**©** by George Jelliss, begun January 2001, latest revision November 2012.

Sections on this page: — Rook Tours — The Clockwork Mouse — Magic Rook Tours — Bishop Tours — Queen Tours (including magic tours)

The chess-piece called a **rook** makes a single move that is like a series of wazir steps in a straight line.
Some problems presented as rook tours are really wazir tours, that is single-step rook tours, this is the case if a
cell is regarded as 'visited' by the rook even though it does not stop there but only rides through. Any non-intersecting
(and non-overlapping) rook path is also a wazir path, except that the moves in the path are counted in terms of the
straight sections of path rather than in unit lengths. For rook tours of this special type see the
Wazir tours section.

By a rook tour we mean a tour in which each move from one cell to the next visited in the tour is a rook move of any length; the cells passed through by the rook en-route from one stopping place to another are not considered to be 'visited' during that move; they will have already been visited or will be visited later in the tour. Rook moves longer than a single step may overlap with others along the same line, so to distinguish them it is necessary to draw the longer moves as curved lines, otherwise they all merge into one straight line. These curves recognise the difference between the passage of the rider through intermediate cells at speed and its stationarity at the end points of the move.

The only tour by a wazir on a 1×n board is an open tour consisting of a straight walk from one end to the other, and a closed tour is possible only on the 1×2 board. A rook, however, can start on any cell and can enter the cells in any order. There are n! ways of numbering the cells of a 1×n board in sequence 1 to n, since there are n choices for the first cell, n−1 for the second, n−2 for the third and so on. Thus the number of open rook tour diagrams, each of which can be numbered from either end, is T(n) = n!/2. All of these open tours are reentrant, since the two end cells are always a rook move apart. The diagrams show the generic open tours of sizes 2 to 4.

Every one-rank tour has the rank, trivially, as an axis of symmetry, so strictly speaking all the tours are symmetric. However, we count as symmetric only those tours that have a vertical axis of symmetry when we draw the moves as curves above the horizontal line. When n is even the number of symmetric open tours is S(n) = [n.(n−2).(n−4)...6.4.2]/2 since there are n choices for first cell, but this also fixes the last cell, then there are n−2 choices for second cell, but this also fixes the next-to-last cell, and so on, until we reach 2 choices for the cell numbered n/2. When n is odd the centre cell of the board must be the centre cell of the symmetric open tour; taking this out we get the same formula for S(n) but with n−1 in place of n. If h = integral part of (n/2) we have S(n) = 2^(h−1)×h! in each case. The number of generic open tours is G(n) = [T(n) + S(n)]/2.

From one closed tour we can derive n open tours by deleting one move. So the closed tours total T(n) = (n−1)!/2 for n > 2. The diagrams show the generic closed tours of sizes 2 to 5.

The above diagrams show the symmetric tours of size 6.

The number of symmetric closed tours on an odd-length board is the same as the number of symmetric open tours on the same board, since they are formed by joining the ends of the open tours; this move being the only symmetric move in the closed tour: S(2h+1) = 2^(h−1)×h!. The number on an even-length board is n/2 + 1 times the total for the previous odd board: S(2h) = 2^(h−1)×(h+1)!. The added '1' in the factor counts the tours with no symmetric move, while the 'n/2' part counts those with two symmetric.

The generic tours are calculated as usual: G(n) = [T(n) + S(n)]/2.

These have not been studied to any extent. Obviously, in a closed tour the number of 'vertical' links must be even.

On the 3×3 board the rook can make 7 geometrically distinct closed tours. The first two illustrated below differ only in the stop at the centre being made on the vertical move or the horizontal move through the centre.

If we regard a cell passed through in two different directions by a rider as 'visited' then we find some distinctive rider tours become possible. The clockwork mouse tour is a particularly distinctive rook crossover tour (one of my earliest tour discoveries, on 23 July 1973). A clockwork mouse (otherwise known as a rotating directed wazir) faces in a definite direction (North, South, East or West) and moves one step at a time directly forward, or turns where it stands to face 90° to the right or left. This rotation counts as a move. These moves can be shown simply by stating the new direction that the piece faces after the turn.

