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On December 01 2012 00:08 CptZouglou 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?
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 ?
Yes, but they can't hear the other's guess so it doesn't really matter. Let's say the cells are really far apart, in different buildings.
On December 01 2012 00:59 Oshuy wrote:
With 3 pirates and seeing the 2 others, its trivial (identical to gnomes bellow), still have trouble with only 2.
Yeah, the way it's worded is intentionally a bit misleading but it's actually quite different from the hat problems.
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
Yup! I used gnomes just because it's a hat problem :p
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On December 01 2012 05:45 starfries wrote:Show nested quote +On December 01 2012 00:08 CptZouglou wrote: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?
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 ? Yes, but they can't hear the other's guess so it doesn't really matter. Let's say the cells are really far apart, in different buildings. Show nested quote +On December 01 2012 00:59 Oshuy wrote:
With 3 pirates and seeing the 2 others, its trivial (identical to gnomes bellow), still have trouble with only 2.
Yeah, the way it's worded is intentionally a bit misleading but it's actually quite different from the hat problems. Show nested quote +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 Yup! I used gnomes just because it's a hat problem :p
The pirates in the cell is the exact same as the 7 different color hats with 7 gnomes. This is just the case of 2 instead of 7. + Show Spoiler +And with 2 seeing your coin is the exact same amount of information as seeing all the other individuals' results except yours. Also you can word the solution differently with 2 without needing to invoke mod. One pirate always guesses what's on his coin, the other always guesses the opposite of his coin. They last indefinitely.
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On April 18 2012 22:58 Captain Mayhem wrote:Show nested quote +On April 18 2012 22:47 yeaR wrote:
If Pinochio said my nose will grow right now, what would happen ? Short solution: Pinocchio is God. Long solution: + Show Spoiler +
Which leads us to another riddle: if pinochio by virtue of said argument is omnipotent, what happens if he says "either my nose will grow, or my nose growing mechanism never existed!"?
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I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
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On December 11 2012 12:49 ziggurat wrote:
I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?"
Got it.
+ Show Spoiler +4. There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4. edit: Not sure how clear the above reasoning was. Maybe this will be better. + Show Spoiler +Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker... that seems less clear if anything.
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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?
I feel as if I am on the right track.
+ Show Spoiler +
Weigh 3 balls vs. 3 balls. If they're uneven, then the odd one is in the set of 6 you weighed. If they're even, then the odd one is in the set of 6 you didn't weigh.
Then weigh 3 of the balls in the set of normal ones vs. 3 in the set you know the odd one is in. If they're uneven, you know the odd one is in one of the 3 you weighed. If they're even, then they're in the 3 you didn't weigh.
So now you have 3 balls you know the odd one is in. 3 is the lowest amount I've managed to reduce the chances to in 2 weighs.
Just to be clear, the solution doesn't involve chopping balls in half, does it?
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On December 11 2012 16:23 dabom88 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? I feel as if I am on the right track. + Show Spoiler +
Weigh 3 balls vs. 3 balls. If they're uneven, then the odd one is in the set of 6 you weighed. If they're even, then the odd one is in the set of 6 you didn't weigh.
Then weigh 3 of the balls in the set of normal ones vs. 3 in the set you know the odd one is in. If they're uneven, you know the odd one is in one of the 3 you weighed. If they're even, then they're in the 3 you didn't weigh.
So now you have 3 balls you know the odd one is in. 3 is the lowest amount I've managed to reduce the chances to in 2 weighs.
Just to be clear, the solution doesn't involve chopping balls in half, does it?
hint
+ Show Spoiler +
must keep track of individual balls
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On December 11 2012 13:06 frogrubdown wrote:Show nested quote +On December 11 2012 12:49 ziggurat wrote:
I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?" Got it. + Show Spoiler +4. There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4. edit: Not sure how clear the above reasoning was. Maybe this will be better. + Show Spoiler +Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker... that seems less clear if anything.
I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is
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On November 30 2012 00:41 aseq wrote:
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?
+ Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
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On December 11 2012 21:39 eluv wrote:Show nested quote +On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
dunno if this is right lol.
+ Show Spoiler +
if you swim in a zigzag line away from the wolf's starting position, wouldnt the wolf get stuck oscillating back and fourth across his initial position?
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On December 11 2012 21:39 eluv wrote:Show nested quote +On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
+ Show Spoiler +Assume the lake is 100m across. Its circumference is thus (via C/100m = PI) about 314m, so the wolf can reach any point from any other point in at most 157m. Your goal is thus to get within 39m of the shore while the wolf is exactly opposite your current position.
Swim toward the wolf until you get 39m from the shore. Now swim in a circle such that you stay 39m from the shore at all times. Your path is describing a circle that is 22m in diameter and thus about 69m in circumference, which means you're swimming about 14% faster than the wolf in terms of degrees. Swim until you're exactly opposite the wolf, then swim to shore.
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On December 11 2012 20:49 Neino wrote:Show nested quote +On December 11 2012 13:06 frogrubdown wrote:On December 11 2012 12:49 ziggurat wrote:
I saw this many years ago, hopefully you find it interesting....
