An Algebraic-Combinatorial Proof Technique for the GM-MDS Conjecture

# An Algebraic-Combinatorial Proof Technique for the GM-MDS Conjecture

Texas A&M University, College Station, TX 77843 USA
###### Abstract

This paper considers the problem of designing maximum distance separable (MDS) codes over small fields with constraints on the support of their generator matrices. For any given binary matrix , the GM-MDS conjecture, due to Dau et al., states that if satisfies the so-called MDS condition, then for any field of size , there exists an MDS code whose generator matrix , with entries in , fits (i.e., is the support matrix of ). Despite all the attempts by the coding theory community, this conjecture remains still open in general. It was shown, independently by Yan et al. and Dau et al., that the GM-MDS conjecture holds if the following conjecture, referred to as the TM-MDS conjecture, holds: if satisfies the MDS condition, then the determinant of a transformation matrix , such that fits , is not identically zero, where is a Vandermonde matrix with distinct parameters. In this work, we generalize the TM-MDS conjecture, and present an algebraic-combinatorial approach based on polynomial-degree reduction for proving this conjecture. Our proof technique’s strength is based primarily on reducing inherent combinatorics in the proof. We demonstrate the strength of our technique by proving the TM-MDS conjecture for the cases where the number of rows () of is upper bounded by . For this class of special cases of where the only additional constraint is on , only cases with were previously proven theoretically, and the previously used proof techniques are not applicable to cases with .

## I Introduction

In recent years, there has been a growing interest in designing maximum distance separable (MDS) codes with constraints on the support of the codes’ generator matrix [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Such constraints arise in wireless network coding and distributed storage scenarios where each user/server has access to only a subset of the information symbols. Two examples of such scenarios are cooperative data exchange in the presence of an eavesdropper [1, 5], and simple multiple access networks with link/relay errors [6, 8].

Given an binary (support) matrix and a field of size , the problem is to design an MDS code with a generator matrix , (i.e., all sub-matrices of are full-rank) fitting (i.e., if , then ). Note that for some , there exists no MDS code whose generator matrix fits (i.e., is not completable to MDS). Nevertheless, there is a simple condition, known as the MDS condition, which characterizes all matrices that are completable to MDS for sufficiently large fields [4, 5]. A matrix satisfies the MDS condition if:

 |∪i∈Isupp(Mi)|≥n−m+|I|,∀I⊆{1,…,m},I≠∅,

where is the th row of , and is the support of . The existence of MDS codes (over sufficiently large fields) whose generator matrix’s support satisfies the MDS condition was shown, e.g., in [8], via Edmonds matrix and Hall’s marriage theorem. The following conjecture, due to Dau et al. [4], aims for generalizing this result for small fields.

###### Conjecture 1 (GM-MDS Conjecture)

If the matrix satisfies the MDS condition, then for any field of size , there exists an MDS code whose generator matrix with entries in fits the matrix .

Notwithstanding all the efforts by the coding theory community, the GM-MDS conjecture remains still open in general. The GM-MDS conjecture and a simplified version of this conjecture where the supports of rows of have the same size were shown in [8] to be equivalent (using a generalized version of Hall’s theorem). Despite this simplification, there are only three classes of special cases for which this conjecture is theoretically proven: (i) the rows of are divided into three groups, and the rows in each group have the same support [6]; (ii) the size of intersection of the supports of every two rows of is upper bounded by [8]; and (iii) the number of rows of is upper bounded by [11]. More importantly, the previously used proof techniques are not applicable to more general cases due to the combinatorial explosion.

One possible approach to find a completion of to MDS is to leverage the structure of Generalized Reed-Solomon (GRS) codes [5, 4] which are known to be MDS. Let be the set of independent indeterminates . Let be an binary matrix whose rows’ supports have the same size, and let be a generic Vandermonde matrix with parameters . Let be a generic transformation matrix such that fits , and let . If the evaluations of are distinct, then every sub-matrix of is non-singular (i.e., is a generator matrix of a GRS code with evaluation points ) so long as is non-singular. That is, if is not generically singular (i.e., the determinant of as a multivariate polynomial in variables is not identically zero), then for any field of size , there exists such that is a generator matrix of an MDS code, and fits . Thus, the GM-MDS conjecture holds if the following conjecture, proposed independently by Dau et al. [4] and Yan et al. [5], holds:

###### Conjecture 2 (TM-MDS Conjecture)

If satisfies the MDS condition, then is not generically singular.

