Tanya Khovanova's Math Blog

We recently wrote a blog post on how to generalize the game of SET and promised to continue. Here we are. But first, a reminder of what the game of SET is.

In the game of SET, we have 81 cards, each containing one, two, or three of the same object. The object is green, red, or purple, the shape is squiggly, oval, or diamond, and the shading is empty, full, or stripped. Three cards form a set if, for every feature, the attributes are all the same or all different. An example of a set with all features different is shown below. By the way, such sets are usually more difficult to spot. In the game, you need to find sets as fast as you can.

If we assign each attribute value a number 0, 1, or 2, we get an equivalent definition of a set. Three cards form a set if and only if the values for each feature sum to zero modulo 3. Thus, we can see our cards as vectors in the space F₃⁴. Three vectors form a set if they sum up to 0.

The generalizations we described in the previous post, were the following. We pick a different group and define a set as a few cards that might need to be in a specific order that multiply to the group’s identity.

However, there is a different way to generalize sets to groups. Three cards that form a set in a classical game of SET, taken in any order, form an arithmetic progression. In other words, if a, b, and c form a set, then vectors b−a and c−b are the same. We can check this. We have c−b = c−(c+b+a)−b =−2b−a = b−a.

Thus, we can generalize the game of SET differently. Suppose cards are vectors in some space. We say three of them, a, b, and c, form a set if and only if b−a = c−b. Now, the order becomes important, similar to our previous generalization. We do not need to use commutative groups like vector spaces. For any group, our condition is equivalent to ba^-1 = cb^-1. Thus, c = ba^-1b.

Interestingly, we do not care much about the identity card, meaning the card deck is a torsor. We introduced the notion of a torsor before, which informally is a group that forgets about its identity. Now, let’s check possible examples.

Suppose the values of one attribute correspond to Z₄. This game is not very inspiring as two values, a=0 and b=2, can be completed to a set with the third card, which equals c=(0,0,0) and is already used. The next interesting example is Z₅. Here, values a=0 and b=2 can be completed to a third value c=4. We will leave it to the reader to check that for any two cards, a and b, the third card, c, differs from both a and b.

To make it more visual, we can use a pentagon with one marked direction. We fix the pentagon in space. In this case, three cards form a set if and only if the directions of the first and the third card are symmetric with respect to the direction of the second card. If we want to use three pentagons, we can reuse the cards from the game C53T, we described in our previous post. To make this game more visual, we can use three pentagons. We can mark each coordinate with a direction on the corresponding pentagon. The colors are to allow players to visually process the cards faster. For theoretical purposes, the colors can be ignored. Also, the pentagons themselves have different intensities of gray to emphasize that the game is played on each of them separately and also help choose the top of the card.

Let us go back and calculate when it is possible that the element that completes a set is already used. If our initial elements are a and b, then we need ba^-1b to complete a set. If this element is equal to b, then a = b, which contradicts the assumption that we start with two different cards. Suppose ba^-1b = a, then, equivalently (ba^-1)² = 1. The new element can be the one that is already used if and only if the group contains elements of order 2.

Can we use other decks we described in the previous post? Consider the game ProSet/Socks. The group is commutative, and every element is of order 2, which means ba^-1b is always a. We can’t use the deck at all! What about the EvenQuads deck? It can be viewed as Z₄³, so it is possible that we can’t complete any two cards to a set. However, there is a bigger problem with the deck. To play with it, we need to actually assign values to colors and shapes. We are saying that even if we decide to use the group, we should make different cards! For example, we can style the cards as squares like the pentagons above.

We also mentioned in the previous post the group, which is the wreath product of S₂ and S₃. Consider the following example of two cards that were screen-printed from the Numberphile video on the variations of the game of SET.

The first card, a, is the inverse of itself, so the third card we are looking for is described as the product bab, which we can visualize as the following. If we ignore the beads, the card that completes the set is a. Luckily, if we do not ignore the beads, it is not a; we need to add two beads to a.

The product of permutations is difficult to visualize, so playing this game with the cards in the Numberphile video might be difficult. The good news is that this group can be visualized in many different ways:

• The group of symmetries of a cube, meaning its rotations and reflections;
• The group of symmetries of an octahedron, meaning its rotations and reflection;
• The wreath product of S₂ and S₃;
• The direct product of S₄ and S₂.

We want to show you a beautiful deck from the tsetse website that allows you to use any one of the four definitions to play the game. The game is called OCTA Set, as the underlying group is called an octahedral group. An example of a set is in the image below, where the top and bottom shapes represent the same element of the group. Moreover, the deck is a torsor: there is no the identity card.

