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 0 | ] ^ (n-2) * | F(2) |
| F(n-1) | [ | 1 0 0 | ] | F(1) |
| F(n-2) | [ | 0 1 0 | ] | F(0) |
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] = {{1,1,0}, {1,0,0}, {0,1,0}}; // Multiply it with initial values i.e with // F(0) = 0, F(1) = 1, F(2) = 1 if (n==1) return F[0][0] + F[0][1]; power(F, n/2); multiply(F, F); if (n%2 != 0) multiply(F, M); // 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) = 1int findNthTerm(int n){ int F[3][3] = {{1,1,0}, {1,0,0}, {0,1,0}} ; //Base cases if(n==0) return 0; if(n==1 || n==2) return 1; return power(F, n-2);}// Driver codeint 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) = 1import 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++) // Updating our matrix a[i][j] = mul[i][j]; } // 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 if (n == 1) return F[0][0] + F[0][1]; power(F, n / 2); multiply(F, F); if (n%2 != 0) multiply(F, M); // 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) { int F[][] = {{1, 1, 1}, {1, 0, 0}, {0, 1, 0}} ; 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. |
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 for x in range(3)] for y 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 return a;# 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 if (n == 1): return F[0][0] + F[0][1]; power(F, int(n / 2)); F = multiply(F, F); if (n % 2 != 0): F = multiply(F, M); # 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) = 1def findNthTerm(n): F = [[1, 1, 1], [1, 0, 0], [0, 1, 0]]; return power(F, n - 2);# Driver coden = 5;print("F(5) is", findNthTerm(n));# This code is contributed by mits |
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) = 1using 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++) // Updating our matrix a[i, j] = mul[i, j]; } // 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 if (n == 1) return F[0, 0] + F[0, 1]; power(F, n / 2); multiply(F, F); if (n % 2 != 0) multiply(F, M); // 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) { int[, ] F = { { 1, 1, 1 }, { 1, 0, 0 }, { 0, 1, 0 } }; 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
<script>// javascript 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 bfunction multiply(a , b){ // Creating an auxiliary matrix to // store elements of the // multiplication matrix var mul = Array(3).fill(0).map(x => Array(3).fill(0)); for (i = 0; i < 3; i++) { for (j = 0; j < 3; j++) { mul[i][j] = 0; for (k = 0; k < 3; k++) mul[i][j] += a[i][k] * b[k][j]; } } // storing the multiplication // result in a for (i = 0; i < 3; i++) for (j = 0; j < 3; j++) // Updating our matrix a[i][j] = mul[i][j]; }// Function to compute F raise to// power n-2.function power(F , n){ var 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 if (n == 1) return F[0][0] + F[0][1]; power(F, parseInt(n / 2)); multiply(F, F); if (n % 2 != 0) multiply(F, M); // 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) = 1function findNthTerm(n){ var F = [[1, 1, 1], [1, 0, 0], [0, 1, 0]] ; return power(F, n-2);} // Driver code var n = 5; document.write("F(5) is " + findNthTerm(n));// This code is contributed by Princi Singh </script> |
PHP
<?php// PHP 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[][]function multiply(&$a, &$b){ // Creating an auxiliary matrix to store // elements of the multiplication matrix $mul = array_fill(0, 3, array_fill(0, 3, 0)); for ($i = 0; $i < 3; $i++) { for ($j = 0; $j < 3; $j++) { $mul[$i][$j] = 0; for ($k = 0; $k < 3; $k++) $mul[$i][$j] += $a[$i][$k] * $b[$k][$j]; } } // storing the multiplication result in a[][] for ($i = 0; $i < 3; $i++) for ($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){ $M = array(array(1, 1, 1), array(1, 0, 0), array(0, 1, 0)); // Multiply it with initial values i.e with // F(0) = 0, F(1) = 1, F(2) = 1 if ($n == 1) return $F[0][0] + $F[0][1]; power($F, (int)($n / 2)); multiply($F, $F); if ($n % 2 != 0) multiply($F, $M); // Multiply it with initial values i.e with // F(0) = 0, F(1) = 1, F(2) = 1 return $F[0][0] + $F[0][1] ;}// Return n'th term of a series defined // using below recurrence relation.// f(n) is defined as// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3// Base Cases :// f(0) = 0, f(1) = 1, f(2) = 1function findNthTerm($n){ $F = array(array(1, 1, 1), array(1, 0, 0), array(0, 1, 0)); return power($F, $n - 2);}// Driver code$n = 5;echo "F(5) is " . findNthTerm($n);// This code is contributed by mits?> |
Output
F(5) is 7
Time Complexity: O(logN)
Auxiliary Space: O(logN)
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