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Topic: distinguishability - in context, according to definitions
Replies: 43   Last Post: Feb 22, 2013 10:04 AM

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 fom Posts: 1,968 Registered: 12/4/12
Re: distinguishability - in context, according to definitions
Posted: Feb 15, 2013 6:57 AM
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On 2/14/2013 12:40 PM, fom wrote:
> On 2/14/2013 9:32 AM, Shmuel (Seymour J.) Metz wrote:
>> In <qImdnYCz5tRmvITMnZ2dnUVZ_oWdnZ2d@giganews.com>, on 02/11/2013
>> at 10:53 AM, fom <fomJUNK@nyms.net> said:
>>
>> You really need to step back, separate out the philosophy from the
>> mathematics and define any terms that you aren't uisng in accordance
>> with standard practice.

http://finitegeometry.org/sc/24/MOG.html

The Miracle Octad Generator
is a two-dimensional array
with 24 loci. Its standard
labeling is given by the
quadratic residues modulo 23
and 0, that is, the set

Q={0,1,2,3,4,6,8,9,12,13,16,18}

with remaining labels based on
the involution

z:=>(-1/z)

and located according to an involution
on the 24-element array.

a_(0,0)=0
a_(0,1)=oo=-1/0
a_(0,2)=1
a_(0,3)=11=-1/2
a_(0,4)=2
a_(0,5)=22=-1/1
a_(1,0)=19=-1/6
a_(1,1)=3
a_(1,2)=20=-1/8
a_(1,3)=4
a_(1,4)=10=-1/16
a_(1,5)=18
a_(2,0)=15=-1/3
a_(2,1)=6
a_(2,2)=14=-1/18
a_(2,3)=16
a_(2,4)=17=-1/4
a_(2,5)=8
a_(3,0)=5=-1/9
a_(3,1)=9
a_(3,2)=21=-1/12
a_(3,3)=13
a_(3,4)=7=-1/13
a_(3,5)=12

With a distinguished 3-set, the
MOG array partitions into the
given 3-set and a 21-set interpretable
as a 21-point projective plane
relative to the S{5,8,24} Witt design.

For this design, each 5-set of symbols
occurs in some 8-set block with a
multiplicity of 1 -- that is, the
5-sets occur uniquely.

When the MOG is partitioned into
its 35 standard sextets, the loci
may be labelled with 3 Roman numerals
and normalized coordinates over the
Galois field on 4 elements.

Given, {0,1,[],{}}

Let,

1+[]={}
1+{}=[]
{}+[]={}*[]=1
[]^2={}
{}^2=[]
{}^3=[]^3=1
1+1=0
{}+{}=0
[]+[]=0

then

a_(0,0)=(0,1,0)
a_(0,1)=(1,0,0)
a_(0,2)=(0,0,1)
a_(0,3)=(1,0,1)
a_(0,4)=([],0,1)
a_(0,5)=({},0,1)
a_(1,0)=I(Roman)
a_(1,1)=(1,1,0)
a_(1,2)=(0,1,1)
a_(1,3)=(1,1,1)
a_(1,4)=([],1,1)
a_(1,5)=({},1,1)
a_(2,0)=II(Roman)
a_(2,1)=(1,[],0)
a_(2,2)=(0,[],1)
a_(2,3)=(1,[],1)
a_(2,4)=([],[],1)
a_(2,5)=({},[],1)
a_(3,0)=(III)(Roman)
a_(3,1)=(1,{},0)
a_(3,2)=(0,{},1)
a_(3,3)=(1,{},1)
a_(3,4)=([],{},1)
a_(3,5)=({},{}.1)

For the purposes of this construction,
the names occurring as second coordinates
for the alphabet on the free orthomodular
lattice on 2 generators need to be
placed in their corresponding positions.

a_(0,0)=SOME
a_(0,1)=ALL
a_(0,2)=NTRU
a_(0,3)=NOR
a_(0,4)=NIMP
a_(0,5)=DENY
a_(1,0)=I(Roman)
a_(1,1)=NOT
a_(1,2)=AND
a_(1,3)=LEQ
a_(1,4)=FIX
a_(1,5)=IF
a_(2,0)=II(Roman)
a_(2,1)=NO
a_(2,2)=NIF
a_(2,3)=FLIP
a_(2,4)=XOR
a_(2,5)=NAND
a_(3,0)=III(Roman)
a_(3,1)=OTHER
a_(3,2)=LET
a_(3,3)=IMP
a_(3,4)=OR
a_(3,5)=TRU

