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Matrix Exponentiation

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This is one of the most used techniques in competitive programming. Let us first consider below simple question.

What is the minimum time complexity to find n’th Fibonacci Number? 
We can find n’th Fibonacci Number in O(Log n) time using Matrix Exponentiation. Refer method 4 of this for details. In this post, a general implementation of Matrix Exponentiation is discussed. 

For solving the matrix exponentiation we are assuming a
linear recurrence equation like below:
F(n) = a*F(n-1) + b*F(n-2) + c*F(n-3) for n >= 3
. . . . . Equation (1)
where a, b and c are constants.
For this recurrence relation, it depends on three previous values.
Now we will try to represent Equation (1) in terms of the matrix.
[First Matrix] = [Second matrix] * [Third Matrix]
| F(n) | = Matrix 'C' * | F(n-1) |
| F(n-1) | | F(n-2) |
| F(n-2) | | F(n-3) |

Dimension of the first matrix is 3 x 1 .
Dimension of the third matrix is also 3 x 1.
So the dimension of the second matrix must be 3 x 3
[For multiplication rule to be satisfied.]
Now we need to fill the Matrix 'C'.
So according to our equation.
F(n) = a*F(n-1) + b*F(n-2) + c*F(n-3)
F(n-1) = F(n-1)
F(n-2) = F(n-2)
C = [a b c
1 0 0
0 1 0]
Now the relation between matrix becomes :
[First Matrix] [Second matrix] [Third Matrix]
| F(n) | = | a b c | * | F(n-1) |
| F(n-1) | | 1 0 0 | | F(n-2) |
| F(n-2) | | 0 1 0 | | F(n-3) |
Lets assume the initial values for this case :-
F(0) = 0
F(1) = 1
F(2) = 1
So, we need to get F(n) in terms of these values.
So, for n = 3 Equation (1) changes to
| F(3) | = | a b c | * | F(2) |
| F(2) | | 1 0 0 | | F(1) |
| F(1) | | 0 1 0 | | F(0) |
Now similarly for n = 4
| F(4) | = | a b c | * | F(3) |
| F(3) | | 1 0 0 | | F(2) |
| F(2) | | 0 1 0 | | F(1) |
- - - - 2 times - - -
| F(4) | = | a b c | * | a b c | * | F(2) |
| F(3) | | 1 0 0 | | 1 0 0 | | F(1) |
| F(2) | | 0 1 0 | | 0 1 0 | | F(0) |
So for n, the Equation (1) changes to
- - - - - - - - n -2 times - - - - -
| F(n) | = | a b c | * | a b c | * ... * | a b c | * | F(2) |
| F(n-1) | | 1 0 0 | | 1 0 0 | | 1 0 0 | | F(1) |
| F(n-2) | | 0 1 0 | | 0 1 0 | | 0 1 0 | | F(0) |
| F(n) | = [ | a b c | ] ^ (n-2) * | F(2) |
| F(n-1) | [ | 1 0 0 | ] | F(1) |
| F(n-2) | [ | 0 1 0 | ] | F(0) |

So we can simply multiply our Second matrix n-2 times and then multiply it with the third matrix to get the result. Multiplication can be done in (log n) time using Divide and Conquer algorithm for power (See this or this)

Let us consider the problem of finding n’th term of a series defined using below recurrence. 

n'th term,
F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
Base Cases :
F(0) = 0, F(1) = 1, F(2) = 1

We can find n’th term using following : 

Putting a = 1, b = 1 and c = 1 in above formula
| F(n) | = [ | 1 1 1 | ] ^ (n-2) * | F(2) |
| F(n-1) | [ | 1 0 0 | ] | F(1) |
| F(n-2) | [ | 0 1 0 | ] | F(0) |

Recommended Practice

Below is the implementation of above idea.

C++




// C++ program to find value of f(n) where f(n)
// is defined as
// F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
#include <bits/stdc++.h>
using namespace std;
 
// A utility function to multiply two matrices
// a[][] and b[][]. Multiplication result is
// stored back in b[][]
void multiply(int a[3][3], int b[3][3])
{
    // Creating an auxiliary matrix to store elements
    // of the multiplication matrix
    int mul[3][3];
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            mul[i][j] = 0;
            for (int k = 0; k < 3; k++)
                mul[i][j] += a[i][k] * b[k][j];
        }
    }
 
    // storing the multiplication result in a[][]
    for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++)
            a[i][j] = mul[i][j]; // Updating our matrix
}
 
// Function to compute F raise to power n-2.
int power(int F[3][3], int n)
{
    int M[3][3] = { { 1, 1, 1 }, { 1, 0, 0 }, { 0, 1, 0 } };
 
