Articulation Points in Graph in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the adjacency list for the graph.

DSA C++

vector<int> adj[[1]];

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

Agraph

Bedges

Cn

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'edges' instead of 'n' for adjacency list size.

Using 'visited' which is unrelated here.

The adjacency list is declared with size 'n', the number of vertices.

2fill in blank

medium

Complete the code to initialize the discovery time of a node.

DSA C++

disc[u] = [1]++;

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Atimer

Bcount

Ctime

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'time' which may be a reserved word or undefined.

Using 'count' or 'index' which are unrelated.

The discovery time is set using the 'timer' variable which increments after assignment.

3fill in blank

hard

Fix the error in the condition to check if a node is an articulation point for root.

DSA C++

if (parent == -1 && children > [1]) articulation[u] = true;

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

A3

B2

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 which incorrectly marks root nodes with fewer children.

Using 3 which is too strict.

A root node is an articulation point if it has more than 1 child, so children > 1 means 2 or more children.

4fill in blank

hard

Fill both blanks to update the low time after visiting a neighbor.

DSA C++

low[u] = min(low[u], [1][[2]]);

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Adisc

Blow

Cparent

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using parent or visited arrays which are unrelated here.

Swapping disc and low arrays incorrectly.

We update low[u] with the minimum of low[u] and disc[v] to track earliest reachable ancestor.

5fill in blank

hard

Fill all three blanks to complete the DFS function signature and recursive call.

DSA C++

void dfs(int [1], int [2], vector<int> adj[], vector<bool> &[3]) {
    visited[u] = true;
    // rest of DFS code
}

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

Au

Bparent

Cvisited

Dtimer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'timer' as a parameter which is unrelated here.

Mixing parameter order or names.

The DFS function takes current node 'u', its 'parent', adjacency list, and visited array by reference.

Practice

1. What is an articulation point in a graph?

easy

A. A vertex with the highest degree in the graph

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

C. A vertex connected to all other vertices

D. A vertex that forms a cycle with its neighbors

Solution

1. Step 1: Understand the role of articulation points

   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 describes this property correctly; others describe different vertex properties.

3. Final Answer:

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

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 forms cycles
• Assuming articulation point connects to all vertices

2. Which of the following is the correct way to declare a vector of vectors for adjacency list in C++?

easy

A. vector adj;

B. vector adj<int>;

C. vector<int> adj[];

D. vector<vector> adj<int>;

Solution

1. Step 1: Recall correct syntax for 2D vector declaration

   In C++, a vector of vectors is declared as vector name; to represent adjacency lists.

2. Step 2: Check each option

   vector adj; matches correct syntax; A, B and D have syntax errors.

3. Final Answer:

   vector adj; -> Option A

4. Quick Check:

   2D vector syntax = vector [OK]

Hint: Use vector for adjacency lists in C++ [OK]

Common Mistakes:

• Using array syntax instead of vector
• Incorrect template syntax
• Missing angle brackets for nested vectors

3. Consider the following code snippet for finding articulation points in a graph with 4 vertices (0 to 3):

int timer = 0;
vector<int> disc(4, -1), low(4, -1);
vector<bool> visited(4, false), isAP(4, false);
vector adj = {{1, 2}, {0, 2}, {0, 1, 3}, {2}};

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

dfs(0, -1);

vector<int> result;
for (int i = 0; i < 4; i++) {
    if (isAP[i]) result.push_back(i);
}
for (int x : result) cout << x << " ";

What will be the output?

medium

A. 2

B. 0 1 2

C. 1 2

D. 0 2

Solution

1. Step 1: Analyze the graph structure and DFS traversal

   The graph edges are: 0-1, 0-2, 1-2, 2-3 (triangle 0-1-2 + pendant 3 on 2). DFS from 0 visits 1 then 2 then 3; when back to 0, 2 already visited so root children=1 (not AP).

2. Step 2: Check articulation point conditions in code

   Root children=1 (<=1, not AP). For non-root u=2 (parent=1), low[3]=3 >= disc[2]=2 so isAP[2]=true. No others. Output is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Root children=1 and vertex 2 disconnects 3 = 2 [OK]

Hint: Root with >1 child and low[v]>=disc[u] mark AP [OK]

Common Mistakes:

• Missing root articulation point condition
• Confusing low and disc arrays
• Ignoring parent check in DFS

4. The following code snippet is intended to find articulation points but has a bug:

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

What is the bug in this code?

medium

A. Not marking visited[u] as true at start

B. Incorrect update of low[u] inside else if block

C. Not incrementing timer before assigning disc[u]

D. Missing check for parent != -1 before marking isAP[u]

Solution

1. Step 1: Understand articulation point condition for non-root nodes

   For non-root nodes, isAP[u] should be set only if parent != -1 and low[v] >= disc[u]. The code misses the parent != -1 check before setting isAP[u].

2. Step 2: Verify other parts of code

   Timer is incremented correctly, visited[u] is set true, low[u] updated properly. So only missing parent check is the bug.

3. Final Answer:

   Missing check for parent != -1 before marking isAP[u] -> Option D

4. Quick Check:

   Parent check needed before marking articulation point [OK]

Hint: Add parent != -1 check before marking articulation point [OK]

Common Mistakes:

• Marking root as articulation point incorrectly
• Confusing order of visited and timer increment
• Ignoring parent in back edge condition

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

hard

A. 1

B. 0 1 3

C. 1 3

D. 3 4

Solution

1. Step 1: Analyze graph connectivity and removal effects

   Vertices 0,1,2 form a triangle (cycle), so none of them alone disconnects graph except vertex 1 which connects to 3 and 4. Removing 1 disconnects 3 and 4. Removing 3 disconnects 4. So 1 and 3 are articulation points.

2. Step 2: Confirm articulation points

   Vertex 1 connects two parts: cycle and chain (3-4). Vertex 3 connects 4 only. Removing 1 or 3 increases components. Vertex 0, 2, 4 are not articulation points.

3. Final Answer:

   1 3 -> Option C

4. Quick Check:

   Removing 1 or 3 disconnects graph = 1 3 [OK]

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

Common Mistakes:

• Including vertices in cycles as articulation points
• Missing leaf node articulation points
• Confusing articulation points with bridges