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Germany149 Posts
On April 19 2012 10:51 FeUerFlieGe wrote:Show nested quote +On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
- The only possible eyecolours are green and blue.
+ Show Spoiler +I would say this one is easily solved by simply scaling down the numbers. 200 people is a lot for the mind to wrap it's head around so lets just say there are 2 people on the island, one blue eyed one green eyed. If the green eyed guy sees the other guy has blue eyes then he will leave on the first day since he knows at least one of them has green eyes.
Now how about 2 green eyed people and 2 blue eyed people? The first day each green eyed person will see 2 blue eyed people and 1 green eyed person. They know that if they don't have green eyes then the other green eyed person will leave on the first night, because he would not see any other green eyes. So on the second night when it is confirmed that the other didn't leave they will both leave. Then you just work your way up until on the 100th day all 100 of the leave.
I think you don't even have to consider the blue eyed people, as the riddle will work without them.
+ Show Spoiler +Since logic tells us that you can only get off the island eventually if you're green eyed, shouldn't everyone just state their eye color is green on the first day? all of the green eyed will be saved, and the others are screwed either way
edit: does everyone know "not green" means blue? in that case i retract my statement
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On April 19 2012 11:43 Alvin853 wrote:Show nested quote +On April 19 2012 10:51 FeUerFlieGe wrote:On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
- The only possible eyecolours are green and blue.
+ Show Spoiler +I would say this one is easily solved by simply scaling down the numbers. 200 people is a lot for the mind to wrap it's head around so lets just say there are 2 people on the island, one blue eyed one green eyed. If the green eyed guy sees the other guy has blue eyes then he will leave on the first day since he knows at least one of them has green eyes.
Now how about 2 green eyed people and 2 blue eyed people? The first day each green eyed person will see 2 blue eyed people and 1 green eyed person. They know that if they don't have green eyes then the other green eyed person will leave on the first night, because he would not see any other green eyes. So on the second night when it is confirmed that the other didn't leave they will both leave. Then you just work your way up until on the 100th day all 100 of the leave.
I think you don't even have to consider the blue eyed people, as the riddle will work without them. + Show Spoiler +Since logic tells us that you can only get off the island eventually if you're green eyed, shouldn't everyone just state their eye color is green on the first day? all of the green eyed will be saved, and the others are screwed either way
edit: does everyone know "not green" means blue? in that case i retract my statement
Once all of the blue eyes leave, the green eyes will know they have green eyes and will leave as well.
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There is a missing part of the question I think. Why not just everyone say "i have green eyes" on the first day and "I have blue eyes the second day" and everyone is off after two days.
Also, do they KNOW that it's 100 and 100? it doesn't state that in the problem, which has massive implications.
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On April 19 2012 14:17 BluePanther wrote:
There is a missing part of the question I think. Why not just everyone say "i have green eyes" on the first day and "I have blue eyes the second day" and everyone is off after two days.
Also, do they KNOW that it's 100 and 100? it doesn't state that in the problem, which has massive implications.
I think its because you have to know for certain, if you guess green, even if you have green eyes you do not know for certain you have green eyes so they wont let you go. Just because you don't get off the first night doesn't mean you have the other color, it just means you weren't sure and can still have green or blue eyes.
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There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
- The only possible eyecolours are green and blue.
Dont know if this is right.
+ Show Spoiler +You just count how many green eyes there are and how many blue eyes there are, your eye color is the one that has 99 people in it.
or
You get 3 people, one person with blue eyes and one with green and a random one. You stare at the guy until he puts you with a partner.
or
You split them into two halves, blue and green eyes, if they dont trus you just wink with your eye, that is a sign of trust so they should let you split them, then you ask a friend and a person you trust to put you on one half.
or
You go to the captain and see if you can go if he says yes then you have green eyes if he says no you have blue eyes.
