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1 5. Kirkman s Schoolgirl Problem In a boarding school there are fifteen schoolgirls who always take their daily walks in row of threes. How can it be arranged so that each schoolgirl walks in the same row with every other schoolgirl exactly once a week? This extraordinary problem was posed in the Lady s and Gentleman s Diary for 1850, by the English mathematician T.P. Kirkman. We give two solutions of the many that have been found. One is by the English minister Andrew Frost ("General Solution and Extension of the Problem of the 15 Schoolgirls", Quarterly Journal of Pure and Applied Mathematics, vol. XI, 1871) and the other is that of B. Pierce ("Cyclic Solutions of the Schoolgirl Puzzle", The Astronomical Journal, vol. VI, ). (Dörrie asserts that Sylvester thought Pierce s solution the best, but does not state how many solutions he examined.) Frost s Solution The problem consists of arranging the 15 elements x, a 1,a 2,b 1,b 2,c 1, c 2,d 1,d 2, e 1,e 2, f 1,f 2, g 1,g 2 in seven columns of five triplets each in such a way that any two elements always occur in one and only one of the 35 triplets. We shall select xa 1 a 2 xb 1 b 2 xc 1 c 2 xd 1 d 2 xe 1 e 2 xf 1 f 2 xg 1 g 2 as the initial triplets of the seven columns. Then we only have to distribute the 14 elements a 1, a 2, b 1, b 2,...,g 1, g 2 over the other four lines of our system. Using the seven letters a,b,c, d, e,f, we form groups of triplets in which each pair of letters occurs exactly once: abc ade afg bdf beg cdg cef From this group we can take exactly four triplets for each column that contain all the letters except for those in the first line of the column. If we put the triplets in alphabetical order in each column, we get the following preliminary arrangement: bdf ade ade abc abc abc abc beg afg afg afg afg ade ade cdg cdg bdf beg bdf beg bdf cef cef beg cef cdg cdg cef Now we have to index the triplets bdf,beg,cdg,cef,ade,afg,abc, i.e., provide them with indices of 1 or 2. We index them in the order just mentioned, i.e., first all triplets bdf, then 1

2 all triplets beg, etc., observing the following three rules: 1. When a letter in one column has been indexed, the next time that letter occurs in the same column, it gets the other index number. 2. If two letters of a triplet have already been indexed, these two index numbers must not be used in the same sequence for the same letters in other triplets. 3. If the index number of a letter is not determined by the first two rules, the letter is assigned the index number 1. The letters will be indexed in three steps: First step. The triplets bdf, beg,cdg,cef and all the letters aside from a that can be indexed by rules 1, 2 and 3 are successively indexed. (Note: ALL bdfs in the order b 1 d 1 f 1, b 1 d 2 f 2, b 2 d 1 f 2, b 2 d 2 f 1 are done first, then all begs, etc.) The result is: b 1 d 1 f 1 ad 2 e 2 ad 1 e 1 ab 2 c 2 ab 1 c 1 ab 2 c 1 ab 1 c 2 b 2 e 1 g 1 af 2 g 2 af 1 g 1 af 2 g 1 af 1 g 2 ad 2 e 1 ad 1 e 2 (The reader is advised to do this on his/her own.) Second step. The missing indices for a in the triplets ade and afg and for the last two as in line 2 are assigned. The result is: b 1 d 1 f 1 a 1 d 2 e 2 a 1 d 1 e 1 ab 2 c 2 ab 1 c 1 a 1 b 2 c 1 a 1 b 1 c 2 b 2 e 1 g 1 a 2 f 2 g 2 a 2 f 1 g 1 af 2 g 1 af 1 g 2 a 2 d 2 e 1 a 2 d 1 e 2 Third step. The still missing indices on a in columns 4 and 5 are inserted, in accordance with the rules above. The final result is: 2

