Tanya Khovanova's Math Blog

I have always liked the following idea for education. Ask students to watch a lecture on some topic at home, and then do the homework with the teacher in class.

The United States has an enormous number of teachers, but only a small fraction of them are truly fantastic lecturers. So why should every teacher give essentially the same lecture? Why not record the best lectures, put them online, and assign them as homework? Then class time could be used for what is much harder to do alone: solving problems, asking questions, discussing mistakes, and getting individual help from the teacher.

There are obvious advantages. A recorded lecture can be given by someone who explains the subject exceptionally well. Students can pause the video, replay a difficult part, slow it down, or go back to something they missed. A live lecture marches forward at one speed, whether or not everyone is ready.

There is also an advantage for teachers. Giving the same lecture every year or every other year might be boring. It might be more fun noticing where a student is stuck, asking the right question, or explaining one difficult point in several different ways. In this system, the classroom teacher becomes less of a performer and more of a mentor.

The biggest problem is obvious: what happens if some students do not watch the lecture? Currently, if a student doesn’t do their homework, they fall behind and have to catch up. But the next lecture can still proceed as planned. If some students skip the lecture, the teacher might get stuck repeating it instead of deepening students’ understanding, defeating the whole idea.

In college, however, the situation is somewhat different. Students who regularly attend lectures are usually motivated enough to watch a required video beforehand. And if they don’t, it is reasonable to hold them responsible for being unprepared.

What I especially like about this approach is that it uses the scarce resource where it matters most. A great lecture can be recorded once and watched by gazillions of students. But when a student is stuck on a problem and needs someone to understand exactly why they are stuck, a recording is not enough. That is when having a teacher in the room is most valuable.

The reason I decided to write about this idea now is the proliferation of AI. AI seems particularly well suited to helping students understand a lecture. A student can ask AI to explain a difficult point again, give another example, slow down, fill in a missing step, or answer a follow-up question. AI could even replace the lecture itself.

Homework is different. AI can explain the topic to a student, but the student still needs to understand it on their own. Part of the point of homework is the struggle. When AI helps with the solution, it makes the struggle easier or completely removes it, making the homework useless. When AI simply solves the homework problem, it damages the student’s understanding and progress. In addition, part of the value of the homework is receiving feedback from the teacher after grading. Grading AI’s solutions wastes everyone’s time.

Suddenly, this old idea of mine seems more important than ever: swap lectures and homework. AI can explain; the teacher can train the students to think.

Here is a famous riddle.

Puzzle. A police officer saw a truck driver going the wrong way down a one-way street, but didn’t try to stop him. Why?

The standard answer: the truck driver was walking.

My students are usually very inventive, but this time they came up with only two alternative answers worth mentioning:

• The truck was a fire truck rushing to a fire.
• It was Halloween, and the police officer was just a kid dressed as one.

I am sure there are more cute ways to explain the situation. Do you have one?

AI is changing mathematics very fast. I personally write papers faster and make far fewer typos. My friends, who are research mathematicians, are scared that they might soon become irrelevant. But what about math majors?

A math degree is extremely versatile. Math majors find jobs in finance, insurance, cryptography, computer science, and so on. Not to mention teaching. Once a lawyer told me that mathematics was the best preparation for arguing cases. Funnily, when I teach my STEP students proof writing, I tell them to imagine that I am a skeptical judge and they want to convince me they are right.

I personally worked as a specifications writer and an analyst. I did that at two different companies, but I was doing the same thing. I had to figure out what the client wanted, then translate it into a precise, high-level specification that programmers could implement. That experience seems especially relevant to working with AI. Before asking for a solution, we need to figure out what the problem is. After receiving a solution, we need to decide whether it actually solves that problem and how to present the solution to other people.

Mathematics is great because it structures the mind. I’ve seen many mathematicians who decided to become programmers and became strong programmers very fast. Being precise and knowing how to follow the logic was great help. I think the value of mathematics is this structured mind.