The puzzle is to place a clockwork mouse __anywhere__ on the chessboard, wind it up and allow it to wander
round the board in such a way that it spends the same amount of time (i.e. number of moves) in every cell, and
ends up where it started (facing in the same direction). Surprisingly, the path is uniquely determinate.
This is a generalised type of tour.

The first magic rook tour I know of is this diagonal magic one by Joachim Brügge, published in the German
chess problem magazine *Die Schwalbe* August 1985. It uses 6 lengths of rook move (1, 2, 3, 4, 5, 6).

This inspired me to further research and I have a note that I found the tour on the right on 20 May 1986, and followed this with the other examples shown below on 21 May 1986, but then seem to have forgotten about this work, probably due to big changes in my life that took place later that year. My tours use rook moves of only 4 lengths (1, 2, 3, 4), and each quarter forms a bisatin of one circuit, using two cells in each rank and file. The circles mark the ends of the quarters where the numbers 1, 16, 17, 32, 33, 48, 49, 64 occur. These tours are all diagonally magic.

These biaxial tours remain magic in ranks and files when the origin of numbering is shifted to the next quarter, i.e. 17 becomes 1, 18 becomes 2 and so on. The three tours shown below also remain diagonally magic when renumbered in this way, but the tour above loses the diagonal property; this is because it has more than two numbers £ 16 in a diagonal.

An enumeration of the possible bisatins that could be used in this type of tour would be the next step, if anyone with computing ability would like to take it on.

The following items appeared in *The Games and Puzzles Journal*
issue #31, Jan-Feb 2004 having previously been presented as puzzles for solution in issue #30, Nov-Dec 2003.

Siep Korteling e-mailed on 10 May 2003: "I maintain a website http://www.dorinde.nl/steenslag on International Checkers problems. In that game, played on a 10 by 10 board, a solitary king behaves just like a bishop. I am looking for the shortest path for a king (bishop) to travel all the black fields in a 10 by 10 checkerboard, where moving over a field counts for 'hitting' a field. Could you point me the way?" In subsequent correspondence we found that the answer depends on how a 'path' and its 'length' are defined and what counts as 'visiting' a cell. It is impossible to make a 'tour' that enters or passes through each cell once: some cells must necessarily be visited twice. We have found solutions for four distinct cases.

Find the shortest path for a bishop to visit all the black cells on (A) an 8 by 8 and (B) a 10 by 10 board, for the following cases: (a) minimise number of bishop moves, (b) minimise distance travelled by the bishop, (c) minimise number of moves times distance, (d) minimise number of moves when 'visiting' a cell means either stopping there or passing through it twice. In each case we require that the bishop completes each straight section of the path before turning, and that there be only one route along the path under this condition.

**SOLUTIONS (A)**
**(Aa)** Solution by Siep Korteling.
11 - 88 - 77 - 86 - 31 - 13 - 68 - 57 - 84 - 51 - 15 - 48 - 37 - 82 - 71 - 17 - 28 which takes only 16 moves (geometric length 49√2).
**(Ab)** Solution by George Jelliss.
11 - 33 - 15 - 26 - 17 - 28 - 37 - 48 - 57 - 68 - 86 - 75 - 84 - 73 - 82 - 71 62 - 51 - 42 - 31 - 13 - 46 - 64 - 53 - 44 - 88. Length 35√2 (25 moves).
**(Ac)** Solution by Siep Korteling.
11 - 44 - 17 - 28 - 82 - 71 - 53 - 31 - 13 - 68 - 86 - 75 - 84 - 51 - 15 - 48 - 66 - 88 Moves 17, Length 43√2
(product = 731&radic2) as compared with 784√2 in (a) or 875√2 in (b).
**(Ad)** Solution by H. E. Dudeney.
11 - 55 - 82 - 71 - 17 - 28 - 46 - 13 - 31 - 86 - 68 - 57 - 48 - 15 - 51 - 84 - 66 - 88.
This solution to a bishop tour of the black cells of the 8 by 8 board in 17 moves (geometric length 45√2)
was given by H. E. Dudeney in *The Tribune* (Dec.3 1906), and in his *Amusements in Mathematics* (1917 problem 325),
and is reported in W.W.Rouse Ball, *Mathematical Recreations and Essays*, (1956 edition page 187).
However the above results show this route is not minimal in the number of moves (as seems to be implied by the context of Ball's work)
nor is it the geometrically shortest length. In his 1917 text Dudeney specifies that "You can visit any squares more than once,
but you are not allowed to move twice between the same two adjoining squares." This shows how carefully problems must sometimes
be phrased to eliminate alternative solutions. If it takes the bishop one second to pass through a cell and one second to make a
right-angled turn then, apart from the initial and final cells, he spends the same amount of time in every cell.

