# Logic problem: Identifying poisoned wines out of a sample, minimizing test subjects with constraints

I just got out from my Math and Logic class with my friend. During the lecture, a well-known math/logic puzzle was presented:

The King has $$1000$$ wines, $$1$$ of which is poisoned. He needs to identify the poisoned wine as soon as possible, and with the least resources, so he hires the protagonist, a Mathematician. The king offers you his expendable servants to help you test which wine is poisoned.

The poisoned wine is very potent, so much that one molecule of the wine will cause anyone who drinks it to die. However, it is slow-acting. The nature of the slow-acting poison means that there is only time to test one "drink" per servant. (A drink may be a mixture of any number of wines) (Assume that the King needs to know within an hour, and that any poison in the drink takes an hour to show any symptoms)

What is the minimum amount of servants you would need to identify the poisoned wine?

With enough time and reasoning, one can eventually see that this requires at most ten ($$10$$) servants (in fact, you could test 24 more wines on top of that 1000 before requiring an eleventh servant). The proof/procedure is left to the reader.

My friend and I, however, was not content with resting upon this answer. My friend added the question:

What would be different if there were $$2$$ wines that were poisoned out of the 1000? What is the new minimum then?

We eventually generalized the problem to this:

Given $$N$$ bottles of wine ($$N \gt 1$$) and, of those, $$k$$ poisoned wines ($$0 \lt k \lt N$$), what is the optimum method to identify the all of the poisoned wines, and how many servants are required ($$s(N,k)$$)?

After some mathsing, my friend and I managed to find some (possibly unhelpful) lower and upper bounds:

$$log_2 {N \choose k} \le s(N,k) \le N-1$$

This is because $$log_2 {N \choose k}$$ is the minimum number of servants to uniquely identify the $$N \choose k$$ possible configurations of $$k$$ poisoned wines in $$N$$ total wines.

Can anyone help us find an optimum strategy? Besides the trivial one requiring $$N-1$$ servants. How about a possible approach to start?

Would this problem be any different if you were only required to find a strategy that would for sure find a wine that is not poisoned, instead of identifying all poisoned wines? (other than the slightly trivial solution of $$k$$ servants)

• It would be better to change the title of your question to say something about the problem itself. Something like "Math/Logic problem - 1000 cups of wine, one of which is poisoned". This will help people who know the answer look at the question, as well as make it easier to find later. Jul 24, 2010 at 12:24
• I have a slight problem with the reasoning here. It all checks out, apart from the fact the poison is slow acting. I can't see a way to manage to test all wines with just 10 servants when it takes the entire time-period for symptoms to show. Jul 24, 2010 at 13:38
• @workmad3: Number the wines from 0 to 999 and number the servants from 1 to 10. Feed servant k all the wines with a 1 in the kth binary position of their number. Wait an hour. If D is the set of dead servants, the number of the poisoned wine is $\sum_{i \in D} 2^{i-1}$. Jul 24, 2010 at 13:51
• @Justin: When someone else starts typing in the same question, the duplicate matcher will display your question title for them. Consider what title would make them most likely to click on it. "Can someone help me with a logic problem" will not do a good job of convincing them their exact question has already been asked. Jul 24, 2010 at 23:28
• @97832123 @Kaestur - I've edited my title; is it more helpful? =) Jul 24, 2010 at 23:31

I asked this question on MathOverflow and got a great answer there.

For $k = 2$ I can do it with ${\lceil \log_2 N \rceil + 2 \choose 2} - 1$ servants. In particular for $N = 1000$ I can do it with $65$ servants. The proof is somewhat long, so I don't want to post it until I've thought about the problem more.

I haven't been able to improve on the above result. Here's how it works. Let $n = \lceil \log_2 N \rceil$. Let me go through the algorithm for $k = 1$ so we're all on the same page. Number the wines and assign each of them the binary expansion of their number, which consists of $n$ bits. Find $n$ servants, and have servant $i$ drink all the wines whose $i^{th}$ bit is $1$. Then the set of servants that die tells you the binary expansion of the poisoned wine.

For $k = 2$ we need to find $n$ butlers, $n$ maids, and ${n \choose 2}$ cooks. The cooks will be named $(i, j)$ for some positive integers $1 \le i < j \le n$. Have butler $i$ drink all the wines whose $i^{th}$ bit is $1$, have maid $i$ drink all the wines whose $i^{th}$ bit is $0$, and have cook $(i, j)$ drink all the wines such that the sum of the $i^{th}$ bit through the $j^{th}$ bit, inclusive, mod 2, is $1$. This is how the casework breaks down for butlers and maids.

