No, that's not what I meant, but maybe I didn't understand the question.
What I suggested would involve sorting y, not x: "sort the *distances*".
If you want to minimize the sd of a subset of numbers, you sort the numbers and 
find a subset that is clumped together.
If the numbers are a function of pairs, you compute the function for all pairs 
of numbers, and find a subset that's clumped together.
Anyway, it's an idea, not a theorem, so proof is left as an exercise for the 
esteemed reader.

-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Hans W Borchers
Sent: Monday, February 28, 2011 2:17 PM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] Finding pairs with least magnitude difference from mean

 <rex.dwyer <at> syngenta.com> writes:

> James,
> It seems the 2*mean(x) term is irrelevant if you are seeking to
> minimize sd. Then you want to sort the distances from smallest to
> largest. Then it seems clear that your five values will be adjacent in
> the list, since if you have a set of five adjacent values, exchanging
> any of them for one further away in the list will increase the sd. The
> only problem I see with this is that you can't use a number more than
> once. In any case, you need to compute the best five pairs beginning
> at position i in the sorted list, for 1<=i<=choose(n,2), then take the
> max over all i.
> There no R in my answer such as you'd notice, but I hope it helps just
> the same.
> Rex

You probably mean something like the following:

    x <- rnorm(10)
    y <- outer(x, x, "+") - (2 * mean(x))

    o <- order(x)
    sd(c(y[o[1],o[10]], y[o[2],o[9]], y[o[3],o[8]], y[o[4],o[7]], y[o[5],o[6]]))

This seems reasonable, though you would have to supply a more stringent
argument. I did two tests and it works alright.

--Hans Werner

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