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We know from a theorem by Schwenk that for any (m x n) chess board with $m \leq n$ it is always possible to create a knight's tour unless one or more of these three conditions are met:

  1. m and n are both odd
  2. m = 1,2 or 4
  3. m = 3 and n = 4, 6 or 8

When I was about 15 years old, I played a game that I had "invented" myself. Years later, I realised this game was actually a generalisation of the Knight's tour problem. The generalisation is that the knight can jump of the edge, and reappear on the opposite side of the board. From there, it can complete its knight move.

Here is an example of what I mean. The red numbers indicate the trajectory of a knight that traverses in the conventional manner. The knight begins at tile 1 and ends in tile 3. The green trajectory indicates the manner in which the knight can travel in the new generalised Knight's Tour problem. Again, it begins at tile 1 and ends at tile 3.

examples of knight moves

I was wondering whether, with these new abilities, the knight can make a Knight's Tour under the conditions mentioned in Schwenk's Theorem.

Please note that the knight can only reappear on the other side of the board once every move. Furthermore, it can only do it when it is moving in a direction perpendicular to the edge of the board.

Max Lonysa Muller
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    The problem you pose is that of a Knights tour on the Torus you get by identifying opposite edges of the board. There are papers on the subject e.g. http://www.math.oregonstate.edu/~math_reu/proceedings/REU_Proceedings/Proceedings1994/1994Seibel.pdf which you can read online and/or http://www.jstor.org/discover/10.2307/2691257?uid=3738032&uid=2129&uid=2&uid=70&uid=4&sid=21103720877127 – Mark Bennet Mar 23 '14 at 17:31
  • Is it the Torus, or perhaps just a cylinder (identifying along only one pair of edges)? – hardmath Mar 24 '14 at 01:03
  • @MarkBennet and hardmath: ah I guess it is the Torus, the two pairs of opposing edges are identified with one another. Thank you for the links to the paper, Mark. I guess you can pose them as an answer if you want. – Max Lonysa Muller Mar 24 '14 at 11:30

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Let me modify an argument that I gave in this MSE-post:

On an infinite board represented by $\Bbb Z \times \Bbb Z$ the positions the knight can visit starting from $(0,0)$ form a group. This group $P$ is generated by the moves represented by $(1,2),(-1,2) , (2,1),(2,-1)$. $$P=\langle (1,2),(-1,2) , (2,1),(2,-1) \rangle = \Bbb Z \times \Bbb Z.$$ Now you are identifying every $n$-th column and every $m$-th row, which is, algebraically speaking, to factor out $n \Bbb Z \times m \Bbb Z$. Thus the possible visiting positions are given by the image of $P= \Bbb Z \times \Bbb Z$ in $\Bbb Z /n\Bbb Z \times \Bbb Z /m\Bbb Z$, which is $\Bbb Z /n\Bbb Z \times \Bbb Z /m\Bbb Z$, i.e. the whole board.

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benh
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