On 1 March 2024 09:07:26 GMT, 'Martin R' via sage-devel
<sage-devel@googlegroups.com> wrote:
>I'd be OK with raising an exception or with -oo, but it should be uniform,
>and I think it should be the same for polynomials, Laurent polynomials and
>in the same spirit for degree and valuation.
>
>It might be best to raise an exception, because this ensures that the zero
>polynomial gets special treatment.
Exceptions are expensive, performance-wise, and using them as a regular means
of controlling the flow of the algorithm execution is not a good idea.
A simple if/then/else is much cheaper.
Dima
>
>Martin
>On Thursday 29 February 2024 at 22:54:20 UTC+1 Nils Bruin wrote:
>
>> On Thursday 29 February 2024 at 11:15:21 UTC-8 Dima Pasechnik wrote:
>>
>> How about using something like https://github.com/NeilGirdhar/extended_int
>> ?
>> (Even better, do a PEP to have such a thing in Python proper...)
>> In old, totally duck-typed, Python this didn't really matter, but nowadays
>> it does make
>> a perfect sense.
>>
>> At the moment, I think most degree functions do their best to return sage
>> Integer objects; mainly so that coercion works well with them. So whatever
>> solution we use should probably be based on objects that naturally live in
>> the sage hierarchy. We do have an infinity object in sage and it already
>> gets used for valuations.
>>
>> Incidentally:
>>
>> sage: R.<x>=LaurentSeriesRing(QQ)
>> sage: z=R(0)
>> sage: z.valuation()
>> +Infinity
>> sage: z.degree()
>> -1
>>
>> I don't quite know why laurent series have a degree defined at all, but
>> they're keeping to the deg(0)=-1 convention.
>>
>> Incidentally:
>>
>> sage: A.<x>=QQ[]
>> sage: B.<y>=LaurentPolynomialRing(QQ)
>> sage: x.valuation(oo)
>> -1
>> sage: y.valuation(oo)
>> 1
>> so polynomial rings have a valuation (that will return +oo when
>> appropriate), but on LaurentPolynomialRing this gets silently broken: the
>> argument simply gets ignored and the valuation at 0 is returned. So I guess
>> you can get a well-behaving degree with
>>
>> f=0*y
>> -f(1/y).valuation()
>>
>
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