Hello Jürgen,
SVN 969 on my platform (Fedora 25, Intel(R) Core(TM) i7-6700
CPU)
5J3 | 14J5 1J4 ¯4J1
gives
1J4 ¯4J1 ¯4J1
not
¯1J4 ¯4J1 ¯4J1
Regards,
Fred
On Fri, 2017-06-23 at 17:38 +0200, Juergen Sauermann wrote:
> Hi,
>
>
>
> I have changed A∣B to literally follow the paper pointed
> out by Jay.
>
> The complex floor itself was already implemented like described
> in
> the paper,
>
> but A∣B was not.
>
>
>
> Now (SVN 969) the complex A∣B is computed as
>
>
>
> Z←B-A×⌊B÷A+A=0
>
>
>
> without any attempts to improve the performance of the
> operation.
>
>
>
> The result in the 5J3 modulus are now the same as in IBM
> APL2 (and I suppose also in J)
>
>
>
> 5J3 ∣ 14J5
> 1J4 ¯4J1
>
> ¯4J1 ¯4J1 ¯4J1
>
>
>
> I hope this finally fixed it. Thanks a lot to all that helped
> fixing this bug.
>
>
>
> /// Jürgen
>
>
>
>
>
> On 06/23/2017 09:34 AM, Jay Foad wrote:
>
>
>
> > I urge you to read Eugene McDonnell's Complex
> > Floor, which also discusses Residue. I believe the design
> > he comes up with in this paper was adopted more or less
> > verbatim
> > in APL. Also bear in mind that Floor and Residue in APL
> > have to
> > work well on all complex numbers, not just the Gaussian
> > integers.
> >
> >
> >
> > Jay.
> >
> >
> >
> > On 23 June 2017 at 01:42, Frederick
> > Pitts <fred.pi...@comcast.net>
> > wrote:
> >
> >
> > >
> > > Hello Jürgen,
> > >
> > >
> > >
> > > Some observations:
> > >
> > >
> > >
> > > 1) When performing a residue calculation on
> > > positive
> > > integers, a straight-forward integer division
> > > with
> > > remainder calculation
> > > suffices. For example, 5 ∣ 13 is computed with 13 /
> > > 5
> > > = 2 r 3 and so 5 ∣ 13 = 3 where 3 is in the
> > > complete
> > > residue system
> > > { 0, 1, 2, 3, 4 }. When performing the calculation
> > > on
> > > negative integers, one has to take advantage of
> > > the fact
> > > that the
> > > integer division quotient and remainder are not
> > > unique in order to compute a residue that is in
> > > the
> > > complete residue system.
> > > For 5 ∣ ¯13, ¯13 / 5 = ¯2 r ¯3 where ¯3 is not in
> > > the
> > > CRS. However, ¯13 / 5 = ¯3 r 2 where 3 is in the
> > > CSR.
> > > The same concept applies Gaussian integers.
> > >
> > >
> > >
> > > 2) I suspect the decision to have the APL2 floor
> > > function round toward negative infinity, instead
> > > of
> > > toward zero,
> > > was made based on the desire to save cpu cycles and
> > > memory in the residue function code.
> > >
> > >
> > >
> > > 3) I read at least one math literature article
> > > discussing Gaussian integer Euclidean division
> > > algorithms, that recommended
> > > rounding down to the nearest real and imaginary
> > > part
> > > toward negative infinity. Unfortunately I cannot
> > > find
> > > the article right now. I will continue to look for
> > > it. None of the articles discussed using a
> > > complex
> > > integer floor function.
> > >
> > >
> > >
> > > 4) The reason MOD_TEST.apl shows total disagreement
> > > MODJ and the builtin residue function is that the
> > > complex floor function code change in SVN 965
> > > relocated
> > > the CRS's on complex plane. Attached are
> > > CRS0-CRS1-6J-6-SVN964.out
> > > CRS0-CRS1-6J-6-SVN965.out. The first file contains
> > > a
> > > CRS map for modulus ¯6J¯6 produced with the
> > > residue
> > > function
> > > followed by a map for the same modulus produced
> > > with
> > > MODJ using SVN 964. The second file contains the
> > > same
> > > maps
> > > using SVN 965. Observe that for SVN 964 the residue
> > > function CRS is in the bottom half of the complex
> > > plane,
> > > but for SVN 965 it is in the top half. The CRS
> > > for the
> > > MODJ function is in the bottom half in both SVN
> > > cases.
