From mboxrd@z Thu Jan  1 00:00:00 1970
From: "H. Peter Anvin" <hpa@zytor.com>
Subject: Re: Triple parity and beyond
Date: Wed, 20 Nov 2013 11:10:11 -0800
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On 11/20/2013 11:05 AM, Andrea Mazzoleni wrote:
> 
> For the first row with j=0, I use xi = 2^-i and y0 = 0, that results in:
> 

How can xi = 2^-i if x is supposed to be constant?

That doesn't mean that your approach isn't valid, of course, but it
might not be a Cauchy matrix and thus needs additional analysis.

> row j=0 -> 1/(xi+y0) = 1/(2^-i + 0) = 2^i (RAID-6 coefficients)
> 
> For the next rows with j>0, I use yj = 2^j, resulting in:
> 
> rows j>0 -> 1/(xi+yj) = 1/(2^-i + 2^j)

Even more so here... 2^-i and 2^j don't seem to be of the form xi and yj
respectively.

	-hpa