# Iterated Logarithm log*(n)

• Difficulty Level : Medium
• Last Updated : 19 Aug, 2022

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 the 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

Time Complexity: O(logn)

Auxiliary Space: O(logn) due to recursive stack space
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

## Javascript

 

Time Complexity: O(logn)

Auxiliary Space: O(1)

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