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

On 2025-05-27 13:34, [email protected] wrote:
>> 
>> 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.
> 
I found the reason for no difference — real(b) and imag(b) were relative 
prime.
If not relative prime, there are differences:

? b*=2
6 + 4*I
? 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)
[-3 - 2*I, -1 - 5*I, 2 - 3*I]
? setminus(myS,S)
[-2 + 3*I, 1 + 5*I, 3 + 2*I]
?

Regards,

Hermann.