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# Iterated Logarithm log*(n)

• Difficulty Level : Medium
• Last Updated : 26 Apr, 2021

Iterated Logarithm or Log*(n) is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1. Applications: It is used in analysis of algorithms (Refer Wiki for details)

## C++

 // Recursive CPP program to find value of// Iterated Logarithm#include using namespace std; int _log(double x, double base){    return (int)(log(x) / log(base));} double recursiveLogStar(double n, double b){    if (n > 1.0)        return 1.0 + recursiveLogStar(_log(n, b), b);    else        return 0;} // Driver codeint main(){    int n = 100, base = 5;    cout << "Log*(" << n << ") = "         << recursiveLogStar(n, base) << "\n";    return 0;}

## Java

 // Recursive Java program to// find value of Iterated Logarithmimport java.io.*; class GFG{static int _log(double x,                double base){    return (int)(Math.log(x) /                 Math.log(base));} static double recursiveLogStar(double n,                               double b){    if (n > 1.0)        return 1.0 +               recursiveLogStar(_log(n,                                 b), b);    else        return 0;} // Driver codepublic static void main (String[] args){    int n = 100, base = 5;    System.out.println("Log*(" + n + ") = " +                  recursiveLogStar(n, base));}} // This code is contributed by jit_t

## Python3

 # Recursive Python3 program to find value of# Iterated Logarithmimport math def _log(x, base):     return (int)(math.log(x) / math.log(base)) def recursiveLogStar(n, b):     if(n > 1.0):        return 1.0 + recursiveLogStar(_log(n, b), b)    else:        return 0  # Driver codeif __name__=='__main__':    n = 100    base = 5    print("Log*(", n, ") = ", recursiveLogStar(n, base)) # This code is contributed by# Sanjit_Prasad

## C#

 // Recursive C# program to// find value of Iterated Logarithm using System; public class GFG{static int _log(double x, double baset){    return (int)(Math.Log(x) /                Math.Log(baset));} static double recursiveLogStar(double n,                            double b){    if (n > 1.0)        return 1.0 +            recursiveLogStar(_log(n,                                b), b);    else        return 0;} // Driver code    static public void Main (){         int n = 100, baset = 5;    Console.WriteLine("Log*(" + n + ") = " +                recursiveLogStar(n, baset));}} // This code is contributed by ajit.

## PHP

  1.0)        return 1.0 +               recursiveLogStar(_log($n, $b), $b); else return 0;} // Driver code$n = 100; $base = 5;echo "Log*(" , $n , ")"," = ",recursiveLogStar($n, $base), "\n"; // This code is contributed by ajit?>

## Javascript

 

Output :

Log*(100) = 2

Iterative Implementation :

## C++

 // Iterative CPP function to find value of// Iterated Logarithmint iterativeLogStar(double n, double b){    int count = 0;    while (n >= 1) {        n = _log(n, b);        count++;    }    return count;}

## Java

 // Iterative Java function to find value of// Iterated Logarithmpublic static int iterativeLogStar(double n, double b){    int count = 0;    while (n >= 1) {        n = _log(n, b);        count++;    }    return count;} // This code is contributed by pratham76

## Python3

 # Iterative Python function to find value of# Iterated Logarithm  def iterativeLogStar(n, b):     count = 0    while(n >= 1):        n = _log(n, b)        count = count + 1     return count # This code is contributed by# Sanjit_Prasad

## C#

 // Iterative C# function to find value of// Iterated Logarithmstatic int iterativeLogStar(double n, double b){    int count = 0;    while (n >= 1)    {        n = _log(n, b);        count++;    }    return count;} // This code is contributed by rutvik_56

This article is contributed by Abhishek rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.