Re: Wrong (negative) degree returned by poldegree.

[Karim BELABAS <Karim.Belabas@math.u-psud.fr>]

On Wed, 29 Jan 2003, Alexander Shumakovitch wrote:
> I've recently came across the following problem with Pari 2.2.4:
>   (13:16) gp > p1 = t^2 / q^4
>   %1 = 1/q^4*t^2
>   (13:16) gp > poldegree(p1, q)
>   %2 = -1
>
> On the other hand,
>   (13:16) gp > p2 = p1 / t^7
>   %3 = 1/(q^4*t^5)
>   (13:17) gp > poldegree(p2, q)
>   %4 = -4
>
> poldegree obviously assumes that all exponents in a polynomial are positive
> (it takes a maximum with -1 (basemath/gen3.c: 215), if one asks for a degree
> with respect to a secondary variable. But this is not the case for p1:
>   (13:17) gp > Vec(p1)
>   %5 = [1/q^4, 0, 0]
>
> Does the problem lie with the internal simplifications of (Laurent)
> polynomials in Pari? The easiest solution would be to define poldegree(0) as
> -VERYBIGINT (one can't have _that_ many monomials anyway). The the code would
> work as it is. To keep the old convention intact, one could rename poldegree
> to, say, poldegree0 (since it's recursive) and define poldegree as
> poldegree0 == -VERYBIGINT ? -1 : poldegree0

The old convention didn't make much sense, since -1 is a perfectly legitimate
result. I _think_ it was only chosen because internally the degree of a
non-zero polynomial x is lgef(x) - 3, which gives -1 when x is a zero
polynomial, so a uniform formula could be used...

-VERYBIGINT is a good approximation to -oo, and consistent with
valuation(0, p) [ = VERYBIGINT ], and a few other functions needing to
represent infinity (padicprec, newtonpoly).

I've committed the corresponding patch [ and fixed the docs ].

Thanks,

    Karim.
-- 
Karim Belabas                    Tel: (+33) (0)1 69 15 57 48
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Université Paris-Sud             Email: Karim.Belabas@math.u-psud.fr
F-91405 Orsay (France)           http://www.math.u-psud.fr/~belabas/
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