Is it correct to use JavaScript Array.sort() method for shuffling?

Question

I was helping somebody out with his JavaScript code and my eyes were caught by a section that looked like that:

function randOrd(){
  return (Math.round(Math.random())-0.5);
}
coords.sort(randOrd);
alert(coords);

My first though was: hey, this can't possibly work! But then I did some experimenting and found that it indeed at least seems to provide nicely randomized results.

Then I did some web search and almost at the top found an article from which this code was most ceartanly copied. Looked like a pretty respectable site and author...

But my gut feeling tells me, that this must be wrong. Especially as the sorting algorithm is not specified by ECMA standard. I think different sorting algoritms will result in different non-uniform shuffles. Some sorting algorithms may probably even loop infinitely...

But what do you think?

And as another question... how would I now go and measure how random the results of this shuffling technique are?

update: I did some measurements and posted the results below as one of the answers.

Answer 1

After Jon has already covered the theory, here's an implementation:

function shuffle(array) {
    var tmp, current, top = array.length;

    if(top) while(--top) {
        current = Math.floor(Math.random() * (top + 1));
        tmp = array[current];
        array[current] = array[top];
        array[top] = tmp;
    }

    return array;
}

The algorithm is O(n), whereas sorting should be O(n log n). Depending on the overhead of executing JS code compared to the native sort() function, this might lead to a noticable difference in performance which should increase with array sizes.

---

In the comments to bobobobo's answer, I stated that the algorithm in question might not produce evenly distributed probabilities (depending on the implementation of sort()).

My argument goes along these lines: A sorting algorithm requires a certain number c of comparisons, eg c = n(n-1)/2 for Bubblesort. Our random comparison function makes the outcome of each comparison equally likely, ie there are 2^c equally probable results. Now, each result has to correspond to one of the n! permutations of the array's entries, which makes an even distribution impossible in the general case. (This is a simplification, as the actual number of comparisons neeeded depends on the input array, but the assertion should still hold.)

As Jon pointed out, this alone is no reason to prefer Fisher-Yates over using sort(), as the random number generator will also map a finite number of pseudo-random values to the n! permutations. But the results of Fisher-Yates should still be better:

Math.random() produces a pseudo-random number in the range [0;1[. As JS uses double-precision floating point values, this corresponds to 2^x possible values where 52 ≤ x ≤ 63 (I'm too lazy to find the actual number). A probability distribution generated using Math.random() will stop behaving well if the number of atomic events is of the same order of magnitude.

When using Fisher-Yates, the relevant parameter is the size of the array, which should never approach 2^52 due to practical limitations.

When sorting with a random comparision function, the function basically only cares if the return value is positive or negative, so this will never be a problem. But there is a similar one: Because the comparison function is well-behaved, the 2^c possible results are, as stated, equally probable. If c ~ n log n then 2^c ~ n^(a·n) where a = const, which makes it at least possible that 2^c is of same magnitude as (or even less than) n! and thus leading to an uneven distribution, even if the sorting algorithm where to map onto the permutaions evenly. If this has any practical impact is beyond me.

The real problem is that the sorting algorithms are not guaranteed to map onto the permutations evenly. It's easy to see that Mergesort does as it's symmetric, but reasoning about something like Bubblesort or, more importantly, Quicksort or Heapsort, is not.

---

The bottom line: As long as sort() uses Mergesort, you should be reasonably safe except in corner cases (at least I'm hoping that 2^c ≤ n! is a corner case), if not, all bets are off.

Answer 2

It's never been my favourite way of shuffling, partly because it is implementation-specific as you say. In particular, I seem to remember that the standard library sorting from either Java or .NET (not sure which) can often detect if you end up with an inconsistent comparison between some elements (e.g. you first claim A < B and B < C, but then C < A).

It also ends up as a more complex (in terms of execution time) shuffle than you really need.

I prefer the shuffle algorithm which effectively partitions the collection into "shuffled" (at the start of the collection, initially empty) and "unshuffled" (the rest of the collection). At each step of the algorithm, pick a random unshuffled element (which could be the first one) and swap it with the first unshuffled element - then treat it as shuffled (i.e. mentally move the partition to include it).

This is O(n) and only requires n-1 calls to the random number generator, which is nice. It also produces a genuine shuffle - any element has a 1/n chance of ending up in each space, regardless of its original position (assuming a reasonable RNG). The sorted version approximates to an even distribution (assuming that the random number generator doesn't pick the same value twice, which is highly unlikely if it's returning random doubles) but I find it easier to reason about the shuffle version :)

This approach is called a Fisher-Yates shuffle.

I would regard it as a best practice to code up this shuffle once and reuse it everywhere you need to shuffle items. Then you don't need to worry about sort implementations in terms of reliability or complexity. It's only a few lines of code (which I won't attempt in JavaScript!)

The Wikipedia article on shuffling (and in particular the shuffle algorithms section) talks about sorting a random projection - it's worth reading the section on poor implementations of shuffling in general, so you know what to avoid.

Answer 3

I did some measurements of how random the results of this random sort are...

