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Re: Fisher-Yates theory... does this prove that it is invalid?by BrowserUk (Patriarch) |
on Jul 25, 2003 at 02:04 UTC ( [id://277774]=note: print w/replies, xml ) | Need Help?? |
The problem is, your shuffle routine is not an implementation of a Fisher-Yates shuffle. This line
is no way equivalent to this line from the FAQ implementation
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.
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