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In the link
http://plato.stanford.edu/entries/qt-quantlog/#1
you will find the remark
"Whereas logicians have usually assumed that
properties … of negation were the ones least
able to withstand a critical analysis, the study of
mechanics points to the distributive identities …
as the weakest link in the algebra of logic.
[1937, p. 839]"
There is nothing particularly authoritative about the link. It is just
that I have heard this particular statement about negation before and
wished to provide some source.
If you scroll down to where you find quasigroup products such as
5*0=4
| 1 | | | | 0 |
| | | N | | |
| 0 | | | | 1 |
| | * | O | = | |
| 0 | | | | 1 |
| | | T | | |
| 1 | | | | 0 |
you can decide if you are interested in this post.
I am tired of arguing.
:-)
mitch
---
I will start this post presenting the incidence relations for the
trivial affine geometry consisting of 4 points and 6 lines.
[begin fixed width]
a b c d e f
A x x x
B x x x
C x x x
D x x x
[end fixed width]
This starting point was explained somewhat in
news:<[EMAIL PROTECTED]>. The explanations will be
revisited as needed in this post.
As was done at that time, comparison is made to the presentation,
[begin fixed width]
a b c d e f
(1,1) 1 0 0 1 1 0
(1,0) 1 1 0 0 0 1
(0,1) 0 1 1 0 1 0
(0,0) 0 0 1 1 0 1
[end fixed width]
So that it can be seen that a fixed specification for six connectives of
the sixteen propositional connectives relate to one another in a manner
comparable to the components of the geometric model.
As discussed in news:<[EMAIL PROTECTED]>, however, the
actual intuitions involved here come from rotational symmetries in solid
geometry. The 2.3.4 symmetry family consists of six 2-fold symmetries,
four 3-fold symmetries, and three 4-fold symmetries.
Relative to a cube, a 2-fold symmetry is a 180 degree rotation. For a
given 2-fold rotation, the axis of symmetry is the line from the
midpoint of a given edge to the midpoint of the opposite edge. The six
2-fold symmetries segregate into pairs according to their relation to
the three 4-fold symmetries.
To see this, consider the diagram,
[begin fixed width]
/|
/ |
------/--|----------
/ / | /
/ / | /
/ / / /
/ | / /
/ | / /
--------|---/--------
| /
| /
|/
[end fixed width]
The line corresponding to the intersection of the two planes would
correspond to one of the 4-fold symmetries. The following diagram
attempts to show axes for the 4-fold symmetries. The '*' symbol
represents the intersection of the three axes in the center of the cube
and the '+' symbols represent the center of the three faces through
which these axes pass.
Obviously, the diagram lacks perspective given the nature of the
presentation in ASCII.
[begin fixed width]
|
|
-----|------------------
/ | /|
/ | / |
/ | / |
/ + / |
/ | / |
/ | / |
/ | / |
------------------------ |
| | | |
| | | |
| |/ | /
| --*---------|--+---------
| /| | /
| / | /
| + | /
| / | /
| / |/
-------/----------------
/
/
[end fixed width]
So, each 4-fold symmetry corresponds with four of the 2-fold symmetries.
The reason the 3-dimensional perspective becomes important is that there
are 24 possible ways to orient a truth table using columns chosen from
among
[begin fixed width]
1 0 0 1 1 0
1 1 0 0 0 1
0 1 1 0 1 0
0 0 1 1 0 1
[end fixed width]
A specific selection of two vectors that accomplishes an orientation
must be chosen so that they are not complementary--that is, such a
selection must result in 4 distinct rows when the two columns are placed
into a 4x2 matrix. So, while
[begin fixed width]
a b c d e f
(1,1) 1 0 0 1 1 0
(1,0) 1 1 0 0 0 1
(0,1) 0 1 1 0 1 0
(0,0) 0 0 1 1 0 1
[end fixed width]
had been used to begin this discussion, it is certainly comparable to
something like
[begin fixed width]
a b c d e f
(1,0) 1 0 0 1 1 0
(0,0) 1 1 0 0 0 1
(0,1) 0 1 1 0 1 0
(1,1) 0 0 1 1 0 1
[end fixed width]
where columns d and c are used rather than colums a and e. That is, the
resultant interpretation still yields the projection connectives, their
complements, logical equivalence, and exclusive disjunction.
