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Find Nth term (A matrix exponentiation example)
• Difficulty Level : Hard
• Last Updated : 16 May, 2019

We are given a recursive function that describes Nth terms in form of other terms. In this article we have taken specific example.

Now you are given n, and you have to find out nth term using above formula.

Examples:

Input : n = 2
Output : 5

Input : n = 3
Output :13


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisite :

Basic Approach:This problem can be solved by simply just iterating over the n terms. Every time you find a term, using this term find next one and so on. But time complexity of this problem is of order O(n).

Optimized Approach
All such problem where a term is a function of other terms in linear fashion. Then these can be solved using Matrix (Please refer : Matrix Exponentiation ). First we make transformation matrix and then just use matrix exponentiation to find Nth term.
Step by Step method includes:
Step 1. Determine k the number of terms on which T(i) depends.
In our example T(i) depends on two terms.so, k = 2

Step 2. Determine initial values

Step 3. Determine TM, the transformation matrix.
This is the most important step in solving recurrence relation. In this step, we have to make matrix of dimension k*k.
Such that
T(i)=TM*(initial value vector)
Here initial value vector is vector that contains intial value.we name this vector as initial.

Below is the program to implement above approach

## C++

 // CPP program to find n-th term of a recursive// function using matrix exponentiation.#include using namespace std;#define MOD 1000000009  #define ll long long int  ll power(ll n){    if (n <= 1)        return 1;      // This power function returns first row of    // {Transformation Matrix}^n-1*Initial Vector    n--;      // This is an identity matrix.    ll res[2][2] = { 1, 0, 0, 1 };      // this is Transformation matrix.    ll tMat[2][2] = { 2, 3, 1, 0 };      // Matrix exponentiation to calculate power of {tMat}^n-1    // store res in "res" matrix.    while (n) {          if (n & 1) {            ll tmp[2][2];            tmp[0][0] = (res[0][0] * tMat[0][0] + res[0][1] * tMat[1][0]) % MOD;            tmp[0][1] = (res[0][0] * tMat[0][1] + res[0][1] * tMat[1][1]) % MOD;            tmp[1][0] = (res[1][0] * tMat[0][0] + res[1][1] * tMat[1][0]) % MOD;            tmp[1][1] = (res[1][0] * tMat[0][1] + res[1][1] * tMat[1][1]) % MOD;            res[0][0] = tmp[0][0];            res[0][1] = tmp[0][1];            res[1][0] = tmp[1][0];            res[1][1] = tmp[1][1];        }        n = n / 2;        ll tmp[2][2];        tmp[0][0] = (tMat[0][0] * tMat[0][0] + tMat[0][1] * tMat[1][0]) % MOD;        tmp[0][1] = (tMat[0][0] * tMat[0][1] + tMat[0][1] * tMat[1][1]) % MOD;        tmp[1][0] = (tMat[1][0] * tMat[0][0] + tMat[1][1] * tMat[1][0]) % MOD;        tmp[1][1] = (tMat[1][0] * tMat[0][1] + tMat[1][1] * tMat[1][1]) % MOD;        tMat[0][0] = tmp[0][0];        tMat[0][1] = tmp[0][1];        tMat[1][0] = tmp[1][0];        tMat[1][1] = tmp[1][1];    }      // res store {Transformation matrix}^n-1    // hence will be first row of res*Initial Vector.    return (res[0][0] * 1 + res[0][1] * 1) % MOD;}  // Driver codeint main(){    ll n = 3;    cout << power(n);    return 0;}

## Java

 // Java program to find n-th term of a recursive// function using matrix exponentiation.class GfG {      static int MAX = 100;    static int MOD = 1000000009;    static int power(int n)    {        if (n <= 1) {            return 1;        }          // This power function returns first row of        // {Transformation Matrix}^n-1*Initial Vector        n--;          // This is an identity matrix.        int res[][] = { { 1, 0 }, { 0, 1 } };          // this is Transformation matrix.        int tMat[][] = { { 2, 3 }, { 1, 0 } };          // Matrix exponentiation to calculate power of {tMat}^n-1        // store res in "res" matrix.        while (n > 0) {              if (n % 2 == 1) {                int tmp[][] = new int[2][2];                tmp[0][0] = (res[0][0] * tMat[0][0]                             + res[0][1] * tMat[1][0])                            % MOD;                tmp[0][1] = (res[0][0] * tMat[0][1]                             + res[0][1] * tMat[1][1])                            % MOD;                tmp[1][0] = (res[1][0] * tMat[0][0]                             + res[1][1] * tMat[1][0])                            % MOD;                tmp[1][1] = (res[1][0] * tMat[0][1]                             + res[1][1] * tMat[1][1])                            % MOD;                res[0][0] = tmp[0][0];                res[0][1] = tmp[0][1];                res[1][0] = tmp[1][0];                res[1][1] = tmp[1][1];            }              n = n / 2;            int tmp[][] = new int[2][2];            tmp[0][0] = (tMat[0][0] * tMat[0][0]                         + tMat[0][1] * tMat[1][0])                        % MOD;            tmp[0][1] = (tMat[0][0] * tMat[0][1]                         + tMat[0][1] * tMat[1][1])                        % MOD;            tmp[1][0] = (tMat[1][0] * tMat[0][0]                         + tMat[1][1] * tMat[1][0])                        % MOD;            tmp[1][1] = (tMat[1][0] * tMat[0][1]                         + tMat[1][1] * tMat[1][1])                        % MOD;            tMat[0][0] = tmp[0][0];            tMat[0][1] = tmp[0][1];            tMat[1][0] = tmp[1][0];            tMat[1][1] = tmp[1][1];        }          // res store {Transformation matrix}^n-1        // hence wiint be first row of res*Initial Vector.        return (res[0][0] * 1 + res[0][1] * 1) % MOD;    }      // Driver code    public static void main(String[] args)    {        int n = 3;        System.out.println(power(n));    }}  // This code contributed by Rajput-Ji

