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[PDF page 66] 50 SOME GEOMETRICAL QUESTIONS. [CH. II a regular hexagon and the three points of intersection of opposite sides form such a group, and therefore any projection of that figure will give a solution. Thus at present it is not possible to say what is the maximum number of rows of three which can be formed fromn counters placed on a plane. Extension to p-in-a-row. The problem mentioned above at once suggests the extension of placingn counters so as to form as many rows as possible, each of which shall containp and only p counters. Such problems can be often solved immediately by placing at infinity the points of intersection of some of the lines, and (if it is so desired) sub- sequentlyprojectingthediagramthusformedsoastobringthesepoints to a finite distance. One instance of such a solution is given above. As easy examples I may give the arrangement of16 counters in15 rows1, each containing4 counters; and the arrangement of19 counters in 10 rows, each containing 5 counters. A solution of the second of these problems can be obtained by placing counters at the19 points of intersection of the10 lines x = ±a, x = ±b, y = ±a, y = ±b, y = ±x: of these points two are at infinity. The first problem I leave to the ingenuity of my readers. Tesselation. Another of these statical recreations is known as tes- selation and consists in the formation of geometrical designs or mosaics by means of tesselated tiles. To those who have never looked into the matter it may be surpris- ing that patterns formed by the use of square tiles (of which one-half bounded by a diagonal is white and the other half black) should be subject to mathematical analysis. In view of the discussion of this subject by Montucla*, Lucas†, and other writers it would be hard to refuse to call the formation of such patterns a mathematical amuse- ment, but the treatment is (perhaps necessarily) somewhat empirical, and though there are some interesting puzzles of this kind, I do not propose to describe them here. * See Ozanam, 1803 edition, vol. i, p. 100; 1840 edition, p. 46. † Lucas, Récréations Mathématiques, Paris, 1882–3, vol.ii, part 4: hereafter I shall refer to this work by the name of the author. 1. ‘13’ corrected to ’15’ as per errata sheet

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[PDF page 67] CH. II] COLOUR-CUBE PROBLEM. 51 Sylvester* proposed a modified tesselation problem which consists in forming anallagmatic squares, that is, squares such that in every row and every column the number of changes of colour or the number of permanences is constant, the tiles used being square white tiles and square black tiles. If more than two colours are used, the problems become increas- ingly difficult. As a simple instance of this class of problems I may mention one, sent to me by a correspondent who termed itCross-Fours, wherein sixteen square counters are used, the upper half of each being yellow, red, pink, or blue, and the lower half being gold, green, black, or white, no two counters being coloured alike. Such counters can be arranged in the form of a square so that in each vertical, horizontal, and diagonal line there shall be 8 colours and no more: they can be also arranged so that in each of these ten lines there shall be 6 colours and no more, or 5 colours and no more, or 4 colours and no more. Puzzles of this kind are but little known; they are however not uninstructive. Colour-Cube Problem. As an example of a recreation analogous to tesselation I will mention the colour-cube problem; I select this partly because it is one of the most difficult of such puzzles, but chiefly because it has been subjected† to mathematical analysis. Stripped of mathematical technicalities the problem may be enun- ciated as follows. A cube has six faces, and if six colours are chosen we can paint each face with a different colour. By permuting the or- der of the colours we can obtain thirty such cubes, no two of which are coloured alike. Take any one of these cubes,K, then it is desired to select eight out of the remaining twenty-nine cubes, such that they can be arranged in the form of a cube (whose linear dimensions are double those of any of the separate cubes) coloured like the cubeK, and placed so that where any two cubes touch each other the faces in contact are coloured alike. Only one collection of eight cubes can be found to satisfy these conditions. To pick out these eight cubes empirically would be out of the question, but the mathematical analysis enables us to select them by the following rule. Take any face of the cubeK: it has four * Ex. gr. see the Educational Times Reprints, London, 1868, vol. x, pp. 74–76, 112: see also vol.xl v, p. 127; vol.l vi, pp. 97–99. † By Major MacMahon; an abstract of his paper, read before the London Mathe- matical Society on Feb. 9, 1893, was given inNature, Feb. 23, 1893, vol.xl vii, p. 406.

