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.

## Riddler Express

Earlier this year, Dakota Jones used a crystal key to gain access to a hidden temple, deep in the Riddlerian Jungle. According to an ancient text, the crystal had exactly six edges, five of which were 1 inch long. Also, the key was the *largest* such polyhedron (by volume) with these edge lengths.

However, after consulting an expert, Jones realized she had the wrong translation. Instead of definitively having *five* edges that were 1 inch long, the crystal only needed to have *four* edges that were 1 inch long. In other words, five edges *could* have been 1 inch (or all six for that matter), but the crystal definitely had *at least four* edges that were 1 inch long.

The translator confirmed that the key was indeed the *largest* such polyhedron (by volume) with these edge lengths.

Once again, Jones needs your help. *Now* what is the volume of the crystal key?

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

From Phil DeOrsey comes a matter that is anything but cut and dried:

One morning, Phil was playing with his daughter, who loves to cut paper with her safety scissors. She especially likes cutting paper into “strips,” which are rectangular pieces of paper whose shorter sides are at most 1 inch long.

Whenever Phil gives her a piece of standard printer paper (8.5 inches by 11 inches), she picks one of the four sides at random and then cuts a 1-inch wide strip parallel to that side. Next, she discards the strip and repeats the process, picking another side at random and cutting the strip. Eventually, she is left with nothing but strips.

On average, how many cuts will she make before she is left only with strips?

*Extra credit:* Instead of 8.5 by 11-inch paper, what if the paper measures *m* by *n* inches? (And for a special case of this, what if the paper is square?)

The solution to this Riddler Classic can be found in the following column.

## Last week’s Riddler

Congratulations to 👑 Bill Neagle 👑 of Springfield, Missouri, winner of last week’s Riddler and the new ruler of Riddler Nation!

Last week was the sixth 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.

Unlike previous iterations of this challenge, you were allowed to split up phalanxes into *tenths*. In other words, the number of phalanxes you assigned to any given castle could go to one decimal place. For example, you could have assigned five phalanxes to a castle, but 4.9 and 5.1 were also valid assignments. But remember — your total across all 10 castles still had to add up to 100.^{2}

I received a total of 426 battle plans,^{3} of which 413 were valid, with each number of phalanxes extending to at most one decimal place and a sum that was at most 100.

Next, I ran all 85,078 one-on-one matchups, awarding one victory to each victor. In the event of a tie, both warlords were granted half a victory. Bill was the overall winner, tallying 320 wins against just 90 losses and two ties. Here’s a rundown of the 10 strongest warlords, along with how many soldiers they deployed to each castle:^{4}

##### Who were Riddler Nation’s strongest warlords?

The top 10 finishers in FiveThirtyEight’s sixth Battle for Riddler Nation, with their distribution of phalanxes for each castle and overall record

Soldiers per castle, 1-5 | ||||||
---|---|---|---|---|---|---|

Name | 1 | 2 | 3 | 4 | 5 | |

1 | Bill Neagle | 0.1 | 0.2 | 0.3 | 12.4 | 18 |

2 | Sampsa Lahtonen | 0.3 | 0.4 | 10.2 | 0.4 | 0.6 |

3 | Joseph Poirier | 0.3 | 0.1 | 0.3 | 10.6 | 19.6 |

4 | Franco Baseggio | 0.2 | 0.2 | 7 | 11.4 | 12.1 |

5 | Jordan Thomas | 0.1 | 0.2 | 0.3 | 12.2 | 19.2 |

6 | Jesus Petry | 0 | 0 | 0.1 | 12.3 | 19.2 |

8 | Ben Knox | 0.2 | 0.3 | 0.4 | 13.2 | 17.6 |

7 | Paul Heller | 0.2 | 0.3 | 0.5 | 11.4 | 19.6 |

9 | Randi Goldman | 0.7 | 0.6 | 0.7 | 12.1 | 19.2 |

10 | Vince Vatter | 0.2 | 5.4 | 7.3 | 0.2 | 16.1 |

Soldiers per castle, 6-10 | ||||||

Name | 6 | 7 | 8 | 9 | 10 | |

1 | Bill Neagle | 2.1 | 3.2 | 3.3 | 34.4 | 26 |

2 | Sampsa Lahtonen | 23.6 | 0.6 | 3.3 | 34.3 | 26.3 |

3 | Joseph Poirier | 2.1 | 2.2 | 3.6 | 35.6 | 25.6 |

4 | Franco Baseggio | 22.5 | 3.1 | 3.8 | 5.8 | 33.9 |

5 | Jordan Thomas | 2.2 | 2.2 | 3.2 | 34.2 | 26.2 |

6 | Jesus Petry | 1.2 | 2.2 | 2.3 | 35.6 | 27.1 |

8 | Ben Knox | 2.2 | 2.3 | 2.3 | 35.3 | 26.2 |

7 | Paul Heller | 0 | 3.5 | 3.5 | 35.5 | 25.5 |

9 | Randi Goldman | 0.5 | 3 | 3.2 | 33.6 | 26.4 |

10 | Vince Vatter | 1.2 | 4 | 27.1 | 3.3 | 35.2 |

Record | ||||||

Name | Wins | Ties | Losses | |||

1 | Bill Neagle | 320 | 2 | 90 | ||

2 | Sampsa Lahtonen | 319 | 0 | 93 | ||

3 | Joseph Poirier | 316 | 1 | 95 | ||

4 | Franco Baseggio | 315 | 2 | 95 | ||

5 | Jordan Thomas | 315 | 1 | 96 | ||

6 | Jesus Petry | 313 | 3 | 96 | ||

8 | Ben Knox | 314 | 0 | 98 | ||

7 | Paul Heller | 311 | 2 | 99 | ||

9 | Randi Goldman | 311 | 1 | 100 | ||

10 | Vince Vatter | 310 | 2 | 100 |

Like most previous battles, there were 10 castles, with a total of 55 points in play (the sum of the whole numbers from 1 to 10). To win any given war, you needed at least half of the points, meaning you needed at least 28 points.

As you can see from the table above, the optimal strategy was once again to focus most of your phalanxes on castles that earned you *exactly* 28 points. Bill, third-place finisher Joseph Poirier and five others who finished in the top 10 all focused their attention on Castles 4, 5, 9 and 10. Meanwhile, second-place finisher Sampsa Lahtonen opted for Castles 3, 6, 9 and 10, and Franco Baseggio and Vince Vatter found other ways to achieve 28 points.

While the addition of a single decimal place had little effect on overall strategy, the *tactics* — that is, precisely how many phalanxes to place at each of the castles that got you 28 points — was a new front in the battle. Bill’s placement of 12.4 phalanxes at Castle 4, 18.0 at Castle 5, 34.4 at Castle 9 and 26.0 at Castle 10 proved to be the winning combination. One-tenth of a phalanx shifted to any other castle would have spelled likely doom and a different champion.

The complete data set of strategies will be posted in the coming weeks. In the meantime, the following graph summarizes all the strategies. Each column represents a different castle, while each row is a strategy, with the strongest performers on top and the weakest on the bottom. The shading of a cell indicates the number of soldiers placed. It’s a lot to take in, but at the very least you might see a few “bands” — for example, the 4-5-9-10 strategies are clustered together near the top, since they were similarly (somewhat) successful.

Congratulations to everyone who participated! Barring an uprising in Riddler Nation, the next battle will occur next year. If you have an idea for a new set of rules, be sure to send them my way.

## Want more riddles?

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.