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On November 30 2012 03:46 betaflame wrote: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.
We already had this one with gnomes !
Here's another then, almost like the indians one:
7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always
- The gnomes can discuss strategy before getting their hats
- The gnomes cannot communicate after
- Gnomes are smart
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Sanya12364 Posts
On November 30 2012 03:46 betaflame wrote: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.
What's the downside to guessing wrong?
The way I see it, the first person guesses immediately on receiving a bandana. As the person with no information and a flat 50% chance, the first person must take that chance before the other two hostages use their information advantage to reach a conclusion.
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On November 30 2012 04:05 CptZouglou wrote:Show nested quote +On November 30 2012 03:46 betaflame wrote: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. We already had this one with gnomes ! Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ? Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart
+ Show Spoiler +
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
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On November 30 2012 04:21 Geiko wrote:Show nested quote +On November 30 2012 04:05 CptZouglou wrote:On November 30 2012 03:46 betaflame wrote: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. We already had this one with gnomes ! Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ? Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart + Show Spoiler +
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer!
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I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written.
You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
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On November 30 2012 04:29 CptZouglou wrote:Show nested quote +On November 30 2012 04:21 Geiko wrote:On November 30 2012 04:05 CptZouglou wrote:On November 30 2012 03:46 betaflame wrote: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. We already had this one with gnomes ! Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ? Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart + Show Spoiler +
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer!
I understand the answer, but I don't understand how it saves the gnomes.
+ Show Spoiler +
If the gnomes already know which seven colors are present, then each gnome can deduce the answer trivially; they can see the other six colors, so they have the color they don't see. If they just know that some arbitrary seven colors are used, then they have no way of assigning the available colors to numbers in order for Geiko's solution to be applied.
What am I missing?
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On November 30 2012 05:05 AmericanUmlaut wrote:Show nested quote +On November 30 2012 04:29 CptZouglou wrote:On November 30 2012 04:21 Geiko wrote:On November 30 2012 04:05 CptZouglou wrote:On November 30 2012 03:46 betaflame wrote: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. We already had this one with gnomes ! Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ? Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart + Show Spoiler +
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer! I understand the answer, but I don't understand how it saves the gnomes. + Show Spoiler +
If the gnomes already know which seven colors are present, then each gnome can deduce the answer trivially; they can see the other six colors, so they have the color they don't see. If they just know that some arbitrary seven colors are used, then they have no way of assigning the available colors to numbers in order for Geiko's solution to be applied.
What am I missing?
The colors could be anything (there could be 3 red, 4 blue for instance), they have no information on the color distribution. But to save all the gnomes, only one good answer is required.
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On November 30 2012 05:14 CptZouglou wrote:Show nested quote +On November 30 2012 05:05 AmericanUmlaut wrote:On November 30 2012 04:29 CptZouglou wrote:On November 30 2012 04:21 Geiko wrote:On November 30 2012 04:05 CptZouglou wrote:On November 30 2012 03:46 betaflame wrote: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. We already had this one with gnomes ! Here's another then, almost like the indians one: 7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ? Some info as always - The gnomes can discuss strategy before getting their hats - The gnomes cannot communicate after - Gnomes are smart + Show Spoiler +
Gnomes assign numbers from 0 to 6 to colors. Each gnome is assigned a different number from 0 to 6. Gnomes guess their hat color by assuming that the sum of all hats is equal to their number (mod 7) so at least one of them gets it right.
Nice and clean answer! I understand the answer, but I don't understand how it saves the gnomes. + Show Spoiler +
If the gnomes already know which seven colors are present, then each gnome can deduce the answer trivially; they can see the other six colors, so they have the color they don't see. If they just know that some arbitrary seven colors are used, then they have no way of assigning the available colors to numbers in order for Geiko's solution to be applied.
What am I missing?
The colors could be anything (there could be 3 red, 4 blue for instance), they have no information on the color distribution. But to save all the gnomes, only one good answer is required.
So the gnomes meet, and they decide on the following codes:
0 - red
1 - blue
2 - green
3 - yellow
4 - black
5 - white
Now I'm a gnome, and I look around at my six buddies. I see two orange hats, one pink, one grey, and two purple. What do I answer?
Edit: Wait, I get it. The "out of seven different colors" in the question is meant to imply that the gnomes know the set of available colors. I withdraw my objection.
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On November 30 2012 04:05 CptZouglou wrote:
Here's another then, almost like the indians one:
7 gnomes, each one will be given a magic, colored hat (any color out of 7 different colors) . The gnomes can see all the other hats, but not theirs (gnomes are very smart, but god they have trouble taking off their hats). All the gnomes secretly write down a color on a paper, and if at least one found out his hat color, all the gnomes are freed. Is there a way to ensure that all gnomes are saved ?
