EPTyou are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem."
Brainteasers/Math problems - Page 7
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cmpcmp
84 Posts
you are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem." | ||
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lololol
5198 Posts
On April 10 2011 05:20 shadowy wrote: Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely. + Show Spoiler + If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word. The whole grouping thing serves no purpose whatsoever. If they knew the totals(and it clearly says that they don't) then it wouldn't be a puzzle at all, since the difference between the colors of everyone else and the totals is obviously their own eye color. + Show Spoiler [hint] + The only new information they can gain is whether someone has left the island or not, so obviously the answer is based around that. | ||
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cmpcmp
84 Posts
Sorry my bad, but that doesn't change the answer in a meaningful way because that question can be asked in a "yes or no" manner. I edited the answer to reflect this. | ||
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Murderotica
Vatican City State2594 Posts
On April 10 2011 05:20 shadowy wrote: Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely. + Show Spoiler + If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word. "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." Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks. | ||
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ixi.genocide
United States981 Posts
On April 10 2011 05:51 cmpcmp wrote: @genocide you are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem." Yeah, the puzzle becomes obvious when you can ask the middle daughter something about the other daughters. Once you think of a question that never points to the middle daughter when asking the other daughters you win! | ||
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ixi.genocide
United States981 Posts
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Wonderballs
Canada253 Posts
+ Show Spoiler + I (think) understand everybody's logic... and it makes sense if you follow those strict mathematical theorms. What I don't understand is why they are concluding they have blue eyes when they can see other people with blue eyes. If everybody with blue eyes can see someone with blue eyes, nobody will ever leave the island. I don't see how after 100 days that suddenly means they have blue eyes. If the ferry takes somebody away every night... and on the first night after the guru speaks nobody leaves... does that not break the condition of somebody had to have left? Which means for 99 days you are counting 99 (or 100) blue eyes on the island. On the 100th day you realize, wow nobody is leaving.... I must have blue eyes? But wait, theres still 99 people with blue eyes here why aren't they leaving? What is so special about 100? This is probably all jumbled up so don't be too hard on me... I can see why it works if you follow those definitions from the solution of the website.... But only if you assume those theorems are true. They have not given "proof" mind you. Something mathematical textbooks ALWAYS put in a coloured box after a theorem like that. EDIT: reread the part about people leaving; they don't have to leave.... Here's my train of thought... (I have blue eyes) Me:"Hmm, 201 people (200 - guru), 100 brown eyes, 99 blue eyes, hmm, I wonder what colour mine are?" Guru:"SOMEBODY HAS BLUE EYES" Me: "Yup, this is true, what colour are mine?" I don't believe this ground up induction works considering you don't start with 1 person. Therefore cannot assume that on the Nth night N people will leave. AHHHH wtf man stupid ass problem... I see why everyone understands it but it seems like theres is assumptions made by the logicians. + Show Spoiler + 1 person /w blue --> understandable 2 people /w blue---> understandable >3 people /w blue ---> ???? The 1 person example; no other blue eyes, obviously. The 2 people example; the 2nd person knows the first person can see a 2nd set of blue eyes. easy peazy. The 3 people example; the 3rd person knows the other two can see each others eyes. Why would anybody leave? | ||
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shadowy
Bulgaria305 Posts
On April 10 2011 06:01 ixi.genocide wrote: Yeah, the puzzle becomes obvious when you can ask the middle daughter something about the other daughters. Once you think of a question that never points to the middle daughter when asking the other daughters you win! I am sorry, I still don't see, how there could be a question, that will evade the middle sister problem with a YES/NO question, since her answer will always be random, hence there can no be solution. I personally, can not think about @cmpcmp Yeah, you changed the question, but this does not fix the explanations. I am still confused about this one. | ||
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Frigo
Hungary1023 Posts
1. (accessible to everyone) you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol) + Show Spoiler + Pour 3l water into the 5l jug, pour the 3l jug full, pour 2l from it into the 5l jug (till it is full), empty the 5l jug, pour the remaining 1l into it, pour another 3l into it. 2. Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing? I don't think so. You need ceil(log2(10)) measurements even [...] Goddamn why did I read the spoilers. 2.b. Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing? + Show Spoiler + Same as with 2., except you need a superincreasing sequence of number of coins from the bags, so you can't confuse weights with sums of weights. 1a+2b+4c+8d will do nicely. 3. (accessible) You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass. + Show Spoiler + Easy, start both hourglasses at the same time. Flip the 7 minute one at the 7th minute. Flip the 7 minute one again at the 11th minute. You have 4 minutes till it runs out. 4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide? + Show Spoiler + Around 1:05:27 (12/11 * 60 minute). PM of course. 5. (my favorite ) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course. Not necessarily. Based on area alone, they could. But since 4 disk of radius 1/2 can't cover 1 disk of radius 1, they might miss some rectangles as well. Can't provide solid proof. Hehe nice solution by the others. 