• The top shape represents an octahedron where the opposite faces are the same color. Thus, the top shape shows a unique rotation of an octahedron. All the swirls are the same in the image, but in other cards, the swirl could be white. Two different swirls mean that the octahedron needs to be reflected to get from one shape to the other. In our example above, no reflection is involved, so all the swirls are black.
• We can ignore that the top shape represents an octahedron and only look at the choice of colors. The changes of colors represent a permutation in S₄ and a swirl represents an element in S₂.
• The bottom shape represents a cube where the opposite faces are the same color. One of the faces is solid, and the opposite face is hollow. This way, one can reconstruct the complete shape of a cube by the top three faces.
• We can ignore that the top shape represents a cube and only look at the choice of colors. The colors form a permutation, and the beads correspond to changing the hollowness of a color. When viewing the cube this way, the permutation acting on the colors is S₃, and each color has its hollowness associated with an element of S₂. This exactly corresponds to the wreath product of S₃ and S₂.

Let us prove that this is a set. Consider the bottom cube shape. Comparing the first two cards, the top face doesn’t change. We can see that the symmetry of the cube is the 90-degree clockwise rotation around the line that goes through the centers of green faces. In such a rotation, the left face on the second card keeps the color from the first card, while the right face takes the color from the left face on the first card and swaps hollowness. We see that the third cube completes the set.

For another proof, let us look at the top shape and discuss what happens with the permutation of colors when changing from the first card to the second. The left color moves to the bottom, the bottom color to the right, the right color to the center, and the center color to the left. Not surprisingly, we got a cyclic permutation of order 4, similar to a 90-degree rotation being of order 4. The same thing happens when moving from the left to the right. The swirl stays the same.

As we mentioned, when you play this game and pick two cards then calculate what card completes the set, you might discover that it is one of the cards you picked. The probability that two cards in a specific order can’t be completed into a set is the same as the probability of picking a random element in our group and discovering that it has order 2. The symmetric group S₄ has 9 elements of order 2. Thus, the direct product with S₂ has 19 elements of order 2, giving a probability of 19/48. For completeness, this group also has 8 elements of order 3, 12 elements of order 4, 8 elements of order 6, not to mention the identity of order 1.

If ba^-1 is an element of order three, then the cards a, b, and the card c that completes the set form a set when they are taken in any order. As a tradition in mathematical writing, we leave it to the reader to check that fact. Just a reminder that in the game of SET, the group element ba^-1 always has order 3.

Notably, there was nothing special or extraordinary about the group discussed above. It has a pretty visualization as a cube or octahedron, but is not otherwise particularly interesting. The reason why this group allowed for these two platonic solids to be used to visualize it is because the cube is dual to the octahedron. But we could have similarly used any group to play set! One such example might consider using the group of symmetries of the other pair of dual platonic solids, the icosahedron and the dodecahedron. This group is actually equivalent to A₅, also known as the alternating group of order 5, which consists of all even permutations of five elements. The tsetse website we mentioned above contains an implementation of such a game called A5SET (pronounced “asset”). The design of the site, games, and cards was done by Andrew Tockman and Della Hendrickson.

The world is full of groups and symmetries. Any group can be turned into a game of SET!

Foams are cool mathematical objects studied by my brother, Mikhail Khovanov. I already wrote about them in my previous blog posts, Foams Made out of Felt and Tesseracts and Foams. Here, I would like to explain why foams are so cool, but first, I need to remind you of their definition. Foams are finite 2-dimensional CW-complexes, such that each point’s neighborhood must be homeomorphic to one of the three objects below.

• An open disc. Such points are called regular points.
• The product of a tripod and an open interval. Such points are called seam points.
• The cone over the 1-skeleton of a tetrahedron. Such points are called singular vertices.

Foams are cool: they are 2-dimensional CW-complexes embedded in 3-space, with singularities only of the most generic kind, which makes them relatively simple. Moreover, they are combinatorially defined, which makes them easier to work with than with many other geometric objects.

My two previous blog posts have some pictures, but now, I just want to discuss a generic planar cross-section of a foam, which is a planar graph. In the cross-section, seams become vertices, and faces (regular points) become edges. The tripod condition above implies that the resulting graph is trivalent: each vertex has degree 3.

The most interesting foams are tricolarble: foams where their faces can be colored in three colors, so that each face has its own color, and, at the seams, three faces of three different colors meet. The cross-section of such a foam makes a tricolorable trivalent graph. This coloring is called Tait coloring. The cool thing is the Tait’s theorem connects the Tait coloring to the 4-color theorem.

Tait’s theorem. The following two statements are equivalent.

• Every planar graph is 4-colorable.
• The edges of every planar bridgeless trivalent graph are 3-colorable.

I won’t discuss the proof here, but I will explain how to color the edges of a graph in three colors when the faces are colored in four, and vice versa.

Assume that the four colors of the faces form a group of four elements, called the Klein group. Let’s say that gray is the identity, and red, blue, and green are the rest. Then, the product of gray and x is x. The product of any two non-gray colors is the third non-gray color.

Given a trivalent graph G whose edges are colored in three colors, we can color the faces of that graph in the following manner. Color one of the faces a random color. Then, calculate the colors of the other faces so that each edge’s color is the product of the colors of neighboring faces.

