CH. X] KIRKMAN'S SCHOOL-GIRLS PROBLEM 203 



all cases except those where n is of the form 24m + 21. These 

 I have failed to bring under analogous rules, but we can solve 

 them by recourse to the three-step cycles next described. 



Three-Step Cycles. The fact that certain cases are soluble 

 by one-step cycles, and others by two-step cycles, suggests 

 the use of three-step cycles, and the fact that n is a multiple 

 of 3 points to the same conclusion. On the other hand, if 

 we denote the n girls by 1, 2, 3, ... n, and make a cyclical 

 permutation of three steps at a time (or if we denote 

 the girls by Oj, b lt c u a a , b u c 2 , ..., and make a cyclical 

 permutation of the suffixes one step at a time), we cannot 

 get arrangements for more than n/3 days. Hence there will 

 remain (n — l)/2 — n/3 days, that is, (n — 3)/6 days, for which 

 we have to find other arrangements. In fact, however, we 

 can arrange the work so that in addition to the cyclical 

 arrangements for n/3 days we can find (n — 3)/6 single triplets 

 from each of which by a cyclical permutation of the numbers 

 or suffixes an arrangement for one of these remaining days can 

 be obtained ; other methods are also sometimes available. 



For instance take the case of 21 girls. An arrangement 

 for the first day is (1. 4. 10), (2. 5. 11), (3. 6. 12), (7. 14. 18), 

 (8. 15. 16), (9. 13. 17), (19. 20. 21). From this by cyclical 

 permutations of the numbers three steps at a time, we can 

 get arrangements for 7 days in all. The arrangement for the 

 8th day can be got from the triplet (1. 6. 11) by a three-step 

 cyclical permutation of the numbers in it. Similarly the 

 arrangement for the 9th day can be got from the triplet 

 (2. 4. 12), and that for the 10th day from the triplet (3. 5. 10), 

 by three-step cyclical permutations. 



This method was first used by A. Bray in 1883, and was 

 subsequently developed by Dudeney and Eckenstein. It gives 

 a solution for every value of n except 15, but it is not so easy 

 to use as the methods already described, partly because the 

 solution is in two parts, and partly because the treatment 

 varies according as n is of the form 18m + 3, or 18m + 9, or 

 18m + 15. Most of the difficulties in using it arise in the case 

 when n is of the form 18m + 15. 



