This is a clever brainteaser that someone gave me a couple weeks ago that I thought would be fun to share. So not to ruin it for everyone, I’ll post the answer as a comment a bit later :]
You are throwing a magnificent party in your awesome castle and have acquired 1000 bottles of wine for the occasion. The night before the party, some devious person (blue group coordinator) poisons exactly one bottle. The amount of poison is so small as to make differentiation by mass or volume or whatever impossible. Yet the poison is so potent that it will kill anything that drinks it in 9.5 hours, regardless of how diluted it has become. Fortunately, your castle basement has a PETA-approved Acme Humane Rat Trap ™. In it are 10 rats that you may test the wine on. But, the party begins in only 10 hours! In other words, you only have time to conduct one round of testing before the party starts. How can you test the wine on the rats (not PETA approved) in such a way that you can determine exactly which bottle was poisoned? Keep in mind that because of the potency of the poison any mixture containing some poisoned wine will kill the rat.
I think there might be multiple solutions, but one of them is actually quite elegant!
I think I’ve got an elegant solution, but the whole business of mixing little bits of wine together is going to take more than the 0.5 hr grace period! There are 1000 bottles: just opening them will take longer than that! But perhaps in my awesome castle I also have 1000 awesome handmaidens to help.
By: Prof. Kate on November 12, 2008
at 12:19 am
… and pretty soon some very drunk rats.
By: Prof. Kate on November 12, 2008
at 12:28 am
I got this brain teaser in an interview–it has a very elegant solution. I also got a follow-up: what if there were TWO poisoned bottles? The solution to this is ridiculously difficult, and I don’t really remember it.
P.S. I know I blogged about this before, but if you love these kinds of riddles, you should check out the wu::riddles page/answer forum:
http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi
This riddle is called the “Criminal Cupbearers” riddle: http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml#criminalCupbearers
By: rfgarcia on November 12, 2008
at 4:04 pm
So I’m going to make a guess that the elegant answer has to do with binomial coefficients
and poisoning the rats in different combinations, so that when the specific group of rat(s) dies, you can trace it to a unique treatment bottle.
However, the first thing that came to my mind is actually quite inelegant, but effective if the poison is indeed lethal in exactly 9.5 hours…
Since you have 1000 bottles and 10 rats, have your 10 magical maids each take charge of 100 bottles and one rat. The party is in ten hours, so you have a 30-minute grace period. Have each maid dose her rat with one of the 100 bottles every 18 seconds (30 minutes/100 bottles = 18 seconds/bottle). Then, if you can time precisely when the ill-fated rat falls dead, you know exactly which bottle is to blame.
By: justpeachie on November 12, 2008
at 10:33 pm
This calls for a binary readout by rat. Ten rats each with their own dish would allow you to vet up to 2^10 bottles = 1024 > 1000.
One dish must be able to hold 500 drops of wine, and each subsequent dish half that of the previous.
I’ll leave the rest to the next commenter.
By: Ted on November 13, 2008
at 1:38 am
What if there are two poisoned bottles?
Well, there are 1000*999/2 = 499500 ways to pick two bottles out of 1000. But if you only do one round and see how many rats died, there are 2^10 = 1024 combinations of dead rats. So you can’t find out which two bottles are poisoned (there are 499500 possibilities) just by giving each rat some combination of bottles and seeing who dies (there are 1024 possible outcomes).
So what’s the trick then?
Maybe the question asks you to find some number of safe bottles (so you don’t need to know exactly which are poisoned), or something like that. Or maybe it involves timing, like justpeachie suggests? Do you remember, rfgarcia?
By: Admin on November 15, 2008
at 7:30 pm
So what’s the answer? =)
By: Brain Training on November 17, 2008
at 5:10 am
The (elegant) answer, at long last!:
This hinges on familiarity with binary numbers. Number each bottle of wine from 0 to 999 in binary (0000000000 – 1110011111). Assign each rat a digit: rat n is assigned to the nth digit of the binary number. The binary number determines which rats drink which wine, 1=drink 0=do not drink. Therefore rat 1 drinks all bottles whose binary number starts with 1, rat 2 drinks all bottles whose second digit is 1, and so on. Then, if 1=dead and 0=alive, write down the status of the rats. So for example, for rats 1 3 and 9 dead:
1010000010
This number corresponds exactly to the bottle of wine that was poisoned.
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at 9:41 pm
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