Justin Pombrio

Algebra and Data Types

Addition, multiplication, and exponentiation model data types.

In math class you’ve done algebra, with addition and multiplication and exponentiation and polynomials like 1 + x + x². And while programming, you’ve worked with enums and structs and functions and lists. You probably thought these things were unrelated.

Surprise! They’re deeply related, and by the end of this post you’ll see how to use algebra to refactor your data types. The crowning example in this post will be finding an equivalent representation of red-black trees.

Table of contents:

Algebraic Data Types

If you already know why algebraic data types are called “algebraic”, and how they relate to addition and multiplication, feel free to skip this section and jump to “Refactoring with Algebra”. Otherwise, don’t be afraid: algebraic data types are very simple, and I’ll show examples below. All the code in this post will be in Rust.

To have algebraic data types, you need two things: “product types” and “sum types”.

Product Types

structs in Rust are product types:

// Structs are product types
struct Rectangle {
    x: i32,
    y: i32,
    width: u32,
    height: u32,

Rectangle is called a product type because to calculate the number of possible Rectangles, you multiply the number of possible values of its fields. There are 2^32 possible values for x, for y, for width, and for height, so there are 2^32 * 2^32 * 2^32 * 2^32 = 2^128 possible Rectangles. Let’s call this number, 2^128, the cardinality of Rectangle.

The same thing is true for tuples: the number of possible values of a tuple is the product of the number of possible values of its elements. So tuples are also considered product types:

// Also a product type
type Pos = (i32, i32);

Sum Types

We used a product type for Pos because it contains an x-coordinate and a y-coordinate. On the other hand, you use a sum type when you have one thing or another thing. So Rust enums are sum types.

For example, a vending machine has multiple states, and needs to store different information depending on which state it is in:

// This is a sum type
enum VendingMachineState {
    /// Just sitting around
    /// Someone has put money in.
    /// Store the total number of cents inserted so far.
    /// Someone bought an item, and we're dispensing it.
    /// Store the letter label of the item they bought.

(This is not meant to be a full featured implementation of a vending machine. For example, it doesn’t handle “someone put too much money in” or “help I’m out of quarters”.)

VendingMachineState is called a sum type because its number of possible values is the sum of the number of possible values of each of its options. There is 1 Idle value, 2^32 possible “amounts of money” in the MoneyInserted state, and 2^32 possible chars in the Dispensing state (because unlike in C, a Rust char is 4 bytes). So altogether, VendingMachineState has a cardinality of 1 + 2^32 + 2^32 = 1 + 2^33.

Notice that this is not a count of the number of plausible values of VendingMachineState. For example, the char will probably only ever be an ascii letter, and the “amount of money inserted in cents” should always be much less than u32::MAX. Instead, we are counting the number of possible values allowed by the type system.

Result and Option

Let’s look at the cardinality of two common built-in sum types in Rust: Result and Option.

enum Result<T, E> {

A Result<T, E> contains either a value of type T or a value of type E. Thus its cardinality is the sum of the cardinalities of T and E, which I’ll write as T + E.

enum Option<T> {

Since an Option<T> contains either a T or no data, its cardinality is 1 + T: all the possible values of T, plus the extra None option.

The Baffling Lack of Sum Types

To reiterate:


How, then, are you supposed to represent X or Y in a language that lacks sum types?

There are several different workarounds, depending on the language. Here are a couple:

As a programming languages person, this drives me bonkers. Do you know how long we’ve had sum types? Since the 70s! They are very easy to implement and to type check. And they’re both safer and ergonomically nicer than the alternatives.

If you’re ever designing a language, please, I beg you, give it sum types.


That’s my only rant this post, I promise.

Refactoring with Algebra

Now for the magic!

Just as you can refactor code by rewriting it in a way that looks different but does the same thing, you can refactor data type(s) by arranging their contents in a different way. For example, we could replace the x and y in our Rectangle type with a Pos tuple:

struct Rectangle {
    x: i32,
    y: i32,
    width: u32,
    height: u32,
type Pos = (i32, i32);

struct NewRectangle {
    pos: Pos,
    width: u32,
    height: u32,

This probably isn’t news to you. What you may not know is that you can use algebra to verify that this refactoring is correct!

