Last week’s puzzle was the “two switch” puzzle. To solve it, the group of prisoners must identify one person, the counter. Ignore switch B, it is there just to give people something to toggle if they don’t want to toggle switch A. Everyone except the counter follows the rule: if switch A is up, and you’ve never previously put it down, then put it down. The counter follows the rule, if switch A is down, then put it up and add one to the count. If the count matches the number of prisoners (less one to take account of the counter), announce that everyone has visited the room.
With these rules, each prisoner (except the counter) will put switch A down exactly once, and the counter will eventually notice, increment the count and reset the switch.
It is possible to adapt these rules slightly for the more difficult puzzle where the initial positions of switch A is unknown, so as to avoid the off-by-one error that results from the counter thinking there is one extra person if switch A starts off down. Exercise for the reader.
Today something a bit more electrical, since this is an EDA blog. What is the resistance between diagonally opposite corners of a cube of identical resistors?