• If both butler $i$ and maid $i$ die, then one of the poisoned wines has $i^{th}$ bit $0$ and the other has $i^{th}$ bit $1$.
• If only butler $i$ dies, then both of the poisoned wines have $i^{th}$ bit $1$.
• If only maid $i$ dies, then both of the poisoned wines have $i^{th}$ bit $0$.

The second two cases are great. The problem with case 1 is that if it occurs more than once, there's still ambiguity about which wine has which bit. (The worst scenario is if all the butlers and maids die.) To fix the issue with case 1, we use the cooks.

Let $i_1 < ... < i_m$ be the set of bits where case 1 occurs. We'll say that the poisoned wine whose $(i_1)^{th}$ bit is $1$ is wine A, and the other one is wine B. Notice that the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine A mod 2 is the same as the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine B mod 2, and we can determine what this sum is by looking at whether cook $(i_1, i_2)$ died. The value of this sum determines whether the $(i_2)^{th}$ bit of wine A is 1 or 0 (and the same for wine B). Similarly, looking at whether cook $(i_j, i_{j+1})$ died tells us the remaining bits of wine A, hence of wine B.

One last comment for now. The lower bound is not best possible when $k$ is large compared to $N$; for example, when $k = N-1$ it takes $N-1$ servants. The reason is that any servant who drinks more than one wine automatically dies, hence gives you no information.

• By the way, I don't know how I feel about this phrase "the astute mathematics student." It seems kind of demeaning. Jul 29, 2010 at 3:50
• @Qiaochu - fixed :) i await your solution; a rather large group of my friends are about to give up and go with brute-force using a computer programmer. Jul 29, 2010 at 5:15
• Some thoughts. If you want to approach the lower bound, you need to be very careful about how many wines each servant drinks. For fixed N and k there is an optimum number of wines which gets you as close to 1 bit of information as possible, but I have no idea how I would decide what sets of wines to drink other than randomly. The probabilistic method might be capable of showing that a strategy which is close to optimal exists with positive probability, but I don't know much about these matters. Aug 1, 2010 at 5:33
• Some more thoughts. The above strategy generalizes to a strategy which requires kn + k {n \choose 2} servants, where n = log_k N, but I can't actually prove that it works yet... Aug 3, 2010 at 0:26
• This is the best answer I'm probably going to get, and I concluded that this question/puzzle is not one that will be solvable any cleaner than this. Thanks for all your help Sep 7, 2010 at 20:25

### Solution with 42 servants (also 41):

This gives a solution, for 2 poisons, using 42 servants, for 2187 (=3^7) wines. Represent the wines using trinary number system. You will get 7 Trits. We need to find out the trits of the two poisons.

    A-B-C-D-E-F-G


(A) For Trit A, give wines with values 0,1,2 to 3 unique persons respectively. Do similar tests for rest of the Trits. There will be 7*3=21 such tests.

(B) Mix wines with AB values {00,11,22}. Give it to 1 unique person. Do similar tests on all 2-combination of Trits. There will be C(7,2)=21 such tests.

In Step (A), for a given Trit, at the most, 2 persons will die. If only one person dies, that Trit is resolved. Let's say that some Trits (B,C,E) didn't get resolved.

    1-(0|1)-(1|2)-0-(1|2)-0-0


To link B&C, check if person for BC died in Step(2). If so, BC are {02,11}; else {01,12}.
To link C&E, check if person for CE died in Step(2). If so, CE are {11,22}; else {12,21}.

Total # of tests = 21+21 = 42

### Explanation:

The idea is that when comparing Trits (in Step-B), we need to consider only two broad categories. One with only one common value. One with two common values (Not having a common value is impossible). Let's assume that trits A&D did not get resolved.

Follg shows them as having only one common value (0)

    A   D
=====
0   0
1   []
[]  2


Follg shows them as having two common values (0,1).

    A   D
=====
0   0
1   1
[]  []


Either way, to link them, it is enough to know if at least one of {00,11,22} died. That resolves the values of both the trits.

Note: A solution for 1458 wines, 2 poisons, can be formulated with 41 servants, by using 7 digits (6 Trits + 1 Bit). This trick doesn't add much worth; it might find its use in testing higher number of poisons.

