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:

http://www.scribd.com/doc/64349616/O-Connor-A-Historical-Note-on-the-Fisher-Yates-Durstenfeld-Shuffle-Algorithms

 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
			set
	}

	// 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).

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
ABC,
-BCA,
–CAB
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?

” 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 ;-)

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.

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.

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.