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On April 19 2012 06:03 Odoakar wrote:Show nested quote +On April 19 2012 05:35 TheJizWiz 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. Everyone can leave on day one. + Show Spoiler +They just have to stand in a line facing one direction. The last guy of the line comes to the front and walks along the line untill he sees two persons with different eyecolors standing next to eachother, in which case he steps between them. They have to repeat this till the one who was first is at the end of the line again. He and the one after him have to go a second time. This will seperate the first group of eycolors from the rest. This group can leave.
This process can be repeated untill someone walks along the whole line whithout stepping between two persons. If this is the case he either has the same eyecolor as everyone in his group or is the only one with a different one and will never know.
If this counts as communicating then I dont know. I'm puzzled by this 'there is at least one person that has green eyes'...aren't there 100 persons with green eyes, like stated above in the text? On the first day, first person looks into eyes of every other and counts how many green eyes he saw. If he saw 100, he has blue eyes. If he saw 99, he has green eyes. He leaves. Next person knows that the person who left has green eyes ("Everyone knows the eyecolour of every OTHER inhabitant at all times"). He counts green eyes, if he sees 99, he knows he has blue eyes. If he sees 98, he knows he has green eyes. He leaves. And so on, everyone leaves on the first day.
The people don't know that there are 100 people with green eyes and 100 people with blue eyes at the start. They only know that there are 200 people total and all of them have either green eyes or blue eyes.
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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 +after 199 days everybody leaves the island on the same day
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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.
Oh cool, a math problem!
+ Show Spoiler +Solution: on day 100 everyone knows, half will leave, the rest the next day.
Proof by induction:
If only one person has green eyes, he leaves the first night, as he knows the rest have brown eyes. As he knew he was the only one, the rest can deduce that they have brown eyes, and leave the next day.
Let's assume that with X people with green eyes, those X people will leave on the Xth day, the rest the next. Then, if there are X+1 people with green eyes, they would know for a fact that there are more than X people at day X+1, but they can only see X others, and can each deduce that they must have green eyes. The next day, the rest will leave, as they are perfect logicians and understand why the X+1 left.
Hence, this is true for any X that is a natural number, in this case, 100.
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I have a good one.
You have twelve coins. One of them is slightly different in mass, either too light or too heavy.
You have a two pan balance.
How can you find out which coin is different and whether it's too light or too heavy in just three weighing?
here is the answer:
http://mathforum.org/library/drmath/view/55618.html
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I have two good ones that i bugged my mind with for quite some time (credits to my math teacher Bo)
First problem, you have 9 identical small stones, an old fashioned weight and nothing else. One of the stones are heavier, you have to weightings to figure out which one. How? (easy one)
second problem (a lot harder)
You have 12 identical stones, and the same setup as before. One stone is odd, but can be either heavier or lighter, you have three weightings to figure out which one, AND if it is in fact heavier or lighter. good luck! :D
- Rune
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The following puzzle was posed to me and a group of co-workers (we're software engineers):-
You and 20 other 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 will flip exactly one lever one time and then be escorted back to your cell. 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."
There is one hour for you and the other prisoners to devise a strategy to surivive, but there is no way to communicate once you are escorted to your solitary confinement cells. How do you survive? (Solution will follow, I'll keep any eye on this thread and answer questions if needed though).
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LOL diavlo how is that even possible! hahahaha, the odds of us both asking the same riddle at the same time is hilarious!
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Here are two good ones.
First one from the mighty Djini Kalah:
#1 A princess is as old as the prince will be, when the princess is twice as old as the prince was when the princess' age was half the sum of their present age. How old are they now?
There is more than one correct answer to this question, you don't need to guess which one, if you get any of the correct combinations you win ^^.
+ Show Spoiler [hint 1] +Find all the data you can use. The riddle tells you one more constant to work with, that isn't mentioned
+ Show Spoiler [hint 2] +That constant is their age diffirence
+ Show Spoiler [hint 3] +
+ Show Spoiler [solution] +
Prince age = x
Princess age = y
Age diffirence = a
x=y-a
y=x+a
princess was ½(x+y) when prince was ½(x+y)-a
princess age will be 2[½(x+y)-a] so the princess age is now 2[½(x+y)-a]-a
now having y=2[½(x+y)-a]-a we just solve the equation
(x+y-2a)-a=y
y=4a
x=3a
so the possible solutions are all that keep the proportions right: Prince is 3 and princess is 4 years old, or they may be 30 and 40 years old etc.
