Keep track of how many times a recursive function was called

Answer 1

This is an almost purely academic variant, but you can use a modified fixed point combinator for this purpose.

Lets shorten and improve your original function a bit:

function singleDigit(n) {
    let digitProduct = [...(n + '')].reduce((x, y) => x * y, 1);
    return digitProduct <= 9 ? digitProduct : singleDigit(digitProduct);
}

// singleDigit(123234234) == 0

From this variant, we can factor out and curry the recursive call:

function singleDigitF(recur) {
    return function (n) {
        let digitProduct = [...(n + '')].reduce((x, y) => x * y, 1);
        return digitProduct <= 9 ? digitProduct : recur()(digitProduct);
    };
}

This function can now be used with a fixed point combinator; specifically I implemented a Y combinator adapted for (strict) JavaScript as follows:

function Ynormal(f, ...args) {
    let Y = (g) => g(() => Y(g));
    return Y(f)(...args);
}

where we have Ynormal(singleDigitF, 123234234) == 0.

Now comes the trick. Since we have factored out the recursion to the Y combinator, we can count the number of recursions within it:

function Ycount(f, ...args) {
    let count = 1;
    let Y = (g) => g(() => {count += 1; return Y(g);});
    return [Y(f)(...args), count];
}

A quick check in the Node REPL gives:

> Ycount(singleDigitF, 123234234)
[ 0, 3 ]
> let digitProduct = (n) => [...(n + '')].reduce((x, y) => x * y, 1)
undefined
> digitProduct(123234234)
3456
> digitProduct(3456)
360
> digitProduct(360)
0
> Ycount(singleDigitF, 39)
[ 4, 3 ]

This combinator will now work for counting the number of calls in any recursive function written in the style of singleDigitF.

(Note that there's two sources of getting zero as a very frequent answer: numeric overflow (123345456999999999 becoming 123345457000000000 etc.), and the fact that you will almost surely get zero as an intermediate value somewhere, when the size of the input is growing.)