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Doing the Microsoft Shuffle: Algorithm Fail in Browser Ballot

March 6th Update:  Microsoft appears to have updated the www.browserchoice.eu website and corrected the error I describe in this post.  More details on the fix can be found in The New & Improved Microsoft Shuffle.  However, I think you will still find the following analysis interesting.



The story first hit in last week on the Slovakian tech site DSL.sk.  Since I am not linguistically equipped to follow the Slovakian tech scene, I didn’t hear about the story until it was brought up in English on TechCrunch.  The gist of these reports is this: DSL.sk did a test of the “ballot” screen at www.browserchoice.eu, used in Microsoft Windows 7 to prompt the user to install a browser.  It was a Microsoft concession to the EU, to provide a randomized ballot screen for users to select a browser.  However, the DSL.sk test suggested that the ordering of the browsers was far from random.

But this wasn’t a simple case of Internet Explorer showing up more in the first position.  The non-randomness was pronounced, but more complicated.  For example, Chrome was more likely to show up in one of the first 3 positions.  And Internet Explorer showed up 50% of the time in the last position.  This isn’t just a minor case of it being slightly random.  Try this test yourself: Load www.browserchoice.eu, in Internet Explorer, and press refresh 20 times.  Count how many times the Internet Explorer choice is on the far right.   Can this be right?

The DLS.sk findings have lead to various theories, made on the likely mistaken theory that this is an intentional non-randomness.  Does Microsoft have secret research showing that the 5th position is actually chosen more often?  Is the Internet Explorer random number generator not random?  There were also comments asserting that the tests proved nothing, and the results were just chance, and others saying that the results are expected to be non-random because computers can only make pseudo-random numbers, not genuinely random numbers.

Maybe there was cogent technical analysis of this issue posted someplace, but if there was, I could not find it.  So I’m providing my own analysis here, a little statistics and a little algorithms 101.  I’ll tell you what went wrong, and how Microsoft can fix it.  In the end it is a rookie mistake in the code, but it is an interesting mistake that we can learn from, so I’ll examine it in some depth.

Are the results random?

The ordering of the browser choices is determined by JavaScript code on the BrowserChoice.eu web site.  You can see the core function in the GenerateBrowserOrder function.  I took that function and supporting functions,  put it into my own HTML file, added some test driver code and ran it for 10,000 iterations on Internet Explorer.  The results are as follows:

Internet Explorer raw counts
Position I.E. Firefox Opera Chrome Safari
1 1304 2099 2132 2595 1870
2 1325 2161 2036 2565 1913
3 1105 2244 1374 3679 1598
4 1232 2248 1916 590 4014
5 5034 1248 2542 571 605
Internet Explorer fraction of total
Position I.E. Firefox Opera Chrome Safari
1 0.1304 0.2099 0.2132 0.2595 0.1870
2 0.1325 0.2161 0.2036 0.2565 0.1913
3 0.1105 0.2244 0.1374 0.3679 0.1598
4 0.1232 0.2248 0.1916 0.0590 0.4014
5 0.5034 0.1248 0.2542 0.0571 0.0605

This confirms the DSL.sk results.  Chrome appears more often in one of the first 3 positions and I.E. is most likely to be in the 5th position.

You can also see this graphically in a 3D bar chart:

But is this a statistically significant result?  I think most  of us have an intuitive feeling that results are more significant if many tests are run, and if the results also vary much from an even distribution of positions.  On the other hand, we also know that a finite run of even a perfectly random algorithm will not give a perfectly uniform distribution.  It would be quite unusual if every cell in the above table was exactly 2,000.

This is not a question one answers with debate.  To go beyond intuition you need to perform a statistical test.  In this case, a good test is Pearson’s Chi-square test, which tests how well observed results match a specified distribution.  In this test we assume the null-hypothesis that the observed data is taken from a uniform distribution.  The test then tells us the probability that the observed results can be explained by chance.  In other words, what is the probability that the difference between observation and a uniform distribution was just the luck of the draw?  If that probability is very small, say less than 1%, then we can say with high confidence, say 99% confidence, that the positions are not uniformly distributed.   However, if the test returns a larger number, then we cannot disprove our null-hypothesis.  That doesn’t mean the null-hypothesis is true.  It just means we can’t disprove it.  In the end we can never prove the null hypothesis.  We can only try to disprove it.

Note also that having a uniform distribution is not the same as having uniformly distributed random positions.  There are ways of getting a uniform distribution that are not random, for example, by treating the order as a circular buffer and rotating through the list on each invocation.  Whether or not randomization is needed is ultimately dictated by the architectural assumptions of your application.  If you determine the order on a central server and then serve out that order on each innovation, then you can use non-random solutions, like the rotating circular buffer.  But if the ordering is determined independently on each client, for each invocation, then you need some source of randomness on each client to achieve a uniform distribution overall.  But regardless of how you attempt to achieve a uniform distribution the way to test it is the same, using the Chi-square test.

Using the open source statistical package R, I ran the  chisq.test() routine on the above data.  The results are:

X-squared = 13340.23, df = 16, p-value < 2.2e-16

The p-value is much, much less than 1%.  So, we can say with high confidence that the results are not random.

Repeating the same test on Firefox is also non-random, but in a different way:

Firefox raw counts
Position I.E. Firefox Opera Chrome Safari
1 2494 2489 1612 947 2458
2 2892 2820 1909 1111 1268
3 2398 2435 2643 1891 633
4 1628 1638 2632 3779 323
5 588 618 1204 2272 5318
Firefox fraction of total
Position I.E. Firefox Opera Chrome Safari
1 0.2494 0.2489 0.1612 0.0947 0.2458
2 0.2892 0.2820 0.1909 0.1111 0.1268
3 0.2398 0.2435 0.2643 0.1891 0.0633
4 0.1628 0.1638 0.2632 0.3779 0.0323
5 0.0588 0.0618 0.1204 0.2272 0.5318

On Firefox, Internet Explorer is more frequently in one of the first 3 positions, while Safari is most often in last position.  Strange.  The same code, but vastly different results.

The results here are also highly significant:

X-squared = 14831.41, df = 16, p-value < 2.2e-16

So given the above, we know two things:  1) The problem is real.  2) The problem is not related to a flaw only in Internet Explorer.

In the next section we look at the algorithm and show what the real problem is, and how to fix it.

Random shuffles

The browser choice screen requires what we call a “random shuffle”.  You start with an array of values and return those same values, but in a randomized order. This computational problem has been known since the earliest days of computing.  There are 4 well-known approaches: 2 good solutions, 1 acceptable (“good enough”) solution that is slower than necessary, and 1 bad approach that doesn’t really work.  Microsoft appears to have picked the bad approach. But I do not believe there is some nefarious intent to this bug.  It is more in the nature of a “naive” algorithm”, like the bubble sort, that inexperienced programmers  inevitably will fall upon when solving a given problem.  I bet if we gave this same problem to 100 freshmen computer science majors, at least one of them would make the same mistake.  But with education and experience, one learns about these things.  And one of the things one learns early on is to reach for Knuth.

The Art of Computer Programming, Vol. 2, section 3.4.2 “Random sampling and shuffling” describes two solutions:

  1. If the number of items to sort is small, then simply put all possible orderings in a table and select one ordering at random.  In our case, with 5 browsers, the table would need 5! = 120 rows.
  2. “Algorithm P” which Knuth attributes to Moses and Oakford (1963), but is now known to have been anticipated by Fisher and Yates (1938) so it is now called the Fisher-Yates Shuffle.

Another solution, one I use when I need a random shuffle in a database or spreadsheet, is to add a new column, fill that column with random numbers and then sort by that column.  This is very easy to implement in those environments. However, sorting is an O(N log N)  operation where the Fisher-Yates algorithm is O(N), so you need to keep that in mind if performance is critical.

Microsoft used none of these well-known solutions in their random solution.  Instead they fell for the well-known trap.  What they did is sort the array, but with a custom-defined comparison function or “comparator”.  JavaScript, like many other programming languages, allows a custom comparator function to be specified.  In the case of JavaScript, this function takes two indexes into the value array and returns a value which is:

  • <0 if the value at the first index should be sorted before the value at the second index
  • 0 if the values at the first index and the second index are equal, which is to say you are indifferent as to what order they are sorted
  • >0 if the value at the first index should be sorted after the value at the second index

This is a very flexible approach, and allows the programmer to handle all sorts of sorting tasks, from making case-insensitive sorts to defining locale-specific collation orders, etc..

In this case Microsoft gave the following comparison function:

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

Since Math.random() should return a random number chosen uniformly between 0 and 1, the RandomSort() function will return a random value between -0.5 and 0.5.  If you know anything about sorting, you can see the problem here.  Sorting requires a self-consistent definition of ordering. The following assertions must be true if sorting is to make any sense at all:

  1. If a<b then b>a
  2. If a>b then b<a
  3. If a=b then b=a
  4. if a<b and b<c then a<c
  5. If a>b and b>c then a>c
  6. If a=b and b=c then a=c

All of these statements are violated by the Microsoft comparison function.  Since the comparison function returns random results, a sort routine that depends on any of these logical implications would receive inconsistent information regarding the progress of the sort.  Given that, the fact that the results were non-random is hardly surprising.  Depending on the exact search algorithm used, it may just do a few exchanges operations and then prematurely stop.  Or, it could be worse.  It could lead to an infinite loop.

Fixing the Microsoft Shuffle

The simplest approach is to adopt a well-known and respected algorithm like the Fisher-Yates Shuffle, which has been known since 1938.  I tested with that algorithm, using a JavaScript implementation taken from the Fisher-Yates Wikpedia page, with the following results for 10,000 iterations in Internet Explorer:

Internet Explorer raw counts
Position I.E. Firefox Opera Chrome Safari
1 2023 1996 2007 1944 2030
2 1906 2052 1986 2036 2020
3 2023 1988 1981 1984 2024
4 2065 1985 1934 2019 1997
5 1983 1979 2092 2017 1929
Internet Explorer fraction of total
Position I.E. Firefox Opera Chrome Safari
1 0.2023 0.1996 0.2007 0.1944 0.2030
2 0.1906 0.2052 0.1986 0.2036 0.2020
3 0.2023 0.1988 0.1981 0.1984 0.2024
4 0.2065 0.1985 0.1934 0.2019 0.1997
5 0.1983 0.1979 0.2092 0.2017 0.1929

Applying Pearson’s Chi-square test we see:

X-squared = 21.814, df = 16, p-value = 0.1493

In other words, these results are not significantly different than a truly random distribution of positions.  This is good.  This is what we want to see.

Here it is, in graphical form, to the same scale as the “Microsoft Shuffle” chart earlier:


The lesson here is that getting randomness on a computer cannot be left to chance.  You cannot just throw Math.random() at a problem and stir the pot, and expect good results.  Random is not the same as being casual.  Getting random results on a deterministic computer is one of the hardest things you can do with a computer and requires deliberate effort, including avoiding known traps.  But it also requires testing.  Where serious money is on the line, such as with online gambling sites, random number generators and shuffling algorithms are audited, tested and subject to inspection.  I suspect that the stakes involved in the browser market are no less significant.  Although I commend DSL.sk for finding this issue in the first place, I am astonished that the bug got as far as it did.  This should have been caught far earlier, by Microsoft, before this ballot screen was ever made public.  And if the EC is not already demanding a periodic audit of the aggregate browser presentation orderings, I think that would be a prudent thing to do.

If anyone is interested, you can take a look at the file I used for running the tests.  You type in an iteration count and press the execute button.  After a (hopefully) short delay you will get a table of results, using the Microsoft Shuffle as well as the Fisher-Yates Shuffle.  With 10,000 iterations you will get results in around 5 seconds.  Since all execution is in the browser, use larger numbers at your own risk.  At some large value you will presumably run out of memory, time out, hang, or otherwise get an unsatisfactory experience.

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{ 133 comments… add one }

  • anon e moose 2010/02/28, 10:55

    Actually, Fisher-Yates might not be the simplest. Another approach that works well in languages with associative arrays is to do this:

    1. Make an associative array whose keys are the elements you wish to shuffle, and whose values are random numbers.

    2. Sort the keys by their values.

    My Javascript is rusty, so I’ll give an example in Perl, and leave it to someone else to show how it would be done in Javascript:

    sub randomize_array
    my %hash = map {$_ => rand} @_;
    return sort {$hash{$a} $hash{$b}} keys %hash;

    That takes an array, and returns a shuffled version of that array.

  • Rob 2010/02/28, 11:48

    @moose, that is essentially the solution I described as an “acceptable solution that is slower than necessary”. That’s how an end user can do a random shuffle in a spreadsheet: insert a new column parallel to your data column(s). Copy the formula “=rand()” into each cell in that column. Select the entire range, including your existing data and the data in the new column. Sort, using the new random column as your sort key. Same concept.

    But the interesting thing is that doing a sort is not necessary at all to do a random shuffle. Of course, with only 5 elements, the fact that sorting is asymptotically slower than Fisher-Yates isn’t going to matter. Your solution would work just fine at that scale.

  • Josh D. 2010/02/28, 11:51

    “Never attribute to malice that which can be adequately explained by stupidity.” –Hanlon’s razor

  • Darren Kopp 2010/02/28, 12:19

    why do you want equal distribution rather than random? this screen will only appear to the user once, and will be completely random. also, to determine that your results are in fact right, you are going to need 10,000 iterations of your test that takes 10,000 sorted rankings. this will show you how you don’t understand what “random” means.

  • Rob 2010/02/28, 13:36

    In fact, your first p-value is even lower… 2.2E-16 is the lowest result R’s chisq.test can produce :P

  • Rob 2010/02/28, 13:46

    @Darren, In this post I’m not getting into the economic or competitive reasons of “why” a random ordering is desired here. Let’s just take it from the point where Microsoft and the EC agreed to settle the outstanding anti-trust suit, and where as part of the settlement Microsoft agreed to randomize the the browser ballot screen. So take that as a requirement given to a programmer. As was noted originally by DSL.sk, and confirmed by my tests, the delivered ballot screen is not random.

    The statistical analysis here is pretty basic. You don’t need to do 10,000 repetitions of my test to show — to a very high confidence level — that it is not random. The p-value itself showed that the distribution was so uneven that the chances were 1 in 50,000,000,000,000,000 that these results were drawn from a process that was actually producing random orderings. Analysis of the algorithm shows why the results are non-random.

