# Find minimum number of Log value needed to calculate Log upto N

Last Updated : 25 Feb, 2023

Given an integer N. The task is to find the minimum number of log values needed to calculate all the log values from 1 to N using properties of the logarithm.
Examples

Input : N = 6
Output : 3
Value of log1 is already know, i.e. 0.
Except this the three log values needed are,
log2, log3, log5.

Input : N = 4
Output : 2

One of the properties of log function is:

log(x.y) = log(x) + log(y)

Hence, to calculate log(x.y), we must know log values of x and y. Let us understand the concept using an example, for N = 6. Let ans denotes the number of log values needed to find all log values from 1 to 6.

• log(1)=0 (implicit).
• To calculate log(2), we must know its value prior, we can’t find this using property.so, ans become 1.
• To calculate log(3), we must know its value prior, we can’t find this using property.so, ans become 2.
• To calculate log(4), we can use property, log(4)=log(2.2)=log(2)+log(2).As we already find log(2) hence ans remains 2.
• To calculate log(5), we must know its value prior, we can’t find this using property.so, ans become 3.
• To calculate log(6), we can use property, log(6)=log(2.3)=log(2)+log(3).As we already find log(2) and log(3), hence ans remains 3.

The idea is very simple, on observing carefully you will find that you can’t calculate log values of prime number as it has no divisor(other than 1 and itself). So, the task reduces to find all prime numbers from 1 to N.

Algorithm :

Step 1: start
Step 2: Initialize a variable int MAX with the value 1000005.
Step 3: Create a vector name prime of the length MAX.
Step 4: Now make a function that stores the true value at all index in prime with the help of for loop.
Step 5: Now create another function that takes an integer value as input and then stores false at index 0 and 1 because 0 & 1 are                not prime numbers.
Now Check if prime[i] is true by using a for loop that goes from 2 to N.Prime[ij] should be set to false for every j if prime[i]              is true since ij is not a prime. If prime[i] is true, use another for loop that goes from 2 to N/i.
Step 6: Make another function countLogNeeded which takes an integer value as a parameter in the function initialize count to 0                  and then call the above function which we created to find prime numbers.
Now run a for loop that counts all the prime numbers from 1 to N then return count.
Step 7: End

Below is the implementation of the above approach:

## C++

 // C++ program to find number of log values// needed to calculate all the log values// from 1 to N #include using namespace std; #define MAX 1000005 // In this vector prime[i] will store true// if prime[i] is prime, else store falsevector<bool> prime(MAX, true); // Using sieve of Eratosthenes to find // all prime upto Nvoid sieve(int N){    prime[0] = prime[1] = false;         for (int i = 2; i <= N; i++) {        if (prime[i]) {            for (int j = 2; i * j <= N; j++)                prime[i * j] = false;        }    }} // Function to find number of log values needed // to calculate all the log values from 1 to Nint countLogNeeded(int N){    int count = 0;         // calculate primes upto N    sieve(N);         for (int i = 1; i <= N; i++) {        if (prime[i])            count++;    }         return count;} // Driver codeint main(){    int N = 6;         cout<

## Java

 // Java program to find number of log values// needed to calculate all the log values// from 1 to Nimport java.util.*; class GFG {     static int MAX = 1000005;     // In this vector prime[i] will store true    // if prime[i] is prime, else store false    static Vector prime = new Vector<>(MAX);     static void vecIni()     {        for (int i = 0; i < MAX; i++)         {            prime.add(i, true);        }    }     // Using sieve of Eratosthenes to find     // all prime upto N    static void sieve(int N)     {        prime.add(0, false);        prime.add(1, false);         for (int i = 2; i <= N; i++)         {            if (prime.get(i))            {                for (int j = 2; i * j <= N; j++)                {                    prime.add(i * j, false);                }            }        }    }     // Function to find number of log values needed     // to calculate all the log values from 1 to N    static int countLogNeeded(int N)    {        int count = 0;         // calculate primes upto N        sieve(N);         for (int i = 1; i <= N; i++)         {            if (prime.get(i))             {                count++;            }        }         return count;    }     // Driver code    public static void main(String[] args)     {        vecIni();        int N = 6;        System.out.println(countLogNeeded(N));    }} /* This code contributed by PrinciRaj1992 */

## Python3

 # Python3 program to find number of log values# needed to calculate all the log values# from 1 to N MAX = 1000005 # In this list prime[i] will store true# if prime[i] is prime, else store falseprime = [True for i in range(MAX)] # Using sieve of Eratosthenes to find# all prime upto Ndef sieve(N):     prime[0], prime[1] = False, False     for i in range(2, N + 1):        if(prime[i]):            for j in range(2, N + 1):                if(i * j > N):                    break                prime[i * j] = False  # Function to find number of log values needed# to calculate all the log values from 1 to Ndef countLogNeeded(N):     count = 0     # calculate primes upto N    sieve(N)     for i in range(1, N + 1):        if(prime[i]):            count = count + 1     return count # Driver codeif __name__=='__main__':    N = 6    print(countLogNeeded(N)) # This code is contributed by# Sanjit_Prasad

## C#

 // C# program to find number of log values // needed to calculate all the log values // from 1 to N using System;using System.Collections.Generic;using System.Linq; class GFG {      static int MAX = 1000005;      // In this vector prime[i] will store true     // if prime[i] is prime, else store false     static List prime = new List(MAX);      static void vecIni()     {         for (int i = 0; i < MAX; i++)         {             prime.Add(true);         }     }      // Using sieve of Eratosthenes to find     // all prime upto N     static void sieve(int N)     {         prime.Insert(0, false);         prime.Insert(1, false);          for (int i = 2; i <= N; i++)         {             if (prime[i])            {                 for (int j = 2; i * j <= N; j++)                 {                     prime.Insert(i * j, false);                 }             }         }     }      // Function to find number of log values needed     // to calculate all the log values from 1 to N     static int countLogNeeded(int N)     {         int count = 0;          // calculate primes upto N         sieve(N);          for (int i = 1; i <= N; i++)         {             if (prime[i])             {                 count++;             }         }          return count;     }      // Driver code     public static void Main()     {         vecIni();         int N = 6;         Console.Write(countLogNeeded(N));     } }  /* This code contributed by Mohit kumar */

## Javascript

 

Output:
3`

Time Complexity:
Auxiliary Space: O(N)