Re: How to determine Mod(a,b) with t_COMPLEX b?

[hermann@stamm-wilbrandt.de]

On 2025-05-27 01:59, Karim Belabas wrote:
> * hermann@stamm-wilbrandt.de [2025-05-27 00:01]:
> [...]
> > The minimal residue of 1+4*I modulo 3+2*I is the yellow point -I in 
> > the
> > example:
> > https://en.wikipedia.org/wiki/Gaussian_integer#Describing_residue_classes
> >
> > How can minimal residue of an input gaussian integer modulo a gaussian
> > integer be computed in PARI/GP?
>
> ? a = 1+4*I; b = 3+2*I;
> ? a - round(a/b)*b
> %2 = -I
>
> This is not exacly the same normalization as in the Wikipedia article,
> because ties are rounded up (= floor(x+1/2)), not down (= ceil(x-1/2)),
> but it has the same properties (defines a Euclidean division with 
> unique
> quotient and remainder).
>
> If you insist on the same (awkward) normalization, then you must use
> something like
>
>   myround(z) = ceil(real(z)-1/2) + I * ceil(imag(z)-1/2);
>   a - myround(a/b)*b
>
> instead.
>
> ? round(1/2 + I/2)
> %3 = 1 + I
> ? myround(1/2 + I/2)
> %4 = 0
>
> Cheers,
>
>     K.B.
Thank you for that approach of using t_COMPLEX (t_POL in Bill's 
approach).

I cannot find a difference in set of minimal residues for both 
normalizations:

$ gp -q
? a = 1+4*I; b = 3+2*I;
? myround(z) = ceil(real(z)-1/2) + I * ceil(imag(z)-1/2);
? S=Set([a - round(a/b)*b | 
r<-[-real(b)..real(b)];i<-[-imag(b)..imag(b)];a<-[r+i*I]]);
? myS=Set([a - myround(a/b)*b | 
r<-[-real(b)..real(b)];i<-[-imag(b)..imag(b)];a<-[r+i*I]]);
? #S==norml2(b)&&#myS==norml2(b)
1
? setminus(S,myS)
[]
? setminus(myS,S)
[]
?

Regards,

Hermann.