The contributions of this work are as follows. First, we present a generalization of the TM-MDS conjecture for the cases where the supports of rows of have arbitrary sizes. Then, we present an algebraic-combinatorial approach based on polynomial-degree reduction for proving this conjecture. Our technique’s strength is primarily due to reducing the inherent combinatorics in the proof. Specifically, we demonstrate this strength by proving the TM-MDS conjecture for the cases where the number of rows of is upper bounded by .

## Ii Basic Notations and Definitions

Let be a field. For , let be independent indeterminates. Let be a ring of multivariate polynomials in variables with coefficients in , and let be a module of univariate polynomials in variable with coefficients in . Fix and such that . For , denote by . Define polynomials of degrees in :

 Pi(α)≜∏γ∈Ni(α−γ),∀i∈[m], (1)

where , the set of roots of , is a (proper) subset of of size . (Note that the roots of are indeterminates.) For and , .

Note that , where are polynomials in . Define . Note that is a polynomial in .

###### Definition 1

A polynomial in is identically zero, denoted by , if the coefficients of all monomials in the polynomial expansion of are zero.

###### Definition 2

A set of polynomials of degree has rectangular property (RP) if, for some , there exist at least polynomials in with at least common roots. Otherwise, has non-rectangular property (NRP).

###### Definition 3

A set of polynomials of degrees has generalized RP (GRP) if, for some and , there exist at least polynomials of degrees at most in with at least common roots. Otherwise, has generalized NRP (GNRP).

## Iii Main Conjectures and Theorems

The following conjecture is equivalent to the TM-MDS conjecture (Conjecture 2).

###### Conjecture 3

Let be polynomials of degree in . If , then has RP.

The sketch of proof of the equivalency between the TM-MDS conjecture and Conjecture 3 follows. Let be an binary matrix. Let be the th row of and let be the support of . Let , where are independent indeterminates. Suppose that all have the same size. Defining () as in (1), it follows that the matrix satisfies the MDS condition iff has NRP (refer to this as Fact 1). Let be a generic generator matrix of a Generalized Reed-Solomon (GRS) code with evaluation points such that fits (i.e., if , then ). Let be a generic Vandermonde matrix with parameters , and let be a generic transformation matrix such that . Taking and , where , it follows that (for more details, see [5]). Since , then (i.e., is not generically singular) iff (refer to this as Fact 2). By Facts 1 and 2, the TM-MDS conjecture and Conjecture 3 are equivalent.

In the following, we propose a new conjecture which generalizes Conjecture 3 for the cases where the degrees of polynomials are arbitrary. (Conjecture 4 is equivalent to a generalized version of the TM-MDS conjecture where the supports of rows of have arbitrary sizes.)

###### Conjecture 4

Let be polynomials of arbitrary degrees in . If , then has GRP.

If for all , then Conjecture 4 holds trivially: (i) since for all , and (ii) has GRP since for and , there exist polynomials of degrees at most in with at least common roots. Hereafter, w.l.o.g., we assume for some .

The following theorems, which are our main results, prove the GM-MDS conjecture for . More specifically, Theorems 12, and 3 settle Conjecture 4 (and so Conjecture 3) for , and Theorem 4 settles Conjecture 3 for .

###### Theorem 1

For any such that , if , then has GRP.

###### Theorem 2

For any such that , if , then has GRP.

###### Theorem 3

For any such that , if , then has GRP.

###### Theorem 4

For any such that , if , then has GRP.

## Iv Main Ideas and Lemmas

In this section, we explain the main ideas and state the useful lemmas for the proofs of our main results.

Consider an arbitrary set () of polynomials (with the sets of roots ) such that for all , and for some . Define a class of reduction processes over , where any process in this class is associated with a unique reduction set , and it reduces to . Let . Note that . Restrict your attention to those reduction sets such that for some , and for all . Such are referred to as acceptable. For any acceptable reduction set, w.l.o.g., assume that for all and (and so, since ). For any (acceptable) reduction set, the following result holds.

