Given a number N, the task is to evaluate below expression. Expected time complexity is O(1).
f(n-1)*f(n+1) - f(n)*f(n)
Where f(n) is the n-th Fibonacci number with n >= 1. First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, ………..i.e. (considering 0 as 0th Fibonacci number) Examples :
Input : n = 5 Output : -1 f(5-1=4) = 3 f(5+1=6) = 8 f(5)*f(5)= 5*5 = 25 f(4)*f(6)- f(5)*f(5)= 24-25= -1
Although the task is simple i.e. find n-1th, nth and (n+1)-th Fibonacci numbers. Evaluate the expression and display the result. But this can be done in O(1) time using Cassini’s Identity which states that:
f(n-1)*f(n+1) - f(n*n) = (-1)^n
So, we don’t need to calculate any Fibonacci term,the only thing is to check whether n is even or odd. How does above formula work? The formula is based on matrix representation of Fibonacci numbers. 
C/C++
C++
// C++ implementation to demonstrate working // of Cassini’s Identity #include <bits/stdc++.h> using namespace std; // Returns (-1)^n int cassini(int n) { return (n & 1) != 0 ? -1 : 1; } // Driver Method int main() { int n = 5; cout << (cassini(n)); return 0; } // This code is contributed by phasing17 |
Java
// Java implementation to demonstrate working // of Cassini’s Identity class Gfg { // Returns (-1)^n static int cassini(int n) { return (n & 1) != 0 ? -1 : 1; } // Driver method public static void main(String args[]) { int n = 5; System.out.println(cassini(n)); } } |
Python3
# Python implementation # to demonstrate working # of Cassini’s Identity # Returns (-1)^n def cassini(n): return -1 if (n & 1) else 1 # Driver program n = 5print(cassini(n)) # This code is contributed # by Anant Agarwal. |
C#
// C# implementation to demonstrate // working of Cassini’s Identity using System; class GFG { // Returns (-1) ^ n static int cassini(int n) { return (n & 1) != 0 ? -1 : 1; } // Driver Code public static void Main() { int n = 5; Console.Write(cassini(n)); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP implementation to // demonstrate working of // Cassini’s Identity // Returns (-1)^n function cassini($n) { return ($n & 1) ? -1 : 1; } // Driver Code $n = 5; echo(cassini($n)); // This code is contributed by Ajit. ?> |
JavaScript
<script> // Javascript implementation to // demonstrate working of // Cassini’s Identity // Returns (-1)^n function cassini(n) { return (n & 1) ? -1 : 1; } // Driver Code let n = 5; document.write(cassini(n)); // This code is contributed by _saurabh_jaiswal. </script> |
Output :
-1
Time complexity: O(1) since only constant operations are performed
Auxiliary Space: O(1)
Reference : https://en.wikipedia.org/wiki/Cassini_and_Catalan_identities This article is contributed by Sahil Chhabra. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen 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.
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