The correspondence for each Boolean
block of the free orthomodular
lattice on 2 generators is given
as

DENY:=>

a_(0,0)=null
a_(0,1)=null
a_(0,2)=01
a_(0,3)=02
a_(0,4)=05
a_(0,5)=08
a_(1,0)=null
a_(1,1)=null
a_(1,2)=04
a_(1,3)=07
a_(1,4)=11
a_(1,5)=14
a_(2,0)=null
a_(2,1)=null
a_(2,2)=03
a_(2,3)=06
a_(2,4)=10
a_(2,5)=13
a_(3,0)=null
a_(3,1)=null
a_(3,2)=09
a_(3,3)=12
a_(3,4)=15
a_(3,5)=16

FLIP:=>

a_(0,0)=null
a_(0,1)=null
a_(0,2)=17
a_(0,3)=18
a_(0,4)=21
a_(0,5)=24
a_(1,0)=null
a_(1,1)=null
a_(1,2)=20
a_(1,3)=23
a_(1,4)=27
a_(1,5)=30
a_(2,0)=null
a_(2,1)=null
a_(2,2)=19
a_(2,3)=22
a_(2,4)=26
a_(2,5)=29
a_(3,0)=null
a_(3,1)=null
a_(3,2)=25
a_(3,3)=28
a_(3,4)=31
a_(3,5)=32

LEQ:=>

a_(0,0)=null
a_(0,1)=null
a_(0,2)=33
a_(0,3)=34
a_(0,4)=37
a_(0,5)=40
a_(1,0)=null
a_(1,1)=null
a_(1,2)=36
a_(1,3)=39
a_(1,4)=43
a_(1,5)=46
a_(2,0)=null
a_(2,1)=null
a_(2,2)=35
a_(2,3)=38
a_(2,4)=42
a_(2,5)=45
a_(3,0)=null
a_(3,1)=null
a_(3,2)=41
a_(3,3)=44
a_(3,4)=47
a_(3,5)=48

XOR:=>

a_(0,0)=null
a_(0,1)=null
a_(0,2)=49
a_(0,3)=50
a_(0,4)=53
a_(0,5)=56
a_(1,0)=null
a_(1,1)=null
a_(1,2)=52
a_(1,3)=55
a_(1,4)=59
a_(1,5)=62
a_(2,0)=null
a_(2,1)=null
a_(2,2)=51
a_(2,3)=54
a_(2,4)=58
a_(2,5)=61
a_(3,0)=null
a_(3,1)=null
a_(3,2)=57
a_(3,3)=60
a_(3,4)=63
a_(3,5)=64

FIX:=>

a_(0,0)=null
a_(0,1)=null
a_(0,2)=65
a_(0,3)=66
a_(0,4)=69
a_(0,5)=72
a_(1,0)=null
a_(1,1)=null
a_(1,2)=68
a_(1,3)=71
a_(1,4)=75
a_(1,5)=78
a_(2,0)=null
a_(2,1)=null
a_(2,2)=67
a_(2,3)=70
a_(2,4)=74
a_(2,5)=77
a_(3,0)=null
a_(3,1)=null
a_(3,2)=73
a_(3,3)=76
a_(3,4)=79
a_(3,5)=80

LET:=>

a_(0,0)=null
a_(0,1)=null
a_(0,2)=81
a_(0,3)=82
a_(0,4)=85
a_(0,5)=88
a_(1,0)=null
a_(1,1)=null
a_(1,2)=84
a_(1,3)=87
a_(1,4)=91
a_(1,5)=94
a_(2,0)=null
a_(2,1)=null
a_(2,2)=83
a_(2,3)=86
a_(2,4)=90
a_(2,5)=93
a_(3,0)=null
a_(3,1)=null
a_(3,2)=89
a_(3,3)=92
a_(3,4)=95
a_(3,5)=96

As the projections into the MOG
array are interpretable as states
for a Turing machine reading and
writing to toroidal Karnaugh maps,
the 8 loci associated with the
standard octad, that is, the
array elements

a_(0,0)=
a_(0,1)=
a_(1,0)=
a_(1,1)=
a_(2,0)=
a_(2,1)=
a_(3,0)=
a_(3,1)=

may be seen as controlling bits.
There are only 5 visual patterns
generated by the mapping of the
alphabet letters through their
difference sets.

