EPT[SFW] Riddles / Puzzles / Brain Teasers - Page 12
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Tanukki
Finland579 Posts
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Cocacooh
Norway1510 Posts
On April 19 2012 20:54 Tanukki wrote: The easier a riddle is, the more pleasant the feeling when you manage to beat someone with it. Guess I didn't get that today :p It sure is  | ||
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jjhchsc2
Korea (South)2393 Posts
+ Show Spoiler + is there a pattern or something or is it because of the fact the ''rd'' is there | ||
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Iranon
United States983 Posts
On April 19 2012 18:04 ]343[ wrote: Haha, I have a vaguely related one, but it's not so much a brain teaser as a math problem (and I think I've posted it here before). I think it's way more counterintuitive >< Infinitely many prisoners, labeled 1, 2, 3, 4, ... are standing in a line so that person N can see everyone with smaller label (and not himself). Each prisoner k is assigned a hat labeled with a [integer, rational number, or real number; doesn't actually matter (!)], c_k. Now, starting from prisoner 1 and going up, each prisoner will try to guess the number on their own hat; if he fails, he is shot, and if he succeeds, he lives. Show that, if the prisoners are given (an infinite amount of) time beforehand to formulate a strategy, they can ensure that only finitely many prisoners are shot. Nice -- but you don't mean an infinite amount of time, you mean an arbitrarily large finite amount of time. If they had an infinite amount of time to prepare, you could ensure that everyone lived by planning forever, and that's a stupid linguistic loophole. I'll have to think on that a bit more to get it down to cofinitely many survivors, but arbitrarily high proportions of survivors is easy (see spoiler as food for thought). + Show Spoiler + You can code lists of numbers into a single number in any number of ways -- Gödel codes, pairing functions, a bijection from N to N^k (or Q, or C, or whatever), etc. Suppose you want at most one person out of 10,000 to die. Then when you're planning, line everyone up, and have the first person memorize the number that encodes the numbers of the next 9,999 people, and have the 10,001st person do the same thing for the 9,999 people after him, and so on with the (10000k+1)th people for all natural numbers k. Then those people announce that they think their number is the relevant code, and are almost certainly killed, since that would be a ridiculous coincidence, and the next 9,999 people can immediately calculate their numbers with just arithmetic, and so on and so on. Obviously moving from "arbitrarily close to 100% survival rate" to "all but finitely many survivors" is a huge gap that requires a whole new technique to bridge, but I think it's neat that it's so simple to get that much. I'll get my mental hamsters on their wheels and come back later with a full solution. | ||
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Tanukki
Finland579 Posts
On April 19 2012 21:13 jjhchsc2 wrote: + Show Spoiler + is there a pattern or something or is it because of the fact the ''rd'' is there + Show Spoiler + No pattern, just wordplay. Only the 3rd object in the sequence is referred to with "-rd" | ||
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mathilol
8 Posts
+ Show Spoiler + Ask one of the guards: Which doors shouldn't i take? The liar will point out one door: the right one. The other guard will point out 2 wrong doors. | ||
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ThePlayer33
Australia2378 Posts
On April 19 2012 20:50 Tanukki wrote: There's a 7 floor building with each floor dedicated to a single type of facility unique to that floor. On floor 7 there's offices. On floor 6 there's server rooms. On floor 4 there's an arcade. A library is on the __rd floor. Which floor is that? + Show Spoiler + nerd floor? | ||
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Josh-FiveO
Denmark174 Posts
+ Show Spoiler + Obviously 3rd floor. Since 3 is the only number up to 7 that ends with 'rd' | ||
TanGeng
Sanya12364 Posts
On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms. The team is adjudged to win if at least one of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses. You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it? | ||
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Bahamuth
134 Posts
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy. On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms. The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses. You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it? Just to clarify, I'm guessing players don't have to press their buttons simultaneously, and that other players can see what button the others pressed? Am I allowed to use some notion of time? As in, the strategy is to do something under a certain circumstance. If no player does anything for a time x, proceed to the next step? | ||
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VayeshMoru
201 Posts
On April 20 2012 01:22 TanGeng wrote: Teams of three players are invited to engage in a guessing game and told the rules of the game ahead of the time and allowed time to come up with a strategy. On the set, there are three soundproof rooms. Above each room, there are two lights, one blue, one red. The rooms are windowed such that someone inside can only see the lights above other the rooms and nothing else. Inside the room, there are three buttons, a red button, a blue button, and a white button labeled pass. At the start of the game, each player will be led into a separate room, and one of the two lights will be turned on above each rooms. The light that is turned on is determined by a flip of a fair coin independently for all three rooms. The game ends when all players have pressed a button inside their respective rooms. The team is adjudged to win if at least of them pressed the button of the matching color of the light turn on above their room and none of them pressed the button of the non-matching color. A press of the white pass button counts as neither matching nor non-matching. In all other cases, the team loses. You lead a team of three. Do you come up with a strategy to maximize your chances of winning? What is it? Only two lights can ever be on, red or blue. As you know each person can see the other two peoples lights and not their own you set it up as followed. If you can see one of each colour you hit the white button. If you see two of the same colour you hit the opposite button. IE if you see two red lights above your fellows you will hit the blue button. There is a chance obviously that all three people have the same colour, however to maximize your chances of winning I believe this is the best strategy. | ||
