text:
\section{Introduction}

\PARstart{c}onstant-composition codes (CCCs) are a special type of
constant-weight codes  (CWCs) which  are important in coding theory. The
class of constant-composition codes includes the important
permutation codes and have attracted recent interest due to their
numerous applications, such as in determining the zero error
decision feedback capacity of discrete memoryless channels
\cite{Telatar:1990}, multiple-access communications
\cite{D'yachkov:1984}, spherical codes for modulation
\cite{Ericson:1995}, DNA codes
\cite{King:2003,Milenkovic:2006,Chee:2008}, powerline communications
\cite{Chu:2004,Colbourn:2004}, frequency hopping \cite{Chu:2006}, frequency permutation arrays \cite{Huczynska:2006}, and
coding for bandwidth-limited channels \cite{Costello:2007}.

Systematic study began in late 1990's \cite{Bogdanova:1998,Svanstrom:1999,Bogdanova_Ocetarova:1998}. Today, various methods have been applied to
the problem of determining the maximum size of a
constant-composition code, such as computer search methods \cite{Bogdanova:2003}, packing designs
\cite{Ding:2005(2),Chu:2006,Ding:2006,Yan:2008,YinTang:2008,Wen:2009,Yan:2009,Huczynska:2010},
tournament designs \cite{YinYan:2008}, polynomials and nonlinear
functions
\cite{Ding:2005,Ding:2005(3),Chu:2006,Ding:2008,YDing:2008},
difference triangle sets \cite{Chee:2010}, PBD-closure methods
\cite{Yeow:2007,Yeow:2008} and some other methods \cite{Svanstrom:2000,Luo:2003}.

In the paper of Svanstr{\"o}m et al. \cite{Svanstrom:2002}, some
methods for providing upper and lower bounds on the maximum size
$A_3(n,d,\overline{w})$ of a ternary code with length $n$, minimum
Hamming distance $d$, and constant composition $\overline{w}$ were
presented. The sizes of optimal ternary constant-composition codes
with weight three have been determined completely by Chee, Ge and
Ling in \cite{Yeow:2008}. The sizes of optimal ternary
constant-composition codes with  weight four and distance five
have been determined completely by Gao and Ge in \cite{Gao:2010}.

The sizes of optimal quaternary  constant-composition codes with
weight three have been determined almost completely by Chee, Ge
and Ling in \cite{Yeow:2008} with  four lengths in doubt. Recently,
 the problem of determining the sizes for optimal quaternary constant-composition codes with  weight four and
distance seven has been settled by Chee, Dau, Ling and Ling in
\cite{Chee:2010}.  In this paper, we will concentrate our
attention on quaternary CCCs  with   weight four and  distances
five or six. The problem is solved leaving only five lengths
undetermined for the case of distance five. Previously, the
results on the sizes of such quaternary constant-composition codes
were scarce.

\section{Preliminaries}
\vskip 10pt

\subsection{Definitions and Notations}
The set of integers $\{i,i+1,\ldots,j\}$ is denoted by $[i,j]$. The
ring $\bbZ/q\bbZ$ is denoted by $\bbZ_q$. The notation $\Lbag \cdot
\Rbag$ is used for multisets.

All sets considered in this paper are finite if not obviously
infinite. If $X$ and $R$ are finite sets, $R^X$ denotes the set of
vectors of length $|X|$, where each component of a vector $\vu\in
R^X$ has value in $R$ and is indexed by an element of $X$, that is,
$\vu=(\vu_x)_{x\in X}$, and $\vu_x\in R$ for each $x\in X$. A
{\it $q$-ary} code of {\it length} $n$ is a set $\C \subseteq \bbZ_q^X$ for some
$X$ with size $n$. The elements of $\C$ are called {\it codewords}. The
{\it Hamming norm} or the {\it Hamming weight} of a vector $\vu\in\bbZ_q^X$ is
defined as $\|\vu\|=| \{x\in X: \vu_x\not=0\}|$. The distance
induced by this norm is called the {\it Hamming distance}, denoted $d_H$,
so that $d_H(\vu,\vv)=\| \vu-\vv \|$, for $\vu,\vv\in\bbZ_q^X$. The
{\it composition} of a vector $\vu\in \bbZ_q^X$ is the tuple
$\overline{w}=[w_1,\ldots,w_{q-1}]$, where $w_j=|\{x\in X:
\vu_x=j\}|$. For any two vectors $\vu,\vv\in\bbZ_q^X$, define their
{\it support} as $\supp(\vu,\vv)=\{x\in X:\vu_x\not=\vv_x\}$. We write
$\supp(\vu)$ instead of $\supp(\vu,\vzero)$ and also call
$\supp(\vu)$ the support of $\vu$.

A code $\C$ is said to have minimum distance $d$ if $d_H(\vu,\vv)\geq d$ for
all distinct $\vu,\vv\in\C$. If $\|\vu\|=w$ for every codeword $\vu\in\C$,
then $\C$ is said to be of (constant) {\it weight $w$}. A $q$-ary code
$\C$ has {\it constant composition $\overline{w}$} if every codeword in
$\C$ has composition $\overline{w}$. A $q$-ary code of length $n$,
distance $d$, and constant composition $\overline{w}$ is referred to
as an $(n,d,\overline{w})_q$-code. The maximum size of an
$(n,d,\overline{w})_q$-code is denoted as $A_q(n,d,\overline{w})$ and
the $(n,d,\overline{w})_q$-codes achieving this size are called
{\it optimal}. Note that the following operations do not affect distance
and weight properties of an $(n,d,\overline{w})_q$-code:
\begin{enumerate}[(i)]
\item reordering the components of $\overline{w}$, and
\item deleting zero components of $\overline{w}$.
\end{enumerate}
Consequently, throughout this paper, we restrict our attention to
those compositions $\overline{w}=[w_1,\ldots,w_{q-1}]$, where
$w_1\geq \cdots\geq w_{q-1}\geq 1$.

Suppose $\vu\in\bbZ_q^X$ is a codeword of an
$(n,d,\overline{w})_q$-code, where
$\overline{w}=[w_1,\ldots,w_{q-1}]$. Let $w=\sum_{i=1}^{q-1} w_i$.
We can represent $\vu$ equivalently as a $w$-tuple $\langle
a_1, a_2, \ldots, a_w\rangle\in X^w$, where
\begin{align*}
\vu_{a_1}=\cdots=\vu_{a_{w_1}} &=1, \\
\vu_{a_{w_1+1}}=\cdots=\vu_{a_{w_1+w_2}} &=2, \\
\vdots \\
\vu_{a_{\sum_{i=1}^{q-2} w_i+1}} =\cdots=\vu_w &= q-1.
\end{align*}
Throughout this paper, we shall often represent codewords of
constant-composition codes in this form. This has the advantage of
being more succinct and more flexible in manipulation.

\subsection{General Bounds}
\vskip 10pt

\begin{lemma}[Chee et al. \cite{Yeow:2008}]\label{bound1}
\begin{equation*}
\begin{split}
& A_q(n,d,[w_1,\ldots,w_{q-1}])=\\
& \begin{cases} \binom{n}{\sum_{i=1}^{q-1}w_i}
\binom{\sum_{i=1}^{q-1}w_i}{w_1,\ldots,w_{q-1}}, & \text{if $d \leq 2$}\\
\left\lfloor \frac{n}{\sum_{i=1}^{q-1}w_i} \right\rfloor, & \text{if
$d=2
\sum_{i=1}^{q-1}w_i$}\\
1, & \text{if $d \geq 2\sum_{i=1}^{q-1}w_i+1$}.
\end{cases}
\end{split}
\end{equation*}
\end{lemma}
\vskip 10pt

The following Johnson-type bound has been proven for
constant-composition codes.

\vskip 10pt
\begin{lemma}[Svanstr{\"o}m et al. \cite{Svanstrom:2002}]\label{bound2}
\begin{equation*}
A_q(n,d,[w_1,\ldots,w_{q-1}]) \leq \frac{n}{w_1}A_q(n-1,d,[w_1-1,
\ldots ,w_{q-1}]).
\end{equation*}
\end{lemma}
\vskip 10pt

Moreover, we have the following results.

\vskip 10pt
\begin{lemma}[Chee et al. \cite{Yeow:2007}] \label{bound3}
\begin{equation*}
\begin{split}
& A_q(n,d,[w_1,\ldots,w_{q-1}]) \leq\\
& \begin{cases} \left\lfloor \frac{n}{w_1} \left\lfloor
\frac{n-1}{\sum_{i=1}^{q-1}w_i-1} \right\rfloor \right\rfloor &
\text{if
$d=2\sum_{i=1}^{q-1}w_i-3$}\\
\left\lfloor \frac{n}{w_1} \left\lfloor \frac{n-1}{w_1-1}
\right\rfloor \right\rfloor & \text{if $d=2\sum_{i=1}^{q-1}w_i-2$}.\\
\end{cases}
\end{split}
\end{equation*}
\end{lemma}
\vskip 10pt

\begin{corollary}
\begin{equation*}
\begin{split}
A_4(n,5,[2,1,1]) \leq n \left\lfloor \frac{n-1}{2}\right\rfloor\\
\end{split}
\end{equation*}
\begin{equation*}
A_4(n,6,[2,1,1]) \leq \left\lfloor \frac{n}{2} \left\lfloor
\frac{n-1}{3} \right\rfloor \right\rfloor.
\end{equation*}
\end{corollary}
\begin{proof}
The first equation follows by Lemma \ref{bound2} and Lemma
\ref{bound1} where $w_1=1, w_2=2$ and $w_3=1$.

The second equation follows by Lemma \ref{bound3} and Lemma
\ref{bound1} where $w_1=2, w_2=1$ and $w_3=1$.
\end{proof}
\vskip 10pt

In the following, we denote $U(n,5,[2,1,1])=n \left\lfloor
\frac{n-1}{2}\right\rfloor$ and $U(n,6,[2,1,1])=\left\lfloor
\frac{n}{2} \left\lfloor \frac{n-1}{3} \right\rfloor \right\rfloor$.

\subsection{Designs}
\vskip 10pt

Our recursive construction is based on some combinatorial structures
in design theory. The most important tools are pairwise balanced
designs (PBDs) and group divisible designs (GDDs).

Let $K$ be a subset of positive integers and let $\lambda$ be a
positive integer. A {\it pairwise balanced design} (PBD($v,K,\lambda$) or
($K,\lambda$)-PBD) of order $v$ with block sizes from $K$ is a pair
($\V,\B$), where $\V$ is a finite set (the point set) of cardinality
$v$ and $\B$ is a family of subsets (blocks) of $\V$ that satisfy
(1) if $B\in \B$, then $|B|\in K$ and (2) every pair of distinct
elements of $\V$ occurs in exactly $\lambda$ blocks of $\B$. The
integer $\lambda$ is the index of the PBD. The notations PBD($v,K$)
and $K$-PBD of order $v$ are often used when $\lambda=1$. If an
element $k\in K$ is ``starred'' (written $k^\star$), it means that
the PBD has exactly one block with size $k$.

\vskip 10pt
\begin{lemma}[\cite{HBPBD3}]
\label{PBD4-6} For any integer $v\geq 10$, a $(v,\{4,5,6\},1)$-PBD
exists with exceptions $v\in \{7,8,9,10,$ $11,12,14,15,18,19,23\}$.
\end{lemma}
\vskip 10pt

\begin{lemma}[Ling, Zhu, Colbourn, Mullin \cite{Ling:1997}]
\label{PBD5-9} For any integer $v\geq 10$, a
$(v,\{5,6,7,8,9\},1)$-PBD exists with exceptions $v\in
[10,20]\cup[22,24]\cup[27,29]\cup[32,34]$.
\end{lemma}
\vskip 10pt

\begin{lemma}[Rees, Stinson \cite{Rees-Stinson:1989}] \label{PBD4*} There exists a
$(v,\{4,w^\star\},1)$-PBD with $v>w$ if and only if $v\geq 3w+1$,
and:
\begin{enumerate}
\item[(i)] $v\equiv 1$ or $4\pmod{12}$ and $w\equiv 1$ or $4\pmod{12}$;  or
\item[(ii)] $v\equiv 7$ or $10\pmod{12}$ and $w\equiv 7$ or $10\pmod{12}$.
\end{enumerate}
\end{lemma}
\vskip 10pt

Let $K$ and $G$ be sets of positive integers and let $\lambda$ be a
positive integer. A {\it group divisible design} of index $\lambda$ and
order $v$ (($K,\lambda$)-GDD) is a triple ($\V,\G,\B$), where $\V$
is a finite set of cardinality $v$, $\G$ is a partition of $\V$ into
parts (groups) whose sizes lie in $G$, and $\B$ is a family of
subsets (blocks) of $\V$ that satisfy (1) if $B\in \B$ then $|B|\in
K$, (2) every pair of distinct elements of $\V$ occurs in exactly
$\lambda$ blocks or one group, but not both, and (3) $|\G|>1$. If
$v=a_1g_1+a_2g_2+ \dots +a_sg_s$, and if there are $a_i$ groups with
size $g_i$, $i=1, 2, \ldots, s$, then the $(K,\lambda)$-GDD is of
type $g_1^{a_1}g_2^{a_2}\dots g_s^{a_s}$. This is exponential
notation for the group type. If $K=\{k\}$, then the
($K,\lambda$)-GDD is a ($k,\lambda$)-GDD. If $\lambda=1$, the GDD is
a $K$-GDD. Furthermore, a ($\{k\},1$)-GDD is a $k$-GDD. A {\it parallel
class} or {\it resolution class} is a collection of blocks that partition
the point-set of the design. A GDD is {\it resolvable} if the blocks of
the design can be partitioned into parallel classes. A resolvable
GDD is denoted by RGDD.

