This is a new and interesting companion puzzle to the "Fifteen
Schoolgirls" (see solution of No. 269), and even in the simplest
possible form in which I present it there are unquestionable
difficulties. Nine schoolboys walk out in triplets on the six week days
so that no boy ever walks _side by side_ with any other boy more than
once. How would you arrange them?
If we represent them by the first nine letters of the alphabet, they
might be grouped on the first day as follows:--
A B C
D E F
G H I
Then A can never walk again side by side with B, or B with C, or D with
E, and so on. But A can, of course, walk side by side with C. It is here
not a question of being together in the same triplet, but of walking
side by side in a triplet. Under these conditions they can walk out on
six days; under the "Schoolgirls" conditions they can only walk on four
days.