text:
\section{Introduction}
\label{section1}

Quantum error correcting codes (QECCs) can be constructed by means of classical linear codes with the use of the CSS method~\cite{Calderbank,Gottesman,Ketkar}. QECCs over $\mathbb{F}_q$, the finite field with $q$ elements, are constructed from self-orthogonal classical linear codes over $\mathbb{F}_q$ if we consider the Euclidean metric, and over $\mathbb{F}_{q^2}$ if we consider the Hermitian metric. In general the Hermitian metric produces QECCs with better parameters since we may consider linear codes over a bigger field, the downside being, however, that the computations are more involved.

Brun et al. proposed in \cite{Brun} a pre-sharing entanglement protocol between encoder and decoder which increases the communication capacity and simplifies the theory of  quantum error-correction since it is not restricted to self-orthogonal classical linear codes. These quantum codes are known as  {\em entanglement-assisted quantum error-correcting codes} (EAQECCs). They were introduced over the binary field and then extended for an arbitrary finite field in \cite{Galindo:2019}. As for QECCs, we may consider classical linear codes over $\mathbb{F}_q$ if we consider the Euclidean metric, and over $\mathbb{F}_{q^2}$ if we consider the Hermitian metric. EAQECCs have four fundamental parameters $[[n,k,d;c]]$: length, dimension, minimum distance and the number of required pairs of maximally entangled quantum states, respectively.

Among the classical codes used to construct quantum codes by means of the CSS construction, codes arising from algebraic geometry resources and tools (the so-called algebraic geometry codes or AG codes for short) have revealed themselves as good candidates to obtain quantum codes with good parameters \cite{Ashikhmin,Bartoli,Kim,Matsumoto}. In particular, we should highlight AG codes from Castle curves (the so-called Castle codes) \cite{Fer,MTT}, whose prototype is the Hermitian curve, which has traditionally played an important role in the theory of AG codes. Therefore, it is natural to consider the Hermitian curve to obtain EAQECCs with good parameters.

EAQECCs from AG codes have been addressed in \cite{Francisco1,Francisco2}, where the Euclidean inner product is mainly considered. In this work, we consider Hermitian codes \cite{HLP} -- that is, one-point AG codes from the Hermitian curve over $\mathbb{F}_{q^2}$ -- to obtain EAQECCs considering the Hermitian inner product. The length, dimension, and minimum distance of the EAQECCs from the Hermitian curve are known. Only one parameter remains: the number of required maximally entangled quantum states, $c$. Unlike the other parameters, however, $c$ is unknown if we consider the Hermitian inner product. Thus, the main task of this article is to compute $c$.

We may compute $c$ as the rank of a product of generator matrices, specifically the generator matrix of the classical linear code and the generator matrix of its Hermitian dual  (see \cite[Proposition 3]{Galindo:2019}). However, the Hermitian dual of an AG code (and of a Hermitian code) is unknown in general and, hence, there is no formula for computing $c$ in the Hermitian metric case. In this work, we present an algorithmic approach for computing $c$. The algorithm is based on a careful analysis of reduced polynomials modulo the Hermitian curve equation and on the valuation given by the pole order at the infinity point. This allows us to compute the parameters of EAQECCs from the Hermitian curve and illustrate that they have excellent parameters. The computational complexity of our algorithm is $\mathcal{O}(q^5)$, which is significantly better than computing a rank of a matrix using methods from linear algebra since the length of a Hermitian code is $q^3$.

Finally, by considering $q$-adic expansions and Lucas' theorem, we are able to reduce the number of iterations in our algorithm by a careful analysis of the $q$-th powers of Hermitian codes. The proofs are different for $q$ prime and non-prime, but they are developed in a parallel way. Naturally, the proof for the non-prime case is more involved and technical.

The structure of this paper is as follows. In Section \ref{section2}, we present some standard results on classical and quantum codes.
In Section \ref{Sect3}, we review Hermitian codes and introduce the idea of reduced polynomials that is used in the following section.
Section \ref{Sect4} shows the first main result: we describe an algorithmic approach to the problem of determining the parameters of EAQECCs derived from a one-point Hermitian code. Moreover, we provide codes with excellent parameters at the end of the section. Finally, in Section \ref{Sect5}, we analyze the $q$-th powers of Hermitian codes, which allow us to improve the number of iterations of the algorithm in Section \ref{Sect4}. We also compute the computational complexity of our algorithm and show, computationally, that the parameters of most of the EAQECCs in this family exceed the Gilbert-Varshamov bound \cite{Galindo:2019}.

The ideas and methods developed in this article can serve as a paradigm to  study EAQECCs coming from of other AG codes obtained from different curves, and especially from Castle-type curves.

\section{Preliminaries}
\label{section2}

In this section we recall some preliminary and well known results about linear vector spaces over a finite field, the Frobenius map and entan\-glement-assisted quantum codes.
A more complete study of these topics can be found in \cite{Galindo:2019,LaGuardia,Ketkar,LidlNiederreiter:Book}.

\subsection{The Frobenius map and the Hermitian dual of a subspace}
\label{section2.1}

Let $q=p^r$ be a prime power, and let $\mathbb{F}_{q^2}$ be the finite field with $q^2$ elements.
Additionally, let $\sigma:\mathbb{F}_{q^2}\rightarrow\mathbb{F}_{q^2}$ be the Frobenius map, $\sigma(x)=x^q$.
It is well known that $\sigma$ is an $\mathbb{F}_q$-automorphism of $\mathbb{F}_{q^2}$ and an involution,
that is $\sigma^2=id$. Furthermore, considering the same notation, $\sigma$ may be extended to the map
$\sigma:\mathbb{F}_{q^2}^n\rightarrow\mathbb{F}_{q^2}^n$, by $\sigma(\mathbf{v})=(v_1^q,\dots,v_n^q)$.
This map is $\sigma$-semilinear and bijective, thus it preserves  intersections and linear dependence: given
$V,W\subset \mathbb{F}_{q^2}^n$, we have  $\sigma(V\cap W)=\sigma(V)\cap\sigma(W)$ and
$\dim (\langle V\rangle)=\dim (\langle \sigma(V)\rangle)$, where $\langle V\rangle$
denotes the $\mathbb{F}_{q^2}$-linear subspace spanned by $V$. Following the usual notation in linear coding theory,
we often write $\mathbf{v}^q=\sigma(\mathbf{v})$ and $V^q=\sigma(V)=\{\mathbf{v}^q \ : \ \mathbf{v}\in V\}$.

The Euclidean and Hermitian inner products in $\mathbb{F}_{q^2}^n$ are defined as follows:
given $\mathbf{v}, \mathbf{w}\in \mathbb{F}_{q^2}^n$,
\[
  \mathbf{v} \cdot \mathbf{w}= v_1w_1+\cdots+ v_nw_n \; \mbox{ and } \;
  \mathbf{v} \cdot_H \mathbf{w}=\sigma(\mathbf{v}) \cdot \mathbf{w}.
\]
Given a linear subspace $V\subset \mathbb{F}_{q^2}^n$, its Euclidean and Hermitian duals are, respectively,
\begin{eqnarray*}
V^{\perp}&=&\{\mathbf{x}\in\mathbb{F}_{q^2}^n \ : \ \mathbf{v} \cdot \mathbf{x}=0 \mbox{ for all $\mathbf{v}\in V$} \}, \\
V^{\perp_H}&=&\{\mathbf{x}\in\mathbb{F}_{q^2}^n \ : \ \mathbf{v} \cdot_H \mathbf{x}=0 \mbox{ for all $\mathbf{v}\in V$} \}.
\end{eqnarray*}
We define $\Delta(V)=\dim(V\cap V^{\perp_H})$.

\begin{proposition} \label{propDelta(V)}
Let $V\subset \mathbb{F}_{q^2}^n$ be a linear subspace. We have \newline
(a) $V^{\perp_H}=(V^q)^{\perp}=(V^{\perp})^q$;\newline
(b) $\Delta(V)=\dim (V^q\cap V^{\perp})$; \newline
(c) $\Delta(V)=\Delta(V^{\perp})$.
\end{proposition}
\begin{proof}
(a) The first equality is clear. The second one follows from the fact that
$\mathbf{v}^q \cdot \mathbf{x}=0$ if and only if $\mathbf{v} \cdot \mathbf{x}^q=0$. Write $\mathbf{y}=\mathbf{x}^q$.
Then $V^{\perp_H}=\{\mathbf{y}^q\in\mathbb{F}_{q^2}^n \ : \ \mathbf{v} \cdot \mathbf{y}=0 \mbox{ for all $\mathbf{v}\in V$} \}=(V^{\perp})^q$,
and the second equality of (a) holds. (b) Applying $\sigma$ we have
$\Delta(V)=\dim(V\cap (V^{\perp})^q)=\dim(V^q \cap V^{\perp})$ which proves the statement in (b).
(c) Since $(V^{\perp})^{\perp}=V$ we get
$\Delta(V^{\perp})=\dim((V^{\perp})^q\cap (V^{\perp})^{\perp})=\Delta(V)$.
\end{proof}

In what follows, we use the expression given by Proposition \ref{propDelta(V)}(b) to compute $\Delta(V)$.

\subsection{Entanglement-Assisted Quantum Error-Correcting Codes}
\label{Sect2.2}

A quantum code $\mathcal{Q}$ is called an $[[n,k,d;c]]_q$ {\em entanglement-assisted quantum error-correcting code} (EAQECC) if it encodes $k$ logical qudits into $n$ physical qudits using $c$ copies of maximally entangled quantum  states
and it can correct $\lfloor(d-1)/2\rfloor$ quantum errors. The {\em rate} of an EAQECC is  $k/n$, its {\em relative distance} is $d/n$, and its {\em entanglement-assisted rate} is  $c/n$. When  $c = n-k$ then the EAQECC is said to have {\em maximal entanglement}.

\begin{proposition}\cite[Proposition 3 and Corollary 1]{Galindo:2019} \label{Prep:WildeHerm}
Let $\mathcal{C}$ be a linear code over $\mathbb{F}_{q^2}$ with parameters $[n,k,d]_{q^2}$.
Then there is an EAQECC $\mathcal{Q}$ with parameters
$[[n,2k-n+c, d'; c]]_q$, where $d'$
is the minimum Hamming weight of a vector in the set $\mathcal{C}\setminus(\mathcal{C}\cap \mathcal{C}^{\perp_H})$, and
\begin{equation}\label{formulac}
  c = \dim \mathcal{C}^{\perp_H} - \dim (\mathcal{C}^{\perp_H}\cap \mathcal{C})
\end{equation}
is the number of required maximally entangled quantum states.
\end{proposition}

From Propositions~\ref{propDelta(V)} and \ref{Prep:WildeHerm}, the amount of entanglement of a code $\mathcal{Q}$ given by Eq. (\ref{formulac}) can be computed as
$c = \dim (\mathcal{C^\perp}) - \Delta(\mathcal{C})=n-\dim (\mathcal{C}) - \Delta(\mathcal{C})$. In this paper we are going to use this
formula to determine $c$. Furthermore, we clearly have that $d'\ge d(\mathcal{C})$.

In order to show the performance of the EAQECCs given in this article we will consider the Gilbert-Varshamov bound \cite[Theorem 5]{Galindo:2019} for EAQECCs considering the Hermitian inner product.

\begin{theorem}\label{thmgv}
Assume the existence of positive integers $n$, $k\leq n$, $d$, $c\leq (n-k)/2$
such that
  \begin{equation}
    \frac{q^{n+k}-q^{n-k-2c}}{q^{2n}-1}\sum_{i=1}^{d - 1}
         {n \choose i}(q^2-1)^i < 1. \label{eqgv}
  \end{equation}
Then there exists an EAQECC $\mathcal{Q}$ with parameters
$[[n,k-c, d; c]]_q$.
\end{theorem}

Furthermore, we will consider the entanglement-assisted analog of the Singleton bound given in \cite[Theorem A.5]{Singleton}: if there exists an EAQECC with parameters $[[n,k,d;c]]_q$ and $d \le (n+2)/2$, then $2d \le n +2 - k  +c$. Note that the previous Singleton type bound does not hold for all range of parameters of an EAQECC. In the light of this bound, the following defect has been used to evaluate the performance of EAQECCs \cite[Notation 3.7]{Singleton}.

\begin{definition}
Let $\mathcal{Q}$ be an $[[n, k, d; c]]_q$ EAQECC. Then the  {\em entanglement-assisted quantum Singleton defect of $\mathcal{Q}$} is the integer \begin{equation}n + 2 - k + c - 2d.\end{equation}
\end{definition}

Note that the entanglement-assisted quantum Singleton defect is non-negative if $d \le (n+2)/2$, and otherwise it may be negative.