\vskip 10pt
\begin{lemma} [\cite{HBGDD}]
\label{4RGDD38} There exists a $4$-RGDD of type $g^u$ for each $(g,u)\in
\{(3,8),(4,7)\}$.
\end{lemma}

\vskip 10pt
\begin{lemma} [\cite{HBGDD}]
\label{4GDD} There exists a $4$-GDD of type $g^u m^1$ for each
$(g,u,m)\in \{(3,5,0),$ $(4,6,7),$ $(12,4,18),$ $(12,5,18),$
$(15,4,21)\}$.
\end{lemma}
\vskip 10pt

A $\{k\}\text{-GDD}$ of type $n^k$ is also called a {\em transversal
design} and denoted by ${\rm TD}(k,n)$.

\vskip 10pt
\begin{lemma}[\cite{Abel:2007}]
\label{TD} Let $n$ be a positive integer. Then:
\begin{enumerate}
\item[(i)] a TD$(5,n)$ exists if $n\not\in \{2,3,6,10\}$;
\item[(ii)] a TD$(6,n)$ exists if $n\not\in \{2,3,4,6,10,14,18,22$\};
\item[(iii)] a TD$(7,n)$ exists if $n\not\in \{2,3,4,5,6,10,14,15,18,20,$ $22,26,30,34,38,46,60\}$.
\item[(iv)] a TD$(8,9)$ exists.
\end{enumerate}
\end{lemma}
\vskip 10pt

\subsection{Group Divisible Codes}

Given $\vu \in \bbZ_q^X$ and $Y \subseteq X$, the
\emph{restriction of $\vu$ to $Y$}, written $\vu\mid_Y$, is the
vector $\vv \in \bbZ_q^X$ such that
$$\vv_x=
\begin{cases}
\vu_x, & \text{if $x \in Y$} \\
0, & \text{if $x \in X \backslash Y$}. \\
\end{cases}$$

A {\it group divisible code} (GDC) of distance $d$ is a triple
($X,\G,\C$), where $\G=\{G_1,\ldots,G_t\}$ is a partition of $X$
with cardinality $|X|=n$ and $\C\subseteq \bbZ_q^X$ is a $q$-ary
code of length $n$, such that $d_H(u,v) \ge d$ for each distinct
$u,v\in \C$, and $\|u|_{G_i}\|\le 1$ for each $u \in \C$, $1\le i \le
t$. Elements of $\G$ are called groups. We denote a GDC($X,\G,\C$)
of distance $d$ as $w$-GDC($d$) if $\C$ is of constant weight $w$.
If we want to emphasize the composition of the codewords, we denote
the GDC as $\overline{w}$-GDC($d$) when every $u\in \C$ has
composition $\overline{w}$. The type of a GDC($X,\G,\C$) is the
multiset $\Lbag |G|: G\in \G \Rbag$. As in the case of GDDs, the
exponential notation is used to describe the type of a GDC. The size
of a GDC($X,\G,\C$) is $|\C|$. Note that an
$(n,d,\overline{w})_q$-code with size $s$ is equivalent to a
$\overline{w}$-GDC($d$) of type $1^n$ with size $s$.

Constant-composition codes of larger orders can often be obtained from
GDCs via the following two constructions.

\vskip 10pt
\begin{construction}[(Filling in Groups) \cite{Yeow:2008}]
\label{FillGroups} Let $d\leq 2(w-1)$. Suppose there exists a
$w$-${\rm GDC}(d)$ $(X,\G,\C)$ of type $g_1^{t_1}\cdots g_s^{t_s}$
with size $a$. Suppose further that for each $i$, $1\leq i\leq s$,
there exists a $(g_i,d,w)_q$-code $\C_i$ with size $b_i$, then there
exists a $(\sum_{i=1}^s t_ig_i,d,w)_q$-code $\C'$ with size
$a+\sum_{i=1}^s t_ib_i$. In particular, if $\C$ and $\C_i$, $1\leq
i\leq s$, are of constant composition $\overline{w}$, then $\C'$ is
also of constant composition $\overline{w}$.
\end{construction}
\vskip 10pt

\begin{construction}[(Adjoining $y$ Points) \cite{Yeow:2008}]
\label{AdjoinPoints} Let $y\in\bbZ_{\geq 0}$. Suppose there exists a
(master) $w$-${\rm GDC}(d)$ of type $g_1^{t_1}\cdots g_s^{t_s}$ with size $a$, and suppose the following (ingredients) also exist:
\begin{enumerate}[(i)]
\item a $(g_1+y,d,w)_q$-code with size $b$,
\item a $w$-${\rm GDC}(d)$ of type $1^{g_i} y^1$ with size $c_i$ for each $2\leq i\leq s$,
\item a $w$-${\rm GDC}(d)$ of type $1^{g_1} y^1$ with size $c_1$ if $t_1\geq 2$.
\end{enumerate}
Then, there exists a $(y+\sum_{i=1}^s t_ig_i,d,w)_q$-code with size
\begin{equation*}
a+b+(t_1-1)c_1+\sum_{i=2}^{s} t_ic_i.
\end{equation*}
Furthermore, if the master and ingredient codes are of constant
composition, then so is the resulting code.
\end{construction}
\vskip 10pt

The  following two  constructions are useful for generating  GDCs of larger orders from smaller ones.

\begin{construction}[(Fundamental Construction) \cite{Yeow:2008}]
\label{FundCtr} Let $d\le 2(w-1)$, $\D=(X,\G,\A)$ be a (master) GDD,
and $\omega: X\rightarrow \bbZ_{\ge0}$ be a weight function. Suppose
that for each $A\in \A$, there exists an (ingredient) $w$-GDC($d$)
of type $\Lbag \omega(a): a\in A \Rbag$. Then there exists a
$w$-GDC($d$) $\D^{*}$ of type $\Lbag \sum_{x\in G}{\omega(x)}:G\in
\G \Rbag$. Furthermore, if the ingredient GDCs are of constant
composition $\overline{w}$, then $\D^{*}$ is also of constant
composition $\overline{w}$.
\end{construction}

\vskip 10pt
\begin{construction}
\label{Inflation} Suppose there exists a $w$-GDC($d$) of type
$g_1^{t_1}\dots g_s^{t_s}$ with size $a$. Suppose further that there
exists a TD($w,m$), then there exists a $w$-GDC($d$) of type
$(mg_1)^{t_1}\dots (mg_s)^{t_s}$ with size $am^2$. If the original
GDC is of constant composition $\overline{w}$, then so is the
derived GDC.
\end{construction}
\vskip 10pt

\section{Determining the Value of $A_4(n,5,[2,1,1])$}
\vskip 10pt

\subsection{Room Square Construction}
Let $S$ be a set of $n+1$ elements (symbols). A {\it Room square} of side
$n$ (on symbol set $S$), RS($n$), is an $n \times n$ array, $F$,
that satisfies the following properties:
\begin{enumerate}
\item every cell of $F$ either is empty or contains an unordered
pair of symbols from $S$,
\item each symbol of $S$ occurs once in each row and column of $F$,
\item every unordered pair of symbols occurs in precisely one cell
of $F$.
\end{enumerate}

\vskip 10pt
\begin{lemma} [\cite{skewRS}]
\label{RS} A Room square of side $n$ exists if and only if $n$ is
odd and $n\neq 3$ or $5$.
\end{lemma}
\vskip 10pt

   From  each filled cell ($r,c$) of an RS($n$),  $R$, one can obtain an underlying  $4$-subset $\{i,j,r,c\}$, where  $\{i,j\}$ occurs in column $c$ and row $r$ of $R$. A Room
square  of side $n$ is called {\it super-simple} (denoted  by SSRS($n$)),  if for any two filled cells  ($r_1,c_1$) and ($r_2,c_2$) containing the  symbols $\{i_1,j_1\}$ and
$\{i_2,j_2\}$ respectively,  the underlying $4$-subsets $\{i_1,j_1,r_1,c_1\}$ and $\{i_2,j_2,r_2,c_2\}$  share at most two common elements.

\vskip 10pt
\begin{theorem}
\label{SSRStoCodes} For an odd integer $n$, suppose there exists an
SSRS($n$), then there exists an  optimal $(n,5,[2,1,1])_4$-code with size
$n(n-1)/2=U(n,5,[2,1,1])$.
\end{theorem}
\begin{proof}
For each filled cell $(r,c)$ of the given SSRS($n$), $R$, we form a codeword $\langle
i,j,r,c\rangle$ of type $[2,1,1]$, where  $\{i,j\}$ occurs in column $c$ and row $r$ of $R$.  We have in total of  $n(n-1)/2$ such codewords. Now, we  prove that these $n(n-1)/2$ codewords form an optimal code of length $n$.

Because $R$ is super-simple, so any two codewords intersect in at
most two coordinates. If the distance between any two codewords, $\langle
i_1,j_1,r_1,c_1\rangle$ and $\langle i_2,j_2,r_2,c_2\rangle$, is
less than $5$, then one of the following five properties  must be satisfied:
\begin{enumerate}
\item $i_1=i_2$ and $j_1=j_2$,
\item $i_1=j_2$ and $i_2=j_1$,
\item $i_l=j_m$ and $r_1=r_2$ where $l,m\in \{1,2\}$,
\item $i_l=j_m$ and $c_1=c_2$ where $l,m\in \{1,2\}$, or
\item $r_1=r_2$ and $c_1=c_2$.
\end{enumerate}
But any of these five properties  has conflicts with the properties of a Room square.
This means  the distance between any two codewords is greater than or
equal to $5$. So these codewords form an optimal code of length $n$
with size $n(n-1)/2=U(n,5,[2,1,1])$.
\end{proof}
\vskip 10pt

If $\{S_1,\ldots,S_n\}$ is a partition of a set $S$, an
{\it $\{S_1,\ldots,S_n\}$-Room frame} is an $|S|\times |S|$ array, $F$,
indexed by $S$, satisfying:
\begin{enumerate}
\item  every cell of $F$ either is empty or contains an unordered
pari of symbols of $S$,
\item  the subarrays $S_i\times S_i$ are empty, for each $1\le i\le
n$ (these subarrays are holes),
\item  each symbol $x\notin S_i$ occurs once in row (or column) s
for each $s\in S_i$, and
\item  the pairs occurring in $F$ are those $\{s,t\}$, where $(s,t)\in (S\times S)\setminus \cup_{i=1}^{n}{(S_i\times S_i)}
$.
\end{enumerate}

A Room square of side $n$ is equivalent to a Room frame of type
$1^n$.

We can define a super-simple Room frame in the same way as a
super-simple Room square.

\vskip 10pt
\begin{theorem}
\label{SSRFtoCodes}
Suppose there exists a super-simple Room frame
of type $2^t$, then there exists an  optimal $(2t,5,[2,1,1])_4$-code with size
$2t(t-1)=U(2t,5,[2,1,1])$.
\end{theorem}
\begin{proof}
For each filled cell $(r,c)$ of the super-simple Room frame of type $2^t$, $R$, we form a codeword $\langle
i,j,r,c\rangle$ of type $[2,1,1]$, where  $\{i,j\}$ occurs in column $c$ and row $r$ of $R$. There are
$2\times 2t(t-1)/2=2t(t-1)$ filled cells in $R$, which equals $U(2t,5,[2,1,1])$.  Hence, we have in total of  $U(2t,5,[2,1,1])$ such codewords. The rest of the
proof is similar to that of Theorem \ref{SSRStoCodes}.
\end{proof}
\vskip 10pt

A {\it starter} in the  abelian group $G$ of odd order (written additively),
where $|G|=g$ is a set of unordered pairs $S=\{\{s_i,t_i\}:1 \le i
\le (g-1)/2 \}$ that satisfies:
\begin{enumerate}
\item $\{s_i:1 \le i \le (g-1)/2\} \cup \{t_i:1 \le i \le
(g-1)/2\}=G\setminus \{0\}$, and
\item $\{\pm (s_i-t_i): 1 \le i \le (g-1)/2\}=G\setminus  \{0\}$.
\end{enumerate}

A {\it strong starter} is a starter $S=\{\{s_i,t_i\}\}$ in the abelian
group $G$ with the additional property that $s_i+t_i=s_j+t_j$
implies $i=j$, and for each $i$, $s_i+t_i \ne 0$. Let
$S=\{\{s_i,t_i\}: 1 \le i \le (g-1)/2\}$ and $T=\{\{u_i,v_i\}: 1 \le
i \le (g-1)/2\}$ be two starters in $G$. Without loss of generality,
assume that $s_i-t_i=u_i-v_i$, for all $i$. Then $S$ and $T$ are called
{\it orthogonal} starters if $u_i-s_i=u_j-s_j$ implies $i=j$, and if $u_i
\ne s_i$ for all $i$. An {\it adder} for the starter $S=\{\{s_i,t_i\}:1
\le i \le (g-1)/2\}$ is an ordered set
$A_S=\{a_1, a_2, \ldots, a_{(g-1)/2}\}$ of $(g-1)/2$ distinct nonzero
elements from $G$ such that the set $T=\{\{s_i+a_i,t_i+a_i\}:1 \le i
\le (g-1)/2\}$ is also a starter in the group $G$. These two
starters $S$ and $T$ are orthogonal starters.

