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Brandon Rozek

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

Deep Recursion

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3 minute reading time

In functional programming, we often look at a list in terms of its head (first-element) and tail (rest-of-list). This allows us to define operations on a list recursively. For example, how do we sum a list of integers such as [1, 2, 3, 4]?

def sum(l : List[Int]): Int =
    if l.size == 0 then
        0
    else if l.size == 1 then
        l.head
    else
        l.head + sum(l.tail)

We later learn that the fold version is more compact.

l.foldLeft(0)(_ + _)

The big question though, is how do we write this function if we allow lists to be arbitrarily nested? One example of this is the list [[1, 2, [3, 4]], 5, [[6, 7], 8]]

Deep Recursion

To accomplish this, we need to make use of deep recursion. At its essence, we change the previous program so that it also recurses on the head of the list as well since that may be a list.

def deep_sum(l: Int | Matchable): Int =
    if l.isInstanceOf[Int] then
        l.asInstanceOf[Int]
    else
        val ll = l.asInstanceOf[List[Int | Matchable]]
        if ll.size == 0 then
            0
        else if ll.size == 1 then
            deep_sum(ll.head)
        else
            deep_sum(ll.head) + deep_sum(ll.tail)

Lets trace through an example [[1], 2]

deep_sum([[1], 2])
deep_sum([1]) + deep_sum([2])
deep_sum(1) + deep_sum([2])
1 + deep_sum([2])
1 + deep_sum(2)
1 + 2
3

Deep Recursion via Fold

Similar to shallow recursion, we can use the foldLeft function to help clean up the code a little:

def deep_sum(l : Int | Matchable): Int =
    if l.isInstanceOf[Int] then
        l.asInstanceOf[Int]
    else
        val ll = l.asInstanceOf[List[Int | Matchable]]
        ll.foldLeft(0)((c, n) => c + deep_sum(n))

In the above fold, c contains the current partial result (of type Int) which we can then add the recursive result of the next element of the list.

Let’s trace through an example [[1], 2]

deep_sum([[1], 2])
[[1], 2].foldLeft(0)((c, n) => c + deep_sum(n))
(0 + deep_sum([1])) + deep_sum(2)
(0 + [1].foldLeft(0)((c1, n1) => c1 + deep_sum(n1))) + deep_sum(2)
(0 + (0 + deep_sum(1))) + deep_sum(2)
(0 + (0 + 1)) + deep_sum(2)
(0 + 1) + deep_sum(2)
1 + deep_sum(2)
1 + 2
3

Deep Recursion via Fold/Map

In the prior example, the deep recursion and the reduction logic were combined within the same anonymous function. We can separate this out by making use of map.

def deep_sum(l: Int | Matchable): Int = 
    if l.isInstanceOf[Int] then
        l.asInstanceOf[Int]
    else
        val ll = l.asInstanceOf[List[Int | Matchable]]
        l.map(deep_sum).foldLeft(_ + _)

Intuitively, the map will apply deep_sum to each element of the list and returns an Int for each element as that’s the return type of deep_sum. Once we have our list of integers, we can perform the fold to reduce it to a single sum.

Lets trace through an example [[1], 2]

deep_sum([[1], 2])
[deep_sum([1]), deep_sum(2)].foldLeft(0)(_ + _)
[[deep_sum(1)].foldLeft(0)(_ + _), deep_sum(2)].foldLeft(0)(_ + _)
[[1].foldLeft(0)(_ + _), deep_sum(2)].foldLeft(0)(_ + _)
[(0 + 1), deep_sum(2)].foldLeft(0)(_ + _)
[1, deep_sum(2)].foldLeft(0)(_ + _)
[1, 2].foldLeft(0)(_ + _)
(0 + 1) + 2
1 + 2
3
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