CH. x] kirkman's school-girls problem 201 



a time, we get a solution of the problem for the seven days 

 required. Each of these solutions by reflection and inversion 

 gives rise to three others. 



Next, if we take (k. 1. 2) for one triplet on the first day we 

 shall have the points 3, 4,... 14 for the vertices of the four 

 triangles denoting the other triplets on that day. The sides 

 must be of the lengths [1], [2], ... [7], of which [2], ... [6] must 

 be used not more than twice and [1], [7] must be used 

 only once. The [1] used must start from an even number, 

 for otherwise the chord denoted by it would, when the system 

 was rotated, occupy the position joining the points 1 and 2, 

 which has been already used. The only possible groupings are 

 [2, 4, 6], [2, 3, 5], [3, 4, 7], [1, 5, 6] ; or [2, 4, 6], [2, 5, 7], [1, 4, 5], 

 [3, 3, 6] ; or [2, 4, 6], [2, 5, 7], [1, 3, 4], [3, 5, 6]. Each of these 

 groupings gives rise to various solutions. For instance the first 

 grouping gives a set of triplets (k. 1. 2), (3. 7. 10), (4. 5. 13), 

 (6. 9. 11), (8. 12. 14). From this by a cyclical two-step per- 

 mutation we get a solution. This solution is represented in 

 figure iii. If we take (k. 1. 4) or (k. 1. 6) as one triplet on the 

 first day, we get other sets of solutions. 



Solutions involving the triplets (k. 1. 2), (k. 1. 4), (k. 1. 6), 

 (Ic. 1. 8), and other analogous solutions, can be obtained from 

 the solutions illustrated in the above diagrams by re-arranging 

 the symbols denoting the girls. For instance, if in figure iii, 

 where all the triangles used are scalene, we replace the numbers 

 2, 8, 3, 13, 4, 6, 5, 11, 7, 9, 10, 14 by 8, 2, 13, 3, 6, 4, 11, 5, 9, 

 7, 14, 10, we get the solution (k. 1. 8), (4. 5. 7), etc., given above. 

 Again, if in figure iii we replace the symbols 1, 2, 3, 4, 5, 6, 

 7, 8, 9, 10, 11, 12, 13, 14, h by 12, 2, 10, 6, 1, 14, 5, 13, 9, 

 15, 4, 3, 11, 8, 7, we obtain a solution equivalent to that given 

 by T. H. Gill and printed in the fourth edition of this book. In 

 this the arrangement on the first day is (1. 6. 11), (2. 7. 12), 

 (3. 8. 13), (4. 9. 14), (5. 10. 15). The arrangements on the other 

 days are obtained as before by rotating the system so de- 

 lineated round 7 as a centre two steps at a time. Gill's 

 arrangement is thus presented in its canonical form as a 

 central two-step cyclical solution. 



