Number of Triangles in an Undirected Graph - GeeksforGeeks

18 Apr, 2022
Given an Undirected simple graph, We need to find how many triangles it can have. For example below graph have 2 triangles in it. 
Let A[][] be the adjacency matrix representation of the graph. If we calculate A3, then the number of triangles in Undirected Graph is equal to trace(A3) / 6. Where trace(A) is the sum of the elements on the main diagonal of matrix A.  
Trace of a graph represented as adjacency matrix A[V][V] is,
trace(A[V][V]) = A[0][0] + A[1][1] + .... + A[V-1][V-1]

Count of triangles = trace(A3) / 6
Below is the implementation of the above formula. 
C++
Java
Python3
C#
Javascript
// A C++ program for finding// number of triangles in an// Undirected Graph. The program// is for adjacency matrix// representation of the graph#include <bits/stdc++.h>using namespace std; // Number of vertices in the graph#define V 4 //  Utility function for matrix// multiplicationvoid multiply(int A[][V], int B[][V], int C[][V]){    for (int i = 0; i < V; i++)    {        for (int j = 0; j < V; j++)        {            C[i][j] = 0;            for (int k = 0; k < V; k++)                C[i][j] += A[i][k]*B[k][j];        }    }} // Utility function to calculate// trace of a matrix (sum of// diagonal elements)int getTrace(int graph[][V]){    int trace = 0;    for (int i = 0; i < V; i++)        trace += graph[i][i];    return trace;} //  Utility function for calculating// number of triangles in graphint triangleInGraph(int graph[][V]){    // To Store graph^2    int aux2[V][V];     // To Store graph^3    int aux3[V][V];     //  Initialising aux    // matrices with 0    for (int i = 0; i < V; ++i)        for (int j = 0; j < V; ++j)            aux2[i][j] = aux3[i][j] = 0;     // aux2 is graph^2 now  printMatrix(aux2);    multiply(graph, graph, aux2);     // after this multiplication aux3 is    // graph^3 printMatrix(aux3);    multiply(graph, aux2, aux3);     int trace = getTrace(aux3);    return trace / 6;} // driver codeint main(){     int graph[V][V] = {{0, 1, 1, 0},                       {1, 0, 1, 1},                       {1, 1, 0, 1},                       {0, 1, 1, 0}                      };     printf("Total number of Triangle in Graph : %d\n",            triangleInGraph(graph));    return 0;}
// Java program to find number// of triangles in an Undirected// Graph. The program is for// adjacency matrix representation// of the graphimport java.io.*; class Directed{    // Number of vertices in    // the graph    int V = 4;     //  Utility function for   // matrix multiplication   void multiply(int A[][], int B[][],                            int C[][])   {       for (int i = 0; i < V; i++)       {           for (int j = 0; j < V; j++)           {               C[i][j] = 0;               for (int k = 0; k < V;                                   k++)               {                   C[i][j] += A[i][k]*                              B[k][j];               }           }       }   }     // Utility function to calculate   // trace of a matrix (sum of   // diagonal elements)   int getTrace(int graph[][])   {       int trace = 0;        for (int i = 0; i < V; i++)       {           trace += graph[i][i];       }       return trace;   }     // Utility function for   // calculating number of   // triangles in graph   int triangleInGraph(int graph[][])   {       // To Store graph^2       int[][] aux2 = new int[V][V];         // To Store graph^3       int[][] aux3 = new int[V][V];         // Initialising aux matrices       // with 0       for (int i = 0; i < V; ++i)       {           for (int j = 0; j < V; ++j)           {               aux2[i][j] = aux3[i][j] = 0;           }       }         // aux2 is graph^2 now       // printMatrix(aux2)       multiply(graph, graph, aux2);         // after this multiplication aux3       // is graph^3 printMatrix(aux3)       multiply(graph, aux2, aux3);         int trace = getTrace(aux3);        return trace / 6;   }     // Driver code   public static void main(String args[])   {       Directed obj = new Directed();               int graph[][] = { {0, 1, 1, 0},                         {1, 0, 1, 1},                         {1, 1, 0, 1},                         {0, 1, 1, 0}                       };        System.out.println("Total number of Triangle in Graph : "+              obj.triangleInGraph(graph));   }} // This code is contributed by Anshika Goyal.