> > > 5)The complex floor code change did not help with
> > > the
> > > issue that the builtin residue function is not
> > > idempotent for all possible arguments and
> > > consequently
> > > generates too many residues. See attached
> > > CRSOTST0-SVN965.out. For a grid
> > > of Gaussian integers with real and imaginary parts
> > > ranging from ¯15 to 15, using every value with
> > > every
> > > other value as modulus and second argument, there
> > > were
> > > 40 case where the order of CSR exceeded the
> > > modulus
> > > norm. I think that
> > > was the failure count with the previous SVN.
> > >
> > >
> > >
> > > Sincerely, I think the complex floor and ceiling
> > > functions should not be used by other functions
> > > even if
> > > IBM and ISO
> > > imply they are in their documentations. I'm not
> > > seeing them used in the Gaussian integer
> > > literature.
> > > Again, please correct me if I'm wrong.
> > >
> > >
> > >
> > > Regards,
> > >
> > >
> > >
> > > Fred
> > >
> > >
> > >
> > >
> > >
> > > On Thu, 2017-06-22 at 18:08 +0200, Juergen
> > > Sauermann wrote:
> > >
> > > > Hi again,
> > > >
> > > >
> > > >
> > > > sorry a small typo below. Lines 19/20
> > > > should read:
> > > >
> > > >
> > > >
> > > > (¯6J¯5 - 0J¯11) ÷ ¯6J¯6
> > > >
> > > > 0J¯1
> > > >
> > > >
> > > >
> > > > /// Jürgen
> > > >
> > > >
> > > >
> > > > On
> > > > 06/22/2017 05:44 PM, Juergen Sauermann
> > > > wrote:
> > > >
> > > >
> > > >
> > > > > Hi Fred at al.,
> > > > >
> > > > >
> > > > >
> > > > > I have made another attempt to fix
> > > > > the residue
> > > > > function, SVN 965.
> > > > >
> > > > >
> > > > >
> > > > > For complex m∣b It now rounds down
> > > > > the
> > > > > real() and imag() parts of the
> > > > > quotient q←b÷m
> > > > > and returns b-q.
> > > > >
> > > > > Instead of always rounding towards 0
> > > > > or
> > > > > -infinity, the rounding direction is
> > > > > now
> > > > > (compared to the previous
> > > > >
> > > > > attempt) determined by the quadrant
> > > > > in which the
> > > > > modulus m lies.
> > > > >
> > > > >
> > > > >
> > > > > There are still differences to the
> > > > > results
> > > > > displayed by MOD_test.apl, but I
> > > > > suppose
> > > > > they are
> > > > >
> > > > > caused by that program. For example,
> > > > > the first
> > > > > line of MOD_test.apl, says:
> > > > >
> > > > >
> > > > >
> > > > > LA RA MODJ |
> > > > >
> > > > > ¯6J¯6 ¯6J¯5 0J¯11 0J1
> > > > >
> > > > >
> > > > >
> > > > > We have:
> > > > >
> > > > >
> > > > >
> > > > > (¯6J¯5 - 0J1) ÷ ¯6J¯6
> > > > >
> > > > > 1
> > > > >
> > > > > (0J¯11 - 0J1) ÷ ¯6J¯6
> > > > >
> > > > > 1J1
> > > > >
> > > > >
> > > > >
> > > > > so both 0J¯11 and 0J1 are valid
> > > > > remainders modulo ¯6J¯6. However, the
> > > > >
> > > > > magnitude of 0J¯11 (= 11)
> > > > > is larger than the magnitude of the
> > > > > divisor ¯6J¯6 (= around
> > > > > 8.4).
> > > > >
> > > > > I suppose most people expect the
> > > > > remainder of
> > > > > a division to be in some sense
> > > > >
> > > > > smaller than the divisor.
