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 single1in 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()