Second one comes from a certain dragon in the High Forrest of Faerun if i remember correctly, it goes as fallows:
2# Old arab was near his death. He had two sons, but he decided that he won't split his land, but rather will give it to one of them. He gave them a challange. An unusual race, because the son whose camel would arrive last at the gates of neighboring city would get the land of his father. However, they weren't allowed to ride away from their destination, only towards it.
After long day, on the backs of their camels. Exhausted from moving at turtle's pace and being exposed to the buring sun of the dessert they decided to hold the challange and go to the inn. They told their story to innkeeper, and he listened carefully. Then he said only two words of advice to them.
The very morning of the next day, people could see both sons riding on their camels fast as if they were chased by the devil. Kicking and screaming at their mounts to hurry them up. They were heading towards the neighboring city.
What did the innkeeper say?
+ Show Spoiler [hint 1] +What exactly did the father say?
+ Show Spoiler [solution] +
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On April 19 2012 06:44 TWIX_Heaven wrote:
LOL diavlo how is that even possible! hahahaha, the odds of us both asking the same riddle at the same time is hilarious!
We didn't. I posted a shitty riddle one minute before you posted yours. Though yours was better, stole it and typed it back in my earlier message to take all the credit.
I'm really super fast.... and kind of a dick 
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On April 19 2012 06:44 keplersfolly wrote:
The following puzzle was posed to me and a group of co-workers (we're software engineers):-
You and 20 other 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 will flip exactly one lever one time and then be escorted back to your cell. 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."
There is one hour for you and the other prisoners to devise a strategy to surivive, but there is no way to communicate once you are escorted to your solitary confinement cells. How do you survive? (Solution will follow, I'll keep any eye on this thread and answer questions if needed though).
Hmm, isn't this puzzle usually with one lever? Am I misinterpreting it or is there a reason you need two:
+ Show Spoiler +
We only work with the first lever.
We designate a "leader".
Leader algorithm: If lever is down, move lever up. Add one to a mental counter "x".
Everyone else: You keep a mental variable "y" which starts at 2. If y > 0 and the lever is up, move the lever down and subtract one from y.
When x >= 40, the leader knows that everyone has been in the room. x counts the number of times the lever has been pulled up (It's off by at most one, which is why we need to start y at 2 instead of 1. It might be off by one because the lever might have started down, which would make x 1 larger than the number of times the lever has been pulled up.) Once x >= 40, we then know that the lever has been pulled up at least 39 times. Then, if not everyone else has been in the room, there are at most 19 non-leaders who have entered the room, each who could have pulled up the lever at most 2 times, so the lever could have been pulled up at most 38 times.
I've seen a version where everyone must have the same strategy, and here you need two levers; is this the version you want? (And if so, do you want the version where we know the starting position of the two levers or we don't?)
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On April 19 2012 06:44 keplersfolly wrote:
The following puzzle was posed to me and a group of co-workers (we're software engineers):-
You and 20 other 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 will flip exactly one lever one time and then be escorted back to your cell. 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."
There is one hour for you and the other prisoners to devise a strategy to surivive, but there is no way to communicate once you are escorted to your solitary confinement cells. How do you survive? (Solution will follow, I'll keep any eye on this thread and answer questions if needed though).
+ Show Spoiler +The prisoners will assign one person as a counter. The FIRST time a prisoner (non-counter) enters the lever room they put it downward. Nobody will then change the lever until the counter enters the room and puts the lever back in the upward position. The next person who enters the lever room will put the lever downwards again (unless it is someone who already put the lever downwards)
This process will repeat itself in which every person is only allowed to put the lever downwards once. Otherwise they will just let the lever be. If the counter has seen the lever down 20 times (and he has put the lever back upwards 20 times), he will know that every other prisoner has been in the lever room. Therefore he can tell the guards that everyone has been in the room.
I don't know what is up with the 'Pair' of levers. If i'm missing something please tell me! But I guess that if you have had your turn earlier you can just switcheroo the one that is not being used, if you are FORCED to change direction of any of the levers (you decide which lever you're going to use beforehand for the counting process then.)
Hope this is clear enough. My wording is a bit complicated at times.
NINJA EDIT: I saw someone already posted a better result (with 40x instead of 20x because of the possible downward direction of the lever at the start. clever!
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Doh! Fixed my original riddle on page 2.
Ok:
If a blue house is made of blue bricks, an orange house is made of orange bricks, and a red house is made of red bricks, what is a green house made of?
