text:
\section{Introduction}

Recently a new kind of unique decoding algorithm of algebraic geometry codes appeared \cite{kwankyu2011}. The algorithm decodes the primal AG code that consists of codewords obtained by evaluation of functions at rational points of an algebraic curve, unlike the classical syndrome decoding algorithm that decodes the dual code. Based on Gr\"obner bases of modules over a univariate polynomial ring, the algorithm has a regular data and control structure that is suitable for parallel hardware implementation, like K\"otter's algorithm for the syndrome decoding \cite{koetter1998}.

In this paper, we reformulate the previous algorithm, using the theory of Gr\"obner bases of modules on the coordinate ring of the base curve. This approach eliminates the technical complexity of the previous algorithm in a large degree, and results in a conceptually clean description of the algorithm which would contribute for better understanding the majority voting procedure, which plays a central role in the interpolation-based unique decoding. The new algorithm resembles the Berlekamp-Massey-Sakata algorithm for the syndrome decoding \cite{sakata1995}.

In Section \ref{mckwd}, we review the theory of the Gr\"obner bases of modules over the coordinate rings of algebraic curves, and outline the interpolation decoding algorithm based on it.  The algorithm operates by iterating two core steps, the Gr\"obner basis computation step and the message guessing step. Sections \ref{jksadde} and \ref{kleflso} are devoted for each step. In Section \ref{ckdwdd}, we demonstrate the algorithm with Hermitian codes. In the remainder of this introduction, we briefly review basic facts about AG codes. Like the previous algorithm in \cite{kwankyu2011} and the BMS decoding algorithm, the new algorithm is formulated for the AG codes from the Miura-Kamiya curves \cite{miura2007}, which include Hermitian curves as prominent special cases.

A Miura-Kamiya curve $X$ is an irreducible plane curve defined by the equation
\[
	y^a+\sum_{ai+bj<ab}c_{i,j}x^iy^j+dx^b=0
\]
over a field $\mathbb{F}$ with $\gcd(a,b)=1$ and $0\neq d\in\mathbb{F}$.  It is well known that $X$ has a unique point $P_\infty$ at infinity and has a unique valuation $v_{P_\infty}$ associated with it. Let $\delta(f)=-v_{P_\infty}(f)$ for $f$ in the coordinate ring $R$ of $X$. Then $\delta(x)=a$ and $\delta(y)=b$. By the equation of the curve, a function in the coordinate ring $R=\mathbb{F}[x,y]$ can be written as a unique $\mathbb{F}$-linear combination of monomials $x^iy^j$ with $i\ge 0$ and $0\le j<a$, which we call monomials of $R$. The numerical semigroup of $R$ at $P_\infty$,
\[
	\begin{split}
	S&=\set{\delta(f)\mid f\in R}=\set{\delta(x^iy^j)\mid i\ge 0,0\le j<a}\\
	&=\set{ai+bj\mid i\ge 0,0\le j<a}=\ideal{a,b}
	\end{split}
\]
is a subset of the Weierstrass semigroup at $P_\infty$. As $\gcd(a,b)=1$, there is an integer $b'$ such that $b'b\equiv 1\pmod{a}$. If $s=ai+bj$ is a nongap, then $b's\bmod a=j$, $(s-bj)/a=i$, and therefore $i$ and $j$ are uniquely determined. Hence the monomials of $R$ are in one-to-one correspondence with nongaps in $S$. For a nongap $s$, let $\phi_s$ be the unique monomial with $\delta(\phi_s)=s$.

Let $\mathcal{P}=\set{P_1,P_2,\dots,P_n}$ be a set of nonsingular rational points of $X$. The evaluation $\mathrm{ev}$ from $R$ to the Hamming space $\mathbb{F}^n$ defined by
\[
	\phi\mapsto(\phi(P_1),\phi(P_2),\dots,\phi(P_n))
\]
is a linear map over $\mathbb{F}$. Let $u$ be a fixed positive integer less than $n$ and define
\[
	L_u=\set{f\in R\mid \delta(f)\le u}=\langle\phi_s\mid s\in S, s\le u\rangle,
\]
where brackets denote the linear span over $\mathbb{F}$. Then the AG code $C_u$ is defined as the image of $L_u$ under $\mathrm{ev}$. As $u<n$, the evaluation is one-to-one on $L_u$. Therefore the dimension of the linear code $C_u$ equals $\dim_\mathbb{F} L_u=|\set{s\in S\mid s\le u}|$.

\section{Interpolation decoding}\label{mckwd}

We assume a codeword $c$ in $C_u$ is sent through a noisy communication channel and $v\in\mathbb{F}^n$ is the vector  received from the channel. Let $v=c+e$ with the error vector $e$. Then $c=\mathrm{ev}(\mu)$ for a unique
\[
	\mu=\sum_{s\in S,s\le u}\omega_s\phi_s\in L_u,\qquad \omega_s\in\mathbb{F}
\]
We assume encoding by evaluation, and the vector $(\omega_s\mid s\in S,s\le u)$ is the message encoded into the codeword $c$. The decoding problem is essentially to find $\omega_s$ for all nongap $s\le u$ from the given $v$.

For $s\ge u$, let $v^{(s)}=v$, $c^{(s)}=c$, and $\mu^{(s)}=\mu$. For nongap $s\le u$, let
\[
\begin{aligned}
	\mu^{(s-1)}&=\mu^{(s)}-\omega_s\phi_s,\\
	c^{(s-1)}&=c^{(s)}-\mathrm{ev}(\omega_s\phi_s),\\
	v^{(s-1)}&=v^{(s)}-\mathrm{ev}(\omega_s\phi_s),
\end{aligned}
\]
and for gap $s\le u$, let $v^{(s-1)}=v^{(s)}$, $c^{(s-1)}=c^{(s)}$, and $\mu^{(s-1)}=\mu^{(s)}$. Note that
\[
	\mu^{(s)}\in L_s,\quad c^{(s)}=\mathrm{ev}(\mu^{(s)})\in C_s,\quad v^{(s)}=c^{(s)}+e
\]
for all $s$. Hence we can find $\omega_s$ iteratively.

A polynomial in $R[z]$ defines a function on the product surface of $X$ and the line $\mathbb{A}_\mathbb{F}^1$, and can be evaluated at a point $(P,\alpha)$ with $P\in X,\alpha\in\mathbb{F}$. Hence we can define the \emph{interpolation module}
\[
	I_v=\set{ f \in Rz\oplus R\mid f(P_i,v_i)=0, 1\le i\le n}
\]
for $v$ and similarly for $v^{(s)}$. These interpolation modules are indeed modules over $R$, and finite-dimensional vector space over $\mathbb{F}$. Note that
\begin{equation}\label{ckskw}
	I_v=R(z-h_v)+J
\end{equation}
where
\[
	J=\bigcap_{1\le i\le n}\mathfrak{m}_i,
	\qquad
	\mathrm{ev}(h_v)=v,
\]
and $\mathfrak{m}_i=\ideal{x-\alpha_i,y-\beta_i}$ is the maximal ideal of $R$ associated with $P_i=(\alpha_i,\beta_i)$. Recall that by Lagrange interpolation, $h_v$ can be computed fast from $v$. We will see that the key to find $\omega_s$ is the Gr\"obner basis of $I_{v^{(s)}}$ with respect to a monomial order $>_s$, which is defined in the following.

Let $s$ be an integer. The monomial $x^iy^jz^k$  of $R[z]$ is given the weight $\delta(x^iy^j)+sk$. In particular, the weighted degrees of the monomials $x^iy^jz$ and $x^iy^j$ of $Rz\oplus R$ are $ai+bj+s$ and $ai+bj$, respectively. The monomial order $>_s$ on $Rz\oplus R$ orders the monomials by their weighted degrees, and breaks the tie with higher $z$-degree. For $f$ in $Rz\oplus R$, the notations $\mathrm{lt}_s(f)$, $\mathrm{lm}_s(f)$, and $\mathrm{lc}_s(f)$ denote the leading term, the leading monomial, and the leading coefficient of $f$, respectively, with respect to $>_s$. As $f=f^Uz+f^D$ with unique $f^U,f^D\in R$\footnote{The superscripts $U$ and $D$ may be read ``upstairs'' and ``downstairs'', respectively (with $z$ being the staircase).}, note that
\[
	\mathrm{lm}_s(f)\in Rz \iff \delta(f^U)+s\ge \delta(f^D),
\]
where equality holds if and only if $\mathrm{lm}_s(f)\in Rz$ and $\mathrm{lm}_{s-1}(f)\in R$.

Now let $M$ be a submodule of $Rz\oplus R$. A subset $B$ of $M$ is called a \emph{Gr\"obner basis} with respect to $>_s$ if the leading term of every element of $M$ is divided by the leading term of some element of $B$. We will write
 \[
	B=\set{G_i,F_j}
\]
with $i$ and $j$ in some implicit index sets, where the leading term of $G_i$ is in $R$ while that of $F_j$ is in $Rz$. The \emph{sigma set} $\Sigma_s$ or $\Sigma_s(M)$ of $M$ is the set of all leading monomials of polynomials in $M$ with respect to $>_s$. The \emph{delta set} $\Delta_s$ or $\Delta_s(M)$ of $M$ is the complement of $\Sigma_s$ in the set of all monomials of $Rz\oplus R$. We note that
\[
	\begin{aligned}
	\Delta_s&=(Rz\cap\Delta_s)\sqcup(R\cap\Delta_s),\\
	\Sigma_s&=(Rz\cap\Sigma_s)\sqcup(R\cap\Sigma_s).
	\end{aligned}
\]
where $\sqcup$ denotes disjoint union. For the case that $M$ is an ideal of $R$, we may omit the superfluous $s$ from the above notations, and denote $>_s$ simply by $>$ in particular. Note that if $\mathrm{lm}_s(f)\in Rz$, then $\mathrm{lm}_s(f)=\mathrm{lm}(f^U)z$, and if $\mathrm{lm}_s(f)\in R$, then $\mathrm{lm}_s(f)=\mathrm{lm}(f^D)$. It is easy to see by the definition of Gr\"obner bases that
\[
	\begin{split}
	\dim_\mathbb{F}(Rz\oplus R/M)&=|\Delta_s|=|\Delta_s\cap Rz|+|\Delta_s\cap R|\\
	&=|\Delta(\set{F_j^U})|+|\Delta(\set{G_i^D})|,
	\end{split}
\]
where $\Sigma(T)$, $\Delta(T)$ with a set of polynomials in $R$ have natural definitions.

As $J$ is an ideal of $R$, it has a Gr\"obner basis $\set{\eta_i}$ with respect to $>$, and
\begin{equation}\label{camdzs}
	\dim_\mathbb{F} R/J=|\Delta(J)|=|\Delta(\set{\eta_i})|=n
\end{equation}
since $J$ is the ideal associated with the sum of $n$ rational points on $X$. By \eqref{ckskw}, we see that $\dim_\mathbb{F}(Rz\oplus R/I_v)=\dim_\mathbb{F}(R/J)=n$. Let $N=\delta(h_v)$. The set $\set{\eta_i}\cup\set{z-h_v}$ is then a Gr\"obner basis of $I_v$ with respect to $>_N$. Let us denote a Gr\"obner basis of $I_{v^{(s)}}$ with respect to $>_s$ by $B^{(s)}=\set{G_i,F_j}$. Observe that if $s$ is a nongap $\le u$, then the set $\tilde{B}=\set{G_{i}(z+\omega_s\phi_s),F_j(z+\omega_s\phi_s)}$ is still a Gr\"obner basis of $I_{v^{(s-1)}}$ with respect to $>_s$, but not with respect to $>_{s-1}$ in general. These observations lead to the following interpolation decoding algorithm.

\begin{IDT}
Let $v$ be the received vector.
\begin{description}
\item[Initialize]
Compute $h_v$. Let $B^{(N)}=\set{\eta_i}\cup\set{z-h_v}$ where $N=\delta(h_v)$.
\item[Main]
Repeat the following for $s$ from $N$ to $0$.
\begin{description}
\item[M1]
If $s$ is a nongap $\le u$, then make a guess $w^{(s)}$ for $\omega_s$, and let $\tilde{B}=\set{G_{i}(z+w^{(s)}\phi_s),F_j(z+w^{(s)}\phi_s)}$. Otherwise, let $\tilde{B}=B^{(s)}$.
\item[M2]
Compute $B^{(s-1)}$ from $\tilde{B}$.
\end{description}
\item[Finalize]
Output $(w^{(s)}\mid \text{nongap $s\le u$})$.
\end{description}
\end{IDT}

In the next section, we will elaborate on the step \textbf{M2}. The results in the section will lay a foundation for Section \ref{kleflso}, in which we give details of the main steps \textbf{M1} and \textbf{M2}.

