bbchallenge - #bbxy

<@465186903847731210> Just to make sure, have you already adapted the cert for https://bbchallenge.org/1RB2LA0LA_2LC---2RA_0RA2RC1LC to its cousin https://bbchallenge.org/1RB1LC---_0LC2RB1LB_2LA0RC1RC ?

The two have a pretty simple relationship: From the first, replace start configs `A0|1B0|C12|2C2` with `B0|C02|0A2|A00|1B0|C12|1C1`. Exchange `1` and `2`, and left and right. Finally, rotate the states, replacing `A` with `C`, `B` with `A`, and `C` with `B`. Then you have the second.

Alright, I've gotten the forward behavior of https://bbchallenge.org/1RB2LA0LA_2LC---2RA_0RA2RC1LC fully (?) characterized: ``` C(a,b,c) = 0^∞ (12)^a 2^b C> 202 (202)^c 000 (0^∞ A> 0^∞) -> C(1,5,1) 0^∞ C(0,b,c) -> C(b+2,1,c-1) [c ≥ 1] C(1,b,c) -> C(b+5,1,c-1) [c ≥ 1] C(2,b,0) -> C(1,5,2) 0^(b-2) 202 [b ≥ 2] C(a,b,0) -> C(1,5,a-3) 0^b 202 [a ≥ 3] C(a,b,c) -> C(a-2,b+7,c-1) [a ≥ 2, c ≥ 1] C(0,b,0) 0^5 -> C(1,5,b+1) C(1,b,0) 0^5 -> C(1,5,b+4) ``` **EDIT:** The original version of this comment had two incorrect rules (shown below), which have been fixed above: ``` C(2,b,0) -> C(1,5,2) 0^(b+1) 202 C(a,b,0) -> C(1,5,a-3) 0^(b+3) 202 [a ≥ 3] ``` It's hard to determine whether the last two cases can ever occur at all, unless there's some modular invariant I'm not seeing. Also, if they do occur, I believe it's impossible for it to eat another `0^5` before a new `0^(b+1)` or `0^(b+3)` is added, but I don't have a formal proof of this.

OK, I think I can simplify these rules a bit: ``` (0, b, 2c) -> (2b+2, 1, 2c) (1, b, 0) -> Halt (1, 2b, 2c) -> (0, 1, 2(b+c+1)) (1, 2b+1, 2c) -> (2, 1, 2(b+c+3)) (a, 2b, 0) -> (a-2, 2b+3, 0) (2, 2b+1, 0) -> (0, 1, 2b+6) (a, 2b+1, 0) -> (a-3, 1, 2b+4) ``` Notice that c is always even now.

OK, I have a little bit of an idea of what's going on with these long streaks of consistent differences between `c` values. I think we can prove the following behavior: ``` 1RB2LC1RC_2LC---2RB_2LA0LB0RA Let f(n) = 3n+4 h(n) = f^n(1) + 1 = 3^{n+1} - 1 g(n) = sum_{k=0}^{n-1} h(k) = 3/2 (3^n - 1) - n = h(n)/2 - (n+1) h(n) = 2 (g(n) + n + 1) if g(n) < c, then (0, 1, 2c) -> (0, f^n(1), 2 (c-g(n))) -> (2 h(n), 1, 2 (c-g(n))) Let C(n) = (0, 1, 2n) = 0^inf 1 <A2 22 (20)^2n 0^inf C(g(n) + 8k+1) -> C(g(n) + 8k+1 + n+9) for all k: (h(n) - 45)/65 < k < (h(n) - 22)/38 Notably, when 8 divides (n+1) then this rule can potentially be applied repeatedly. Ex: if n = 7, then we get: C(3273 + 8k) -> C(3273 + 8(k+2)) when k in [101, 172] And we see this starting with C(4137) = C(3273 + 8*108) which repeats this rule until we get to C(4665) = C(3273 + 8*174). And as n gets way bigger, these ranges of repeat will increase exponentially. ```

new BBi(8) champ: `1RB1LA------_1RC3LB1RB---_2LA2LC---0LC` leaves `~10^783.13207 = ((-11 + 3^((7 + 3^8)/4))/2)` marks. not sure how many steps!