    // Multiply it with initial values i.e with
    // F(0) = 0, F(1) = 1, F(2) = 1
    while (n) {
        if (n & 1) {
            multiply(F, M);
        }
        multiply(M, M);
        n >>= 1;
    }
 
    // Multiply it with initial values i.e with
    // F(0) = 0, F(1) = 1, F(2) = 1
    return F[0][0] + F[0][1];
}
 
// Return n'th term of a series defined using below
// recurrence relation.
// f(n) is defined as
// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
// Base Cases :
// f(0) = 0, f(1) = 1, f(2) = 1
int findNthTerm(int n)
{
    // Instead of taking F[][] as a 3X1 2D array we took
    // F[][] as 3X3 array with only the first column filled
    // and the rest cells are filled with zeros
    int F[3][3] = { { 1, 0, 0 }, { 1, 0, 0 }, { 0, 0, 0 } };
 
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return 1;
 
    return power(F, n - 2);
}
 
// Driver code
int main()
{
    int n = 5;
 
    cout << "F(5) is " << findNthTerm(n);
 
    return 0;
}


Java




// JAVA program to find value of f(n) where
// f(n) is defined as
// F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
import java.io.*;
 
class GFG {
 
    // A utility function to multiply two matrices
    // a[][] and b[][]. Multiplication result is
    // stored back in b[][]
    static void multiply(int a[][], int b[][])
    {
        // Creating an auxiliary matrix to store elements
        // of the multiplication matrix
        int[][] mul = new int[3][3];
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                mul[i][j] = 0;
                for (int k = 0; k < 3; k++)
                    mul[i][j] += a[i][k] * b[k][j];
            }
        }
 
        // storing the multiplication result in a[][]
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++)
                a[i][j] = mul[i][j]; // Updating our matrix
    }
 
    // Function to compute F raise to power n-2.
    static int power(int F[][], int n)
    {
        int M[][]
            = { { 1, 1, 1 }, { 1, 0, 0 }, { 0, 1, 0 } };
 
        // Multiply it with initial values i.e with
        // F(0) = 0, F(1) = 1, F(2) = 1
        while (n > 0) {
            if (n % 2 == 1) {
                multiply(F, M);
            }
            multiply(M, M);
            n >>= 1;
        }
 
        // Multiply it with initial values i.e with
        // F(0) = 0, F(1) = 1, F(2) = 1
        return F[0][0] + F[0][1];
    }
 
    // Return n'th term of a series defined using below
    // recurrence relation.
    // f(n) is defined as
    // f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
    // Base Cases :
    // f(0) = 0, f(1) = 1, f(2) = 1
    static int findNthTerm(int n)
    {
        // Instead of taking F[][] as a 3X1 2D array we took
        // F[][] as 3X3 array with only the first column
        // filled and the rest cells are filled with zeros
        int F[][]
            = { { 1, 0, 0 }, { 1, 0, 0 }, { 0, 0, 0 } };
 
        // Base cases
        if (n == 0)
            return 0;
        if (n == 1 || n == 2)
            return 1;
 
        return power(F, n - 2);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
 
        System.out.println("F(5) is " + findNthTerm(n));
    }
}
 
// This code is contributed by vt_m.


C#




// C# program to find value of f(n) where
// f(n) is defined as
// F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
using System;
 
class GFG {
 
    // A utility function to multiply two matrices
    // a[][] and b[][]. Multiplication result is
    // stored back in b[][]
    static void Multiply(int[, ] a, int[, ] b)
    {
        // Creating an auxiliary matrix to store elements
        // of the multiplication matrix
        int[, ] mul = new int[3, 3];
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                mul[i, j] = 0;
                for (int k = 0; k < 3; k++)
                    mul[i, j] += a[i, k] * b[k, j];
            }
        }
 
        // storing the multiplication result in a[][]
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++)
                a[i, j] = mul[i, j]; // Updating our matrix
    }
 
    // Function to compute F raise to power n-2.
    static int Power(int[, ] F, int n)
    {
        int[, ] M
            = { { 1, 1, 1 }, { 1, 0, 0 }, { 0, 1, 0 } };
 
        // Multiply it with initial values i.e with
        // F(0) = 0, F(1) = 1, F(2) = 1
        while (n > 0) {
            if (n % 2 == 1) {
                Multiply(F, M);
            }
            Multiply(M, M);
            n >>= 1;
        }
 
        // Multiply it with initial values i.e with
        // F(0) = 0, F(1) = 1, F(2) = 1
        return F[0, 0] + F[0, 1];
    }
 
    // Return n'th term of a series defined using below
    // recurrence relation.
    // f(n) is defined as
    // f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
    // Base Cases :
    // f(0) = 0, f(1) = 1, f(2) = 1
    static int findNthTerm(int n)
    {
        // Instead of taking F[,] as a 3X1 2D array we took
        // F[,] as 3X3 array with only the first column
        // filled and the rest cells are filled with zeros
        int[, ] F
            = { { 1, 0, 0 }, { 1, 0, 0 }, { 0, 0, 0 } };
 
        // Base cases
        if (n == 0)
            return 0;
        if (n == 1 || n == 2)
            return 1;
 
        return Power(F, n - 2);
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
 
        Console.WriteLine("F(5) is " + findNthTerm(n));
    }
}
 
// This code is contributed by vt_m.


