Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.
Riddler Express
From Dan Waterbury, a pizza problem:
You and your two older siblings are sharing two extra-large pizzas and decide to cut them in an unusual way. You overlap the pizzas so that the crust of one touches the center of the other (and vice versa since they are the same size). You then slice both pizzas around the area of overlap. Two of you will each get one of the crescent-shaped pieces, and the third will get both of the football-shaped cutouts.
Which should you choose to get more pizza: one crescent or two footballs?
Riddler Classic
From Dave Moran, a pricy postseason puzzler:
Congratulations! The Acme Axegrinders, which you own, are the regular season champions of the National Squishyball League (NSL). Your team will now play a championship series against the Boondocks Barbarians, which had the second-best regular season record. You feel good about Acme’s chances in the series because Acme won exactly 60 percent of the hundreds of games it played against Boondocks this season. (The NSL has an incredibly long regular season.) The NSL has two special rules for the playoffs:
- The owner of the top-seeded team (i.e., you) gets to select the length of the championship series in advance of the first game, so you could decide to play a single game, a best two out of three series, a three out of five series, etc., all the way up to a 50 out of 99 series.
- The owner of the winning team gets $1 million minus $10,000 for each of the victories required to win the series, regardless of how many games the series lasts in total. Thus, if the top-seeded team’s owner selects a single-game championship, the winning owner will collect $990,000. If he or she selects a 4 out of 7 series, the winning team’s owner will collect $960,000. The owner of the losing team gets nothing.
Since Acme has a 60 percent chance of winning any individual game against Boondocks, Rule 1 encourages you to opt for a very long series to improve Acme’s chances of winning the series. But Rule 2 means that a long series will mean less winnings for you if Acme does take the series.
How long a series should you select in order to maximize your expected winnings? And how much money do you expect to win?
Solution to last week’s Riddler Express
Congratulations to 👏 Ethan Cravener 👏 of Cleveland, Ohio, winner of the previous Express puzzle!
Two long-distance swimmers are standing on a beach, right on the water’s edge. They begin 100 yards away from one another on the shore, which is a straight line of sand. Both swimmers swim at exactly the same speed. Swimmer A heads straight out to sea, directly perpendicular to the shore. At the same time, Swimmer B also heads out, swimming exactly in the direction of Swimmer A at all times. Over time, Swimmer B will approach a position directly in Swimmer A’s wake, where he will follow her at a fixed distance. What is that distance?
It is 50 yards.
Why? To start us off, Chris Ketelsen illustrated what the swimmers’ paths through the water look like:
Swimmer B’s path toward A is not, in fact, a quarter circle, as many solvers supposed. (If it were, the distance would be about 57 yards, which was a commonly submitted answer.) Rather, it is a type of pursuit curve. These cropped up in a Riddler column last summer about an angry, fenced-in ram.
The puzzle’s submitter, Scott Cardell, offered a nifty geometric way of approaching the problem:
Let the y-axis be the shore, and the x-axis be Swimmer A’s path. Have Swimmer B start at (0,100). Let \(d\) be the distance between the swimmers and let \(f\) be the difference in the swimmers’ x-axis values. Draw the triangle connecting these three points: Swimmer A’s position, Swimmer B’s position and the x-axis point closest to Swimmer B.
The numbers \(d\) and \(f\) are the lengths of two of the sides of this triangle. Finally, let \(\theta\) be the angle between those two sides. Swimmer A’s motion is increasing \(f\) at the rate \(v\) (his velocity) while Swimmer B’s motion is reducing \(f\) at the rate \(v\cdot \cos(\theta)\). (The cosine is the ratio of the length of a triangle’s adjacent side to its hypotenuse.) Swimmer B is decreasing \(d\) at the rate \(v\) while Swimmer A is increasing \(d\) at the rate \(v\cdot \cos(\theta)\). Thus \(f+d\) is changing at the rate \(v-v\cdot \cos(\theta)-v+v\cdot \cos(\theta)=0\). Therefore, \(f+d\) is constant. Initially, \(d=100\), \(f=0\) and \(f+d=100\), so \(f+d\) is fixed at 100. When swimmer B is following directly behind Swimmer A, \(f=d\) which means \(f=d=50\) yards.
Tyler Barron animated the paths for hypothetical swimmers that start at different distances down the shore:
In each case, Swimmer B winds up behind Swimmer A at half the distance that separated the two down the beach.
Solution to last week’s Riddler Classic
Congratulations to 👏 Chris Thornett 👏 of Brooklyn, New York, winner of the previous Classic puzzle!
You run a film magazine 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. 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?
Depending on what your critics are capable of during the festival, you need to send either nine or 21 reviewers.
If your critics are not able to coordinate, and you’ve got to plan out the assignments for the entire festival in advance, then you’ll need 21 reviewers. Vivek M. charted one method to assign screens to the 21 critics, with each colorful grid depicting one reviewer’s assignment and the letters on top of them their assigned screens:
Hector Pefo provided a thorough explanation of the 21-reviewer approach. There doesn’t seem to be a silver-bullet mathematical approach here, and most solvers turned to brute force or trial and error. As Josh Simpson explained in his submission: I used one person to watch every movie on each screen. From this I can identify the best movie on each individual screen. After that you need to make sure each of those three are compared to the other two. I did this with a simple pattern where three critics go “clockwise,” viewing screen A then B then C, and starting on different screens. Three also go “counterclockwise.” This covers the first three showings completely, but fails for the fourth showing. At that point, I added some more critics following the same clockwise or counterclockwise pattern, but watching each given theater three times in a row. This gets us through nine showings per theater. For the tenth and final showing I had to extend the pattern, adding six more critics who watch a screen nine times before moving. That’s a total of 21 reviewers.
If your critics are able to coordinate between the films that they review, and share their notes on rankings to update their plans, then you’ll only need to send nine reviewers. As Victor Bible explained in his submission, each pair of films that are not simultaneously shown must be seen by at least one reviewer — the top two films could not be separated otherwise. There are \((30/2) \cdot 27 = 405\) pairs. Each critic can cover \({{10} \choose {2}} = 45\) pairs, using the choose function. So sending 405/45 = 9 critics will work, and any fewer won’t cover enough pairs.
Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.