Going back, if we have a planar trivalent graph with faces colored in four colors, we can assign an edge a color that is the product of the colors of neighboring faces. As neighboring faces have different colors, the product of those colors is never gray (the identity). Thus, the edges will be colored in three colors. I leave it to the reader to check that three edges incident to a vertex must be colored in different colors.

Kronheimer-Mrowka homology theory of graphs states that the Kronheimer-Mrowka homology of a trivalent graph is non-zero if and only if the graph has no bridge. If one can prove that the rank of the homology group is the number of 3-colorings of the edges (or at least that the non-zero homology implies the existence of the tricoloring of that graph), then the Four-color theorem would follow from Tait’s theorem.

Foams are cool by themselves, but there is hope that they might provide a conceptual proof of the Four-color theorem, making them awesome!

Do you know that 2025 is a composite, deficient, evil, odd, square, and powerful number? I collect properties of numbers at my Number Gossip website, where you can also find detailed definitions of these terms. Provocatively, 2025 is also an apocalyptic power, meaning that 2 to the power of 2025 contains 666 as a substring.

Recently, Tamas Fleischer sent me an email discussing additional fascinating properties of 2025. While I am slowly deciding whether to add them to my database, there is some urgency in posting these properties in anticipation of the coming year. Here’s the material from Tamas, retold in my own words.

Out of the properties mentioned earlier, the square property is the only rare one. On my website, I define a property as rare if fewer than 100 numbers below 10,000 possess it. Square numbers barely make the cut. But 2025 is not just a square number—it is the square of a triangular number. If you remember the formula for the sum of cubes of the first n natural numbers, the result is (n(n+1)/2)^{2, which is the square of the nth triangular number. Thus, 2025 is the sum of the cubes of all one-digit numbers.}

Additionally, 2025 is the product of 25 and 81. My website notes an intriguing property shared by 25 and 2025: both remain square numbers when all their digits are incremented by 1. For example, 25 becomes 36, and 2025 becomes 3136, both of which are squares. Moreover, 25 is the smallest such number, and 2025 is the second smallest. What my website does not mention is that their square roots exhibit a similar pattern. The square roots of 25 and 2025 are 5 and 45, respectively. When their digits are incremented by 1, the results are 6 and 56, the square roots of 36 and 3136, respectively. The original and incremented squares and their square roots are tied together in a surprising way.

2025 also shares an interesting property with 81. Both are square numbers with an even number of digits and if you split the digits in half and sum the halves, the result is the square root of the original number. For 81, splitting into 8 and 1 gives 8 + 1 = 9, which is the square root of 81. Similarly, for 2025, splitting into 20 and 25 gives 20 + 25 = 45, the square root of 2025. Intriguingly, 81 is the smallest number with this property, and 2025 is the second smallest.

Thank you, Tamas, and Happy New 2025 to everyone!

The deck for the game of SET consists of 81 cards. Each card has 1, 2, or 3 identical objects. Each object is an oval, a diamond, or a squiggly, colored in red, green, or purple, and the shading can be solid, striped, or empty. Three cards form a set if, for each feature (number, shape, color, shading), all three are either the same or all different. For example, one red striped squiggly, two red striped ovals, and three red striped diamonds form a set. The numbers are all different (1, 2, and 3); the shapes are all different (squiggly, oval, diamond); the color is the same (red), and the shading is the same (stripped). Another example of a set with all features different is shown in the picture.

For every feature, we can assign values of 0, 1, 2, modulo 3. The fact that the values are the same or all different is equivalent to saying that the values sum up to zero modulo 3. Thus, we can view the cards as points in a 4D vector space over the field F₃, namely F₃⁴. Three cards form a set if and only if the corresponding vectors sum up to the zero vector. Sets have a very useful property: for any two cards in a deck, there exists a third card that completes the given two cards to a set.

Before we generalize the game of SET, let’s talk about notation. Above, we looked at the values we assign for one feature as the field F₃. If we want to talk about vectors, since technically a vector space must be over a field, we could use this definition. However we do not use its multiplicative properties, so we can say that values for one attribute are in Z₃. If we want to forget about the exact values, and just use the group structure, we can use notation C₃.

Let’s see in what ways we can extend this definition. People have tried to generalize the game of SET to a group. Suppose each card corresponds to an element in a group. Then, we want to arrange three cards so that the corresponding three elements multiply to the identity. If the group is non-commutative, then the order matters. That means the players do not just pick out the three cards, they have to arrange them in a line so that the corresponding product is the identity. However, we still retain a useful property. If our card deck contains all possible elements of the group, then for any two cards in a given order, there exists a card that completes them to a set so that the product is the identity. However, another problem is that this card might be one of the cards we already have. On the plus side, we can rearrange the order of the cards, getting more options. We can also say that a set can consist of any number of cards, not just 3, as long as the product is the identity.

