Puzzle thread? Puzzle thread.

[Xei]

Okay, you're getting too good at these. Here's a couple of tricky ones to stump you for a while. The first one is really amazing. I worked it out after a bit of effort, but took a long time to convince myself of my own answer. Unfortunately I was given the solution to the second one before I could work it out myself, but I have no idea how long it'd've taken me. The solution is simple and awesome but kinda has the appearance of being pulled out of a hat. I imagine I'd've been stuck for a long while. Blue Eyes [wording by xkcd.com, though he didn't invent it] 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 it instantly. No one knows the color of their 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." And lastly, the answer is not "no one leaves." The Buttons and the Barrel You and I play a game where we take it in turns to place buttons, all circular and of the same size, on the top of an upright barrel. Once placed, a button cannot be moved. The first person who can't fit any more buttons flat on the barrel loses. You can chose if you want to go first or second. Can you always win?

[Arra]

I got this one really quickly and I have no idea how. Can you remember anything about your thought processes prior to the insight, Di? Interesting that you went for a smoke. The guy told me the puzzle at the pub and I think I went to pee. I recall realising how to do the prisoner one in the same circumstance. That is odd. :/ Spoiler for mundane thought process: It was nothing special, just chance. I randomly thought about the fact that there are an even number of prisoners (if it wasn't random I don't remember what led to this thought), then thought about all the possible combinations of whether the white/black hats are even or odd. Either there are odd numbers of both, or even numbers of both, but they can't be different. Then I considered that maybe that knowledge might be used to solve it somehow (not really believing it would). The back person knows the colors of those in front of him... but that knowledge is of 59 prisoners, an odd number, so one color must be even and the other odd. What if the back person just shouted black if black was odd and white if white was odd? Would the second person know his color? *epiphany moment* yes he would, and after he said his color the next one would know too. EDIT: The island one sounds interesting. It doesn't seem anyone can leave... The guru saying anything before anyone has left will do nothing, since everyone will see multiple blue-eyed and brown-eyed people so it's not like one could realize "everyone else has brown eyes so I must have blue eyes." But before she says anything no one can know either. They have absolutely no information about their own eye color. I'm assuming they can't communicate beforehand to plan out that the guru will speak on the nth day where n is the number of blue-eyed people, or something like that. Unless they're all supposed to figure out that that will give them all the best chances. But I doubt that's the answer, and it's hard to believe there is an answer.

[SnowyCat]

Alright, I think I solved the button problem. If this is wrong it'll be really embarrassing. Spoiler for The Buttons and the Barrel: So I solved this one mathematically. Let's call the radius of the barrel "x", and the number of buttons that can be placed in a straight line on that radius "b". "b" will always be a whole number, since the game doesn't allow for the buttons to hang off the edge. Using these variables, we can show that the radius of a single button is half of the number of buttons that can be lined up on the radius, or 1/2 * "b per x", or x/2b. Therefore, the area of a single button is pi(x/2b)^2 , or (pi*x^2)/(4b^2). Also, the area of the barrel is clearly pi*x^2. Just putting that out there for future reference. Since we need to know how many buttons will fit on the barrel, we need to take the area of the barrel over the area of a button. The number of buttons (the solution to that division problem) is essentially the number of turns in our game, since one button will be placed per turn. So let's divide those two expressions we set up earlier. Area Barrel/Area Button = #Turns (pi * x^2)/((pi *x^2)/(4b^2)) = #Turns ... Simplify... ... 4b^2 = #Turns Well! That worked out nicely. Since an even whole number (4) times any whole number will always be even, we know that every time there will be an even number of turns in the game no matter what. Now let's find out what turn to play on. In this example, b = 2, so there will be 16 turns. We'll play on the second turn, and see where that puts us. Turn Player 1 Xei 2 Snowy 3 X 4 S 5 X 6 S 7 X 8 S 9 X 10 S 11 X 12 S 13 X 14 S 15 X 16 S Hey! Would ya look at that, I win. Since every game will have an even number of turns, you can win every time by playing second.

