15 Oct
2007
15 Oct
'07
6:16 p.m.
For p an odd prime, I give a solution for the size N = 2p submarine problem in 2N-1 = 4p-1 steps using 2N-2 shots. At t = 0, 2, ... , N-2 shoot square 0. At t = 1, 3, ... , N-3 shoot square t+1. At t = N-1 shoot your foot (this shot is not needed for the submarines). At t = N, N+2, ... , 2N-2 shoot square p = N/2. At t = N+1, N+3, ... , 2N-3 shoot square (t+1+p) mod N. This gives a more clearly structured solution for the size 10 problem than my previous message, and corrects a minor typo in that message. It's known that the size 6 problem can be massaged down to 10 steps. I don't know if size 2p for p > 3 can massage out the unused shot. Best, - Scott