> > > > >
> > > > >
> > > > >
> > > > > Regarding Jay's idempotency
> > > > > requirement we now
> > > > > have:
> > > > >
> > > > >
> > > > >
> > > > > f←{6J6|⍵}
> > > > >
> > > > > f ¯3 ¯2 ¯3 ¯1 0 1 2 3
> > > > >
> > > > > 3J6 4J6 3J6 5J6 0 ¯5J6 ¯4J6
> > > > > ¯3J6
> > > > >
> > > > > f f ¯3 ¯2 ¯3 ¯1 0 1 2 3
> > > > >
> > > > > 3J6 4J6 3J6 5J6 0 ¯5J6 ¯4J6
> > > > > ¯3J6
> > > > >
> > > > >
> > > > >
> > > > > f←{5J3|⍵}
> > > > >
> > > > > f ¯3 ¯2 ¯3 ¯1 0 1 2 3
> > > > >
> > > > > 2J3 3J3 2J3 4J3 0 ¯2J5 ¯1J5 0J5
> > > > >
> > > > > f f ¯3 ¯2 ¯3 ¯1 0 1 2 3
> > > > >
> > > > > 2J3 3J3 2J3 4J3 0 ¯2J5 ¯1J5 0J5
> > > > >
> > > > >
> > > > >
> > > > > so at least the first modulus seems
> > > > > to work as
> > > > > well. The result is still different
> > > > >
> > > > > from APL2 as reported by Jay, but I
> > > > > can't tell
> > > > > why:
> > > > >
> > > > >
> > > > >
> > > > > IBM APL2:
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > 5J3 ∣ 14J5 1J4 ¯4J1
> > > > >
> > > > > ¯4J1
> > > > > ¯4J1
> > > > > ¯4J1
> > > > >
> > > > >
> > > > >
> > > > > GNU APL:
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > 5J3 ∣ 14J5 1J4 ¯4J1
> > > > >
> > > > > 1J4 1J4 1J4
> > > > >
> > > > >
> > > > >
> > > > > But both 1J4 and ¯4J1
> > > > > are valid remainders. Interestingly
> > > > > Jay's
> > > > > idempotency requirement seems to
> > > > >
> > > > > be fulfilled by both the GNU APL
> > > > > and by IBM
> > > > > APL2, so that that requirement
> > > > > alone does not
> > > > > suffice
> > > > >
> > > > > to tell which result is correct.
> > > > >
> > > > >
> > > > >
> > > > > On the other hand this matter seems
> > > > > to be like
> > > > > discussing if the square root of 4
> > > > > is 2 or -2
> > > > > with
> > > > >
> > > > > both answers being correct.
> > > > >
> > > > >
> > > > >
> > > > > Best Regards,
> > > > >
> > > > > Jürgen Sauermann
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > On
> > > > > 06/21/2017 10:25 PM, Frederick Pitts
> > > > > wrote:
> > > > >
> > > > >
> > > > >
> > > > > > Jürgen,
> > > > > >
> > > > > >
> > > > > >
> > > > > > The proposed change to DIVJ does
> > > > > > not work
> > > > > > because 'q1' is a complex number,
> > > > > > so the '×'
> > > > > > in '× q1' is the complex
> > > > > > complement function,
> > > > > > not the sign function. I tried
> > > > > > the proposed
> > > > > > change and every test fails.
> > > > > >
> > > > > >
> > > > > >
> > > > > > I will try to hack DIVJ to use a
> > > > > > floor
> > > > > > towards zero instead of towards
> > > > > > minus infinity
> > > > > > for the real and imaginary
> > > > > > parts of the quotient and see what
> > > > > > happens.
> > > > > >
> > > > > >
> > > > > >
> > > > > > Correct me if I am wrong, but my
> > > > > > mind set
> > > > > > is that the APL residue function
> > > > > > has to
> > > > > > satisfy the following invariants:
> > > > > > 1) The order of the complete
> > > > > > residue system
> > > > > > (residue count) for a given
> > > > > > modulo 'n' has to
> > > > > > equal the norm of 'n'.
