Suppose the parameters/inputs of the computation include the time at which the configuration of a particular deterministic chaotic system needs to be computed.

Say, for instance, as input we have a start configuration of the chaotic Rule 30 cellular automata and a time-step $t$, and as output, we require the configuration at time-step $t$.

What can be said about the computational complexity of such a computation?


Wolfram's rule 30 is a one dimensional cellular automaton. If you want to know the state of the CA at step $t$, given the initial configuration, you just need to run the rule for $t$ steps. At every step the computation consist in no more than a bunch of if's, so the time complexity of such CA is linear in $t$.

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  • $\begingroup$ Ahhh yes of course, can we do better? $\endgroup$ – ramseysdream111 Oct 18 '19 at 11:08
  • $\begingroup$ @pkwssis I strongly doubt it, but I don't know if anyone ever prooved it formally. If you want to expand your knowledge on cellular automatons, I strongly recommend that you read the book "A New Kind of Science" written by Wolfram himself. $\endgroup$ – Yamar69 Oct 18 '19 at 11:13
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    $\begingroup$ This answer is not quite right; the running time is actually quadratic in $t$, as explained in the link in my comment above. The reason is that you need to simulate up to $t$ cells. Thus, simulating a single step takes $O(t)$ time, not $O(1)$ time. Multiply that by $t$ steps, and you get $O(t^2)$ time. As explained in the link I provided, it is an open question whether we can do substantially better. $\endgroup$ – D.W. Oct 18 '19 at 17:59

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