Puzzle thread? Puzzle thread.

As anybody who frequents chat knows, I freakin' love puzzles. Everybody loves puzzles! But there's a lot of crap out there, and there's nothing more annoying than spending hours steaming your noodles only to find that it's some kind of lame trick question. So I thought it'd be a good idea to create a thread where we can share really good ones and tackle them together. Puzzles involving creative or clever thinking are what we want. There should be one answer, and you should know it when you get it. The best ones should seem impenetrable, until you receive a sudden flash of inspiration. Word riddles are fine, as long as they satisfy the above. I'll post a few favourites off the top of my head to get the ball rolling. Please don't spoil any you've seen before.


The field and the farmer

You love sweetcorn. You're sat in the middle of a circular sweetcorn field, eating some sweetcorn. There is a path around the circular sweetcorn field, upon which rides an angry old sweetcorn farmer with a pitchfork, on a tricycle. The tricycle moves 4 times as fast as you do, but it cannot go off-road. Can you definitely escape?


The prisoners and the light

There are 20 prisoners in a prison. The warden of the prison is slightly insane, so one day, he takes all the prisoners outside into the yard and tells them the following. From now on, once every day, he will select a random prisoner, and take them from their cell to a disused bathroom. In the room is a switch; this switch turns the room's light on and off. Then he'll take the prisoner back to their cell. At any time, any prisoner may say to the warden, "we have all been to the bathroom now". If that prisoner is right, they're all freed. If the prisoner is wrong, they're all fed to the prison dogs. They all have an hour to talk to each other and form a plan, before they're taken back inside the prison, where they can't communicate. What do they do?


The coins and the table

You are blindfolded in a room with a table in the middle. Scattered across the floor are several thick gold coins. You're told that half are heads up and half are tails up, but you don't know and can't tell which each individual coin is. To win the coins, you must put some of them on the table, in such a way that the number of heads on the table is the same as the number of heads on the floor. Can you do it?

First one.

Spoiler for ewrwe:

No, I don't think so.
If I start at one edge and run to the other (maximising the distance he must go, while also minimising the distance I must go), say it is 10m diameter.
He will have to go (c= pi X d)/2 which is 15.7m and I will go 10m, obviously.
At 1m p/s it will take me 10 seconds to get to the other side. At 4m p/s it will take him 3.9 seconds.

Quote:

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Originally Posted by tommo

First one.

Spoiler for ewrwe:

No, I don't think so.
If I start at one edge and run to the other (maximising the distance he must go, while also minimising the distance I must go), say it is 10m diameter.
He will have to go (c= pi X d)/2 which is 15.7m and I will go 10m, obviously.
At 1m p/s it will take me 10 seconds to get to the other side. At 4m p/s it will take him 3.9 seconds.

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Yeah, but you're discounting alternatives. Like if you run to almost the end, and he follows you, then at the last moment you run another direction.

I haven't figured this one out yet, but I have a feeling it has an answer.

Oh, and, for the second two, I do have 'answers' but I think they're likely bs. I started typing them out once and gave up.

For the prison one, how about every person just make a scratch on the wall when they go in? Then when someone sees 19 scratches, he'll know that he's the last.

For the coin one, just put all the coins on the ground in the same area, then turn the table upside down onto those coins. Then all the coins will be on the table and on the ground. (Or gather up all the coins, and then 0 will be on the table and on the ground, but that may not be allowed since you said 'you must put some of them on the table'.

Hah, the table thing's genius. Though if that was the answer, I'd've thought the person asking it was a bit of an asshole, 'cos that's kinda a trick question. No, it's too heavy to lift or whatever, and the coins have to either be on the table or be on the floor. Moving them about is all you can do. ;)

If that was the answer to the prisoner one I'd definitely be an asshole. No, you can't make any marks or anything. It's all about the light. Light on and light off is the only information involved.

Quote:

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Originally Posted by tommo

First one.

Spoiler for ewrwe:

No, I don't think so.
If I start at one edge and run to the other (maximising the distance he must go, while also minimising the distance I must go), say it is 10m diameter.
He will have to go (c= pi X d)/2 which is 15.7m and I will go 10m, obviously.
At 1m p/s it will take me 10 seconds to get to the other side. At 4m p/s it will take him 3.9 seconds.

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Good try, but like Dianeva says, there are alternatives. Though it seems sensible, it turns out that the assumption that the shortest path is the best you can possibly isn't right. :)

If anybody knows any good ones please share, I don't want this to be a "Xei knows the answer and you are dumb" thread. :P

Well, the question for the last one says, "can you do it." So, yes, by luck, puzzle solved.

My guesses:

Field and Farmer: Go to the edge where the farmer comes close to you, then run the opposite direction he is facing so he has to go around the entire circle to catch up...

20 prisoners: If they are in the same block, they could count each time they hear the doors to their cells open.

Coins on the table: Flip all the coins...put half on the table and half on the floor...

Random Riddle from the intarwebs...


"A farmer challenges an engineer, a physicist, and a mathematician to fence off the largest amount of area using the least amount of fence.

The engineer made his fence in a circle and said it was the most efficient.

The physicist made a long line and said that the length was infinite. Then he said that fencing half of the Earth was the best.

The mathematician laughed at the others and with his design, beat the others. What did he do?"

Quote:

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Originally Posted by Dianeva

Yeah, but you're discounting alternatives. Like if you run to almost the end, and he follows you, then at the last moment you run another direction.

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Yes, I was thinking this as I was going to sleep last night.
But.... still I don't know if that would work.

It really depends how he follows you.
I mean if you run in a tiny circle, and he constantly tries to keep the shortest distance to you by going around the entire circle, then yeah, you could maybe get out while he's still trying to catch up. But that would just stupid. So assuming it's a real person, that for some reason can't go inside the circle, and not an automaton, then you can't.

I mean even if I take a 1cm step in one direction, he goes around to the side I'm closest to, and then I run in the other direction, it will still take me 5 seconds and him 3.9 seconds.

