Puzzle thread? Puzzle thread.

[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 ) 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.

[Arra]

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.

[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.

[Xei]

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!

[Arra]

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.

[SnowyCat]

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

[Alyzarin]

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

[Arra]

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.

[Xei]

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.

[Nelzi]

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 :/

[Arra]

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.

[Nelzi]

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

[Xei]

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?

[Darkmatters]

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.

[Xei]

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.

[Darkmatters]

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.

[Arra]

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.

[Alyzarin]

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. That's cool to know. 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....

[Alyzarin]

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.

[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. Actually you did say they both took some time Sure, but an unspecified amount. They mull it over to the best of their ability, that's all it means. 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. 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. Also of course my answer depends on them knowing how many red and blue hats were in the box, which you never stated. Naturally that's implicit. If they didn't know what was in the box it'd obviously be totally impossible. 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. 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.

[Darkmatters]

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

[Alyzarin]

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. Ah, alright then. Hmm, I'm going to need some more time on that then.... 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. Thanks! And I was just trying to say that there are multiple possibilities, so it takes away the certainty.

[Arra]

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.

[Alyzarin]

^^ 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.

[Xei]

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.