The key insight is that for the refactoring to be correct, the information contained in the NewRectangle type must be the same as the information contained in the old Rectangle type. Therefore, the total number of possible values must remain the same. We can verify this with algebra:

= i32 * i32 * u32 * u32
= (i32 * i32) * u32 * u32
= Pos * u32 * u32
= NewRectangle

In general, two data types contain the same information if and only if the algebraic expressions for those data types are equal. There are a couple ways to make use of this:

In the rest of this post, we’ll work through the refactorings implied by a couple dozen laws of algebra, and see many applications of them.


The commutative law for multiplication says A * B = B * A. Remember that A * B as a data type is a pair (A, B) or alternatively a struct with fields of type A and B. Thus commutativity suggests that you can refactor a type by re-ordering its elements (thus winning the contest for most boring refactoring ever):

(String, DateTime) = (DateTime, String)
struct Pos {
    x: u32,
    y: u32,
struct Pos {
    y: u32,
    x: u32,

There’s also a commutative law for addition: A + B = B + A. Remember that A + B is an enum with two variants, one containing an A and one containing a B. So this law says, for example:

Result<T, E> = Result<E, T>

(Note that this isn’t good programming practice: semantically, the Ok and Err variants of a Result are not symmetric. For example, the ? operator treats them differently, and programmers expect that if there’s an error it goes in the Err variant and not in the Ok variant. However, the algebra only cares about the information content, and flipping a Result does keep the same information content.)


enum Color { Red, Yellow } = enum Color { Yellow, Red }


There are also associative laws. One for multiplication:

(A * B) * C = A * (B * C) = A * B * C

Which gives some equivalences between tuples:

((A, B), C) = (A, (B, C)) = (A, B, C)

And one for addition:

(A + B) + C = A + (B + C) = A + B + C

Which gives some equivalences between enums:

enum StopLight { Green, Yellow, Red }


enum DontGoColor { Yellow, Red }
enum StopLight { Green, DontGo(DontGoColor) }


enum MaybeGoColor { Green, Yellow }
enum StopLight { Red, MaybeGo(MaybeGoColor) }


So far we’ve only looked at boring algebraic laws that involved addition or multiplication. But distributivity involves both:

A * (B + C) = (A * B) + (A * C)

In terms of data types, we get a law involving both tuples/structs and enums:

(A, Result<T, E>) = Result<(A, T), (A, E)>

Or a more realistic example:

// A binary tree with data on both leaves and branches.
// A leaf has no children, and a branch has two children.

struct BinaryTree<Data> {
    data: Data,
    children: Children,
enum Children {
    Branch(Box<BinaryTree<Data>>, Box<BinaryTree<Data>>),


enum BinaryTree<Data> {
    Branch(Data, Box<BinaryTree<Data>>, Box<BinaryTree<Data>>),

This is a common refactoring. It is typically better to use the first form, since it doesn’t logically duplicate Data.

Subtraction as a Constraint

One interesting thing we can do is use subtraction to constrain a type.

A Go function that produces an answer T or an error E will return a pair (Option<T>, Option<E>). (I’m writing Go’s nil-able values as Rust-like Options for convenience.) In algebra:

  (Option<T>, Option<E>)
= (1 + T) * (1 + E)

However, programmers are expected to maintain the convention that exactly one of these two Options should be filled. Thus these two cases are invalid:

Let’s start with our full type and subtract out these two invalid cases, to see what’s left:

  (1 + T) * (1 + E) - 1 - T*E
= 1 + T + E + T*E - 1 - T*E
= T + E

This is the cardinality of Rust’s Result<T, E>! So Rust’s Result type allows exactly the valid Go results and no more, using the type system to enforce what is only a convention in Go.


So far we’ve seen big numbers like i32 = 2^32. But what about small numbers?

Counting Down: Two

2 would be the number for a type that has two possible values. Either one thing, or another thing, but nothing else. You know a type like this! It’s the friendly boolean:

enum Bool {

What can you say in algebra about the number two? One thing is A + A = 2A. This means:

Result<A, A> = (bool, A)

This is relevant to Rust’s standard library binary search function:

pub fn binary_search(&self, x: &T) -> Result<usize, usize>

The docs for this say:

If the value is found then Result::Ok is returned, containing the index of the matching element. […] If the value is not found then Result::Err is returned, containing the index where a matching element could be inserted while maintaining sorted order.

Since Result<usize, usize> is equivalent to (bool, usize), this function could also have returned both an index and a boolean specifying the meaning of that index. Though in this case I think the Result type is more clear.

Counting Down: One

1 would be the cardinality of a type that has one possible value.

