Fibonacci Power

Given a number n. You are required to find ( Fib(n) ^ Fib(n) ) % 10^9 + 7, where Fib(n) is the nth fibonacci number.

Examples :

Input : n = 4
Output : 27
4th fibonacci number is 3
[ fib(4) ^ fib(4) ] % 10^9 + 7 = ( 3 ^ 3 ) % 10^9 + 7 = 27



Input : n = 3
Output : 4
3th fibonacci number is 2
[ fib(3) ^ fib(3) ] % 10^9 + 7 = ( 2 ^ 2 ) % 10^9 + 7 = 4

If n is large, fib(n) will be huge and fib(n) ^ fib(n) is not only difficult to calculate but its storage is impossible.

Approach:
( a ^ (p-1) ) % p = 1 where p is prime number (using Fermat Little Theorem).
( a ^ a ) % m can be written as ( ( a % m ) ^ a ) % m
It is also possible to write any number ‘a’ as a = k * ( m – 1 ) + r (Using Division Algorithm)
where ‘k’ is quotient and ‘r’ is remainder. We can say that r = a % (m-1)
So, Steps to reduce our calculation, lets suppose a = fib(n)

  ( a ^ a ) % m 
= ( ( a % m ) ^ a ) % m 
= ( ( a % m ) ^ ( k * ( m - 1 ) + r ) ) % m 
= ( ( ( a % m ) ^ ( m-1 ) ) ^ k  * ( a % m ) ^ r ) % m
= ( (1 ^ k) * ( a % m ) ^ r ) % m
= ( ( a % m ) ^ r ) % m   
= ( ( a % m ) ^ (a % (m-1) ) % m

a % m and a % (m-1) are easy to calculate and easy to store.
and we can calculate ( x ^ y ) % m using this GFG article.
We can find nth fibonacci number in log(n) time using this GFG article.

Below is the implementation of above approach :

C++

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#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
   
// Iterative function to compute modular power
long long modularexpo(long long x, long long y, long long p)
{
    // Initialize result
    long long res = 1; 
    // Update x if it is more than or
    // equal to p
    x = x % p; 
   
    while (y > 0) {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
   
        // y must be even now
        y = y >> 1; 
        x = (x * x) % p;
    }
    return res;
}
  
// Helper function that multiplies 2 matrices
// F and M of size 2*2, and puts the
// multiplication result back to F[][]
void multiply(long long F[2][2], long long M[2][2], long long m)
{
    long long x = ((F[0][0] * M[0][0]) % m + 
                   (F[0][1] * M[1][0]) % m) % m;
  
    long long y = ((F[0][0] * M[0][1]) % m + 
                   (F[0][1] * M[1][1]) % m) % m;
  
    long long z = ((F[1][0] * M[0][0]) % m + 
                   (F[1][1] * M[1][0]) % m) % m;
  
    long long w = ((F[1][0] * M[0][1]) % m + 
                   (F[1][1] * M[1][1]) % m) % m;
   
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
  
// Helper function that calculates F[][] raise to 
// the power n and puts the  result in F[][] 
// Note that this function is designed only for fib() 
// and won't work as general power function 
void power(long long F[2][2], long long n, long long m)
{
    if (n == 0 || n == 1)
        return;
    long long M[2][2] = { { 1, 1 }, { 1, 0 } };
   
    power(F, n / 2, m);
    multiply(F, F, m);
   
    if (n % 2 != 0)
        multiply(F, M, m);
}
  
// Function that returns nth Fibonacci number 
long long fib(long long n, long long m)
{
    long long F[2][2] = { { 1, 1 }, { 1, 0 } };
    if (n == 0)
        return 0;
    power(F, n - 1, m);
    return F[0][0];
}
  
// Driver Code
int main()
{
    long long n = 4;
    long long base = fib(n, mod) % mod;
    long long expo = fib(n, mod - 1) % (mod - 1);
    long long result = modularexpo(base, expo, mod) % mod;
    cout << result << endl;
}

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class GFG  


{ 


static int mod = 1000000007; 


  


// Iterative function to compute modular power 


static long modularexpo(long x, long y, long p) 


{ 


    // Initialize result 


    long res = 1;  


      


    // Update x if it is more than or 


    // equal to p 


    x = x % p;  


  


    while (y > 0) 


    { 


        // If y is odd, multiply x with result 


        if (y % 2 == 1) 


            res = (res * x) % p; 


  


        // y must be even now 


        y = y >> 1;  


        x = (x * x) % p; 


    } 


    return res; 


} 


  


// Helper function that multiplies 2 matrices 


// F and M of size 2*2, and puts the 


// multiplication result back to F[][] 


static void multiply(long F[][],  


                     long M[][], long m) 


{ 


    long x = ((F[0][0] * M[0][0]) % m +  


              (F[0][1] * M[1][0]) % m) % m; 


  


    long y = ((F[0][0] * M[0][1]) % m +  


              (F[0][1] * M[1][1]) % m) % m; 


  


    long z = ((F[1][0] * M[0][0]) % m +  


              (F[1][1] * M[1][0]) % m) % m; 


  


    long w = ((F[1][0] * M[0][1]) % m +  


              (F[1][1] * M[1][1]) % m) % m; 


  


    F[0][0] = x; 


    F[0][1] = y; 


    F[1][0] = z; 


    F[1][1] = w; 


} 


  


// Helper function that calculates F[][] raise to  


// the power n and puts the result in F[][]  


// Note that this function is designed only for fib()  


// and won't work as general power function  


static void power(long F[][], long n, long m) 


{ 


    if (n == 0 || n == 1) 


        return; 


    long M[][] = { { 1, 1 }, { 1, 0 } }; 


