This study is simply a catalogue of tours on odd square boards from the small to the large.
On the 3×3 board the knight has no move when placed on the centre cell, so there is no tour, but it does have a circuit of the 8 edge cells. This is the simplest example of a knight's tour on a board with holes and will be found in our study of tours on Shaped and Holey Boards. Further analysis of tours on 3×n rectangles will be found in our study of tours on Oblong Boards.
The 5×5 board of 25 cells is the smallest knight-tourable square board. There are 112 geometrically distinct tours, all open (since the board has an odd number of cells) of which 8 are symmetric.
Euler (1759) was the first to publish such tours. §36-40: He notes that, since odd and even numbered cells follow alternately, a reentrant tour is not possible on any board with an odd number of cells, since the numbers on the first and last cells in the tour are odd. On the 25-cell board he notes that every tour must commence or finish at a corner cell. He diagrams one corner-to-centre tour and gives formulae for 34 others, including all eight symmetric cases (all corner-to-corner type), which are also given separate diagrams (in numerical form). It happens incidentally that the first of these tours, numbered from a1, has the arithmetic progression 1, 7, 13, 29, 25 along the main diagonal, making it a figured tour.
If you prefer larger numbers and wish to count the different diagrams possible, then each symmetric tour can be viewed in 4 distinguishable orientations, and each asymmetric tour in 8, thus the total becomes: 8×4 + (112 – 8)×8 = 864.
If further you wish to present the tours in numerical form this total has to be doubled since each tour can be numbered from either end, giving the total 1728. This was the total given by Charles Planck in his "Chessboard Puzzles" series in Chess Amateur (puzzles 25 and 26, December 1908 p.83 and February 1909 p.147). This is the earliest reference I have found to the complete enumeration of the 5×5 tours.
Every 5×5 tour must have one end on a corner cell, since if the knight passes through all four corners it makes a short circuit of 8 moves. The other end can be any other cell of the same colour. Thus, classified by separation of end-points there are 8 types of 5×5 tour. The numbers of each type are: {0,2} 14, {0,4} 30, {1,1} 8, {1,3} 14, {2,2} 8, {2,4} 14, {3,3} 6, {4,4} 18.
The moves through the corners form three different patterns: (A) two pieces of 4 cells, (B) pieces of 6 and 2 cells, (C) one path of 8 cells. The moves through the other cells, excluding the centre, form a short circuit of 16 moves, and this is either toured in two pieces of (1) 8+8 cells, (2) 6+10 cells, (3) 4+12 cells (4) 2+14 cells or (5) one piece of 16 cells. Combining these classifications we get 13 classes: A1, A2, A3, A4; B1, B2, B3, B4; C1, C2, C3, C4, C5. The classes (1) to (5) are also characterised by the pattern at the centre cell: (1) the moves go straight through, (2) they form an acute angle, (3) a right angle, (4) an obtuse angle and (5) the tour terminates at the centre.
Classes A1–A4 comprise the 18 tours from corner to opposite corner. Class A1 consists of the eight symmetric tours, with straight move through centre. The maximum number of two-unit lines in a 5×5 tour is three. Murray (1942) noted that there are two such tours, both symmetric (diagrams 2 and 4 here). Class A2 consists of four with acute centre, A3 two with right-angled centre, A4 four with obtuse centre.
Classes B1–B4 comprise 30 tours from corner to adjacent corner, with B1 straight centre 2 tours, B2 acute centre 8 tours, B3 right-angled centre 12 tours, and B4 obtuse centre 8 tours.
Class C1 has 8 tours with straight move through centre. Class C2 has 16 tours with acute centre.
Class C3 has 16 tours with right-angled centre.
Class C4 has 16 tours with obtuse angle at centre. Class C5 has 8 tours terminating at the centre.
Puzzles: Complete the following completely defined 5×5 tours: (1) a1 ... a4-c3. (2) a1-c2 ... b2 with central moves forming (a) diagonal-acute angle, (b) orthogonal-acute angle, (c) right angle containing b2, (d) right angle excluding b2, (e) orthogonal-obtuse angle, (f) diagonal obtuse angle, (g) straight line. (3) a1-c2 ... d4 (a)–(f) as in (2). (4) a1 ... e3 with centre angle (a) orthogonal-obtuse, (b) diagonal-obtuse. (5) a1 ... c1 with centre angle (a) diagonal-acute, (b) orthogonal-acute. (6) Identify which are 'tours of inspection', i.e. they 'pass through' every unit square — not only the squares of the board but also the squares whose corners are the centres of the cells.
If the board is chequered, then a tour must start and finish in a cell of the majority colour, that is the colour of the corner cells. On this board a tour between any two such cells is possible. To show all geometrically distinct cases of end-point placement would require how many diagrams: 50?.
We show some centrosymmetric open tours. The Bergholt tour shows the maximum amount of oblique quaternary symmetry in a centrosymmetric open tour. (The linkage a5-b3-d4-f5-g3, by inserting or deleting the moves along this path, converts it into a centreless tour with oblique quaternary symmetry.) The last two were composed so that examples of all five symmetric end-point positions are shown.
Here are some asymmetric example tours 7×7. The Willis example is simply irregular. The Kraitchik example is compartmental, combining 4×7 with 3×7. The Dudeney and Dickins examples show an obtuse octagon (and acute octagon).
The Dudeney examples are cited in Murray 1942 but without source.
Border method. The border braid consists of two strands of unequal length, one of 16 and the other of 40 cells. Symmetric open tours of course require boards with an odd number of cells. This case was studied by Lucas (1894), including this symmetric corner to opposite corner example, and the three asymmetric tours joining corner to centre, corner to adjacent corner and corner to near-corner. These tours all incorporate a central 5×5 tour.
The method of the first example is easily extended to 13×13, 17×17 and so on.
Border method. The border braid consists of two unequal strands.
See the 17×17 example below.
See the 19×19 example below.
This tour simply extends the Lucas 9×9 example to the 13×13 and 17×17 boards.
Method given by Murray (1942) for "expanding tour on the 4n + 3 boards" (i.e. sides 7, 11, 15, 19 and so on, starting from the Dudeney 7×7 example).
