CaptainEd
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo fireflies xxx foresters xxxoo screeches xxxxo shredders xxxxo streetcar xxxxo screamers xxxx
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo fireflies xxx foresters xxxoo screeches xxxxo shredders xxxxo streetcar xxxxo
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo fireflies xxx foresters xxxoo screeches xxxxo shredders xxxxo
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo fireflies xxx foresters xxxoo screeches xxxxo
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I'll be back in an hour
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo fireflies xxx foresters xxxoo
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo fireflies xxx
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo envelopes xxo
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo barbecues xxo
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo saxophone xooo Did I drop it through the cracks? sorry...
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elephants xoooo agreeable xxo sparkling xooo daredevil xx sometimes xxxo
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elephants xoooo agreeable xxo sparkling xooo
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elephants xoooo agreeable xxo
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elephants xoooo
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I'm available, I have a 9-letter word.
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Yes, that one coin difference is needed to make the further decision of whether the bad coin is light or heavy. Thank you, and thank you for the puzzle. It was fun to understand the whole area and clarify my thoughts. For one thing, I learned about how general the restricted case is. While we all know that, if the bad coin is known to be light, we can find it in n weighings from 3^n coins, I hadn't realized the more general result (which I now proved) that it doesn't matter whether the bad coin is light or heavy, as long as we know which ones could only be light, and which ones could only be heavy.
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Yet another typo, sigh! In the proof for restricted coins, the n>1 case, the 2) clause should read 2) 0 < #+ < (3^n)/2 (in other words, assume the number of "+" is less than half the number of coins. Same argument applies if the number of "-" is less than half)
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Please disregard my previous posts--that proof was sloppy, anecdotal, wrong, and CHEAP!
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Thanks to EventHorizon for putting us out of our misery (manual backtracking got REAL OLD...) and showing us an interesting pattern, to boot. Thanks to Bonanova for coming up with yet another interesting, non-trivial, and yet possible problem.
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What a fascinating kind of problem! An easy problem for backtracking, but I know I haven't the time to debug a program...