3 b 1 d 1 f 1 a 1 d 2 e 2 a 1 d 1 e 1 a 2 b 2 c 2 a 2 b 1 c 1 a 1 b 2 c 1 a 1 b 1 c 2 b 2 e 1 g 1 a 2 f 2 g 2 a 2 f 1 g 1 a 1 f 2 g 1 a 1 f 1 g 2 a 2 d 2 e 1 a 2 d 1 e 2 Pierce s Solution Designate one girl as ', whol walks in the middle of the same row all seven days of the week; divide the other girls into two groups of 7, the girls in the first group designated by 1, 2, 3, 4,5, 6, 7 or by a,b, c,d, e, f, g and the second group designated by I,II,III, IV,V, VI, VII or by A, B,C,D, E,F, G. We also designate the days of the week Sunday, Monday,..., Saturday by 0, 1, 2,...,6. Let the Sunday arrangement have the form: a ) A b * B c + C d ' D E F G Add the same number r R, e.g. 1 and I, 2 and II, etc. to each number mod 7 to get a r ) r A R b r * r B R c r + r C R d r ' D R E R F R G R for the r th weekday. The arrangements so obtained provide a solution to the problem if the following three conditions are satisfied: 1. ) " a 1, * " b 2 and + " c The seven differences A " a, A " ), B " b, B " *, C " c, C " + and D " d mod7 equal 0,1, 2,3, 4, 5, 6 in some order. 3. F " E 1, G " F 2 and G " E 3. Proof. 1. We show first that every girl x of the first group walks with every other girl y of the first group. By 1., x " y q oÿa " ), oÿb " *, or oÿc " + mod 7, and just one of them, say x " y q * " bmod 7, or x " * q y " b q rmod7, with r 0,1,...,6. Then 3

4 x q * rmod 7 and y q b rmod 7 so girls x and y walk in the same row on day r. 2. Next we show that every girl x of the first group walks with every girl X of the second group. By 2., X " x is congruent mod 7 to just one of A " a, A " ), B " b, B " *, C " c, C " + or D " d, say X " x q C " + mod 7, or X " C q x " + q smod7 with s 0, 1,...,6. Then X q C Smod 7 and x q + smod7 so girls X and x walk in the same row on day s. (Here S is the Roman numeral for s.) 3. Finally we show that every girl X of the second group walks exactly once with every other girl Y of the second group. By 3., X " Y q oÿf " E, oÿg " F, or oÿg " E mod 7, and just one of them, say X " Y q G " Fmod 7, or X " G q Y " F q Rmod7, with R VII,I,...,VI. Then X q G Rmod7 and Y q F R mod7 so girls X and Y walk in the same row on day R. R Thus, we need only satisfy conditions 1,2 and 3 to obtain the Sunday arrangement. a 1, b 3, c 4, d 6, ) 2, * 5, + 7 and A I, B VI,C II,D III,E IV, F V and G VII satisfy all the conditions. The differences in 2. are 0, "1, 3, 1, "2, "5, "3 q 0, 6,3, 1, 5,2,4mod 7.b The Sunday arrangement is therefore and the weekday rows are: 1 2 I 3 5 VI 4 7 II 6 ' III IV V VII 2 3 II 3 4 III 4 5 V 4 6 VII 5 7 I 6 1 II 5 1 III 6 2 IV 7 3 V 7 ' IV 1 ' V 2 ' VI V VI I, VI VII II, VII I III, 4

5 5 6 V 6 7 VI 7 1 VII 7 2 III 1 3 IV 2 4 V 1 4 VI 2 5 VII 3 6 I 3 ' V 4 ' 1 5 ' II I II IV, II III V, III IV VI. [Kirkman s schoolgirl problem is an example of a problem in combinatorial design theory. The solution is an example of a resolvable Ÿ35, 15,7, 3, 1 design. See for example Introductory Combinatorics ( ) by Kenneth P. Bogart, Harcourt, 2000.] 5

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