A puzzle for you!

A group of six men (A, B, C, D, E, and F) and six women (a, b, c, d, e, and f) are chosen so that everybody has a perfect match of the opposite gender. They are given the following clues:

- E and e are a match.
- A and c are a match.
- The pairing (A,b), (B,a), (C,c), (D,d), (E,e), (F,f) contains exactly 2 matches.
- The pairing (A,c), (B,d), (C,a), (D,b), (E,e), (F,f) contains exactly 2 matches.
- The pairing (A,c), (B,f), (C,b), (D,d), (E,e), (F,a) contains exactly 3 matches.

If they can figure out everybody’s perfect matches, the group will get a million dollars to share! Can they figure it out?

If you figure that one out, then here’s the next level, with ten pairs and twelve clues!

A group of ten men (ABCDEFGHIJ) and ten women (abcdefghij) are chosen so that everybody has a perfect match of the opposite gender. They are given the following twelve clues:

- F and f are not a match
- J and h are not a match
- B and e are not a match
- D and d are a match
- H and c are a match
- The pairing (Ai Bb Ca Dd Ee Fc Gj Hg If Jh) contains exactly 2 matches.
- The pairing (Af Be Cg Dj Eh Fd Gi Hc Ia Jb) contains exactly 4 matches.
- The pairing (Af Be Ca Dd Ej Fh Gi Hb Ig Jc) contains exactly 2 matches.
- The pairing (Aa Bc Ci Dd Eg Fj Gf Hb Ih Je) contains exactly 2 matches.
- The pairing (Af Bi Ce Dd Eh Fg Gj Hc Ia Jb) contains exactly 5 matches.
- The pairing (Af Ba Cb Dd Eh Fi Gj Hc Ie Jg) contains exactly 5 matches.
- The pairing (Af Bi Ch Dd Eb Fg Gj Hc Ia Je) contains exactly 7 matches.

Can you help them get their million dollars?

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Some background on the puzzle:

I didn’t actually come up with it out of nowhere, I encountered in the wild! A few days ago I started watching a new Netflix reality show called “Are You The One?” The premise of the show: ten men and ten women are paired up via a matchmaking algorithm, and if they can all figure out their other halves after living together for a month, then they win one million dollars. It’s an extremely corny show, but there was one aspect of it which I found pretty interesting: it’s a perfect setting for Bayesian experimental design! Me being me, I spent the whole show thinking about the math of how contestants get information about their potential “perfect matches”.

Let’s start with a very basic look at the underlying math. Ten pairs, chosen from two groups of ten, gives ten factorial possible matchings. 10! is 3,628,800, which is a *lot* of possibilities. If you were to randomly choose the ten matches, with no background information, you’d have a .000028% chance of getting everything right. This is about as likely as you are to correctly guess the outcome of 22 tosses of an unbiased coin in a row!

Of course, the show isn’t so cruel as to force them to just guess with no information. For one thing, they have a whole month to get to know each other and gather lots of that subtle tricky-to-quantify social evidence. But more importantly, every episode they get two crucial pieces of information:

First, they get to choose one pair (a man and a woman) to go into the “Truth Booth”. The group is then informed whether these two are a genuine match or not.

And second, at the end of each episode the entire group gathers together, everybody pairs off with one another, and they are told how many pairs they got right (though not which ones). On the final episode, the matching they choose determines whether they get the million dollars or not.

I call the first type of test the individual pair test and the second the group test. And this is where the two types of clues in the above puzzles come from! The clues for the second puzzle are actually just a translation of the tests that the group decided to do in the first seven episodes. (So if you successfully solved it, then feel good about yourself for being the type of person that would have won the million.) Interestingly, it turns out that by the seventh episode they could had already figured out everybody’s perfect match, but it took them three more episodes to get to that point! Silly non-logically-omniscient humans.

Putting aside the puzzle we started with, the show setup naturally lends itself to some interesting questions. First of all, **is there a strategy that guarantees that the group will get the million dollars by the tenth episode?** In other words, is ten pair tests and nine group tests sufficient to go from 3,628,800 possibilities to 1?

I am not yet sure of the answer. The fact that the group in season one managed to narrow it down to a single possible world with only seven episodes seems like evidence that yes, those 19 tests do provide enough evidence. In addition, there have been eight seasons and in only one did the group fail to find the correct matches. (And in the season whose group failed, they started with eleven pairs – multiplying the number of possible worlds by 11 – and used two fewer pair tests than previous seasons.)

However, it’s worth keeping in mind that while the group of twenty individuals was far from logically omniscient, they did have a great wealth of additional social evidence, and that evidence may have allowed them to make choices for their tests that yielded much more expected information than the optimal test in the absence of social information. (I’m also suspicious that the producers might be a little more involved in the process than they seem. There are a few seasons where the group is consistently getting 2 or 3 matches until the last episode where they suddenly get everything right. This happens in season 3, and by my calculations there were still four possible worlds consistent with all their information by the time of the last matching ceremony!)

We can also ask the Bayesian experimental design question. **What’s the optimal pair test/group test to perform, given the set of remaining possible worlds?**

We can solve this fairly easily using a brute force approach. Consider the problem of finding the optimal pair test. First we actually generate the set of all possible worlds (where each world is a specification of all ten matches). Let N be the size of this set. Then we look through all possible pairs of individuals (i, j), and for each pair calculate the number of worlds in which this pair is a match. Call this quantity n_{ij}. Then the expected number of worlds remaining after performing a pair test on (i, j) is:

n_{ij} Pr(i and j are a match) + (N – n_{ij}) Pr(i and j are not a match) = n_{ij}^{2}/N+ (N – n_{ij})^{2}/N

So we simply search for the pair (i, j) that minimizes n_{ij}^{2} + (N – n_{ij})^{2}. This is equivalent to maximizing n_{ij} (N – n_{ij}): the product of the number of worlds where (i, j) is a match and the number of worlds where (i, j) is not a match.

We do pretty much the same thing for the group test. But here’s where we run into a bit of trouble: though our algorithm is guaranteed to return an optimal test, the brute force approach has to search through 10!^{10!} possibilities, and this is just not feasible. The time taken to solve the problem grows exponentially in the number of worlds, which grows exponentially in the number of individuals.

So we have another interesting question on our hands: **Is there an efficient algorithm for calculating the optimal pair/group test?** In the more general setting, where the tests are not restricted to just being pair tests or group tests, but can be a test of ANY boolean expression, this is the question of whether SAT can be solved efficiently. And given that SAT is known to be NP-complete (it was in fact the first problem to be proven NP-complete!), this more general question ends up being equivalent to whether P = NP!