Javascript




// A utility function to multiply two matrices
// a[][] and b[][]. Multiplication result is
// stored back in b[][]
function multiply(a, b) {
    // Creating an auxiliary matrix to store elements
    // of the multiplication matrix
    let mul = Array.from({ length: 3 }, () => Array(3).fill(0));
    for (let i = 0; i < 3; i++) {
        for (let j = 0; j < 3; j++) {
            mul[i][j] = 0;
            for (let k = 0; k < 3; k++) {
                mul[i][j] += a[i][k] * b[k][j];
            }
        }
    }
 
    // storing the multiplication result in a[][]
    for (let i = 0; i < 3; i++) {
        for (let j = 0; j < 3; j++) {
            a[i][j] = mul[i][j]; // Updating our matrix
        }
    }
}
 
// Function to compute F raise to power n-2.
function power(F, n) {
    let M = [[1, 1, 1], [1, 0, 0], [0, 1, 0]];
 
    // Multiply it with initial values i.e with
    // F(0) = 0, F(1) = 1, F(2) = 1
    while (n > 0) {
        if (n % 2 === 1) {
            multiply(F, M);
        }
        multiply(M, M);
        n >>= 1;
    }
 
    // Multiply it with initial values i.e with
    // F(0) = 0, F(1) = 1, F(2) = 1
    return F[0][0] + F[0][1];
}
 
// Return n'th term of a series defined using below
// recurrence relation.
// f(n) is defined as
// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
// Base Cases :
// f(0) = 0, f(1) = 1, f(2) = 1
function findNthTerm(n) {
    // Instead of taking F[][] as a 3X1 2D array we took
    // F[][] as 3X3 array with only the first column filled
    // and the rest cells are filled with zeros
    let F = [[1, 0, 0], [1, 0, 0], [0, 0, 0]];
 
    // Base cases
    if (n === 0) {
        return 0;
    }
    if (n === 1 || n === 2) {
        return 1;
    }
 
    return power(F, n - 2);
}
 
// Driver code
let n = 5;
console.log(`F(5) is ${findNthTerm(n)}`);


Python3




# Python3 program to find value of f(n)
# where f(n) is defined as
# F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
# Base Cases :
# F(0) = 0, F(1) = 1, F(2) = 1
 
# A utility function to multiply two matrices
# a[][] and b[][]. Multiplication result is
# stored back in b[][]
 
 
def multiply(a, b):
    # Creating an auxiliary matrix to store elements
    # of the multiplication matrix
    mul = [[0, 0, 0] for _ in range(3)]
    for i in range(3):
        for j in range(3):
            mul[i][j] = 0
            for k in range(3):
                mul[i][j] += a[i][k] * b[k][j]
 
    # storing the multiplication result in a[][]
    for i in range(3):
        for j in range(3):
            a[i][j] = mul[i][j]  # Updating our matrix
 
 
# Function to compute F raise to power n-2.
def power(F, n):
    M = [[1, 1, 1], [1, 0, 0], [0, 1, 0]]
 
    # Multiply it with initial values i.e with
    # F(0) = 0, F(1) = 1, F(2) = 1
    while n > 0:
        if n % 2 == 1:
            multiply(F, M)
        multiply(M, M)
        n >>= 1
 
    # Multiply it with initial values i.e with
    # F(0) = 0, F(1) = 1, F(2) = 1
    return F[0][0] + F[0][1]
 
 
# Return n'th term of a series defined using below
# recurrence relation.
# f(n) is defined as
# f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
# Base Cases :
# f(0) = 0, f(1) = 1, f(2) = 1
def findNthTerm(n):
    # Instead of taking F[][] as a 3X1 2D array we took
    # F[][] as 3X3 array with only the first column filled
    # and the rest cells are filled with zeros
    F = [[1, 0, 0], [1, 0, 0], [0, 0, 0]]
 
    # Base cases
    if n == 0:
        return 0
    if n == 1 or n == 2:
        return 1
 
    return power(F, n - 2)
 
 
# Driver code
n = 5
 
print("F(5) is",
      findNthTerm(n))
 
# This code is contributed by mits


Output

F(5) is 7

Time Complexity: O(logN)
Auxiliary Space: O(logN) 



Last Updated : 11 Mar, 2024
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