Ok,
thanks a lot, everyone!
Yes, I should rather have started thinking a bit more myself,
before going the easy route to R-help....
Anyway, the most obvious algorithm,
just putting things into place by swapping elements,
and counting how many times you have to swap, is easy and
quite efficient.
I'll post R code later, being busy for the next few hours.
Martin
>>>>> "MM" == Martin Maechler <[EMAIL PROTECTED]>
>>>>> on Tue, 15 Apr 2008 18:13:43 +0200 writes:
MM> I am looking for an algorithm (written in R (preferably) or C,
MM> but even pseudo-code in a text book maybe fine)
MM> to determine the sign of a permutation.
MM> What is that? Well, a permutation is either even or odd, the
MM> sign is +1 or -1, respectively, see, e.g.,
MM> http://en.wikipedia.org/wiki/Signature_of_a_permutation
MM> which also says
>>> In practice, in order to determine whether a given permutation
>>> is even or odd, one writes the permutation as a product of
>>> disjoint cycles. The permutation is odd if and only if this
>>> factorization contains an odd number of even-length cycles.
MM> but I would not know how to algorithmically
MM> "write the permutation as a product of disjoint cycles"
MM> If you start looking at R code,
MM> let's assume the permutation {\pi(i)}_{i=1..n} is simply given
MM> as the (integer) vector (\pi(1), \pi(2), ..., \pi(n))
MM> {or equivalently, a random permutation is simply found by 'sample(n)'}
MM> Thank you in advance for further pointers,
MM> or even working R code.
MM> Best regards,
MM> Martin Maechler, ETH Zurich
MM> ______________________________________________
MM> R-help@r-project.org mailing list
MM> https://stat.ethz.ch/mailman/listinfo/r-help
MM> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
MM> and provide commented, minimal, self-contained, reproducible code.
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