The smallest knight-tourable boards are the two 7-cell boards shown below, each of which is tourable by both wazir and knight uniquely. The tours have the same symmetry as the boards, namely direct binary symmetry, lateral or diagonal. The U-shaped case appears in the earliest Arabian chess manuscripts (al-'Adli, c.840) in the form of a puzzle where you are to place seven pawns round three edges of the 3×3 board, leaving only the top square vacant, each new pawn being placed at a knight's move from the preceding one (Murray, A History of Chess 1913 p.337). The trick is to start at a knight's move from the vacant cell.
The mediaeval manuscripts also contain the puzzle of the four knights, often attributed to the later writer Guarini (1512), where white and black knights placed in the corners of the 3×3 board are to be interchanged. The knights go in procession round the star-shaped closed tour of the eight edge squares. The 3×3 closed tour is the first example of a tour with octonary symmetry. Also shown is a symmetric open tour of a loosely connected board of 8 cells. This type of board is excluded from the rest of this study, but this is not to say that there may not be interesting results to be found.
There are 10 boards of 8 cells (octominoes) that are knight-tourable, with open tours. Three of these boards have a hole. Only the centreless 3×3 board admits a reentrant or closed tour. The two other holed boards, and one of the other shaped boards, can each be toured in two ways. One open tour, the last diagram here, is centrosymmetric.
The collages below show the open tours of 8-cell boards, with or without holes, joined to form a single open tour. The first, excluding the 3×3 case, fits them within an 11×11 area, the second, including the 3×3 case, takes a 10×13 board. The holes are shaded in.
The 10 symmetric tours and other selected cases are shown:
After a further check (29 August 2009) of my enumeration of the 9-cell tours I now find 57 tourable boards with 94 tours. Of these boards, 1 can be toured in 6 ways, 3 in 4 ways, 4 in 3 ways, 15 in 2 ways, and the remaining 34 uniquely. There are 13 boards with a hole, and 10 of the tours are symmetric (2 oblique, 8 direct).
Here is a chart of all the 9-cell boards in reduced form:
For a closed tour of course the board must have an even number of cells. There are 12 different patterns of 10-cell boards, without holes, on which a knight may play a closed tour. This result was given by O. E. Vinje of Baltimore (Fairy Chess Review, problem 8149, July 1949 p.44 and August 1949 p.53).
Four of the Vinje tours are symmetric, one each of the Bergholtian, Murraian, Sulian and Eulerian types. They are the first four in the top row. The Bergholtian tour, on an H-shaped board, has biaxial symmetry.
In the remarkable collage below, these boards are fitted together within a 13×13 area to form a larger board, also without holes, in which the tours join together to form a closed tour of the composite board. This is due to A. W. Baillie (Fairy Chess Review, problem 8531 November 1949 p.84 and June 1950 p.110).
T. W. Marlow (1995) has noted that there are six further 10-ominoes with holes that have closed knight tours, to be addded to Vinje's enumeration.
On 10 cells there are 9 centro-symmetric open tours (using 6 shapes of board). Note that one of these tours is reentrant. The asymmetric open tours have not been enumerated.
On boards of 11 cells there is 1 tour with lateral axis, 8 centro-symmetric tours (using 5 board shapes), and a holey board with three tours which have a diagonal axis:
Enumeration of the 12-cell boards with closed knight tours is a considerably stiffer problem. I sent a summary of my results to Tom Marlow (27 October 1995) writing: "I think I have found all those up to 4×5 minimum containing rectangle, by a process of systematically removing squares from the rectangle, while ensuring the tourability of the remainder. However, for the larger sizes this method becomes inefficient. The 4×6 cases have mostly been found by 'folding out' a pair of moves from the 4×5 cases, and permuting the different shapes removed from the corners. The 5×5 cases, which I'm sure are very incomplete, resulted mostly from an examination of 12-cell tours that include a 2×3 rectangle. The 5×6 was 'folded out' from a 4×6 case. The particularly nice 6×6, where there seems to be the only one solution, was found by pure intuition before I started the systematic count. I haven't proved to my satisfaction that no cases with a dimension greater than 6 can occur, but this seems probable."
Mr Marlow responded (24 November 1995): "I am finding the polyomino tour problem quite addictive." He gave the 10-cell tours with holes as diagrammed above and diagrams of twenty boards (with 23 tours) not included in my list. He wrote: "My method has been to start from the opposite direction to you and look for 12 step knight loops within a rectangle. Then I see if the squares form a connected shape and reach to all edges. I found that I could easily adapt a magic tour computer programme to search for 12 step loops and then worked by hand. It seems clear that neither dimension can be greater than six."
I revisited the subject again in September 2003 trying a slightly different method, working in from the edges and corners of the rectangles. This provides a proof of the dimension proposition:
THEOREM: A 12-cell closed tour cannot have an enclosing rectangle of side greater than six.