+ Show Spoiler +
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On April 19 2012 05:48 Nehsb wrote:Here's a followup riddle (Spoils the answer, so don't look at if you haven't done the original yet): + Show Spoiler +Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given?
@spoilered question
Does it really matter?
+ Show Spoiler +Since everyone knows that there's either 99 g.e. persons and 100 b.e. persons or 100 g.e. persons and 99 b.e. persons, can't they (by the same method of induction) conclude that since no one leaves on the 99th day, their eye colour must be the same as the one of the 99 blue/green eyed persons that they see?
Edit:
+ Show Spoiler [More thoughts] +
This would work only if the islanders know that there are only blue and green coloured people on the island. Otherwise it'd screw with the base case. Right?
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On April 19 2012 07:05 Shraft wrote:Show nested quote +On April 19 2012 05:48 Nehsb wrote:Here's a followup riddle (Spoils the answer, so don't look at if you haven't done the original yet): + Show Spoiler +Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given? @spoilered question Does it really matter? + Show Spoiler +Since everyone knows that there's either 99 g.e. persons and 100 b.e. persons or 100 g.e. persons and 99 b.e. persons, can't they (by the same method of induction) conclude that since no one leaves on the 99th day, their eye colour must be the same as the one of the 99 blue/green eyed persons that they see?
Edit: + Show Spoiler [More thoughts] +
This would work only if the islanders know that there are only blue and green coloured people on the island. Otherwise it'd screw with the base case. Right?
There actually is a key difference even if the villagers know that there are only blue/green eyes:
+ Show Spoiler +http://en.wikipedia.org/wiki/Common_knowledge_(logic)The idea is that, at the start, everyone knows that there is at least one person with green eyes. Everyone also knows that everyone knows that there is at least one person with green eyes. (Because if someone else didn't see anyone with green eyes, there could be at most 1 person with green eyes, so everyone knows that everyone knows that there is at least one person with green eyes.) And we can take this even further: Everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that there is at least one person with green eyes. But the issue is, without the message, if you take this chain too far (around 100 "everyone knows", maybe 99 or 98 or 101), the statement no longer becomes true. But with the message, the statement is true no matter how many "everyone knows" there are. An example with lower numbers: with 198 blue-eyed people, and 2 green-eyed people, everyone knows that there is at least one green-eyed person on the island. But if you are one of the green-eyed people, you don't know that you're green-eyed, so there's a possibility that the other green-eyed person sees only blue-eyed people. So everyone knows that there is a green-eyed person, but somebody doesn't realize that everybody knows that there is a green-eyed person. Once you add on more "everyone knows", the statement becomes harder and harder to process, but the same basic idea holds.
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Here is a really simple, really stupid riddle.
"At the local deli there is a butcher with size 13 shoes who is 6 feet tall. What does he weigh?"
+ Show Spoiler +
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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 +All will leave. On Night 1 (N1) No one will leave because everyone sees 99 other green eyes, this continues until N99 and on N100 No one will have left and everyone will figure out that there are only 100 green eyes and leave.
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Anne, Bill and Cath each have a positive natural number written on their foreheads. They can only see the foreheads of others. One of the numbers is the sum of the other two. All the previous is common knowledge. The following truthful conversation takes place:
Anne: I don't know my number.
Bill: I don't know my number.
Cath: I don't know my number.
Anne: I now know my number, and it is 50.
What are the numbers of Bill and Cath?
Just to give credit, reportedly it's from the November 2004 issue of Math Horizons (whatever that is).
+ Show Spoiler [Solution] +Bill's number is 20, Cath's is 30.
Why? Let's see what happens:
1. A sees 20 and 30. She could have either 10 or 50.
2. B sees 50 and 30. Also he knows that his number is not equal to 30 (C's number), because then A would've known for sure that her number was 60 (0 is not allowed). He could have either 20 or 80.
3. C sees 50 and 20 and she knows that her number is neither of them. Still, she could have 30 or 70.
4. How does A now know that it's 50 and not 10?
Let's see what would've happened if it was 10 instead:
1. A sees 20 and 30. Same stuff, 10 or 50.
2. B sees 10 and 30. Again, he knows it's not 30. This time it could be 20 or 40.
3. C sees 10 and 20. It could be 10 or 30, but in this situation she knows that it's not 10 and she would've screamed "I know! I know!".
Since she didn't, A's number can't be 10 and the only other option is 50.