\section{Gr\"obner basis computation}\label{jksadde}

First we review the concept of the \emph{lcm}, least common multiple, for the monomials of $R$. For two monomials $\phi_s$ and $\phi_t$, we say $\phi_s$ \emph{divides} $\phi_t$ if there exists a unique monomial $\lambda$ such that
\[
	\delta(\phi_t-\lambda\phi_s)<\delta(\phi_t).
\]
The unique monomial $\lambda$ will be denoted by the quotient $\phi_t/\phi_s$. Note that $\phi_s$ divides $\phi_t$ if and only if $t-s$ is a nongap, and in this case, actually $\lambda=\phi_{t-s}$. We will also simply say $s$ \emph{divides} $t$ if $t-s$ is a nongap.

\begin{proposition}
Let $s$ and $t$ be nongaps that do not divide each other. Then there are unique nongaps $l_1$ and $l_2$ such that
$l_1$ and $l_2$ are both divisible by $s$ and $t$, and if a nongap $c$ is divisible by $s$ and $t$, then $l_1$ or $l_2$ divides $c$.
\end{proposition}

\begin{proof}
Let $s=as_1+bs_2$ and $t=at_1+bt_2$. Without loss of generality, we may assume $s_1<t_1$ and $s_2>t_2$. Since $s$ divides $a(s_1+b)$, we also have $t_1<s_1+b$.

Now let $l_1=at_1+bs_2$ and $l_2=a(s_1+b)+bt_2$. It is easily verified that $l_1$ and $l_2$ are divisible by $s$ and $t$. Suppose a nongap $c=ac_1+bc_2$ is divisible by $s$ and $t$. Then $c-s=a(c_1-s_1)+b(c_2-s_2)$ is a nongap as $s$ divides $c$. Note that
\[
	b'(c-s)\bmod a=\begin{cases}
	c_2-s_2 & \text{if $c_2\ge s_2$,}\\
	c_2-s_2+a & \text{if $c_2<s_2$.}
	\end{cases}
\]
Therefore if $c_2\ge s_2$, then $c_1\ge s_1$ while if $c_2<s_2$, then $c_1\ge s_1+b$. Similarly, as $t$ divides $c$, if $c_2\ge t_2$, then $c_1\ge t_1$ while if $c_2<t_2$, then $c_1\ge t_1+b$. So in any case, we have at least $c_1\ge t_1$. Now let us check that $c$ is divisible either by $l_1$ or $l_2$. Assume $l_1$ does not divide $c$. Then $c_1<t_1+b$ as $c_1\ge t_1+b$ contradicts our assumption. Therefore $c_2\ge t_2$. If $c_1<s_1+b$, then $c_2\ge s_2$, which also contradicts the assumption. Therefore $c_1\ge s_1+b$. Then $l_2$ divides $c$.
\end{proof}

We will call $\phi_{l_1}$ and $\phi_{l_2}$ the lcms of $\phi_s$ and $\phi_t$. In the case when $\phi_s$ divides $\phi_t$, we will call $\phi_t$ the lcm of $\phi_s$ and $\phi_t$.

Let $B=\set{G_i,F_j}$ be a Gr\"obner basis of a submodule $M$ of $Rz\oplus R$ with respect to $>_s$. We want to compute a Gr\"obner basis of the same module $M$ with respect to $>_{s-1}$ from $B$. Note that while $\mathrm{lm}_{s-1}(G_i)=\mathrm{lm}_s(G_i)\in R$, we may have either $\mathrm{lm}_{s-1}(F_j)=\mathrm{lm}_s(F_j)\in Rz$ or $\mathrm{lm}_{s-1}(F_j)\in R$. Let $\Sigma_s$ and $\Delta_s$ denote the sigma set and the delta set of $M$ with respect to $>_s$, respectively. Observe that
\[
	\begin{gathered}
	Rz\cap\Sigma_{s-1} \subset Rz\cap\Sigma_s,\quad
	R\cap\Sigma_{s-1} \supset R\cap\Sigma_s,\\
	Rz\cap\Delta_{s-1} \supset Rz\cap\Delta_s,\quad
	R\cap\Delta_{s-1} \subset R\cap\Delta_s.
	\end{gathered}
\]
For those $j$ such that $\mathrm{lm}_{s-1}(F_j)=\mathrm{lm}_s(F_j)\in Rz$, define
\[
	\mathrm{spoly}(F_j)=\set{F_j}.
\]
If $\mathrm{lm}_{s-1}(F_j)\in R\cap\Sigma_s$, then there is an $i$ such that $\mathrm{lm}_s(G_i)|\mathrm{lm}_{s-1}(F_j)$, and then, with one such $i$, define
\[
	\mathrm{spoly}(F_j)=\set{\frac{1}{\mathrm{lc}_{s-1}(F_j)}F_j-\frac{\mathrm{lm}_{s-1}(F_j)}{\mathrm{lt}_s(G_i)}G_i}.
\]
Finally, if $\mathrm{lm}_{s-1}(F_j)\in R\cap\Delta_s$, then define
\[
	\mathrm{spoly}(F_j)=\set{\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i\mid\text{$\psi$ is an lcm of $\mathrm{lm}_{s-1}(F_j)$ and $\mathrm{lm}_s(G_i)$}}.
\]

\begin{proposition}\label{kcmald}
For every $f\in\mathrm{spoly}(F_j)$, $\mathrm{lm}_{s-1}(f)$ is in $Rz$.
\end{proposition}

\begin{proof}
Recall that $\mathrm{lm}_s(F_j)\in Rz$. Suppose $\mathrm{lm}_{s-1}(F_j)\in R$, and let $\psi$ be an lcm of $\mathrm{lm}_{s-1}(F_j)$ and $\mathrm{lm}_s(G_i)$ for any $i$. Then
\[
	\begin{gathered}
	\delta(\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j^U)=\delta(\psi)-\delta(F_j^D)+\delta(F_j^U)=\delta(\psi)-s,\\
	\delta(\frac{\psi}{\mathrm{lt}_s(G_i)}G_i^U)=\delta(\psi)-\delta(G_i^D)+\delta(G_i^U)<\delta(\psi)-s.
	\end{gathered}
\]
Therefore
\[
	\delta((\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i)^U)=\delta(\psi)-s.
\]
On the other hand,
\[
	\begin{gathered}
	\delta(\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j^D)=\delta(\psi)-\delta(F_j^D)+\delta(F_j^D)=\delta(\psi),\\
	\delta(\frac{\psi}{\mathrm{lt}_s(G_i)}G_i^D)=\delta(\psi)-\delta(G_i^D)+\delta(G_i^D)=\delta(\psi).
	\end{gathered}
\]
As the monic terms cancel each other, we have
\[
	\delta((\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i)^D)<\delta(\psi).
\]
Therefore
\[
	\delta((\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i)^U)+s-1\ge\delta((\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i)^D),
\]
and hence
\begin{equation}\label{jfjwa}
	\mathrm{lm}_{s-1}(\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i)=\mathrm{lm}(\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j^U)z\in Rz.
\end{equation}
For the case when $\mathrm{lm}_{s-1}(F_j)\in R\cap\Sigma_s$, notice that $\mathrm{lm}_{s-1}(F_j)$ is the lcm.
\end{proof}

\begin{proposition}\label{qhkxa}
A monomial $\phi$ is in $R\cap\Sigma_{s-1}$ if and only if there exists an $i$ such that $\mathrm{lm}_{s-1}(G_i)|\phi$ or there exists a $j$ such that $\mathrm{lm}_{s-1}(F_j)\in R\cap\Delta_s$ and $\mathrm{lm}_{s-1}(F_j)|\phi$.
\end{proposition}

\begin{proof}
Both $\mathrm{lm}_{s-1}(G_i)|\phi$ and $\mathrm{lm}_{s-1}(F_j)|\phi$ imply $\phi\in R\cap\Sigma_{s-1}$. Let us show the converse. If $\phi\in R\cap\Sigma_s$, then $\mathrm{lm}_{s}(G_i)|\phi$ for some $i$, and therefore $\mathrm{lm}_{s-1}(G_i)|\phi$.
As $R\cap\Sigma_{s-1}\supset R\cap \Sigma_s$, it remains to consider the case when $\phi\in R\cap (\Sigma_{s-1}\backslash\Sigma_s)$.

Suppose $f\in M$ is such that $\phi=\mathrm{lm}_{s-1}(f)\in R\cap (\Sigma_{s-1}\backslash\Sigma_s)$. Since $\phi\notin R\cap\Sigma_s$, we must have $\mathrm{lm}_s(f)\in Rz$, and hence
\[
	\delta(f^U)+s=\delta(f^D)=\delta(\phi).
\]
Then $\mathrm{lm}_s(F_j)|\mathrm{lm}_s(f)$ for some $j$. As $\mathrm{lm}_s(F_j)\in Rz$, we have $\delta(F_j^U)+s \ge \delta(F_j^D)$, where actually equality holds as we will show now. Assume the contrary, that is,
\[
	\delta(F_j^U)+s>\delta(F_j^D).
\]
Then
\[
	\begin{gathered}
	\delta(\frac{\mathrm{lt}_s(f)}{\mathrm{lt}_s(F_j)}F_j^D)=\delta(f^U)-\delta(F_j^U)+\delta(F_j^D)<\delta(f^U)+s=\delta(f^D),\\
	\delta(\frac{\mathrm{lt}_s(f)}{\mathrm{lt}_s(F_j)}F_j^U)=\delta(f^U)-\delta(F_j^U)+\delta(F_j^U)=\delta(f^U).
	\end{gathered}
\]
These imply
\[
	\mathrm{lm}_s(f-\frac{\mathrm{lt}_s(f)}{\mathrm{lt}_s(F_j)}F)=\mathrm{lm}(f^D)=\mathrm{lm}_{s-1}(f)=\phi,
\]
contradictory to the assumption $\phi\notin R\cap\Sigma_s$. Hence $\delta(F_j^U)+s=\delta(F_j^D)$, and
\[
	\delta(\frac{\mathrm{lm}_s(f)}{\mathrm{lm}_s(F_j)}\mathrm{lm}_{s-1}(F_j))
	=\delta(f^U)-\delta(F_j^U)+\delta(F_j^D)=\delta(f^U)+s=\delta(\phi).
\]
Therefore $\mathrm{lm}_{s-1}(F_j)|\phi$, and $\mathrm{lm}_{s-1}(F_j)\in R\cap\Delta_s$.
\end{proof}

\begin{proposition}
A monomial $\phi$ is in $Rz\cap\Sigma_{s-1}$ if and only if there exists a $j$ such that $\mathrm{lm}_{s-1}(f)|\phi$ for some $f\in\mathrm{spoly}(F_j)$.
\end{proposition}

\begin{proof}
By Proposition \ref{kcmald}, the converse is clear. Let us assume $\phi\in Rz\cap\Sigma_{s-1}$. Suppose $\phi=\mathrm{lm}_{s-1}(f)$ for some $f\in M$. Then $\phi=\mathrm{lm}_s(f)$, and there exists some $j$ such that $\mathrm{lm}_s(F_j)|\phi$.
If $\mathrm{lm}_{s-1}(F_j)\in Rz$, then $F_j\in\mathrm{spoly}(F_j)$ and $\mathrm{lm}_{s-1}(F_j)=\mathrm{lm}_s(F_j)|\phi$.

