Arbitrary precision integer arithmetic: ceiling?
Steven D'Aprano
steve at REMOVE-THIS-cybersource.com.au
Sun Mar 9 03:16:39 CET 2008
On Sat, 08 Mar 2008 15:26:35 -0800, Paul Rubin wrote:
> Alasdair <amca01 at gmail.com> writes:
>> What is the best way of finding a ceiling of a quotient of arbitrary
>> sized integers?
>
> ceiling(a/b) = (a+b-1)//b
Unfortunately that doesn't work reliably.
>>> a, b = 9007199254741000.0, -3.0
>>> a/b
-3002399751580333.5
>>> (a+b-1)//b # should be -3002399751580333
-3002399751580332.0
I make no claim that this is the most efficient way to do it, but this
should work:
def splitfloat(f):
"""Return the integer and fraction part of a float."""
fp = abs(f) % 1.0
ip = abs(f) - fp
if f < 0:
ip, fp = -ip, -fp
return (ip, fp)
def ceil(f):
ip, fp = splitfloat(f)
if fp == 0:
return ip
elif f < 0:
return ip
else:
return ip + 1
>>> a, b = 9007199254741000.0, -3.0
>>> ceil(a/b)
-3002399751580333.0
>>> a, b = 9007199254741000.0, 3.0
>>> ceil(a/b)
3002399751580334.0
It even works for infinities, if supported by your platform:
>>> ceil(1.0/inf)
0.0
(Disclaimer: if you consider that 1.0/inf is a tiny bit more than zero,
and therefore you want ceil(1.0/inf) to give 1.0, then you will disagree
with me.)
--
Steven
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