\section{Reduced polynomials for Hermitian codes}
\label{Sect3}

In this section we recall some basic facts about the Hermitian curve, the classical codes derived from it, Hermitian codes, and some tools we use in subsequent computations. For a complete treatment see \cite{HLP}.

\subsection{Hermitian curves and codes}
\label{Sect3.1}

Let $\mathcal{X}$ be the Hermitian curve defined over $\mathbb{F}_{q^2}$ by the affine equation
$x^{q+1} = y^q + y$. This is a nonsingular plane curve of genus $g=q(q-1)/2$,  with $n = q^3$
rational affine points, $P_1, \ldots, P_n$, plus one point at infinity, $Q$.
Let $\nu_Q$ be the valuation of $\mathbb{F}_{q^2}(\mathcal{X})$ given by the order at $Q$ and $\nu=-\nu_Q$.
From the equation of $\mathcal{X}$ we have $\nu(x)=q$ and $\nu(y)=q+1$.
The following property of $\nu$  will be widely used in what follows.

\begin{proposition}\cite[Theorem 2.16(iii)]{HLP}\label{Lemma_0}
Given functions $f_1,f_2\in\mathbb{F}_{q^2}(\mathcal{X})^*$, we have $\nu(f_1+f_2)\leq \max\{\nu(f_1),\nu(f_2)\}$
with equality if $\nu(f_1)\neq\nu(f_2)$.
\end{proposition}

For $m\geq 0$, we consider the Riemann-Roch space
\[
  \mathcal{L}(m) = \mathcal{L}(mQ) = \{f\in \mathbb{F}_{q^2}(\mathcal{X}) : f = 0 \text{ or  div}(f)\ge -mQ  \}.
\]
This is a linear space, whose dimension is denoted by $\ell(m)$. Recall that $\ell(m) = m+1-g$ when
$m\geq 2g-1$, according to the Riemann-Roch theorem~\cite{HLP}.
Let $\mathcal{L}(\infty) = \cup_{m=0}^\infty \mathcal{L}(m)$ and denote by $S$ the Weierstrass semigroup of $Q$, then $S=\{ \nu(f) : f\in \mathcal{L}(\infty), f\neq 0 \}= \langle q, q+1\rangle$.
Since $\mathcal{L}(\infty)$ is a finitely generated $\mathbb{F}_{q^2}$-algebra, we have that
$\mathcal{L}(\infty) = \mathbb{F}_{q^2}[x,y]$. Abusing the notation, the elements of $\mathcal{L}(\infty)$ will be called {\em polynomials}.

\begin{lemma} \label{Lemma_2}
If $f_1, \dots, f_t\in \mathcal{L}(\infty)$ are nonzero polynomials of pairwise different orders,
then they are linearly independent.
\end{lemma}
\begin{proof}
Suppose $\lambda_1 f_1 + \cdots + \lambda_t f_t = 0$, then Proposition \ref{Lemma_0} ensures that
$\nu(0) = \nu(\lambda_1 f_1 + \cdots + \lambda_t f_t) = \max\{\nu(f_i) : \lambda_i\neq 0\}$. Hence
$\lambda_1 = \cdots = \lambda_t = 0$.
\end{proof}

Consider the evaluation map $ev:\mathcal{L}(\infty) \rightarrow \mathbb{F}_{q^2}^n$,
$ev(f)=(f(P_1),\dots,f(P_n))$. The algebraic geometry code $\mathcal{C}(m)$ is defined as
$\mathcal{C}(m) = ev(\mathcal{L}(m))$.  Since $\mathcal{C}(n+2g-1) = \mathbb{F}_{q^2}^n$ -- see eg.~\cite{Stichtenoth} --
we can restrict to $0\leq m \leq n+2g-1$. In this range, we can consider the quantum code $\mathcal{Q}(m)$
over $\mathbb{C}^q$ obtained from $\mathcal{C}(m)$ by the construction described in Proposition \ref{Prep:WildeHerm}. The code $\mathcal{Q}(m)$ has parameters $[[n,2k-n+c,\ge d;c]]_q$, where $n=q^3,k=k(m)$ and $d=d(m)$ are, respectively, the length,
dimension and minimum distance of $\mathcal{C}(m)$, and $c= n-k(m)-\Delta (\mathcal{C}(m))$ is the minimum number of maximally entangled quantum states
consumed by the code. Recall that the dimension of $\mathcal{C}(m)$ is $k(m)=\ell(m)-\ell(m-n)$, and its exact minimum distance $d(m)$ has been determined in \cite{YK}.
Therefore, our main task is to compute $\Delta(m)=\Delta(\mathcal{C}(m))=\dim (\mathcal{C}(m)^q\cap \mathcal{C}(m)^{\perp})$, which is the only unknown value among the previous parameters. Please note that the dual code $\mathcal{C}(m)^{\perp}$ satisfies the well known relation $\mathcal{C}(m)^{\perp}=\mathcal{C}(m^{\perp})$, where $m^{\perp}=n+2g-2-m$, see \cite{HLP}.

\subsection{Reduced polynomials}
\label{Sect3.2}
A monomial $x^a y^b$ is called \emph{reduced} if $0\leq a<q^2$, $0\leq b<q$. Thus, the number of
reduced monomials is at most $n=q^3$. A polynomial $f\in\mathcal{L}(\infty)$ is \emph{reduced}  if $f=0$ or it is the sum of reduced monomials.  We denote by $\mathcal{R}$ the set of all reduced polynomials and
$\mathcal{R}^* = \mathcal{R}\setminus\{0\}$.  Consider the reduction
$\mathfrak{r}': \mathcal{L}(\infty) \rightarrow \mathcal{R}$ as follows. Given a polynomial $f$,
$\mathfrak{r}'(f)$ is obtained by performing the substitutions
\[
  \mbox{\bf (R1) } \mathfrak{r}'(y^q)= x^{q+1} - y \hspace*{5mm}  \mbox{ and }   \hspace*{5mm}  \mbox{\bf (R2) } \mathfrak{r}'(x^{q^2})= x
\]
as many times as possible. In other words, $\mathfrak{r}'(f)$ is the unique (up to multiplication by a constant $\lambda\in\mathbb{F}_{q^2}^*$) reduced polynomial in the coset of $f$ modulo the ideal $( x^{q+1}-y^q-y, x^{q^2} - x )\subset  \mathcal{L}(\infty)$. This comes from considering the ideal generated from the Hermitian equation, the field equations $x^{q^2}-x$ and $y^{q^2}-y$, and the lexicographical ordering with $y>x$.

The next lemma gives some properties of reduced polynomials that we shall use later.

\begin{lemma} \label{Lemma_1}
The following statements hold.\newline
(a) Two distinct reduced monomials  are linearly independent;\newline
(b) For any $f\in \mathcal{L}(\infty)$, we have $ev(f)=ev(\mathfrak{r}'(f))$;\newline
(c) The evaluation map $ev: \mathcal{R}\rightarrow\mathbb{F}_{q^2}^n$ is an isomorphism of vector spaces.
Thus, any code $\mathcal{C}\subseteq \mathbb{F}_{q^2}^n$ can be obtained by evaluating reduced polynomials.
\end{lemma}
\begin{proof}
(a) follows from the fact that $\nu(x^a y^b)=aq+b(q+1)$ and Lemma \ref{Lemma_2}. (b) is a consequence of $y^q= x^{q+1} - y$ in $ \mathcal{L}(\infty)$ and
$ev(x^{q^2})=ev(x)$. For (c), note that the order of a reduced polynomial is at most $q^3+q^2-q-1=n+2g-1$ and that $ev:\mathcal{L}(q^3+q^2-q-1)\rightarrow \mathbb{F}_{q^2}^n$ is surjective. From this, it follows that $ev: \mathcal{R}\rightarrow\mathbb{F}_{q^2}^n$ must be surjective as well.  Since the dimension of $\mathcal{R}$ is at most $n$,
we get the isomorphism.
\end{proof}

In view of property (a) in Lemma \ref{Lemma_1}, any nonzero reduced polynomial has a unique leading monomial with respect to $\nu$. We define the normalization of $f\in\mathcal{R}$ as
$\mathfrak{n}(f)=0$ if $f=0$, and $\mathfrak{n}(f)=\lambda f$, where $\lambda\in\mathbb{F}_{q^2}$ is chosen
such that  $\mathfrak{n}(f)$ has leading coefficient equal to 1 (with respect to $\nu$), if $f\in\mathcal{R}^*$.
The normalized reduction of  $f\in \mathcal{L}(\infty)$ will be $\mathfrak{r}(f)=\mathfrak{n}(\mathfrak{r}'(f))$.

Let $\mathcal{M}=\{ 1,x,y,\dots\}=\{ f_1,f_2,\dots,f_n\}$ be the set of reduced monomials ordered
according to the values of $\nu$ and let  $\mathcal{M}_m=\{ f_1,\dots,f_{\ell(m)}\}$ for  $0\le m \le q^3+q^2-q-1$.  From Lemma \ref{Lemma_1} it holds that
$\{  ev(f_i) : f_i \in\mathcal{M}\}$ is a basis of $\mathcal{C}(q^3+q^2-q-1) \cong \mathbb{F}_{q^2}^n$ and
$\{  ev(f_i) : f_i \in\mathcal{M}_m\}$ is a basis of $\mathcal{C}(m)$.
In what follows, we deal with codes in terms of reduced polynomials.

\begin{example}
As mentioned before, the Hermitian curve over $\mathbb{F}_{q^2}$ has $q^3$ rational affine points and one point at infinity, which is also rational. The genus of the curve is $g = q(q-1)/2$.
    A basis for $\mathcal{L}(m)$ is given by the set $\{x^i y^j| i\geq 0, 0\leq j\leq q-1, iq+j(q+1)\leq m\}$. Now, assume that $q = 3$ and $m = 22$.
    Then,
    \begin{equation*}
    B = \{1, x, y, x^2, xy, y^2, x^3, x^2y, xy^2, x^4, x^3y, x^2y^2, x^5, x^4y, x^3y^2,x^6, x^5y,x^4y^2, x^7, x^6y\}
  \end{equation*}
  is a basis for $\mathcal{L}(m)$. Consider $B^q = \{f^q\colon f\in B\}$. Since $q = 3$, the reductions are given by $\mathfrak{r}'(x^9) = x$ and $\mathfrak{r}'(y^3) = x^{4}-y$. Thus, the normalized reduction $\mathfrak{r}$ on $B^q$ gives the polynomials listed in Table~\ref{tab:reductions}.
    \begin{table}[h]
      \centering
      \setlength{\tabcolsep}{12pt}
      \begin{tabular}{rccc}
        $i$  & $f_i$    & $\mathfrak{r}'(f_i^q)$                   & $\mathfrak{r}(f_i^q)=\mathfrak{n}(\mathfrak{r}(f_i^q))$\\
        \hline
        $1$  & $1$      & $1$                           & $1$                           \\
        $2$  & $x$      & $x^3$                         & $x^3$                         \\
        $3$  & $y$      & $x^4$ $+$ $2y$                & $x^4$ $+$ $2y$                \\
        $4$  & $x^2$    & $x^6$                         & $x^6$                         \\
        $5$  & $xy$     & $x^7$ $+$ $2x^3y$             & $x^7$ $+$ $2x^3y$             \\
        $6$  & $y^2$    & $x^8$ $+$ $x^4y$ $+$ $y^2$    & $x^8$ $+$ $x^4y$ $+$ $y^2$    \\
        $7$  & $x^3$    & $x$                           & $x$                           \\
        $8$  & $x^2y$   & $2x^6y$ $+$ $x^2$             & $x^6y$ $+$ $2x^2$             \\
        $9$  & $xy^2$   & $x^7y$ $+$ $x^3y^2$ $+$ $x^3$ & $x^7y$ $+$ $x^3y^2$ $+$ $x^3$ \\
        $10$ & $x^4$    & $x^4$                         & $x^4$                         \\
        $11$ & $x^3y$   & $x^5$ $+$ $2xy$               & $x^5$ $+$ $2xy$               \\
        $12$ & $x^2y^2$ & $x^6y^2$ $+$ $x^6$ $+$ $x^2y$ & $x^6y^2$ $+$ $x^6$ $+$ $x^2y$ \\
        $13$ & $x^5$    & $x^7$                         & $x^7$                         \\
        $14$ & $x^4y$   & $x^8$ $+$ $2x^4y$             & $x^8$ $+$ $2x^4y$             \\
        $15$ & $x^3y^2$ & $x^5y$ $+$ $xy^2$ $+$ $x$     & $x^5y$ $+$ $xy^2$ $+$ $x$     \\
        $16$ & $x^6$    & $x^2$                         & $x^2$                         \\
        $17$ & $x^5y$   & $2x^7y$ $+$ $x^3$             & $x^7y$ $+$ $2x^3$             \\
        $18$ & $x^4y^2$ & $x^8y$ $+$ $x^4y^2$ $+$ $x^4$ & $x^8y$ $+$ $x^4y^2$ $+$ $x^4$ \\
        $19$ & $x^7$    & $x^5$                         & $x^5$                         \\
        $20$ & $x^6y$   & $x^6$ $+$ $2x^2y$             & $x^6$ $+$ $2x^2y$             \\
        \hline
      \end{tabular}
      \caption{Normalized reductions for $q=3$ and $m=22$}
      \label{tab:reductions}
    \end{table}
    Lastly, one can check that $\langle ev(B^q)\rangle = \langle ev(\mathfrak{r}(B^q))\rangle $.
\end{example}

