Finding number of digits in n’th Fibonacci number

Given a number n, find number of digits in n’th Fibonacci Numbers. First few Fibonacci 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) = (?n - ?-n) / ?5
where
? = (1 + ?5) / 2
? = (1 - ?5) / 2
The above formula can be simplified, 
fib(n) = round(?n / ?5) 
Here round function indicates nearest integer.
Count of digits in Fib(n) = Log10Fib(n)
                          = Log10(?n / ?5)
                          = n*Log10(?) - Log10?5
                          = n*Log10(?) - (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.

Javascript

<script>// Javascript 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
    let d = (n * Math.log10(1.6180339887498948)) -
                        ((Math.log10(5)) / 2);

    return Math.ceil(d);
}

    // Driver Code
    let i;
    for (let i = 1; i <= 10; i++)
    document.write(`Number of Digits in F(${i}) - ${numberOfDigits(i)} <br>`);

// This code is contributed by _saurabh_jaiswal
</script>

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), "\n";

// 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

Time Complexity: O(1)
Auxiliary Space: O(1)

Another Approach(Using fact that Fibonacci numbers are periodic):

Fibonacci sequence is periodic modulo any integer, with period equal to 60 (known as Pisano’s period). This means that we can calculate the nth Fibonacci number modulo 10^k for some large k, and then use the periodicity to compute the number of digits. For example, we can compute F_n modulo 10^10 and count the number of digits:

F_n_mod = F_n % 10**10
digits = floor(log10(F_n_mod)) + 1

Below is the implementation of above approach:

C++

#include<bits/stdc++.h>
using namespace std;

long long numberOfDigits(long long n){
    int k = 10; // module 10^k
    int phi = (1 + sqrt(5)) / 2; //golden ratio

    // compute the n-th Fibonacci number modulo 10^k
    int a = 0, b = 1;
    for (int i = 2; i <= n; i++) {
        int c = (a + b) % int(pow(10, k));
        a = b;
        b = c;
    }
    int F_n_mod = b;

    // compute the number of digits in F_n_mod
    int digits = 1;
    while (F_n_mod >= 10) {
        F_n_mod /= 10;
        digits++;
    }
    return digits;
}

int main(){
    long long i;
    for (i = 1; i <= 10; i++)
    cout << "Number of Digits in F("
         << i <<") - "
         << numberOfDigits(i) << "\n";
    return 0;
}

// This code is contributed by Yash Agarwal(yashagarwal2852002)

Java

import java.util.*;

public class GFG {
    public static long numberOfDigits(long n) {
        int k = 10; // module 10^k
        double phi = (1 + Math.sqrt(5)) / 2; //golden ratio

        // compute the n-th Fibonacci number modulo 10^k
        int a = 0, b = 1;
        for (int i = 2; i <= n; i++) {
            int c = (a + b) % (int) Math.pow(10, k);
            a = b;
            b = c;
        }
        int F_n_mod = b;

        // compute the number of digits in F_n_mod
        int digits = 1;
        while (F_n_mod >= 10) {
            F_n_mod /= 10;
            digits++;
        }
        return digits;
    }

    public static void main(String[] args) {
        long i;
        for (i = 1; i <= 10; i++)
            System.out.println("Number of Digits in F(" + i + ") - " + numberOfDigits(i));
    }
}

Python3

import math


def numberOfDigits(n):
    k = 10
    # Golden ratio (approximately 1.618033988749895)
    phi = (1 + math.sqrt(5)) / 2

    # Compute the n-th Fibonacci number modulo 10^k
    a, b = 0, 1
    # Start the loop from 2, as we already have F(0) and F(1)
    for i in range(2, n + 1):
        c = (a + b) % pow(10, k)
        # Update the previous Fibonacci numbers for the next iteration
        a = b
        b = c
    F_n_mod = b

    # Compute the number of digits in F_n_mod
    # Initialize the digit counter to 1 (as any number has at least one digit)
    digits = 1
    # Keep dividing F_n_mod by 10 until it becomes less than 10
    while F_n_mod >= 10:
        F_n_mod = F_n_mod // 10
        # Increment the digit counter
        digits += 1
# Return the number of digits in the n-th Fibonacci number modulo 10^k
    return digits


# Driver code
for i in range(1, 11):
    # Calculate and print the number of digits in F(i) modulo 10^10
    print("Number of Digits in F(" + str(i) + ") - " + str(numberOfDigits(i)))

# THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

C#

using System;

class GFG
{
    static int NumberOfDigits(long n)
    {
        int k = 10; // modulo 10^k
        // Compute the n-th Fibonacci number modulo 10^k
        int a = 0, b = 1;
        for (int i = 2; i <= n; i++)
        {
            int c = (a + b) % (int)Math.Pow(10, k);
            a = b;
            b = c;
        }
        int F_n_mod = b;
        // Compute the number of digits in F_n_mod
        int digits = 1;
        while (F_n_mod >= 10)
        {
            F_n_mod /= 10;
            digits++;
        }
        return digits;
    }
    static void Main(string[] args)
    {
        for (long i = 1; i <= 10; i++)
        {
            Console.WriteLine($"Number of Digits in F({i}) - {NumberOfDigits(i)}");
        }
    }
}

Javascript

function numberOfDigits(n) {
  let k = 10; // module 10^k
  let phi = (1 + Math.sqrt(5)) / 2; // golden ratio

  // compute the n-th Fibonacci number modulo 10^k
  let a = 0,
    b = 1;
  for (let i = 2; i <= n; i++) {
    let c = (a + b) % Math.pow(10, k);
    a = b;
    b = c;
  }
  let F_n_mod = b;

  // compute the number of digits in F_n_mod
  let digits = 1;
  while (F_n_mod >= 10) {
    F_n_mod = Math.floor(F_n_mod / 10);
    digits++;
  }
  return digits;
}

// main function
let i;
for (i = 1; i <= 10; i++)
  console.log("Number of Digits in F(" + i + ") - " + numberOfDigits(i));

// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

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

Time Complexity: O(nk)

Auxiliary Space: O(1)

References:
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section2
https://en.wikipedia.org/wiki/Fibonacci_number