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Total number of Spanning Trees in a Graph

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If a graph is a complete graph with n vertices, then total number of spanning trees is n(n-2) where n is the number of nodes in the graph. In complete graph, the task is equal to counting different labeled trees with n nodes for which have Cayley’s formula.

What if graph is not complete? 

Follow the given procedure:

  • STEP 1: Create Adjacency Matrix for the given graph. 
  • STEP 2: Replace all the diagonal elements with the degree of nodes. For eg. element at (1,1) position of adjacency matrix will be replaced by the degree of node 1, element at (2,2) position of adjacency matrix will be replaced by the degree of node 2, and so on. 
  • STEP 3: Replace all non-diagonal 1’s with -1. 
  • STEP 4: Calculate co-factor for any element. 
  • STEP 5: The cofactor that you get is the total number of spanning tree for that graph. 

Consider the following graph:

 kirchoff-formula 

Adjacency Matrix for the above graph will be as follows:

 kirchoff-matrix

 After applying STEP 2 and STEP 3, adjacency matrix will look like

 kirchoff-theorem 

The co-factor for (1, 1) is 8. Hence total no. of spanning tree that can be formed is 8. 

NOTE: Co-factor for all the elements will be same. Hence we can compute co-factor for any element of the matrix. This method is also known as Kirchhoff’s Theorem. It can be applied to complete graphs also.

let’s see another example to solve these problems by making use of the Laplacian matrix.

A Laplacian matrix L,
where L[i, i] is the degree of node i and L[i, j] = -1 if there is an edge between nodes i and j, and otherwise L[i, j] = 0.

Kirchhoff’s theorem provides a way to calculate the number of spanning trees for a given graph as a determinant of a special matrix.

consider the following graph,

the undirected graph

All possible spanning trees are as follows

three spanning trees

In order to calculate the number of spanning trees, construct a Laplacian matrix L, where L[i, i] is the degree of node i and L[i, j] = ?1 if there is an edge between nodes i and j, and otherwise L[i, j] = 0. 

for the above graph, The Laplacian matrix will look like this

 

The number of spanning trees equals the determinant of a matrix.

The Determinant of a matrix that can be obtained when we remove any row and any column from L.

For example, if we remove the first row and column, the result will be,

 

The determinant is always the same, regardless of which row and column we remove from L.

C++




#include<bits/stdc++.h> 
using namespace std; 
  
// C++ program to find number of spanning 
// trees in a graph using Matrix 
// Chain Multiplication 
#define MAX 100 
#define MOD 1000000007 
  
// Matrix Multiplication 
void multiply(int A[MAX][MAX], int B[MAX][MAX], int C[MAX][MAX]) 
    for (int i = 0; i < MAX; i++) 
        for (int j = 0; j < MAX; j++) 
            for (int k = 0; k < MAX; k++) 
                C[i][j] = (C[i][j] + (A[i][k] * B[k][j])%MOD)%MOD;     
  
// Function to find Nth power of A 
void power(int A[MAX][MAX], int N, int result[MAX][MAX]) 
    int temp[MAX][MAX]; 
    for (int i = 0; i < MAX; i++) 
        for (int j = 0; j < MAX; j++) 
            result[i][j] = (i == j); 
  
    while (N>0) 
    
        if (N%2 == 1) 
        
            multiply(A, result, temp); 
            for (int i=0; i<MAX; i++) 
                for (int j=0; j<MAX; j++) 
                    result[i][j] = temp[i][j]; 
        
  
        N = N/2; 
        multiply(A, A, temp); 
        for (int i=0; i<MAX; i++) 
            for (int j=0; j<MAX; j++) 
                A[i][j] = temp[i][j]; 
    
  
// Function to find number of Spanning 
// Trees in a Graph using Matrix Chain 
// Multiplication. 
int numOfSpanningTree(int graph[][MAX], int V) 
    int result[MAX][MAX] = {0}; 
    int temp[MAX][MAX] = {0}; 
  
    // Create a copy of graph as the 
    // Adjacency matrix will be changed 
    // during the process 
    for (int i = 0; i < V; i++) 
        for (int j = 0; j < V; j++) 
            temp[i][j] = graph[i][j]; 
  