But, the reason for the 3-dimensional perspective is that such a
selection constitutes a deselection of a complementary pair. For the
introductory example, that pair would be
[begin fixed width]
1 0
0 1
0 1
1 0
[end fixed width]
and for the second example it would be
[begin fixed width]
1 0
0 1
1 0
0 1
[end fixed width]
In both cases, the deselected pair ends up being interpreted as logical
equivalence and exclusive disjunction.
It is this deselection that corresponds to the relationship with the
4-fold rotational symmetry. One manifestation of this relationship can
be seen in
http://citeseer.nj.nec.com/feigelson97forbidden.html
where Lemma 4 concludes with
"Therefore, there are precisely two assignments
that are justifying for both x and y"
followed by the footnote,
"Before proving Lemma 4, note that it is stated only
for |V|>2. If |V|=2 then there are only four possible
assignments to the variables of V and f must be either
XOR or ~XOR. In either case, all four assignments
are justifying for both x and y. Thus the conclusion of
the lemma does not hold for |V|=2."
But, it has proven somewhat fruitless to attempt establishing this
4-fold relationship using the recognition problem for unate switching
functions.
Hopefully, the above remarks suffice to show reasonable cause for
investigating the relation between logic, the trivial affine geometry,
and the symmetry families in solid geometry.
I would like to now direct attention to the diagram in
http://phil240.tamu.edu/LectureNotes/6.3.pdf
As noted in a prior posting, this paper shows a picture of a Venn
diagram labeled with 8 regions. The paper discusses the use of Venn
diagrams for testing the validity of categorical syllogisms. The method
described in the paper is vaguely comparable to what is being discussed
here to the extent that representing the conclusion involves only two of
the three terms. Our purpose is somewhat more subtle.
In a prior post, we noted that the mapping
[begin fixed width]
1 -> 3
2 -> 4
3 -> 6
4 -> 2
5 -> 0
6 -> 1
7 -> 5
8 -> #
[end fixed width]
would align the diagram with the multiplication table,
[begin fixed width]
* | 0 1 2 3 4 5 6
--|----------------------------
0 | 0 3 6 1 5 4 2
|
1 | 3 1 4 0 2 6 5
|
2 | 6 4 2 5 1 3 0
|
3 | 1 0 5 3 6 2 4
|
4 | 5 2 1 6 4 0 3
|
5 | 4 6 3 2 0 5 1
|
6 | 2 5 0 4 3 1 6
[end fixed width]
For the moment ignore region 8 and its mapping to the octothorpe
(Previously, I had used an ampersand; but, 8 and octo- seem to go
together better.).
The multiplication given here is for a Steiner quasigroup. By
definition, a Steiner quasigroup is a commutative quasigroup whose
binary quasigroup satisfies
x*x=x
(x*y)*y=x
The multiplication table given above is for the Steiner quasigroup of
order seven.
With respect to the incidence matrixes given above, we need to extend
our mapping to a new set of labels. Once again, the first number comes
from the diagram in
http://phil240.tamu.edu/LectureNotes/6.3.pdf
while the second number is associated with the multiplication table
above. So, the new labels are given by
[begin fixed width]
| 1 |
| |
| 1 |
1 -> 3 -> | |
| 0 |
| |
| 0 |
| 0 |
| |
| 1 |
2 -> 4 -> | |
| 1 |
| |
| 0 |
| 1 |
| |
| 0 |
3 -> 6 -> | |
| 1 |
| |
| 0 |
| 0 |
| |
| 1 |
4 -> 2 -> | |
| 0 |
| |
| 1 |
| |
| N |
| |
5 -> 0 -> | O |
| |
| T |
| |
| 0 |
| |
| 0 |
6 -> 1 -> | |
| 1 |
| |
| 1 |
| 1 |
| |
| 0 |
7 -> 5 -> | |
| 0 |
| |
| 1 |
[end fixed width]
What now follows is a long listing verifying the sense of the quasigroup
products for these labels. We ignore the idempotence of the products,
x*x=x
The first grouping involves interpretation of the quasigroup 0 as the
complementation operation.