Proof (sketch): Each edge must contain at least one cell that is part of the tour, with two cells
reachable from it (say a3, b5, c4 or a3, c2, c4), and at least two other cells (say b3, b4 or b3, c3) to connect
these. These in turn generate two other cells each, but it will be found that none can coincide with the previous
five. Two such batches of nine against opposite edges separated by seven units must have an overlap of at least
six to give a 12-cell tour (3 + 6 + 3), since an overlap of 5 uses 13 (4 + 5 + 4) and of 4 uses 14 (5 + 4 + 5).
The only possible arrangements to do this result in lozenge-shaped short circuits of four moves. (For example a3,
b3, c2, c3, c4, d2, d4, e2, e3, e4, f3, g3.)
We now give a catalogue with diagrams of all the 27 symmetric closed tours of 12 cells. There are two 12-cell tours with quaternary symmetry, both of the type with two diagonal axes. The one within the 4×4 frame was found by Euler (1759), the one within the 6×6 is my own discovery. There are 10 other closed tours of 12 cells with centrosymmetry (i.e. 180 degree rotational symmetry). Of these, 6 are Bergholtian (i.e. passing through the centre), one within the 6×3 frame and five within the 4×5 frame, and 4 are Eulerian (i.e. not passing through the centre) three within the 4×6 frame and one on the 5×5, with hole):
The remaining symmetric tours have a single axis of symmetry; one tour 4×6 has a lateral axis, the o thers have a diagonal axis, namely four 4×4 and ten 5×5:
Rather than diagram all the asymmetric tours we show here all the closed-tourable board shapes of 12 cells in reduced size. The tours are easily reconstructed since only two moves are available at many of the cells. Where more than one geometrically distinct tour is possible the number is stated below the diagram. S = symmetric board, H = board with hole(s). The diagrams are arranged with (a) shaped boards preceding holey boards, (b) symmetric boards before asymmetric, (c) multiply toured boards before those with single tours. The boards are oriented with the longer side (if any) of the containing rectangle horizontal, and according to the binary number method, by which the cut away cells (black) count as 0s and the board cells (white) count as 1s, and the board is oriented so as to give the smallest number.
3×5, 3×6 and 4×4 cases:
4×5 shaped cases:
4×5 holey cases:
4×6 cases:
5×5 cases:
5×6 and 6×6 cases:
The following table summarises the results we have obtained.
| Frame size | Boards | Tours | Symmetric | With hole |
| 3×5 | 1 | 1 | 0 | 0 |
| 3×6 | 2 | 2 | 1 | 0 |
| 4×4 | 10 | 14 | 3B/5T | 2 |
| 4×5 | 83 | 104 | 8B/6T | 30 |
| 4×6 | 22 | 27 | 2B/3T | 8 |
| 5×5> | 67 | 83 | 12B/11T | 16 |
| 5×6 | 5 | 6 | 0 | 2 |
| 6×6 | 1 | 1 | 1 | 0 |
| Totals | 191 | 238 | 27B/27T | 58 |
Thus for example within the 4×5 frame there are 83 different 12-square boards, of which 30 have a hole. Since some of these can be toured in two or more ways they give 104 tours, (67 on the shaped boards and 37 on the holey boards). There are 8 symmetric boards but only 6 symmetric tours since two of the symmetric boards only have asymmetric tours (another has two tours, but only one of them is symmetric).
In the 5×5 frame there are 12 symmetric boards but one of these, the one with lateral axis of symmetry, has only an asymmetric tour, so there are only 11 symmetric tours. Again, one of the symmetric boards has two tours but only one of these is symmetric.
Shown are the tour within the smallest frame 3×5, and the asymmetric tour within the 3×6 frame, and one within the 5×5 frame that includes three two-move knight lines (and also cycles round the centre in one cnsistent direction):
Next we show the five cases of asymmetric tours of symmetric boards. The 4×5 and 5×5 cases with lateral axis differ only in having a pair of moves, and the associated cell, folded up or down.
Finally we also show the 5×5 boards with 4 and 6 tours. The 4×5 board with 4 tours, not shown, is related to the 5×5, as above, by the top cell and its associated two moves being folded down to the bottom.
There are 49 geometrically distinct 12-cell symmetric open tours. These consist of: 2 rectangular (3×4 board), 12 reentrant (derived from the 6 bergholtian tours above by deleting one of the central moves), 27 other shaped-board tours, and 8 with holes. The 27 shaped-board tours use 18 shapes of board (1 with 3 tours, 7 with 2 tours, and 10 with 1 tour).
There are 2 with 3×6 frame, 17 with 4×5 frame, 1 with 4×7 frame, and 7 with 5×6 frame:
There are also eight 12-cell centrosymmetric open tours with holes. Of these 7 are on the same board with two single-cell holes, and 1 is on a board with a two-cell hole.
The 12-cell asymmetric open tours have not been counted. There is only one closed tour within the 3×5 frame, and it is asymmetric, so produces 12 reentrant tours. In a partial enumeration not yet checked, I found 39 non-reentrant tours within this frame. These tours use 10 different board shapes. Removing the minority color square from a2 the other two cells removed can be a1, b2; a1, c3; a1, d2; a1, e1; a1, e3; b2, e1; d2, e1; e1, e3; and with b1 as the minority cell, the others are a3, d2 or a3, e1.
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