The Numberphile video The Game of Set (and some variations) shows examples of such games. One of the decks they show represents the group S₃: permutations of 3 elements. The picture shows a screenshot from the video with a set in a different group, but if you ignore the blue beads, you will get a set in S₃. The red dots at the bottom mark odd permutations. This helps see sets faster: each set has to have an even number of red dots. This deck is a toy example of what is possible with only 6 elements in the group.

They also show a similar deck corresponding to group S₄: permutations of 4 elements. Thus, the deck size is 24. Despite the size not being very big, the video claims that the game is tough to play. One more deck represents two permutations of 3 elements, and one must get to the identity on both of them. Thus, the deck size is 36, and the underlying group is S₃². These games are sometimes called Non-abelian sets. A third deck is a deck with three crossing lines with beads. The crossing lines correspond to permutations of three elements, and each string either has a bead or not. To get to the identity, you need an even number of beads on each line. Thus, the deck size is 48 and corresponds to the group that is called the wreath product of C₂ and S₃, where C_n is the cyclic group with n elements. The wreath product is used because a copy of C₂, each of the beads, is associated with each of the three lines that the first group, S₃, is acting on.

The screenshot from the video we mentioned corresponds to this group, showing four cards forming a set. The red dot at the bottom signals the parity of a permutation, and there has to be an even number of them, but you can ignore them. However, the meaningful dots are small blue dots on some lines, and there has to be an even number of them on each line.

In practice, one card in each deck will contain the group identity, so this card is typically removed from such games. Leaving it in would allow anyone to add it to any other set they found, getting a free point (if the score is the number of cards for each set). Thus, the actual deck size contains 1 fewer card than the corresponding group.

One famous example of such an extension of SET that wasn’t shown in the Numberfile video is often called “ProSet” (short for Projective Set). It is usually marketed under other names, for example, “Socks”. It is played on the group F₂⁶ with the identity card removed, making the deck 63 cards. The cards each contain some non-empty subset of 6 colored dots/socks. A set is any group of cards where each color appears an even number of times. This is equivalent to adding vectors in F₂⁶ and having them add to the identity: the zero vector. Since the group operation is addition, which is commutative, the sets found do not need to be in a particular order. The picture shows a set in the game of Socks.

However, there is a fundamental difference between the classic game of SET and the extensions we have talked about. No card in the SET deck obviously corresponded to the identity and had to be removed. It is almost as if the SET group “forgot” about its identity element. A group that has “forgotten” about its identity can be informally called a torsor. A torsor has a more sophisticated definition, but this casual one will do for our purposes. We also note that SET sets always contain exactly three cards. The sets have the following property: if we had chosen a particular assignment of the values of 0, 1, and 2 to each property, as in the Numberphile video, we would find that changing this assignment would have the following effect: if the initial set has all attributes different, the set will not change; if the attributes are all the same, each value would be shifted by the same shift s. The sum would thus be shifted by 3s = 0, as we look at everything mod 3.

Another affine game is a game of EvenQuads played on a deck of size 64. Each card has 3 attributes with 4 values each. Each card has symbols of a color (red, yellow, green, or blue), a shape (spiral, polyhedron, circle, square), and a quantity (from 1 to 4). The set, called quad in this game, consists of four cards such that for each feature, one of three things is true: 1) all of the cards are the same, 2) all of the cards are different, and 3) the card’s values are divided fifty-fifty. The picture shows the set in this game.

Can we extend the notion of a torsor to ProSet? The answer is yes! EvenQuads is almost equivalent to the game of ProSet: you just need to pick the card corresponding to the origin and only allow sets with four cards. The four values of each attribute in EvenQuads represent all possible values of 2 different socks in ProSet/Socks. With 3 attributes, we can express all possible values of 6 dots/socks. This is more restrictive than the set that we could choose in ProSet, but the advantage is that our deck is now completely free of a choice of a particular identity value. ProSet can be used to play EvenQuads and vice versa (with the exception of the identity card); the “torsor-ness” of EvenQuads makes the definition of a set cleaner. Indeed, a set in EvenQuads is not any set of cards that forms a set in ProSet. See more about how different famous card games are equivalent to each other in the paper Card Games Unveiled: Exploring the Underlying Linear Algebra.

At this point, it is worth noting that nothing about the games of SET, ProSet, or EvenQuads requires the exact dimension chosen. The game of SET uses C₃ as an underlying group and chooses to use 4 copies of it. Both ProSet and EvenQuads use 6 copies of C₂. The choice of dimension, the number of copies, is usually made with practical considerations in mind to make the number of cards in the deck, 81 and 64, respectively, to be of a reasonable quantity. However, to design a new game, we only have to verify that the underlying group is satisfactory, and only then do we need to fix a dimension.

