So what do you think these should return?  I'm assuming you think that
at least 0/q should return 0/1 regardless of what include is set to.
What about p/0?  Should p.cancel(Poly(0, t)) work or raise some kind
of exception?

Aaron Meurer

On Tue, Jul 19, 2011 at 10:33 PM, Aaron Meurer <[email protected]> wrote:
> If you look at the code in dmp_cancel(), it is special-cased for f ==
> 0 or g == 0, in which case it just returns f, g for include=True or
> K.one, K.one, f, g for include=False.  So there is no bug, per se,
> just a design flaw.
>
> I think you wrote this code.  Is there a reason you special-cased
> this?  Perhaps for speed?  Or maybe some stuff fails when g == 0 (I
> didn't try it)?
>
> Aaron Meurer
>
> On Tue, Jul 19, 2011 at 9:10 PM, Mateusz Paprocki <[email protected]> wrote:
>> Hi,
>>
>> On 20 July 2011 05:02, Aaron Meurer <[email protected]> wrote:
>>>
>>> Hi.
>>>
>>> This question is mainly for Mateusz, but I encourage others to chime
>>> in too if they have an opinion.
>>>
>>> I need the following behavior in the polys:
>>>
>>> >>> Poly(0, t).cancel(Poly(t, t), include=True)
>>> (Poly(0, t), Poly(1, t))
>>>
>>> Whereas it currently does the following:
>>>
>>> >>> Poly(0, t).cancel(Poly(t, t), include=True)
>>> (Poly(0, t), Poly(t, t))
>>>
>>> (notice that the denominator is reduced to 1 in the first case and
>>> left alone in the second).  I need the former to make it easier to
>>> deal with zero rational functions in my Risch code.
>>>
>>> Now, this is one of four possibilities in cancel.  They are:
>>>
>>> 1. Poly(0, t).cancel(Poly(t, t), include=True) # 0/q include=True;
>>> this is the one I use
>>> 2. Poly(0, t).cancel(Poly(t, t), include=False) # 0/q include=False
>>> 3. Poly(t, t).cancel(Poly(0, t), include=True) # p/0 include=True
>>> 4. Poly(t, t).cancel(Poly(0, t), include=False) # p/0 include=False
>>>
>>> I did not change 2-4 in my branch yet because I only ever use the
>>> first idiom in my code, but I think we should try to make it
>>> consistant.  So my questions are:
>>>
>>> - Should 2 return a one denominator too?  I don't actually care if it
>>> does, but it seems like it should to be consistant with 1.
>>> - What should the returned coefficient from 2 be?  I never actually
>>> use include=False, so I don't really have a feel for this.
>>> - Should 3 and 4 even work?  To me they should raise an exception, but
>>> maybe it isn't necessary to think of f.cancel(g) as f/g (though that
>>> is what the docstring says).
>>
>> I didn't yet analyze all the cases pointed out above, but it seems there is
>> a bug in cancel():
>> In [1]: gcd(0, t)
>> Out[1]: t
>> In [2]: Poly(0, t).cancel(Poly(t, t))
>> Out[2]: (1, Poly(0, t, domain='ZZ'), Poly(t, t, domain='ZZ'))
>> gcd(0, something) always returns something, thus no matter what `include` is
>> set to, cancel() whould give Poly(1, t) as the denominator. Note that
>> cofactors() works correctly:
>> In [3]: Poly(0, t).cofactors(Poly(t, t))
>> Out[3]: (Poly(t, t, domain='ZZ'), Poly(0, t, domain='ZZ'), Poly(1, t,
>> domain='ZZ'))
>> (it gives GCD, cofactor(lhs), cofactor(rhs)).
>>
>>>
>>> Aaron Meurer
>>>
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>>
>> Mateusz
>>
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>

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