\vskip 10pt
\begin{theorem}[\cite{HBRS}]
If there exist two orthogonal starters in a group of order $n$, then
there exists a Room square of side $n$. If the group is $\bbZ_n$,
then the resulting Room square is cyclic.
\end{theorem}
\vskip 10pt

Let $G$ be an additive abelian group of order $g$, and let $H$ be a
subgroup of order $h$ of $G$, where $g-h$ even. An $h$-{\it frame starter}
of order $g/h$ in $G\setminus H$ is a set of unordered pairs
$S=\{\{s_i,t_i\},1\le i \le (g-h)/2\}$ such that
\begin{enumerate}
\item $\{s_i : 1\le i \le (g-h)/2\} \cup \{t_i : 1\le i \le (g-h)/2\}=G\setminus
H$, and
\item $\{\pm (s_i-t_i) : 1\le i \le (g-h)/2\}=G\setminus H$.
\end{enumerate}
A frame starter $A=\{\{s_i,t_i\}\}$ is strong if $s_i+t_i=s_j+t_j$
implies $i=j$, and $s_i+t_i\notin H$ for all $i$. Let
$A=\{\{s_i,t_i\}\}$ and $B=\{\{u_i,v_i\}\}$ be two frame starters.
We may assume that $t_i-s_i=v_i-u_i$, for each $1\le i \le (g-t)/2$. We
say that $A$ and $B$ are orthogonal frame starters if
$u_i-s_i=u_j-s_j$ implies $i=j$, and $u_i-s_i\notin H$ for all $i$.

\vskip 10pt
\begin{lemma}[\cite{Dinitz:1980}]
\label{strongframestarter} If $A=\{\{s_i,t_i\}\}$ is a strong
frame starter then $A$ and $-A=\{\{-s_i,-t_i\}\}$ are orthogonal
frame starters.
\end{lemma}
\vskip 10pt

\begin{lemma}[\cite{Dinitz:1980}]
\label{framestarter} If there exists a pair of orthogonal $t$-frame
starters in $G\setminus H$ with $|G|=g$ and $|H|=t$, then there
exists a Room frame of type $t^u$, where $u=g/t$.
\end{lemma}
\vskip 10pt

So if we have a strong starter in a group of order $n$ which can
generate an SSRS(n), then we get an optimal $(n,5,[2,1,1])_4$-code with size $U(n,5,[2,1,1])$. Similarly, if we  have  a strong frame starter
in $G\setminus H$ with $|G|=g$ and $|H|=2$ which can generate a
super-simple Room frame of type $2^{g/2}$, then we get an optimal
$(g,5,[2,1,1])_4$-code with size $U(g,5,[2,1,1])$.

\subsection{Some Small $[2,1,1]$-GDC$(5)$s and Optimal Codes with Distance $5$}
\vskip 10pt

In the sequel, we construct some small $[2,1,1]$-GDC$(5)$s and optimal
codes via computer search. The constructions are based on the familiar difference method,
where a finite group (mostly abelian group $\bbZ_u$) will be
utilized to generate all the codewords of a code or a GDC.
Thus, instead of listing all the codewords, we list a set of base
codewords and generate the others by an additive group and
perhaps some further automorphisms.  Mostly, the set of base
codewords  are divided into two parts, $P$ and $R$, where
each codeword of $P$ will be multiplied by $m^i$ for each $0 \le i \le
s-1$ to generate $s$ codewords, and $R$ is the set of the remaining
base codewords. The desired codes are generated by developing the
base codewords $+M$ modulo $n$. Then, we just need to list $n$, $m$, $s$, $M$, $P$ and $R$
for each code. Sometimes, $R$ may be empty, which will be omitted.

\vskip 10pt
\begin{proposition}
\label{GDC(5)gt}There exists a $[2,1,1]$-GDC$(5)$ of type $g^t$ with size $\frac{g^2t(t-1)}{2}$ for the following parameters:
\begin{enumerate}
\item $g=2$, $t \in \{8, 9, 11\}$,
\item $g=3$, $t \in \{7, 9\}$,
\item $g=4$, $t \in \{5, 6, 7, 8, 9, 11\}$,
\item $g=6$, $t \in \{5\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
For each given pair $\{g, t\}$, let $X_{\{g, t\}}=\bbZ_{gt}$,
$\G_{\{g, t\}}=\{\{i,t+i, \ldots, (g-1)t+i\}:i \in \bbZ_{t}\}$ and  $\C_{\{g, t\}}$ be the set of cyclic (or quasi-cyclic) shifts of the vectors generated by the following vectors respectively. Then $(X_{\{g, t\}},\G_{\{g, t\}},\C_{\{g, t\}})$ is a
$[2,1,1]$-GDC$(5)$ of type $g^t$ with size $\frac{g^2t(t-1)}{2}$. In order to save space, we list only one example here. Other cases can be found in Propositions 6.1--6.4. For $g=2$ and $t=8$, we have $n=16$, $m=11$, $s=2$, $M=2$ and
\begin{equation*}
\footnotesize
\begin{array}{cccccc}
P:& \langle 0, 13, 15, 6\rangle\\
R:& \langle 1, 11, 0, 13\rangle & \langle 0, 1, 4, 14\rangle & \langle 0, 11, 9, 15\rangle & \langle 0, 3, 7, 10\rangle\\
&\langle 0, 6, 2, 1\rangle &\langle 0, 2, 13, 7\rangle & \langle 1, 5, 2, 11\rangle & \langle 0, 7, 14, 12\rangle\\
&\langle 0, 5, 1, 3\rangle & \langle 0, 9, 3, 4\rangle & \langle 0, 12, 6, 9\rangle & \langle 1, 3, 12, 2\rangle.\\
\end{array}
\end{equation*}
\end{proof}
\vskip 10pt

\begin{proposition}
\label{GDC(5)4t21}There exists a $[2,1,1]$-GDC$(5)$ of type $4^t2^1$
with size $8t^2$ for each $t \in \{5,6\}$.
\end{proposition}
\begin{proof}
Detailed constructions can be found in Propositions 6.5.
\end{proof}
\vskip 10pt

\begin{proposition}
\label{CCC(5)}$A_4(n,5,[2,1,1])$  =  $U(n,5,[2,1,1])$ for each $n \in
\{12,14,15,17,20,28,30,36,38,44,46,52,54,62,$
$68,70,76,78,86,92,94,110,126,134\}$.
\end{proposition}
\begin{proof}
Detailed constructions can be found in Propositions 6.6.
\end{proof}
\vskip 10pt

\begin{proposition}
\label{CCC(5s)}$A_4(n,5,[2,1,1])$  =  $U(n,5,[2,1,1])$ for each $n \in
\{19,21,23,24,25,26,27,29,31,32,33,34,35,$
$37,39,40,41,42,43,45,47,48,49,50,51,53,55,56,57,58,$
$59,61,63,64,65,66,67,69,71,73,74,75,77,79,83,85,87,$
$88,89,93,95,99,103,104,106,107,109,111,123,125,127,$
$131,133,138,139\}$.
\end{proposition}
\begin{proof}
All these optimal codes are constructed by strong starters or strong
frame starters with n odd or even respectively. By Theorems \ref{SSRStoCodes} and \ref{SSRFtoCodes}, there
exists an optimal code of such length $n$. The starters are given in a similar way as the codewords in the above propositions.
In order to save space, we list only one example here. Other cases can be found in Proposition 6.7. For  $n=19$, we have $m=4$, $s=9$, $M=1$ and $P: \{1, 3\}$.
\end{proof}
\vskip 10pt

\subsection{The Case of Length $n \equiv 0 \pmod{4}$}
\vskip 10pt

\begin{lemma}
\label{CCC(5)4 8} $A_4(4,5,[2,1,1])=1$ and $A_4(8,5,[2,1,1]) \ge
18$.
\end{lemma}
\begin{proof}
For $n=4$, the one required codeword is $\langle 0, 1, 2, 3\rangle$. For $n=8$, the $18$  required codewords can be found in Proposition 6.8.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC4t} There exists an optimal $(4t,5,[2,1,1])_4$-code with
size $4t(2t-1)$ for all $t \ge 3$.
\end{theorem}
\begin{proof}
For each $3\le t\le 14$ or $t\in \{16,17,19,22,23,$ $26\}$, there exists an
optimal $(4t,5,[2,1,1])_4$-code by Propositions \ref{GDC(5)gt},
\ref{CCC(5)} and \ref{CCC(5s)}.

For each $u\in \{5,6,7,9,11\}$, there exists a $[2,1,1]$-GDC(5) of type
$4^u$. Apply Construction \ref{Inflation} with a TD($4,3$) (which
exists by Lemma \ref{TD}) to get a $[2,1,1]$-GDC(5) of type $12^u$. Filling in the groups of this GDC with an
optimal $(12,5,[2,1,1])_4$-code coming from Proposition
\ref{CCC(5)}, the result is an optimal $(4t,5,[2,1,1])_4$-code for each
$t\in \{15,18,21,27,33\}$.

For each $u\in \{7,8\}$, there exists a $[2,1,1]$-GDC(5) of type $4^u$,
and apply Construction \ref{Inflation} with a TD($4,4$) to get a
$[2,1,1]$-GDC(5) of type $16^u$ for each $u\in \{7,8\}$. Filling in the
groups of this GDC with an optimal $(16,5,[2,1,1])_4$-code (see
 Proposition \ref{GDC(5)gt}), the result is an optimal
$(4t,5,[2,1,1])_4$-code for each $t\in \{28,32\}$.

For $t=31$, take a $4$-RGDD of type $3^8$ (see Lemma~\ref{4RGDD38}). There are $7$ parallel
classes in this $4$-RGDD. Add one ideal point to each of these  $7$ parallel
classes to complete them. The result is a $5$-GDD of type $3^8 7^1$.  Apply Construction \ref{FundCtr} with
weight $4$ to each point of this $5$-GDD and fill in the groups with optimal
codes of lengths $12$ and $28$ to obtain an optimal
($124,5,[2,1,1]$)-code. Here, the input design, a $[2,1,1]$-GDC(5) of
type $4^5$, exists by Proposition \ref{GDC(5)gt}.

For all $t\geq 34$ or $t\in \{20,24,25,29,30\}$, take a
$(t+1,\{5,6,7,8,9\},1)$-PBD from Lemma \ref{PBD5-9} and remove one
point to obtain a $\{5,6,7,8,9\}$-GDD of type $4^i 5^j 6^k 7^l 8^m$
with $4i+5j+6k+7l+8m=t$. Apply Construction \ref{FundCtr} with
weight $4$ using  $[2,1,1]$-GDC(5)s of types $4^t$ for  $t\in
\{5,6,7,8,9\}$  (Proposition \ref{GDC(5)gt}) as input ingredients to obtain a
$[2,1,1]$-GDC(5) of type $16^i 20^j 24^k 28^l 32^m$ and length $4t$. Filling in the groups of this GDC with optimal
$(n,5,[2,1,1])_4$-codes for   $n\in \{16,20,24,28,32\}$ by Propositions
\ref{GDC(5)gt}, \ref{CCC(5)} and \ref{CCC(5s)}, the result is an
optimal $(4t,5,[2,1,1])_4$-code for all $t\geq 34$ or $t\in
\{20,24,25,29,30\}$.
\end{proof}

\subsection{The Case of Length $n \equiv 1 \pmod{4}$}
\vskip 10pt

\begin{lemma}
\label{CCC(5)5913} $A_4(5,5,[2,1,1])=2$, $A_4(9,5,[2,1,1])\ge 27$
and $A_4(13,5,[2,1,1])\ge 72$.
\end{lemma}
\begin{proof}
All the required codewords can be found in Proposition 6.9.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC4t+1}  There exists an optimal $(4t+1,5,[2,1,1])_4$-code
with size $2t(4t+1)$ for all $t \ge 4$.
\end{theorem}
\begin{proof}
For each $4\le t\le 19$ or $t\in \{21,22,23,27,$ $31,33\}$, there exists an
optimal $(4t+1,5,[2,1,1])_4$-code by Propositions \ref{CCC(5)} and
\ref{CCC(5s)}.

For $t=26$, there exists a $[2,1,1]$-GDC(5) of type $3^7$. Apply
Construction \ref{Inflation} with a TD($4,5$) to get a
$[2,1,1]$-GDC(5) of type $15^7$.  Fill in the groups with an
optimal $(15,5,[2,1,1])_4$-code to obtain an optimal
$(4t+1,5,[2,1,1])_4$-code.

For each $u\in \{7,8\}$, there exists a $[2,1,1]$-GDC(5) of type $4^u$.
Apply Construction \ref{Inflation} with a TD($4,4$) to get a
$[2,1,1]$-GDC(5) of type $16^u$. Adjoin one ideal
point to this GDC, and fill in the groups together with the extra point with an optimal
$(17,5,[2,1,1])_4$-code to obtain an optimal
$(4t+1,5,[2,1,1])_4$-code for each $t\in \{28,32\}$.

