```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.htmlThe Miracle Octad Generatoris a two-dimensional arraywith 24 loci.  Its standardlabeling is given by thequadratic residues modulo 23and 0, that is, the setQ={0,1,2,3,4,6,8,9,12,13,16,18}with remaining labels based onthe involutionz:=>(-1/z)and located according to an involutionon the 24-element array.a_(0,0)=0a_(0,1)=oo=-1/0a_(0,2)=1a_(0,3)=11=-1/2a_(0,4)=2a_(0,5)=22=-1/1a_(1,0)=19=-1/6a_(1,1)=3a_(1,2)=20=-1/8a_(1,3)=4a_(1,4)=10=-1/16a_(1,5)=18a_(2,0)=15=-1/3a_(2,1)=6a_(2,2)=14=-1/18a_(2,3)=16a_(2,4)=17=-1/4a_(2,5)=8a_(3,0)=5=-1/9a_(3,1)=9a_(3,2)=21=-1/12a_(3,3)=13a_(3,4)=7=-1/13a_(3,5)=12With a distinguished 3-set, theMOG array partitions into thegiven 3-set and a 21-set interpretableas a 21-point projective planerelative to the S{5,8,24} Witt design.For this design, each 5-set of symbolsoccurs in some 8-set block with amultiplicity of 1 -- that is, the5-sets occur uniquely.When the MOG is partitioned intoits 35 standard sextets, the locimay be labelled with 3 Roman numeralsand normalized coordinates over theGalois field on 4 elements.Given, {0,1,[],{}}Let,1+[]={}1+{}=[]{}+[]={}*[]=1[]^2={}{}^2=[]{}^3=[]^3=11+1=0{}+{}=0[]+[]=0thena_(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 coordinatesfor the alphabet on the free orthomodularlattice on 2 generators need to beplaced in their corresponding positions.a_(0,0)=SOMEa_(0,1)=ALLa_(0,2)=NTRUa_(0,3)=NORa_(0,4)=NIMPa_(0,5)=DENYa_(1,0)=I(Roman)a_(1,1)=NOTa_(1,2)=ANDa_(1,3)=LEQa_(1,4)=FIXa_(1,5)=IFa_(2,0)=II(Roman)a_(2,1)=NOa_(2,2)=NIFa_(2,3)=FLIPa_(2,4)=XORa_(2,5)=NANDa_(3,0)=III(Roman)a_(3,1)=OTHERa_(3,2)=LETa_(3,3)=IMPa_(3,4)=ORa_(3,5)=TRUThe correspondence for each Booleanblock of the free orthomodularlattice on 2 generators is givenasDENY:=>a_(0,0)=nulla_(0,1)=nulla_(0,2)=01a_(0,3)=02a_(0,4)=05a_(0,5)=08a_(1,0)=nulla_(1,1)=nulla_(1,2)=04a_(1,3)=07a_(1,4)=11a_(1,5)=14a_(2,0)=nulla_(2,1)=nulla_(2,2)=03a_(2,3)=06a_(2,4)=10a_(2,5)=13a_(3,0)=nulla_(3,1)=nulla_(3,2)=09a_(3,3)=12a_(3,4)=15a_(3,5)=16FLIP:=>a_(0,0)=nulla_(0,1)=nulla_(0,2)=17a_(0,3)=18a_(0,4)=21a_(0,5)=24a_(1,0)=nulla_(1,1)=nulla_(1,2)=20a_(1,3)=23a_(1,4)=27a_(1,5)=30a_(2,0)=nulla_(2,1)=nulla_(2,2)=19a_(2,3)=22a_(2,4)=26a_(2,5)=29a_(3,0)=nulla_(3,1)=nulla_(3,2)=25a_(3,3)=28a_(3,4)=31a_(3,5)=32LEQ:=>a_(0,0)=nulla_(0,1)=nulla_(0,2)=33a_(0,3)=34a_(0,4)=37a_(0,5)=40a_(1,0)=nulla_(1,1)=nulla_(1,2)=36a_(1,3)=39a_(1,4)=43a_(1,5)=46a_(2,0)=nulla_(2,1)=nulla_(2,2)=35a_(2,3)=38a_(2,4)=42a_(2,5)=45a_(3,0)=nulla_(3,1)=nulla_(3,2)=41a_(3,3)=44a_(3,4)=47a_(3,5)=48XOR:=>a_(0,0)=nulla_(0,1)=nulla_(0,2)=49a_(0,3)=50a_(0,4)=53a_(0,5)=56a_(1,0)=nulla_(1,1)=nulla_(1,2)=52a_(1,3)=55a_(1,4)=59a_(1,5)=62a_(2,0)=nulla_(2,1)=nulla_(2,2)=51a_(2,3)=54a_(2,4)=58a_(2,5)=61a_(3,0)=nulla_(3,1)=nulla_(3,2)=57a_(3,3)=60a_(3,4)=63a_(3,5)=64FIX:=>a_(0,0)=nulla_(0,1)=nulla_(0,2)=65a_(0,3)=66a_(0,4)=69a_(0,5)=72a_(1,0)=nulla_(1,1)=nulla_(1,2)=68a_(1,3)=71a_(1,4)=75a_(1,5)=78a_(2,0)=nulla_(2,1)=nulla_(2,2)=67a_(2,3)=70a_(2,4)=74a_(2,5)=77a_(3,0)=nulla_(3,1)=nulla_(3,2)=73a_(3,3)=76a_(3,4)=79a_(3,5)=80LET:=>a_(0,0)=nulla_(0,1)=nulla_(0,2)=81a_(0,3)=82a_(0,4)=85a_(0,5)=88a_(1,0)=nulla_(1,1)=nulla_(1,2)=84a_(1,3)=87a_(1,4)=91a_(1,5)=94a_(2,0)=nulla_(2,1)=nulla_(2,2)=83a_(2,3)=86a_(2,4)=90a_(2,5)=93a_(3,0)=nulla_(3,1)=nulla_(3,2)=89a_(3,3)=92a_(3,4)=95a_(3,5)=96As the projections into the MOGarray are interpretable as statesfor a Turing machine reading andwriting to toroidal Karnaugh maps,the 8 loci associated with thestandard octad, that is, thearray elementsa_(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 