High dimensional affine codes whose square has a designed minimum distance

Abstract

Given a linear code ${\mathcal {C}}$, its square code ${\mathcal {C}}^{(2)}$ is the span of all component-wise products of two elements of ${\mathcal {C}}$. Motivated by applications in multi-party computation, our purpose with this work is to answer the following question: which families of affine variety codes have simultaneously high dimension $k({\mathcal {C}})$ and high minimum distance of ${\mathcal {C}}^{(2)}$, $d({\mathcal {C}}^{(2)})$? More precisely, given a designed minimum distance d we compute an affine variety code ${\mathcal {C}}$ such that $d({\mathcal {C}}^{(2)})\ge d$ and the dimension of ${\mathcal {C}}$ is high. The best constructions we propose mostly come from hyperbolic codes. Nevertheless, for small values of d, they come from weighted Reed–Muller codes.

Appendices

A For which affine codes ${\mathcal {C}}_A$ is it verified that ${\mathrm {FB}}({\mathcal {C}}_A) = d({\mathcal {C}}_A)$?

Let $A\subseteq [\![0,q-1]\!]^m$ and consider the code ${\mathcal {C}}_A$ as the affine variety code C(I, L) with $I=(0)$ and $L={\mathbb {F}}_q[A]$. Then, we know that the length of ${\mathcal {C}}_A$ is $q^m$ and its dimension coincides with the cardinality of the set A. Moreover its minimum distance, denoted as $d({\mathcal {C}}_A)$, satisfies that $d({\mathcal {C}}_A)\ge {\mathrm {FB}}({\mathcal {C}}_A)$. In this section we will study when these two values coincide. More concretely, we provide sufficient conditions to have the equality $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.

Lemma 6

Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-\alpha _1)\cdots (q-\alpha _m)$. Then $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$ if all the elements $\beta = (\beta _1, \ldots , \beta _m)$ with $0\le \beta _i \le \alpha _i$ belong to the set A.

Proof

First, to simplify the proof let us suppose that $m=2$. Let ${\mathcal {P}} = \left\{ P_1, \ldots , P_n\right\}$ be the ordered enumeration of the $q^2$ different points of ${\mathbb {F}}_q^2$. Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-\alpha _1)(q-\alpha _2)$. Now we can define the polynomial

$$\begin{aligned} f(x) = (X_1-P_1) \cdots (X_1-P_{\alpha _1}) \cdot (X_2-P_1) \cdots (X_2-P_{\alpha _2}). \end{aligned}$$

Take notice that by hypothesis $f(X_1, X_2) \in {\mathbb {F}}_q[A]$ since all the elements $\beta = (\beta _1, \beta _2)$ with $0\le \beta _1 \le \alpha _1$ and $0\le \beta _2 \le \alpha _2$ belongs to the set A. Moreover, the ${\mathbb {F}}_q$-roots of f are all the points of form:

$$\begin{aligned} \begin{array}{cccc} (P_i, z_2)&\hbox { and }&(z_1, P_j)&\hbox { with } i \in \{1, \ldots , \alpha _1\} \hbox { , } j \in \{1, \ldots , \alpha _2\} \hbox { and } z_1, z_2\in {\mathbb {F}}_q. \end{array} \end{aligned}$$

That is, the number of ${\mathbb {F}}_q$-roots of f(x) is $(\alpha _1 + \alpha _2) q - \alpha _1 \alpha _2$. Therefore, we have found a codeword ${\mathbf {c}} = {\mathrm {ev}}_{{\mathcal {P}}} (f)\in {\mathcal {C}}_A$ of weight $q^2 - (\alpha _1 + \alpha _2) q - \alpha _1 \alpha _2 = {\mathrm {FB}}({\mathcal {C}}_A)$. Hence the minimum distance of ${\mathcal {C}}_A$ is ${\mathrm {FB}}({\mathcal {C}}_A)$.

