Check if a number is power of 8 or not

Given a number check whether it is power of 8 or not.

Examples :

Input : n = 64
Output : Yes

Input : 75
Output : No

First solution

We calculate log8(n) of the number if it is an integer, then n is in power of 8. We use trunc(n) function that finds closest integer for a double value.

C++



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// CPP program to check if a number is power of 8
#include <cmath>
#include <iostream>
using namespace std;
  
/* function to check if power of 8 */
bool checkPowerof8(int n)
{
    /* calculate log8(n) */
    double i = log(n) / log(8);
  
    /* check if i is an integer or not */
    return (i - trunc(i) < 0.000001);
}
  
/* driver function */
int main()
{
    int n = 65;
    checkPowerof8(n) ? cout << "Yes" : cout << "No";
    return 0;
}

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Java

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// Java program to check if
// a number is power of 8
  
class GFG {
  
    // function to check
    // if power of 8
    static boolean checkPowerof8(int n)
    {
        /* calculate log8(n) */
        double i = Math.log(n) / Math.log(8);
  
        /* check if i is an integer or not */
        return (i - Math.floor(i) < 0.000001);
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int n = 65;
        if (checkPowerof8(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
  
// This code is contributed by Sam007

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C#

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// C#  program to check if
// a number is power of 8
using System;
  
class GFG {
  
    // function to check
    // if power of 8
    static bool checkPowerof8(int n)
    {
  
        // calculate log8(n) */
        double i = Math.Log(n) / Math.Log(8);
  
        // check if i is an integer or not */
        return (i - Math.Floor(i) < 0.000001);
    }
  
    // Driver Code
    static public void Main()
    {
        int n = 65;
  
        if (checkPowerof8(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by akt_mit

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PHP

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<?php
// PHP program to check if 
// a number is power of 8 
  
// function to check 
// if power of 8 
function checkPowerof8($n
    /* calculate log8(n) */
    $i = log($n) / log(8); 
  
    /* check if i is an integer or not */
    return ($i - floor($i) < 0.000001); 
  
// Driver Code 
$n = 65; 
if(checkPowerof8($n)) 
    echo "Yes"
else
    echo "No"
  
// This code is contributed 
// by Sach_Code
?>

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Output :

No

Second solution

A number is power of 8 if following conditions are satisfied.

  1. The number is power of two. A number is power of two if it has only one set bit, i.e., bitwise and of n and n-1 is 0.
  2. The number has its only set bit at position 0 or 3 or 6 or …. 30 [For a 32 bit number]. To check the position of its set bit we can use a mask (0xB6DB6DB6)16 = (10110110110110110110110110110110)2.

Below is implementation of above idea.

C++

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// C++ program to check if a number is power of 8
// using bit mask.
#include <bits/stdc++.h>
using namespace std;
  
/*function to check if power of 8*/
bool checkPowerof8(int n)
{
    return (n && !(n & (n - 1)) && !(n & 0xB6DB6DB6));
}
  
/*driver function*/
int main()
{
    int n = 65;
    checkPowerof8(n) ? cout << "Yes" : cout << "No";
  
    return 0;
}

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Python3

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# Python3 program to check if a number 
# is power of 8
  
# function to check if power of 8
def checkPowerof8(n):
  
    return (n and not (n & (n - 1)) and 
                  not (n & 0xB6DB6DB6))
  
# Driver Code
if __name__ == "__main__":
  
    n = 65
      
if checkPowerof8(n):
    print ("Yes")  
else:
    print ("No")
  
# This code is contributed by ita_c

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PHP

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<?php
// PHP program to check if a number 
// is power of 8 using bit mask. 
  
// function to check if power of 8
function checkPowerof8($n
    $t = ($n && !($n & ($n - 1)) && 
                !($n & 0xB6DB6DB6));
    return $t
  
// Driver Code
$n = 65; 
if(checkPowerof8($n))
    echo "Yes" ;
else
    echo "No"
  
// This code is contributed by Sach 
?>

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Output :

No

One simple observation that can be made here is that if a number is power of 8 then it has only one bit set and that bit is at positions 1, 4, 7, 10, …
Thus we can just check if the only bit set in the number is at one of these positions then it is a power of 8 otherwise not.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if n is power of 8
bool checkPowerof8(int n)
{
    // Variable i will denote the bit
    // that we are currently at
    int i = 0;
    unsigned long long l = 1;
    while (i <= 63) {
        l <<= i;
  
        // If only set bit in n
        // is at position i
        if (l == n)
            return true;
  
        // Get to next valid bit position
        i += 3;
        l = 1;
    }
    return false;
}
  
// Driver code
int main()
{
    int n = 65;
    if (checkPowerof8(n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Output :

No

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