patternsgenerated by the mapping of thealphabet letters through theirdifference sets.The first two patterns are"reflected forms," one oddand one evenXOXOOXX0XOXOOXOXThe next pattern simply shiftsthe odd pattern above in onecolumnXOXOOOXXThe last two patterns have a formreminiscent of truth tablesXOOOOXXXXXOXXOOOThe last pattern is special in thatit is related to the pattern of involutionby which the quadratic residues modulo 23are used to label the array.In these representations, the Romansare the lowest triple in the first column.The sense in which the proposed Turingmachine outputs its data is throughas sequence of triples for each dimension.The actual idea is that Roman II is acontrol bit and that Romans I and III playa counting game with wins, losses, and ties.The outcomes would determine whether a pureI fragment is concatenated, a pure III fragmentis concatenated or a combined modulo 2 additionof the two sequences is concatenated because ofa tie.The idea would be to find a start position interms of a locus in some complementary squareand a start configuration on the Karnaugh mapsthat would lead to "winning percentages"in the approximate range of neutrino mixingangle probabilities.For example, because the "given letter" canalways have its Romans interpreted with aconstant value, a string fragment might looklikeXOXOXXOXXXXOXOOOOOOOOOOXXXXXXOXOXXXOWith the two triples at the end taken asdelimiters that are not counted.  The lastsymbol is from the distinguished pattern,XXOXXOOOand is taken to be a controlled not gatethat halts the processing in that dimensionin the sense of a delay in a machine circuit.Taken as a game, the sequence abovehas the scoreI:=>7III:=>8and appends the stringXXXXOXOXXXto the output for the Turing machine.  Thedelimiting characters have been truncated.The transitions involving the "given letter"involve toroidal transforms of the data fieldaccording to the canonical enumeration andthen "presenting" the configuration of thesuccessor 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 LEQThis has to do with coloring the complete graphon 6 symbols.  There are 15 edges in that graph,and, in any 2-edge coloring, there is at leastone monochromatic triangle.The second idiosyncracy is that NTRU exchanges theparity of every symbol in the Karnaugh map beforemaking its "presentation" of its successor, AND.This additional action reflects the fact that NTRUis bound with the name for NOT by the name manglingassociated with line names and point names.  It alsocharacterizes the toroidal surface as a non-orientablesurface with non-orientability sensed when the "curve"closes at an NTRU locus.By "presentation" it is meant that the Karnaughmap has the parity of those symbols determined bythe associaated line elements reversed.  In fact,this is true of every Karnaugh map for all sixdimensions.  Thus, the Karnaugh map data is evolvingwith the state transitions for all six dimensions.So, first NTRU will reverse the parity of everysymbol. And, then it will reverse the parity forthe line elements of AND.Also, with regard to the five Boolean blocksinvolved with mapping from the difference set,the 5-set is always completed by the locus,a_(1,1)=NOTThe transformation of line namesto line elements is given by thefollowing.  