Bonus points if you can prove that it's the only solution.
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On April 19 2012 07:13 Nehsb wrote:Show nested quote +On April 19 2012 07:05 Shraft wrote:On April 19 2012 05:48 Nehsb wrote:Here's a followup riddle (Spoils the answer, so don't look at if you haven't done the original yet): + Show Spoiler +Everyone knew that there was at least one green eyed person at the start, because they could see everyone else's eye color. So why does it matter that the message was given? @spoilered question Does it really matter? + Show Spoiler +Since everyone knows that there's either 99 g.e. persons and 100 b.e. persons or 100 g.e. persons and 99 b.e. persons, can't they (by the same method of induction) conclude that since no one leaves on the 99th day, their eye colour must be the same as the one of the 99 blue/green eyed persons that they see?
Edit: + Show Spoiler [More thoughts] +
This would work only if the islanders know that there are only blue and green coloured people on the island. Otherwise it'd screw with the base case. Right?
There actually is a key difference even if the villagers know that there are only blue/green eyes: + Show Spoiler +http://en.wikipedia.org/wiki/Common_knowledge_(logic)The idea is that, at the start, everyone knows that there is at least one person with green eyes. Everyone also knows that everyone knows that there is at least one person with green eyes. (Because if someone else didn't see anyone with green eyes, there could be at most 1 person with green eyes, so everyone knows that everyone knows that there is at least one person with green eyes.) And we can take this even further: Everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that there is at least one person with green eyes. But the issue is, without the message, if you take this chain too far (around 100 "everyone knows", maybe 99 or 98 or 101), the statement no longer becomes true. But with the message, the statement is true no matter how many "everyone knows" there are. An example with lower numbers: with 198 blue-eyed people, and 2 green-eyed people, everyone knows that there is at least one green-eyed person on the island. But if you are one of the green-eyed people, you don't know that you're green-eyed, so there's a possibility that the other green-eyed person sees only blue-eyed people. So everyone knows that there is a green-eyed person, but somebody doesn't realize that everybody knows that there is a green-eyed person. Once you add on more "everyone knows", the statement becomes harder and harder to process, but the same basic idea holds.
I see the difference now. This stuff is very mind-boggling (and also very interesting). Thanks for the explanation.
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Ever play the game 24? Use the numbers 1, 5, 5, 5 and the four basic math operations ( + / - *) to get the answer 24. You can only use each number once. The operations can be used as many times as you want.
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On April 19 2012 06:43 TWIX_Heaven wrote:
second problem (a lot harder)
You have 12 identical stones, and the same setup as before. One stone is odd, but can be either heavier or lighter, you have three weightings to figure out which one, AND if it is in fact heavier or lighter. good luck! :D
- Rune
Just one question is there a method to figure this out with 100% accurancy or can you get occurences where you have bad "luck" and can't figure it out in three goes?
+ Show Spoiler +
So I think I've figured this out now or at least some of the scenarios. Had to find pen and paper to do it though:D
PS: I'm dividing the stones into 3 bulks of four :A B C, in A you have Aa Ab Ac Ad, the same with B and C. I started off writing out my thoughts but it just made the whole thing messy. X-Y is how I write my scale.
1:
You divide your stones into A B C - 4 stones in each block. The benefit of this is if you weigh A - B and its equal you know that C is the where the light or heavy one is. Lets say that happens.
You then know that the 8 stones on the scale are the same.
You then take 2 of the Cs on one side of the scale and on the other you take 1C and 1 from A.
*Ca Cb - Cc A is equal, you now know the Cd is the odd stone.
Cd - A is you last move. Cd goes up when its light and down when its heavy.
2
We continue on 1, this time though:
*CaCb - Cc A is not equal.
You also get to know that IF the odd one is on the left and the left side go up that the odd one is light, IF its on the right its heavy. (visaversa if right side go up.) Next move.
Ca-Cb. If they are equal you know that Cc is the odd one and its either heavy or light depending on the result of the previous weighing if Cc went up its light, down its heavy.
Ca-Cb are unequal. You have to look on the 2nd result to know wich is the odd one. CaCb was lighter then CcA you know that the light one in this weighing will be the odd one. Similar story if its heavy in weighing number 2. Sick wall of text.
3:
A-B are unequal.
A goes down B up.
Either the odd one is in A and is heavy or its in B and is light.
Edit: Gave up
Update: so I have figured out some of the scenarios.
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EPT
