How to strip the trailing zeros from the bit representation of a number

Question

This is the python version of the same C++ question.

Given a number, num, what is the fastest way to strip off the trailing zeros from its binary representation?

For example, let num = 232. We have bin(num) equal to 0b11101000 and we would like to strip the trailing zeros, which would produce 0b11101. This can be done via string manipulation, but it'd probably be faster via bit manipulation. So far, I have thought of something using num & -num

Assuming num != 0, num & -num produces the binary 0b1<trailing zeros>. For example,

num   0b11101000
-num  0b00011000
&         0b1000

If we have a dict having powers of two as keys and the powers as values, we could use that to know by how much to right bit shift num in order to strip just the trailing zeros:

#        0b1     0b10     0b100     0b1000
POW2s = {  1: 0,    2: 1,     4: 2,      8: 3, ... }

def stripTrailingZeros(num):
  pow2 = num & -num
  pow_ = POW2s[pow2]  # equivalent to math.log2(pow2), but hopefully faster
  return num >> pow_

The use of dictionary POW2s trades space for speed - the alternative is to use math.log2(pow2).

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Is there a faster way?

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Perhaps another useful tidbit is num ^ (num - 1) which produces 0b1!<trailing zeros> where !<trailing zeros> means take the trailing zeros and flip them into ones. For example,

num    0b11101000
num-1  0b11100111
^          0b1111

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Yet another alternative is to use a while loop

def stripTrailingZeros_iterative(num):
  while num & 0b1 == 0:  # equivalent to `num % 2 == 0`
    num >>= 1
  return num

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Ultimately, I need to execute this function on a big list of numbers. Once I do that, I want the maximum. So if I have [64, 38, 22, 20] to begin with, I would have [1, 19, 11, 5] after performing the stripping. Then I would want the maximum of that, which is 19.

Answer 1

You say you "Ultimately, [..] execute this function on a big list of numbers to get odd numbers and find the maximum of said odd numbers."

So why not simply:

from random import randint


numbers = [randint(0, 10000) for _ in range(5000)]


odd_numbers = [n for n in numbers if n & 1]
max_odd = max(odd_numbers)
print(max_odd)

To do what you say you want to do ultimately, there seems to be little point in performing the "shift right until the result is odd" operation? Unless you want the maximum of the result of that operation performed on all elements, which is not what you stated?

I agree with @TimPeters answer, but if you put Python through its paces and actually generate some data sets and try the various solutions proposed, they maintain their spread for any number of integer size when using Python ints, so your best option is integer division for numbers up to 32-bits, after that see the chart below:

from pandas import DataFrame
from timeit import timeit
import math
from random import randint


def reduce0(ns):
    return [n // (n & -n)
            for n in ns]


def reduce1(ns, d):
    return [n >> d[n & -n]
            for n in ns]


def reduce2(ns):
    return [n >> int(math.log2(n & -n))
            for n in ns]


def reduce3(ns, t):
    return [n >> t.index(n & -n)
            for n in ns]


def reduce4(ns):
    return [n if n & 1 else n >> ((n & -n).bit_length() - 1)
            for n in ns]


def single5(n):
    while (n & 0xffffffff) == 0:
        n >>= 32
    if (n & 0xffff) == 0:
        n >>= 16
    if (n & 0xff) == 0:
        n >>= 8
    if (n & 0xf) == 0:
        n >>= 4
    if (n & 0x3) == 0:
        n >>= 2
    if (n & 0x1) == 0:
        n >>= 1
    return n


def reduce5(ns):
    return [single5(n)
            for n in ns]


numbers = [randint(1, 2 ** 16 - 1) for _ in range(5000)]
d = {2 ** n: n for n in range(16)}
t = tuple(2 ** n for n in range(16))
assert(reduce0(numbers) == reduce1(numbers, d) == reduce2(numbers) == reduce3(numbers, t) == reduce4(numbers) == reduce5(numbers))

df = DataFrame([{}, {}, {}, {}, {}, {}])
for p in range(1, 16):
    p = 2 ** p
    numbers = [randint(1, 2 ** p - 1) for _ in range(4096)]

    d = {2**n: n for n in range(p)}
    t = tuple(2 ** n for n in range(p))

    df[p] = [
        timeit(lambda: reduce0(numbers), number=100),
        timeit(lambda: reduce1(numbers, d), number=100),
        timeit(lambda: reduce2(numbers), number=100),
        timeit(lambda: reduce3(numbers, t), number=100),
        timeit(lambda: reduce4(numbers), number=100),
        timeit(lambda: reduce5(numbers), number=100)
    ]
    print(f'Complete for {p} bit numbers.')


print(df)
df.to_csv('test_results.csv')

Result (when plotted in Excel):

Note that the plot that was previously here was wrong! The code and data were not though. The code has been updated to include @MarkRansom's solution, since it turns out to be the optimal solution for very large numbers (over 4k-bit numbers).

Answer 2

There's really no answer to questions like this in the absence of specifying the expected distribution of inputs. For example, if all inputs are in range(256), you can't beat a single indexed lookup into a precomputed list of the 256 possible cases.

If inputs can be two bytes, but you don't want to burn the space for 2**16 precomputed results, it's hard to beat (assuming that_table[i] gives the count of trailing zeroes in i):

low = i & 0xff
result = that_table[low] if low else 8 + that_table[i >> 8]

And so on.

You do not want to rely on log2(). The accuracy of that is entirely up to the C library on the platform CPython is compiled for.

What I actually use, in a context where ints can be up to hundreds of millions of bits:

    assert d

    if d & 1 == 0:
        ntz = (d & -d).bit_length() - 1
        d >>= ntz

A while loop would be a disaster in this context, taking time quadratic in the number of bits shifted off. Even one needless shift in that context would be a significant expense, which is why the code above first checks to see that at least one bit needs to shifted off. But if ints "are much smaller", that check would probably cost more than it saves. "No answer in the absence of specifying the expected distribution of inputs."

Answer 3

while (num & 0xffffffff) == 0:
    num >>= 32
if (num & 0xffff) == 0:
    num >>= 16
if (num & 0xff) == 0:
    num >>= 8
if (num & 0xf) == 0:
    num >>= 4
if (num & 0x3) == 0:
    num >>= 2
if (num & 0x1) == 0:
    num >>= 1

The idea here is to perform as few shifts as possible. The initial while loop handles numbers that are over 32 bits long, which I consider unlikely but it has to be provided for completeness. After that each statement shifts half as many bits; if you can't shift by 16, then the most you could shift is 15 which is (8+4+2+1). All possible cases are covered by those 5 if statements.

Answer 4

On my computer, a simple integer divide is fastest:

import timeit
timeit.timeit(setup='num=232', stmt='num // (num & -num)')
0.1088077999993402
timeit.timeit(setup='d = { 1: 0, 2 : 1, 4: 2, 8 : 3, 16 : 4, 32 : 5 }; num=232', stmt='num >> d[num & -num]')
0.13014470000052825
timeit.timeit(setup='import math; num=232', stmt='num >> int(math.log2(num & -num))')
0.2980690999993385