With Pith

Ethan Petuchowski

A Functional Union of Two Sets

I’m in the 3rd week of Functional Programming Principles in Scala course on Coursera, taught by Martin Odersky, the creator of Scala. The class has been surprisingly awesome so far, but one thing above all struck me as impossible: the “functional” implementation of the union of two immutable binary search trees.

def union(other: IntSet) = ((left union right) union other) incl elem


After an intensive investigation, I eventually determined that this operation performs what I’m calling a ‘post-order insertion’ of the elements of this (non-empty) IntSet into other.

Ok, you need some context. Odersky defines our IntSet binary search tree as follows:


trait IntSet {
    def contains(x: Int): Boolean
    def incl(x: Int): IntSet
    def union(other: IntSet): IntSet

object Empty extends IntSet {
    def contains(x: Int) = false
    def incl(x: Int) = new NonEmpty(x, Empty, Empty)
    def union(other: IntSet) = other

class NonEmpty(val elem: Int, val left: IntSet, val right: IntSet) extends IntSet {

    def contains(x: Int) =
        if (x < elem) left contains x
        else if (x > elem) right contains x
        else true

    def incl(x: Int) =
        if (x < elem) new NonEmpty(elem, left incl x, right)
        else if (x > elem) new NonEmpty(elem, left, right incl x)
        else this

    def union(other: IntSet) = ((left union right) union other) incl elem

The only thing that’s non-obvious is the union method I mentioned above. The reader should make sure she understands the other methods before continuing.

At first I wondered, “What sort of tree is produced by the union of two IntSet?” Given fact that the contains method relies on the tree being a Binary Search Tree, I wondered how this method could possibly retain the sortedness. The short answer is that each element of this is passed through incl to be inserted into other. In the process, I noticed a number of properties of this operation.

The best explanation I’ve got

Some preliminary properties of (A union B)

By post-order insertion, I mean we do a post-order traversal of A, inserting each element into B.

First of all, let “int_set” be an arbitrary IntSet.

Prop 1.a:

(Empty union int_set) == int_set

This is simply the definition of union on Empty.

Prop 1.b:

(int_set union Empty) does not neccessarily == int_set

Counterexample: the IntSet produced by

(((new NonEmpty(2, Empty, Empty)) incl 1) incl 3)

Prop 2:

Inserting nodes from left into right with the incl method will produce a tree where every element of left is hanging off the left edge of the leftmost element of right.

We know every element of right is greater than every element of left. Therefore each element being added to right from left will travel strictly leftwards through the right tree before being deposited.

Prop 3:

other is left fully intact in the returned IntSet

Recall that other is the IntSet operand on the right side of the union operator. This is explained below.

Watch the operation unfold (this is the meat)

Let’s think of the union operator in terms of how it get’s rewritten.

  • a U b
  • ((a.left U a.right) U b) incl a.elem, a substitution of the formula
  • (a1 U b) incl a.head, a rewrite, letting a.left U a.right := a1
  • (((a1.left U a1.right) U b) incl a1.elem) incl a.elem, by substituting the formula
  • ((a2 U b) incl a1.elem) incl a.elem, letting a1.left U a1.right := a2
  • repeat as necessary

So from these few steps we can already see quite a few important things:

  1. We keep removing one element from IntSet a, and prepending it to a list of single elem to incl into IntSet b.
  2. So now it is clear that what this function does, is one-by-one incl elem of this into other. This gives us Prop 3 above.

What remains to be determined is the order in which these incl occur.

  1. a.elem will be the last element included
  2. a1.elem will be the second to last element included
  3. By Prop 3 which was demonstrated just a minute ago, a1.elem is the root node of a.right
  4. Now the post-order algorithm goes left, right, root; and we’ve got ???, right, root.
  5. Instead of spending forever demonstrating why ??? shall be filled in with left, I’ll simply point out that:
    1. In our counterexample demonstration of Prop 1.b above, the root follows the right follows the left.
    2. There’s simply no other element to choose from in a NonEmpty IntSet.

Short example of union in action

Perhaps the notation of this example is a bit hard to parse. Well ain’t it darn fortuitous that I’m here to explain it.

  1. At the very top we have a restatement of the formula to be applied upon taking the union of two IntSet.

  2. Next we have the two IntSet we’d like to find the union of, named A and B. A references a 3-element Binary Tree containing (3, 5, 7), and B references a Tree containing (6, 7, 8).

  3. Next we substitute A U B into the equation at the top to give us a problem of reduction with three components. If we were to define the operators to associate to the left we could remove the parentheses, but I don’t know if that’s legal, in Scala or in Math.

  4. By following the order of operations, we first determine the result of the inner-most parentheses, and proceed outwards, which leaves us the desired result.