\section{The parameters of EAQECCs from Hermitian codes}
\label{Sect4}

In this section we shall compute the parameters of the EAQECC $\mathcal{Q}(m)$ obtained  from the Hermitian code $\mathcal{C}(m)$. Let us remember that this problem leads us to the computation of
$\Delta(m)=\dim(\mathcal{C}(m)^{\perp_H} \cap \mathcal{C}(m))= \dim(\mathcal{C}(m)^{q} \cap \mathcal{C}(m^{\perp}))$, where $m^{\perp}=n+2g-2-m$.
In some cases, this number is easy to determine, see \cite{SK}.

\begin{proposition}  \label{Proposition_4}
If $m\leq q^2 - 2$, then $\Delta(m) = \ell(m)$.
\end{proposition}
\begin{proof}
Note that $\mathcal{L}(m)^q\subseteq \mathcal{L}(qm)$, where $\mathcal{L}(m)^q = \{f^q :  f\in\mathcal{L}(m)\}$.
If $m\leq q^2-2$, then $qm\leq m^\perp$ and thus $\mathcal{C}(m)^q\subseteq \mathcal{C}(m^\perp)$, so $\Delta(m)= \dim(\mathcal{C}(m)^q)=\dim(\mathcal{C}(m))$.
\end{proof}

For $m< q^3$ the evaluation map $ev:\mathcal{L}(m)^q\rightarrow \mathbb{F}_{q^2}^n$ is injective since $ev(f^q)=ev(f)^q$ and $ev(\mathcal{L}(m))$ is one-to-one. For $m\geq q^3$ this is no longer true. Thus, in this case we cannot expect a formula for $\Delta(m)$ in terms of
$\ell(m)$, as in Proposition~\ref{Proposition_4}. In the sequel, we develop a procedure that performs such a computation.
We can restrict to $m\geq q^2 -1$ since Proposition~\ref{Proposition_4} covers the remaining cases.
On the other hand, from Proposition \ref{propDelta(V)}(c), we can also assume $m\le m^{\perp}$, that is,
we can restrict to $m\le m^*=\lfloor n/2+g-1\rfloor$ \cite{Tiersma}.
We  write $\mathcal{M}_*=\mathcal{M}_{m^*}$.

Let $m\le m^*$. The code $\mathcal{C}(m)^q$ can be obtained by
\[
\mathcal{C}(m)^q = ev(\langle f_1^q, \ldots, f_\ell^q\rangle) = ev(\langle \mathfrak{r}(f_1^q), \ldots, \mathfrak{r}(f_\ell^q)\rangle),
\]
where $\ell=\ell (m)$ (since $m<q^3=n$). Notice that the functions $\mathfrak{r}(f_1^q), \ldots, \mathfrak{r}(f_\ell^q)$ are linearly independent, as
$\dim(\mathcal{C}(m)^q)=\dim(\mathcal{C}(m))$. In general, however, they do not have pairwise different orders.
We shall construct a new set of functions $\phi_1,\dots,\phi_\ell$ such that
$\langle \phi_1, \ldots, \phi_\ell\rangle = \langle \mathfrak{r}(f_1^q), \ldots, \mathfrak{r}(f_\ell^q)\rangle$
and all $\phi_i$'s have different orders. To that end, we define a second reduction on the $\mathfrak{r}(f_i^q)$'s, denoted by $\mathfrak{s}$ and described iteratively via the rule
\[
  \mathfrak{s}(\mathfrak{r}(f_1^q)):=1
  \quad\text{and}\quad
  \mathfrak{s}(\mathfrak{r}(f_i^q)) := \mathfrak{r}(f_i^q)  \bmod \big\langle \mathfrak{s}(\mathfrak{r}(f_1^q)), \dots, \mathfrak{s}(\mathfrak{r}(f_{i-1}^q))\big\rangle, \; i=2,\dots,\ell .
\]
We then set $\phi_i=\mathfrak{s}(\mathfrak{r}(f_i^q))$ for $ i=1,\dots,\ell $.

This reduction may be described step-by-step as follows.
Define $\phi_1 := 1$. Since $f_1=1$, it is obvious that
$ \langle\phi_1\rangle=\langle\mathfrak{r}(f_1^q)\rangle$. Once normalized reductions
$\phi_1, \ldots, \phi_{t-1}$ such that  $\langle\phi_1, \dots, \phi_{t-1}\rangle=\langle \mathfrak{r}(f_1^q), \dots,  \mathfrak{r}(f_{t-1}^q) \rangle $ are computed, set $\phi =  \mathfrak{r}(f_t^q)$ and do:
\begin{description}
\item[(S1)] If $\nu(\phi) \neq \nu(\phi_i)$ for all  $i<t$, then set $\phi_t \leftarrow  \phi$. End.
\item[(S2)] If $\nu(\phi) = \nu(\phi_i)$ for some  $i<t$, then set $\phi \leftarrow \mathfrak{n}(\phi - \phi_i)$.
Repeat from (S1).
\end{description}
Note that when $\nu(\phi) = \nu(\phi_i)$ for some $i<t$, then we have  $\langle\phi_1, \dots, \phi_{t-1},\mathfrak{n}(\phi - \phi_i)\rangle=\langle\phi_1, \dots, \phi_{t-1},\phi \rangle$. Then, by the initial choice  $\phi =  \mathfrak{r}(f_t^q)$  and the induction hypothesis $\langle\phi_1, \dots, \phi_{t-1}\rangle=\langle \mathfrak{r}(f_1^q), \dots,  \mathfrak{r}(f_{t-1}^q) \rangle $, in each step of the previous procedure it holds that
$\langle\phi_1, \dots, \phi_{t-1},\phi\rangle=\langle \mathfrak{r}(f_1^q), \dots,  \mathfrak{r}(f_t^q) \rangle$. In particular, the  functions $\mathfrak{r}(f_1^q), \ldots, \mathfrak{r}(f_\ell^q)$ being linearly independent, we deduce that $\phi\neq 0$. Note also that, in the case  $\nu(\phi) = \nu(\phi_i)$ for some $i<t$,
the leading terms of $\phi$ and $\phi_i$ coincide  by  Lemma~\ref{Lemma_1}. Hence, $\nu(\phi)$ decreases in each iteration, and after at most $t$ such
iterations, we obtain $\phi_t$.\footnote{We are showing in the following section that we can reduce the number of iterations. It is achieved by a careful analysis of the reduced monomials created in the present process.}
Furthermore, the $\phi_i$'s are reduced since they are linear combinations of reduced polynomials.
Thus, finally we obtain reduced functions  $\phi_1, \ldots, \phi_{t}$, of  pairwise different orders, such that
$\langle\phi_1, \ldots, \phi_{t} \rangle) = \langle \mathfrak{r}(f_1^q), \dots,  \mathfrak{r}(f_t^q) \rangle$.

Define the set $\Phi(m) = \{\phi_i  : 1\leq i\leq \ell(m) \}$. The above properties can be summarized as follows:

\begin{proposition} \label{Lemma_5}
$\Phi(m)$ is a set of linearly independent reduced polynomials with pairwise different orders and
 $\mathcal{C}(m)^q = ev(\langle\Phi(m)\rangle)$.
\end{proposition}

The set $\Phi(m)$ provides an efficient way to compute $\Delta(m)$.

\begin{lemma} \label{Lemma_6}
Let $m, m'$ be two nonnegative integers with $m \leq m^*$. The set
$\{\phi\in\langle\Phi(m)\rangle : \nu(\phi)\leq m'\}$ is a linear space whose basis is
$\Phi(m,m')=\{\phi_i\in\Phi(m) : \nu(\phi_i)\leq m'\}$.
\end{lemma}
\begin{proof}
It is a consequence of Proposition \ref{Lemma_0} and Lemma \ref{Lemma_2}, taking into account
that all functions of $\Phi(m)$ have different orders.
\end{proof}

\begin{proposition} \label{Proposition_7}
Let $0\leq m \leq m^*$. We have $\Delta(m) = \#\{\phi_i\in\Phi(m) : \nu(\phi_i)\leq m^\perp\} = \#\Phi(m,m^\perp)$.
\end{proposition}
\begin{proof}
Let $\mathbf{x}\in\mathcal{C}(m)^q\cap\mathcal{C}(m^\perp)$. There exists a reduced polynomial
$f\in \mathcal{R}$ such that $\nu(f)\leq m^\perp$ and $ev(f) = \mathbf{x}$. Furthermore, by Proposition~\ref{Lemma_5},
there exists $\phi\in\langle\Phi(m)\rangle$ such that $\mathbf{x}=ev(\phi)$. Now, since both $f$ and $\phi$ are reduced, Lemma \ref{Lemma_1} gives $\phi = f$. Thus $\nu(\phi)\leq m^\perp$. Then the  number of vectors $\mathbf{x}$ in $\mathcal{C}(m)^q\cap\mathcal{C}(m^\perp)$ is exactly the number of $\phi$'s in $\mathcal{C}(m)^q$ such that $\nu(\phi)\leq m^{\perp}$.
The conclusion follows from Lemma~\ref{Lemma_6}.
\end{proof}

The previous arguments lead directly to an algorithmic way of computing $\Delta(m)$. Note that $-\nu$ is the valuation given by the point at infinity and $\mbox{LC}$ stands for leading coefficient (with respect to $\nu$).

\newcommand{\ensuremath{\mbox{\sc A basis for $\Delta(m)$ (for computing $c$)}}}{\ensuremath{\mbox{\sc A basis for $\Delta(m)$ (for computing $c$)}}}
\begin{algorithm}[h!]
 \caption{\ensuremath{\mbox{\sc A basis for $\Delta(m)$ (for computing $c$)}}}
 \algsetup{indent=2em}
 \begin{algorithmic}[1]\label{alg:dec1}
    \REQUIRE $m$ with  $q^2 -1 \le m \le m^*=\lfloor n/2+g-1\rfloor=\lfloor q^3/2+(q^2-q)/2-1\rfloor$;
                    $\mathcal{M}_m = \{f_1, \ldots, f_\ell \}$, where $\ell=m+1-(q^2-q)/2$.
    \ENSURE A basis for $\Delta(m)$.
    \medskip
    \FOR{$j=1, \ldots, \ell$}
         \STATE $\phi_j \gets f_j^q$;
         \WHILE{$\deg_y (\phi_j)\ge q$ or $\deg_x (\phi_j) \ge q^2$}
             \STATE In $\phi_j$, substitute $y^q$ for $x^{q+1}-y$ and $x^{q^2}$ for $x$;
         \ENDWHILE
         \STATE $\phi_j \gets \phi_j/\mbox{LC}(\phi_j)$;
    \ENDFOR
    \FOR{$j=2, \ldots , \ell$}
         \STATE $i=1$;
         \WHILE{$i<j$}
            \IF{$\nu(\phi_j) = \nu(\phi_i)$}
            	\STATE $\phi_j \gets \phi_j - \phi_i$;\label{alg:reduction1}
                \STATE $\phi_j \gets \phi_j/\mbox{LC}(\phi_j)$;\label{alg:reduction2}
                \STATE $i \gets 1$;
            \ELSE
            	\STATE $i \gets i + 1$;
            \ENDIF
         \ENDWHILE
    \ENDFOR
\STATE $\Delta(m)\gets\{\phi_1 , \ldots , \phi_\ell \}$;
\RETURN $\Delta(m)$;
\end{algorithmic}
\end{algorithm}

The next remark clarifies some properties of Algorithm~\ref{alg:dec1}.