Quote:

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Good try, but like Dianeva says, there are alternatives. Though it seems sensible, it turns out that the assumption that the shortest path is the best you can possibly isn't right.

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Well, the assumption was the shortest path while maximising the length of his path.


Prison light one
"What do they do?"
Kill the guard when he takes the first guy to the bathroom.

Spoiler for seriously though it's quite simple:

Assuming he can't take the same prisoner twice.
After twenty days the prisoner taken on that day can go to the bathroom and then tell him they've all been there

This is so frustrating. I can't stop thinking about these puzzles. I was just playing a 2 hour game of l4d and was terrible because I was just thinking about that fucking light switch one the whole time.

For the first, I've realized it's hard to come up with a solution without knowing the algorithm the farmer is following. Does he always stay as close to you as possible? Or if he sees you running in a direction, will he assume you're headed for the other side in a straight line and head for the other end? Or are you just supposed to assume he could go anywhere? If so, there can't be a solution using any definite path, as he could always just place himself at your planned point of exit, from the beginning. So the solution may be some algorithm that will depend on where he goes, which would get complicated.

The prisoners. I may have somewhat of a solution which is probably inadequate.

Spoiler for Prisoners solution:

They consider time in segments of some number of days. The greater the number of days, the longer it will take, but it will be more likely that it will end up working eventually. I'll pretend it's a month, but it should probably be more than that.

If this is the first time the person has been to the bathroom with the light off when he entered, he leaves the light on for the next person.
Otherwise, the person leaves the light unchanged.
At the end of the month, the last person turns the light off.

This way, each month, if someone new has been to the bathroom, everyone who went after he did will know because they saw the light on. So each person just has to wait until they've witnessed the light on in 19 separate months. But this isn't a perfect solution. Not only would it take a really long time (19 months minimum), but it isn't guaranteed to work. Like if the warden chose to focus on the same group of 10 people for one month, then the next 10 the next month, and keep alternating, then neither group could be aware of the other being there.

EDIT: !!! This is getting so fucking complicated! I've realized that might not really work well unless someone near the beginning of the month turns the light on. Because if the person on the last day or so turns the light on, there will be hardly anyone to see it before it's turned off again. But there's no way to signal that the whole process should start over if the light isn't turned on early enough.... maybe each prisoner should turn the light on his first time in the bathroom with the light off, only if it's in the first few days of the month. Otherwise he should save his statement for a time he's chosen earlier in the month, so more people will see it. I suppose you could use math to come up with some span of days that will make it extremely likely that at least one person will witness all 19 switches being turned on... I don't feel like doing that now though.

And the coin one... I don't see how there can be a solution to that. No matter what you do to the coins, you can never know which ones are heads. I can't even think of some indirect way to do it. If you pick up more than half of them, you know that you have at least one heads and one tails. That's about as far as I can imagine knowing something that might seem unintuitive. I don't know where you could go from there though.

Quote:

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Originally Posted by Chimpertainment

"A farmer challenges an engineer, a physicist, and a mathematician to fence off the largest amount of area using the least amount of fence.

The engineer made his fence in a circle and said it was the most efficient.

The physicist made a long line and said that the length was infinite. Then he said that fencing half of the Earth was the best.

The mathematician laughed at the others and with his design, beat the others. What did he do?"

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Don't do anything, say the fenced area is infinitely small but it's really the outside and everything outside of it (ie. all space) is being fenced off?

Actually just realised a possible solution to the coin one as well

Spoiler for efwe:

put all the coins on their edges like _O_ that, standing up, there are 0 heads on the table and 0 on the floor.

I'm glad you're enjoying them. The light switch one pestered me for about two days and nights, lol. Though I simplified the puzzle slightly.

Quote:

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Originally Posted by Chimpertainment

My guesses:

Field and Farmer: Go to the edge where the farmer comes close to you, then run the opposite direction he is facing so he has to go around the entire circle to catch up...

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The farmer can turn around. :P

He can also slow down and stop of course, as can you. 4 times your max speed is his max speed.

Quote:

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20 prisoners: If they are in the same block, they could count each time they hear the doors to their cells open.

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But they might just hear the same person going 20 times.

The light is their only means of communication.

Quote:

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Coins on the table: Flip all the coins...put half on the table and half on the floor...

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Half are heads and half are tails. If you flip them all this is still the case. But the half you put on the table could all be heads, for instance. Then all the heads would be on the table, and no heads on the floor.

Quote:

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Random Riddle from the intarwebs...

"A farmer challenges an engineer, a physicist, and a mathematician to fence off the largest amount of area using the least amount of fence.

The engineer made his fence in a circle and said it was the most efficient.

The physicist made a long line and said that the length was infinite. Then he said that fencing half of the Earth was the best.

The mathematician laughed at the others and with his design, beat the others. What did he do?"

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I can't really tell what the physicist is supposed to have done... and I don't know what "fence off the largest about of area using the least amount of fence" means. Who wins if one person has a larger area but lots of fence and another has a small area but a small amount of fence?

My vague guess at the moment is basically what Dianeva said.

Quote:

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Originally Posted by tommo

I mean even if I take a 1cm step in one direction, he goes around to the side I'm closest to, and then I run in the other direction, it will still take me 5 seconds and him 3.9 seconds.

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That's all true, and you're thinking along the right lines, but perhaps there is a better technique. :)

Quote:

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After twenty days the prisoner taken on that day can go to the bathroom and then tell him they've all been there.

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The prisoners are selected at random - so repeats are allowed.

Though you can be sure that they'll all be picked eventually. It may take a very long time... but you can assume that the prisoners don't die by then. :P

Quote:

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put all the coins on their edges like _O_ that, standing up, there are 0 heads on the table and 0 on the floor.

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Dang, that's another good answer. I shouldn't have mentioned that the coins are thick, it's actually irrelevant. So... yeah, say that have curved edges or whatever and can't be balanced. Heads or tails only, sorry.

I promise it is possible with a few simple steps. It's probably a good idea to focus on this one first, actually.

