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May 10, 2019

# Operating on Infinite Lists

Today we're going to step through one of my favorite exercises in composing functions - building and operating on infinite streams of data.

The question we'll be answering is: How might you represent the following operations?

nums
// =>  1, 2, 3, 4, ...
double(nums)
// =>  2, 4, 6, 8, ...
fib
// =>  0, 1, 1, 2, 3, 5, 8, ...

This post contains code written in JavaScript, but the ideas here apply to any language with functions and lists.

## The Finite

We can build a finite list in JavaScript like so:

const ls = [1, 2, 3, 4, 5]
// => [1, 2, 3, 4, 5]

We can access any element inside of it as well as its length:

ls[0]
// => 1
ls[10]
// => undefined
ls.length
// => 5

Alternatively, we can self-referentially represent lists as an element, followed by a list. This is commonly referred to as a "linked list." (You're welcome to scroll to the next section if this is familiar to you. You're welcome either way I guess, I'm not a cop.)

In this data structure we'll use an object with a first field for the element and a rest field for the rest of the list. We'll use null for the empty list - where our list ends.

const ll = {
first: 0,
rest: {
first: 1,
rest: {
first: 2,
rest: null,
},
},
}

This is conceptually simpler, but pretty awkward (imagine if you had to use this instead of [...]!). Still, we can access elements inside of it:

ll.first
// => 0
ll.rest.first
// => 1
ll.rest.rest.first
// => 2

And we can compute its length using recursion (where our function is defined in terms of itself):

function length(ll) {
if (ll === null) {
// empty list has length 0
return 0
} else {
// its length is 1 (the item)
// plus the length of the rest
return 1 + length(ll.rest)
}
}
length(ll)
// => 3

## The Infinite

Conceptually, an infinite list is a lot like a finite list. We'll want a way to get an item, and a way to get the rest. Let's kick things off by building a stream of ones:

1, 1, 1, 1, ...

Our first will be a single piece of data in our stream - in this case, the number 1.

const ones = {
first: 1,
rest: ???,
}

In our linked list above, our rest was also a linked list. Similarly, the rest of our stream will be... a stream!

const ones = {
first: 1,
rest: ones,
}

You'll quickly find however, that we can't just define a variable in terms of itself. In JavaScript, we'll receive the following from our console:

ReferenceError: ones is not defined

What can we define in terms of itself? Functions! Remember length from above?

function length(ll) {
if (ll === null) {
return 0
} else {
return 1 + length(ll.rest)
}
}

So...let's switch ones to be a function:

function ones() {
return {
first: 1,
rest: ones(),
}
}

But, we're not in the clear just yet:

> ones()
RangeError: Maximum call stack size exceeded
at ones (repl:1:14)
at ones (repl:4:11)
at ones (repl:4:11)

Uh oh. When creating our stream, ones is eagerly making subsequent calls to itself over and over again - never coming up for air before our JavaScript console lets us know we probably messed up.

Instead, let's make our implementation lazier by returning the ones function and calling it when we need it.

function ones() {
return {
first: 1,
rest: ones,
}
}

ones()
// => { first: 1, rest: [Function: ones] }

No infinite loops! We can create a ones stream, and gather elements from it like so:

ones().first
// => 1
ones().rest().first
// => 1
ones()
.rest()
.rest().first
// => 1

## Defining and Building Streams

Taking a step back, we can define a stream as a function which returns two things:

(1) an item
(2) a stream

We can define a twos stream similarly to our ones:

function twos() {
return {
first: 2,
rest: twos,
}
}

twos().first
// => 2
twos().rest().first
// => 2
twos()
.rest()
.rest().first
// => 2

Still, these streams don't seem particularly useful. One potential smell is our streams are functions, but so far they haven't taken any arguments. What if they did?