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[PDF page 68] 52 SOME GEOMETRICAL QUESTIONS. [CH. II angles, and at each angle three colours meet. By permuting the colours cyclically we can obtain from each angle two other cubes, and the eight cubes so obtained are those required. Forinstancesupposethatthesixcoloursareindicatedbytheletters a, b, c, d, e, f. Let the cubeK be put on a table, and to fix our ideas suppose that the face colouredf is at the bottom, the face coloureda is at the top, and the faces colouredb, c, d, and e front respectively the east, north, west, and south points of the compass. I may denote such an arrangement by (f; a; b, c, d, e). One cyclical permutation of the colours which meet at the north-east corner of the top face gives the cube (f; c; a, b, d, e), and a second cyclical permutation gives the cube (f; b; c, a, d, e). Similarly cyclical permutations of the colours which meet at the north-west corner of the top face ofK give the cubes (f; d; b, a, c, e) and (f; c; b, d, a, e). Similarly from the top south-west corner of K we get the cubes (f; e; b, c, a, d) and (f; d; b, c, e, a): and from the top south-east corner we get the cubes (f; e; a, c, d, b) and (f; b; e, c, d, a). The eight cubes being thus determined it is not difficult to arrange them in the form of a cube coloured similarly toK, and subject to the condition that faces in contact are coloured alike; in fact they can be arranged in two ways to satisfy these conditions. One such way, taking the cubes in the numerical order given above, is to put the cubes3, 6, 8, and 2 at the SE, NE, NW, and SW corners of the bottom face; of course each placed with the colourf at the bottom, while3 and 6 have the colour b to the east, and2 and 8 have the colourd to the west: the cubes 7, 1, 4, and5 will then form the SE, NE, NW, and SW corners of the top face; of course each placed with the coloura at the top, while7 and 1 have the colourb to the east, and5 and 4 have the colourd to the west. If howeverK is not given, then, without the aid of mathematical analysis, it is a difficult puzzle to arrange the eight cubes in the form of a cube coloured similarly to one of the other twenty-two cubes and subject to the condition that faces in contact are coloured alike. It is easy to make similar puzzles in two dimensions which are fairly difficult; it is somewhat surprising that none are to be bought, but I have never seen any except those that I have made myself. Dynamical Games of Position. Games which are played by moving pieces on boards of various shapes—such as merrilees, fox and geese, solitaire, backgammon, draughts, and chess—present more

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[PDF page 69] CH. II] SHUNTING PROBLEMS. 53 interest. In general, however, they permit of so many movements of the pieces that any mathematical analysis of them becomes too intricate to follow out completely. Probably this is obvious, but it may emphasize the impossibility of discussing such games effectively if I add that it has been shown that in a game of chess there may be as many as197299 waysofplayingthefirstfourmoves, andnearly 72000differentpositions at the end of the first four moves (two on each side), of which16556 arise when the players move pawns only*. Games in which the possible movements are very limited may be susceptible of mathematical treatment. One or two of these are given in the next chapter: here I shall confine myself mainly to puzzles and simple amusements. Shunting Problems. The first I will mention is a little puzzle which I bought some years ago and which was described as the “Great North- ern puzzle.” It is typical of a good many problems connected with the shunting of trains, and though it rests on a most improbable hypoth- esis, I give it as a specimen of its kind. A BC DE F P Q R The puzzle shows a railway, DEF, with two sidings, DBA and F CA, connected atA. The portion of the rails atA which is common to the two sidings is long enough to permit of a single wagon, likeP or Q, running in or out of it; but is too short to contain the whole of an engine, like R. Hence, if an engine runs up one siding, such asDBA, it must come back the same way. * L’Intermédiaire des mathématiciens, Paris, December, 1903, vol. x, pp. 305– 308: alsoRoyal Engineers Journal, London, August–November, 1889; orBritish Association Transactions, 1890, p. 745.