[begin fixed width]
0*1=3
| | | 0 | | 1 |
| N | | | | |
| | | 0 | | 1 |
| O | * | | = | |
| | | 1 | | 0 |
| T | | | | |
| | | 1 | | 0 |
3*1=0
| 1 | | 0 | | |
| | | | | N |
| 1 | | 0 | | |
| | * | | = | O |
| 0 | | 1 | | |
| | | | | T |
| 0 | | 1 | | |
3*0=1
| 1 | | | | 0 |
| | | N | | |
| 1 | | | | 0 |
| | * | O | = | |
| 0 | | | | 1 |
| | | T | | |
| 0 | | | | 1 |
1*0=3
| 0 | | | | 1 |
| | | N | | |
| 0 | | | | 1 |
| | * | O | = | |
| 1 | | | | 0 |
| | | T | | |
| 1 | | | | 0 |
1*3=0
| 0 | | 1 | | |
| | | | | N |
| 0 | | 1 | | |
| | * | | = | O |
| 1 | | 0 | | |
| | | | | T |
| 1 | | 0 | | |
0*3=1
| | | 1 | | 0 |
| N | | | | |
| | | 1 | | 0 |
| O | * | | = | |
| | | 0 | | 1 |
| T | | | | |
| | | 0 | | 1 |
0*2=6
| | | 0 | | 1 |
| N | | | | |
| | | 1 | | 0 |
| O | * | | = | |
| | | 0 | | 1 |
| T | | | | |
| | | 1 | | 0 |
6*2=0
| 1 | | 0 | | |
| | | | | N |
| 0 | | 1 | | |
| | * | | = | O |
| 1 | | 0 | | |
| | | | | T |
| 0 | | 1 | | |
6*0=2
| 1 | | | | 0 |
| | | N | | |
| 0 | | | | 1 |
| | * | O | = | |
| 1 | | | | 0 |
| | | T | | |
| 0 | | | | 1 |
2*0=6
| 0 | | | | 1 |
| | | N | | |
| 1 | | | | 0 |
| | * | O | = | |
| 0 | | | | 1 |
| | | T | | |
| 1 | | | | 0 |
2*6=0
| 0 | | 1 | | |
| | | | | N |
| 1 | | 0 | | |
| | * | | = | O |
| 0 | | 1 | | |
| | | | | T |
| 1 | | 0 | | |
0*6=2
| | | 1 | | 0 |
| N | | | | |
| | | 0 | | 1 |
| O | * | | = | |
| | | 1 | | 0 |
| T | | | | |
| | | 0 | | 1 |
0*5=4
| | | 1 | | 0 |
| N | | | | |
| | | 0 | | 1 |
| O | * | | = | |
| | | 0 | | 1 |
| T | | | | |
| | | 1 | | 0 |
4*5=0
| 0 | | 1 | | |
| | | | | N |
| 1 | | 0 | | |
| | * | | = | O |
| 1 | | 0 | | |
| | | | | T |
| 0 | | 1 | | |
4*0=5
| 0 | | | | 1 |
| | | N | | |
| 1 | | | | 0 |
| | * | O | = | |
| 1 | | | | 0 |
| | | T | | |
| 0 | | | | 1 |
5*0=4
| 1 | | | | 0 |
| | | N | | |
| 0 | | | | 1 |
| | * | O | = | |
| 0 | | | | 1 |
| | | T | | |
| 1 | | | | 0 |
5*4=0
| 1 | | 0 | | |
| | | | | N |
| 0 | | 1 | | |
| | * | | = | O |
| 0 | | 1 | | |
| | | | | T |
| 1 | | 0 | | |
0*4=5
| | | 0 | | 1 |
| N | | | | |
| | | 1 | | 0 |
| O | * | | = | |
| | | 1 | | 0 |
| T | | | | |
| | | 0 | | 1 |
[end fixed width]
The remainder of this listing will give products whose factors could be
interpreted as orienting columns for a truth table. The reader is asked
to observe the consistency with which the constant rows are matched to
the same value (0).