6. What's your eyes color? (hard problem) + Show Spoiler + God I hate that problem. Took me a whole day to understand it, even though I knew the answer. In the queen + 100 wives + 100 husbands version, every husband dies at the 100th day. 7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale? + Show Spoiler + The trick here is we are comparing THREE groups of coins at once. Two on the scale, one in our hands. It is quite clear which group the counterfeit coin is in, be the scale balanced or unbalanced. So we take 3 coins in our hands and compare 3 coins against 3 coins. We know which group the counterfeit coin is in. We take 1 coin in our hands and compare 1 coin against 1 coin. We know which one is the counterfeit. 8. (Day9 wants you to do this one) On April 10 2011 00:29 Munk-E wrote: Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes? Note: the string possibly burns unevenly. + Show Spoiler + Light one string on one end, the other string on both ends. Once the other strings burns out, light the other end of the first string as well. 9. (if you know what prime numbers are) When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not? + Show Spoiler + It sure as hell can't be divisible by 2, 3 or 5. That leaves 7, 11, 13, 17, 19, 23, 29 - all primes. 10. On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)? + Show Spoiler + All of them see the hats of those later in the line, the first one in line sees them all, except his. He communicates the parity of the sum of the binary representation of hats (0 or 1) by saying the respective color of the parity (it doesn't matter which color is 0 and which is 1). The guy next in the line can deduce his hat color based on this parity and what he sees. The others can deduce their hat color from the parity, and what others have said before him. Only the first in the line might die, the others surely live. Unless one of them fucks up  11. Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.) No idea. 12. (Monty Hall?) On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch? + Show Spoiler + Yes, this is a very famous problem. You should switch, your chances increase to 2/3 from 1/3 or so. 13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)? + Show Spoiler + Yes, they can be. And it is unique. It's simply the binary representation of the numbers. 14. On April 10 2011 02:14 ghrur wrote: Question: What is the maximum number of times 10 lines can intersect in the same plane? paraphrased: What is the maximum number of intersection points between 10 distinct lines on a plane? + Show Spoiler + Looks like it is always possible to construct n lines so every line intersect the others. So it is 10 choose 2 = 45. 15. Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7? + Show Spoiler + In other words: Can you select 15 integers from 100 distinct integers such that all of them are in the same equivalence class mod 7? The answer is yes, if there are only 14 integers from each equivalence class, that is only 14 * 7 = 98, so it is guaranteed there are at least one equivalence class with 15 integers falling into it. | ||
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azn_dude1
162 Posts
On April 10 2011 06:16 shadowy wrote: I am sorry, I still don't see, how there could be a question, that will evade the middle sister problem with a YES/NO question, since her answer will always be random, hence there can no be solution. I personally, can not think about @cmpcmp Yeah, you changed the question, but this does not fix the explanations. I am still confused about this one. + Show Spoiler + I've heard this question before but I forgot the answer and I'm too lazy to think about it now but it has to do with asking one of the sisters if another will tell the truth or is more likely to tell the truth. I forgot. Edit: my bad it's like the same thing so nevermind | ||
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shadowy
Bulgaria305 Posts
On April 10 2011 05:53 Murderotica wrote: "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." Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks. As I said already, shall we enforce the "no communication" rule strictly, my solution it's not possible - in this case I have already provided the answer + Show Spoiler + n+1 for blues, n+2 for browns. I was asking if there is otherwise is any flaw in my logic, which allows everybody on the island to figure out the color of their eyes by simple grouping/ordering, without anybody actually communicating, but only taking action based on information from visually inspection others. | ||
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cmpcmp
84 Posts
Further Explanation + Show Spoiler + Essentially, my solution picks a sister at random (which you must do) and asks her a question. The trick is that you NEVER pick the sister that you asked the question to. Thus, if you ask the middle sister the question it doesn't matter what she answers because you can't possibly pick her. So: 1. Ask oldest sister: you pick the youngest sister (she told the truth) 2. Ask middle sister: she answers randomly and you pick the oldest sister 1/2 of the time and the youngest sister 1/2 of the time. 3. Ask the youngest sister: you pick the oldest sister (she lied about which one was younger) | ||
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tomnov
Israel148 Posts
the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler + * a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer how do they split the money and how many pirates die? | ||
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Murderotica
Vatican City State2594 Posts
On April 10 2011 06:22 shadowy wrote: As I said already, shall we enforce the "no communication" rule strictly, my solution it's not possible - in this case I have already provided the answer + Show Spoiler + n+1 for blues, n+2 for browns. I was asking if there is otherwise is any flaw in my logic, which allows everybody on the island to figure out the color of their eyes by simple grouping/ordering, without anybody actually communicating, but only taking action based on information from visually inspection others. You are still completely wrong because brown eyed people would not have any indication that they can leave, ever. Also, the whole premise of your solution being that we don't stick to the premise of the problem... That just sounds retarded. Hey, let's do 1+2 except let's assume that + isn't actually + because if we bend the rules a little it can be -. Oh wait, shall we enforce the RULE of + SOOOO STRICTLY? Jesus dude. | ||
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lololol
5198 Posts
