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.
PHP
<?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($n))); }   // Driver Code if(isPowerOfTwo(31)) echo "Yes\n"; elseecho "No\n";   if(isPowerOfTwo(64)) echo "Yes\n"; elseecho "No\n";       // This code is contributed // by Sam007 ?> |
No Yes
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.
PHP
<?php   // Function to check if // x is power of 2 function isPowerOfTwo($n) { if ($n == 0)     return 0; while ($n != 1) {     if ($n % 2 != 0)         return 0;     $n = $n / 2; } return 1; }   // Driver Code if(isPowerOfTwo(31))     echo "Yes\n"; else    echo "No\n";   if(isPowerOfTwo(64))     echo "Yes\n"; else    echo "No\n";   // This code is contributed // by Sam007 ?> |
No Yes
3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. Please see Count set bits in an integer for counting set bits.
4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.
For example for 4 ( 100) and 16(10000), we get following after subtracting 1
3 –> 011
15 –> 01111
So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
PHP
<?php // PHP program to efficiently // check for power for 2   // Function to check if // x is power of 2 function isPowerOfTwo ($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 ?> |
No Yes
Please refer complete article on Program to find whether a no is power of two for more details!
