Hello Jürgen,
I get
¯11J¯1 | ¯10J¯12modulus (A) is: (-11,-1)⎕CT is: 1e-13A=0 is:
(0,0)A+A=0 is: (-11,-1)B÷A+A=0 is: (1,1)bif_floor(): area c: x=1,
y=1⌊B÷A+A=0 is: (1,0)A×⌊B÷A+A=0 is: (-11,-1)B-A×⌊B÷A+A=0 is: (1,-
11)1J¯11
According to the diagram in Eugene's article, ⌊ 1J1 is 1J0, not
1J1.
Regards,
Fred
On Sat, 2017-06-24 at 23:12 +0200, Juergen Sauermann wrote:
> Hi Fred,
>
>
>
> interesting. I get:
>
>
>
> ¯11J¯1 |
> ¯10J¯12
>
> modulus (A) is: (-11,-1)
>
> ⎕CT is: 1e-13
>
> A=0 is: (0,0)
>
> A+A=0 is: (-11,-1)
>
> B÷A+A=0 is: (1,1)
>
> bif_floor(): area a/d: x=0, y=0
>
> ⌊B÷A+A=0 is: (1,1)
>
> A×⌊B÷A+A=0 is: (-10,-12)
>
> B-A×⌊B÷A+A=0 is: (0,0)
>
> 0
>
>
>
> Attached is a debug variant using SVN 761.
>
>
>
> Best Regards,
>
> Jürgen
>
>
>
>
>
> On 06/24/2017 09:27 PM, Frederick Pitts
> wrote:
>
>
>
> >
> > Jürgen,
> >
> >
> >
> > Using SVN 971 and testing all possible parings of Gaussian
> > integers where the real and imaginary parts range
> > independently
> > from ¯15 to 15, the residue function yields the following
> > anomalous results
> >
> >
> >
> > ¯11J¯1 | ¯10J¯12 1J¯11
> > 1J¯11 0
> > ¯11J1 | ¯12J¯10 ¯1J¯11
> > ¯1J¯11 0
> > ¯1J¯11 | 10J¯12 11J¯1
> > 11J¯1 0
> > 1J¯11 | 12J¯10 11J1
> > 11J1 0
> > ¯1J11 | ¯12J10 ¯11J¯1
> > ¯11J¯1 0
> > 1J11 | ¯10J12 ¯11J1
> > ¯11J1 0
> > 11J¯1 | 12J10 1J11
> > 1J11 0
> > 11J1 | 10J12 ¯1J11
> > ¯1J11 0
> >
> >
> >
> > The first item in each test result line is not in the
> > complete residue system for the given modulus as evidenced
> > by
> > the second item on the line.
> >
> >
> >
> > The same test using McDonnell's APL implementations of floor
> > and residue yields no errors.
> >
> >
> >
> > Regards,
> >
> >
> >
> > Fred
> >
> >
> >
> >
> >
> >
> > On Sat, 2017-06-24 at 20:21 +0200, Juergen Sauermann wrote:
> >
> > > Hi Fred,
> > >
> > >
> > >
> > > I am glad to hear that. It is in SVN 971 now. It was
> > > Jay who moved us into the right
> > >
> > > direction, thanks for that. I had used the Donell paper
> > > earlier (when designing complex
> > >
> > > floor) but the borderline cases (i.e. when ⎕CT makes a
> > > difference) were not considered
> > >
> > > in the paper, and the descriptions in both ISO and the
> > > APL2
> > > language reference are
> > >
> > > entirely misleading in that respect.
> > >
> > >
> > >
> > > Have a nice weekend,
> > >
> > >
> > >
> > > Best Regards,
> > >
> > > /// Jürgen
> > >
> > >
> > >
> > >
> > >
> > > On 06/24/2017 07:55 PM, Frederick
> > > Pitts wrote:
> > >
> > >
> > >
> > > >
> > > > Hello Jürgen,
> > > >
> > > >
> > > >
> > > >
> > > > SUCCESS.
> > > >
> > > >
> > > >
> > > >
> > > > The cut-and-paste below from my platform is
> > > > identical to
> > > > yours
> > > >
> > > >
> > > >
> > > > 5J3 | ¯7J6
> > > > ⎕CT is: 1e-13
> > > > modulus (A) is: (5,3)
> > > > A=0 is: (0,0)
> > > > A+A=0 is: (5,3)
> > > > B÷A+A=0 is: (-0.5,1.5)
> > > > ⌊B÷A+A=0 is: (0,1)
> > > > A×⌊B÷A+A=0 is: (-3,5)
> > > > B-A×⌊B÷A+A=0 is: (-4,1)
> > > > ¯4J1
> > > >
> > > >
> > > >
> > > > and 5J3 | 4J¯1 ¯4J1 give the correct answer too.
> > > >
> > > >
> > > >
> > > > If you want, I can patch the undebugged version of
> > > > Complex.cc and run a battery of tests. If not, I
> > > > will wait
> > > > and run the tests on the next SVN version.
> > > >
> > > >
> > > >
> > > > I think i need to find something useful to do with
> > > > Gaussian integers.
> > > >
> > > >
> > > >
> > > > Regards,
> > > >
> > > >
> > > >
> > > > Fred
> > > >
> > >
> > >
> > >
> > >
> >
> >
>
>
>
>
>