For all $t\geq 34$ or $t\in \{20,24,25,29,30\}$, take a
$(t+1,\{5,6,7,8,9\},1)$-PBD from Lemma \ref{PBD5-9} and remove one
point to obtain a $\{5,6,7,8,9\}$-GDD of type $4^i 5^j 6^k 7^l 8^m$
with $4i+5j+6k+7l+8m=t$. Apply construction \ref{FundCtr} with
weight 4 using $[2,1,1]$-GDC(5)s of types $4^t$  for  $t\in
\{5,6,7,8,9\}$ (Lemma \ref{GDC(5)gt})  as input ingredients to obtain a $[2,1,1]$-GDC(5)
of type $16^i 20^j 24^k 28^l 32^m$ and length $4t$. Adjoin one
ideal point to this GDC, and fill in the groups  together with the extra point with optimal
$(n,5,[2,1,1])_4$-codes for  $n\in \{17,21,25,29,33\}$ (Propositions
\ref{CCC(5)} and \ref{CCC(5s)}) to obtain an optimal
$(4t+1,5,[2,1,1])_4$-code for all $t\geq 34$ or $t\in
\{20,24,25,29,30\}$.
\end{proof}

\subsection{The Case of Length $n \equiv 2 \pmod{4}$}
\vskip 10pt

\begin{lemma}
\label{CCC(5)610} $A_4(6,5,[2,1,1])=6$, $A_4(10,5,[2,1,1]) \ge 36$.
\end{lemma}
\begin{proof}
All the required codewords can be found in Proposition 6.10.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC4t+2(1)} There exists an optimal $(4t+2,5,[2,1,1])_4$-code
with size $2t(4t+2)$ for each $3\le t \le 37$.
\end{theorem}

\begin{proof}
For each $3\le t\le 34$ and $t\notin \{20,22,24,25,$ $28,29,30,32\}$, there exists an
optimal $(4t+2,5,[2,1,1])_4$-code with  size $2t(4t+2)$ by Propositions
\ref{GDC(5)gt}, \ref{CCC(5)} and \ref{CCC(5s)}.

For each $t\in\{20,24,28,32,36\}$, there exists a $[2,1,1]$-GDC(5) of type $4^u$ with $u\in\{5,6,7,8,9\}$. Apply
Construction \ref{Inflation} with a TD($4,4$) to get a
$[2,1,1]$-GDC(5) of type $16^u$ with $u\in\{5,6,7,8,9\}$. Adjoin two additional points and fill in the groups together with the two extra points with a $[2,1,1]$-GDC(5) of type $2^9$ to obtain an optimal $[2,1,1]$-GDC(5) of type $2^{8u+1}$ and lengths $82,98,114,130$ or $146$.

For each $t\in \{22, 37\}$, there exists a $[2,1,1]$-GDC(5) of type $6^5$. Let $t_1=(4t+2)/30$. Apply
Construction \ref{Inflation} with a TD($4,t_1$) to get a
$[2,1,1]$-GDC(5) of type $(6t_1)^5$. Fill in the groups with an
optimal $(6t_1,5,[2,1,1])_4$-code to obtain an optimal code of length
$4t+2$.

For each $t\in\{25,30,35\}$, there exists a $[2,1,1]$-GDC(5) of type $4^u$ with $u\in\{5,6,7\}$. Apply
Construction \ref{Inflation} with a TD($4,5$) to get a
$[2,1,1]$-GDC(5) of type $20^u$ with $u\in\{5,6,7\}$. Adjoin two additional points and fill in the groups  together with the two extra points with a $[2,1,1]$-GDC(5) of type $2^{11}$ to obtain an  optimal  $[2,1,1]$-GDC(5) of type $2^{10u+1}$ and  lengths $102,122$ or $142$.

For $t=29$, take a TD($6,5$) from Lemma \ref{TD}. Apply
Construction \ref{FundCtr} with weight $4$ to the points in the
first five groups and $4$ points in the last group,  and  weight $2$ to
the other $1$ points in the last group. All the remaining points are given weight $0$.
Note that there exist $[2,1,1]$-GDC(5)s of types $4^5$, $4^6$ and
$4^5 2^1$ by Propositions \ref{GDC(5)gt} and \ref{GDC(5)4t21}. The result is a $[2,1,1]$-GDC(5) of type $20^5
 18^1$. Filling in the groups with optimal
codes of lengths $18$ or $20$, the results are optimal codes
of length $118=4\times 29+2$.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC4t+2(2)} There exists an optimal $(4t+2,5,[2,1,1])_4$-code
with size $2t(4t+2)$ for all $t \ge 38$.
\end{theorem}

\begin{proof}
Take a TD$(7,r)$ for  $r \ge 7$ and $r\not\in \{10,14,15,18,$ $20, 22,26,30,34,38,46,60\}$
  from Lemma \ref{TD}. Apply Construction
\ref{FundCtr} with weight $4$ to the points in the first five
groups, $x$ points in the sixth group and $y$ points in the last
group, and  weight $2$ to the other $z$ points in the last group. All the remaining points are given weight $0$. Here,   we require that
$x\ge 4$, $4y+2z\ge 14$ and $z$ odd.  Note that there exist $[2,1,1]$-GDC(5)s
of types $4^5$, $4^6$, $4^7$, $4^5 2^1$, and $4^6 2^1$ by Propositions
\ref{GDC(5)gt} and \ref{GDC(5)4t21}. The result is a
$[2,1,1]$-GDC(5) of type $(4r)^5 (4x)^1 (4y+2z)^1$. Fill in the groups with optimal codes of
lengths $4u$ with $u\ge 4$ (which exist by Theorem \ref{CCC4t}) or
$4v+2$ with $3\le v \le 37$ (which exist by Theorem
\ref{CCC4t+2(1)}). The result is an optimal
$(20r+4x+4y+2z,5,[2,1,1])_4$-code, where $20r+4x+4y+2z$ can take any
value $n$ with $n \equiv 2 \pmod{4}$ and $n \ge 4\times 38+2=154$.
\end{proof}

\subsection{The Case of Length $n \equiv 3 \pmod{4}$}
\vskip 10pt

\begin{lemma}
\label{CCC(5)711} $A_4(7,5,[2,1,1])=10$, $A_4(11,5,[2,1,1])\ge 48$.
\end{lemma}
\begin{proof}
All the required codewords can be found in Proposition 6.11.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC4t+3(1)} There exists an optimal $(4t+3,5,[2,1,1])_4$-code
with size $(2t+1)(4t+3)$ for each $3\le t \le 37$.
\end{theorem}

\begin{proof}
For each $3\le t\le 34$ and $t\notin \{22,28,29,33\}$, there exists an
optimal $(4t+3,5,[2,1,1])_4$-code with size $(2t+1)(4t+3)$ by
Propositions \ref{CCC(5)} and \ref{CCC(5s)}.

For $t=22$, there exists a $[2,1,1]$-GDC(5) of type $6^5$. Apply
Construction \ref{Inflation} with a TD($4,3$) to get a
$[2,1,1]$-GDC(5) of type $18^5$. Adjoin one ideal point and fill in
the groups together with the extra point with an optimal $(19,5,[2,1,1])_4$-code to obtain an
optimal code of length $91=4\times 22+3$.

For each $t\in\{28,29\}$, take a TD($6,5$) from Lemma \ref{TD}. Apply
Construction \ref{FundCtr} with weight $4$ to the points in the
first five groups and  $x$ points in the last group, and weight $2$ to
the other $y$ points in the last group.  All the remaining points are given weight $0$.
Note that there exist $[2,1,1]$-GDC(5)s of types $4^5$, $4^6$ and
$4^5 2^1$ by Propositions \ref{GDC(5)gt} and \ref{GDC(5)4t21}. The result is a $[2,1,1]$-GDC(5) of type $20^5
 (4x+2y)^1$. Here, $4x+2y$ can take $14$ or $18$ when $x=3, y=1$ or
 $x=4, y=1$.  Adjoining one ideal point and filling in the groups together with the extra point with optimal codes of lengths $15$, $19$ or $21$, the results are optimal codes
of lengths $115=4\times 28+3$ or $119=4\times 29+3$.

For $t=33$, there exists a $[2,1,1]$-GDC(5) of type $3^9$ by
Proposition \ref{GDC(5)gt}. Apply Construction \ref{Inflation}
with a TD($4,5$) to get a $[2,1,1]$-GDC(5) of type $15^9$. Fill
in the groups with an optimal $(15,5,[2,1,1])_4$-code to obtain an
optimal code of length $135=4\times 33+3$.

For $t=35$, take a TD($9,8$) from Lemma~\ref{TD}. Apply Construction \ref{FundCtr}
with weight $2$ to the points in the first eight groups and $7$
points in the last group. The other points are given weight 0. Note
that there exist $[2,1,1]$-GDC(5)s of types $2^8$, $2^9$  by Proposition
\ref{GDC(5)gt}. The result is a $[2,1,1]$-GDC(5) of type $16^8
14^1$. Adjoining one ideal point and filling in the group together with the extra point with
optimal codes of lengths $15$ or $17$, the result is an optimal code of
length $143=4\times 35+3$.

For $t=36$, there exists a $[2,1,1]$-GDC(5) of type $3^7$ by
Proposition \ref{GDC(5)gt}. Apply Construction \ref{Inflation}
with a TD($4,7$) to get a $[2,1,1]$-GDC(5) of type $21^7$. Fill
in the groups with an optimal $(21,5,[2,1,1])_4$-code to obtain an
optimal code of length $147=4\times 36+3$.

For $t=37$, there exists a $[2,1,1]$-GDC(5) of type $6^5$. Apply
Construction \ref{Inflation} with a TD($4,5$) to get a
$[2,1,1]$-GDC(5) of type $30^5$. Adjoin one ideal point and fill in
the groups together with the extra point with an optimal $(31,6,[2,1,1])_4$-code to obtain an
optimal code of length $151=4\times 37+3$.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC4t+3(2)} There exists an optimal $(4t+3,5,[2,1,1])_4$-code
with size $2t(4t+3)$ for all $t \ge 38$.
\end{theorem}

\begin{proof}
Take a TD$(7,r)$ for  $r \ge 7$ and $r\not\in \{10,14,15,18,$ $20, 22,26,30,34,38,46,60\}$
 from Lemma \ref{TD}. Apply Construction
\ref{FundCtr} with weight $4$ to the points in the first five
groups, $x$ points in the sixth group and $y$ points in the last
group, and weight $2$ to the other $z$ points in the last group.  All the remaining points are given weight $0$. We require that
$x\ge 4$ and $4y+2z\ge 14$. Note that there exist $[2,1,1]$-GDC(5)s
of types $4^5$, $4^6$, $4^7$, $4^5 2^1$, and $4^6 2^1$ by Propositions
\ref{GDC(5)gt} and \ref{GDC(5)4t21}. The result is a
$[2,1,1]$-GDC(5) of type $(4r)^5 (4x)^1 (4y+2z)^1$. Adjoin one ideal point and fill in the
groups together with the extra point with optimal codes of lengths $4u+1$ with $u\ge 4$ (which exist
by Theorem \ref{CCC4t}) or $4v+3$ with $3\le v \le 37$ (which
exist by Theorem \ref{CCC4t+3(1)}). The result is an optimal
$(20r+4x+4y+2z+1,5,[2,1,1])_4$-code, where $20r+4x+4y+2z+1$ can take
any value of $4t+3$ with $t$ greater than $38$.
\end{proof}
\vskip 10pt

\section{Determining the Value of $A_4(n,6,[2,1,1])$}
\vskip 10pt

\subsection{Some Small $[2,1,1]$-GDC$($6$)$ and Optimal Codes with Distance $6$}
\vskip 10pt

First, we construct some small $[2,1,1]$-GDC$(6)$s and optimal
codes via computer search. In the codes with infinite points, the subscripts on the elements
$x_0\in \{x\}\times \bbZ_u$ for $x\in \{a,b,c,d,e\}$ are developed modulo the unique subgroup
in the abelian group $\bbZ_{n}$ of order $u$.
\vskip 10pt

\begin{proposition}
\label{GDC(6)gt}There exists a $[2,1,1]$-GDC$(6)$ of type $g^t$ with size $\frac{g^2t(t-1)}{6}$ for the following parameters:
\begin{enumerate}
\item $g=2$, $t \in \{10, 13, 16, 19, 22, 25, 28, 34\}$,
\item $g=3$, $t \in \{5, 7\}$,
\item $g=4$, $t \in \{4, 7\}$,
\item $g=6$, $t \in \{4, 5, 6, 7\}$,
\item $g\in \{7, 10, 13, 22\}$, $t \in \{4\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Detailed constructions can be found in Propositions 7.1--7.6, 7.9 and 7.10.
\end{proof}
\vskip 10pt

\begin{proposition}
\label{GDC(6)12tu1}There exists a $[2,1,1]$-GDC$(6)$ of type
$12^tu^1$ with size $12t(2t+\frac{u}{3}-2)$ for the following parameters:
\begin{enumerate}
\item $u=9$, $t \in \{4,5,\ldots,15,17,18,19, 23\}$,
\item $u=15$, $t \in \{7, 8, \ldots, 15\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Detailed constructions can be found in Propositions 7.7 and 7.8.
\end{proof}
\vskip 10pt

\begin{proposition}
\label{GDC(6)14_2}There exists a $[2,1,1]$-GDC$(6)$ of type
$1^{12}2^1$ with size $28$.
\end{proposition}
\begin{proof}
Detailed construction can be found in Proposition 7.11.
\end{proof}
\vskip 10pt

\begin{proposition}
\label{GDC(6)1t111}There exists a $[2,1,1]$-GDC$(6)$ of type
$1^t11^1$ with size $6u^2+20u$, where $6u=t$ for each $t \in \{30,36,54,$
$66,78\}$.
\end{proposition}
\begin{proof}
Detailed constructions can be found in Proposition 7.12.
\end{proof}
\vskip 10pt

\begin{proposition}
\label{CCC(6)}$A_4(n,6,[2,1,1])$  =  $U(n,6,[2,1,1])$ for each $n
\in \{6, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 25, 28, 31, 34,$ $
35, 37, 43, 55, 67, 79, 103\}$.
\end{proposition}
\begin{proof}
Detailed constructions can be found in Proposition 7.13.
\end{proof}
\vskip 10pt

