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astine
I can honestly say that that isn't the algorithm that would have occured to me. In CL, this is the what I would have thought to do:

  (defun butnth (index list)
    (append (subseq list 0 index)
  	    (subseq list (1+ index))))

  (defun shuffle (list)
    (unless (null list)
      (let ((index (random (list-length list))))
        (cons (nth index list)
  	      (shuffle (butnth index list))))))
Building a second sequence by randomly selecting nodes from the old sequnece and inserting them into the new sequence seems more intuitive to me. It certainly works better with lists. If I where working with arays, the OPs method would be more efficient.
duncanj
I think you're doing the same thing as the "good" knuth algorithm which, as I read it, always exchanges the random card with the one at the decreasing index from the end of the deck.
joe_the_user
Interesting.

It makes me realize how much whether one begins with lists which allow cheap insertions or not affects both one's choice of algorithm and one's thinking.

crux_
Won't that explode somewhat performance-wise once you have more than a handful of cards? It looks like something along the lines of O(n^2)...
astine
As it's written I'll blow the call stack before I see a performance decrease. :)

Actually, the algorithm is O(n), it's essentially a list reversal with a call to random inserted. In psuedocode:

  oldlist = [...]
  newlist = []
  
  if oldlist not empty
  do
    x = remove random from oldlist
   insert x into newlist
  repeat
Of course, I'm ignoring the complexity of 'butnth' in that assesment, but this was just a 'first instinct' sort of deal.
kmcgivney
I didn't like the butnth. I get an allergic reaction to append. Blame Ken Tilton.

  (defun shuffle-in-place (l)
    (cond ((null l) nil)
          (t (rotatef (nth 0 l)
                      (nth (random (length l)) l))
             (shuffle-in-place (rest l))))
    l)

  (defun shuffle (l)
    (shuffle-in-place (copy-list l)))
astine
Blame Ken Tilton.

I do. Always.

One could use nconc instead of append, but that's not where the processor cycles are being lost as far as I can tell.

crux_
It's not O(n): both 'nth' and 'butnth' are themselves O(n), and get called 'n' times. I'm fairly certain you'll end up with O(n^2) unless you introduce mutation. I'm also fairly certain there is no purely functional O(n) algorithm. ;)
astine
You're right of course; I'm not looking very deeply.

I'm also fairly certain there is no purely functional O(n) algorithm.

  (defun nth (index list)
    (if (zerop index)
        (first index)
        (nth (1- index) (rest list))))
Tichy
It annoyed me when I heard that some guys slapped their name on that algorithm (having "invented" it independently for a game). It seems like another of those not patent worthy obvious things. Why is it obvious: because you should realise that there is a random number generator in your programming language. Therefore shuffling is silly. You want to assign the cards to slots determined by the random number generator directly. From there on it is pretty straightforward. </brag>
scott_s
http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
Tichy
Yes? That is what I described - I left the swapping part as an exercise to the reader (since it is already described in the article submission). It seems to be the only logical way (I have not tried to think of another way, though - but none has been mentioned yet on HN, either).
scott_s
They didn't "slap their name on it." Other people refer to it as their algorithm because they came up with it first. It's obvious to you that someone would want an algorithm for a random shuffle of a sequence. It wasn't obvious at the time.
Tichy
What I was trying to say is that the algorithm is the obvious consequence of wanting to assign a sequence to random places in an array. I wanted it because I was programming a game. I think the same must have happened to many people. Anyway, no biggie - of course the world doesn't end because now the algorithm has a name.
scott_s
And what you're missing is that they realized people might want to randomly shuffle a sequence before there was obvious applications for it.
Tichy
I guess I did not describe it fully enough - to assign randomly, you end up noticing that you don't want to waste time checking for slots you already assigned. So the logical solution is to shrink the array by swapping out the randomly chosen index - basically the "good knuth". If you don't have an uninterrupted streak of cells to assign, you end up doing wasteful checks. Can't think of another way, off the top of my head.
allenbrunson
I just happen to have implemented something like this myself recently, which seems easier to me. here it is: generate a large random number for each card, then sort the list according to the random numbers. Can anybody here poke holes in that one?
l0stman
A problem can occur if you're shuffling a large number of cards (greater than the maximum of the random number generator) because each random number should be unique.
fhars
It's O(N * log(N)) instead of O(N).

And of course it has the usual problem of generating permutations using a pseudo random number generater: Even for moderately sized collections, the number of permutations is far larger than the number of differnt initialization states of many common PRNGs, so even if an algorithm looks correct, it may not be able to generate all permutations. For example, a 63 bit linear congruential shift generator has 2^63 = 9x10^18 possible sequences, while there are about 52! = 80x10^66 permutations of 52 cards.

allenbrunson
i'm not really worried about the complexity. it's plenty fast enough for my needs. but i am worried about getting decent results. i suck at anything that smells like math, but i think you're saying that the distribution of random numbers might not be random enough to get decent results, right?

is there some random number generator in a C-like language i should investigate, as opposed to using srand() and rand()?

astine
distribution of random numbers might not be random enough to get decent results, right?