  • Antti 2010/02/28, 14:00

    Actually, the tag-and-sort algorithm is not a perfect shuffle, whereas Fisher-Yates is. See http://okmij.org/ftp/Haskell/perfect-shuffle.txt for an explanation.

  • Albert A. Freeman 2010/02/28, 14:00

    What’s more interesting is the question: does it benefit Microsoft? It would appear to actually work against them, since competing browsers appear more often at the front of the list.

    Seems to me that would perfectly satisfy the aims of people behind the Browser Choice List. It seems like they had a hidden goal of reducing Microsoft’s presence, and this will do it, ever so slightly.

  • ben 2010/02/28, 14:18

    I think by “equal distribution” he means an “equal probability of appearing in a given spot” which is what you’d expect if a number was truly random.

  • Rob 2010/02/28, 14:24

    @Antii, you are certainly correct. But I think that if you have a small number of elements to shuffle, you should avoid the kind of key collisions that would cause a problem. And at that point the performance liabilities would also tend to lead one away from a sort-dependent solution and toward Fisher-Yates.

    It would be interesting to see at what point the key collisions cause a problem. How many elements do you need for it to fail the chi-square test with p-value < 1% ?

  • Anthony 2010/02/28, 14:43

    An alternate shuffle, that you could compute in N time, assuming all elements are already in an array:

    for each element in the array:
    generate a random value between [1 and size of array] (assumption = 1 is the index of the first value)
    swap current element with the element in the randomly generated position

    Every element in the array has at least 1 chance to get moved to a random position.

  • amnon 2010/02/28, 14:50

    Many years ago (early 80′s) I saw in some magazine an elegant and fast O(N) solution for doing this, as follows:

    for (i=0; i<N; i++) A[i] = i;
    for (i=0; i<N; i++) swap( A[i], A[ rand(N) ] );

    I never analysed the result to see if mathematically it is identical to a true random permutation, but intuitively it looks that way and definitely was good enough for the games I wrote back then… :-)

  • Jeremy 2010/02/28, 14:53

    Neither script seems to use more memory for more iterations. Running the test in firefox on my machine, there was no difference in memory usage between 50,000 iterations and 5,000,000 anyway. It just took a lot longer to finish.

  • Matt 2010/02/28, 15:03

    Using Chrome “ unknown (38071)” which is latest non-dev according to the About screen, your script executes, displays the results, and then immediately flashes back to the input, blanking the results. It’s rather inconvenient.

    Is there something I’m doing wrong?

  • Carlos 2010/02/28, 15:12

    Nice article and analysis – it’s always easy to forget about the theory behind the code once you’ve been though years of rushed projects.

    For the third possible return case in the compare function possibilities section above:
    ■<0 if the value at the first index should be sorted before the value at the second index
    ■0″ instead of the first and third both being “<0" cases.

  • Rob 2010/02/28, 15:15

    @Carlos, good catch. I’ll correct.

    @Matt, I don’t have Chrome installed. But it could be error in my script, or compatibility issue. Anyone else seeing this behavior?

  • Jo Hermans 2010/02/28, 15:25

    amnon: you solution is basically Fisher-Yates, except for a mistake in the rand function (you remembered it incorrectly). Compare with the solution on the Wikipedia page.

  • Tim 2010/02/28, 15:29

    Just curious, what’s the third good solution?

  • steve 2010/02/28, 15:32

    I don’t see how this code could cause an infinite loop. Please elaborate?

  • Nick 2010/02/28, 15:36

    When localizing educational computer games from a US manufacturer for the German market, I’ve seen my fair share of failed shuffle implementations. The typical implementation was

    for (i=0; i<n; i++) {
    r = random(n);
    swap(a[i], a[r]);

    This is very similar to Fisher-Yates but doesn't result in an equal distribution.

  • Brandt 2010/02/28, 15:44

    The problem in Chrome is that you’re submitting the form, which causes the page to re-render. Easiest workaround is to change the button from “type=submit” to “type=button”.

  • Jonas B. 2010/02/28, 15:54

    That’s not the naive solution. That’s just … wrong. I don’t think anyone would expect the outcome to be random if you sort by a weird changing comparator function. That outcome will depend on the sorting algorithm used and just be weird.

    The naive solution would be to assign each element a random element in the output array. I think that would work too, but be slower and involve floating point divisions and stuff that better algorithms can dispose of.

  • Michael Clark 2010/02/28, 15:56

    Solutions to creating software answer are flawed, the software is running on hardware incapable of producing a random number. No matter how complex the software it’s limited to the incapability/capability of the hardware.

    You need to use a non random hardware input such as giga readings from a radioactive source – {which is a random scientific event’.


  • Rob 2010/02/28, 16:21

    @Brandt, thanks. I changed the form. Hopefully it works better with Chrome. Would be interest in how the results are there, and Opera and Safari as well.

    @Nick, do you see the problem with that algorithm? The items at the end the array can be swapped out less frequently, though there are more opportunities for values to be swapped in there. In other words, elements in the front have a greater chance of being put back in the front from later exchanges. So it is biased with respect to the original sort order.

    @Steve, for example a bubble sort could easily end up in an infinite loop. Remember, that sort will continue making passes over the entire array until everything is pairwise ordered. So with a random comparator, the chance of any pass finding all elements sorted is 1 in 2^N, regardless of the number of passes previously made. OK. Not infinite, but can get quite long quite fast. With a 100 element array, I doubt it would finish in our lifetime. Of course, one hopes that bubble sorts would not be used in Microsoft code. But one also would have thought that a broken random shuffle algorithm would not be used either.

    @Tim — Re: the third good solution… That is simply bad editing on my part. I was going to go into the original Yates-Fisher versus modern variations. But I never got there. I’ll clean that up.

  • Sinetheta 2010/02/28, 16:23


    You’re assuming that the last position on the list is the least desirable; unfortunately you seem to have missed out on an important lesson here.

  • Anon 2010/02/28, 16:24

    You mix up the concepts. The distribution of browsers is random but non-uniform. They are completely different things. I can even be pedantic and say EU only required randomness, not a uniform distribution, but I don’t have access to the agreement between EU and MS and therefore will not say that.

  • Tom 2010/02/28, 16:32

    Nick, what is your evidence that swapping(i, rand(n)) for i = 0..n doesn’t make for equal distribution? Can you prove it with test runs? Every element is given a shot and being swapped with any other element at least once. Some will be swapped multiple times, but that’s ok, since everything is symmetrical.

    What would be problematic is if one swapped (rand(n), rand(n)) n times, which is what I saw people do when I used to ask this as an interview question.

  • Shay 2010/02/28, 16:44

    @Darren Kopp: If the order didn’t matter, you’d just show the same order to every user and be done with it. Presumably the order is considered to exert some influence, thus they want each browser to fill each spot a roughly equal number of times in the grand scheme of things, so that the differences even out overall. Also, there’s not necessarily only one spot that’s best.

  • Gunstick 2010/02/28, 16:46

    “Where serious money is on the line, such as with online gambling sites, random number generators and shuffling algorithms are audited, tested and subject to inspection.”

    Well real gambling sites should use the dice-o-matic. That’s real random :-)
    http://www.youtube.com/watch?v=7n8LNxGbZbs Amazing…

  • zak 2010/02/28, 17:17

    It is possible to use solution 1 (n!) without computing all possibilities using a recursive-type algorithm. Here is a python illustration of what this might look like (in c it is done better recursively):

    while m > 1:
    m = m – 1
    yeild A.pop( r / m! )
    r = r % m!
    yeild A.pop(r)
    yield A.pop()

  • Dave 2010/02/28, 17:26

    Tom, regarding why swapping (i, rand(n)) for i = 0..n isn’t a good solution: Consider how this algorithm would run on a set of three items. There are 6 possible permutations of 3 elements, and the desired goal is a uniform distribution – each permutation having equally likely probability.

    Each iteration through the loop there are 3 possible outcomes, each equally likely. And this is done 3 times, so you have 3^3 = 27 different paths, each with a probability of 1/27. Now count how many of those paths lead to each permutation. If the distribution is uniform then there should be the same number of paths to each permutation, but there is no way to divide 27 by 6 evenly, so some permutations will have more paths than others, and thus the distribution will not be uniform.

    Note that decreasing the range of the random() by 1 each iteration solves this particular problem because there will be N! paths (instead of N^N), and the number of permutations is N! – each permutation has exactly one path (and all paths are equally likely).

  • Stan 2010/02/28, 18:00

    @Sinetheta: The algorithm is flawed, but in this application, who cares. The concept of randomness in this application is to prevent bias towards Microsoft’s own browser and nothing more.

  • Lucian Wischik 2010/02/28, 18:08

    You write “Depending on the exact search algorithm used, it could have done a few exchanges operations and then prematurely stopped. It could have been worse. It could have lead to an infinite loop.”

    To be precise, ECMA 262 (the Javascript specification) says that if a non-stable comparison function is used, then the results are implementation-defined. And the Microsoft implementation (at least of JScript) specifies that a non-stable comparison function *will* yield a random order. http://msdn.microsoft.com/en-us/library/k4h76zbx(VS.85).aspx

    So: no infinite loop. (but I think they should rewrite their specification to say “will yield an unspecified order”.)

  • Bill 2010/02/28, 18:15

    You can do it O(1), also. Just create a permutation generator which permutes the set of array indices. This is easiest and most efficient when dealing with sets with a power-of-2 order.

    I use this to “shuffle” records (DNS, etc) with an iterator that uses a callback much like at issue here. The iterator simply uses a counter, i. For i = 1..N it just returns array[permute(i)] at each step. By the time i is N you’ve returned each and every record, but out-of-order. No intermediate step was necessary. permute(i) is O(1).

    How do you create a permutation generator? You can create a simple 2^N-bit generator using a single random seed in combination with exclusive-ors, shifts, and optionally substitution boxes (those operations simply “mix” the bits; other operations might not guarantee a one-to-one mapping of input to output). There are other ways to do it, I’m sure, but this is what I’m most familiar with. This is the basis for many cryptographic algorithms. If all you wish to achieve is some degree of statistically random distribution, you don’t need to be an expert. Just hack together an algorithm and test to verify the strength of the distribution. Indeed, if you’re doing anything other than a straight Fisher-Yates shuffle with a known good PRNG, then you want to test anyhow.

    Of course, O(1) might still take more wall time than the other algorithms for smaller sets. But even so, I typically use it because it doesn’t require any extra data structures other than two integers for the seed and counter. The simplicity extends to the actual code, as well, which I find most important. Plus, there are many occasions when you wish to randomize a set but then only retrieve a subset. With this method, there’s no wasted work.

  • Steve Weller 2010/02/28, 18:20

    A simpler solution is this two-step process.

    1. Microsoft, the browser makers, and the EC to sit down together ONCE in a room, draw the browser names from a hat and write down the order in which they appear.

    2. Then when those browsers are presented to the user, they are shown always in the same order as chosen in 1), but with a randomly-added shift applied. This involves a single, simple random() call that is hard to get wrong.

    For example, if the order the browser names were drawn from the hat was Safari, Explorer, Firefox, Opera, the user would only ever see one of the following lists presented on the screen:

    Safari, Explorer, Firefox, Opera

    Explorer, Firefox, Opera, Safari

    Firefox, Opera, Safari, Explorer

    Opera, Safari, Explorer, Firefox

    It’s hard to screw up the implementation, and it’s demonstrably fair.

  • Mits 2010/02/28, 18:28

    @Anon 4:24 – You hit the nail in the head. I doubt any of the EU or MS officials or programmers know the difference. Being pedantic like you however, I’d reverse the requirement and say that EU should have required just a uniform distribution. In my humble opinion, “randomness” is NOT required. Here’s an example:

    Imagine this scenario: The computer upon Windows installation contacts a MS site that uses a global installation counter – each new installation would increase the counter from N to (N+1) and then present a browser order according to (N modulo 5!). This is a totally deterministic process, with no randomness at all (statistical tests for randomness would fail because of the autocorrelation), which however would lead to perfect uniformity: at any given time instant, each browser would have been placed in each of the 5 positions with a percentage of precisely 20%, as required !!!

    The same kind of uniformity could be produced by using the installation serial number (licence) of Windows: since the licence key space is well-defined, the order of browsers could be also well (uniformly) defined from the serial number itself. There might be a problem with volume licences, but VLKs are a small percentage of total installations.

    However, on a single offline computer, with no knowledge of history (what ballot was presented globally) or without a licence key, programmers have to resort to mathematics in order to produce uniform (not necessarily random) distributions. This is an application of the law of large numbers: if the ballot is uniform on the same computer, it will be uniform globally.

    In conclusion, the only requirement should be that the browser ballot should produce uniform results. If the ballot operation is going to be performed online, this is an easy task. Otherwise, the only way to uniformity is through randomness – and as the fine articles have shown, this is not a trivial problem.

  • but does it matter ? 2010/02/28, 19:18

    You know you talk only about one set of random positions.
    But besides that i think there is another random part in this game
    (well statistics is origin is gaming math after-all)..
    The fact that also unknowing people (should) choose randomly.

    However people more often these days look at how much bling bling valeu something has.
    And simply this results in a graphical evaluation of icons.
    Yep i know its not great, but that psychology at work, and so i think Firefox is better off…

    To get a purely random option for a web browser people should gues a number and get a unknown browser based on that number. At least that would be the correct way. Or for example create a random number based on mouse movement.

    Anyway i think you and I know that any math based shuffle technique is always a pseudo random generator, unless you get some external random data, like quantum noice random engines. Well some CPU’s these days have such a function, but not all of the home pc have such options.

    Anyway i think the best result of this EU is that know Microsoft creates their backend applications to work with Firefox and some others, you no longer are forced to use IExplorer because your exchange servers only works with it, no these days their new server product are (finaly) keeping respect to other web standards dont get me wrong microsoft isnt against standards, but most of them they copyright (but not all)) Now as a result finally Microsoft has stopped doing this on the web browser market.
    Which is pretty nice if your a web designer, and cool if your a consumer who doesnt like to be binded to Iexplorer.

    (now if only SAP went to firefox too… and some others..)

