Test deep equality *with sharing* of JavaScript objects

Question

Much ink has been spilled on the subject of testing two objects for deep equality in JavaScript. None, however, seem to care about distinguishing the following two objects:<

Answer 1

Version with no ES6 features that runs in quadratic time:

function deepGraphEqual(a, b) {
    var left = [], right = [], has = Object.prototype.hasOwnProperty;
    function visit(a, b) {
        var i, k;
        if (typeof a !== 'object' || typeof b !== 'object' || a === null || b === null)
            return a === b;
        if (Object.getPrototypeOf(a) !== Object.getPrototypeOf(b))
            return false;
        for (i = 0; i < left.length; i++) {
            if (a === left[i])
                return b === right[i];
            if (b === right[i])
                return a === left[i];
        }
        for (k in a)
            if (has.call(a, k) && !has.call(b, k))
                return false;
        for (k in b)
            if (has.call(b, k) && !has.call(a, k))
                return false;
        left.push(a);
        right.push(b);
        for (k in a)
            if (has.call(a, k) && !visit(a[k], b[k]))
                return false;
        return true;
    }
    return visit(a, b);
}

Version with ES6 Map that runs in linear time:

function deepGraphEqual(a, b) {
    let left = new Map(), right = new Map(), has = Object.prototype.hasOwnProperty;
    function visit(a, b) {
        if (typeof a !== 'object' || typeof b !== 'object' || a === null || b === null)
            return a === b;
        if (Object.getPrototypeOf(a) !== Object.getPrototypeOf(b))
            return false;
        if (left.has(a))
            return left.get(a) === b
        if (right.has(b))
            return right.get(b) === a
        for (let k in a)
            if (has.call(a, k) && !has.call(b, k))
                return false;
        for (let k in b)
            if (has.call(b, k) && !has.call(a, k))
                return false;
        left.set(a, b);
        right.set(b, a);
        for (let k in a)
            if (has.call(a, k) && !visit(a[k], b[k]))
                return false;
        return true;
    }
    return visit(a, b);
}

Answer 2

how to improve Anders Kaseorg's answer:

If you use the algorithm on extremely large data structures, you can get a stack overflow error. That happens for example with a complete graph with 5,000 nodes. So I wrote a non-recursive version, which uses breadth first search instead of depth first search, since that seemed way easier to implement (when not using recursion). The iterative version works fine for the complete graph with 5,000 nodes (takes a whopping 6 seconds, though, on my machine). Here it is:

function deepEqual(item1, item2){
    var EQUAL_ATOM = 1, UNEQUAL = 2, OBJECT = 3;
    function compareSimple(first, second){
        var ty1 = typeof first, ty2 = typeof second;
        if (ty1!==ty2) return UNEQUAL;
        if (ty1!=='object'){
            if (first===second) return EQUAL_ATOM;
            if ((ty1==='number')&&isNaN(first)&&isNaN(second)) return EQUAL_ATOM;
            return UNEQUAL;
        }
        if (first===null) return (second===null) ? EQUAL_ATOM : UNEQUAL;
        if (second===null) return UNEQUAL;
        if (Object.getPrototypeOf(first) !== Object.getPrototypeOf(second)) return UNEQUAL;
        return OBJECT;
    }
    var c = compareSimple(item1, item2);
    if (c !== OBJECT) { return (c===EQUAL_ATOM); }
    var stack1 = [], stack2 = [], inverse1 = new Map(), inverse2 = new Map();
    stack1.push(item1); stack2.push(item2);
    inverse1.set(item1, 0); inverse2.set(item2, 0);
    var currentIdx = 0;
    var firstItem, secondItem, i, own, has1, has2, key, kid1, kid2, itemCount;
    while (currentIdx < stack1.length){
        firstItem = stack1[currentIdx]; secondItem = stack2[currentIdx];
        own = {};
        for (key in firstItem){
            has1 = firstItem.hasOwnProperty(key);
            has2 = secondItem.hasOwnProperty(key);
            if (has1 !== has2) return false;
            if (has1) { own[key] = null; }
        }
        for (key in secondItem){
            if (!(key in own)){
                has1 = firstItem.hasOwnProperty(key);
                has2 = secondItem.hasOwnProperty(key);
                if (has1 !== has2) return false;
                if (has1) { own[key] = null; }
            }
        }
        for (key in own){
            kid1 = firstItem[key];
            kid2 = secondItem[key];
            c = compareSimple(kid1, kid2);
            if (c === UNEQUAL) return false;
            if (c === OBJECT){
                has1 = inverse1.has(kid1);
                has2 = inverse2.has(kid2);
                if (has1 !== has2) return false;
                if (has1){
                    if (inverse1.get(kid1) !== inverse2.get(kid2)) { return false; }
                } else {
                    itemCount = stack1.length;
                    stack1.push(kid1); stack2.push(kid2);
                    inverse1.set(kid1, itemCount); inverse2.set(kid2, itemCount);
                }
            }
        }
        ++currentIdx;
    }
    return true;
}

I added some speed tests on the jsperf.com website. Interestingly, depending on the data structure, sometimes Anders's recursive version is faster, and sometimes my iterative version is faster, with the average being more in Anders's favor.

Here are the links to the tests on jsperf:

nephews example

cycle free real world JSON from reddit example

uncles example

complete graph with 2K nodes

complete graph with 5K nodes

Moreover, builtin objects aren't handled in the way you'll probably want. Many or most builtin objects "hide" their keys. If you do Object.keys(...), you'll just get an empty array.

now = new Date();
keys = Object.keys(now);  // result: []

Hence, for example, any 2 Dates are deepGraphEqual to each other, also any 2 RegExps. That's most probably not what you want. I don't have a "catch all" for all those, and going through all existing "builtin" objects would take really long. But as for Dates and RegExps, here is how you can put in something more reasonable, using .toString() to compare them instead.