Puzzle thread? Puzzle thread.

[Arra]

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. "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?" 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?

[Xei]

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. 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... The farmer can turn around. He can also slow down and stop of course, as can you. 4 times your max speed is his max speed. 20 prisoners: If they are in the same block, they could count each time they hear the doors to their cells open. But they might just hear the same person going 20 times. The light is their only means of communication. Coins on the table: Flip all the coins...put half on the table and half on the floor... 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. 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?" 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. 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. That's all true, and you're thinking along the right lines, but perhaps there is a better technique. After twenty days the prisoner taken on that day can go to the bathroom and then tell him they've all been there. 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. put all the coins on their edges like _O_ that, standing up, there are 0 heads on the table and 0 on the floor. 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. 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. 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. 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. 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.