Category:3-ary Boolean functions; cube permutations

permuting 3 arguments

original order or arguments

permuting 2 arguments

Permutations of a cube (or a vector of length 8) corresponding to 3-ary Boolean functions in a big equivalence class (compare this category)

The 48 permutations are the elements of the full octahedral group (FOG).
The 8×6 matrix on the right shows that each of these permutations can be written as ${\displaystyle fog(m,n)}$.
Each file corresponds to a permutation of the 3 arguments - and to a column of that matrix.

SVG files edit

The data shown in these images can be found in here.
The SVG files have been created with Python using this script and these templates.

The permutations (maps from ${\displaystyle \{0...7\}}$ to itself) are shown in the upper half of each rectangle and on the left in the footer of each file.
The results of applying them to the sequence ${\displaystyle (0...7)}$ are shown in the lower half of each rectangle and on the right in the footer.
The results are always shown with a gray background. (Dark gray for numbers with odd binary digit sum.)

All the gray 8×8 matrices have the pattern of the XOR table in file 0. So replacing a set of numbers with 1s and the others with 0s gives a SEC matrix.

Mathematical description edit

Let ${\displaystyle f}$ be a 3-ary Boolean function, ${\displaystyle neg(m)}$ a negator pattern and ${\displaystyle perm(n)}$ a permutation of 3 elements.
Then the truth table of ${\displaystyle f}$ with ${\displaystyle neg(m)}$ and then the inverse of ${\displaystyle perm(n)}$ applied to its arguments is the initial truth table permuted by ${\displaystyle fog(m,n)}$: ${\displaystyle f\circ perm(n)^{-1}\circ neg(m)\circ (A,B,C)~~=~~fog(m,n)\circ f\circ (A,B,C)}$         (Some parentheses omitted for readability.)

Finding the position edit

The position of ${\displaystyle f({\overline {C}},A,B)}$ among the 48 permutations is to be found.

${\displaystyle perm(n)\circ ({\overline {C}},A,B)=(A,B,{\overline {C}})~~\iff ~~perm(n)={\begin{pmatrix}0&1&2\\2&0&1\end{pmatrix}}~~\iff ~~n=3}$         (Compare Permutation notation)

${\displaystyle neg(m)\circ (A,B,C)=(A,B,{\overline {C}})~~\iff ~~neg(m)=(0,0,1)~~\iff ~~m=4}$

So ${\displaystyle f({\overline {C}},A,B)}$ is at position in cube , and its truth table is that of ${\displaystyle f(A,B,C)}$ permuted by ${\displaystyle fog(4,3)}$.

${\displaystyle fog(4,3)={\begin{pmatrix}0&1&2&3&4&5&6&7\\4&0&5&1&6&2&7&3\end{pmatrix}}}$         (Compare File:Cube permutation 4 3.svg)

For the sake of simplicity the truth table of the initial function ${\displaystyle f}$ is identified (${\displaystyle \cong }$) with the integers 0..7:

${\displaystyle f(A,B,C)~~=~~({\overline {A}}{\overline {B}}{\overline {C}},A{\overline {B}}{\overline {C}},{\overline {A}}B{\overline {C}},AB{\overline {C}},{\overline {A}}{\overline {B}}C,A{\overline {B}}C,{\overline {A}}BC,ABC)~~\cong ~~(0,1,2,3,4,5,6,7)}$

${\displaystyle f({\overline {C}},A,B)~~=~~fog(4,3)\circ f(A,B,C)~~=~~(A{\overline {B}}{\overline {C}},AB{\overline {C}},A{\overline {B}}C,ABC,{\overline {A}}{\overline {B}}{\overline {C}},{\overline {A}}B{\overline {C}},{\overline {A}}{\overline {B}}C,{\overline {A}}BC)~~\cong ~~(1,3,5,7,0,2,4,6)}$

(As usual when dealing with permutations that are not self-inverse, there is the danger of confusing a permutation with the result of applying it to the natural order.
That is why both the permutations and their results are shown in these graphics.)

Specific Boolean function edit

${\displaystyle f(A,B,C)={\overline {(A\land B)\lor C}}=(1,1,1,0,0,0,0,0)=}$ .      It is at position in this cube ( with unrotated functions).
${\displaystyle f({\overline {C}},A,B)={\overline {({\overline {C}}\land A)\lor B}}=(1,0,0,0,1,1,0,0)=}$ .      It is at position in this cube ( with rotations to the left).

${\displaystyle fog(4,3)\circ (1,1,1,0,0,0,0,0)=(1,0,0,0,1,1,0,0)}$

While the positions of general formulas like ${\displaystyle f(A,B,C)}$ or ${\displaystyle f({\overline {C}},A,B)}$ are unique, for specific ${\displaystyle f}$ there is always at least one duplicate,
because the maximum number of functions in a big equivalence class (of 3-ary Boolean functions) is 24.

For the ${\displaystyle f}$ in the above example ${\displaystyle f(A,B,C)=f(B,A,C)}$ and ${\displaystyle f(A,{\overline {C}},B)=f({\overline {C}},A,B)}$.

That means that ${\displaystyle fog(0,0)}$ and ${\displaystyle fog(0,1)}$ have no effect and ${\displaystyle fog(4,2)}$ and ${\displaystyle fog(4,3)}$ have the same effect on the truth table of ${\displaystyle f}$.

Compare File:Hyperoctahedral group 2; active postfix.svg with permutations of a square