  • Thomas B. 2010/02/28, 19:20

    They should just put the list in a top-right to bottom-left diagonal orientation. This is how simultaneous lead billing problems are solved in Hollywood.

  • Tom 2010/02/28, 19:36


    I could not have asked for a better explanation of why that doesn’t work! Apparently, I should not hire myself. I also confirmed by experiment that the distribution was not even close to uniform for larger orders.

    This javascript seems to work well after experimentation. It uses your idea of decreasing the random set on each iteration.

    function shuffleArrayInPlace(items) {
    for (var i = items.length; i > 1; ) {
    var j = Math.floor(Math.random() * i);
    var t = items[--i];
    items[i] = items[j];
    items[j] = t;

  • Anon 2010/02/28, 19:43

    @Rob Weir

    You say you used the Fisher-Yates implementation from Wikipedia… but the Javascript implementation on that page has a comment in the code stating that the way the random number is derived will introduce modulo bias into the results. And looking at your code, I see the same comment. You couldn’t have missed it while copying and pasting, so I’m curious to know how you missed this caveat in your article since the whole point of the article was to lay bare (inadvertent) bias.

  • daniel 2010/02/28, 20:16

    You make a very good point, but I think that you miss your own point in the summary. This issue isn’t about throwing a random number at something, its about sorting with a variable comparator. You point to this with an example of a database, just assign a random number to each element and then sort on that number rather than generating a random number in the comparator. For all the fuss on slashdot, this really is just a simple error, but one that should have been picked up in any code review. Great article though.

  • Rob 2010/02/28, 20:22

    @Anon, You need to worry about modulo bias when the number of possible combinations of your elements is an appreciable fraction of the number of random numbers your generator can produce. In this case we have only 5! = 120 possible orderings, so we’re fine, presuming that Internet Explorer has a non-pathological PRNG. The test of the results with the Fisher-Yates Shuffle suggests this is the case. But if you have a larger array, say representing a deck of 52 cards, then modulo bias can be more of a problem. Note that this is not a specific weakness of the shuffling algorithm. It occurs with the other shuffling algorithms as well. It is simply a statement that the PRNG is inadequate to represent the number of possible states in the problem.

  • Firoze Lafeer 2010/02/28, 20:27

    The agreement that was agreed to with the EU just says that the top 5 browsers should appear “randomly”. So the requirement could have been more precise. But still, that’s not an excuse for such terrible code.

  • Firoze Lafeer 2010/02/28, 20:28
  • RickRussellTX 2010/02/28, 20:44

    “The lesson here is that getting randomness on a computer cannot be left to chance.”

    Oh ho, Mr. Algorithm Man, you just made my day.

  • David 2010/02/28, 21:29

    For the lazy among us could you show how to 1) assign the data to a variable and 2) how to run the chisq.test() against it in R?

  • Rob 2010/02/28, 22:02

    @David, You do the chi-square test on the raw counts. You first get the table into a CSV file on disk. So headers in quotes, separated by comma, and the rows, with numbers separated by comas. You don’t want the “Position” column. You just want the counts.

    Then, in R, try this:

    data <- read.csv(“/path/data.csv”)

  • RickRussellTX 2010/02/28, 22:34

    FYI, Excel has a Chi-Squared test. Look up CHITEST in the help. I believe it returns the P-value.

  • Joshua 2010/02/28, 22:40

    I can prove that the qsort() function I have on hand as source code terminates on an ill-behaved comparison function. But yeah order by random() isn’t going to be very uniform for it.

  • Bugstomper 2010/02/28, 23:08

    It is ironic that Rob was able to analyze the bug and the solution so thoroughly because the code was in an open source javascript file.

    If the code had been developed in the first place in typical open source fashion, the bug may have never made it into production.

    By the way, I think another solution that only requires generating one random number is to use the factoradic representation of the permutations. See Wikipedia entry for Factoradic in the subsection on permutations. It is an algorithm for finding the Nth permutation given an integer N. That would let you generate a random integer N from 0 to 119 and compute from that the Nth permutation of the set {0, 1, 2, 3, 4}. That is elegant in that you only need one random number, but if you squint the right way at the algorithm for finding the Nth permutation it looks an awful lot like what you go through for Fisher-Yates.

    Actually, with only five items to shuffle, just about any way of doing other than what Microsoft did would work. You could even generate one random number and compute it mod 5, mod 4, mod 3, and mod 2, to determine which of the five to choose first, then which of the remaining 4, remaining 3, remaining 2, and the one that’s left.

  • Steve Weissman 2010/02/28, 23:22

    IE, with its “world’s most widely used browser” culmination, is most likely to be placed last, and it is kept away from the long wall of Opera descriptive text, while most likely to be next to the short and unthreatening (also in market terms) Safari. UI designers will have tested enough people to find out what they are most likely to choose and how the choice can be influenced; the “naive” algorithm has taken advantage of this while providing an excuse to the powers that saw political advantage in forcing Microsoft to accommodate for its competitors.

    The only naive thing here is the geek so overwhelmed by his understanding of first-year statistics and computer science that he forgets that business is the application of technology to profit, not the perfection (by some Platonic ideal, I assume) of technology itself. You are being forced to do something you don’t want to do and you can afford the smartest team on your side — why wouldn’t you employ them to resist this force?

    (Reading the MSDN documentation, of course, a Microsoft JScript sort abused like this returns items in a random order. Since reading the documentation was the first thing that came to mind for me, before performing a statistical test on the implementation and making proud guesses as to what went “wrong”, I couldn’t help but give wry smile at in the evident sense of humour that the ballot developer has.)

  • Joseph 2010/02/28, 23:43

    Nice article, but there is a major mistake!

    In the case of this problem. A unique user/pc will get this script to run once. In order to repeat your test, you have to get 10,000 machines at 10,000 different time spread over some time. By doing so, each random call will be “independent” of the other calls. In you case, each call is somehow dependent on the previous CPU clock , instruction … thus never random.

    In the real world case, where the Microsoft script is running, You will get real random distribution.

    So my suggestion to you is this:

    Add the script to this page, and let it RUN and report back the result . After 10,000 users visiting this page, you can then POST the new results. I would love to see them. You can limit one run per IP.


  • Dorian 2010/03/01, 01:10

    Although I am a scientist I am not a statistician and a very mediocre programmer. Can someone describe for me in laymans terms why the solution Microsoft’s programmer used doesn’t work? To my eye it seems that the code above should produce a nice array of random numbers which could subsequently have been sorted.

    In addition, why did the programmer use “0.5 – rand()” instead of just “rand()”?


  • Fwip 2010/03/01, 01:11

    Bill – Isn’t your algorithm O(n)? It seems that you do one step for each of n elements, which requires at least n operations. If your algorithm was O(1), it would take the same amount of time to shuffle 5 elements as 5 million elements, right?

    Not meaning to be a jerk, just checking to see that I understand.

  • Kamil 2010/03/01, 01:41

    Interestingly there was very little bias for a 10k test using Chrome

    Position I.E. Firefox Opera Chrome Safari
    1 2446 2477 1646 2195 1236
    2 2462 2439 1658 2207 1234
    3 1879 1977 1697 1885 2562
    4 1776 1689 2205 1902 2428
    5 1437 1418 2794 1811 2540

  • Sid 2010/03/01, 02:50

    If I am not wrong… On The ability to generate random numbers… UNIX has a /dev/rand something that is supposed to be as old as linux and it picks up random electrical disturbances to generate random numbers… Why/do they? we have something like that in Windows?? That’d be truly random no?

  • Bill Parker 2010/03/01, 03:46

    PHP shuffle()

    Can someone please tell me how PHP implements shuffle()? I can find plenty of hits explaining how to use it (umm, it couldn’t be simpler really) but none explaining what lies behind it? But I’m inclined to bet it’s not based on Fisher-Yates. So what exactly IS it?

  • Peter Bierman 2010/03/01, 04:26

    I’m surprised no one has checked which position is actually the most favorable. Everyone has assumed it’s the first slot. Are there behavioral studies to support that assumption? What if disinterested users typically click the last option?

  • Erik 2010/03/01, 05:40

    User generated randomness!
    I think the simplest solution in this case to achieve effective randomness and a uniform distribution is to just use system time to loop through all 120 possibilities. The screen would just change every second and since no-one can predict when a user is going to request the webpage: voila you have perfect randomness plus uniform distribution. Just not when you use a script, but then that wasn’t really the question…

  • Martin 2010/03/01, 06:18

    @Mits: Sorry for being pedantic, too, but: I would agree that a truly random distribution is PROBABLY not required — but I believe it is harder to take a non-random with some bias and then prove that this bias does not matter, than just take a uniform random distribution which cannot by definition have any bias. For example, if we follow your proposal and just enumerate all the possible permutations and present them one-by-one, the browser in the first position stays there for quite some while. That would e.g. enable an attack where somebody who hates Firefox could issue a lot of simultaneous requests to “push Firefox off the first position”. Okay, maybe this is a bad example, but it illustrates the kind of things you have to think about when deliberately NOT taking a random distribution.

  • Martin 2010/03/01, 06:27


    I doubt that O(1) calls to a truly random “n-element permutation generator” are really faster than O(n) calls to a random number generator. Your “permutation generator” would have to do something comparable to Fisher-Yates inside, otherwise it is presumably not correct.

    I can solve the Travelling Salesman Problem in O(1), given a solve_tsp() function.

    Also, you cannot “verify” the randomness of a process by running statistical tests, only fail to falsify it. “Testing can never show the absence of bugs, only their presence” (Dijkstra). If the randomness of that process is critical to the security of your application, I would hesitate to use some “hacked-together algorithm” using some swaps. Please use some tested-and-proven standard implementation.

  • Anon 2010/03/01, 08:00

    Considering most people who view the browser choice screen will just want to keep browsing “like they used to” and will, after speaking to their technical friends, make sure that IE is still installed, it’s all a bit academic really. But great article and analysis.

    (source: my own experiences over the last couple of days)

  • Rob 2010/03/01, 08:33

    @Steve Weissman, I’m very aware of the business, legal and political context this ballot screen is in. Certainly, it does not require “platonic perfection”. I never suggested it did . That’s a red herring. But I doubt it is satisfactory that some browsers are put in the first place twice as often as others. And this just isn’t about Internet Explorer. Look at the disparity between Chrome and Safari as well. In the end, the degree this does or does not meet Microsoft’s business, legal and political needs will be illustrated by how quickly they fix the bug, now that they are aware of it. Come back in a few days and tell me that this was not important enough to warrant correction.

    @Dorian, The problem appears to be the way the custom comparator function interacts with their sort() routine. Without seeing the source code for Internet Explorer, it is hard to say more. But typically sort() routines are written to give fast results given a fully-ordered comparator function. How they behave when given a unstable comparator function is going to be implementation dependent.

    @Joseph, are you suggesting that these results are not uniform because of non-randomness in the PRNG caused by the tests not being run on a single machine and not “spread out over some time” and that if I used 10,00 different machines with 10,000 different IP addresses I’d get truly random results? If that were true, then my tests would show biased results when running the Fisher-Yates algorithm, since that relies on calls to Math.random() as well. But that algorithm works fine. So this suggests the problem is in the choice of algorithm not a problem in the PRNG.

  • Jeff Walden 2010/03/01, 08:36

    Lucian, Microsoft’s documentation doesn’t say that their sort function shuffles if the compare function isn’t consistent; that’s community-contributed content akin to the comments on PHP’s method documentation pages, not normative statements.

  • John 2010/03/01, 08:58

    Awesome job with your use of R and explanation of sorting algorithms. What you failed to explain, however, is the relevance of your analysis to the end-user’s choice. This isn’t the Monty-Hall problem; the browser isn’t hidden behind the door. Presumably, the end-user is acquainted with one or more of the choices, and will select their *previously-chosen* preferred-browser. Where’s the evidence that browser share will be affected by ordering? Do you have *any*? Meh, this barely rises to the level of tempest in a teapot.

  • Rob 2010/03/01, 09:27

    @John, It was not the purpose of this post to analyze the reasons for an unbiased browser ballot screen. I’m taking that as a requirement which a programmer needs to implement in their code. In this case the requirement comes from an agreement made between Microsoft and the EU to settle their anti-trust case. It would be a fascinating discussion, I am sure, to debate why other browser vendors insisted on having a randomized ballot, and why Microsoft agreed to do this, but that would only make a too-long post even longer. But if you want to go down that route, I’d start with a read of Jakob Nielsen’s “The Power of Defaults”.

  • mnicky 2010/03/01, 09:52

    A couple of years ago, I saw a research showing, that when man opens a newspaper, he is attracted mostly by the titles of the right upper part of the newspaper (with probability of 36%; where normally it should be 25%).

    And we must also take the fact, that user visiting BrowserChoice can also scroll to the right (and then the first places are not visible), into account…

  • Erick G 2010/03/01, 09:55

    Am I suppose to care about this issue?When someone makes me an offer, I don’t know for you, but for me I have other things in mind than the algorithm used in order to present me a choice. I am not a normal tech, not nerd enough I think…

  • Wol 2010/03/01, 09:59

    Rob – you said you replicated DSL’s results, and then asked “is that statistically significant?”. You shouldn’t have needed to run your chi-squared test!

    It’s reasonably well known in the world of statistics that someone ran a test and concluded that “Israeli fighter pilots have more girl children than boys”. Whoops. That’s what past data says. Only if future data also agrees can you decide that the first study meant anything (and if the second study *does* produce the *same* result, then you can conclude with a high degree of certainty that you have picked up something of significance.

    I don’t know what the chances were of DSL producing the results they did were. But the chances of your test coming up with the *same* results by chance are pretty miniscule. Just because you’ve won the lottery once, it doesn’t increase your chances of doing it again. The mere fact that you could reproduce DSL’s results said that MS’s “random order” wasn’t random, without having to run any other tests at all.

    (though proving it for sure with chi-squared wasn’t a bad idea…)


    • Rob 2010/03/01, 10:27

      @Wol, if the DSL.sk test was based on only 10 iterations, and my test was based on only 10 iterations, then if the results matched, you might still doubt the significance. When I first saw the DSL.sk results I was very skeptical. It looked very much a like they were cherry picking their results. That’s what lead me to try to reproduce it. If you have a 5×5 table of outcomes, the chances of all of them being 2,000 is a less than any given one of them being 2,000. Ditto for having them all be in an arbitrary range, say 1800-2200. My initial concern was that, short of doing a Chi-square test our intuition fails us. There is a hindsight bias if you run the test and then decide which combinations should be tested for significance. The Chi-square test neatly takes this into account, looking at the whole of the table.

