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CTK Exchange
DdrgL
Charter Member
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Nov-30-00, 09:57 AM (EST) |
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"Coin puzzle"
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Hi, This puzzle seems to be very esy at first. But it turns out to be tough. Lets try! You have 12 coins that look identical. But, 1 coin is bogus and is either lighter or heavier than the others (!) You have a balance scale that can see if its two sides are equal. Find the bogus coin and whether it is lighter or heavier than the others in only 3 weighings.
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jon wegener (Guest)
guest
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Dec-02-00, 00:55 AM (EST) |
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2. "RE: Coin puzzle"
In response to message #0
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i finally got the solution for 12 coins after agonizing for a day, but I dont have much insight into this amazing fact that you need only 3 weighings other than the influence of two facts: (1) the information gained at some steps is usually trinary (+, =, -) not binary (yes, no) becasue the scale can remain unbalanced, become balaneced, or flip and (2) also if the removed coins have been determined to be normal they provide a reference state which again provides 3 pieces of info (normal, heavy, or light) when used to compare uncertain coins on the balance. My question is this: I generalized this to n coins to see where the divide between 2 and 3 weighings occurs. It'seems solvable with 2 weighs only up to 4 coins (one bogus), but did I miss a way to solve it with only 2 weighings with say 5 or 6?? If 4 is correct, it is also amazing, because it means that 3 weighings works over a very large range, up to 12 at least!! |
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