The problem is, your shuffle routine is not an implementation of a Fisher-Yates shuffle.

This line 

    my $j = shift @$rand;

[download]

is no way equivalent to this line from the FAQ implementation 

     my $j = int rand ($i+1);

[download]

The latter picks a swap partner for the current value of $i, randomly between 0 and $i-1. I can't quite wrap my brain around what your code is doing here, but it isn't even vaguely equivalent.

Therefore you are not testing a Fisher-Yates shuffle, but some shuffle algorithm of your own invention, which you succeed in proving isn't as good as the Fisher-Yates.

You might find this post Re: When the Best Solution Isn't that does a statistical analysis of several shuffle routines, a Fisher-Yates amongst them, including frequency and standard deviation interesting.

Examine what is said, not who speaks.
"Efficiency is intelligent laziness." -David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller

In reply to Re: Fisher-Yates theory... does this prove that it is invalid? by BrowserUk
in thread Fisher-Yates theory by Anonymous Monk

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