This end of year issue consists of ten puzzle topics (most with several puzzle questions) that have been sent to me over the past two years, or have arisen from my own researches. The best set of solutions sent to me (preferably in text form without diagrams) before the end of January 2004 will earn the PRIZE: bound sets of volumes 1 and 2 of The Games and Puzzles Journal.

The 1- 2- 5- 10- ... cent values of a typical coinage display some curious arithmetical features. In particular, the sum of 10 cents cannot be realised in three coins, though it can be realised in any other number of coins from 1 to 10 inclusive. In the three little puzzles below, it is assumed that the coins available have cent values 1, 2, 5, 10, 20, 50, and 100. The first two problems have unique solutions, but the third has three solutions.

PROBLEM 56(A). Alicia has twice as much money as Bella; Bella has twice as much as Celia; Celia has twice as much as Delia. Each has two coins. Which coins does each of them hold?

PROBLEM 56(B). Alicia, Bella, Celia, Delia, and Ella each have the same amount of money, but Alicia has one coin, Bella has two, Celia has three, Delia has four, and Ella has five. Which coins does each of them hold?

PROBLEM 56(C). Alicia has three times as much money as Bella; Bella has three times as much as Celia. Each has three coins. Which coins does each of them hold?

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: (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.

PROBLEMS 57(A) and (B): Solve these cases for (A) the 8 by 8 board and (B) the 10 by 10 board. Thus there are eight cases: Aa, Ab, Ac, Ad, Ba, Bb, Bc, Bd. 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.

Juha reported (19 May 2003) that he "examined some tours with Popeye during 2 days", and found the following results which are offered here for solution. The maximummer condition means that at each move the bishop travels the maximum possible distance without passing through a cell already visited. If there is a choice of moves of equal length the bishop takes the one that best fulfills the other specified conditions. If two choices lead to equally good solutions there is said to be a dual.

On 8 by 8 chessboard:

PROBLEM 58(A): Longest maximummer bishop's tour.

PROBLEM 58(B): Longest dual-free maximummer bishop's tour.

PROBLEM 58(C): Minimal dual-free maximummer bishop's tour.

PROBLEM 58(D): Minimal maximummer bishop's tour.

Also in the e-mail (19 May 2003) Juha gave some results using the fairy chess pieces Mao and Moa. The Mao is the knight used in Chinese Chess (Xiangqi). It moves just like an ordinary knight but in two steps, a {0,1) wazir move followed by a {1,1} fers move, and the intermediate cell must be clear. In the case of a Mao tour this means the cell passed through must not have been visited previously. The Moa, invented by the late Dr Werner Speckmann, is similar but makes the fers move first and the wazir move second. Juha has found Mao and Moa tours of maximum length on the 5 by 5 and 6 by 6 boards.

PROBLEM 59(A): Maximal Mao tour in 5 by 5:

PROBLEM 59(B): Maximal Moa tour in 5 by 5:

PROBLEM 59(C): Maximal Mao tour in 6 by 6:

PROBLEM 59(D): Maximal Moa tour in 6 by 6:

Results on 7 by 7 and 8 by 8 would also be of interest.

PROBLEM 60(A) by John Beasley.

White Knight c1, Black King e8, White and Black play alternately. White, who is to make the first move, seeks to invade the palace (e8). Black's defence is to sow mines; specifically, at each turn he may lob a mine onto any unoccupied square outside the palace, and this square becomes permanently unavailable to the knight.

PROBLEM 60(B) by John Beasley.

On a large chessboard of 15 by 20 squares, White has a fiveleaper at h1, Black a palace occupied by a king at h20. Black and White play alternately. Black, to play, may sow a mine on any unoccupied square. White's aim is to invade the palace. A fiveleaper is a knight-like piece that makes moves of types {3,4} or {0,5}. It takes five moves to get from h1 to h20, for example h1 - h6 - l9 - i13 - e16 - h20.

PROBLEM 60(C) by George Jelliss.

A chessboard represents an Emperor's palace. The Emperor reclines in his throne room (which can be any room he chooses). Each room of the palace has access to six other rooms which are reached by rook moves of length 1, 2 or 4 steps. A 4-step can be made freely, but a 2-step must be within a 4 by 4 quarter of the board, and a 1-step must be within a 2 by 2 sixteenth of the board. A visiting Mandarin is taken to the Entrance room of the palace, which is the uniquely determined room that is farthest from the throne room (finding this is the first "move"). The sounding of a gong indicates that the Emperor has shut one of the six doors leading to the audience rooms, and that the Mandarin must make an approach move. How should the Mandarin move to ensure he enters one of the six audience rooms with access to the throne room before all the doors are shut against him?

These problems are extracted from a 40-page dissertation on the subject of Pythagorean and Rectangle-Square Dissections that Chris sent me in October 2002.

PROBLEM 61(A): By cuts parallel to the edges, no pieces being turned over, transform two squares of sides 3 and 4 into one square of side 5 using 4 pieces, one of which is a unit square. There are two distinct solutions.

Given a rectangle of sides 64 by 81 (= 8² by 9²) we can dissect it into two pieces that can be moved without rotation or turning over to form a square of side 72 (= 8×9) by 15 cuts parallel to the edges. The cuts must be in staircase formation as shown here, where the small component cells are of size 8×9. This is a general method that works for any rectangle q² by (q+1)² to form a square of side q(q+1) in two pieces using 2q – 1 cuts in staircase formation.

PROBLEM 61(B): By cuts parallel to the edges, no pieces being rotated or turned over, make the transformation of rectangle 64×81 to square 72×72 using 3 pieces formed with only 7 cuts.

