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 ÑÑâÐ Greg Burnham ÑÑâÐ of Brooklyn and ÑÑâÐ Daniel Ferguson ÑÑâÐ of Navarre, Florida, our respective Express and Classic winners. You can find solutions to the previous Riddlers at the bottom of this post.
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?
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.)
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, classist, classists
la, lap, laps, lapse, elapse, relapse, relapser, relapsers
pi, pin, ping, aping, raping, craping, scraping, scrapings
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:
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.)
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:
- Calculating the cosmos [The New York Times]
- Puzzles about circles [The Wall Street Journal]
- “Two tantalising teasers from the prince of puzzles” [The Guardian]
- Three puzzles involving time [The Players’ Tribune]
- Halloween puzzles! [Expii]
- Can you solve the puzzle Dyson uses to hire its staff? [The Telegraph]
Have an excellent weekend!