Proving the existence of Euclidean knight's tours on $n \times n \times \cdots \times n$ chessboards for $n<4$

The Knight's Tour problem consists of finding a Hamiltonian path for the knight on a given set of points so that the knight can visit exactly once every vertex of the mentioned set. In the present paper, we provide a $5$-dimensional alternative to the well-known statement that it is not ever possible for a knight to visit once every vertex of $C(3,k) := \{0,1,2\}^k$ by performing a sequence of $3^k-1$ jumps of standard length, since the most accurate answer to the original question actually depends on which mathematical assumptions we are making at the beginning of the game, when we decide to extend a planar chess piece to the third dimension and above. Our counterintuitive outcome follows from the observation that we can alternatively define a $2$D knight as a piece that moves from one square to another on the chessboard by covering a fixed Euclidean distance of $\sqrt{5}$ so that also the statement of Theorem~3 in [Erde, J., Gol{\'e}nia, B., \&Gol{\'e}nia, S. (2012), The closed knight tour problem in higher dimensions, The Electronic Journal of Combinatorics, 19(4), \#P9] does not hold anymore for such a Euclidean knight, as long as a $2 \times 2 \times \cdots \times 2$ chessboard with at least $2^6$ cells is given. Moreover, we show a classical closed knight's tour on $C(3,4)-\{(1,1,1,1)\}$ whose arrival is at a distance of $2$ from $(1,1,1,1)$, and we finally construct closed Euclidean knight's tours on $\{0,1\}^k$ for each integer $k \geq 6$.


Introduction
Given a set C ⊆ R k consisting of m ∈ Z + points, it is commonly agreed that every knight's tour is a sequence of m − 1 knight jumps, of Euclidean length √ 5 chessboard units each (where one chessboard unit is the distance between the centers of adjacent squares of the chessboard), that let the knight visit exactly once all the m vertices of C. In particular, we say that the knight's tour is closed if and only if the m-th visited vertex of C (including the starting vertex) is at a unit knight-distance from the beginning point, otherwise we have an open knight's tour on C.
The origins of the Knight's Tour problem are lost in the centuries, this being a thousands of years old puzzle [2] that lists among its contributors some very big names in mathematics, such as Abraham de Moivre, Alexandre-Théophile Vandermonde, Adrien-Marie Legendre, and Leonard Euler himself, who found one solution for the planar 8 × 8 configuration in 1759 [7,18].Euler's solution is an open knight's tour since the center of the last square visited by the knight is not at a distance of √ 2 2 + 1 2 chessboard units from the center of its starting square (in the most common sense, for arbitrary k, being the beginning vertex at a Euclidean distance of √ 5 from the arrival would represent a necessary but not sufficient condition for having a closed knight's tour).Now, if we agree that the Euclidean √ 5-rule (see [8], Article 3.6, that uses the superlative of near as a criterion for the official knight move rule) defines also the knight metric for any k-dimensional n × n × • • • × n chessboard (while a customizable definition of the discrete knight pattern is at the bottom of the generalized knight's tour problem in two and three dimensions, as described in [3] and [1], respectively), we trivially have that a k-knight is a mathematical object whose move rule consists of performing only jumps having Euclidean length equal to √ 5 chessboard units [9], from a cell of the given chessboard to one of the remaining n k − 1 cells.Thus, we can move our favorite chess piece from the vertex V 1 , identified by the k-tuple of Cartesian coordinates (x 1 , x 2 , . . ., x k ) : Hence, (1) can be compactly rewritten as It is easily possible to show that, although our Euclidean knight produces a metric for every pair (n, k) that allows the usual knight to do so (i.e., n ≥ 4 ∧ k ≥ 2 represents a sufficient condition [15]), it also induces a metric space on any given 3 × 3 × • • • × 3 chessboard with at least 3 5 cells (Section 2) and even on every 2 × 2 × • • • × 2 chessboard consisting of at least 2 6 cells (see Section 4, Theorem 4.1).
Furthermore, we construct an open Euclidean knight's tour on {{0, Accordingly, Section 2 is devoted to proving the existence of Euclidean knight's tours when n = 3 is given, and then a pair of corollaries will follow, while Section 3 proposes a variation of the main problem [16] by removing the central vertex from any grid {{0, [12], under the additional constraint of ending the path in a vertex that is at a Euclidean distance of
Finally, Section 4 entirely covers the n = 2 case.
