That is a very strange logic puzzle, i had to look at the solution, and it makes almost perfect sense, it's just confusing...
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OK, if there were two people. 1 brown, 1 blue, then the blue would leave that night. (trivial). If the only other possible eye colour is known to be brown, then the brown would leave the night after, as he knows that his non-blue eyes are needed to help the blue. Otherwise, he stays for eternity.
If there were three people. 1 brown, 2 blue, then nobody leaves on the first night. But now, each of the blues knows that they are blue, as had they been another colour, the other blue would have been able to work out their blueness on the first night. So here, on the second night, the 2 blues leave. On the third night, depending, the brown leaves.
Now, the number of browns is irrelevant. If they are blue, then the blues will leave later, so as soon as the blues leave, all of the browns know they are brown, if brown is the only other possible colour.
Thus a position of x browns and y blues = a position of 1 brown and y blues
Whenever you add a blue to the scenario (if I were to explain out what would happen again with 3 blues) the basic pattern is that it follows the scenario with y-1 blues, but if they haven't left by then, this newly added blue knows he must be blue (as no other colour changes anything) and so they all leave on the day after.
When y=1, they leave on the first day, so with y blues, they leave on the yth day, and the browns leave on the (y+1)th, depending.
In the scenario given, blues leave on the 100th day, and browns either stay forever, or leave on the 101th.
Here's a nice puzzle:
You have a shuffled deck of 52 cards, 10 of which are randomly face up. Deal them out into two piles so that each pile has the same number of face up cards. You are blind.
bugoobiga wrote:
OK, here's one:
There are two hills, side by side, with a stream that runs between.
On top of one hill is a tiny house, in which a woman and her very young daughter live.
One day, the house caught on fire, so the mother grabbed her daughter's hand and they ran out the door. They ran down the hill, crossed the stream and ran to the top of the other hill.
While sitting safely on the hill, looking across at their burning house, the young daughter said, "Thank God the three of us made it out of there."
???
Is it a grandmother, a mother, and her daughter?
doodinthemood wrote:
OK, if there were two people. 1 brown, 1 blue, then the blue would leave that night. (trivial). If the only other possible eye colour is known to be brown, then the brown would leave the night after, as he knows that his non-blue eyes are needed to help the blue. Otherwise, he stays for eternity.
If there were three people. 1 brown, 2 blue, then nobody leaves on the first night. But now, each of the blues knows that they are blue, as had they been another colour, the other blue would have been able to work out their blueness on the first night. So here, on the second night, the 2 blues leave. On the third night, depending, the brown leaves.
Now, the number of browns is irrelevant. If they are blue, then the blues will leave later, so as soon as the blues leave, all of the browns know they are brown, if brown is the only other possible colour.
Thus a position of x browns and y blues = a position of 1 brown and y blues
Whenever you add a blue to the scenario (if I were to explain out what would happen again with 3 blues) the basic pattern is that it follows the scenario with y-1 blues, but if they haven't left by then, this newly added blue knows he must be blue (as no other colour changes anything) and so they all leave on the day after.
When y=1, they leave on the first day, so with y blues, they leave on the yth day, and the browns leave on the (y+1)th, depending.
In the scenario given, blues leave on the 100th day, and browns either stay forever, or leave on the 101th.
Here's a nice puzzle:
You have a shuffled deck of 52 cards, 10 of which are randomly face up. Deal them out into two piles so that each pile has the same number of face up cards. You are blind.
Very nicely done, as that puzzle is very difficult for most people. It seems odd because when the guru announces she sees somebody with blue eyes, she doesn't seem to give them any information--they already knew there was somebody there with blue eyes. However, they didn't know that everybody knew that everybody knew that . . . (x 100) . . . that there was at least one person with blue eyes, which is key to the puzzle. Note that the first paragraph is carefully phrased to be self-referencing, because otherwise I would need to recurse "everybody knows that everybody knows that . . . everybody knows the above rules," or again the puzzle would not work.
