Nth Fibonacci number using Pell’s equation
Last Updated :
19 Oct, 2022
Given an integer N, the task is to find the Nth Fibonacci number.
Examples:
Input: N = 13
Output: 144
Input: N = 19
Output: 2584
Approach: The Nth Fibonacci number can be found using the roots of the pell’s equation. Pells equation is generally of the form (x2) – n(y2) = |1|.
Here, consider y2 = x, n = 1. Also, taken positive (+1) in the right-hand side.
Now the equation becomes x2 – x = 1 which is same as x2 – x – 1 = 0.
Here, {x = (pi – qi) / (p – q)} is termed as Nth term of the fibonacci series where i = n – 1 and (p, q) are the roots of the pell’s equation.
To find roots of general quadratic equation (a*x2 + b*x + c = 0).
x1 = [-b + math.sqrt(b2 – 4*a*c)] / 2*a
x2 = [-b – math.sqrt(b2 – 4*a*c)] / 2*a
i.e.
p = (1 + math.sqrt(5)) / 2
q = (1 – math.sqrt(5)) / 2
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int fib( int n)
{
double p = ((1 + sqrt (5)) / 2);
double q = ((1 - sqrt (5)) / 2);
int i = n - 1;
int x = ( int ) (( pow (p, i) -
pow (q, i)) / (p - q));
return x;
}
int main()
{
int n = 5;
cout << fib(n);
}
|
Java
class GFG
{
static double p = (( 1 + Math.sqrt( 5 )) / 2 );
static double q = (( 1 - Math.sqrt( 5 )) / 2 );
static int fib( int n)
{
int i = n - 1 ;
int x = ( int ) ((Math.pow(p, i) -
Math.pow(q, i)) / (p - q));
return x;
}
public static void main(String[] args)
{
int n = 5 ;
System.out.println(fib(n));
}
}
|
Python3
import math
p = ( 1 + math.sqrt( 5 )) / 2
q = ( 1 - math.sqrt( 5 )) / 2
def fib(n):
i = n - 1
x = (p * * i - q * * i) / (p - q)
return int (x)
n = 5
print (fib(n))
|
C#
using System;
class GFG
{
static double p = ((1 + Math.Sqrt(5)) / 2);
static double q = ((1 - Math.Sqrt(5)) / 2);
static int fib( int n)
{
int i = n - 1;
int x = ( int ) ((Math.Pow(p, i) -
Math.Pow(q, i)) / (p - q));
return x;
}
static public void Main ()
{
int n = 5;
Console.Write(fib(n));
}
}
|
Javascript
<script>
function fib(n)
{
let p = ((1 + Math.sqrt(5)) / 2);
let q = ((1 - Math.sqrt(5)) / 2);
let i = n - 1;
let x = parseInt((Math.pow(p, i) -
Math.pow(q, i)) / (p - q));
return x;
}
let n = 5;
document.write(fib(n));
</script>
|
Time complexity: O(logn)
Auxiliary Space: O(1)
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