# A Python3 program for finding number of# triangles in an Undirected Graph. The# program is for adjacency matrix# representation of the graph # Utility function for matrix# multiplicationdef multiply(A, B, C):    global V    for i in range(V):        for j in range(V):            C[i][j] = 0            for k in range(V):                C[i][j] += A[i][k] * B[k][j] # Utility function to calculate# trace of a matrix (sum of# diagonal elements)def getTrace(graph):    global V    trace = 0    for i in range(V):        trace += graph[i][i]    return trace # Utility function for calculating# number of triangles in graphdef triangleInGraph(graph):    global V         # To Store graph^2    aux2 = [[None] * V for i in range(V)]     # To Store graph^3    aux3 = [[None] * V for i in range(V)]     # Initialising aux    # matrices with 0    for i in range(V):        for j in range(V):            aux2[i][j] = aux3[i][j] = 0     # aux2 is graph^2 now printMatrix(aux2)    multiply(graph, graph, aux2)     # after this multiplication aux3 is    # graph^3 printMatrix(aux3)    multiply(graph, aux2, aux3)     trace = getTrace(aux3)    return trace // 6 # Driver Code # Number of vertices in the graphV = 4graph = [[0, 1, 1, 0],         [1, 0, 1, 1],         [1, 1, 0, 1],         [0, 1, 1, 0]] print("Total number of Triangle in Graph :",                    triangleInGraph(graph)) # This code is contributed by PranchalK
// C# program to find number// of triangles in an Undirected// Graph. The program is for// adjacency matrix representation// of the graphusing System; class GFG{// Number of vertices// in the graphint V = 4; // Utility function for// matrix multiplicationvoid multiply(int [,]A, int [,]B,                        int [,]C){    for (int i = 0; i < V; i++)    {        for (int j = 0; j < V; j++)        {            C[i, j] = 0;            for (int k = 0; k < V;                              k++)            {                C[i, j] += A[i, k]*                           B[k, j];            }        }    }} // Utility function to// calculate trace of// a matrix (sum of// diagonal elements)int getTrace(int [,]graph){    int trace = 0;     for (int i = 0; i < V; i++)    {        trace += graph[i, i];    }    return trace;} // Utility function for// calculating number of// triangles in graphint triangleInGraph(int [,]graph){    // To Store graph^2    int[,] aux2 = new int[V, V];     // To Store graph^3    int[,] aux3 = new int[V, V];     // Initialising aux matrices    // with 0    for (int i = 0; i < V; ++i)    {        for (int j = 0; j < V; ++j)        {            aux2[i, j] = aux3[i, j] = 0;        }    }     // aux2 is graph^2 now    // printMatrix(aux2)    multiply(graph, graph, aux2);     // after this multiplication aux3    // is graph^3 printMatrix(aux3)    multiply(graph, aux2, aux3);     int trace = getTrace(aux3);     return trace / 6;} // Driver codepublic static void Main(){    GFG obj = new GFG();             int [,]graph = {{0, 1, 1, 0},                    {1, 0, 1, 1},                    {1, 1, 0, 1},                    {0, 1, 1, 0}};     Console.WriteLine("Total number of " +                   "Triangle in Graph : "+              obj.triangleInGraph(graph));}} // This code is contributed by anuj_67.