**SOLUTIONS (B)**

So that each cell on the 10 by 10 board is represented by two digits we write 10 = a.
**(Ba)** Solution by Siep Korteling.
11 - aa - 99 - a8 - 31 - 13 - 8a - 79 - a6 - 51 - 15 - 6a - 59 - a4 - 71 - 17 - 4a - 39 - a2 - 91 - 19 - 2a. Taking 21 moves (geometric length 81√2).
**(Bb)** Solution by George Jelliss.
11 - 33 - 15 - 26 - 17 - 28 - 19 - 2a - 39 - 4a - 59 - 6a - 79 - 8a - a8 - 97 - a6 - 95 - a4 - 93 - a2 - 91 - 82 - 71 - 62 - 51 - 42 - 31 - 13 - 46 - 37 - 48 - 57 - 68 - 86 - 75 - 84 - 73 - 64 - 53 - 44 - aa.
Length 53√2 (41 moves).
**(Bc)** Solution by Siep Korteling.
11 - 33 - 15 - 37 - 19 - 2a - 48 - 6a - a6 - 84 - a2 - 91 - 73 - 51 - 42 - 31 - 13 - 8a - a8 - 53 - 71 - a4 - 4a - 17 - 44 - aa.
Taking 25 moves, with geometric length 67√2. The product is 1675√2 as compared with 1701√2 for (a) and 2173√2 for (b).
**(Bd)** Solution by George Jelliss.
11 - 44 - 17 - 4a - 68 - 13 - 31 - 53 - 71 - a4 - 77 - 55 - 19 - 2a - a2 - 91 - 64 - a8 - 8a - 79 - 6a - 15 - 51 - a6 - 88 - aa.
Taking 25-moves (geometric length 73√2.

Comments: Here are some generalisations to boards of side 2h.

On a board of side 2h the bishop lines form h components, consisting of the diagonal path and h−1 rectangular paths,
comprising 4h−3 moves, of total geometric length (2h−1)^{2}√2.

It is possible to make a 'unicursal tour' of these lines (i.e. to draw them in a continuous pencil stroke) since there are only two odd nodes (the ends of the diagonal). If we try to minimise the number of straight moves of the pencil in this tracing each rectangular circuit takes at least 5 moves and to connect each of these to the diagonal requires h−1 breaks in the diagonal, making h diagonal moves, giving the minimum 6(h−1) + 1. This sequence runs: 1, 7, 13, 19, 25, 31, 37, ...

(Ca) If we have to visit all cells but not all branches we can reduce the number of moves for each circuit to 4 by making the breaks at their corners.
The procedure as for (Aa) and (Ba) then leads to a solution in (5h−4) moves (i.e. the 4h−3 plus h−1 switchbacks), with total geometric
length still (2h−1)^{2}√2, since each of the h−1 circuits has one √2 step removed.

(Cb) To minimise the geometric length. The problem as stated is soluble on boards with h = 4k or 4k + 1 (including the trivial 2 by 2 board), but to become solvable, in a uniquely defined path, on boards with h = 4k + 2 or 4k + 3 switchbacks must be allowed.

(Cc) This is solved by a method similar to (Cd) but making the deleted rectangles as large as possible.