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DarkPlasmaBall
United States44391 Posts
On April 20 2012 01:04 Josh-FiveO wrote: + Show Spoiler + Obviously 3rd floor. Since 3 is the only number up to 7 that ends with 'rd' + Show Spoiler + Agreed. Wasn't sure if he messed up his ending (as a math educator, I've seen many instances of this in the past with people getting sloppy trying to generalize sequence numbers) or if this was the trick. If it was done on purpose, then it's definitely 3rd for the reason you cited. Otherwise, I have no idea. | ||
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Umpteen
United Kingdom1570 Posts
That said, I think even the xkcd version isn't strict enough to mandate the 'right' answer. Here's what I think the intended correct solution is: + Show Spoiler + Upon being told that at least one of them has green eyes, anyone who can see only blue eyes can leave on the first ferry. Thus if only one person had green eyes, they would leave on day 1. If two people had green eyes, each would think 'I can see one person with green eyes, so it might be him and not me', and not leave the first night. However, upon waking the next day to find the other green-eyed person still around, and knowing that they themselves are the only candidate for having the second pair of green eyes, they can go down to the ferry and leave on the second night. If three people had green eyes, each would look at the other two, follow the logic above, and think 'Since It takes two green-eyed people two days to figure out they can leave, if I wait two days and they're still here, there must be a third - and the only candidate is me." Thus the three can leave on the third day. If four people had green eyes, each would see three greens, follow the logic above and conclude that the three greens would leave on the third day unless there was a fourth - them. Thus, in general, if there are N green-eyed people, each green-eyed one can see N-1. He knows the N-1 would take N-1 days to figure out they could leave if he himself didn't have green eyes. The remaining blue-eyed ones, able to see N greens and wondering if there are N+1, need to wait an extra day and will be disappointed. However, I think there's another answer that gets everyone off the island in just two days without the need for a note or message: EDIT: Strike all that. I think there might be a way to do it without the note/message but the idea I had wasn't right. | ||
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XiGua
Sweden3085 Posts
I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? ![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg) | ||
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SFX
United States89 Posts
On April 20 2012 02:33 XiGua wrote: + Show Spoiler + I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? I think someone already listed this riddle. | ||
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Crownlol
United States3726 Posts
On April 19 2012 04:10 Go1den wrote: Here are a few, they aren't too hard: 1. Forwards I'm heavy, backwards I'm not. What am I? 2. A black dog is standing in the middle of an intersection in a town painted black. A storm has caused a power outage, meaning all of the town lights are not functioning properly. A car with two broken headlights drives towards the dog, but turns in time to avoid hitting him. How could the driver have seen the dog in time? 3. How much dirt is in a hole 3 feet wide, 3 feet long, and 3 feet deep? 4. How can you rearrange the letters in "new door" to make one word? 5. How many times can you subtract the number 5 from the number 25? + Show Spoiler + Ton! + Show Spoiler + It's the daytime + Show Spoiler + None. It's a hole. + Show Spoiler + It IS "one word"! + Show Spoiler + Once. If you do it again, it's 5 from 20. | ||
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DarkPlasmaBall
United States44391 Posts
Yup. Please read the thread first  | ||
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Taekwon
United States8155 Posts
Girl A said: "I didn't eat it!" Boy B said: "Neither A nor D ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!" One of these children is lying. Who is it? | ||
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Crownlol
United States3726 Posts
On April 20 2012 02:33 XiGua wrote: I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? Ok, I'm not sure, but here ya go: + Show Spoiler + The group designates a single person the "counter". This person is the only one who will reset switch A to the bottom position. A starts in the down position. If it is your first visit to the room, you move switch A to up. If switch A is in the up position, you don't touch it, you just move B (also, if you have been there before you move B). If the counter finds the switch in the up position, he moves it down and adds +1 to his count. When the counter's count reaches 22, he frees himself and all his co-prisoners! | ||
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SFX
United States89 Posts
On April 20 2012 03:36 Taekwon wrote: A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!" Girl A said: "I didn't eat it!" Boy B said: "Neither A nor B ate it." Boy C said: "I swear on my momma's grave I didn't eat it!" Girl D said: "C is telling the truth!" One of these children is lying. Who is it? At least one, or only one? | ||
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