Maximal Clique Problem | Recursive Solution

Given a small graph with N nodes and E edges, the task is to find the maximum clique in the given graph. A clique is a complete subgraph of a given graph. This means that all nodes in the said subgraph are directly connected to each other, or there is an edge between any two nodes in the subgraph. The maximal clique is the complete subgraph of a given graph which contains the maximum number of nodes.
Examples:

Input: N = 4, edges[][] = {{1, 2}, {2, 3}, {3, 1}, {4, 3}, {4, 1}, {4, 2}} 
Output: 4
Input: N = 5, edges[][] = {{1, 2}, {2, 3}, {3, 1}, {4, 3}, {4, 5}, {5, 3}} 
Output: 3 

Approach: The idea is to use recursion to solve the problem. 

• When an edge is added to the present list, check that if by adding that edge to the present list, does it still form a clique or not.
• The vertices are added until the list does not form a clique. Then, the list is backtracked to find a larger subset which forms a clique.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int MAX = 100;

// Stores the vertices
int store[MAX], n;

// Graph
int graph[MAX][MAX];

// Degree of the vertices
int d[MAX];

// Function to check if the given set of
// vertices in store array is a clique or not
bool is_clique(int b)
{

    // Run a loop for all set of edges
    for (int i = 1; i < b; i++) {
        for (int j = i + 1; j < b; j++)

            // If any edge is missing
            if (graph[store[i]][store[j]] == 0)
                return false;
    }
    return true;
}

// Function to find all the sizes
// of maximal cliques
int maxCliques(int i, int l)
{
    // Maximal clique size
    int max_ = 0;

    // Check if any vertices from i+1
    // can be inserted
    for (int j = i + 1; j <= n; j++) {

        // Add the vertex to store
        store[l] = j;

        // If the graph is not a clique of size k then
        // it cannot be a clique by adding another edge
        if (is_clique(l + 1)) {

            // Update max
            max_ = max(max_, l);

            // Check if another edge can be added
            max_ = max(max_, maxCliques(j, l + 1));
        }
    }
    return max_;
}

// Driver code
int main()
{
    int edges[][2] = { { 1, 2 }, { 2, 3 }, { 3, 1 },
                       { 4, 3 }, { 4, 1 }, { 4, 2 } };
    int size = sizeof(edges) / sizeof(edges[0]);
    n = 4;

    for (int i = 0; i < size; i++) {
        graph[edges[i][0]][edges[i][1]] = 1;
        graph[edges[i][1]][edges[i][0]] = 1;
        d[edges[i][0]]++;
        d[edges[i][1]]++;
    }

    cout << maxCliques(0, 1);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

static int MAX = 100, n;

// Stores the vertices
static int []store = new int[MAX];

// Graph
static int [][]graph = new int[MAX][MAX];

// Degree of the vertices
static int []d = new int[MAX];

// Function to check if the given set of
// vertices in store array is a clique or not
static boolean is_clique(int b)
{

    // Run a loop for all set of edges
    for (int i = 1; i < b; i++)
    {
        for (int j = i + 1; j < b; j++)

            // If any edge is missing
            if (graph[store[i]][store[j]] == 0)
                return false;
    }
    return true;
}

// Function to find all the sizes
// of maximal cliques
static int maxCliques(int i, int l)
{
    // Maximal clique size
    int max_ = 0;

    // Check if any vertices from i+1
    // can be inserted
    for (int j = i + 1; j <= n; j++)
    {

        // Add the vertex to store
        store[l] = j;

        // If the graph is not a clique of size k then
        // it cannot be a clique by adding another edge
        if (is_clique(l + 1))
        {

            // Update max
            max_ = Math.max(max_, l);

            // Check if another edge can be added
            max_ = Math.max(max_, maxCliques(j, l + 1));
        }
    }
    return max_;
}

// Driver code
public static void main(String[] args)
{
    int [][]edges = { { 1, 2 }, { 2, 3 }, { 3, 1 },
                    { 4, 3 }, { 4, 1 }, { 4, 2 } };
    int size = edges.length;
    n = 4;

    for (int i = 0; i < size; i++)
    {
        graph[edges[i][0]][edges[i][1]] = 1;
        graph[edges[i][1]][edges[i][0]] = 1;
        d[edges[i][0]]++;
        d[edges[i][1]]++;
    }
    System.out.print(maxCliques(0, 1));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
MAX = 100;
n = 0;

# Stores the vertices
store = [0] * MAX;

# Graph
graph = [[0 for i in range(MAX)] for j in range(MAX)];

# Degree of the vertices
d = [0] * MAX;

# Function to check if the given set of
# vertices in store array is a clique or not
def is_clique(b):

    # Run a loop for all set of edges
    for i in range(1, b):
        for j in range(i + 1, b):

            # If any edge is missing
            if (graph[store[i]][store[j]] == 0):
                return False;

    return True;

# Function to find all the sizes
# of maximal cliques
def maxCliques(i, l):

    # Maximal clique size
    max_ = 0;

    # Check if any vertices from i+1
    # can be inserted
    for j in range(i + 1, n + 1):

        # Add the vertex to store
        store[l] = j;

        # If the graph is not a clique of size k then
        # it cannot be a clique by adding another edge
        if (is_clique(l + 1)):

