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++ 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 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));
}
} |
# 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 = 5
print (cassini(n))
# This code is contributed # by Anant Agarwal. |
// 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 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. ?> |
<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