      So I set out to show that the results indeed were random, but when I ran the tests, I found that the DSL.sk observations were accurate, and that the results were indeed biased. Thus this blog post.

  • Rob 2010/03/01, 11:30

    Some things to try, if anyone feels like pursing this further:

    1) Repeat the same tests, on Internet Explorer, but instead of using Math.random(), use a source of true random numbers, such as from ww.randomnumbers.info. You probably need around 100,000 randum numbers for 10,000 iterations of the test. Put them into a Javascript array and then iterate through them. Demonstrate that the algorithm used is flawed even when given that as input.

    2) Or alternatively, dump out the pseudo random numbers actually used by I.E. and run them through the Diehard test battery: http://www.phy.duke.edu/~rgb/General/dieharder.php

  • Tim 2010/03/01, 12:37

    A thought on the comparison function.

    The effect of this comparison function is rather mysterious given that the sorting algorithm used by javascript differs for various browsers. This can have rather nasty effects. Any algorithm that checks whether the array is sorted before terminating is quite likely to go into a loop with nondeterministic running time, i.e. there exist incredibly improbably but never terminating sequences of computation. This can happen with say bubble sort checking that no swap has been made (http://en.wikipedia.org/wiki/Bubble_sort#Pseudocode_implementation).

    But is it possible for this to run just fine? I am somewhat confident that the sequence of comparisons that quicksort would make would satisfy rules 1-6. (To handwave, this is because no pair of elements is compared twice, and once a pair has been separated by a pivot they can never be compared.) It is not necessarily going to give a random permuatation. If the pivot is selected a certain way, say always from the back of the list, the positions of that element is going to follow a binomial distribution within its range. It will appear after k elements where k is the number of things the comparison sort said it was greater than in a range containing n elements. (This is the easiest to see with the first pivot.)

    Now here is the stumper, well at least it stumped me. If quicksort is done with uniform random pivot selection, do we have a “good” random permutation algorithm? My suspicion is that we do, but I would love to know anybody else’s thoughts on this.

  • Daniel Elstner 2010/03/01, 13:29

    Houston, we’ve got a problem: http://www.javascriptkit.com/javatutors/arraysort.shtml
    Scroll to the bottom…

  • hark 2010/03/01, 14:24

    So what if the results are rigged to show IE on the right side on a horizontal list of 5 or 6 items? Which side will people pick most often? The answer is the one on the right. See this link here: http://www.newsweek.com/id/210194

    It seems that microsoft was using some carefully planned algorithms to benefit them, at least so it would seem in a small list where the majority of right-handers will pick the items on the right side of the screen, rather than the better browsers on the left. Luckily, I use my left hand, so I am immune until such a time as they can figure out my handedness. This would be my guess as to why MS would want to show up last on the list, especially if there is a menu on a horizontal plane. :) Cool, huh?

  • Renee Marie Jones 2010/03/01, 14:52

    What is really instructive here is that Microsoft — some of the “best” programmers on earch, according to their proponents — are too stupid to compute a random permutation. And, apparently, too arrogant to even look up the correct algorithm.

    Lots of Microsoft bugs are the result of this arrogance. This kind of carelessness and arrogance is, I think, the major reason why all Microsoft products are such junk. They are successfull — in a profit sense — only because of a combination of illegal marketing activities, abuse of the leagl and patent system, and an almost unlimited ability to spread lie-based propaganda. It only works because the general public is so terribly ignorant of science and mathematics.

    • stroyde 2013/12/29, 15:35

      @Renee Marie Jones

      Ridiculous comment. The method in question was “not invented here”
      as seen from Daniel Elstner’s post:
      “Houston, we’ve got a problem: http://www.javascriptkit.com/javatutors/arraysort.shtml
      Scroll to the bottom…”

      Yes, Microsoft just went with a recommended method, and anti-Microsoft zealots are no less vulnerable, including the ones handing out blanket condemnations (which would probably be the whole lot).

  • Rob 2010/03/01, 14:59

    There have been a few suggestions that my test is biased because it requests random numbers in a tight loop, and all on one machine, rather than on different machines, running at different times. So I modified my test to dump out a list of all of the random numbers retrieved in the calls to Math.random() during a 10,000 iteration run. You are welcome to download then and apply whatever tests you wish to them. I don’t see anything here out of the ordinary.

    The interesting thing was how many times Math.random() was actually called. It was called 272,991 times for 10,000 iterations of the Microsoft Shuffle. So that tells us that every sort operation of the 5 element array required around 27 calls to their comparator function. Interesting. But as anyone who has seen a magician do a card trick knows, the intensity of the shuffle says nothing about the randomness of the results.

  • Daniel Elstner 2010/03/01, 15:33

    @Rob: Is it always the same number of calls? I’d expect the number of calls to the comparison function to be random as well.

    Also, the whole discussion about bias in the analysis completely misses the point. This is broken code, plain and simple. An API is being misused, the result is undefined behavior. It’s purely by chance that they even got something that looks like a random permutation at all.

    • Rob 2010/03/01, 16:13

      @Daniel, the number of calls varies, but averages out to close to 27.

      @Bugstomper, there is nothing in the EcmaScript standard that requires the use of quicksort to implement the sort() function. So it is not a safe way to code a random shuffle in Javascript. But still that is interesting to know.

  • Bugstomper 2010/03/01, 15:34

    Thinking about Tim’s comment on quicksort, I came up with a slightly less hand-wavy argument why the Microsoft Shuffle should work ok if the sort algorithm is quicksort – You can write out the steps of any one execution of quicksort as a series of binary choices that progress through a binary tree of all possible runs. Since the algorithm never compares the same pair of elements twice, then making a random choice at each comparison could only have the effect of selecting one of the possible execution paths, with all being possible and all being equally likely. It would have the effect of choosing a permutation at random, sorting it with quicksort, and using the result to let you know what the initial random permutation was. It would require as many calls to random() as you have comparisons, which with quick_sort is O(n log(n)).

    But paraphrasing Knuth, the above is only a proof of correctness, so we don’t know if it will work without testing. I coded up a variation of the Javascript quicksort code from http://en.literateprograms.org/Quicksort_(JavaScript) in which I added an argument to provide the comparison function, and added to the test here a copy of Microsoft Shuffle that call quick_sort(array, RandomSort) instead of array.sort(RandomSort).

    I also added calculation of the time in each one, var startTime = (new Date).getTime(); before the loop that iterates the requested number of times, and outputting ((new Date).getTime() – startTime1) before the table of results.

    The Microsoft Shuffle using quicksort produced the same statistically even distribution as the Fisher-Yates algorithm. The times were interesting. On my MacBook using Firefox 3.6 for 100,000 iterations the incorrect Microsoft Shuffle took 2.28 seconds, the quicksort Microsoft Shuffle took 4.61 seconds (but got correct results), and the Fisher-Yates one took just 0.185 seconds.

  • Sid 2010/03/01, 16:08

    @Rob I would like to ask you one thing and that is that both the Javascript version which is incorrect and The fisher algorithm use Math.random(). Now you were able to empirically proove(using chi square) that fishers thingy gives a uniformly distributed permutation, but since Math.random() is behind them both its hard to see how even n(n-1)/2 shuffles of the javascript function would not produce a uniformly random result? I mean if Math.random() is the only important thing here then the fisher algorithm would also fail right?
    I mean the empirical evaluation is fine but is there an analytical answer as to why the average distribution is not random in the MS version?

    • Rob 2010/03/01, 16:35

      @Sid, Since we don’t have Microsoft’s Internet Explorer code to inspect, there is no way to know. Who knows what IE is really doing behind the scenes? It is possible that they are not even using a single sort routine. The could be branching to one routine for long arrays, another optimized for short arrays and maybe even some special logic to “unroll” the sort logic for very short arrays of 2-5 elements.

      Firefox gives non-uniform output as well. Maybe we can look at the sort() implementation there? Anyone have a direct link to that source file?

  • Sid 2010/03/01, 16:14

    @Bugstomper So does that not mean that the only explanation here is that there is something other than just the sort function being called… Its not just quick sort or any sort for that matter… Forget the incorrectness… This has just to do with a generating a random permutation of N elements. That java script function might have been defined
    function randomresult(a,b){
    return 0.5-Math,random();
    for all we care…
    And by calling O(NlogN) or O(N^2) calls to any comparison sort, the results should be uniformly random…
    Some thing else is obviously wrong…

  • Stefan W. 2010/03/01, 17:26

    I’m sorry, but the pagelayout does not allow to increase the font-size in firefox with ctrl-+, and reading the text without scrolling left and right. That is far more annoying than a poorly done browser-choice – who need’s that?

    Well – I would go for a simple solution: Pick a browser from 5 by random, from the rest pick one, pick one from 3, pick one from 2, and you’re done. 5 calls to random while waiting on a user reaction is nothing worth considering performance issues.

    val r = new java.util.Random

    def sort (n : List[Int]) : List [Int] = n match {
    case Nil => Nil
    case _ => val x = r.nextInt (n.size)
    n (x) :: sort (n - n(x))
    for (i <- (1 to 10)) println (sort (List(1, 2, 3, 4, 5)).mkString ("\t"))
    5 2 4 3 1
    4 3 1 2 5
    2 4 3 1 5
    3 2 5 1 4
    5 3 4 1 2
    1 2 5 4 3
    5 3 2 4 1
    4 5 2 1 3
    5 1 3 2 4
    1 4 3 2 5

    It’s just how you would do it by hand – like a lottery-machine works.

  • commenter 2010/03/01, 17:41

    Rob, nice analysis. :)

    Sadly, I can imagine making this kind of schoolboy error.

    My implementation wouldn’t bother with any sorting:

    * take a random item from the list of browsers B
    * add it to the end of the displayed list.
    * keep doing this till B is empty.

    I think that’d work.

  • Anon 2010/03/01, 17:48

    Nice work!

    I can’t believe such an elementary mistake was committed, but I’m sure it will make a great example in an introductory algorithms course :-).

    If in doubt, consult Knuth.

  • Mits 2010/03/01, 18:29

    New glorious opportunity to run checks for “randomness”: Microsoft offers browser choices to Europeans (http://news.bbc.co.uk/2/hi/technology/8537763.stm) . The candidate alternative browsers now are 12 !!!

    “The Opera, Firefox, Chrome, Safari and Internet Explorer browsers are randomly ordered on the first section of this screen. Another seven browsers, namely Sleipnir, Green Browser, Maxthon, Avant, Flock, K-meleon, and Slim, will be randomly ordered on the rest of the screen. They can be viewed by scrolling sideways. ”

    I think one does not need a PhD to see why IE does not appear in the “second section of the screen”!

  • Bugstomper 2010/03/01, 19:46

    @Sid, the explanation as to what is going on with different sort algorithms can best be expressed with the old saying “Garbage in, garbage out.” The specification for the EcmaScript sort function is that the second argument is a function of two arguments, call them a and b, which will be passed two elements of the arry being sorted and must return -1, 0, 0r +1 depending on whether a comes before, is the same, or comes after b in the sort order that you want.

    The call to Math.random violates the terms of the specification, giving the sort algorithm garbage information about whether a comes before or after b. In the case of quicksort, the algorithm happens to act on the results of each comparison and not only never revisits the same pair, it separates the elements of the pair into two sections of the array that are then sorted separately. The bogus result of the compare function jumbles things up but doesn’t come back to haunt the rest of the processing.

    Contrast that to a sort algorithm that does something like this three element example, calling the elements of the array a, b, and c:

    compare(a,b), and compare(b,c). If the results are a<b and bb and b>c then the answer is c b a, again with a 1/4 probability of getting this result

    If the results are ac then you need to look at compare(a,c) and get either a c b or c b a each with 1/8 probability.

    If the results are a>b and b<c then you also need to look at compare(a,c) and get either b c a or b a c, again each with 1/8 probability.

    Result: Two of the six permutations each occur 1/4 of the time, the remaining four permutations each occur 1/8 of the time.

    Yet a different sort algorithm that chooses what to compare in a different way would produce yet different inconsistent results that are not necessarily a uniform distribution even if compare() has a uniform random distribution.

  • Bill 2010/03/01, 23:14

    Fwip, Martin:

    It would be O(N) if you wanted to loop over every index, true. I said O(1) for the following reason: If you have 128 elements and want to take 3 uniformly at random, using Fisher-Yates you would have to first shuffle all 128 elements in O(N) time, and then take, say, the top 3 items (O(N) again). Using the generator method, the first step isn’t needed. The “shuffle” is O(1)–initializing the generator–though subsequently taking 3 items is indeed O(N). So, yes, very strictly speaking whatever you’re doing is asymptotically O(N) either way, but a better way of looking at it is that a particular Fisher-Yates-employing method is O(N) + O(M) while the method I described is O(1) + O(M). The distinction can be useful.

    As for “hacked-together algorithm”: For my DNS library I implemented a 16-bit permutation generator (for ports and QIDs) using a Luby-Rackoff Feistel construction where the strength of the construction is provably a function of the strength of the mixing function F. F needn’t be invertible–you can use SHA-512 if desired–but I used TEA. So, you can use this method with similar confidence as a Fisher-Yates shuffle. (You wouldn’t want to use TEA to protect your passwords, but as the mixing function for a 16-bit block cipher you can bet the resulting distribution will be quite indistinguishable from random.)

    But, for “randomizing” my resource record iterator (e.g. to contact nameservers in non-linear order, or to “shuffle” SRV records with the same priority) I indeed hacked together a naive permutation box. Horses for courses.

  • Bugstomper 2010/03/01, 23:28

    Sorry about the bolding in the previous comment. I didn’t realize that the blog software would interpret the inequalities with variable b as the HTML bold tag :). Where it turns bold the HTML ate that I said “if a<b and b>c then”

    Anyway, I think I located the source code for Firefox’s Javascript sort function in

    It uses the Merge Sort algorithm, except it has an optimization that preprocesses the array by sorting each chunk of four elements using an insertion sort. The Merge Sort would then split the array into two chunks, of size 3 and 2 or 2 and 3 (I didn’t figure out which it is doing), sort each, and then merge the results.