  


    power(F, n / 2, m); 


    multiply(F, F, m); 


  


    if (n % 2 != 0) 


        multiply(F, M, m); 


} 


  


// Function that returns nth Fibonacci number  


static long fib(long n, long m) 


{ 


    long F[][] = { { 1, 1 }, { 1, 0 } }; 


    if (n == 0) 


        return 0; 


    power(F, n - 1, m); 


    return F[0][0]; 


} 


  


// Driver Code 


public static void main(String[] args) 


{ 


    long n = 4; 


    long base = fib(n, mod) % mod; 


    long expo = fib(n, mod - 1) % (mod - 1); 


    long result = modularexpo(base, expo, mod) % mod; 


    System.out.println(result); 


} 


} 


  


// This code is contributed by PrinciRaj1992 



















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mod = 1000000007; 


  


# Iterative function to compute modular power 


def modularexpo(x, y, p): 


  


    # Initialize result 


    res = 1;  


      


    # Update x if it is more than or 


    # equal to p 


    x = x % p;  


  


    while (y > 0):  


          


        # If y is odd, multiply x with result 


        if (y & 1): 


            res = (res * x) % p; 


  


        # y must be even now 


        y = y >> 1;  


        x = (x * x) % p; 


      


    return res; 


  


# Helper function that multiplies 2 matrices 


# F and M of size 2*2, and puts the 


# multiplication result back to F[][] 


def multiply(F, M, m): 


  


    x = ((F[0][0] * M[0][0]) % m +


         (F[0][1] * M[1][0]) % m) % m; 


  


    y = ((F[0][0] * M[0][1]) % m +


         (F[0][1] * M[1][1]) % m) % m; 


  


    z = ((F[1][0] * M[0][0]) % m +


         (F[1][1] * M[1][0]) % m) % m; 


  


    w = ((F[1][0] * M[0][1]) % m +


         (F[1][1] * M[1][1]) % m) % m; 


  


    F[0][0] = x; 


    F[0][1] = y; 


    F[1][0] = z; 


    F[1][1] = w; 


  


# Helper function that calculates F[][] raise to  


# the power n and puts the result in F[][]  


# Note that this function is designed only for fib()  


# and won't work as general power function  


def power(F, n, m): 


  


    if (n == 0 or n == 1): 


        return; 


    M = [[ 1, 1 ], [ 1, 0 ]]; 


  


    power(F, n // 2, m); 


    multiply(F, F, m); 


  


    if (n % 2 != 0): 


        multiply(F, M, m); 


  


# Function that returns nth Fibonacci number  


def fib(n, m): 


  


    F = [[ 1, 1 ], [ 1, 0 ]]; 


    if (n == 0): 


        return 0; 


    power(F, n - 1, m); 


    return F[0][0]; 


  


# Driver Code 


if __name__ == '__main__': 


    n = 4; 


    base = fib(n, mod) % mod; 


    expo = fib(n, mod - 1) % (mod - 1); 


    result = modularexpo(base, expo, mod) % mod; 


    print(result); 


  


# This code is contributed by 29AjayKumar 



















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using System; 


  


class GFG  


{ 


static int mod = 1000000007; 


  


// Iterative function to compute modular power 


static long modularexpo(long x, long y, long p) 


{ 


    // Initialize result 


    long res = 1;  


      


    // Update x if it is more than or 


    // equal to p 


    x = x % p;  


  


    while (y > 0) 


    { 


        // If y is odd, multiply x with result 


        if (y % 2 == 1) 


            res = (res * x) % p; 


  


        // y must be even now 


        y = y >> 1;  


        x = (x * x) % p; 


    } 


    return res; 


} 


  


// Helper function that multiplies 2 matrices 


// F and M of size 2*2, and puts the 


// multiplication result back to F[,] 


static void multiply(long [,]F,  


                     long [,]M, long m) 


{ 


    long x = ((F[0, 0] * M[0, 0]) % m +  


              (F[0, 1] * M[1, 0]) % m) % m; 


  


    long y = ((F[0, 0] * M[0, 1]) % m +  


              (F[0, 1] * M[1, 1]) % m) % m; 


  


    long z = ((F[1, 0] * M[0, 0]) % m +  


              (F[1, 1] * M[1, 0]) % m) % m; 


  


    long w = ((F[1, 0] * M[0, 1]) % m +  


              (F[1, 1] * M[1, 1]) % m) % m; 


  


    F[0, 0] = x; 


    F[0, 1] = y; 


    F[1, 0] = z; 


    F[1, 1] = w; 


} 


  


// Helper function that calculates F[,] raise to  


// the power n and puts the result in F[,]  


// Note that this function is designed only for fib()  


// and won't work as general power function  


static void power(long [,]F, long n, long m) 


{ 


    if (n == 0 || n == 1) 


        return; 


    long [,]M = { { 1, 1 }, { 1, 0 } }; 


  


    power(F, n / 2, m); 


    multiply(F, F, m); 


  


    if (n % 2 != 0) 


        multiply(F, M, m); 


} 


  


// Function that returns nth Fibonacci number  


static long fib(long n, long m) 


{ 


    long [,]F = { { 1, 1 }, { 1, 0 } }; 


    if (n == 0) 


        return 0; 


    power(F, n - 1, m); 


    return F[0, 0]; 


} 


  


// Driver Code 


public static void Main(String[] args) 


{ 


    long n = 4; 


    long Base = fib(n, mod) % mod; 


    long expo = fib(n, mod - 1) % (mod - 1); 


    long result = modularexpo(Base, expo, mod) % mod; 


    Console.WriteLine(result); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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Output:


27





Time Complexity: log(n)



          
          
          

          

    

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