticulationPoints(graph: number[][]): void

C. function findArticulationPoints(graph: string): number[]

D. function findArticulationPoints(graph: Map): number[]

Solution

1. Step 1: Identify the graph representation type

   An adjacency list is best represented as a Map from node number to array of connected nodes.

2. Step 2: Check function signature compatibility

   function findArticulationPoints(graph: Map): number[] uses Map and returns number[] which fits the problem.

3. Final Answer:

   function findArticulationPoints(graph: Map<number, number[]>): number[] -> Option D

4. Quick Check:

   Adjacency list as Map with number arrays returns number array [OK]

Hint: Use Map for adjacency list in TypeScript [OK]

Common Mistakes:

• Using string type for graph input
• Returning void instead of list of articulation points
• Passing graph as a single number

3. Given the following TypeScript code snippet to find articulation points, what will be the output for the graph edges: [[0,1],[1,2],[2,0],[1,3]]?

const graph = new Map();
graph.set(0, [1, 2]);
graph.set(1, [0, 2, 3]);
graph.set(2, [0, 1]);
graph.set(3, [1]);

const articulationPoints = findArticulationPoints(graph);
console.log(articulationPoints);

medium

A. [0, 1]

B. [3]

C. [1]

D. []

Solution

1. Step 1: Analyze the graph structure

   The graph has a triangle between 0,1,2 and a leaf node 3 connected to 1.

2. Step 2: Identify articulation points

   Removing node 1 disconnects node 3 from the rest, so 1 is an articulation point. Nodes 0 and 2 are part of a cycle, so not articulation points. Node 3 is a leaf, not articulation point.

3. Final Answer:

   [1] -> Option C

4. Quick Check:

   Only node 1 disconnects graph when removed [OK]

Hint: Check if removing node splits graph into more parts [OK]

Common Mistakes:

• Thinking nodes in cycle are articulation points
• Including leaf nodes as articulation points
• Ignoring connectivity after removal

4. The following TypeScript code attempts to find articulation points but has a bug. What is the bug?

function dfs(u: number, parent: number) {
  visited[u] = true;
  disc[u] = time;
  low[u] = time;
  time++;
  let children = 0;
  for (const v of graph.get(u) || []) {
    if (!visited[v]) {
      children++;
      dfs(v, u);
      low[u] = Math.min(low[u], low[v]);
      if (parent !== -1 && low[v] >= disc[u]) {
        articulationPoints.add(u);
      }
    } else if (v !== parent) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
  if (parent === -1 && children > 1) {
    articulationPoints.add(u);
  }
}

medium

A. The 'visited' array is not initialized before DFS

B. The root node condition for articulation point is checked after the loop

C. The variable 'time' is not incremented correctly

D. The condition 'low[v] >= disc[u]' should be 'low[v] > disc[u]'

Solution

1. Step 1: Review DFS variables initialization

   The code uses 'visited' array but does not show its initialization before DFS calls.

2. Step 2: Identify missing initialization impact

   If 'visited' is not initialized to false for all nodes, DFS may skip nodes or behave incorrectly.

3. Final Answer:

   The 'visited' array is not initialized before DFS -> Option A

4. Quick Check:

   Uninitialized visited array causes DFS errors [OK]

Hint: Always initialize visited array before DFS traversal [OK]

Common Mistakes:

• Assuming time increments incorrectly
• Misunderstanding root node articulation condition
• Changing condition from >= to > incorrectly

5. You have a network of computers represented as an undirected graph. You want to find all critical computers whose failure would split the network. The graph has 7 nodes with edges: [[0,1],[1,2],[2,0],[1,3],[3,4],[4,5],[5,3],[4,6]]. Which nodes are articulation points?

hard

A. [0, 2, 5]

B. [1, 3, 4]

C. [3, 5, 6]

D. [1, 4, 6]

Solution

1. Step 1: Analyze graph components and cycles

   Nodes 0,1,2 form a triangle (cycle). Nodes 3,4,5 form another cycle. Node 6 is connected only to 4.

2. Step 2: Identify articulation points

   Removing 1 disconnects node 3 and its cycle from 0,2. Removing 3 disconnects 4,5,6 from 1. Removing 4 disconnects node 6 from rest. So articulation points are 1, 3, and 4.

3. Final Answer:

   [1, 3, 4] -> Option B

4. Quick Check:

   Nodes connecting cycles and leaf nodes are articulation points [OK]

Hint: Look for nodes connecting cycles or leaf nodes [OK]

Common Mistakes:

• Including nodes inside cycles as articulation points
• Missing nodes connecting leaf nodes
• Confusing leaf nodes as articulation points