[begin fixed width]
2*1=4
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
| | * | | = | |
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
4*1=2
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
| | * | | = | |
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
2*4=1
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 1 | | 0 |
| | * | | = | |
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 0 | | 1 |
1*4=2
| 0 | | 0 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
| | * | | = | |
| 1 | | 1 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
4*2=1
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 1 | | 0 |
| | * | | = | |
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 1 | | 1 |
1*2=4
| 0 | | 0 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
| | * | | = | |
| 1 | | 0 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
6*1=5
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
| | * | | = | |
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
5*1=6
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
| | * | | = | |
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
6*5=1
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 0 | | 0 |
| | * | | = | |
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 1 | | 1 |
1*5=6
| 0 | | 1 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
| | * | | = | |
| 1 | | 0 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
1*6=5
| 0 | | 1 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
| | * | | = | |
| 1 | | 1 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
5*6=1
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 0 | | 0 |
| | * | | = | |
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 0 | | 1 |
3*2=5
| 1 | | 0 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
| | * | | = | |
| 0 | | 0 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
5*2=3
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 1 | | 1 |
| | * | | = | |
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 1 | | 0 |
3*5=2
| 1 | | 1 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
| | * | | = | |
| 0 | | 0 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
2*5=3
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 0 | | 1 |
| | * | | = | |
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 1 | | 0 |
2*3=5
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
| | * | | = | |
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
5*3=2
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
| | * | | = | |
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
6*3=4
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
| | * | | = | |
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
4*3=6
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
| | * | | = | |
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
6*4=3
| 1 | | 0 | | 1 |
| | | | | |
| 0 | | 1 | | 1 |
| | * | | = | |
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 0 | | 0 |
3*4=6
| 1 | | 0 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
| | * | | = | |
| 0 | | 1 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
3*6=4
| 1 | | 1 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
| | * | | = | |
| 0 | | 1 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
4*6=3
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 0 | | 1 |
| | * | | = | |
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 0 | | 0 |
[end fixed width]
Once again, we shall eventually discuss region 8 from our diagram (our
octothorpe). However, at this point we consider what has been done so
far. The exposition began by discussing how the list of columns
[begin fixed width]
1 0 0 1 1 0
1 1 0 0 0 1
0 1 1 0 1 0
0 0 1 1 0 1
[end fixed width]
relate to the manner by which the system of propositional connectives in
logic associate with one another via truth tables. In this analysis, we
noted that the orientation choices segregate logical equivalence and
exclusive disjunction from the projections and their complements. This
aspect of the orientation was related to the 2.3.4 family of rotational
symmetries from solid geometry, and, an algebraic system has been
presented so that context specific labels relate complements to one
another while simultaneously associating labels so that possible
orientations of the propositional system enforce the segregating
relationship with logical equivalence and exclusive disjunction.
Now we can look at what all of this has accomplished with respect to
invariance.
The Steiner triple system that results from the quasigroup above
consists of the triples,
[begin fixed width]
{0,1,3}
| | | 0 | | 1 |
| N | | | | |
| | | 0 | | 1 |
| O | | | | |
| | | 1 | | 0 |
| T | | | | |
| | | 1 | | 0 |
{0,2,6}
| | | 0 | | 1 |
| N | | | | |
| | | 1 | | 0 |
| O | | | | |
| | | 0 | | 1 |
| T | | | | |
| | | 1 | | 0 |
{0,4,5}
| | | 0 | | 1 |
| N | | | | |
| | | 1 | | 0 |
| O | | | | |
| | | 1 | | 0 |
| T | | | | |
| | | 0 | | 1 |
{1,2,4}
| 0 | | 0 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 0 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
{1,6,5}
| 0 | | 1 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 1 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
{2,3,5}
| 0 | | 1 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 0 | | 0 |
| | | | | |
| 1 | | 0 | | 1 |
{3,4,6}
| 1 | | 0 | | 1 |
| | | | | |
| 1 | | 1 | | 0 |
| | | | | |
| 0 | | 1 | | 1 |
| | | | | |
| 0 | | 0 | | 0 |
[end fixed width]
If we recall our original incidence matrix,
[begin fixed width]
a b c d e f
A x x x
B x x x
C x x x
D x x x
[end fixed width]
we may compare it with the relabeling,
[begin fixed width]
a/1 b/5 c/3 d/4 e/2 f/6
{1,2,4} x x x
{1,6,5} x x x
{2,3,5} x x x
{3,4,6} x x x
[end fixed width]
It should be obvious what is going on here:
a corresponds with 1 because {1}={1,2,4} cap {1,6,5}
b corresponds with 5 because {5}={2,3,5} cap {1,6,5}
c corresponds with 3 because {3}={3,4,6} cap {2,3,5}
d corresponds with 4 because {4}={1,2,4} cap {3,4,6}
e corresponds with 2 because {2}={1,2,4} cap {2,3,5}
f corresponds with 6 because {6}={3,4,6} cap {1,6,5}
It should also be clear that our construction has reversed the sense of
our original comparison. Whereas we translated the original incidence
matrix as
[begin fixed width]
a b c d e f
A 1 0 0 1 1 0
B 1 1 0 0 0 1
C 0 1 1 0 1 0
D 0 0 1 1 0 1
[end fixed width]
this relabeling corresponds with
[begin fixed width]
a b c d e f
A 0 1 1 0 0 1
B 0 0 1 1 1 0
C 1 0 0 1 0 1
D 1 1 0 0 1 0
[end fixed width]
But, we note that our quasigroup was formulated without any semantic
interpretation of 0 and 1, whence the juxtaposition is an artifact of
the introductory exposition rather than any failure on the part of the
formal structure being discussed.