• I think your solution with 40 servants might be in error, or I misunderstood your description. Consider the two cases: (i) the bottles labeled 00111 and 22333 are poisoned and (ii) the bottles labeled 00333 and 22111 are poisioned. If I understand your proposed solution correctly, then the same servants die in both cases (in A the servants [AB] and [CDE], in B the servants [AB][AB] and [CDE][CDE] and in C the servants [AB][CDE]). May 7, 2014 at 14:35
• @Kallus Tx for insightful comment! Step-C fails to handle ur case. There is an ambiguity. May be 42 is the optimum. I will post again if I get new ideas. Otherwise, I will delete the the solution for 40. May 8, 2014 at 23:16
• @Kallus I removed the solution with 4-base-system. An edge between two 4-base-digits has 16 combinations. If two tests reduce it twice (16/2/2=4), that would still remain ambiguous. The best, I could do was, to solve it with 41 servants (by replacing one Trit with one Bit). Tx for your comments! May 10, 2014 at 17:19
• That was so clever! According to Qiaochu Yuan's MathOverflow link, 2187 wines can't be tested with fewer than 42 servants.
– Eric
Mar 8, 2017 at 10:22
• @blackpen , can you adapt this method of yours to solve this question : puzzling.stackexchange.com/questions/12770/… Sep 27, 2020 at 1:37

### Solution with 40 servants:

Represent the wines using base-4 number system. You will get 5 digits (for 4^5=1024 wines)

    A-B-C-D-E


(A) For digit A, give wines with values 0,1,2,3 to four unique persons respectively. Do similar tests for rest of the digits. There will be 5*4=20 such tests.

(B) Mix wines with AB values {00,11,22,33}. Give it to one unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (blue graph).

(C) Mix wines with AB values {10,21,32,03}. Give it to one unique person. Do similar tests on all 2-combination of digits. There will be C(5,2)=10 such tests (green graph). In Step (A), for a given digit, at the most, 2 persons will die. If only one person dies that digit is resolved. If 2 persons die, it means it has two values (one for each poison).

Let's say two digits D&E got two values each. To link them up, you need to look if the person in Step(B) and Step(C) died for D&E. That is enough to link up the digits.

### Explanation:

Look at the two graphs.

Assume D has values (1,3). Together, they connect with 0,1,2,3 (all possible values of E). This is the best case. No matter which two values of E you choose, you can uniquely connect them with values of D.

Assume D has values (1,2). Together, they connect with 0,1,2. One value of E (=3) remains unconnected. This is the worst case. Still, no matter which two values of E you select, at least one of them remains connected with values of D. That is enough for us.

I have a solution that uses just 28 slaves. The idea is to use a 28x1000 2-separable testing matrix $$M$$, where $$M_{ij}$$ is 1 if slave $$i$$ drinks bottle $$j$$ and 0 if he doesn't drink it. A matrix is 2-separable if the boolean OR of any of its two columns is unique. You can read more about such matrices here: https://en.wikipedia.org/wiki/Disjunct_matrix

You can find the actual matrix $$M$$ here: https://pastebin.com/4yDd1XvG

When I find some free time, I plan to write a short report describing the method I used to find $$M$$.

UPDATE: A 27x1065 2-separable matrix has been found so it is possible to solve this problem with 27 slaves. See https://oeis.org/A054961

• Are you still around? Could you please write out the proof of your solution?
– Hans
Mar 29, 2018 at 7:32
• Hi Hans. The proof is in the matrix that I provided. If you take the boolean OR of any of its two columns then you will find that it is unique. This means that we can use this matrix to identify the location of any two poisoned bottles out of 1000. Mar 30, 2018 at 9:02
• Yes, Zhaohui Du actually posted about the 27 solution here as well: math.stackexchange.com/a/3330768/346149 Aug 11, 2020 at 3:25
• How can one modify this matrix method to solve the following question ? puzzling.stackexchange.com/questions/12770/… Sep 28, 2020 at 22:08
• @HemantAgarwal I believe the problem in your link is EXACTLY the same as the problem here, so the same matrix method would apply to give you 27 slaves. Sep 30, 2020 at 1:08

This gives a solution using 60 servants for 1024 wines.Represent the wines using 10-bit binary numbers. Name the bits from A-J. Group the adjacent bits.

    AB--CD--EF--GH--IJ


What is the value of AB for poisons? Give follg sets of wines to 4 different people.