[Xei]

It's a good try, but it's not right. There's no way of working out the size of the buttons. They could literally be anything - there was no requirement that an exact number fit along the diameter. Your calculation would also only be right if every single bit of area of the barrel was covered. But obviously even if the players packed the coins as densely as possible (which they aren't required to - maybe your opponent uses a different technique), there would still be a lot of left over area between the coins. The answer is really simple (though probably very hard to come up with). No algebra or anything, just cunning.

[SnowyCat]

It's a good try, but it's not right. Your calculation would also only be right if every single bit of area of the barrel was covered. But obviously even if the players packed the coins as densely as possible (which they aren't required to - maybe your opponent uses a different technique), there would still be a lot of left over area between the coins. The answer is really simple (though probably very hard to come up with). No algebra or anything, just cunning. Darn, I didn't think of that. Hmmm...

[Alyzarin]

The Buttons and the Barrel You and I play a game where we take it in turns to place buttons, all circular and of the same size, on the top of an upright barrel. Once placed, a button cannot be moved. The first person who can't fit any more buttons flat on the barrel loses. You can chose if you want to go first or second. Can you always win? Well I just simulated this with quarters and an upside down bowl and was able to win going both first and second lol. So what am I missing here?

[Xei]

The question is whether you can beat me for certain somehow. Obviously if I play badly you can probably beat me either way. The question is whether you can beat me no matter how I play.

[Nelzi]

Heh, I got the barrel. Nice try SnowyCat. I knew it could not be a matter of calculating sizes and how they would fit into a circle, that would be an overkill in complication to a simple question. Spoiler for The Buttons and the Barrel: I choose to have the first turn and place the first button in the center. Now for each button you place, I simply place my next one on the opposite side! I just failed the 'going back to bed part' of my modest attempt for WBTB, so I'll fight with the eye color problem now...

[SnowyCat]

Heh, I got the barrel. Nice try SnowyCat. I knew it could not be a matter of calculating sizes and how they would fit into a circle, that would be an overkill in complication to a simple question. Spoiler for The Buttons and the Barrel: I choose to have the first turn and place the first button in the center. Now for each button you place, I simply place my next one on the opposite side! I just failed the 'going back to bed part' of my modest attempt for WBTB, so I'll fight with the eye color problem now... Oooo, nice one! I thought about doing something kinda like that, but I didn't know how to make sure it would work every time. Care to share you're thought process?

[Arra]

Heh, I got the barrel. Nice try SnowyCat. I knew it could not be a matter of calculating sizes and how they would fit into a circle, that would be an overkill in complication to a simple question. Spoiler for The Buttons and the Barrel: I choose to have the first turn and place the first button in the center. Now for each button you place, I simply place my next one on the opposite side! I just failed the 'going back to bed part' of my modest attempt for WBTB, so I'll fight with the eye color problem now... Nice lol. I barely even thought about this one because I thought it would involve some complicated math which isn't as fun to me. But that's so simple and makes a lot of sense.

[Nelzi]

... Care to share you're thought process? Well it was not a mastermind-beautiful-insight-moment, I just felt right from the beginning that the answer cannot be a complicated calculation. And then when reading 'You can chose if you want to go first or second', I was kinda reminded of the tic-tac-toe game (you know the 9 squares to be filled with O and X), where the middle square is kind of an advantage square.

[SnowyCat]

Well it was not a mastermind-beautiful-insight-moment, I just felt right from the beginning that the answer cannot be a complicated calculation. And then when reading 'You can chose if you want to go first or second', I was kinda reminded of the tic-tac-toe game (you know the 9 squares to be filled with O and X), where the middle square is kind of an advantage square. Huh, makes sense to me. Well done!