Suppose $\mathrm{lm}_{s-1}(F_j)\in R\cap\Sigma_s$. Then there is an $i$ such that $\mathrm{lm}_s(G_i)|\mathrm{lm}_{s-1}(F_j)$ and
\[
	\frac{1}{\mathrm{lc}_{s-1}(F_j)}F_j-\frac{\mathrm{lm}_{s-1}(F_j)}{\mathrm{lt}_s(G_i)}G_i\in\mathrm{spoly}(F_j)
\]
and by \eqref{jfjwa},
\[
	\mathrm{lm}_{s-1}(\frac{1}{\mathrm{lc}_{s-1}(F_j)}F_j-\frac{\mathrm{lm}_{s-1}(F_j)}{\mathrm{lt}_s(G_i)}G_i)=\mathrm{lm}_s(F_j)|\phi.
\]

Suppose $\mathrm{lm}_{s-1}(F_j)\in R\cap\Delta_s$. Note that
\[
	\delta(f^U)+s>\delta(f^D),\quad \delta(F_j^U)+s=\delta(F_j^D),
\]
and hence
\[
	\begin{gathered}
	\delta(\frac{\mathrm{lt}_s(f)}{\mathrm{lt}_s(F_j)}F_j^U)=\delta(f^U)-\delta(F_j^U)+\delta(F_j^U)=\delta(f^U),\\
	\delta(\frac{\mathrm{lt}_s(f)}{\mathrm{lt}_s(F_j)}F_j^D)=\delta(f^U)-\delta(F_j^U)+\delta(F_j^D)=\delta(f^U)+s>\delta(f^D).
	\end{gathered}
\]
Thus we see that
\[
	\mathrm{lm}_s(f-\frac{\mathrm{lt}_s(f)}{\mathrm{lt}_s(F_j)}F_j)=\frac{\mathrm{lm}_s(f)}{\mathrm{lm}_s(F_j)}\mathrm{lm}_{s-1}(F_j)\in R
\]
and hence there is an $i$ such that
\[
	\mathrm{lm}_s(G_i)|\frac{\mathrm{lm}_s(f)}{\mathrm{lm}_s(F_j)}\mathrm{lm}_{s-1}(F_j).
\]
Now there is an lcm $\psi$ of $\mathrm{lm}_{s-1}(F_j)$ and $\mathrm{lm}_s(G_i)$ such that
\begin{equation}\label{dkwda}
	\psi | \frac{\mathrm{lm}_s(f)}{\mathrm{lm}_s(F_j)}\mathrm{lm}_{s-1}(F_j),
\end{equation}
and
\[
	\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i\in\mathrm{spoly}(F_j).
\]
By \eqref{jfjwa},
\[
	\mathrm{lm}_{s-1}(\frac{\psi}{\mathrm{lt}_{s-1}(F_j)}F_j-\frac{\psi}{\mathrm{lt}_s(G_i)}G_i)
	=\frac{\psi}{\mathrm{lm}_{s-1}(F_j)}\mathrm{lm}_s(F_j)\in Rz
\]
and finally from \eqref{dkwda},
\[
	\frac{\psi}{\mathrm{lm}_{s-1}(F_j)}\mathrm{lm}_s(F_j)|\mathrm{lm}_s(f)=\phi.
\]
\end{proof}

Combining the above results, we see that the set
\[
\set{G_i,F_j\mid \mathrm{lm}_{s-1}(F_j)\in R\cap\Delta_s}\cup\bigcup_{j}\mathrm{spoly}(F_j)
\]
is a Gr\"obner basis of $M$ with respect to $>_{s-1}$. In general, the Gr\"obner basis may contain more polynomials than necessary. Indeed, we can reduce each set in the union by removing polynomials whose leading term is divisible by that of other polynomial in the same set. We will denote the \emph{reduced} Gr\"obner basis of $M$ with respect to $>_{s-1}$ by
 \[
\set{G_i,F_j\mid \mathrm{lm}_{s-1}(F_j)\in R\cap\Delta_s}'\cup\bigcup_{j}'\mathrm{spoly}(F_j).
\]

\section{Message Guessing}\label{kleflso}

The ideal of the error vector $e$ defined by
\[
	J_e=\bigcap_{e_i\neq 0}\mathfrak{m}_i
\]
has a Gr\"obner basis $\set{\epsilon_i}$ with respect to $>$, and
\begin{equation}\label{mdcad}
	\dim_\mathbb{F} R/J_e=|\Delta(J_e)|=\mathrm{wt}(e).
\end{equation}
Recall that $B^{(s)}=\set{G_i,F_j}$ is a Gr\"obner basis of $I_{v^{(s)}}$ with respect to $>_s$. Observe that $J_e(z-\mu^{(s)})\subset I_{v^{(s)}}$, which result in $\Sigma(J_e)z\subset\Sigma_s(I_{v^{(s)}})\cap Rz$, and hence $\Delta_s(I_{v^{(s)}})\cap Rz\subset\Delta(J_e)z$. Therefore
\[
	|\Delta_s(I_{v^{(s)}})\cap Rz|=|\Delta(F_j^U)|\le\mathrm{wt}(e).
\]
Now let $s$ be a nongap $\le u$. Let us consider the module
\[
	\tilde{I}_w=\set{f(z+w\phi_s)\mid f\in I_{v^{(s)}}}\subset Rz\oplus R.
\]
for $w\in\mathbb{F}$. Note that
\[
	\tilde{B}=\set{G_i(z+w\phi_s),F_j(z+w\phi_s)}
\]
is a Gr\"obner basis of $\tilde{I}_w$ with respect to $>_s$ since $\mathrm{lm}_s(f(z+w\phi_s))=\mathrm{lm}_s(f)$ for all $f\in I_{v^{(s)}}$. For the same reason,
\[
	\Sigma_s(\tilde{I}_w)=\Sigma_s(I_{v^{(s)}}),\quad \Delta_s(\tilde{I}_w)=\Delta_s(I_{v^{(s)}}).
\]
Observe that $\tilde{I}_{\omega_s}=I_{v^{(s-1)}}$. Hence
\begin{equation}\label{xjakd}
	|\Delta_{s-1}(\tilde{I}_{\omega_s})\cap Rz|\le\mathrm{wt}(e).
\end{equation}

In Theorem \ref{mxmjw} below, we will see that $\omega_s$ is such a $w$ that makes the value
\[
	|\Delta_{s-1}(\tilde{I}_w)\cap Rz|
\]
smallest, provided that $\mathrm{wt}(e)$ is not too large. First note that
\[
	\begin{gathered}
	|\Delta_{s-1}(\tilde{I}_w)\cap Rz|+|\Delta_{s-1}(\tilde{I}_w)\cap R|=|\Delta_{s-1}(\tilde{I}_w)|=n,\\
	|\Delta_s(\tilde{I}_w)\cap R|+|\Delta_s(\tilde{I}_w)\cap Rz|=|\Delta_s(\tilde{I}_w)|=n.
	\end{gathered}
\]

\begin{lemma}
For $w\neq \omega_s$,
\[
	|\Delta_{s-1}(\tilde{I}_w)\cap Rz|\ge n-|\Delta(J_e\phi_s)\cap\Delta(J)|.
\]
\end{lemma}

\begin{proof}
Observe that $J_e(z-(\omega_s-w)\phi_s-\mu^{(s-1)})\subset\tilde{I}_w$ and $J\subset\tilde{I}_w$. Therefore $\Sigma(J_e\phi_s)\cup\Sigma(J)\subset\Sigma_{s-1}(\tilde{I}_w)\cap R$, that is
\[
	\Delta_{s-1}(\tilde{I}_w)\cap R\subset\Delta(J_e\phi_s)\cap\Delta(J).
\]
Hence $|\Delta_{s-1}(\tilde{I}_w)\cap R|\le |\Delta(J_e\phi_s)\cap\Delta(J)|$, equivalent to the second equality.
\end{proof}

\begin{lemma}
$|\Delta(J_e\phi_s)|=\mathrm{wt}(e)+s$.
\end{lemma}

\begin{proof}
Note that
\[
	\begin{split}
	|\Delta(J_e\phi_s)|&=|\Sigma(R)\backslash\Sigma(J_e\phi_s)|=|\Delta(J_e)|+|\Sigma(R)\backslash\Sigma(R\phi_s)|\\
	&=\mathrm{wt}(e)+|S\backslash(s+S)|=\mathrm{wt}(e)+s.
	\end{split}
\]
The equality $|S\backslash(s+S)|=s$ holds for any numerical semigroup and can be proved by induction on the Frobenius number.
\end{proof}

\begin{theorem}\label{mxmjw}
The value $|\Delta_{s-1}(\tilde{I}_w)\cap Rz|$ is smallest for $w=\omega_s$, provided that
\[
	|\Delta(J)\cup\Delta(R\phi_s)|-s>2\mathrm{wt}(e).
\]
\end{theorem}

\begin{proof}
We need to show that for $w\neq\omega_s$,
\[
	|\Delta_{s-1}(\tilde{I}_w)\cap Rz|>|\Delta_{s-1}(\tilde{I}_{\omega_s})\cap Rz|.
\]
By \eqref{xjakd} and the previous lemmas, a sufficient condition for the above is
\[
	\begin{split}
	&n-|\Delta(J)\cap\Delta(J_e\phi_s)|>\mathrm{wt}(e)\\
	&\iff n-|\Delta(J)|-|\Delta(J_e\phi_s)|+|\Delta(J)\cup\Delta(J_e\phi_s)|>\mathrm{wt}(e)\\
	&\iff |\Delta(J)\cup\Delta(J_e\phi_s)|-s>2\mathrm{wt}(e)
	\end{split}
\]
since $|\Delta(J)|=n$. Finally note that $|\Delta(J)\cup\Delta(J_e\phi_s)| \ge |\Delta(J)\cup\Delta(R\phi_s)|$.
\end{proof}

Note that $|\Delta_{s-1}(\tilde{I}_w)\cap Rz|$ is smallest when so is
\[
	\begin{split}
	|\Delta_{s-1}(\tilde{I}_w)\cap Rz|-|\Delta_s(\tilde{I}_w)\cap Rz|
	&=|\Delta_s(\tilde{I}_w)\cap R|-|\Delta_{s-1}(\tilde{I}_w)\cap R|\\
	&=|(\Delta_s(\tilde{I}_w)\backslash\Delta_{s-1}(\tilde{I}_w))\cap R)|\\
	&=|\Sigma_{s-1}(\tilde{I}_w)\cap\Delta_s(\tilde{I}_w)\cap R|.
	\end{split}
\]
since $|\Delta_s(\tilde{I}_w)\cap Rz|=|\Delta_s(I_{v^{(s)}})\cap Rz|$ is independent of $w$. The value
\[
	|\Sigma_{s-1}(\tilde{I}_w)\cap\Delta_s(\tilde{I}_w)\cap R|
\]
can be computed using the Gr\"obner bases of $\tilde{I}_w$ with respect to $>_s$ and $>_{s-1}$. As we saw in Section \ref{jksadde}, the Gr\"obner basis of $\tilde{I}_w$ with respect to $>_{s-1}$ is determined from $\tilde{B}$, the Gr\"obner bases of $\tilde{I}_w$ with respect to $>_s$. Precisely, according to Proposition \ref{qhkxa}, the set
\[
	\Sigma_{s-1}(\tilde{I}_w)\cap\Delta_s(\tilde{I}_w)\cap R
\]
is determined by $\mathrm{lm}_{s-1}(F_j(z+w\phi_s))$ that lies in $\Delta_s(\tilde{I}_w)\cap R$. We note that for each $j$, there is a unique $w_j\in\mathbb{F}$ such that
\[
	\mathrm{lm}_{s-1}(F_j(z+w_j\phi_s))\in Rz,
\]
and $\mathrm{lm}_{s-1}(F_j(z+w\phi_s))=\mathrm{lm}(F_j^U\phi_s)\in R$ if and only if $w\neq w_j$. In fact,
\[
	w_j=-\frac{d}{\mathrm{lc}(F_j^U)},
\]
where $d$ is the coefficient of the monomial $\mathrm{lm}(F_j^U\phi_s)$ in $F_j^D$.