In Rust, this is the unit type, written (). It’s like a tuple with 0 elements. It’s the most boring type. It only has one possible value, which is also written (). Rust stores it in literally 0 bytes:

println!("{}", std::mem::size_of::<()>());
// prints 0

Multiplicative Identity

One is the multiplicative identity. This is a fancy way of saying:

1 * A = A

In terms of types, this means:

((), A) = A

In other words, there’s no reason to include the (). It doesn’t add any information.

(Why, then, does Rust even have ()? One reason is that it’s the return type for functions that “don’t return anything”. In C or Java this would be written “void”.)

Counting Down: Zero?

What about zero? Zero is the sum of no things. It’s an enum with no variants:

enum Zero {}

Since there are no variants for it, you cannot construct a Zero. It has 0 possible values.

In Rust, this type is written ! and pronounced “the never type”.

Additive Identity

In algebra, zero is the additive identity of multiplication:

A + 0 = A

In data types, this means:

Result<A, !> = A

As explained in the Rust docs:

Since the Err variant contains a !, it can never occur. If the exhaustive_patterns feature is present this means we can exhaustively match on Result<T, !> by just taking the Ok variant. This illustrates another behaviour of ! - it can be used to “delete” certain enum variants from generic types like Result.

Multiplicative Absorbative

Zero plays a second role in algebra too. It’s the “absorbative” element of multiplication (aren’t these names great?):

A * 0 = 0

In data types, this means:

(A, !) = !

In other words, ! is contagious. There’s no way to construct it, so there’s no way to construct a tuple or struct that contains it.


We just saw some very small numbers: 2, 1, 0. And we’ve seen intermediate numbers, like char = 2^32. How about very large numbers?

Let’s take String. String has infinitely many possible values. We could say that the cardinality of String is , but isn’t really a number, and our algebra would fall apart if we tried to use it.

Instead we’re going to just… not simplify String. In the algebra, we’ll continue to call it String, and treat it as an indeterminate, never replacing it with a concrete number.

In fact, it’s useful to do this for more than just String. For example, you may want to avoid treating char and u32 as identical, even if they have the same number of possible values, since one is meant to store a unicode character, and the other a number. You can achieve this just by leaving char as char and not setting it equal to 2^32.

Summing up the numbers we’ve seen:

Data Type   Algebraic Expression
!           0
()          1
bool        2
u8          2^8, or just u8
char        2^32, or just char
String      String


What’s the algebraic expression for an array? Well, the array [A; n] contains n As. So its expression is A * A * ... * A = A^n.

Exponentiation Laws

And now we can get some refactoring rules for arrays from the laws of exponents!

Exponentiation is Repeated Multiplication

An array can be stored in a product type instead:

A^3 = A * A * A
// An array of three u8s
type Color = [u8; 3];

// Can also be represented as a struct with 3 fields:
struct Color {
    red: u8,
    green: u8,
    blue: u8,

Matrix Storage

(A^m)^n = (A^n)^m

This gives the equivalence between row-major and column-major storage for matrices:

type RowMajorMatrix<A, const M: usize, const N: usize>
  = [[A; N]; M];
// access with `matrix[row][col]`

type ColMajorMatrix<A, const M: usize, const N: usize>
  = [[A; M]; N];
// access with `matrix[col][row]`

Array Flattening

Of course, you can always just jam your matrix into one big array:

(A^m)^n = A^(m*n)
type FlatMatrix<A, const M: usize, const N: usize>
  = [A; M * N];
// access with `matrix[N * row + col]`

(I probably got some of these indices backwards… you get the idea, though.)

Array Splitting

If you have a long array, you can split it into a first part and a second part:

A^(m+n) = A^m * A^n
[A; 6] = ([A; 2], [A; 4])

More Array Splitting

An array of pairs is equivalent to a pair of arrays:

(A*B)^n = A^n * B^n
[(A, B); n] = ([A; n], [B; n])

This is the famous array of structs vs. struct of arrays equivalence.

Exponentiation with Small Numbers

We can find more rules by asking what happens if either the base or the exponent is a small number.

Short Arrays

An array of length 1 might as well not be an array:

A^1 = A
[A; 1] = A

Shorter Arrays

An array of length 0 contains no information at all:

A^0 = 1
[A; 0] = ()

The Rust compiler confirms:

println!("{}", std::mem::size_of::<[usize; 0]>());
// prints 0

Array of Units

An array of units also contains no information:

1^n = 1
[(); n] = ()

Rustc confirms again:

println!("{}", std::mem::size_of::<[(); 100]>());
// prints 0


There’s one last data type we will model: functions!