> > > > > > 2) And as Jay Foad so succinctly
> > > > > > expressed
> > > > > > it, the residue function has to
> > > > > > be idempotent
> > > > > > with respect to its right
> > > > > > argument,
> > > > > > e.g., ( n | m ) = n | n | m .
> > > > > > regardless of the implementation of
> > > > > > the
> > > > > > residue function.
> > > > > >
> > > > > >
> > > > > >
> > > > > > Later,
> > > > > >
> > > > > >
> > > > > >
> > > > > > Fred
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > On Wed, 2017-06-21 at 19:46 +0200,
> > > > > > Juergen
> > > > > > Sauermann wrote:
> > > > > >
> > > > > > > Hi Fred,
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > I have a question about the
> > > > > > > MOD_test.apl
> > > > > > > that you kindly provided.
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > In function DIVJ on line 57
> > > > > > > ff it
> > > > > > > says:
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > z
> > > > > > > ← q , a - b × q ← CMPLX ⌊
> > > > > > > ( 9 11 ) ○ a ÷
> > > > > > > b
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > so the quotient is rounded
> > > > > > > down towards
> > > > > > > minus infinity.
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > I wonder if that should be
> > > > > > > something like
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > z
> > > > > > > ← q , (× q1) × a - b × q
> > > > > > > ← CMPLX ⌊ ∣ 9
> > > > > > > 11 ○ q1 ← a ÷ b
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > so that the quotient is
> > > > > > > rounded towards 0?
> > > > > > > Interestingly IBM and ISO
> > > > > > >
> > > > > > > give different definitions
> > > > > > > for the residue
> > > > > > > in terms of APL:
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > IBM (language reference, page
> > > > > > > 227):
> > > > > > >
> > > > > > > Z←L∣R
> > > > > > >
> > > > > > > Z is R-L×⌊ R÷L+L=0
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > ISO (chapter 7.2.9 Residue):
> > > > > > >
> > > > > > > R←Q∣P
> > > > > > >
> > > > > > > R←P-(×P)×|Q×⌊|P÷Q
> > > > > > >
> > > > > > > and return R if (×R)=×Q,
> > > > > > > or R+Q
> > > > > > >
> > > > > > > otherwise.
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > That suggest that IBM rounds
> > > > > > > the quotient down
> > > > > > > towards minus infinity while
> > > > > > > ISO rounds
> > > > > > >
> > > > > > > towards 0.
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > My naive view on remainder is
> > > > > > > that the nearest
> > > > > > > integer quotient shall be
> > > > > > > smaller in
> > > > > > >
> > > > > > > magnitude and not smaller in
> > > > > > > value. Regarding
> > > > > > > your proposal (which is
> > > > > > > different from
> > > > > > >
> > > > > > > both IBM and ISO) my concern is
> > > > > > > that may lead
> > > > > > > to different results for modulo
> > > > > > > N and
> > > > > > >
> > > > > > > modulo N×1J0
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > Best Regards,
> > > > > > >
> > > > > > > Jürgen Sauermann
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > On
> > > > > > > 06/21/2017 03:08 AM,
> > > > > > > Frederick Pitts wrote:
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > > Jürgen,
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > This
> > > > > > > > message is being resent
> > > > > > > > because last
> > > > > > > > minute changes I made
> > > > > > > > to CRS0.apl and
> > > > > > > > CRS1.apl do not output
> > > > > > > > the
> > > > > > > > data I
> > > > > > > > intended. This message
> > > > > > > > has corrected
> > > > > > > > versions of those files
> > > > > > > > attached.
> > > > > > > > Please discard the old
> > > > > > > > CRS0.apl and
> > > > > > > > CRS1.apl files. The
> > > > > > > > first line of
> > > > > > > > output is the modulo
> > > > > > > > basis, the second
> > > > > > > > line is the calculated
> > > > > > > > complete residue
> > > > > > > > system values and the
> > > > > > > > third line is the
> > > > > > > > number of residues in
> > > > > > > > the CRS on the
> > > > > > > > previous line.
