Rig The Election … With Math!

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from many top-notch puzzle folks around the world — including you!

Recently, we started something new: Riddler Express problems. These are bite-size puzzles that don’t take as much fancy math or computational power to solve. For those of you in the slow-puzzle movement, worry not — we still feature our classic, more challenging Riddler.

You can mull both over on your commute, dissect them on your lunch break and argue about them with your friends and lovers. When you’re ready, submit your answer(s) using the links below. I’ll reveal the solutions next week, and a correct submission (chosen at random) will earn a shoutout in this column.1

Before we get to the new puzzles, let’s reveal the winners of last week’s. Congratulations to 👏 Greg Burnham 👏 of Brooklyn and 👏 Daniel Ferguson 👏 of Navarre, Florida, our respective Express and Classic winners. You can find solutions to the previous Riddlers at the bottom of this post.

Now, two gerrymandering redistricting puzzles from Eli Ross of Brilliant.org — an online community for students and professionals looking for tantalizing math, physics and computer science challenges — just in time for the election.

First, in Riddler Express, learn how to gerrymander:

Imagine your job is to draw districts and you happen to be a member of the Blue Party. The grid below gives the locations of 25 voters in a region, which you must divide into five districts with five voters each. In each district, the party with the most votes will win. The districts must be non-overlapping and contiguous (that is, each square in a district must share an edge with at least one other square in the district). Can you draw the districts such that the Blue Party wins more districts than the Red Party?

If you need a hint, you can try asking me nicely. Want to submit a new Riddler Express? Email me.

Now, in Riddler Classic, gerrymander a whole state!

In the real world, of course, there aren’t just 25 voters. Even if you can group neighborhoods together, the grid of voters in an entire state is going to be much larger, meaning that a computer program will probably be necessary to optimally gerrymander. Below is a rough approximation of Colorado’s voter preferences, based on county-level results from the 2012 presidential election, in a 14-by-10 grid. Colorado has seven districts, so each would have 20 voters in this model. What is the most districts that the Red Party could win if you get to draw the districts with the same rules as above? What about the Blue Party? (Assume ties within a district are considered wins for the party of your choice.)

If you need a hint, you can try asking me nicely. Want to submit a new Riddler Classic? Email me.

Here’s the solution to last week’s Riddler Express, which asked you to build the longest Scrabble word, one letter at a time, where each step along the way is also a valid word. Due to the peculiarities of the Scrabble dictionary and the rules of this challenge, the longest possible word is nine letters — and there are a few equally correct ways to build one that long.

as, ass, lass, lassi, lassis, classis, classism, classisms

as, ass, lass, lassi, lassis, classis, classist, classists

as, ass, lass, lassi, lassie, glassie, glassies, glassiest

la, lap, laps, lapse, elapse, relapse, relapser, relapsers

pi, pin, ping, aping, raping, craping, scraping, scrapings

at, eat, eath, heath, sheath, sheathe, sheather, sheathers

ai, rai, rais, raise, praise, upraise, upraiser, upraisers

li, lin, blin, bling, blings, ablings, tablings, stablings

Within those paths are variations, too. To build “upraisers,” for example, you could also start with “is, ais, rais …”

Much to my delight, your approaches to this problem varied widely. Bob Maucher used a spreadsheet where he entered all the words, organized by word length, and wrote a formula to see which longer words built off shorter ones. Paul M., on the other hand, found “relapsers” using “just a guess.” He’d seen a similar string built up in a game of Words With Friends. Others turned to code in Python and R.

Your possible answers also depend on which dictionary you use. If you use the North American Scrabble dictionary, you can build the strings above. If you use the public-domain ENABLE list, you miss out on a few (“lassi,” for example, an Indian beverage, doesn’t appear in ENABLE). And if you use the Collins Scrabble dictionary, used for play outside North America, you can add a few more nine-letter words, including

in, pin, ping, oping, ooping, hooping, whooping, whoopings

“Ooping” and “whoopings” — alas — aren’t valid in North American Scrabble. And Evans Clinchy, who finished seventh in the 2015 Scrabble world championship, pointed out that you can actually build a 10-letter word using the newest version of the international dictionary:

hi, hin, hing, ching, eching, eeching, reeching, breeching, breechings!

And here’s the solution to last week’s Riddler Classic, concerning the most point-rich Boggle board possible.

Zach Wissner-Gross was able to find an 1,818-point board, its scoring neatly animated here:

Michael Weiss found a 4,100-point board with a heuristic “placing letters that had more appearances in 2 and 3 letter substrings of words (e.g. ing, er) on a board more frequently.”

```SEAS
RTND
AIEE
MPSR```

(He scored it using this handy Boggle solver. There are some minor scoring discrepancies that, again, are a result of the specific word list being used. Zach’s creation, for example, scores 2,380 using the Boggle solver.)

Eshan Mitra found a 4,391-point board using a Metropolis algorithm, which he ran for eight hours and 3.5 million simulation iterations.

```SEGS
RNTR
EIAE
TSLP```

It turns out this is a very tough computational problem. The Riddler with the highest score — Daniel Ferguson — matched the highest score I was able to find online, using the board generated (in a non-Riddler context) by Chuong Do. According to the Boggle solver, they both contain 4,600 points.

```SERS
PATG
LINE
SERS```

And this is Daniel’s:

```SLPS
EIAE
RNTR
SEGS```

My name, including an honorific and middle initials, scores just 277 points.

```DROL
IVER
KTRO
EDER```

Elsewhere in the puzzling world:

Have an excellent weekend!

## Footnotes

1. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday. Have a great weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.

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