[Xei]

Nice lol. I barely even thought about this one because I thought it would involve some complicated math which isn't as fun to me. But that's so simple and makes a lot of sense. Ye of little faith! When I say there is little to no maths, I mean it. I am able to guage these things without vastly overestimating how much most people love algebra. :p Although really, solving these kinds of puzzle is really what proper maths is all about, not the boring stuff you have to do at school. Maybe Nelzi secretly has a PhD, he's awesome. Sorry about your WBTB! You should try to find a zinger of a puzzle to get your revenge. By the way, I think the "mastermind-beautiful-insight-moment" is probably a pop-culture fiction. There's no such thing as an intellect which somehow manipulates some kind of pure logical structure with total comprehension, insight and purpose. Fundamentally, everybody just stumbles about in the dark until they hit something. Some people such as yourself just have a knack for stumbling.

[tommo]

I didn't think this thread would be so popular, damn all you people figuring them out! This is a cool idea lol. I haven't read through all the posts yet but I definitely agree with Nelzi's response to the coin one (that one was neat ) and I'm assuming the farmer one is right too, not that I've checked the math. But here's my solution to The prisoners and the light: Tell 19 of the prisoners: if you go into the bathroom, the light is off, and it's your first time seeing the light off, turn it on. The last prisoner's only job is to turn off the light if he goes in there and it's on, and to keep track of how many times he had to turn the light off. Since each of the first prisoners only turns on the light once and only the last prisoner turns it off, they basically just have to endure going back and forth until his count reaches 19 he'll know that they've all been in there. Goddamit! I thought of this and quickly dismissed it coz I just didn't think it would work. Makes perfect sense though, well done. Spoiler for Solution to the field and the farmer: The basic trick is to realise that if you run around a small circle centred on the centre of the field, you can outpace the farmer. In fact, if the circle is a quarter of the distance from the edge, the circumference will be 4 times less, and you and the farmer will take the same time to do a lap - so on a circle a tiny bit smaller than this one, you can outpace the farmer. So run around this circle until you're on the opposite side of the centre to him. Then just run in a straight line to where the path is closest. The farmer has to travel half of the way around the circular field to meet you there, and it's easy to check that this is more than 4 times as far. So you can always make it. Again, I thought about this, but dismissed it, but I still don't agree with this. Maybe I'm not giving it enough thought, but the farmer can just slow down or turn around, right? You can't really overlap him. If you start to get too far ahead he can just wait for you to come back around, or turn around and "catch up". Nevermind, makes sense. I think I need to think through my ideas more Blue Eyes [wording by xkcd.com, though he didn't invent it] I read this on xkcd a while ago, spent hours on it, read the answer and still don't understand it. It's convoluted as hell IIRC.

[Xei]

Hah. I wouldn't call it convoluted per se... I don't think the answer is complex. I would say it's very counter-intuitive though, and hard to think of, unless you stumble upon the right approach. With the farmer thing, the point is, on a small concentric circle, you run a lap faster than the farmer does. So if he tries his best to stay level with you, he will fall behind, and ultimately you can get to the furthest point on your circle from him (so that you both lie on a line through the centre). If he turns around, well, even better. In that case he's going in the 'wrong' direction and you don't even have to do anything to get opposite to him.

[tommo]

Haven't looked at the answer for this, but it seems like a puzzle that suits the type of ones here so: A giant and an elf live on an island. This island has seven poisoned wells, numbered 1 to 7. If you drink from a well, you can only save yourself by drinking from a higher numbered well. Well 7 is located at the top of a high mountain, so only the giant can reach it. One day they decide that the island isn’t big enough for the two of them, and they have a duel. Each of them brings a glass of water to the duel, they exchange glasses, and drink. After the duel, the elf lives and the giant dies. Why did the elf live? Why did the giant die?

[Xei]

You can get that result if you assume Giant only thinks one step ahead but Elf can think two. Giant thinks the strongest poison will surely kill Elf. Elf knows this so he drinks the weakest poison beforehand, and is cured by Giant's poison. Giant also thinks the strongest poison will surely cure Elf's. Elf knows this so he gives Giant water. If they were able to think further or even arbitrarily far ahead, things'd get complicated.

