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[PDF page 112] 96 MISCELLANEOUS MATHEMATICAL RECREATIONS. [CH. IV e e e e e e e 1101000 Figure i. e e e e e e e 1101001 Figure ii. e e e e e e e 1100111 Figure iii. every ring on the bar gives a variation (that is,1 to 0 or 0 to 1) and every ring off the bar gives a continuation, the effect of a step by which a ring is taken off or put on the bar is either to subtract unity from this number or to add unity to it. For example, the number denoting the position of the rings in figure ii is obtained from the number denoting that in figure i by adding unity to it. Similarly the number denoting the position of the rings in figure iii is obtained from the number denoting that in figure i by subtracting unity from it. The position when all the seven rings are off the bar is denoted by the number 0000000: when all of them are on the bar by the number 1010101. Hence to change from one position to the other requires a number of steps equal to the difference between these two numbers in the binary scale. The first of these numbers is0: the second is equal to 26 + 24 + 22 + 1, that is, to 85. Therefore 85 steps are required. In a similar way we may show that to put on a set of2n + 1 rings requires (1 + 21 + 22 + . . . + 22n) steps, that is, 1 3(22n+2 − 1) steps; and to put on a set of2n rings requires (2 + 23 + . . . + 22n−1) steps, that is, 1 3(2n+1 − 2) steps. I append a table indicating the steps necessary to take off the first four rings from a set of five rings. The diagrams in the middle column show the successive position of the rings after each step. The number following each diagram indicates that position, each number being ob- tained from the one above it by the addition of unity. The steps which are bracketed together can be made in one movement, and, if thus ef- fected, the whole process is completed in7 movements instead of 10 steps: this is in accordance with the formula given above. M. Gros asserted that it is possible to take from64 to 80 steps a minute, which in my experience is a rather high estimate. If we accept the lower of these numbers, it would be possible to take off10 rings in less than 8 minutes; to take off25 rings would require more than 582 days, each of ten hours work; and to take off60 rings would

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[PDF page 113] CH. IV] THE EIGHT QUEENS PROBLEM. 97 Initial position d d d d d 10101 After 1st step d d d d d 10110 }" 2nd " d d d d d 10111 " 3rd " d d d d d 11000 " 4th " d d d d d 11001 }" 5th " d d d d d 11010 " 6th " d d d d d 11011 " 7th " d d d d d 11100 " 8th " d d d d d 11101 " 9th " d d d d d 11110 }" 10th " d d d d d 11111 necessitatenolessthan 768614, 336404, 564650steps, andwouldrequire nearly 55000, 000000 years work—assuming of course that no mistakes were made. The Eight Queens Problem *. The determination of the number of ways in which eight queens can be placed on a chess- board—or more generally, in which n queens can be placed on a board of n2 cells—so that no queen can take any other was proposed originally by Nauck in 1850. In 1874 Dr S. Günther† suggested a method of solution by means of determinants. For, if each symbol represents the corresponding cell of the board, the possible solutions for a board ofn2 cells are given by * On the history of this problem see W. Ahrens, Mathematische Unterhaltun- gen und Spiele, Leipzig, 1901, chap.ix—a work issued subsequent to the third edition of this book. † Grunert’sArchiv der Mathematik und Physik, 1874, vol.l vipp. 281–292.