\begin{remark}
(a) Note that all polynomials involved in the computations described above have coefficients in
$\mathbb{F}_p$, where $p$ is the characteristic of $\mathbb{F}_{q^2}$. Therefore, the algorithm runs over
$\mathbb{F}_p$, although both the Hermitian curve $\mathcal{X}$ and AG codes obtained from them
are defined over $\mathbb{F}_{q^2}$. \newline
(b) In view of Proposition \ref{propDelta(V)}(c), we have treated only the case  $m\le m^*$. For larger
values of $m$, one can use the identity $\Delta(m)=\Delta(m^{\perp})=\#\Phi(m^\perp,m)$. \newline
(c) For $m\ge n+2g-2-(q^2-2)=q^3-q$, we have $m^\perp\leq q^2-2$. By Proposition~\ref{propDelta(V)} and Proposition~\ref{Proposition_4} this implies $\Delta(m)=\Delta(m^\perp)=\ell(m^\perp)$. Hence, the EAQECC constructed from $\mathcal{C}(m)$ has entanglement $c=\ell(m^\perp)-\ell(m^\perp-n)-\Delta(m)=0$, meaning that it is a standard quantum code.
Similarly, if $m\le  q^2-2$ we have $\Delta(m)=\ell(m)$ and $\mathcal{C}(m)$ has dimension $k(m)=\ell(m)$. Thus, the resulting EAQECC has dimension $0$.
All these codes  can be discarded if one is only interested in EAQECCs with $c>0$ and $k>0$.
\end{remark}

\begin{example}
Let $q=3$. Here $g=3$ and  $m^*=15$. Algorithm~\ref{alg:dec1} gives the data shown in Table~\ref{tab:tabla1}.
\small
\begin{table}[h!]
\begin{center}
\begin{tabular}{cccccc}
$f_i$    & $\nu(f_i)$ & $\mathfrak{r}(f_i^q)$ & $\nu(\mathfrak{r}(f_i^q))$ & $\phi_i$          & $\nu(\phi_i)$ \\ \hline
$1$      & 0          & $1$                   & 0                          & $1$               & 0   \\
$x$      & 3          & $x^3$                 & 9                          & $x^3$             & 9   \\
$y$      & 4          & $x^4+2y$              & 12                         & $x^4+2y$          & 12  \\
$x^2$    & 6          & $x^6$                 & 18                         & $x^6$             & 18  \\
$xy$     & 7          & $x^7+2x^3y$           & 21                         & $x^7+2x^3y$       & 21  \\
$y^2$    & 8          & $x^8+x^4y+y^2$        & 24                         & $x^8+x^4y+y^2$    & 24  \\
$x^3$    & 9          & $x$                   & 3                          & $x$               & 3   \\
$x^2y$   & 10         & $x^6y+2x^2$           & 22                         & $x^6y+2x^2$       & 22  \\
$xy^2$   & 11         & $x^7y+x^3y^2+x^3$     & 25                         & $x^7y+x^3y^2+x^3$ & 25  \\
$x^4$    & 12         & $x^4$                 & 12                          & $y$               & 4   \\
$x^3y$   & 13         & $x^5+2xy$             & 15                         & $x^5+2xy$         & 15  \\
$x^2y^2$ & 14         & $x^6y^2+x^6+x^2y$     & 26                         & $x^6y^2+x^6+x^2y$ & 26  \\
$x^5$    & 15         & $x^7$                 & 21                         & $x^3y$            & 13  \\
\hline
\end{tabular}
\caption{Algorithm~\ref{alg:dec1} for $q=3$.}
\label{tab:tabla1}
\end{center}
\end{table}

In the first column we list the monomials in $\mathcal{M}_*$.
The second and sixth columns of this table allows us to compute $\Delta(m)$ for all values of $m$.
For example, $\Delta(10)=\#\Phi(10,21)=6$. If $m>15$ we apply the duality property; for example
$\Delta(21)=\#\Phi(21,10)=\#\Phi(10,21)=6$.

We obtain quantum codes over $\mathbb{C}^3$ with parameters
$[[27, 1, 19; 16]]_3$, $[[27, 4, 16; 13]]_3$,  $[[27, 6, 13; 9]]_3$,  $[[27, 9, 10; 6]]_3$,  $[[27, 13, 7; 4]]_3$,
and $[[27, 16, 4; 1]]_3$. All previous codes have entanglement-assisted quantum Singleton defect equal to 6. An important feature of the first three examples above is
that there is no (quantum) stabilizer code attaining the respective parameters.

Continuing the analysis for EAQECCs over $\mathbb{C}^4$, we derive the codes
$[[64, 1, 49; 45]]_4$, $[[64, 5, 42; 35]]_4$, $[[64, 16, 30; 22]]_4$,
$[[64, 24, 21; 12]]_4$, $[[64, 33, 14; 7]]_4$, $[[64, 35, 12; 3]]_4$, and\linebreak[4]
$[[64, 39, 8; 1]]_4$. They have entanglement-assisted quantum Singleton defect equal to
$12$, $12$, $12$, $12$, $12$, $10$, and $12$.
Lastly, considering EAQECCs over $\mathbb{C}^5$, the codes have parameters equal to
$[[125, 1, 101; 96]]_5$, $[[125, 9, 91; 84]]_5$, $[[125, 15, 81; 70]]_5$, $[[125, 36, 56; 41]]_5$,
$[[125, 54, 41; 29]]_5$, $[[125, 70, 26; 15]]_5$, and $[[125, 90, 10; 1]]_5$, which have
entanglement-assisted quantum Singleton quantum defect equal to
$20, 20, 20, 20, 20, 20$, and $18$.  Regarding parameters attainability using stabilizer codes, we have that
the first three and the first four EAQECCs over $\mathbb{C}^4$ and over $\mathbb{C}^5$, respectively,
are unique. We summarize these results in Table \ref{tab:tabla2}, where GV stands for the Gilbert-Varshamov bound \cite{Galindo:2019}.

\begin{table}[h!]
\begin{center}
\begin{tabular}{ccccc}
Parameters              & Singleton defect & Exceeding GV \\ \hline
$[[27, 1, 19; 16]]_3$   & 6                               & \checkmark                        \\
$[[27, 4, 16; 13]]_3$   & 6                               & \checkmark                        \\
$[[27, 13, 7; 4]]_3$    & 6                               & \checkmark                        \\
$[[27, 16, 4; 1]]_3$    & 6                               & \checkmark                        \\
$[[64, 5, 42; 35]]_4$   & 12                               & \checkmark                        \\
$[[64, 16, 30; 22]]_4$  & 12                               & \checkmark                        \\
$[[64, 35, 12; 3]]_4$   & 10                               & \checkmark                        \\
$[[64, 39, 8; 1]]_4$    & 12                               & \checkmark                        \\
$[[125, 1, 101; 96]]_5$ & 20                             & \checkmark                        \\
$[[125, 9, 91; 84]]_5$  & 20                             & \checkmark                        \\
$[[125, 36, 56; 41]]_5$ & 20                             & \checkmark                        \\
$[[125, 70, 26; 15]]_5$ & 20                             & \checkmark                        \\
 $[[125, 90, 10; 1]]_5$ & 18                               & \checkmark                        \\
\hline
\end{tabular}
\caption{Examples of code's parameters and comparative analysis by means of coding bounds.}\label{tab:tabla2}
\end{center}
\end{table}

\end{example}

\section{$q$-th powers of Hermitian codes}
\label{Sect5}

The computation of $\Delta(m)$ in the previous algorithm relies on the comparison of $\nu(f_i)$ and
$\nu(\phi_i)$ for all $f_i\in\mathcal{M}_*$.
The hardest part of this algorithm is the computation (and storage) of all reductions $\mathfrak{r}(f_i^q)$, $1\le i\le m^*$.
However, in practice, the algorithm only requires the knowledge of $\mathfrak{r}(f_i^q)$ when $f_i$ is involved in some  reduction $\mathfrak{s}$, which does not happen for all the values of $i$ (see for example the case $ q = 3 $ in Table \ref{tab:tabla1}).  Otherwise (if $f_i$ is not involved in any reduction $\mathfrak{s}$), it is enough to know $\nu(\mathfrak{r}(f_i^q))$. In this section we show how this value can be determined directly, without computing $\mathfrak{r}(f_i^q)$ through {\bf (R1)} and  {\bf (R2)}.
We start by giving a formula for the reduction  $\mathfrak{r}'(f_i^q)$ that allows us a fast computation of this data when necessary.

\subsection{Reducing $q$-th powers}
\label{Sect5.1}

Let us recall that the reduction $\mathfrak{r}'(f)$ of a polynomial $f\in \mathcal{L}(\infty)=\mathbb{F}_{q^2}[x,y]$  is obtained by performing the substitutions
\[
  \mbox{\bf (R1) } \mathfrak{r}'(y^q)= x^{q+1} - y \hspace*{5mm}  \mbox{ and }   \hspace*{5mm}  \mbox{\bf (R2) } \mathfrak{r}'(x^{q^2})= x
\]
as many times as possible. In other words, $\mathfrak{r}'(f)$ is the unique (up to multiplication by a constant $\lambda\in\mathbb{F}_{q^2}^*$) reduced polynomial in the coset of $f$ modulo the ideal $( x^{q+1}-y^q-y, x^{q^2} - x )\subset  \mathcal{L}(\infty)$, with respect to $\nu$. The next lemma gives some properties of $\mathfrak{r}'$.

\begin{lemma} \label{propreduc}
Let $g_1,g_2\in \mathbb{F}_{q^2}[x,y]$. The following properties hold. \newline
(a) $\mathfrak{r}'(g_1+g_2)=\mathfrak{r}'(g_1)+\mathfrak{r}'(g_2)$; \newline
(b) $\mathfrak{r}'(g_1g_2)=\mathfrak{r}'(\mathfrak{r}'(g_1) \mathfrak{r}'(g_2))$.
\end{lemma}
\begin{proof}
Both properties follow from the fact that the natural map $\mathbb{F}_{q^2}[x,y] \rightarrow \mathbb{F}_{q^2}[x,y]/( x^{q+1}-y^q-y, x^{q^2} - x )$ is a ring homomorphism.
\end{proof}

As a notation, given  a non negative integer $\delta$, we write
\[
  \mathfrak{m}(\delta)= \left\{
    \begin{array}{ll}
      0                     & \mbox{if $\delta=0$}                         \\
      \delta \bmod{(q^2-1)} & \mbox{if $q^2-1\nmid \delta$}                \\
      q^2-1                 & \mbox{if $q^2-1|\delta$ and $\delta\neq 0$.} \\
    \end{array}
  \right.
\]
That is, for $\delta\neq 0$, $\mathfrak{m}(\delta)$ is the remainder of $\delta$ modulo $q^2-1$ in the interval $[1, q^2-1]$ rather than $[0,q^2-2]$ as usual. In particular   $\mathfrak{m}(\delta)\le q^2-1$.

\begin{lemma}\label{sumarq^2+1}
Let $\delta_1,\delta_2$ be nonnegative integers.  The following properties hold. \newline
(a)  $\mathfrak{m}(\delta_1+\delta_2)=\mathfrak{m}(\mathfrak{m}(\delta_1)+\delta_2)=\mathfrak{m}(\mathfrak{m}(\delta_1)+\mathfrak{m}(\delta_2))$.\newline
(b) If $\delta_2>0$ then $\mathfrak{m}(q^2-1+\delta_2)=\mathfrak{m}(\delta_2)$.
\end{lemma}

Given nonnegative integers $\delta,\mu$, we denote the binomial coefficient modulo $p$ as
\[
  \binom{\mu}{\delta}_{\!\!  p} = \binom{\mu}{\delta} \;  \bmod{p}
\]
where $p$ is the characteristic of $\mathbb{F}_q$.

\begin{proposition}\label{mathfrakr'}
Let $f=x^ay^b$ with $0\le a$ and $0\le b<q$.  Then we have
\[
  \mathfrak{r}'(f^q)=\sum_{j=0}^b (-1)^j\binom{b}{j}_{\!\! p} x^{\mathfrak{m}(\nu(f)-j(q+1))} y^j.
\]
\end{proposition}
\begin{proof}
If $b=0$ then only the reduction (R2) is involved in $\mathfrak{r}'$.  If $a=0$,  according to the Newton's binomial formula and Lemma \ref{propreduc}  we have
\begin{align*}
\mathfrak{r}'((y^b)^q)= \mathfrak{r}'((x^{q+1}-y)^b) &=\mathfrak{r}'\left(\sum_{j=0}^b (-1)^j\binom{b}{j}_{\!\! p} x^{(b-j)(q+1)} y^j\right)\\
                                                     &=\sum_{j=0}^b (-1)^j\binom{b}{j}_{\!\! p} x^{(b-j)(q+1)} y^j.
\end{align*}
If $ab>0$, according to Lemma \ref{propreduc} and the previous computations  we have
\begin{align*}
\mathfrak{r}'((x^ay^b)^q) &=\sum_{i=0}^b (-1)^j\binom{b}{j}_{\!\! p} \mathfrak{r}'(x^{\mathfrak{m}(aq)+(b-j)(q+1)}) y^j\\
&=\sum_{j=0}^b (-1)^j\binom{b}{i}_{\!\! p} x^{\mathfrak{m}(aq+(b-j)(q+1))} y^j
\end{align*}
since all summands in the last expression are reduced monomials.
\end{proof}

\subsection{Computing  $\nu(\mathfrak{r}(f_i^q))$. Case $q$ prime}
\label{Sect5.2}

From Proposition \ref{mathfrakr'} we can deduce the values $\nu(\mathfrak{r}(f_i^q))$, for $f_i\in \mathcal{M}_*$ we need to run our algorithm.
When $q=p$ is a prime number, then all binomial coefficients in the formula of Proposition \ref{mathfrakr'} are  non-zero. When $q$ is not a prime then some of these binomial coefficients may vanish, and the description of $\nu(\mathfrak{r}(f_i^q))$ becomes more involved.  We first study the case in which $q$ is a prime number. The case $q$  not a prime will be treated in the next subsection.

