Date: Feb 15, 2013 6:57 AM Author: fom Subject: Re: distinguishability - in context, according to definitions 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)=