CH. x] kirkman's school-girls problem 209 



For 51 girls, an arrangement in the first part ia (17. 34. 51), (49. 16. 10), 

 (11. 44. 50), (15. 33. 27), (19. 4. 46), (2. 38. 29), (21. 36. 45), (22. 25. 24), (5. 8. 7), 

 (39. 42. 41), (43. 13. 20), (26. 47. 3), (9. 30. 37), (1. 14. 6), (35. 48. 40), (18. 31. 23), 

 (28. 32. 12). The triplets in the second part are (29. 33. 13), (30. 34. 14), 

 (1. 26. 12), (2. 27. 13), (3. 28. 14), (1. 11. 30), (2. 12. 31), (3. 13. 32). 



For 69 girls, an arrangement in the first part is (23. 46. 69), (31. 34. 43), 

 (11. 8. 68), (54. 57. 66), (58. 52. 28), (35. 29. 5), (45. 51. 6), (16. 1. 49), 

 (62. 47. 26), (39. 24. 3), (4. 55. 42), (50. 32. 19), (27. 9. 65), (40. 67. 18), 

 (17. 44. 64), (63. 21. 41), (10. 14. 15), (33. 37. 38), (56. 60. 61), (25. 53. 36), 

 (2. 30. 13), (48. 7. 59), (22. 20. 12). The triplets in the second part are 

 (23. 21. 13), (24. 22. 14), (1. 8. 33), (2. 9. 34), (3. 10. 35), (1. 17. 36), (2. 18. 37), 

 (3. 19. 38), (1. 41. 15), (2. 42. 16), (3. 43. 17). 



For 87 girls, an arrangement in the first part is (29. 68. 87), (76. 82. 73), 

 (47. 53. 44), (15. 9. 18), (70. 1. 13), (41. 59. 71), (12. 30. 42), (4. 67. 31), 

 (2. 26. 62), (60. 84. 33), (40. 79. 25), (11. 50. 83), (69. 21. 54), (43. 64. 63), 

 (14. 35. 34), (72. 6. 5), (61. 16. 77), (32. 74. 48), (3. 45. 19), (28. 68. 66), 

 (37. 86. 39), (10. 8. 57), (46. 80. 36), (22. 65. 75), (7. 17. 51), (85. 23. 78), 

 (52. 20. 27), (49. 56. 81), (55. 38. 24). The triplets in the second part are 

 (56. 39. 25), (57. 40. 26), (1. 5. 42), (2. 6. 43), (3. 7. 44), (1. 29. 6), (2. 30. 7), 

 (3. 31. 8), (1. 9. 20), (2. 10. 21), (3. 11. 22), (1. 14. 36), (2. 15. 37), (3. 16. 38). 



Before proceeding to the consideration of other methods I should add that 

 it is also possible to obtain irregular solutions of cases where n is of any 

 of these three forms. As an instance I give a three-step solution of 33 girls. 

 A possible arrangement in the first part is (1. 4. 10), (14. 23. 26), (9. 15. 30), 

 (3. 28. 83), (2. 6. 8), (11. 18. 27), (13. 24. 25), (5. 16. 20), (7. 17. 22), (21. 29. 31), 

 (12. 19. 32). The triplets in the second part are (1. 2. 21), (1. 8. 30), (1. 3. 26), 

 (1. 15. 20), (1. 17. 18). I describe this solution as irregular, since all, saye one, 

 of the triads used are different. 



The Focal Method. Another method of attacking the 

 problem, comparatively easy to use in practice, is applicable 

 when n is of the form 24m + 3p, where p = 6q + 3. It is due 

 to Eckenstein. Here it is convenient to use a geometrical 

 representation by denoting 24m + 2p girls by equidistant 

 numbered points on the circumference of a circle, and the 

 remaining p girls by lettered points placed inside the circle; 

 these p points are termed foci. The solution is in two parts. 

 In the first part, we obtain an order from which the arrange- 

 ments for 12m +p days are deducible by a two-step cycle 

 of the numbers: in none of these triplets does more than 

 one focus appear. In the second part, we find the arrange- 

 ments for the remaining 3q + 1 days ; here the foci and the 

 numbered points are treated separately, the former being 



B- b. 14 