My technique was to take a small array [1,2,3,4] and create all (4! = 24) permutations of it. Then I would apply the shuffling function to the array a large number of times and count how many times each permutation is generated. A good shuffling algoritm would distribute the results quite evenly over all the permutations, while a bad one would not create that uniform result.

Using the code below I tested in Firefox, Opera, Chrome, IE6/7/8.

Surprisingly for me, the random sort and the real shuffle both created equally uniform distributions. So it seems that (as many have suggested) the main browsers are using merge sort. This of course doesn't mean, that there can't be a browser out there, that does differently, but I would say it means, that this random-sort-method is reliable enough to use in practice.

EDIT: This test didn't really measured correctly the randomness or lack thereof. See the other answer I posted.

But on the performance side the shuffle function given by Cristoph was a clear winner. Even for small four-element arrays the real shuffle performed about twice as fast as random-sort!

// The shuffle function posted by Cristoph.
var shuffle = function(array) {
    var tmp, current, top = array.length;

    if(top) while(--top) {
        current = Math.floor(Math.random() * (top + 1));
        tmp = array[current];
        array[current] = array[top];
        array[top] = tmp;
    }

    return array;
};

// the random sort function
var rnd = function() {
  return Math.round(Math.random())-0.5;
};
var randSort = function(A) {
  return A.sort(rnd);
};

var permutations = function(A) {
  if (A.length == 1) {
    return [A];
  }
  else {
    var perms = [];
    for (var i=0; i<A.length; i++) {
      var x = A.slice(i, i+1);
      var xs = A.slice(0, i).concat(A.slice(i+1));
      var subperms = permutations(xs);
      for (var j=0; j<subperms.length; j++) {
        perms.push(x.concat(subperms[j]));
      }
    }
    return perms;
  }
};

var test = function(A, iterations, func) {
  // init permutations
  var stats = {};
  var perms = permutations(A);
  for (var i in perms){
    stats[""+perms[i]] = 0;
  }

  // shuffle many times and gather stats
  var start=new Date();
  for (var i=0; i<iterations; i++) {
    var shuffled = func(A);
    stats[""+shuffled]++;
  }
  var end=new Date();

  // format result
  var arr=[];
  for (var i in stats) {
    arr.push(i+" "+stats[i]);
  }
  return arr.join("\n")+"\n\nTime taken: " + ((end - start)/1000) + " seconds.";
};

alert("random sort: " + test([1,2,3,4], 100000, randSort));
alert("shuffle: " + test([1,2,3,4], 100000, shuffle));

Answer 4

Interestingly, Microsoft used the same technique in their pick-random-browser-page.

They used a slightly different comparison function:

function RandomSort(a,b) {
    return (0.5 - Math.random());
}

Looks almost the same to me, but it turned out to be not so random...

So I made some testruns again with the same methodology used in the linked article, and indeed - turned out that the random-sorting-method produced flawed results. New test code here:

function shuffle(arr) {
  arr.sort(function(a,b) {
    return (0.5 - Math.random());
  });
}

function shuffle2(arr) {
  arr.sort(function(a,b) {
    return (Math.round(Math.random())-0.5);
  });
}

function shuffle3(array) {
  var tmp, current, top = array.length;

  if(top) while(--top) {
    current = Math.floor(Math.random() * (top + 1));
    tmp = array[current];
    array[current] = array[top];
    array[top] = tmp;
  }

  return array;
}

var counts = [
  [0,0,0,0,0],
  [0,0,0,0,0],
  [0,0,0,0,0],
  [0,0,0,0,0],
  [0,0,0,0,0]
];

var arr;
for (var i=0; i<100000; i++) {
  arr = [0,1,2,3,4];
  shuffle3(arr);
  arr.forEach(function(x, i){ counts[x][i]++;});
}

alert(counts.map(function(a){return a.join(", ");}).join("\n"));

Answer 5

I have placed a simple test page on my website showing the bias of your current browser versus other popular browsers using different methods to shuffle. It shows the terrible bias of just using Math.random()-0.5, another 'random' shuffle that isn't biased, and the Fisher-Yates method mentioned above.

You can see that on some browsers there is as high as a 50% chance that certain elements will not change place at all during the 'shuffle'!

Note: you can make the implementation of the Fisher-Yates shuffle by @Christoph slightly faster for Safari by changing the code to:

function shuffle(array) {
  for (var tmp, cur, top=array.length; top--;){
    cur = (Math.random() * (top + 1)) << 0;
    tmp = array[cur]; array[cur] = array[top]; array[top] = tmp;
  }
  return array;
}

Test results: http://jsperf.com/optimized-fisher-yates

Answer 6

I think it's fine for cases where you're not picky about distribution and you want the source code to be small.

In JavaScript (where the source is transmitted constantly), small makes a difference in bandwidth costs.

Answer 7

It's been four years, but I'd like to point out that the random comparator method won't be correctly distributed, no matter what sorting algorithm you use.