Some info as always
- The gnomes can discuss strategy before getting their hats
- The gnomes cannot communicate after
- Gnomes are smart
The Gnomes could just decide to write the color of one specific gnomes hat, before they get their hats. So they would have 6 times the color of his hat writen down and he just says that color.
-> one gnome knows his hatcolor
And if every gnome needs to know the color of his hat to be freed they could number themselfs 1-7 and decide that gnome 1 writes the color of gnome 2's hat (2->3, 3->4, ...). Therefore they would have every hatcolor writen down and every gnome just needs to sort out the colors that he sees and the one that is left is the color of his hat.
-> every gnome knows his hatcolor
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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 +Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase)
Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
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On November 30 2012 04:39 CptZouglou wrote:I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written. Show nested quote +You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?
+ Show Spoiler + This can only be solved if you know the odd ball is lighter or heavier.
Lets say its heavier.
1st weigh: weigh 4 balls against 4 balls. If one set of 4 balls is heavier, the heavy ball is in that group. If scales are equal, heavy ball is in the leftover group of 4.
2nd weigh: From the 4 ball group which includes the heavy ball, weigh 1 ball against 1 ball. If 1 ball is heavier, its that ball. Game over. If scales are equal, the heavy ball is one of the two remaining balls.
3rd weigh: weigh the two remaining balls to find the heavier one. Game over.
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On November 30 2012 18:52 rangi wrote:Show nested quote +On November 30 2012 04:39 CptZouglou wrote:I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written. You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it? + Show Spoiler + This can only be solved if you know the odd ball is lighter or heavier.
Lets say its heavier.
1st weigh: weigh 4 balls against 4 balls. If one set of 4 balls is heavier, the heavy ball is in that group. If scales are equal, heavy ball is in the leftover group of 4.
2nd weigh: From the 4 ball group which includes the heavy ball, weigh 1 ball against 1 ball. If 1 ball is heavier, its that ball. Game over. If scales are equal, the heavy ball is one of the two remaining balls.
3rd weigh: weigh the two remaining balls to find the heavier one. Game over.
No ! There's a way to find the odd ball and determine if it's lighter or heavier, but it's not trivial !
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On November 30 2012 17:33 starfries 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 +Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
+ Show Spoiler +Solution supposes you work in N and not Z (gnome line has a start), therefore there is always a finite number of gnomes behind the one answering.
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase)
Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average?
edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
If both answer at the same time, basically 1/4 of both wrong at each try and an average survival would be of 3 days ? Once again if there is no communication it's answering on random independant events, which doesn't leave much room for strategy ...
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On November 30 2012 19:37 Oshuy wrote:Show nested quote +On November 30 2012 17:33 starfries 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 +Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
+ Show Spoiler +Solution supposes you work in N and not Z (gnome line has a start), therefore there is always a finite number of gnomes behind the one answering.
Oh, that explains it. I wonder if it's unsolvable in Z then?
If both answer at the same time, basically 1/4 of both wrong at each try and an average survival would be of 3 days ? Once again if there is no communication it's answering on random independant events, which doesn't leave much room for strategy ...
There is a way to do better than chance ^.^
Actually, while I was looking for where I first saw this problem I found another nice one (credit goes to mindyourdecisions.com). Hopefully it hasn't been posted, there are a lot of hat riddles here...
Three gnomes are in a room and a red or blue hat (with equal probability) is placed on each one's head. They must simultaneously either guess the colour of their hat or pass. There is only one round. The gnomes win if at least one gnome guesses correctly and no one guesses incorrectly. What's the best strategy?
standard stuff about gnomes being smart and having planning time applies
Hint:
+ Show Spoiler +AFAIK the best you can do is a 75% chance of winning. It gets veeery interesting with larger numbers of players.
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On November 30 2012 17:33 starfries 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 +Is being able to see the hats in front enough to determine which equivalence class the set of all gnomes belongs to? Couldn't you have two sets which don't differ in front of one gnome but differ at an infinite number of points behind and are therefore in different classes?
A relatively easy puzzle (I don't remember the exact wording so I'll paraphrase) Two pirates are captured and kept in separate cells with no communication between them. Every day, two coins are flipped - one in front of each pirate, where he can see it land but not influence it. Each pirate then states their guess for the other pirate's coin. If they're both wrong, they're put to death. How long will they last on average? edit: they had a chance to discuss strategy before they were captured just in case something like this happened, and both pirates are smart
Is the guess simultaneous ?