So, how would we generalize further? If each attribute was an element of Z₄ instead of Z₂², would that work? Recall that in the game of SET, we require three cards to sum to zero in Z₃². In ProSet and EvenQuads, we also require the cards to sum to zero in Z₂⁶, where in ProSet, we use any number of cards, and in EvenQuads, we use four cards. Coming back to Z₄, suppose we announce that x cards form a set if their sum is zero. If we want the cards that are all the same in each feature to form a set, we need x to be divisible by 4. If we choose x to be 4, we will lose the ability to select “all different” as a valid combination of attributes since 0+1+2+3 = 2 mod 4. Any other x that is divisible by 4 is too big to play. And in any case, we lose symmetry between different values. For example, when the attributes are divided fifty-fifty, sometimes they sum to zero, as in 1+1+3+3, and sometimes not, as in 1+1+2+2.

Is this it? Should we stop there? Let’s not. Let’s check Z₅. To allow all cards to be the same, we will claim that sets must have 5 cards. The good news is that 0+1+2+3+4 = 0 mod 5. Suppose we define the set as 5 cards that sum to zero modulo 5. The big question is, can we define the rule without assigning the exact values to each attribute? Let’s look at an example of a set: 0,0,0,2,3. We can’t claim a set to be three of the same values, and two others are different. However, if we specify a relative order of our elements and allow 0,0,0,2,3 to be a set, then together with it, all shifted sequences correspond to a set; for example, shifting by 1 produces 1,1,1,3,4.

In particular, to make a game based on Z₅, we can demand that our sets consist of 5 cards that add to the identity. The fact that each set has 5 cards means that we can use a torsor and do not need to explicitly specify the identity. If we represent our 5 attributes as points on a pentagon or lines pointing in one of 5 directions, we observe a remarkable pattern: the 5 values add to 0 exactly when there is a line of symmetry in the chosen attributes! Below, we see all possible ways to add 5 values mod 5 and arrive at 0, up to rotations. As an example, if we chose 3+2+1+2+2=0 mod 5, we would see that this corresponds to the second symmetry in the picture, which we can view as drawing an axis of symmetry “through the 2”.

This is a property unique to the number 5, and also for numbers smaller than 4. This is not true for larger sizes of groups; for example, 0+0+0+1+2+3=6 has no symmetries in Z₆. Number 4 represents an interesting case. If we represent our 4 attributes as points on a square or lines pointing in one of 4 directions, we observe that when the 4 values add to 0, there is a line of symmetry in the chosen attributes. However, in the case of values 0, 0, 3, 3, or, 0, 1, 2, 3, there is a line of symmetry, but the values do not sum to zero.

So we have shown that a torsor of C₅, the cyclic group of 5 elements, might make for a convenient group on which to play a variant of SET. A reasonable choice of dimension might be 3, making a deck of size 125. This game would then be played on a C₅³-torsor. This inspired the name of this game as, this has been shortened to C53T, pronounced “C-Set”. To illustrate a card, we use pentagons with an indicated vertex to denote an element of C₅. We need 3 pentagons, one for each dimension. To help players work around the fact that cards often get turned around, each direction on each pentagon is assigned a color to help players tell them apart better. Additionally, the three pentagons are shaded to clarify which pentagon corresponds to each dimension. In the image below, we see a C53T set, where the symmetric axes in each of the three dimensions go through the red, orange, and “any” axes from top to bottom.

This and many other games can be found on tsetse website. The website has a description, and also allows for a solitaire play.

Finding sets in C53T is extremely hard, so this is not a game for casual players, but it does serve to illustrate what really makes a set a set. In everything discussed in this blog post, we either used a group with a clear identity (such as in the Numberphile video) or a commutative operation (such as in C53T). Is there any way to have a non-abelian group torsor to play a variant of SET? The answer is yes, but that is a story for another day.

A foam is a finite 2-dimensional CW-complex with extra properties. This one opening sentence is already more advanced than any of my usual blog posts. Let me define a finite 2-dimensional CW-complex in layman’s terms.

To construct such a CW-complex, we can start with a bunch of discrete points. This will be the 0-dimensional part of our future CW-complex. To continue, we glue-in segments between some pairs of points, making the whole thing into a 1-dimensional CW-complex. We can view such a complex as a graph. Now, what we have left to do is glue some disks in. There are two ways to do it. First, we can take the disk’s border and attach it to one of the points. Second, we can glue the border of the disc to a cycle in a graph.

The previous paragraph explained how to construct a finite 2-dimensional CW-complex. Foams have additional properties. Given a point in the CW-complex, its neighborhood needs to be homeomorphic to one of three objects:

1. An open disc. Such points are called regular points. These are any points from inside of the discs.
2. The product of a tripod and an open interval. Such points are called seam points. These are any points from inside of the segments. To meet the tripod condition, all the segments have to attach themselves to three discs.
3. The cone over the 1-skeleton of a tetrahedron. Such points are called singular vertices. This means our starting vertices can’t remain unattached. The underlying 1-skeleton of our complex has to be a 4-regular graph (each vertex has to have degree 4). Moreover, any two edges coming from the same vertex must be on a disc’s border.

