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. |
PHP
<?php // PHP Program to find // GCD of Fib(M) and Fib(N) $MAX = 1000; // Create an array for memoization $f = array_fill (0, $MAX , 0); // Returns n'th Fibonacci number // using table f[]. Refer method // 6 of below post for details. function fib( $n ) { global $f ; // 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) ? ( $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 function gcd( $M , $N ) { if ( $M == 0) return $N ; return gcd( $N % $M , $M ); } // Returns GCD of Fib(M) and Fib(N) function findGCDofFibMFibN( $M , $N ) { return fib(gcd( $M , $N )); } // Driver code $M = 3; $N = 12; print (findGCDofFibMFibN( $M , $N )) // This code is contributed // by mits ?> |
Output:
2
This article is contributed by Shubham Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.