There has been a recent interest in finding an MDS code with a generator matrix constrained on the support. This problem appears in many areas such as distributed coding and storage, multiple access networks, where each relay nodes has access to a subset of the sources , , and weakly secure data exchange, where users have a subset of the data packets and want to exchange them without revealing information to eavesdroppers , .
For a linear code with length and dimension , the singleton bound on the minimum distance is . To achieve this bound, namely for MDS codes, any columns of the generator matrix should be linearly independent. Let be the set of positions of the zeros in the th row of . Then, for any subset , the columns indexed in will have zeros in all their entries in .111 represents the set Since those columns need to be linearly independent, for any nonempty ,
is a necessary condition for the code to be MDS.
It is not hard to show that generating a random matrix with a constrained support satisfying (1) will result in an MDS code with high probability if the field size is large enough . Nonetheless, since random codes are not easy to decode, it is more preferable to design structured codes like Reed-Solomon codes, which have efficient decoders. The GM-MDS conjecture stated by Dau et al.  describes (1) as also a sufficient condition for the existence of a Reed-Solomon code whose generator matrix satisfies the support constraints.
Although the conjecture has many equivalent versions and partial proofs have been proposed in ,,,, it has not been proven yet in general. Heidarzadeh et al.  proved it for . Halbawi et al.  proved the statement for if there are distinct support sets on the rows of the generator matrix. In , , the result is proven when the generator matrix is sparsest and balanced. Yan et al.  give a partial induction step, a way to reduce the problem from to if one of the inequalities in (1) holds with equality for some such that .
We should mention that there is a related problem where, given a support constraint on the generator matrix, one would like to find a code with the largest minimum distance. This is because not every support constraint will admit an MDS code. Again it can be shown that, for a large enough field size, a random generator matrix satisfying the support constraints achieves the maximum minimum distance with high probability. In , it has been shown that the existence of Reed-Solomon codes that achieve the maximum minimum distance is equivalent to the GM-MDS conjecture studied in this paper. Finally, we should mention that the dual problem where the support constraint is on the parity check matrix, is of interest in locally-repairable codes , .
In this paper, we will group the rows with the same support constraint and prove the GM-MDS conjecture for , which improves all the previous results except the sparsest and balanced case. Furthermore, we will give a more extensive way for reducing the problem, which covers any equality case, not limited to .
The rest of the paper is organized as follows. In the next section we formulate the problem and, in fact, introduce a slightly more general problem including multisets which will be of use in our proof. The proof of the main result appears in Section 3 and the paper concludes with Section 4.
Ii Problem Setup
Ii-a GM-MDS conjecture
Consider the generator matrix of a generalized Reed-Solomon code:
for distinct . For a nonsingular , define the generator matrix
The GM-MDS conjecture  is
If satisfy for any nonempty ,
then, there exists and a nonsingular such that for all and where .
Ii-B Grouping equal subsets
We will group the equal subsets and represent the position of zeros in the rows of the generator matrix by the sets with multiplicities where . That is, the first rows of will have zeros at positions in , the next rows will have zeros at positions in and so on. Then, the condition (4) on these sets becomes
for any nonempty .
Suppose satisfies condition (5). Let for and be any partition of such that where . Note that due to (5) for . Then, will also satisfy the condition (5). Define similarly by introducing new variables . If is a solution for , then will be a solution for .
After this assumption, the span of the grouped rows in will be uniquely determined. Therefore, we can analyze the singularity of one example:
which is partitioned into blocks where the th block has rows formed by the coefficients of the polynomials , , , where . The precondition
ensures that the matrix is square. Furthermore, in the multiplication , the rows will consist of the substitution of ’s in the polynomials of the form , which will have zeros at the desired positions.
As a result, we end up with an equivalent conjecture to creftypecap 1:
For , let such that , and for any nonempty ,
Then, (which is a multivariate polynomial of ’s) is not identically zero,
where is given by (6).
From now on, we will assume that are indeterminates and write or to indicate that the determinant is identically zero or nonzero, respectively.
Ii-C Extension to multisets
Multisets are the generalization of the sets where multiple instances of the set elements are allowed . The multiset extension for the sets will be useful later in the proof of our main results. Although multisets have no meaning regarding the positions of the zeros in the generator matrix, we can still define the matrix in (6) for ’s being multisets, in which case, the polynomials may have multiple roots. We will not write the conjecture for multisets; however, the fact that the condition (8) is necessary can be extended for multisets as well:
Let be multisets in the universe such that and . Let be the polynomials defined as for all .