\begin{proposition}\label{kaqqe}
\[
	\begin{split}
	\Sigma_{s-1}(\tilde{I}_w)\cap\Delta_s(\tilde{I}_w)\cap R
	&=\bigcup_{w_j\neq w}\Sigma_{s-1}(F_j(z+w\phi_s))\cap\Delta_s(\tilde{I}_w)\\
	&=\bigsqcup_{c\neq w}\bigcup_{w_j=c}\Sigma_{s-1}(F_j(z+w\phi_s))\cap\Delta_s(\tilde{I}_w)
	\end{split}
\]
where $\sqcup$ denotes disjoint union.
\end{proposition}

\begin{proof}
The first equality follows from Proposition \ref{qhkxa}. It remains to show that the second union is disjoint. Assume that for $c_1\neq c_2$, there is a monomial $\phi\in R$ such that $\phi$ is in the intersection of
\[
	\bigcup_{w_j=c_1}\Sigma_{s-1}(F_j(z+w\phi_s))\cap\Delta_s(\tilde{I}_w)
\]
and
\[
	\bigcup_{w_j=c_2}\Sigma_{s-1}(F_j(z+w\phi_s))\cap\Delta_s(\tilde{I}_w).
\]
Let
\[
	\phi=\psi \mathrm{lm}_{s-1}(F_{j_1}(z+w\phi_s))
	=\chi \mathrm{lm}_{s-1}(F_{j_2}(z+w\phi_s))
\]
with $w_{j_1}=c_1$, $w_{j_2}=c_2$, and monomials $\psi$, $\chi$. Then we will show that
\begin{equation}\label{cjqdd}
	\mathrm{lm}_s(\frac{\psi}{\mathrm{lc}(F_{j_1}^U)}F_{j_1}(z+w\phi_s)-\frac{\chi}{\mathrm{lc}(F_{j_2}^U)} F_{j_2}(z+w\phi_s))=\phi,
\end{equation}
contradicting the assumption that $\phi\in\Delta_s(\tilde{I}_w)$. Indeed notice that $\phi=\mathrm{lm}(\psi F_{j_1}^U\phi_s)=\mathrm{lm}(\chi F_{j_2}^U\phi_s)$. Hence the coefficient of the monomial $\phi$ in the first term of the polynomial in \eqref{cjqdd} is
\[
	\frac{1}{\mathrm{lc}(F_{j_1}^U)}(w+d_1)
\]
where $d_1$ is the coefficient of the monomial $\mathrm{lm}( F_{j_1}^U\phi_s)$ in $F_{j_1}^D$. In the same way, the coefficient of the monomial $\phi$ in the second term after the minus in \eqref{cjqdd} is
\[
	\frac{1}{\mathrm{lc}(F_{j_2}^U)}(w+d_2)
\]
where $d_2$ is the coefficient of the monomial $\mathrm{lm}(F_{j_2}^U\phi_s)$ in $F_{j_2}^D$. These two coefficients are different because we assumed
\[
	w_{j_1}=-\frac{d_1}{\mathrm{lc}(F_{j_1}^U)}\neq w_{j_2}=-\frac{d_2}{\mathrm{lc}(F_{j_2}^U)}.
\]
Hence \eqref{cjqdd} follows.
\end{proof}

We observe that for $c,w\in\mathbb{F}$ with $w\neq c$,
\[
	\bigcup_{w_j=c}\Sigma_{s-1}(F_j(z+w\phi_s))\cap\Delta_s(\tilde{I}_w)
	=\bigcup_{w_j=c}\Sigma(F_j^U\phi_s)\cap\Delta(\set{G_i^D}).
\]
Therefore this set is independent of $w$, and is determined by $B^{(s)}$. Let
\[
	d_c=\Bigl|\bigcup_{w_j=c}\Sigma(F_j^U\phi_s)\cap\Delta(\set{G_i^D})\Bigr|.
\]
Then Proposition \ref{kaqqe} implies
\[
	|\Delta_{s-1}(\tilde{I}_w)\cap Rz|-|\Delta_s(\tilde{I}_w)\cap Rz|=\sum_{c\neq w}d_c
\]
is smallest when $w=c$ with $d_c$ largest. Now we can elaborate the main steps of the interpolation decoding algorithm as follows:

\begin{description}
\item[M1] If $s$ is a nongap $\le u$, then do the following, but otherwise let $\tilde{B}=\set{G_i,F_j}$.
\begin{description}
\item[M1.1] Compute the set $W=\set{w_j}$, where
\[
	w_j=-\frac{d}{\mathrm{lc}(F_j^U)},
\]
and $d$ is the coefficient of the monomial $\mathrm{lm}(F_j^U\phi_s)$ in $F_j^D$.
\item[M1.2] For each $c\in W$, compute the value
\[
	d_c=\Bigl|\bigcup_{w_j=c}\Sigma(F_j^U\phi_s)\cap\Delta(\set{G_i^D})\Bigr|.
\]
\item[M1.3] Let $w^{(s)}=c$ with largest $d_c$, and let
\[
	\tilde{B}=\set{G_{i}(z+w^{(s)}\phi_s),F_j(z+w^{(s)}\phi_s)}.
\]
\end{description}
\item[M2] Suppose $\tilde{B}=\set{\tilde{G}_i,\tilde{F}_j}$. Let
\[
	B^{(s-1)}=\set{\tilde{G}_i,\tilde{F}_j\mid \mathrm{lm}_{s-1}(\tilde{F}_j)\in R\cap\Delta_s(\set{\tilde{G}_i})}'\cup\bigcup_{j}'\mathrm{spoly}(\tilde{F}_j).
\]
\end{description}

\begin{theorem}
The algorithm outputs $w^{(s)}=\omega_s$ for all $s\in S,s\le u$ if
\[
	d_u=\min_{s\in S,s\le u}\nu(s)>2\mathrm{wt}(e),
\]
where $\nu(s)=|\Delta(J)\cup\Delta(R\phi_s)|-s$ for $s\in S$. Moreover $d_u\ge n-u$.
\end{theorem}

\begin{proof}
By Theorem \ref{mxmjw}, the condition $d_u>2\mathrm{wt}(e)$ implies that the algorithm computes $w^{(s)}=\omega_s$ for each iteration for nongap $s$ from $u$ to $0$. To see $d_u\ge n-u$, notice that $|\Delta(J)\cup\Delta(R\phi_s)|\ge |\Delta(J)|=n$.
\end{proof}

\section{Decoding Hermitian Codes}\label{ckdwdd}

In this section, we demonstrate the decoding algorithm on the Hermitian codes defined on Hermitian curves with equation
\[
	y^q+y-x^{q+1}=0
\]
over $\mathbb{F}_{q^2}$. There are $q^3$ rational points on the Hermitian curve, and $J=\ideal{x^{q^2}-x}$. We now determine the performance of the decoding algorithm for the Hermitian code $C_u$.

\begin{theorem}
For nongap $u<q^3$,
\[
	\begin{aligned}
	d_u=\begin{cases}
	q^3-aq & b \le a+q-q^2\\
	q^3-u & b> a+q-q^2
	\end{cases}
	\end{aligned}
\]
if $u=aq+b$, $0\le b<q$.
\end{theorem}

\begin{proof}
We first compute $\nu(s)$ for nongap $s=qs_1+s_2<q^3$. As
\[
	\begin{split}
	|\Delta(J)\cup\Delta(R\phi_s)|&=|\Sigma(J)\cap\Delta(R\phi_s)|+|\Delta(J)|\\
	&=|\set{t\in S\mid q^3+t\notin s+S}|+q^3.
	\end{split}
\]
we have $\nu(s)=|\set{t\in S\mid q^3+t-s\notin S}|+q^3-s$. Note that
\[
	q^3+t-s=q(q^2+t_1-s_1)+t_2-s_2
\]
with $t=qt_1+t_2$. Therefore $q^3+t-s\notin S$ if and only if
\[
	t_2-s_2\ge 0,\quad q^2+t_1-s_1<t_2-s_2
\]
or
\[
	t_2-s_2<0,\quad q^2+t_1-s_1<q+1+t_2-s_2.
\]
The first case is actually impossible since $s_1<q^2$. Hence
\[
	|\set{t\in S\mid q^3+t-s\notin S}|=s_2\max\set{s_1-s_2+q+1-q^2,0}.
\]
Thus
\[
	\nu(s)=s_2\max\set{s_1-s_2+q+1-q^2,0}+q^3-s
\]
for $s=qs_1+s_2<q^3$. If $a-b+q-q^2\ge 0$, then the minimum is attained at $s=aq$, and hence
$d_u=q^3-aq$ while if $a-b+q-q^2<0$, then the minimum is attained at $s=u$, and hence $d_u=q^3-u$.
\end{proof}

\begin{figure}
\begin{center}
\begin{tikzpicture}[scale=2,x=3pt,y=2pt,
	declare function={
		q = 3;
		b(\u) = mod(\u,q);
		a(\u) = div(\u,q);
		d(\u) = b(\u) <= a(\u) - (q^2 - q) ? q^3 - a(\u)*q : q^3 - \u;
		s(\u)= a(\u) - b(\u) < 0 ? a(\u)*q+a(\u) : \u;
		del(\u)= d(s(\u));
	}]
	\draw[samples at={0,1,...,27},mark=ball,mark size=.7pt,ball color=black,only marks,variable=\u]
		plot (\u,{del(\u)});
	\draw (0,0) -- (28,0);
	\draw (0,0) -- (0,27);
	\foreach \x in {0,10,...,27}
		\draw (\x,0) -- (\x,3pt);
	\foreach \x in {0,5,...,27}
		\draw (\x,0) -- (\x,2pt);
	\foreach \x in {10,20,...,27}
		\draw (\x,0) node[anchor=north] {\x};
	\foreach \y in {0,10,...,27}
		\draw (0,\y) -- (3pt,\y);
	\foreach \y in {0,5,...,27}
		\draw (0,\y) -- (2pt,\y);
	\foreach \y in {0,10,...,27}
		\draw (0,\y) node[anchor=east] {\y};
\end{tikzpicture}
\caption{Decoding performance of Hermitian codes of length $27$}\label{ckqe}
\label{default}
\end{center}
\end{figure}

Figure \ref{ckqe} shows the decoding performance $d_u$ of $C_u$ over the Hermitian curve $y^3+y-x^4=0$ over $\mathbb{F}_9$, where $\mathbb{F}_9=\mathbb{F}_3(\alpha)$ with $\alpha^2-\alpha-1=0$. The $27$ rational points on the curve are
\[
	\begin{gathered}
	( 0, 0 ), ( 0, \alpha^2 ), ( 0, \alpha^6 ), ( 1, 2 ), ( 1, \alpha ), ( 1, \alpha^3 ), ( 2, 2 ), ( 2, \alpha ), ( 2, \alpha^3 ),\\
 	( \alpha, 1 ), ( \alpha, \alpha^7 ), ( \alpha, \alpha^5 ), ( \alpha^2, 2 ), ( \alpha^2, \alpha ), ( \alpha^2, \alpha^3 ), ( \alpha^7, 1 ), ( \alpha^7, \alpha^7 ), ( \alpha^7, \alpha^5 ),\\
	 ( \alpha^5, 1 ), ( \alpha^5, \alpha^7 ), ( \alpha^5, \alpha^5 ), ( \alpha^3, 1 ), ( \alpha^3, \alpha^7 ), ( \alpha^3, \alpha^5 ), ( \alpha^6, 2 ), ( \alpha^6, \alpha ), ( \alpha^6, \alpha^3 ),
	\end{gathered}
\]
and there is a unique point $P_\infty$ at infinity. As $\delta(x)=3$ and $\delta(y)=4$, the numerical semigroup of the coordinate ring $R$ is
\[
	S=\tuple{3,4}=\set{0,3,4,6,7,8,9,10,\dots}.
\]
Note that $S$ has three gaps $1$, $2$, and $5$. The monomials of $R$ correspond to nongaps in $S$ and are displayed in the diagram
\[
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  	\draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
	\draw (0,0) grid[step=4mm] +(11,3);
	\draw (0,0) +(.5,.5) node[scale=.6] {$1^{\vphantom{1}}$};
	\draw (1,0) +(.5,.5) node[scale=.6] {$x^{\vphantom{1}}$};
	\draw (10,0) +(.5,.5) node[scale=.6] {$\cdots$};
	\foreach \x in {2,...,9}
	  		\draw (\x,0) +(.5,.5) node[scale=.6] {$x^{\x}$};
	\draw (0,1) +(.5,.5) node[scale=.6] {$y^{\vphantom{1}}$};
	\draw (1,1) +(.5,.5) node[scale=.6] {$xy^{\vphantom{1}}$};
	\draw (10,1) +(.5,.5) node[scale=.6] {$\cdots$};
	\foreach \x in {2,...,9}
	  		\draw (\x,1) +(.5,.5) node[scale=.6] {$x^{\x}y$};
	\draw (0,2) +(.5,.5) node[scale=.6] {$y^2$};
	\draw (1,2) +(.5,.5) node[scale=.6] {$xy^2$};
	\draw (10,2) +(.5,.5) node[scale=.6] {$\cdots$};
	\foreach \x in {2,...,9}
	  	\draw (\x,2) +(.5,.5) node[scale=.6] {$x^{\x}y^2$};
\end{tikzpicture}
\]

Let $u=16$. Then the Hermitian code $C_{16}$ has dimension $14$ and minimum distance $11$, and the decoding algorithm can correct up to $5$ errors.
Suppose we received the vector
\[
	v=(0,0,0,0,0,\alpha^2,2,0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \alpha^3, 0, 0, \alpha^7, 0, 0, 2, 0)
\]
from a noisy channel. We now follow the steps of the decoding algorithm.

