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].

é (16) A Self-Referential Sentence.

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.

é (17) Trisected Triangles.

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:

order:	1	2	3	4	5	6	7	8	9
solutions:	0	1	1	2	5	20	56	276	2,136
order:	10	11	12	13		14		15
solutions:	13,756	148,352	2,727,448	41,044,816		1,056,334,024		46,033,137,324

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.

é (18) Dissected Hexagons.

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:

order:	1	2	3	4	5	6	7	8
sixths:	1	2	12	173	5,429	392,544	66,961,869	27,094,069,322
trisections:	1	5	116	15,785	11,599,297	47,212,453,928
halves:	1	8	731	982,648	16,305,532,683

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.

é (19) Polycubes.

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.

é (20) Magic and Semi-magic Knight Tours.

(a) Diagonally magic square on 8x8 torus

This was published in Variant Chess in 1992 (issue 7 page 91).

06	59	44	25	12	15	38	61
43	28	05	58	37	64	11	14
54	45	24	03	18	51	36	29
21	04	55	46	33	30	19	52
48	53	02	23	50	17	32	35
01	22	47	56	31	34	49	20
60	07	26	41	16	09	62	39
27	42	57	08	63	40	13	10

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.

(b) Diagonally magic knight tour on the vertical cylinder

This property is retained if the tour is renumbered upwards starting at 9, 17, 33, 41 or 49.

10	27	34	49	32	07	54	47
55	48	29	06	35	52	09	26
28	11	50	33	08	31	46	53
45	56	05	30	51	36	25	12
04	21	44	19	62	13	60	37
57	38	63	16	41	18	03	24
22	01	20	43	14	61	40	59
39	58	15	64	17	42	23	02

The geometrical diagram is based on a repeated 4x4 pattern.

(c) Semi-magic knight tours on the 6×6 board

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
29	18	15	22	07	20		08	31	34	15	06	17
16	23	30	19	14	09		33	14	07	18	35	04
31	28	17	08	21	06		30	09	32	05	16	19
24	03	26	35	10	13		13	22	11	26	03	36
27	32	01	12	05	34		10	29	24	01	20	27
02	25	04	33	36	11		23	12	21	28	25	02

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
01	10	35	28	25	12		02	25	28	09	36	11		02	27	18	31	04	29
34	29	02	11	08	27		27	06	01	12	29	34		17	12	03	28	19	32
03	36	09	26	13	24		24	03	26	35	10	13		26	01	36	13	30	05
30	33	20	05	16	07		07	16	05	20	33	30		11	16	09	22	33	20
21	04	31	18	23	14		04	23	18	31	14	21		08	25	14	35	06	23
32	19	22	15	06	17		17	06	15	22	19	32		15	10	07	24	21	34

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.