Some of the coolest foams are tricolorable. A foam is called tricolorable if it is possible to color its faces each in its own color by using three colors total so that at the seams, the faces of all three colors meet.

I first heard about foams during my brother Mikhail Khovanov’s lecture. He made a fascinating claim about tricolorable foams. If you remove regular points of a particular color from a foam, then the neighborhood of each point is an open disc. I was so curious, I decided to make a physical model. My readers know I hate crocheting, so I thought making a model out of felt would be easier. Thus, I made a tricolored neighborhood of a singular point. The two pictures show the same model from different angles.

Then, I needed to check my brother’s statement and made the same model with one color attached to the foam by zippers. This way, I can actually unzip one color and see the result. This color in real life is neon-green but looks yellowish in the pictures.

Guess what? The result was a smooth neighborhood isomorphic to an open disc. My brother was right!

The term fibonometry was coined by John Conway and Alex Ryba in their paper titled, you guessed it, “Fibonometry”. The term describes a freaky parallel between trigonometric formulas and formulas with Fibonacci (F_n) and Lucas (L_n) numbers. For example, the formula sin(2a) = 2sin(a)cos(a) is very similar to the formula F_2n = F_nL_n. The rule is simple: replace angles with indices, replace sin with F (Fibonacci) and cosine with L (Lucas), and adjust coefficients according to some other rule, which is not too complicated, but I am too lazy to reproduce it. For example, the Pythegorian identity sin²a + cos²a = 1 corresponds to the famous identity L_n² – 5F_n² = 4(-1)ⁿ.

My last year’s PRIMES STEP senior group, students in grades 7 to 9, decided to generalize fibonometry to general Lucas sequences for their research. When the paper was almost ready, we discovered that this generalization is known. Our paper was well-written, and we decided to rearrange it as an expository paper, Fibonometry and Beyond. We posted it at the arXiv and submitted it to a journal. I hope the journal likes it too.

Consider the following Fibonacci trick. Ask your friends to choose any two integers, a and b, and then, starting with a and b, ask them to write down 10 terms of a Fibonacci-like sequence by summing up the previous two terms. To start, the next (third) term will be a+b, followed by a+2b. Before your friends even finish, shout out the sum of the ten terms, impressing them with your lightning-fast addition skills. The secret is that the seventh term is 5a+8b, and the sum of the ten terms is 55a+88b. Thus, to calculate the sum, you just need to multiply the 7th term of their sequence by 11.

If you remember, I run a program for students in grades 7 through 9 called PRIMES STEP, where we do research in mathematics. Last year, my STEP senior group decided to generalize the Fibonacci trick for their research and were able to extend it. If n=4k+2, then the sum of the first n terms of any Fibonacci-like sequence is divisible by the term number 2k+3, and the result of this division is the Lucas number with index 2k+1. For example, the sum of the first 10 terms is the 7th term times 11. Wait, this is the original trick. Okay, something else: the sum of the first 6 terms is the 5th term times 4. For a more difficult example, the sum of the first 14 terms of a Fibonacci-like sequence is the 9th term times 29.

My students decided to look at the sum of the first n Fibonacci numbers and find the largest Fibonacci number that divides the sum. We know that the sum of the first n Fibonacci numbers is F_n+2 – 1. Finding a Fibonacci number that divides the sum is easy. There are tons of cute formulas to help. For example, we have a famous inequality F_4k+3 – 1 = F_2k+2L_2k+1. Thus, the sum of the first 4k+1 Fibonacci numbers is divisible by F_2k+2. The difficult part was to prove that this was the largest Fibonacci number that divides the sum. My students found the largest Fibonacci number that divides the sum of the first n Fibonacci numbers for any n. Then, they showed that the divisibility can be extended to any Fibonacci-like sequence if and only if n = 3 or n has remainder 2 when divided by 4. The case of n=3 is trivial; the rest corresponds to the abovementioned trick.

They also studied other Lucas sequences. For example, they showed that a common trick for all Jacobsthal-like sequences does not exist. However, there is a trick for Pell-like sequences: the sum of the first 4k terms (starting from index 1) of such a sequence is the (2k + 1)st term times 2P_2k, where P_n denotes an nth Pell number.

You can check out all the tricks in our paper Fibonacci Partial Sums Tricks posted at the arXiv.

The famous 5-card trick begins with the audience choosing 5 cards from a standard deck. The magician’s assistant then hides one of the chosen cards and arranges the remaining four cards in a row, face up. Upon entering the room, the magician can deduce the hidden card by inspecting the arrangement. To eliminate the possibility of any secret signals between the assistant and the magician, the magician doesn’t even have to enter the room — an audience member read out the row of cards.

