Can You Grow a Hibiscus Hedge?

You’re planting a row of colorful flowers. If you want the colors to appear random-ish, what’s the longest row you can plant?

Apr 11, 2025

∙ Paid

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. 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

If you enjoyed last week’s puzzles, then you’re in for a treat! From Dean Ballard comes a colorful conundrum:

Dean has three colors of the hibiscus: red, orange, and yellow. He wants to plant them in a straight hedge of shrubs (each of which is one color) so that the order appears somewhat random, but not truly random. More specifically, he wants the following to be true:

No two adjacent shrubs have the same color.
No ordering of three consecutive shrubs appears more than once in the hedge. (But a prior ordering can appear in reverse. For example, ROYOR is an acceptable hedge, but ROYROY is not.)

What is the greatest number of shrubs Dean’s hedge can contain?

Submit your answer

This Week’s Extra Credit

From Dean also comes some Extra Credit:

In addition to red, orange, and yellow hibiscus flowers, Dean now includes a fourth color: pink. Again, he wants to plant a straight hedge of shrubs that appears somewhat random. Here are the rules for ordering the shrubs this time:

No two adjacent shrubs have the same color.
No ordering of four consecutive shrubs appears more than once in the hedge. (Again, a prior ordering can appear in reverse.)
Among any group of four consecutive shrubs, at least three distinct colors are represented.

What is the greatest number of shrubs Dean’s hedge can contain?

Submit your answer

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 a “trivia question” from LearnedLeague that is in fact a math puzzle.

LearnedLeague is an online trivia platform with four seasons per calendar year that consist of daily general trivia questions. There are also subject-specific “MiniLeagues” between these seasons. Right now, there happens to be a MiniLeague specific to mathematics! (I am 100 percent participating in this, and am doing fairly well so far.)

As you may recall, LearnedLeague came up in a previous Fiddler, titled “Can You Defend Your Trivia Knowledge?” Here’s the relevant portion from that puzzle on the mechanics of daily play:

Every day, you and your opponent for the day are presented with the same six trivia questions. You each do your best to answer these, and you assign point values for your opponent, without knowing (until the following day) which questions your opponent answered correctly. You must assign point values of 0, 1, 1, 2, 2, and 3 to the six questions.

Without further ado, here was one of yesterday’s questions from the MiniLeague:

Suppose on a given LearnedLeague match day you get 1 correct answer and your opponent gets 3. If you both play defense uniformly at random, what is the probability that you win? Express your answer as a fraction a/b in lowest terms.

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.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. (I’m starting to automate some of this tabulation. If you think you see a mistake in the standings, kindly let me know.)

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 David Cohen 🎻 from Silver Spring, Maryland. I received 63 timely submissions, of which 49 were correct—good for a 78 percent solve rate.

Last week, a teacher was handing out candy to his students, of which there were at least four. He abided by the following rules:

He handed out candy to groups of three students (i.e., “trios”) at a time. Each member of the trio got one piece of candy.
Each unique trio could ask for candy, but that same trio couldn’t come back for seconds. If students in the trio wanted more candy, they had to return as part of a different trio.
When a trio got candy, the next trio couldn't contain any students from that previous trio.

It turned out that every possible trio could get a helping of candy. What was the smallest class size for which this was possible?

When I first posted this puzzle last week, I failed to specify that there were at least four students. Without this specification, there was a trivial answer of three. The single trio received their candy, at which point all trios had done so.

But if three was an acceptable answer, why not two, or one, or zero? In this case, there were no possible trios. But it was still technically true that “every possible trio” got a helping of candy—all zero of the trios.

In my opinion, such interpretations were clearly not the intent of the puzzle, and I applaud the readers who reached out to clarify. I updated the post within hours to specify that there were indeed “at least four” students in the class. Therefore, I will not be accepting solutions of zero, one, two, or three. If you ever find yourself submitting a trivial solution (and it’s not April Fools’ Day), I strongly encourage you to reach out and seek clarification.

Okay, so there were at least four students in the class. If there were exactly four students, was it possible for every trio to get candy while abiding by the rules? No, it was not! To see why, suppose the four students were named A, B, C, and D. Without loss of generality, suppose the first trio were A, B, and C. The second trio to receive candy could include D … and no one else, since all the others had just received their candy as the prior trio.

Similarly, things didn’t work when there were five students in the class. After the first trio got their candy, the next trio consisted of at most the two remaining students, meaning it couldn’t have been a trio at all.

With six students, it was now possible to have two distinct trios. Huzzah! Among six students, there were 6 choose 3, or 20 distinct trios of students. Let’s call the students A, B, C, D, E, and F. Without loss of generality, suppose the first trio were A, B, and C. The next trio to receive candy couldn’t include A, B, or C, so it had to be D, E, and F. And the next trio to receive candy after that couldn’t include D, E, or F, so it had to be … A, B, and C again. But that trio already received candy! Of the 20 possible trios, at most two trios could get their candy. So the class needed more than six students.

Now suppose there were seven students. This time, there were 7 choose 3, or 35 distinct trios of students. Let’s call the students A, B, C, D, E, F, and G. Without loss of generality, suppose the first trio were A, B, and C, and the second trio were D, E, and F. From there, you knew the third trio had to include G, as well as two students among A, B, and C (for a total of three possible trios). Beyond the third trio, there were even more possibilities.

So it seemed plausible that every trio could get candy when there were seven students. To convince yourself, you could list the 35 trios in order, so that no two consecutive trios had any students in common. Here was solver Alain Camus’ list:

Inspecting this list, sure enough, all 35 trios were there—and no two consecutive trios had a student in common. So the smallest possible class size was indeed seven!

It turned out there was some deeper mathematics at play here, but I’ll save that for the Extra Credit. For now, I’ll share Andy Quick’s neat result that were a whopping 1,419,264 ways the 35 groups could have been ordered, including the one shown above.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Eric Widdison 🎻 from Kaysville, Utah. I received 39 timely submissions, of which 24 were correct—good for a 62 percent solve rate.

Instead of trios of students, now there were groups of 10 students coming up to get candy. This time, there were at least 11 students in the class. As before:

Each member of the group of 10 got one piece of candy per visit.
Each unique group of 10 could ask for candy, but the exact same group of 10 couldn’t come back for seconds. If students in the group wanted more candy, they had to return as part of a different group.
When a group of 10 got candy, the next group of 10 couldn’t contain any students from the previous group of 10.

Suppose the class size was the minimum that allowed every possible group of 10 to get a helping of candy. How many pieces of candy did each student receive?

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