In this regard, let us observe how Qing and Watkins indicated a different way to extend in 3D the planar knight's move pattern by proposing, in [13], pp.45-46, that every knight jump has to mandatorily change all the k = 3 Cartesian coordinates of its starting vertex (by 2 0 , 2 1 , and 2 2 ).Although this personal interpretation of Article 3.6 of [8] is obviously not compatible with the existence of any knight's tour on 3 × 3 × • • • × 3 grids (since the Euclidean distance between (1, 1, . . ., 1) and (0, 0, . . ., 0) is equal to √ k, which is clearly smaller than for any k > 1), the underlying idea of a k-knight that can change (or has to mandatorily change) the values of all its k Cartesian coordinates by performing a single move is fascinating and useful [11].Now we are ready to prove that an open Euclidean knight's tour actually exists if n = 3 and k is set at 5, so the trivial consideration that the knight graph is not connected for any 3 × 3 × • • • × 3 board does not apply anymore, as the k-knight is a Euclidean k-knight.
Accordingly, let us constructively prove Theorem 2.1 by simply providing the sequence of the 243 Cartesian coordinates that describe an open knight's tour on the set C(3, 5) (see Figure 1 for a graphical proof), since every vertex is visited exactly once and the Euclidean length of each knight jump is equal to √ 5.Then, the polygonal chain , even if this time the taxicab length [10,17] of our knight jump is equal to 5, instead of 3 as for any other knight move), while the Euclidean distance between its starting point and ending point is 2, so the knight tour is open.Therefore, a knight tour exists for the set {{0, Proof.In order to prove Corollary 2.1, let us introduce the well-known parity argument [13].
Then, we have that the generic vertex (x 1 , x 2 , . . ., x k ) of {0, 1, 2, . . ., n − 1} k is a dark vertex if and only if k j=1 x j ≡ 0 (mod 2), whereas any light vertex is such that k j=1 x j ≡ 1 (mod 2) (i.e., if we add together the k Cartesian coordinates of a dark vertex of C the result is always an even number, otherwise the given vertex is a light vertex).Figure 2 shows how to consistently represent the set C(3, 5) as a 5D chessboard.Thus, we invoke the parity argument by observing that the knight (including the Euclidean knight, which is subject to the same constraint by construction since |1|+|1|+|1|+|1|+|1| is odd as |1| + |2|) can only move from a dark square to a light square and vice versa (see [4], Figure 2).For this purpose, let us observe that the taxicab length [17] of any Euclidean knight jump has to necessarily be 3 or 5 since all the Diophantine equations of the form 2)) admit only two types of solutions, one having only 2 non-zero terms (i.e., a ±1 term and a ±2 term) and a second set of solutions which is characterized by 5 non-zero terms that are all elements of {−1, 1}.
Hence, a necessary (but not sufficient) condition for having a closed (possibly Euclidean) knight's tour on C(n, k) is that the number of light vertices is equal to the number of dark vertices [13], and this is impossible if n = 3 since 3 k is odd for any k.
Corollary 2.2.A Euclidean knight can visit every vertex of C(3, 5) and then return to the vertex on which it began by performing no more than 3 5 + 1 jumps.
In detail, the mentioned Theorem 3 assumes k ≥ 3 and states that a k-dimensional rectangular grid of the form admits a closed knight's tour if and only if the following three conditions hold: Proof.In order to state the necessary and sufficient condition for the existence of Euclidean knight's tours on 2 × 2 × • • • × 2 chessboards, let us preliminary point out that a Euclidean knight can move also on a 2 × 2 × • • • × 2 chessboard, but this is possible only if k ≥ 5, since here is mandatory that a Euclidean knight changes by one 5 Cartesian coordinates at any jump, given the fact that n < 3 does not let the knight make its well-known L-shaped move.Additionally, it is trivial to observe that only one move is possible on C(2, 5) since the maximum distance between two vertices of a k-cube corresponds to the distance between opposite pairs of corners, which is always equal to and then none of the unvisited vertices of C(2, 5) is far enough to allow a second move.
