Karim Belabas on Wed, 21 Jan 2015 21:49:22 +0100 |
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Re: Mixing variables in Mod expressions |
* Karim Belabas [2015-01-20 11:51]: > * Pascal Molin [2015-01-20 11:39]: > > What suprises me is that the moduli is removed in the result. [...] > The moduli is not removed, but indeed not printed: [...] > A 0 t_POLMOD is printed as 0 and omitted in polynomial coefficients. > For t_INTMOD, 0 is still explicitly written as Mod(0, N) [ but still omitted > when a polynomial coefficient ] > > I don't see any rationale for this. I can fix the discrepancy, and > explicitly write Mod(0, x) above instead of 0. Done in master. Now we have \\ 0 by itself is printed verbosely (21:26) gp > Mod(x,x^2-3) + Mod(x,x^2-5) %1 = Mod(0, 1) (21:26) gp > Mod(1,2)+Mod(1,3) %2 = Mod(0, 1) \\ ... but not as polynomial coefficients (21:26) gp > (x^100+1)*Mod(1,2) %3 = Mod(1, 2)*x^100 + Mod(1, 2) (21:26) gp > (x^100+1)*Mod(1,y) %4 = Mod(1, y)*x^100 + Mod(1, y) * Aurel.Page@math.u-bordeaux1.fr [2015-01-20 11:21]: > What about a warning when having to take gcd of moduli ? It would keep the > current behaviour but a student should understand he is doing something > wrong. If we consider an operation as legitimate (and here I do, PARI philosophy...) there should be no warning. OTOH it's *probably* a mistake when the "base ring" suddenly changes in this way. Maybe a warning when 'debug' is non-zero ? As in (21:32) gp > f(x)=0; (21:32) gp > \g1 debug = 1 (21:32) gp > f(y) *** Warning: compiler generates copy for `y'. %1 = 0 <aside> Here the copy optimizer tells us it had to generate a copy of 'y' when calling user function 'f', although it would probably be alright not to. The user can then rewrite his code as e.g. my(y=y);f(y) thereby killing the warning and producing faster bytecode: ? y=vector(10^5); ? for(i=1,1000,f(y)) *** Warning: compiler generates copy for `y'. time = 884 ms. ? my(y=y);for(i=1,1000,f(y)) time = 12 ms. </aside> Something like ? Mod(1,2)+Mod(1,3) %1 = Mod(0, 1) ? \g1 debug = 1 ? Mod(1,2)+Mod(1,3) *** _+_: Warning: coercing quotient rings; moduli 2 and 3 -> 1. %2 = Mod(0, 1) Cheers, K.B. -- Karim Belabas, IMB (UMR 5251) Tel: (+33) (0)5 40 00 26 17 Universite de Bordeaux Fax: (+33) (0)5 40 00 69 50 351, cours de la Liberation http://www.math.u-bordeaux.fr/~kbelabas/ F-33405 Talence (France) http://pari.math.u-bordeaux.fr/ [PARI/GP] `