Then, if and only if there exists some polynomials , not all zero, such that
Iii Main Results
Due to the lack of a complete proof for creftypecap 2, we will apply the minimal counterexample method in order to present all our findings and to show that the conjecture holds for all . If creftypecap 2 is not true, then, there will be at least one counterexample which satisfies the conditions in creftypecap 2 but for which, . Among these counterexamples, there will be one (or many) that is minimal with regards to the parameters when considered in lexicographical order. In Lemma 2, some necessary conditions are listed for a minimal counterexample. Note that these conditions are not necessary for any counterexample but for a minimal counterexample.
It will be needed in the statement of Lemma 2 to define a new collection of sets where .
If creftypecap 2 is not true and is a counterexample such that is the smallest possible in the lexicographical order222It turns out that Lemma 2 is also true for different orderings of ., then, the following333Although there are eight conditions listed, the last five are consequences of the first three. must be true:
For any nonempty such that ,
For any ,
For any ,
For any , there exists such that and (Equivalently, ).
For any , .
For any ,
Furthermore, since , .
There exists such that .
If for some , then, for any , .
Since , by Proposition 1, there exist polynomials , not all zero, such that and . Assume the contrary. Then, there exists some with such that
Assume the contrary. Hence, there exists such that and . W.l.o.g. assume that and . For all , define the sets
Since , if , then yielding . Hence, we can define for ’s and
Since , . Since , for any nonempty ,
Therefore, for ,
Hence, is also a counterexample with parameters . Contradiction.
Assume the contrary. W.l.o.g. assume that . For all , define the multisets
where is the multiset summation. Then, similar to (ii), . We have that . Denote by the multiplicity of in . Then, for any , , , and for . Then, for any nonempty ,
(21) (23) (24)
Therefore, for ,
So, satisfies the conditions in creftypecap 2 except they are multisets. However, we can apply Lemma 1 by defining and . Note that ’s are normal sets i.e. they do not contain any element with multiplicity more than one. Then, by Lemma 1, is also a counterexample with parameters . Contradiction.
Assume the contrary. Then, there exists such that . By (i),
yielding . Contradiction.
Assume that . Then, . Apply Lemma 1. Hence, is also a counter example with smaller parameters. Contradiction.
Assume that . Then, by (iii), for any , . Then, either meaning or meaning . Hence, there is no containing but not . Contradiction due to (iv).
Assume that . Then, by (ii), for either or . Hence, we can partition the set into
If , then , which implies . Hence, . Therefore,
where we use the fact that for since .
By (i), . Hence, .
Assume the contrary. Then, for all , . By (v) and (vi), . Then, by (ii), for any , ; so, either or . W.l.o.g. assume that . Then, by (iv), for any , there exists such that and . Then, . Then, for any , . Then, there is no containing but not . Contradiction due to (iv).
Corollary of (ii) and (v).
Iii-a Consequences of Lemma 2
Firstly, Lemma 2 allows us to make the assumptions listed from (i) to (viii) when proving creftypecap 2. If creftypecap 2 is true under these assumptions, then it must be also true without these assumptions; otherwise, it will lead to a contradiction for the minimal counterexample. For example, the condition (i) implies that “it is enough to prove creftypecap 2 only for the case where all the inequalities in (8) are strict”.
Secondly, if one of the conditions listed in Lemma 2 does not hold, then, the problem can be reduced to the one with a smaller parameter or . The way in which it is reduced can be found in the proof of Lemma 2 for parts (i)-(iii).
Thirdly, it can help us to solve the problem for small parameters. The conditions (v) and (vii) already imply that creftypecap 2 is true for because by condition (vii), there exists a set with size at least , whose size is upper bounded by in condition (v). By a little more work, we can also solve as shown in Theorem 2:
creftypecap 2 is true for .
Assume the contrary. Then, there exists a minimal counterexample such that . By (v) and (vii), . Hence, . By (v) and (vi), for all . If and are size , then by (iii). There will be three cases:
Case 1. . Assume that and are two different sets of size . By (viii), their intersection must have exactly one element. W.l.o.g. assume that and . By (viii), for any , and ; hence, either or . For any , there is a set containing but not , which has to be . Let . Let be the set containing but not . Then, . Since , we have , which means . Contradiction.
Case 2. . W.l.o.g. let . If , then and by (viii) either or . Let be the set containing but not . Then, wlog. . Let be the set containing but not . Then, . Let be the set containing but not . Then, for some . Since , . Then, . Contradiction.
Case 3. For all , . Let . Let contain but not . Then, wlog . Let contain but not . Then, . Let contain but not . Then, . Let . Then, has at least one element from . Wlog assume that . Then, . . Contradiction.
We have established the correctness of the GM-MDS conjecture of Dau et al. for , where is the number of distinct support sets defined on the rows of the generator matrix. The result subsumes all earlier known results on the GM-MDS conjecture except for those pertaining to sparsest and balanced generator matrices. Our results followed has a careful study of properties that must hold for any minimal counterexample to the conjecture. It remains to be seen whether this approach can be extended to prove the conjecture for values of beyond .
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