Proof:

1. For an array of n elements, there are exactly n! permutations (i.e. possible shuffles).
2. Every comparison during a shuffle is a choice between two sets of permutations. For a random comparator, there is a 1/2 chance of choosing each set.
3. Thus, for each permutation p, the chance of ending up with permutation p is a fraction with denominator 2^k (for some k), because it is a sum of such fractions (e.g. 1/8 + 1/16 = 3/16).
4. For n = 3, there are six equally-likely permutations. The chance of each permutation, then, is 1/6. 1/6 can't be expressed as a fraction with a power of 2 as its denominator.
5. Therefore, the coin flip sort will never result in a fair distribution of shuffles.

The only sizes that could possibly be correctly distributed are n=0,1,2.

---

As an exercise, try drawing out the decision tree of different sort algorithms for n=3.

---

There is a gap in the proof: If a sort algorithm depends on the consistency of the comparator, and has unbounded runtime with an inconsistent comparator, it can have an infinite sum of probabilities, which is allowed to add up to 1/6 even if every denominator in the sum is a power of 2. Try to find one.

Also, if a comparator has a fixed chance of giving either answer (e.g. (Math.random() < P)*2 - 1, for constant P), the above proof holds. If the comparator instead changes its odds based on previous answers, it may be possible to generate fair results. Finding such a comparator for a given sorting algorithm could be a research paper.

Answer 8

No, it is not correct. As other answers have noted, it will lead to a non-uniform shuffle and the quality of the shuffle will also depend on which sorting algorithm the browser uses.

Now, that might not sound too bad to you, because even if theoretically the distribution is not uniform, in practice it's probably nearly uniform, right? Well, no, not even close. The following charts show heat-maps of which indices each element gets shuffled to, in Chrome and Firefox respectively: if the pixel (i, j) is green, it means the element at index i gets shuffled to index j too often, and if it's red then it gets shuffled there too rarely.

These screenshots are taken from Mike Bostock's page on this subject.

As you can see, shuffling using a random comparator is severely biased in Chrome and even more so in Firefox. In particular, both have a lot of green along the diagonal, meaning that too many elements get "shuffled" somewhere very close to where they were in the original sequence. In comparison, a similar chart for an unbiased shuffle (e.g. using the Fisher-Yates algorithm) would be all pale yellow with just a small amount of random noise.

Answer 9

It is a hack, certainly. In practice, an infinitely looping algorithm is not likely. If you're sorting objects, you could loop through the coords array and do something like:

for (var i = 0; i < coords.length; i++)
    coords[i].sortValue = Math.random();

coords.sort(useSortValue)

function useSortValue(a, b)
{
  return a.sortValue - b.sortValue;
}

(and then loop through them again to remove the sortValue)

Still a hack though. If you want to do it nicely, you have to do it the hard way :)

Answer 10

If you're using D3 there is a built-in shuffle function (using Fisher-Yates):

var days = ['Lundi','Mardi','Mercredi','Jeudi','Vendredi','Samedi','Dimanche'];
d3.shuffle(days);

And here is Mike going into details about it:

http://bost.ocks.org/mike/shuffle/

Answer 11

Here's an approach that uses a single array:

The basic logic is:

Starting with an array of n elements

Remove a random element from the array and push it onto the array

Remove a random element from the first n - 1 elements of the array and push it onto the array

Remove a random element from the first n - 2 elements of the array and push it onto the array

...

Remove the first element of the array and push it onto the array

Code:

for(i=a.length;i--;) a.push(a.splice(Math.floor(Math.random() * (i + 1)),1)[0]);

Answer 12

Can you use the Array.sort() function to shuffle an array – Yes.

Are the results random enough – No.

Consider the following code snippet:

/*
 * The following code sample shuffles an array using Math.random() trick
 * After shuffling, the new position of each item is recorded
 * The process is repeated 100 times
 * The result is printed out, listing each item and the number of times
 * it appeared on a given position after shuffling
 */
var array = ["a", "b", "c", "d", "e"];
var stats = {};
array.forEach(function(v) {
  stats[v] = Array(array.length).fill(0);
});
var i, clone;
for (i = 0; i < 100; i++) {
  clone = array.slice();
  clone.sort(function() {
    return Math.random() - 0.5;
  });
  clone.forEach(function(v, i) {
    stats[v][i]++;
  });
}
Object.keys(stats).forEach(function(v, i) {
  console.log(v + ": [" + stats[v].join(", ") + "]");
});

Sample output:

a: [29, 38, 20,  6,  7]
b: [29, 33, 22, 11,  5]
c: [17, 14, 32, 17, 20]
d: [16,  9, 17, 35, 23]
e: [ 9,  6,  9, 31, 45]

Ideally, the counts should be evenly distributed (for the above example, all counts should be around 20). But they are not. Apparently, the distribution depends on what sorting algorithm is implemented by the browser and how it iterates the array items for sorting.

Answer 13

There is nothing wrong with it.

The function you pass to .sort() usually looks something like

function sortingFunc( first, second )
{
  // example:
  return first - second ;
}

Your job in sortingFunc is to return:

• a negative number if first goes before second
• a positive number if first should go after second
• and 0 if they are completely equal

The above sorting function puts things in order.

If you return -'s and +'s randomly as what you have, you get a random ordering.

Like in MySQL:

SELECT * from table ORDER BY rand()