\subsection{The Case of Length $n \equiv 0, 1 \pmod{6}$}
\vskip 10pt

\begin{lemma}
\label{CCC(6)7} $A_4(7,6,[2,1,1])=4$.
\end{lemma}
\begin{proof}
All the required codewords can be found in Proposition 7.14.
\end{proof}
\vskip 10pt

\begin{lemma}
\label{CCC6t} For any positive integer $t$, if
$A_4(6t+1,6,[2,1,1])=U(6t+1,6,[2,1,1])$ then
$A_4(6t,6,[2,1,1])=U(6t,6,[2,1,1])$.
\end{lemma}
\begin{proof}
In an optimal $(6t+1,6,[2,1,1])_4$-code, every coordinate has exactly $2t+2t=4t$ non-zero
elements. Fix a  coordinate $x$ and remove all the $4t$ codewords containing   non-zero
elements in this  coordinate $x$. Shorten all the remaining codewords by deleting the element $0$ in   coordinate $x$ from them.  The resultant
codewords form an optimal $(6t,6,[2,1,1])_4$-code with size
$U(6t+1,6,[2,1,1])-4t=6t^2-3t=U(6t,6,[2,1,1])$.
\end{proof}
\vskip 10pt

\begin{lemma}
\label{GDC(6)12t} There exists a $[2,1,1]$-GDC(6) of type $12^t$ for
all $t\ge 4$.
\end{lemma}
\begin{proof}
When $t \equiv 0$ or $1\pmod{4}$ and $t\geq 4$, there exists a
$(3t+1,\{4\},1)$-PBD   by Lemma \ref{PBD4*}. Deleting one
point from the point set gives a $\{4\}$-GDD of type $3^t$. When
$t\equiv 2$ or $3\pmod{4}$ and  $t\geq 7$, there exists a
$(3t+1,\{4,7^\star\},1)$-PBD  by Lemma \ref{PBD4*}.  Remove one
point from this PBD which is not in the unique block with size
$7$ to  obtain  a
$\{4,7^\star\}$-GDD of type $3^t$. Hence, we always have a
$\{4,7\}$-GDD of type $3^t$ for all $t\geq 4$ and $t\neq 6$.

Apply Construction \ref{FundCtr} with weight $4$ to obtain a
$[2,1,1]$-GDC($6$) of type $12^t$ for all $t\geq 4$ and $t\neq 6$.
Here, the input $[2,1,1]$-GDC($6$)s of types $4^4$ and $4^7$ exist by
Proposition \ref{GDC(6)gt}.

For $t=6$, take a $\{5\}$-GDD of type $4^6$ (see \cite{GeLing:2005})
and apply Construction \ref{FundCtr} with weight $3$ to obtain a
$[2,1,1]$-GDC($6$) of type $12^6$. Here, the input
$[2,1,1]$-GDC($6$) of type $3^5$ exists by Proposition
\ref{GDC(6)gt}.
\end{proof}

\vskip 10pt
\begin{theorem}
\label{CCC6t+1(1)} There exists an optimal
$(12t+1,6,[2,1,1])_4$-code with size $24t^2+2t$ for all $t\ge 1$.
\end{theorem}
\begin{proof}
For all $t\ge 4$, adjoin one ideal point to a $[2,1,1]$-GDC($6$) of
type $12^t$ (Lemma \ref{GDC(6)12t}) and fill in the groups together with the extra point with an
optimal $(13,6,[2,1,1])_4$-code (which exists by Proposition
\ref{CCC(6)}) to obtain an optimal $(12t+1,6,[2,1,1])_4$-code with size
$24t^2+2t$.

For each $t\in \{1,2,3\}$, there exists an optimal
$(12t+1,6,[2,1,1])_4$-code with size $24t^2+2t$ by Proposition
\ref{CCC(6)}.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC6t+1(2)} There exists an optimal
$(12t+7,6,[2,1,1])_4$-code with size $(12t+7)(2t+1)$ for each $1 \le
t\le 16$.
\end{theorem}
\begin{proof}
For each $t\in \{1,2,3,4,5,6,8\}$, there exists an optimal
$(12t+7,6,[2,1,1])_4$-code with size $(12t+7)(2t+1)$ by Proposition
\ref{CCC(6)}.

For the other $9$ values of $t$, we first construct $9$  GDCs  of types $18^5$, $24^4 18^1$,  $24^4 30^1$,  $30^4 18^1$, $30^5$,  $30^5 12^1$,  $30^5 24^1$, $24^7 18^1$ and $24^7 30^1$ as follows: For $t=7$, take a $\{4\}$-GDD of type $3^5$ (which exists by Lemma
\ref{4GDD}). Apply Construction \ref{FundCtr} with weight $6$ to
obtain a $[2,1,1]$-GDC($6$) of type $18^5$. For $t=9$, take a TD($5,4$) from Lemma \ref{TD}. Remove one point
from a group to obtain a $\{4,5\}$-GDD of type $4^4 3^1$. Apply
Construction \ref{FundCtr} with weight $6$ to obtain a
$[2,1,1]$-GDC($6$) of type $24^4 18^1$.
For $t=10$, take a TD($5,5$) from Lemma \ref{TD}. Remove $4$ points
from a block to obtain a $\{4,5\}$-GDD of type $4^4 5^1$. Apply
Construction \ref{FundCtr} with weight $6$ to obtain a
$[2,1,1]$-GDC($6$) of type $24^4 30^1$.
For $t=11$, take a TD($5,5$) from Lemma \ref{TD}. Remove $2$ points
from a group to obtain a $\{4,5\}$-GDD of type $5^4 3^1$. Apply
Construction \ref{FundCtr} with weight $6$ to obtain a
$[2,1,1]$-GDC($6$) of type $30^4 18^1$.
For $t=12$, take a TD($5,5$) from Lemma \ref{TD}. Apply Construction
\ref{FundCtr} with weight $6$ to obtain a $[2,1,1]$-GDC($6$) of type
$30^5$.
For each $t\in \{13,14\}$, take a TD($6,5$) from Lemma \ref{TD}. Remove
$3$ or $1$ points from a group to obtain a $\{4,5\}$-GDD of types
$5^5 2^1$ or $5^5 4^1$. Apply Construction
\ref{FundCtr} with weight $6$ to obtain a $[2,1,1]$-GDC($6$) of types
$30^5 12^1$ or $30^5 24^1$.
For each $t\in \{15,16\}$, take a $4$-RGDD of type $4^7$ (see Lemma~\ref{4RGDD38}). There are $8$ parallel
classes in this $4$-RGDD. Add one ideal point to each of the $u$ parallel
classes for $u\in \{3, 5\}$ to complete them. The result is a $\{4,5\}$-GDD of type
$4^7 u^1$.  Apply Construction \ref{FundCtr} with weight $6$ to obtain a $[2,1,1]$-GDC($6$) of types $24^7 18^1$ or $24^7 30^1$.
Here, the input $[2,1,1]$-GDC($6$)s of types $6^4$ and $6^5$ exist by Proposition \ref{GDC(6)gt}.
Adjoining one ideal point to each of the above GDCs and filling in the groups together with the extra point with an optimal code of length $n_1 \in \{13, 19, 25, 31\}$, the result is an optimal code of length $12t+7$ with $t\in \{7, 9, 10, 11, 12, 13, 14, 15, 16\}$ as desired.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC6t+1(3)} There exists an optimal
$(12t+7,6,[2,1,1])_4$-code with size $(12t+7)(2t+1)$ for all $t\ge
17$.
\end{theorem}
\begin{proof}
Take a TD$(6,2t)$ from Lemma \ref{TD}. Apply Construction
\ref{FundCtr} with weight $6$ to the points in the first $4$ groups,
$2x$ points in the fifth group, and $y$ points in the last group.
The other points are given weight 0. Note that there exist
$[2,1,1]$-GDC($6$)s of types $6^4$, $6^5$, $6^6$ by Proposition
\ref{GDC(6)gt}. The result is a $[2,1,1]$-GDC($6$) of type $(12t)^4
(12x)^1 (6y)^1$. We require that $y\ge 3$. Adjoin one ideal point
and fill in the groups together with the extra point with optimal codes of lengths $12t+1$ with
$t\ge 1$ (which exist by Theorem \ref{CCC6t+1(1)}) or
$12u+7$ with $1\le u \le 16$ (which exist by Theorem
\ref{CCC6t+1(2)}). The result is an optimal
$(48t+12x+6y+1,6,[2,1,1])_4$-code, where $48t+12x+6y+1$ can take any
value greater than $12\times 17+7=211$ except for the case of
$295=12\times 24+7$.

For $t=24$, take a TD$(7,8)$. Apply Construction  \ref{FundCtr}
with weight $6$ to the points in the first $5$ groups, $6$ points in
the sixth group, and $3$ points in the last group. The other points
are given weight 0. Note that there exist $[2,1,1]$-GDC($6$)s of
types $6^5$, $6^6$, $6^7$ by Proposition \ref{GDC(6)gt}. The result
is a $[2,1,1]$-GDC($6$) of type $48^5 36^1 18^1$. Adjoin one
ideal point and fill in the groups together with the extra point with optimal codes of lengths
$37$, $49$ (both exist by Theorem \ref{CCC6t+1(1)}) or $19$ (which exists by Theorem \ref{CCC6t+1(2)}). The result is an optimal $(295,6,[2,1,1])_4$-code.
\end{proof}
\vskip 10pt

\begin{theorem} There exists an optimal $(6t,6,[2,1,1])_4$-code with size $6t^2-3t$ for all $t\ge
1$.
\end{theorem}
\begin{proof}
The result follows by combining Lemma \ref{CCC6t}, Theorems
\ref{CCC6t+1(1)}--\ref{CCC6t+1(3)}
and the fact that there exists an optimal $(6,6,[2,1,1])_4$-code with size $3$ (see Proposition \ref{CCC(6)}).
\end{proof}
\vskip 10pt

\subsection{The Case of Length $n \equiv 2 \pmod{6}$}
\vskip 10pt

\begin{lemma}
\label{CCC6t+2(1)} Every $[2,1,1]$-GDC($6$) of type $2^{3t+1}$ is an
optimal $(6t+2,6,[2,1,1])_4$-code with size $6t^2+2t$.
\end{lemma}
\begin{proof}
The size of a $[2,1,1]$-GDC($6$) of type $2^{3t+1}$ is $6t^2+2t$
which meets the upper bound of an optimal $(6t+2,6,[2,1,1])_4$-code.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC6t+2(2)} There exists an optimal $(6t+2,6,[2,1,1])_4$-code
with size $6t^2+2t$ for each $t\in \{1,2,\ldots,11\} \cup
\{14,17,18,22\}$.
\end{theorem}
\begin{proof}
For each $t\in \{1,2\}$, there exists an optimal
$(6t+2,6,[2,1,1])_4$-code with size $6t^2+2t$ by Proposition
\ref{CCC(6)}.

For each $t\in \{3,4,5,6,7,8,9,11\}$, there exists a $[2,1,1]$-GDC($6$)
of type $2^{3t+1}$ by Proposition \ref{GDC(6)gt}.

For each $t\in \{10,14,18,22\}$, take a $[2,1,1]$-GDC($6$) of type
$12^{t/2}$ (which exists by Lemma \ref{GDC(6)12t}). Adjoin two ideal
points and fill in the groups together with the two extra points with a $[2,1,1]$-GDC($6$) of type
$1^{12} 2^1$ (which exists by Proposition \ref{GDC(6)14_2}). The
result is an optimal $(6t+2,6,[2,1,1])_4$-code with size $6t^2+2t$.

For $t=13$, take an optimal $(10,6,[2,1,1])_4$-code with size $10$
which  exists by Proposition \ref{CCC(6)}. This code can also be
regarded as a $[2,1,1]$-GDC($6$) of type $1^{10}$. Apply Construction
\ref{Inflation} with a TD($4,8$) to get a $[2,1,1]$-GDC($6$) of type
$8^{10}$. Fill in the groups with an optimal
$(8,6,[2,1,1])_4$-code. The result is an optimal code of length
$80=6\times 13+2$.

