Why does sorting a JS array of numbers with < work?

Question

When sorting an array of numbers in JavaScript, I accidentally used < instead of the usual - -- but it still works. I wonder why?

Example:

var a  = [1,3,2,4]
a.sort(function(n1, n2){
    return n1<n2
})
// result is correct: [4,3,2,1]

And an example array for which this does not work (thanks for Nicolas's example):

[1,2,1,2,1,2,1,2,1,2,1,2]

Answer 1

This sort works on your input array due to its small size and the current implementation of sort in Chrome V8 (and, likely, other browsers).

The return value of the comparator function is defined in the documentation:

• If compareFunction(a, b) is less than 0, sort a to an index lower than b, i.e. a comes first.
• If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements.
• If compareFunction(a, b) is greater than 0, sort b to an index lower than a, i.e. b comes first.

However, your function returns binary true or false, which evaluate to 1 or 0 respectively when compared to a number. This effectively lumps comparisons where n1 < n2 in with n1 === n2, claiming both to be even. If n1 is 9 and n2 is 3, 9 < 3 === false or 0. In other words, your sort leaves 9 and 3 "unchanged with respect to each other" rather than insisting "sort 9 to an index lower than 3".

If your array is shorter than 11 elements, Chrome V8's sort routine switches immediately to an insertion sort and performs no quicksort steps:

// Insertion sort is faster for short arrays.
if (to - from <= 10) {
  InsertionSort(a, from, to);
  return;
}

V8's insertion sort implementation only cares if the comparator function reports b as greater than a, taking the same else branch for both 0 and < 0 comparator returns:

var order = comparefn(tmp, element);
if (order > 0) {
  a[j + 1] = tmp;
} else {
  break;
}

Quicksort's implementation, however, relies on all three comparisons both in choosing a pivot and in partitioning:

var order = comparefn(element, pivot);
if (order < 0) {
  // ...
} else if (order > 0) {
  // ...
}
// move on to the next iteration of the partition loop

This guarantees an accurate sort on arrays such as [1,3,2,4], and dooms arrays with more than 10 elements to at least one almost certainly inaccurate step of quicksort.

---

Update 7/19/19: Since the version of V8 (6) discussed in this answer, implementation of V8's array sort moved to Torque/Timsort in 7.0 as discussed in this blog post and insertion sort is called on arrays of length 22 or less.

The article linked above describes the historical situation of V8 sorting as it existed at the time of the question:

Array.prototype.sort and TypedArray.prototype.sort relied on the same Quicksort implementation written in JavaScript. The sorting algorithm itself is rather straightforward: The basis is a Quicksort with an Insertion Sort fall-back for shorter arrays (length < 10). The Insertion Sort fall-back was also used when Quicksort recursion reached a sub-array length of 10. Insertion Sort is more efficient for smaller arrays. This is because Quicksort gets called recursively twice after partitioning. Each such recursive call had the overhead of creating (and discarding) a stack frame.

Regardless of any changes in the implementation details, if the sort comparator adheres to standard, the code will sort predictably, but if the comparator doesn't fulfill the contract, all bets are off.

Answer 2

After my initial comment, I wondered a little bit about how easy it is to find arrays for which this sorting method fails.

I ran an exhaustive search on arrays of length up to 8 (on an alphabet of size the size of the array), and found nothing. Since my (admittedly shitty) algorithm started to be too slow, I changed it to an alphabet of size 2 and found that binary arrays of length up to 10 are all sorted properly. However, for binary arrays of length 11, many are improperly sorted, for instance [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0].

// Check if 'array' is properly sorted using the "<" comparator function
function sortWorks(array) {
    let sorted_array = array.sort(function(n1, n2) {
        return n1 < n2;
    });
    for (let i=0; i<sorted_array.length-1; i++) {
        if (sorted_array[i] < sorted_array[i+1]) return false;
    }
    return true;
}

// Get a list of all arrays of length 'count' on an alphabet of size 'max_n'.
// I know it's awful, don't yell at me :'(
var arraysGenerator;
arraysGenerator = function (max_n, count) {
    if (count === 0) return [[]];
    let arrays = arraysGenerator(max_n, count-1);
    let result = [];
    for (let array of arrays) {
        for (let i = 0; i<max_n; i++) {
            result.push([i].concat(array));
        }
    }
    return result;
}

// Check if there are binary arrays of size k which are improperly sorted,
// and if so, log them
function testArraysOfSize(k) {
    let arrays = arraysGenerator(2, k);
    for (let array of arrays) {
        if (!sortWorks(array)) {
            console.log(array);
        }
    }
}

I'm getting some weird false-positives for some reason though, not sure where my mistake is.

---

EDIT:

After checking for a little while, here's a partial explanation on why OP's "wrong" sorting method works for lengths <=10 and for lengths >=11: it looks like (at least some) javascript implementations use InsertionSort if the array length is short (length <= 10) and QuickSort otherwise. It looks like QuickSort actively uses the "-1" outputs of the compare function while InsertionSort does not and relies only on the "1" outputs.

Source: here, all thanks to the original author.

Answer 3

If we analyze what's being done, it seems that this is mostly luck as in this case, 3 and 2 are considered to be "the same" and should be interchangeable. I suppose in such cases, the JS engines keep the original order for any values that have been deemed equal:

let a = [1, 3, 2, 4];
a.sort((n1, n2) => {
  const result = n1 < n2;
  if (result < 0) {
    console.log(`${n1} comes before ${n2}`);
  } else if (result > 0) {
    console.log(`${n2} comes before ${n1}`);
  } else {
    console.log(`${n1} comes same position as ${n2}`);
  }
  return result;
})

console.log(a);

As pointed out in the comments, this isn't guaranteed to work ([1,2,1,2,1,2,1,2,1,2,1,2] being a counter-example).