Can You Squeeze the Particles Into the Box?

Three repelling particles are confined to a square. How will they arrange so as to minimize their overall energy? What if there are nine particles?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

You have three particles inside a unit square that all repel one another. The energy between each pair of particles is 1/r, where r is the distance between them. To be clear, the particles can be anywhere inside the square or on its perimeter. The total energy of the system is the sum of the three pairwise energies among the particles.

What is the minimum energy of this system, and what arrangement of the particles produces it?

This Week’s Extra Credit

Instead of three particles, now you have nine. Again, the energy between each pair of particles is 1/r, where r is the distance between them. The total energy is the sum of the 36 pairwise energies among the particles.

What is the minimum energy of this system, and what arrangement of the particles produces it?

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing FrontierMath, “a benchmark of hundreds of original, exceptionally challenging mathematics problems crafted and vetted by expert mathematicians.”

Why was this benchmark created? Because apparently high school- and college-level contest problems are now too easy for AIs to solve, and they need something harder to test themselves against. As the linked abstract states, “Current state-of-the-art AI models solve under 2% of [FrontierMath’s] problems, revealing a vast gap between AI capabilities and the prowess of the mathematical community. As AI systems advance toward expert-level mathematical abilities, FrontierMath offers a rigorous testbed that quantifies their progress.”

If you’re curious, Appendix A of the PDF includes several sample problems and solutions, ranging from “low difficulty” to “high difficulty.” (For the record, I’m still muddling my way through one of the “low difficulty” items.)

As always, feel free to discuss this puzzle and your approach in the comments below.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jon Moon 🎻 from Sunnyvale, California. I received 76 timely submissions, of which 73 were correct—good for a 96 percent solve rate.

Last week, you analyzed a recent xkcd webcomic by Randall Munroe:

The question was, very simply: Was this comic mathematically correct?

In other words, let p represent the probability that if you randomly selected two arrows from a group of 10 (five of which were cursed), neither arrow was cursed. And let q represent the probability that if you rolled three d6 dice (cubes with faces numbered 1 through 6) and one d4 die (a tetrahedron with faces numbered 1 through 4), the sum of the rolls was at least 16.

Did p equal q? If so, you had to determine their common value. If not, you had to determine the value of each.

Of the two probabilities, p was easier to compute, so let’s start there. The probability that the first arrow you picked wasn’t cursed was 5/10, since five arrows were cursed and five were not. Assuming you were successful, five of the remaining nine arrows were cursed. So the probability the second arrow was not cursed (assuming the first arrow was also not cursed) was 4/9. The probability of both these events occurring was the product of these probabilities: (5/10)·(4/9), or 2/9. So p was equal to 2/9.

But what about q? That was a fair bit more work to check. One way was to analyze the four dice two at a time. The probability distribution for the sum of two d6 (or standard dice) is fairly well known: there was a 1-in-36 chance of rolling a two, a 2-in-36 chance of rolling a three, and so on. The likeliest roll was a six, which had a 6-in-36 chance.

You could produce a similar distribution for rolling a d4 and a d6 together, with the sums ranging from two to 10. There was a 1-in-24 chance of rolling a two (due to the 1-in-4 and 1-in-6 chances of rolling a one on each die), a 2-in-24 chance rolling a three (you could roll a one or two on the d4 and then a two or one on the d6), and so on. This time, the most likely rolls were five, six, and seven, which each had a 4-in-24 chance of occurring.

The columns of the table below show the distribution for d6+d6 and the rows sow the distribution for d4+d6. The probability of any given combination d6+d6 and d4+d6 was the product of their respective probabilities. For example, the probability that d6+d6 was seven and d4+d6 was nine was (1/6)·(1/12), or 1/72.

The outlined region in the bottom right of the table contained all the cases where the sum of the four dice was at least 16. Adding up the probabilities therein gave a total of 192/864, which simplified to 2/9. This was q, the total probability the sum of the four dice was at least 16.

Surprise, surprise: Both the arrow selection and the roll of the dice had the same probability of 2/9, which meant xkcd was right. (If you were at all familiar with xkcd, it really shouldn’t have been a surprise.)

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Riley Thompson 🎻 from Burlington, Ontario, Canada. I received 35 timely submissions, of which 32 were correct—good for a 91 percent solve rate. (That was the most solvers for an Extra Credit since June of this year. Great work, everyone!)

A typical set of dice from Dungeons & Dragons (D&D) includes a d4, a d6, a d8, a d10, a d12, and d20. Here, “dN” indicates a die with N faces numbered 1 through N that are all equally likely to come up.

Again, let p represent the probability that if you randomly selected two arrows from a group of 10 (five of which were cursed), neither arrow was cursed.

You wanted to simulate this event by rolling up to four D&D dice and having the sum be at least some specific value. These dice could all have been different from each other, they could all have been the same kind of die (say, d4), or anywhere in between, with a few that were the same and others that were different.

How many ways could you have simulated picking the arrows, an event with probability p? And what were these ways? For each way, you had to specify the dice as well as the value their sum should have been greater than or equal to.