Re: [Bug-apl] Modulo or residue function is not generating consistent complete residue systems for some arguments

[Juergen Sauermann]

Hi, please disregard my comment on bif_floor() below. The proper line should be this (note the - instead of +): if (D < 1.0 - Workspace::get_CT())   return zv(Z, fr, fi); With that change my testcases (which contain the examples of both the ISO standard and the IBM language reference) are now properly working again. Fixed in SVN 970. /// Jürgen On 06/23/2017 09:02 PM, Juergen Sauermann wrote: Hi Frederick, I suppose that this is caused by a rounding error just before the complex floor. I have attached a debug version of ComplexCell.cc that prints every partial result of the computation. If I run it then I get:       5J3 | 14J5 modulus (A) is: (5,3) A=0 is: (0,0) A+A=0 is: (5,3) B÷A+A=0 is: (2.5,-0.5) ⌊B÷A+A=0 is: (3,-1) A×⌊B÷A+A=0 is: (18,4) B-A×⌊B÷A+A=0 is: (-4,1) ¯4J1 However if I use a slightly smaller B (kind of simulating a rounding error) then I get:       5J3 | 14J4.999999999 modulus (A) is: (5,3) A=0 is: (0,0) A+A=0 is: (5,3) B÷A+A=0 is: (2.4999999999118,-0.50000000014706) ⌊B÷A+A=0 is: (2,-1) A×⌊B÷A+A=0 is: (13,1) B-A×⌊B÷A+A=0 is: (1,3.999999999) 1J3.999999999 I suppose that is what happens on your machine but not on mine. You may also want to try out the following. In ComplexCell::bif_floor() around line 231 ff it says:    // ISO: if D is tolerantly less than 1 return fr + 0J1×fi    // IBM: if D is            less than 1 return fr + 0J1×fi    // However, ISO examples follow IBM (and so do we)    // // if (D < 1.0 + Workspace::get_CT())   return zv(Z, fr, fi);   // ISO    if (D < 1.0)    return zv(Z, fr, fi);   // IBM and examples in ISO If you uncomment the first if statement then you get the ISO variant of bif_floor() while otherwise you get the IBM variant (which is also the one described in the McDonnel paper). It could be that the ISO variant works better for the bif_residue() function, but it causes problems in other places. If uncommenting the if line should fix your problem, then I can create two variants of bif_floor(): one for bif_residue() and one for the rest. Best Regards, Jürgen Sauermann On 06/23/2017 07:31 PM, Frederick Pitts wrote: 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