It is important to understand what is going on here. First of all,
there is nothing particularly special about the triples
{1,2,4}
{1,6,5}
{2,3,5}
{3,4,6}
Relative to the triple system with which they are associated, they form
a Pasch configuration (also known as a fragment or a quadrilateral).
Such a configuration is characterized by the relations
{a,w,x}
{a,y,z}
{b,w,z}
{b,x,y}
so that, for example,
{0,1,3}
{0,2,6}
{5,1,6}
{5,3,2}
is also a Pasch configuration for our system. What is important,
however, is that the Pasch configuration used to label the incidence
matrix is the single Pasch configuration whose triples are composed only
of those labels referring to the columns.
Unfortunately, we are not yet in a position to understand this labeling
as independent from the assignment
[begin fixed width]
| |
| N |
| |
0 = | O |
| |
| T |
| |
[end fixed width]
The reason for this is that a Steiner triple system specified by
{0,1,3}
{0,2,6}
{0,4,5}
{1,2,4}
{1,6,5}
{2,3,5}
{3,4,6}
is nested in a (7,4,2)-design given by
[begin fixed width]
|-------|---|
| 0 1 3 | 6 |
| | |
| 1 2 4 | 0 |
| | |
| 2 3 5 | 1 |
| | |
| 3 4 6 | 2 |
| | |
| 4 5 0 | 3 |
| | |
| 5 6 1 | 4 |
| | |
| 6 0 2 | 5 |
|-------|---|
[end fixed width]
By a (7,4,2)-design we mean that 7 symbols can be described by blocks of
4 symbols so that any given pair of symbols will occur in 2 blocks. The
Steiner triple system of our quasigroup is a (7,3,1)-design.
To understand why this (7,4,2)-design interferes with our objectives,
first consider the triple system that is of interest in the sense that
it only refers to labels for the columns in our incidence matrix:
{1,2,3}
{1,3,4}
{1,4,5}
{1,5,6}
{1,2,6}
{2,4,6}
{2,3,5}
{3,4,6}
{2,4,5}
{3,5,6}
This is a (6,3,2)-design--or, in other words, a TS(6,2) triple system
since the block size is 3. Interest in this design is specifically
motivated by the desire to recover a purely geometric interpretation for
the incidence matrix.
To see how the (7,4,2)-design interferes with the geometric
intepretation, we recast the explanation of the newly labeled incidence
matrix to see what is being obscured by the set-theoretic intersection
operation. The presentation is given in outline form so that the
relationship between the (7,4,2)-design and the (6,3,2)-design is made
clear. Also, for similar reasons, there is an abuse of notation
involving ordered pairs.
So, now we write:
[begin fixed width]
I. a corresponds with 1 because {1}={1,2,4} cap {1,6,5}
A. (7,4,2)-design
1. {<1,4>,0,2}
2. {<1,4>,5,6}
B. (6,3,2)-design
1. {<1,4>,3}
2. {<1,4>,5}
II. b corresponds with 5 because {5}={2,3,5} cap {1,6,5}
A. (7,4,2)-design
1. {<1,5>,2,3}
2. {<1,5>,4,6}
B. (6,3,2)-design
1. {<1,5>,4}
2. {<1,5>,6}
III. c corresponds with 3 because {3}={3,4,6} cap {2,3,5}
A. (7,4,2)-design
1. {<2,3>,4,6}
2. {<2,3>,1,5}
B. (6,3,2)-design
1. {<2,3>,1}
2. {<2,3>,5}
IV. d corresponds with 4 because {4}={1,2,4} cap {3,4,6}
A. (7,4,2)-design
1. {<2,4>,0,1}
2. {<2,4>,3,6}
B. (6,3,2)-design
1. {<2,4>,5}
2. {<2,4>,6}
V. e corresponds with 2 because {2}={1,2,4} cap {2,3,5}
A. (7,4,2)-design
1. {<1,2>,0,4}
2. {<1,2>,3,5}
B. (6,3,2)-design
1. {<1,2>,3}
2. {<1,2>,6}
VI. f corresponds with 6 because {6}={3,4,6} cap {1,6,5}
A. (7,4,2)-design
1. {<4,6>,1,5}
2. {<4,6>,3,2}
B. (6,3,2)-design
1. {<4,6>,2}
2. {<4,6>,3}
[end fixed width]
Note that the (6,3,2)-design triples do not convey the properties of the
geometric interpretation directly. Rather, the outline above is
presented to emphasize the fact that the (7,4,2)-design emulates the
(6,3,2)-design triples in a definite way. Specifically, the relation
between the designs is such that a naive use of set-theoretic operations
obtains the same result.
