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Mar 30, 2018 at 13:35 vote accept Morteza Azad
Mar 30, 2018 at 12:37 comment added Robin Houston Watkins also gives the following elegant argument: “Since you can remove four vertices from the knight’s graph of the $4\times 4$ chessboard and have the graph fall into six pieces, there could not possible have been a Hamiltonian path (you can’t make four cuts in a single length of string and end up with six pieces of string–well, most people can’t, maybe Ricky Jay could do it).”
Mar 30, 2018 at 12:34 comment added Robin Houston There is no $4\times 4$ Knight’s tour. For a proof, see Solution 3.4 of Across the Board: The Mathematics of Chessboard Problems by John J. Watkins. So 5 is indeed minimal.
Mar 30, 2018 at 12:09 history edited Joel David Hamkins CC BY-SA 3.0
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Mar 30, 2018 at 2:32 comment added paul garrett Continuing my idle free-association, the latter point you raise vaguely reminds me of the (craziness of the) Stone-Cech compactification of innocuous spaces like $\mathbb R^2$... suggesting that, as you say, the obvious tame notion of "board" cannot accommodate the inevitable limits... but perhaps something fancier might make sense of it. (Again, a lovely response to the question above!)
Mar 30, 2018 at 2:11 comment added Joel David Hamkins @paulgarrett Some people have looked at transfinite boards, with bishops, rooks and queens able to bridge the gap by following lines obeying their defining movements. I'm not sure, however, how much the result is like chess. For a knight's tour, one would want to define the limit ordinal time position of the knight on such a board, but this would seem difficult if the knight had been following the spiral pattern suggested in this answer, for example. Where would it be at time $\omega$?
Mar 30, 2018 at 1:58 comment added paul garrett How could even-larger infinite chessboards be described to make sense of a similar question (and possibly extrapolation of this very nice answer)? E.g., how can knights move to "cross" the thresh-hold of limit cardinals/ordinals? Of course, this idle extrapolation may be boring...
Mar 30, 2018 at 1:50 history edited Joel David Hamkins CC BY-SA 3.0
Improved the argument about tiling
Mar 30, 2018 at 1:21 history edited Joel David Hamkins CC BY-SA 3.0
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Mar 30, 2018 at 1:16 history edited Joel David Hamkins CC BY-SA 3.0
added 5 characters in body
Mar 30, 2018 at 1:02 comment added Joel David Hamkins Thanks! A $3\times 3$ board doesn't work since the middle square is unreachable, and I tried with $4\times 4$ for a while but didn't find anything that worked, so I moved on to $5\times 5$. At first I tried to start in the middle of a side, intending to have one "going straight" pattern and one "turning right" pattern, as you suggested, but then I realized that starting in a corner meant you could do both patterns at once.
Mar 30, 2018 at 1:00 comment added Morteza Azad (+1) Beautiful construction, Joel! I wonder if 5 is minimum $n$ in which one may carry out the spiral argument for an infinite knight's tour?
Mar 30, 2018 at 0:51 history edited Joel David Hamkins CC BY-SA 3.0
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Mar 30, 2018 at 0:37 history answered Joel David Hamkins CC BY-SA 3.0