Functional Pearl: Implementing trace in JavaScript

Question

Ross Paterson: Arrows and Computation introduces the trace function (on page 11):

trace :: ((a, c) -> (b, c)) -> a -> b
trace f a = let (b, c) = f (a, c) in b

The trace function is useful for modularizing the magic feedback step in circular programs. For example, consider Richard Bird's famous repmin function which finds the minimum leaf value of a tree and creates an identical tree with every leaf value replaced by the minimum leaf value, both in a single pass by making clever use of lazy evaluation and local recursion (as provided by letrec):

data Tree = Leaf Int | Node Tree Tree deriving Show

repmin :: Tree -> Tree
repmin = trace repmin'

repmin' :: (Tree, Int) -> (Tree, Int)
-- put the minimum value m into the leaf and return the old value n as the minimum
repmin' (Leaf n, m) = (Leaf m, n)
-- copy the minimum value m into both the left and right subtrees and
-- set the minimum value m to the minimum of both the left and right subtrees
repmin' (Node l r, m) = let (l', lmin) = repmin' l m in
                        let (r', rmin) = repmin' r m in
                        (Node l' r', lmin `min` rmin)

Anyway, I was wondering how to implement the trace function in JavaScript such that we can implement repmin as follows:

const Leaf = (value) => ({ tag: "Leaf", value });

const Node = (left, right) => ({ tag: "Node", left, right });

const repmin = trace(function repmin(tree, min) {
    switch (tree.tag) {
    case "Leaf":
        return [Leaf(min), tree.value];
    case "Node":
        const [left,  lmin] = repmin(tree.left,  min);
        const [right, rmin] = repmin(tree.right, min);
        return [Node(left, right), Math.min(lmin, rmin)];
    }
});

In order to implement trace we need local recursion as provided by letrec so that we can write something like:

const trace = (f) => (a) => {
    const [b, c] = f(a, c);
    return b;
};

I originally thought of making c a promise. However, that changes the semantics of trace. So, can you think of a way to implement trace in JavaScript without changing its semantics?

Answer 1

Actually, you only need lazy evaluation because assignments in JavaScript are like letrec. Lazy evaluation is generally implemented using thunks. Hence, you can implement trace as follows:

const trace = (f) => (a) => {
    const [b, c] = f(a, () => c);
    return b;
};

Using this definition of trace, the repmin function can remain the same:

const repmin = trace(function repmin(tree, min) {
    switch (tree.tag) {
    case "Leaf":
        return [Leaf(min), tree.value];
    case "Node":
        const [left,  lmin] = repmin(tree.left,  min);
        const [right, rmin] = repmin(tree.right, min);
        return [Node(left, right), Math.min(lmin, rmin)];
    }
});

However, you'd want to make your data constructors possibly lazy using getters:

const descOf = (value) =>
    typeof value === "function" && value.length === 0 ?
        { enumerable: true, get: value } :
        { enumerable: true, value };

const Leaf = (value) => Object.defineProperties({ tag: "Leaf" }, {
    value: descOf(value),
});

const Node = (left, right) => Object.defineProperties({ tag: "Node" }, {
    left: descOf(left),
    right: descOf(right),
});

Putting it all together:

const trace = (f) => (a) => {
    const [b, c] = f(a, () => c);
    return b;
};

const descOf = (value) =>
    typeof value === "function" && value.length === 0 ?
        { enumerable: true, get: value } :
        { enumerable: true, value };

const Leaf = (value) => Object.defineProperties({ tag: "Leaf" }, {
    value: descOf(value),
});

const Node = (left, right) => Object.defineProperties({ tag: "Node" }, {
    left: descOf(left),
    right: descOf(right),
});

const repmin = trace(function repmin(tree, min) {
    switch (tree.tag) {
    case "Leaf":
        return [Leaf(min), tree.value];
    case "Node":
        const [left,  lmin] = repmin(tree.left,  min);
        const [right, rmin] = repmin(tree.right, min);
        return [Node(left, right), Math.min(lmin, rmin)];
    }
});

const show = (tree) => {
    switch (tree.tag) {
    case "Leaf": return `Leaf(${tree.value})`;
    case "Node": return `Node(${show(tree.left)}, ${show(tree.right)})`;
    }
}

const tree = Node(Node(Leaf(1), Leaf(2)), Node(Leaf(3), Leaf(4)));

console.log("Input: ", show(tree));

console.log("Output:", show(repmin(tree)));

The only problem is that you won't be able to write functions like:

const sqr = trace((x, y) => [y * y, x]);

This is because the * operator is not lazy. Hence, you would have to define a lazy mul function:

const trace = (f) => (a) => {
    const [b, c] = f(a, () => c);
    return b;
};

const evaluate = (value) =>
    typeof value === "function" && value.length === 0 ?
        value() : value;

const mul = (a, b) => () => evaluate(a) * evaluate(b);

const sqr = trace((x, y) => [mul(y, y), x]);

console.log(evaluate(sqr(10)));

Hope that helps.