Articulation Points in Graph in DSA C - Build from Scratch

Articulation Points in Graph

📖 Scenario: You are working as a network engineer. You want to find critical computers (nodes) in a network that, if removed, would break the network into disconnected parts. These critical nodes are called articulation points.We will represent the network as a graph using adjacency lists.

🎯 Goal: Build a program in C to find all articulation points in an undirected graph using DFS and low-link values.

📋 What You'll Learn

Create an adjacency list representation of the graph with 5 nodes and given edges

Add arrays to track discovery times, low values, and articulation points

Implement DFS to find articulation points using Tarjan's algorithm

Print all articulation points found in the graph

💡 Why This Matters

🌍 Real World

Finding articulation points helps identify critical nodes in computer networks, social networks, or transportation systems that can cause disconnection if removed.

💼 Career

Network engineers, system architects, and software developers use articulation point detection to improve network reliability and fault tolerance.

1

Create the graph adjacency list

Create an integer V with value 5 for number of vertices. Create an adjacency list array called adj of size V where each element is a pointer to an integer array. Initialize the adjacency list with these edges: 0-1, 0-2, 1-2, 1-3, 3-4.

DSA C

# Define number of vertices and adjacency list
// Your code here

Hint

Use an array of pointers for adjacency list. Use a helper function add_edge to add edges both ways.

2

Add arrays for DFS tracking

Add integer arrays disc, low, and parent of size V. Also add a boolean array ap of size V to mark articulation points. Initialize all disc and low values to -1, all parent values to -1, and all ap values to 0.

DSA C

#include 
#include 

int V = 5;
int* adj[5];
int adj_size[5] = {0};

void add_edge(int u, int v) {
    adj[u][adj_size[u]++] = v;
    adj[v][adj_size[v]++] = u;
}

int main() {
    for (int i = 0; i < V; i++) {
        adj[i] = (int*)malloc(5 * sizeof(int));
    }

    add_edge(0, 1);
    add_edge(0, 2);
    add_edge(1, 2);
    add_edge(1, 3);
    add_edge(3, 4);

    // Add arrays disc, low, parent, ap and initialize
    // Your code here

    return 0;
}

Hint

Use arrays of size V to track discovery time, low values, parents, and articulation points. Initialize them properly.

3

Implement DFS to find articulation points

Write a recursive function APUtil with parameters int u, int* time. Inside, set disc[u] and low[u] to ++(*time). Use a variable children to count DFS tree children. For each adjacent vertex v of u, if disc[v] is -1, set parent[v] = u, recursively call APUtil(v, time), update low[u] with minimum of low[u] and low[v]. Mark ap[u] = 1 if u is root with more than one child or if low[v] >= disc[u] for non-root. If disc[v] is not -1 and v != parent[u], update low[u] with minimum of low[u] and disc[v]. Call APUtil for all vertices with disc[i] == -1 in main.

DSA C

#include 
#include 

int V = 5;
int* adj[5];
int adj_size[5] = {0};

void add_edge(int u, int v) {
    adj[u][adj_size[u]++] = v;
    adj[v][adj_size[v]++] = u;
}

int disc[5];
int low[5];
int parent[5];
int ap[5];

// Implement APUtil function here
// Your code here

int main() {
    for (int i = 0; i < V; i++) {
        adj[i] = (int*)malloc(5 * sizeof(int));
    }

    add_edge(0, 1);
    add_edge(0, 2);
    add_edge(1, 2);
    add_edge(1, 3);
    add_edge(3, 4);

    for (int i = 0; i < V; i++) {
        disc[i] = -1;
        low[i] = -1;
        parent[i] = -1;
        ap[i] = 0;
    }

    int time = 0;
    for (int i = 0; i < V; i++) {
        if (disc[i] == -1) {
            APUtil(i, &time);
        }
    }

    return 0;
}

Hint

Use DFS recursion with discovery and low times. Update articulation points based on conditions for root and non-root nodes.

4

Print all articulation points

Add a for loop in main to print all vertices i where ap[i] == 1. Print them in the format: "Articulation Point: i\n".

DSA C

#include 
#include 

int V = 5;
int* adj[5];
int adj_size[5] = {0};

void add_edge(int u, int v) {
    adj[u][adj_size[u]++] = v;
    adj[v][adj_size[v]++] = u;
}

int disc[5];
int low[5];
int parent[5];
int ap[5];

void APUtil(int u, int* time) {
    disc[u] = low[u] = ++(*time);
    int children = 0;

    for (int i = 0; i  1)
                ap[u] = 1;

            if (parent[u] != -1 && low[v] >= disc[u])
                ap[u] = 1;
        } else if (v != parent[u]) {
            if (disc[v] < low[u])
                low[u] = disc[v];
        }
    }
}

int main() {
    for (int i = 0; i < V; i++) {
        adj[i] = (int*)malloc(5 * sizeof(int));
    }

    add_edge(0, 1);
    add_edge(0, 2);
    add_edge(1, 2);
    add_edge(1, 3);
    add_edge(3, 4);

    for (int i = 0; i < V; i++) {
        disc[i] = -1;
        low[i] = -1;
        parent[i] = -1;
        ap[i] = 0;
    }

    int time = 0;
    for (int i = 0; i < V; i++) {
        if (disc[i] == -1) {
            APUtil(i, &time);
        }
    }

    // Print articulation points
    // Your code here

    return 0;
}

Hint

Loop through ap array and print all indices where value is 1.