Functions are totally data. You can stick one in a variable:

fn main() {
    fn func(n: u8) -> bool { n > 0 }
    let var: fn(u8) -> bool = func;

(There is a variety of function types in Rust, due to its ownership model. Let’s completely ignore this fact.)

The whole trick of this blog post is to count the number of possible values of a data type. So how many distinct functions from u8 to bool are there?

There are infinitely many. For example:

This is, uh, not really what we were looking for. The u8-to-bool-ness is getting buried beneath the side effects.

The better question to ask is, how many pure, terminating functions from u8 to bool are there?

This is easier to answer: to specify such a function, you have to say what boolean it returns on input 0 and what boolean it returns on input 1and what boolean it returns on input 255. Altogether, that’s 2*2*...*2 = 2^256 possible functions.

In general, the cardinality of a function from A to B is B^A.

Exponentiation: now for Functions

Let’s go through some of the algebraic laws for exponentiation again, this time applying them to functions rather than arrays.


First off:

A^n = A^n

Well no duh. What am I getting at? While the two A^ns look the same, I mean the first one to be interpreted as the cardinality of fn(n) -> A, and the second one as the cardinality of [A; n]. This is saying that a function whose input has n possible values and whose output is A, can be represented as an array of n As.

This was the answer to a Google interview question! The question was to find the fastest way to count the number of set bits in a byte (i.e., the number of 1s in its binary representation). The answer was to realize that:

  fn count_ones(byte: u8) -> u8
= u8^u8
= u8^256
= [u8; 256]

In other words: if you really care about the speed of a function, you can store all 256 of its possible return values in a lookup table. Memory is cheap these days!


Currying is the idea that a function that takes two arguments can also be expressed as a function that takes the first argument and returns a function that takes the second argument:

A^(m*n) = (A^m)^n
// A function from `A` and `B` to `C`:
fn(A, B) -> C

// Can also be expressed as a function from `A` to a function from `B` to `C`:
fn(A) -> (fn(B) -> C)

Function Splitting

If you have a function that takes an enum (like Result), you can split it into functions that each handle one of the variants of the enum (like one for the Ok and one for the Err):

C^(A+B) = C^A * C^B
fn(Result<A, B>) -> C
(fn(A) -> C, fn(B) -> C)

(This is analogous to the way you can emulate sum types in OOP: instead of having a single function that acts on the sum A + B, you have two methods—one for handling A, and one for handling B—that live in different classes.)

More Function Splitting

A function that returns a pair is equivalent to a pair of functions:

(B*C)^A = B^A * C^A
fn(A) -> (B, C)
(fn(A) -> B, fn(A) -> C)

Exponentiation with Small Numbers

Again, we can find more rules by asking what happens if either the base or the exponent is a small number.

Short Functions

A function that just takes the unit value (or equivalently no arguments) might as well not be a function:

A^1 = A
fn(()) -> A = A

Remember that exponentiation models pure, terminating functions. So this law is saying that a function that takes a unit argument can only be useful (beyond its single return value) if it has a side effect.

Shorter Functions

A function that takes in a “never” argument is equivalent to unit:

A^0 = 1
fn(!) -> A  =  ()

[UPDATE] I wasn’t really sure about this, but a friend clarified it. We’re classifying functions based on their behavior. How many functions are there from ! to A, counting functions as equal if they have the same behavior? Why, there’s just one. They’re all the same, because you can’t call such a function.

Void Functions

A function that doesn’t return anything is sometimes called a void function (because it returns “void” in C). Void functions can’t do anything unless they have a side effect:

1^A = 1
fn(A) -> ()  =  ()


We’ve looked at arrays, but how about lists? Unfortunately, there’s a lot of confusing terminology around lists and their representations in various languages, so let me be clear. By “list”, I mean a finite but arbitrarily long (and resizable) sequence of elements of the same type. Rust’s most commonly used list type is called Vec<A>. It’s backed by an array that’s re-allocated as needed, but that’s an implementation detail.

