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A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
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On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
Don't you have to specify how many white/black hats there are (or some other relation betwee the hats)? Cause this way, there is no correlation between what hat a gnome is wearing and the hats other gnomes wear right?
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Bahamuth November 30 2012 00:14. Posts 70 PM Profile Report Quote #
On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
Don't you have to specify how many white/black hats there are (or some other relation betwee the hats)? Cause this way, there is no correlation between what hat a gnome is wearing and the hats other gnomes wear right?
No correlation is needed between the hats' colours.
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gnomes are smarter than indians. I could see how an infinite number of gnomes could find their hats, but not how only a finite number of gnome answer wrong...
They all answer at the same time: this means their only available information is the hats before them ?
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Hmm, them having to answer at the same time makes it rather impossible, it seems.
There is a variation to this problem where the gnomes can see all the hats except their own (no front or back watching). They're told to form a line, but have all the white and hats next to each other and all the red hats next to each other. They can't say anything at all (but have a strategy). How do they do that?
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If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true.
If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true.
It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
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I think you are right Oshuy. I knew there was some kind of trick involved =)
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On November 30 2012 00:46 Oshuy wrote:
If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true.
If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true.
It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
That's right for the infinite amount of right, but the finite wrong is much trickier one ^^
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A variant of the hat puzzle, from game theory:
A holy man on his pilgrimage stumbles upon a small town where 100 couples live. This village has a unusual cerimony: every night all the men gather around the town square and each publically either praises his wife imensely if he thinks she's been faithful in their marriage or curses her for her infidelity.
While spending his day there, the holy man notices that all 100 wives are cheating on their husbands, and that every man knows about every single infidelity, except his own! Though every man notices that his companions are being cheated by their respective wives, he assumes with absolute certainty that his wife is and always has been faithful and also knows that all other men think the same way about their own respective wives.
In the holy man's 1st night there, before the men begin to praise their wives, he declares: "one woman in this village is being unfaithful" and leaves the village. None of the men look surprised and every single one begins to praise his wife. The same thing happens for another 98 nights.
In the 100th night, however, they all know with certainty that they've been cheated and forevermore curse their wives.
How did this happen, if the holy man said something that everyone already knew (and better yet, knew that everybody knew)?
Assume that each man only thinks over what happened in the night after the cerimony is over (in other words, if he is to realise something from the cerimony, he will only do so after it's over). Also assume everyone is smart in game theory, etc.
Solution:
+ Show Spoiler +There's a mathematical explanation for this, but honestly I don't remember it all to well, so here's a simplified explanation:
Since each man knew about every single infidelity but his own, he assumed that on the 99th night everyone except himself would curse their wives. Since this didn't happen, he knows he was being cheated on.
It's easier to explain with fewer people:
Imagine there are only 2 men in the same situation, and that only one of them are being cheated on (John is being cheated and Mike isn't). Once the holy man says there is one wife cheating on her husband, John will be sure it is his because he doesn't know about any other wife cheating on her husband.
In turn, if both men are being cheated, then John expects Mike to curse his wife on the first night and Mike expects John to do the same. Therefore, once the night is over and neither wife has been cursed, the both realise that there must be another cheating wife, and that it must be their own. Thus, on night 2 they would curse their wifes.
Imagine now with 3 men, all being cheated (John, Mike and Paul). From Paul's point of view, there are only two women cheating and their husbands don't know about it, thus he expects it to play out like it did above: with Mike and John cursing their wives on night 2, since he thinks that both John and Mike only see one cheating wife: each other's. His surprise comes on night 2 when they don't curse their wives, indicating that there is a third cheating wife, which must be his.
This keeps going until you have 100 men, cursing on the 100th night.
The funny thing is that the holy man said something everyone already knew: that there was a cheating wife in their mist (everyone knew about 99 of them in fact)! What changed was that at the end of the chain of expectations is always someone who is expected to think there are no cheating wives. In the 2 man example, John thinks Mike thinks there are no cheating women. In the 3 man example, John thinks Mike thinks Paul thinks there are no cheating women. This is impossible once someone declares there is a single cheating wife.
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On November 30 2012 01:18 Sienionelain wrote:Show nested quote +On November 30 2012 00:46 Oshuy wrote:
If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true.
If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true.
It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent. That's right for the infinite amount of right, but the finite wrong is much trickier one ^^
There are lots of tricks that can be pulled : buying a lot of black paint would also work 
If we put a hat on each gnome by tossing a coin to decide the color, each draw is independant and the probability of white or black for any given gnome are both 1/2, whatever they see. So they need a form of communication.