Let $f\in \mathcal{M}_*$. We will write the order $\nu(f)$ as $uq^2+sq+t$ with $0\leq s,t<q$ and  $0\le u\leq (q+1)/2$.  Note that this representation is unique.

\begin{proposition} \label{prop:qthOrders}
Let $q=p$ be a prime, and let $f\in\mathcal{M}_*$. Write $\nu(f)=uq^2+sq+t$ with $0\leq u\leq (q+1)/2$ and $0\leq s,t<q$. Then
  \[
    \nu(\mathfrak{r}(f^q))=
    \begin{cases}
      sq^2+(u+t)q & \mbox{\rm if $s\ge t$ and $u+t<q^2-sq$;} \\
      q^3-2q^2+(u+t)q+1 & \mbox{\rm if $s\ge t$ and $u+t\ge q^2-sq$;} \\
      q^3-q^2+(u+t-1)q+s+1 & \mbox{\rm if $s< t$ and $u+t\le q+s+1$;} \\
      q^3-2q^2+(u+t-1)q+s+2 & \mbox{\rm if $s< t$ and $u+t> q+s+1$.}
    \end{cases}
  \]
\end{proposition}
\begin{proof}
The assumption $\nu(f)=uq^2+sq+t$ implies $f=x^{uq+s-t}y^t$ and therefore
\begin{equation} \label{prooforderfq}
\mathfrak{r}'(f^q)=\sum_{j=0}^t (-1)^j\binom{t}{j}_{\!\! p} x^{\mathfrak{m}(uq^2+sq+t-j(q+1))} y^j
\end{equation}
from Proposition \ref{mathfrakr'}. Since $t<q=p$, all binomial coefficients are nonzero modulo $p$, so it is enough to find the highest order among all the summands in the previous expression.
To do this, we use the following observation. According to Lemma \ref{sumarq^2+1} (b),  when the condition $(S): sq+u+t>j(q+1)$ holds, then  $\mathfrak{m}(uq^2+sq+t-j(q+1))=\mathfrak{m}(sq+t+u-j(q+1))$. In particular $(S)$ is always satisfied for $j=0$, and for all values of $j$ when $s\ge t$.
Let us distinguish three separate cases.\newline
{\em Case 1:} If $s\ge t$ and $sq+u+t<q^2$, as above we have $\mathfrak{m}(sq+u+t-j(q+1))=sq+u+t-j(q+1)$. The maximum order among the summands of $\mathfrak{r}'(f^q)$  is $sq^2+(u+t)q$, obtained for $j=0$. \newline
{\em Case 2:} If $s\ge t$ and $sq+u+t\ge q^2$, then $s=q-1, t\ge 1$ and $sq+u+t\le q^2+q$, hence  $\mathfrak{m}(sq+u+t-j(q+1))=sq+u+t-(q^2-1)$ if $j=0$ and $\mathfrak{m}(sq+u+t-j(q+1))=sq+u+t-j(q+1)$ if $j>0$. Thus the maximum order in $\mathfrak{r}'(f^q)$ is  $q^3-2q^2+(u+t)q+1$, which is obtained for $j=1$.
\newline
{\em Case 3:} If  $s< t$, then $u> 0$. Note that the condition $(S)$ is satisfied for all $j\le s$. Since $sq+u+t< q^2$,  in this range we have $\mathfrak{m}(sq+u+t-j(q+1))=sq+u+t-j(q+1)$. So  the maximum order among the summands of $\mathfrak{r}'(f^q)$ corresponding to $j\le s$ is $sq^2+(u+t)q$, which is obtained for $j=0$. Consider now the summand corresponding to $j=t$. Since $uq^2+sq+t-t(q+1)= (u-1)(q^2-1)+(q+s-t)q +u-1$, we have $\mathfrak{m}(uq^2+sq+t-t(q+1))=(q+s-t)q +u-1$ and the corresponding monomial has order $(q+s-t)q^2 +(u-1)q+t(q+1)$, which is bigger than the orders we have obtained for $j\le s$.
Finally let $j=t-h>s$. According to Lemma \ref{sumarq^2+1}(a) and the computation made for $j=t$, we have
$\mathfrak{m}(uq^2+sq+t-j(q+1))=\mathfrak{m}((q+s-t)q+(u-1)+h(q+1))=\mathfrak{m}((q+s-t+h)q+(u+h-1))$,
hence the maximum order is obtained when $h$ is as large as possible satisfying the condition $(q+s-t+h)q+(u+h-1)\le q^2-1$, or equivalently $(t-s-h)q \ge u+h$. It is easy to check that such largest value of $h$ is
$h=t-s-1$ (that is $j=s+1$) when $q \ge u+t-s-1$, and $h=t-s-2$ (that is $j=s+2$) when $q < u+t-s-1$. A straightforward computation gives the maximum order among the summands of  \eqref{prooforderfq} in both cases, which is $q^3-q^2+(u+t-1)q+s+1$ if  $q \ge u+t-s-1$ and $q^3-2q^2+(u+t-1)q+s+2$ if  $q< u+t-s-1$.
\end{proof}
Proposition~\ref{prop:qthOrders} gives us the $\nu$-value of the leading monomial in $f^q$, it can be used to determine when $\nu(\mathfrak{r}(f^q))= \nu(\mathfrak{r}(f'^q))$ holds for two functions $f, f'\in\mathcal{M}_*$. However, we will postpone this until after Proposition~\ref{prop:monomialSupport}, which concerns the $\nu$-values of \emph{all} monomials in $f^q$.
To state this proposition, we use $\Supp f$ to denote the monomial support of $f$. That is, if $f=\sum c_{a,b}x^ay^b\in\mathbb{F}_{q^2}[x,y]$, we define $\Supp f=\{x^ay^b\mid c_{a,b}\neq 0\}$.

\begin{proposition}\label{prop:monomialSupport}
Let $f\in\mathcal{M}$. Then
\begin{equation*}
\Supp \mathfrak{r}(f^q) \subseteq \left\{f_j\in \mathcal{M} : \nu(f_j)\equiv q\nu(f)\pmod{q^2-1}\right\}.
\end{equation*}
Furthermore, if $q+1$ divides $\nu(f)$, then $\nu(f)\equiv q\nu(f)\pmod{q^2-1}$.
\end{proposition}
\begin{proof}
  If $f=1$ then both results are clear. Let us assume $f\neq 1$.
  We have $\nu(f^q)=q\nu(f)$, so in particular, $\nu(f^q)\equiv q\nu(f)\pmod{q^2-1}$. Thus, it suffices to show that when applying each of the reductions \textbf{(R1)} and \textbf{(R2)} to $f^q$, the orders of all resulting monomials remain in the original equivalence class.

  Hence, let $f^q=x^ay^b$. If $a<q^2$ and $b<q$, then the result is immediate since $\mathfrak{r}(x^ay^b)=x^ay^b$. If $a\geq q^2$, we can apply \textbf{(R2)} once to obtain the monomial $x^{a-(q^2-1)}y^b$. This has order $\nu(x^ay^b)-q(q^2-1)$, meaning that the equivalence class modulo $q^2-1$ is preserved. If $b\geq q$, applying \textbf{(R1)} gives two monomials $x^{a+(q+1)}y^{b-q}$ and $x^ay^{b-(q-1)}$. These have orders $\nu(x^ay^b)$ and $\nu(x^ay^b)-(q^2-1)$, respectively. Again, the resulting orders are in the same equivalence class as $\nu(x^ay^b)$. This proves the first claim of the proposition.

  For the second claim, one has that $q+1\mid\nu(f)$ implies $(q-1)\nu(f)\equiv 0\pmod{q^2-1}$, which can be rearranged to obtain the result.
\end{proof}
\begin{corollary}\label{psi}
Let $f,f'\in\mathcal{M}_*$. If $\nu(\mathfrak{r}(f^q))= \nu(\mathfrak{r}(f'^q))$ then  $\nu(f)\equiv\nu(f') \pmod{q^2-1}$.
\end{corollary}
\begin{proof}
  Proposition~\ref{prop:monomialSupport} tells us that $\nu(\mathfrak{r}(f^q))\equiv q\nu(f)\pmod{q^2-1}$, and similarly for $\mathfrak{r}(f'^q)$. Combining these equivalences gives the result.
\end{proof}
We now have a necessary condition for two functions $f, f'\in\mathcal{M}_*$ to satisfy
$\nu(\mathfrak{r}(f^q))= \nu(\mathfrak{r}(f'^q))$. This can be used to bound the number of reductions in our algorithm in Section~\ref{Sect4}, since the reduction $\mathfrak{s}$ is only applied to the monomials $f_i$ for which there exists $f_j\in \mathcal{M}_*$ with $j<i$ and $\nu(\mathfrak{r}(f_j^q))= \nu(\mathfrak{r}(f_i^q))$. The number of such $f_j$'s should be moderate, as the following proposition shows that the map $f\mapsto\nu(f)\bmod{(q^2-1)}$ is quite uniformly distributed.

\begin{proposition} \label{papendice}
(a) The map $\mathcal{M}_*\rightarrow \mathbb{Z}/(q^2-1)\mathbb{Z}$ given by $f\mapsto \nu(f)\bmod{(q^2-1)}$ is surjective. \newline
(b) For $0\leq k<q^2-1$, the number of monomials $f\in\mathcal{M}$ with $\nu(f)\equiv k\pmod{q^2-1}$ is
\begin{itemize}
  \item $q+2$ when $k=0$
  \item $q+1$ when $k\neq 0$ and $q+1\mid k$
  \item $q$ when $k\neq 0$ and $q+1\nmid k$
\end{itemize}
\end{proposition}

The proof of this result can be found in the Appendix.
To illustrate the preceding results, we give the following example.
\begin{example}
  Let $q=5$, and consider $f=f_{24}=x^3y^3$ with $\nu(f_{24})=33=q^2+q+3$. Thus, $\nu(f_{24})\bmod{(q^2-1)}=9$.
  Turning to the $q$'th power, we see that $\mathfrak{r}(f_{24}^q)=3x^{21}y^2 + 4x^{15}y^3 + x^9 + 2x^3y$. Table~\ref{tab:exampleSupport} lists the orders of the monomials in $\Supp\mathfrak{r}(f_{24}^q)$ along with their orders modulo $q^2-1$. In each case, this remainder is $21$, which is exactly $q\nu(f_{24})\bmod{(q^2-1)}$.

  The remaining monomials $f_j$ with $\nu(f_j)\equiv 9\pmod{q^2-1}$ are $f_{48}=x^9y^2$, $f_{72}=x^{15}y$, $f_{96}=x^{21}$, and $f_{119}=x^{21}y^4$. Thus, there are a total of $5$ such monomials as predicted by Proposition~\ref{papendice}. If we consider the support of their $q$'th powers, we obtain the four monomials in Table~\ref{tab:exampleSupport} and the monomial $f_{60}=x^9y^4$.