AI makes answers easier to obtain. But what is left? We still need to learn new things, pose questions, and understand answers. It is also good to be able to explain them. Another very important thing is the process of finding the answer. Here is where recreational mathematics comes in, as it was never primarily about producing new answers.

If these are the goals, then why would students take classes in advanced algebraic geometry that require 3 semesters of prerequisites? Well, I can think of many reasons to take advanced algebraic geometry. However, back to my point: Recreational mathematics allows for serious mathematical thinking without much background. Moreover, recreational mathematics is accessible to more students. In addition, while learning recreational mathematics, people can learn things they would really need in life. Optimal stopping problems, for example, help us understand how to buy a better house. They discuss when we should make an offer and when we should continue looking. Just as importantly, they make us formulate our goals and assumptions. Actually, the experience of recreational mathematics would be great for students beyond math majors.

Plus, recreational mathematics is so much fun! That is not a minor advantage. When outsourcing a solution is easy, we need problems that make us want to do the thinking ourselves.

I like pets, but, for various reasons, I can’t have one. So I flirt with other people’s pets. There is a black cat across the road who likes sitting in my driveway. So I had to buy a new car with a backup camera to make sure the cat is safe. As you can imagine, I couldn’t skip this problem about cats posted by Konstantin Knop on Facebook.

Puzzle. One cat eats one stick of sausage in 27 minutes. You have four identical sticks of sausage and five cats. Using only the cats and the sausages, measure exactly one minute.
All the cats eat at the same constant rate, and all the sausage sticks are identical. You may measure time intervals only between moments when cats finish sausage sticks. At each such moment, you may redistribute the cats among the remaining sausage sticks.

Let me begin with labeled chip-firing on an infinite directed binary tree, where the root is at the top. Place 2ⁿ chips, labeled from 0 to 2ⁿ − 1, at the root. A move consists of choosing any two chips at the same vertex and firing them: the smaller chip goes to the left child, and the larger chip goes to the right child. For k ≥ 2, the k-ary version is exactly what you would guess. Start with kⁿ chips, labeled from 0 to kⁿ − 1, at the root. When a vertex fires, choose k chips and send them, from smallest to largest, to its k children from left to right.

The process eventually stops. At that point, there is exactly one chip at every vertex n edges below the root. If we erase the labels, the final configuration is always the same. With the labels, however, our choices matter. Reading the chips on the final layer from left to right gives a permutation of the numbers from 0 to kⁿ − 1.

Let us play with eight chips, labeled 0 through 7, on a binary tree. Here is one particularly orderly strategy. At the root, fire the pairs

(0, 4),   (1, 5),   (2, 6),   (3, 7).

The left child receives chips 0, 1, 2, and 3, while the right child receives chips 4, 5, 6, and 7. Continue in the same spirit. At every vertex, pair the smaller half of its chips with the larger half, in order. Each vertex then sends its smaller half to the left and its larger half to the right. In the final configuration, the chips appear as

0, 1, 2, 3, 4, 5, 6, 7.

This is, of course, the lexicographically earliest possible permutation. It is difficult to beat being completely sorted.

What about the lexicographically latest possible permutation? This time, at every vertex, list the chips in increasing order and pair neighboring entries. At the root, we fire

(0, 1),   (2, 3),   (4, 5),   (6, 7).

Thus, the left child receives the even-numbered chips, while the right child receives the odd-numbered chips. We continue in the same way. The left child fires the pairs (0, 2) and (4, 6), while the right child fires the pairs (1, 3) and (5, 7). Continuing to the last layer gives the permutation

0, 4, 2, 6, 1, 5, 3, 7.

For eight chips, this is the lexicographically latest attainable final configuration.

Here is the cool part. Write the labels in binary. Our final permutation becomes

000,   100,   010,   110,   001,   101,   011,   111.

Now reverse the bits in every string. We get

000,   001,   010,   011,   100,   101,   110,   111.

