fom
Posts:
1,969
Registered:
12/4/12


Re: distinguishability  in context, according to definitions
Posted:
Feb 15, 2013 6:57 AM


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 twodimensional 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 24element 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 3set, the MOG array partitions into the given 3set and a 21set interpretable as a 21point projective plane relative to the S{5,8,24} Witt design.
For this design, each 5set of symbols occurs in some 8set block with a multiplicity of 1  that is, the 5sets 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 2edge 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 nonorientable surface with nonorientability 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 5set 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)=