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[PDF page 70] 54 SOME GEOMETRICAL QUESTIONS. [CH. II Initially a small block of wood,P, coloured to represent a wagon, is placed at B; a similar block,Q, is placed atC; and a longer block of wood, R, representing an engine, is placed atE. The problem is to use the engineR to interchange the wagonsP and Q, without allowing any flying shunts. This is effected thus. (i)R pushes P into A. (ii) R returns, pushes Q up to P in A, couples Q to P, draws them both out toF, and then pushes them toE. (iii) P is now uncoupled,R takes Q back toA, and leaves it there. (iv) R returns to P, pulls P back to C, and leaves it there. (v) R running successively throughF, D, B comes to A, draws Q out, and leaves it at B. A somewhat similar puzzle, on sale in the streets in 1905, is made as follows. A loop-lineBGE connects two pointsB and E on a railway track AF, which is supposed blocked at both ends, as shown in the diagram. In the model, the trackAF is 9 inches long,AB = EF = 1 5 6 inches, and AH = F K = BC = DE = 1 4 inch. On the track and loop are eight wagons, numbered successively1 to 8, each one inch long and AB CD EF G HK one-quarter of an inch broad, and an engine of the same dimensions. Originally the wagons are on the track fromA to F and in the order 1, 2, 3, 4, 5, 6, 7, 8, and the engine is on the loop. The construction and the initial arrangement ensure that at any one time there cannot be more than eight vehicles on the track. Also if eight vehicles are on it only the penultimate vehicle at either end can be moved on to the loop, but if less than eight are on the track then the last two vehicles at either end can be moved on to the loop. If the points at each end of the loop-line are clear, it will hold four, but not more than four, vehicles. The object is to reverse the order of the wagons on the track, so that fromA to F they will be numbered successively8 to 1; and to do this by means which will involve as few transferences of the engine or a wagon to or from the loop as is possible. Other shunting problems are not uncommon, but these two exam- ples will suffice.

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[PDF page 71] CH. II] FERRY-BOAT PROBLEMS. 55 Ferry-Boat Problems. Everybody is familiar with the story of the showman who was travelling with a wolf, a goat, and a basket of cab- bages; and for obvious reasons was unable to leave the wolf alone with the goat, or the goat alone with the cabbages. The only means of transporting them across a river was a boat so small that he could take in it only one of them at a time. The problem is to show how the passage could be effected*. A similar problem, given by Alcuin, Tartaglia, and others, is as follows†. Three beautiful ladies have for husbands three men, who are as jealous as they are young, handsome, and gallant. The party are travelling, and find on the bank of a river, over which they have to pass, a small boat which can hold no more than two persons. How can they pass, it being agreed that, in order to avoid scandal, no woman shall be left in the society of a man unless her husband is present? The method of transportation to be used in the above cases is obvious, and can be illustrated practically by using six court cards out of a pack. Another problem similar to the one last mentioned is the case of n married couples who have to cross a river by means of a boat which can be rowed by one person and will carryn − 1 people, but not more, with the condition that no woman is to be in the society of a man unless her husband is present. Alcuin’s problem is the case ofn = 3. Let y denote the number of passages from one bank to the other which will be necessary. Then it has been shown that ifn = 3, y = 11; if n = 4, y = 9; and ifn > 4, y = 7; the demonstration presents no difficulty. The following analogous problem is due to the late Prof. Lucas‡. To find the smallest numberx of persons that a boat must be able to carry in order that n married couples may by its aid cross a river in such a manner that no woman shall remain in the company of any man unless her husband is present; it being assumed that the boat can be rowed by one person only. Also to find the least number of passages, say y, from one bank to the other which will be required. M. Delannoy has shown that ifn = 2, then x = 2, and y = 5. If n = 3, then x = 2, and y = 11. If n = 4, then x = 3, and y = 9. If n = 5, then x = 3, and y = 11. And finally if n > 5, then x = 4, and y = 2n − 1. M. De Fonteney has remarked that, if there was an island in the * Ozanam, 1803 edition, vol. i, p. 171; 1840 edition, p. 77. † Bachet, Appendix, problemiv, p. 212. ‡ Lucas, vol. i, pp. 15–18, 237–238.

