Erasure resilient MDS code with four redundant packets

Erasure resilient MDS code with four redundant packets

Emin Gabrielyan

EPFL / Switzernet Sàrl

2005-11-03

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We are trying to build (11, 7), (10, 6) and (9, 5) MDS codes

Given are:

- 7, 6 or 5 information packets

- 4 redundant packets

- Packet sizes are identical and are divisible by 3, minimum 3 bits

- Information must be retrieved if number of losses does not exceed 4

We must check for which numbers of information packets we can build an MDS code.

Let (a,b,c) be a packet where a, b and c are its first, second and third portions.

Let f be a function which applied to a packet (a,b,c) forms another packet of the same size, whose first, second and third elements are XOR results of some subsets given from {a,b,c}. Example: f(x, y, z) = (x+y, z, x+z), where operation + is XOR.

We are interested in only invertible functions. There are 168 such functions producing 168 invertible packets. Each function can be represented by a binary 3 by 3 matrix.

Four redundant packets are constructed as follows

where k is the number of information packets, i.e. is equal to 7, 6 or 5.

f, g and h are vectors whose elements are from the list of 168 invertible functions

Restoring two information packets from (1, f), (1, g) and (1, h)

In case, two information packets i, j are lost and we received the first and the second redundant packet (the other two are also lost). Then if + is invertible for any pair of i and j we can restore and successively . Similarly for the cases when two information packets must be restored from the first and third (g-redundant) packets or from the first and fourth (h-redundant) packets.

In the set of 168 invertible functions, there are 4032 subsets of the size of 5-functions, 1344 subsets of the size of 6-functions and 192 subsets of the size of 7-functions, for which the above condition ( + 1 is invertible) holds for any pair of the subset.

Restoring two information packets from (f, g)

Let us now examine valid combinations of f and g vectors. For the sizes of 5, 6 and 7 functions there are 403240325!, 134413446! and 1921927! possible pairs of f and g vectors. An (f, g) pair is valid only if for any two i and j (the lost packets) the following function:

is invertible

Restoring three information packets from (1, f, g)

Additionally, an (f, g)-pair is valid only if it can retrieve, together with the first redundant packet, any three lost information packets.

For that the following function must be invertible for any i, j and k:

(+1(+1) + (+1(+1)

Instead of examining all possible pairs of vectors

403240325! – for 5 information packets (codeword length = 9)

134413446! – for 6 information packets (codeword length = 10)

1921927! – for 7 information packets (codeword length = 11)

We fixed the f-vector on the first candidate

(11,73,140,167,198) – for 5 information packets

(11,73,140,167,198,292) – for 6

(11,73,140,167,198,292,323) – and for 7

Thus we limited our choice by the following number of pairs

40325! – for 5 information packets

13446! – for 6

1927! – and for 7

for 7 information packets we have found 1680 valid (f, g)-pairs

for 6 information packets we have found 1680 valid (f, g)-pairs as well

and for 5 information packets also we have found 1680 valid (f, g)-pairs

Thus (10, 7)-code exists, which is an MDS code.

Choosing h-redundant packet, restoring two information packets from (g, h) and three information packets from (1, g, h)

For any of 1680 valid (f, g)-pairs we must examine a valid (f, h)-pair, thus there are 1680(1680–1)/2 possible (f, g, h) combinations to examine.

(g, h)-pair is valid only if:

+ is invertible for any two i and j (the case when two information packets must be retrieved from the g and h-redundant packets)

and if:

(+1(+1) + (+1(+1) is also invertible for any three lost information packets i, j and k (the case when three information packets must be retrieved from the first redundant packets and from the g and h-redundant packets).

there are 28224 valid (g, h)-pairs for 7 information packets

there are also 28224 valid (g, h)-pairs with 6 information packets

and there are 56448 valid (g, h)-pairs with 5 information packets

Restoring three information packets from (f, g, h) and four information packets from (1, f, g, h)

Three lost information packets can be retrieved from f, g and h-redundant packets if the following function is invertible

(+(+) + (+(+)

for any three lost information packets i, j and k

Among 28224 (g, h)-pairs with 7 information packets and 28224 (g, h)-pairs with 6 information packets there were none, satisfying the above constraint, thus:

(11, 7) MDS code does not exist and

(10, 6) MDS code does not exist (at least with this method)

Additionally vector h is valid only if we can also restore any four i, j, k and l lost information packets from the four redundant packets. From the four redundant packets we can obtain these three (by eliminating corposant)

(+1)+

(+1)

(+1)+

(+1)

(+1)+

(+1)

From them we can obtain these two by eliminating the composant:

((+1 (+1) + (+1 (+1)) +

((+1 (+1) + (+1 (+1))

From the above two, we can eliminate and obtain the below function applied to . If this function is invertible then we can retrieve and consecutively all other information packets.

( (+1 (+1) + (+1 (+1)

((+1 (+1) + (+1 (+1))

( (+1 (+1) + (+1 (+1)

((+1 (+1) + (+1 (+1))