(Cd) Looking at a diagram of the Dudeney tour (Ad) I noted that it is based on a line 11 - 88 from corner to corner and three rectangular circuits, from each of which one single-step bishop move has been deleted. These four deleted moves form a small circuit, namely 55 - 46 - 57 - 66 - 55. The same method doesn't quite work on the 10 by 10 board since there is one diagonal 11 - aa and four rectangular circuits, so that at least five deletions are needed. My 10 by 10 solution uses eight deletions forming two small circuits. We can make 4 by 4 and 6 by 6 solutions possible by allowing switchbacks. On a 32 by 32 board a solution is possible by deleting five small squares: {(3,3), (4,4), (5,3), (4,2)}, {(11,3), (12,4), (13,3), (12,2)}, {(19,3), (20,4), (21,3), (20,2)}, {(27,3), (28,4), (29,3), (28,2)}, {(15,9), (16,10), (17,9), (16,8)}. The first four join up the rectangles and diagonal in sets of four, and the last joins up the four resulting paths. (81 moves).

Any king tour is of course also a queen tour, restricted to single lateral or diagonal steps. These are studied in the separate King Tour section. The queen tours studied here all use longer moves.

According to my notes this was first presented by Sam Loyd in *Le Sphinx* March 1867, one month before
the *Illustrated London News* example. The *ILN* solution is the same as that given by Dudeney
(*Amusements in Mathematics* p.99), but rotated 180°. I give diagrams of three solutions:

These 'Queen Tours' are not strictly speaking 'Tours' in the same sense as Knight Tours or King Tours since they visit some squares twice (e.g. de56 in all the above examples). My solution has no K-type junctions which are both visited and passed through.

The simple 15-move non-intersecting open Queen Tour from c6 to f3 was used in a puzzle by H.E.Dudeney in
*The Tribune* 3 Oct 1906, and later problem 70 in *The Canterbury Puzzles*.

There are two diagonally magic queen tours on the 4×4 board. They are numbers 2 and 619 in the list
of 4×4 diagonally magic squares given by B. Frénicle in 1693 (as reproduced in *New Recreations
with Magic Squares* by W.H.Benson and O.Jacoby, 1976).

1604 05090416050915 03 10 06 03 15 10 06 02 14 07 11 14 02 07 110113 120813011208

Three non-diagonal magic queen tours with biaxial symmetry on the 4×4 board, using only two types of move,
are also possible (G.P.Jelliss, *Chessics*, #26, p.119, 1986). The frst two use wazir and alfil moves, the
third uses fers and three-leaper moves.

15 05 04 10 07 13 12 02 041609051606 030906160903 15 03 06 100111 140811010814 02 14 11 07 02 12 13 07 10 04 05 15 13010812

Here are some 8×8 magic queen tours. The first tour here by Joachim Brügge, *Die Schwalbe* 1985,
is said to be based on Durer's 4×4 square. The second is a slight modification of a non-diagonal tour given
by Brügge, it is based on spirals (G.P.Jelliss, *Chessics* #30, p.163, 1987).

36 15 23 58 37 10 18 631746 52 08 25 45 511614 35 59 22 11 38 62 19 47 53 09 26 07 24 44 50 24 60 34161761 39 09 54 28 27 34 63 06 05 43 57 21 13336420 12 40 10 29 35336442 04 23 25 53 45013252 44 08 55 36 30320103 61 42 56 26 04484931 05 41 11 37 38 31 02 59 60 22 46 03 27 54 43 06 30 51 18 12 56 39 58 41 21 15 02 47 55 28 07 32 50 294819 13 57 40 20 1449

These two tours (G.P.Jelliss, *Chessics* #26, p.120, 1986), show mainly diagonal and lateral moves
respectively.

2364254948083310 10 11 61 62 35 36 22 23 63 24 26 47 50 07 09 34 09 12 60 63 34 31 21 24 22 62 46 27 06 51 35 11 08 13 59643338 20 21 21 45 61 05 28 36 52 12 07 14 50494847 19 26 44 20 04 60 37 29 13 53 58 51 15161718 46 39 43 03 19 38 59 14 30 54 57 52 06013227 45 40 02 41 39 18 15 58 56 31 56 53 05 02 31 28 44 41 4201401617573255 55 54 04 03 30 29 43 42

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