It would be possible to change complex mod without changing floor. Floor is consistent on reals and complex numbers, while complex mod is not, so it doesn't necessarily need to be fixed.
The problem with complex mod is that for left argument b it that it ensures that the result is in the box defined by corners -:(j./~ _0.5 0.5)*b . That means that it subtracts the closest Gaussian integer multiple of b from the right argument. This is inconsistent with mod for the reals, which subtracts the floor under division by b. Marshall On Sun, Jan 22, 2012 at 1:28 PM, Henry Rich <henryhr...@nc.rr.com> wrote: > Well, I know complex floor was something Gene McDonnell was proud of, > and I expect it's embedded in some applications. You'll need to produce > a compelling argument for changing it. > > Henry Rich > > On 1/22/2012 12:38 PM, Marshall Lochbaum wrote: > > I'm pretty sure I understand the definition, but I still don't get the > > rationale. Under my scheme the "obvious" complex floor would be (- 1&|), > > which also has the properties of convexity, translatability, and > > compatibility listed in the dictionary. > > > > Marshall > > > > On Sun, Jan 22, 2012 at 12:08 PM, Henry Rich<henryhr...@nc.rr.com> > wrote: > > > >> As Don said, make sure you understand complex floor before you start > >> coding. > >> > >> Henry Rich > >> > >> On 1/22/2012 11:38 AM, Marshall Lochbaum wrote: > >>> The theory of moduli is based on the quotient group of the integers by > a > >>> subgroup. For instance, the integers (mod 2) are produced by taking all > >> the > >>> integers and identifying all the ones that are even, as well as all the > >>> ones that are odd. Then we get a two-element group which we can preform > >>> addition on: even+even=even, even+odd=odd, etc. > >>> > >>> To reduce a number in a particular modulus, we need to find a canonical > >>> representation for that number. For positive numbers n the choice is > >> fairly > >>> simple: n|l gives the l' such that 0<=l'<n. In the complex plane, a > >> number > >>> generates a grid by taking its product with the Gaussian integers; try > >>> 'dot; pensize 2' plot , 1j2 * j./~i:10 > >>> to see what I mean. Then what we want is a canonical form for what > >> happens > >>> when we identify all those points together. We're allowed to "shift" by > >> any > >>> Gaussian integer times the modulus. > >>> > >>> Based on this, I think a good way to calculate the modulus is to get > the > >>> number into the square that lies counterclockwise of the modulus > number. > >>> Practically, this means we decompose a complex number y into (a j.b)*x, > >> and > >>> then return (1|a)j.(1|b) . > >>> > >>> I'll see if I can get around to editing this. I have a working copy of > >> the > >>> source, but I haven't made sense of it entirely. > >>> > >>> Marshall > >>> > >>> On Sun, Jan 22, 2012 at 10:45 AM, Raul Miller<rauldmil...@gmail.com> > >> wrote: > >>> > >>>> Yes, this is a bug. > >>>> > >>>> Someone should fix it. > >>>> > >>>> J is open source. (Though distributed sources do not compile for me, > >>>> and I keep getting sidetracked when I investigate forks that might > >>>> compile.) > >>>> > >>>> -- > >>>> Raul > >>>> ---------------------------------------------------------------------- > >>>> For information about J forums see > http://www.jsoftware.com/forums.htm > >>>> > >>> ---------------------------------------------------------------------- > >>> For information about J forums see http://www.jsoftware.com/forums.htm > >>> > >> ---------------------------------------------------------------------- > >> For information about J forums see http://www.jsoftware.com/forums.htm > >> > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