The trick was introduced by Fitch Cheney in 1950. Here is the strategy. With five cards, you are guaranteed to have at least two of the same suit. Suppose this suit is spades. The assistant then hides one of the spades and starts the row with the other one, thus signaling that the suit of the hidden card is spades. Now, the assistant needs to signal the value of the card. The assistant has three other cards than can be arranged in 6 different ways. So, the magician and the assistant can agree on how to signal any number from 1 to 6. This is not enough to signal any random card.

But wait! There is another beautiful idea in this strategy — the assistant can choose which spade to hide. Suppose the two spades have values X and Y. We can assume that these are distinct numbers from 1 to 13. Suppose, for example, Y = X+5. In that case, the assistant hides card Y and signals the number 5, meaning that the magician needs to add 5 to the value of the leftmost card X. To ensure that this method always works, we assume that the cards’ values wrap around. For example, king (number 13) plus 1 is ace. You can check that given any two spades, we can always find one that is at most 6 away from the other. Say, the assistant gets a queen of spades and a 3 of spades. The 3 of spades is 4 away from the queen (king, ace, two, three). So the assistant would hide the 3 and use the remaining three cards to signal the number 4.

I skipped some details about how permutations of three cards correspond to numbers. But it doesn’t matter — the assistant and the magician just need to agree on some correspondence. Magically, the standard deck of cards is the largest deck with which one can perform this trick with the above strategy.

Later, a more advanced strategy for the same trick was introduced by Michael Kleber in his paper The Best Card Trick. The new strategy allows the magician and the assistant to perform this trick with a much larger deck, namely a deck of 124 cards. But this particular essay is not about the best strategy, it is about the Cheney strategy. So I won’t discuss the advanced strategy, but I will redirect you to my essay The 5-Card Trick and Information, jointly with Alexey Radul.

63 years later, the 4-card trick appeared in Colm Mulcahy’s book Mathematical Card Magic: Fifty-Two New Effects. Here the audience chooses not 5 but 4 cards from the standard deck and gives them to the magician’s assistant. The assistant hides one of them and arranges the rest in a row. Unlike in the 5-card trick, in the 4-card trick, the assistant is allowed to put some cards face down. As before, the magician uses the description of how the cards are placed in a row to guess the hidden card.

The strategy for this trick is similar to Cheney’s strategy. First, we assign one particular card that the magician would guess if all the cards are face down. We now can assume that the deck consists of 51 cards and at least one of the cards in the row is face up. We can imagine that our 51-card deck consists of three suits with 17 cards in each suit. Then, the assistant is guaranteed to receive at least two cards of the same imaginary suit. Similar to Cheney’s strategy, the leftmost face-up card will signal the imaginary suit, and the rest of the cards will signal a number. I will leave it to the reader to check that signaling a number from 1 to 8 is possible. Similar to Cheney’s strategy, the assistant has an extra choice: which card of the two cards of the same imaginary suit to hide. As before, the assistant chooses to hide the card so that the value of the hidden card is not more than the value of the leftmost face-up card plus 8. It follows that the maximum number of cards the imaginary suit can have is 17. Magically, the largest possible deck size for performing this trick is 52, the standard deck of cards.

Last academic year, my PRIMES STEP junior group decided to dive deeper into these tricks. We invented many new tricks and calculated their maximum deck sizes. Our cutest trick is a 3-card trick. It is similar to both the 5-card trick and the 4-card trick. In our trick, the audience chooses not 5, not 4, but 3 cards from the standard deck and gives them to the magician’s assistant. The assistant hides one of them and arranges the other two in a row. The assistant is allowed to put some cards face down, as in the 4-card trick, and, on top of that, is also allowed to rotate the cards in two ways: by putting each card vertically or horizontally.

We calculated the maximum deck size for the 3-card trick, which is not 52, as for the 5- and 4-card trick, but rather 54. Still, this means the 3-card trick can be performed with the standard deck. The details of this trick and other tricks, as well as some theory, can be found in our paper Card Tricks and Information.

The famous 5-card trick begins with the audience choosing 5 cards from a standard deck. The magician’s assistant then hides one of these cards and arranges the remaining four cards in a row, face up. On entering the room, the magician can deduce the hidden card by inspecting the arrangement. To eliminate the possibility of secret signals between the assistant and the magician, the magician needn’t even enter the room — an audience member can call them and read out the row of cards.

We will not delve into the mechanics of the trick, which are widely available online. Instead, we will explore the information theory underlying it. Michael Kleber’s paper, The Best Card Trick, provides an information-theoretic argument that works as follows:

For a deck of N cards, the number of different messages the magician can receive is N(N-1)(N-2)(N-3). The magician must guess the hidden card, which is equivalent to determining the set of five cards chosen by the audience. The number of such sets is N choose 5. For the trick to work, the number of messages must not exceed the number of possible answers, leading to the inequality: (N choose 5) ≤ N(N-1)(N-2)(N-3). After some manipulation, we get that (N-4)/120 doesn’t exceed 1. This implies that the deck can have at most 124 cards. The bound turns out to be tight: as discussed in Kleber’s paper, the trick can still be performed with such a huge deck. The paper expands this argument to a trick with K instead of 5 cards and shows that the maximum deck size for such a trick is K! + K – 1.

