Matrix Exponentiation is a technique used to calculate a matrix raised to a power efficiently, that is in logN time. It is mostly used for solving problems related to linear recurrences.
Idea behind Matrix Exponentiation:
Similar to Binary Exponentiation which is used to calculate a number raised to a power, Matrix Exponentiation is used to calculate a matrix raised to a power efficiently.
Let us understand Matrix Exponentiation with the help of an example:
We can calculate matrix M^(N – 2) in logN time using Matrix Exponentiation. The idea is same as Binary Exponentiation:
When we are calculating (MN), we can have 3 possible positive values of N:
- Case 1: If N = 0, whatever be the value of M, our result will be Identity Matrix I.
- Case 2: If N is an even number, then instead of calculating (MN), we can calculate ((M2)N/2) and the result will be same.
- Case 3: If N is an odd number, then instead of calculating (MN), we can calculate (M * (M(N – 1)/2)2).
Use Cases of Matrix Exponentiation:
Finding nth Fibonacci Number:
The recurrence relation for Fibonacci Sequence is F(n) = F(n – 1) + F(n – 2) starting with F(0) = 0 and F(1) = 1.
Below is the implementation of above idea:
C++ // C++ Program to find the Nth fibonacci number using // Matrix Exponentiation #include <bits/stdc++.h> using namespace std; int MOD = 1e9 + 7; // function to multiply two 2x2 Matrices void multiply(vector<vector<long long> >& A, vector<vector<long long> >& B) { // Matrix to store the result vector<vector<long long> > C(2, vector<long long>(2)); // Matrix Multiply C[0][0] = (A[0][0] * B[0][0] + A[0][1] * B[1][0]) % MOD; C[0][1] = (A[0][0] * B[0][1] + A[0][1] * B[1][1]) % MOD; C[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % MOD; C[1][1] = (A[1][0] * B[0][1] + A[1][1] * B[1][1]) % MOD; // Copy the result back to the first matrix A[0][0] = C[0][0]; A[0][1] = C[0][1]; A[1][0] = C[1][0]; A[1][1] = C[1][1]; } // Function to find (Matrix M ^ expo) vector<vector<long long> > power(vector<vector<long long> > M, int expo) { // Initialize result with identity matrix vector<vector<long long> > ans = { { 1, 0 }, { 0, 1 } }; // Fast Exponentiation while (expo) { if (expo & 1) multiply(ans, M); multiply(M, M); expo >>= 1; } return ans; } // function to find the nth fibonacci number int nthFibonacci(int n) { // base case if (n == 0 || n == 1) return 1; vector<vector<long long> > M = { { 1, 1 }, { 1, 0 } }; // Matrix F = {{f(0), 0}, {f(1), 0}}, where f(0) and // f(1) are first two terms of fibonacci sequence vector<vector<long long> > F = { { 1, 0 }, { 0, 0 } }; // Multiply matrix M (n - 1) times vector<vector<long long> > res = power(M, n - 1); // Multiply Resultant with Matrix F multiply(res, F); return res[0][0] % MOD; } int main() { // Sample Input int n = 3; // Print nth fibonacci number cout << nthFibonacci(n) << endl; } // Java Program to find the Nth fibonacci number using // Matrix Exponentiation import java.util.*; public class GFG { static final int MOD = 1000000007; // Function to multiply two 2x2 matrices public static void multiply(long[][] A, long[][] B) { // Matrix to store the result long[][] C = new long[2][2]; // Matrix multiplication C[0][0] = (A[0][0] * B[0][0] + A[0][1] * B[1][0]) % MOD; C[0][1] = (A[0][0] * B[0][1] + A[0][1] * B[1][1]) % MOD; C[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % MOD; C[1][1] = (A[1][0] * B[0][1] + A[1][1] * B[1][1]) % MOD; // Copy the result back to the first matrix for (int i = 0; i < 2; i++) { for (int j = 0; j < 2; j++) { A[i][j] = C[i][j]; } } } // Function to find (Matrix M ^ expo) public static long[][] power(long[][] M, int expo) { // Initialize result with identity matrix long[][] ans = { { 1, 0 }, { 0, 1 } }; // Fast exponentiation while (expo > 0) { if ((expo & 1) != 0) { multiply(ans, M); } multiply(M, M); expo >>= 1; } return ans; } // Function to find the nth Fibonacci number public static int nthFibonacci(int n) { // Base case if (n == 0 || n == 1) { return 1; } long[][] M = { { 1, 1 }, { 1, 0 } }; // F(0) = 1, F(1) = 1 long[][] F = { { 1, 0 }, { 0, 0 } }; // Multiply matrix M (n - 1) times long[][] res = power(M, n - 1); // Multiply resultant with matrix F multiply(res, F); return (int)((res[0][0]) % MOD); } public static void main(String[] args) { // Sample input int n = 3; // Print nth Fibonacci number System.out.println(nthFibonacci(n)); } } # Python Program to find the Nth fibonacci number using # Matrix Exponentiation MOD = 10**9 + 7 # function to multiply two 2x2 Matrices def multiply(A, B): # Matrix to store the result C = [[0, 0], [0, 0]] # Matrix Multiply C[0][0] = (A[0][0] * B[0][0] + A[0][1] * B[1][0]) % MOD C[0][1] = (A[0][0] * B[0][1] + A[0][1] * B[1][1]) % MOD C[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % MOD C[1][1] = (A[1][0] * B[0][1] + A[1][1] * B[1][1]) % MOD # Copy the result back to the first matrix A[0][0] = C[0][0] A[0][1] = C[0][1] A[1][0] = C[1][0] A[1][1] = C[1][1] # Function to find (Matrix M ^ expo) def power(M, expo): # Initialize result with identity matrix ans = [[1, 0], [0, 1]] # Fast Exponentiation while expo: if expo & 1: multiply(ans, M) multiply(M, M) expo >>= 1 return ans def nthFibonacci(n): # Base case if n == 0 or n == 1: return 1 M = [[1, 1], [1, 0]] # F(0) = 0, F(1) = 1 F = [[1, 0], [0, 0]] # Multiply matrix M (n - 1) times res = power(M, n - 1) # Multiply Resultant with Matrix F multiply(res, F) return res[0][0] % MOD # Sample Input n = 3 # Print the nth fibonacci number print(nthFibonacci(n))
// C# Program to find the Nth fibonacci number using // Matrix Exponentiation using System; using System.Collections.Generic; public class GFG { static int MOD = 1000000007; // function to multiply two 2x2 Matrices public static void Multiply(long[][] A, long[][] B) { // Matrix to store the result long[][] C = new long[2][] { new long[2], new long[2] }; // Matrix Multiply C[0][0] = (A[0][0] * B[0][0] + A[0][1] * B[1][0]) % MOD; C[0][1] = (A[0][0] * B[0][1] + A[0][1] * B[1][1]) % MOD; C[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % MOD; C[1][1] = (A[1][0] * B[0][1] + A[1][1] * B[1][1]) % MOD; // Copy the result back to the first matrix A[0][0] = C[0][0]; A[0][1] = C[0][1]; A[1][0] = C[1][0]; A[1][1] = C[1][1]; } // Function to find (Matrix M ^ expo) public static long[][] Power(long[][] M, int expo) { // Initialize result with identity matrix long[][] ans = new long[2][] { new long[] { 1, 0 }, new long[] { 0, 1 } }; // Fast Exponentiation while (expo > 0) { if ((expo & 1) > 0) Multiply(ans, M); Multiply(M, M); expo >>= 1; } return ans; } // function to find the nth fibonacci number public static int NthFibonacci(int n) { // base case if (n == 0 || n == 1) return 1; long[][] M = new long[2][] { new long[] { 1, 1 }, new long[] { 1, 0 } }; // F(0) = 0, F(1) = 1 long[][] F = new long[2][] { new long[] { 1, 0 }, new long[] { 0, 0 } }; // Multiply matrix M (n - 1) times long[][] res = Power(M, n - 1); // Multiply Resultant with Matrix F Multiply(res, F); return (int)((res[0][0] % MOD)); } public static void Main(string[] args) { // Sample Input int n = 3; // Print nth fibonacci number Console.WriteLine(NthFibonacci(n)); } } const MOD = 1e9 + 7; // Function to multiply two 2x2 matrices function multiply(A, B) { // Matrix to store the result const C = [ [0, 0], [0, 0] ]; // Matrix Multiply C[0][0] = (A[0][0] * B[0][0] + A[0][1] * B[1][0]) % MOD; C[0][1] = (A[0][0] * B[0][1] + A[0][1] * B[1][1]) % MOD; C[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % MOD; C[1][1] = (A[1][0] * B[0][1] + A[1][1] * B[1][1]) % MOD; // Copy the result back to the first matrix A[0][0] = C[0][0]; A[0][1] = C[0][1]; A[1][0] = C[1][0]; A[1][1] = C[1][1]; } // Function to find (Matrix M ^ expo) function power(M, expo) { // Initialize result with identity matrix const ans = [ [1, 0], [0, 1] ]; // Fast Exponentiation while (expo) { if (expo & 1) multiply(ans, M); multiply(M, M); expo >>= 1; } return ans; } // Function to find the nth fibonacci number function nthFibonacci(n) { // Base case if (n === 0 || n === 1) return 1; const M = [ [1, 1], [1, 0] ]; // F(0) = 0, F(1) = 1 const F = [ [1, 0], [0, 0] ]; // Multiply matrix M (n - 1) times const res = power(M, n - 1); // Multiply Resultant with Matrix F multiply(res, F); return res[0][0] % MOD; } // Sample Input const n = 3; // Print nth fibonacci number console.log(nthFibonacci(n)); Time Complexity: O(logN), because fast exponentiation takes O(logN) time.
Auxiliary Space: O(1)
Finding nth Tribonacci Number:
The recurrence relation for Tribonacci Sequence is T(n) = T(n – 1) + T(n – 2) + T(n – 3) starting with T(0) = 0, T(1) = 1 and T(2) = 1.
Below is the implementation of above idea:
C++ // C++ Program to find the nth tribonacci number #include <bits/stdc++.h> using namespace std; // Function to multiply two 3x3 matrices void multiply(vector<vector<long long> >& A, vector<vector<long long> >& B) { // Matrix to store the result vector<vector<long long> > C(3, vector<long long>(3)); for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { for (int k = 0; k < 3; k++) { C[i][j] = (C[i][j] + ((A[i][k]) * (B[k][j]))); } } } // Copy the result back to the first matrix for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { A[i][j] = C[i][j]; } } } // Function to calculate (Matrix M) ^ expo vector<vector<long long> > power(vector<vector<long long> > M, int expo) { // Initialize result with identity matrix vector<vector<long long> > ans = { { 1, 0, 0 }, { 0, 1, 0 }, { 0, 0, 1 } }; // Fast Exponentiation while (expo) { if (expo & 1) multiply(ans, M); multiply(M, M); expo >>= 1; } return ans; } // function to return the Nth tribonacci number long long tribonacci(int n) { // base condition if (n == 0 || n == 1) return n; // Matrix M to generate the next tribonacci number vector<vector<long long> > M = { { 1, 1, 1 }, { 1, 0, 0 }, { 0, 1, 0 } }; // Since first 3 number of tribonacci series are: // trib(0) = 0 // trib(1) = 1 // trib(2) = 1 // F = {{trib(2), 0, 0}, {trib(1), 0, 0}, {trib(0), 0, // 0}} vector<vector<long long> > F = { { 1, 0, 0 }, { 1, 0, 0 }, { 0, 0, 0 } }; vector<vector<long long> > res = power(M, n - 2); multiply(res, F); return res[0][0]; } int main() { // Sample Input int n = 4; // Function call cout << tribonacci(n); return 0; } // Java program to find nth tribonacci number import java.util.