2 ms·
A rule R is reversible if there exists a rule R^-1 such that for all a, b, c, d, e, f, g we see the following evolution (where . stands for "don't care"):
by orlp 2y ago
A rule R is reversible if there exists a rule R^-1 such that for all a, b, c, d, e, f, g we see the following evolution (where . stands for "don't care"):
a b c d e f g
Apply rule R
. b' c' d' e' f' .
Apply rule R^-1
. . c d e . .
In equations, it means there must exist some R^-1 such that for all a, b, c, d, e, f, g the following holds:
c = R^-1(R(a, b, c), R(b, c, d), R(c, d, e))
d = R^-1(R(b, c, d), R(c, d, e), R(d, e, f))
e = R^-1(R(c, d, e), R(d, e, f), R(e, f, g))
Writing a quick program that checks all possible combinations of R and R^-1 (https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=e677edf47d45067cfa4988b8680c5a79 https://play.rust-lang.org/?version=stable&mode=debug&editio...) we find the following inverse pairs:
0x33 <-> 0x33
0x55 <-> 0x0f
0xcc <-> 0xcc
0xf0 <-> 0xaa
That is, the following two rules are self-inverses (they're the NOT and IDENTITY gates on the center cell respectively):
111 110 101 100 011 010 001 000
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
We have the left <-> right moving pair you identified:
111 110 101 100 011 010 001 000
1 0 1 0 1 0 1 0
111 110 101 100 011 010 001 000
1 1 1 1 0 0 0 0
And there's just one more, which is the same as the above but it also inverts the output (move left and invert is the inverse of move right and invert):
111 110 101 100 011 010 001 000
0 1 0 1 0 1 0 1
111 110 101 100 011 010 001 000
0 0 0 0 1 1 1 1
- Y_Y 2y agoVery neat, thanks for working this out. (Why would it not be sufficient to start with just five symbols?) The solution is like a restricted version of Hilbert's Hotel, we're still in a group of invertible maps on binary sequences {0,1}^N but we aren't allowed to do anything non-local. https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Gra...