The algorithm first compute the Lagrange interpolation of $v$,
\[
	\begin{split}
	h_v&=\alpha^3x^8y^2 + x^7y^2 + \alpha^6x^8y + \alpha^7x^6y^2 + x^7y + 2x^8 + \alpha^5x^5y^2 \\
	&\quad + x^6y + \alpha x^7 + \alpha^3x^4y^2 + \alpha x^5y + \alpha^6x^6 + \alpha^6x^3y^2 + \alpha^6x^4y\\
	&\quad + 2x^2y^2 + \alpha^7x^3y + 2x^4 + \alpha^2xy^2 + \alpha^2x^3 + \alpha^3xy + x.
	\end{split}
\]
The algorithm iterates the main steps for $s$ from $N=\delta(h_v)=32$ to $0$. The ideal $J$ has Gr\"obner basis $\set{\eta_1=x^9-x}$. Hence the Gr\"obner basis of $I_{v^{(32)}}=I_v$ is
\[
B^{(32)}=\left\{\begin{array}{rcrcl}
G_1&=&                          0\,z&+&x^9-x\\
F_1&=&                           1\,z&+&\alpha^7x^8y^2+\cdots\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (0,0) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (0.5,0.5) node[scale=.6] {$1$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9.5,0.5) node[scale=.6] {$x^9$};
\end{tikzpicture}
\end{center}
Here the left diagram exhibits the monomials in $\Sigma_{32}(I_{v^{(32)}})\cap Rz$ omitting the common $z$ variable, while the right diagram shows the monomials in $\Sigma_{32}(I_{v^{(32)}})\cap R$. The leading terms of the polynomials in the Gr\"obner basis are also shown.

For $s\ge u=16$ or a gap $s$, as $\tilde{B}=\set{\tilde{G}_i,\tilde{F}_j}=B^{(s)}=\set{G_i,F_j}$ in the step \textbf{M1}, we will omit the tilde in the following. In the step \textbf{M2}, $\mathrm{lm}_{31}(F_1)=x^8y^2\in R\cap \Delta_{32}(G_1)$, and the lcms of $\mathrm{lm}_{31}(F_1)=x^8y^2$ and $\mathrm{lm}_{32}(G_1)=x^9$ are $x^9y^2$ and $x^{12}$. Hence $\mathrm{spoly}(F_1)=\set{\alpha xz+\alpha^5x^8y^2 + \dots+ \alpha^5x^2,\alpha yz+\alpha^5x^{11} + \dots + \alpha^2xy}$. Then the Gr\"obner basis of $I_{v^{(31)}}$ is
\[
B^{(31)}=\left\{\begin{array}{rcrcl}
G_1&=&                          0\,z&+&x^9+\cdots\\
G_2&=&                          1\,z&+&\alpha^7x^8y^2+\cdots\\
F_1&=&                     \alpha x\,z&+&\alpha^5x^8y^2+\cdots\\
F_2&=&                      \alpha y\,z&+&\alpha^5x^{11}+\cdots\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (1,0) |- (0,1) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (1.5,0.5) node[scale=.6] {$x$};
  \draw (0.5,1.5) node[scale=.6] {$y$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (8,2) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9.5,0.5) node[scale=.6] {$x^9$};
  \draw (8.5,2.5) node[scale=.6] {$x^8y^2$};
\end{tikzpicture}
\end{center}
As $\mathrm{lm}_s(F_1),\mathrm{lm}_s(F_2)\in Rz$ for  $s=31,30$, there is no change in the Gr\"obner basis. So we get to the unaltered Gr\"obner basis of $I_{v^{(29)}}$
\[
B^{(29)}=\left\{\begin{array}{rcrcl}
G_1&=&                          0\,z&+&x^9+\cdots\\
G_2&=&                          1\,z&+&\alpha^7x^8y^2+\cdots\\
F_1&=&                     \alpha x\,z&+&\alpha^5x^8y^2+\cdots\\
F_2&=&                     \alpha y\,z&+&\alpha^5x^{11}+\cdots\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (1,0) |- (0,1) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (1,0) +(.5,.5) node[scale=.6] {$x$};
  \draw (0,1) +(.5,.5) node[scale=.6] {$y$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (8,2) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9,0) +(.5,.5) node[scale=.6] {$x^9$};
  \draw (8,2) +(.5,.5) node[scale=.6] {$x^8y^2$};
\end{tikzpicture}
\end{center}
Now since $\mathrm{lm}_{29}(G_2)=x^8y^2$ divides $\mathrm{lm}_{28}(F_1)=x^8y^2$ and $\mathrm{lm}_{29}(G_1)=x^9$ divides $\mathrm{lm}_{28}(F_2)=x^{11}$, both $\mathrm{lm}_{28}(F_1)$ and $\mathrm{lm}_{28}(F_2)$ are in $R\cap\Sigma_{29}(G_1,G_2)$ Thus
\[
	\begin{aligned}
	\mathrm{spoly}(F_1)&=\set{2xz + \alpha^5z+\alpha^6x^9y + \dots + \alpha x},\\
	\mathrm{spoly}(F_2)&=\set{2yz+\alpha^6x^8y^2 + \dots+ \alpha^5xy}.
	\end{aligned}
\]
Hence the Gr\"obner basis of $I_{v^{(28)}}$ is
\[
B^{(28)}=\left\{\begin{array}{rcrcl}
G_1&=&                          0\,z&+&x^9+\cdots\\
G_2&=&                          1\,z&+&\alpha^7x^8y^2+\cdots\\
F_1&=&                     (2 x+\alpha^5)\,z&+&\alpha^6 x^9y+\cdots\\
F_2&=&                     2 y\,z&+&\alpha^6 x^8y^2+\cdots\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (1,0) |- (0,1) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (1,0) +(.5,.5) node[scale=.6] {$x$};
  \draw (0,1) +(.5,.5) node[scale=.6] {$y$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (8,2) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9,0) +(.5,.5) node[scale=.6] {$x^9$};
  \draw (8,2) +(.5,.5) node[scale=.6] {$x^8y^2$};
\end{tikzpicture}
\end{center}
Similar steps are iterated. Eventually, we get to the Gr\"obner basis of $I_{v^{(16)}}$,
\[
B^{(16)}=\left\{\begin{array}{rcrcl}
G_1&=&                          0\,z&+&x^9+\cdots\\
G_2&=&                     (\alpha^2 xy+\dots)\,z&+&\alpha^2 x^7y+\cdots\\
F_1&=&(\alpha^2x^2+\cdots)\,z&+&0\\
F_2&=&            (\alpha^5y^2+\cdots)\,z&+&x^8+\cdots\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (2,0) |- (0,2) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (2,0) +(.5,.5) node[scale=.6] {$x^2$};
  \draw (0,2) +(.5,.5) node[scale=.6] {$y^2$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (7,1) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9,0) +(.5,.5) node[scale=.6] {$x^9$};
  \draw (7,1) +(.5,.5) node[scale=.6] {$x^7y$};
\end{tikzpicture}
\end{center}
Now $s=16$ is a nongap  and $\le u=16$. So in the step \textbf{M1}, we proceed to guess $\omega_{16}$ for the monomial $\phi_{16}=x^4y$. The leading coefficient of $F_1$ is $\alpha^2$ and the coefficient of the monomial $x^6y$ in $F_1$ is $0$, where $x^6y$ is the leading monomial of $x^2\phi_{16}$. Hence $w_1=-(0/\alpha^2)=0$. The leading coefficient of $F_2$ is $\alpha^5$ and the coefficient of the monomial $x^8$ in $F_2$ is $1$, where $x^8$ is the leading monomial of $y^2\phi_{16}$. Hence $w_2=-(1/\alpha^5)=\alpha^7$. So $W=\set{0,\alpha^7}$. The shape of
\[
	\bigcup_{w_j=0}\Sigma(F_j^U\phi_{16})\cap\Delta(\set{G_i^D})=\Sigma(x^6y)\cap\Delta(x^9,x^7y)
\]
is
\[
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (7,1) |- (0,3) |- (11,3);
  \draw[pattern=crosshatch] (6,1) |- (7,3) |- (6,1);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9,0) +(.5,.5) node[scale=.6] {$x^9$};
  \draw (7,1) +(.5,.5) node[scale=.6] {$x^7y$};
\end{tikzpicture}
\]
and thus $d_0=2$. On the other hand, the shape of
\[
	\bigcup_{w_j=\alpha^7}\Sigma(F_j^U\phi_{16})\cap\Delta(\set{G_i^D})=\Sigma(x^8)\cap\Delta(x^9,x^7y)
\]
is
\[
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (9,0) |- (7,1) |- (0,3) |- (11,3);
  \draw[pattern=crosshatch] (8,0) |- (9,1) |- (8,0);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (9,0) +(.5,.5) node[scale=.6] {$x^9$};
  \draw (7,1) +(.5,.5) node[scale=.6] {$x^7y$};
\end{tikzpicture}
\]
and thus $d_{\alpha^7}=1$. Hence we take $w^{(16)}=0$. Then
\[
\tilde{B}=\left\{\begin{array}{rcrcl}
\tilde{G}_1&=&                          0\,z&+&x^9+\cdots\\
\tilde{G}_2&=&                     (\alpha^2 xy+\dots)\,z&+&\alpha^2 x^7y+\cdots\\
\tilde{F}_1&=&(\alpha^2x^2+\cdots)\,z&+&0\\
\tilde{F}_2&=&            (\alpha^5y^2+\cdots)\,z&+&x^8+\cdots\\
\end{array}\right\}
\]
In the step \textbf{M2}, $\mathrm{lm}_{15}(\tilde{F}_1)=x^2z\in Rz$ and $\mathrm{lm}_{15}(\tilde{F}_2)=x^8\in R\cap\Delta_{15}(\tilde{G}_1,\tilde{G}_2)$. So $\mathrm{spoly}(\tilde{F}_1)=\set{\tilde{F}_1}$, and since the lcm of $\mathrm{lm}_{15}(\tilde{F}_2)=x^8$ and $\mathrm{lm}_{15}(\tilde{G}_1)=x^9$ is $x^9$, and the lcms of $\mathrm{lm}_{15}(\tilde{F}_2)$ and $\mathrm{lm}_{15}(\tilde{G}_2)=x^7y$ are $x^8y$ and $x^{11}$,
\[
	\mathrm{spoly}(\tilde{F}_2)=\set{\alpha^5xy^2z+\dots+x,\alpha^5x^4z+\dots+x^2,\alpha^5x^3y^2z+\dots+\alpha^5xy}.
\]
Removing redundant polynomials, we have
\[
	\begin{gathered}
	\set{\tilde{G}_1,\tilde{G}_2,\tilde{F}_2}'=\set{\tilde{G}_2,\tilde{F}_2},\\
	(\mathrm{spoly}(\tilde{F}_1)\cup\mathrm{spoly}{\tilde{F}_2})'=\set{\tilde{F}_1,\alpha^5xy^2z+\dots+x}.
	\end{gathered}
\]
Thus the Gr\"obner basis of $I_{v^{(15)}}$ is
\[
B^{(15)}=\left\{\begin{array}{rcrcl}
G_1&=&(\alpha^2 xy+\cdots)\,z&+&\alpha^2x^7y+\cdots\\
G_2&=&(\alpha^5y^2+\cdots)\,z&+&x^8+\cdots\\
F_1&=&(\alpha^2x^2+\cdots)\,z&+&0\\
F_2&=&            (\alpha^5xy^2+\cdots)\,z&+&2 x^7y+\cdots\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (2,0) |- (1,2) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (2,0) +(.5,.5) node[scale=.6] {$x^2$};
  \draw (1,2) +(.5,.5) node[scale=.6] {$xy^2$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (8,0) |- (7,1) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (8,0) +(.5,.5) node[scale=.6] {$x^8$};
  \draw (7,1) +(.5,.5) node[scale=.6] {$x^7y$};
\end{tikzpicture}
\end{center}
Continuing in this way, after the last iteration for $s=0$, we get to the Gr\"obner basis of $I_{v^{(-1)}}$,
\[
B^{(-1)}=\left\{\begin{array}{rcrcl}
G_1&=&(\alpha^2xy+\cdots)\,z&+&\alpha^2x^7y+\cdots\\
G_2&=&(\alpha^5y^2+\cdots)\,z&+&x^8+\cdots\\
F_1&=&(\alpha^2x^2+\cdots)\,z&+&0\\
F_2&=&            (xy^2+\cdots)\,z&+&0\\
\end{array}\right\}
\]
\begin{center}
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (2,0) |- (1,2) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (2,0) +(.5,.5) node[scale=.6] {$x^2$};
  \draw (1,2) +(.5,.5) node[scale=.6] {$xy^2$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=1.2,x=4mm,y=4mm,baseline=(current bounding box)]
  \draw[dimgray,fill] (0,0) |- (11,3) |- (0,0);
  \draw[lightgray,fill] (11,3) |- (8,0) |- (7,1) |- (0,3) |- (11,3);
  \draw (0,0) grid[step=4mm] +(11,3);
  \draw (8,0) +(.5,.5) node[scale=.6] {$x^8$};
  \draw (7,1) +(.5,.5) node[scale=.6] {$x^7y$};
\end{tikzpicture}
\end{center}
Finally, the algorithm output $(w^{(s)}=0\mid\text{nongap $s\le 16$})$.

