How to check if a given number is a power of two?

Bit Manipulations

One approach would be to use bit manipulations:

(n & (n-1) == 0) and n != 0

Explanation: every power of 2 has exactly 1 bit set to 1 (the bit in that number’s log base-2 index). So when subtracting 1 from it, that bit flips to 0 and all preceding bits flip to 1. That makes these 2 numbers the inverse of each other so when AND-ing them, we will get 0 as the result.

For example:

                    n = 8

decimal |   8 = 2**3   |  8 - 1 = 7   |   8 & 7 = 0
        |          ^   |              |
binary  |   1 0 0 0    |   0 1 1 1    |    1 0 0 0
        |   ^          |              |  & 0 1 1 1
index   |   3 2 1 0    |              |    -------
                                           0 0 0 0
-----------------------------------------------------
                    n = 5

decimal | 5 = 2**2 + 1 |  5 - 1 = 4   |   5 & 4 = 4
        |              |              |
binary  |    1 0 1     |    1 0 0     |    1 0 1
        |              |              |  & 1 0 0
index   |    2 1 0     |              |    ------
                                           1 0 0

So, in conclusion, whenever we subtract one from a number, AND the result with the number itself, and that becomes 0 – that number is a power of 2!

Of course, AND-ing anything with 0 will give 0, so we add the check for n != 0.

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math functions

You could always use math functions, but notice that using them without care could cause incorrect results:

• math.log(x[, base]) with base=2:

  import math
  
  math.log(n, 2).is_integer()
  

• math.log2(x):

  math.log2(n).is_integer()
  

Worth noting that for any n <= 0, both functions will throw a ValueError as it is mathematically undefined (and therefore shouldn’t present a logical problem).

• math.frexp(x):

  abs(math.frexp(n)[0]) == 0.5
  

As noted above, for some numbers these functions are not accurate and actually give FALSE RESULTS:

• math.log(2**29, 2).is_integer() will give False
• math.log2(2**49-1).is_integer() will give True
• math.frexp(2**53+1)[0] == 0.5 will give True!!

This is because math functions use floats, and those have an inherent accuracy problem.

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(Expanded) Timing

Some time has passed since this question was asked and some new answers came up with the years. I decided to expand the timing to include all of them.

According to the math docs, the log with a given base, actually calculates log(x)/log(base) which is obviously slow. log2 is said to be more accurate, and probably more efficient. Bit manipulations are simple operations, not calling any functions.

So the results are:

Ev: 0.28 sec

log with base=2: 0.26 sec

count_1: 0.21 sec

check_1: 0.2 sec

frexp: 0.19 sec

log2: 0.1 sec

bit ops: 0.08 sec

The code I used for these measures can be recreated in this REPL (forked from this one).