    The merging process has the same kind of uneveness as in the example at the end of my previous comment. The merge algorithm goes like this:

    merge(array_left, array_right) {

    while there are elements in both arrays {

    compare the first element of array_left with the first element of array_right

    pop the smaller one off its array and append it to the result


    if any elements are left in one of the arrays append them to the end of the result


    Let’s say that the sort is almost complete and it is ready for the final merge of array_left of length 2 containing elements a b, and array_right of length 3 containing c d e. There is a 1/4 chance that the next two random comparisons will select a and then b as the smaller, resulting in the output being a b c d e. Similarly there is a 1/8 chance of that resulting in a c b d e and a 1/8 chance it would result in a c d b e. I haven’t worked out all the probabilities of different results with the initial insertion sort and every way you get merges, but just from that you can see nonuniform results from the merge sort algorithm.

  • Rob 2010/03/02, 00:03

    @Bugstomper, good work! I’m thinking that one could calculate the exact expected distribution. With a 5-element shuffle, the state space is small enough to do by hand. Root node is the initial state. Each call to the comparator generates one random number, and you get one child node for each of the two possibilities. The state space can’t be too big. On the other hand, I’m seeing 27 comparator calls per iteration in Internet Explorer….

    @Stefan W., Your “lottery” approach is an interesting one. (@Commenter suggested essentially the same approach). Note that it does require an auxiliary array equal in size to the original one. So it is not doing an in-place shuffle. But that’s fine since we only have 5 elements. And it certainly meets the requirements.

    To all, I just thought of another solution. Requires 2N space for N elements, but only one call to random. Distribution is uniform. Any guesses?

  • Adam Williamson 2010/03/02, 00:44

    ““Never attribute to malice that which can be adequately explained by stupidity.” –Hanlon’s razor”

    It’s a fine sentiment, but somewhat balanced out by its obvious vulnerability to exploitation by the malicious.

    Google’s very good at that. ‘What, us, want to expose all your contacts to everyone else? Nooo, noo, that was just a *mistake*. We’re sorry. We’ll fix it next month sometime. Maybe.”

  • Pepe 2010/03/02, 02:04

    The method Microsoft used is the one provided by the well-regarded javascript tutorial site, JavaScriptKit.com. See http://www.javascriptkit.com/javatutors/arraysort.shtml

    As I write this on 3/1/2010 (they may change the code because of this now highly publicized flaw), here is the code used to perform a “random sort” of and array (basically passing in the “random” comparison function into a standard sorting function):

    “Shuffling (randomizing) the order of an array
    To randomize the order of the elements within an array, what we need is the body of our sortfunction to return a number that is randomly 0, irrespective to the relationship between “a” and “b”. The below will do the trick:
    //Randomize the order of the array:
    var myarray=[25, 8, "George", "John"]
    myarray.sort(function() {return 0.5 – Math.random()}) //Array elements now scrambled”

    A slashdot comment indicated that many javascript coders consult javascriptkit.com for examples on how to code desired functions, and there’s a good chance that many websites are using this random sort implementation. Regardless of that, Microsoft should change the code anyway, so as not to be accused of anything.

  • Sid 2010/03/02, 03:07

    @bugstomper I ll try to solve this and see… Thanks for that Insight apparently all comparison sorts are not going to be the same distribution!

  • Sid 2010/03/02, 03:11

    @bugStomper Thanks for that not so trivial explanation. This is a good exercise to do… I will get on it. Apparently all comparison sorts are not going to yield the uniform distribution of permutations… Hmm definitely not intuitive…

  • Isaac 2010/03/02, 03:55

    In case anyone’s curious, mid-last-week I bashed out 1 million trial runs of the shuffle method used in the browser ballot with several different sorting algorithms and posted the distributions.

  • Bugstomper 2010/03/02, 04:42

    I just coded up javascript to run the test using the Microsoft Shuffle with a set of sorting algorithms, including the Firefox internal sort (your original Microsoft Shuffle test), Merge Sort, Merge Sort preprocessed with length four insertion sort like Mozilla does, Insertion Sort, Heap Sort, Quick Sort, and then the Fisher-Yates algorithm. When the page loads it generates all 120 permutations of {0,1,2,3,4} and sorts them all with every one of the sort functions to verify that I don’t have any bugs in the sort functions themselves.

    While tracking down why my early results in the merge sort were different from the Mozilla sort I noticed another optimization that they have in their code. When going to do a merge, the merge sort has two arrays that have each been sorted already which are to be merged. Also the two arrays are actually two contiguous sections of the array being sorted. If the last element of the left array is not greater than the first element of the right array, then you don’t have to do any merging, the whole combined sequence of those two array segments is already in sorted order. When the comparison function is random it means that each merge step has a 50% chance of being skipped.

    That explains how the sort is so skewed: The items that end up in the first partition at the start of the algorithm (IE and Firefox in this case) have a 50% chance of never being merged into the second half of the array right off the bat.

    Adding that optimization to my Merge Sort code gives me close but not identical results to the Mozilla version.

    My full results are too long to post in a comment. Rob, would you want me to email you the Javascript?

  • Rob 2010/03/02, 07:31

    @Bugstomper. Interesting. One thing I tried last night was to increase the size of the array. So instead of doing a Microsoft shuffle of 5 elements, I did a Microsoft shuffle of 25. With the caveat that this is not at all what was used in the browser ballot, the results were much starker. There was a clear diagonal band of excess counts. The effect seems to be magnified with larger arrays. If you want to send the Javascript file, my contact info is on my about page.

    @Isaac, very good. I think what we’re suspecting now is that some browsers make small optimizations to the textbook sort algorithms and that these optimizations can introduce additional ways for random comparators to lead to biased results.

    @Pepe, are you telling me that someone is wrong on the internet? ;-) If you look around you’ll also see bad financial advice, bad medical advice, bad nutrition advice, etc. It is not surprising that there is bad programming advice out there was well. As I said in the post, this is a well-known error, a trap that novice programmers tend to fall into.

    @Adam, I’ve never seen a team made up of perfect programmers. We all make mistakes, and rookies make more mistakes. What separates high quality code from more buggy code is not the presence of super-human programmers, but the presence of quality control, at all levels. A bug in released code may have been written by one person. But many other things need to go wrong for it it make it out into high-profile released code. In other words, the modern large software organization has processes that tolerate the existence of novice programmers and puts such reviews and tests in place to ensure that the output is of sufficient quality. I say “sufficient” not “perfect” because generally the consumer neither needs nor can afford perfect software.

    So, although I’m fretting about the micro-details of the algorithm choice here, because I find that interesting in my own geeky way, that doesn’t mean that the best way to prevent such problems in the future is to make all novice programmers memorize Knuth. Sure, that would not be a bad solution, IMHO. But it is not a realistic solution and probably is not the most economical solution, in terms of the effort needed to achieve a given level of quality. Remember, you could book up on Knuth all you want, but then implement the algorithm incorrectly, or introduce a typo, or off-by one error, or fail to check in the right version of your code, or any of a dozen other mistakes that even more experienced programmers make. To prevent this kind of thing in practice you need a quality process, both at the person level and organizational level. It is part culture, part science. part process. So if I were sitting in Redmond, I’d avoid beating up on the programmer who made the coding error and instead ask questions like: 1) Where are the test cases for this ballot? 2) Who reviewed an signed off on the test cases? 3)Where is the evidence that the test cases were actually run and passed?

    To all: Here is the other solution I can up with last night. Create an array of length 2N and put the browser choices in twice, like this: {0,1,2,3,4,0,1,2,3,4}. Then take a single random number from 0 to 4. Start from that index in the array, use 5 browser choices starting from there. So, if your random number was 3, you would use the sub-array {2,3,4,0,1}. This may be the fastest uniform solution out there, since it requires only one call to Math.random() and no exchanges. It returns a uniform distribution of positions. But it clearly is not random, since the relative positions are highly correlated.

  • Jose_X 2010/03/02, 08:31

    >> Where serious money is on the line, such as with online gambling sites, random number generators and shuffling algorithms are audited, tested and subject to inspection. I suspect that the stakes involved in the browser market are no less significant.

    Microsoft doesn’t need to play games creating situations they can later exploit through tiny innocent updates to IE+Windows; however, what one must wonder is if Microsoft is making enough money to give their dealings with the EU sufficient attention and resources. [Clearly someone in Q&A dropped the ball here.]

    To address this potential deficiency, has the EU considered raising a tax on computer users, a tax which can be directed to Microsoft so that Microsoft has sufficient money to better effect what the EU asks of them?

    If the EU authorities need ideas, may I suggest that they adopt Microsoft software for use in government in such a way that computer users are forced to buy Microsoft software in order to preserve interoperability? This way the “tax” on computer users can even be efficiently set and collected for Microsoft by Microsoft themselves without creating yet another layer of middlemen.

    Could I further suggest that the EU mandate the use of OOXML?

    Lastly, Microsoft is a major supporter of software patents. While it’s true that software patents, where possible within the EU, are likely subjected to a higher standard than they are in the US, the EU could try to ensure that software patents unambiguously become as permanent a part of EU law as we can expect of any law. Monopolies, especially those that can be bought in large numbers and which have wide scope by definition, provide yet another efficient way for taxes to be raised on all EU software users for the benefit of a company like Microsoft. You can’t get much cleverer than a tax on thinking and sharing. [All Microsoft has to do is to be the first to write down general ideas of where technology is almost surely to head and they will gain entitlement to 20 years of monopoly.] And we know open source developers like to think and share products that are becoming more popular by the day. Ha! Whoever thought they could bypass Microsoft taxes by creating their own software and sharing it for free was not thinking the morning they hatched out their plot.

  • Jose_X 2010/03/02, 08:57

    Some of those less familiar with open source software now may have a(nother) reason for liking open source software: Even when one company slips up, open source code (javascript in this case) means that others can catch the mistakes and provide fixes. As a byproduct, if the reason for the screw-up was bad intentions, it comes out in the wash just the same.

    Now, if we can only get Microsoft to open source, not just bits of javascript, but all the tens of millions of lines of code that constitute IE+Windows+MSOffice+MSservers+…, we’d be in business.. and a whole lot safer from malware and what not.

  • Bugstomper 2010/03/02, 09:13

    @Rob, I took another closer look at the Mozilla code to try to understand the remaining differences in my results and I realized something remarkable about it. Even though the function names and comments indicate that they are using Merge Sort, it is not the textbook algorithm. In every place that I’ve seen Merge Sort explained, the algorithm goes like this: split the array in half; recursively call merge_sort on each half; merge the results; There is of course something to end the recursion when the array is small enough. However the Mozilla code appears to be running the algorithm bottom up. It goes something like this:

    Preprocess the array by running through doing an in-place insertion sort of each run of 4 elements. Now the array consists of a series of sorted length 4 sequences, with the obvious slight special case at the end of the array.

    Run across the array doing a merge of each successive pair of runs of 4 elements, resulting in a series of sorted length 8 sequences. Note that the merge has the optimization that if the last element of the left array is less than or equal to the first element of the right array then they are already fully in order and the merge is a no-op. That optimization is the step that makes it very sensitive to using the random compare function.

    Repeat for 8, 16, 32, … until the entire array is sorted

    Clearly this is the same algorithm as the standard Merge Sort, but with the usual recursion unrolled to make it bottom-up. I modified my code to do the same thing when the array size is less than 8 (not wanting to take the time for now to rewrite it to the exact behavior when I’m only testing length 5 arrays) and the result is much closer to the Mozilla one, though still not identical.

    I found an interesting page that analyzes the Javascript sort algorithms in different browsers How Does Your Browser Sort? by instrumenting the comparison function to keep count of the number of comparisons. It occurs to me that you could use that same idea, of having the comparison function log what it is doing, to record the results of every possible sequence of the binary choices of the random compare function. For example, have the compare function return -1 every time and see how many times it is called when using it to sort {0, 1,2,3,4} and what the resulting sort order is, then do the same for returning 1 followed by all -1′s, etc. Unless the sort algorithm takes too many comparisons to make it practical you should be able to run an exhaustive test and count how many times each of the 120 different permutations occurs. That would give an exact map of the results when using the Microsoft Shuffle with that sort function, even when, as in the case of IE, you don’t have knowledge of the specific algorithm that the sort function is using.

  • Alex T 2010/03/02, 09:50

    It just seems embarrassing that Microsoft a company so many laymen think of as the best software writer there is should screw up so simply. Shame none of them will understand it to have their faith pranged.

  • Jag 2010/03/02, 09:59

    Strange to do it on the client side… I have JS deactivated (and thus randomness) and the order is : IE, FF, Opera, Chrome, Safari.

  • James Gregory 2010/03/02, 11:41

    @weissman has been enjoying too many spy stories and conspiracy books.

    I personally would assume that the way this mistake happened was along the lines of:

    1. EU forces directive on Microsoft
    2. Microsoft doesn’t like the directive but realises it has no choice, so implements it in the cheapest fashion possible
    3. Microsoft therefore assigns an intern or a low-skill low-pay web front-end monkey or some other programmer who doesn’t cost too much to write the Javascript code for the choice page
    4. Hence the simple mistake

    I’m sure there are lots of good programmers at Microsoft, but they will be doing things that are either a) challenging and hence require lots of programming skill (like working on the NT kernel or whatever) or b) making improvements to software that actually makes Microsoft lots of money (like Office and Visual Studio). Microsoft isn’t going to assign its best and most expensive programmers to writing a short piece of Javascript they’d rather wasn’t written at all.

  • Jose_X 2010/03/02, 12:39

    James Gregory, apparently you disagree with Rob that getting this correct is very important.

    Microsoft probably thinks similarly (call it the Monopolious Arrogancia) and may help explain why they would have been fined over a billion euros.

    To think of how many PHDs Microsoft could have hired and put to the task of writing this sort routine if they had those 1 billion euros in their pocket!

  • Bernard Swiss 2010/03/02, 17:08

    Any psychologists in the house? Poli-sci?

    As I understand, in many such “ballot” choice situations, first and last on the list are both considered to be advantageous/privileged positions.

    If you can’t put yourself first on the list (and sometime even if you can), last on the list may be the smart way to go.