This is a problem, but it will be a problem no matter what algorithm you choose. You can always use something like http://www.random.org/ or special hardware, but I'd just look to /dev/random and whatever your default psudo-random generator is for your platform.

jrp
You probably have something like 2^32 ways to seed your RNG. But there are N! possible outcomes of the shuffle, so at most 2^32 of those will appear (one per seed).

2^32 = 4294967296

100! is about 10^158

So for just 100 items, you're already missing most of your outcome space. You might switch to a generator with a bigger seed. For this example with 100 items, you need at least

log(100 !) / log(2) = 524.764993

bits in your seed. For N items to be sorted, you need about NlogN bits in your seed. So if you're trying to actually make each outcome possible, then no matter which clever algorithm you use to go from random numbers to a shuffled sequence, you've already lost.

allenbrunson
hrm. never thought of it that way before. but assuming I seed the prng only once at the beginning of a program run, i could hit every possible shuffle on subsequent deals, couldn't I?
fhars
No, you couldn't, that's what this subthread is about. A PRNG with 32 bit entropy will only ever produce 2^32 different sequences of random numbers and so only 2^32 different shuffles. If you want more, you must introduce more entropy, either by using a larger PRNG, or using a hardware RNG. Calling the same PRNG fro diffferent threads (don't forget your synchronization primitives!) might also work as a cheap source of entropy but will likely have either bad statistical properties or horrible performance: for it to make a real difference, every single shuffle of a deck must be interspersed with many calls to the PRNG from other threads. If calls from other threads do only happen between shuffles, all you gain from introducing shared state concurrency into your program is that the same 2^32 permutations are generated in a different order.

But, as I remarked earlier, if this is just for a game that is for fun where opponents can't cheat you or other players out of large amounts of money by predicting which permutations may or may not occurr, you may get away with whatever PRNG you have at hand and only swap is out once many players start to complain that they had this very hand one or two billon games ago.

gjm11
That problem (not enough entropy in your PRNG state) isn't specific to the particular way of shuffling: it applies just the same if you do a Fisher-Yates shuffle, for instance.
crux_
Besides the performance difference, the main issue is the ideal of a perfect shuffle: each of the N! sorts ought to be equally likely.

(N.b.: this post ended up being mostly the same point as fhars, only phrased differently.)

However, this is tricky to achieve. It requires either:

- an infinite source of random numbers

- a psuedo random number generator whose pool size (the # of random numbers it can generate) is an even multiple of N! (If not, then it have bias towards certain shuffles almost by definition.)

Here's what happens when someone does it all wrong -- the wrong shuffle algorithm and, far more devastatingly, a poor PRNG choice:

http://www.cigital.com/papers/download/developer_gambling.ph...

(Consider that if your PRNG takes a 32-bit seed, there will be at most be 2^32 possible shuffles, even with a correctly perfect shuffling algorithm --- few enough for an attacker to pre-calculate every possible shuffle to use against you.)

fhars
As far as I can tell, the "correct" shuffle in that post is wrong, int(rand() * $i) is always smaller than $i, so it will never generate a permutation where an element stays at the same position.

It is obvious from the numbers, too: the loop goes from $length - 1 to 1, so it has $length - 1 iterations, in the first iteration there are $length - 1 possible values for $r, in the second $length - 2, and the last iteration will always swap elements 0 and 1, for a total of (N-1)! different paths through the program, which are clearly not sufficient to generate N! different permutations.

[Edit: I had an off-by-one in my calculations, too. I hope it is correct now.]

[Edit: Actually, there still is an error in the first sentence: the algoritm will never leave the current element in its place, so the only element that is guaranteed not to stay at the same position is the last one. Run this:

  for my $i (1..1000000) {
    my @a = (1..10);
    my @b = shuffle(@a);
    if ($b[10] == 10) {
	die "Criticism is wrong $i";
    }
  }
]
fhars
After a good night's sleep I see that my second edit is in fact wrong and my intuition in the first sentence was right (and the code in my edit fell victim to the same off-by-one as the code in the article).

Final proof:

  for my $i (1..1000000) {
    my @a = (0..10);
    my @b = shuffle(@a);
    for $j (0..10) {
	if ($b[$j] == $j) {
	    die "Criticism is wrong $i";
	}
    }
  }
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jrockway
The best way to get this right in Perl:

    use List::Util qw(shuffle);
    shuffle(@list);
astine
But that misses the point of the exercise. :)
rflrob
So presumably if the naive interpretation is wrong, then the bias should be predictable. Any ideas how to go about determining which permutations are more likely than they ought to be?
songism
This is how python's shuffle works:

  def shuffle(x):
    for i in reversed(xrange(1, len(x))):
      j = int(random() * (i+1))
      x[i], x[j] = x[j], x[i]
In the first loop the last element in the list 'x' has a chance to be exchanged with itself or any of the other elements. Then the second to last element has a chance to be exchanged with itself or any other element.

All the way down to the second to last element, which has the chance to be exchanged with itself or the first element.

This is exactly how the good Knuth shuffle algorithm works, right?

thisrod
I'm surprised that the median job candidate gets that wrong. It isn't an especially hard question, and I would have assumed professional programers had a pretty good instinct for combinatorics.