    AB=00; AB=01; AB=10; AB=11


If only one of them die, AB is completely resolved! If not, we get partial information. Do similar tests on remaining 4 groups.

    00--0(0|1)--11--(0|1)1--1(0|1)


Consider CD & GH. If we know if set DG=00 died, we can link them. Since we dont know which test would be required, we run all 4-tests:

    CG=00; CH=00; DG=00; DH=00


Similarly for GH and JK. If GJ=00 didnt die, the poisons will be:

    00--00--11--01--11
00--01--11--11--10


Total # of tests/people needed: 5*4 + C(5,2)*4 = 60

PS: If you use groups of 3 (rather than 2), it doesn't improve. If you use groups of 4 or more, it worsens.

Technically, a molecule of wine can't contain a molecule of poison, but I'm splitting hairs. I know what you mean.

I consider this problem impossible to solve because of the time constraints. Any time taken to find the correct bottle is taken away from the time necessary for the poison to kill the servant.

If one's objective were to protect the king, it could be done with a single servant. Simply give wine to the servant an hour before the king is ready for his. If the servant dies, pick another bottle. It could take up to 999 tries to find the right bottle, but the king would be safe, and only one servant's life is put at risk.

I don't know about you, but it would probably take me the whole hour to number a thousand bottles. Mixing wines would seem equally time consuming.

There are two solutions Binary and Dimensional.

The binary solution:

We give each bottle a binary number corresponding to its numerical position in the line of bottles. We need enough columns to accommodate the given number of bottles. For instance, if we have 8 bottles :

0 -> 000

1 -> 001

2 -> 010

3 -> 011

4 -> 100

5 -> 101

6 -> 110

7 -> 111

(Note: the first bottle has position 0)

We can test with three columns. Three doses are made one for each column, taking one drop from each bottle if and only if it has a 1 in that column. For instance, the first sample has drops from bottles 4,5,6 and 7.

Each dose is given to one person. We just note which of them die. If none die, the first bottle has poison. If all die, the eight bottle (number 7, binary) is poisoned. If the first and third die, the sixth bottle (binary 5).

The binary solution results in an uncertain number of deaths, averaging 1.5, or 50%. But it uses the smallest number of tasters.

The Dimensional solution:

In the simplest case, treating the bottles as points in a line, we need 1000 servants and 1 dies.

Making a square, we need 32 rows and 32 columns. Some spaces are empty, it won't matter, since empty spaces have the same effect as unpoisoned bottles.

Each dose has drops from 1 row. 64 persons take the dose, two die, indicating a row and a column, the poison bottle is at the intersection.

Making a cube, we have 10 x 10 x 10 bottles. We make 3 doses with a drop from each square, on the x y and z axis. Again, the poisoned bottle is at the intersection.

The progression should now be clear:

Arrangement: Line

Consists of: Points

Test using : Points

Number used: 1000

Arrangement: Square

Consists of: Lines

Test using : Lines

Number used: 32 x 2 = 64

Arrangement: Cube

Consists of: Squares

Test using : Squares

Number used: 10 x 3 = 30

For any given number of bottles, to arrange in n dimensions we take the nth root and round of to the next higher integer. At 5 dimensions, we use 20 testers and 5 die.

Increasing the dimensions can go on and on.

• I assume this solution is only for "1-poison" problem and not for "2-poisons". The "dimensional solution" reaches optimum value when you use log_2(N) dimensions (and thus it becomes the "binary solution"). Aug 22, 2015 at 0:08
• Sorry blackpen, I checked the site for a few days and then thought no one was active on this problem! Sep 2, 2016 at 15:26
• The dimensional solution works for any number of poisoned bottles, and they can be arranged in any order. It will not give a false negative. However, it will give a false positive, if an unpoisoned bottle is surrounded in all dimensions by poisoned bottles. So the King and his guests may keep aside some perfectly good bottles, but they will not ever drink poisoned wine. Sep 2, 2016 at 15:29
• Chris, Consider 2-dimensional solution with 2-poisons. Each point is a single wine; each line/row stands for a mixture of those wines. You are testing rows/columns. If you found that 4 persons died (row=1, row=2, col=1, col=2), you don't know which diogonal has the two-poisons : [(1,1),(2,2)] OR [(1,2),(2,1)]. In 2-dimensions, you have 2-diagonals that you need to resolve. In general (dims,diagonals) grow as (1,0),(2,2),(3,4),(4,8), ..($n$,$2^{n-1}$). Sep 2, 2016 at 21:11

Look at Identifying 2 poisoned wines out of 2^n wines for a easier, guided explanation on how it can be solved using $6n$ tests given $2^n$ wines.