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[PDF page 72] 56 SOME GEOMETRICAL QUESTIONS. [CH. II middle of the river, the passage might be always effected by the aid of a boat which could carry only two persons. If there are only two or only three couples the island is unnecessary, and the case is covered by the preceding method. Ifn > 3 then the least number of passages from land to land which will be required is8(n − 1). His solution is as follows. The first nine passages will be the same, no matter how many couples there may be: the result is to transfer one couple to the island and one couple to the second bank. The result of the next eight passages is to transfer one couple from the first bank to the second bank: this series of eight operations must be repeated as often as necessary until there is left only one couple on the first bank, only one couple on the island, and all the rest on the second bank. The result of the last seven passages is to transfer all the couples to the second bank. The solution for the case when there are four couples may be rep- resented as follows. LetA and a, B and b, C and c, D and d, be the four couples. The letters in the successive lines indicate the positions of the men and their respective wives after different passages of the boat. First Bank Island Second Bank Initially ABCD abcd . . . . .... . . . . .... After 1st passage ABCD .. cd . . . . ab.. . . . . .... " 2nd " ABCD . bcd . . . . a... . . . . .... " 3rd " ABCD ... d . . . . abc. . . . . .... " 4th " ABCD .. cd . . . . ab.. . . . . .... " 5th " . . CD .. cd AB . . ab.. . . . . .... " 6th " . . CD .. cd AB . . .... . . . . ab.. " 7th " . . CD .. cd AB . . . b.. . . . . a... " 8th " . . CD .. cd . . . . . b.. AB . . a... " 9th " . . CD .. cd . B . . . b.. A . . . a... " 10th " . BCD .. cd . . . . . b.. A . . . a... " 11th " . BCD .... . . . . . bcd A . . . a... " 12th " . BCD ... d . . . . . bc. A . . . a... " 13th " . . . D ... d . BC . . bc. A . . . a... " 14th " . . . D ... d . . . . . bc. ABC . a... " 15th " . . . D ... d . . . . abc. ABC . .... " 16th " . . . D ... d . . . . . b.. ABC . a.c. " 17th " . . . D ... d . B . . . b.. A . C . a.c. " 18th " . B . D ... d . . . . . b.. A . C . a.c.

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[PDF page 73] CH. II] GEODESICS.. 57 First Bank Island Second Bank After19th passage . . . . ... d . B . D . b.. A . C . a.c. " 20th " . . . . ... d . . . . . b.. ABCD a .c. " 21st " . . . . ... d . . . . . bc. ABCD a ... " 22nd " . . . . ... d . . . . .... ABCD abc . " 23rd " . . . . .. cd . . . . .... ABCD ab .. " 24th " . . . . .... . . . . .... ABCD abcd Prof. G. Tarry has suggested an extension of the problem, which still further complicates its solution. He supposes that each husband travels with a harem of m wives or concubines; moreover, as Mo- hammedan women are brought up in seclusion, it is reasonable to sup- pose that they would be unable to row a boat by themselves without the aid of a man. But perhaps the difficulties attendant on the trav- els of one wife may be deemed sufficient for Christians, and I content myself with merely mentioning the increased anxieties experienced by Mohammedans in similar circumstances. Geodesics. Geometrical problems connected with finding the shortest routes from one point to another on a curved surface are often difficult, but geodesics on a flat surface or flat surfaces are in general readily determinable. I append an instance*, but I should have hesitated to do so had not experience shown that some readers do not readily see the solution. It is as follows: A room is30 feet long,12 feet wide, and12 feet high. On the middle line of one of the smaller side walls and one foot from the ceiling is a wasp. On the middle line of the opposite wall and11 feet from the ceiling is a fly. The wasp catches the fly by crawling all the way to it: the fly, paralysed by fear, remaining still. The problem is to find the shortest route that the wasp can follow. To obtain a solution we observe that we can cut a sheet of paper so that, when folded properly, it will make a model to scale of the room. This can be done in several ways. If, when the paper is again spread out flat, we can join the points representing the wasp and the fly by a straight line lying wholly on the paper we shall obtain a geodesic route between them. Thus the problem is reduced to finding the way of cutting out the paper which gives the shortest route of the kind. * I heard a similar question propounded at Cambridge in 1903, but the only place where I have seen it in print is theDaily Mail, London, February 1, 1905.

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[PDF page 74] 58 SOME GEOMETRICAL QUESTIONS. [CH. II A B C D W F Here is the diagram corresponding to a solution of the above ques- tion, where A represents the floor, B and D the longer side-walls, C the ceiling, and W and F the positions on the two smaller side-walls occupied initially by the wasp and fly. In the diagram the square of the distance betweenW and F is (32)2+(24)2; hence the distance is40 feet. Problems with Counters placed in a row. Numerous dynamical problems and puzzles may be illustrated with a box of counters, espe- cially if there are counters of two colours. Of course coins or pawns or cards will serve equally well. I proceed to enumerate a few of these played with counters placed in a row. First Problem with Counters. The following problem must be fa- miliar to many of my readers. Ten counters (or coins) are placed in a row. Any counter may be moved over two of those adjacent to it on the counter next beyond them. It is required to move the coun- ters according to the above rule so that they shall be arranged in five equidistant couples. If we denote the counters in their initial positions by the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, we proceed as follows. Put7 on 10, then5 on 2, then 3 on 8, then 1 on 4, and lastly9 on 6. Thus they are arranged in pairs on the places originally occupied by the counters2, 4, 6, 8, 10. Similarly by putting4 on 1, then 6 on 9, then 8 on 3, then 10 on

