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The latest introduction of MDS conjecture, from a french sicentistla: Introduction-The Singleton Bound
Singleton Bound. k< n-d+1, for any [n, k, d] code
Proof Any 2 codewords disagree in the first n-d+l coordinates
somewhere, so there are gn-d+i in total L
Linear codes achieving equality are called Maximum Distance Sepa
rable(MDs)codes. a general question given d and k, what is the
greatest length n of an MDs code?
1b: MDS codes and generic subsets of Fk
List k rows generating mds code c as a k x n matrix A
Claim. The columns of a give rise to a set s of n vectors, such that
any k=n-d+l of them are Ll. We call such an s generic
Proof. any k-dependence in columns
→∑cA6=0(K|=k)
6∈K
→)x6=0 V rows x of a(→x∈C)
6∈K
So not all gn-d+i choices for the n-d+1 such(xs sek appear in
C. Contradiction!
Conversely, any generic s gives rise to an MDs code of length S
Ic:(Supposedly) best example
We have a correspondence
Length of MDS code +Si
ize of generic SCFg
Try RM codes! We know they meet the Singleton Bound
Enc: fb(f(a1),., f(an))as an rS encoder needs to use distinct
i,so we can obtain a length of up to n
usIn
g all possible
elements of
Under this correspondence, the generic s obtained is the normal ra
tiona| curve“{(1,t,t
t E Fqg-any k such form a VDM
matrix, hence are LI! We can add (0, .. 0, 1) to this S, reaching
n=q+1
MDS Conjecture:
Hfk≤q, then a generic|S≤q+1. We prove case k≤p(=q)
2a: Segre's Tangent Function
Say S C fk is generic. Then, if Z C s has Z=k-2, consider
the codimension-1 hyperplanes 2) z with normal vectors vy. We
define a variable polynomia
7z(X):=Ⅱ<,X>
∑∩S=Z
Then, if x,y, zJUY is a basis, we have
TYux (y)Truly (z) Trufz(x
(-1)17y
YUx(zT
(x)Ty{z}()
where t=p+k-1-s
2b: Interpolating T
For E=a1,., at+2 and Y=k-2 disjoint in
0=∑7(a)I
∈E
b∈E\a
But all we needed was that{b,a}∪ Y was a basis va≠b∈E.We
never sp
olit
up
Idea: exchange elements of E and r. More generally, if r
T、(a)I
∈E
0;+1(y)
∈EUy(0rU{ar}
t (ar, z, 0r)
Here B;=(al, .. ai-1, yi,..., yk-2), as a set and a tuple
2c: Using Segre's lemma to simplify the
interpolation equation
Any order of a1,..., ar give the same term in the sum
So, we have
0(a
0=r!
T、(a)II
det(ar, z, 8r)
k-l, we use 0. which only has k-2 entries. But
t=q+k-1-|5
If sp, that S< q+k+1-min k, p
But we cannot hope to replace p by g in this proof because of the
final step: p!=0 in Fo
Nevertheless, in a follow-up paper, Ball relaxed the condition from
k< p to k 2p-2, for mds to hold
Iso when p
Als
2 and k=3 or g-l, the conjecture isn t quite
true-insteadS
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