The generalization to m variables is straightforward. Let ${\mathcal {P}} = \left\{ P_1, \ldots , P_n\right\}$ be the ordered enumeration of the $q^m$ different points of ${\mathbb {F}}_q^m$. Then, using all the hypothesis we can define the following polynomial in ${\mathbb {F}}_q[A]$:

$$\begin{aligned} f(X_1, \ldots , X_m) = \prod _{i=1}^m (X_i - P_1) \cdots (X_i - P_{\alpha _i}) \in {\mathbb {F}}_q[A]. \end{aligned}$$

Thus, we have found a codeword of ${\mathcal {C}}_A$ of weight ${\mathrm {FB}}({\mathcal {C}}_A)$, hence $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.$\square$

The following result shows that if l is a divisor of $q-1$ then, there exists a polynomial $f(x) = X^l-\alpha \in {\mathbb {F}}_q[X]$ with small support but a large number of ${\mathbb {F}}_q$-roots. This result will be useful for computing the minimum distance of codes of type ${\mathcal {C}}_A$ by just checking that a very small number of points belongs to the set A.

Lemma 7

Let $\alpha$ be a primitive element of ${\mathbb {F}}_q^*$. Consider the polynomial $f(X) = X^l-\alpha ^j \in {\mathbb {F}}_q[X]$. Then $X^l-\alpha ^j$ has at least one root in ${\mathbb {F}}_q$ if and only if $\gcd (l, q-1)$ divides j. In such case, the exactly number of ${\mathbb {F}}_q$-roots of f(X) is $\gcd (l,q-1)$.

Proof

Suppose that $\alpha ^i$ is an ${\mathbb {F}}_q$-root of f(X), then $f(\alpha ^i) = 0$, that is $\alpha ^{il} = \alpha ^j$ which implies that the order of $\alpha$, which is $q-1$, divides $il-1$. In other words, there exists an integer x such that $x(q-1) + il = j$. Take notice that such x exists if and only if $\gcd (l,q-1)$ divides j.

In such case, if (x, y) is a solution of the equation $x(q-1) + yl = j$. Then, all solutions of this equations has the form:

$$\begin{aligned} \left( x-\lambda \frac{l}{\gcd (l,q-1)}, y + \lambda \frac{q-1}{\gcd (l,q-1)} \right) \hbox { with } \lambda \in {\mathbb {Z}} \end{aligned}$$

Therefore, if f(X) has at least one root in ${\mathbb {F}}_q$, then it will have exactly $\gcd (l,q-1)$${\mathbb {F}}_q$-roots.$\square$

Corollary 1

Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-l)q^{m-1}$ with l a divisor of $q-1$. Then, $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$ if $\{1, X_i^l\}\subseteq {\mathbb {F}}_q[A]$ for some $i\in \{1, \ldots , m\}$.

Proof

By hypothesis we can define the following polynomial in ${\mathbb {F}}_q[A]$:

$$\begin{aligned} f(X) = X_i^l - \beta \hbox { for certain }\beta \in {\mathbb {F}}_q. \end{aligned}$$

Then, by Lemma 7, f(X) has $lq^{m-1}$${\mathbb {F}}_q$-roots. That is, we have found a codeword of ${\mathcal {C}}_A$ of weight ${\mathrm {FB}}({\mathcal {C}}_A)$, hence $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.$\square$

Lemma 8

Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-kl)q^{m-1}$ with l a divisor of $q-1$. Then, $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$ if $\{1, X_i^l, X_i^{2l}, \ldots , X_i^{kl}\}\subseteq {\mathbb {F}}_q[A]$ for some $i\in \{1, \ldots , m\}$.

Proof

By hypothesis we can define the following polynomial in ${\mathbb {F}}_q[A]$:

$$\begin{aligned} f(X) = (X_i^l - \beta ) (X_i^{2l}-\beta ^2) \cdots (X_i^{kl} - \beta ^k) \hbox { for certain }\beta \in {\mathbb {F}}_q. \end{aligned}$$

Then, by Lemma 7, f(X) has $klq^{m-1}$${\mathbb {F}}_q$-roots. That is, we have found a codeword of ${\mathcal {C}}_A$ of weight ${\mathrm {FB}}({\mathcal {C}}_A)$, hence $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.$\square$

Lemma 9

Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-l_1)\cdots (q-l_m)$ with $l_i$ a divisor of $q-1$. Then, $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$ if $\{1, X_1^{l_1}, \cdots , X_m^{l_m}\}\subseteq {\mathbb {F}}_q[A]$.

