To investigate this type of symmetry, as noted in the section on symmetry in general, I use the method of numbering the cells from the centre outwards, as illustrated here for boards up to 8×8.
Apart from the star-shaped circuit on the 3×3 board, the existence of other tours with octonary symmetry seems to have been overlooked hitherto. They all require boards with a containing square of odd side and omit the centre cell. Thus no octonary tour is possible on a board without holes. Other cells may be omitted in groups of four on the diagonals or medians, or of eight elsewhere.
Note that for the cells to constitute a board, as opposed to being merely an arrangement of cells, they must be wazir-connected, i.e. it must be possible to get from any cell to any other in a series of single-step rook moves.
Every octonary tour consists of eight equal paths, each path having one end on a median cell and the other end on a diagonal cell. Tours with octonary symmetry thus all use a multiple of 8 cells (4 diagonal, 4 medial, each other position in sets of 8).
There is one octonary tour with 8 or 16 cells and there are two with 24 cells.
Within a 3×3 area the knight's moves are between the cells coded 1 and 2, so we denote the 8-move star by 1.2, this being one eighth of the octonary tour. Within a 5×5 area we find a tour 3.4.2 of a cross-shaped board with a central cross-shaped hole. Within a 7×7 area the shortest quaternary tours have the formula 18.104.22.168, where the 7.4 move can be taken in two different ways we write it 7:4 or 7;4 according as the move crosses a median or diagonal.
I find 6 within the 7×7 area.
And 8 within the 9×9 area. The cells form ever more tenuously connected board shapes.
Note: The last tour above has the peculiar property that all the angles between successive pairs of knight's moves are obtuse (including straight). An unsolved problem is whether a tour of this type is possible without holes. A solution is possible with all angles greater than or equal to 90 degrees (i.e. allowing right angles). See the Quaternary Symmetry pages for an example.
These five examples are all within a 9×9 frame.
Here are three within a 9×9 frame.
Here are examples with 56, 64, 72 and 80 cells (maximum possible) within the 11×11 frame.
Theorem. Within a square frame of odd side greater than 3 the maximum number of moves (or cells) in an octonary tour is 16(h^2 h + 1) when the side is 4h + 1 and 16(h^2 + 1) when the side is 4h + 3.
Sketch Proof: Assume the board chequered with white cells on the diagonals. Then the number of white cells in the triangle bounded by a diagonal, a median and the frame edge is either 2 + 4 + 6 + ... + 2(h 1) = h(h 1) or else is 1 + 3 + 5 + ... + (2h 1) = h^2. In a knight's path these must alternate with black cells, and for the longest possible path we must have a black cell at each end, since the black cells are in the majority in these triangles. So if there are w white cells we add w + 1 black cells and 1 more to account for the links to the cells on diagonal, giving 8(2w + 2) = 16(w + 1) for the whole tour.
This sequence of maximum values therefore runs; 8, 16, 32, 48, 80, 112, 160, ...or, dividing by 8 to give the size of each octant: 1, 2, 4, 6, 10, 14, 20, ...
Back to top