One strange part of the riddle is that you state the gnomes cannot see the hats behind them. Even if they see those all other hats, it doesn't provide them with any meaningful information if they all answer at the same time.
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On November 30 2012 01:18 Sbrubbles wrote:A variant of the hat puzzle, from game theory: A holy man on his pilgrimage stumbles upon a small town where 100 couples live. This village has a unusual cerimony: every night all the men gather around the town square and each publically either praises his wife imensely if he thinks she's been faithful in their marriage or curses her for her infidelity. While spending his day there, the holy man notices that all 100 wives are cheating on their husbands, and that every man knows about every single infidelity, except his own! Though every man notices that his companions are being cheated by their respective wives, he assumes with absolute certainty that his wife is and always has been faithful and also knows that all other men think the same way about their own respective wives. In the holy man's 1st night there, before the men begin to praise their wives, he declares: "one woman in this village is being unfaithful" and leaves the village. None of the men look surprised and every single one begins to praise his wife. The same thing happens for another 98 nights. In the 100th night, however, they all know with certainty that they've been cheated and forevermore curse their wives. How did this happen, if the holy man said something that everyone already knew (and better yet, knew that everybody knew)? Assume that each man only thinks over what happened in the night after the cerimony is over (in other words, if he is to realise something from the cerimony, he will only do so after it's over). Also assume everyone is smart in game theory, etc. Solution: + Show Spoiler +There's a mathematical explanation for this, but honestly I don't remember it all to well, so here's a simplified explanation:
Since each man knew about every single infidelity but his own, he assumed that on the 99th night everyone except himself would curse their wives. Since this didn't happen, he knows he was being cheated on.
It's easier to explain with fewer people:
Imagine there are only 2 men in the same situation, and that only one of them are being cheated on (John is being cheated and Mike isn't). Once the holy man says there is one wife cheating on her husband, John will be sure it is his because he doesn't know about any other wife cheating on her husband.
In turn, if both men are being cheated, then John expects Mike to curse his wife on the first night and Mike expects John to do the same. Therefore, once the night is over and neither wife has been cursed, the both realise that there must be another cheating wife, and that it must be their own. Thus, on night 2 they would curse their wifes.
Imagine now with 3 men, all being cheated (John, Mike and Paul). From Paul's point of view, there are only two women cheating and their husbands don't know about it, thus he expects it to play out like it did above: with Mike and John cursing their wives on night 2, since he thinks that both John and Mike only see one cheating wife: each other's. His surprise comes on night 2 when they don't curse their wives, indicating that there is a third cheating wife, which must be his.
This keeps going until you have 100 men, cursing on the 100th night.
The funny thing is that the holy man said something everyone already knew: that there was a cheating wife in their mist (everyone knew about 99 of them in fact)! What changed was that at the end of the chain of expectations is always someone who is expected to think there are no cheating wives. In the 2 man example, John thinks Mike thinks there are no cheating women. In the 3 man example, John thinks Mike thinks Paul thinks there are no cheating women. This is impossible once someone declares there is a single cheating wife.
No offense, but I'm familiar with the actual problem and you seriously misdescribed both it and the mechanism of its solution to the point of it making no sense. Shame given how cool the original is.
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On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution:
+ Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
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On November 30 2012 00:46 Oshuy wrote:
If there is a finite number of whites ahead, either all answer white to get an infinite false or all answer black to get infinite true.
If there is a finite number of blacks ahead, either all answer black to get an infinite false or all answer white to get infinite true.
It both are infinite, all answer white by convention to get both an infinite true and an infinite false.
Err... There must be something I'm not getting here.
Edit: ok - was finite false. No solution then appart from all gnomes staying silent.
Infinite correct:
+ Show Spoiler +
Also I would add since the problem doesn't specify, if there is a finite number of both in front of you (ie the line extends infinitely behind you but not in front) it's impossible to guarantee infinite correct. Stated simply though, guess a color you see an infinite amount of in front of you and there will be infinite correct.
Although I also have a question which has to do with the theory of random numbers, which I dont' really understand. If each gnome is able to produce an actual unpredictable random guess wouldn't they get infinite correct even in the line only stretches infinitely behind you and not in front of you case? There would be an infinitely small chance that the hats were assigned in a direct contrary pattern to what they end up guessing?
Finite wrong:
This seems impossible to guarantee in almost all cases. Unless the line only extends infinitely to the front and there's a finite number of one of the colors. I want to hear the solution to this.
Edit: reading above.
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On November 30 2012 02:07 frogrubdown wrote:Show nested quote +On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution: + Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial.
