# Identifying 2 poisoned wines out of 2^n wines

This problem has been asked and discussed in following posts:

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 = $2*n$ + $2*(2*n)$ = $6n$.

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.

• 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? Nov 21, 2015 at 14:06
• 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. Nov 21, 2015 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? Nov 21, 2015 at 15:40
• Sorry I made an error. I will try to update my post with another example later. Nov 21, 2015 at 19:32
• So your explantion was wrong, but the part about the circle and line explains why your test works. Nov 21, 2015 at 23:09

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 $M$.