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > CRSOTST0.apl and
> > > > > > > > CRSOTST1.apl are
> > > > > > > > unchanged.
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > Also
> > > > > > > > please find attached
> > > > > > > > MOD_TEST.apl which
> > > > > > > > compares the residues
> > > > > > > > calculated by MODJ
> > > > > > > > and the builtin residue
> > > > > > > > function and
> > > > > > > > reports discrepancies.
> > > > > > > > The first column
> > > > > > > > of output is the modulo
> > > > > > > > basis, the
> > > > > > > > second column the right
> > > > > > > > argument to the
> > > > > > > > functions, the third
> > > > > > > > column the MODJ
> > > > > > > > result and the fourth
> > > > > > > > column is the
> > > > > > > > builtin residue
> > > > > > > > function result.
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > Regards
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > Fred
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > Hello Jürgen,
> > > > > > > > SVN 964 moved us in
> > > > > > > > the right direction but not completely out of the
> > > > > > > > woods. SVN 964 still
> > > > > > > > exhibits errors. For instance
> > > > > > > > 2J6 | 5J5
> > > > > > > > ¯1J7
> > > > > > > > 2J6 | ¯1J7
> > > > > > > > ¯3J1
> > > > > > > > 2J6 | ¯3J1
> > > > > > > > ¯3J1
> > > > > > > > I found this and
> > > > > > > > previous residue function errors using the attached APL
> > > > > > > > code files. The files with
> > > > > > > > base name ending in '0' use the builtin residue
> > > > > > > > function. Those with base
> > > > > > > > name ending in '1' use a residue function implemented
> > > > > > > > in APL. The files with
> > > > > > > > base name beginning with 'CRSOTST' test if the order of
> > > > > > > > the complete residue system
> > > > > > > > (CRS) equals the norm of the modulo basis. That
> > > > > > > > test fails for several
> > > > > > > > modulo bases, 2J6 being one of them, using the builtin
> > > > > > > > residue function. No errors
> > > > > > > > are detected with the APL implementation. The other
> > > > > > > > files
> > > > > > > > can be used to plot the CRS
> > > > > > > > for a given modulo basis where 'a' and 'b' in
> > > > > > > > 'a + b * i' are limited to
> > > > > > > > +15 to -15 range. A full screen terminal window is
> > > > > > > > needed to see the plot.
> > > > > > > > My APL
> > > > > > > > implementation of the residue function is very close to
> > > > > > > > what you
> > > > > > > > described in your previous
> > > > > > > > email. Maybe comparing the two implementations will
> > > > > > > > give insight into why the
> > > > > > > > builtin residue function fails for some modulo bases.
> > > > > > > > I make no assertion
> > > > > > > > that my implementation is correct in all
> > > > > > > > aspects.
> > > > > > > > Regards,
> > > > > > > > Fred
> > > > > > > > On Tue, 2017-06-20 at 14:14
> > > > > > > > +0200,
> > > > > > > > Juergen Sauermann wrote:
> > > > > > > >
> > > > > > > > > Hi
> > > > > > > > > Frederick,
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > the algorithm for A ∣
> > > > > > > > > B used in
> > > > > > > > > GNU APL is this:
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > - compute the
> > > > > > > > > quotient Q←B÷A,
> > > > > > > > >
> > > > > > > > > - "round down" Q to
> > > > > > > > > the next
> > > > > > > > > (complex) integer
> > > > > > > > > Q1,
> > > > > > > > >
> > > > > > > > > - return B - Q1×A
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > Now the problem seems
> > > > > > > > > to be what is
> > > > > > > > > meant by "round
> > > > > > > > > down". There are two
> > > > > > > > > candidates:
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > Q1 ← ⌊
> > > > > > > > > Q
> > > > > > > > >
> > > > > > > > > i.e. use APL
> > > > > > > > > floor to
> > > > > > > > > round down Q
> > > > > > > > >
> > > > > > > > > Q1 ← Complex(
> > > > > > > > > floor(Q.real(),
> > > > > > > > > floor(Q.imag())
> > > > > > > > > ) i,e, use
> > > > > > > > > C/C++ floor()
> > > > > > > > > to round down Q.
