2

Not a trivial question: How would you define the knight move rule on a d-dimensional chessboard (say an 8 X 8 X ... X 8 board) by extending the official knight move rule that states how a knight moves on a regular 8 X 8 chessboard?

The most strict and consistent proposal I have found is to take the Euclidean distance between the centers of adjacent cells (as an example, just consider the distance between (0,0,...,0) and (1,0,...,0) or between (0,0,...,0) and (0,1,0,...,0), generalizing the distance between the centers of a1 and b1 or between the centers of a1 and a2 on a planar chessboard) and state that our knight can only move from a cell of our chessboard to another one of the same chessboard, under the constraint that this distance have to be equal to sqrt(5).

I used the Euclidean distance in the definition above, since Article 3.6, Section E, of the FIDE Handbook defines the 2D knight move rule 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". Now, if we do not accept my generalization, we should assume that a 3-dimensional knight can move from the (1,1,1) to (0,0,1) (or (2,0,1) or (2,2,1) and so forth...) and I do not think that this would be reasonable.

What do you think, is this the best way to describe a multidimensional knight?

3
  • 2
    Shouldn't you first determine where a multidimensional Knight can move to before worrying about the best way to describe it? The 2D Knight moves to a square 2 away in one dimension, and 1 away in the other, that is, it's a (2, 1) leaper. It's not obvious to me how a 3D Knight moves. ((3, 2, 1)? (2, 1, 1)? (2, 2, 1)?)
    – Abigail
    Sep 22, 2023 at 13:06
  • 1
    I don't think there's one clearly correct generalization, in comparison to rook.
    – qwr
    Sep 22, 2023 at 14:47
  • 1
    I am asking this since I am working on a long preprint about metric spaces in chess and you are right, there is not a unique way to generalize some chess pieces in multiple dimensions. Anyway, I discuss the knight move rule in this recent preprint on arXiv: arxiv.org/abs/2309.09639 The outcome is quite surprising in 5 dimensions and above (see Theorem 2.1 and Theorem 4.1). Sep 23, 2023 at 11:17

2 Answers 2

1

Im not sure if this is an answer, but it is too long for a comment.

I have seen it noted that a Knight, on the center of a 5x5 set of squares, covers all of the squares that would not be covered by the pieces (Bishop and Rook) that move in straight lines. This would certainly generalize, but might not be best.

I think the problem starts with the Bishops. Even with three dimensions, there are two sorts of diagonal, so we have two sorts of Bishop. Once the Rooks and Bishops are sorted out, the Knights could take whatever is left.

In $d$ dimensions I think there are $d-1$ kinds of diagonal. Together with the Rooks, they give a choice of $4d$ destinations to the linear pieces. This leaves the Knights with $5^d-4d-1$ destinations, which eventually becomes most of the $5^d$ hypercube. In very high dimensions, the Knight would have access to a fraction $(5/8)^d$ of the board.

4
  • 5^d - 4d - 1 equals 16 if d == 2. But a Knight in the center of a standard board can reach only 8 squares. The assumption there are d-1 kinds of diagonal is wrong. If correct, there would be just one diagonal if d == 2, but there are two diagonals. A d dimensional cube has 2^d vertices, which means there are 2^{d-1} diagonals. Rooks can move in d directions. Which leaves 5^d - 4*(2^{d-1} + d) - 1 squares for the Knight to move to. This will make the Knight by far the most powerful piece on the board for higher dimensions.
    – Abigail
    Sep 22, 2023 at 23:28
  • Dear Philip Roe, you are abosolutely right. I've already solved this problem in the manuscript I am writing about metric spaces in chess and the solution comes from the parity argument of the chessboard, but it is too long to explain in a comment (I will post the DOI of the preprint when it will be ready). In this question let us forget about everything and just focus ourselves on the knight definition given by the FIDE in the Handbook, since it is related to this preprint that I've just released about knight's tours: arxiv.org/abs/2309.09639 Sep 23, 2023 at 11:21
  • P.S. The generalized knight that I define in the preprint I am working on is very powerful, but not so powerful as the queen, which covers the union of the cells covered by the rook and by the bishop at any move... even if in some cases the knight can reach a square earlier than the queen, this is very tricky and my generalized pawn definition takes into account a wide set of arising constraints, in order to let it produce a metric space, letting me calculate also the eccentricity of every cell (and thus the diameter and the radius of the induced graph). Sep 23, 2023 at 11:26
  • @Abigail My formula was wrong. But by kinds of diagonal in d=3, I mean edge to edge and vertex to vertex diagonals, There are therefore two kinds of diagonal and three of each. If we count the Rooks as face-to face diagonal there are d^2 "diagonals". I think this is general and my formula should be $5^d-4d^2-1$. The Knight is only the most powerful piece over a short range. The other pieces have longer range. Not so very different from 2D. For large d the formula is dominated by the exponential term however you define the Bishops.
    – Philip Roe
    Sep 23, 2023 at 21:06
0

What do you think, is this the best way to describe a multidimensional knight?

I'd use a more precise fairy chess notation to describe multi-dimensional chess movement. There are a few different notations, but they are all more precise than the FIDE description. For example, the extended Partlett notation for a knight is ~1/2 which means "this piece jumps over any obstacles to land and capture on its destination location, which is 1 unit away in a dimension and 2 units away in another." Another term for this is a (1,2) leaper.

In 2D, this yields the 8 regular knight moves. In more than one dimension, the same analogy would apply (so a 3D knight could move 2 units "upward" and 1 unit "inward").

If instead you wanted to translate the confusing FIDE definition into multiple dimensions, you are certainly welcome to try. But the terms "rank," "file," and "diagonal" will need more precise definition in higher dimensions. And depending on how you define those terms, you may just end up with a regular old (1,2) leaper after all the work.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.