    // Find (V-2)th power of Adjacency 
    // matrix 
    power(temp, V-2, result); 
  
    int ans = 0; 
  
    // Find sum of all elements in (V-2)th 
    // power 
    for (int i = 0; i < V; i++) 
        for (int j = 0; j < V; j++) 
            ans = (ans + result[i][j])%MOD; 
  
    return ans; 
  
  
  
// Driver program 
int main() 
    // Let us create the following graph 
    // 2 <-> 3 
    // | | 
    // 0 <-1-> 1 
    int V = 4; // Number of vertices in graph 
    int E = 5; // Number of edges in graph 
    int graph[][MAX] = { 
                        {0, 1, 1, 1}, 
                        {1, 0, 1, 1}, 
                        {1, 1, 0, 1}, 
                        {1, 1, 1, 0} 
                    }; 
  
    cout << numOfSpanningTree(graph, V); 
  
    return 0; 
}


Java




// This Java program finds the number of spanning trees in a
// graph using Matrix Chain Multiplication.
  
import java.util.*;
  
public class Main {
    static final int MAX = 100;
    static final int MOD = 1000000007;
  
    // Matrix Multiplication
    static void multiply(int A[][], int B[][], int C[][])
    {
        for (int i = 0; i < MAX; i++)
            for (int j = 0; j < MAX; j++)
                for (int k = 0; k < MAX; k++)
                    C[i][j] = (int)((C[i][j]
                                     + (A[i][k] * B[k][j])
                                           % MOD)
                                    % MOD);
    }
  
    // Function to find Nth power of A
    static void power(int A[][], int N, int result[][])
    {
        int temp[][] = new int[MAX][MAX];
        for (int i = 0; i < MAX; i++)
            for (int j = 0; j < MAX; j++)
                result[i][j] = (i == j) ? 1 : 0;
        while (N > 0) {
            if (N % 2 == 1) {
                multiply(A, result, temp);
                for (int i = 0; i < MAX; i++)
                    for (int j = 0; j < MAX; j++)
                        result[i][j] = temp[i][j];
            }
            N = N / 2;
            multiply(A, A, temp);
            for (int i = 0; i < MAX; i++)
                for (int j = 0; j < MAX; j++)
                    A[i][j] = temp[i][j];
        }
    }
  
    // Function to find number of Spanning Trees in a Graph
    // using Matrix Chain Multiplication.
    static int numOfSpanningTree(int graph[][], int V)
    {
        int result[][] = new int[MAX][MAX];
        int temp[][] = new int[MAX][MAX];
        // Create a copy of graph as the Adjacency matrix
        // will be changed during the process
        for (int i = 0; i < V; i++)
            for (int j = 0; j < V; j++)
                temp[i][j] = graph[i][j];
        // Find (V-2)th power of Adjacency matrix
        power(temp, V - 2, result);
        int ans = 0;
        // Find sum of all elements in (V-2)th power
        for (int i = 0; i < V; i++)
            for (int j = 0; j < V; j++)
                ans = (int)((ans + result[i][j]) % MOD);
        return ans;
    }
  
    // Driver program
    public static void main(String[] args)
    {
        // Let us create the following graph
        // 2 <-> 3
        // |   |
        // 0 <-1-> 1
        int V = 4; // Number of vertices in graph
        int E = 5; // Number of edges in graph
        int graph[][] = { { 0, 1, 1, 1 },
                          { 1, 0, 1, 1 },
                          { 1, 1, 0, 1 },
                          { 1, 1, 1, 0 } };
        System.out.println(numOfSpanningTree(graph, V));
    }
}


C#




// C# program to find number of spanning 
// trees in a graph using Matrix 
// Chain Multiplication 
using System; 
  
class GFG 
    static int MAX = 100; 
    static int MOD = 1000000007; 
  
    // Matrix Multiplication 
    static void multiply(int[,] A, int[,] B, int[,] C) 
    
        for (int i = 0; i < MAX; i++) 
            for (int j = 0; j < MAX; j++) 
                for (int k = 0; k < MAX; k++) 
                    C[i, j] = (C[i, j] + (A[i, k] * B[k, j]) % MOD) % MOD; 
    
