Wednesday, July 15, 2015

Checking a number as it is Fibonacci number or not

Hello Friends, today I want to give you a basic information that is very useful in letting know whether the number is a fibonacci number just by simple formula.

Here it goes,

Let us suppose we have to check whether the number n is a fibonacci number or not, then the condition 5n^2 + 4 or 5n^2 - 4 should be a perfect square.

 // A utility function that returns true if x is perfect square  
 bool isPerfectSquare(int x)  
   int s = sqrt(x);  
   return (s*s == x);  
 // Returns true if n is a Fibinacci Number, else false  
 bool isFibonacci(int n)  
   // n is Fibinacci if one of 5*n*n + 4 or 5*n*n - 4 or both  
   // is a perferct square  
   return isPerfectSquare(5*n*n + 4) ||  
       isPerfectSquare(5*n*n - 4);  
 int main()  
  for (int i = 1; i <= 10; i++)  
    isFibonacci(i)? cout << i << " is a Fibonacci Number \n":  
            cout << i << " is a not Fibonacci Number \n" ;  
  return 0;  

Now once you pass the number to the function it will tell you whether or not the number is a part of fibonacci series or not.


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