Proof

By hypothesis we can define the following polynomial in ${\mathbb {F}}_q[A]$:

$$\begin{aligned} f(X) = \prod _{i=1}^m( X_i^{l_i} - \beta _i) \hbox { for certain }\beta _1, \ldots , \beta _m \in {\mathbb {F}}_q. \end{aligned}$$

Then, by Lemma 7, f(X) has

$$\begin{aligned} (l_1 + \ldots l_m) q^{m-1} - \sum _{1\le i< j\le m} l_i l_j q^{m-2} - \sum _{1\le i< j<k\le m} l_i l_jl_k q^{m-3} - \ldots - l_1 \cdots l_m \end{aligned}$$

${\mathbb {F}}_q$-roots. That is, we have found a codeword of ${\mathcal {C}}_A$ of weight ${\mathrm {FB}}({\mathcal {C}}_A)$, hence $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.$\square$

Lemma 10

Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-k_1l_1)\cdots (q-k_ml_m)$ with $l_i$ a divisor of $q-1$. Then, $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$ if $\{1, X_1^{l_1}, \ldots , X_1^{ml_1}, \cdots , X_m^{l_m}, \ldots , X_m^{k_ml_m}\}\subseteq {\mathbb {F}}_q[A]$.

Proof

By hypothesis we can define the following polynomial in ${\mathbb {F}}_q[A]$:

$$\begin{aligned} f(X) = \prod _{i=1}^m( X_i^{l_i} - \beta _i) (X_i^{2l_i}-\beta _i^2) \cdots (X_i^{k_il_i} - \beta _i^{k_i}) \hbox { for certain }\beta _1, \ldots , \beta _m \in {\mathbb {F}}_q. \end{aligned}$$

Then, by Lemma 7, we have found a codeword of ${\mathcal {C}}_A$ of weight ${\mathrm {FB}}({\mathcal {C}}_A)$, hence $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.$\square$

Lemma 11

Suppose that ${\mathrm {FB}}({\mathcal {C}}_A) = (q-l)q^{m-1}$ with $l-1$ a divisor of $q-1$. Then, $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$ if $\{X_i, X_i^l\}\subseteq {\mathbb {F}}_q[A]$ for some $i\in \{1, \ldots , m\}$.

Proof

By hypothesis we can define the following polynomial in ${\mathbb {F}}_q[A]$:

$$\begin{aligned} f(X) = (X_i^l - \beta X_i) = X_i (X_i^{l-1}-\beta ) \hbox { for certain }\beta \in {\mathbb {F}}_q. \end{aligned}$$

Then, by Lemma 7, f(X) has $lq^{m-1}$${\mathbb {F}}_q$-roots. That is, we have found a codeword of ${\mathcal {C}}_A$ of weight ${\mathrm {FB}}({\mathcal {C}}_A)$, hence $d({\mathcal {C}}_A) = {\mathrm {FB}}({\mathcal {C}}_A)$.$\square$

Lemma 12

Let $A\subseteq [\![0,q-1]\!]^m$ and $s\in [\![0,q-1]\!]$. If for all $f\in {\mathbb {F}}_q[A]$ we have that $X_1^s$ is a divisor of f(X), then $d({\mathcal {C}}_A) = d({\mathcal {C}}_B)$ with

$$\begin{aligned} B=\{ (i_1-s-1, i_2,\ldots , i_m)\mid (i_1, \ldots , i_m) \in A\}. \end{aligned}$$

The result can be generalized to any other coordinate $X_i$ with $i=2, \ldots , m$.

Proof

By hypothesis every polynomial $f\in {\mathbb {F}}_q[A]$ can be written as $f=X_1^s g$ with $g\in {\mathbb {F}}_q[B]$. And both polynomials f and g have exactly the same number of ${\mathbb {F}}_q$-roots.$\square$

B Proof of Theorem 3 when d is even

We consider now the case of Theorem 3 when the minimum distance d is even.

Theorem 4

Let ${\mathbb {F}}_q$ be a finite field and $d \in {\mathbb {Z}}^+$ be an even integer with $d < q$. If ${\mathcal {C}}$ is a weighted Reed–Muller code over ${\mathbb {F}}_q$ with $d({\mathcal {C}}^{\, (2)}) \ge d$, then $k({\mathcal {C}}) \le k({\mathcal {C}}_B),$ where ${\mathcal {C}}_B$ is any of the weighted Reed–Muller codes described in Lemma 5.

