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

From Dean Ballard comes something quite “fun”:

If you use each of the counting numbers 1, 2 and 3 at most once, and you can add or multiply them in any order you choose, you can make all the numbers from 1 to 9:

1. 1
2. 2
3. 3
4. 1+3
5. 2+3
6. 2×3
7. 2×3+1
8. (3+1)×2
9. (2+1)×3

However, there is no way to make the number 10. Therefore, 10 is the First Unmakeable Number, or “FUN,” for the first three counting numbers.

What is the FUN for the first four counting numbers?

Extra credit: How many other “FUN numbers” can you find?

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

Riddler Classic

From Jason Zimba comes a surprisingly sandy puzzle:

In the Great Riddlerian Desert, there is a single oasis that is straight and narrow. There are N travelers who are trapped at the oasis, and one day, they agree that they will all leave. They independently pick a random location in the oasis from which to start and a random direction in which to travel. Once their supplies are packed, they all head out.

What is the probability that none of their paths will intersect, in terms of N? (For the purposes of this puzzle, assume the oasis is a line segment, while the desert is an infinite Cartesian plane.)

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

Solution to last week’s Riddler Express

Congratulations to 👏 Ken Butler  👏 of Franklin, Massachusetts, winner of last week’s Riddler Express.

Last week, you were playing a game of chess, but the enemy king was invisible! You had no idea where he was on the chessboard, while your opponent knew precisely where his king was at all times. Nevertheless, you took turns making one move at a time. If at any point you checkmated the enemy king, he revealed himself and you won the game.

However, if your king ever moved to a square that is adjacent to the enemy king, then be warned, for he would have captured your king and won the game. Meanwhile, the enemy king never moved into check or checkmate.

Your king began in the bottom left corner of the board, flanked by a rook on either side, as shown below. You also knew the enemy king was not currently in check.

It was your move. How could you guarantee a victory for yourself without losing a piece?

There were several ways to solve this using what is known as a ladder checkmate, moving the rooks one row at a time and reducing the number of allowable positions for the enemy king. But because you don’t know precisely where that enemy king is, you must keep those rooks together, guarding them with the king when necessary.

Solver Lucas Yan animated one such solution:

Here's a visualization I made for Riddler Express. pic.twitter.com/2NQEVIaiJP

— Lucas Yan (@smartfuse) May 30, 2022

While Lucas had the king leading the charge, Kiera Jones and Brad Dittmer both had the king lag behind the two rooks. Brad simulated the scenario using Stockfish, a chess-playing engine that has also previously appeared on FiveThirtyEight:

As it turned out, even an invisible king could be captured.

Solution to last week’s Riddler Classic

Congratulations to 👏Peter Ji 👏 of Madison, Wisconsin, winner of last week’s Riddler Classic.

Last week, the enemy king had gone missing again, but this time he was no chess piece. He was somewhere in his castle’s square courtyard. It was poorly lit, so you carried around a lantern that helped you spot him as long as he was within 10 meters of your current position in any direction.

The king was wily and tried to avoid detection. Both you and he moved at the exact same speed at all times.

You found yourself in the center of the courtyard. Since you could not see the king, you knew he was somewhere outside your 10-meter circle of visibility, but beyond that, you had no idea where he was. After some mental math, you realized that you have no hope of ever finding him, no matter where he started from.

What was the smallest possible side length of the square courtyard?

There were three regions for the king’s initial position that you had to consider: (1) when the king was very close to the circle of light, (2) when the king was near one of the edges or corners of the courtyard and (3) when the king was somewhere in between.

Let’s look at the second case first — when the king started near an edge or corner. Let’s also try to find the largest possible courtyard where it was possible to trap the king for at least one initial position. This occurred when the king started in one of the four corners. Of course, you had a one-in-four chance of guessing the correct corner — but if you guessed correctly, at least it was possible to trap the king.

Reader Nikhil Tilak sketched this out (assuming a radius of 5 meters rather than 10 meters).

With a little trigonometry, when the radius of the circle of light was 10 meters, the king could not escape when the side length of the square courtyard was at most 20+10√2, or about 34.14 meters.

Now, in the third case, where the king wasn’t initially near an edge or corner and not particularly close to the circle of light, he could gauge which direction you started along and always stay away from you.

But I admittedly failed to closely consider that first case when crafting this puzzle. (Hey, mistakes happen!) Several solvers, including Starvind and Izumihara Ryoma, considered what happened when the king started very close to the circle of light, and you luckily decided to move in the direction of the king. Most of the king’s motion had to be directly away from you — otherwise, he would have been caught by the circle — leaving a relatively small component of his motion in a tangential direction. After a sufficiently long pursuit, these tangential components would have compounded and the king would have eventually sneaked around the circumference undetected. As you considered initial positions that were closer and closer to the light, it took longer and longer for the king to escape around the circle.

That meant that for any finite courtyard, if the king started sufficiently close to the circle, it was possible to catch him. And so the smallest courtyard from which he was guaranteed escape was … infinitely large!

Solver Al Shaheen reimagined the puzzle to make it a little more interesting. Suppose the king did indeed start in one of the four corners and that you correctly guessed which corner he was in. How much faster than you would the king need to be to guarantee an escape?

In this case, you again might make a beeline for the corner to trap the king at one of two tangent points. From there, you’d have to guess which of the two tangent points the king was at and then approach him as he worked his way around the circumference of the circle of light.

Based on last week’s puzzle, it’s clear that I’m intrigued by these sorts of pursuit puzzles where one party (the pursuer or the pursued) is missing information about the other. I hope to “circle” back to this again with another puzzle in the not-to-distant future.

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.