I had a go at creating rules for two of these patterns. I tried to minimise the population that the corresponding metacells would have. In the metacell the rule is stored as a lookup table on a glider tape where each possible neighbourhood corresponds to a location on the tape and there are n gliders in that location if the metacell should go to state n. So we should try to use as few transitions as possible and try to prefer low numbered states.
Of course this is just scraping a few cells off a population of millions, but it's still satisfying.
Calcyman, do you know approximately at which point a metacell would achieve its minimum population? Presumably the population is always changing slightly as the tape runs through its snark loop. Would it be practical to run a metacell over the possible time interval in which the minimum population might be achieved in order to calculate its exact minimum population? (On the other hand we could say that we only care about bounding box rather than population, since the reverse caber-tosser gives us quadratic replicators, RROs and SMOSes with population 143. What's the minimal bounding box of the metacell? Does it depend on the rule?)
Quadratic replicatorCode: Select all
x = 1, y = 1, rule = MAPACAAAIAAAAAAAAAAAAAAAIAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
o!
I think this (or one of its rotations) is the minimum population quadratic replicator. The rule says that you become live if you have a cell to your north only, a cell to your west only, or cells to your south and east only. It only uses one state and three transitions. You need at least three transitions to produce quadratic growth, and this is the only choice of three transitions that does so (the other possibility is the rule where a cell becomes live if it has one cell to its west, north or east, but this rule sierpinskis in such a way that its population returns infinitely often to three cells, meaning I'm not sure if it counts as a quadratic replicator).
SMOS
Code: Select all
@RULE metacell-SMOS0
@TABLE
n_states:4
neighborhood:vonNeumann
symmetries:none
var a={0,1,2,3}
var b={0,1,2,3}
var c={0,1,2,3}
var d={0,1,2,3}
var e={0,1,2,3}
var f={1,2,3}
f,a,b,c,d,0
0,2,0,0,1,3
0,3,0,0,0,1
0,0,0,0,3,2
0,2,0,0,0,2
0,0,0,0,1,1
a,b,c,d,e,0
Code: Select all
x = 5, y = 5, rule = metacell-SMOS0
4.B4$A!
This is a one-cell SMOS using only five transitions. I suspect this is minimal but I can't prove it. I'm also not quite sure if this is the minimum-population way to number the states in this rule. I have the one-cell state being state 3 and the two ships being states 1 and 2, which only puts 9 gliders on the rule tape. But the metacell must store its own state somewhere. Depending on how efficient this storage is it might be better to have the one-cell state be state 1 and the other two states be 2 and 3, even though this would put 11 gliders on the tape.
RRO
This seems really tricky, I'm stumped. I can't even write down a super inefficient one using one of
2718281828's isotropic RROs, because I can't get four (or even two) of those to orbit together in the way I described above.