Proof

This proof will follow the same ideas in Theorem 3. Let ${\mathcal {C}}$ be a weighted Reed–Muller code over ${\mathbb {F}}_q$ with $d({\mathcal {C}}^{\, (2)}) \ge d$. We assume without loss of generality that ${\mathcal {C}}= {\mathrm {WRM}}_q(\lambda ,2,\{w_1,1\})$ for some $\lambda , w_1 > 0$. Taking

$$\begin{aligned} A := \{(i,j) \in [\![0, q-1 ]\!]\, \vert \, w_1 i + j \le \lambda \} \end{aligned}$$

we have that ${\mathcal {C}}= {\mathcal {C}}_A$.

In this proof we denote $a:= (q-1)/2$ and $b := (q-d+1)/2$; and observe that either $(a,b) \in {\mathbb {N}}^2$ or both $(a-\frac{1}{2}, b + \frac{1}{2}), (a+\frac{1}{2}, b - \frac{1}{2}) \in {\mathbb {N}}^2$. We divide the proof in two cases depending on the value of $\lambda$.

Case I:$\lambda \le a + b$. We take $B = B_1$ as in Lemma 5. To prove that $|A| = k({\mathcal {C}}) \le k({\mathcal {C}}_B) = |B|$ we are going to prove that either $A \subseteq B$, or

$$\begin{aligned} \begin{array}{lccl} \varphi : &{} A - B &{}\ \longrightarrow \ &{} B - A \\ &{} (\alpha ,\beta ) &{} \mapsto &{} (2a-\alpha , 2b - \beta ) \end{array} \end{aligned}$$

is an injective map (see Fig. 9 for a graphic representation of this idea).

Fig. 9

Figure illustrating the proof of Theorem 4 for $d=6$, $q=11$, $(a,b)=(5,3)$, $A=\left\{ (i,j)\in {\mathbb {N}}^2 \mid 0.4 i + j \le 4 \right\}$ and $B=\left\{ (i,j) \in {\mathbb {N}}^2 \mid i+j< 8\right\} \cup \left\{ (i,j) \in {\mathbb {N}}^2 \mid i+j = 8 \hbox { and } j<3\right\}$

Since the injectivity of $\varphi$ is easy to check, we are showing that $\varphi$ is well defined in three steps:

• (a) if $(\alpha ,\beta ) \in A$, then $(2a-\alpha ,2b-\beta ) \notin A$,

• (b) if $(\alpha ,\beta ) \in A - B$, then $(2a-\alpha ,2b-\beta ) \in {\mathbb {N}}^2$, and

• (c) if $(\alpha ,\beta ) \in A - B$, then $(2a-\alpha ,2b-\beta ) \in B$.

If (a) does not hold, then both $(\alpha ,\beta )$ and $(2a-\alpha ,2b-\beta ) \in A$. Hence, $(2a,2b) = (\alpha ,\beta ) + (2a-\alpha ,2b-\beta ) \in A + A$ and ${\mathcal {C}}_A^{\, (2)} = {\mathcal {C}}_{A+A}$. Since ${\mathcal {C}}_A$ is a weighted Reed–Muller code, by Lemma 4 we have that $d \le d({\mathcal {C}}^{\, (2)}) = {\mathrm {FB}}({\mathcal {C}}^{\, (2)}) \le (q-2a)(q-2b) = d-1$, a contradiction.

We observe that $(2a-\alpha ,2b-\beta ) \in {\mathbb {Z}}^2$ and that $\alpha \le q-1 = 2a$, so to prove (b) we just need to see that $2b - \beta \ge 0$. Assume that $2b < \beta$ and let us prove that

• (b.1) $P = (a,b) \in A$ if q is odd, or

• (b.2) $Q_1 = (a - \frac{1}{2}, b + \frac{1}{2}),\ Q_2 = (a + \frac{1}{2}, b- \frac{1}{2}) \in A$ if q is even.

If $\alpha > a$, then $\alpha \ge a+\frac{1}{2}$ since $\beta \ge 2b + 1 > b + \frac{1}{2}$ we have that $P \in A$ in case (b.1) and $Q_1, Q_2 \in A$ in case (b.2). If $\alpha \le a$, from one side we have that $(\alpha , \beta ) \notin B$, so

$$\begin{aligned} \alpha + \beta \ge a+b \end{aligned}$$

(4)

and, if we have equality, then $\beta \ge b$. From the other side we have that $(\alpha ,\beta ) \in A$, which implies that

$$\begin{aligned} w_1 \alpha + \beta \le \lambda . \end{aligned}$$

(5)

