To check whether an integer is a power of two, I’ve deployed hacks like this:

def is_power_of_two(x: int) -> bool:
    return x > 0 and hex(x)[-1] in ("0", "2", "4", "8")

While this works1, I’ve never liked explaining the pattern matching hack that’s going on here.

Today, I came across this tweet2 by Raymond Hettinger where he proposed an elegant solution to the problem. Here’s how it goes:

def is_power_of_two(x: int) -> bool:
    return x > 0 and x.bit_count() == 1

This is neat as there’s no hack and it uses a mathematical invariant to check whether an integer is a power of 2 or not. Also, it’s a tad bit faster.

Explanation

Any integer that’s a power of 2, will only contain a single 1 in its binary representation.

For example:

>>> bin(2)
'0b10'
>>> bin(4)
'0b100'
>>> bin(8)
'0b1000'
>>> bin(16)
'0b10000'
>>>

The .bit_count() function checks how many on-bits (1) are there in the binary representation of an integer.

Complete example with tests

import unittest


def is_power_of_two(number: int) -> bool:
    return number > 0 and number.bit_count() == 1


class IsPowerofTwoTest(unittest.TestCase):
    def setUp(self):
        self.power_of_twos = [2**x for x in range(2, 25_000)]
        self.not_power_of_twos = [3**x for x in range(2, 25_000)]

    def test_is_power_of_two(self):
        for x, y in zip(self.power_of_twos, self.not_power_of_twos):
            self.assertIs(is_power_of_two(x), True)
            self.assertIs(is_power_of_two(y), False)


if __name__ == "__main__":
    unittest.main()

  1. My tweet on the hack ↩︎

  2. A better solution by Raymond Hettinger ↩︎

Recent posts