0
$\begingroup$

This question is inspired by the following challenge from CodeGolf.SE: https://codegolf.stackexchange.com/q/251510/88765.

Given positive integer $N$, we can consider a version of Conway’s game of life restricted to the $N\times N$ grid. Namely, given a state $s$, viewed as a function from $[N]^2$ to $\{0,1\}$, we define its updated state $s’= T(s)$ so that $s’(v) =1$ if there are exactly three $u$ at $L_\infty$ distance $1$ from $v$ with $s(u)=1$ or $s(v)=1$ and there are exactly two $u$ at $L_\infty$ distance $1$ from $v$ with $s(u)=1$.

Anyways, we can define a directed graph $G_N$ on the set of these states, with an edge from each state $s$ to its updated state $T(s)$. Let $L(N)$ denote the length of the longest directed path in $G_N$. Let $L’(N)$ be the length of longest directed path that doesn’t intersect a directed cycle.

Is the asymptotic growth of $L(N),L’(N)$ known? Also, is it possible for $L(N)$ to equal $L(N+1)$ or $L’(N)$ to equal $L’(N+1)$? In the latter case, this suggests the aforementioned Code Golf challenge can’t go on forever with perfect play (though they work with a slightly different variant I believe).

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ I don't have an answer other than "the asymptotic growth is really, really, fast". See this forum post for a demonstration of the fact that $L(111) \geq {}^{15}10$ (that's tetration, not exponentiation). I haven't done a proper analysis, but the construction used there could probably be used to show something like $L(N) \geq {}^{N/8}10$ for large $N$. This all assumes that, like in the CodeGolf challenge, the pattern is allowed to evolve outside of the $N \times N$ grid after the first timestep. $\endgroup$ Commented Sep 10, 2022 at 19:01
  • $\begingroup$ If the pattern is not allowed to evolve outside of the original $N \times N$ grid then of course you can't have this sort of blowup: you trivially get $L(N) \leq 2^{N^2}$ since there are only that many configurations possible. $\endgroup$ Commented Sep 10, 2022 at 19:03
  • $\begingroup$ hm, well since it grows very fast, I’m reasonably convinced the second case shouldn’t occur. getting good bounds on the growth still seems interesting me. I am in fact forbidding evolution outside the $N \times N$ grid, which means thing can be calculated in finite time (no halting problem/busy beaver shenanigans). $\endgroup$ Commented Sep 10, 2022 at 19:44
  • $\begingroup$ If you want to restrict to the $N \times N$ square, then I'm pretty sure that the asymptotic growth is $\Theta(2^{N^2})$. I already argued the upper bound. For a lower bound, you could construct a pattern like O(sqrt(log(t)) that is programmed to count in binary in a 2D region of the $N \times N$ square and then self-destruct once it is full. That wiki page is light on details, but this construction is done in more detail in Section 9.7.2 of our book. $\endgroup$ Commented Sep 10, 2022 at 21:18
  • 1
    $\begingroup$ @NathanielJohnston: Are you sure it is $\Theta(2^{N^2})$ and not $2^{\Theta(N^2)}$? $\endgroup$ Commented Sep 11, 2022 at 6:23

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.