Can You Capture The Invisible Kings?
Illustration by Guillaume Kurkdjian
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
You are playing a game of chess, but the enemy king is invisible! You have no idea where he is on the chessboard, while your opponent knows precisely where his king is at all times. Nevertheless, you take turns making one move at a time. If at any point you checkmate the enemy king, he reveals himself and you win the game.
However, if your king ever moves to a square that is adjacent to the enemy king, then be warned, for he will capture your king and win the game. Meanwhile, the enemy king will never move into check or checkmate. (If a stalemate is available, he may very well go for it.)
Your king begins in the bottom left corner of the board, flanked by a rook on either side, as shown below. You also know the enemy king is not currently in check.
It’s your move. How can you guarantee a victory for yourself without losing a piece?
The solution to this Riddler Express can be found in the following column.
Riddler Classic
The enemy king has gone missing again, but this time he is no chess piece. He is somewhere in his castle’s square courtyard. It’s poorly lit, so you carry around a lantern that will help you spot him as long as he is within 10 meters of your current position in any direction.
The king is wily and tries to avoid detection. Both you and he move at the exact same speed at all times.
You find yourself in the center of the courtyard. Since you cannot see the king, you know he is somewhere outside your 10-meter circle of visibility, but beyond that you have no idea where he is. After some mental math, you realize that you have no hope of ever finding him, no matter where he starts from.
What is the smallest possible side length of the square courtyard?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Steve Schaefer 👏 of Carlsbad, California, winner of last week’s Riddler Express.
Last week, Amare the ant was trying to escape from a spider web.
Starting at the very center of the web, he then crawled 1 inch radially away in a random direction. From there, he randomly chose between crawling in the radial or tangential directions, with a 50 percent chance of each. When he crawled radially, he moved another inch from his current position directly away from the center of the web. When he crawled tangentially, he moved another inch in one of the two directions perpendicular to the radial direction. (To be clear, when Amare moved tangentially, he was still moving farther away from the center of the web, at least a little bit.)
After Amare had crawled a total of 4 inches, how far away from the center of the web was he, on average?
Suppose Amare was currently a distance x from the center of the web. After a radial move, he would then be a distance x+1 from the center. But after a tangential move (in either of the two directions), his distance from the center would be √(x2+1) thanks to the Pythagorean theorem. For larger values of x, these tangential moves had less of an effect on Amare’s distance.
Now for this problem, you knew that Amare’s first move was radial. After that, solver Harshil Mehta recognized that there were eight equally likely possibilities. Let’s list them out in descending order of distance from the center of the web, with an R for radial and a T for tangential moves:
- RRR: 4
- TRR: 2+√2, or about 3.414
- RTR: 1+√5, or about 3.236
- RRT: √10, or about 3.162
- TTR: 1+√3, or about 2.732
- TRT: √(4+2√2), or about 2.613
- RTT: √6, or about 2.449
- TTT: 2
The average of these eight values was approximately 2.951, which meant Amare was, on average, 2.951 inches from the center of the web.
Solver Joao Coelho plotted the distribution of Amare’s final distances if he had moved a total of 7 inches:
Solver Rohan Lewis further animated Amare’s possible positions over the course of 10 moves. As Amare got farther away from the center of the web, his tangential moves had less of an effect on his distance from the center. So after N moves, for sufficiently large values of N, his average distance was approximately N/2, since about half his moves were in the radial direction.
In any case, I hope Amare successfully evaded any spiders.
Solution to last week’s Riddler Classic
Congratulations to 👏 Tom Knief 👏 of Des Moines, Iowa, winner of last week’s Riddler Classic.
Last week, you (hopefully) marveled at the never-before-seen image of the black hole — Sagittarius A* — at the center of our galaxy:
This image was generated using data from the Event Horizon Telescope. One of the most striking things about the image was how clearly we could make out the black hole’s shadow. That was because the plane of its accretion disk was nearly perpendicular to the vector between us and the black hole. Was this likely to occur, or was it just a cosmic coincidence?
Assuming the accretion disk was equally likely to be in any plane, what was the probability of it being within 10 degrees of perpendicular to us, thereby resulting in a spectacular image?
Instead of thinking about the angle of the plane of the accretion disk, solver Mari Fujioka instead focused on a vector that was normal (i.e., perpendicular) to the plane. If this vector pointed within 10 degrees of Earth, then it was within 10 degrees of being perpendicular.
At this point, the problem was more or less an exercise in multivariable calculus. What fraction of a sphere’s surface area is within 10 degrees of a single point? Many solvers computed this by integrating the surface element of a unit sphere — sin𝜃d𝜃d𝜙, where 𝜃 is the polar angle and 𝜙 is the azimuthal angle — for polar angles between 80 and 90 degrees and azimuthal angles between 0 and 360 degrees.
If the maximum angle of the normal vector is 𝛼, then the area of the region on the unit sphere within angle 𝛼 of a single point is given by the formula 2𝜋(1−cos𝛼). Meanwhile, the surface area of a unit sphere is 4𝜋. And so the fraction of the surface that is within the region of interest is (1−cos𝛼)/2. Plugging in 10 degrees for 𝛼 gave you a value of about 0.0076, meaning there was a 0.76 percent chance of generating a spectacular image of the black hole.
But wait a minute, that wasn’t right! Because of the symmetry in the problem, a similar image would have been generated had the normal vector been pointing 180 degrees away from you. This meant you had to double the previous answer, and that there was a 1.52 percent chance of generating a spectacular image.
While a 10 degree difference makes for beautiful galactic artwork, what if we had relaxed this constraint so that the plane was merely within 30 degrees of being perpendicular? If the orientation of the accretion disk had truly been random, then there was still just a 13.4 percent chance of this happening.
That’s all pretty remarkable, especially when you consider that the previously imaged black hole at the center of the supergiant galaxy Messier 87 was similarly spectacular.
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.
