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

Question

The code below isn't working right for some inputs.

a, i = set(), 1
while i <= 10000:
    a.add(i)
    i <<= 1

N = int(input())
if N in a:
    print("True")
else:
    print("False")

My initial idea was to check for each input if it's a power of 2 by starting from 1 and multiplying by 2 until exceeding the input number, comparing at each step. Instead, I store all the powers of 2 in a set beforehand, in order to check a given input in O(1). How can this be improved?

Answer 1

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).

Answer 2

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.

Answer 3

In binary representation, a power of 2 is a 1 (one) followed by zeros. So if the binary representation of the number has a single 1, then it's a power of 2. No need here to check num != 0:

print(1 == bin(num).count("1")) 

Answer 4

The bin builtin returns a string "0b1[01]?" (regex notation) for every strictly positive integer (if system memory suffices, that is), so that we can write the Boolean expression

'1' not in bin(abs(n))[3:]

that yields True for n that equals 0, 1 and 2**k.

1 is 2**0 so it is unquestionably a power of two, but 0 is not, unless you take into account the limit of x=2**k for k → -∞. Under the second assumption we can write simply

check0 = lambda n: '1' not in bin(abs(n))[3:]

and under the first one (excluding 0)

check1 = lambda n: '1' not in bin(abs(n))[3:] and n != 0

^{_{Of course the solution here proposed is just one of the many possible ones that
you can use to check if a number is a power of two... and for sure not the most
efficient one but I'm posting it in the sake of completeness :-)}}