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I wonder if Knight's Tours have been explored in higher dimensions, using the following definition of a knight move.

  • In dimension $d=2$, the knight moves left/right and forward/back one step and two steps in either order. So $\{ \pm 1, \pm 2 \}$ steps. This leads to $8$ moves.

  • In dimension $d=3$, the knight moves $\{ \pm 1, \pm 2, \pm 3 \}$ steps in any order along $x,y,z$ axes. This leads to $48$ moves; see below.

  • In dimension $d$, the knight moves $\{ \pm 1, \pm 2, \ldots \pm d \}$ steps in any order along the axes. This leads to $d!\, 2^d$ moves.

I have found literature on knight's tours in 3D but under a different (planar) definition of move.1

It would be interesting to learn which (if any!) $n \times n \times n$ cube chess boards allow a knight's tour.

KnightMoves3D

1 Awani Kumar. "Magic Knight's Tours in Higher Dimensions." (arXiv link)


**Addendum**. *BenBarber* found much more relevant references for planar knight moves in higher dimensions: "with $n$ even, there is always a knight's tour provided that $n$ is sufficiently large." *TMA* made an incisive observation that answered my explicit question: There is no knight's tour for $d=3,4$, due to parity considerations. So, what remains is: (a) $d \ge 5$ and other odd triangular numbers, or (b) knight's tours of the even-coordinate-sum ("black") cells in $d=3,4$.
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    $\begingroup$ Some more references on the usual problem: the $n$-dimensional chessboards admitting a (conventional) Knight's tour were classified by Josh Erde and independently by Bruno and Sylvain Golenia. $\endgroup$
    – Ben Barber
    Commented May 11, 2014 at 14:03
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    $\begingroup$ Note for when the dth triangular number is even (d is 3 or 4 mod 4), the given move does not change parity, so only half the board can be covered in those dimensions. $\endgroup$ Commented May 11, 2014 at 14:14
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    $\begingroup$ Even coordinates is one subset...however, the idea of every other cell comes from taking the sum of the coordinates to be even. These are the lattices $D_n.$ $\endgroup$
    – Will Jagy
    Commented May 11, 2014 at 23:24
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    $\begingroup$ @WillJagy: That's what I meant by "even coordinates," although I now realize how misleading was that phrasing. Thanks for your correction. $\endgroup$ Commented May 11, 2014 at 23:45
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    $\begingroup$ I think this can be extended to multiples of $6$ in the 3 dimensional case, and to the 4 dimensional case, with $n=8$, by doing half of the black cells in each pair 1-5, 2-6, etc., then instead of returning to $(1,1,1,1)$, switch the sign of the $1$ or $3$ movement to return to a square with sum $2$ (mod $4$), and do it all over again. It will take some more care to make the path closed. $\endgroup$ Commented May 12, 2014 at 5:28

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In my recent preprint, available on arXiv, I show that a more natural $d$-dimensional generalization of the planar knight move rule can be deduced from Article 3.6 of Section E of the well-known FIDE hanbook (see FIDE Handbook).
Basically, since the aforementioned Article 3.6 refers to the Euclidean distance between the starting cell and the ending cell of a knight jump, we can take as a rule that this Euclidean distance has to be fixed at any jump and it is clearly equal to $\sqrt{5}$ chessboard units. Then, the arising diophantine equation admits only two kinds of solutions for any $d \geq 5$, one describing the well-known L-shaped move (whose taxicab length is $3$) and one changing by one $5$ out of the $d$ Cartesian coordinates of the starting cell.

As a result, by Theorem 4.1, we have that closed (Euclidean) knight's tours esist also on any $2 \times 2 \times \cdots \times 2 \subseteq \mathbb{Z}^d$ chessboard, as long as $d > 6$ is given.

Theorem 2.1 constructively proves the existence of open Euclidean knight's tours on a $3 \times 3 \times 3 \times 3 \times 3 \subseteq \mathbb{Z}^5$ chessboard, while there cannot be any closed Euclidean knight's tour on $3 \times 3 \times \cdots \times 3$ chessboards due to the parity argument.

Now, if we still wish to extend the knight move rule by following the pattern $\{\pm 1, \pm 2, \pm 3, \dots, \pm d\}$ instead of the more reasonable $\{\pm 2^0, \pm 2^1, \pm 2^2, \dots, \pm 2^d\}$, we should consider that this violates the light-cell/dark-cell rule so that such a knight would jump from a light-cell to another light-cell even on three dimensional chessboard (in 2006, Qing&Watkins clearly explained this fact in their paper entitled Knigh's Tours for Cubes and Boxes, p. 46). Figure 2 of my preprint shows how to properly color a $k$-dimensional chessboard according to the (widely accepted) parity argument and consequently the $1,2,3,\dots,d$ pattern is not coeherent with the light-square/dark-square jumping principle.

I honestly do not thing that this is a minor issue for the given problem and my proposal of considering the $\sqrt{5}$ constraint in order to extend the knight move rule from $d=2$ to $d=3$ and above could open a new perspective on this topic, avoiding to guess the continuation of a short integer sequence as $1, 2, \dots, ?$.

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    $\begingroup$ The FIDE definition says that "The knight may move to one of the squares nearest to that on which it stands but not on the same rank, file or diagonal". How this generalizes to higher dimensions depends on what meaning is then given to "rank, file or diagonal". It seems you have chosen some meaning, but there could be other possibilities. In particular, one could say that a "diagonal" is a direction where any two or more coordinates change by the same amount. Under that interpretation, changing five coordinates by $\pm 1$ would not be a knight move. $\endgroup$ Commented Sep 22, 2023 at 13:48
  • $\begingroup$ As far as I know, it's generally agreed that a diagonal of a $k$-cube is a segment that joins two corners, but if we say so, also every edge should be considered as a diagonal. On the other hand, it seems to me fair obvious that $rank \cup file$ may refer, in $k$-dimensions, to the $k$ orthogonal axes centered in the given vertex. Thus, it wouldn't be very reasonable to assume that the generalized FIDE move rule includes any "diagonal" as the considered set of diagonals, but only the major diagonals. Anyway, a 3D knight cannot move from (1,1,1) to (2,2,1), two of the "nearest" vertices, right? $\endgroup$ Commented Sep 23, 2023 at 11:36
  • $\begingroup$ In the last sentence of my comment above, I assumed that our $3$D chessboard is described as the set $\{(0,1,2)\times(0,1,2)\times(0,1,2)\}$ of $3^3$ vertices. According to Article 3.6, we would have that the vertices $(1,1,1)$ and $(2,2,1)$ (or $(0,0,1)$ or $(0,2,1)$ and so forth) do not lay on the same file, rank, minor diagonal, major diagonal, or even a Z-axis going from (1,1,0) to (1,1,2), and their distance is only $\sqrt{2}$, so $(2,2,1)$ is one vertex "near" to $(1,1,1)$. Thus, our FIDE knight may be allowed to jump from $(1,1,1)$ to $(2,2,1)$... otherwise it is a Euclidean knight. $\endgroup$ Commented Sep 23, 2023 at 12:12
  • $\begingroup$ Of course, I meant a Euclidean knight that couldn't be allowed to move along the major diagonal of the given $k$-cube... but the proof of Theorem 4.1 is enough to provide infinitely many closed knight's tours on $n \times n \times \cdots \times n$ chessboards, where $n \in \{2,3\}$ $\endgroup$ Commented Sep 23, 2023 at 12:23

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