Quote:

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Originally Posted by Dianeva

For the first, I've realized it's hard to come up with a solution without knowing the algorithm the farmer is following. Does he always stay as close to you as possible? Or if he sees you running in a direction, will he assume you're headed for the other side in a straight line and head for the other end? Or are you just supposed to assume he could go anywhere? If so, there can't be a solution using any definite path, as he could always just place himself at your planned point of exit, from the beginning. So the solution may be some algorithm that will depend on where he goes, which would get complicated.

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Yeah these are all good questions and insights. Basically the farmer may use any of those plans. But there may be a way to beat him no matter what he does. That's what the question's asking for. Can you always beat him no matter what he does? Or can he always beat you no matter what you do? I promise the answer doesn't use a complicated algorithm. Just a simple idea.

Quote:

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The prisoners. I may have somewhat of a solution which is probably inadequate.

Spoiler for Prisoners solution:

They consider time in segments of some number of days. The greater the number of days, the longer it will take, but it will be more likely that it will end up working eventually. I'll pretend it's a month, but it should probably be more than that.

If this is the first time the person has been to the bathroom with the light off when he entered, he leaves the light on for the next person.
Otherwise, the person leaves the light unchanged.
At the end of the month, the last person turns the light off.

This way, each month, if someone new has been to the bathroom, everyone who went after he did will know because they saw the light on. So each person just has to wait until they've witnessed the light on in 19 separate months. But this isn't a perfect solution. Not only would it take a really long time (19 months minimum), but it isn't guaranteed to work. Like if the warden chose to focus on the same group of 10 people for one month, then the next 10 the next month, and keep alternating, then neither group could be aware of the other being there.

EDIT: !!! This is getting so fucking complicated! I've realized that might not really work well unless someone near the beginning of the month turns the light on. Because if the person on the last day or so turns the light on, there will be hardly anyone to see it before it's turned off again. But there's no way to signal that the whole process should start over if the light isn't turned on early enough.... maybe each prisoner should turn the light on his first time in the bathroom with the light off, only if it's in the first few days of the month. Otherwise he should save his statement for a time he's chosen earlier in the month, so more people will see it. I suppose you could use math to come up with some span of days that will make it extremely likely that at least one person will witness all 19 switches being turned on... I don't feel like doing that now though.

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Yes, the really big problem with this method is that, even if a new prisoner turned the light on at the start of the month, it seems quite unlikely that any one prisoner would get to visit the bathroom every single month (it's around 1% I think), and so there would be at least one prisoner they didn't count. In any case, it's definitely not certain. Even if you changed it from a month to a year, you still couldn't be certain. The answer uses quite a simple idea which will certainly work, but it took me a long while to stumble across it. But it looks like you're thinking along the right lines.

Hm.. is the farmer one,

Spoiler for possible solution:

No, you cannot escape because r > pi*r/4 => 1 > ~0.785.
This assumes you start in the center and dart opposite the direction of the farmer's initial position.

Right?

Or,

Spoiler for ...:

would the optimal path be a spiral? Though, you still probably wouldn't be able to escape, but don't ask to prove this; I've forgotten my polar maths.

Quote:

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Originally Posted by reci

Hm.. is the farmer one,

Spoiler for possible solution:

No, you cannot escape because r > pi*r/4 => 1 > ~0.785.
This assumes you start in the center and dart opposite the direction of the farmer's initial position.

Right?

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I don't know the answer but I know this isn't it, since you can't make assumptions. Unless it really is impossible.

Quote:

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Originally Posted by reci

Or,

Spoiler for ...:

would the optimal path be a spiral? Though, you still probably wouldn't be able to escape, but don't ask to prove this; I've forgotten my polar maths.

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I was thinking that too but that wouldn't work either. If the farmer could do anything he could just go to your point of exit, so no planned path would work.

For the coins: just place half of them on the table and flip them (only those on the table). I liked it, had a big smile on my face, when I got that insight :)

Wow the farmer is cool too :D:
lets say the radius is 100 m, then:
-> the farmer's circle has a length of 628,3 m
-> go out 22 m from the center. You are now on a smaller circle with total length 138,2 m, which is less than one fourth of the farmers total length. So you can get on his opposite side by following your smaller, inner circle, even though he is 4 times as fast!
-> now your shortest path to freedom is 78 m, while the farmer still has to travel half his circle to reach that point. While you run the 78 m, he will travel 78 x 4 = 312 m on his path, which is about 2 m less than half his circle!
-> hope his pitchfork is less than 2 m long! :D

The prisoner lights will keep me awake though (why post insomnia inducing riddles in a dreaming forum btw?). I enjoy this thread, will try to find riddles also...

Farmer problem

My first thought was that you could not escape the farmer, because he can outpace you by a factor of pi or greater. So he can ride further than the circumference of a semicircle in the same time that it takes you to run the radius of that same circle. If you simply run in a straight line, he will always catch you.

But then I scrolled down and read Xei's hint about running in a straight line not necessarily being the best strategy. After considering this hint, I believe I have a solution. The solution is quite easy to visualize, but turned out to be slightly long and complicated to explain in words, or at least to do so in any kind of thorough way. Some of this could probably be simplified, but it's late...

Spoiler for farmer solution:

TL, DR: The path that you end up running to escape the farmer involves running an outward-spiraling curve from the center to just past pi/4 radius lengths from the center, at which point you can turn and dash in a straight line directly away from the center and out of the corn field.

We have already established that running in a straight line away from the farmer's initial position will not work. However, if you had a "head start" such that you did not have to run the full radius of the circle, but instead only had to run less than pi/4 of the radius of the circle, then from there you could escape the farmer by running in a straight line. So if we can show that there is a way to get yourself into the position of being pi/4 radius lengths from the edge of the circle, and with the farmer on the edge directly opposite the center of the circle from you, then we will have shown that you can escape the farmer. Let's call this position the "safe zone."

