Peter, I think I understand it differently. Concrete example in GF(256) for k=6, m=4:
First, create a 3 by 6 cauchy matrix, using x_i = 2^-i, and y_i = 0 for i=0, and y_i = 2^i for other i. In this case: x = { 1, 142, 71, 173, 216, 108 } y = { 0, 2, 4). The cauchy matrix is:
1 2 4 8 16 32
244 83 78 183 118 47
167 39 213 59 153 82
Divide row 2 by 244 and row 3 by 167. Then extend it with a row of ones on top and it's still MDS, and that's the code for m=4, with RAID-6 as a subset. Very nice!
Jim
----------
On Nov 20, 2013, at 2:10 PM, H. Peter Anvin wrote:
> 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
>
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