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| Acmlm's Board - I3 Archive - Brain Teasers - Teh Box O' Brain Hurtrz |
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SamuraiX Broom Hatter                  Since: 11-19-05 Last post: 6285 days Last view: 6286 days |
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| Method 1: The first person sees if green or red hats are more dominant, and then chooses the most common one. This guarantees that the number of people saved are [50,100].
50~100% save rate. And a note, I'm pretty sure it would be a column, not a row, that you're talking about. Rows are horizontal, and would imply that each person is not facing towards another person. If I'm right, they should be looking at the people in front of them, correct? |
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Uki-Ki-rby Poppy Bros. Jr    Since: 01-27-07 From: New York City Last post: 6285 days Last view: 6284 days |
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| true but see we do not live in a 2 dimensional world | |||
Kejardon Shyguy  Since: 05-21-06 Last post: 6285 days Last view: 6284 days |
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| A person has 2 options: If he knows his own color, say it, else give information on the next color(s). I'm going to assume that information may only be binary - green or red (which may be code for something else).
Let's try something simple to start with. Person 1 says if the next to colors are the same or different. Person 2 and 3 are then guaranteed to live, and the loop restarts with the 4th person. 66 people live, minimum. There is a 'key' for the information, based on the actual line of people. If you think carefully, you can encode a lot of information based on this key. Now then, how can you best compress information in this way? Red = 0, Green = 1 Person 1 looks ahead at the next 3. 1: If 000, say 0. If 001, say 1. If 010, say 1. If 011, say 0. Invert all answers if 1XX. 2 looks ahead at the next 2 (XX) and remembers what 1 said (Y). (YXX) 2: If 000, say 0. If 101, say 0. If 110, say 0. If 111, say 0. All others, say 1. 3 looks at 4 (X) and remembers what 1 and 2 said (YY). (YYX) 3: If 000, say 0. If 100, say 1. If 101, say 0. If 001, say 1. If 010, say 1. If 110, say 0. If 111, say 1. If 011, say 0. 4 remembers what 1, 2, and 3 said (YYY). (YYY) 4: If 000, say 0. If - Screw it. You get the pattern. Eventually, while writing that all up, I realized all you need to do is have the first person say whether or not there is an odd number of, let's say, green. Person 2 then sees if there is an odd number of greens in front of him, and if so, he's red. Else he's green. If he's red, person 3 sees if there are an odd number of greens, if so, he's green, else red. If 2 is green, 3 sees if there is an *even* number of greens, if so, he's green, else red. Swap between odd and even every time a green pops up. Wala, 99/100 people live. Just don't draw the short straw. Alternatively, answer in a language in which the words for green and red are identical, thus insuring 100% accuracy. Hah. |
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Uki-Ki-rby Poppy Bros. Jr Since: 01-27-07 From: New York City Last post: 6285 days Last view: 6284 days |
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| Yea but you cannot know your own color but you could have 99% living | |||
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SamuraiX Broom Hatter Since: 11-19-05 Last post: 6285 days Last view: 6286 days |
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Originally posted by Uki-Ki-rby 95.5% actually, since there's a half-chance that the first person's color is the same as the color which is used to inform the 2nd person. |
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Uki-Ki-rby Poppy Bros. Jr Since: 01-27-07 From: New York City Last post: 6285 days Last view: 6284 days |
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| Oh yea but its TRULY 90 if they can only say 1 thing | |||
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SamuraiX Broom Hatter Since: 11-19-05 Last post: 6285 days Last view: 6286 days |
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Originally posted by Uki-Ki-rby Originally posted by Kejardon I think I know which statement here is the correct one. And I thought it was "voil…," not wala. =\ |
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Uki-Ki-rby Poppy Bros. Jr Since: 01-27-07 From: New York City Last post: 6285 days Last view: 6284 days |
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AAAAAAAaannny way I think this chat qualifies as answering it  |
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ziratha Koopa Since: 11-19-05 Last post: 6302 days Last view: 6285 days |
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| And the correct answer is 99 out of 100.
ps. You could also save 93 of them by having the first 7 prisoners give a binary representation of the number of hats of a given color. Let me see... do I know any more? Ok heres another one: There are 100 prisoners. The are put in jail in solitary confinement. Once each day, in the morning, the warden will select a random prisoner and put him in a common room, and return him to his cell at the end of the day. The common room has a lightbulb, that's all. While in the common room, the prisoner may turn the lightbulb off it it is on, or turn it on if it is off, or leave it as it is. The prisoner may not mark the wall or do anything except interact with the lightbulb. The lightbulb does not burn out and will not be turned off or on except by the prisoners. While in the common room, the prisoner may also declare that all 100 prisoners have been in the common room. If he declares correctly, all prisoners will be let go. If he declares incorrectly they will all be executed. The prisoners are allowed to assemble beforehand and agree on a strategy. What strategy can the prisoners implement so as to be sure all prisoners have been in the room? (edited by ziratha on 01-29-07 03:17 PM) |
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Uki-Ki-rby Poppy Bros. Jr Since: 01-27-07 From: New York City Last post: 6285 days Last view: 6284 days |
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| man what is with the death sentences? anyway my guess is to pull the lightbulb chain once per guy and count the amount of times the others see the lightbulb be changed therefore at the 99th change the 100th prisoner could declare they have all been in there via the sight of the lights.
