PHP Program to find whether a no is power of two

Given a positive integer, write a function to find if it is a power of two or not.

Examples :

Input : n = 4
Output : Yes
22 = 4

Input : n = 7
Output : No

Input : n = 32
Output : Yes
25 = 32

1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2.

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<?php
// PHP Program to find 
// whether a no is 
// power of two
  
// Function to check
// Log base 2
function Log2($x)
{
    return (log10($x) / 
            log10(2));
}
  
  
// Function to check
// if x is power of 2
function isPowerOfTwo($n)
{
    return (ceil(Log2($n)) == 
            floor(Log2($n)));
}
  
// Driver Code
if(isPowerOfTwo(31))
echo "Yes\n";
else
echo "No\n";
  
if(isPowerOfTwo(64))
echo "Yes\n";
else
echo "No\n";
      
// This code is contributed 
// by Sam007
?>

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Output:

No
Yes

2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.

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<?php
  
// Function to check if
// x is power of 2
function isPowerOfTwo($n)
{
if ($n == 0)
    return 0;
while ($n != 1)
{
    if ($n % 2 != 0)
        return 0;
    $n = $n / 2;
}
return 1;
}
  
// Driver Code
if(isPowerOfTwo(31))
    echo "Yes\n";
else
    echo "No\n";
  
if(isPowerOfTwo(64))
    echo "Yes\n";
else
    echo "No\n";
  
// This code is contributed 
// by Sam007
?>

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Output:

No
Yes

3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. Please see Count set bits in an integer for counting set bits.

4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.

For example for 4 ( 100) and 16(10000), we get following after subtracting 1
3 –> 011
15 –> 01111

So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.

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<?php
// PHP program to efficiently 
// check for power for 2
  
// Function to check if
// x is power of 2
function isPowerOfTwo ($x)
{
// First x in the below expression
// is for the case when x is 0 
return $x && (!($x & ($x - 1)));
}
  
// Driver Code
if(isPowerOfTwo(31))
    echo "Yes\n" ;
else
    echo "No\n";
  
if(isPowerOfTwo(64))
    echo "Yes\n" ;
else
    echo "No\n";
          
// This code is contributed by Sam007
?>

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Output:

No
Yes

Please refer complete article on Program to find whether a no is power of two for more details!



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