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And now for something completely different
+ Show Spoiler + haha took me ten minutes I think. I counted the number of non-prime factors of each number in each series, and added that up., In the case where there were no non-prime factors and a zero, increased the count by 1 for each zero (There were two numbers). Still worked
^^
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Show nested quote +
If both answer at the same time, basically 1/4 of both wrong at each try and an average survival would be of 3 days ? Once again if there is no communication it's answering on random independant events, which doesn't leave much room for strategy ...
There is a way to do better than chance ^.^
With 3 pirates and seeing the 2 others, its trivial (identical to gnomes bellow), still have trouble with only 2.
Three gnomes are in a room and a red or blue hat (with equal probability) is placed on each one's head. They must simultaneously either guess the colour of their hat or pass. There is only one round. The gnomes win if at least one gnome guesses correctly and no one guesses incorrectly. What's the best strategy? standard stuff about gnomes being smart and having planning time applies Hint: + Show Spoiler +AFAIK the best you can do is a 75% chance of winning. It gets veeery interesting with larger numbers of players.
Haven't seen the gnome version. A few similar ones generated a bit of traffic a while back ^^
+ Show Spoiler +Name opposite color if I see 2 hats of the same color, pass if 2 different
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On December 01 2012 00:51 Arachne wrote:And now for something completely different + Show Spoiler + haha took me ten minutes I think. I counted the number of non-prime factors of each number in each series, and added that up., In the case where there were no non-prime factors and a zero, increased the count by 1 for each zero (There were two numbers). Still worked ^^
I had to use the hint to solve it. XD
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On November 30 2012 19:12 CptZouglou wrote:Show nested quote +On November 30 2012 18:52 rangi wrote:On November 30 2012 04:39 CptZouglou wrote:I don't know if this one has been posted already, but I like it. Instead of writing it myself, I found it on the internet so that it's better written. You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).
You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).
You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it? + Show Spoiler + This can only be solved if you know the odd ball is lighter or heavier.
Lets say its heavier.
1st weigh: weigh 4 balls against 4 balls. If one set of 4 balls is heavier, the heavy ball is in that group. If scales are equal, heavy ball is in the leftover group of 4.
2nd weigh: From the 4 ball group which includes the heavy ball, weigh 1 ball against 1 ball. If 1 ball is heavier, its that ball. Game over. If scales are equal, the heavy ball is one of the two remaining balls.
3rd weigh: weigh the two remaining balls to find the heavier one. Game over. No ! There's a way to find the odd ball and determine if it's lighter or heavier, but it's not trivial !
Solution:
+ Show Spoiler +Split the twelve balls into three piles of four. You weigh two of the three piles with each other. Now their are two different progressions depending on the result.
If they're equal you weigh three of the now known balls with three of the unknown ones. If the scales are equal then the ball is the last remaining and you need to weigh it with a known ball in order to determine if it's heavier or lighter. If the scales are unequal then you swap a ball with one from the other side and then remove a different ball from each side, you then weigh the scales one last time, if a different side is now heavier than the odd is the unknown ball you swapped, if the sides are now equal than it's the removed ball, otherwise it's the unknown ball you didn't do anything to. In order to determine whether it was lighter or heavier than the rest of the balls you simply need to check which side of the uneven scales it was on.
If they're unequal you swap three of the balls with the other side (the third by adding one of the balls you know isn't odd) and remove three balls (the third is once again substituted with a known ball). Then you weigh the two sides again, if a different side is now heavier than the odd ball is one of the balls you swapped, if the sides are now equal than it's among the removed balls, otherwise it's one of the two remaining unknown balls. After that you replace one of the balls and swap it with one of the others and weigh it a final time. If the scales are equal the odd ball was the replaced one, if a different side is now heavier the odd ball was the swapped one. If nothing happened with the weight distribution the odd ball was the one remaining unknown ball you didn't do anything to. In order to determine whether it was lighter or heavier than the rest of the balls you simply need to check which side of the uneven scales it was on.
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On December 01 2012 00:51 Arachne wrote:And now for something completely different + Show Spoiler + haha took me ten minutes I think. I counted the number of non-prime factors of each number in each series, and added that up., In the case where there were no non-prime factors and a zero, increased the count by 1 for each zero (There were two numbers). Still worked ^^
That is a funny coincidence, that + Show Spoiler +the number of non-prime factors for every digit is equal to the number of enclosed spaces they have, other than zero which is a special case anyway cause it doesn't have factors. Well in writing 2's and 4's sometimes are written differently though.
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EPT![[image loading]](http://i.imgur.com/LYebe.png)