    Note: I would be surprised if certain segments of Microsoft were unacquainted with this phenomenon. So the question of how such “sloppy” programming of a rather simple routine got through, is of some interest.

  • Jose_X 2010/03/02, 19:26


    You posted an algorithm that is like Fisher–Yates except approximately that in each cycle a random number between 1 and n is used instead of the correct 1 and i range:

    for (i=0; i> This is very similar to Fisher-Yates but doesn’t result in an equal distribution.

    Yes, below I calculate the distributions for sizes of 1, 2, and 3.

    What I found was that: for size 1, these are the same; for size 2, they are the same; and for size 3, they differ slightly.

    I break the data into 3 major sets corresponding (in order) to the sizes 1, 2, and 3. Each such grouping is further broken into 2 parts: the first part covers Fisher-Yates (F-Y) and the following part covers the variation you presented (algorithm B or “B”). Each such part consists of a listing of the final and intermediate permutations (of the intial set) for all possible series of random numbers that could occur during a run of the algorithm.

    For example, Fisher-Yates and algorithm B each have only one possible series of random numbers for a set of size 1: the only random chosen in the single step is 1. For size 2, F-Y has 2!=2 possibilities for the 2 steps: we can get 1 followed by 1 or 1 followed by 2. Meanwhile, B has 2^2=4 possibilities for the 2 steps: 1 and 1, 1 and 2, 2 and 1, or 2 and 2. For case 3 steps, F-Y has 3!=6 possibilities while B has 3^2=27 possibilities.

    Here is an example of how to read the data. The line is taken from Algorithm B.


    “312 ->cba ->bca ->bac”

    In this hypothetical scenario (of the 27 possible ones), we have 3 items we want shuffled. The initial ordering is abc. There are 3 steps. In step 1, the random value is 3. This is read as: we swap the first element with the 3rd element. Since we start with abc, after this swap we have cba. In step 2, random value is 1, so we swap the second element with the 1st element. Since we currently have cba, we end up with bca after the swap. In final step 3, random value is 2. We swap third element with 2nd, so the current order bca now becomes bac. Thus, if the series of random values that came up for an Algorithm B sort of 3 elements was 3 followed by 1 followed by 2, we would have swapped abc into bac. All of these steps are summarized by the example line above.

    Another example would be (also from the section of Algorithm B on size 3):


    “131 ->abc ->acb ->bca”

    Briefly.. we get random value 1, so we turn abc into abc (first value is swapped with the first value => no change in order). Then random value 3 turns current abc into acb. Finally, random value 1 turns current acb into bca.

    The F-Y lines are read identically as these 2 examples because the two algorithms are identical (except that the random values possible at each step are more constrained in the F-Y case).

    Finally, we see why we must have different distributions for size 3. 1! divides 1^2, 2! divides 2^2, but 3! does not divide 3^2. For the latter case, we can see that Algorithm B turns abc into abc, cba, or cab, each in exactly 4 of the 27 scenarios. Meanwhile, bca, bac, and acb appear as the ending result, each, in exactly 5 of the 27 scenarios. 4+4+4+5+5+5=27. 6 does not divide 27 evenly but only 4 times with remainder 3 (the remainder accounts for the 5,5,5 cases). This distribution is almost even but not quite, we note, so, if we used this broken Algorithm B on a group of size 3, we would get somewhat close to a random shuffle. The bias would be towards producing any of the final orderings bca, bac, and acb 25% more frequently than the others. In other words, any of the aforementioned 3 ending results will tend to appear 5 times for every 4 times that any of abc, cba, or cab appear. If we ran a small number of trials, we might not catch this tendency, but, if we ran something like 1000 trials, the favorable cases would tend to appear around 185 times each while each of the less favorable cases would likely appear near 148 times. In contrast, F-Y would have each of these 6 cases equally tend towards appearing 167 times.

    One other note is that each of the scenarios described below (ie, all 1! + 1^1 + 2! + 2^2 + 3! + 3^2 cases) has the same chances of happening as any other within its subpart. Eg1, we will find that for every 3! times we run F-Y on size 3, 1 followed by 1 followed by 3 will tend to occur 1/6th of the time. Eg2, we will find that for every 3^2 times we run B on size 3, 1 followed by 1 followed by 3 (or 3,1,2 or 3,3,3 or 1,1,1 or any other) will tend to occur 1/27th of the time. If this weren’t the case (in other words, if the random function was weighted more towards producing some series of random values over other series, eg, if its distribution function was not constant across all values at all times), then the results below would have to be weighted accordingly; however, the model is of a perfect random function, and each case below counts as a single distinct case as does any other. This means that to determine if a distribution is even we merely have to verify that each final permutation occurs in the same number of cases as any other permutation.

    Another interesting point is that, for the F-Y case, the first iteration is always the identity iteration because the initial value is necessarily swapped with itself. [This is why F-Y can be described as skipping this initial step and instead starting the swapping with the second element.] This is shown below: for a, ab, and abc, the first step of F-Y is always labeled ->a, ->ab, or ->abc, respectively, while only a fraction of the first steps of the B cases are similarly labelled.


    1 ->a (1)

    1 ->a


    11 ->ab ->ba (2)
    12 ->ab ->ab (2)

    11 ->ab ->ba
    12 ->ab ->ab
    21 ->ba ->ba
    22 ->ba ->ab


    111 ->abc ->bac ->cab (4)
    112 ->abc ->bac ->bca (5)
    113 ->abc ->bac ->bac (5)
    121 ->abc ->abc ->cba (4)
    122 ->abc ->abc ->acb (5)
    123 ->abc ->abc ->abc (4)

    111 ->abc ->bac ->cab
    112 ->abc ->bac ->bca
    113 ->abc ->bac ->bac
    121 ->abc ->abc ->cba
    122 ->abc ->abc ->acb
    123 ->abc ->abc ->abc
    131 ->abc ->acb ->bca
    132 ->abc ->acb ->abc
    133 ->abc ->acb ->acb
    211 ->bac ->abc ->cba
    212 ->bac ->abc ->acb
    213 ->bac ->abc ->abc
    221 ->bac ->bac ->cab
    222 ->bac ->bac ->bca
    223 ->bac ->bac ->bac
    231 ->bac ->bca ->acb
    232 ->bac ->bca ->bac
    233 ->bac ->bca ->bca
    311 ->cba ->bca ->acb
    312 ->cba ->bca ->bac
    313 ->cba ->bca ->bca
    321 ->cba ->cba ->abc
    322 ->cba ->cba ->cab
    323 ->cba ->cba ->cba
    331 ->cba ->cab ->bac
    332 ->cba ->cab ->cba
    333 ->cba ->cab ->cab

  • Jose_X 2010/03/02, 19:44

    Bernard Swiss, hark addressed your question with an interesting link http://www.newsweek.com/id/210194 “Right is Right … Except for Lefties”

    However, we should note that there are enough variables in this picture that, without changing the javascript, Microsoft can adjust Internet Explorer or else change the order of the initial listings on the webpage (html) or do something else tomorrow so as to achieve whatever result they want to achieve from IE+Windows tomorrow. Meanwhile, the other browsers’ source can be changed so that, for any given javascript/html combination, they get a new set of results. Naturally, the http://www.browserchoice.eu webpage can change without notice in a way to achieve effect X for browser Y. This would hold at least until browser Y’s behavior was changed to counteract.

    Simply, to prevent problems, whether through javascript or on the server or otherwise, Microsoft should have tried to ensure each browser ordering was equally likely as the next. I think that was the deal/goal. To test this approximately, we would refresh the page many many times and measure the results to see if randomness was suggested (this is effectively what Rob did with the simulated repeat trials and Chi-squared test). [We assume neither IP address nor anything else would affect the outcome.]

  • Jose_X 2010/03/02, 22:21

    Dorian, Rob replied to you, but let me see if I can add a “few” more details.

    We can’t do very much in terms of predictions and analysis when we don’t know everything the browser and the OS are doing and when a software interface contract is broken. Had the comparator function/sort routine contract not been violated, we could run tests, and these would have to fulfill certain criteria we could check.

    Are you interested in the distribution or orderings achievable for a merge-sort when we use .5-rand()? This we can try to answer, but we don’t currently publicly know what sorting routine IE+Windows will be running when the time comes to load up that page. [However, we likely can know the distribution of orderings for something like Firefox+Linux because these are open source.]

    Note, that the Microsoft javascript code did not use the random values to build an array or list and then sort that. It used the random values essentially to create a 50% chance that during a given comparison cycle of the sort routine (of the javascript sort routine implemented differently by each browser) either of the two values compared during that step might be chosen as being “first” for the purposes of any potential swapping during that step. Basically, this code simply breaks the contract between the sort routine and that comparison function. Broken contract means the results are implementation-defined. [I don't think the javascript does what you attributed to it, but you may want to clarify (for my sake) precisely what you think is happening.]

    See http://www.devguru.com/Technologies/ecmascript/quickref/sort.html for two examples of a proper definition of a comparison function (“reverseSort” and “sortNumbers”). You will note that for any given a and b values, each of those two comparison functions always returns the same result. That is part of the contract. Rob summarized this within the main posting. The Microsoft javascript code broke this contract, so what is the final ordering after such a broken sort is anyone’s guess (anyone outside Microsoft). The final ordering depends on the details of the sort algorithm within the browser.

    An example.

    Sorting (1,2,3) for a given sorting routine might return 3,1,2 more frequently than any other ordering if we assume that any given comparison can go either way with equal probability. [I construct such an algorithm next.] To know the distribution, we have to know the details of the routine. One routine might even try to remember past comparison values and behave oddly if future test of the same pairs yielded a different value. This is why I asked if you wanted to analyze a specific algorithm or (implementation).

    [Now, for the algorithm with lopsided distribution...]

    Eg, let’s say the sorting algorithm is designed to work on 3 element sets exclusively (or is a special case of a more general routine). It might compare elements 1 and 2, then 1 and 3, then 1 and 2 again redundantly, then finally 2 and 3. It might then use a lookup value to map the 4-tuple result into a specific permutation. Then it might apply that permutation to the original array being sorted. Also, that lookup table is initialized as if to sort e1,e2,e3 into e3,e1,e2 (ie, every value in that lookup table is initialized with the correct value for the 3,1,2 case). After this initialization just described, enough lookup values needed to always solve the 3-sort problem correctly are properly set. The remaining values are left in the initialized state.

    For a correct comparison function passed into this sort routine, this algorithm is assumed to be correct; however, if the two 1,2 comparison checks don’t match on each of the two times its done, the table would be looked up in places that still point to the original initialization value of 3,1,2. I’ll step through this a little more carefully below after I list the actual table mappings.

    Note, that this crazy algorithm might work well with specially constrained hardware or the table might be overlapped with other data within the program (perhaps to allow for a very compact implementation) of some type (???). The point is that, for proper comparison functions passed to this sort algorithm, it always produces the correct sorting, but, for bad comparison functions, all bets are off.

    [Now, for the 2^4=16 table rows... These rows are keyed by a 4-tuple, and a row essentially holds a mapping. I am assuming each comparison result is either + or - (ignore 0).]

    e1?e2; e1?e3; e1?e2; e2?e3:

    +; +; +; + -> e3,e2,e1
    +; +; +; – -> e2,e3,e1
    +; -; +; + -> //this case is impossible since we cannot have (+) element 2 precede element 1, (-) element 1 precede element 3, yet (+) element 3 precede element 2. The value here is the initialization value of e3,e1,e2.
    +; -; +; – -> e2,e1,e3 //note that the 2,3 check (-) would have been redundant
    -; +; -; + -> e3,e1,e2 //note that the 2,3 check (+) would have been redundant
    -; +; -; – -> //this case is impossible since e1 cannot precede e2, e3 precede e1, yet e2 precede e3. The value here is the initialization value of e3,e1,e2.
    -; -; -; + -> e1,e3,e2
    -; -; -; – -> e1,e2,e3

    +; +; -; + -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    +; +; -; – -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    +; -; -; + -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    +; -; -; – -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    -; +; +; + -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    -; +; +; – -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    -; -; +; + -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match
    -; -; +; – -> e3,e1,e2 //this case is impossible since each of the two 1,2 checks must match

    Note that all the cases where the two 1,2 checks don’t match remain pointing to the e3,e1,e2 “solution”.

    Note that, for any correct comparison function, we get correct results. This was incorporated into the table purposely by analyzing the e?e above logically by hand. Exactly 6 of the entries are needed to be correct (and are). Every other case is an impossibility logically. Eg, if e1<e2, e2<e3, then e3e2. Another way to put this is that for a correct comparison function, each of these 6 entries has 1/6th chance of being the one looked up for any given initial array ordering, while each of the other 10 entries (all of which point to e3,e1,e2) has exactly 0% chance of being looked up. Note 6 = 3!.

    [Now, for a description of what happens when we put in a comparison function that returns 0.5-rand(). Note that the breaking of the contract between sort routine and comparison function (that comparison function is not consistent across all invocations of the same a,b value) implies what we get in this case cannot be trusted to be anything resembling any sorting, and we also cannot expect any type of distribution. The distribution of "orderings" produced will depend on the implementation details. In our case, we will see that the 3,1,2 ordering will happen with high frequency in many cases, in particular, in the case of 0.5-rand().]

    Now, let’s look at the distribution for the case where there is a random decision at each comparison check, where 50% of the time we get + and 50% of the time we get -: [Note, I have assumed 0 doesn't happen in order to keep this analysis manageable in size. Where the 0 case might be used, we can still get the correct result by picking either + or -, since two similar elements can be resorted amongst each other (for our purposes) with no substantial consequence. Also, if rand() picks a "real" number between 0 and 1 with equal probability, then in the limiting case, the single exact value 0.5 is chosen 0.00% of the time.]

    Here we see that the 6 correct cases happen with equal probability and with the same probability each as each of the other 10 entries. In other words, each of the 16 possibilities from the lookup table above has an equal likelihood of being chosen. The reason is that now, unlike for correct comparison functions, each +,- 4-tuple is equally likely to occur when we make our 4 measurements.

