Given a number n, find number of digits in n’th Fibonacci Numbers. First few Fibinacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ….
Examples:
Input : n = 6 Output : 1 6'th Fibonacci number is 8 and it has 1 digit. Input : n = 12 Output : 3 12'th Fibonacci number is 144 and it has 3 digits.
A simple solution is to find n’th Fibonacci Number and then count number of digits in it. This solution may lead to overflow problems for large values of n.
A direct way is to count number of digits in the nth Fibonacci number using below Binet’s Formula.
fib(n) = (Φn - Ψ-n) / √5
where
Φ = (1 + √5) / 2
Ψ = (1 - √5) / 2
The above formula can be simplified,
fib(n) = round(Φn / ? 5)
Here round function indicates nearest integer.
Count of digits in Fib(n) = Log10Fib(n)
= Log10(Φn / √5)
= n*Log10(Φ) - Log10√5
= n*Log10(Φ) - (Log105)/2
As mentioned in this G-Fact, this formula doesn’t seem to work and produce correct Fibonacci numbers due to limitations of floating point arithmetic. However, it looks viable to use this formula to find count of digits in n’th Fibonacci number.
Below is the implementation of above idea :
C++
/* C++ program to find number of digits in nth Fibonacci number */#include<bits/stdc++.h> using namespace std; // This function returns the number of digits // in nth Fibonacci number after ceiling it // Formula used (n * log(phi) - (log 5) / 2) long long numberOfDigits(long long n) { if (n == 1) return 1; // using phi = 1.6180339887498948 long double d = (n * log10(1.6180339887498948)) - ((log10(5)) / 2); return ceil(d); } // Driver program to test the above function int main() { long long i; for (i = 1; i <= 10; i++) cout << "Number of Digits in F(" << i <<") - " << numberOfDigits(i) << "n"; return 0; } |
Java
// Java program to find number of digits in nth // Fibonacci number class GFG { // This function returns the number of digits // in nth Fibonacci number after ceiling it // Formula used (n * log(phi) - (log 5) / 2) static double numberOfDigits(double n) { if (n == 1) return 1; // using phi = 1.6180339887498948 double d = (n * Math.log10(1.6180339887498948)) - ((Math.log10(5)) / 2); return Math.ceil(d); } // Driver code public static void main (String[] args) { double i; for (i = 1; i <= 10; i++) System.out.println("Number of Digits in F("+i+") - " +numberOfDigits(i)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to find # number of digits in nth # Fibonacci number import math # storing value of # golden ratio aka phi phi = (1 + 5**.5) / 2 # function to find number # of digits in F(n) This # function returns the number # of digitsin nth Fibonacci # number after ceiling it # Formula used (n * log(phi) - # (log 5) / 2) def numberOfDig (n) : if n == 1 : return 1 return math.ceil((n * math.log10(phi) - .5 * math.log10(5))) // Driver Code for i in range(1, 11) : print("Number of Digits in F(" + str(i) + ") - " + str(numberOfDig(i))) # This code is contributed by SujanDutta |
C#
// C# program to find number of // digits in nth Fibonacci number using System; class GFG { // This function returns the number of digits // in nth Fibonacci number after ceiling it // Formula used (n * log(phi) - (log 5) / 2) static double numberOfDigits(double n) { if (n == 1) return 1; // using phi = 1.6180339887498948 double d = (n * Math.Log10(1.6180339887498948)) - ((Math.Log10(5)) / 2); return Math.Ceiling(d); } // Driver code public static void Main () { double i; for (i = 1; i <= 10; i++) Console.WriteLine("Number of Digits in F("+ i +") - " + numberOfDigits(i)); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP program to find number of // digits in nth Fibonacci number // This function returns the // number of digits in nth // Fibonacci number after // ceiling it Formula used // (n * log(phi) - (log 5) / 2) function numberOfDigits($n) { if ($n == 1) return 1; // using phi = 1.6180339887498948 $d = ($n * log10(1.6180339887498948)) - ((log10(5)) / 2); return ceil($d); } // Driver Code $i; for ($i = 1; $i <= 10; $i++) echo "Number of Digits in F($i) - " , numberOfDigits($i), "
"; // This code is contributed by nitin mittal ?> |
Output:
Number of Digits in F(1) - 1 Number of Digits in F(2) - 1 Number of Digits in F(3) - 1 Number of Digits in F(4) - 1 Number of Digits in F(5) - 1 Number of Digits in F(6) - 1 Number of Digits in F(7) - 2 Number of Digits in F(8) - 2 Number of Digits in F(9) - 2 Number of Digits in F(10) - 2
References:
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section2
https://en.wikipedia.org/wiki/Fibonacci_number
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