Can You Cover The Baking Sheet With Cookies?
Illustration by Guillaume Kurkdjian
Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. 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 for either,1 and you may get a shoutout in the next column. 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.
Riddler Express
From Mike Strong comes a brief return to last week’s battle:
In last week’s Battle for Riddler Nation, you had to assign 100 phalanxes of soldiers to 10 castles, each worth a distinct number of points. For example, you could assign all 100 phalanxes to a single castle (and none to the others), split them evenly so that there were 10 phalanxes at every castle or arrange them in some other way.
What was the total possible number of strategies you could have submitted?
Riddler Classic
On social media, I recently saw an image of cookies, which had presumably been circular when they were placed in the oven, emerging as expanded hexagons.
This made me wonder more about how circles could overlap to fill up a rectangular tray.
Suppose you have a unit square (i.e., with side length 1). If you also have four identical circles that can overlap, they would need to have a radius of 0.25·√2 to completely cover the square, as shown below:
Now suppose that, instead of four identical circles, you have five identical circles that can overlap. What is the minimum radius they would need to completely cover a unit square?
Extra credit: Suppose you have six identical circles that can overlap. What is the minimum radius they would need to completely cover a unit square?
Last week’s Riddler
Congratulations to ÑбâÑб÷âбâбâ¤âбâб⤠David Hacker ÑбâÑб÷âбâбâ¤âбâб⤠of Chevy Chase, Maryland, winner of last week’s Riddler and the new ruler of Riddler Nation!
Last week was the seventh Battle for Riddler Nation, and once again things were a little different this time around.
In a distant, war-torn land, there were 10 castles and 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 phalanxes of soldiers to distribute, any way you liked, to fight at any of the 10 castles. Whoever sent more phalanxes to a given castle conquered that castle and won its victory points. If you each sent the same number of phalanxes, you split the points. You didn’t know what distribution of forces your enemy had chosen until the battles began. Whoever won the most points won the war.
As in previous years, I adjudicated all the possible one-on-one matchups. But instead of declaring the winner at this point, I used the results of all these matchups to seed the teams in a single-elimination tournament. In the end, the optimal type of strategy no longer only preyed on weaker strategies but also defeated similarly strong strategies.
This year, I received a whopping 3,238 strategies before the midnight deadline. However, just one person submitted 2,770 of those strategies, in an apparent attempt to “flood the zone.” While their dedication was admirable, it would have been unfair to the remaining strategists to consider every entry. I therefore took only the last strategy submitted by each warlord, for a total of 432 strategies.
Next, I had them all compete against each other in a grand total of 93,096 one-on-one matchups. As I have done in previous years, I generated a heat map (with darker orange representing more phalanxes) with all 432 strategies, organized by how well each approach fared in this preliminary stage:
In particular, the top 10 one-on-one finishers (counting a tie as half of a win) are shown below, along with two other strategies that did well in the tournament — but more on those later.
Who were Riddler Nation’s strongest warlords?
The top strategies in FiveThirtyEight’s Seventh Battle for Riddler Nation, with their distribution of phalanxes for each castle and overall record
| Seed | Name | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | W | T | L |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | Temple Kirk | 1 | 3 | 8 | 13 | 11 | 24 | 5 | 28 | 3 | 4 | 304 | 15 | 112 |
| 2 | Betts Slingluff | 0 | 0 | 7 | 11 | 12 | 23 | 3 | 4 | 6 | 34 | 306 | 10 | 115 |
| 3 | Jim Crimmins | 0 | 0 | 8 | 11 | 13 | 22 | 3 | 4 | 5 | 34 | 306 | 10 | 115 |
| 4 | Alex Conant | 0 | 1 | 8 | 11 | 16 | 21 | 2 | 4 | 4 | 33 | 306 | 9 | 116 |
| 5 | Brian Petrone | 3 | 5 | 6 | 7 | 12 | 1 | 27 | 31 | 4 | 4 | 308 | 1 | 122 |
| 6 | Chuck McKnight | 0 | 6 | 7 | 11 | 12 | 22 | 3 | 26 | 6 | 7 | 306 | 5 | 120 |
| 8 | Chris Payne | 0 | 1 | 11 | 0 | 1 | 1 | 23 | 27 | 1 | 35 | 305 | 4 | 122 |
| 7 | Mark Conkle | 1 | 1 | 2 | 14 | 3 | 3 | 22 | 24 | 26 | 4 | 303 | 6 | 122 |
| 9 | Andrew Jiang | 2 | 1 | 8 | 1 | 1 | 1 | 25 | 27 | 1 | 33 | 300 | 5 | 126 |
| 10 | Jenny Tai | 0 | 0 | 8 | 12 | 14 | 22 | 3 | 3 | 6 | 32 | 297 | 10 | 124 |
| 112 | Graydon Snider | 1 | 1 | 9 | 15 | 12 | 16 | 8 | 7 | 8 | 23 | 255 | 7 | 169 |
| 266 | David Hacker | 5 | 0 | 14 | 19 | 5 | 6 | 8 | 8 | 18 | 17 | 203 | 3 | 225 |
In previous Battles for Riddler Nation, the strongest strategies clearly attempted to win at least half — i.e., 28 — of the total 55 points up for grabs. This time around, fewer of the top one-on-one finishers deployed the “go for 28” strategy. For example, Temple Kirk’s top one-on-one strategy sent at least eight phalanxes to castles 3, 4, 5, 6 and 8, for a total of 26 points. But with clever allocations to a few of the other castles, Temple came out on top — at least in the one-on-one matchups.
Nevertheless, the battle was far from over. All 432 strategies then squared off in a single-elimination tournament, a new feature of this year’s battle. Of the 432 strategies, 352 played in a total of 176 “play-in” games, while the remaining 80 received a bye. That resulted in 256 — a power of 2 — for the remainder of the tournament.
There were quite a few surprises, as many of the top seeds that had vanquished the weaker strategies one-on-one failed to advance deep into the tournament. Here was how the final 16 teams fared in the tournament:
David Hacker, the new ruler of Riddler Nation, began as the 266th seed and had a losing record in the one-on-one matches. David won the play-in game as an underdog and continued to win as the underdog in every single match the rest of the way — a truly incredible feat. Along the way, David defeated the Nos. 247, 10, 138, 55, 87, 103, 47, 83 and 112 seeds — a string of nine consecutive victories.
How did David beat the odds? By assigning phalanxes to defeat the “4-5-9-10” and “3-6-9-10” strategies of battles past. This required only 75 phalanxes that David described as a “sharpened weapon against top strategies,” but this distribution still lost to the average distributions from some of the previous years’ battles. The good news for him was that there were still 25 phalanxes remaining. After placing those in various castles, the end result was more than ready to punch above its weight in the tournament.
It was immensely fun to see this tournament play out — thanks again to all who entered, and congratulations to David! If you have ideas for a twist to put on next year’s Battle for Riddler Nation, let me know.
Want more puzzles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at riddlercolumn@gmail.com.