    So this means:
    an e3,e2,e1 sort will happen 1/16th of the time (and not 1/6th).
    an e2,e3,e1 sort will happen 1/16th of the time.
    an e2,e1,e3 sort will happen 1/16th of the time.
    an e1,e3,e2 sort will happen 1/16th of the time.
    an e1,e2,e3 sort will happen 1/16th of the time.
    an e3,e1,e2 sort will happen 11/16th of the time.

    Thus, we see that, for a given initial ordering, e1,e2,e3, there will be a huge chance (11/16th) that the resulting output value (the “ordering”) becomes e3,e1,e2 whenever the comparison function was the version returning “0.5-rand()”.

    [Beware of pun ahead] This is the sort of result Rob and others got: for the Microsoft javascript page, the different final browser orderings were not likely to be achieved each with equal probability (at least not if we trust the Chi-square test and the accuracy of the repeat trial results). The reason had to do with how the implementation of the sort routine interacted with the broken comparison function that returned an inconsistent/improper value of “0.5-rand()”.

    Note, that Rob did not list out all the 120 permutations with respective frequencies (as was done for the 3-case in the made-up example just described), but we can infer these permutations did not each occur with equal likelihood (as they should have), as, otherwise, each of the five browsers would have landed at each spot about 20% of the time.

    Finally, if instead of “0.5-rand()” the returned value had been something like “0.25-rand()”, then we can go back to that table and assign a different corresponding probability to each of the 16 rows, but, in general, we have no reason to expect this will result in an constant distribution for our made-up algorithm [remember that the e3,e1,e2 mapping is over-represented]. As for what would be the distribution for a different but unknown algorithm, that is anyone’s guess. The point of this posting was simply to show that we could construct an example algorithm where the distribution was not constant when “0.5-rand()” was used as the comparison function.

  • Jose_X 2010/03/02, 22:59

    The short version of the prior comment to Dorian is:

    If we spec a sort routine (javascript.sort) such that we place certain restrictions on the comparison function that will be used to make the checks between 2 values at each step, then we expect those constraints to be met. If they are met, then we implement our sorting algorithm to guarantee a correct result; however, if the restrictions are not met, all bets are off. We are designing to spec, not to a non-spec.

    In the long prior comment, I gave an example of a sort algorithm that uses a lookup table to find the ending mapping to apply to the initial array so as to “order” it properly. The algorithm only applies to arrays of size 3 (but the idea can be generalized).The table it uses has 16 keyed values. Exactly 6 of these (16) table rows are used to sort all possible 3-element arrays correctly whenever we get a comparison function that fulfills the requirements on comparison functions, namely, that all calls to that function return the same value, say + or -, for every invocation of the same 2 values; however, if the comparison function does not fulfill its requirement, we essentially get garbage out as our “ordering”, with no reason to expect one type of distribution in orderings over another.

    I specifically constructed the 16 rows of that table so that, for a comparison function returning “0.5-rand()”, as is the case with the Microsoft-coded javascript that is the topic of this thread, the hypothetical sort function would produce one specific ordering 11/16th of the time. [Notice how the "0.5-rand()" comparison function violates the contract.] The reason for the 11/16th bias towards a single mapping was because this mapping value occupied most of the table rows. Despite this, the algorithm worked properly when the comparison function was good because only 6 rows of the table were used; however, for a broken comparison function, the sort generally relied on all 16 rows to produce its garbage.

    I took the time to write up the long example only because I know that it might appear unintuitive that a “balanced” randomization comparison can fail so badly within a sort. The reason it can is because the sorting was not engineered (was not spec’d) for a case such as 0.5-rand().

    Moral of the story: If you don’t engineer a box to hold kitties, no matter how fluffy and cute kitties are, you should not expect the kitties to be held. Some boxes will do a fine job despite not being designed to hold kitties, but others will not.

    ***** Note that by Microsoft relying on behavior outside of the spec, they help create the interoperability failures that are so common between their dominant software and third party competing software. Is the EU watching? “Microsoft specs” are worthless if Microsoft dominates a particular market and their software doesn’t follow the corresponding specs or these specs are ambiguous, underspecified, or inconsistent. *****

  • Bugstomper 2010/03/03, 00:28

    @Rob, I was able to code up something that finds the exact probability distribution of results when using the browser’s Javascript sort without having to know the algorithm being used or seeing the code. The results when I try it in Firefox and in IE 8 match up very well to the random tests that you’ve done.
    @Jose_X, this is ironic, given what you said at the beginning of your comment about how you can’t do much analysis when you can’t see the code and the contract is broken. I thought so to and then had a flash of inspiration.

    Here’s what I did, which I started thinking about at the end of my last comment: The Microsoft Shuffle runs the browser’s sort function on an array {0,1,2,3,4}. Without knowing the underlying sort algorithm or implementation we can say that in any given run it will call the randomized comparison function some number N times, and in that run we can call the return values of the comparison function r0, r1, r2, …rN where each r is either -1 or +1. Unless the sort function is written with some really strange probabilistic part, it’s pretty safe to assume that every time the comparison function returns the same sequence r0 … rN of results, the sort function will produce the same output using the same number of comparisons.

    That means that the behavior of the Microsoft Shuffle with a particular sort can be completely represented as a binary tree, the two branches of the root node of the tree representing the value returned by the first comparison r0, then the next level in the tree for r1, and so on. Every possible run of the sort is represented as one path from the root to that path’s leaf node.

    How big is the tree? Is it finite? Is it of a size that would be practical to traverse in a program?

    Well, it is conceivable, given that the comparison function violates the contract of the sort function, that there might be some set of values that would cause the sort function to run endlessly. In that case we would have an infinite length branch in the tree and probably would not want to try to traverse it to the end. But we know from running these tests that we haven’t seen such infinite loops, and not even an extremely long run time for the sorts.

    If the comparison function were well behaved we would expect the sort to run with something like O(n log n) comparisons or worst case O(n^2). Given that we have just 5 elements we can expect around 11 to 15 comparisons, or maybe as much as 25. Where it is in that range does make a difference. Limiting the tree to branches of length 11 means that we don’t have to follow paths to more than 2048 leaves. At 25 there are around 32 million, which might take too long for a javascript program. At this point in the analysis it does seem likely that it will prove practical to traverse the entire tree.

    Each leaf node of the tree represents the end point of a particular sort in which the comparison function output a particular r0 … rN. We can find out what the sort function will produce corresponding to that leaf by calling the sort function on {0,1,2,3,4} passing it a comparison function that is hard-wired to produce that specific r0 … rN sequence. We also know that the probability of the random comparison function producing that sequence is precisely 1/2^(N+1).

    That gives us a good way of dealing with the possibility that there are some sequences of output of the random comparison that would lead to an infinite loop. The longer a branch is, the less likely it is to occur. Unless there are a near infinite number of them which collectively have a high probability, it should be safe to ignore those hypothetical very long branches.

    The algorithm I came up with, then, includes setting an arbitrary cutoff point at which it stops following a branch.

    I have a bit string R of length N that represents the path r0 … rN where ri==0 means the comparison function will return a -1 the ith time it is called, and ri==1 means the return will be 1. N is small enough so that I can use an integer for the bit string. The bit string integer is initialized to 0 at the start, which represents the first path to follow.

    I then call the browser’s Javascript sort function passing it the array {0,1,2,3,4} and a comparison function that is instrumented to use the bits of R to determine its results. Of course it has to keep count of how many times it is called in order to make its way through R. If it is called more than N times then what I should do is throw an exception to abort the sort. I didn’t do that because I have never worked with Javascript exceptions before and didn’t want to mess with it if I didn’t have to. I did have it not try to compile statistics on that branch and keep track of the largest branch it saw and output that at the end so I could see if I wanted to adjust my limit to suit that browser.

    After the sort is called, I update the statistics that I keep on how often each browser ended up in each position. The results are added into the same 5×5 resultsMatrix array that Rob uses in his code. There is one crucial detail: Instead of incrementing by one, I add a number proportional to the probability of getting this particular result by chance. That probability is 1/2 raised to the power of the number of comparisons that were done by this call to the sort function. To make the counters be integers, I increment the counter by 2 to the power (N – number of comparisons).

    Let’s say, for example’s sake, that the first sort using the path bit string 000…0 performed its sort using 6 calls to the comparison function, i.e., 000000. Then we are all done with any path that begins with 000000 and the next path to try is the first one beginning with 000001 which is 00000100…0 (N bits long). Each time we perform a sort, the path used for the next sort is found by setting all the bits after the last one that was used to 0 and add 1 to the integer at the bit position that was used by the last comparison.

    We know that the entire tree has been traversed when we add that 1 and the result overflows the N bit length of the string, i.e., the integer being used to represent the N bit string is no longer less than 2^N.

    Since the results are stored in the same matrix data structure that Rob used to compile the results of the thousands of shuffles, I can re-use his code to display the tables of results. On my MacBook in Firefox 3.6 the program only required 144 sorts with a maximum path length of 11 comparisons, and did it in 3 msec. In Windows XP under VirtualBox on my MacBook running IE 8 it required 1120 sorts with a maximum path length of 12 comparisons and took 75 msec. I wanted to see how much of a difference running on a virtual machine under VirtualBox made, so I also tried it with Firefox in the Windows XP virtual machine. That took 15msec, which I thought indicated that the virtualization was slowing things down. Then I noticed that I was still running Firefox 3.5.6 there, whereas I have Firefox 3.6 on the native MacBook. When I updated Firefox in Windows to 3.6 and tried again it also ran the test in 3 msec.

    Here are the frequency tables generated by the test, as bet as I can figure out how to paste them in to the comment box. First for Firefox 3.6, then for IE 8.


    Position I.E. Firefox Opera Chrome Safari

    1 0.2461 0.2461 0.1641 0.0938 0.2500
    2 0.2871 0.2871 0.1914 0.1094 0.1250
    3 0.2461 0.2461 0.2578 0.1875 0.0625
    4 0.1621 0.1621 0.2695 0.3750 0.0313
    5 0.0586 0.0586 0.1172 0.2344 0.5313

    Now IE 8

    Position I.E. Firefox Opera Chrome Safari

    1 0.1328 0.2168 0.2109 0.2527 0.1868
    2 0.1328 0.2168 0.2109 0.2527 0.1868
    3 0.1094 0.2227 0.1406 0.3735 0.1538
    4 0.1250 0.2188 0.1875 0.0586 0.4102
    5 0.5000 0.1250 0.2500 0.0625 0.0625

  • Norm 2010/03/03, 02:17

    Bugstomper (#100),

    The sort algorithm that you encountered in Mozilla’s code sounds a lot like something known as the “Merge-Insertion Sort”. Dig into Knuth vol. 3 for a description of this algorithm and see how it compares. It’s a somewhat fiddly and complicated sorting algorithm that is designed to be optimal at minimising comparisons. This doesn’t make it particularly *fast* in comparison with other sorting algorithms unless the comparisons are very costly relative to the other operations involved in sorting. Still, an interesting method — worth looking up!

  • Michael Geary 2010/03/03, 03:49
  • Bugstomper 2010/03/03, 05:45

    Using my instrumented compare function and the exhaustive test I described I figured out why IE ends up in 5th place 50% of the time when you use IE’s Javascript sort, and in fact I found the reason for the exact percentages for 5th place for all the other items.

    It turns out that if the fourth call to the comparison function returns -1 (“less than”), then the first element of the original array always ends up in the last place of the sort result. Since any one comparison has a 50/50 chance of going either way, the first element, which in the menu is IE, ends up in 5th place half the time.

    How can so much depend on just one comparison operation? This exact behavior may only happen to be true when sorting a five-element array (I didn’t test other size arrays), and I don’t know what exact algorithm Microsoft is using, but the effect is as follows:

    The first steps in the sort establish which element goes last. At least when the array is as small as five, this is done by comparing the fifth element with the fourth, swapping them if the fifth element is smaller; then comparing what is now the fifth element with the third, swapping them if it is smaller, then what is now the fifth element with the second, swapping them if it is smaller, and then that crucial fourth comparison of what is now the fifth element with the first, swapping them if it is smaller.

    At this point with a real comparison function the last array element contains the largest element of the array and doesn’t have to be touched again. With the random compare function, no matter what happened in the first three comparisons, the fourth one has a 50% chance of sticking the first element a the end where it will not be touched again.

    You can see that the original second array element has exactly 1/4 chance of ending up in the last place, the third element has 1/8 chance, and the remaining two each have a 1/16 chance of ending up last. And that is exactly the distribution we see in all the tests when they are run in IE.

    Would this hold true for any size array or is it part of something that Microsoft does when the array is very small? That would be easy to test by repeatedly sorting a large array using the random compare function and seeing if the first element ends up in the last position half the time. I’ll leave that as an exercise for someone who has not already wasted as much time on this as I have :)

    I can think of a reason for doing this even for large arrays. Microsoft appears to be using a kind of treesort algorithm. I can imagine that there could be a variation of treesort in which as a first step you anchor the tree by doing an initial O(n) scan of the array to find the largest element and lock it in place at the end. But that is pure speculation on my part.

  • John 2010/03/03, 08:34

    The code has been changed now. Should be better.

  • Calum 2010/03/03, 08:39

    So, instead of choosing an O(N) algorithm or an O(N log N) algorithm, they chose an O(random) algorithm, where the random number may be between 1 and infinity…

  • Jose_X 2010/03/03, 10:47

    Bugstomper, hopefully today I will spend more time with your very interesting comments. What I said (that you called ironic) requires clarification. You can always attempt to analyze something a lot. Depending on the context you may be able to say very much yet still cover less than 1 millionth of a percent. 1 trillion bits of info are tiny compared to 1 trillion times that or to the infinite. We can say a whole lot about some software, yet be entirely wrong on all counts tomorrow after a software update or after the software runs on the next trillion bit data set. In your analysis, you found yourself many times making assumption x or y. Anyway, I think you understand what I was talking about now. It’s a coin toss, really. When we have no choice and if we have time and inclination, we try to get lots of milage and say interesting things (especially when there is application to other places), but there is always that looming asterisk and the chance we will entirely miss the boat when the show starts. We should be particularly careful when the performers have put special effort to create a show that will defy the statistics. If you have many hundreds of billions on the line and some experience and desire, you might just try to create such a show. Trust me, magicians can wow alright and not by accident.

