|The Games and Puzzles Journal Issue 22, January-April 2002|
This issue is devoted to items from Tom Marlow, some dating back 15 years, but not previously published. Work by Awani Kumar on 6×6 semi-magic knight tours is also summarised in the last section. Both authors sent updated results in May 2002 after this issue first appeared, so this version is now a second edition (something not normally possible with a printed paper journal). A full account of Mr Kumar's enumeration of semi-magic tours, and major results by Mr Marlow on magic fiveleaper tours appear on two new pages on the Knight's Tour Notes website [now part of this mayhematics site].
The following self-referential sentence appeared on Tom Marlow's Christmas / New Year card for (1987-88) and seems to be the first he produced:
"This sentence contains three A's, one B, three C's, two D's, thirty three E's, seven F's, one G, six H's, ten I's, one J, one K, one L, one M, twenty four N's, sixteen O's, one P, one Q, nine R's, twenty six S's, seventeen T's, five U's, five V's, four W's, four X's, four Y's and one Z."
A similar one beginning "This sentence created for GPJ contains four A's, one B, ..." was published in the Games and Puzzles Journal issue 5+6 p.92.
The following results were sent on 13 September 1989. They follow on from the related square problem which was discussed in the special issue of Chessics on "Chessboard Dissections". Note that in the diagrams, the triangles are not accurately equilateral, since they have been drawn by joining dots on a square grid. The component triangles have base and height the same. By looking at them from a suitable angle below, perspective should give an impression of equilaterals.
Chessics 28, page 149, dealt with the problem of dividing a square of side n into either two or four equal and congruent parts by cuts along the grid lines that divide it into n^2 cells. A count was given of the number of solutions, rotations and reflections not being considered as distinct. This included the case where n is odd by treating the central cell as a hole after which the remaining cells number a multiple of four. A similar process can be applied to an equilateral triangle by cutting along the isometric grid lines that divide it into triangular cells.
A triangle of order n has n^2 cells. If n is of the form 3x then n^2 is exactly divisible by three, but if n = 3x ± 1 then n^2 is of the form 3y + 1. Consequently it is necessary in these cases to treat a central cell as a hole as was done for odd ordered squares. If n = 3x + 1 then the hole is oriented in the same way as the outer triangle, but for n = 3x 1 it is inverted.
The diagrams for lower orders give an outline of how the count of solutions can be made and the known results are:
The results up to order nine were found by hand and confirmed by a computer count which went on to produce the remainder. So far as I know this dissection has not been considered before so an independent check is desirable. Even better would be a formula for a general solution. (The figures for 14 and 15 added 6 May 2002.)
Diagram of cases 2 to 5.
Diagram of the 20 trisections in case 6.
These results also date from 13 September 1989. As before the diagrams are based on component triangles that have base and height equal, so are not accurately equilateral.
The process of trisecting equilateral triangles can be extended to regular hexagons which can be considered as constructed from triangular cells. A hexagon of order n contains 6n^2 cells. hence it can be divided exactly into two, three or six parts of equal size, and because of its sixfold symmetry these parts can be congruent. Diagrams are given for the solutions of the lower orders and the known, computer compiled, results are as follows:
All counts that number no more than three digits have been checked by hand. It will be seen that in each case the count grows so rapidly that it soon outpaces the computer! (Figures in darker boxes added 6 May 2002.)
Hexagons of sides 1 and 2. The bisections, trisections and six-sections.
Hexagons, side 3, the 12 six-sections.
Hexagons, side 3, the 116 trisections. (a) 21 cases.
Hexagons, side 3, the 116 trisections. (b) 42 cases.
Hexagons, side 3, the 116 trisections. (c) 53 cases.
Hexagons, side 4, the 173 sixths. (a) 49 cases.
Hexagons, side 4, sixths (b) 27 cases.
Hexagons, side 4, sixths (c) 35 cases.
Hexagons, side 4, sixths (d) 35 cases.
Hexagons, side 4, sixths (e) 27 cases.
By T. W. Marlow 14 December 1994
You may be interested in some work I have been doing on packing rectangular boxes with like copies of a single solid pentacube.
The diagram below shows the only result so far (apart from multiples) for the shape (A) shown. Note rotational symmetry in the plane of the diagram and vertically.
Even if the mirror image shape is used as well only 6×5×3 results. On the other hand for piece (B) I have solutions for 5×2×2, 5×5×4, 5×5×5, 6×5×3 and several others. There is no obvious reason why this shape should be so much more helpful.
This was published in Variant Chess in 1992 (issue 7 page 91).
The broken diagonals a2-g8,h1 and b1-h7,a8 also sum to 260. Also all rectangles (a1,b4), (c1,d4) etc. and some horizontal rectangles. (a2,d3), (a6,d7) etc.
This property is retained if the tour is renumbered upwards starting at 9, 17, 33, 41 or 49.
The geometrical diagram is based on a repeated 4x4 pattern.
The following tables show the only two earlier examples published. (a) One of the five tours in quaternary symmetry is semi-magic when numbered from a suitable cell. This result was noted by Maurice Kraitchik in Le Problème du Cavalier 1927 p.36. This is #16 in Mr Kumar's list (inverted). Since it is a symmetric tour the numbers in diametrally opposite cells differ by the constant value 18 (i.e. 6×6/2). (b) An example in which the numbers in diametrally opposite cells differ by 6 (#19 in the list) was given in my note on Piece-wise Symmetric Tours in the Journal of Recreational Mathematics (problem vol.28 no.1, 1996-7, pp.63-64; solution vol.29 no.1, 1998 pp.69-71; diagram on p.70).
|(a) Kraitchik||(b) Jelliss|
The tables below show tours #1, #17 and #20 from Mr Kumar's list. Tour #17 is an alternative numbering of the same tour as used in my example shown above, but because of the choice of initial cell the differences between numbers in diametrally opposite cells are now not all 6, some are 30 (i.e. 36 6).
|Kumar #1||Kumar #17||Kumar #20|
The 88 semi-magic tours found by Mr Kumar occur in 44 pairs that are reverse-numberings of each other. Of these, 13 pairs are reentrant tours (i.e. 1 and 36 are a knight move apart) and 31 pairs are nonreentrant. The 13 pairs of reentrant tours derive from 4 geometrically distinct closed tours. The quaternary tour mentioned above gives two numerical forms, while three other asymmetric tours each give eight numerical forms.
The open tour pair #20 and #25 in Mr Kumar's list is the only one in which the diagonal sums are 111 plus or minus 1, i.e. as near as possible to the magic constant.
There are 50 of the 88 tours in which the sum of both diagonals is 222 (twice the magic constant). These consist of 24 reentrant and 36 nonreentrant. This group includes all the reentrant tours except the tour #1 and its reverse #53.
In his original e-mail Mr Kumar concluded: A careful study of 6*6 semi-magic tours will help us in discovering magic tours on other singly-even square boards. I wish to know about magic or semi-magic tours on singly-even boards, that is 10*10, 14*14 etc. Have they been found? Please suggest literature on it. What about magic tours of fairy chess pieces? Have they been found? As an answer to the last question see the new Knight's Tour Notes page containing the work by Tom Marlow on Fiveleaper tours.