Articulation Points in Graph in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the discovery time array for all vertices.

DSA Java

int[] disc = new int[[1]];

Drag options to blanks, or click blank then click option'

AE

B0

Cgraph.length

DV

Attempts:

3 left

💡 Hint

Common Mistakes

Using the number of edges instead of vertices.

Using zero instead of the array size.

The discovery time array should be initialized with the number of vertices V.

2fill in blank

medium

Complete the code to mark a vertex as visited.

DSA Java

visited[[1]] = true;

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Au

Bparent

Ctime

Ddisc[u]

Attempts:

3 left

💡 Hint

Common Mistakes

Marking the parent vertex instead of the current vertex.

Using time or discovery time as index.

The vertex u should be marked as visited when it is first discovered.

3fill in blank

hard

Fix the error in updating the low time of vertex u.

DSA Java

low[u] = Math.min(low[u], [1]);

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Alow[v]

Btime

Cdisc[v]

Ddisc[u]

Attempts:

3 left

💡 Hint

Common Mistakes

Using discovery time instead of low time.

Using time or discovery time of u incorrectly.

The low time of u should be updated with the low time of v to track back edges.

4fill in blank

hard

Fill both blanks to check if u is an articulation point when v is not visited.

DSA Java

if (disc[u] == 0 && low[v] >= [1]) {
    ap[u] = true;
}
if (disc[u] > 0 && low[v] >= [2]) {
    ap[u] = true;
}

Drag options to blanks, or click blank then click option'

Adisc[u]

Btime

Cdisc[v]

Dlow[u]

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing discovery and low times incorrectly.

Using time variable instead of disc or low arrays.

For root vertex, check if low[v] >= disc[u]. For other vertices, check low[v] >= low[u].

5fill in blank

hard

Fill all three blanks to implement the DFS call for articulation points.

DSA Java

for (int v : graph[[1]]) {
    if (!visited[v]) {
        parent[v] = [2];
        dfs(v, [3], graph, visited, disc, low, parent, ap);
    }
}

Drag options to blanks, or click blank then click option'

Au

Bv

Ctime

Dparent[u]

Attempts:

3 left

💡 Hint

Common Mistakes

Using v instead of u for adjacency.

Setting parent incorrectly.

Passing wrong parameters to dfs.

Iterate neighbors of u, set parent[v] = u, and call dfs with v and current time.

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