The first two patterns are
"reflected forms," one odd
and one even

XO
XO
OX
X0

XO
XO
OX
OX

The next pattern simply shifts
the odd pattern above in one
column

XO
XO
OO
XX

The last two patterns have a form
reminiscent of truth tables

XO
OO
OX
XX

XX
OX
XO
OO

The last pattern is special in that
it is related to the pattern of involution
by which the quadratic residues modulo 23
are used to label the array.

In these representations, the Romans
are the lowest triple in the first column.
The sense in which the proposed Turing
machine outputs its data is through
as sequence of triples for each dimension.
The actual idea is that Roman II is a
control bit and that Romans I and III play
a counting game with wins, losses, and ties.
The outcomes would determine whether a pure
I fragment is concatenated, a pure III fragment
is concatenated or a combined modulo 2 addition
of the two sequences is concatenated because of
a tie.

The idea would be to find a start position in
terms of a locus in some complementary square
and a start configuration on the Karnaugh maps
that would lead to "winning percentages"
in the approximate range of neutrino mixing
angle probabilities.

For example, because the "given letter" can
always have its Romans interpreted with a
constant value, a string fragment might look
like

XOXOXXOXXXXO
XOOOOOOOOOOX
XXXXXOXOXXXO

With the two triples at the end taken as
delimiters that are not counted. The last
symbol is from the distinguished pattern,

XX
OX
XO
OO

and is taken to be a controlled not gate
that halts the processing in that dimension
in the sense of a delay in a machine circuit.

Taken as a game, the sequence above
has the score

I:=>7
III:=>8

and appends the string

XXXXOXOXXX

to the output for the Turing machine. The
delimiting characters have been truncated.

The transitions involving the "given letter"
involve toroidal transforms of the data field
according to the canonical enumeration and
then "presenting" the configuration of the
successor in that Boolean block.

<LEQ, OR, DENY, FLIP, NIF, NTRU, AND, NIMP, XOR, IMP, NAND, TRU, IF,
FIX, LET, NOR>

There are two idiosyncracies to this mechanism.

First, NOR cycles to OR and not to LEQ

This has to do with coloring the complete graph
on 6 symbols. There are 15 edges in that graph,
and, in any 2-edge coloring, there is at least
one monochromatic triangle.

The second idiosyncracy is that NTRU exchanges the
parity of every symbol in the Karnaugh map before
making its "presentation" of its successor, AND.
This additional action reflects the fact that NTRU
is bound with the name for NOT by the name mangling
associated with line names and point names. It also
characterizes the toroidal surface as a non-orientable
surface with non-orientability sensed when the "curve"
closes at an NTRU locus.

By "presentation" it is meant that the Karnaugh
map has the parity of those symbols determined by
the associaated line elements reversed. In fact,
this is true of every Karnaugh map for all six
dimensions. Thus, the Karnaugh map data is evolving
with the state transitions for all six dimensions.

So, first NTRU will reverse the parity of every
symbol. And, then it will reverse the parity for
the line elements of AND.

Also, with regard to the five Boolean blocks
involved with mapping from the difference set,
the 5-set is always completed by the locus,

a_(1,1)=NOT

The transformation of line names
to line elements is given by the
following. The 8 element blocks
are completed with the 3 Romans.

LEQ:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=NIMP
a_(0,5)=
a_(1,0)=
a_(1,1)=NOT
a_(1,2)=
a_(1,3)=
a_(1,4)=
a_(1,5)=IF
a_(2,0)=
a_(2,1)=
a_(2,2)=NIF
a_(2,3)=
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=IMP
a_(3,4)=
a_(3,5)=

OR:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=DENY
a_(1,0)=
a_(1,1)=
a_(1,2)=AND
a_(1,3)=
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=NO
a_(2,2)=
a_(2,3)=
a_(2,4)=XOR
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=IMP
a_(3,4)=
a_(3,5)=

DENY:=>

a_(0,0)=
a_(0,1)=ALL
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=LET
a_(3,3)=IMP
a_(3,4)=OR
a_(3,5)=TRU

FLIP:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=NOR
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=
a_(1,4)=FIX
a_(1,5)=
a_(2,0)=
a_(2,1)=NO
a_(2,2)=NIF
a_(2,3)=
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=
a_(3,5)=TRU

NIF:=>

a_(0,0)=SOME
a_(0,1)=
a_(0,2)=
a_(0,3)=NOR
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=LEQ
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=FLIP
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=IMP
a_(3,4)=
a_(3,5)=

NTRU(NOT):=>

a_(0,0)=
a_(0,1)=
a_(0,2)=NTRU
a_(0,3)=
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=NOT
a_(1,2)=
a_(1,3)=LEQ
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=XOR
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=
a_(3,5)=TRU