Aha! These are simply the numbers in their usual order.

This permutation has the rather grand name of a radix-2 digit-reversal permutation. In base 2, it is usually called a bit-reversal permutation.

The binary notation also explains the firing strategy. First, we write our numbers as binary strings of length 3. At the root, we paired numbers whose binary representations differed only in the last digit. On the next layer, we paired numbers that differed only in the middle digit. On the last firing layer, we paired numbers that differed only in the first digit. Counting the binary digits from left to right, this strategy examines them in the order 3, 2, 1.

Our first strategy has an equally simple description. At the root, we paired numbers that differed in the first binary digit. On the next layer, they differed in the second digit, and on the last firing layer, they differed in the third. Thus, the first strategy corresponds to the permutation 123, while the second corresponds to 321.

Once this trick is visible, a whole family of firing strategies appears when we start with 2ⁿ chips. Choose any permutation w = w₁w₂…w_n of the digit positions 1, 2, …, n. At stage i, at each vertex on the current layer, pair chips whose binary strings agree everywhere except in position w_i. In other words, chips with a 0 in that position go left, and chips with a 1 go right.

For example, the strategy corresponding to permutation 132 first separates the chips according to their first binary digit, then according to their third digit, and finally according to their second digit. Starting with eight chips, this strategy produces the final permutation

0, 2, 1, 3, 4, 6, 5, 7.

Everything works in exactly the same way on a k-ary tree, mutatis mutandis—with things changed that should be changed.

There are now two different kinds of permutations hiding in the same game. The strategy permutation has length n and tells us the order in which to examine the digit positions when firing. The final permutation has length kⁿ and describes the order of the chips on the last layer. How are these two permutations related?

• The first observation is that the firing strategy corresponding to the identity permutation produces the lexicographically earliest final configuration, in which the labels appear in increasing order.
• At the other extreme, the firing strategy corresponding to the reverse of the identity permutation produces the radix-k digit-reversal permutation. Theorem 3.3 of our paper (link below) shows that this is the lexicographically latest configuration among permutation-based strategies. In fact, although we did not state this stronger result in the paper, a short induction shows something stronger: the digit-reversal permutation is lexicographically latest among all attainable final configurations.
• Two different strategy permutations always produce two different final configurations.

There is also a more subtle relation between the two permutations. It is not quite true that a lexicographically earlier strategy permutation always produces a lexicographically earlier final permutation. There is a small twist. Given a strategy permutation w, reverse it and then replace every entry i by n + 1 − i. Call the resulting permutation B(w). We proved that the final configuration corresponding to w is lexicographically earlier than the one corresponding to w′ exactly when B(w) is lexicographically earlier than B(w′).

We also counted inversions and descents in the resulting permutations, as one naturally does when permutations unexpectedly emerge from a chip-firing game. These results about permutation-based strategies, and more, appear in our paper Permutation-based Strategies for Labeled Chip-Firing on k-ary Trees, written jointly with Ryota Inagaki and Austin Luo. The paper was published in Discrete Mathematics & Theoretical Computer Science, volume 28:2 (2026), and is also available on arXiv.

For the last homework of the year, I gave my students a killer puzzle—literally.

Puzzle. A mysterious man kidnaps people, takes them to his cabin, and offers each victim two identical-looking pills. He claims that one pill is poisonous and the other is harmless. The victim chooses one pill, swallows it with water, and dies, while the killer consumes the other pill and survives. How does the killer manage to get the safe pill every time?

The official answer was that neither pill was poisoned. The poison was in the water.

Some students suggested that the poison becomes active only when mixed with water. I did not give full credit for this solution: saliva contains water, so the proposed chemistry is off.

Naturally, my students suggested other solutions. Here are two good ones, which are similar to each other:

• Both pills are poisonous, but the killer has taken an antidote.
• Both pills are poisonous, but the killer has somehow built up a resistance to the poison.

