# Knight's Tours with Octonary Symmetry

by George Jelliss

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. ## Octonary Tours

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).

### Octonary tours with 8, 16 or 24 cells

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 1.7.4.2, 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.

### Octonary tours with 32 cells

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.

### Octonary tours with 40 cells

These five examples are all within a 9×9 frame. ### Octonary tours with 48 cells

Here are three within a 9×9 frame. ### Larger Octonary Tours

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, ...

Note: Material previously on this page has now been moved to the Octonarian Tours page.