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.

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.

(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).


Refer to the excellent and detailed answer to "How to check if a number is a power of 2" — for C#. The equivalent Python implementation, also using the "bitwise and" operator &, is this:

def is_power_of_two(n):
    return (n != 0) and (n & (n-1) == 0)

As Python has arbitrary-precision integers, this works for any integer n as long as it fits into memory.

To summarize briefly the answer cited above: The first term, before the logical and operator, simply checks if n isn't 0 — and hence not a power of 2. The second term checks if it's a power of 2 by making sure that all bits after that bitwise & operation are 0. The bitwise operation is designed to be only True for powers of 2 — with one exception: if n (and thus all of its bits) were 0 to begin with.

To add to this: As the logical and "short-circuits" the evaluation of the two terms, it would be more efficient to reverse their order if, in a particular use case, it is less likely that a given n be 0 than it being a power of 2.

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