It turns out there is a way to successfully get to the safe zone. The trick is to run not in a straight line, but instead to run along a path such that, at any given instant, you are headed directly away from the farmer's current or instantaneous position. This leads to you running a curved, outward-spiraling sort of path. (The curve arises because the farmer will presumably be racing around the edge of the cornfield trying to get closer to you, rather than just sitting still. Obviously the problem is trivial if he just sits still.) But of course, you can't keep up this spiraling sort of path the entire way out to the edge of the circle. Because the farmer's pace exceeds yours, there will necessarily be a point in this spiral path at which you will no longer be able to approach the edge of the circle, at any angle, and still also be running directly away from the farmer's instantaneous location. Basically, there is a smaller, inner circle within which you can run on such a path, and outside of which you are not physically able to do so. Let's call this inner circle the "spiral zone," for lack of a more creative name.

If there is any overlap between the spiral zone and the safe zone, then you can always reach the safe zone by using this spiraling strategy, and therefore you can always escape the farmer. Below I show that the spiral zone and the safe zone do in fact overlap.

Letting r = the radius of the circular corn field, the distance from the center of the circle to the safe zone is r - pi*r/4 = r(1 - pi/4).

The circle defining the spiral zone has radius r/4. We know this because the farmer can outpace you by a factor of 4, so in the time it takes him to travel the circumference of his circle (or any fraction thereof), you can in that same time travel the circumference of a circle (or any fraction thereof) that has at most 1/4 the radius of his circle. That smaller circle, of course, is what defines the spiral zone. So anyway, this all means that the distance from the center of the circle to the edge of the spiral zone is r/4.

We can rewrite r/4 as r(1 - 3/4), which makes it easy to compare to r(1 - pi/4), which was the distance from the center of the circle to the safe zone. Because pi > 3, the distance from the center to the edge of the spiral zone exceeds the distance from the center to the safe zone, therefore the safe zone and the spiral zone overlap (barely in this case), therefore you can always escape the farmer.

The path that you end up running to do so entails running an outward-spiraling curve from the center to just past pi/4 from the center, at which point you run in a straight line away from the center and out of the corn field.

I will probably look at the others later. As of now I haven't even looked, to prevent myself from spending time trying to solve them right now... I am supposed to be pulling an all-nighter getting important work done, not solving silly puzzles :P

So much math for the farmer problem o.o
How about this for the farmer one:

Spoiler for The Answer!:

What if you take the sweetcorn and lay it across the path. He wouldn't be able to go over it because he can't go offroad. So just make a path for yourself to walk on that is thick enough so you don't get stabbed by his pitchfork.

Quote:

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Originally Posted by Nelzi

For the coins: just place half of them on the table and flip them (only those on the table). I liked it, had a big smile on my face, when I got that insight

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Kind of mad at myself for not getting this one earlier. The answer was so simple.

Just looked at Nelzi's farmer solution. I didn't check the exact figures but it appears to be basically the same idea that I elaborated. I guess you won't be surprised to hear that I think it's missing some rather important details, but then maybe I am just overly verbose :whyme:.

Maybe I'm missing something but I don't see how Nelzi's coin solution does the trick. It clearly works in expectation -- that is, the most likely number of heads remaining on the floor after you pick up half of the coins is 1/2 of the original total -- and likewise the most likely number of heads on the table after you flip that half of the coins is again 1/2 of the total -- but for any one particular attempt there is no guarantee that the two numbers of heads will be equal. In fact, if the total number of coins is large, the strategy will very rarely work on any given attempt.

My feeling right now is that anything involving flipping the coins is not going to be a good way to go, because it introduces additional uncertainty into the situation. But I'll think on it some more.

Edit: I see that reci hit on one of my ideas for the farmer solution as well.

I've got the coin one.

Spoiler for coin solution:

Take each coin and place it on the table, without turning it over, one at a time so that, one by one, you eventually transfer every coin from the floor to the table. No matter which coins you pick in which order, at some point the number of heads on the table has to equal the number of heads on the floor, assuming that there are an even number of heads (which I guess we can assume, because the task is impossible if there are an odd number of heads).

Of course, you won't be aware at exactly what point you have satisfied the task, since you can't see the coins...

Edit: ....actually this solution doesn't work in the general case. It works under the assumption that I state, but I see now that my reasoning in justifying the assumption is wrong, and the assumption need not hold.

^ I think you're overcomplicating it. No matter how I figure it, putting half the coins on the table and flipping them works.

Let's say you have 20 coins in all - 10 heads, 10 tails on the floor to begin with.

Pick up 1 head and 9 tails, leaving 9 heads and 1 tail on the floor. Of course, when you flip the ones you picked up, you have the same on the table.

Pick up 2 heads and 8 tails - same thing happens. It leaves 8 heads and 2 tails on the floor, and that's what you get on the table when you flip them.

Without actually running through it for every possible combination, it seems clear this will happen every time.

Oh, I see, I was misunderstanding what you meant by "flip." I assumed you meant to "toss" the coin (i.e., so that it lands with an equal chance of head or tails), but I guess you actually just mean to "turn it over" to the other side.

Well, in that case, I like Nelzi's solution better than mine :P

Hahaha yeah, if you had to literally flip them that would complicate things!

I also just figured out that my proposed coin solution is flawed... :( I edited the post.

I will take all of this as a sign that I need to put the puzzles down and get back to work...

I want to figure out the light switch solution. If anyone posts that one I'm not going to look at it.

LOL guys... funny how you are still guessing on solutions after reading mine, because they actually work. To convince yourself of the coin solution, just pick some examples, just like Darkmatters did. And the farmer example I gave does also work, no matter what the farmer does... and yes it is basically the same idea as DuB had. For clarity (and without numbers - for GooseMan), the important insight is that you can find a small circle inside the field (with the same center as the field, think this is the 'spiral zone' suggested by DuB) with a circumference less than than one fourth of the field's circumference. So your speed in angle/time is greater than the farmers angle/time and you can get a head start equal to the inner circle's radius, which is enough to make it in the shortest path.