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[PDF page 114] 98 MISCELLANEOUS MATHEMATICAL RECREATIONS. [CH. IV those terms, if any, of the determinant a1 b2 c3 d4 . . . . . . . . . . . . β2 a3 b4 c5 . . . . . . . . . . . . γ3 β4 a5 b6 . . . . . . . . . . . . δ4 γ5 β6 a7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a 2n−3 b2n−2 . . . . . . . . . . . . . . β 2n−2 a2n−1 in which no letter and no suffix appears more than once. The reason is obvious. Every term in a determinant contains one and only one element out of every row and out of every column: hence any term will indicate a position on the board in which the queens cannot take one another by moves rook-wise. Again in the above de- terminant the letters and suffixes are so arranged that all the same letters and all the same suffixes lie along bishop’s paths: hence, if we retain only those terms in each of which all the letters and all the suf- fixes are different, they will denote positions in which the queens cannot take one another by moves bishop-wise. It is clear that the signs of the terms are immaterial. In the case of an ordinary chess-board the determinant is of the8th order, and therefore contains8!, that is,40320 terms, so that it would be out of the question to use this method for the usual chess-board of 64 cells or for a board of larger size unless some way of picking out the required terms could be discovered. A way of effecting this was suggested by Dr J.W.L. Glaisher* in 1874, and as far as I am aware the theory remains as he left it. He showed that if all the solutions ofn queens on a board ofn2 cells were known, then all the solutions of a certain type forn + 1 queens on a board of(n+1) 2 cells could be deduced, and that all the other solutions of n + 1 queens on a board of(n + 1)2 cells could be obtained without difficulty. The method will be sufficiently illustrated by one instance of its application. It is easily seen that there are no solutions whenn = 2 and n = 3. If n = 4 there are two terms in the determinant which give solutions, namely, b2c5γ3β6 and c3β2b6γ5. To find the solutions when n = 5 , * Philosophical Magazine, London, December, 1874, series 4, vol.xl viii, pp. 457– 467.

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[PDF page 115] CH. IV] THE EIGHT QUEENS PROBLEM. 99 Glaisher proceeded thus. In this case Günther’s determinant is a1 b2 c3 d4 e5 β2 a3 b4 c5 d6 γ3 β4 a5 b6 c7 δ4 γ5 β6 a7 b8 ε5 δ6 γ7 β8 a9 To obtain those solutions (if any) which involvea9 it is sufficient to ap- pend a9 to such of the solutions for a board of16 cells as do not involve a. As neither of those given above involves ana we thus get two solu- tions, namely, b2c5γ3β6a9 and c3β2b6γ5a9. The solutions which involve a1, e5 and ε5 can be written down by symmetry. The eight solutions thus obtained are all distinct; we may call them of the first type. The above are the only solutions which can involve elements in the corner squares of the determinant. Hence the remaining solutions are obtainable from the determinant 0 b2 c3 d4 0 β2 a3 b4 c5 d6 γ3 β4 a5 b6 c7 δ4 γ5 β6 a7 b8 0 δ6 γ7 β8 0 If, in this, we take the minor ofb2 and in it replace by zero every term involving the letterb or the suffix2 we shall get all solutions involving b2. But in this case the minor at once reduces tod6a5δ4β8. We thus get one solution, namely,b2d6a5δ4β8. The solutions which involveβ2, δ4, δ6, β8, b8, d6, and d4 can be obtained by symmetry. Of these eight solutions it is easily seen that only two are distinct: these may be called solutions of the second type. Similarly the remaining solutions must be obtained from the de- terminant 0 0 c3 0 0 0 a3 b4 c5 0 γ3 β4 a5 b6 c7 0 γ5 β6 a7 0 0 0 γ7 0 0 If, in this, we take the minor ofc3, and in it replace by zero every term involving the letterc or the suffix3, we shall get all the solutions