  One may also note that there is a duality between these two sets of monomials. Namely, the monomial support of $\{f_{12}^q,f_{36}^q,f_{60}^q,f_{84}^q,f_{108}^q\}$ is exactly $\{f_{24},f_{48},\allowbreak f_{72},f_{96},f_{119}\}$.
\end{example}
\begin{table}[ht]
    \centering
    \begin{tabular}{cccc}
      $j$ & $f_j$ & $\nu(f_j)$ & $\nu(f_j)\bmod{(q^2-1)}$\\
      \hline
      $108$ & $x^{21}y^2$ & $117$ & $21$ \\
      $84$ & $x^{15}y^3$ & $93$ & $21$ \\
      $36$ & $x^9$ & $45$ & $21$ \\
      $12$ & $x^3y$ & $21$ & $21$ \\
      \hline
    \end{tabular}
    \caption{Monomials in the support of $\mathfrak{r}(f_{24}^q)$ for $q=5$}
	\label{tab:exampleSupport}
\end{table}

\subsection{Computing  $\nu(\mathfrak{r}(f_i^q))$. Case $q$ non-prime}
\label{sec:non-prime-q}

In this section we describe $\nu(\mathfrak{r}(f^q))$ when $q=p^r$ is not a prime, analogous to what was done in Proposition \ref{prop:qthOrders} when $q$ is a prime number. Our study will rely on Lucas' theorem, which relates the binomial coefficient $\binom{t}{j}\bmod{p}$ to the $p$-ary representations of $t$ and $j$. In \cite{GKS13}, the concept of a \emph{$p$-shadow} is defined as follows. Let $t=\sum_{i=0}^{r-1} t_i p^i$ and $j=\sum_{i=0}^{r-1} j_i p^i$ be the $p$-ary representations of $t$ and $j$, where $0\le j\le t<q$. If $j_i\leq t_i$ for all $i$, then $j$ is said to be in the $p$-shadow of $t$, and we write $j\leq_p t$. As a corollary to Lucas' theorem \cite{Fine47,lucas1891}, we  have
\begin{equation}\label{eq:lucasTheorem}
  \binom{t}{j}_{\!\!  p}\neq  0  \quad\mbox{ if and only if }\quad j\leq_p t.
\end{equation}
Inspired by the $p$-shadow, we introduce the \emph{$p$-illumination}.

\begin{definition}\label{defi:illumination}
Let $q=p^r$ be a prime power and let $0\leq j\leq t<q$ be integers. If $t=\sum_{i=0}^{r-1} t_i p^i$ and $j=\sum_{i=0}^{r-1} j_i p^i$ are the $p$-ary representations of $t$ and $j$, we let
\[
    \illum{t}{j}=\sum_{i=0}^{i^\ast} (j_i-t_i)p^i,
\]
where $i^\ast$ is the largest index such that $j_{i^\ast}>t_{i^\ast}$. If no such $i^\ast$ exists, we let $\illum{t}{j}=0$. We call $\illum{t}{j}$ the {\em $p$-illumination} of $j$ with respect to $t$.
\end{definition}

It is easy to verify that $\illum{t}{j}$ is a non-negative integer. Additionally, $j-\illum{t}{j}$ is in the $p$-shadow of $t$, and it is the largest integer less than or equal to $j$ that satisfies this property. This follows from the observation that $j-\illum{t}{j}$ has the same $p$-ary digits as $t$ for indices $1,2,\ldots,i^\ast$. Each increment up to $j$ will have at least one of these digits greater than the corresponding $p$-digit of $t$. These considerations yield the following lemma.

\begin{lemma}\label{lem:binomials}
  Let $q=p^r$ be a prime power, $t<q$ and $0\leq j\leq t$. If $\illum{t}{j}=0$, then $\binom{t}{j}_{\!\!  p}\neq 0$. Otherwise, if $\illum{t}{j}>0$, then
  \[
    \binom{t}{j}_{\!\!  p}= \binom{t}{j-1}_{\!\!  p}= \cdots =\binom{t}{j-\illum{t}{j}+1}_{\!\!  p}= 0 \;
    \mbox { and }
    \quad \binom{t}{j-\illum{t}{j}}_{\!\!  p}\neq 0.
  \]
\end{lemma}
\begin{proof}
  By \eqref{eq:lucasTheorem}, $\binom{t}{j'}_{\!\!  p}$ is non-zero if and only if $j'\leq_p t$. The observations immediately below Definition~\ref{defi:illumination} imply that  the first such $j'$ below $j$ is $j-\illum{t}{j}$, proving the lemma.
\end{proof}

In other  words, the largest integer $j^*\le j$ satisfying $\binom{t}{j^*}_{\!\!  p}\neq 0$ is $j^*=j-\rho_t(j)$. In addition, we can apply Lemma~\ref{lem:binomials} to $t-j$ and use the symmetry of the binomial coefficient, $\binom{t}{i}=\binom{t}{t-i}$, to infer that $\binom{t}{j}_{\!\! p}=\cdots=\binom{t}{j+\rho_t(t-j)-1}_{\!\! p}=0$ and $\binom{t}{j+\rho_t(t-j)}\neq 0$. That is, the smallest integer $j^*\ge j$ satisfying $\binom{t}{j^*}_{\!\!  p}\neq 0$ is $j^*=j+\rho_t(t-j)$.

\begin{proposition}\label{qthordersnoprime}
  Let $q=p^r$ be a prime power, and let $f\in\mathcal{M}_*$. Write $\nu(f)=uq^2+sq+t$ with $0\leq u\leq (q+1)/2$ and $0\leq s,t<q$. Then $ \nu(\mathfrak{r}(f^q))$ equals
\[
    \begin{cases}
      sq^2+(u+t)q & \mbox{\rm if $s\ge t$, $u+t<q^2-sq$;} \\
      q^3-2q^2+(u+t)q+1-(q^2-1)\rho_t(t-1) & \mbox{\rm if $s\ge t$, $u+t\ge q^2-sq$;} \\
      q^3-q^2+(u+t-1)q+s+1-(q^2-1)\rho_t(t-s-1) & \mbox{\rm if $s< t$, $u+t\le q+s+1$;} \\
      q^3-2q^2+(u+t-1)q+s+2-(q^2-1)\rho_t(t-s-2) & \mbox{\rm if $s< t$, $u+t> q+s+1$.}
    \end{cases}
  \]
\end{proposition}

\begin{proof}
The proof of this result is similar to that of Proposition \ref{prop:qthOrders}, in which we sought the summand $j$ providing the highest order in  the  writing of $\mathfrak{r}'(f^q)$ given in the Proposition \ref{mathfrakr'}.
Let $j^*$ be the index of the summand providing the maximum order in our case $q$ not a prime.
Following the proof of Proposition  \ref{prop:qthOrders}, in case~1 such maximum order is obtained for $j=0$. Since $\binom{t}{0}_{\!\!  p}\neq 0$, we have $j^*=0$.  In case~2, the maximum order comes from the summand corresponding to the smallest index $j^*\ge 1$ with $\binom{t}{j^*}_{\!\!  p}\neq 0$, that is for $j^*=1+\rho_t(t-1)$ according to Lemma \ref{lem:binomials}.
In case~3, the maximum order comes as well from the summand corresponding to the smallest index $j^*\ge j$ with $\binom{t}{j^*}_{\!\!  p}\neq 0$,  where $j=s+1$ when  $u+t\le q+s+1$ and $j=s+2$ when  $u+t> q+s+1$. Such indices are $j^*=s+1+\rho_t(t-s-1)$ and $j^*=s+2+\rho_t(t-s-2)$ respectively.
In all cases, it is enough to compute the orders of the summands corresponding to these $j^*$'s to obtain the stated formula.
\end{proof}

Note that when $\nu(f)\ge q^2$, the values  $ \nu(\mathfrak{r}(f^q))$ are, in most cases, bigger when $q$ is a prime number than when it is not, $q=p^r$ with $r>0$. Therefore, also $\Delta$ increases (and so the entanglement $c$ decreases) when $r$ increases.

\subsection{Bounding the complexity of Algorithm~\ref{alg:dec1}}
Having described the $q$'th powers given in the previous sections, we now use those results to bound the computational complexity of Algorithm~\ref{alg:dec1}. Since the most computationally costly part are the reductions $\mathfrak{s}$ -- that is, lines~\ref{alg:reduction1} and \ref{alg:reduction2} --  we focus on bounding the number of such reductions as well as bounding the cost of each reduction.

\begin{proposition}\label{prop:algComplexity}
  The total number of reductions $\mathfrak{s}$ in Algorithm~\ref{alg:dec1} is $\mathcal{O}(q^4)$, and each reduction requires $\mathcal{O}(q)$ field operations.
\end{proposition}
\begin{proof}
  First recall that $\phi_i=\mathfrak{r}(f_i^q)$. Thus, Proposition~\ref{prop:monomialSupport} ensures that all monomials $f_k$ in $\phi_i$ satisfy $\nu(f_k)\equiv q\nu(f_i)\pmod{q^2-1}$.
  If we consider some $\phi_j$ during the algorithm, similar arguments as in Corollary~\ref{psi} show that the reduction $\phi_j=\phi_j-\phi_i$ can happen only if $\nu(f_j)\equiv\nu(f_i)\pmod{q^2-1}$. Thus, we bound the total number of reductions by an amortized analysis, grouping polynomials $\phi_i$ based on the equivalence class of $\nu(f_i)$.

  Fix a $k\in\{0,1,\ldots,q^2-2\}$, and consider all polynomials $\phi_i$ such that $\nu(f_i)\equiv k\pmod{q^2-1}$. By Proposition~\ref{papendice}, there can be at most $q+2$ such monomials. The first time the algorithm encounters such a $\phi_i$ no reduction will happen. The second time such a $\phi_i$ is found, there will be at most one reduction. The third time at most two and so forth.
  Thus, the total number of reductions required to process the $\phi_i$ with $\nu(f_i)\equiv k\pmod{q^2-1}$ cannot exceed
  \[
    \sum_{j=1}^{q+2}(j-1)=\frac{q(q+1)}{2}.
  \]
  Because there are $q^2-1$ possible values of $k$, the overall number of reductions during the algorithm is at most $\frac{1}{2}q(q+1)(q^{2}-1)$, which is $\mathcal{O}(q^4)$.

  To prove the claim on the number of field operations per reduction, note that each $\phi_i$ can contain at most $q+2$ monomials by Propositions~\ref{prop:monomialSupport} and \ref{papendice}.
\end{proof}
Since the reductions $\mathfrak{s}$ are what dominates the complexity of Algorithm~\ref{alg:dec1}, Proposition~\ref{prop:algComplexity} implies that the number of field operations used during the algorithm is $\mathcal{O}(q^5)$, which is significantly better than computing $c$ as in Proposition \ref{Prep:WildeHerm} using linear algebra methods. Namely, it is significantly better than computing the rank of $\mathbf{G}\mathbf{G}^*$, where $\mathbf{G}$ is a generator matrix of the linear code and $\mathbf{G}^*$ is the $q$-th power of the transpose matrix of $\mathbf{G}$ (see \cite[Proposition 3]{Galindo:2019}).

We can optimize the algorithm even further by using the values of $\nu(\mathfrak{r}(f^q))$ found in Sections~\ref{Sect5.2} and \ref{sec:non-prime-q}.
Namely, the algorithm only requires the knowledge of $\mathfrak{r}(f_i^q)$ when $f_i$ is involved in some  reduction $\mathfrak{s}$, which does not happen for all the values of $i$ (see for example the case $ q = 3 $ in Table \ref{tab:tabla1}).

Therefore, we can modify the algorithm as follows: we first compute the orders $\nu(\mathfrak{r}(f_i^q))$ from Propositions  \ref{prop:qthOrders} and \ref{qthordersnoprime}. If $f_i$ is not involved in any  reduction $\mathfrak{s}$, then we will not calculate $\mathfrak{r}(f_i^q)$. Otherwise, when it is necessary to know this data, it is computed from Proposition~\ref{mathfrakr'}, and the corresponding polynomials $\phi$'s from the reduction $\mathfrak{s}$.
The obtained result is shown in Table \ref{tab:tabla4}  for $q=3$.  The reader can compare this table with Table
\ref{tab:tabla1}, in which the unmodified process is shown.

\begin{table}[h!]
\centering
\begin{tabular}{cccccc}
$f_i$    & $\nu(f_i)$ & $\mathfrak{r}(f_i^q)$ & $\nu(\mathfrak{r}(f_i^q))$ & $\phi_i$    & $\nu(\phi_i)$ \\ \hline
$1$      & 0          &                       & 0                          &             & 0   \\
$x$      & 3          &                       & 9                          &             & 9   \\
$y$      & 4          &   $x^4+2y$       & 12                         &  $x^4+2y$    & 12  \\
$x^2$    & 6          &                       & 18                         &             & 18  \\
$xy$     & 7          & $x^7+2x^3y$           & 21                    & $x^7+2x^3y$ & 21  \\
$y^2$    & 8          &                       & 24                         &             & 24  \\
$x^3$    & 9          &                       & 3                          &             & 3   \\
$x^2y$    & 10         &                       & 22                         &             & 22  \\
$xy^2$   & 11         &                       & 25                         &             & 25  \\
$x^4$    & 12         &  $x^4$                & 12                         &  $y$        & 4   \\
$x^3y$   & 13         &                       & 15                         &             & 15  \\
$x^2y^2$ & 14         &                       & 26                         &             & 26  \\
$x^5$    & 15         & $x^7$                 & 21                         & $x^3y$      & 13  \\
\hline
\end{tabular}
\caption{Results of the modified algorithm for $q=3$.}\label{tab:tabla4}
\end{table}

\subsection{Computational results}
\label{reults}
We have computed the parameters of all Hermitian EAQECCs up to field size $16$ using our algorithm. These lists of parameters reveal that many of the resulting EAQECCs exceed the Gilbert-Varshamov bound~\cite{Galindo:2019}. Once $\mathcal{C}(m)$ is large enough to produce an EAQECC of non-zero dimension, all following values of $m$ seem to produce codes exceeding the bound as well until the resulting entanglement $c$ is `too small' compared to the code length. More specifically, we have verified that the parameters of the EAQECCs exceed the Gilbert-Varshamov bound \cite{Galindo:2019} for the values of $c$ specified in Table~\ref{tab:tabla6}. For lower values of $c$, it is sometimes possible to find codes that exceed the bound, but it will not be true for all codes with this entanglement.
\begin{table}[h!]
  \centering
    \begin{tabular}{ccc}
      $q$  & $n$ & $c$   \\ \hline
      2 & 8 & 0--3 \\
      3 & 27 & 1--16 \\
      4 & 64 &  3--45 \\
      5 & 125 & 4--96 \\
      7 & 343 & 10--288 \\
      8 & 512 & 9--441\\
      9 & 729 & 14--640 \\
      11 & 1331 & 38--1200 \\
      13 & 2197 & 51--2016\\
      16 & 4096 & 45--3825\\
      \hline
    \end{tabular}
    \caption{Hermitian EAQECCs exceding the GV bound.}
    \label{tab:tabla6}
\end{table}

\section*{Acknowledgements}
The authors thank the anonymous reviewers, whose helpful comments have led to a better manuscript.