How many possible values does a Vec<A> have? This is easiest to answer if you think of a Vec<A> like a linked list (it’s the same information content either way). It’s either empty, or it has an element and another list:

Vec<A> = 1 + A*Vec<A>

Let’s substitute in the right hand side of the equation… repeatedly:

Vec<A> = 1 + A*Vec<A>
       = 1 + A*(1 + A*Vec<A>)
       = 1 + A + A^2*Vec<A>
       = 1 + A + A^2*(1 + A*Vec<A>)
       = 1 + A + A^2 + A^3*Vec<A>
       = ...
       = 1 + A + A^2 + A^3 + ...

This is saying that a list of As can have 0 As (because it’s empty), or 1 A (because it has length 1), or 2 As (because it has length 2), etc.

It’s tedious to work with infinite polynomials like this, though, so let’s try to find a closed-form solution:

Vec<A> = 1 + A*Vec<A>
Vec<A> - A*Vec<A> = 1
(1 - A)*Vec<A> = 1
Vec<A> = 1/(1 - A)

There we go! A list of As has 1/(1-A) possible values.

“But Justin,” you say, “what does division mean? You haven’t shown any data type that it models.”

“And Justin,” you continue, “what if A is bool”? Then 1/(1-A) = 1/(1-2) = -1. What are negative numbers supposed to be?”

Shhh child, do not fear. One need not understand the formula for it to be true.

“You don’t know either, do you?”

Fear leads to anger. Anger leads to hate. Hate leads to algebraically manipulating infinite polynomials because you weren’t willing to simplify them.

Short Lists

And the 1/(1-A) list formula does work.

Say we have a short list, with fewer than 3 elements. Equivalently, this is a list subject to the constraint that it does not have 3 or more elements. We can represent all lists with 3 or more elements as A^3 * Vec<A>: you store the first three elements, plus a list for the rest. Therefore our short list has cardinality:

Vec<A> - A^3 * Vec<A>

I.e., it’s a list (Vec<A>) subject to the constraint (-) that it does not have 3 or more elements (A^3 * Vec<A>).

Simplifying with algebra:

  Vec<A> - A^3 * Vec<A>
= 1/(1 - A) - A^3/(1 - A)
= (1 - A^3) / (1 - A)
= (1 - A) * (1 + A + A^2) / (1 - A)
= 1 + A + A^2

In other words, it has length 0, 1, or 2. Not a startling insight, but it shows that the list formula works.


Here’s a more interesting example. The zip function takes two iterators and produces an iterator of pairs, by consuming from both iterators at once until one of them stops producing values. If we assume the iterators are finite (the methods in this blog post don’t work on infinite data structures), then they’re equivalent to lists, so we can use the 1/(1-A) list formula for them.

Since the zip function might not fully consume its input iterators, it pretty clearly throws some information away. Let’s figure out what is lost, exactly. zip(Vec<A>, Vec<B>) returns Vec<A*B>, but if it were lossless, it would have to return some additional information. Let’s call this extra information X. Then Vec<A>*Vec<B> = X*Vec<A*B>. Solving for X and simplifying:

X = Vec<A>*Vec<B> / Vec<A*B>
  = [1/(1-A) * 1/(1-B)] / [1/(1-A*B)] 
  = [1/(1-A)(1-B)] / [1/(1-A*B)] 
  = (1-A*B) / (1-A)(1-B)
  = 1 + A/(1-A) + B/(1-B)   // this step's tricky
  = 1 + A*Vec<A> + B*Vec<B>

So if you wanted zip to be a lossless function, then not only would it have to produce Vec<A*B>, it would also have to preserve the unused elements of its inputs:

Alternating Lists

Here’s another example of refactoring with algebra.

Say we have a list of alternating elements A and B, that could start with either A or B. We could represent that with this Rust type:

enum Alternating {

enum ListA {
    Cons(A, Box<ListB>),

enum ListB {
    Const(B, Box<ListA>),

There’s one subtlety here. I missed it the first time I wrote this section, but the algebra made it visible.

These definitions give two ways to express an empty list: either an empty StartWithA or an empty StartWithB. To avoid having two ways to represent the same information, we should pick one representation, and declare the other one illegal. So let’s say that an empty list should be represented by Alternating::StartWithA(ListA::Empty), and that Alternating::StartWithB(ListB::Empty) is illegal.