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[PDF page 116] 100 MISCELLANEOUS MATHEMATICAL RECREATIONS. [CH. IV which involve c3. But in this case the minor vanishes. Hence there is no solution involvingc3, and therefore by symmetry no solutions which involve γ3, γ7, or c3. Had there been any solutions involving the third element in the first or last row or column of the determinant we should have described them as of the third type. Thus in all there are ten and only ten solutions, namely, eight of the first type, two of the second type, and none of the third type. Similarly, if n = 6, we obtain no solutions of the first type, four solutions of the second type, and no solutions of the third type; that is, four solutions in all. If n = 7, we obtain sixteen solutions of the first type, twenty-four solutions of the second type, no solutions of the third type, and no solutions of the fourth type; that is, forty solutions in all. If n = 8, we obtain sixteen solutions of the first type, fifty-six solutions of the second type, and twenty solutions of the third type, that is, ninety-two solutions in all. It will be noticed that all the solutions of one type are not always distinct. In general, from any solution seven others can be obtained at once. Of these eight solutions, four consist of the initial or fundamen- tal solution and the three similar ones obtained by turning the board through one, two, or three right angles; the other four are the reflex- ions of these in a mirror: but in any particular case it may happen that the reflexions reproduce the originals, or that a rotation through one or two right angles makes no difference. Thus on boards of42, 52, 62, 72, 82, 92, 102 cells there are respectively1, 2, 1, 6, 12, 46, 92 fundamental solutions; while altogether there are respectively2, 10, 4, 40, 92, 352, 724 solutions. The following collection of fundamental solutions may interest the reader. The positions on the board of the queens are indicated by digits: the first digit represents the number of the cell occupied by the queen in the first column reckoned from one end of the column, the second digit the number in the second column, and so on. Thus on a board of 42 cells the solution3142 means that one queen is on the3rd square of the first column, one on the1st square of the second column, one on the4th square of the third column, and one on the2nd square of the fourth column. If a fundamental solution gives rise to only four solutions the number which indicates it is placed in curved brackets, ( ); if it gives rise to only two solutions the number which indicates it is placed in square brackets, [ ]; the other fundamental solutions give rise

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[PDF page 117] CH. IV] THE EIGHT QUEENS PROBLEM. 101 to eight solutions each. On a board of 42 cells there is 1 fundamental solution: namely, [3142]. On a board of52 cells there are2 fundamental solutions: namely, 14253, [25314]. On a board of 62 cells there is 1 fundamental solution: namely, (246135). On a board of72 cells there are6 fundamental solutions: namely, 1357246, 3572461, (5724613), 4613572, 3162574, (2574136). On a board of82 cells there are12 fundamental solutions: namely, 25713864, 57138642, 71386425, 82417536, 68241753, 36824175, 64713528, 36814752, 36815724, 72418536, 26831475, ( 64718253). The arrangement in this order is due to Mr Oram. It will be noticed that the 10th, 11th, and 12th solutions somewhat resemble the 4th, 6th, and7th respectively. The6th solution is the only one in which no three queens are in a straight line. On a board of92 cells there are 46 fundamental solutions; one of them is 248396157. On a board of102 cells there are92 fundamental solutions; these were given by Dr A. Pein*; one of them is2468t13579, where t stands for ten. On a board of112 cells there are 341 funda- mental solutions; these have been given by Dr T.B. Sprague†: one of them is 15926t37e48. I may add that for a board ofn2 cells there is always a symmetrical solution of the form246 . . . n135 . . .(n −1), when n = 6 m or n = 6 m + 4, Also Mr Oram has shown that for a board of n2 cells, when n is a prime, cyclical arrangements of then natural numbers, other than in their natural order, will give solutions; see, for instance, the solution quoted above. The puzzle in the form of a board of 36 squares is sold in the streets of London for a penny, a small wooden board being ruled in the manner shown in the diagram and having holes drilled in it at the points marked by dots. The object is to put six pins into the holes so that no two are connected by a straight line. * Aufstellung von n Königinnen auf einem Schachbrett vonn2 Feldern, Leipzig, 1889. † Proceedings of the Edinburgh Mathematical Society, vol.xvii, 1898–9, pp. 43–68.