## Python3

 # Python3 program to find n-th term of a recursive# function using matrix exponentiation.MOD = 1000000009;  def power(n):    if (n <= 1):        return 1;      # This power function returns first row of    # {Transformation Matrix}^n-1 * Initial Vector    n-= 1;      # This is an identity matrix.    res = [[1, 0], [0, 1]];      # this is Transformation matrix.    tMat = [[2, 3], [1, 0]];      # Matrix exponentiation to calculate     # power of {tMat}^n-1 store res in "res" matrix.    while (n):        if (n & 1):            tmp = [[0 for x in range(2)] for y in range(2)];            tmp[0][0] = (res[0][0] * tMat[0][0] +                         res[0][1] * tMat[1][0]) % MOD;            tmp[0][1] = (res[0][0] * tMat[0][1] +                         res[0][1] * tMat[1][1]) % MOD;            tmp[1][0] = (res[1][0] * tMat[0][0] +                         res[1][1] * tMat[1][0]) % MOD;            tmp[1][1] = (res[1][0] * tMat[0][1] +                        res[1][1] * tMat[1][1]) % MOD;            res[0][0] = tmp[0][0];            res[0][1] = tmp[0][1];            res[1][0] = tmp[1][0];            res[1][1] = tmp[1][1];              n = n // 2;        tmp = [[0 for x in range(2)] for y in range(2)];        tmp[0][0] = (tMat[0][0] * tMat[0][0] +                     tMat[0][1] * tMat[1][0]) % MOD;        tmp[0][1] = (tMat[0][0] * tMat[0][1] +                     tMat[0][1] * tMat[1][1]) % MOD;        tmp[1][0] = (tMat[1][0] * tMat[0][0] +                     tMat[1][1] * tMat[1][0]) % MOD;        tmp[1][1] = (tMat[1][0] * tMat[0][1] +                     tMat[1][1] * tMat[1][1]) % MOD;        tMat[0][0] = tmp[0][0];        tMat[0][1] = tmp[0][1];        tMat[1][0] = tmp[1][0];        tMat[1][1] = tmp[1][1];      # res store {Transformation matrix}^n-1    # hence will be first row of res * Initial Vector.    return (res[0][0] * 1 + res[0][1] * 1) % MOD;  # Driver coden = 3;print(power(n));      # This code is contributed by mits

## C#

 // C# program to find n-th term of a recursive// function using matrix exponentiation.using System;  class GfG {      // static int MAX = 100;    static int MOD = 1000000009;    static int power(int n)    {        if (n <= 1) {            return 1;        }          // This power function returns first row of        // {Transformation Matrix}^n-1*Initial Vector        n--;          // This is an identity matrix.        int[, ] res = { { 1, 0 }, { 0, 1 } };          // this is Transformation matrix.        int[, ] tMat = { { 2, 3 }, { 1, 0 } };          // Matrix exponentiation to calculate power of {tMat}^n-1        // store res in "res" matrix.        while (n > 0) {              if (n % 2 == 1) {                int[, ] tmp = new int[2, 2];                tmp[0, 0] = (res[0, 0] * tMat[0, 0]                             + res[0, 1] * tMat[1, 0])                            % MOD;                tmp[0, 1] = (res[0, 0] * tMat[0, 1]                             + res[0, 1] * tMat[1, 1])                            % MOD;                tmp[1, 0] = (res[1, 0] * tMat[0, 0]                             + res[1, 1] * tMat[1, 0])                            % MOD;                tmp[1, 1] = (res[1, 0] * tMat[0, 1]                             + res[1, 1] * tMat[1, 1])                            % MOD;                res[0, 0] = tmp[0, 0];                res[0, 1] = tmp[0, 1];                res[1, 0] = tmp[1, 0];                res[1, 1] = tmp[1, 1];            }              n = n / 2;            int[, ] tmp1 = new int[2, 2];            tmp1[0, 0] = (tMat[0, 0] * tMat[0, 0]                          + tMat[0, 1] * tMat[1, 0])                         % MOD;            tmp1[0, 1] = (tMat[0, 0] * tMat[0, 1]                          + tMat[0, 1] * tMat[1, 1])                         % MOD;            tmp1[1, 0] = (tMat[1, 0] * tMat[0, 0]                          + tMat[1, 1] * tMat[1, 0])                         % MOD;            tmp1[1, 1] = (tMat[1, 0] * tMat[0, 1]                          + tMat[1, 1] * tMat[1, 1])                         % MOD;            tMat[0, 0] = tmp1[0, 0];            tMat[0, 1] = tmp1[0, 1];            tMat[1, 0] = tmp1[1, 0];            tMat[1, 1] = tmp1[1, 1];        }          // res store {Transformation matrix}^n-1        // hence wiint be first row of res*Initial Vector.        return (res[0, 0] * 1 + res[0, 1] * 1) % MOD;    }      // Driver code    public static void Main()    {        int n = 3;        Console.WriteLine(power(n));    }}  // This code contributed by mits

## PHP

 
Output:
13


Time Complexity : O(Log n)

The same idea is used to find n-th Fibonacci number in O(Log n)

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