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Suppose we have $n$ bottles of water, one of which is poisoned and which we want to identify. A mixture can be made by mixing some number of bottles of water. We have $k$ rabbits, and each day, each of the rabbits can be given a mixture - a rabbit dies if the mixture contains any poison. How many days are necessary to find out which bottle is poisoned?

A similar, but slightly different problem, is discussed here: Logic problem: Identifying poisoned wines out of a sample, minimizing test subjects with constraints

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Micheal
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    You need to carry out $\lceil \log_{2} n \rceil$ tests to identify a single poisoned bottle. The tests can be planned in advance (e.g., you can prepare the mixtures each day without knowing any of the results from earlier days). With $k$ tests per day, you need $\lceil \frac{1}{k} \lceil \log_2{n} \rceil \rceil = \lceil \log_{2^{k}}{n} \rceil$ days. – mjqxxxx Jan 30 '13 at 23:25
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    @mjqxxxx I might be wrong, but I think that your numerical answer is dependent on all the rabbits surviving each day, while your solution assumes that some rabbits may die out. – Calvin Lin Jan 31 '13 at 00:00
  • @CalvinLin: You're right; I was treating it as if you had $k$ rabbits each day. – mjqxxxx Jan 31 '13 at 00:05

2 Answers2

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In $d$ days of testing, each rabbit can experience at most $(d+1)$ outcomes: live for all $d$ days, or die on some day from $1$ to $d$. With $k$ rabbits, then, the number of possible outcomes in $d$ days is at most $(d+1)^{k}$. The maximum number of bottles from which a single poison bottle can be identified must also satisfy $$ N(k,d) \le (d+1)^{k}.$$ We will show that this bound is tight, so $N(k,d)=(d+1)^k$ exactly. This is clearly true for $d=0$. The proof for $d\ge 1$ is inductive on $d$. Suppose we have shown that $N(m,d-1)=d^{m}$ for all $m$. Given $(d+1)^k$ bottles and $k$ rabbits, we proceed as follows. For each bitstring of length $k$ that contains $m$ $0$'s, label $d^m$ bottles with that bitstring. The total number of labeled bottles is $$ \sum_{m=0}^{k}{{k}\choose{m}}d^m=\sum_{m=0}^{k}{{k}\choose{m}}d^m1^{k-m}=(d+1)^k. $$ Now feed a mixture to each rabbit, $i$, that includes water from each bottle with a $1$ in the $i$-th place of its bitstring. At the end of the day, the bitstring with $1$'s in the places corresponding to dead rabbits will be exactly the label on the poisoned bottle. If $k-m$ rabbits have died (which will be the case for some $0\le m\le k$), the selected bitstring has $k-m$ $1$'s and $m$ $0$'s; there are $d^m$ bottles with that bitstring, and we have $m$ surviving rabbits and $d-1$ days left to identify the poison bottle from among them. By the inductive hypothesis, this is enough, which completes the proof.

So, with $k$ rabbits and $d$ days, the maximum number of distinguishable bottles is exactly $(d+1)^k$. Inverting this, we see that for the case with $n$ bottles and $k$ rabbits, we need to be allowed $$ D(n,k)=\left\lceil n^{1/k}-1 \right\rceil $$ days to find the poison.

mjqxxxx
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Let the answer be $S(n, k)$. All fractional number of days should be round up (i'm lazy to type the ceiling function.

If $k=1$, then we can only test 1 wine each day, so we need $n-1$ days. Remember that if there's only 1 wine left, it must be the poisoned one. So $S(n, 1) = n-1$.

If $k=2$, then by dividing up the wine into 3 groups of $ \alpha, \beta, n-\alpha - \beta$ on the first day, and testing the first 2 groups against the 2 rabbits. If both survive, poison is in the last group, and it will take $S(n - \alpha - \beta, 2) + 1$ days. Otherwise, poison is in the first 2 groups and it will take $S(\alpha, 1) + 1$ or $S(\beta, 1) + 1$ days. Hence, in order to minimize the number of days, we want to set these values to be all the same. This gives us $S(\alpha, 1) + 1 = \alpha - 1 + 1 = \alpha$, so $\alpha = \beta = S(n - 2\alpha, 1)+1$.

Since $S(1, 2) = 0$ and $S(2, 2) = 1$, so if we have $\alpha = \beta = 1$, then $n - 2\alpha \leq 1$. This shows that $ S(2, 2)=1, S(3, 2) = 1$ by giving 1 rabbit 1 wine each and that $(4, 2) =2$.

Since $S(3, 2)=1$ and $(4, 2) =2$., so if we have $\alpha = \beta = 2$, then $n - 2\alpha \leq 3$. This shows that $S(4, 2) = 2, S(5, 2) = 2, S(6, 2) = 2, S(7, 2) = 2$ by giving 1 rabbit 2 wines each, and that $S(8, 2) = 3$.

The pattern of the highest value of $n$ such that $S(n, 2) = m-1$ is $1, 3, 7, 13, \ldots$, where we're adding $2m$ to each term, so the terms of the series is $1+(m-1)\times m$. Thus, if $1 + (m-2) \times (m-1) < n < 1 + (m-1)\times m$, then $S(n, 2) = m$.

You can continue this procedure with $k=3$ rabbits. Break up the wines into 4 groups of $\alpha, \beta, \gamma, n-\alpha -\beta - \gamma$, and we test the first 3 groups on the 3 rabbits. We want $\alpha = \beta = \gamma = S(n- \alpha - \beta - \gamma, 2)$. Then, $S(n,3) = 1 + \alpha$.

Sadly, I'm not certain how to generalize this.

Calvin Lin
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