# A toughie... (or maybe it isn't)

kyngj
Posers and Puzzles 16 Jul '03 23:37
1. bbarr
Chief Justice
22 Jul '03 22:38
Originally posted by ddebened
Excellent solution. I am not sure I agree that two visits are required. Would it not work equally well if each person flips the left switch on their first visit and only on their first visit (which would vastly reduce the number of visits required to conclude that all persons had visited the room)?
The two flips are necessary because I have no way of knowing the original position of the left switch, whether it starts up or down. If, on my first visit, I find the left switch in the up position, then it is either the case that the switch started out like that, or that a prisoner been to the room. Thus, it is possible that if everyone flips the left switch only once my count will only proceed to 18, even long after everyone has visited (or, alternatively, it's possible that if I assume on my first visit that the switch in the up position indicates the prior visit of a prisoner, I'll stop counting at 19 with only 18 prisoners having actually visited).
2. Phlabibit
Mystic Meg
23 Jul '03 16:183 edits
Originally posted by bbarr
Sorry for the delay. First, I designate myself as the only one with the authority to inform the guards that every other prisoner has visited the room. I'll do this by counting the number of times the left switch is flipped. I pay no atten ...[text shortened]... times, then I'll know each prisoner has been there at least once.
UG,

That will take forever, but I suppose it will work eventually... if you visit the room that many times.

Yikes!

I need to think about this a bit more also, since someone is going to flip it once, and you won't know if they did or did not. You may also be waiting a long time for visitor 20, and he thinks he already flipped it enough and won't flip it again for you.

Wouldn't that stink if they sent all 20 in a row, sending you last... and decided to never send you again......