0
   

impossible Prison Riddle!!!

 
 
Reply Tue 22 May, 2007 12:20 pm
The following was given to me by a maths professor. This is a hard question.

The governor of a prison is bored. He decides to amuse himself by playing
a game with the prisoners. He sets up a room with two switches in it. Each
switch has 2 positions (on and off ). He tells the prisoners that he is going to
send them to the room one at a time and that they have to move exactly one
of the switches each time they visit. He is free to send them in whatever order
he likes and can send them as often as he likes (perhaps sending one prisoner
100 times before sending some other prisoner, then sending the first one again...
anything!). The only promise that he makes is that if any prisoner picks a
number n and waits long enough, they can be sure to visit the room n times.
The prisoners are kept in solitary and are unable to communicate once the
game has started. They will never know how many other prisoners have visited
the room since they were last there. They are also not told what the initial
position of the switches was.
The governor tells them that if at any stage a prisoner comes to him and
states correctly that all prisoners have now visited the room at least once, all of
the prisoners will be freed. But if a false claim is made, they will all be shot.
The night before the game begins the prisoners meet in a large room to
discuss their tactics. Do they have a strategy that ensures they will win their
freedom?
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Discussion • Score: 0 • Views: 30,928 • Replies: 6
No top replies

 
contrex
 
  2  
Reply Tue 22 May, 2007 01:14 pm
Easy. One prisoner is designated as the "caller." He and only he will make the determination of when everyone's been in the room. Whenever he sees that the left-hand switch is "down," he will move it "up" and add 1 to his count; otherwise, he will flip the right-hand switch and not think anything of it. When the other prisoners enter the room, they have a different strategy. If the left-hand switch is "up," they will switch it "down," but they will only do this the first two times they encounter the left-hand switch "up." If they've moved the left switch twice already, or if the left-hand switch is "down," they will flip the right-hand switch.

When the caller's count is twice the number of prisoners plus one, each prisoner must have been in the room at least once, regardless of the initial configuration of the switches.
betty boo
 
  1  
Reply Tue 22 May, 2007 01:34 pm
cheers for that. got me a limited edition LP Laughing
0 Replies
 
contrex
 
  1  
Reply Tue 22 May, 2007 01:53 pm
I found it like this

http://www.google.co.uk/search?num=100&hl=en&q=prison+switches+riddle&btnG=Search&meta=
0 Replies
 
Quincy
 
  1  
Reply Sun 27 May, 2007 04:43 pm
Wow. That seemed pretty impossible to solve. Who the hell thinks up this stuff anyway!
0 Replies
 
Morlock26
 
  1  
Reply Mon 8 Sep, 2008 09:28 pm
@contrex,
Yea, there are a lot of solutions like that if you can distinguish between left and right switches. No where in the riddle does it say that the prisoners no they will be oriented the same. There could be more then door, they could be dropped in from a door in the ceiling. One switch could be on one wall of the room, the other switch on the other side of the room, the prisoners could be thrown in blind folded completely disoriented, and a door cannot be used as a reference frame because there are two identical doors maybe. There is not enough information about how they enter the room or about the room itself. Doesn't seem to be an easy answer yet, perhaps a solution could be come up for every possible situation the prisoners can come up with. But they might miss a condition, and therefore their freedom not ensured.
0 Replies
 
andbria1
 
  1  
Reply Tue 13 Apr, 2010 06:53 pm
@contrex,
I don't see why each prisoner has to flick the switch twice... unless they want to stay in prison for longer?

I think it should work like this:
One person is the counter.
The counting must start from the first time the counter enters the room.
Every time the counter enters the room he must make sure that switch A is in the on position. If it is not he makes it so and adds one to his count, if it is already in the on position he flicks switch B.

Everyone else must only flick switch A from the on position to the off position and they must only do this once; if these conditions can not be met then they must flick switch B.

When the counter reaches the same count as the number of prisoners he knows that every prisoner has been in the room at least once.
0 Replies
 
 

Related Topics

Alternative Einstein's riddle answer - Discussion by cedor
Urgent !!! Puzzle / Riddle...Plz helpp - Question by zuzusheryl
Bottle - Question by Megha
"The World's Hardest Riddle" - Discussion by maxlovesmarie
Hard Riddle - Question by retsgned
Riddle Time - Question by Teddy Isaiah
riddle me this (easy) - Question by gree012
Riddle - Question by georgio7
Trick Question I think! - Question by sophocles
Answer my riddle - Question by DanDMan52
 
  1. Forums
  2. » impossible Prison Riddle!!!
Copyright © 2024 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 05/20/2024 at 11:11:53