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[PDF page 118] 102 MISCELLANEOUS MATHEMATICAL RECREATIONS. [CH. IV Other Problems with Queens. Captain Turton called my atten- tion to two other problems of a somewhat analogous character, neither of which, as far as I know, has been published elsewhere, or solved otherwise than empirically. The first of these is to place eight queens on a chess-board so as to command the fewest possible squares. Thus, if queens are placed on cells 1 and 2 of the second column, on cell 2 of the sixth column, on cells 1, 3, and 7 of the seventh column, and on cells2 and 7 of the eighth column, eleven cells on the board will not be in check; the same number can be obtained by other arrangements. Is it possible to place the eight queens so as to leave more than eleven cells out of check? I have never succeeded in doing so, nor in showing that it is impossible to do it. The other problem is to place m queens (m being less than 5) on a chess-board so as to command as many cells as possible. For instance, four queens can be placed in several ways on the board so as to command58 cells besides those on which the queens stand, thus leaving only2 cells which are not commanded:ex. gr.this is effected if the queens are placed on cell5 of the third column, cell1 of the fourth column, cell 6 of the seventh column, and cell2 of the eighth column; or on cell1 of the first column, cell7 of the third column, cell3 of the fifth column, and cell5 of the seventh column. A similar problem is to determine the minimum number and the position of queens which can be placed on a board ofn2 cells so as to occupy or command every cell. It would seem that, even with the additional restriction that no queen shall be able to take any other queen, there are no less than ninety-one typical solutions in which five queens can be placed on a chess-board

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[PDF page 119] CH. IV] THE FIFTEEN SCHOOL-GIRLS PROBLEM. 103 so as to command every cell*. Extension to other Chess Pieces.Analogousproblemsmaybepro- posed with other chess-pieces. For instance, questions as to the max- imum number of knights which can be placed on a board ofn2 cells so that no knight can take any other, and the minimum number of knights which can be placed on it so as to occupy or command every cell have been propounded†. Similar problems have also been proposed fork kings placed on a chess-board ofn2 cells‡. It has been asserted that, ifk = 2, the number of ways in which two kings can be placed on a board so that they may not occupy adjacent squares is1 2(n − 1)(n − 2)(n2 + 3n − 2). Similarly, if k = 3, the number of ways in which three kings can be placed on a board so that no two of them occupy adjacent squares is said to be 1 6(n − 1)(n − 2)(n4 + 3n3 − 20n2 − 30n + 132). The Fifteen School-Girls Problem. This problem— which was first enunciated by Mr T.P. Kirkman, and is sometimes known as Kirkman’s problem§—consists in arranging fifteen things in different sets of triplets. It is usually presented in the form that a school-mistress was in the habit of taking her girls for a daily walk. * L’Intermédiaire des mathématiciens, Paris, 1901, vol.viii, p. 88. † Ibid., March, 1896, vol. iii, p. 58; 1897, vol. iv, p. 15, 254; and 1898, vol.v, p. 87. ‡ Ibid., June, 1901, p. 140. § It was published first in the Lady’s and Gentleman’s Diaryfor 1850, p. 48, and has been the subject of numerous memoirs. Among these I may single out the papers by A. Cayley in thePhilosophical Magazine, July, 1850, series 3, vol. xxxvii, pp. 50–53; by T.P. Kirkman in theCambridge and Dublin Math- ematical Journal, 1850, vol. v, p. 260; by R.R. Anstice,Ibid., 1852, vol. vii, pp. 279–292; by B. Pierce,Gould’s Journal, Cambridge, U.S., 1860, vol. vi, pp. 169–174; by T.P. Kirkman,Philosophical Magazine, March, 1862, series 4, vol. xxiii, pp. 198–204; by W.S.B. Woolhouse in theLady’s Diaryfor 1862, pp. 84–88, and for 1863, pp. 79–90, and in theEducational Times Reprints, 1867, vol. viii, pp. 76–83; by J. Power in theQuarterly Journal of Mathemat- ics, 1867, vol.viii, pp. 236–251; by A.H. Frost,Ibid., 1871, vol.xi, pp. 26–37; by E. Carpmael in theProceedings of the London Mathematical Society, 1881, vol. xii, pp. 148–156; by Lucas in hisRécréations, vol.ii, part vi; by A.C. Dixon in theMessenger of Mathematics, Cambridge, October, 1893, vol.xxiii, pp. 88– 89; and by W. Burnside,Ibid., 1894, vol.xxiii, pp. 137–143. It has also, since the issue of my third edition, been discussed by W. Ahrens in hisMathematische Unterhaltungen und Spiele, Leipzig, 1901, chapter xiv.