*; public class Main { // Function to multiply two 3x3 matrices static void multiply(long[][] A, long[][] B) { // Matrix to store the result long[][] C = new long[3][3]; for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { for (int k = 0; k < 3; k++) { C[i][j] += (A[i][k] * B[k][j]); } } } // Copy the result back to the first matrix for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { A[i][j] = C[i][j]; } } } // Function to calculate (Matrix M) ^ expo static long[][] power(long[][] M, int expo) { // Initialize result with identity matrix long[][] ans = { { 1, 0, 0 }, { 0, 1, 0 }, { 0, 0, 1 } }; // Fast Exponentiation while (expo > 0) { if ((expo & 1) == 1) multiply(ans, M); multiply(M, M); expo >>= 1; } return ans; } // function to return the Nth tribonacci number static long tribonacci(int n) { // base condition if (n == 0 || n == 1) return n; // Matrix M to generate the next tribonacci number long[][] M = { { 1, 1, 1 }, { 1, 0, 0 }, { 0, 1, 0 } }; // Since first 3 number of tribonacci series are: // trib(0) = 0 // trib(1) = 1 // trib(2) = 1 // F = {{trib(2), 0, 0}, {trib(1), 0, 0}, {trib(0), // 0, 0}} long[][] F = { { 1, 0, 0 }, { 1, 0, 0 }, { 0, 0, 0 } }; long[][] res = power(M, n - 2); multiply(res, F); return res[0][0]; } public static void main(String[] args) { // Sample Input int n = 4; // Function call System.out.println(tribonacci(n)); } } # Python3 program to find nth tribonacci number # Function to multiply two 3x3 matrices def multiply(A, B): # Matrix to store the result C = [[0]*3 for _ in range(3)] for i in range(3): for j in range(3): for k in range(3): C[i][j] += A[i][k] * B[k][j] # Copy the result back to the first matrix A for i in range(3): for j in range(3): A[i][j] = C[i][j] # Function to calculate (Matrix M) ^ expo def power(M, expo): # Initialize result with identity matrix ans = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] # Fast Exponentiation while expo > 0: if expo & 1: multiply(ans, M) multiply(M, M) expo >>= 1 return ans # Function to return the Nth tribonacci number def tribonacci(n): # Base condition if n == 0 or n == 1: return n # Matrix M to generate the next tribonacci number M = [[1, 1, 1], [1, 0, 0], [0, 1, 0]] # Since first 3 numbers of tribonacci series are: # trib(0) = 0 # trib(1) = 1 # trib(2) = 1 # F = [[trib(2), 0, 0], [trib(1), 0, 0], [trib(0), 0, 0]] F = [[1, 0, 0], [1, 0, 0], [0, 0, 0]] res = power(M, n - 2) multiply(res, F) return res[0][0] # Sample Input n = 4 # Function call print(tribonacci(n))
// C# program to find nth tribonacci number using System; public class MainClass { // Function to multiply two 3x3 matrices static void Multiply(long[][] A, long[][] B) { // Matrix to store the result long[][] C = new long[3][]; for (int i = 0; i < 3; i++) { C[i] = new long[3]; for (int j = 0; j < 3; j++) { for (int k = 0; k < 3; k++) { C[i][j] += A[i][k] * B[k][j]; } } } // Copy the result back to the first matrix A for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { A[i][j] = C[i][j]; } } } // Function to calculate (Matrix M) ^ expo static long[][] Power(long[][] M, int expo) { // Initialize result with identity matrix long[][] ans = new long[3][]; for (int i = 0; i < 3; i++) { ans[i] = new long[3]; ans[i][i] = 1; // Diagonal elements are 1 } // Fast Exponentiation while (expo > 0) { if ((expo & 1) == 1) Multiply(ans, M); Multiply(M, M); expo >>= 1; } return ans; } // Function to return the Nth tribonacci number static long Tribonacci(int n) { // Base condition if (n == 0 || n == 1) return n; // Matrix M to generate the next tribonacci number long[][] M = new long[][] { new long[] { 1, 1, 1 }, new long[] { 1, 0, 0 }, new