Finally, we can repeat the same process to extend the 7-cube solution to the 8-cube, and so forth.We are allowed to do so since any vertex of a k-cube is also a corner, and thus the symmetry is preserved.
Therefore, there exist (many) closed Euclidean knight's tours on each 2×2×• • •×2 chessboard with at least 2 6 cells, whereas no Euclidean knight's tour is possible on C(2, k) for k < 6, and the proof of Theorem 4.1 is complete.

Conclusion
The move pattern of the 2D knight, the piece that also appears in the FIDE logo, is described by Article 3.6 of Reference [8] as follows: "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".Thus, we have defined the knight by assuming the standard Euclidean metric and, consequently, the resulting distance covered by any jump is √ 5.Then, it seems legitimate to assume that a multidimensional knight can also be defined in a more inclusive way than the usual description of a piece that merely moves, from a vertex of C(n, k) to another one of the same set, by adding or subtracting 2 to one of the k Cartesian coordinates of the starting vertex and, simultaneously, adding or subtracting 1 to another of the remaining k−1 elements of the mentioned k-tuple.Accordingly, we have provided the alternative definition of the knight as a piece that is allowed to move from V m ∈ C(n, k) to V m+1 ∈ C(n, k) if and only if the condition d(V m , V m+1 ) = d(V m+1 , V m ) = √ 5 is satisfied (where d(A, B) indicates the Euclidean distance between the point A and the point B, as usual).
Consequently, the present paper has shown that such a Euclidean knight can produce a knight's tour also on k-dimensional grids of the form {{0, 1, 2} × {0, 1, 2} × • • • × {0, 1, 2}}, for some k ≥ 5.In particular, by Corollary 2.2, every Euclidean knight's tour for any 3 × 3 × • • • × 3 chessboard is necessarily an open tour.Now, if this is not enough to fully accept the √ 5 knight metric, since the knight may not be allowed to move along one of the main diagonals of a 5-cube by unimaginatively extending beyond the k = 2 case the semantic meaning of Article 3.6 of [8], then Theorem 4.1 shows that it is possible to construct closed Euclidean knight's tours on any (k-dimensional) 2 × 2 × • • • × 2 chessboard, as long as k ≥ 6, avoiding by construction to move along any main diagonal of the given k-cube.Furthermore, we could invoke the same argument to suggest that, for any k > 4, the central vertex of a 3 × 3 × • • • × 3 k-cube is (Euclidean) knight-connected to any other element of the set C(3, k).
As a result, Theorem 3 of [6] can no longer be invoked on any metric space C(2, k) : k > 5 where the distance between two vertices A ∈ C(n, k) and B ∈ C(n, k) is given by the minimum number of jumps [15] of (Euclidean) length √ 5 that the described Euclidean k-knight requires in order to go from A to B (and vice versa).
Lastly, a related open problem is to prove the existence of a closed (possibly Euclidean) knight's tour also for every C(3, k) − {(1, 1, . . ., 1)} such that k is even (in Section 3, we have verified the correctness of this conjecture for the cases k = 2 and k = 4, but making such inferences in low dimensions can be notoriously misleading).

Corollary 2 . 1 .
and this concludes the proof of the theorem.For any given k ∈ N − {0, 1}, it does not exist any closed Euclidean knight's tour on C(3, k).
Here, we consider the problem of finding (possibly open) Euclidean knight's tours on 3×3×• • •×3 chessboards.Then, in order to constructively show that a Euclidean knight's tour exists only if k > 4, it is sufficient to point out that the (Euclidean) distance between the vertex (1, 1, . . ., 1) and one of the farthest 2 k vertices (corners) of the k 2 Euclidean knight's tours on a 3 × 3 × 3 × 3 × 3 chessboard [6]orem 3 of Reference[6]would imply that C(2, k) does not admit any closed knight tour, but this is no longer true if closed Euclidean knight's tours are included.Theorem 4.1.Euclidean knight's tours exist on a 2 × 2 × • • • × 2 Furthermore, if there is a Euclidean knight's tour on a given 2 × 2 × • • • × 2 chessboard, then closed Euclidean knight's tours exist for the same chessboard.