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > In your 5J3 ∣ 14J5
> > > > > > > > > example, the
> > > > > > > > > quotient is 2.5J¯0.5,
> > > > > > > > > which
> > > > > > > > > gives different
> > > > > > > > > results for the APL
> > > > > > > > > floor ⌊ and the C/C++
> > > > > > > > > floor().
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > The APL floor
> > > > > > > > > ⌊2.5J¯0.5 is
> > > > > > > > > 3J¯1 (a
> > > > > > > > > somewhat dubious
> > > > > > > > > invention in the ISO
> > > > > > > > > standard on page 19,
> > > > > > > > > which I used up to
> > > > > > > > >
> > > > > > > > > including SVN 963),
> > > > > > > > > while the
> > > > > > > > > C/C++ floor() is
> > > > > > > > > 2J¯1.
> > > > > > > > > The difference
> > > > > > > > > between the APL floor and
> > > > > > > > > the C/C++ floor is
> > > > > > > > > 1.0 which,
> > > > > > > > >
> > > > > > > > > multiplied by the
> > > > > > > > > divisor, explains the
> > > > > > > > > differences that we
> > > > > > > > > see.
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > As of SVN 964 I have
> > > > > > > > > changed the
> > > > > > > > > residue function (∣)
> > > > > > > > > to use the
> > > > > > > > > C/C++ floor instead
> > > > > > > > > of the APL floor.
> > > > > > > > > The APL floor and
> > > > > > > > >
> > > > > > > > > Ceiling functions (⌊
> > > > > > > > > and ⌈)
> > > > > > > > > are still using the
> > > > > > > > > apparently broken
> > > > > > > > > definition in the ISO
> > > > > > > > > standard.
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > I hope this works
> > > > > > > > > better for you. At
> > > > > > > > > least I am getting
> > > > > > > > > this in SVN 964:
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > 5J3 | 14J5
> > > > > > > > >
> > > > > > > > > 1J4
> > > > > > > > >
> > > > > > > > > 5J3 | 1J4
> > > > > > > > >
> > > > > > > > > 1J4
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > whereas SVN 963 was
> > > > > > > > > giving:
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > 5J3 | 14J5
> > > > > > > > >
> > > > > > > > > ¯4J1
> > > > > > > > >
> > > > > > > > > 5J3 | 1J4
> > > > > > > > >
> > > > > > > > > ¯4J1
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > Best Regards,
> > > > > > > > >
> > > > > > > > > /// Jürgen
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > On
> > > > > > > > > 06/19/2017 07:03 PM,
> > > > > > > > > Frederick Pitts
> > > > > > > > > wrote:
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > > Jürgen,
> > > > > > > > > >
> > > > > > > > > > With gnu apl (svn 961 on Fedora 25, Intel(R)
> > > > > > > > > > Core(TM) i7-6700
> > > > > > > > > > CPU), the residue function (∣) yields the
> > > > > > > > > > following:
> > > > > > > > > >
> > > > > > > > > > 5J3 ∣ 14J5
> > > > > > > > > > 1J4
> > > > > > > > > > 5J3 | 1J4
> > > > > > > > > > ¯4J1
> > > > > > > > > > 5J3 | ¯4J1
> > > > > > > > > > ¯4J1
> > > > > > > > > > The above result means that two elements in the
> > > > > > > > > > complete residue system
> > > > > > > > > > (CSR) for mod 5J3 are equal, i.e. 1J4 = ¯4J1 mod
> > > > > > > > > > 5J3, which is not
> > > > > > > > > > allowed. None of the elements of a CSR can be
> > > > > > > > > > equal modulo the CSR's
> > > > > > > > > > basis.
> > > > > > > > > >
> > > > > > > > > > Regards,
> > > > > > > > > >
> > > > > > > > > > Fred
> > > > > > > > > >
> > > > > > > > > >
> > > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > >
> > > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > >
> > > >
> > > >
> > > >
> > >
> > >
> > >
> > >
> > >
> >
> >
> >
> >
> >
> >
>
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