Hello Jürgen,
Some observations:
1) When performing a residue calculation on positive integers, a
straight-forward integer division with remainder
calculationsuffices. 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 theinteger 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 recommendedrounding down to
the nearest real and imaginary part toward negative
infinity. Unfortunately I cannot findthe 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.outCRS0-CRS1-6J-6-SVN965.out. The first file contains
a CRS map for modulus ¯6J¯6 produced with the residue functionfollowed
by a map for the same modulus produced with MODJ using SVN 964. The
second file contains the same mapsusing 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
gridof 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 thatwas 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 ISOimply 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
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > >
> > > > >
> > > >
> > > >
> > > >
> > > >
> > >
> > >
> >
> >
> >
> >
>
>
>
>
> ¯6J¯6
¯3J¯3 3J¯8 3J¯7 3J¯6 3J¯5 3J¯4 3J¯3 ¯3J¯8 ¯3J¯7 ¯3J¯6 ¯3J¯5 ¯3J¯4 ¯2J¯3 ¯2J¯2
4J¯7 4J¯6 4J¯5
4J¯4 ¯2J¯9 ¯2J¯8 ¯2J¯7 ¯2J¯6 ¯2J¯5 ¯2J¯4 ¯1J¯3 ¯1J¯2 ¯1J¯1 5J¯6 5J¯5
¯1J¯10 ¯1J¯9 ¯1J¯8
¯1J¯7 ¯1J¯6 ¯1J¯5 ¯1J¯4 0J¯3 0J¯2 0J¯1 0 0J¯11 0J¯10 0J¯9 0J¯8 0J¯7 0J¯6
0J¯5 0J¯4 1J¯3
1J¯2 1J¯1 ¯5J¯6 ¯5J¯5 1J¯10 1J¯9 1J¯8 1J¯7 1J¯6 1J¯5 1J¯4 2J¯3 2J¯2 ¯4J¯7
¯4J¯6 ¯4J¯5
¯4J¯4 2J¯9 2J¯8 2J¯7 2J¯6 2J¯5 2J¯4
72
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- ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
¯6J¯6
¯3J¯3 3J¯8 3J¯7 3J¯6 3J¯5 3J¯4 3J¯3 ¯3J¯8 ¯3J¯7 ¯3J¯6 ¯3J¯5 ¯3J¯4 ¯2J¯3 ¯2J¯2
4J¯7 4J¯6 4J¯5
4J¯4 ¯2J¯9 ¯2J¯8 ¯2J¯7 ¯2J¯6 ¯2J¯5 ¯2J¯4 ¯1J¯3 ¯1J¯2 ¯1J¯1 5J¯6 5J¯5
¯1J¯10 ¯1J¯9 ¯1J¯8
¯1J¯7 ¯1J¯6 ¯1J¯5 ¯1J¯4 0J¯3 0J¯2 0J¯1 0 0J¯11 0J¯10 0J¯9 0J¯8 0J¯7 0J¯6
0J¯5 0J¯4 1J¯3
1J¯2 1J¯1 ¯5J¯6 ¯5J¯5 1J¯10 1J¯9 1J¯8 1J¯7 1J¯6 1J¯5 1J¯4 2J¯3 2J¯2 ¯4J¯7
¯4J¯6 ¯4J¯5
¯4J¯4 2J¯9 2J¯8 2J¯7 2J¯6 2J¯5 2J¯4
72
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
- ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
CLEAR WS
¯6J¯6
3J3 3J4 3J5 3J6 3J7 3J8 ¯3J3 ¯3J4 ¯3J5 ¯3J6 ¯3J7 ¯3J8 ¯2J9 4J4 4J5 4J6 4J7 ¯2J2
¯2J3 ¯2J4 ¯2J5
¯2J6 ¯2J7 ¯2J8 ¯1J9 ¯1J10 5J5 5J6 ¯1J1 ¯1J2 ¯1J3 ¯1J4 ¯1J5 ¯1J6 ¯1J7 ¯1J8
0J9 0J10 0J11
0 0J1 0J2 0J3 0J4 0J5 0J6 0J7 0J8 1J9 1J10 ¯5J5 ¯5J6 1J1 1J2 1J3 1J4 1J5
1J6 1J7 1J8 2J9
¯4J4 ¯4J5 ¯4J6 ¯4J7 2J2 2J3 2J4 2J5 2J6 2J7 2J8
72
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
- ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
¯6J¯6
¯3J¯3 3J¯8 3J¯7 3J¯6 3J¯5 3J¯4 3J¯3 ¯3J¯8 ¯3J¯7 ¯3J¯6 ¯3J¯5 ¯3J¯4 ¯2J¯3 ¯2J¯2
4J¯7 4J¯6 4J¯5
4J¯4 ¯2J¯9 ¯2J¯8 ¯2J¯7 ¯2J¯6 ¯2J¯5 ¯2J¯4 ¯1J¯3 ¯1J¯2 ¯1J¯1 5J¯6 5J¯5
¯1J¯10 ¯1J¯9 ¯1J¯8
¯1J¯7 ¯1J¯6 ¯1J¯5 ¯1J¯4 0J¯3 0J¯2 0J¯1 0 0J¯11 0J¯10 0J¯9 0J¯8 0J¯7 0J¯6
0J¯5 0J¯4 1J¯3
1J¯2 1J¯1 ¯5J¯6 ¯5J¯5 1J¯10 1J¯9 1J¯8 1J¯7 1J¯6 1J¯5 1J¯4 2J¯3 2J¯2 ¯4J¯7
¯4J¯6 ¯4J¯5
¯4J¯4 2J¯9 2J¯8 2J¯7 2J¯6 2J¯5 2J¯4
72
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
- ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘
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∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘
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∘ ∘ ∘ ∘ ∘
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∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ⎕ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ⎕ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
∘ ∘ ∘ ∘ ∘
modulo basis = ¯15J¯5 basis norm = 250 residue count = 252
modulo basis = ¯15J5 basis norm = 250 residue count = 252
modulo basis = ¯14J¯2 basis norm = 200 residue count = 202
modulo basis = ¯14J2 basis norm = 200 residue count = 202
modulo basis = ¯12J¯4 basis norm = 160 residue count = 166
modulo basis = ¯12J4 basis norm = 160 residue count = 166
modulo basis = ¯6J¯4 basis norm = 52 residue count = 54
modulo basis = ¯6J¯2 basis norm = 40 residue count = 42
modulo basis = ¯6J2 basis norm = 40 residue count = 42
modulo basis = ¯6J4 basis norm = 52 residue count = 54
modulo basis = ¯5J¯15 basis norm = 250 residue count = 252
modulo basis = ¯5J15 basis norm = 250 residue count = 252
modulo basis = ¯4J¯12 basis norm = 160 residue count = 166
modulo basis = ¯4J¯6 basis norm = 52 residue count = 54
modulo basis = ¯4J6 basis norm = 52 residue count = 54
modulo basis = ¯4J12 basis norm = 160 residue count = 166
modulo basis = ¯2J¯14 basis norm = 200 residue count = 202
modulo basis = ¯2J¯6 basis norm = 40 residue count = 42
modulo basis = ¯2J6 basis norm = 40 residue count = 42
modulo basis = ¯2J14 basis norm = 200 residue count = 202
modulo basis = 2J¯14 basis norm = 200 residue count = 202
modulo basis = 2J¯6 basis norm = 40 residue count = 42
modulo basis = 2J6 basis norm = 40 residue count = 42
modulo basis = 2J14 basis norm = 200 residue count = 202
modulo basis = 4J¯12 basis norm = 160 residue count = 166
modulo basis = 4J¯6 basis norm = 52 residue count = 54
modulo basis = 4J6 basis norm = 52 residue count = 54
modulo basis = 4J12 basis norm = 160 residue count = 166
modulo basis = 5J¯15 basis norm = 250 residue count = 252
modulo basis = 5J15 basis norm = 250 residue count = 252
modulo basis = 6J¯4 basis norm = 52 residue count = 54
modulo basis = 6J¯2 basis norm = 40 residue count = 42
modulo basis = 6J2 basis norm = 40 residue count = 42
modulo basis = 6J4 basis norm = 52 residue count = 54
modulo basis = 12J¯4 basis norm = 160 residue count = 166
modulo basis = 12J4 basis norm = 160 residue count = 166
modulo basis = 14J¯2 basis norm = 200 residue count = 202
modulo basis = 14J2 basis norm = 200 residue count = 202
modulo basis = 15J¯5 basis norm = 250 residue count = 252
modulo basis = 15J5 basis norm = 250 residue count = 252
1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 1 1
1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 0
1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1
1 0 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1
1 1 1 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1
1 1 1 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1
1 0 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 0
1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1
1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1
1 1 1 1 1