\section{Remarks}

We presented an interpolation-based unique decoding algorithm for the primal algebraic geometry codes. The algorithm iteratively finds each entry of the sent message by a majority voting procedure. We showed that the majority voting is successful if the codeword encoding the message is close enough to the received vector.

redpajama_set_name: RedPajamaArXiv


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text:
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redpajama_set_name: RedPajamaCommonCrawl


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text:
Q: Docker-compose - build with maven that re-uses the maven repository When building my Spring-boot image using Maven I now do this inside the Dockerfile. Maven will download all dependencies and then compile my files. This takes quite some time.
How can I specify that the build process via docker-compose (Dockerfile) re-uses my "Windows10 Home" Maven repository? So, the number of (new) downloads is minimal. My dev context: I use the Docker quickstart terminal, so using a docker-machine.
The is is a part of my docker-compose.yml file:
version: '3'
services:
  spring-boot-app:
    image: spring-boot-image
    build:
      context: ./
      dockerfile: Dockerfile
    depends_on:
      - mysql-docker-container
    ports:
      - 8087:8080
    volumes:
      - $HOME/.m2/repository:/root/.m2/repository
      - /data/spring-boot-app

My Dockerfile is:
FROM java:8
FROM maven:alpine
WORKDIR /app
COPY . /app
RUN mvn -v
RUN mvn clean install -DskipTests
EXPOSE 8080
LABEL maintainer="xyz@holland.nl"
ADD ./target/spring-boot-example-0.0.1-SNAPSHOT.jar /developments/
ENTRYPOINT ["java","-jar","/developments/spring-boot-example-0.0.1-SNAPSHOT.jar"]

A: @Jack Gore - thank you very much for pointing me the way. I had seen the post, but there was not a definitive answer yet. After diving in all seperate answers, the info provided me both an answer to the symptom as well with insights how to solve similar questions.
ANSWER: Via the Dockerfile you can build the final images via 1 or more subsequent image layers. To prevent the step from re-downloading the dependencies each time, you can make an image layer for the 'downloading the world' of dependencies. Re-downloading will only be done when a dependency changes in the pom.xml file.
To do so, you split the Dockerfile in 2 build steps: first copy the pom.xml file and build it. That will create an image layer with all dependencies. As a second step you build the application.
FROM java:8
FROM maven:alpine

# image layer
WORKDIR /app
ADD pom.xml /app
RUN mvn verify clean --fail-never

# Image layer: with the application
COPY . /app
RUN mvn -v
RUN mvn clean install -DskipTests
EXPOSE 8080
ADD ./target/your.jar /developments/
ENTRYPOINT ["java","-jar","/developments/your.jar"]

Then you get the following build scenario's:

*

*The first time you build this (docker build .)  the dependencies are downloaded and as step 2 the application jar is build.

*When you rebuild immediately the dependencies (pom.xml) and the application sources were not changed. So, the image layers don't need to be changed. The build is ready in no time.

*If you change 1 of your application source files, only a few downloads are downloaded and the application is build. So you are NOT downloading the world.

*If you change the pom.xml file, thus changing the dependencies, then all dependency downloads are done.

The impact of seperating image layers is shown via a number of very short videos.
The disadvantage of this way of building is of course that the final (production) image contains more than the appication. Not only the JAR is in it, but also a number of dependencies.
How to solve this iamge being far too big:

*

*Reduce the image size via a trick. Use the option: --squash. This is explained in this post.

*Seperate the build process (here: maven and/or ng build --prod) as a step before the docker build step. Then execute the docker build and put ONLY the jar file in final image.

*Use an CI/CD environment with e.g. a Jenkinsfile (pipeline). In the Jenkins pipeline you first build the image. All dependencies are already there. You only rebuild the application. The you perform a docker build, etc.

For my case this is the best option possible. You automate the process AND keep the image size low.

redpajama_set_name: RedPajamaStackExchange


text:
Strupiechów  is a village in the administrative district of Gmina Wierzbno, within Węgrów County, Masovian Voivodeship, in east-central Poland. It lies approximately  north-east of Wierzbno,  south-west of Węgrów, and  east of Warsaw.

The village has an approximate population of 190.

References

Villages in Węgrów County

redpajama_set_name: RedPajamaWikipedia


One of the worst sins that a restaurant owner could commit would be to serve food that they knew to be stale. One of the worst sins that a policeman could commit would be to ignore the distressed 911 call of someone in trouble, just so that they could watch the end of the football match on television. One of the worst sins that a government official could commit would be to steal from public funds for their own personal use.
Human beings are naturally flawed and sometimes we do make mistakes. People can be and should be forgiven for their mistakes, but there are some errors or results that occur because of sloppy work that musn't be forgiven. PPC Managers that commit the following PPC Management Sins should be dumped by the way-side immediately.
If you notice any tell-tale signs that the following PPC Management Sins are being committed by your PPC Manager, it is probably time to find a new one. If you are a PPC Manager and you are committing the following sins then you really do need to stop. Take professional pride in your work and make sure that you cannot be accused of making the following unforgiveable errors.
There is no excuse for not checking the PPC Campaign Settings of every campaign that you manage. The PPC Campaign Settings that you elect are incredibly important and can have a huge effect on any campaign that you manage.
It is important to be aware of the fact that there are automatically set PPC Campaign Settings configured to a series of default choices by Google. These default settings are there so that everyone and anyone is able to run any kind of basic PPC Campaign if they so wish without much guidance.
However, as a PPC Specialist, you should know that these default settings are not going to fit the needs and demands of every single PPC Campaign that you manage. For instance, if you do not want to advertise nationally, if you are a local company and you only want to advertise in Chicago, the default setting that Google sets to "national" is a huge waste of time and money.
Your budget may run out over the course of the day before anyone in Chicago has even seen your advertisement because the default "national" setting was not adjusted accordingly. This is merely one example, but it gives you an idea of what we are trying to highlight here.
PPC Management is all about analyzing performance and then optimizing for efficiency. How can you expect to be analyzing performance of the PPC Campaign you are managing thoroughly enough if you haven't set up conversion tracking on those websites you are promoting?
If you haven't taken the time to check to see whether the conversion tracking tool has been put in place on the corresponding website and you begin to run your PPC Campaign regardless, you are offering PPC Management which is sloppy and below average. In short, you are committing a PPC Management Sin that really cannot be excused.
A PPC Manager must be analytical and must make use of those tools available in order to do that analysis to their best of their ability. There is no excuse for a PPC Campaign that lacks conversion tracking.
Lazy PPC Managers adopt the all or nothing approach to their job and it is simply unforgiveable. The all or nothing approach describes PPC Managers who select only broad match keywords for instance. They are the kind of PPC Manager who does little research and spends even less time in developing effective keywords. They completely ignore the opportunity they have to work with a combination of broad match, phrase match and exact match keywords, for example.
The all or nothing approach is adopted by those PPC Ma nagers who are always aiming for the number one spot on the Google SERP because they don't consider that a number two or three spot might actually be more affordable and more profitable for that particular PPC advertising campaign.
In short, the all or nothing PPC Manager is someone who is not doing any work whatsoever and who knows hardly anything about PPC Management. He or she is a PPC fraudster. The best advice is to stay away from PPC Managers who exhibit these kinds of qualities at all times. You want someone who is going to create a PPC Campaign which is tailor-made to your business and advertising needs, not someone who is going to set up the same campaign for you that they set up for everyone else.


text:
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<document>
  <properties>
    <title>Apache Receivers Companion for log4j 1.2.</title>
  </properties>
  <body>
    <release version="1.0" date="TDB" description="First release">
      <action action="update">Upgraded to Apache Maven 3</action>
      <action issue="42102" type="add">
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redpajama_set_name: RedPajamaGithub


