Okay. So, I have another riddle for you. This one is not really physics based, but still it can be solved with logic alone (no math). I’ll give it to you in storybook form just for fun.
Once upon a time there was a forest, and in this forest there lived a sufficiently large number of gnomes (the exact number of gnomes doesn’t matter). They were extremely logical beings and valued the needs of the group far above their own individual needs, so much so that no mirrors existed in their village, so as to keep the focus of their attention away from themselves. They valued homogeneity, which was just as well since all of them were identical… well, except for one thing: their hats. For some inexplicable reason, while most of the gnomes had red hats, there were a certain number of them (say “N”) who had blue hats. These blue hatted gnomes didn’t themselves know that they had blue hats (for lack of mirrors, you see) and it was a taboo subject of the highest degree. No gnome would EVER give any indication — verbal or otherwise — as to the color of another gnome’s hat.
One day, at one of the gnome village meetings, where all the gnomes gathered to discuss serious matters, they decided as a group that because they valued homogeneity so much, it would be better for the village if all of the blue hatted gnomes left and lived elsewhere. Nothing more was discussed. No gnomes were singled out as having blue hats. The blue hatted gnomes were simply expected to leave as soon as they knew they had blue hats.
How many village meetings passed before all of the blue hatted gnomes left?
This is a really tricky riddle. Just remember, the solution has nothing to do with the gnomes using sign language to tell other gnomes about their hats, or using spoons as mirrors to see themselves. It’s a much more elegant and logical solution. If you haven’t heard it before, feel free to bounce ideas back and forth in the comments.
The gnomes don’t know that there are N blue hatters do they?
The gnomes only know what they can see for themselves. That said, they can’t see the colour of their own hat, but they can see the colours of the other gnomes hats (it’s just a taboo subject).
I thought about this for a while and couldn’t figure any way possible, but now I have some thoughts. Lemme know if I’m on the right track. I’m going to try to work in reverse order, from the last blue-hat to leave back.
Say there’s only 1 blue-hat left. Then at the next meeting, he will see that there are no other blue-hats, but the fact that they’re having a meeting at all means that there’s at least one left which he can conclude is him. He leaves.
Say there’s two left. A blue-hat ‘A’ will see one other, call him blue-hat B. From A’s perspective, if A had a red hat, then he should know that B should be able to figure out that he has to go by the reasoning for one blue-hat remaining. If A sees that B does NOT leave, then there must be at least 2 left and concludes that he’s blue, and leaves.
For three left, we have A, B, and C. From any one’s perspective, he sees only 2, and if they were the only 2 left then one should leave by the reasoning above. If one does NOT, then the third can conclude that he’s blue, and leaves.
This can then carry backward in reasoning until we hit N.
So, I’m not sure how the number of meetings works out though. If only one gnome leaves per meeting than it’s N meetings. If more can leave at once, then since the chain above carries back for all N and at each level the reasoning doesn’t have a preference of which gnome’s perspective we consider, then I don’t see why they wouldn’t ALL leave at meeting one.
Do the gnomes know how many of them have blue hats? If not, It would take one meeting, wouldn’t it? Any gnome with a blue hat would see N-1 blue hats and realize he was wearing a blue hat and leave? The same goes for all the other N-1 Gnomes
If so, rather.
If so, that’s not a very interesting riddle.
Nigel’s comment makes sense. If you’re a gnome, you see N blue hats, so you know that the total is either N or N+1.
The gnomes know that all the other gnomes are perfectly rational, so they know they’ll figure it out the same way.
If there’s one blue, he’ll leave the first day, knowing he’s the only one. If no one leaves the first day, you know there must be more than one blue hat. If you can only see 1, that means the other one is yours. So you both leave on day 2.
So if there are N blue hats, they all leave on day N.
Did their discussion (that the blue hats should leave) also include the information that “at least one gnome has a blue hat”?
That’s one of the elements I find so fascinating about this problem. If N>1, this statement doesn’t include any information that the gnomes didn’t already know, yet without the statement, I don’t believe the blue-hatted gnomes will be able to leave.
