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.

Start with the Naive Modulator:

a          =          [ hz ]

b          =          [ gz ]

c          =          [ ad ]

                    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 “Naïve 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.