I urge you to read Eugene McDonnell's Complex Floor
<http://www.jsoftware.com/papers/eem/complexfloor.htm>, 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
>
>
>
>
>
>
>