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[PDF page 75] CH. II] PROBLEMS WITH COUNTERS.. 59 7, and lastly2 on 5, they are arranged in pairs on the places originally occupied by the counters 1, 3, 5, 7, 9. If two superposed counters are reckoned as only one, solutions anal- ogous to those given above will be obtained by putting7 on 10, then5 on 2, then 3 on 8, then 1 on 6, and lastly9 on 4; or by putting4 on 1, then 6 on 9, then 8 on 3, then 10 on 5, and lastly2 on 7*. There is a somewhat similar game played with eight counters, but in this case the four couples finally formed are not equidistant. Here the transformation will be effected if we move5 on 2, then3 on 7, then 4 on 1, and lastly6 on 8. This form of the game is applicable equally to (8 + 2n) counters, for if we move4 on 1 we have left on one side of this couple a row of(8 + 2n − 2) counters. This again can be reduced to one of(8 + 2n − 4) counters, and in this way finally we have left8 counters which can be moved in the way explained above. A more complete generalization would be the case ofn counters, where each counter might be moved over them counters adjacent to it on to the one beyond them. Second Problem with Counters. Another problem of a somewhat similar kind is due to Tait†. Place four florins (or white counters) and four halfpence (or black counters) alternately in a line in contact with one another. It is required in four moves, each of a pair of two contiguous pieces, without altering the relative position of the pair, to form a continuous line of four halfpence followed by four florins. His solution is as follows. Let a florin be denoted by a and a halfpenny byb, and let×× denote two contiguous blank spaces. Then the successive positions of the pieces may be represented thus: Initially ................. × × a b a b a b a b . After the first move ..... b a a b a b a × × b . After the second move .. b a a b × × a a b b . After the third move .... b × × b a a a a b b . After the fourth move ... b b b b a a a a × × . The operation is conducted according to the following rule. Sup- pose the pieces to be arranged originally in circular order, with two contiguous blank spaces, then we always move to the blank space for * Note by J. Fitzpatrick to a French translation of the third edition of this work, Paris, 1898. † Philosophical Magazine, London, January, 1884, series 5, vol.xvii, p. 39.

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[PDF page 76] 60 SOME GEOMETRICAL QUESTIONS. [CH. II the time being that pair of coins which occupies the places next but one and next but two to the blank space on one assigned side of it. A similar problem with2n counters—n of them being white and n black—will at once suggest itself, and, ifn is greater than4, it can be solved inn moves. I have however failed to find a simple rule which covers all cases alike, but solutions, due to M. Delannoy, have been given* for the four cases wheren is of the form4m, 4m + 2, 4m + 1, or 4m+3; in the first two cases the first1 2 n moves are of pairs of dissimilar counters and the last1 2 n moves are of pairs of similar counters; in the last two cases, the first move is similar to that given above, namely, of the penultimate and antepenultimate counters to the beginning of the row, the next 1 2(n − 1) moves are of pairs of dissimilar counters, and the final 1 2(n − 1) moves are of similar counters. The problem is also capable of solution if we substitute the restric- tion that at each move the pair of counters taken up must be moved to one of the two ends of the row instead of the condition that the final arrangement is to be continuous. Tait suggested a variation of the problem by making it a condition that the two coins to be moved shall also be made to interchange places; in this form it would seem that5 moves are required; or, in the general case, n + 1 moves are required. Problems on a Chess-board with Counters or Pawns.The follow- ing three problems require the use of a chess-board as well as of counters or pieces of two colours. It is more convenient to move a pawn than a counter, and if therefore I describe them as played with pawns it is only as a matter of convenience and not that they have any connection with chess. The first is characterized by the fact that in every position not more than two moves are possible; in the second and third problems not more than four moves are possible in any position. With these lim- itations, analysis is possible. I shall not discuss the similar problems in which more moves are possible. First Problem with Pawns†. On a row of seven squares on a chess- board 3 white pawns (or counters), denoted in the diagram by “a”s, are placed on the3 squares at one end, and3 black pawns (or counters), denoted by “b”s, are placed on the3 squares at the other end—the middle square being left vacant. Each piece can move only in one * La Nature, June, 1887, p. 10. † Lucas, vol. ii, part 5, pp. 141-143.