Let's kick things off with a stream of the natural numbers.

function nums(n) {
return {
first: n,
// Remember, rest is a stream,
// and streams are functions.
rest: () => nums(n + 1),
}
}

nums(1).first
// => 1
nums(1).rest().first
// => 2
nums(1)
.rest()
.rest().first
// => 3
nums(999).first
// => 999

An initial argument for nums doesn't seem particuarly useful, so let's use a default.

function nums(n = 1) {
return {
first: n,
// Remember, rest is a stream,
// and streams are functions.
rest: () => nums(n + 1),
}
}

nums().first
// => 1

All these .rest()/.first chains are becoming cumbersome, so let's take a second and define a function to grab the first n elements of a stream.

function take(stream, n) {
if (n <= 0) {
return []
} else {
const { first, rest } = stream()
return [first].concat(take(rest, n - 1))
}
}

take(nums, 5)
// => [1, 2, 3, 4, 5]

Exercise: Write a function to grab the nth item of a stream

nth(nums, 100)
// => 101
Solution
function nth(stream, n) {
if (n < 0) return null

const {first, rest} = stream()
if (n === 0) {
return first
} else {
return nth(rest, n - 1)
}
}


At this point we have a stream with some interesting data in it.

Now it's your turn. Try the following examples on your own machine. The solutions can be expanded if you get stuck (and it's okay if you do!)

Define a stream which returns the odd numbers.

take(odds, 5)
// => [1, 3, 5, 7, 9]
Solution
function odds(n = 0) {
return {
first: 2 * n + 1,
rest: () => odds(n + 1),
}
}


Define a stream which returns the square numbers.

take(squares, 5)
// => [1, 4, 9, 16, 25]
Solution
function squares(n = 1) {
return {
first: n * n,
rest: () => squares(n + 1),
}
}


Define a stream which loops between 0 and 1.

take(flipFlop, 5)
// => [0, 1, 0, 1, 0]
Solution
function flipFlop(n = 0) {
return {
first: n % 2,
rest: () => flipFlop(n + 1),
}
}


Right on. Feel free to take a break, refill your water, and we'll continue onto the next section - writing functions to operate on our streams.

## Streams Out / Streams In

Now that we're comfortable creating streams, let's start writing functions that can create streams for us.

To kick things off, let's write a function that takes in a number and returns a stream of that number.

function fixed(n) {
// Streams are functions
return () => ({
// Which return an item...
first: n,

// ...and a stream
rest: fixed(n),
})
}

const sevens = fixed(7)
take(sevens, 5)
// => [7, 7, 7, 7, 7]

Our pattern here is a little different than the ones and twos from earlier - in particular we're returning a function instead of an object with first and rest.

Additionally, our value for rest is a little simpler since fixed(n) returns a function (we don't need to make a new one inline like before).

For our next example, let's recall our definitions for odds, squares, and flipFlop. Specifically, just the first and rest parts of each.

odds
first: 2 * n + 1,
rest: () => odds(n + 1),

squares
first: n * n,
rest: () => squares(n + 1),

flipFlop
first: n % 2,
rest: () => flipFlop(n + 1),

Interestingly, our three streams share the same rest! And the first is just a function of n.

Let's materialize this some more, by building a function funcStream to generalize our three examples.

function funcStream(f, n = 0) {
return () => ({
first: f(n),
rest: funcStream(f, n + 1),
})
}

take(funcStream(n => 2 * n + 1), 5)
// => [1, 3, 5, 7, 9]
take(funcStream(n => n * n), 5)
// => [0, 1, 4, 9, 16]
take(funcStream(n => n % 2), 5)
// => [0, 1, 0, 1, 0]

Not bad - but we can do better. funcStream still needs to keep track of its own n and update it appropriately, but nums can do that for us. What if our function operated on nums, or any stream for that matter?

We can think of this as a "map" equivalent for streams. Here's how we might do it:

function map(f, stream) {
return () => {
const { first, rest } = stream()

return {
first: f(first),
rest: map(f, rest),
}
}
}

take(map(n => 2 * n, nums), 5)
// => [0, 2, 4, 6, 8]

Using this, we can define functions to operate on streams:

const double = stream => map(n => 2 * n, stream)
take(double(nums), 5)
// => [0, 2, 4, 6, 8]

Better yet, these functions can compose.

take(double(double(nums)), 5)
// => [0, 4, 8, 12, 16]

Let's take stock. nums is an infinite stream of numbers, and we're able to apply a function to it. Under the hood, this function is lazily invoked when we need it - such as when we display the output with take.

But mapping isn't the only thing we do with finite lists, so it shouldn't be the only thing we do with streams. What if we wanted to filter a stream?

take(filter(n => n % 3 > 0, nums), 6)
// => [1, 2, 4, 5, 7, 8]

When filtering a stream, we'll grab it's first and rest, then test first. If the test passes (and the function passed into filter returns true), we'll make sure to include that in the stream.