Later on, we will have cause to return to the assignment,
[begin fixed width]
| |
| N |
| |
0 = | O |
| |
| T |
| |
[end fixed width]
so that we can understand how it chromatically generates the Steiner
triple system of our quasigroup.
However, we are now in a position to begin discussing region 8 from our
diagram and the octothorpe symbol to which we mapped it. Recall the
multiplication table for our Steiner quasigroup,
[begin fixed width]
* | 0 1 2 3 4 5 6
--|----------------------------
0 | 0 3 6 1 5 4 2
|
1 | 3 1 4 0 2 6 5
|
2 | 6 4 2 5 1 3 0
|
3 | 1 0 5 3 6 2 4
|
4 | 5 2 1 6 4 0 3
|
5 | 4 6 3 2 0 5 1
|
6 | 2 5 0 4 3 1 6
[end fixed width]
Now, given a Steiner triple system, it is possible to extend it to a
totally symmetric loop. This loop is called a Steiner loop.
A totally symmetric loop is obtained from a Steiner triple system by
extending the symbol set by one symbol,
<0,1,2,3,4,5,6> -> <0,1,2,3,4,5,6,#>
The symbols are given a product satisfying:
1. #*x = x*# = x
2. x*y=z whenever {x,y,z} is a block of the triple system
It follows that the new symbol set is a totally symmetric loop because
the following axioms are satified,
1. x*y = y*x
2. #*x = x
3. x*x = #
4. x*(x*y) = y
where # is the loop identity.
These products are expressed in the multiplication table for the
unipotent quasigroup,
[begin fixed width]
* | 0 1 2 3 4 5 6 #
--|--------------------------------
0 | # 3 6 1 5 4 2 0
|
1 | 3 # 4 0 2 6 5 1
|
2 | 6 4 # 5 1 3 0 2
|
3 | 1 0 5 # 6 2 4 3
|
4 | 5 2 1 6 # 0 3 4
|
5 | 4 6 3 2 0 # 1 5
|
6 | 2 5 0 4 3 1 # 6
|
# | 0 1 2 3 4 5 6 #
[end fixed width]
The importance of this extension comes from factorization.
Let us begin with the original quasigroup. For a symbol set V, a
Steiner near-1-factorization is defined as
{{{x,y}: x,y e V, ~(x=y), x*y=z} : z e V}
So, the near-1-factorization obtained for the labeling of our
quasigroup, for example, is given by
{{1,3},{2,6},{4,5}}
{{0,3},{2,4},{6,5}}
{{0,6},{1,4},{3,5}}
{{0,1},{2,5},{4,6}}
{{0,5},{1,2},{3,6}}
{{0,4},{2,3},{1,6}}
{{0,2},{3,4},{1,5}}
For the Steiner loop and its associated unipotent quasigroup we can
define the Steiner 1-factorization,
{{{#,z}}cup {{x,y}: x,y e V, ~(x=y), x*y=z} : z e V}
so that our corresponding Steiner 1-factorization is
{{#,0},{1,3},{2,6},{4,5}}
{{#,1},{0,3},{2,4},{6,5}}
{{#,2},{0,6},{1,4},{3,5}}
{{#,3},{0,1},{2,5},{4,6}}
{{#,4},{0,5},{1,2},{3,6}}
{{#,5},{0,4},{2,3},{1,6}}
{{#,6},{0,2},{3,4},{1,5}}
---
I ended here. It is just too long.
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