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My solution:
+ Show Spoiler +People will have to get a lot of babies and see their eye colours. EDIT: eye colour schematic picBut it looks like there's still some uncertainty, so it's theoretically possible to mate with blue+blue and get 10 children with green eyes :D not very likely, but possible
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On April 19 2012 14:17 BluePanther wrote:
There is a missing part of the question I think. Why not just everyone say "i have green eyes" on the first day and "I have blue eyes the second day" and everyone is off after two days.
Also, do they KNOW that it's 100 and 100? it doesn't state that in the problem, which has massive implications. They would still not know their own eye colour with certainty on day 1.
+ Show Spoiler +This problem can be solved with any mix of blue and green eyed people.
If you only see blue eyed people, you would be the only one with green eyes and leave on day 1.
If you see only 1 with green eyes, he would leave on day 1 if you have blue eyes.
If he doesn't leave on day 1, it means he sees someone else with green eyes which can only be you and you'll both leave on day 2.
If you see 2 green eyed people, they'll leave on day 2 if you have blue eyes. If they don't, it means you have green eyes and You'll leave on day 3.
If you see 3 green eyed people, they'll leave on day 3 if you have blue eyes. If they don't, it means you have green eyes and You'll leave on day 4.
This pattern is repeating and end result is: If there's X number of green eyed people, they'll leave on day X, and the blue eyed people 1 day later.
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Sweden45 Posts
On April 19 2012 00:38 Cocacooh wrote:
A family inherit 6 million dollars from a friend of the family who has passed away. The family consist of two fathers and two sons. The family decides to spilt the money so that they each get 2 millions. How is that possible?
+ Show Spoiler +I would say that there are 2 fathers and 3 sons, the grandfather was someones son aswell
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About the Island with blue and green eyes:
+ Show Spoiler +
Make two people stand next to eachother, then another one (w/e the color of his eyes) will come stand between you if you have the same color of eyes. Otherwise he will stand next to them.You do this till no one is left. This way everyone will know what color their eyes are. Example:
1: gg
2: Bgg
3: bBgg
3: bbGgg
4: bbGggg
5: ...
199: B x 100 and G x 100
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On April 19 2012 16:46 mathilol wrote:About the Island with blue and green eyes: + Show Spoiler +
Make two people stand next to eachother, then another one (w/e the color of his eyes) will come stand between you if you have the same color of eyes. Otherwise he will stand next to them.You do this till no one is left. This way everyone will know what color their eyes are. Example:
1: gg
2: Bgg
3: bBgg
3: bbGgg
4: bbGggg
5: ...
199: B x 100 and G x 100
+ Show Spoiler +
Nice solution. The only problem I see is that the last guy has no way to know what colour he has.
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There is no form of communication possible, so most of you have the wrong approach.
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On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
- The only possible eyecolours are green and blue.
We did this in #bg9 some time ago:
+ Show Spoiler +After 100 days, if no one leaves, you have green eyes, and so on. I believe the riddle is supposed to include an announcement about eye color daily, but I could be wrong.
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United States10328 Posts
On April 19 2012 03:24 Bahamuth wrote:
green eyes problem
Haha, I have a vaguely related one, but it's not so much a brain teaser as a math problem (and I think I've posted it here before). I think it's way more counterintuitive ><
Infinitely many prisoners, labeled 1, 2, 3, 4, ... are standing in a line so that person N can see everyone with smaller label (and not himself). Each prisoner k is assigned a hat labeled with a [integer, rational number, or real number; doesn't actually matter (!)], c_k. Now, starting from prisoner 1 and going up, each prisoner will try to guess the number on their own hat; if he fails, he is shot, and if he succeeds, he lives.
Show that, if the prisoners are given (an infinite amount of) time beforehand to formulate a strategy, they can ensure that only finitely many prisoners are shot.
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On April 18 2012 23:42 mentallyafk wrote:Show nested quote +On April 18 2012 23:33 sc2system wrote:you are presented with 2 doors. one has millions of dollars behind it and the other has a lion that will eat you behind it. there are 2 guards in front of the doors that know what is behind the doors. one of them always tells the truth and one of them always lies, but you don't know which is which. You can only ask one question. What do you ask? Took me like 2 minutes. Solution: + Show Spoiler + You ask "What would your brother say if I asked him behind which door the million dollars are?".