For $t=17$, take a $[2,1,1]$-GDC($6$) of type $2^{13}$ (which exists
by Proposition \ref{GDC(6)gt}). Apply Construction \ref{Inflation}
with a TD($4,4$) to get a $[2,1,1]$-GDC($6$) of type $8^{13}$.
Fill in the groups with an optimal $(8,6,[2,1,1])_4$-code. The
result is an optimal code of length $104=6\times 17+2$.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC6t+2(3)} There exists an optimal $(6t+2,6,[2,1,1])_4$-code
with size $6t^2+2t$ for all $t\ge 23$ and $t\in\{12,15,16,$
$19,20,21\}$.
\end{theorem}
\begin{proof}
For each $t\geq 23$ or $t\in \{12,15,16,19,20,21\}$, take a
$(t+1,\{4,5,6\},1)$-PBD from Lemma \ref{PBD4-6} and remove one point
to obtain a $\{4,5,6\}$-GDD of type $3^i 4^j 5^k$ with $3i+4j+5k=t$.
Apply construction \ref{FundCtr} with weight $6$ and input
$[2,1,1]$-GDC($6$)s of types $6^t$ for $t\in \{4,5,6\}$ (Lemma
\ref{GDC(6)gt}) to obtain a $[2,1,1]$-GDC($6$) of type $18^i 24^j
30^k$ and length $6t$. Adjoining two ideal points and filling
in the groups together with the two extra points with $[2,1,1]$-GDC($6$)s of types $2^u$ for
$u\in \{10,13,16\}$ (see Proposition \ref{GDC(6)gt}), the result is a
$[2,1,1]$-GDC($6$) of type $2^{3t+1}$ for all $t\geq 23$ or $t\in
\{12,15,16,19,20,21\}$. By Lemma \ref{CCC6t+2(1)}, the resultant $[2,1,1]$-GDC($6$) is an
optimal $(6t+2,6,[2,1,1])_4$-code with size $6t^2+2t$.
\end{proof}
\vskip 10pt

\subsection{The Case of Length $n \equiv 3, 4 \pmod{6}$}
\vskip 10pt

\begin{lemma}
\label{CCC(6)4} $A_4(4,6,[2,1,1])=1$.
\end{lemma}
\begin{proof}
The one required codeword is $\langle 0, 1, 2, 3\rangle$.
\end{proof}

\vskip 10pt

\begin{lemma}
\label{CCC6t3} For any positive integer $t$, if
$A_4(6t+4,6,[2,1,1])=U(6t+4,6,[2,1,1])$ then
$A_4(6t+3,6,[2,1,1])=U(6t+3,6,[2,1,1])$.
\end{lemma}
\begin{proof}
In an optimal $(6t+4,6,[2,1,1])_4$-code, every coordinate has exactly $2t+1+2t+1=4t+2$ non-zero
elements. Fix a  coordinate $x$ and remove all the $4t+2$ codewords containing   non-zero
elements in this  coordinate $x$. Shorten all the remaining codewords by deleting the element $0$ in   coordinate $x$ from them.  The resultant
codewords form an optimal $(6t+3,6,[2,1,1])_4$-code with size
$U(6t+4,6,[2,1,1])-4t-2=6t^2+3t=U(6t+3,6,[2,1,1])$.
\end{proof}
\vskip 10pt

\begin{lemma}
\label{GDC12t91(1)} There exists a $[2,1,1]$-GDC($6$) of type $12^t
9^1$ for all $t\ge 4$.
\end{lemma}
\begin{proof}
For each $t\in \{4,5,\ldots,15\} \cup \{17,18,19,23\}$, a
$[2,1,1]$-GDC($6$) of type $12^t 9^1$ exists by Proposition
\ref{GDC(6)12tu1}.

For $t\in \{16, 21, 22, 26, 27, 28, 31, 32, 33\}$, we first construct $9$  GDCs  of types $48^4$, $48^4 60^1$, $48^4 72^1$, $48^5 72^1$, $60^4 84^1$, $84^4$, $48^6 84^1$, $84^4 48^1$ and  $84^4 60^1$ as follows: For $t=16$, take a TD($4,8$) from Lemma \ref{TD}. Apply Construction
\ref{FundCtr} with weight $6$ to obtain a $[2,1,1]$-GDC($6$) of type
$48^4$. For $t=21$, take a TD($5,5$) from Lemma \ref{TD}. Remove $4$ points
from a block to obtain a $\{4,5\}$-GDD of type $4^4 5^1$. Apply
Construction \ref{FundCtr} with weight $12$ to obtain a
$[2,1,1]$-GDC($6$) of type $48^4 60^1$.
For each $t\in\{22,26\}$, take a $4$-GDD of type $12^u 18^1$ for
$u\in\{4,5\}$ from Lemma \ref{4GDD}. Apply Construction \ref{FundCtr}
with weight $4$ to obtain a $[2,1,1]$-GDC($6$) of type $48^u 72^1$.
For $t=27$, take a $4$-GDD of type $15^4 21^1$ from Lemma \ref{4GDD}.
Apply Construction \ref{FundCtr} with weight $4$ to obtain a
$[2,1,1]$-GDC($6$) of type $60^4 84^1$.
For $t=28$, take a TD($4,14$) from  Lemma \ref{TD}. Apply Construction
\ref{FundCtr} with weight $6$ to obtain a $[2,1,1]$-GDC($6$) of type
$84^4$.
For $t=31$, take a $4$-GDD of type $4^6 7^1$ from  Lemma \ref{4GDD}.
Apply Construction \ref{FundCtr} with weight $12$ to obtain a
$[2,1,1]$-GDC($6$) of type $48^6 84^1$.
For each $t\in \{32,33\}$, take a TD($5,14$) from  Lemma \ref{TD}. Delete
$6$ or $4$ points from a group to get a $\{4,5\}$-GDD of types $14^4
8^1$ or $14^4 10^1$. Apply Construction \ref{FundCtr}
with weight $6$ to obtain a $[2,1,1]$-GDC($6$) of types $84^4 48^1$
or $84^4 60^1$.
Here, the input $[2,1,1]$-GDC($6$)s of types $4^4, 6^4, 12^4$ and $12^5$ exist by
Proposition \ref{GDC(6)gt}. Now, adjoin $9$ ideal points to each of the above GDCs and fill in
the groups together with the nine extra points with a $[2,1,1]$-GDC($6$) of type $12^v 9^1$ with $v \in \{4, 5, 6, 7\}$ (which
all exist by Proposition \ref{GDC(6)12tu1}) to obtain a
$[2,1,1]$-GDC($6$) of type $12^t 9^1$ as desired.

For all $t\geq 34$ or $t\in \{20,24,25,29,30\}$, take a
$(t+1,\{5,6,7,8,9\},1)$-PBD from Lemma \ref{PBD5-9} and remove one
point to obtain a $\{5,6,7,8,9\}$-GDD of type $4^i 5^j 6^k 7^l 8^m$
with $4i+5j+6k+7l+8m=t$. Apply construction \ref{FundCtr} with
weight $12$ and input $[2,1,1]$-GDC($6$)s of types $12^t$ for $t\in
\{5,6,7,8,9\}$ (Lemma \ref{GDC(6)12t}) to obtain a
$[2,1,1]$-GDC($6$) of type $48^i 60^j 72^k 84^l 96^m$ and length
$12t$. Adjoin $9$ ideal points and fill in the groups together with the nine extra points with
$[2,1,1]$-GDC($6$)s of types $12^u 9^1$ for $u\in \{4,5,6,7,8\}$
(which exist by Proposition \ref{GDC(6)12tu1}) to obtain a
$[2,1,1]$-GDC($6$) of type $12^t 9^1$.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC12t+10(1)} There exists an optimal
$(12t+10,6,[2,1,1])_4$-code with size $(4t+3)(6t+5)$ for all $t \ge
0$.
\end{theorem}
\begin{proof}
For each $t\in\{0,1,2\}$, there exists an optimal
$(12t+10,6,[2,1,1])_4$-code with size $(4t+3)(6t+5)$ by Proposition
\ref{CCC(6)}.

For $t=3$, take a $[2,1,1]$-GDC($6$) of type $3^5$ (which exists by
Proposition \ref{GDC(6)gt}). Apply Construction \ref{Inflation} with
a TD($4,3$) to get a $[2,1,1]$-GDC($6$) of type $9^5$.  Adjoin one
ideal point and fill in the groups together with the extra point with an optimal
$(10,6,[2,1,1])_4$-code. The result is an optimal code of length
$46=12\times 3+10$.

For all $t\ge 4$, there exists a $[2,1,1]$-GDC($6$) of type $12^t 9^1$.
Adjoin one ideal point to this GDC and fill in the groups together with the extra point with
optimal codes of lengths $10$ or $13$ to obtain an optimal
$(12t+10,6,[2,1,1])_4$-code with size $(4t+3)(6t+5)$.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC12t+16(1)} There exists an optimal
$(12t+16,6,[2,1,1])_4$-code with size $(4t+5)(6t+8)$ for each $0 \le t
\le 15$.
\end{theorem}
\begin{proof}
For each $t\in \{0,1\}$, there exists an optimal
$(12t+16,6,[2,1,1])_4$-code with size $(4t+5)(6t+8)$ by Proposition
\ref{CCC(6)}.

For $t=2$, take a $[2,1,1]$-GDC($6$) of type $10^4$ (which exists by
Proposition \ref{GDC(6)gt}). Fill in the groups with an optimal
$(10,6,[2,1,1])_4$-code. The result is an optimal code of length
$40=12\times 2+16$.

For $t=3$, take a $[2,1,1]$-GDC($6$) of type $13^4$ (which exists by
Proposition \ref{GDC(6)gt}). Fill in the groups with an optimal
$(13,6,[2,1,1])_4$-code. The result is an optimal code of length
$52=12\times 3+16$.

For $t=4$, take a $[2,1,1]$-GDC($6$) of type $3^7$ (which exists by
Proposition \ref{GDC(6)gt}). Apply Construction \ref{Inflation} with
a TD($4,3$) to get a $[2,1,1]$-GDC($6$) of type $9^7$. Adjoin one
ideal point and fill in the groups together with the extra point with an optimal
$(10,6,[2,1,1])_4$-code. The result is an optimal code of length
$64=12\times 4+16$.

For $t=5$, take a $[2,1,1]$-GDC($6$) of type $3^5$ (which exists by
Proposition \ref{GDC(6)gt}). Apply Construction \ref{Inflation} with
a TD($4,5$) to get a $[2,1,1]$-GDC($6$) of type $15^5$. Adjoin one
ideal point and fill in the groups together with the extra point with an optimal
$(16,6,[2,1,1])_4$-code. The result is an optimal code of length
$76=12\times 5+16$.

For $t=6$, take a $[2,1,1]$-GDC($6$) of type $22^4$ (which exists by
Proposition \ref{GDC(6)gt}). Fill in the groups with an optimal
$(22,6,[2,1,1])_4$-code. The result is an optimal code of length
$88=12\times 6+16$.

For each $t\in \{7,8,\ldots,15\}$, there exists a $[2,1,1]$-GDC(6) of
type $12^t 15^1$ by Proposition \ref{GDC(6)12tu1}. Adjoining one
ideal point and filling in the groups together with the extra point with optimal codes of lengths
$13$ or $16$, the result is an optimal $(12t+16,6,[2,1,1])_4$-code.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC12t+16(2)} There exists an optimal
$(12t+16,6,[2,1,1])_4$-code with size $(4t+5)(6t+8)$ for all $t \ge
16$.
\end{theorem}
\begin{proof}
For all $t\ge 16$, take a TD($6,2r$)  for $r\ge 4$ and $r\not\in \{5,7,9,11\}$  from Lemma \ref{TD}. Apply
Construction  \ref{FundCtr} with weight $6$ to the points in the
first $4$ groups, $2x$ points in the fifth group, and $y$ points in
the last group. We require that $x=0$ or $x\ge4$ and $1\le y \le 31$, $y$ odd. The other points are given weight 0. Note that there
exist $[2,1,1]$-GDC($6$)s of types $6^4$, $6^5$ and $6^6$ by Proposition
\ref{GDC(6)gt}. The result is a $[2,1,1]$-GDC($6$) of type $(12r)^4
(12x)^1 (6y)^1$.
Adjoining $9$ ideal points and filling in the groups together with the nine extra points with
$[2,1,1]$-GDC(6)s of types $12^u 9^1$ for all $u\ge 4$, the result is a
$[2,1,1]$-GDC(6) of type $12^{4r+x} (6y+9)^1$. Adjoin one more
ideal point and fill in the groups together with the extra point with optimal codes of lengths $13$
or $6y+10$ for $1\le y \le 31$ and $y$ odd (which exist
by Theorem \ref{CCC12t+16(1)}). The result is an optimal
$(48r+12x+6y+10,6,[2,1,1])_4$-code, where $48r+12x+6y+10$ can take
any value of $12t+16$ with $t$ greater than $16$.
\end{proof}
\vskip 10pt

\begin{theorem} There exists an optimal $(6t+3,6,[2,1,1])_4$-code with size $6t^2+3t$ for all $t\ge
1$.
\end{theorem}
\begin{proof}
The result follows by combining Lemma \ref{CCC6t3} and Theorems
\ref{CCC12t+10(1)}--\ref{CCC12t+16(2)}.
\end{proof}
\vskip 10pt

\subsection{The Case of Length $n \equiv 5 \pmod{6}$}
\vskip 10pt

\begin{lemma}
\label{CCC(6)5} $A_4(5,6,[2,1,1])=1$.
\end{lemma}
\begin{proof}
The one required codeword is $\langle 0, 1, 2, 3\rangle$.
\end{proof}

\vskip 10pt

\begin{theorem}
\label{CCC12t+11(1)} There exists an optimal
$(12t+11,6,[2,1,1])_4$-code with size $24t^2+40t+16$ for all $t \ge
0$.
\end{theorem}
\begin{proof}
For each $t\in \{0,1,2\}$, there exists an optimal
$(12t+11,6,[2,1,1])_4$-code with size $24t^2+40t+16$ by Proposition
\ref{CCC(6)}.