text:
Raiders |
Live updates: Mistake-prone Raiders beaten by…
Dieter Kurtenbach
Live updates: Mistake-prone Raiders beaten by Chargers 26-10
Quarterback Derek Carr #4 of the Oakland Raiders faces pressure by defensive end Melvin Ingram #54 of the Los Angeles Chargers in the second quarter at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Los Angeles Chargers running back Austin Ekeler (30) celebrates his touchdown with teammate wide receiver Mike Williams during the first half of an NFL football game against the Oakland Raiders Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Mark J. Terrill)
Melvin Gordon #28 of the Los Angeles Chargers is hit short of the goal line by Marquel Lee #55 and Marcus Gilchrist #31 of the Oakland Raiders during the second quarter at StubHub Center on October 7, 2018 in Carson, California. (Photo by Harry How/Getty Images)
Austin Ekeler's 44-yard catch-and-run touchdown gave the Chargers a lead they never relinquished. (Photo by Harry How/Getty Images)
Defensive end Melvin Ingram #54 of the Los Angeles Chargers celebrates a sack with defensive end Isaac Rochell #98 and outside linebacker Jatavis Brown #57 in the first quarter against the Oakland Raiders at StubHub Center on October 7, 2018 in Carson, California. (Photo by Harry How/Getty Images)
Tight end Antonio Gates #85 of the Los Angeles Chargers runs with the ball in front of defensive back Dominique Rodgers-Cromartie #45 of the Oakland Raiders and defensive back Erik Harris #25 at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Quarterback Derek Carr #4 of the Oakland Raiders looks to pass in the second quarter against the Los Angeles Chargers at StubHub Center on October 7, 2018 in Carson, California. (Photo by Harry How/Getty Images)
Running back Austin Ekeler #30 of the Los Angeles Chargers scores a touchdown in the second quarter in front of strong safety Marcus Gilchrist #31 of the Oakland Raiders at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Strong safety Jahleel Addae #37 of the Los Angeles Chargers recovers the Oakland Raiders' fumble in front of wide receiver Martavis Bryant #12 of the Oakland Raiders in the second quarter at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Los Angeles Chargers quarterback Philip Rivers prepares to throw a pass during the first half of an NFL football game against the Oakland Raiders, Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Mark J. Terrill)
Wide receiver Martavis Bryant #12 of the Oakland Raiders fumbles in front of strong safety Jahleel Addae #37 of the Los Angeles Chargers in the second quarter at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Wide receiver Keenan Allen #13 of the Los Angeles Chargers is tackled by free safety Reggie Nelson #27 of the Oakland Raiders in the second quarter at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Marshawn Lynch faces his former team for the first time on Sunday. (Photo by Harry How/Getty Images)
Running back Melvin Gordon #28 of the Los Angeles Chargers makes a 34 yard run in the second quarter against the Oakland Raiders at StubHub Center on October 7, 2018 in Carson, California. (Photo by Sean M. Haffey/Getty Images)
Oakland Raiders punt returner Dwayne Harris, right, is hauled down by the Los Angeles Chargers' Austin Ekeler during the first half of an NFL football game Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Mark J. Terrill)
Oakland Raiders quarterback Derek Carr prepares to throw a pass during the first half of an NFL football game against the Los Angeles Chargers, Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Jae C. Hong)
Los Angeles Chargers running back Melvin Gordon, center, is upended by Oakland Raiders free safety Reggie Nelson (27) and linebacker Marquel Lee, second from right, during the first half of an NFL football game Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Mark J. Terrill)
Los Angeles Chargers cornerback Trevor Williams, right, can't hold on to the ball as he breaks up a end-zone pass intended for Oakland Raiders tight end Jared Cook with Los Angeles Chargers strong safety Jahleel Addae, left, during the first half of an NFL football game Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Jae C. Hong)
An Oakland Raiders fan poses while tailgating before an NFL football game against the Los Angeles Chargers on Sunday, Oct. 7, 2018, in Carson, Calif. (AP Photo/Jae C. Hong)
By Jon Becker | Bay Area News Group
PUBLISHED: October 7, 2018 at 4:30 a.m. | UPDATED: October 7, 2018 at 4:07 p.m.
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Get live updates, news and analysis Sunday afternoon as the Raiders take on the Chargers in Los Angeles beginning at 1:05 p.m.
Fresh off their first win of the season, the Raiders take on their AFC West rival Chargers, led by the Raiders' archenemy Philip Rivers. L.A. beat the 49ers 29-27 last Sunday on the strength of Rivers' three TD passes.
In the process of knocking off the 49ers, Rivers climbed past John Elway into eighth place on the all-time passing list with 51,504 career yards.
The Raiders will have to deal not only with Rivers but his top-notch weapons, led by ex-Cal receiver Keenan Allen as well as running back Melvin Gordon.
Meanwhile, the Chargers will be facing a Raiders offense fresh off a 45-point performance against the Browns. Derek Carr threw for 437 yards and four touchdowns against Cleveland, which also had trouble stopping Marshawn Lynch (130 yards rushing).
Oakland's offense line bears watching a rookie right tackle Brandon Parker, its third-round pick, steps in as the starter with Donald Penn injuring his groin and getting placed on injured reserve. With Parker on the line with Kolton Miller on the left, the Raiders have bookend 6-foot-8 rookie tackles.
The Raiders' next game will be across the pond when they face the Seattle Seahawks at London's Wembley Stadium on Oct. 14.
To view the live updates feed on your mobile device click here.
Like our Oakland Raiders Facebook page for more Raiders news, commentary and conversation.
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Jon Becker | Online producer
Jon Becker is a digital producer for the Bay Area News Group.
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Senate panel rejects attempt to overturn labor board ruling
By Erik Wasson - 06/14/12 03:36 PM EDT
Democrats on the Senate Appropriations Committee on Thursday blocked a Republican attempt to overturn a National Labor Relations Board ruling that would allow unions to form smaller bargaining units.
The amendment to the 2013 Labor, Health and Human Services measure was defeated on a 15-to-15 vote after a heated clash between Sen. Lindsey Graham Lindsey Olin GrahamMcConnell: I'm going to give Biden's Supreme Court nominee 'a fair look' The Hill's Morning Report - Who will replace Justice Breyer? Breyer retirement throws curveball into midterms MORE (R-S.C.), who was pushing for the overturn, and Sen. Mary Landrieu Mary Loretta LandrieuDemocratic ex-senators join pro-gas organization 11 former Democratic senators call for 'meaningful reform to Senate rules' 10 Democrats who could run in 2024 if Biden doesn't MORE (D-La.).
Sen. Mark Pryor Mark Lunsford Pryor11 former Democratic senators call for 'meaningful reform to Senate rules' Kyrsten Sinema is less of a political enigma than she is a strategic policymaker Bottom line MORE (R-Ark.), whose state is home to Wal-Mart, was the only Democrat voting in favor of the ban.
Graham was attempting to overturn an August 2011 NLRB ruling, known as Specialty Healthcare, that he and other critics argue allows the creation of "micro-unions." The NLRB said their ruling should only apply to certain industries, but critics complain it is being used elsewhere.
Business groups have challenged the ruling in court but have also lobbied Congress to stop it. Legislation passed in the House last year would void the ruling, but that bill has not moved in the Senate.
Arguing for the ban, Graham said in the hearing Thursday that the ruling "throws the whole business model in chaos" in many sectors.
Landrieu argued on behalf of smaller unions, pointing out that the ban would hurt, for example, shoe saleswomen in department stores. She recounted her own experiences selling shoes and the grueling demands of customers who demand to try on 20 shoes for every pair purchased.
"Every time you go to a shoe store or your wives go to a shoe store just know who you are voting against," she said.
She also said Graham's amendment was an "insult" to the working class and, referring to the ruling allowing nurses in a nursing home to form a union without the participation of other employees, noted nurses deserve breaks and to work with clean bedpans.
"It is sickening to me that you would sit here and take their rights away," she said.
Graham heatedly responded that the NLRB is a "disaster."
"Oh, please!" Landrieu exclaimed.
Graham said President Obama is giving political payback to union leaders and that the NLRB is undermining the economic recovery. He suggested Obama will not be "my president" for long.
Landrieu, Pryor and Sens. Barbara Mikulski Barbara Ann MikulskiTwo women could lead a powerful Senate spending panel for first time in history Harris invites every female senator to dinner next week Will the real Lee Hamiltons and Olympia Snowes please stand up? MORE (D-Md.) and Ben Nelson (D-Neb.) later voted for a GOP amendment that would delay for a year a Labor Department regulation meant to restrict the use of immigrant workers.
Mikulski and Landrieu said that seafood industry would suffer from the ruling. At one point Mikulski talked at length about the life cycle of the blue crab.
That amendment, sponsored by Sen. Richard Shelby (R-Ala.), was approved.
— Kevin Bogardus contributed
This story was last updated at 12:16 p.m.
Tags Barbara Mikulski Lindsey Graham Mary Landrieu Mark Pryor


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Wellbeing Check Ins
If you or one of your colleagues are feeling unsure as to whether you need some support, why not email us for a free, confidential wellbeing check in? We are at the end of the phone to talk through any struggles or worries you might have or just to talk through the options available to you. We are here to support your mental health and wellbeing, whether you are struggling with anxiety, stress, low mood or anything else, our wellbeing check in offer can signpost you to the support you might need.
New free wellbeing offers for everyone - New Year focus on keeping well!
Addressing your own wellbeing in busy and stretched roles can often feel overwhelming, but taking small and simple steps can help to move the dial on your emotional and physical health. Our trusted partners HALSA and the Cellar Trust have now released their free Health and Wellbeing Programmes for 2022 which are open to everyone working in the West Yorkshire health and care system – including volunteers and people who work in local authorities.
Between February and April HALSA is offering 23 different sessions with more than 45 different time and date slots to choose from. You can find out more about the HALSA programme by viewing and downloading this flyer. You can book onto the HALSA events here.
These sessions will be:
Delivered by specialist external trainers from Hälsa Wellbeing
Live Webinar Sessions (log in from anywhere)
Focused on fast and practical tips.
The Cellar Trust e-learning modules are free for anyone in West Yorkshire. Register for any of these great offers on Living Well Training – Training for a healthy mind:
Stress Buster
This course focuses on understanding the effects of stress and how stress can influence the way we think, feel and behave. Advice and tips on how to manage and control stress in addition to providing information about further help and support.
This course provides guides for wellbeing at work aimed at employees, managers and senior leaders. The aim is to improve employee understanding of wellbeing in the workplace and provide helpful guidance and advice on a wide range of topics. These include facts and figures, impact of work on our wellbeing, signs of stress, identifying workplace stressors and promoting wellbeing at work. The brief guide for employers and managers provides additional information for supporting and managing staff wellbeing within the workplace including spotting warning signs of poor mental wellbeing and outlining roles and responsibilities for mental wellbeing within organisations.
Skills for Being Well in Adversity
This course provides skills, knowledge and information to help manage and improve physical and mental wellbeing with a range of practical tools/strategies to help individuals gain insight into their own wellbeing. Topics covered include managing stress and anxiety, identifying triggers, self-care planning and dealing with difficult feelings, thinking styles and beliefs.
For teams and individuals
Leading Mental Wellbeing Together webinar
Leading mental wellbeing is everyone's responsibility and this webinar provides all the skills and knowledge to manage mental wellbeing. This two-hour interactive facilitated webinar explores the concept of mental wellbeing within the context of leadership and the ongoing response to the Covid-19 pandemic. The course provides leaders with information and guidance on managing wellbeing in the workplace, using up-to-date research evidence and best practice. Topics covered during the session include recognising the signs of poor mental wellbeing, self-care, leading through adversity, as well as practical wellbeing tools which can be used with colleagues.
To book an individual place or a bespoke session for your team email bdcccg.wyh.mentalwellbeinghub@nhs.net or for more information visit Living Well Training – Training for a healthy mind. (Please note- places are limited).
February dates- Thurs 3 Feb (10am-12noon), Tues 8 Feb (1pm-3pm), Thurs 10 Feb (10am-12noon), Tues 15 Feb (1pm-3pm), Thurs 17 Feb (10am-12noon).
As someone who works or volunteers in health and care services, you know that you and your team have been at the forefront of the Covid19 pandemic. Your efforts may have come at a personal cost to your mental health and wellbeing.
If you are worried about your mental or emotional wellbeing, you can use the self-referral option into our psychological assessment and therapy service, specially set up to help you.
The self-referral process asks you to complete a couple of questionnaires that might help you decide if such support would be useful for you at this time. If you choose to proceed, you can fill in a referral form that asks you to share a bit of information with us so that we can match your needs with a therapist that is best placed to help you.
You must complete the form yourself as it asks about your personal experience. The information will be forwarded securely to our therapy team who will seek to be in touch with you within 72 hours (excluding bank holidays) with next steps.
If you would prefer to have an initial conversation with a trained listener about your situation or would like help to complete the referral form please call 0808 196 3833 or click here to chat online. Telephone and chat are available 8am – 8pm every day.
The MindEd COVID-19 Resilience Hub has content designed to be rapidly accessible and provide quick help when needed.
Try the Every Mind Matters Quiz - 5 questions in this interactive quiz to get top tips and advice for you.
Improving Access to Psychological Therapies (IAPT) services
Most people won't need to access psychological therapies, but it is important to get help if you need it. If you have been exposed to highly stressful, traumatic or frightening experiences at work you are more likely to develop post-traumatic stress disorder (PTSD).
Symptoms include reliving what has happened through vivid dreams or flashbacks, feeling very anxious, angry, irritable and guilty. Some people will feel low in mood, have trouble sleeping and have physical symptoms.
For many people these symptoms will improve over time, but if they don't improve after one month, you can be referred, or self-refer for psychological therapy. Services will also work with you if you are suffering from depression or anxiety.
Find out more about IAPT and your nearest IAPT service here.
If you need help now, you can get in touch phoning the helpline or using the chat function. The service is open from 8am to 8pm, every day, including weekends and bank holidays.
Talk to us now anonymously
Or call our freephone number 0808 1963833.
Outside of these hours, you can get help from the Samaritans, on freephone 116 123.
Support for families and carers
Our Working Carers section has been designed to help working carers look after the person they care for, and themselves, during these unprecedented times. The areas covered are:
Money worries;
Your rights as an employee;
Looking after your mental health;
Looking after your physical health;
Healthy eating;
Looking after the person you care for;
Looking after children's mental and physical health; and
Trying to get back to 'normal'.
Parenting during the pandemic guide
Relationship Matters - When two people share responsibility for a child, no matter whether they are still together or separated, the relationship between them has a direct impact on the mental health of their children.
A national guide for key worker parents and carers when talking to children about Covid-19
Children's guide to coronavirus - A guide for parents and carers to share with their children. It answers questions about coronavirus, explains to children how to stay safe and protect others, and helps them to make the best of their time at home.
Domestic violence - The stress of lockdown means that domestic violence is on the rise. If this is happening to you a friend or a family member it's important to know that you are not alone please this Domestic Violence Information Sheet for support and further help.
Home-schooling - A resource for families with children of different ages and areas of interest
How to maintain your wellbeing as a carer (during the COVID-19 pandemic) - guide and useful links
Think Ninja is a free app which will bring self-help knowledge and skills to children and young people (10-18 years old) who may be experiencing increased anxiety and stress during the crisis.
To help you think about the different ways and people that can help you in an emergency, if you look after someone who couldn't manage without your support.
Samaritans building resilience video (below)
The Hub is offering a new support group as part of our Critical Incident Staff Support Pathway (CrISSP). We know that events at work can impact on home and personal relationships. If this describes you or colleagues you know and you would be interested in attending a small group with other staff who have been impacted in this way please have a look at this poster and register your interest with us at bdcccg.wyh.mentalwellbeinghub@nhs.net
Resources for colleagues from ethnic minority backgrounds
As part of the national health and wellbeing programme for our NHS people, and working in collaboration with our staff networks, we have developed a number of bespoke support offers for NHS colleagues from ethnic minority backgrounds. Click through to find out more.
BAME Network Podcast:
'Can You Hear Me?' is a new podcast giving a voice to the diverse talent working to improve health and care for people in West Yorkshire and Harrogate, presented by our Black, Asian and Minority Ethnic Staff Network.
Episode 1 is presented by network member Fatima Khan-Shah, Programme Lead for the Unpaid Carers and Personalised Care Programmes.
Fatima was joined by some fantastic people to get the series underway, including Dr Sal Uka, Medical Lead for The West Yorkshire Association of Acute Trusts; Dr Habib Naqvi; Deputy Director for the NHS Workforce Race Equality Standard at NHS England; Yvonne Coghill, Director for WRES Implementation at NHS England, and Nyoka Fothergill, Leeds City Council.
Listen to episode 1 here.
Episode 2 is about carers, ahead of Carers Week which starts on
It features presenter Mo Sattar, a GP in Leeds and member of the West Yorkshire and Harrogate Health and Care Partnership's BAME Staff Network, in conversation with Owen Williams OBE, Chief Executive at Calderdale and Huddersfield NHS Foundation Trust, and Sherone Phillips from the Palliative and End of Life Team within the Personalised Care Group at NHS England, who is also a carer herself.
Episode 3 is about white privilege
Usha Kaushal, Lead Pharmacist and an Operational Lead at Yorkshire Ambulance Service, leads the discussion with Samantha Allen, Chief Executive of Sussex Partnership NHS Foundation Trust, Wallace Sampson, Chief Executive of Harrogate Borough Council and Rob Webster, Lead CEO for West Yorkshire and Harrogate Health and Care Partnership and Chief Executive at South West Yorkshire Partnership Foundation Trust.
Episode 4 is around workforce issues
Delphine Arinze from The Leeds Teaching Hospitals NHS Trust talks with Uduak Archibong, Professor of Diversity at the University of Bradford, Kez Hayat from Bradford Teaching Hospitals NHS Foundation Trust and Sayed Ahmed from Leeds and York Partnership NHS Foundation Trust. Conversation on this episode was centred around workforce issues and the experience of being from a Black, Asian and minority ethnic community in the health and care workforce.
For further information, guidance and training on risk assessment please see the following resources:
Covid-19 - staff risk assessment training support
Additional management guidance for risk assessment
Risk assessment - at risk staff (Covid-19)
Managing your mental and physical health and wellbeing with Long Covid
Dr Sari Harenwall, Senior Clinical Psychologist and COVID Rehab Clinical Lead for the Primary Care Wellbeing Service at Bradford District Care NHS Foundation Trust and colleagues in Bradford have recorded a series of videos on how to manage your physical and mental health and wellbeing with Long Covid. These videos are also available to watch on YouTube.
Supporting your recovery after COVID-19
As you find yourself recovering from COVID-19 you may still be coming to terms with the impact the virus has had on both your body and mind.
These changes should get better over time, some may take longer than others, but there are things you can do to help.
Your COVID Recovery helps you to understand what has happened and what you might expect as part of your recovery. Click here to find out more.
Information for Care Home workers
Staff in care homes, whether Council or privately owned, are part of our West Yorkshire and Harrogate Health and Care Partnership. The COVID 19 pandemic has been especially challenging for those working and volunteering in care homes, whatever their role - from the lack of PPE, to anxieties around protecting loved ones and those they care for, the relentless pressure on keeping residents safe has taken its toll. Whilst we are hopefully coming out the other side of the pandemic, we know that this brings rise to other concerns such as anxieties around vaccinations, impact of losing people they have cared for or struggled with the level of care they have been able to provide.
The Staff Mental Health and Wellbeing Hub has produced a newsletter for care homes which you can read here together with a covering letter from our consultant psychologists.
We have also recorded a podcast about the issues and challenges facing care home staff as a result of the pandemic. You can listen to that here.
You can find more resources for people who work as carers and you can get confidential help every day from trained listeners on our free support line 0808 196 3833, open 8am -8pm or online chat here.
Information for people working in Primary Care
Primary Care includes everyone working or volunteering in general practice, community pharmacy, dental practices and opticians' practices, whether NHS or privately owned. The COVID 19 pandemic has been especially challenging for all of you, whatever your role - from the lack of PPE, to anxieties around protecting patients and service-users, the relentless pressure on keeping everyone safe has taken its toll. Whilst we are hopefully coming out the other side of the pandemic, we know that this brings rise to other concerns such as increasing workload, abuse from people you are trying to help, staff leaving and pressure throughout the system.
The Staff Mental Health and Wellbeing Hub has produced a newsletter for Primary Care which you can read here together with a covering letter from our consultant psychologists.
We are also making a podcast about the issues and challenges facing Primary Care staff as a result of the pandemic.
If you are a manager or leader working in Primary Care, you can find resources to help you support your team here.
Confidential help is available every day from trained listeners on our free support line 0808 196 3833, open 8am -8pm or online chat here. You can self-refer for an assessment by our psychological therapy service here.
West Yorkshire Health and Care Partnership @WYpartnership
Swap after school cupboard attacks for healthier snacks with the NHS Food Scanner app. Download it for free today:… twitter.com/i/web/status/1…
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The plan is very easy and straight forward. With 2 full examinations of your pet per year you can feel assured that they are fit and healthy. With once a month easy treatments your pet and family are protected against ALL parasites. The plan is also perfect for puppies and kittens and includes their primary vaccinations.
With our plans you save 25% on preventative health care compared to pay as you go. Also you can spread the cost with monthly direct debit payments.