You will get the door where the dragon is.
you are presented with 3 doors and 3 gaurds. same story, 1 door has money, other 2 dont. 1 guard always tells the truth and the other 2 always lie. what do you ask?
+ Show Spoiler +
Heh this is interesting, I'll write what I think while I try to solve this.
The basic premise of the riddle is simple enough, especially when you know the answer to the version with 2 doors and 2 guards. You need to phrase the question in a way that will get you the same answer regardless of which guard you happen to ask. Phrasing the problem as a pseudo math equation makes it even simpler. 1 (truth) * -1 (lie) = -1 and of course -1 * 1 = -1 as well, so regardless which guard you get, you will always get -1, ie the wrong answer.
Now, if this problem was just 3 guards but still 2 doors, the answer would be similar and very simple. You would just ask guard 1 'what would guard 2 say that guard 3 would say if I asked him where the money was?'. The equation would turn into 1 * -1 * -1 = 1 and you would always get the correct answer.
The problem here is that we need to eliminate 2 doors, not one and that adds some ambiguity. If we asked the same question as above, the liars would always have 2 possible lies, not just 1. This would both make different outcomes possible and it would make it so that none of the guards would actually know which of the 2 lies the lying guards would pick, so they could not accurately tell the truth or lie about which door the lying guards would pick.
As an example: Guard 1 liar, guard 2 truth-teller, guard 3 liar. You ask guard 1: 'What would guard 2 say that guard 3 would say if I asked him where the money was? Guard 3 would say one of the two doors without money behind it. Guard 2 wouldn't know which one so his answer would have to be 'I don't know' or 'one of these two'. Even if we assume that guard 2 knows which lie guard 3 would pick, guard 1 would again have two lies available. He could either choose the right door or the wrong door not picked by guard 3.
The answer then has to be that you go to guard 1 and you ask: 'If I asked guard 2 what guard 3 would never say if I asked him where the money is, which door would never be the answer?'. (yay negatives) The point and difference being that with the never included, it removes the ambiguity of the liars being able to lie in two different ways.
Let's go through the 3 different options:
Option 1 (guard 1 truth-teller, 2 and 3 liars): Guard 3 would never pick the door with the money behind it, because he's a liar. Because guard 2 is also a liar, he would lie and say the answer is one of the two doors with no money behind it (there's still ambiguity, but it doesn't matter here). Because guard 1 knows this, he knows the door with the money behind it would never be the answer, so he will give that as the answer. So we pick that door.
Option 2 (guard 1 liar guard 2 truth-teller, guard 3 liar): Again, guard 3 would never pick the door with the money behind it. The truth-teller knows this so his answer would be the right door. Because guard 1 is a liar, he will also give us the right door as the answer (because we're asking him which door won't be the answer).
Option 3 (guards 1 and 2 liras, 3 truth-teller): Guard 3, the truth-teller, would never say either of the two doors without the money behind it. Because guard 2 is a liar, he will give us the door with the money behind it as the answer. Similarly to option 2, guard 1 will also give us this answer, as the question is what door wouldn't be the answer.
Fun!
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On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain.
- The only possible eyecolours are green and blue.
+ Show Spoiler +
Seems pretty straightforward. Let's imagine for a second that there are 199 with blue eyes and only 1 with green. Since the guy with the green eyes knows there's someone with green eyes and he knows noone else has green eyes, he has to be that one, so he could go home on the first night.
Similarly, if there were 2 people with green eyes, they would see one person with green eyes. If they themselves didn't have green eyes, that person would leave day 1, since he would see noone else with green eyes. Since that person doesn't leave day 1, they can deduce that they themselves must have green eyes too and they could leave day 2.
And on and on. If there were 3 people with green eyes they'd leave on day 3, if 10 they'd leave on day 10. Since there's no communication possible, this is the only way they can figure out what eye colour they have.
In this example, all the people with green eyes will leave on day 100. On day 99 they expect all the people with green eyes to go home if they themselves didn't have green eyes. On day 100 they know they in fact do.
