Welcome to The Riddler. Every week over the past eight years, we have offered up problems related to the things we hold dear around here: math, logic and probability. Usually, two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer,¹ and you may get a shoutout. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.

All good things …

The time has come. This is indeed the final column for The Riddler here at FiveThirtyEight. On behalf of myself and Oliver Roeder, as well as current and former FiveThirtyEight staff (in particular, my editors: Chadwick Matlin, Santul Nerkar and Maya Sweedler), I want to thank you, dear readers, for all your riddle ideas, solutions and everything in between. Writing and solving math puzzles is good fun — but what’s even better has been taking part in this awesome puzzling community. Thank you, Riddler Nation!

That said, this isn’t quite the end.

The puzzling will continue on Substack via my brand-new newsletter, Fiddler on the Proof, affectionately known as “The Fiddler.” The first puzzle (as well as the solution to this week’s riddle) will drop on July 7. Sign up (for free!) to keep the puzzle goodness going!

The Final Riddler

As a graduate student in the Riddler Research Lab, I have been tasked with aligning a laser beam inside a perfectly circular mirror. The beam starts at one point on the circumference of the circle and bounces around inside the circle many times before returning to that very same point.

Upon further analysis, I find that the beam creates a 184-gon (i.e., a polygon with 184 sides) inscribed in the circle — but not a regular 184-gon, mind you. The head of the lab has asked me to further measure the angles the beam makes as it bounces around, which honestly seems like a lot of busywork. To lighten the load, I grab a labmate and we decide to split the work in half. We each will measure every other angle. That is, if the points around the polygon are labeled ABCDEFGH …, I will measure angles ABC, CDE, EFG, etc., while my labmate will measure BCD, DEF, FGH, etc.

In total, I’m responsible for measuring 92 angles. I find that the first 89 angles each measure precisely 178 degrees. What is the sum of the final three angles I’m supposed to measure?

Submit your answer

Important note: As this is the final column for The Riddler, the solution to this puzzle will appear next Friday morning at thefiddler.substack.com.

Last Week’s Riddler

Congratulations to Ñб─Ñб÷âб─б≤âб─б≤ Tom Hanrahan Ñб─Ñб÷âб─б≤âб─б≤ of Lexington Park, Maryland, winner of last week’s Riddler and the final ruler of Riddler Nation!

Last week was the eighth and final Battle for Riddler Nation, and things were a little different this time around.

In a distant, war-torn land, there were 10 castles. There were two warlords: you and your archenemy. Each castle had its own strategic value for a would-be conqueror. Specifically, the castles were worth 1, 2, 3, … , 9 and 10 victory points. You and your enemy each had 100 soldiers to distribute, any way you liked, to fight at any of the 10 castles. Whoever sent more soldiers to a given castle conquered that castle and won its points. If you each sent the same number of troops, you split the points. You don’t know what distribution of forces your enemy has chosen until the battles begin. Whoever won the most points won the war.

As in previous years, I adjudicated all the possible one-on-one matchups. A victory was worth one “victory point,” while a tie was worth 0.5 victory points. Instead of declaring the winner after this round robin, I eliminated the team with the fewest victory points and repeated the entire process with one fewer competitor.²

This year, I received 394 strategies. After eliminating duplicate submissions and unfair strategies (e.g., placing more than 100 total soldiers or effectively placing more than 100 by allocating negative numbers to some castles), 364 strategies remained.

As always, I generated a heat map (with darker orange representing more soldiers), organized by how well each approach fared in the initial round robin, shown on the left below.

With the initial round robin out of the way, the elimination rounds began. One round at a time, I eliminated the weakest-performing strategy in a round robin of the remaining teams, until only one warlord ruled supreme. A similar heat map showing these final rankings appears above on the right.

A few trends are apparent at the tops of these heat maps. In the initial round robin, the strongest warlords clustered their soldiers in castles worth 8, 6, 5, 4, 3 and 2 points. This was one of many ways to earn a total of 28 points, which was sufficient to guarantee a victory (there were 55 total points at stake, so 28 was more than half). In prior years, similar strategies have prevailed, such as clustering in castles worth 10, 9, 5 and 4 points, or in castles worth 10, 9, 6 and 3 points.

That said, these heat maps can be difficult to interpret. Friend-of-The-Riddler Vince Vatter (also a former ruler around here) did an impressive job reconstructing the strategies from the seventh Battle for Riddler Nation by analyzing a similar heat map from last year. (Cheers to Vince for using all the resources at his disposal!)

So let’s take a deeper dive into the strategies that stood atop the final rankings:

This time around, a few “conventional” strategies (which won prior battles) did fairly well. Ben Knox made a play for castles 4, 5, 9 and 10, while Michael DeHaye and Martin Stearne both went for castles 3, 6, 9 and 10. The three of them did well in the initial round robin and ultimately came in seventh, eighth and ninth. Meanwhile, Matthieu and JD (who came in second and third, respectively) still targeted castles worth 28 points in total, but in combinations that didn’t fare as well in prior battles.

But, in the end, many of the best strategies didn’t do quite so well at the beginning. To win in this new format, you had to make it through every elimination round, even if it wasn’t always pretty. More specifically, you wanted a strategy that merely survived against the weaker strategies early on, but that was strong against the better strategies later on.

And so, at the very top, Tom Hanrahan only made a strong play for castles 6 and 9 (worth a combined 15 points), while leaving a smattering of soldiers at the remaining eight castles. Meanwhile, Tabitha Togersen, who came in fourth, only made a strong play for castle 10. And David Kuplic, who came in fifth, spread soldiers all over the place, with less than 20 at every castle.

While this may have been the Final Battle for Riddler Nation, it was refreshing to see a new set of strategies rise to the top. Indeed, in this format, none of the final top six warlords had placed in the top 40 after the initial round robin. Here’s a graph showing how each warlord’s initial and final rankings compared:

In the top right, a really bad strategy (e.g., putting 100 soldiers at castle 1) was bad no matter the format of the battle. But beyond those low performers, things got noisy. Put another way, the correlation between initial and final rank was surprisingly low (to me, at least).

Looking at the bottom right of this graph, I have to award special kudos to “John The Warload Winner” (I presume there’s a typo in there), who was the most improved between the initial and final rankings. In the initial round robin, John ranked 306th. But John was a scrappy warlord, avoiding hundreds of eliminations to ultimately place 16th in the final standings.

Congratulations to the winners, and to everyone who participated in this memorable, final edition of the Battle for Riddler Nation!

Want more puzzles?

Well, aren’t you lucky? While this may be the final column of The Riddler here at FiveThirtyEight, the puzzling continues over at Fiddler on the Proof!

Want to submit a riddle?

Email Zach Wissner-Gross at thefiddler@substack.com.