A more general poisonous water puzzle The simple version of the poisonous water problem can be described as follows: Given 1000 bottle with water with one bottle with poisonous water among 1000 bottles. You have several animal that can test the water. Once they drink the poisonous one(no matter how much) they will die 24 hours later. Now you want to find the poisonous one after 24 hours. Then the minimum amount of animal we need is 10 (log(1000)). Note that without the time limit it can be viewed as a binary search problem. 
   Now the general form of this problem is what happens if we have m poisonous bottles instead of one? What's the answer of the minimum amount of animal we need? Theoretically I would say that the lower bound is not hard by taking the log of all the possibilities. But to find out the real value and algorithm is difficult. I will be grateful if someone can answer the general form of the problem
Update: I heard someone using m-(de)cimal but I don't see how this will leads to a simple answer.
 A: This problem is similar to but not identical to the $m-1$-error correction code problem ($m-1$ because the $\log_2(N)$ animals you would feed if $m=1$ correspond to the main $\log_2(N)$ bits you are correcting). The difference is that in this problem, the additional questions you ask (experiments you do, animals you feed) cannot themselves give erroneous results; in the correction code case, they can.
The other difference is that here, you know in advance there are precisely $m$ "errors."
At any rate, while there are in the literature nice methods for single and double error correction, and there are complex methods (see BCH theory) for dealing with multiple corrections, there do not seem to be bounds on the number of bits that can be corrected using $k$ additional bits, with those bounds constructively proven by demonstrating algorithms.
For your problem, we can show that there are cases where the "information theory" lower bound of 
$$\left\lceil \log_2\binom{N}{m}\right\rceil$$ does not suffice.  Consider, for example, ten bottles, three of which are poisoned, which has $120$ possible answers.  If you give any animal water from one bottle, then the answer splits the cases into $36$ if the animal dies and $84 > 64$ if the animal lives, so that leads to at least one extra trial compared to that bound.  Giving any animal water from two bottles splits the cases into $64$ and $56$, but the $64$ cases in which the animal dies don't permit a splitting into $32|32$ on the next animal.  Giving any animal water from more than two bottles splits the cases into $k>64$ and $120-k$.
The actual required number for general $N,m$ appears to be a hard problem.
A: If there are $m$ bottles there are ${N \choose m}$ possible combinations.  An animal lives or dies (binary) so you need $\log_2{N \choose m}$ animals to match all possibilities.