I wrote a code that searches for such an arragement with less than 40 servants. I randomized 1000 groups of 12 servants, representing the servants that drink each bottle. Afterwords, I found two couples of wine bottles that have the same union of servants who drink them. I rerandomized the servants that drink each of the four(or three) bottles in the unions.

That approach found a solution for as much as 36 servants.

– Hans
Mar 29, 2018 at 7:33

# Two solutions with 20 tests

I am about $$10=\lceil\log_21000\rceil$$ years late to the party (weird coincidence since it's the solution to $$T_*(1,1000)$$), yet I am surprised no one has mentioned a solution with $$20$$ steps for the $$T_*(2,1000)$$ problem. (The notation I am using is that $$T_*(k,N)$$ represents the maximum number of steps required to find exactly $$k$$ poisoned wine bottles in exactly $$N$$ wine bottles by algorithm $$*$$ .)

There is an algorithm known as the Generalised Binary Splitting, developed by Hwang, and for the sake of completeness, I shall reproduce that here later.

But first, here's an algorithm for the specific case of $$T_s(k=2,N)$$ which we claim can be solved in at most $$4\lceil\log_4N\rceil$$ tests, which gives, for $$N=1000$$, $$T_s(2,1000)=4\lceil\log_41000\rceil=20$$ tests.

(Side note : None of these algorithms are proven to be optimal, although both are near optimal, as one can check how much the answers differ from the information lower bound, which is an obvious lower bound, but is in general unachievable).

Here's a proof of the claim.We proceed via induction on $$N$$. We first see that for $$2\le N\le4$$, we do need to check every single one of them for identification, thus the number of tests is $$N\le 4$$, and we also have $$T_s(2,N)=4\lceil\log_4N\rceil=4$$ and thus the result holds in these cases.

Take any of them as the base case and assume that the statement is true for all number of items $$\le N-1$$. Then for the case of $$N$$ items, consider the following algorithm:

1. If we are in any base case, the number of tests required is at most $$4$$, so add a $$4$$ and stop.

2. If not, make $$4$$ groups of bottles such that there are $$n_1$$ groups of $$2^{\lceil \log_2(N/4)\rceil}=\lambda$$ (say) and another $$n_2$$ groups of $$b\le \lambda$$ bottles. This gives us 2 equations: $$n_1\lambda+n_2b=N$$ $$n_1+n_2=4$$ Solving simultaneously gives: $$n_2(b-\lambda)=4\lambda-N$$ This gives us solutions for the pair $$(n_2,b)\in \mathbb Z^2$$ and we only consider those solutions which satisfy $$0\le n_2\le 4$$ and $$0\le b\le \lambda$$. Choose the pair such that $$n_2$$ is maximum (this minimizes $$n_1$$, but any solution will work. Just for the sake of a concrete algorithm, we are considering this). Observe that \begin{align}\lceil \log_2 b\rceil\le\lceil\log_2\lambda\rceil\end{align} We label the groups with their sizes ($$\lambda$$ and $$b$$). Now we test each group, which requires $$t_1=4$$ tests. Consider the possibilities. If just $$1$$ group signals poisoned, then we can have a $$\lambda$$ group signalling or a $$b$$ group signalling. If $$2$$ groups signal poisoned, then the possibilities are $$2\ \lambda$$ groups signalling or $$2\ b$$ groups signalling or a $$\lambda$$ and a $$b$$ group signalling. These scenarios are further referred to as $$A,B,C,D$$ and $$E$$ respectively.

We shall use the result that $$1$$ poisoned bottle in $$n$$ bottles can be found in at most $$\lceil \log_2 n\rceil$$ tests.