Here, we want to present a more direct, intuitive argument. We will make the argument for the 5-card trick, which is easily generalizable to the K-card trick. The assistant has 5 ways to choose which card to hide and 24 ways to arrange the remaining four cards, so they only have 120 actions in any given situation. Ergo, the magician should only be able to extract 120 alternatives’ worth of information from knowing what action the assistant would take.

This is a bit fishy, because of course even with N > 120, the trick could happen to work sometimes. That is, if the magician tells the assistant the strategy by which they will guess the missing card, the assistant may, for some sets of 5 cards drawn even from a large deck, manage to show an arrangement of four that will lead the magician to guess correctly.

The crux of formalizing the argument is to move to the global view, but we can do that without additional computations. Consider the space of all states reachable by any strategy of the assistant. In our case, this is equivalent to ordered sequences of five cards, with the last face down. There are obviously (N-4)M of these, where M is the number of states the magician observes (four-card sequences, in our case), however many of those there are. When the assistant and the magician choose a strategy for the assistant, they make most of these impossible. Indeed, since the assistant always has exactly 120 options, after they have chosen one to take in each situation, we have exactly (N-4)M/120 states that remain possible with that strategy. For the trick to always work, this last expression must be no more than M; M cancels, saving us the trouble of computing it, and we are left with N-4 ≤ 120 as desired.

By the way, one of the authors of this essay, Tanya Khovanova, taught this trick to her PRIMES STEP students, who were students in grades 7 through 9. They found and studied interesting generalizations of this trick and wrote the paper Card Tricks and Information available at the arXiv. They studied many variations of the trick, including the ones where the assistant is allowed to put the cards face down. This interesting variation is outside the scope of this essay.

We would like to use as an example one of the tricks described in the paper: the K-card trick, where the assistant hides one card and arranges the rest in a circle. The implication is that when the audience member describes the arrangement to the magician, they describe the circle clockwise in any order. Our argument works here as follows. We count the number of the assistant’s actions: K ways to choose the hidden card and (K-2)! ways to arrange the cards in different circles up to rotation. Thus, the number of different actions is K(K-2)!. Hence, the deck size doesn’t exceed K(K-2)! + K – 1, as we can exclude the K-1 cards in the circle, as they aren’t hidden. Not surprisingly, this is the same formula as in the paper.

The puzzle In the Details by Derek Kisman is one of my favorite MIT Mystery Hunt puzzles of all times. I even wrote a blog post advertising it and another post with comments on the solution. This puzzle type became known as fractal word search.

In the standard word search, you have a grid of letters and a list of words. You need to find the words written in the grid in all eight directions. The unused letters provide the answer, or a clue to the answer, of the word search puzzle.

In the fractal word search, you have a grid of letters and a list of words. It looks like a regular word search, but you will not find all the words inside the given grid. The given grid is only a snapshot of the whole grid on some particular level k. To go to level k+1, you have to use replacement rules: each letter is replaced by a 2-by-2 block of letters. This creates a much bigger grid where you might find more words from the given list.

An interesting question is, where do you find the replacement rules? In Derek’s puzzle, the rules were not given, but the initial grid was level 2. So you could notice that this grid can be decomposed into 2-by-2 squares, and there are only 26 different squares, implying that each square corresponds to a letter. Assuming that replacing the 2-by-2 squares with correct letters will allow you to find more words from the list, you can decipher the replacement rules. This will allow you to get to level 1 as well as any other level. Small levels are easy to search, but on large levels, the grid gets so huge that it might not fit in the memory of a computer, or a million computers. That is why this puzzle was presented at the MIT mystery hunt, but not at any other puzzle hunt. It is quite difficult: one of the given words is hiding on level 86.

I liked the puzzle so much, I included it in one of my lectures. After I gave my talk at Brown University, a student, Klára Churá, approached me. She got as fascinated with the puzzle as I was. We ended up collaborating on the paper Fractal Word Search: How Deep to Delve. As the title suggests, we focused on finding the upper bound of the level where we could find a word of a given length. We had two parameters: the size of the alphabet n and the block size b used in the replacement rules.

For different reasons, the most interesting case is words of length 3. I will leave it to the reader to figure out why this is the most interesting case, or the reader can check our paper. We showed that any word of length 3 appears no later than level n³ + n² + 1.

When we posted the paper, I sent the link to Derek. He immediately wrote a program and showed that our bound is fairly tight. His code is available at GitHub. He created a configuration that puts a 3-letter word at depth LCM(a,b,c)+1, where a,b,c ≤ n-3. If n is even, this gives us a lower bound of (n-3)(n-4)(n-5) + 1. If n is odd, this gives us a lower bound of (n-4)(n-5)(n-6) + 1. In any case, asymptotically, it is very close to our upper bound.