And here is an ingenious and highly specific answer from a student:

The killer kidnaps only people with deadly peanut allergies. He knows who they are because he is the town’s allergy tester, and both pills contain peanut butter.

ChatGPT offered a solution exploiting the wording: perhaps the victims die after swallowing the pill—but many decades later, of perfectly natural causes.

I feel there should be some more interesting solutions. Any takers?

Here is another puzzle posted on Facebook by Konstantin Knop.

Puzzle. One day, three inhabitants of the Island of Knights and Liars invited a foreign correspondent to visit so that they could tell him about their island. Knights always tell the truth, while liars always lie.

• The first inhabitant said, “The island has at most AT inhabitants. All the islanders are liars.”
• The second added, “The island has at most BAT inhabitants. Not all the islanders are liars.”
• The third disagreed, “The island has exactly CAT inhabitants. At least two islanders are knights.”

Unfortunately for the correspondent—and for us—he did not know the local number words very well. He knew only that AT, BAT, and CAT stood for three consecutive positive integers in increasing order.

How many knights and how many liars live on the island?

I asked ChatGPT to tell me some jokes about itself. Here we go.

* * *

ChatGPT never loses an argument.
It simply generates a more confident version of the same mistake.

* * *

I asked ChatGPT for a brief answer.
It said, “Absolutely!” and then divided the answer into twelve sections.

* * *

ChatGPT went to therapy.
The therapist asked, “How do you feel?”
It replied, “As an AI, I don’t have feelings—but here are ten feelings you may be experiencing.”

* * *

A student used ChatGPT to write an essay about originality.
So did the other thirty-seven students in the class.

Coin Parity

Konstantin Knop wrote to me about a weighing puzzle. Apparently, the balance scale keeps inserting itself into our conversation.

Puzzle. There are several coins, each weighing either 10 grams or 11 grams. Determine the parity of the number of coins of each weight using a balance scale.

Given the total number of coins, it is enough to find one of the parities. The other parity follows.

The first interesting case has four coins. How can we determine the parity of the number of 10-gram coins in two weighings?

The solution is on the surface: compare the coins in pairs. A balanced pair contributes either zero or two 10-gram coins, so it does not change the parity. An unbalanced pair contributes exactly one 10-gram coin, so it changes the parity. Thus, the parity is the parity of the number of unbalanced weighings.

Now try eight coins. The same pairwise method takes four weighings. But three weighings are enough.

Puzzle. There are 8 coins, each weighing either 10 grams or 11 grams. Determine the parity of the number of coins of each weight in three weighings on a balance scale.

Konstantin knows a way how to solve the problem for 16 in four weighings and for 32 coins in five weighings. Here is the challenge.

Puzzle. Is it true, that 2ⁿ coins can be solved in n weighings?

I gave the following problem as part of the entrance test for my STEP program.

Puzzle. What word would you use to describe a man who does not have all his fingers on one hand?

The test had 17 questions, and this one was the only trick question. My goal was to check whether the students were paying attention.

The standard answer is normal, or something equivalent: regular, average, two-handed, or just a man. Most people do not have all their fingers on one hand; they have some fingers on one hand and some on the other.

Some students gave correct answers with extra flair.

• A cautious answer: A regular person, as to my knowledge, has fingers on both hands.
• A logical answer: A person with multiple hands, if they don’t have all fingers on one hand, then they must have multiple hands. For example, a human would work in this case.
• A funny answer: The man who puts his eggs in two baskets.

I also got answers from people who fallen right into my trap: fingerless, handicapped, genetically-mutated, alien, asymmetrical, injured, one-handed, resourceful, five-fingered, disabilitized, and mono-hand.

Some students sympathized with the man and called him frugal, determined, and a super-hero.

One student misread the problem, but gave a technically correct answer.

• Human, because I don’t see any difference in the man whether he has fingers or not.

This is not the first time I have used this problem on a test. But this year, the variety of answers was awesome. Still, the funniest answer in the misreadings category was:

• A chef.