The 8 element blocksare completed with the 3 Romans.LEQ:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=NIMPa_(0,5)=a_(1,0)=a_(1,1)=NOTa_(1,2)=a_(1,3)=a_(1,4)=a_(1,5)=IFa_(2,0)=a_(2,1)=a_(2,2)=NIFa_(2,3)=a_(2,4)=a_(2,5)=a_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=IMPa_(3,4)=a_(3,5)=OR:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=a_(0,5)=DENYa_(1,0)=a_(1,1)=a_(1,2)=ANDa_(1,3)=a_(1,4)=a_(1,5)=a_(2,0)=a_(2,1)=NOa_(2,2)=a_(2,3)=a_(2,4)=XORa_(2,5)=a_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=IMPa_(3,4)=a_(3,5)=DENY:=>a_(0,0)=a_(0,1)=ALLa_(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)=LETa_(3,3)=IMPa_(3,4)=ORa_(3,5)=TRUFLIP:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=NORa_(0,4)=a_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=a_(1,4)=FIXa_(1,5)=a_(2,0)=a_(2,1)=NOa_(2,2)=NIFa_(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)=TRUNIF:=>a_(0,0)=SOMEa_(0,1)=a_(0,2)=a_(0,3)=NORa_(0,4)=a_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=LEQa_(1,4)=a_(1,5)=a_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=FLIPa_(2,4)=a_(2,5)=a_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=IMPa_(3,4)=a_(3,5)=NTRU(NOT):=>a_(0,0)=a_(0,1)=a_(0,2)=NTRUa_(0,3)=a_(0,4)=a_(0,5)=a_(1,0)=a_(1,1)=NOTa_(1,2)=a_(1,3)=LEQa_(1,4)=a_(1,5)=a_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=a_(2,4)=XORa_(2,5)=a_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=a_(3,4)=a_(3,5)=TRUAND:=>a_(0,0)=SOMEa_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=NIMPa_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=a_(1,4)=FIXa_(1,5)=a_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=a_(2,4)=XORa_(2,5)=a_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=a_(3,4)=ORa_(3,5)=NIMP:=>a_(0,0)=a_(0,1)=ALLa_(0,2)=a_(0,3)=a_(0,4)=a_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=ANDa_(1,3)=LEQa_(1,4)=FIXa_(1,5)=IFa_(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)=NORa_(0,4)=a_(0,5)=a_(1,0)=a_(1,1)=NOTa_(1,2)=ANDa_(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)=NANDa_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=a_(3,4)=ORa_(3,5)=IMP:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=a_(0,5)=DENYa_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=LEQa_(1,4)=a_(1,5)=a_(2,0)=a_(2,1)=a_(2,2)=NIFa_(2,3)=a_(2,4)=a_(2,5)=a_(3,0)=a_(3,1)=OTHERa_(3,2)=a_(3,3)=a_(3,4)=ORa_(3,5)=NAND:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=NORa_(0,4)=a_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=a_(1,4)=a_(1,5)=IFa_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=a_(2,4)=XORa_(2,5)=a_(3,0)=a_(3,1)=OTHERa_(3,2)=a_(3,3)=LETa_(3,4)=a_(3,5)=TRU:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=a_(0,5)=DENYa_(1,0)=a_(1,1)=NOTa_(1,2)=a_(1,3)=a_(1,4)=FIXa_(1,5)=a_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=FLIPa_(2,4)=a_(2,5)=a_(3,0)=a_(3,1)=a_(3,2)=LETa_(3,3)=a_(3,4)=a_(3,5)=IF:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=NIMPa_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=LEQa_(1,4)=a_(1,5)=a_(2,0)=a_(2,1)=NOa_(2,2)=a_(2,3)=a_(2,4)=a_(2,5)=NANDa_(3,0)=a_(3,1)=a_(3,2)=LETa_(3,3)=a_(3,4)=a_(3,5)=FIX:=>a_(0,0)=a_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=NIMPa_(0,5)=a_(1,0)=a_(1,1)=a_(1,2)=ANDa_(1,3)=a_(1,4)=a_(1,5)=a_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=FLIPa_(2,4)=a_(2,5)=a_(3,0)=a_(3,1)=OTHERa_(3,2)=a_(3,3)=a_(3,4)=a_(3,5)=TRULET:=>a_(0,0)=SOMEa_(0,1)=a_(0,2)=a_(0,3)=a_(0,4)=a_(0,5)=DENYa_(1,0)=a_(1,1)=a_(1,2)=a_(1,3)=a_(1,4)=a_(1,5)=IFa_(2,0)=a_(2,1)=a_(2,2)=a_(2,3)=a_(2,4)=a_(2,5)=NANDa_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=a_(3,4)=a_(3,5)=TRUNOR:=>a_(0,0)=a_(0,1)=ALLa_(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)=NIFa_(2,3)=FLIPa_(2,4)=XORa_(2,5)=NANDa_(3,0)=a_(3,1)=a_(3,2)=a_(3,3)=a_(3,4)=a_(3,5)=
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