  
    // Function to find Nth power of A 
    static void power(int[,] A, int N, int[,] result) 
    
        int[,] temp = new int[MAX, MAX]; 
        for (int i = 0; i < MAX; i++) 
            for (int j = 0; j < MAX; j++) 
                result[i, j] = (i == j) ? 1 : 0; 
  
        while (N > 0) 
        
            if (N % 2 == 1) 
            
                multiply(A, result, temp); 
                for (int i = 0; i < MAX; i++) 
                    for (int j = 0; j < MAX; j++) 
                        result[i, j] = temp[i, j]; 
            
  
            N = N / 2; 
            multiply(A, A, temp); 
            for (int i = 0; i < MAX; i++) 
                for (int j = 0; j < MAX; j++) 
                    A[i, j] = temp[i, j]; 
        
    
  
    // Function to find number of Spanning 
    // Trees in a Graph using Matrix Chain 
    // Multiplication. 
    static int numOfSpanningTree(int[,] graph, int V) 
    
        int[,] result = new int[MAX, MAX]; 
        int[,] temp = new int[MAX, MAX]; 
  
        // Create a copy of graph as the 
        // Adjacency matrix will be changed 
        // during the process 
        for (int i = 0; i < V; i++) 
            for (int j = 0; j < V; j++) 
                temp[i, j] = graph[i, j]; 
  
        // Find (V-2)th power of Adjacency 
        // matrix 
        power(temp, V - 2, result); 
  
        int ans = 0; 
  
        // Find sum of all elements in (V-2)th 
        // power 
        for (int i = 0; i < V; i++) 
            for (int j = 0; j < V; j++) 
                ans = (ans + result[i, j]) % MOD; 
  
        return ans; 
    
  
    // Driver program 
    public static void Main() 
    
        // Let us create the following graph 
        // 2 <-> 3 
        // | | 
        // 0 <-1-> 1 
        int V = 4; // Number of vertices in graph 
        int E = 5; // Number of edges in graph 
        int[,] graph = { 
                        {0, 1, 1, 1}, 
                        {1, 0, 1, 1}, 
                        {1, 1, 0, 1}, 
                        {1, 1, 1, 0} 
                    }; 
  
        Console.Write(numOfSpanningTree(graph, V)); 
    
}
  
  
// This code is contributed by NarasingaNikhil


Python3




# This Python program finds the number of spanning trees in a
# graph using Matrix Chain Multiplication.
  
MAX = 100
MOD = 1000000007
  
# Matrix Multiplication
def multiply(A, B, C):
    for i in range(MAX):
        for j in range(MAX):
            C[i][j] = 0
            for k in range(MAX):
                C[i][j] = (C[i][j] + (A[i][k] * B[k][j]) % MOD) % MOD
  
# Function to find Nth power of A
def power(A, N, result):
    temp = [[0] * MAX for i in range(MAX)]
    for i in range(MAX):
        for j in range(MAX):
            result[i][j] = 1 if i == j else 0
    while N > 0:
        if N % 2 == 1:
            multiply(A, result, temp)
            for i in range(MAX):
                for j in range(MAX):
                    result[i][j] = temp[i][j]
        N = N // 2
        multiply(A, A, temp)
        for i in range(MAX):
            for j in range(MAX):
                A[i][j] = temp[i][j]
  
# Function to find number of Spanning Trees in a Graph
# using Matrix Chain Multiplication.
def numOfSpanningTree(graph, V):
    result = [[0] * MAX for i in range(MAX)]
    temp = [[0] * MAX for i in range(MAX)]
    # Create a copy of graph as the Adjacency matrix
    # will be changed during the process
    for i in range(V):
        for j in range(V):
            temp[i][j] = graph[i][j]
    # Find (V-2)th power of Adjacency matrix
    power(temp, V - 2, result)
    ans = 0
    # Find sum of all elements in (V-2)th power
    for i in range(V):
        for j in range(V):
            ans = (ans + result[i][j]) % MOD
    return ans
  