You said that the only possible eye colours are green and blue. If the people on the island don't know that, the people with blue eyes are fucked. They can never know for sure what eye colour they have. If they do know, they will all leave on day 101 (or day 100 if they can immediately react to all the green eyed people leaving). All along, they saw 100 people with green eyes. Since all of them left on day 100, they can deduce that they themselves didn't have green eyes and therefore must have blue eyes.
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On April 19 2012 10:20 SwizzY wrote:
When does yesterday come after today?
The day after tomorrow
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On April 19 2012 08:44 Nehsb wrote:Show nested quote +On April 19 2012 07:54 keplersfolly wrote:I'm not a regular on this forum so apologies if I'm quoting wrong :D Solution discussion for Nehsb and Miicro: + Show Spoiler +I'm impressed :D Was that off the top of your heads or had you heard it before? It took me and my colleagues a while to work it out from a position of total ignorance. Nehsb is more correct though as he doesn't assume the starting state of the levers and he hints towards fully understanding the issue (but falls a little short). Basically there are two related facets:-
You MUST flip a lever exactly once + There are 2 levers.
What this effectively means is that the prisoners decide that the left lever is the "noop" lever. So the counter only cares about the right lever and you follow the logic you both described, so if a non-counter enters the room and the right lever is flipped to indicate someone needs counting then they throw away their flip on the left lever. You can think of this as changing the puzzle conditions to be having just one lever but flipping it is optional. Still, congrats if you worked this out from scratch! Ninja Edit: Nehsb's solution accounts for the unknown starting state - good job! Miicro's solution accounts for the fact there are two levers - good job! I'd heard it before. There's also a really hard version that I'm curious if anyone has an easier solution to: Show nested quote +20 prisoners are in jail who are all gathered in the main hall and addressed by the warden who explains the following. "In 1 hour you will all be placed in solitary confinement and never see each other again. You will individually be escorted by the guards to the "Lever Room" in which the are a pair of levers that have no function except being able to be flipped either Up or Down. You can flip any number of levers, including all of them or none of them. We make no guarantees about the order you will be called to the Lever Room in, and we may take any of you there more than once. If any of you prisoners tell the guard "We have all been to the lever room" and this is true you will all be released, otherwise you will all be shot."You do not know the starting position of the levers. You are allowed to write down a strategy and give it to all of the prisoners, but the strategy each prisoner gets must be the same; you are not allowed to "pick a leader" or anything like that. The changes are in bold: - Still 2 levers - You can now flip any number of levers - Every prisoner must have the same strategy - You do not know the starting states of the levers The problem gets much easier if you do know the starting state of the levers.
I have a question : do you know how often people go to the lever room, that is, do you have a way to keep track of the time ?
If not, then I don't know the answer yet. If so, then I have a much much harder version of your riddle, which can be stated as follows :
The basic set up is the same. Each day, a prisoner is brought at random to the lever room, where he has to switch exactly one lever. However, the prisoners don't know in advance how the lever room will look, that is, they can't agree on which lever is the left lever, and which one is the right lever, and even more, they can't agree on what is down position and what is up position. They just know that, when they will eventually get there, they will be able to switch exactly one of two devices which can have two distinct states. They don't know the initial position of the levers.
Then they will get freed not only if they are able to say that everyone went to the room at least one, but that this happened in the last 20 days too.
They are still allowed to make the strategy they want beforehand. They are not forced to all have the same strategy.
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There's a 7 floor building with each floor dedicated to a single type of facility unique to that floor.
On floor 7 there's offices.
On floor 6 there's server rooms.
On floor 4 there's an arcade.
A library is on the __rd floor. Which floor is that?
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On April 19 2012 20:50 Tanukki wrote:
There's a 7 floor building with each floor dedicated to a single type of facility unique to that floor.
On floor 7 there's offices.
On floor 6 there's server rooms.
On floor 4 there's an arcade.
A library is on the __rd floor. Which floor is that?
+ Show Spoiler +
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EPT