I hope that was right...  |
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ziratha Koopa Since: 11-19-05 Last post: 6302 days Last view: 6285 days |
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| The prisoners are all in solitary confinement, so they cannot talk to each other, Also, assume they can't see into the common room, cuz then they could just visually watch and count manually.
The ONLY information that any prisoner has is the state of the lightbulb when he enters the room and the stategy they agreed on beforehand. |
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Uki-Ki-rby Poppy Bros. Jr Since: 01-27-07 From: New York City Last post: 6285 days Last view: 6284 days |
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D'oh! Oh well... My loophole worked until that was fixed.. |
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Kejardon Shyguy Since: 05-21-06 Last post: 6285 days Last view: 6284 days |
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I can think of an easy solution, though there might be a much faster method. But you didn't ask for the fastest method. 
One person will be the designated counter. He will turn the light bulb on every time he's in the room. Everyone else will turn the light bulb off *once*, and never do anything else. When the one person has turned the light bulb on 100 times, everyone's been in the room at least once. Assuming random chance of picking someone, this will take on average, a bit more than 10000 days, or somewhere around 30 years. Is there a faster method I should try to figure out? |
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ziratha Koopa Since: 11-19-05 Last post: 6302 days Last view: 6285 days |
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| No, that seems to work. That is essentially the solution I was thinking of except off for on and visa versa. I don't really think there is a quicker solution.
Let's see.... what other riddles do I know??? Ok, here's some: You have a scale. The scale is the kind of scale where it has two arms with a platform on each arm. The arm with the most weight will go down, the other arm goes up. 1) Given 8 standard metal weights and the knowledge that one of the weights is heavier than the others, how can you find the heavier ball given only 3 weighings? 2) Given 27 standard metal weights, and the knowledge that one of the weights is heavier than the others, how can you find the heavier ball given only 3 weighings? 3) Given 12 standard metal weights, and the knowledge that one of the weights is either heavier, or lighter than the others. How can you find the odd weight given only 3 weighings, as well as determine whether the odd weight is heavier or lighter? |
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Skreename Giant Red Paratroopa Since: 11-18-05 Last post: 6290 days Last view: 6284 days |
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Originally posted by zirathaEasy. Put 4 on each, weigh them. Remove the lighter set; take the heavier one and split it into 2 sets of 2, and weigh them. Take the heavier set, split it into 1 and 1, then weigh them. The heavier one there is the heavier weight. |
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NSNick Gohma IF ALL ELSE FAILS USE BOOZE  Since: 11-17-05 From: Last post: 6286 days Last view: 6286 days |
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Originally posted by ziratha
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ziratha Koopa Since: 11-19-05 Last post: 6302 days Last view: 6285 days |
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| Good, good. nsnick has the one and skreename has the other. Can anybody get the third, way hard one?
reposted for your convienence! 3) Given 12 standard metal weights, and the knowledge that one of the weights is either heavier, or lighter than the others. How can you find the odd weight given only 3 weighings, as well as determine whether the odd weight is heavier or lighter? And another one, for HER pleasure! You find yourself in a room with three gods. Somehow you know, that one is the god of the past, one is the god of the present, and one is the god of the future (maybe it's written on a sign or something). You may ask three yes or no questions, which all gods will answer at the same time, however there is a catch. The god's speak a different language, In this language, they will answer either "ya" or "da", you do not know which means yes and which means no. THis is a further complication. The god of the past answers the last question asked, the god of the future answers the next question to be asked, and the god of the present answers the current question. Also, assume that there was someone in the room before you who asked questions and there will be someone after you in the room who asks questions( so the god of the past will not remain silent on the first question.) Given all this, How can you determine who isthe god of the past, present, and future as well as whether "ya" and "da" mean yes or no? |
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SamuraiX Broom Hatter Since: 11-19-05 Last post: 6285 days Last view: 6286 days |
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| K. To question 3.
Divide into 3 groups of four, namely groups A, B, C. 1. Weigh A and B. If one side is heavier, that side has the heavier weight. Else, the heavy weight is in group C. 2. Weigh two of them (D, E), if one is heavier, it is the heavy weight. Else go on to step 3 in the case that D and E are of equal weight. 3. Weigh the two others (F, G) and the heavier one is the heavy weight. And for the questions, ask "Is 1+1=2?" Then, ask "Is the answer to my next question "yes"?" For your last question ask "Was the answer to my last question "yes"?" (edited by SamuraiX on 01-30-07 09:24 PM) |
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ziratha Koopa Since: 11-19-05 Last post: 6302 days Last view: 6285 days |
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| erk. to #3, You don't know one of the weights is heavyier, you know it is heavier OR lighter, You have to find out which during the courses of weighings. so if you weight weight 'a' and ball 'b', and a is seen as heavier, it could be that 'a' is the odd ball and is heavier, or it could be that 'b' is the odd ball and is lighter.
As for: And for the questions, ask "Is 1+1=2?" Then, ask "Is the answer to my next question "yes"?" For your last question ask "Was the answer to my last question "yes"?" My brain hurts, don't do that! |
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SamuraiX Broom Hatter Since: 11-19-05 Last post: 6285 days Last view: 6286 days |
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Originally posted by ziratha Just replace ever instance of heavier with heavier/lighter, and it still works out correctly. |
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