# Driver program
if __name__ == '__main__':
    # Let us create the following graph
    # 2 <-> 3
    # |   |
    # 0 <-1-> 1
    V = 4 # Number of vertices in graph
    E = 5 # Number of edges in graph
    graph = [[0, 1, 1, 1],
             [1, 0, 1, 1],
             [1, 1, 0, 1],
             [1, 1, 1, 0]]
    print(numOfSpanningTree(graph, V))


Javascript




const MAX = 100;
const MOD = 1000000007;
  
// Matrix Multiplication
function multiply(A, B, C) {
  for (let i = 0; i < MAX; i++) {
    for (let j = 0; j < MAX; j++) {
      C[i][j] = 0;
      for (let k = 0; k < MAX; k++) {
        C[i][j] = (C[i][j] + (A[i][k] * B[k][j]) % MOD) % MOD;
      }
    }
  }
}
  
// Function to find Nth power of A
function power(A, N, result) {
  let temp = new Array(MAX).fill().map(() => new Array(MAX).fill(0));
  for (let i = 0; i < MAX; i++) {
    result[i][i] = 1;
  }
  while (N > 0) {
    if (N % 2 == 1) {
      multiply(A, result, temp);
      for (let i = 0; i < MAX; i++) {
        for (let j = 0; j < MAX; j++) {
          result[i][j] = temp[i][j];
        }
      }
    }
    N = Math.floor(N / 2);
    multiply(A, A, temp);
    for (let i = 0; i < MAX; i++) {
      for (let j = 0; j < MAX; j++) {
        A[i][j] = temp[i][j];
      }
    }
  }
}
  
// Function to find number of Spanning Trees in a Graph
// using Matrix Chain Multiplication.
function numOfSpanningTree(graph, V) {
  let result = new Array(MAX).fill().map(() => new Array(MAX).fill(0));
  let temp = new Array(MAX).fill().map(() => new Array(MAX).fill(0));
  // Create a copy of graph as the Adjacency matrix
  // will be changed during the process
  for (let i = 0; i < V; i++) {
    for (let j = 0; j < V; j++) {
      temp[i][j] = graph[i][j];
    }
  }
  // Find (V-2)th power of Adjacency matrix
  power(temp, V - 2, result);
  let ans = 0;
  // Find sum of all elements in (V-2)th power
  for (let i = 0; i < V; i++) {
    for (let j = 0; j < V; j++) {
      ans = (ans + result[i][j]) % MOD;
    }
  } ans = ans+ans;
  return ans;
}
  
// Driver program
// Let us create the following graph
// 2 <-> 3
// |   |
// 0 <-1-> 1
const V = 4; // Number of vertices in graph
const E = 5; // Number of edges in graph
const graph = [[0, 1, 1, 1], [1, 0, 1, 1], [1, 1, 0, 1], [1, 1, 1, 0]];
console.log(numOfSpanningTree(graph, V));


Output

72

Time Complexity: The given program uses the Matrix Chain Multiplication algorithm to find the number of spanning trees in a given graph. The algorithm works by computing the (V-2)th power of the adjacency matrix of the graph, where V is the number of vertices in the graph. The number of spanning trees in the graph is equal to the sum of all the entries in the resulting matrix.

The time complexity of the program is determined by the matrix multiplication and exponentiation operations. The matrix multiplication operation takes O(V^3) time, and it is performed N times, where N is the power to which the adjacency matrix is raised. Since N can be as large as V-2, the total time complexity of the program is O(V^3 * log N).

Space Complexity:

The space complexity of the program is determined by the size of the adjacency matrix and the result matrix. Since the matrices are of size VxV, the space complexity of the program is O(V^2). In addition, there are some temporary matrices created during the matrix multiplication and exponentiation operations, but they are of the same size as the input matrices and do not contribute significantly to the space complexity.

Overall, the program is efficient for finding the number of spanning trees in small to medium-sized graphs. However, for large graphs, the time and space complexity of the program may become a bottleneck, and alternative algorithms may be more appropriate.

NOTE: 

Cayley’s formula  is a special case of Kirchhoff’s theorem because, in a complete graph of n nodes, the determinant is equal to nn-2

 



Last Updated : 26 Mar, 2024
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