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This problem has been asked and discussed in following posts:

  1. Logic problem: Identifying poisoned wines out of a sample, minimizing test subjects with constraints,
  2. Finding 2 poisoned bottles of wine out of a 1000
  3. Identifying poisoned wines

There are better optimized solutions to this problem. But, for easy comprehension, I formulated following solution with 6n tests (for $2^n$ wines). Are there any loop holes in it?

Part-I
Let's say there are $2^5$ (=32) wines. Represent them using binary system using 5 bits:

     A-B-C-D-E

Give wines with A=$0$ to one person. Give wines with A=$1$ to another person. Do such tests for all bits (10 tests). Let's say you got a result like the following:

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

Bits B&C have fixed values (B=$0$. C=$1$). Bits A,D,E have two probable values each. The solution must be one from the following enumerations (Two solutions if you count both (a,b) and (b,a)). We represent the solutions in binary/decimal forms for clarity. It is easy to prove that they are just pairs of numbers with a constant sum (=27 in our case).

     (00100, 10111) = (4,23)
     (00101, 10110) = (5,22)
     (00110, 10101) = (6,21)
     (00111, 10100) = (7,20)
     (10100, 00111) = (20,7)
     (10101, 00110) = (21,6)
     (10110, 00101) = (22,5)
     (10111, 00100) = (23,4)

Part-II
Renumber the wines from (0,1,2,3,4,5 ...) to (0,1,4,9,16,25 ...) and put them into a bigger set of size $(2^5)^2$=$2^{10}$=1024 (We are squaring the numbers). Only 32 of the 1024 will be actual wines. Fill the rest of them with water.

Repeat the tests similar to that ones from Part-I on this bigger set (2*10=20 tests). This will give another set of solutions. Ignore all the solutions which contain water in one or both the bottles. Only the pair that matches with the corresponding pair from Part-I will be the actual solution (Two solutions if you consider duplicates).

If you get these from Part-I: $$(a,b),(c,d),(e,f),(f,e),(d,c),(b,a)$$ and if you get these from Part-II: $$(a^2,b^2),(g^2,h^2),(c^2,i^2),(i^2,c^2),(h^2,g^2),(b^2,a^2)$$

then the solution must be (a,b).

The reason is: Part-I satisfies $x+y=c1$, and Part-II satisfies $x^2+y^2=c2$. Given that x>=0 & y>=0, there is only one solution that satisfies these conditions (two solutions if you consider duplicates).

Number of tests required for $2^5$ wines with 2 poisons = $2*5 + 2*(2*5)$ = 30.
Number of tests required for $2^n$ wines with 2 poisons = $2*n$ + $2*(2*n)$ = $6n$.

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blackpen
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2 Answers2

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If $$a+b=c+d$$ and $$a^2+b^2=c^2+d^2$$ then $a=c$ and $b=d$ is not always true. Let $$a=1\quad b=2 \quad c=2 \quad d=1$$ and note that $a\neq c$ and $b\neq d$.

So your test may work, but not for the reason you specified. For example, let there be $4$ wines labelled $0,1,2,3$ and let the wines $0,3$ be poisoned. In binary this is $(00,11)$. Your first test yields the solutions $$00,11\quad 01,10$$ Squaring, we have the wines labelled $0,1,4,9$ with the wines $0,9$ poisoned. In binary this is $(0000,1001)$. Your second test yields the solutions $$0000,1001\quad 0001,1000$$ which is $0,9$ and $1,8$. The last pairs are not perfect squares, yielding the solution $0,9$.

Squaring uniquely identifies which solution is correct because only two pairs of integers can satisfy $x+y=k_1$ and $x^2+y^2=k_2$, as these are the equations of a line and circle and can only intersect at two points. The two pairs are always mirrored solutions $(a,b)$ and $(b,a)$. These are the same two poisoned wines, so the test works.

cheesyfluff
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  • I agree that mirrored/duplicate/flipped solutions of the kind (a,b), (b,a) lead to cancelling out zeroes on both sides, thus the absurdity.

    1+2=2+1 => 1-1=2-2 and 1^2+2^2=2^2+1^2 => 1^2-1^2=2^2-2^2 => (1+1)(1-1)=(2+2)(2-2) => 2=4

    Observe that, we strike off such cases in Part-I. May be I should have said, "Only consider solutions (a,b) where a<b"? Or does this problem completely invalidates my solution?

     – blackpen Nov 21 '15 at 14:06
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    Mirrored solutions are the only type of solutions for these two equations. Note that $x+y=k_1$ is a line and $x^2+y^2=k_2$ is a circle, which can only intersect at two real points corresponding to $(a,b)$ and $(b,a)$. I've updated the answer to further illustrate why your method does not work. – cheesyfluff Nov 21 '15 at 14:43
  • @cheesystuff, Nice observation (circle & line)! I am fighting with formatting my comment. Observe that when (0,3) are the poisons, Part-I gives test outcome as (0|1)-(0|1) that results into pairs of probable solutions (both binary/decimal included for clarity): $$(00,11)=(0,3)$$ $$(01,10)=(1,2)$$ Part-II gives the outcome as (0|1)-0-0-(0|1) that results into following pairs of probable solutions: $$(0000,1001)=(0,9)$$ $$(0001,1000)=(1,8)$$ The above ones have constant sums (=9). It is the very nature of the test that produces the constant sums. Does that look reasonable? – blackpen Nov 21 '15 at 15:40
  • Sorry I made an error. I will try to update my post with another example later. – cheesyfluff Nov 21 '15 at 19:32
  • So your explantion was wrong, but the part about the circle and line explains why your test works. – cheesyfluff Nov 21 '15 at 23:09
  • cheesyfluff, Your circle/line idea brought contrast to the explanation. Since I eliminated the duplicates in Part-I, I never considered their impact on "zero/zero cancellation" in Part-II. With the elimination of duplicates, the equations in Part-II still work. Right? – blackpen Nov 22 '15 at 05:24
  • Apart from the exception of "duplicates", does the scheme looks Ok? I posted the question, because, this problem proved particularly tricky. Everytime, I thought I had a solution, I ended up disproving it myself. – blackpen Nov 22 '15 at 05:30
  • Your explanation of part 2 does not work because the solutions that part 2 gives are not the same solution. Note my example. Part 2 can give multiple solutions, but only one pair of wines satisfies both the first part and the second part, for the reasons I stated. You can eliminate pairs of wines that don't make sense based on what you get from both tests. – cheesyfluff Nov 22 '15 at 05:31
  • I understand your objection about "the solutions in Part-I and Part-II are not the same". Part-II, in worst case, will come out with as many as 2^n pairs of solutions, all of which (except just one) will turn out to be ignorable (because they contain water in one or both the bottles). That is why I do not start with Part-II. I start with Part-I. I use Part-II, strictly for narrowing down the solutions from Part-I. – blackpen Nov 22 '15 at 05:52
  • I considered a few other optimal formulas in place of squaring, but gave them up in favour of easy comprehension. For example, $f(x)={(x(x+1))} / 2 $ is slightly more optimal. – blackpen Nov 22 '15 at 05:53
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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

I plan to write a short report describing the method I used to find $M$.

Dmitry Kamenetsky
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