<script> // Javascript program to find number// of triangles in an Undirected// Graph. The program is for// adjacency matrix representation// of the graph // Number of vertices in the graphlet V = 4; //  Utility function for matrix// multiplicationfunction multiply(A, B, C){    for(let i = 0; i < V; i++)    {        for(let j = 0; j < V; j++)        {            C[i][j] = 0;            for(let k = 0; k < V; k++)                C[i][j] += A[i][k] * B[k][j];        }    }} // Utility function to calculate// trace of a matrix (sum of// diagonal elements)function getTrace(graph){    let trace = 0;    for(let i = 0; i < V; i++)        trace += graph[i][i];             return trace;} //  Utility function for calculating// number of triangles in graphfunction triangleInGraph(graph){         // To Store graph^2    let aux2 = new Array(V);     // To Store graph^3    let aux3 = new Array(V);     // Initialising aux    // matrices with 0    for(let i = 0; i < V; ++i)    {        aux2[i] = new Array(V);        aux3[i] = new Array(V);        for(let j = 0; j < V; ++j)        {            aux2[i][j] = aux3[i][j] = 0;        }    }           // aux2 is graph^2 now  printMatrix(aux2);    multiply(graph, graph, aux2);     // After this multiplication aux3 is    // graph^3 printMatrix(aux3);    multiply(graph, aux2, aux3);     let trace = getTrace(aux3);    return (trace / 6);} // Driver codelet graph = [ [ 0, 1, 1, 0 ],              [ 1, 0, 1, 1 ],              [ 1, 1, 0, 1 ],              [ 0, 1, 1, 0 ] ]; document.write("Total number of Triangle in Graph : " +               triangleInGraph(graph));                // This code is contributed by divyesh072019     </script>
Output: 
Total number of Triangle in Graph : 2
How does this work? If we compute An for an adjacency matrix representation of the graph, then a value An[i][j] represents the number of distinct walks between vertex i to j in the graph. In A3, we get all distinct paths of length 3 between every pair of vertices.A triangle is a cyclic path of length three, i.e. begins and ends at the same vertex. So A3[i][i] represents a triangle beginning and ending with vertex i. Since a triangle has three vertices and it is counted for every vertex, we need to divide the result by 3. Furthermore, since the graph is undirected, every triangle twice as i-p-q-j and i-q-p-j, so we divide by 2 also. Therefore, the number of triangles is trace(A3) / 6.
Time Complexity: The time complexity of above algorithm is O(V3) where V is number of vertices in the graph, we can improve the performance to O(V2.8074) using Strassen’s matrix multiplication algorithm. 
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Another approach: Using Bitsets as adjacency lists.
For each node in the graph compute the corresponding adjacency list as a bitmask.
If two nodes, i & j, are adjacent compute the number of nodes that are adjacent to i & j and add it to the answer.
In the end, divide the answer by 6 to avoid duplicates.
In order to compute the number of nodes adjacent to two nodes, i & j, we use the bitwise operation & (and) on the adjacency list of i and j, then we count the number of ones.
Below is the implementation of the above approach:
C++
#include<iostream>#include<string>#include<algorithm>#include<cstring>#include<vector>#include<bitset> using namespace std; #define V 4 int main(){    // Graph represented as adjacency matrix    int graph[][V] = {{0, 1, 1, 0},                      {1, 0, 1, 1},                      {1, 1, 0, 1},                      {0, 1, 1, 0}};     // create the adjacency list of the graph (as bit masks)    // set the bits at positions [i][j] & [j][i] to 1, if there is an undirected edge between i and j    vector<bitset<V>> Bitset_Adj_List(V);    for (int i = 0; i < V;i++)        for (int j = 0; j < V;j++)            if(graph[i][j])                Bitset_Adj_List[i][j] = 1,                Bitset_Adj_List[j][i] = 1;     int ans = 0;     for (int i = 0; i < V;i++)        for (int j = 0; j < V;j++)                         // if i & j are adjacent            // compute the number of nodes that are adjacent to i & j            if(Bitset_Adj_List[i][j] == 1 && i != j){                bitset<V> Mask = Bitset_Adj_List[i] & Bitset_Adj_List[j];                ans += Mask.count();            }    // divide the answer by 6 to avoid duplicates   ans /= 6;    cout << "The number of Triangles in the Graph is : " << ans;       // This code is contributed    // by Gatea David}
The number of Triangles in the Graph is : 2
Time Complexity: First we have the two for nested loops O(V2) flowed by Bitset operations & and count, both have a time complexity of O(V / Word RAM), where V = number of nodes in the graph and Word RAM is usually 32 or 64. So the final time complexity is O(V2 * V / 32) or O(V3).
References:
http://www.d.umn.edu/math/Technical%20Reports/Technical%20Reports%202007-/TR%202012/yang.pdf
Number of Triangles in Directed and Undirected Graphs. 
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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