I suspect that there must also be some requirements about the gnomes only seeing each other during the meetings. You need some sort of discretization of time and events for this to work.
The trivial answer is that each one takes off his hat and looks at it to see its color. But that’s boring.
The above answer makes sense – each individual leaves on day N, knowing that if he had a red hat, then there would be only N-1 blue hats, so they would all have left the previous day.
Consider the total number is M and assume M/3=X (X is integer), randomely divide the gnomes every three persons, thus only X times meeting are enough
Wow! So much brainstorming and so many answers. So what is the actual solutions then.
Ok so… the right answer was that stated by excitedstate (and pioneered by Nigel).
@Anonymous Coward
I don’t think any requirement that the gnomes only see each other during meetings is needed. Just that once the gnomes realize what colour hat they have they leave after the meeting is over.
Great thinking, everyone!
@Jasper
I disagree. You need some way of discretizing time.
Try the following example: monthly meetings and only two blue-hatted dudes which we will call A & B. If A sees B still in town on the Tuesday following the meeting, he will realize that his own hat must also be blue (since B didn’t take off), and skip town. If B didn’t notice A seeing him, he will happily stay. But if B sees A (also in town a week later), he will leave too. This will screw up extrapolation to higher numbers.
I still maintain that, for similar reasons, it must be announced at the first meeting that “at least one dude has a blue hat”. Even though this information is redundant with what the dudes all know (if N>1), it is essential.
But perhaps I should revise my position entirely, as I now think that Mc’s “trivial” answer is far and away the best answer. I mean, why wouldn’t “extremely logical beings” (who wanted to leave as soon as possible) take off their hats to look at them? It would seem a bit illogical not to, no?
@AC
Jasper and I talked a lot about the discretized time problem. There’s a nifty bit of a paradox that can occur if the gnomes leave DURING the meeting, and what can happen based on time steps. (Depending on how you discretize time, the number of meetings needed is reduced by at least 1) This can all be avoided by saying that once gnomes are certain that they are blue, they leave immediately AFTER that meeting, and this is the only time when they would leave.
This clears up the problem in the first part of your post entirely. As for the second part; come on. They obviously know that at least one is blue, otherwise there’d be no point to this. Finally, as for your last point, well now you’re just being pedantic. Stick to the spirit of the problem.
@Nigel
Re: the only times gnomes can leave is “immediately” following a meeting.
I think that fixes it.
Re: “They obviously know that at least one is blue”
Admittedly, this is unlikely to be a problem in any practical situation, but as you point out in your dismissal of the “removing one’s own hat” method, the joy of this problem is not about what would happen in a practical situation. Whether or not this must be specified is – to me at least – the most interesting thing about the problem.
If you imagine a situation where the gnome overlord decrees “I’m not saying whether any of you have blue hats, but if you do you must leave”, then if there is only one blue-hatted guy, the gnomes will never be able to figure it out. On the other hand, if the gnome overlord says “At least one of you has a blue hat, and if you do you must leave”, the solitary blue gnome will be able to leave.
So what happens with larger numbers? If you have 2 blue hats, the head gnome’s statement that “At least one of you has a blue hat” is redundant with what all the gnomes already know. But since that piece of information affects the 1-gnome case, it affects the gnomes’s logical extrapolation to higher numbers and whether they will be able to leave.
This is what I like so much about this problem: the seemingly paradoxical situation that telling the gnomes something they already know affects the outcome.
Or am I missing something?
So I know that I’m several months late on this, but after reading everyone’s ideas, I feel like after the first meeting every gnome (red hats and blue hats) will leave. If each member places the whole above himself, each gnome would assume that he has a blue hat and leave.
@myfriendlyscience
But that isn’t a logical solution for the gnomes to take, since if everyone leaves, no one will benefit.
@Jasper
Very true, everyone leaving would not benefit the whole.
It’s hard to distinguish which values come above others. On the one hand, each individual might assume that he wears a blue hat, and leave. But if an individual saw a red hat leaving, would he tell him not to leave? Or would he not mention it because hats are taboo?
That would still be a breach of taboo.