For $t=3$, there exists a $[2,1,1]$-GDC(6) of type $1^{36} 11^1$ by
Proposition \ref{GDC(6)1t111}. Fill in the groups with an optimal
$(11,6,[2,1,1])_4$-code to obtain an optimal
$(47,6,[2,1,1])_4$-code.

For all $t\ge 4$, there exists a $[2,1,1]$-GDC(6) of type $12^t 9^1$ by
Lemma \ref{GDC12t91(1)}. Adjoining $2$ ideal points and
filling in the groups together with the two extra points with a $[2,1,1]$-GDC(6) of type $1^{12} 2^1$
(which exists by Proposition \ref{GDC(6)14_2}) and an optimal
$(11,6,[2,1,1])_4$-code, the result is an optimal
$(12t+11,6,[2,1,1])_4$-code with size $24t^2+40t+16$.
\end{proof}
\vskip 10pt

\begin{theorem}
\label{CCC12t+17(1)} There exists an optimal
$(12t+17,6,[2,1,1])_4$-code with size $24t^2+64t+42$ for all $t \ge
0$.
\end{theorem}
\begin{proof}
For $t=0$, there exists an optimal $(17,6,[2,1,1])_4$-code with size
$42$ by Proposition \ref{CCC(6)}.

For $t=1$, take a $[2,1,1]$-GDC($6$) of type $7^4$ (which exists by
Proposition \ref{GDC(6)gt}). Adjoin one ideal point and fill in the
groups together with the extra point with an optimal $(8,6,[2,1,1])_4$-code. The result is an
optimal code of length $29=12\times 1+17$.

For each $t\in \{2,4,5,6\}$, there exists a $[2,1,1]$-GDC(6) of type
$1^{12t+6} 11^1$ by Proposition \ref{GDC(6)1t111}. Fill in the
groups with an optimal $(11,6,[2,1,1])_4$-code to obtain an optimal
$(12t+17,6,[2,1,1])_4$-code.

For $t=3$, take a $[2,1,1]$-GDC($6$) of type $13^4$ (which exists by
Proposition \ref{GDC(6)gt}). Adjoin one ideal point and fill in the
groups together with the extra point with an optimal $(14,6,[2,1,1])_4$-code. The result is an
optimal code of length $53=12\times 3+17$.

For each $t\in \{7,8,\ldots,15\}$, there exists a $[2,1,1]$-GDC(6) of
type $12^t 15^1$ by Proposition \ref{GDC(6)12tu1}. Adjoining $2$
ideal points and filling in the groups together with the two extra points with a $[2,1,1]$-GDC(6) of
type $1^{12} 2^1$ and an optimal code of length $17$, the result is
an optimal $(12t+17,6,[2,1,1])_4$-code.

For all $t\ge 16$, take a TD($6,2r$) for $r\ge 4$ and $r\not\in \{5,7,9,11\}$ from Lemma \ref{TD}. Apply
Construction  \ref{FundCtr} with weight $6$ to the points in the
first $4$ groups, $2x$ points in the fifth group, and $y$ points in
the last group. We require that $x=0$ or $x\ge 4$ and $1\le y
\le 31$, $y$ odd. The other points are given weight 0. Note that there
exist $[2,1,1]$-GDC($6$)s of types $6^4$, $6^5$ and $6^6$ by Proposition
\ref{GDC(6)gt}. The result is a $[2,1,1]$-GDC($6$) of type $(12r)^4
(12x)^1 (6y)^1$.
Adjoining $9$ ideal points and filling in the groups together with the nine extra points with
$[2,1,1]$-GDC(6)s of types $12^u 9^1$ for all $u\ge 4$, the result is a
$[2,1,1]$-GDC(6) of type $12^{4r+x} (6y+9)^1$. Adjoin $2$ more
ideal points and fill in the groups together with the two extra points with a $[2,1,1]$-GDC(6) of type
$1^{12} 2^1$ and an optimal code of length $6y+11$ with $1\le y
\le 31$ and $y$ odd (which exists by Theorem \ref{CCC12t+16(1)}).
The result is an optimal $(48r+12x+6y+11,6,[2,1,1])_4$-code, where
$48r+12x+6y+11$ can take any value of $12t+17$ with $t$
greater than $16$.
\end{proof}
\vskip 10pt

\section{Conclusion}
\vskip 10pt

In this paper, we determine almost completely the spectrum with
sizes for optimal quaternary constant-composition codes with
weight four and minimum distances five or six. We summarize our
main results of this paper as follows:

\vskip 10pt
\begin{theorem} For any integer $n \ge 4$
\begin{equation*}
A_4(n,5,[2,1,1])=\begin{cases}
1,&\text{if $n=4$} \\
2,&\text{if $n=5$} \\
6,&\text{if $n=6$} \\
10,&\text{if $n=7$} \\
n \left\lfloor \frac{n-1}{2}\right\rfloor,&\text{if  $n\ge 12$ and
$n \ne 13$.}\end{cases}
\end{equation*}
\begin{equation*}
A_4(8,5,[2,1,1])\ge 18,
\end{equation*}
\begin{equation*}
A_4(9,5,[2,1,1])\ge 27,
\end{equation*}
\begin{equation*}
A_4(10,5,[2,1,1])\ge 36,
\end{equation*}
\begin{equation*}
A_4(11,5,[2,1,1])\ge 48,
\end{equation*}
\begin{equation*}
A_4(13,5,[2,1,1])\ge 72.
\end{equation*}

\end{theorem}

\vskip 10pt
\begin{theorem} For any integer $n \ge 4$
\begin{equation*}
A_4(n,6,[2,1,1])=\begin{cases}
1,&\text{if $n=4,5$} \\
4,&\text{if $n=7$} \\
\left\lfloor \frac{n}{2} \left\lfloor \frac{n-1}{3} \right\rfloor
\right\rfloor, &\text{if  $n\ge 6$ and $n \ne 7$.}\end{cases}
\end{equation*}
\end{theorem}
\vskip 5pt

\section*{Acknowledgments}

The authors thank Prof. Olgica Milenkovic, the Associate Editor, Prof.
Helmut B{\"o}lcskei, the Editor-in-Chief of IEEE
Transactions on Information Theory, and the two anonymous
referees for their constructive comments and suggestions that
greatly improved the readability of this article.

\vskip 10pt
\bibliographystyle{IEEEtran}

redpajama_set_name: RedPajamaArXiv


text:
HomeCelebrityJamie Lynn Spears Insists She's Britney's 'Biggest Supporter' Amid Alleged Rift
Jamie Lynn Spears Insists She's Britney's 'Biggest Supporter' Amid Alleged Rift
13/01/2022 Computer generated check source link below Celebrity 0
Jamie Lynn Spears responded to Britney Spears' clams about their relationship, stating in an emotional interview that she remains her sister's "biggest supporter" amid reports of a family rift.
Last year, Britney Spears tore into Jamie Lynn Spears on social media, suggesting that her younger sister played a role in the 13-year, court-ordered conservatorship that granted Spears limited authority over her personal and professional affairs. Earlier this month, she reportedly unfollowed Jamie Lynn Spears on Instagram.
Appearing on "Good Morning America" on Thursday, Jamie Lynn Spears said she "took no steps" to be involved with the conservatorship, which was formally terminated in November.
"I've always been my sister's biggest supporter," she said. "So when she needed help, I set up ways to do so, went out of my way to make sure that she had the contacts she needed to possibly go ahead and end this conservatorship and just end this all for our family. If it's going to cause this much discord, why continue it?"
"Everyone has a voice and it should be heard," she continued. "So if she wanted to talk to other people, then I set that up. I even spoke to her [previous] legal team, and that did not end well in my favor. I did take the steps to help, but … she has to walk through the door."
Watch Jamie Lynn Spears' "Good Morning America" interview below.
Later in the interview, Spears teared up when asked to comment on what the future of her relationship with her sister might entail.
"I love my sister," she said. "I've only ever loved and supported her and done what's right by her, and she knows that. So I don't know why we're in this position right now."
HuffPost reached out to a representative for Britney Spears but did not immediately hear back.
Jamie Lynn Spears, who is currently appearing on Netflix's "Sweet Magnolias," also published a memoir, "Things I Should Have Said." Released Tuesday, the book also touches on the media scrutiny she endured after becoming pregnant at 16 while starring on the Nickelodeon series "Zoey 101," as well as her father's alcoholism.
"Things I Should Have Said," however, has already weathered its share of pre-release controversy. Though Spears had planned to donate proceeds from the book to the mental health nonprofit This Is My Brave, the group opted to reject the contribution following a flurry of criticism from Britney Spears fans.
Source link Celebrity
Alleged
Britneys
insists
Maria Ewing, multifaceted opera singer, dies at 71
32 Things Everyone Used To Have In Their Home That Literally No One Has Anymore
Prince Harry, Oprah Continue Mental Health Series With Surprise Event
LOS ANGELES (AP) — Oprah Winfrey and Prince Harry are continuing their mental health series with a virtual town hall featuring Lady Gaga, Glenn Close and others. Apple TV+ announced Wednesday that it will release […]
Best Bachelor Season Episode 2 Twitter Reactions
Who else wants Victoria to go home, like, right now? Let's all say it together: GO HOME, VICTORIA. I know she's the early season villain, but my god she's annoying. Is she the least relatable […]
Best Celebrity Interview Moments From TikTok
"I did lose weight for this role…and I think it's a stupid conversation." —Carrie Fisher Recently, TikTok user @missbeifong went viral for sharing her favorite moment during a celebrity interview and asking others to share […]

redpajama_set_name: RedPajamaCommonCrawl


Catch up on today's top five analyst upgrades with this list compiled by The Fly: 1. ConocoPhillips (COP) upgraded to Hold from Underperform at Jefferies with analyst Jason Gammel citing valuation and improving financial metrics. 2. DineEquity (DIN) upgraded to Overweight from Sector Weight at KeyBanc with analyst Chris O'Cull saying the stock has reached a good entry point, while its risk/reward ratio is "compelling." 3. Gaming and Leisure Properties (GLPI) upgraded to Overweight at Morgan Stanley. 4. Aviva (AV) upgraded to Buy from Neutral at UBS with the firm saying the company is positioned to be the risk-adjusted dividend leader in the sector. 5. Ligand (LGND) upgraded to Hold from Sell at Deutsche Bank. This list is just a portion of The Fly's full analyst coverage. To see The Fly's full Street Research coverage, click here.
H.C. Wainwright analyst Carol Ann Werther raised her price target for Ligand Pharmaceuticals to $160 citing three recent positive events, namely positive Phase II focal segmental glomerulosclerosis data from partner Retrophin, the beginning of a Phase II clinical trial with LGD-6972 for the treatment of type 2 diabetes and the licensing of several therapeutic programs to Seelos. Werther keeps a Buy rating on Ligand.
Roth Capital analyst Joseph Pantginis raised his price target on Ligand to $154 from $150 after partner Lundbeck announced that the FDA approved Carnexiv injection as a short-term replacement therapy for oral carbamazepine formulations in adults with certain seizure types. Ligand will receive a $1.25M milestone payment this quarter and is entitled to receive a 2.75% royalty on sales of the drug, noted Pantginis, who keeps a Buy rating on Ligand shares.
Stephens analyst Drew Jones lowered his price target for Ligand to $125 from $150 on valuation. The analyst notes that the company reported mixed Q3 results as revenue softness led to lowered guidance for the year. Nonetheless, he reiterates an Overweight rating on the shares.


