GCD and Fibonacci Numbers

You are given two positive numbers M and N. The task is to print greatest common divisor of M’th and N’th Fibonacci Numbers.

The first few Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ….
Note that 0 is considered as 0’th Fibonacci Number.

Examples:

Input  : M = 3, N = 6
Output :  2
Fib(3) = 2, Fib(6) = 8
GCD of above two numbers is 2

Input  : M = 8, N = 12
Output :  3
Fib(8) = 21, Fib(12) = 144
GCD of above two numbers is 3



A Simple Solution is to follow below steps.
1) Find M’th Fibonacci Number.
2) Find N’th Fibonacci Number.
3) Return GCD of two numbers.

A Better Solution is based on below identity

GCD(Fib(M), Fib(N)) = Fib(GCD(M, N))

The above property holds because Fibonacci Numbers follow
Divisibility Sequence, i.e., if M divides N, then Fib(M)
also divides N. For example, Fib(3) = 2 and every third
third Fibonacci Number is even.

Source : Wiki

The steps are:
1) Find GCD of M and N. Let GCD be g.
2) Return Fib(g).

Below are implementations of above idea.

C++

// C++ Program to find GCD of Fib(M) and Fib(N)
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1000;
  
// Create an array for memoization
int f[MAX] = {0};
  
// Returns n'th Fibonacci number using table f[]. 
// Refer method 6 of below post for details.
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
  
    // If fib(n) is already computed
    if (f[n])
        return f[n];
  
    int k = (n & 1)? (n+1)/2 : n/2;
  
    // Applying recursive formula [Note value n&1 is 1
    // if n is odd, else 0.
    f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
  
    return f[n];
}
  
// Function to return gcd of a and b
int gcd(int M, int N)
{
    if (M == 0)
        return N;
    return gcd(N%M, M);
}
  
// Returns GCD of Fib(M) and Fib(N)
int findGCDofFibMFibN(int M,  int N)
{
    return fib(gcd(M, N));
}
  
// Driver code
int main()
{
   int M = 3, N = 12;
   cout << findGCDofFibMFibN(M, N);
   return 0;
}

Java

// Java Program to find GCD of Fib(M) and Fib(N)
class gcdOfFibonacci
{
    static final int MAX = 1000;
    static int[] f;
  
    gcdOfFibonacci()  // Constructor
    {
        // Create an array for memoization
        f = new int[MAX];
    }
  
    // Returns n'th Fibonacci number using table f[].
    // Refer method 6 of below post for details.
    private static int fib(int n)
    {
        // Base cases
        if (n == 0)
            return 0;
        if (n == 1 || n == 2)
            return (f[n] = 1);
  
        // If fib(n) is already computed
        if (f[n]!=0)
            return f[n];
  
        int k = ((n & 1)==1)? (n+1)/2 : n/2;
  
        // Applying recursive formula [Note value n&1 is 1
        // if n is odd, else 0.
        f[n] = ((n & 1)==1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
               : (2*fib(k-1) + fib(k))*fib(k);
  
        return f[n];
    }
  
    // Function to return gcd of a and b
    private static int gcd(int M, int N)
    {
        if (M == 0)
            return N;
        return gcd(N%M, M);
    }
  
    // This method returns GCD of Fib(M) and Fib(N)
    static int findGCDofFibMFibN(int M,  int N)
    {
        return fib(gcd(M, N));
    }
  
    // Driver method
    public static void main(String[] args)
    {
        // Returns GCD of Fib(M) and Fib(N)
        gcdOfFibonacci obj = new gcdOfFibonacci();
        int M = 3, N = 12;
        System.out.println(findGCDofFibMFibN(M, N));
    }
}
// This code is contributed by Pankaj Kumar

Python3

# Python Program to find
# GCD of Fib(M) and Fib(N)
  
MAX = 1000
   
# Create an array for memoization
f=[0 for i in range(MAX)]
   
# Returns n'th Fibonacci
# number using table f[]. 
# Refer method 6 of below
# post for details.
def fib(n):
  
    # Base cases
    if (n == 0):
        return 0
    if (n == 1 or n == 2):
        f[n] = 1
   
    # If fib(n) is already computed
    if (f[n]):
        return f[n]
   
    k = (n+1)//2 if(n & 1) else n//2
   
    # Applying recursive
    # formula [Note value n&1 is 1
    # if n is odd, else 0.
    f[n] = (fib(k)*fib(k) + fib(k-1)*fib(k-1)) if(n & 1) else ((2*
           fib(k-1) + fib(k))*fib(k))
   
    return f[n]
  
   
# Function to return
# gcd of a and b
def gcd(M, N):
  
    if (M == 0):
        return N
    return gcd(N % M, M)
  
   
# Returns GCD of
# Fib(M) and Fib(N)
def findGCDofFibMFibN(M, N):
  
    return fib(gcd(M, N))
  
   
# Driver code
  
M = 3
N = 12
  
print(findGCDofFibMFibN(M, N))
  
# This code is contributed
# by Anant Agarwal.

C#

// C# Program to find GCD of 
// Fib(M) and Fib(N)
using System;
  
class gcdOfFibonacci {
      
    static int MAX = 1000;
    static int []f;
  
    // Constructor
    gcdOfFibonacci() 
    {
        // Create an array
        // for memoization
        f = new int[MAX];
    }
  
    // Returns n'th Fibonacci number
    // using table f[]. Refer method 
    // 6 of below post for details.
    private static int fib(int n)
    {
        // Base cases
        if (n == 0)
            return 0;
        if (n == 1 || n == 2)
            return (f[n] = 1);
  
        // If fib(n) is 
        // already computed
        if (f[n]!=0)
            return f[n];
  
        int k = ((n & 1)==1)? (n+1)/2 : n/2;
  
        // Applying recursive formula 
        // [Note value n&1 is 1
        // if n is odd, else 0.
        f[n] = ((n & 1) == 1) ? (fib(k) * fib(k) +
               fib(k - 1) * fib(k - 1)) : 
               (2 * fib(k - 1) + fib(k)) * fib(k);
  
        return f[n];
    }
  
    // Function to return gcd of a and b
    private static int gcd(int M, int N)
    {
        if (M == 0)
            return N;
        return gcd(N%M, M);
    }
  
    // This method returns GCD of
    // Fib(M) and Fib(N)
    static int findGCDofFibMFibN(int M, int N)
    {
        return fib(gcd(M, N));
    }
  
    // Driver method
    public static void Main()
    {
        // Returns GCD of Fib(M) and Fib(N)
        new gcdOfFibonacci();
        int M = 3, N = 12;
        Console.Write(findGCDofFibMFibN(M, N));
    }
}
  
// This code is contributed by nitin mittal.


Output:

2

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Improved By : nitin mittal



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