This post follows this other post about times cover by Langton's ant of $n$ by $n$ gridded torus.

For $n$ by $n$ gridded torus, I've checked for $n \le 1000$ that the ant covers all. This fact needs to be proved for every $n$, but this should be not surprising.

Contrariwise, we can easily checked that for the $n$ by $2$ gridded torus and $n \ge 3$, the ant enters into a cycle before covering the torus:

Idem for the $n$ by $4$ gridded torus and $n \ge 5$:

Now for $n$ by $6$ (with $n \ge 5$) there is a problem.

For $n=50$, it is not yet covered after around $10^4$ steps:

and it is not yet covered after around $10^5$ steps:

in fact I've checked that it's also not covered after $10^7$ steps.

So there are two possibilities:

- for $n$ by $6$ gridded torus, the covering needs a long time, but finite (checked for $n \in [5, 60]$).
- $\exists n_0$ such that the ant covers $n_0$ by $6$ but
**not**$n$ by $6$ for $n > n_0$.

**Question**: Does Langton's ant cover every n by 6 gridded torus ($n \ge 5$)?

[If yes, what's the asymptotic of the covering time? Else, what's $n_0$?]

*If no:* i.e. if the ant enters into a cyclic pattern of width $n_0$ then we need *a priori* to go at least up to the step $4 \cdot 6 \cdot n_0 \cdot 2^{6 \cdot n_0}$ for checking that, which could become out of computation.

*If yes:* let $s^m_n$ be the number of steps Langton's ant needs for covering the $n$ by $m$ torus.

$\scriptsize{ \begin{array}{c|c} n &10&20&30&40&50&60&70&80 \newline \hline s^6_n &977&17623&87113&3135267&19563171&331665879&4738404219&44105120036 \newline \hline s^n_6 &240&6708&166542&2027581&32781038&220780386&4820615241&76325278885 \newline \hline n/ln(s^6_n) &1.45&2.04&2.6373&2.6741&2.9781&3.0581&3.1419&3.2639 \newline \hline n/ln(s^n_6) &1.82&2.26&2.4952&2.7543&2.8892&3.1229&3.1395 &3.1925 \end{array} } $

This new data (sage computation) suggests that: $s^6_n \sim s^n_6 \sim e^{n/c}$, with $c \sim 3$.

*Remark*: we can checked that there is no such a problem for the $n$ by $3$ or $5$ or $7$ $\dots$ gridded torus, but there is the same problem for the $n$ by $8$ or $10$ or $12$ $\dots$ gridded torus (at least up to $18$).

The pictures was computed online on http://www.turnerbohlen.com/langtonsant/