If , then .

###### Proof 1

Consider the resulting from for an arbitrary (acceptable) reduction set . Let . For any , let be the number of polynomials in such that , and let . Let be the resulting polynomial from by taking derivative times with respect to each variable . (Since , and is the sum of monomials for some (depending on ), then the derivatives of with respect to any variable are independent of .) Note that (by using the Leibniz formula for determinant), and (since and for all , and and , where ). Since (by assumption), then . Thus, .

Lemma 1 enables us to use an inductive argument towards the proof of Conjecture 4 as follows. Suppose that Conjecture 4 holds for any , i.e., for any such that for all and for some , if , then has GRP. We need to prove that for any such that for all and for some , if , then has GRP. The proof follows by contradiction. Assume that and does not have GRP. Consider the resulting from for an (acceptable) reduction set such that has GNRP. By definition, for all . Since (by assumption), then (by Lemma 1), and so, has GRP (by the induction hypothesis), yielding a contradiction. Our goal is thus to devise an (acceptable) reduction process such that if has GNRP, then so does . The problem of designing such a process is still open in general. In the following, we propose a simple yet powerful reduction process which solves this problem for and for all , and for and for all .

From now on, we assume that () is a set of polynomials (with the sets of roots ) such that for all , and .

###### Definition 4

A subset is an -subset in a subset of if belongs to polynomials in (i.e., there exist polynomials in such that ), and . Moreover, an -subset has higher order than an -subset if , or and .

The following lemma gives the intuition behind the definition of -subsets.

###### Lemma 2

If has GNRP, then there exists no -subset in such that .

###### Proof 2

The proof is straightforward and follows from the definitions (and hence omitted).

Intuitively, for any (acceptable) reduction set , any highest-order -subset , if not broken (i.e., ), is the most likely to cause rectangularity in for any with non-rectangular property. This is the main idea of the proposed reduction process.

###### Definition 5

An element of a subset is removable if is a root of some but not all polynomials of degree .

###### Definition 6

A subset is weakly reducible if belongs to a polynomial of degree , and has a removable element.

###### Definition 7

A weakly reducible -subset is strongly reducible if no other weakly reducible -subset has higher order than .

### Proposed Reduction Process

Given , choose an arbitrary strongly reducible subset in , and choose an arbitrary removable element of , say , such that no other removable element of , when compared to , belongs to more polynomials of degree in . Break via removing from the sets of roots of all polynomials , and update all polynomials via replacing by . Repeat this process (in rounds) over the resulting if there exist more than one polynomial of degree . Otherwise, terminate the process, and return the resulting denoted by .

Note that if has GNRP initially, then (i) in each round of the process, such exists, and (ii) the process terminates eventually. Otherwise, there must exist two (or more) identical polynomials of degree in (and hence has GRP), which is a contradiction.

Consider an arbitrary run of the reduction process over and its corresponding . Let be the set of the roots that the reduction process removes over the rounds. (Note that, due to the arbitrary choices in the reduction process, may or may not be unique.) Hereafter, for any such , assume, w.l.o.g., that the (initial) indexing of polynomials in is such that () and , and denote by .

The proofs of our main theorems rely on the following properties of the proposed reduction process.

###### Lemma 3

If has GNRP, then any -subset in such that belongs to a polynomial in such that .

###### Proof 3

Let be an arbitrary -subset in such that . Let be the set of all polynomials in such that . Note that is a set of at most polynomials of degrees at most . Since has GNRP (by assumption), then (and hence ) has GNRP. Thus, there exists no -subset in such that (by Lemma 2). Suppose that belongs to no polynomial in . Then, is an -subset in such that . This is, however, a contradiction. Thus, belongs to a polynomial in .

###### Lemma 4

If has GNRP, then any -subset in such that is weakly reducible.

###### Proof 4

Let be an arbitrary -subset in such that . Note that belongs to a polynomial in (and so ) of degree (by Lemma 3). Note, also, that has degree . Thus, if there exists such that , then is weakly reducible since is removable (by definition). Otherwise, if , then is an -subset in . Since , then has GRP (by Lemma 2), yielding a contradiction.