The first two diagrams below show the best known dissections of two squares to form a third. They are due to Henry Perigal (1873) and Henry Dudeney (1917). They work for any sizes of the squares and use 5 pieces (except when the two squares are the same size in which case Perigal's method reduces to 4 pieces). Each dissection can be regarded as formed by superposing a tessellation of the large square over a combined tessellation of the two smaller squares. The corners of the large squares coincide in Perigal's case with corners of the smaller squares, and in Dudeney's case with the centres of the middle-sized squares.

If the corners of the larger squares are made to coincide with the centres of the smallest squares, as in the third diagram, the dissection resulting has quaternary symmetry like Dudeney's but uses the maximum of 9 pieces.

PROBLEM 61(C): Find positions of the large squares that (a) dissect both smaller squares symmetrically, (b) dissect both smaller squares similarly (in the geometrical sense).

PROBLEM 61(D). Given squares of sides 1 and Ö3 units, with sides parallel, dissect the two squares into five pieces that can be combined, without rotation or turning over, to form a square of side 2. (a) with two pieces the same (i.e. congruent), (b) with all pieces different.

PROBLEM 61(E). Given squares of sides 4 and 2 + 3Ö2 dissect them into five pieces that can be combined to form a square of side 6 + Ö2, including two equal triangles.

This problem is from an email sent in May 2003.

PROBLEM 62. A regular tetrahedron is a solid with four faces, each an equilateral triangle. Divide each face into three congruent segments as the diagram. Then use one of three different colours to colour each segment under the rule that segments that meet along a line or edge cannot be the same colour. The problem then is to find how many distinguishable tetrahedra can be made.

This was proposed by Nick at a meeting of the Outlanders (8 August 2002). He first saw it in a source, some ten years before, that he could not recall.

PROBLEM 63. We start with a cube, each side being 3 inches. Straight cylindrical holes of 1 inch diameter are bored from the centre of each side through to the centre of the opposite side. What is the volume of the material removed from the holes?

One way of calculating the volume would be to weigh the cube before and after drilling the holes, but that would be cheating. The shape of the central volume common to the three cylinders is also puzzling.

PROBLEMS 64(A) and (B) by Jean-Charles Meyrignac: Find longest non-intersecting knight paths on (A) the English 33-cell and (B) the French 37-cell Solitaire Boards.

Neither the English nor the French Solitaire Board admits a closed knight's tour because the number of cells is odd in each case. The French board clearly does not admit an open tour because, when chequered, there are 21 cells of one colour and only 16 of the other. In the case of the English Board the distribution is 17 to 16 so it might seem that an open tour should be possible, but in fact this is not the case.

PROBLEMS 64(C), (D) and (E) by George Jelliss: Construct 31-move open tours on the 33-cell board with the following properties: (C) the two end cells and the omitted cell are all adjacent, and all moves are within the border, (D) the centre cell is omitted and the tour goes outside the border once, (E) the centre cell is omitted and, when the cells visited by the knight are numbered in sequence, the numbers in the diametrally opposite pairs of cells differ by the constant value 4.

PROBLEM 64(F) by George Jelliss: The Snake Pit: On this 82-cell board pass from one side to the other, joining the two marked cells by a non-intersecting knight path of maximum length.

The best I have been able to achieve is 61 moves (62 cells), omitting 20 cells.

Simple cases of the problem of forming "plantations" of T trees in L lines with (density) D per line were investigated in the Puzzle Questions in Games and Puzzles Journal issues 13 - 16. Here we turn to the question of fitting such patterns onto a rectangular array as small as possible. Some sample cases are illustrated below. We list them first according to D, the number of trees in each line, and secondly according to T, the total number of trees planted. The number L of lines is not necessarily the greatest achievable.

6 in 4 lines of 3: board 4 by 5 (two cases shown).

7 in 6 lines of 3: board 5 by 5.

9 in 10 lines of 3: board 3 by 5 (Attributed by Dudeney to Isaac Newton no less).

10 in 5 lines of 4: board 7 by 7 (Dudeney, A in M 1917 problem 210).

16 in 12 lines of 4: board 7 by 10 (Dudeney, Canterbury Puzzles 1907, problem 21 gives a solution on a grid of triangles which I have distorted to fit a grid of squares. He also shows 16 in 15 lines of 4, based on a construction of concentric pentagons, which I have not been able to fit to a lattice of squares.)

22 in 21 lines of 4: board 7 by 7 (Dudeney, A in M 1917 problem 212).

19 in 9 lines of 5: board 13 by 13 (T.R.Dawson Chess Amateur 1923).

Here are three results for solving. The solutions I have require much larger boards than the examples above, the first two being 61 by 61, though it may be possible to do better.

PROBLEM 65(A) Plant 11 trees in 6 lines of 4 on a grid as small as possible.

PROBLEM 65(B) Plant 21 trees in 12 lines of 5 on a grid as small as possible.

PROBLEM 60(C) Plant 13 trees in 9 lines of 4 on a grid as small as possible.

John Beasley reports that he was amused to see his "3NT against any defence" bridge deal (G&PJ Volume 1, 1988, pp 52 and 68) quoted in "The Guardian" last autumn as a sensational new discovery by computer. Apparently one Thomas Andrews had recently examined all the possible rotationally symmetric deals by computer, and had highlighted this one on his web site as particularly interesting. John says that both Andrews and "The Guardian" unhesitatingly and generously acknowledged the prior work as soon as it was pointed out to them.

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