            # Update max
            max_ = max(max_, l);

            # Check if another edge can be added
            max_ = max(max_, maxCliques(j, l + 1));

    return max_;

# Driver code
if __name__ == '__main__':
    edges = [[ 1, 2 ],[ 2, 3 ],[ 3, 1 ],
           [ 4, 3 ],[ 4, 1 ],[ 4, 2 ]];
    size = len(edges);
    n = 4;

    for i in range(size):
        graph[edges[i][0]][edges[i][1]] = 1;
        graph[edges[i][1]][edges[i][0]] = 1;
        d[edges[i][0]] += 1;
        d[edges[i][1]] += 1;

    print(maxCliques(0, 1));

# This code is contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;

class GFG
{

static int MAX = 100, n;

// Stores the vertices
static int []store = new int[MAX];

// Graph
static int [,]graph = new int[MAX,MAX];

// Degree of the vertices
static int []d = new int[MAX];

// Function to check if the given set of
// vertices in store array is a clique or not
static bool is_clique(int b)
{

    // Run a loop for all set of edges
    for (int i = 1; i < b; i++)
    {
        for (int j = i + 1; j < b; j++)

            // If any edge is missing
            if (graph[store[i],store[j]] == 0)
                return false;
    }
    return true;
}

// Function to find all the sizes
// of maximal cliques
static int maxCliques(int i, int l)
{
    // Maximal clique size
    int max_ = 0;

    // Check if any vertices from i+1
    // can be inserted
    for (int j = i + 1; j <= n; j++)
    {

        // Add the vertex to store
        store[l] = j;

        // If the graph is not a clique of size k then
        // it cannot be a clique by adding another edge
        if (is_clique(l + 1))
        {

            // Update max
            max_ = Math.Max(max_, l);

            // Check if another edge can be added
            max_ = Math.Max(max_, maxCliques(j, l + 1));
        }
    }
    return max_;
}

// Driver code
public static void Main(String[] args)
{
    int [,]edges = { { 1, 2 }, { 2, 3 }, { 3, 1 },
                    { 4, 3 }, { 4, 1 }, { 4, 2 } };
    int size = edges.GetLength(0);
    n = 4;

    for (int i = 0; i < size; i++)
    {
        graph[edges[i, 0], edges[i, 1]] = 1;
        graph[edges[i, 1], edges[i, 0]] = 1;
        d[edges[i, 0]]++;
        d[edges[i, 1]]++;
    }
    Console.Write(maxCliques(0, 1));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
    // Javascript implementation of the approach

    let MAX = 100, n;

    // Stores the vertices
    let store = new Array(MAX);
    store.fill(0);

    // Graph
    let graph = new Array(MAX);
    for(let i = 0; i < MAX; i++)
    {
        graph[i] = new Array(MAX);
        for(let j = 0; j < MAX; j++)
        {
            graph[i][j] = 0;
        }
    }

    // Degree of the vertices
    let d = new Array(MAX);
    d.fill(0);

    // Function to check if the given set of
    // vertices in store array is a clique or not
    function is_clique(b)
    {

        // Run a loop for all set of edges
        for (let i = 1; i < b; i++)
        {
            for (let j = i + 1; j < b; j++)

                // If any edge is missing
                if (graph[store[i]][store[j]] == 0)
                    return false;
        }
        return true;
    }

    // Function to find all the sizes
    // of maximal cliques
    function maxCliques(i, l)
    {
        // Maximal clique size
        let max_ = 0;

        // Check if any vertices from i+1
        // can be inserted
        for (let j = i + 1; j <= n; j++)
        {

            // Add the vertex to store
            store[l] = j;

            // If the graph is not a clique of size k then
            // it cannot be a clique by adding another edge
            if (is_clique(l + 1))
            {

                // Update max
                max_ = Math.max(max_, l);

                // Check if another edge can be added
                max_ = Math.max(max_, maxCliques(j, l + 1));
            }
        }
        return max_;
    }

    let edges = [ [ 1, 2 ], [ 2, 3 ], [ 3, 1 ],
                    [ 4, 3 ], [ 4, 1 ], [ 4, 2 ] ];
    let size = edges.length;
    n = 4;

    for (let i = 0; i < size; i++)
    {
        graph[edges[i][0]][edges[i][1]] = 1;
        graph[edges[i][1]][edges[i][0]] = 1;
        d[edges[i][0]]++;
        d[edges[i][1]]++;
    }
    document.write(maxCliques(0, 1));

// This code is contributed by suresh07.
</script>

Output:

4

Time complexity: O(2^n * n^2)
The time complexity of this algorithm is O(2^n * n^2), where ‘n’ is the number of vertices in the graph. This is because ‘maxCliques’ is a recursive function that calculates all possible subsets of the vertices and calls ‘is_clique’ function to check if the given set of vertices form a clique or not. The ‘is_clique’ function takes O(n^2) time as it needs to check for all the edges in the graph.

Space complexity: O(n^2 + n)
The space complexity of this algorithm is O(n^2 + n), where ‘n’ is the number of vertices in the graph. This is because we need to store the graph in the form of an adjacency matrix which requires O(n^2) space. Apart from this, we need to store the vertices in ‘store’ array which takes O(n) space.