    Let me fix a mistake I made in a few parts. In the comment to Nick detailing “Algorithm B” (ie, Fisher-Yates-like algorithm using rand(N) instead of rand(i)), whenever I said 1^2 or 3^2, the correct expression should have been 1^1 and 3^3, resp.

  • Jose_X 2010/03/03, 11:24

    Bugstomper, I gave a twisted example (that same long comment to Dorian, discussing a lookup table) that shows how you can clearly present an interface to outsiders that takes on a correct and simplified form (the equivalent of 6 lookup lines) most of the time yet is several (or many times) more complex below the surface (in the example I gave, there were actually 16 lookup lines of key complexity). An outsider might do a great job poking around and discovering the details of those 6 lines possibly without ever cracking below that layer. In fact, apparent cracks might become anomalies or things the outsider would prefer to skip (avoid, write off) this time around by adding a particular assumption. This is done in order to keep exploring and further detailing what you might consider to be tractable. An example in physics: it’s easier to consider various bits of data and deviations from expected value to be noise or to model it statistically at least so that we can continue using simplified models we understand; however, at some point in time, we usually do decide to take a leap and then find that the next best level of modeling might be much more complex than the simpler model we just left (look at how the simple, long-lasting, and elegant Newtonian mechanics/dynamics eventually gives way to many more complex models). In mathematics and perhaps in just about all scientific fields, we strive to say as much as possible with as few assumptions as possible, but, generally, it’s by making assumptions and setting constraints that we are able to say the most number of definite and (arguably) interesting things about something .. Maybe I can think of other analogies or examples, but I really think you know what I mean. Maybe Dorian and others also see what I mean. I think keeping in mind the magician analogy I used in the earlier comment will be useful. [Magicians make their money from indirection and from pulling off the improbable, and they enjoy what they do and find those types of challenges to be what motivates them.] To some degree, it’s an academic exercise to ask at any point in time if any given model has in fact simply discovered “the 6 lines” or if it has gone further. We are the audience. They are some sort of magician, but how good are they and how much do they want to stay in the spotlight?

    Alright, well, this said, I still find these efforts interesting and revealing. I really respect magicians who control the show, but….

  • Bugstomper 2010/03/03, 14:27

    I said I would stop wasting time on this, but I did have to try one more thing :)

    I coded up a function that repeats 1000 times sorting a length 100 array of the numbers 0 … 99 using the random comparison function (i.e., the Microsoft Shuffle) and counts how often after the sort the last element of the array contains the 0 that started out in the first element of the array.

    If I do this using Fisher-Yates to shuffle the array, I get the expected results that approximately 1/100 of the time the 0 ends up in last place.

    On IE8 the Microsoft Shuffle put the 0 in the last place 0.481 and 0.485 of the time on the two runs I did, which appears close enough that we could guess that their algorithm does start out with that pass from end to beginning to find the last element as I described in my last comment, even with large arrays. I also got the same result with a length 1000 array.

    Firefox 3.6 produced the interesting result that the first element never ended up last place.I think that is because their bottom up algorithm splits the array into 25 pieces providing a huge number of opportunities for one comparison to indicate that a subsequence can be locked down as being before the following one and once that happens no element in that first subsequence can ever end up in last place.

    Ok, now I think I really am done with this issue :)

  • Andrew 2010/03/04, 00:30
  • Daniel 2010/03/08, 09:19

    I prefer this Ballot Screen: http://www.realbrowserchoice.eu

    Very funny!

  • funkspock 2010/03/09, 12:43

    By comparing 1 million runs to 10 million runs we see that:
    1. the fractions of microsoft shuffle get worse
    2. the fractions of Fisher-Yates improve

    see output below from Opera 10.50, Build 3296, Platform Win32, System Windows XP

    1 milllion runs

    Using the Microsoft Shuffle
    Raw countsPosition I.E. Firefox Opera Chrome Safari
    1 374760 374695 187781 31303 31461
    2 374734 374808 109642 70143 70673
    3 93955 93789 491185 160812 160259
    4 78185 78099 106319 369003 368394
    5 78366 78609 105073 368739 369213

    Fraction of totalPosition I.E. Firefox Opera Chrome Safari
    1 0.3748 0.3747 0.1878 0.0313 0.0315
    2 0.3747 0.3748 0.1096 0.0701 0.0707
    3 0.0940 0.0938 0.4912 0.1608 0.1603
    4 0.0782 0.0781 0.1063 0.3690 0.3684
    5 0.0784 0.0786 0.1051 0.3687 0.3692

    Using the Fisher-Yates algorithm
    Raw countsPosition I.E. Firefox Opera Chrome Safari
    1 200254 199442 200135 199652 200517
    2 199799 200570 200337 199580 199714
    3 200651 199409 199779 200235 199926
    4 199910 200326 199875 200690 199199
    5 199386 200253 199874 199843 200644

    Fraction of totalPosition I.E. Firefox Opera Chrome Safari
    1 0.2003 0.1994 0.2001 0.1997 0.2005
    2 0.1998 0.2006 0.2003 0.1996 0.1997
    3 0.2007 0.1994 0.1998 0.2002 0.1999
    4 0.1999 0.2003 0.1999 0.2007 0.1992
    5 0.1994 0.2003 0.1999 0.1998 0.2006

    10 million runs

    Using the Microsoft Shuffle
    Raw countsPosition I.E. Firefox Opera Chrome Safari
    1 3748313 3750186 1875945 312652 312904
    2 3750780 3750064 1092376 703531 703249
    3 936538 936404 4922050 1601583 1603425
    4 781766 781039 1056028 3690080 3691087
    5 782603 782307 1053601 3692154 3689335

    Fraction of totalPosition I.E. Firefox Opera Chrome Safari
    1 0.3748 0.3750 0.1876 0.0313 0.0313
    2 0.3751 0.3750 0.1092 0.0704 0.0703
    3 0.0937 0.0936 0.4922 0.1602 0.1603
    4 0.0782 0.0781 0.1056 0.3690 0.3691
    5 0.0783 0.0782 0.1054 0.3692 0.3689

    Using the Fisher-Yates algorithm
    Raw countsPosition I.E. Firefox Opera Chrome Safari
    1 1999795 1998960 1998794 1999788 2002663
    2 1999306 2000813 2000887 1999840 1999154
    3 2001656 1999856 1999453 2000241 1998794
    4 2001929 1998767 2001334 1999646 1998324
    5 1997314 2001604 1999532 2000485 2001065

    Fraction of totalPosition I.E. Firefox Opera Chrome Safari
    1 0.2000 0.1999 0.1999 0.2000 0.2003
    2 0.1999 0.2001 0.2001 0.2000 0.1999
    3 0.2002 0.2000 0.1999 0.2000 0.1999
    4 0.2002 0.1999 0.2001 0.2000 0.1998
    5 0.1997 0.2002 0.2000 0.2000 0.2001

  • kruador 2010/03/09, 13:19

    There’s a lot of conspiracy on here about fifth place. To those I would pose the question: if fifth place was so attractive, why did Opera spend so much effort getting the EU to require it to be done randomly rather than taking the fifth place that was rightfully theirs through usage share?

    Microsoft’s own research for Office 2007 suggests that second place would attract most eyes. This is why the most frequently-used commands are not at the far left-hand end of the Ribbon but in the second group from the left. http://blogs.msdn.com/jensenh/archive/2005/09/29/475296.aspx.

  • Charles 2010/03/24, 18:07

    Ummm… Who cares.

    That was 45min of my life I will never get back. Who cares the positioning of the browsers? If people are to stupid to choose the browser that they want to use then f**k them.

    I hit refresh about 5 times and i got a bunch of random positions. Did it make me want to install Opera or Safari? No. Why? Because they basically suck.

  • Rob 2010/03/28, 09:42

    @Charles (from,

    A word to the wise. If you are going to submit a comment that: 1) contains spelling errors, 2) contains profanity, 3) insults your users, 4) insults your competitors and 5) shows general ignorance of the topic, then you might want to avoid posting it from a Microsoft.com IP address.

  • My name 2010/04/02, 19:17

    Rob February 28, 2010 at 1:46 pm : Good find! Just to be nitpicking a bit now that we are talking statistics ;) But you write:

    ” The p-value itself showed that the distribution was so uneven that the chances were 1 in 50,000,000,000,000,000 that these results were drawn from a process that was actually producing random orderings”.

    I think that statement is wrong. What you probably meant was:

    “The p-value itself showed that the distribution was so uneven that IF the results were drawn from a random process, the chances you would get something deviating at least as much from the most likely distribution is 1 in 50,000,000,000,000,000″.

    See the difference? You can’t say “The results show that the probability is x% their algorithm is correct [i.e. null hypothesis true]” (which is essentially what your statement is saying). Maybe if you had some a priori idea on what the probability the algorithm was correct to begin with is, and took that into account in the calculation, you could say something like that. The intuition is that if the a priori probability that Microsoft could such an embarassing mistake is only 10^-42 then clearly it is still very likely to be correct even after your observation what you did. ;-) There’s also some more technical correction in the 2nd part of the statement but you can’t really untie it from the overall error in the statement ;-)

  • Rob 2010/04/05, 10:50

    @My name, I like the way you put it better. Thanks.

  • Stefan W. 2010/04/15, 11:30

    I’m not sure whether I understood all comments from Jose_X especially. I guess I’m just repeating in my own words.

    Rob, if you have two browsers AB and choose from ABAB with index 0 to 2, the possibilities are AB, BA and AB again.
    So it should just be ABA with index (0,1) => AB, BA.

    Now with 3 Browsers
    we could form ABCAB wiht index (0 to 2). This would contain every browser at every position, but not every permutation. Maybe Icon B looks pretty inviting if placed left to C? This is a very academical question, but let’s try to solve it:
    We need ACB, CBA and BAC – right?

    So let’s choose a digit from 0 to 2 to find a startindex in [ABC], and a direction (0,1) which means left-or-right.
    We multiply 3*2 choices and get 6. So we need an Int from 0 to 5 and take – for example 4. 4/3 >= 1 so we walk to the right in the array.
    4%3 == 1 so startindex is 1.. We proceed over the border in [ABC] from B to C and reenter the set to A.

    Unfortunately a set of 3 elements is the last one where this works. For 4 elements we would need another approach.
    The idea of crossing the border could be used for every number of browsers (lynx, elinks come to mind :) ).

    We could enumerate every possible combination, and just choose the enumerated combination.
    e browsers numbered map
    0 ABC 012
    1 ACB 021
    2 BAC 102
    3 BCA 120
    4 CAB 201
    5 CBA 210
    you see, the numbering is increasing, RAND=> 2 would lead to choice CAB. I can’t give an adhoc Implementation of how to produce such an enumeration in 2 or 3 lines of code. The number of possibilities is n!, which is for 5, 5!=120 Possibilities.

    My solution from above could do the work in place too, but scala is a mixed approach, functional and OOP, and the functional idea is more about immutability. It would be possible to switch a choosen value from the current position with that at the current position, and increase the current position by 1 per step. Possible, but not elegant.

    Apropos browser: In firefox (linux) this page is rendered hardly readable – the text is vanishing at the right, while I increased the font size. Can there something be done, please?

  • Stefan W. 2010/04/15, 16:59

    Well – the idea didn’t let me sleep, how to enumerat a permutation in such a way, that you needn’t do all permutations, to get a reproduceable element.
    If you have 3 browsers, we have 3*2=6 possible combinations, with 4 browsers 4*6, with 5 browsers 5! permutations. If we have a way to enumerate those, we can choose a x=random.nextInt (n!) and generate the appropriate Permutation. (again, scala-code)
    [code] object Code2Perm {

    val r = new java.util.Random

    def code2perm (code: Int, set: List[String]) : List [String] = {
    val len = set.size
    val pos = code % len
    if (len > 1) {
    val rest = code2perm (code / len, set.tail)
    rest.take (pos) ::: (set.head :: rest.drop (pos))
    } else

    // just for Input > 0, and not too big :)
    def fak (n: Int) : Int = if (n
    println (code2perm (i, set)))
    The proof of the algorithm is, that you can produce every permutation with (0 to fak(n)-1) like demonstratet in ‘permuatateAll’.
    It works by constructing a list (ab) for (0) and (ba) for (1), n=2 elements: n=2! permutations. For more elements the sublist for 2 elements is generated first. Then the sublist for n+1 elements is generated like this:
    The index is the code % (n+1), so for 3 elements and a given sublist (xy), if code=4 then code%3 = 1, so the Position for a in aXY is 1. => XaY.
    The code / (n+1) is the number used to generate the ordering for the last two elements, so 4/3 = 1 (Integer-division) and bc from abc mens, that b is put to position 1, which leads to ba (little endian).
    With 5 elements, max=5!=120. Let’s assume 93 to be our random-number.
    source rest index character mod i div i
    fcsoi 93 0 f 3 18 iscfo
    fcsoi 18 1 c 2 4 isco
    fcsoi 4 2 s 1 1 iso
    fcsoi 1 3 o 1 0 io
    fcsoi 0 4 i 0 0 i
    Ignore the last column until you reach the bottom.
    93 % 5 is 3 rest 18. 3 will be the final position for character ‘f’. 18 is the rest-value for the next line.
    18 % 4 is 2 rest 1. 2 will be the temporary position of ‘c’. 18/4=4, 4 goes to the next line.
    4% 3 is 1 rest 1. 1 will be temp. pos. of ‘s’ and so on. The modulo-value will be the position of that character.

    To construct the final result we start at the bottom, i at Index 0. (from left to right). o at Position 1 => io. But then s occupies Position 1 and pushes the o to the right: iso. C at position 2 => isco, and f at pos. 3 => iscfo.

    The algorithm works in linear time to the size n of permutation, O(n).

  • tibetus 2011/06/07, 14:31

    To randomize the order of the elements within an array, what we need is the body of our sortfunction to return a number that is randomly 0, irrespective to the relationship between “a” and “b”.

  • Derek O'Connor 2011/09/09, 15:01

    Algorithm P in Knuth Vol 2, is Durstenfeld’s Random Permutation algorithm (1964), which is optimal in time and space, i.e., O(n).

    The Fisher-Yates Shuffle algorithm (1938?) is not optimal. It is O(n^2).

    I believe Knuth is wrong to attribute Durstenfeld’s algorithm to Fisher-Yates. See my argument here:


  • sjp 2012/01/27, 21:06
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