Well for the coin one, as DuB said he misunderstood what you meant by 'flip' (gets it now). No one else has tried to re-answer that one, so what are you talking about?

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 :chuckle:) 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.

Spoiler for light solution:

Time is divided into 20-day segments.
Each prisoner is assigned a day of the 20 (ex: one prisoner might be day 3), which everyone else memorizes.

Each time a prisoner enters the bathroom, if it is on his day, he leaves the light on.
Now, the next prisoner will know that that particular prisoner has been to the bathroom.

Once a prisoner knows that another prisoner has been to the bathroom, he can now leave the light on that prisoner's day too to tell others that that prisoner has been to the bathroom.

Ex:
On day 3, [3] is chosen and so he leaves the light on.
On day 4, [12] is chosen, so he doesn't leave the light on.
On the next sequence of 20 days, on day 3 (day 23), the light will be left on if either prisoner [3] or [12] is chosen.
And on the next sequence of 20 days, on day 3 (day 43), prisoners [3], [12], and whoever was chosen on day [4] the last month may leave the light on.

It would get off to a slow start, and I haven't done the math, but the knowledge would build up until eventually someone should get it.

EDIT: Damnit aly! That sounds a lot more simple and probably wouldn't take as long as mine. Nice job.

EDIT 2: I want to make a program to test exactly how long these proposed solutions would take.

Since it looks like I missed out on this batch of puzzles, I figured I would share a new one with y'all. I still haven't found the solution, but I'm also not very good at puzzles. ;)

Concert Puzzle: Four men have a concert that starts in 17 minutes and they must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them.

The flashlight must be walked back and forth. It cannot be thrown and other tricks like that are not needed to solve the problem. The solution is simply a matter of allocating resources in a certain order. Each band member walks at a different speed. A pair must walk together at the rate of the slower man's pace:

John: 1 minute to cross
Harry: 2 minutes to cross
Albert: 5 minutes to cross
Sam: 10 minutes to cross

For example: if John and Sam walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Sam then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission.

Congrats to Nelzi and Dub for getting the farmer one, and to Nelzi for getting the coin one! :)

Their answers basically boil down to this:

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.

Spoiler for Solution to the coins and the table:

Put 10 of the coins on the table. Say for example that 3 of them are heads. This means there are 7 tails on the table, and on the floor there are 7 heads and 3 tails. So just turn over all of the coins on the table! Now 7 heads are on the table and 7 heads are on the floor.

Also well done to Alyzarin for getting the hardest bit of the prisoner puzzle, which is probably the hardest puzzle so far. There's still a little problem, though: what if the light is on at the beginning? Then the counter would turn it off and count it as a prisoner. But then he'd prematurely tell the warden they'd all been.

Dianeva, given the way I phrased the question, I believe that's a fair answer. Clever. The warden could easily break it of course, if he were so inclined, by never selecting one of the prisoners on their day, for instance. But I did say he did it randomly, so you could be certain that eventually he would choose the prisoner on their day. How about a very slightly different puzzle where the warden can do such things; the only thing he promises is that each prisoner can always be sure he'll take them to the bathroom eventually (including if they've been before). Alyzarin's solution goes a long way to solving this but isn't quite there.

Edit: Thanks Snowycat for the new riddle. Unfortunately I've seen it before, but it's a nice one! :)

Quote:

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Originally Posted by SnowyCat

Concert Puzzle: Four men have a concert that starts in 17 minutes and they must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them.

The flashlight must be walked back and forth. It cannot be thrown and other tricks like that are not needed to solve the problem. The solution is simply a matter of allocating resources in a certain order. Each band member walks at a different speed. A pair must walk together at the rate of the slower man's pace:

John: 1 minute to cross
Harry: 2 minutes to cross
Albert: 5 minutes to cross
Sam: 10 minutes to cross

For example: if John and Sam walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Sam then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission.

---

Spoiler for solution:

Albert and Sam go together - 10 mins.
John goes to get the flashlight - 2 mins.
John and Harry go together - 2 mins.

The whole thing only takes 14 minutes. I spent about half an hour thinking about this, writing down all the combinations before realizing it wouldn't work - there must be some type of 'move' I haven't thought to make... I guess it's tricky because it isn't immediately obvious to allow someone to go across just to get the flashlight.

EDIT: Oh, shit. Never mind! lol.... John needs a flashlight to get across. I'll keep my failed solution here for laughs.

Quote:

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Originally Posted by Dianeva

Spoiler for solution:

Albert and Sam go together - 10 mins.
John goes to get the flashlight - 2 mins.
John and Harry go together - 2 mins.

The whole thing only takes 14 minutes. I spent about half an hour thinking about this, writing down all the combinations... I guess it's tricky because it isn't immediately obvious to allow someone to go across just to get the flashlight.

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You can't cross the bridge without the flashlight actually in your hand. Or at least that's how I interpreted it.

EDIT: Lol

Quote:

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Originally Posted by Dianeva

EDIT: Damnit aly! That sounds a lot more simple and probably wouldn't take as long as mine. Nice job.

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Thanks! :) I like yours too, though it may take a long time to work, but in that situation I think anything would sound good haha. I never would have thought of something like that, that's more of a team effort than I usually think of lol.

Quote:

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Originally Posted by SnowyCat

Since it looks like I missed out on this batch of puzzles, I figured I would share a new one with y'all. I still haven't found the solution, but I'm also not very good at puzzles. ;)

Concert Puzzle: Four men have a concert that starts in 17 minutes and they must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them.

The flashlight must be walked back and forth. It cannot be thrown and other tricks like that are not needed to solve the problem. The solution is simply a matter of allocating resources in a certain order. Each band member walks at a different speed. A pair must walk together at the rate of the slower man's pace:

John: 1 minute to cross
Harry: 2 minutes to cross
Albert: 5 minutes to cross
Sam: 10 minutes to cross

For example: if John and Sam walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Sam then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission.

---

This one is bugging me.... May be reporting back.

