Welcome to The Riddler. Every week, I offer up problems 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! There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement.
You can mull them 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 buttons 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 Diana Salsbury ÑÑâÐ of Baltimore and ÑÑâÐ David Kaczér ÑÑâÐ of Leuven, Belgium, our respective Express and Classic winners. You can find solutions to the previous Riddlers at the bottom of this post.
First, for Riddler Express, a very scary bridge puzzle, from Brent Edwards.
Four people have to cross an old, rickety bridge over deep, cold water infested with sharks, alligators, crocodiles and shrieking eels. The bridge is so old that it can hold no more than two people at any given time. It’s the middle of the night, so on every trip across the bridge, the person crossing needs to use the group’s only flashlight to cross safely. The four people who need to cross all walk at different speeds. One of them takes one minute to cross, one takes two minutes, one takes five minutes and one takes 10 minutes. If two people cross together, they need to stay together to share the flashlight, so they cross at the speed of the slower person. For example, the one-minute person can cross with the 10-minute person, and that trip would take 10 minutes.
How quickly can all four people get across the bridge?
If you need a hint, you can try asking me nicely. Want to submit a new Riddler Express? Email me.
Apparently there is an election or something coming up soon, so this week’s Riddler Classic is a voting problem from Andrew Spann, with a hat tip to his colleague Daniel Kane:
You are the only sane voter in a state with two candidates running for Senate. There are N other people in the state, and each of them votes completely randomly! Those voters all act independently and have a 50-50 chance of voting for either candidate. What are the odds that your vote changes the outcome of the election toward your preferred candidate?
More importantly, how do these odds scale with the number of people in the state? For example, if twice as many people lived in the state, how much would your chances of swinging the election change?
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 taught you how to gerrymander a small, imaginary region of voters. Even though Red Party voters outnumber Blue Party voters 16 to nine, you can draw five districts such that the Blue Party wins three and the Red Party only two. Democracy!
There are a few different ways to draw the districts, but there is a common principle among them: “packing.” The idea, if you’re a Blue Party member, is to pack as many Red Party voters as possible into a couple of districts. The Red Party will win those handily but won’t have enough voters in the remaining districts to win them. Here’s one example map, from Chris van B.:
Another arrangement that does the trick, from Steve Caplan:
In both cases, the Red Party wins two districts by a landslide vote of 5 to 0, but the Blue Party wins the remaining three districts, each by a narrow vote of 3 to 2.
And here’s the solution to last week’s Riddler Classic, which asked you to gerrymander a larger hypothetical area — in this case a rough approximation of the state of Colorado. Specifically, you were tasked with finding the largest number of districts, of Colorado’s seven, that either party could win if they had a sympathetic district-drawer. (In this case, Red voters outnumbered Blue voters 89 to 51.)
Given its hefty majority of voters in the state, the Red Party can win all seven of the districts. Here’s a map that accomplishes that, from Vikrant Kulkarni:
With some clever redistricting, the Blue Party can win at most five of the seven districts. (Each district has 20 voters, so it takes at least 10 voters to win one. Blue has 51 voters, so five districts is their ceiling. We assumed that tied districts were won by the party of your choice.) Here’s a map that achieves that outcome, from David Goldfarb:
As in Riddler Express, gobs of Red voters are packed into a couple of districts, which Red wins handily, but their influence is diluted elsewhere. That allows Blue to eke out some narrow victories. Maps matter.
Elsewhere in the puzzling world:
- Farewell to Numberplay’s Gary Antonick [The New York Times]
- Puzzles about Diwali, the Hindu festival of lights [Expii]
- A puzzle about a sum [The Players’ Tribune]
Have a fantastic weekend!