[Nelzi]

... Maybe Nelzi secretly has a PhD, he's awesome. Sorry about your WBTB! You should try to find a zinger of a puzzle to get your revenge. ... Thanks , but the credit must go to you for coming up with these challenges. Yeah I might look for some puzzles... or try to invent some. I don't have a PhD, lol, and I don't want to. While those people are obviously the best within their respective fields, they are often limited by the fact that their perspective is so specific and they defend their own approach so vehemently (even while they have honest intentions). Also, a PhD is no guarantee for a happy and healthy state of being, especially if one feels the need to document the own intellect with that piece of paper and to prove it to other people. I agree with you that true intellect is more a matter of trust in one's own intuition, so it is accessible to everyone who believes in themselves. Back to topic. I had some thoughts on the Blue eyes problem: Spoiler for Blue eyes: What's undoubtedly true for me: - every blue-eyed sees 99 blue-eyed, 100 brown-eyed, 1 green-eyed - every brown-eyed sees 100 blue-eyed, 99 brown-eyed, 1 green-eyed - the brown-eyed ones and the guru can never leave the island, because 'brown' and 'green' is never mentioned, each of them might as well have another, fourth color altogether. So the only question each of the 201 people can ask themselves is: 'do I have blue eyes or not' - there is no distinction between people with the same eye-color and each of them knows that plus each of them knows, that everyone else knows that too . So they treat a group of people with the same eye-color (no matter if they see 99 or 100 of them) as one individual mind. So I'm thinking: The important thing is not how many people there are of a specific color (might as well be 1000, but there has to be more than one, so that there is always at least one observer plus one observed within a color-group), but the important thing is how many different colors there are. With this premise, the problem is reduced in number of people: - a blue-eyed sees one (collective) blue-eyed mind, one (collective) brown-eyed mind, one green-eyed - a brown-eyed sees one (collective) blue-eyed mind, one collective brown-eyed mind, one green-eyed Basically my point is: a blue-eyed sees one other blue eyed 'person' and asks himself, whether that is all of the present blue-eyed ones or whether he is included in that group as well (and he knows that that blue-eyed 'person' asks the same question). If he sees that the blue-eyed 'person' does not leave on the first night, then it must be because they see a blue-eyed person. So all blue eyed persons will know it the second night and leave the island (????????). Hmm funny thing is that each brown-eyed and even the guru will believe the same thing the second night, so they will also claim to have blue eyes, but they are then of course not let off the island... yes I have my doubts about this...

[Alyzarin]

Spoiler for Blue eyes: What's undoubtedly true for me: - every blue-eyed sees 99 blue-eyed, 100 brown-eyed, 1 green-eyed - every brown-eyed sees 100 blue-eyed, 99 brown-eyed, 1 green-eyed - the brown-eyed ones and the guru can never leave the island, because 'brown' and 'green' is never mentioned, each of them might as well have another, fourth color altogether. So the only question each of the 201 people can ask themselves is: 'do I have blue eyes or not' - there is no distinction between people with the same eye-color and each of them knows that plus each of them knows, that everyone else knows that too . So they treat a group of people with the same eye-color (no matter if they see 99 or 100 of them) as one individual mind. So I'm thinking: The important thing is not how many people there are of a specific color (might as well be 1000, but there has to be more than one, so that there is always at least one observer plus one observed within a color-group), but the important thing is how many different colors there are. With this premise, the problem is reduced in number of people: - a blue-eyed sees one (collective) blue-eyed mind, one (collective) brown-eyed mind, one green-eyed - a brown-eyed sees one (collective) blue-eyed mind, one collective brown-eyed mind, one green-eyed Basically my point is: a blue-eyed sees one other blue eyed 'person' and asks himself, whether that is all of the present blue-eyed ones or whether he is included in that group as well (and he knows that that blue-eyed 'person' asks the same question). If he sees that the blue-eyed 'person' does not leave on the first night, then it must be because they see a blue-eyed person. So all blue eyed persons will know it the second night and leave the island (????????). Hmm funny thing is that each brown-eyed and even the guru will believe the same thing the second night, so they will also claim to have blue eyes, but they are then of course not let off the island... yes I have my doubts about this... Yeah, it fails because every single blue-eyed person already knows that every other blue-eyed person already sees another blue-eyed person who is not them, so it really doesn't tell them anything lol.