Analysing $$C,D$$ and $$E$$ first, we note that identifying the $$2$$ poisoned bottles in the groups can be done in $$C\to 2\lceil \log_2 \lambda\rceil$$ $$D\to 2\lceil \log_2 b\rceil\le 2\lceil \log_2 \lambda\rceil$$ $$E\to \lceil \log_2 b\rceil+\lceil \log_2 \lambda\rceil\le 2\lceil \log_2 \lambda\rceil$$ steps, and we see that they are all bounded by $$2\lceil \log_2 \lambda\rceil$$ and thus the total number of steps is bounded by $$T_1=2\lceil \log_2 \lambda\rceil+4$$ steps. Putting in the definition of $$\lambda$$ and simplifying, we get the total number of tests to be $$T_1=2\lceil\log_2N\rceil=2\lceil2\log_4N\rceil\le 4\lceil\log_4N\rceil=T_s(2,N)$$ Now we analyse $$A$$ and $$B$$. If we are in any base case, go to step $$1$$. If not, go to the beginning of step $$2$$ with $$N:=\lambda$$ for $$A$$ or $$N:=b$$ for $$B$$.

By induction, this last step would require either $$4\lceil\log_4\lambda\rceil$$ or $$4\lceil\log_4b\rceil$$ steps to identify the poisoned bottles. We require the maximum of these 2, which is easy to see once we note that $$4\lceil\log_4b\rceil\le 4\lceil\log_4\lambda\rceil$$ and hence $$T_2=4\lceil\log_4\lambda\rceil+4$$. Simplifying again, we have $$T_2=4\left\lceil\frac12\lceil\log_2N\rceil\right\rceil=4\left\lceil\frac12\lceil2\log_4N\rceil\right\rceil\le 4\lceil\log_4N\rceil$$ This completes the proof.$$\square$$

The Generalised Binary Splitting kind of uses the same idea, but is much more general. The algorithm goes as follows:

1. If $$N\le 2k-2$$, test the $$N$$ items individually. If $$N\ge2k-1$$, set $$l=N-k+1$$. Define $$\alpha=\lfloor\log(l/k)\rfloor$$.
2. If $$N>2d-2$$, test a group of size $$2^\alpha$$. If the outcome is negative, the $$2^\alpha$$ items in the group are identified as good . Set $$N:=N-2^\alpha$$ and go to Step 1. If the outcome is positive , use binary splitting to identify one defective and an unspecified number , say $$x$$, of good items. Set $$N:=N-1-x$$ and $$k=k-1$$. Go to Step 1.

This algorithm gives: $$T_G(k,N)=\begin{cases}N\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for}\ N\le 2k-2\\ (\alpha+2)d+p-1\ \text{for}\ N\ge 2k-1\end{cases}$$ where $$p is a non-negative integer uniquely defined in $$l=2^\alpha k+2^\alpha p+\theta,\ 0\le\theta<2^\alpha$$.

The proof goes as follows. The case $$N\le 2k-2$$ is true due to step 1. For $$2k-1\le N\le 3k-2$$,$$\alpha$$ must be $$0$$ and $$G$$ is reduced to individual testing. Furthermore, $$\theta=0$$ and $$p=l-k=N-2k+1$$. Thus $$(\alpha+2)k+p-1=2k+N-2k+1-1=N=T_G(k,N)$$ For $$k=1$$, $$l=N-k+1$$. Note that except for $$N=2^\alpha$$, $$G$$ is reduced to binary splitting and $$(\alpha+2)k+p-1=\alpha+1=\lfloor\log_2N\rfloor+1=\lceil\log_2N\rceil=T_G(1,N)$$ For $$N=2^\alpha$$, $$G$$ spends one more test than binary splitting (by testing the whole set first) and $$(\alpha+2)k+p-1=1+\lceil\log_2N\rceil=T_G(1,N)$$ For the general case $$k\ge 2$$ and $$N\ge 3k-1$$, we proceed via induction on $$k+N$$. From step 2 $$T_G(k,N)=\max\{1+T_G(k,N-2^\alpha),1+\alpha+T_G(k-1,N-1)\}$$ For $$N'=N-2^\alpha$$ and $$k'=k$$, \begin{align*}l'&=N'-d'+1=N-2^\alpha-k+1=l-2^\alpha\\&=\begin{cases}2^\alpha k+2^\alpha(p-1)+\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for}\ p\ge 1 \\2^{\alpha-1}k+2^{\alpha-1}(k-2)+\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for}\ p=0,\ \theta<2^{\alpha-1}\\2^{\alpha-1}k+2^{\alpha-1}(k-1)+(\theta-2^{\alpha-1})\ \text{for}\ p=0,\ \theta\ge 2^{\alpha-1}\end{cases}\end{align*} Hence by induction $$T_G(k,N-2^\alpha)=\begin{cases}(\alpha+2)k+(p-1)-1\ \text{for}\ p\ge 1\\ (\alpha+1)k+(k-2)-1\ \text{for}\ p=0,\ \theta<2^{\alpha-1}\\ (\alpha+1)k+(k-1)-1\ \text{for}\ p=0,\ \theta\ge 2^{\alpha-1}\end{cases}$$ Consequently $$1+T_G(k-N-2^\alpha)=\begin{cases}(\alpha+2)k+p-2\ \text{for}\ p=0,\ \theta<2^{\alpha-1}\\ (\alpha+2)k+p-1\ \text{otherwise}\end{cases}$$ For $$k'=k-1$$ and $$N'=N-1$$, $$l'=N'-k'+1=l=\begin{cases}2^\alpha(k-1)+2^\alpha(p+1)+\theta\ \text{for}\ p\le k-3\\ 2^{\alpha+1}(k-1)+\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for}\ p=k-2\\ 2^{\alpha+1}(k-1)+2^\alpha+\theta\ \ \ \ \ \ \ \ \ \ \text{for}\ p=k-1\end{cases}$$ And hence by induction $$T_G(k-1,N-1)=\begin{cases}(\alpha+2)(k-1)+(p+1)-1\ \text{for}\ p\le k-3\\ (\alpha+3)(k-1)-1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for}\ k-2\le p\le k-1\end{cases}$$ Which gives $$1+\alpha+T_G(k-1,N-1)=\begin{cases}(\alpha+2)k+p-2\ \text{for}\ p=k-1\\(\alpha+2)k+p-1\ \text{otherwise}\end{cases}$$ Since for $$d\ge 2$$, $$p=0$$ and $$p=k-1$$ are mutually exclusive, $$T_G(k,N)=\max\{1+T_G(k,N-2^\alpha),1+\alpha+T_G(k-1,N-1)\}=(\alpha+2)k+p-1\square$$