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[PDF page 120] 104 MISCELLANEOUS MATHEMATICAL RECREATIONS. [CH. IV The girls were fifteen in number, and were arranged in five rows of three each so that each girl might have two companions. The problem is to dispose them so that for seven consecutive days no girl will walk with any of her school-fellows more than once. More generally we may require to arrange3m girls in triplets to walk out for1 2(3m − 1) days, so that no girl will walk with any of her school-fellows more than once. The theory of the formation of all such possible triplets in the case of fifteen girls is not difficult, but the extension to3m girls is, as yet, unsolved. I proceed to describe three methods of solution: these methods are analytical, but I may add that the problem can be also attacked by geometrical methods. Frost’s Method.The first of these solutions is due to Mr Frost. A full exposition of it would occupy a good deal of space, but I hope that the following sketch will make the process intelligible. Denote one of the girls by k. Her companions on each day are different: suppose that on Sunday they area1 and a2, on Monday b1 and b2, and so on, and finally on Saturdayg1 and g2. Hence for each day we have one triplet, and we have to find four others, but in each of the latter no two like letters can occur together, that is, the three letters in any of them must be all different. Let a stand for a1, or a2, b for b1 or b2, and so on. The suffixes1 and 2 are called complementary. Then, since the three letters in each of the triplets we are trying to find must be different, we must make some arrangement such as puttinga with bc, de, andf g; and, if so,b may be associated with d fand eg; and c with dg and ef. Thus there are seven possible triads, such asabc, ade, af g, bd f, beg, cdg, and cef. Moreover each of these may stand for any one of four triplets: for instance, the triad bd fmay stand for any of the tripletsb1d1f1, b1d2f2, b2d1f2, b2d2f1. Thefourtriadswhichdonotinvolve amustbeplacedintheSunday column, the four which do not involveb in the Monday column, and so on. Thus each triad will occur four times. It only remains to insert the proper suffixes. This is done as follows. Take one triad, such asbd f, and insert a different set of suffixes each time that it occurs; for instance, the four sets given above. Next, the other like letters (b, d, or f as the case may be) in these four columns must have the complementary suffixes attached. After this is done, the next triplet in the Sunday column will be b2eg. The triad beg occurs in four columns and includes four possi-

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[PDF page 121] CH. IV] THE FIFTEEN SCHOOL-GIRLS PROBLEM. 105 ble triplets, such as b2e1g1, b2e2g2, b1e1g2, b1e2g1. Insert these, and then give the complementary suffixes to the other like letters in these four columns. In this way the arrangement is constructed gradually, by taking one triad at a time, inserting the proper suffixes to the four triplets included in it, and then the complementary suffixes in the other like letter in the same columns. One final arrangement, thus obtained, is as follows: Sunday Monday Tuesday Wednesday Thursday Friday Saturday ka1a2 kb1b2 kc1c2 kd1d2 ke1e2 kf1f2 kg1g2 b1d1f1 a1d2e2 a1d1e1 a2b2c2 a2b1c1 a1b2c1 a1b1c2 b2e1g1 a2f2g2 a2f1g1 a1f2g1 a1f1g2 a2d2e1 a2d1e2 c1d2g2 c1d1g1 b1d2f2 b1e1g2 b2d1f2 b1e2g1 b2d2f1 c2e2f2 c2e1f1 b2e2g2 c1e2f1 c2d2g1 c2d1g2 c1e1f2 We might obtain other solutions by selecting other seven triads or by choosing other arrangements of the suffixes in each triad (or by merely interchanging letters and suffixes in the above order). By these means Mr Power showed that there are no