"Last night I went to a dinner party with my wife and four other couples (total of 10 people present). As people arrived, handshakes were exchanged. However, not everyone shook hands. Some did and some didn't. No one shook hands with his or her spouse, and no one shook hands with him or her self. After the party I asked all the others how many hands they shook, and to my surprise each person told me a different number.
The puzzle to solve is: how many hands did my wife shake?" Got it. + Show Spoiler +4. There are 9 possible numbers of hands shaken (0-8) for 9 possible shakers (10 minus you). The person that shakes hands with 8 must be married to the person that shakes hands with 0 because everyone outside of this couple in the group is shaken at least once (by the 8-shaker). Similarly, the 7-shaker must be married to the 1-shaker, since, from the prior sentence, we know that the 1-shaker was shaken only by the 8-shaker, and the 7-shaker must shake hands with everyone else except the 0-shaker and 1-shaker to get to 7. Similar reasoning applies to 6-shaker/2-shaker and 5-shaker/3-shaker. Only possibility left is 4. edit: Not sure how clear the above reasoning was. Maybe this will be better. + Show Spoiler +Every shaker position in {0,...,8} is occupied. The 8-shaker must shake hands with everyone not in spouse-group to reach 8. So everyone outside it has at least one, meaning the spouse is the only candidate for 0-shaker. The 7-shaker must shake hands with everyone not in spouse-group or 0-shaker, meaning everyone not in either of these spouse groups is at least a 2-shaker. So the 7-shaker spouse is the only candidate for 1-shaker. The 6-shaker must shake hands with everyone not in spouse-group or 0-shaker or 1-shaker (we already know the 1 person that shakes hands with this person), meaning everyone not in the first 3 spouse-groups shakes hands with at least 3 people. So the 6-shaker spouse is the only candidate for 2-shaker... that seems less clear if anything. I feel like I'm missing something here, but I just woke up so I got an excuse. How do you know the person telling the riddle isn't the person who shook hands with 8 people, and his wife shook hands with 0? I just missed how you determined which couple the story teller is
Due to the line that says he asked and everyone gave a different response.
The groups of couples go
8 0
7 1
6 2
5 3
4 4
As you can see, there are two people who shook 4 peoples hands. As the question asker said that everyone gave different answers, the duplicate must be between him and his wife as he didnt ask himself.
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On December 11 2012 21:39 eluv wrote:Show nested quote +On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
+ Show Spoiler +Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
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On December 12 2012 01:37 helvete wrote:Show nested quote +On December 11 2012 21:39 eluv wrote:On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? + Show Spoiler +Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you.
This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center.
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On December 11 2012 22:33 gameguard wrote:Show nested quote +On December 11 2012 21:39 eluv wrote:On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? dunno if this is right lol. + Show Spoiler +
if you swim in a zigzag line away from the wolf's starting position, wouldnt the wolf get stuck oscillating back and fourth across his initial position?
Once you start moving, the wolf will start moving toward the edge closest to you. If you zig toward any point in the circle ahead of the wolf's position, he'll just keep running that direction. If you zig toward any point in the circle behind his position, he'll turn around when you get closer to that point than to the point you initially swam toward. Every zig and zag will end with the wolf closer to your current position, unless you swim back toward the center, in which case you've gained nothing.
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On December 11 2012 21:39 eluv wrote:Show nested quote +On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten?
+ Show Spoiler +start swimming in the opposite direction of the wolfs starting position. swim slightly circular in the same direction like the wolf. You can have a positional advantage until you are at 0.25*r at which point your relative speed will be the same as the wolves since your circumference is 1/4th of his. Make sure to keep the wolf at the opposite side until you reach 0.25 r. Now make a straight line to the shore.
If r = 1 and human speed = 1m/s it will take the human 0.75s to reach the shore. The wolf needs (pi*r)/(4m/s) which is 0.78s.
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You have a circle with radius 4R. You have unlimited semicircles with radius R. How many semicircles maximum can you place inside the big circle with no overlap? PM for an answer.
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On December 11 2012 20:10 gameguard wrote:Show nested quote +On December 11 2012 16:23 dabom88 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? I feel as if I am on the right track. + Show Spoiler +
Weigh 3 balls vs. 3 balls. If they're uneven, then the odd one is in the set of 6 you weighed. If they're even, then the odd one is in the set of 6 you didn't weigh.
Then weigh 3 of the balls in the set of normal ones vs. 3 in the set you know the odd one is in. If they're uneven, you know the odd one is in one of the 3 you weighed. If they're even, then they're in the 3 you didn't weigh.
So now you have 3 balls you know the odd one is in. 3 is the lowest amount I've managed to reduce the chances to in 2 weighs.
Just to be clear, the solution doesn't involve chopping balls in half, does it? hint + Show Spoiler +
must keep track of individual balls
I remember working this out YEARS ago, in about 2003 or so during high school (ha ha...) Never confirmed if this is true or not. This was a pain to type up in text form; a flowchart or binomial/trinomial tree is much easier!