Quote:

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Originally Posted by Xei

Also well done to Alyzarin for getting the hardest bit of the prisoner puzzle, which is probably the hardest puzzle so far. There's still a little problem, though: what if the light is on at the beginning? Then the counter would turn it off and count it as a prisoner. But then he'd prematurely tell the warden they'd all been.

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Oh hmm.... I guess I just figured that whoever went first would know it and could just count that as them turning on the light, or the counter could ignore it if they went first. Is that not the case?

As an extension to Alyzarin's solution, either discount the first day and tell the first prisoner chosen to turn off the light, or better yet, plan that the first prisoner will be the special one who turns the light off and does the counting. The other prisoners will never know who that prisoner is but that doesn't matter.

Ah, good point, I forgot I said it was one per day. Yes, well done Alyzarin for solving it! :)

What if it wasn't one prisoner per day, but just completely random times? In hindsight that's how I should have phrased the puzzle. That makes stuff clearer and eliminates clever solutions like Dianeva's previous one, and makes the initial light problem more serious.

Hey, thanks for the nice challenging puzzles... I found out the concert puzzle and I found out how to do spoilers lol, in case people don't want to look at it:

Spoiler for Concert:

John + Harry -> total time 2 min
John going back -> total time 3 min
Albert + Sam -> total time 13 min
Harry going back -> total time 15 min
John + Harry -> total time 17 min

Was looking for new puzzles to keep the thread alive, but you named it, Xei, there is a lot of crap out there :/

Results of simulation of both my and Alyzarin's solutions to The Prisoners and the Light:

Spoiler for results:

My solution:
Trials: 100000
Average days: 2379.81
Min days: 1332
Max days: 5663

Alyzarin's solution:
Trials: 100000
Average days: 452.038
Min days: 169
Max days: 947

So her's takes 19% of the time that mine does and so is way better. The lowest number of days for mine doesn't even even match up to the maximum for her's. It was obvious but I just wanted to see, and got to brush up on programming.

Very cool! Seems intuitively plausible... nevertheless, don't discount the fact that you found an alternative and valid solution to the problem.

Okay, looks like we're all out. Here's a couple of classic ones about hats for the time being, but please post more!

The first is probably easier, though less interesting. The second is harder but a lot more fun, and the answer is clever. I think there's only one answer, anyway - the answer I got agreed with the answer of the person who asked me.


The Three Fezes

There is a box in a dark room containing 3 red fezes and 2 blue fezes.

3 men enter this room, take a fez from the box, put it on their head, and exit. They then try to ascertain the colour of their own fez; they can only see the fezes that the others are wearing.

After some time, the first man says that he does not know. Some more time passes, and the second man says that he does not know, either. Finally the third man, who is blind, declares the colour of his fez.

What was it, and why?


The Sixty Sombreros

The insane prison warden is rather irritated by the recent emancipation of twenty of his prisoners, so he plans a new, tougher trial for his inmates.

He gathers 60 of them in the yard, and tells them the following:

"In one hour's time, I will make all of you stand in a queue, so that each of you can only see the prisoners in front of you. If you turn around, I will shoot you and feed you to the prison dogs.

I will then place either a white or a black sombrero upon each of your heads. Don't ask me where I got them. Each of you may then, in whatever order you like, shout out the colour of your own sombrero.

If you get it right, you'll be freed. If you get it wrong, I'll shoot you and feed you to the prison dogs."

The prisoners then come up with a plan which is guaranteed to save at least 59 of them. How?

I've got the answer for the 3 fezes (can't believe I'm the first!) And I'll go one better and say the blind guy knows not only what color his own fez is but the other 2 as well. Pretty good for a blind guy!

Spoiler for the answer:

The blind guy's fez is blue.

1) You can eliminate all 3 of them having red, because if the first two had seen 2 red fezes they would have quickly answered "I don't know". The fact that they both took time to think about it means they each saw one red and one blue fez.

2) You can eliminate either of them seeing 2 blue fezes because anyone seeing 2 blue fezes would instantly know he has red.

3) That means there's got to be 1 blue fez, and since both of the first 2 saw it it must be on the third guy.

Heh, I don't even care to elaborate on all the convoluted scenarios I went through to figure this out, but I will say it involved Larry Moe and a blind Curly...

And interestingly (or maybe not) I could not make any headway until I had slept on it after pondering for longer than I care to admit last night. As soon as I woke up I just had the answer.

That's pretty close but it's not exactly right. I didn't really specify if they answered quickly or otherwise. And I don't really see how the other two hats both being red makes it any easier than if they were red and blue, for the first guy at least.

D'oh!!!

Actually you did say they both took some time: "After some time, the first man says that he does not know. Some more time passes, and the second man says that he does not know, either."

I assumed that was an important detail.

You're right the first guy could only know his if the other 2 are blue, so you can't eliminate all 3 being red until after the 2nd guy answers - I should have said the first guy proves the other 2 aren't both wearing blue and the second proves that there's only 1 blue fez in play.

Also of course my answer depends on them knowing how many red and blue hats were in the box, which you never stated.

Quote:

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Originally Posted by Xei[b

The Sixty Sombreros[/b]

The insane prison warden is rather irritated by the recent emancipation of twenty of his prisoners, so he plans a new, tougher trial for his inmates.

He gathers 60 of them in the yard, and tells them the following:

"In one hour's time, I will make all of you stand in a queue, so that each of you can only see the prisoners in front of you. If you turn around, I will shoot you and feed you to the prison dogs.

I will then place either a white or a black sombrero upon each of your heads. Don't ask me where I got them. Each of you may then, in whatever order you like, shout out the colour of your own sombrero.

If you get it right, you'll be freed. If you get it wrong, I'll shoot you and feed you to the prison dogs."

The prisoners then come up with a plan which is guaranteed to save at least 59 of them. How?