less than15567, 552000 different solutions; but, since the total number of ways in which the school can walk out for a week in triplets is(455)7, the probability that any chance way satisfies Kirkman’s condition is very small. Frost’s method is applicable to the case of22n − 1 girls walking out for 22n−1 − 1 days in triplets. The detailed solution for63 girls walking out for 31 days, which corresponds ton = 3, have been given. Anstice’s Method.Another method of attacking the problem is due to Mr Anstice; it is illustrated by the following elegant solution, by which from the order on Sunday we can obtain the order on the following six days by a cyclical permutation. Let the girls be denoted respectively by the lettersk, a1, a2, a3, a4, a5, a6, a7, b1, b2, b3, b4, b5, b6, b7; and suppose the order on Sunday to be ka1b1, a 2a3a5, a 4b3b6, a 6b2b7, a 7b4b5 . Then, if the suffixes are permuted cyclically, we obtain six other ar- rangements which satisfy the conditions of the problem: the reason being that in the above arrangement the difference of the suffixes of ev- ery pair of like letters—such as either the “a”s or the “b”s—in a triplet

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[PDF page 122] 106 MISCELLANEOUS MATHEMATICAL RECREATIONS. [CH. IV is different for each triplet, as also is the difference of the suffixes of every pair of unlike letters which are in a triplet. Two other arrangements for Sunday, from which those for the re- maining days are obtainable by cyclical permutations can be formed. Theseare ka1b1, a2a3a5, a4b5b7, a6b3b4, a7b2b6; andka1b1, a2a3a5, a4b2b6, a6b5b7, a7b3b4. Anstice’s method is applicable to the case of2p + 1 girls walking out for p days in triplets so that no pair may walk together more than once, provided p is a prime of the form12m + 7. In such a case he showed how to construct a fundamental arrangement for one day from which the arrangements for the remainingp − 1 days can be obtained by cyclical permutations of suffixes. The number of such fundamental arrangements is 3(2m + 1)(3m + 1). The problem of15 girls corresponds tom = 0, and the three fun- damental Anstician arrangements are given above. Ifm = 1 we have the problem of 39 girls. One Anstician arrangement in this case is as follows: ka1b1, a2a8a12, a5a7a10, a6a17a18, a3b10b15, a4b3b5, a9b18b19, a11b8b14, a13b9b17, a14b12b16, a15b4b7, a16b2b11, a19b6b13. If m = 2 we have the problem of63girls, of which Frost has given one solution; and so on. Gill’s Method.Anothermethod ofattackingtheproblemhasbeen suggested to me by Mr T.H. Gill. Representing the girls bya1, a2, a3, . . . , a3m he (i) forms one triplet of the typea1am+1a2m+1, from which, by cyclical permutation of the suffixes1, 2, . . . ,3m he obtainsm triplets which constitute an arrangement for one day, and (ii) forms1 2(m − 1) other triplets such that the three differences of the suffixes are different, from which, by cyclical permutations of the suffixes, the arrangements for the remaining 3 2(m − 1) other days can be obtained. Thus in the case of 15 girls, the triplet a1a6a11 gives, by cyclical permutations of the suffixes, an arrangement for the first day and two triplets such as a1a2a5, a1a3a9 enable us to form30 triplets from which an arrangement for the other six days can be found. Here is a solution thus determined. First Day: 1. 6.11; 2. 7.12; 3. 8.13; 4. 9.14; 5.10.15. Second Day: 1. 2. 5; 3. 4. 7; 8. 9.12; 10.11.14; 13.15. 6. Third Day: 2. 3. 6; 4. 5. 8; 9.10.13; 11.12.15; 14. 1. 7. Fourth Day: 5. 6. 9; 7. 8.11; 12.13. 1; 14.15. 3; 2. 4.10. Fifth Day: 7. 9.15; 8.10. 1; 3. 5.11; 4. 6.12; 13.14. 2. Sixth Day: 9.11. 2; 10.12. 3; 5. 7.13; 6. 8.14; 15. 1. 4. Seventh Day: 11.13. 4; 12.14. 5; 15. 2. 8; 1. 3. 9; 6. 7.10.