[melanieb]

Puzzle-puzzle. While exploring the wild highlands of Ireland, Robert was captured by goblins. Grumpy, the chief of the goblins told him he was allowed one final statement on which would hinge how he would die. If the statement he made was false, he would be boiled in water. If the statement were true, he would be fried in oil. Sine Robert did not like either option, so he made a statement that forced the goblins to release him. What is the one statement he could make to save himself?

[SnowyCat]

Puzzle-puzzle. While exploring the wild highlands of Ireland, Robert was captured by goblins. Grumpy, the chief of the goblins told him he was allowed one final statement on which would hinge how he would die. If the statement he made was false, he would be boiled in water. If the statement were true, he would be fried in oil. Sine Robert did not like either option, so he made a statement that forced the goblins to release him. What is the one statement he could make to save himself? Spoiler for solution: He could say, "This statement is false" or something related. If he was boiled for making a false statement, it would be assumed that his statement being false is false, which would make it true. Therefore, his punishment would be unjust. Similarly, if he were fried for telling the truth, it would be assumed that his statement being false is true, which would require he be boiled instead. It's pretty much the "liar paradox" being practically applied.

[melanieb]

It's something related but the wording is what's important. You need the right wording.

[SnowyCat]

It's something related but the wording is what's important. You need the right wording. Spoiler for ?: Other options include "This statement is untrue" and "I am lying". They all check out logically, though... (or should I say the all fail to check out logically? ) Is there something I'm missing here?

[Xei]

To be fair that's a legit answer... any statement which can't be assigned a truth value answers the question. "I will be boiled in water" is what the question probably wants you to say but it isn't necessary. Also, grumpy Goblins aren't exactly known for their logical fidelity or tolerance of smartass answers, so "I have a vault full of gold" or something similar would probably be a wiser answer.

[Arra]

I'm still trying to figure out the blue eyes one and there doesn't seem to be a solution. Spoiler for thoughts on blue eyes: There are only a few possible actions that might be taken which could be taken as clues about eye color a person has: - The guru says the line - All the blue-eyed people leave - All the brown-eyed people leave Or the opposite of all of those. Ex: The fact that the guru hasn't said the line might hint at something. For example, if it were true that: "if the guru saw 99 blue-eyed people and 101 brown-eyed people she'd say the line," then a blue-eyed person witnessing the fact that she hasn't said that would know that he has blue eyes. But that quoted statement isn't true as far as I can tell. I've also realized that (I guess I thought of this because of the hat problem), for every person, the question "do I have blue eyes?" can be rephrased as "Are there an odd number of an even number of blue-eyed people?" If someone knows the answer to that question, they'll know whether they have blue eyes or not. It seems I've gone down every track already and have come up with nothing. If the guru never says anything, no one will ever know their eye color, so no one can leave. If the guru says something on any day, that still doesn't tell anyone anything. The only way the guru saying anything could lead to someone realizing their eye color is if there were only one blue-eyed person left. But that state can never happen since there is more than one blue-eyed person. Also, individual people can't leave. Either all blue eyed people would have to leave as a group, or all brown-eyed people, or all of them together. (I'm pretty sure the guru can't leave as she could have purple eyes and how would she ever know that?) There just doesn't seem to be an answer. All of the other problems seemed to be solvable before I even heard or thought of the solution. But this one... I feel like I've thought about it from every perspective already, so there just isn't a solution unless I'm misunderstanding the question. The only type of solution I can think of is for them to all individually come up with the simplest possible plan, which I believe would be something like "the guru counts the number of blue-eyed people and waits that many days before saying the line." They don't communicate this plan, but because they're all completely rational they realize that all the other people will be thinking along the same lines as themselves and assume this plan as well. Since that's the most likely way they'll get out, and they know they're all completely rational, it isn't just "the best chance" anymore. Instead it becomes 100% certain that that's what they're all going to do. Then all the blue-eyed people would leave on the 100th night. Is that the answer...?