\bibliographystyle{plainurl}

redpajama_set_name: RedPajamaArXiv


The Planet of the Apes / Green Lantern crossover has been a hit or miss affair in my opinion, suffering from some shaky art issues and a plot whose core bears too much of a resemblance to a recent Green Lanterns story arc. On the other side of that the series does have some interesting elements which keeps me engaged despite the issues I have with it. With that in mind the recently released finale truly represents my feelings about the series as a whole as issue six and I have a love / hate relationship.
On a positive note Justin Jordan's script takes a number of interesting turns since he's allowed to play with the toys in both universes and has no repercussions to face when he decides the story is well serve by breaking the toys. Issue five saw the death of Zira and by the end of the last issue the body count is upped substantially, and these are minor characters we're talking about here. Knowing that nothing's off limits makes the tension in the finale much more real – granted this is essentially an Elseworlds type story and nothing here is in canon – but in experiencing the story the heightened sense that anyone is eligible to be cannon fodder brings a degree of anxiety if you're willing to fully invest in the experience. I also liked some of the surprise plot twists that comprise the last few pages which leaves the door open for a sequel if BOOM! Studios is so inclined to go back for a second effort. To be honest my general disinterest in this story don't leaving me wanting more unless there was a better effort from the art side and a more original plot from the writer.
The downsides lie mainly in the art department this issue. I know that Barnaby Bagenda was highly praised for his artwork on the Omega Men mini-series for DC Comics, but I honestly think that he effort on this limited series has been pretty poor. The artwork is primitive looking and I guess from one perspective you could say it's symbolic of the plot, but I really found it more distracting than anything else. Unless this entire series was finished prior to publication you'd think that someone would've caught the white boot issue and corrected it. In my mind it shows a lack of reverence for the source material and I just can't find a way to excuse it. There are also some glaring continuity issues with this installment – things like Arisia shooting laser blasts out of her non-ring hand, and the gauntlets which protect the ring bearers from the Universal Ring disappearing and reappearing on characters arms. There was even a panel where Arisia's nose is missing! Constructs look dull and uninspired and when Cornelius goes all-powerful this issue the art team decides to convey it by covering him in multicolored dashes and lines. It's all just so hard on the eyes.
Planet of the Apes / Green Lantern #6 is both a triumph and a failure. While the plot manages to crawl out from under the spectre of the Phantom Ring with some surprising plot twists and risky character decisions it's bogged down by some of the worst art I've seen in a long time. I suspect that the enjoyment factor for readers will come from their interest in see the two franchise collide, but for my it's just not enough to make me bananas about it. Four out of ten lanterns.


Art Basel's official annual publication, "Art Basel | Year 45," will be available for purchase in the U.S. in May for $80. It depicts the 2014 shows in Basel, Miami Beach and Hong Kong through an A-Z format that spotlights the 528 participating galleries and offers varied perspective of the global artworld through interviews and essays. Customers from the U.S. can order through artbook|D.A.P. at orders@dapinc.com. For more information, visit artbasel.com/en/About-Art-Basel/Year-45.
Broward Cultural Division's Public Art & Design Program announced a call to artists for design and install of a site-specific artwork at the Dillard Green Space located at the northeast corner of Sunrise Blvd. and NW 27th Ave. in Fort Lauderdale. The application deadline is March 30 and guidelines are outlined here: broward.org/Arts/PublicArt/Documents. For more information, contact Dominique Denis, Public Art & Design Administrator, at 954-357-7236 or ddenis@broward.org.


Emily helps organisations to deliver business critical transformation programmes and to leverage technology to achieve their goals.
She has also worked in more strategic roles to identify how technology can simplify business processes, reduce costs and improve the customer experience.
This has allowed her to gain a holistic view of technology implementation from design through to launch.
Outside of work, Emily plays netball in a recreational league. She enjoys hiking and often spends her weekends exploring the countryside.


Remediation of air pollution can be costly and officially difficult. Many point out and local pollution control companies have discovered that encapsulating contaminated concrete is the most economical and long lasting solution for existing contaminants. 100% sturdy epoxy can seal impurities into the cement with levels of 10 to 100 mils solid. In effect, the encapsulation works as an insect caught in amber. Solids, gasses, and fluids are all captured inside a sound hard capsule. Contaminants are thus in a position to be left set up, with no danger to crews that might experienced to work near remove them.
100% epoxy has no odor, can go on as solid as needed, and comes in colors as well as clear. When making use of floor seals that can be 20 to 500 mils heavy as they repair and level imperfect floor surfaces, it is advisable to consider the height of the finish in addition to the length issues of normal roll-on applications. When normal water is poured on cup, it protrudes above the surface at the edges. Tugging a finger although water, one considers flattening at the accessibility and exit items due to its surface anxiety. When leveling floors, these same ramifications of the epoxy's surface tension must be handled. Quite simply "self-levelling," which is imperfect even for fluids like water, will not produce a smooth top layer of epoxy.
100% epoxy coatings do not reduce and will harden at any height it is left at after being catalyzed. But special methods are needed to be able to assure a flat surface while filling uneven anomalies in the substrate. Experience has motivated that by using a squeegee-like movement with an extremely short nap roller (3/16″) creates a pool of epoxy that may be pulled in your direction to drip into low areas while keeping the surface flat. Care must be used to apply an even coat at the surface, and too much rolling can cause the top to try to contour over the floor's imperfections.
A good way to obtain 100% epoxy floor coatings is Durall Industrial Floor of Minneapolis, MN. Durall provides customized kits specially designed to include every item necessary for the specific project of the customer. Durall gives these kits directly to the job site and provides factory-direct support during the application process. Web tourists can obtain free, job-specific quotations on materials or countrywide turnkey installations by completing a simple questionnaire. This full-service strategy allows anyone to enjoy the benefits associated with industrial strength surfaces.


Juda Store 608.934.5201 2009 CIH 2020 25' head, 3"cut, tracker .
Juda Store 608.934.5201 1020 -30' CIH head 3" – new cutter bar .
Brooklyn Store 608.455.2411 CIH 1020 25' head, 1.5" cut, Johnson stone bar .
Brooklyn Store 608.455.2411 CIH 1020 head, 17.5' , manual reel fore/aft .
Juda Store 608.934.5201 30' Head.
Juda 1987 CIH combine, IH engine -3400 hrs, rocktrap, serviced .
Juda 2010 CIH 6088 combine, duals, chopper, rock trap, GPS, yield monitor, tracker, hours -982 Eng, 730 Sep .
Juda Store 608.934.5201 Unverferth 30' head cart .


text:
COVID-19: Scotland's COVID restrictions tightened - here are the six new rules — Sky News
Nicola Sturgeon has tightened COVID restrictions in Scotland, laying out six changes which will come into effect on Saturday.
apple.news
Scotland tightening their rules from Saturday.
65th changing on the rules in the UK since march
Edit: Reading them, they don't seem overly strict. Unless that click and collect shopping for essentials is like when Wales stopped supermarkets from selling clothes etc. and doesn't mean only from essential shops
Think we have most of those in England already, don't we?
Technically I think the key one is the statutory guidance which basically says if you can work from home before there's no excuse now. I think many companies are arguing under efficiency this time around and telling employees to come in.
Honestly we need a far harder stay at home guidence with proper support for people who can't work from home beyond the the furlough scheme. But Amazon are still allowed to run as with many other firms offering delivery services or the manufacturing sector.
Some thing that get classified as key worker are a joke as well.
Aye that's fair,
I'm lucky in that my work never went back to the office in between lockdowns (my bosses did, but they live local and found it easier to work from there than at home) - but I've seen a lot of people saying their companies are forcing them to go in, saying that they made themselves COVID safe last summer (distancing, sanitizer etc) so it's fine.
It shouldn't be "Work from home where possible" it should be "Go to work if it's impossible to work from home".
Agree on that last part as well - was reading an article about how a shop chain, can't remember which, had a freezer sent to every one of their stores at the start of first lockdown, and a job lot of cheap frozen chips/chicken nuggets etc. so that they could stay open as essential shops.
I honestly don't think that there's a single shop closed in my closest town centre - everyone's found a loophole to stay open.
Coronavirus vaccinations to be available 24/7, Boris Johnson says in major U-turn
Move comes days after Downing Street said there was 'no clamour' for round-the-clock jabs
www.independent.co.uk
Good, should never have been shot down in the first place.
...how many u-turns is this government up to now?
Reactions: The_Blindside
187,645 shots (first and second combined) administered yesterday in England apparently. Getting there, but still edging the required run rate ever higher.
Edit: PsyberAttack (hacker / data analyst on Twitter) things that this is on pace to have everyone properly vaccinated in 18 months. Do we know how long the protection lasts? We could end up in a situation where those who received theirs early are needing another shot / shots while others are waiting for their first.
Largest single casualty list of over 1,500 deaths reported today taking the aggregate of the second wave beyond the first.
New cases are slowing a little and more are being vaccinated which is good, but it doesn't really feel right to be worrying about stuff like rugby in the midst of this.
Read most of the cases are people who contracted it over xmas. Wonder if it was mixing households or just the general melee of supermarket shopping in the festive period
For fear of overanalysing a throwaway comment, sport in general is a welcome distraction for many people at a difficult time. As long as it's not contributing to the problem or diverting resources away from making it better, I think it is important enough to worry about.
Reactions: Yulia
duncanb
Bench Player
Plus add new strain, a trifecta.
Read most of the cases are people who contracted it over xmas.
3 weeks is the length of time generally accepted to go from licking a plaguebearer to winding up in hospital, so it's likely that we're just starting to see the damage that Christmas did.
Wonder if it was mixing households or just the general melee of supermarket shopping in the festive period
There seems to be a widespread enthusiasm for finding a single cause to blame, I guess it's human nature to want to simplify things into easily understandable chunks, but in a problem that is potentially made worse by every single interaction we have, it's never going to be that simple. Surely both were a factor, as were many others. The other big one that comes to mind is the fact that the tier system that was in place at the time wasn't sufficiently robust (or well enforced) to protect against an exponential spread of the UK strain.
Let's also remember the Christmas melee got heavier because of the last minuite cancellations and people had to get their own food suddenly.
Government should absolutely be crucified for the Christmas bungling especially as it went against the scientific advice.
Given that is was the answer to every vaguely awkward question for so many months, the abandoning of the "we're following the science" mantra should be a much bigger scandal. This is the one U-turn that I can think of that has left us travelling in the wrong direction. Who knows, if the government had followed the science and gone for the circuit breaker when it was called for, maybe we wouldn't have grown our own variant. Still, it's better for the economy that we kept it going for a week or two in the autumn and shut it down for multiple months in winter!
The mad rush out of London just before the new tiers, as well (/other big cities, I imagine - but it was the London pictures that stick in your mind)
Train stations absolute crammed full of people - only takes a few asymptomatic people to give it to a hell of a lot more, in the train station, who then spread it around the country
RedruthRFC said:
Yeah I get that and everyone will see it in their own way depending on personal experience, how COVID has directly impacted them, age etc. For me sport's importance / interest has dulled but I accept that it will be the polar opposite for many.
My interest has dulled too, but for me it's more of a product of the lack of an affiliation with a GP team and lash up of a European tournament that have had this effect. I wish I didn't feel that way as I would welcome the distraction from everyday life. I was more responding to what I construed as you meaning that it was morally wrong to be worrying about rugby while the NHS is on its knees and corpses are piling up. I can understand anyone who feels that it's fatuous at the current time, but given that few of us can do anything to help matters (other than limiting personal contact and practicing good hygene / social distancing), I would argue that it's a positive thing for personal wellbeing and not something to feel guilty about.
Theres a reason why that mantra was quietly dropped I think it was about the time of eat out to help out. But scuttlebutt from people in the know basically says the government ditched following the science when they tried to supercharge the economy by exiting the first lockdown as they did.
Whitty and Vallance are going to write one hell of a book someday/give astounding evidence at an inquiry. I assume they are only still working for government out of obligation and knowing resigning means the next people might be even more yes men.
Reactions: RedruthRFC and Which Tyler
Reactions: RedruthRFC