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[PDF page 123] CH. IV] THE FIFTEEN SCHOOL-GIRLS PROBLEM. 107 But, although this method gives triplets with which the problem can be solved, the final arrangement is empirical. A solution of the problem of21 girls for 10 days can be got by the samemethod: a1a8a15 giving 7tripletswhichconstituteanarrangement for one day; anda1a2a6, a1a3a11, a1a4a10 giving 63 triplets from which an arrangement for the other nine days can be formed. Here is the solution thus determined. First Day: 1. 8.15; 2. 9.16; 3.10.17; 4.11.18; 5.12.19; 6.13.20; 7.14.21. Second Day: 1. 2. 6; 4. 5. 9; 7. 8.12; 10.11.15; 13.14.18; 16.17.21; 19.20. 3. Third Day: 7.10.16; 8.11.17; 12.15.21; 18.19. 2; 20. 1. 9; 3. 5.13; 4. 6.14. Fourth Day: 13.16. 1; 14.17. 2; 18.21. 6; 3. 4. 8; 5. 7.15; 9.11.19; 10.12.20. Fifth Day: 4. 7.13; 5. 8.14; 9.12.18; 15.16.20; 17.19. 6; 21. 2.10; 1. 3.11. Sixth Day; 1. 4.10; 2. 5.11; 6. 9.15; 12.13.17; 14.16. 3; 18.20. 7; 19.21. 8. Seventh Day: 2. 3. 7; 5. 6.10; 8. 9.13; 11.12.16; 14.15.19; 17.18. 1; 20.21. 4. Eighth Day: 10.13.19; 11.14.20; 15.18. 3; 21. 1. 5; 2. 4.12; 6. 8.16; 7. 9.17. Ninth Day: 16.19. 4; 17.20. 5; 21. 3. 9; 6. 7.11; 8.10.18; 12.14. 1; 13.15. 2. Tenth Day: 19. 1. 7; 20. 2. 8; 3. 6.12; 9.10.14; 11.13.21; 15.17. 4; 16.18. 5. I should be interested if any of my readers could give me a similar solution of the analogous arrangement of33 girls for 16 days formed from typical triplet suffixes like1, 12, 23; 1, 2, 10; 1, 3, 16; 1, 4, 18; 1, 5, 11; 1, 6, 13; or from other sets of triplets formed in a similar way so that (except in the first triplet) the differences of the suffixes are all different. Walecki’s Theorem. Lastly, Walecki—quoted by Lucas—has shown that, if a solution for the case ofn girls walking out in triplets for 1 2(n − 1) days is known, then a solution for3n girls walking out for 1 2(3n − 1) days can be deduced. For if an arrangement of then girls, a1, a2, . . . , an for 1 2(n − 1) days is known; and also one of then girls, b1, b2, . . . , bn; and also one of the n girls c1, c2, . . . , cn; then an arrangement of these3n girls for 1 2(n − 1) days is known. A set of n triplets for another day will be given by ambm+kcm+2k where m is put equal to 1, 2, . . . , n successively. Here k may have any of then values, 0, 1, 2, . . . ,(n − 1); but, wherever a suffix is greater thann, it is to be divided byn and only the remainder retained. Hence altogether we have an arrangement forn + 1 2(n − 1) days, i.e. for 1 2(3n − 1) days. The arrangement of 3 girls for one day is obvious. Hence, by Walecki’s theorem, we can deduce at once an arrangement of3m girls for 1 2(3m − 1) days. And, generally, as I have given solutions of the problem in the case of3n girls when n = 1, 3, 5, 7, 9, 13, 15, it follows