AND:=>

a_(0,0)=SOME
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=NIMP
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=
a_(1,4)=FIX
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=XOR
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=OR
a_(3,5)=

NIMP:=>

a_(0,0)=
a_(0,1)=ALL
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=AND
a_(1,3)=LEQ
a_(1,4)=FIX
a_(1,5)=IF
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=
a_(3,5)=

XOR:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=NOR
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=NOT
a_(1,2)=AND
a_(1,3)=
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=
a_(2,5)=NAND
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=OR
a_(3,5)=

IMP:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=DENY
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=LEQ
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=NIF
a_(2,3)=
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=OTHER
a_(3,2)=
a_(3,3)=
a_(3,4)=OR
a_(3,5)=

NAND:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=NOR
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=
a_(1,4)=
a_(1,5)=IF
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=XOR
a_(2,5)=
a_(3,0)=
a_(3,1)=OTHER
a_(3,2)=
a_(3,3)=LET
a_(3,4)=
a_(3,5)=

TRU:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=DENY
a_(1,0)=
a_(1,1)=NOT
a_(1,2)=
a_(1,3)=
a_(1,4)=FIX
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=FLIP
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=
a_(3,2)=LET
a_(3,3)=
a_(3,4)=
a_(3,5)=

IF:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=NIMP
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=LEQ
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=NO
a_(2,2)=
a_(2,3)=
a_(2,4)=
a_(2,5)=NAND
a_(3,0)=
a_(3,1)=
a_(3,2)=LET
a_(3,3)=
a_(3,4)=
a_(3,5)=

FIX:=>

a_(0,0)=
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=NIMP
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=AND
a_(1,3)=
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=FLIP
a_(2,4)=
a_(2,5)=
a_(3,0)=
a_(3,1)=OTHER
a_(3,2)=
a_(3,3)=
a_(3,4)=
a_(3,5)=TRU

LET:=>

a_(0,0)=SOME
a_(0,1)=
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=DENY
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=
a_(1,4)=
a_(1,5)=IF
a_(2,0)=
a_(2,1)=
a_(2,2)=
a_(2,3)=
a_(2,4)=
a_(2,5)=NAND
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=
a_(3,5)=TRU

NOR:=>

a_(0,0)=
a_(0,1)=ALL
a_(0,2)=
a_(0,3)=
a_(0,4)=
a_(0,5)=
a_(1,0)=
a_(1,1)=
a_(1,2)=
a_(1,3)=
a_(1,4)=
a_(1,5)=
a_(2,0)=
a_(2,1)=
a_(2,2)=NIF
a_(2,3)=FLIP
a_(2,4)=XOR
a_(2,5)=NAND
a_(3,0)=
a_(3,1)=
a_(3,2)=
a_(3,3)=
a_(3,4)=
a_(3,5)=

Date Subject Author
2/10/13 fom
2/10/13 J. Antonio Perez M.
2/10/13 fom
2/11/13 Shmuel (Seymour J.) Metz
2/11/13 fom
2/14/13 Shmuel (Seymour J.) Metz
2/14/13 fom
2/14/13 fom
2/15/13 fom
2/15/13 Shmuel (Seymour J.) Metz
2/16/13 fom
2/17/13 Shmuel (Seymour J.) Metz
2/19/13 fom
2/21/13 Shmuel (Seymour J.) Metz
2/15/13 fom
2/15/13 fom
2/14/13 fom
2/17/13 Shmuel (Seymour J.) Metz
2/17/13 fom
2/17/13 Barb Knox
2/18/13 fom
2/19/13 Shmuel (Seymour J.) Metz
2/19/13 fom
2/21/13 Shmuel (Seymour J.) Metz
2/19/13 Shmuel (Seymour J.) Metz
2/19/13 fom
2/21/13 Shmuel (Seymour J.) Metz
2/21/13 fom
2/22/13 Shmuel (Seymour J.) Metz
2/15/13 fom
2/17/13 Shmuel (Seymour J.) Metz
2/17/13 fom
2/19/13 Shmuel (Seymour J.) Metz
2/16/13 dan.ms.chaos@gmail.com
2/16/13 fom
2/17/13 dan.ms.chaos@gmail.com
2/17/13 fom
2/17/13 dan.ms.chaos@gmail.com
2/18/13 Shmuel (Seymour J.) Metz
2/20/13 fom
2/21/13 Shmuel (Seymour J.) Metz
2/16/13 fom
2/19/13 Shmuel (Seymour J.) Metz
2/19/13 fom

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