###### Lemma 5

If has GNRP, then the strongly reducible -subsets in such that belong to disjoint subsets of .

###### Proof 5

Let and be two arbitrary strongly reducible -subsets in such that . Let or be the set of polynomials in such that or belongs to , respectively. Note that . First, suppose that . Then, is an -subset in . Since , then has GRP (by Lemma 2), and hence a contradiction. Next, suppose that . We consider two cases. First, suppose that . Then, is a -subset in . Let and . Since , then has GRP (by Lemma 2), which is a contradiction. Next, suppose that . Then, is a -subset in . Let and . Note that . If , then has GRP (by Lemma 2), and hence a contradiction. If , then is weakly reducible (by Lemma 4), and has higher order than and since and . This is also a contradiction since and are strongly reducible (by assumption). Thus, .

###### Lemma 6

For and for all , and for and for all , if has GNRP, then the reduction process breaks any strongly reducible -subset in such that .

###### Proof 6

Let be an arbitrary strongly reducible -subset in such that . Since has GNRP, belongs to a polynomial in of degree (by Lemma 3) and no other strongly reducible -subset in belongs to (by Lemma 5). Moreover, for any and any , there exists no other -subset in such that (otherwise, has GRP). Thus must be broken to reduce .

For and , is either a - or - or -subset in . First, suppose that is a - or -subset in . Since there exists no other - or -subset in (otherwise, has GRP), then must be broken to reduce . Next, suppose that is a -subset in . Let be the set of two polynomials in , say and , such that belongs to both and . Let be an arbitrary (if any) strongly reducible -subset in . If does not exist, then must be broken to reduce both and . If exists, no element of is a common root of both and (otherwise, there exists a strongly reducible -subset in , which is a contradiction since is a strongly reducible -subset). Since has GNRP, then there exists no other strongly reducible -subset in (by Lemma 5), and breaking cannot reduce both and simultaneously. Thus, must be broken to reduce or (or both).

## V Proofs of Main Theorems

In this section, we prove our main theorems. For simplicity, we denote the degree-set of polynomials and by and , respectively.

###### Proof 7 (Proof of Theorem 1)

Assume that . If , then , which is a contradiction. If , then . Thus, , i.e., has RP (and hence GRP).

###### Proof 8 (Proof of Theorem 2)

The proof follows by contradiction. Assume that , and has GNRP. If , then (since the reduction process either reduces both and simultaneously, or it first reduces one, and then reduces the other one). Since (by assumption), then (by Lemma 1). Thus, has GRP (by Theorem 1), i.e., there exists a -subset in . Thus, is a strongly reducible -subset in such that (by Lemmas 4 and 2), and it must have been broken (by Lemma 6), yielding a contradiction.

If , then (since the reduction process must reduce , and reducing may or may not reduce ). If , then has GRP, yielding a contradiction as before. If , then . Thus, (by Lemma 1), which is again a contradiction.

If , then (since the reduction process must reduce , and reducing does not reduce ). Since , then (by Lemma 1), yielding a contradiction. If , then (since the reduction process does not reduce and ). Thus, () has GRP (by the same argument as before), which is a contradiction. If , then , yielding a contradiction. If , then . Thus, has GRP, again a contradiction.

###### Proof 9 (Proof of Theorem 3)

Due to the lack of space, we only give the proofs for the cases of . (The proofs of the rest of the cases follow the exact same lines.) The proof is by way of contradiction. Assume that , and has GNRP. Since (by Lemma 1), then has GRP (by Theorem 2).

First, consider . By the procedure of the reduction process, . Since has GRP, either there exists a -subset , or if does not exist, there exists a -subset , in . Since (or ) is a strongly reducible -subset in such that (by Lemmas 4 and 2), (or ) must have been broken (by Lemma 6), which is a contradiction.

Second, consider . Then, . For any of these cases, by the same arguments as for the previous case, we arrive at a contradiction.

Next, consider . Then, . For the cases of , similar to the previous cases, we reach a contradiction. For the case of