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# 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:

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 using namespace std; const int MAX = 100; // Stores the verticesint store[MAX], n; // Graphint graph[MAX][MAX]; // Degree of the verticesint d[MAX]; // Function to check if the given set of// vertices in store array is a clique or notbool 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 cliquesint 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 codeint 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 approachimport java.util.*; class GFG{ static int MAX = 100, n; // Stores the verticesstatic int []store = new int[MAX]; // Graphstatic int [][]graph = new int[MAX][MAX]; // Degree of the verticesstatic int []d = new int[MAX]; // Function to check if the given set of// vertices in store array is a clique or notstatic 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 cliquesstatic 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 codepublic 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 approachMAX = 100;n = 0; # Stores the verticesstore = [0] * MAX; # Graphgraph = [[0 for i in range(MAX)] for j in range(MAX)]; # Degree of the verticesd = [0] * MAX; # Function to check if the given set of# vertices in store array is a clique or notdef 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 cliquesdef 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 codeif __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 approachusing System; class GFG{ static int MAX = 100, n; // Stores the verticesstatic int []store = new int[MAX]; // Graphstatic int [,]graph = new int[MAX,MAX]; // Degree of the verticesstatic int []d = new int[MAX]; // Function to check if the given set of// vertices in store array is a clique or notstatic 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 cliquesstatic 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 codepublic 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



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.