You all uploaded your thoughts on the best wireless routers this week, and four emerged from the packet to make the finals. Two of them are mesh systems, and two are more traditional routers, so we're going to set up a firewall and have two distinct votes, and two champions.
Google Wifi. Even though my house is a typical '50s ranch style, it has some stubborn dead spots that no previous router has been able to reach. I've gone through top-of-the-range NetGear and Linksys stuff, AirPorts aplenty, and Google's first attempt (Home), all without success.
Then I got a deal on three Google Wifi pods. Setup was simple and scattering them to the corners of the house gave me full coverage at very respectable speeds. My cable connection is 300 Mbps (I regularly get about 340 Mbps due to buying my own cable modem and using the full channel range - up yours, Spectrum). This translates to 250-300 Mbps in the room with the cable modem; 150-200 Mbps in the furthest bedroom (containing one of the satellite pods); and 50-100 Mbps in my office above the detached garage, where the other pod is located.
I compare these numbers with my previous (non mesh) routers, which barely topped 200 Mbps in the room with the cable modem; an intermittent 50ish Mbps in the bedroom (with lots of dropout), and nothing at all in the office.
In addition to providing great connectivity and speed, the Google pods are zero maintenance; I haven't needed to restart them once in a year of operation.
Really love my Netgear Orbi system (RBK50 model). AC3000 WiFi speeds, 4 ethernet ports on EACH satellite. Ability to do Gigabit hardwired backhaul, which is great for large, older houses like mine that have unknown materials in the walls.
I'll agree. I've gone the Asus with Merlin approach, I've gone the Linksys with Tomato approach, Netgears with 15 antennas, I even kicked around a reliable old Airport Extreme for a few years a while back.
But for my money and skill level (know just enough to tweak things without being an IT guy), the Orbi has been the best mix of easy, reliable and strong/consistent coverage. Could I do a "true" mesh network with Ubiquiti or the like? Probably. But I have a larger home, not a hotel and conference center. Nor do I want to be a dedicated IT manager just to make sure that kids can stream Netflix in the basement while I watch DirecTV Now on my upstairs AppleTV.
Editor's note: Yes, you can technically set these up as a mesh network, but they aren't sold in combo packs or typically marketed in that way, and the process isn't as seamless as a true mesh system.
Fast, easy to set up, and extremely strong wireless signal thanks to excellent beam-forming. I own several - one for home use, and several more for controlling sound systems in locations awash with wifi, bluetooth, and 2.4GHz wireless microphones (I don't have control of those).
I've run a number of other routers in each location, and the AC1750 is the only one that's managed to keep a connection once the 2.4GHz wireless mics get turned on.


The greatest declaration in history is, "It is finished!" Jesus' last words speak through eternity: Mission accomplished. The power of the cross is more than head knowledge. The more you meditate on this great work, the more you step into the flow of its blessing and power for your life. There is redemption in the cross. There is healing glory in the cross. There is victory in the cross. There is power over sin and death in the cross. There is resurrection life in the cross.
The Resurrection is the most awesome and glorious moment in history. It is the final word over every situation. Jesus is the resurrection and the life. He is the same yesterday, today, and forever. This is good news for you. He has defeated every power of death and darkness.
God is invisible, but He's not intangible! There is an atmosphere around His throne that we can enter into and carry with us wherever we go. It's what I call the "Glory Bubble". It's where His miracles dwell. It's where His promises reside. It's His Presence, and it surrounds us like a bubble. When you enter into this realm, everything changes in the living substance of His glory.
These are indeed perilous times (2 Timothy 3) but God has a plan for America. He has placed you here in this hour because you are part of His solution. This two-part, 4-disc teaching series by Mahesh and Bonnie Chavda will show you God's prognosis for what ails our country. There is a prescription for America, and the key ingredient is YOU!
The Almighty God, translated as "El Shaddai", is our ultimate reality. The blessings God gave to Abraham are for you and your children. Learn how to strengthen your faith, and go to the next level for miracles as you go from glory to glory.
Through Scripture and dramatic testimonies from his own life and ministry, Mahesh Chavda removes the confusion that often surrounds healing, builds your faith and equips you for miracles.


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Griffith Health Institute
Tackling issues of sexuality among people with dementia
Published March 21, 2013 April 4, 2013 AuthorContributed
Managing the delicate issue of sexual expression amongst people with dementia is the focus of a new education resource produced by Griffith University researcher Dr Cindy Jones.
The first resource of its kind and the subject of funding from the Department of Health and Aging and Queensland Dementia Training and Study Centres (DTSC), Sexualities and Dementia: Education Resource for Health Professionals is aimed at assisting health professionals working across care settings.
Based on national and international literature and research by Dr Jones from Griffith's Centre for Health Practice Innovation, the free eLearning resource aims to increase awareness of intimacy, sexualities and sexual behaviours specific to people with dementia and to guide their carers on how to better support them.
First resource of its kind to address the LGBTI community
It is also the first e-learning resource in Australia to include concerns and issues faced by non-heterosexual people with dementia, including those who identify as lesbian, gay, bisexual, transgender and intersex (LGBTI).
"It became apparent that there was a real gap in the education of health professionals regarding sexualities and dementia, with no other comprehensive resource which specifically addresses the needs of this group," says Dr Jones.
"Our research revealed that when presented with these challenging situations, interventions taken by health professionals are often quite reactive rather than proactive, with many staff simply not knowing what to do.
"The resource provides a framework to guide the development of policy and guidelines to address issues of sexual expression in care settings. There is also a facilitator guide providing workshop information for which 'train the trainer' education can be applied."
"The important message is that expression of sexuality is a part of life no matter what age we are and that older people with dementia therefore also need to be considered," says Dr Jones.
Meredith Gresham, Senior Dementia Consultant – Research & Design, HammondCare was part of the expert panel reviewing the new resource.
"The modules really produced thoughtful discussion. The conclusion points are fabulous. I would like to congratulate Dr Jones on the work," she said.
This resource is available from the National DTSC website
http://dtsc.com.au/new-resource-coming-soon/ as a PDF document to
download, and as a CD-Rom. For inquiries or a hard copy, please contact
Sandra Jeavons on 07 3138 5943 [email protected]
Categories Features, Griffith Health, Griffith Health InstituteTagged Dr Cindy Jones, Griffith Health Institute, sexuality and dementia
Combined probiotics use may help reduce pre-term infant mortality
Probiotic and prebiotic supplementation may help reduce the mortality rate of preterm infants if given in the first few months of life.
Australian Institute for Suicide Research and Prevention
Digital solutions may aid athlete's mental health
Digital mental health platforms including the use of smartphones and online behaviour analysis could help improve the mental health of athletes especially during the COVID-19 pandemic.


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