Find the smallest power of 2 greater than or equal to n in Python

Question

What is the simplest function to return the smallest power of 2 that is greater than or equal to a given non-negative integer in Python?

For example, the smallest po

Answer 1

Always returning 2**(x - 1).bit_length() is incorrect because although it returns 1 for x=1, it returns a non-monotonic 2 for x=0. A simple fix that is monotonically safe for x=0 is:

def next_power_of_2(x):  
    return 1 if x == 0 else 2**(x - 1).bit_length()

Sample outputs:

>>> print(', '.join(f'{x}:{next_power_of_2(x)}' for x in range(10)))
0:1, 1:1, 2:2, 3:4, 4:4, 5:8, 6:8, 7:8, 8:8, 9:16

It can pedantically be argued that x=0 should return 0 (and not 1), since 2**float('-inf') == 0.

Answer 2

Would this work for you:

import math

def next_power_of_2(x):
    return 1 if x == 0 else 2**math.ceil(math.log2(x))

Note that math.log2 is available in Python 3 but not in Python 2. Using it instead of math.log avoids numerical problems with the latter at 2**29 and beyond.

Sample outputs:

>>> print(', '.join(f'{x}:{next_power_of_2(x)}' for x in range(10)))
0:1, 1:1, 2:2, 3:4, 4:4, 5:8, 6:8, 7:8, 8:8, 9:16

It can pedantically be argued that x=0 should return 0 (and not 1), since 2**float('-inf') == 0.

Answer 3

Let's test it:

import collections
import math
import timeit

def power_bit_length(x):
    return 2**(x-1).bit_length()

def shift_bit_length(x):
    return 1<<(x-1).bit_length()

def power_log(x):
    return 2**(math.ceil(math.log(x, 2)))

def test(f):
    collections.deque((f(i) for i in range(1, 1000001)), maxlen=0)

def timetest(f):
    print('{}: {}'.format(timeit.timeit(lambda: test(f), number=10),
                          f.__name__))

timetest(power_bit_length)
timetest(shift_bit_length)
timetest(power_log)

The reason I'm using range(1, 1000001) instead of just range(1000000) is that the power_log version will fail on 0. The reason I'm using a small number of reps over a largeish range instead of lots of reps over a small range is because I expect that different versions will have different performance over different domains. (If you expect to be calling this with huge thousand-bit numbers, of course, you want a test that uses those.)

With Apple Python 2.7.2:

4.38817000389: power_bit_length
3.69475698471: shift_bit_length
7.91623902321: power_log

With Python.org Python 3.3.0:

6.566169916652143: power_bit_length
3.098236607853323: shift_bit_length
9.982460380066186: power_log

With pypy 1.9.0/2.7.2:

2.8580930233: power_bit_length
2.49524712563: shift_bit_length
3.4371240139: power_log

I believe this demonstrates that the 2** is the slow part here; using bit_length instead of log does speed things up, but using 1<< instead of 2** is more important.

Also, I think it's clearer. The OP's version requires you to make a mental context-switch from logarithms to bits, and then back to exponents. Either stay in bits the whole time (shift_bit_length), or stay in logs and exponents (power_log).

Answer 4

We can do this as follows using bit manipulation:

def next_power_of_2(n):
    if n == 0:
        return 1
    if n & (n - 1) == 0:
        return n
    while n & (n - 1) > 0:
        n &= (n - 1)
    return n << 1

Sample outputs:

>>> print(', '.join(f'{x}:{next_power_of_2(x)}' for x in range(10)))
0:1, 1:1, 2:2, 3:4, 4:4, 5:8, 6:8, 7:8, 8:8, 9:16

For further reading, refer to this resource.

Answer 5

v+=(v==0);
v--;
v|=v>>1;
v|=v>>2;
v|=v>>4;
v|=v>>8;
v|=v>>16;
v++;

For a 16-bit integer.