From (4) and (5) we get that

$$\begin{aligned} (w_1 - 1) \alpha + a + b \le (w_1 - 1) \alpha + \alpha + \beta = w_1 \alpha + \beta \le \lambda \le a + b \end{aligned}$$

and, thus, $w_1 \le 1$. Hence, we separate three cases:

Subcase I.I. If $\alpha + \beta > a + b$.

$$\begin{aligned} w_1(a+\tfrac{1}{2}) + b - \tfrac{1}{2} \le w_1 a + b< w_1 a + b + \tfrac{1}{2} < \alpha + \beta + (w_1-1)\alpha = w_1 \alpha + \beta \le \lambda . \end{aligned}$$

So, $P \in A$ if q is odd, or both $Q_1,Q_2 \in A$ if q is even.

Subcase I.II. If $\alpha + \beta = a + b$ and q is odd. Since $\beta \ge b$ and $w_1 < 1$, we have that $w_1(\alpha - a) + \beta - b \ge w_1 (\alpha - a + \beta - b) = 0$. As a consequence,

$$\begin{aligned} w_1 a + b \le w_1 a + b + w_1(\alpha - a) + \beta - b = w_1 \alpha + \beta \le \lambda . \end{aligned}$$

Therefore $P \in A$.

Subcase I.III. If $\alpha + \beta = a + b$ and q is even. Since $\beta \ge b$ and $b \notin {\mathbb {N}}$, then $\beta \ge b + \frac{1}{2}$; moreover, $w_1 < 1$, then we have that $w_1(\alpha - a + \frac{1}{2}) + \beta - b - \frac{1}{2} \ge w_1 (\alpha - a + \frac{1}{2} + \beta - b - \frac{1}{2}) = 0$. As a consequence,

$$\begin{aligned} w_1(a+\tfrac{1}{2}) + b - \tfrac{1}{2}\le & {} w_1(a-\tfrac{1}{2}) + b + \tfrac{1}{2} \\\le & {} w_1(a-\tfrac{1}{2}) + b + \tfrac{1}{2} + w_1(\alpha - a + \tfrac{1}{2}) + \beta - b - \tfrac{1}{2} \\= & {} w_1 \alpha + \beta \le \lambda \end{aligned}$$

and we conclude that $Q_1,Q_2 \in A$.

Moreover, since $P + P = Q_1 + Q_2 = (2a,2b)$, in both cases we obtain that $(2a,2b) \in A+A$ and ${\mathcal {C}}_A^{\, (2)} = {\mathcal {C}}_{A+A}$. Since ${\mathcal {C}}_A$ is a weighted Reed–Muller code, by Lemma 4 we have that $d \le d({\mathcal {C}}^{\, (2)}) \le (q-2a)(q-2b) = d-1$, a contradiction.

Let us prove now (c). Take $(\alpha ,\beta ) \in A-B$, then either

• (c.1) $\alpha + \beta > a+b$, or

• (c.2) $\alpha + \beta = a+b$ and $\beta \ge b$.

In (c.1) we have that $2a-\alpha + 2b-\beta < a+b$, so $(2a-\alpha ,2b-\beta ) \in B$. In (c.2) we observe that $\beta \ne b$ because $(a,b) \notin A$. Then, we have that $2a-\alpha + 2b-\beta = a+b$ and $2b - \beta < b$, so $(2a-\alpha ,2b-\beta ) \in B$.

Case II:$\lambda \ge a + b$. We claim that $\frac{\lambda }{w_1} < a + b$. Otherwise, we have that $P \in A$ if q is odd, or $Q_1,Q_2 \in A$ if q is even. In both cases $(2a,2b) \in A + A$ and ${\mathcal {C}}_A^{\, (2)} = {\mathcal {C}}_{A+A}$. Since ${\mathcal {C}}_A$ is a weighted Reed–Muller code, by Lemma 4 we have that $d \le d({\mathcal {C}}^{\, (2)}) \le (q-2a)(q-2b) = d-1$, a contradiction. Since $\frac{\lambda }{w_1} < a + b$, then $A = \{(i,j) \in {\mathbb {N}}^2 \, \vert \, 0 \le i,j \le q-1 {\text { and }} i + \frac{1}{w_1}{j} \le \frac{\lambda }{w_1}\}$ and a symmetric argument to Case I using $B = B_2$ with $B_2$ as in Lemma 5 applies here.$\square$