---

I don't get what is allowed of the prisoners in this question. Can the prisoners choose their queue order, or is it assigned to them? Also, can they communicate in any way once they have the hats on? I'm guessing not, since if they could everyone could just tell the person in front of them what color hat they have. So the whole answer must have something to do with how they line up... but that doesn't make sense either. If the warden can put hats on anyone then choosing some special lineup with no further communication couldn't tell them anything. Seems like I'm missing some big point of this one.

Quote:

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Originally Posted by Dianeva

Results of simulation of both my and Alyzarin's solutions to The Prisoners and the Light:

Spoiler for results:

My solution:
Trials: 100000
Average days: 2379.81
Min days: 1332
Max days: 5663

Alyzarin's solution:
Trials: 100000
Average days: 452.038
Min days: 169
Max days: 947

So her's takes 19% of the time that mine does and so is way better. The lowest number of days for mine doesn't even even match up to the maximum for her's. It was obvious but I just wanted to see, and got to brush up on programming.

---

Oh wow, you actually did it. :chuckle: That's cool to know. :P

Quote:

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Originally Posted by Xei

The Three Fezes

There is a box in a dark room containing 3 red fezes and 2 blue fezes.

3 men enter this room, take a fez from the box, put it on their head, and exit. They then try to ascertain the colour of their own fez; they can only see the fezes that the others are wearing.

After some time, the first man says that he does not know. Some more time passes, and the second man says that he does not know, either. Finally the third man, who is blind, declares the colour of his fez.

What was it, and why?

---

My solution:

It was red. The first person would have to see either two reds or a red and a blue to not know his own color. But, the second person would hear this and be aware of what both the third and the first person were wearing. If the first person was blue and the first was blue, he was red. If the first was red and the third was blue, he would know that he would have to be red as well because otherwise the first person wouldn't have had to say I don't know. The second person only doesn't know if the first and third person are wearing red. This opens up the possibility that the first person either didn't know because of two reds OR a red on third and a blue on second, so he couldn't be sure about it. Therefore, the third person must know he's wearing red when the second person passes.

Still working on the sombrero one....

Alright, tell me if I'm off here.... I'm not sure if this seems a little too complicated or not, but unless Dianeva is right about there being some information missing this is the first thing I could think of.

My sombrero solution:

The person at the very back of the line well shout out the color of the person in front of them. That gives them a 50/50 chance of survival, which is why you can only save 59 out of the 60. The second person now knows their color, and what they do is if the next person has the same color as them, they yell it out right away. Then the next person knows theirs and can follow these same rules. If the next person's color is different, they hesitate just for a moment. The next person now knows they have the opposite of the previous person's color. If the next person has the same color as them they can say it right away, and then whoever got left behind to hesitate can say their own before the line picks up where it left off. If the next person has the opposite color of them too, then they can also hesitate. Once the moment has gone on long enough, the first person who has to pass can say their own color, and then the person in front of them can say their own color, and then finally the last person can pick up where they left off, and this can go on until the end.

An example would be... the second person has black, so the first person says black and either is set free or dies. Now, the third person has black, so the second person says black and is set free. The fourth person has white though, so the third person hesitates. The fifth person also has white, so the fourth person says white, then the third person says black, and then the fifth person can pick up where they left off. The sixth person has black, so they hesitate, and the seventh person has white, so the sixth hesitates as well. This can signal the fifth person to say their own, white, and then the sixth person can say their own, black, and the seventh person picks it up again. This goes until the 59th person either says their own or hesitates on the 60th, telling them theirs.

Clever solution regarding the sombrero one - it makes sense. No cigar though - let me clarify a couple of things.

The only information the prisoners can convey is a hat colour. The warden calls out "next", and the next person then shouts either 'black' or 'white'. No other words, no pauses, or they're fed to the prison dogs.

Regarding the queue, I don't think it really matters. But for the sake of argument let's say the Warden tells them where to stand.

Quote:

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Originally Posted by Darkmatters

Actually you did say they both took some time

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Sure, but an unspecified amount. They mull it over to the best of their ability, that's all it means.

Quote:

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You're right the first guy could only know his if the other 2 are blue, so you can't eliminate all 3 being red until after the 2nd guy answers - I should have said the first guy proves the other 2 aren't both wearing blue and the second proves that there's only 1 blue fez in play.

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I'm not convinced. How exactly does that follow? Couldn't person 1 and 2 both be wearing blue fezes? Then person 1 and person 2 would both see blue and red.

Quote:

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Also of course my answer depends on them knowing how many red and blue hats were in the box, which you never stated.

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Naturally that's implicit. If they didn't know what was in the box it'd obviously be totally impossible.

Quote:

---

Originally Posted by Alyzarin

My solution:

It was red. The first person would have to see either two reds or a red and a blue to not know his own color. But, the second person would hear this and be aware of what both the third and the first person were wearing. If the first person was blue and the first was blue, he was red. If the first was red and the third was blue, he would know that he would have to be red as well because otherwise the first person wouldn't have had to say I don't know. The second person only doesn't know if the first and third person are wearing red. This opens up the possibility that the first person either didn't know because of two reds OR a red on third and a blue on second, so he couldn't be sure about it. Therefore, the third person must know he's wearing red when the second person passes.

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Nice job again! I'm not exactly sure what you mean by "opening up a possibility", but if you cut that out then the puzzle is basically done already - you've proven that the third man must be wearing a red hat.

Quote:

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Originally Posted by Xei

Couldn't person 1 and 2 both be wearing blue fezes? Then person 1 and person 2 would both see blue and red.

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Oh damn! I've siad it before and I'll say it again - whenever I start thinking I might be smart it usually means I've misunderstood something. :poof:

Quote:

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Originally Posted by Xei

Clever solution regarding the sombrero one - it makes sense. No cigar though - let me clarify a couple of things.

The only information the prisoners can convey is a hat colour. The warden calls out "next", and the next person then shouts either 'black' or 'white'. No other words, no pauses, or they're fed to the prison dogs.

Regarding the queue, I don't think it really matters. But for the sake of argument let's say the Warden tells them where to stand.