If the test does not pass, we'll just filter the tail and ignore the first. Let's turn this into code:

function filter(f, stream) {
const { first, rest } = stream()
if (f(first)) {
// Streams are functions
return () => ({
first: first,
rest: filter(f, rest),
})
} else {
// Ignore first, filter rest
return filter(f, rest)
}
}

And just like that, we now have a filter that operates on an infinite stream of data.

take(filter(n => n % 2 > 0, nums), 5)
// => [1, 3, 5, 7, 9]
take(filter(_ => true, nums), 5)
// => [0, 1, 2, 3, 4]
take(filter(_ => Math.random() < 0.05, nums), 5)
// => [28, 31, 33, 37, 54]

## Closing Exercises

Before you go, try your hand at the following exercises.

Using map, create a FizzBuzz stream.

take(fizzbuzz, 5)
// => [1, 2, 'Fizz', 4, 'Buzz']
Solution
const fizzbuzz = map(n => {
if (n % 15 === 0) {
return 'FizzBuzz'
} else if (n % 3 === 0) {
return 'Fizz'
} else if (n % 5 === 0) {
return 'Buzz'
} else {
return n
}
}, nums)


Using map and nums, write a function that takes two values and produces a stream that toggles between them.

take(toggle("hi", "ho"), 5)
// => ['hi', 'ho', 'hi', 'ho', 'hi']
Solution
function toggle(a, b) {
return map(
// nums starts at 1
n => (n - 1) % 2 === 0 ? a : b,
nums
)
}


Using map and nums, write a function that takes an array of values and produces a stream that cycles between them.

take(cycle(["a", "b", "c"]), 5)
// => ['a', 'b', 'c', 'a', 'b']
Solution
function cycle(items) {
return map(
// nums starts at 1
n => items[(n - 1) % items.length],
nums
)
}


Write a function that takes two streams and interleaves them.

take(interleave(fixed(0), fixed(9)), 5)
// => [0, 9, 0, 9, 0]
Solution
function interleave(a, b) {
const aPair = a()
const bPair = b()

return () => ({
first: aPair.first,
rest: () => ({
first: bPair.first,
rest: interleave(aPair.rest, bPair.rest),
})
})
}


Write a function which accepts a stream and returns a new stream whose values are a running total of the original stream.

For nums: return 1, then 1 + 2, then 1 + 2 + 3, then 1 + 2 + 3 + 4, etc.

take(total(nums), 6)
// => [1, 3, 6, 10, 15, 21]

Hint: give total a second argument that defaults to 0

Solution
function total(stream, value = 0) {
return () => {
const pair = stream()
return {
first: value + pair.first,
rest: total(pair.rest, value + pair.first),
}
}
}


Write reduce for streams.

reduce should take a 2-argument accumulator function, an initial value, and a stream.

take(reduce((acc, value) => acc + value, 0, nums), 6)
// => [1, 3, 6, 10, 15, 21]
take(reduce((acc, value) => acc * value, 1, nums), 6)
// => [1, 2, 6, 24, 120, 720]

Hint: Generalize your solution for total.

Solution
function reduce(fn, init, stream) {
return () => {
const pair = stream()
const newValue = fn(init, pair.first)
return {
first: newValue,
rest: reduce(fn, newValue, pair.rest),
}
}
}


Using map and reduce, create a stream of the Fibonacci sequence.

take(fib, 8)
// => [0, 1, 1, 2, 3, 5, 8, 13]

Hint #1: Your initial value should be the first two values of the sequence.

Hint #2: You do not need to use the input stream's values.

Solution
function fib() {
return {
// reduce will consume the first
// 0, so we'll manually put it
// at the front
first: 0,
rest: map(
pair => pair[0],
reduce(
([a, b], _) => [b, a + b],
[0, 1],
fixed('foobar')
)
)
}
}


## 🎉🎉

Using nothing but functions and some JavaScript objects, we were able to create and modify infinite sequences of data - and we did it all without making our computer cry!

Thanks for reading this post, and an extra special thanks if you took the time to go through the exercises I put together after lots of editing. I really appreciate it. If not, you may want to try them out on a rainy day, I bet you'll learn something!

It would be a mistake to leave you without some links for further reading.

That's all I have for you now. Thanks again for reading, feel free to share on social media, and come work with me at Stripe (we're hiring almost everywhere!)