robert king - 1 month ago 12

Python Question

for given

`x < 10^15`

`p`

`2^p <= x`

Here are some things I've tried:

First I tried this but it's not accurate for large numbers:

`>>> from math import log`

>>> x = 2**3

>>> x

8

>>> p = int(log(x, 2))

>>> 2**p == x

True

>>> x = 2**50

>>> p = int(log(x, 2))

>>> 2**p == x #not accurate for large numbers?

False

I could try something like:

`p = 1`

i = 1

while True:

if i * 2 > n:

break

i *= 2

p += 1

not_p = n - p

Which would take up to 50 operations if p was 50

I could pre-compute all the powers of 2 up until 2^50, and use binary search to find p. This would take around log(50) operations but seems a bit excessive and ugly?

I found this thread for C based solutions: Compute fast log base 2 ceiling

However It seems a bit ugly and I wasn't exactly sure how to convert it to python.

Answer

In Python >= 2.7, you can use the `.bit_length()`

method of integers:

```
def brute(x):
# determine max p such that 2^p <= x
p = 0
while 2**p <= x:
p += 1
return p-1
def easy(x):
return x.bit_length() - 1
```

which gives

```
>>> brute(0), brute(2**3-1), brute(2**3)
(-1, 2, 3)
>>> easy(0), easy(2**3-1), easy(2**3)
(-1, 2, 3)
>>> brute(2**50-1), brute(2**50), brute(2**50+1)
(49, 50, 50)
>>> easy(2**50-1), easy(2**50), easy(2**50+1)
(49, 50, 50)
>>>
>>> all(brute(n) == easy(n) for n in range(10**6))
True
>>> nums = (max(2**x+d, 0) for x in range(200) for d in range(-50, 50))
>>> all(brute(n) == easy(n) for n in nums)
True
```

Source (Stackoverflow)

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