You run a film magazine, Groovy Movies, and you have been invited to attend a new film festival. The festival organizers will screen 30 films evenly distributed across three different screens. Each film will premiere at this festival, and you want to get the scoop on which one was the best. The problem is, though, that because there are three screens, you don’t know which screen will show the best film. You could watch only Screen A, see the best movie there and report on it, but it may not be as good as one of the movies on Screen B or C.
Some more details you know from your many years of experience in the magazine biz:
- Whenever a film is playing on one screen, the other two screens also have a film playing. But there is enough time between each movie that one person can always watch the nth round on one screen and the n+1th round on another screen.
- The best movie on one screen will never play at the same time as the best movie on another screen. However, you don’t know what time slots they will occupy for each theater.
- All of your reviewers are good rankers — they won’t have any disagreement about which movies are better than others that they’ve seen. (They’re ordinal reviewers, in other words.)
- That said, all of your reviewers are terrible raters, so they cannot give an objective measure of how good a movie was (a 9 out of 10, say) and compare it to another reviewer’s measure of how good a different movie was. (To put it another way: They aren’t cardinal reviewers.)
With all that in mind, if you want to know for sure what the best film at the festival was, what is the minimum number of reviewers you would need to send to the festival?
- Extra credit: What if there were more movies shown per screen? What if there were more screens?
There is an ambiguity in the statement of the problem. The static interpretation requires us to assign critics to watch fixed sequences of movies in advance. The dynamic interpretation allows for the critics to base their plans on all the critics’ evaluative judgments before every viewing. While I doubt it makes a difference, I am not sure of that, and I will operate on the assumption that the static interpretation is intended.
First, let’s establish that for every pair of non-simultaneous movies – all $30 \times 27 / 2$, or 405, of them – at least one of the critics must watch both. Suppose some pair of movies is not “co-watched” in that sense. Then if it contains the best and second-best movies, all we will know at the end of the festival is that every critic who has seen either of them rated it higher than every other movie they’ve seen. That provides no way to rank them with respect to each other. So to guarantee that we’ll find the best movie, every pair of movies must be co-watched.
Each movie must be watched by a unique crowd (set of critics), because if two movies are watched by the same crowd, then the earlier of the movies is not co-watched with the two movies shown at the same time as the later.
There have to be at least three critics in each initial crowd to be able to fan out into all the three second-showing movies. Nine critics can form $3\times3\times3$, or 27 unique crowds of three overlapping each of the initial crowds, which is exactly the number of unique crowds needed for the final 9 showings. And nine critics end up watching $9 \times 10 \times 9 / 2$, or 405 pairs of non-simultaneous movies, which is exactly what we need. However, given the need for any two non-simultaneous crowds to overlap, and given the fact that each critic can see only one movie at a time, it turns out that nine critics can cover at most four showings of three movies each.
In general terms, the problem (which is really a problem in set theory and combinatorics) is to determine, for $c$ critics and $s$ screens, the maximum number of distinct single-showing-assignments (partitions) of the $c$ critics into $s$ crowds (subsets) such that any two crowds in any two (distinct) assignments are distinct but overlap. That sounds like a hard problem to me, and I look forward to seeing if the puzzle author (or anyone else) has a nice way into it.
What I do have is a relatively simple and effective strategy that uses a lavish staff of 21 critics.
Let’s represent a critic’s overall assignment of theaters as a ten-character string, such as “ACCBBCAABA”. Then, nine critics can satisfy the desideratum—necessary, but not sufficient to solve the problem—of having movies in consecutive showings co-watched, as follows:
1: AAAAAAAAAA 2: ABABABABAB 3: ACACACACAC 4: BABABABABA 5: BBBBBBBBBB 6: BCBCBCBCBC 7: CACACACACA 8: CBCBCBCBCB 9: CCCCCCCCCC
These assignments actually ensure that every pair of movies containing one from an odd- and one from an an even-numbered showing are co-watched. But many pairs of movies from odd-odd and even-even showings are not co-watched.
We can start to repair that with nine more critics, using the same assignment chart as above but reading it differently, namely by considering the showings to be represented in the order 1, 3, 5, 7, 9, 2, 4, 6, 8, 10. That will cover pairs of movies that are in showings whose numbers are odd and even in this ordering. Reordering, this second set of assignments is:
1: AAAAAAAAAA 2: ABBAABBAAB 3: ACCAACCAAC 4: BAABBAABBA 5: BBBBBBBBBB 6: BCCBBCCBBC 7: CAACCAACCA 8: CBBCCBBCCB 9: CCCCCCCCCC
To finish up, do the same thing yet again, but scramble the ordering as follows: 1,5,9,3,7,2,6,10,4,8. This yields:
1: AAAAAAAAAA 2: ABBABAABAB 3: ACCACAACAC 4: BAABABBABA 5: BBBBBBBBBB 6: BCCBCBBCBC 7: CAACACCACA 8: CBBCBCCBCB 9: CCCCCCCCCC
Now every pair of showings has been odd-even (as concerns our first-chart strategy), and so all pairs of movies are co-watched by the 27 critics we’ve employed. The six constant-theater assignments in the second and third charts are redundant (we already had those assignments covered among our initial nine critics). So in the words of our President, they are fired, and in total we get by with 21 critics.
The first chart has an obvious pattern, which can be repeated for any number of showings without a need for additional critics. And a similar chart can be made for any number $s$ of screens, with $s^2$ lines, and again needing to re-use the chart twice, for a total (after eliminating redundancies) of $3s^2-2s$ critics for any number of showings.
I do doubt that this scheme is optimal, as among 21 critics, $21 \times 10 \times 9 /2 $, or 945, movie pairs are covered—more than half redundantly. But it’s Sunday night, and it’s what I got.