On April 10 2011 06:05 Wonderballs wrote: 6. (discussion) + Show Spoiler + I (think) understand everybody's logic... and it makes sense if you follow those strict mathematical theorms. What I don't understand is why they are concluding they have blue eyes when they can see other people with blue eyes. If everybody with blue eyes can see someone with blue eyes, nobody will ever leave the island. I don't see how after 100 days that suddenly means they have blue eyes. If the ferry takes somebody away every night... and on the first night after the guru speaks nobody leaves... does that not break the condition of somebody had to have left? Which means for 99 days you are counting 99 (or 100) blue eyes on the island. On the 100th day you realize, wow nobody is leaving.... I must have blue eyes? But wait, theres still 99 people with blue eyes here why aren't they leaving? What is so special about 100? This is probably all jumbled up so don't be too hard on me... I can see why it works if you follow those definitions from the solution of the website.... But only if you assume those theorems are true. They have not given "proof" mind you. Something mathematical textbooks ALWAYS put in a coloured box after a theorem like that. + Show Spoiler + Case 1: You see 0 people with blue eyes. That's enough to determine your eye color, so you will leave on the first night(the words of the guru guarantee at least 1 person with blue eyes, so that would be you). Case 2: You see 1 person with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 2.1. Your eyes aren't blue and the other person leaves on the first night(case 1 applies to him) Case 2.2. Your eyes are blue and the other person does not leave on the first night(case 2 applies to him). If your eye color wasn't blue he would've left, so you now know you have blue eyes(the same thing applies to him) and both of you leave on the second night. Case 3: You see 2 people with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 3.1. Your eyes aren't blue and the 2 other people leave on the second night. Case 3.2. Your eyes are blue and the 2 other people don't leave on the second night. If your eye color wasn't blue they would've left, so you now know you have blue eyes(the same thing applies to them) and the three of you leave on the third night. ... Case 100: You see 99 people with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 100.1. Your eyes aren't blue and the 99 people with blue eyes will leave on the 99th night. Case 100.2. Your eyes are blue and the 99 other people don't leave on the 99th night. If your eye color wasn't blue they would've left, so you now know you have blue eyes(the same thing applies to all of them) and all 100 of you leave on the 100th night. | ||
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shadowy
Bulgaria305 Posts
On April 10 2011 06:24 cmpcmp wrote: @shadowy Further Explanation + Show Spoiler + Essentially, my solution picks a sister at random (which you must do) and asks her a question. The trick is that you NEVER pick the sister that you asked the question to. Thus, if you ask the middle sister the question it doesn't matter what she answers because you can't possibly pick her. So: 1. Ask oldest sister: you pick the youngest sister (she told the truth) 2. Ask middle sister: she answers randomly and you pick the oldest sister 1/2 of the time and the youngest sister 1/2 of the time. 3. Ask the youngest sister: you pick the oldest sister (she lied about which one was younger) I finally see it. OMG!, I had to type it, to make sure it all works out. However you missed an important bit of explanation. + Show Spoiler + You have to always skip the sister you ask the question to and choose the younger one from the remaining two, assuming the answer is correct. So, all possible scenarios: 1. You asked the middle one. It doesn't matter. You win. 2. You asked the oldest, if THAT one is younger then the other and pointing at: (and that part is actually random) a. At the youngest. She will tell the true and tell you Yes. You pick her and win. b. At the middle. She will lie and say "No", you pick the youngest and win. 3. You asked the youngest, if THAT one is younger then the other and pointing at: a. At the oldest. She will lie and tell you Yes. You pick her and win. b. At the middle. She will lie and say "No", you pick the youngest and win. | ||
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cmpcmp
84 Posts
That's why it's my favorite puzzle! Explaining is clearly over the internet is also very difficult for this one. | ||
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shadowy
Bulgaria305 Posts
On April 10 2011 06:32 Murderotica wrote: You are still completely wrong because brown eyed people would not have any indication that they can leave, ever. Also, the whole premise of your solution being that we don't stick to the premise of the problem... That just sounds retarded. Hey, let's do 1+2 except let's assume that + isn't actually + because if we bend the rules a little it can be -. Oh wait, shall we enforce the RULE of + SOOOO STRICTLY? Jesus dude. You just being angry right now. Chill, dude. It's just riddles for which correct wording is quite important. Also, regardless, how you solve, the problem, the moment all blue eyes leave, the guru will not confirm there are people with blue eyes anymore on the island, so everybody else will know they have brown eyes, hence they can leave. However I would like to task you with my own version of this riddle. And this one is actually much harder: There is an island where people are either blue-eyed or brown-eyed. There is no other person with any other color of his/her eyes. No exceptions!. On this island there are no reflecting surface or anything that a person may learn his/hers eyes color, unless somebody else tells him/her. However telling/showing anybody the color of their eyes is strictly forbidden as a taboo and will not be violated under any circumstances. Is it possible for all people on the island to figure out the color of their eyes? If so, provide an answer, please. | ||
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thesideshow
930 Posts
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MadVillain
United States402 Posts
+ Show Spoiler + Just put 1 coin from each bag on the scale arranged such that you know which bag the coins came from, simply remove 1 coin at a time lol. I guess that wouldn't count as 1 weighing, though you only are putting something on the scale once so maybe.. | ||
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