Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

LPF Donation via Stripe | LPF Donation - Other Methods

Links below open in new window

ArcticMyst Security by Avery

Riddles

Grix

0
Joined
Dec 9, 2008
Messages
2,190
Points
63
Here's a very hard logic riddle, can anyone solve it? (No googling! :tsk:)


  • So basically you have 50 perfect logicians in a room. All of their foreheads are painted red. They know that the others in the room are perfect logicians as well.
  • The logicians does not know what color their own forehead is painted, but can see the foreheads of all the other logicians.
  • They have all been told that at least one of them has a red-painted forehead.
  • They cannot communicate in any way.
  • After a while the lights are turned off, and they are told that if you know for sure that your forehead is painted red, you must leave the room when the lights are off. Then the lights are turned on and off again until there are no logicians left in the room.

The riddle is: How many times must the lights be turned off before there are no logicians left in the room? (There is a real answer, it's not infinity)

Anyone up for the challenge? :) If you have heard it before or if you couldn't stop yourself from googling it, please don't spoil the answer..
 
Last edited:





Joined
Aug 31, 2010
Messages
176
Points
0
Once, I didn't see anything in the rules stating that they couldn't tell other people what color their forehead was painted :). The only thing I see in there would be that they couldn't see their own forehead directly.

I assume they would also see the relfected color of their forehead in someone else's eyes (assuming their are no mirror like surfaces in the room).
 

Grix

0
Joined
Dec 9, 2008
Messages
2,190
Points
63
I know there was something I forgot to add.. damn. They can not communicate. :)

And no, they have no way of knowing what color their forehead is except looking at other people's foreheads and their actions during lights off.

I'll post hints if noone gets it soon.. :)
 
Joined
Jul 30, 2010
Messages
215
Points
0
They will all leave the first time lights go off: once it's dark they will all touch their heads and realize they're painted.:)
 

HIMNL9

0
Joined
May 26, 2009
Messages
5,318
Points
0
Two times.

The first time, all think that all the others are painted red, except themselves, and stay all in the room ..... the second time (if they are logicians), must understand the fact that all the others stayed cause they see all the others red except our own, understand that, in this case, means that also their own forehead is painted red, and left the room. (at least i see the thing in this way)

(sorry for the bad english, don't know how to explain better my line of think)
 

Grix

0
Joined
Dec 9, 2008
Messages
2,190
Points
63
It's good thinking, but it's wrong. They can assume that the reason noone left is that they are all wondering about the same thing, but perfect logicians don't assume, they KNOW! :p

HINT 1: If you don't know the logic behind it, the result (and the execution of the result) would seem bizarre.
 
Joined
Jul 30, 2010
Messages
215
Points
0
For the case of 50 logicians it would take 50 light switches: everyone would stay and all will leave at the same time after the last flip?
 
Joined
Jul 30, 2010
Messages
215
Points
0
YAY! I am still at work, so I'll have to watch the vid later. Our solution was kinda hard to get your head around all at once, so I'd love to see if these people have a better way to explain it.

I did not come up with this myself, it took me and 5 other engineers the whole lunch hour to figure this out:) First you look at the case of just 2 people, then 3 (this is where it gets harder). We used induction and proof by contradiction for cases where N>2. The credit truly goes to the team!
 

Grix

0
Joined
Dec 9, 2008
Messages
2,190
Points
63
YAY! I am still at work, so I'll have to watch the vid later. Our solution was kinda hard to get your head around all at once, so I'd love to see if these people have a better way to explain it.

I did not come up with this myself, it took me and 5 other engineers the whole lunch hour to figure this out:) First you look at the case of just 2 people, then 3 (this is where it gets harder). We used induction and proof by contradiction for cases where N>2. The credit truly goes to the team!

Sounds about right! Yes, it kinda hurts your head to think about it. :p I was told this has been used as an exam question for computer engineers at MIT, but I don't know if it's true.
 
Joined
Jul 30, 2010
Messages
215
Points
0
Perfect time for that NOW! ;)

EDIT: oh man, I was 4:20 for me, but the forum thinks it's 4:29. oh well, any time is 4.20...
 
Joined
Aug 30, 2008
Messages
6,891
Points
83
Crap......
I just got mind fucked when he started explaining when there are 3 people in the room and then i realized what it would be like with 50.

It was like trying to understand infinity, or witnessing someone divide by 0 :D


speaking of 420...I'm going to snap if i hear someone say "it's 4:20 somewhere" when it's like 1:48pm..... no, it's not 4:20....ANYWHERE!
 
Last edited:




Top