As for your puzzle, it sounds more lateral thinking than logic, because obviously if you are truly blind and have no way of differentiating a face-up from a face-down card, the puzzle is impossible as stated (unless you get really lucky). Perhaps I could cut the cards (literally, with scissors) lengthwise and place half the deck in each pile?
Maybe they are braille cards.
Or maybe you bend every card in half. In this way, half of each card is face-up and half is face-down, so you just divide them evenly.
Or maybe you set the cards up sideways so none of the cards are face-up, in which case it doesn't even matter how you deal them.
Or maybe since the puzzle just says "two piles," not "only two piles," you have three piles: The original, and two other, empty piles, each of which contains zero face-up cards.
There are many possibilities. 
I understand all the mathematical induction involved but I believe that the initial condition makes the induction irrelevent. The guru was talking to x number of people. Therefore you cannot split them up into groups. If there are 2 people that were spoken to it is unknown if both are blue or if neither are blue. Someone in some other group could be blue so it is impossible for that group to determine. This puzzle seems to assume that each group has a number of each color. If for example, a group had 3 blues in it, none of them could leave or determine there own color. (Then again I guess that they could all assume they were blue and leave on day 2 since no one figured it out and left on day 1...) I'm confusing myself now... Perhaps you can split it up like that...
For my card puzzle, there is no trickiness with braille cards or asking someone for help or dealing out more than two piles or tearing up cards or anything.
It is simply dealing the cards into 2 piles, both of which have the same number of face up cards.
It is possible.
For that puzzle with the eyes, if you know induction and want to do it that way, then here ya go:
When the number of people with blue eyes is 1, that person leaves on day 1. This is because he knows there is a blue eyed person but sees none so knows it must be him.
Take the number of days after which all the blues leave as being y when there are x blue eyed people. If there were x+1 blue eyed people, then after y days pass, nobody has left. On the (y+1)th day, everybody knows that there must be more than x blue eyed people. All the blue eyed people only see x other blue eyed people, and so know they must be blue eyed. Thus, when there are x+1 blue eyed people, they leave on the (y+1)th day.
when x=1, y=1, so x=y
In this case, on the 100th day, the blue people all leave.
Fiveofswords wrote:
but after three, eveyone knows that everyone knows that noone is getting new information
In the case of n = 4, while it is true that everyone knows that everyone knows that no one is getting new information (in the sense of knowing that n >= 1), they don't know that everyone knows that everybody knows that no one is getting new information.
In the case of n = 5, everyone knows that everyone knows that everyone knows that no one is getting new information, but they don't know that everyone knows that everyone knows that everyone knows that no one is getting new information.
In the case of n = n, everyone knows that everyone knows that . . . (n) . . . that no one is getting new information, but they don't know that everyone knows that everyone knows that . . . (n) . . . that no one is getting new information.
doodinthemood's solution is a valid induction proof that on the nth night, n people will leave.
But I still don't get that card puzzle.
To be honest, I don't think the card puzzle is all that hard. Simply deal out the top ten cards, and then flip all ten of those cards over. This will give you two piles, each with the same number of face-up cards (though the exact number may vary). For instance, if there happened to be 3 face-up cards in the top 10, then you'd end up with 7 face-up cards in each pile.
May 15, 2010
0
#33
Fiveofswords wrote:
no lol, you cannot itereate the everyone knows more than twice...Its pretty messy and all, but theres only going to be 2 groups of people, those who see n and those who see n-1. So if n = 4, the people who see 3 are the only improtant ones that need to know that everyone is getting no new information. If there actually were people who did see 2, then this 'everyone knows' iteration would work, but there are not.
There is only one relevant group of people: those with blue eyes. The brown eyed people are just random filler who are there to remind you that it is possible you might not have blue eyes. They can never figure out what color their eyes are, because they could be brown or green or red or fuchsia or just about anything else. It is easier to just imagine the puzzle having only blue-eyed people, so let's do that.
