Articulation Points in Graph in DSA Java - 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.

beginner

Why are articulation points important in network design?

Articulation points represent critical nodes whose failure disconnects the network. Identifying them helps improve network reliability by adding redundancy.

intermediate

What two arrays are commonly used in the DFS algorithm to find articulation points?

The 'discovery time' array records when a node is first visited, and the 'low' array records the earliest visited vertex reachable from the subtree rooted at that node.

intermediate

In the DFS tree, under what condition is a root node considered an articulation point?

A root node is an articulation point if it has two or more children in the DFS tree because removing it disconnects those children.

advanced

Explain the role of the 'low' value in detecting articulation points.

The 'low' value of a node shows the earliest discovery time reachable from that node or its descendants. If a child's 'low' value is greater than or equal to the parent's discovery time, the parent is an articulation point.

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

D The graph becomes complete

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

Which algorithm technique is used to find articulation points?

A Dijkstra's Algorithm

B Breadth First Search (BFS)

C Kruskal's Algorithm

D Depth First Search (DFS)

DFS is used to explore nodes and track discovery and low times to find articulation points.

What does the 'discovery time' represent in articulation point detection?

A The order in which nodes are visited in DFS

B The number of edges connected to a node

C The shortest distance from the root

D The degree of the node

Discovery time records when a node is first visited during DFS traversal.

When is a non-root node considered an articulation point?

A If it has a child with low value >= its discovery time

B If it has more than two children

C If it is connected to all other nodes

D If it has the highest degree

A non-root node is an articulation point if a child's low value is not less than the node's discovery time.

What data structure is typically used to represent the graph when finding articulation points?

A Stack

B Adjacency list

C Queue

D Heap

An adjacency list efficiently represents the graph for DFS traversal.

Describe the step-by-step process to find articulation points in a graph using DFS.

Think about how discovery and low times help identify critical nodes.

Explain why the 'low' value is crucial in detecting articulation points and how it is updated during DFS.

Focus on the connection between low values and discovery times.

Practice

1. What is an articulation point in a graph?

easy

A. A vertex that forms a cycle in the graph

B. A vertex with the highest degree in the graph

C. A vertex connected to all other vertices

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

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 which, when removed, increases the number of connected components matches this definition exactly.

3. Final Answer:

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

4. Quick Check:

   Articulation point = vertex breaking connectivity [OK]

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

Common Mistakes:

• Confusing articulation point with highest degree vertex
• Thinking articulation point forms cycles
• Assuming articulation point connects to all vertices

2. Which of the following is the correct way to declare a graph adjacency list in Java for articulation point detection?

easy

A. int graph[][] = new int[];

B. HashMap graph = new HashMap<>();

C. ArrayList graph = new ArrayList<>();

D. List<Integer> graph = new List<>();

Solution

1. Step 1: Understand adjacency list structure

   An adjacency list for a graph is usually a list of lists, where each inner list holds neighbors of a vertex.

2. Step 2: Check Java syntax for adjacency list

   ArrayList graph = new ArrayList<>(); correctly declares an ArrayList of ArrayLists of Integers, suitable for adjacency list.

3. Final Answer:

   ArrayList graph = new ArrayList<>(); -> Option C

4. Quick Check:

   Adjacency list = list of lists [OK]

Hint: Adjacency list uses list of lists in Java [OK]

Common Mistakes:

• Using 2D array without size initialization
• Using HashMap with wrong value type
• Trying to instantiate interface List directly

3. Consider the following Java code snippet for articulation point detection. What will be the output when the graph has edges: 0-1, 1-2, 1-3?

int V = 4;
ArrayList graph = new ArrayList<>();
for (int i = 0; i < V; i++) graph.add(new ArrayList<>());
graph.get(0).add(1);
graph.get(1).add(0);
graph.get(1).add(2);
graph.get(2).add(1);
graph.get(1).add(3);
graph.get(3).add(1);
// Articulation points found: 1

medium

A. []

B. [1]

C. [2, 3]

D. [0, 1]

Solution

1. Step 1: Analyze the graph structure

   The graph has vertices 0,1,2,3 with edges 0-1, 1-2, 1-3. Vertex 1 connects to all others.

2. Step 2: Identify articulation points

   Removing vertex 1 disconnects the graph into three parts, so 1 is an articulation point. Others are not.

3. Final Answer:

   [1] -> Option B

4. Quick Check:

   Only vertex 1 breaks connectivity [OK]

Hint: Check which vertex removal splits graph [OK]

Common Mistakes:

• Assuming leaf nodes are articulation points
• Including vertices with degree 1 as articulation points
• Ignoring connectivity after removal

4. The following Java code snippet is intended to find articulation points but has a bug. What is the error?

void dfs(int u, int parent) {
  visited[u] = true;
  disc[u] = low[u] = time++;
  int children = 0;
  for (int v : 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])
        articulation[u] = true;
    } else if (v != parent) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
  if (parent == -1 && children > 1)
    articulation[u] = true;
}

medium

A. The time variable is not initialized before DFS

B. The condition for root articulation point is incorrect

C. The low array is not updated correctly for back edges

D. The visited array is not marked true for vertex u

Solution

1. Step 1: Check initialization of time variable

   The variable 'time' is used to assign discovery times but must be initialized before DFS starts.

2. Step 2: Verify other parts of DFS

   Conditions for root and back edges are correct; visited[u] is marked true properly.

3. Final Answer:

   The time variable is not initialized before DFS -> Option A

4. Quick Check:

   Uninitialized time causes wrong discovery times [OK]

Hint: Initialize time counter before DFS call [OK]

Common Mistakes:

• Forgetting to initialize time before DFS
• Misusing parent check for root node
• Not updating low values for back edges

5. Given a connected undirected graph with 5 vertices and edges: 0-1, 1-2, 2-0, 1-3, 3-4, which vertices are articulation points?

hard

A. [1, 3]

B. [0, 2]

C. [3]

D. [1]

Solution

1. Step 1: Analyze the graph structure

   Vertices 0,1,2 form a triangle (cycle). Edges 1-3 and 3-4 form a chain attached to vertex 1.

2. Step 2: Identify articulation points

   Removing vertex 1 disconnects 3 and 4 from the cycle, so 1 is an articulation point. Removing 3 disconnects 4, so 3 is also an articulation point. Vertices 0 and 2 are in a cycle, so not articulation points.

3. Final Answer:

   [1, 3] -> Option A

4. Quick Check:

   Vertices breaking connectivity: 1 and 3 [OK]

Hint: Check vertices whose removal splits graph into parts [OK]

Common Mistakes:

• Assuming cycle vertices are articulation points
• Missing vertex 3 as articulation point
• Only considering vertex 1 as articulation point