But would it not be sufficient to count p?
If switch A starts out in the ON position it would equal one 'fake' prisoner. If the counter just counts one extra (which equals p - 1 (himself) + 1 (fake prisoner) = p) it should be enough, no?
If switch A starts out in the ON position it would equal one 'fake' prisoner. If the counter just counts one extra (which equals p - 1 (himself) + 1 (fake prisoner) = p) it should be enough, no?