Let's consider the n = 1 case. As soon as the Guru announces that she sees somebody with blue eyes, you know it must be you, because there is nobody else there. You leave that night.
n=2: As soon as the Guru announces that she sees somebody with blue eyes, you know that if you do not have blue eyes, the n=1 case applies, and the other person will leave that night. When he does not, you know you must have blue eyes, so you leave the next night. By symmetry, so does the other person.
n=3: As soon as the Guru announces that she sees somebody with blue eyes, you know that if you do not have blue eyes, the n=2 case applies, and the other people will leave the next night. When they do not, you know you must have blue eyes, so you leave the night after next. By symmetry, so do the other people.
What makes n=4 fundamentally different? Again, if you do not have blue eyes, you know the n=3 case must apply (because then you will be an irrelevant brown-eyes who is just filler, like I mentioned above). If you find the n=3 case did not apply, you must therefore have blue eyes.
This can clearly be extended to n=100, n>9000, or to any natural number n.
cobra91 wrote:
To be honest, I don't think the card puzzle is all that hard. Simply deal out the top ten cards, and then flip all ten of those cards over. This will give you two piles, each with the same number of face-up cards (though the exact number may vary). For instance, if there happened to be 3 face-up cards in the top 10, then you'd end up with 7 face-up cards in each pile.
Yup! Well done. To most people it seems impossible.
doodinthemood wrote:
For that puzzle with the eyes, if you know induction and want to do it that way, then here ya go:
When the number of people with blue eyes is 1, that person leaves on day 1. This is because he knows there is a blue eyed person but sees none so knows it must be him.
Take the number of days after which all the blues leave as being y when there are x blue eyed people. If there were x+1 blue eyed people, then after y days pass, nobody has left. On the (y+1)th day, everybody knows that there must be more than x blue eyed people. All the blue eyed people only see x other blue eyed people, and so know they must be blue eyed. Thus, when there are x+1 blue eyed people, they leave on the (y+1)th day.
when x=1, y=1, so x=y
In this case, on the 100th day, the blue people all leave.
101st
bjazz wrote:
doodinthemood wrote:
For that puzzle with the eyes, if you know induction and want to do it that way, then here ya go:
When the number of people with blue eyes is 1, that person leaves on day 1. This is because he knows there is a blue eyed person but sees none so knows it must be him.
Take the number of days after which all the blues leave as being y when there are x blue eyed people. If there were x+1 blue eyed people, then after y days pass, nobody has left. On the (y+1)th day, everybody knows that there must be more than x blue eyed people. All the blue eyed people only see x other blue eyed people, and so know they must be blue eyed. Thus, when there are x+1 blue eyed people, they leave on the (y+1)th day.
when x=1, y=1, so x=y
In this case, on the 100th day, the blue people all leave.
101st
Ah yes. Well, the 100th day in which it is available to leave.
Fiveofswords wrote:
it makes it fundamentally different, because if you see 3, then on the first night, you know that nobody is wondering if there is one or two...therefore they return to the exact same situation before anyone said anything, wondering if there is three or four...the amount of days passing does not help, each day you only wonder if there is three or four, like the first day.
consider the perspective of the person who sees 3. He doesnt knwo if other people see 2, or other people see 4. he KNOW noone sees 1. When the first day passes, the people who see 2, if they exist, would still only be wondering if there is 2 or three...you KNOW that noone was ever wondering if there was one or two. So you therefore are only left wonder IF there were people who see 2, or not...exactly the same situation as before the guru spoke.
Consider this, what if the guru spoke at said there are at least zero people with blue eyes...would they sole anything? but a day later? of course everyone knows that noone sees negative one blut eyes lol...but this is exactly as tautological as the person seeing three knows that noone sees one.
I'll try my best to untangle your argument, but it's a confusing one. If the guru said there are at least zero people with blule eyes, then nothing would happen, ever. In a situation with 1 blue eyed person, he would never know.
In a situation with 3 blues. I am a blue and see 2 other blues. If I am blue, then any one other blue sees 2 blues. If I am something else than any one other blue (we'll call our selection Bill) sees 1 blue. At the end of day 1, nobody has gone. This would tell Bill that the blue he sees is not the only one, because if the blue he sees was the only one, that blue would have left at day 1, knowing that the only blue the Guru sees is him. Now Bill knows that the blue he sees is not the only one, but more importantly, I now know that Bill knows this. At the end of day 2, Bill should leave, but he doesn't. What does this tell me? Bill knows that there are more blues than the one he can see, but also isn't sure that his eyes are blue, thus, he must have evidence for a third person with blue eyes, and the only person that could be would be me. So I now leave on day 3.