• This sounds like an adaptive testing strategy. I think the question is about non-adaptive testing ("slow-acting poison" and so forth). Jul 19, 2020 at 9:23
• No this is non adaptive. The algorithm is linear. It's basically the known Binary splitting, except the split is done recursively on splits. Jul 19, 2020 at 9:40
• The procedure says test X, if outcome is negative, do this; if outcome is positive do that. How should I understand this to be non-adaptive? Jul 19, 2020 at 9:50
• It's just like the binary splitting. In the original situation with 1 poisoned wine, you split the set into 2 equal parts, and you say the part that tests positive is good. Repeat the procedure with the bad part. Thus the binary splitting also says different things for good and bad tests. This also uses the same exact argument, except the repeating procedure has one more variable to account for the number of poisoned wine bottles. Jul 19, 2020 at 9:53
• I think the procedure you describe for 1 poisoned wine is also adaptive. The standard non-adaptive procedure is to assign to each wine a distinct binary string of length 10 and each servant $i$ gets a mixture of the wines with a $1$ bit in position $i$ (or equivalent). Jul 19, 2020 at 10:02

You should be able to identify $$k$$ poisoned bottles in at most $$\sum\limits_{i=1}^k (\lceil \textrm{log}_2 N \rceil - \lceil \textrm{log}_2 k \rceil) - 1$$ steps, as follows:
Build a binary tree of subsets as for the $$k = 1$$ case, and follow each path that contains at least one poisoned bottle. After any "round" of tests, you can have at most $$k$$ subsets containing poison, so the largest number of tests occurs when this limit is reached as early as possible and each poisoned bottle is not identified until round $$\lceil \textrm{log}_2 N \rceil$$. However, after $$n$$ rounds there can only be at most $$2^n$$ subsets, so the $$i$$'th poisoned bottle can only be isolated after at least $$\lceil \textrm{log}_2 i \rceil$$ rounds, placing it in $$\lceil \textrm{log}_2 N \rceil - \lceil \textrm{log}_2 i \rceil$$ subsets.
I subtract $$1$$ because I already know the root (the set of all bottles) contains poison and thus I don't need to check it.

• This sounds like an adaptive testing strategy. I think the question is about non-adaptive testing ("slow-acting poison" and so forth). (This means, you wouldn't be able to choose what to test next based on the results of a previous test. All the servants must be given their wine cocktail at the same time.) Jul 19, 2020 at 9:54
• Ah, I must have been thrown off by some of the other adaptive answers. Jul 20, 2020 at 1:06