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