Blog

Brandon Rozek

Photo of Brandon Rozek

PhD Student @ RPI studying Automated Reasoning in AI and Linux Enthusiast.

Corecursion, Unfold and Infinite Sequences

Published on

Updated on

4 minute reading time

Recursion takes a large problem and breaks it down until it reaches some base cases. One popular example, is the factorial function.

def fact(x: Int): Int =
    if x == 0 then
        1
    else if x == 1 then
        1
    else
        x * fact(x - 1)

Though we can similarly arrive at the answer by starting at the base case 1 and multiplying until we reach x. This is called co-recursion.

1 * 2 * ... * x

Unfold allows us to create sequences given some initial state and a function that takes some state and produces a value for the sequence. For the factorial function, we want to keep track of in our state the last factorial computed and the current index. (lastFact, currInd).

Therefore, our initial state is (1, 0).

val fact_seq = () => Iterator.unfold((1, 0))((x, y) => Some(
    x, // currFact
    (x * (y + 1), y + 1) // (nextFact, nextInd)
))

val fact = (x: Int) => fact_seq().take(x + 1).toList.last

Let’s trace an execution of fact(4).

fact(4)
Iterator.unfold((1, 0))((x, y) => Some((x, (x * (y + 1), y + 1)))).take(5).toList.last
States: (1, 0) -> (1, 1) -> (2, 2) -> (6, 3) -> (24, 4).....
[1, 1, 2, 6, 24, ...].take(4).toList.last
[1, 1, 2, 6, 24].last
24

Now why is this useful when maybe the recursive version can seem cleaner? Co-recursion and in turn unfolding can help remove redundancies. Let’s look at the Fibbonaci sequence for an example. The recursive version would be as follows:

def fib(n : Int): Int =
    if n == 0 then
        0
    else if n == 1 then
        1
    else
        fib(n - 1) + fib(n - 2)

Now let’s trace through an execution of fib(4)

fib(4)
fib(3) + fib(2)
(fib(2) + fib(1)) + fib(2)
((fib(1) + fib(0)) + fib(1)) + fib(2)
((1 + fib(0)) + fib(1)) + fib(2)
((1 + 0) + fib(1)) + fib(2)
(1 + fib(1)) + fib(2)
(1 + 1) + fib(2)
2 + fib(2)
2 + (fib(1) + fib(0))
2  + (1 + fib(0))
2 + (1 + 0)
2 + 1
3

Notice how there are many redundant calculations, for example fib(2) is evaluated twice separately in line 3 above.

Now lets look at how unfold helps. For our state, we need to keep track of the last two evaluations. Therefore, we can represent our state as (currentAnswer, nextAnswer).

val fib_seq = () => Iterator.unfold((0, 1))((x, y) => Some(x, (y, x + y)))
val fib: (Int => Int) = (n) => fib_seq.take(n + 1).toList.last

Tracing through fib(4)

fib(4)
Iterator.unfold((0, 1))((x, y) => Some(x,(y, x + y))).take(5).toList.last
State: (0, 1) -> (1, 1) -> (1, 2) -> (2, 3) -> (3, 5)
[0, 1, 1, 2, 3, ...].take(5).toList.last
[0, 1, 1, 2, 3].last
3

Small Unfold Examples

To get a better handle of unfold. Here are three examples:

(1) Build an iterator from start to infinity with a step size of step

Built-in way in Scala:

Iterator.from(start, step)

Using unfold

val fromStep: ((Int, Int) => Iterator[Int]) = (n, step) => Iterator.unfold(n)(x => Some((x, x + step)))

(2) Build an infinite sequence of even numbers

Using from and map:

val evens = Iterator.from(0).filter(_ % 2 == 0)

Using fromStep in (1)

val evens = fromStep(0, 2)

Using unfold

val evens = Iterator.unfold(0)(x => Some((x, x + 2)))

(3) Build a countdown from $n$ to $0$

Notice how the function within unfold needs to return an Option. If the returned option is None then the sequence is terminated.

val countdown = (n: Int) => Iterator.unfold(n)(x => if x == -1 then None else Some((x, x - 1)))

(4) Repeat an element forever

For this example, we don’t need to carry any state throughout hte computation.

val repeat: (Int => Iterator[Int]) = (n) => Iterator.unfold(None)(_ => Some(n, None))

Recursive Sequences

In the past, I’ve written about analyzing sequences from real analysis within Haskell. Within it, I was looking at the following sequence: $$ f(1) = 1, f(2) = 2, f(n) = \frac{1}{2}(f(n - 2) + f(n - 1)) $$ The technique I described in that post is to build out the function f and then map it to the natural numbers. In Scala that would look like:

val f: (Int => Double) = n => if n == 1 then 1.0 else if n == 2 then 2.0 else 0.5 * (f(n - 2) + f(n - 1))
val f_sequence = Iterator.from(1).map(f)

However as mentioned in prior in this post, this methodology is suboptimal since there will be many repeated computations.

Corecursion and unfold comes to the rescue again. For recursive sequences, we can make the state the base cases (1.0, 2.0).

val f_sequence = () => Iterator.unfold((1.0, 2.0))((x1, x2) => Some(
    x1,
    (x2, 0.5 * (x1 + x2))
))

We can get a good guess at where this sequence converges by looking at the $100^{th}$ element.

f_sequence().take(100).toList.last
// 1.6666666666666665

If you want to learn more about unfold or see a different take, then the following two blog posts helped me craft this one:

https://blog.genuine.com/2020/07/scala-unfold/

https://john.cs.olemiss.edu/~hcc/csci555/notes/FPS05/Laziness.html

Reply via Email

Have you published a response to this? :