+ Show Spoiler +
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
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On November 30 2012 02:36 CptZouglou wrote:Show nested quote +On November 30 2012 02:07 frogrubdown wrote:On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution: + Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial. + Show Spoiler +
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
+ Show Spoiler + Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with whatever equivalence class they happen to find themselves in. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
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On November 30 2012 02:45 frogrubdown wrote:Show nested quote +On November 30 2012 02:36 CptZouglou wrote:On November 30 2012 02:07 frogrubdown wrote:On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution: + Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial. + Show Spoiler +
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
+ Show Spoiler + Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large.
+ Show Spoiler +Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
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On November 30 2012 02:49 CptZouglou wrote:Show nested quote +On November 30 2012 02:45 frogrubdown wrote:On November 30 2012 02:36 CptZouglou wrote:On November 30 2012 02:07 frogrubdown wrote:On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution: + Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial. + Show Spoiler +
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
+ Show Spoiler + Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large. + Show Spoiler +Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
+ Show Spoiler +Impressively smart , though to be fair, infinite sight was explicit in the question.
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On November 30 2012 02:49 CptZouglou wrote:Show nested quote +On November 30 2012 02:45 frogrubdown wrote:On November 30 2012 02:36 CptZouglou wrote:On November 30 2012 02:07 frogrubdown wrote:On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution: + Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial. + Show Spoiler +
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
+ Show Spoiler + Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large. + Show Spoiler +Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart!
They must have generations and generations of riddle experience.
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On November 30 2012 02:56 AmericanUmlaut wrote:Show nested quote +On November 30 2012 02:49 CptZouglou wrote:On November 30 2012 02:45 frogrubdown wrote:On November 30 2012 02:36 CptZouglou wrote:On November 30 2012 02:07 frogrubdown wrote:On November 30 2012 00:05 Sienionelain wrote:
A similar to the indian one:
An infinite amount of gnomes are set to a line and after that either a black or a white hat is put on top of them. Gnomes must guess the color of their hat. Every gnome sees the hats which are before him, but not his own hat or the hats of the gnomes behind him.
What is the strategy with which an infinite amount of gnomes answer right?
What is the strategy with which a finite number of gnomes answer wrong?
Rules are almost similar to the indian riddle:
-no communicating in the line (they can discuss the strategy beforehand)
-every gnome answers at the same time !
-gnomes are smart
solution: + Show Spoiler +Define equivalence classes of infinte sequence of black white patterns based on the equivalence relation of only differing in finitely many points in the sequence (it's trivial to show this relation is an equivalence one, might give a proof when not using iphone). Every gnome can see which equivalence class the set of all gnomes is in when they look at the remaining gnomes ahead of them. For each equivalence class, they agree ahead of time to all say the given color for their position from some single member of that class. This guarantees only finite deaths, because the actual sequence cannot differ from their selcted one by more than finitely many places.
Edit: by the way, the reason the equivalence relation thing is important is because equivalnce relations partition sets. The proof of this is also trivial. + Show Spoiler +
So you mean that they prepared an infinite number of classes of infinite sequences of possible patterns, and for each one they have one reference sequence ? So one gnome could actually answer for all the others no ? I don't really get this answer maybe...
+ Show Spoiler + Each gnome knows what all the others will say, so in that sense could answer for them. Of course, the puzzle states that they must each answer with their own hat color, which they will do by selecting the color of their position in the sequence (from that equivalence class) that they agreed to all associate with that equivalence class. This will result in a finite number of deaths, though you can make that finite number arbitrarily large. + Show Spoiler +Ok. That would require an infinite amount of preparation, capacity to memorize infinite number of infinite sequences and infinite sight (to see all the gnomes in front) =) Gnomes are indeed very smart! They must have generations and generations of riddle experience.
If gnomes reproduction follows any known pattern, they have been reproducing for quite some time to reach infinity indeed.
Misleading data for spoilered answer is the definition of "smart". Usually would mean any average gnome can do before answering any task humanly feasible in a finite amount of time. Here they need "a little" more.
I don't like the choice part though.
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There are 3 people who were taken hostage. Their captors bury them each into the ground with only their head sticking out, in a line (column), so that all of them were facing in the same direction (They can't turn their head to look back). The captors then took out 4 bandanas, 2 black, 2 white and said to the hostages that they would put on one of the bandanas onto each hostage's head. If the hostage guesses what colour of bandana is on their head, they will be freed and the rest will be killed. Which hostage got out?
Hint:
+ Show Spoiler +Front person can see nothing, 2nd person can see first person, 3rd person can see everyone.
Solution:
+ Show Spoiler +The 2nd person. He can see the first person and since the 3rd person hasn't called out yet, he knows that he must have a different bandana than the first person.
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EPT