+ Show Spoiler +
Balls are numbered 1-12. First, weigh them 1,2,3,4 // 5,6,7,8
1-a) If original weighing is balanced, weigh 1,9 // 10,11 (I.e.3 new balls, 1 good one.) => We know all 8 in first weighing are good.
1-a-i) If [1-a] balanced, weigh 1 // 12. We know 12 is bad, and weighing it against a good one to see if it's lighter or heavier.
1-a-ii) If [1-a] tips heavier left, weigh 10 // 11. => We know either 9 is heavier, OR 10 or 11 are lighter. Weighing 10 vs 11 lets us know which is lighter, or if they’re the same, then 9 is heavier. (Same logic if weighing tipped other way; flip heavier/lighter in the sentence.)
2-a) If original weighing is not balanced (let’s say heavier left), weigh 1,5,9 // 6,7,2. (I.e. remove 2 from one side, remove 1 from the other side and replace with a good ball, and swap the sides of 2 balls.)
2-a-i) If [2-a] balanced, weigh 3 // 4. => One of the 3 removed balls (3,4,8) is bad.
(2-a-i-1) If [2-a-i] balanced, we know 8, the other ball removed, is the heavier/lighter ball (based on which way original weighing tipped).
(2-a-i-2) if [2-a-i] didn’t balance, we know 3 or 4 are bad. You know original weighing tipped left, meaning it was heavier, so whichever side remains heavier is the heavier ball out. (Same logic if original weighing tipped other way, except you know 3 or 4 are lighter.)
2-a-ii) if [2-a] tips the other way from 1st weighing (in this case, heavier right) weigh 5 // 12. We know one of the swapped balls (2 or 5) are bad, so you weigh one of them against a good ball. If it doesn’t balance, we know it’s bad and if it’s heavier or lighter; if it balances, then you know the other ball is bad, and either heavier or lighter.
2-a-iii) if [2-a] tips same as 1st weighing (in this case, left), weigh 6 // 7. We know either 1, 6, or 7 are bad. If 6 and 7 balance, then 1 is bad and we know which way as well. If 6 vs 7 don't balance, we know if it's heavier or lighter from previous weighing since they were on the same side, so we know the ball that tips the same way is is bad.
TL;DR: Split 4 vs 4. Following this, the harder scenario is when they don’t balance. For the 2nd weighing remove a pair, swap a pair, and replace one ball with a good ball. From there you can always determine the bad ball, and if it’s heavier or lighter, with a 3rd weighing.
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On December 12 2012 01:47 AmericanUmlaut wrote:Show nested quote +On December 12 2012 01:37 helvete wrote:On December 11 2012 21:39 eluv wrote:On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? + Show Spoiler +Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you. This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center.
I don't think you got my solution. It has nothing to do with the center. + Show Spoiler +Swim to anywhere on the edge of the lake. The wolf will get there before you and will be waiting. Now SLOWLY swim towards the other side. As the wolf runs at full speed and you swim slowly he'll get there while you've only covered a tiny distance. You now reverse and swim at full speed. The wolf won't be able to cover the distance in the time it takes you to once again reach the shore since (wolf speed/slow swim speed) > (wolf speed /max swim speed) and the distances are the same.
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On December 12 2012 03:04 helvete wrote:Show nested quote +On December 12 2012 01:47 AmericanUmlaut wrote:On December 12 2012 01:37 helvete wrote:On December 11 2012 21:39 eluv wrote:On November 30 2012 00:41 aseq wrote:
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? + Show Spoiler +
Line up "at-random", just so you have some order. Then first in line starts a new line. Each person in the old line then enters the new line as follows. If the new line consists entirely of 1 color, enter it at one end. If the new line has both color hats already, enter at the "split" between the two.
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? + Show Spoiler +Swim close to the edge of the lake. The wolf will be waiting for you. Swim very slowly towards the opposite side of the lake. When the wolf arrives at the point you're slowly swimming towards, do a 180 and swim full speed to the shore. The wolf won't get there in time to eat you. This doesn't work. The wolf has to run PI*(the lake's radius), to get from one side of the lake to the other. PI is less than 4, so he can always get from any point on the edge of the lake to any other point faster than you can swim there from the center. I don't think you got my solution. It has nothing to do with the center. + Show Spoiler +Swim to anywhere on the edge of the lake. The wolf will get there before you and will be waiting. Now SLOWLY swim towards the other side. As the wolf runs at full speed and you swim slowly he'll get there while you've only covered a tiny distance. You now reverse and swim at full speed. The wolf won't be able to cover the distance in the time it takes you to once again reach the shore.
+ Show Spoiler +"He will always run towards the point on the shore closest to your current position"
The wolf doesn't run to your destination, as your method requires. He runs to the point on the shore closest to your current position. If you're swimming directly away from the wolf, however slowly you do so, that point will remain in the same place until you pass the center of the circle.
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EPT![[image loading]](http://ompldr.org/vN2JkOA/Pinocchio-paradox-solved.jpg)