Then the cardinalities are:

Alternating = ListA + ListB - 1  // minus 1 for the illegal state
ListA = 1 + A*ListB
ListB = 1 + B*ListA

Now to do some algebra, to find another representation! First, substitute ListB into the equation for ListA, so that we can solve for ListA:

ListA = 1 + A*ListB
ListA = 1 + A*(1 + B*ListA)
ListA = 1 + A + A*B*ListA
ListA(1 - A*B) = 1 + A
ListA = (1 + A)/(1 - A*B)

Then use that to solve for Alternating:

  = ListA + ListB - 1
  = ListA + 1 + B*ListA - 1
  = ListA + B*ListA
  = (1 + B) * ListA
  = (1 + B) * (1 + A) / (1 - A*B)
  = (1 + B) * (1 + A) * 1/(1 - A*B)

Wow! What is this type? Remember that 1 + X means Option<X> and 1/(1 - X) means Vec<X>. Putting this together, and helpfully describing the fields, gives:

struct Alternating<A, B> {
    /// If the sequence starts with a B, it's stored here
    initial_b: Option<B>,
    /// If the sequence ends with an A, it's stored here
    final_a: Option<A>,
    /// The rest of the elements, in order:
    middle_elements: Vec<(A, B)>,

That’s a nicer representation, isn’t it? Three types became one, the linked-lists became a Vec, and the illegal state vanished.

Binary Trees

One last magic trick. A Red-black tree is a kind of balanced binary tree. To stay balanced, it considers every node to be either red or black, and maintains these invariants:

Say that the leaves have type X, and the non-leaf nodes have type Y. Then we can capture the first three invariants in Rust types:

enum Node<X, Y> {
    Red(RedNode<X, Y>),
    Black(BlackNode<X, Y>),

enum BlackNode<X, Y> {
    Leaf {
        leaf_data: X,
    Branch {
        branch_data: Y,
        child_0: Box<Node<X, Y>>,
        child_1: Box<Node<X, Y>>,

struct RedNode<X, Y> {
    branch_data: Y,
    child_0: Box<BlackNode<X, Y>>,
    child_1: Box<BlackNode<X, Y>>,

We can use algebra to find an alternate form for red-black trees.

First, translate the Rust types to algebraic equations:

// N: Node
// R: RedNode
// B: BlackNode

N = R + B
B = x + y*N*N
  = x + y*(R + B)*(R + B)
R = y*B*B

But remember that this didn’t capture the last invariant, that “every path from the root to a leaf crosses the same number of black nodes”. In order to capture it, we’ll need to index the types R and B by the number of black nodes crossed. So I’ll write Bn for “the type of a black node, such that every path from it to its leaves crosses n black nodes (including itself). Using this, we can get the full equations for a red-black tree:

B1 = x    // n=1 -> must be a leaf
B{n+1} = y*(Bn + Rn)*(Bn + Rn)
Rn = y*Bn*Bn

Now that the equations are complete, we can simplify them. With a little algebra we can substitute away Rn, so that there’s only one remaining type:

B1 = x  // this stays the same

B{n+1} = y * (Bn + Rn) * (Bn + Rn)
       = y * (Bn + y*Bn*Bn) * (Bn + y*Bn*Bn)
       = y * Bn * (1 + y*Bn) * Bn * (1 + y*Bn)
       = y * Bn * Bn * (1 + y*Bn) * (1 + y*Bn)

We can translate this back into Rust. There’s no good way to keep the n parameter, so we’ll erase it but remember that, per the last equation, the n value of a child is always 1 less than its parent’s. Thus the tree now has constant depth (given by n). So erasing n we have:

enum Node<X, Y> {
    Leaf {
        leaf_data: X,
    Branch {
        branch_data: Y,
        child_0: Box<Node<X, Y>>,
        child_1: Box<Node<X, Y>>,
        child_2: Option<(Y, Box<Node<X, Y>>)>,
        child_3: Option<(Y, Box<Node<X, Y>>)>,

So we’ve found another representation of red-black trees!

Is it useful? It at least warranted a prominent mention on the Wikipedia page for red-black trees. This section starts out:

A red–black tree is similar in structure to a B-tree of order 4, where each node can contain between 1 and 3 values and (accordingly) between 2 and 4 child pointers.

By “values”, it means “data in nodes”, i.e. our Ys. And indeed the Branch variant has 1-3 Ys and (accordingly) 2-4 Node<X, Y>s.

So that’s how you can use algebra to find another representation of a data structure, even one with complicated invariants like a red-black tree.


I don’t know about you, but I was excited to discover just how extensively you can model data types with algebra. This post covered a lot, so here’s a cheat sheet to remember all the algebraic laws and what they say about types:

Cheat Sheet

To read more in this vein, you might want to check out:

Discussion on lobste.rs

March 11, 2021