redpajama_set_name: RedPajamaCommonCrawl


text:
education2u
Joe Biden blames Trump for violence on Capitol Hill as he picks Merrick Garland for attorney general
by education2uPosted on January 7, 2021
U.S. president-elect Joe Biden on Thursday denounced the rioters who stormed the U.S. Capitol as "domestic terrorists" and blamed President Donald Trump for the violence that has shaken the nation's capital and beyond.
The actions of Trump supporters who breached the security of Congress on Wednesday, said Biden, was "not dissent, was not disorder, was not protest. It was chaos."
In solemn tones, Biden said the steps Trump has taken to subvert the nation's democratic institutions throughout his presidency led directly to the mayhem in Washington.
Those who massed on Capitol Hill intending to disrupt a joint session of Congress that was certifying Biden's election victory over Trump "weren't protesters. Don't dare call them protesters. They were a riotous mob — insurrectionists, domestic terrorists. It's that basic," Biden said.
"In the past four years, we've had a president who's made his contempt for our democracy, our constitution, the rule of law clear in everything he has done," Biden said. "He unleashed an all-out assault on our institutions of our democracy from the outset. And yesterday was the culmination of that unrelenting attack."
WATCH | Biden lays the blame for Wednesday's attack on the Capitol on Donald Trump:
U.S. president-elect Joe Biden planted the blame for Wednesday's attack on the U.S. Capitol squarely on President Donald Trump, accusing him of "trying to use a mob to silence the voices of nearly 160 million Americans." 5:32
The mob of hundreds of Trump backers broke into the Capitol and roamed the halls looking for lawmakers, who were forced to halt their deliberations and evacuate to safety. The violent protesters were egged on by Trump himself, who has falsely contended that he lost the election due to voter fraud.
Trump's claims have been repeatedly dismissed in the courts, including the Supreme Court, and by state election officials from both parties, and even some in his own administration.
Merrick Garland for attorney general
Biden made his remarks as he introduced Merrick Garland as his pick for attorney general on Thursday along with three others he has selected for senior Justice Department positions to "restore the independence" of the Justice Department and faith in the rule of law.
Garland is the former chief judge of the U.S. Court of Appeals for the D.C. circuit. If confirmed by the Senate, which is likely, he would take over as the nation's top law enforcement official at a critical moment for the country and the agency.
He would inherit immediate challenges related to civil rights, an ongoing criminal tax investigation into Biden's son Hunter and calls from many Democrats to pursue criminal inquiries into Trump after he leaves office.
Appeals Court Judge Merrick Garland speaks in the Rose Garden of the White House in 2016 after being nominated to the U.S. Supreme Court by President Barack Obama. (Kevin Lamarque/Reuters)
Beyond the specific issues, he will be tasked with repairing the American people's broad distrust in the U.S. Justice Department, among other institutions of democracy undermined by Trump's turbulent presidency.
Biden vowed that Garland's loyalty would rest not with the president, but with the law and Constitution.
"You don't work for me," Biden charged as he introduced Garland.
Black Lives Matter treated very differently, says Biden
Biden used the event to also address what he said was blatant inequality in dealing with the mob on Capitol Hill Wednesday.
"No one can tell me that if it had been a group of Black Lives Matter protesting yesterday, they wouldn't have been treated very, very differently than the mob of thugs who stormed the Capitol."
WATCH | Biden calls out the inequality over how the rioters were treated Wednesday:
U.S. president-elect Joe Biden says it is clear that had the people who stormed the U.S. Capitol been members of Black Lives Matter, they would have been treated very differently by security forces. 1:35
Garland would inherit a Justice Department that has endured a tumultuous four years and abundant criticism from Democrats over what they see as the overpoliticization of law enforcement. The department is expected to dramatically change course under new leadership, including through a different approach to civil rights issues and national policing policies, especially after months of mass protests over the deaths of Black Americans at the hand of law enforcement.
Black and Latino advocates had wanted a Black attorney general or someone with a background in civil rights causes and criminal justice reform. Groups including the NAACP Legal Defense and Educational Fund had championed Garland's Supreme Court nomination, but the extent of support from minority groups for the attorney general job was not immediately clear.
Other justice posts announced
Biden introduced three others for senior Justice Department leadership posts, including Obama administration homeland security adviser Lisa Monaco as deputy attorney general and former Justice Department civil rights chief Vanita Gupta as associate attorney general, the No. 3 official. He also named an assistant attorney general for civil rights, Kristen Clarke, now the president of Lawyers' Committee for Civil Rights Under Law, an advocacy group.
Though Garland is a white man, the selection of Gupta and Clarke, two women with significant experience in civil rights, appeared designed to blunt any concerns and served as a signal that progressive causes would be prioritized in the new administration.
Posted in Arts & Entertainment, Business, Canada, Health, Politics, Top Stories, World
Published by education2u
View all posts by education2u
Prev Elon Musk passes Jeff Bezos to become world's richest person on Bloomberg list
Next U.S. Capitol police chief defends response to storming of the complex
U.S.-bound migrant caravan hunkers down on highway after Guatemala crackdown


text:
Home » Nation
Wage tiff triggers fatal bombing
Source: Xinhua | April 3, 2009, Friday | Print Edition
A DISGRUNTLED former employee killed himself and injured two people yesterday when he blew himself up in an office building in Urumqi, capital of northwest China's Xinjiang Uygur Autonomous Region, police said.
The man, identified as Han Wushun, 42, originally from southwest China's Sichuan Province, arrived at the office building in downtown Urumqi at around 1pm with a black bag.
Han threatened the office workers and demanded that the company, Xinjiang Beixin Road and Bridge Construction Co Ltd, give him 4,500 yuan (US$662) in unpaid wages, police said. He had sued the company for the money but lost the case last year.
Han walked into the first floor of the building and talked to an office worker, saying, "Society is unfair. I don't want to live any more."
Three office workers were trapped in the room. One escaped while Han was not watching him and alerted other occupants of the building to evacuate, police said.
The two trapped men, Yang Tao and Lu Jianzhong, tried to reason with Han in the office, police said. They rushed out of the room when Han tried to connect the wires of the bomb.
Yang sustained injuries to his face and Lu to his hands. Yang was sent to a hospital, but his condition was not life-threatening, police said.


Kindergarten writing paper with lines pdf what is an argumentative research paper research paper on language and culture definition aids essay pdf creative writing courses bible verse about problem solving how to write a methods section of a research paper format assigned risk planet essay library in english example of review of related literature thesis format free teddy bear writing paper model business plan of an npc critical thinking rubrics for discussions critical essay thesis statement examples amcas essay questions. Brian doyle essays driving with my mom examples of analysis essays mla style topics for research papers in education can you start an essay with a question writing short essays for scholarships write an assignment on cancer introduction sample of argumentative essay introduction example university of east anglia creative writing distance learning average amount of homework in high school calendar medical essay lab, sample action research papers apa style how to write an executive summary for an assignment example research proposal methodology term paper structure apa style sample example of a synthesis essay on comics assignment desk editor jobs health insurance research papers homework for toddler printable template does god exist philosophy essay term papers for sale online canada ethical issues essay topics. Ways to improve critical thinking video.
Research paper on language and culture definition, what is an argumentative research paper quoting statements from websites in essays sample essays for scholarships focusing on accomplishments and biography business development performance plan essays on service learning best controversial topics for a research paper pdf consulting firm business plans, assignment of claims far clause free business plan pro serial number graphic organizer for the argument essay dissertation topics in supply chain management job what is a business plan definition literature review on microfinance essay on why i want to be a nurse educator.
Review of literature outline free online small business plan template assignment answers in english mobile app development business plan essay on character traits commercial lease assignment requirements meaning of argumentative essay and example opinion editorial essay example english essay checkers sample essays for scholarships focusing on accomplishments and biography business plan development services san diego research paper topics for graphic design essays on health care reform a working business plan template calligraphy writing on antique paperGroup games for problem solving business development plan template samples admission essay apa format 6th grade essay prompts dna profiling research paper literature review on unemployment good business plans skills of critical thinking how to write a research paper apa style outline template solving business problems using a calculator, an example of a business plan for a new business loan, art of problem solving classes.
Help with essay writing for scholarships. How to find the literature review of an article tobacco essay in kannada language. Assignment 2 informative speech good topic to write a research paper on cloth bag making business plan, writing thesis and dissertation proposals free meaning of assignment in law consumerism essay thesis tobacco essay in kannada language gastro pub business plan problem solving physics grade 8 examples mla essay header homework tips for middle school students meaning of argumentative essay and example, argumentative essay writing steps define a thesis in an essay soccer homework pdf research topic proposal essay sample child day care business plan business plans for nail salons essay on censorship in the united states sample of business continuity plan manual business plan template free download ppt why is the introduction to the constitution essay simple business plan template microsoft word solving for t in half life problems finite math homework helper free business plan template word doc kids homework help on neil armstrong how to cite a research paper mla style is buying research papers legal essays on changes in life what is the 8 step problem solving method what should be on a title page of an essay 5 year business financial plan template solving initial value problems pdf market analysis sample business plan format ideas of a business plan time schedule for research proposal. Dissertation template gcu.
Parent homework meme essay about reading books business plan event space comprehension and critical thinking grade 8 best music to write a paper to how to write a essay about a poem business plan event space problem solving with subtraction for kindergarten homework motivation songs state farm business plan proposal business plan for realtors 2017, studying creative writing examples environmental science topics for research papers business plan points legal research proposal sample template research papers on assisted suicide title assignment 2 informative speech literature review paper example example of research paper powerpoint presentation how to submit an assignment on moodle unhcr during the cold war essay make an essay appear longer mos 25p assignment research paper about business management decision-making and problem solving skills in management examples of topic proposal for research papers examples working at a golf course research paper on capital structure pdf tips on how to write an essay theme essay outline template how to find the literature review of an article patent assignment agreement sample related coursework for business. Help with essay writing for scholarships business restructuring plans writing thesis and dissertation proposals free ideal job essay short theme essay outline template.
Bible verse about problem solving creative nonfiction essay prompts, sample essays for graduate school counseling proposal essay topics research paper how to write medical research proposal how to write a marketing research proposal.
Boat builder business plan, solving speed problems mercantile law masters dissertation topics easy kids homework sheets, smoke shop business plan template sample gantt chart for business plan examples examples of critical thinking in the workplace science students multitasking essay popular essay topics for college students website that will write a paper for you video i need motivation to do my homework.
Example of abstract in research paper objectives financial holding company business plan computer architecture homework solutions login research rationale paper comedy of manners essay for grad school examples creative writing yorkies data mining research papers 2018 download free research papers on mobile computing how to make a persuasive essay good tobacco essay in kannada language. What types of essays are there american revolution topics for research paper.
Middle school essay contest 2020 problem solving with angles best research papers. Possible cbest essay topics, what are some good research paper topics best assignment model movie business plan torrent creative writing courses decision-making and problem solving skills in management how to critique a research paper powerpoint presentation microsoft excel business plan template pdf how to use transitions in an essay help with essay writing for scholarships. Free business plan template word doc developing a research question paper car wash business plan pdf government accounting term paper assignment theme essay outline template good criminal justice research paper topics creative writing job opportunities ideas business development compensation plans 2018 website that will write a paper for you video how to solve basic arithmetic problems sample problem solving in geometry research paper on ruth chapter 2. Montessori writing paper writing a rhetorical analysis essay sample college entrance essay format drunk driving essay subtopics social problem solving activities for teens science research paper template what is essay formatting on microsoft.
Purdue owl paraphrase assignment family narrative essay examples essay on communication example the dog ate my homework series 4 sample problem solving in geometry.


For windows, just download, extract, and continue to use the new extracted files. We are currently working on building an auto installer to the Windows wallet which we will release in the near future. This new feature should making installations on windows a lot easier.
For Linux, please visit https://github.com/VegasCoin/vegascoin for more information on how to compile the wallet. Use git to clone https://github.com/VegasCoin/vegascoin.git. If you already have the repository cloned, all you need to do is checkout the production branch ("git checkout production" on the command line), and git pull to pull in the latest updates. Then you can follow compile instructions to build the wallet. We are currently also looking into building Linux packages to make installation of the Vegas Coin on Linux a breeze as well.
As a small spoiler: We are also preparing an Android wallet, with work on that starting in the upcoming weeks.