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Ah, alright then. Hmm, I'm going to need some more time on that then....

Quote:

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Originally Posted by Xei

Nice job again! I'm not exactly sure what you mean by "opening up a possibility", but if you cut that out then the puzzle is basically done already - you've proven that the third man must be wearing a red hat.

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Thanks! And I was just trying to say that there are multiple possibilities, so it takes away the certainty.

I want to get this sombrero one so fucking badly. Spent the last like 2 hours drawing little dots representing the hats on paper. I've come up with like 10 different algorithms which seemed to work and then ended up failing.
EDIT: Don't bother reading the following spoiler unless you want to know the track I was on before. The second spoiler is the solution.

Spoiler for trying process, don't have to read:

I've come up with a solution if there were only 3 people in the queue. The back prisoner (#1) says 'black' if #2 and #3 have different hat colors, and white if they're the same color. Then #2 knows which hat color he has with this information and by looking at #3. Then #3 knows his hat color based on his knowledge of #2's color (since he just yelled it out) and whether his is the same or different.

I feel like I'm onto something but I don't really know. I tried many variations on this, like #1 yelling if #2 and #4 are different (instead of 2 and 3), or #2 and #60... none of them work out. It gives them all knowledge you wouldn't expect them to have. Ex: If #1 yells out black if #2 and #60 are different, then suddenly prisoners 1-59 all know #1's color even though they can't see it. But I can't seem to get past the first 3.

Some things I've realized:
- #1 has to be the one who can't surely guess his color, as no one can see his hat, so it might as well be invisible.
- #1 must convey #2's hat color with whatever he says, since #1 is the only prisoner who can see #2's hat. If #1 can't convey this then #2 would be in the same situation as #1.
- The solution must have something to do with the order at which they say their colors.

But, for it to have to do with the order, since pauses aren't allowed, at any time after prisoner #1 goes, two prisoners must know their own hat colors at the same time. That's the only way that asking 'which one of us will go?' will make sense. For example I was thinking, as the next step, #2 could go if #4 and #5 have different hat colors, and #3 could go if they're the same, or something like that. But #3 doesn't yet know his hat color until #2 says his, so there's no choice.

@*(#YU*(# HFEOI I Just figured it out!!! Don't bother reading the above spoiler. Damnit that feels soo good after trying for so long! I had just given up too, that's why I wrote all that above in frustration. Then I was out for a smoke and the idea just came to me when I wasn't even trying to think about it anymore. I can't believe how simple it is....

Spoiler for Solution to sombrero problem:

Since there are an odd number of prisoners after prisoner #1, there must be an even number of one color, and an odd number of the other color.

So prisoner #1 says the color of the odd number of hats that he sees in front of him (or the even number for a parallel solution).

With this information, all the others can, in order, infer the colors of their hats.


Ex with 10 prisoners:

>b b w w b b b b b w>

The first prisoner says 'white' because there are an odd number of white hats in front of him.
The second prisoner sees an odd number of white hats in front of him already, so to keep the number odd, his hat must be black.
The third prisoner sees an even number of white hats in front of him. Since he knows there's an odd number of white hats in total, and that #2's hat is black, his hat must be white.
etc.

^^ If that's the solution then I was looking at it wrong, and that makes it a lot easier lol. I was thinking they could only see the person directly in front of them.

Great job Dianeva, I did basically the same thing.

Spoiler for Solution to Sixty Sombreros:

The very first guy counts the number of white hats of the 59 prisoners in front of him. If it's even, he says "white", if it's odd, he says "black". The code could be the other way round, it doesn't matter - point is, he communicates to the other 59 prisoners whether the number of white hats is even or odd. There's a 50% chance that the codeword happens to coincide with the colour of his hat and he isn't shot.

The rest can then easily ensure their survival.

The first guy of the remaining 59 can easily work out the colour of his hat based on the number of white hats he sees in front of him, and whether the total number of white hats for all 59 men is even or odd.

The second guy can also easily work out the colour of his hat based on the number of white hats in front of him, the colour of the first guy's hat, and whether the total number of white hats is even or odd.

And so on.

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. :D

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?

Quote:

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Originally Posted by Xei

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. :D

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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.

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.

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.

Quote:

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Originally Posted by Xei

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.

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Darn, I didn't think of that. Hmmm...

Quote:

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Originally Posted by Xei

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?

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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?

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.

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! :D

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...

Quote:

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Originally Posted by 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! :D

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...

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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?

Quote:

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Originally Posted by 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! :D

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.

Quote:

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Originally Posted by SnowyCat

... Care to share you're thought process?

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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.

Quote:

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Originally Posted by Nelzi

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.

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Huh, makes sense to me. Well done!

Quote:

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Originally Posted by Dianeva

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.

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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.

I didn't think this thread would be so popular, damn all you people figuring them out!

Quote:

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Originally Posted by Alyzarin

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 :chuckle:) 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.

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Goddamit! I thought of this and quickly dismissed it coz I just didn't think it would work. Makes perfect sense though, well done.

Quote:

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Originally Posted by Xei

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.

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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 :D

Quote:

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Originally Posted by Xei

Blue Eyes [wording by xkcd.com, though he didn't invent it]

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http://i.imgur.com/lKwSGou.gif

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.

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.

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?

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.

Quote:

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Originally Posted by Xei

...
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.
...

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Thanks :D, 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 :D. 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...

Quote:

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Originally Posted by Nelzi

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 :D. 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...

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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.

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?

Quote:

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Originally Posted by 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?

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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.

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

You need the right wording.

Quote:

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Originally Posted by melanieb

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

You need the right wording.

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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? :chuckle: ) Is there something I'm missing here?

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.

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...?

No, sorry. There is no tacit mutual "plan" to beat the system. Certainly the Guru does not conspire. Those who leave do so simply because they've figured out their eye colour.

I've seen a few blind stumbles approach the vicinity of the answer. Don't give up.

So how difficult was the Giant one, Xei? Easy as hell? Need to know so I can choose others.