If you imagine me as being any of the other three blue eyed, then the same logic holds, and you can see that with three blues, they leave on day 3.
Fiveofswords wrote:
[1.]it makes it fundamentally different, because if you see 3, then on the first night, you know that nobody is wondering if there is one or two...therefore they return to the exact same situation before anyone said anything, wondering if there is three or four...the amount of days passing does not help, each day you only wonder if there is three or four, like the first day.
[2.]consider the perspective of the person who sees 3. He doesnt knwo if other people see 2, or other people see 4. he KNOW noone sees 1. When the first day passes, the people who see 2, if they exist, would still only be wondering if there is 2 or three...you KNOW that noone was ever wondering if there was one or two. So you therefore are only left wonder IF there were people who see 2, or not...exactly the same situation as before the guru spoke.
[3.]Consider this, what if the guru spoke at said there are at least zero people with blue eyes...would they sole anything? but a day later? of course everyone knows that noone sees negative one blut eyes lol...but this is exactly as tautological as the person seeing three knows that noone sees one.
Off your 1: It is true that for n = 4, I know that everybody knows that there are at least 2 people with blue eyes, but not everybody knows that. It could be that I have brown eyes, in which case the other 3 all see 2 blue-eyed people, and therefore the n = 3 case applies. But if that were the case, they would all leave on day 3. DO YOU AGREE THAT IF THREE PEOPLE HAVE BLUE EYES, THEY WILL ALL LEAVE ON DAY 3? If you do not, please explain why. If you do, then n = 4 must work, because after I see that the other blue-eyed people did not leave after day 3, I know that it must not be the case that only 3 people have blue eyes. Since that is not the case, I must have blue eyes.
That is how I gain new information. I am no longer wondering if there are 3 or 4, because if there were only three, they would have left.
Off your 2: It doesn't matter if anybody was ever wondering whether there were 1 or 2 people with blue eyes; the guru doesn't help with that anyways, only with whether there are 0 or 1 people with blue eyes, so I don't understand this argument at all. If I see three blue-eyed people, I know that they must each see either 2 or 3 blue-eyed people, but I don't know if they know that. Supposing I have brown eyes, they might see 2 blue-eyed people and think "Those other 2 blue-eyed people must be seeing either 1 or 2 blue-eyed people, depending on whether or not I have blue eyes." If they are indeed thinking this, then they will know after 2 days, and with that new knowledge, leave. However, it turns out they don't leave, meaning they were not thinking this. Now that I know they weren't thinking that, I must have blue eyes, so I leave.
Off your 3: No, there is a crucial difference between the scenarios. Not only does everybody already know that there must be a nonnegative number of blue-eyed people on the island, it is impossible for that not to be the case. That the number is nonnegative then is similar to the rules stated in the first paragraph, which everybody automatically knows. That last sentence in the first paragraph is key, because it is self-referential. Not only does everybody know the rules of the first paragraph, but they know everybody knows the rules of the first paragraph, and they know that, and they know that, and they know that, ad infinitum. They know everything said in this paragraph, including this sentence. But such is not the case with whether or not at least one person on the island has blue eyes. Even before the guru speaks, everyone knows at least one person has blue eyes, and everyone knows that, and everyone knows that, but in the case of n = 4, not everyone knows that. In fact, the only way to know that is to wait three days and see.
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I have seen sin x / n = six, but I personally prefer:
women = time * money
time = money
women = money²
money = √(all evil)
women = [√(all evil)]²
women = all evil
Q.E.D.
Anyway, here is a very nice logic puzzle, taken from the xkcd wiki Puzzles page:
A group of people with assorted eye colors live on an island. They are all perfect logicians—if a conclusion can be logically deduced, they will do so instantly. No one knows the color of their own eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
Note: The answer is not "no one leaves."