## Thursday, May 24, 2018

### Diamond Bracket Forms and How to Count to Two; 9 of 10

From "Cybernetics & Human Knowing", Vol. 24 (2017), No. 3-4, pp. 161-188
8. Diffracted Modulators are Rotors

Now let’s analyze these modulators by diffracting them.
a          =          [ hz ]
b          =          [ gz ]
d          =          [ bc ]
e          =          [ c [z] ]
f           =          [ d [z] ]
g          =          [ eh ]
h          =          [ fg ]
Let (A,B) = J(a,b) ; so that A = a/b,  and  B = b/a.
Also let (C,D) = J(c,d) ; (E,F) = J(e,f) ; (G,H) = J(g,h)
Then:
A         =          [ G z ]
B         =          [ H z ]
C          =          [ C B ]
D         =          [ D A ]
E          =          [ C [z] ]
F          =          [ D [z] ]
G         =          [ G E ]
H         =          [ H F ]
z/2       =           G/H

We can write this in a cycle, to get the “NaĂ¯ve Rotor”:
A         =          [ G z ]
D         =          [ A D ]
F          =          [ D [z] ]
H         =          [ F H ]
B         =          [ H z ]
C          =          [ B C ]
E          =          [ C [z] ]
G         =          [ E G ]
z/2        =          G/H

This diagram is of a diamond circuit; gates and wires are doubled, with a twist at each gate.

z          A         D         F          H         B         C          E          G         G/H
--------------------------------------------------------------------------------
1          0          i           i           i           0          j           j           j           0
0          j           j           0          i           i           i           0          j           0
1          0          j           j           j           0          i           i           i           1
0          i           i           0          j           j           j           0          i           1

Now try the First Brownian Modulator:
a          =          [ zd ]
b          =          [ ag ]
c          =          [ db ]  =  z/2
d          =          [ cf ]
e          =          [ cz ]
f           =          [ eh ]
g          =          [ ah ]
h          =          [ eg ]
Let (A,E) = J(a,e) ; so that A = a/e,  and  E = e/a.
Also let (B,F) = J(b,f) ; (C,D) = J(c,d) ; (G,H) = J(g,h)
Then:
A         =          [ C z ]
E          =          [ D z ]
B         =          [ E H ]
F          =          [ A G ]
C          =          [ C F ]
D         =          [ D B ]
G         =          [ G E ]
H         =          [ H A ]
z/2       =          C/D

Written in a cycle, you get a “Brownian Rotor”:
A         =          [ C z ]
H         =          [ A H ]
B         =          [ H E ]
D         =          [ B D ]
E          =          [ D z ]
G         =          [ E G ]
F          =          [ G A ]
C          =          [ F C ]
z/2      =          C/D

z          A         H         B         D         E          G         F          C          C/D
--------------------------------------------------------------------------------
1          0          i           i           i           0          j           j           j           0
0          j           j           0          i           i           i           0          j           0
1          0          j           j           j           0          i           i           i           1
0          i           i           0          j           j           j           0          i           1

Now try Kauffman’s Modulator:
a          =          [ bdz ]
b          =          [ ae ]
c          =          [ df ]
d          =          [ acz ]
e          =          [ af ]  =  z/2
f           =          [ de ]

Let (A,D) = J(a,d) ; so that A = a/d,  and  D = d/a.
Also let (B,C) = J(b,c) ; (E,F) = J(e,f)
Then:

A         =          [ A C z ]
D         =          [ D B z ]
B         =          [ D F ]
C          =          [ A E ]
E          =          [ D E ]
F          =          [ A F ]
z/2      =            E/F

Written in a cycle, you get a “Kauffman Rotor”:
A         =          [ C A z ]
F          =          [ A F ]
B         =          [ F D ]
D         =          [ B D z ]
E          =          [ D E ]
C          =          [ E A ]
z/2      =          E/F

z          A         F          B         D         E          C                      E/F
------------------------------------------------------------------------
1          0          i           i           0          j           j                       0
0          j           j           0          i           i           0                      1
1          0          j           j           0          i           i                       1
0          i           i           0          j           j           0                      0

The paradox values I and J spin around the rotor, bearing phase information.