long[] { 0, 1, 0 } }; // Since first 3 numbers of tribonacci series are: // trib(0) = 0 // trib(1) = 1 // trib(2) = 1 // F = [[trib(2), 0, 0], [trib(1), 0, 0], [trib(0), // 0, 0]] long[][] F = new long[][] { new long[] { 1, 0, 0 }, new long[] { 1, 0, 0 }, new long[] { 0, 0, 0 } }; long[][] res = Power(M, n - 2); Multiply(res, F); return res[0][0]; } public static void Main(string[] args) { // Sample Input int n = 4; // Function call Console.WriteLine(Tribonacci(n)); } } // JavaScript Program to find nth tribonacci number // Function to multiply two 3x3 matrices function multiply(A, B) { // Matrix to store the result let C = [ [0, 0, 0], [0, 0, 0], [0, 0, 0] ]; for (let i = 0; i < 3; i++) { for (let j = 0; j < 3; j++) { for (let k = 0; k < 3; k++) { C[i][j] += A[i][k] * B[k][j]; } } } // Copy the result back to the first matrix A for (let i = 0; i < 3; i++) { for (let j = 0; j < 3; j++) { A[i][j] = C[i][j]; } } } // Function to calculate (Matrix M) ^ expo function power(M, expo) { // Initialize result with identity matrix let ans = [ [1, 0, 0], [0, 1, 0], [0, 0, 1] ]; // Fast Exponentiation while (expo > 0) { if (expo & 1) multiply(ans, M); multiply(M, M); expo >>= 1; } return ans; } // Function to return the Nth tribonacci number function tribonacci(n) { // base condition if (n === 0 || n === 1) return n; // Matrix M to generate the next tribonacci number let M = [ [1, 1, 1], [1, 0, 0], [0, 1, 0] ]; // Since first 3 numbers of tribonacci series are: // trib(0) = 0 // trib(1) = 1 // trib(2) = 1 // F = [[trib(2), 0, 0], [trib(1), 0, 0], [trib(0), 0, 0]] let F = [ [1, 0, 0], [1, 0, 0], [0, 0, 0] ]; let res = power(M, n - 2); multiply(res, F); return res[0][0]; } // Main function function main() { // Sample Input let n = 4; // Function call console.log(tribonacci(n)); } // Call the main function main(); Time Complexity: O(logN), because fast exponentiation takes O(logN) time.
Auxiliary Space: O(1)
Applications of Matrix Exponentiation:
Matrix Exponentiation has a variety of applications. Some of them are:
- Any linear recurrence relation, such as the Fibonacci Sequence, Tribonacci Sequence or linear homogeneous recurrence relations with constant coefficients, can be solved using matrix exponentiation.
- The RSA encryption algorithm involves exponentiation of large numbers, which can be efficiently handled using matrix exponentiation techniques.
- Dynamic programming problems, especially those involving linear recurrence relations, can be optimized using matrix exponentiation to reduce time complexity.
- Matrix exponentiation is used in number theory problems involving modular arithmetic, such as finding large powers of numbers modulo some value efficiently.
Advantages of Matrix Exponentiation:
Advantages of using Matrix Exponentiation are:
- Matrix Exponentiation helps in finding Nth term of linear recurrence relations like Fibonacci or Tribonacci Series in log(N) time. This makes it much faster for large values of N.
- It requires O(1) space if we use the iterative approach as it requires constant amount of extra space.
- Large numbers can be handled without integer overflow using modulo operations.
Disadvantages of Matrix Exponentiation:
Disadvantages of using Matrix Exponentiation are:
- Matrix Exponentiation is more complex than other iterative or recursive methods. Hence, it is harder to debug.
- Initial conditions should be handled carefully to avoid incorrect results.
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