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- Today
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I feel so upset. I cannot solve this problem.
- Yesterday
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Please remember to put answers in spoilers (the picture of the eye next to the B, I, U in the editor).
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I think that the only way to get the child back withouth any risk is stating that if he doesn't return the child he will eat him,then at that point will have no choice but returning him.Or she could say some thing like if it makes it she'll send him to school or buy him new clothes;since the crocodille never said it couldn't be something obvious
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I believe that is an approximation. It is irrational because every trigonometric (and inverse) is irrational at a non-zero rational point. What I meant was "not nice" as in "not nice irrational" i.e. even arccos(-1/3)/pi is irrational.
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? arccos(-1/3) = 109.5 degrees.
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Congratulations! Unfortunately, the answer changes if I grant you this boon, so request denied. But no worries! You've still solved it. The only thing that changes... And so the answer becomes the one you initially gave. I'd also like to give my own explanation of the recursive step for "constrained" coins. Some may find it interesting and/or more intuitive.
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They are on opposite sides of the river (no where does it said that they come to the river together!). One simply uses the boat to cross the river and the other person gets into the boat and takes it back to where it started.
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Conjoined twins?
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My answer:
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Answers in spoilers, please. Two people wish to cross a river. There is only one boat. The boat can only carry one person at a time. A person can not cross the river unless in the boat. The boat can not cross the river without someone in it. Both people cross the river. Can you explain how it happened?
- Last week
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Not sure what "really nice" means.
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Your solution does simplify to something really nice. I will see if anyone sees it before I give it away. Nice work.
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First, I define Restricted and Unrestricted coins, then describe and prove the algorithm for Restricted coins. Next, I request a slight tweak to the OP conditions--one additional coin known to be good. Then comes the definition of the procedure for Unconstrained coins, followed by the edge case (how many coins with only one weighing), and the formula for the Unconstrained case, since the OP requests the Unconstrained formula. Then come the first few values.
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This is a nice one. I really think it should have a nice geometrical solution, but:
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The true King lives in yellow castle.
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Three castles, each occupied. One of them is king, other 2 are not. One always tells the truth, the other 2 always lie. Red and yellow castles are by the lake, green castle is not. The three each make a statement: Green: I am the king. Yellow: I am not the king. Red: The king lives by the lake. The riddle's question: in which castle is the king? Understood. Answer:
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cant do this compactly with iPhone. I'll answer sooon. (1) yes I'll make better demonstration about Restricted approach. (2) yes, I'll describe complete Unrestricted approach, which gives larger number
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This is actually correct! Another way to write it: However, your rationale has a few loose ends. I am interested in seeing your full solution. Equivalently, what is an explicit algorithm?
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Do you understand this riddle?
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for 11 for 22 on one shelf for 22 on 2 shelves
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My first guess would be
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This was answered on the original forum. To see: http://imdb2.freeforums.net/post/548811/thread The answer was:
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araver changed their profile photo
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sorry, my attempt to remove the post failed.
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[spoiler=answer maybe]in N weighings, can distinguish bad coin out of M=Sum(3^i), i=0...N-1 Can't always tell whether heavy or light[/spoiler] [spoiler=Background: restricted vs unrestricted]if you weigh coins in the balance scale, and the Left pan is heavy, then the coins in that pan are "restricted": each one is either Good or Heavy, while the coins in the other pan are restricted as either Good or Light. I abbreviate "heavy or good" as "H", and "light or good" as "L". We can distinguish among 3^N restricted coins in N weighings. [/spoiler] [spoiler=Strategy]However, we are starting with coins we know nothing about (and maybe one coin known to be good) So we will weigh a number of coins (p), leaving some of them (q) unweighed. Once you weigh the first batch, they become restricted, and you can resolve one of them in ceil(log base 3 (p)). If they balanced, then you have N-1 weighings to resolve the remaining q coins. So our initial task when approaching a set of coins is to choose p and q so that the number of weighings for p restricted coins equals number of weighings for q unrestricted coins. [/spoiler]
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