text:
The Free Library > Business and Industry > Business, international > International Business Times - US ed. > December 31, 2012
The Free Library > Communications > News, opinion and commentary > International Business Times - US ed. > December 31, 2012
The Free Library > Date > 2012 > December > 31 > International Business Times - US ed.
How The New Year's Eve Ball Drop In Times Square Works.
<a href="https://www.thefreelibrary.com/How+The+New+Year%27s+Eve+Ball+Drop+In+Times+Square+Works.-a0313522266
MLA style: "How The New Year's Eve Ball Drop In Times Square Works.." The Free Library. 2012 International Business Times 17 Jan. 2021 https://www.thefreelibrary.com/How+The+New+Year%27s+Eve+Ball+Drop+In+Times+Square+Works.-a0313522266
Chicago style: The Free Library. S.v. How The New Year's Eve Ball Drop In Times Square Works.." Retrieved Jan 17 2021 from https://www.thefreelibrary.com/How+The+New+Year%27s+Eve+Ball+Drop+In+Times+Square+Works.-a0313522266
APA style: How The New Year's Eve Ball Drop In Times Square Works.. (n.d.) >The Free Library. (2014). Retrieved Jan 17 2021 from https://www.thefreelibrary.com/How+The+New+Year%27s+Eve+Ball+Drop+In+Times+Square+Works.-a0313522266
Where else but Times Square on New Year's Eve could you get into a free party with Taylor Swift and "Gangnam Style" singer Psy? Of course, you'll be barricaded in with thousands of other people, unable to move your limbs, but it will surely be worth it to celebrate the annual ritual of a large ball moving down a pole. Just when did we start celebrating the new year with this sort of thing, and how is it timed to finish dropping at precisely midnight?
The New Year's Eve ball drop tradition traces its origins back to 19th-century timekeeping. In 1833, the Royal Observatory in Greenwich, England, installed a ball that would drop down a pole at 1 o'clock every afternoon to help ship captains calibrate their instruments. Other places besides Times Square use time-balls throughout the year, like the U.S. Naval Observatory, where a ball drops at noon every day.
But of course, the most famous ball drop these days happens in the city that never sleeps. The ball that will drop in Times Square Monday night is a geodesic sphere 12 feet across, sporting 2,688 Waterford Crystal triangles and illuminated by 32,256 LED lights.
It's a much more elaborate affair than the first New Year's Eve ball, used on the maiden voyage that marked the end of 1907 and the beginning of 1908. That was an iron and wood affair that weighed a comparatively svelte 700 pounds and was lit by 100 light bulbs. Throughout the decades, a succession of balls have made the trip down to herald the new year. A wholly iron ball was used starting in 1920, replaced in 1955 by a lighter aluminum one. In the 1980s, the ball was transformed into a Big Apple with the help of an artificial stem and red light bulbs.
In 1995, the ball, now back to a basic spherical shape, sported rhinestones and strobe lights and was also lowered with the aid of computers. In 2000, Waterford took the helm and invited revelers to gaze into the new millenium with the aid of a wholly crystal New Year's Eve Ball.
Though New York City Mayor Michael Bloomberg will push a button with a minute to go, that gesture doesn't actually initiate movement of the ball. A technician keeps an eye on the official time and, when the moment strikes, hits the real button that lowers the ball electronically.
"What we do is we get a global positioning system signal from the atomic clock in Colorado, which is the main clock the government uses to guide all of its electronic devices, whether it's missiles or planes or anything like that," Tim Tomkins of the Times Square Business Improvement District (http://www.ny1.com/content/36078/technology-helps-times-square-new-year-s-eve-ball-drop-run-smoothly) told NY1 in 2003.
Once the electronic gears are set in motion, the ball will drop 70 feet in 60 seconds to ring in 2013.
COPYRIGHT 2012 International Business Times
International Business Times - US ed.
Geographic Code:
1U2NY
Lady Gaga Takes Her Born This Way Foundation A Step Further.
'Auld Lang Syne' Song Lyrics, Meaning And Everything You Need To Know About The Popular New Year's Eve Song.


text:
Q: Why would square root of 1 equals 1 returns false in java? public static boolean checkSquare(int i){
    return IntStream
            .rangeClosed(1, i/2)
            .anyMatch(x -> Math.sqrt(x) == i);
}

When I enter 1 as an user input, it return false. I don't understand why square root of 1 is not equals to 1. Can anyone tell me is my code correct?

A: If your user input is assigned your i variable, it's clear why
IntStream.rangeClosed(1, i/2).anyMatch(x -> Math.sqrt(x) == i);

returns false when i==1, since 1/2 == 0, so IntStream.rangeClosed(1, 0) is an empty stream.
Change your method to:
public static boolean checkSquare(int i){
    return IntStream
            .rangeClosed(1, i)
            .anyMatch(x -> Math.sqrt(x) == i);
}

or, if you really want to keep the optimization of halving the size of the IntStream:
public static boolean checkSquare(int i) {
    return IntStream
            .rangeClosed(1, Math.max(1,i/2))
            .anyMatch(x -> Math.sqrt(x) == i);
}

redpajama_set_name: RedPajamaStackExchange


Today we announce the Sao Paulo 2017 Qualifier rules and prizes for the Overclocking World Championship 2017. On February 3, 2017, the qualification tournament kicks off with a three-hour qualifier featuring three benchmarks. The top four overclockers return on February 4, 2017, to fight for a spot in the Final and win the prestigious ticket to the Overclocking World Championship at the end of the year.
2017 will go down in history as an important year for Overclocking as it will feature the first ever Overclocking World Championship. The season-long contest is essentially a quest to find the best extreme Overclocker in the world in 2017. Each HWBOT World Tour 2017 event will feature an OCWC Qualifier where local (and possibly even not so local) Overclockers are invited to compete for a place in the Overclocking World Championship Final at the end of the year. Win a OCWC Qualifier, and you will have your seat in the Final come December.
The OCWC Qualifier in Sao Paulo will be spread over two days. Day 1 will act as a preliminary qualification round with the four fastest Overclockers moving on to Day 2 which will involve two Semi-Finals, a Bronze Final and a Grand Final.
On the third day of the HWBOT World Tour – Sao Paulo 2017 event, Brazil's most renowned and respected Overclockers will battle it out across three benchmarks in just three hours. The top four will be invited to return for the Semi-Finals on the following day.
Each overclocker will possess one veto which can be used strategically to avoid an unfavorable benchmark being selected.
The top-3 of the Overclocking World Championship – Sao Paulo 2017 Qualifier go home with an impressive set of prizes.
Monitors in limited supply – please bring your own.
Overclockers are expected to bring their overclocking and competition equipment to the qualifier. This includes but is not limited to thermal flask, insulation, torch or heat gun, peripherals.
You can learn more about the HWBOT World Tour – Sao Paulo 2017 here.
You can purchase a ticket to the Sao Paulo event here.
Renowned and respected Power Supply manufacturer Seasonic is supplying the Official Power Supply of the HWBOT World Tour 2017. The new Snow Silent Power Supply series will power the Overclocking World Championship Qualifiers as well as the amateur workshop systems during next year's World Tour.
As part of its commitment to the HWBOT World Tour 2017, Seasonic has pledged to provide Snow Silent Series PSUs throughout the World Tour. This is fantastic news, not only for the extreme overclockers competing in the Overclocking World Championship Qualifier contests, but also for all the amateur and ambient overclockers attending our OC Workshops. Seasonic Snow Silent PSUs will ensure stable and reliable power delivery to all of this year's World Tour attendees.


text:
Military Rings And Military Gifts
Armed Forces Personnel & Civilians
Marine Rings
K9 Rings
Family Crest Signet Ring
Family Crest Jewelry
Coat of Arms Plaques
O'Sullivan Coat of Arms
Air Force Graduation Gifts
Army Graduation Gifts
US Army Gifts
Family Crest Gift Ideas
Plaques and Coat of Arms
Family Crest Plaques
Gift Ideas For Men In The Military
Gifts for Army Officers
Military Wives
Gifts For Marines
Military Dogs
Navy Graduation Gifts
US Navy Gifts
Veteran Gifts
Gifts For Veterans
Military Clothes
Military dogs are wonderful because they are loyal. Dogs can be trained well and contribute to the safety of our nation. Truly K9 military dogs do hold a special place in the hearts of those who train them and work with them. Especially those who work with them in the United States Military in the defence of this great nation. In fact, these dogs can be trained for various tasks. Some of the dogs become rescue dogs, some sniffer dogs and some become handling dogs., etc.
Loyal Military Dogs
When it comes to the matter of which military dog breeds are used, most dogs tend to be German Shepherds. Dutch Shepherds are also in the mix. In addition, there is the frequent usage of Belgian Malinois. The reason why these breeds are the most prevalent ones used when it comes to military dogs is due to the fact that they hold high levels of intelligence. They are also supremely loyal. Moreover, they are powerfully aggressive and demonstrate real athletic capabilities.
German Shepards
It is noted that German Shepherds hold the highest preference among those who serve in the US Military. This is because these dogs possess a truly unparalleled mix of traits. They are moderately aggressive and easy to train. German Shepherds are predictable, demonstrating the characteristic of dependability. They posses a high level of intelligence and are agile in adapting to most climate conditions. Though some other breeds of dogs may manifest some or many of these desirable characteristics, the German Shepherd is the top breed that showcases all of these highly essential characteristics that are undeniably important when it comes to the service of defending the safety and freedom of the United States of America.
Design The Perfect K9 Military Ring At Online Store
K-9 Training & Military Rings
It will take adjustments for the trainers and dogs during the training process. Also, the ones who will be using the dogs in the service will need to adjust to the dogs. This is a fun time, but it is also a serious endeavor. This takes commitment and dedication. While the dogs are intended for serving the nation to protect its safety and freedom, it is noted that these dogs must be treated humanely and with compassion. However, one cannot take a light approach with these dogs the same way one may take with a family dog.
A sense of seriousness and command must be used with these dogs most of the time, though they must also be shown love and attention as well. When the seriousness and command side is kept in place with these dogs as humans interact with them, the dogs will remain seriously focused on what their aim and purpose is, even when there is the showing of love and attention during more relaxed human interaction. Many of the handlers design K9 Military Rings in honor of their canine friend.
Once trained, the military dogs will perform well. They will do as they are told. They understand their instructions and they follow through. It is not uncommon for a dog to fearlessly put its life in danger knowingly in order to protect the ones it works with. With such a sense of reliability in military dogs, they do grip a deep place in the hearts of all who interact with them. These dogs are honorable in the service that they provide concerning defending the safety and freedom of this nation.
Military Online Shopping / Universal Promotions LLC, Suite 1032 #U139, 332 S Michigan Avenue, Chicago, Illinois, 60604-4434, United States of America. Tel: (1) 888-572-5875.
Family History and Coat of Arms
Deploying To Kuwait
US Army Band
Measure Ring Size Online
© 2022 Military Shopping


G-Block called Kekra-I well some 230-km South West of the Karachi coast. "The companies will be able to collect specimens in March or April for further lab examination in America and Italy, following which the picture will get clear that the well contains gas or oil and in what quantity," official sources told APP.
cost of US $ 75 million.
waiting for a good news about confirmation of gas and oil deposits.
world leading companies in Pakistan to try its luck in offshore explorations.
country was in dire need of a major gas and oil discovery.
but could not find significant reserves matching the country's requirements.
While, existing reserves are depleting gradually, they added.
hydrocarbon reserves in the country.


text:
PUBLIC OFFER CONTRACT
Conditions for the transfer of rights to the breeding use of animals:
These provisions are due to the high breeding value of the animal.
The Agreement on the transfer of breeding rights to an animal from the "Monti Rey" cattery contains the terms of the Agreement on the transfer of the exclusive right to use a breeding animal, which, in accordance with Article 6 of the Civil Code of the Russian Federation (Analogy), is governed by Article 7 of the Federal Law "On Livestock". The use of tribal products (materials) as an object of exclusive rights (intellectual property) is allowed in the manner prescribed by law.
The owner receives the right to breed an animal from the "Monti Rey" cattery with the restrictions stipulated by the terms of this agreement listed in clauses: 1-7. Upon termination of the breeding work and after submitting the photo of sterilization / castration to the "Monti Rey" cattery by mail: laura9@mail.ru, the animal becomes the OWNER"S FULL PROPERTY, and this transfer of the rights to reproduce is terminated. The transfer of the rights to reproduce to an animal from the cattery "Monti Rey" is not intended for commercial gain breeder for breeding is the further breeding work of the above animal in the Owner"s kennel.
Violation of any of the restrictions stipulated by the terms of the agreement (paragraphs: 1,2,3,4,5,6,7) violates the breeder"s rights and damages the cattery "Monti Rey".
1. This animal is transferred to the Owner for breeding work in a private nursery, exclusively for personal use. The owner undertakes NOT to open a cat from the cattery "Monti Rey" for general use and not to breed other cats from third-party kennels.
2. The Owner agrees to knit this male only with those females that directly belong to him (acquired and owned by the Owner or born in the Owner's kennel).
3. The owner undertakes not to sell kittens from the manufacturer to Russia, Ukraine and all former Soviet republics for breeding work from the manufacturer "Monti Rey". Castres can be sold (castration / sterilization operations performed by the Owner before transferring the kitten to a new buyer) as pets in any region without restriction.
4. The owner understands and agrees that the rights to breeding work can not be transferred to third parties !!! The owner undertakes to castrate / sterilize the producer, obtained in the cattery "Monti Rey" after the end of breeding.
Photos castration provided by e-mail: laura9@mail.ru
Then he must find him the best home where the animal will receive care. Or keep it at home
5. In case of non-compliance with any clause of the above conditions, The owner undertakes to return this pet to the breeder of the Monti Rey kennel, not later than one month from the moment of detection of the violation at least one of the conditions (paragraph 1 - 7)using the breeding animal from the cattery "Monti Rey"
6. Having received a breeding animal from the cattery "Monti Rey", the Owner confirms that he understands, agrees and accepts all the terms of this Offer Agreement.
7. The owner has no right to remove claws, apply tattoos, to be cruel with a kitten.


text:
Cationic amino acid transporter 4 is a protein that in humans is encoded by the SLC7A4 gene.

See also

References

Further reading

Solute carrier family

redpajama_set_name: RedPajamaWikipedia


This client is a great dane breeder of AKC champion show dogs. Her kennel name was a play on her own last name and a reference to the Sherlock Holmes novel, "The Hounds of Baskervilles." She wanted it to feature a natural-eared great dane and noted that she LOVED hearts.


text:
Rapper Mac Miller is dead at 26
Miller was set to go on tour next month.
By Alex Abad-Santos Sep 7, 2018, 5:24pm EDT
Share All sharing options for: Rapper Mac Miller is dead at 26
Mac Miller on October 28, 2017.
Malcolm James McCormick, known to his fans and the music industry as Mac Miller, died on Friday at the age of 26, according to TMZ. Miller reportedly died of an apparent drug overdose and was found dead at his home in Studio City, California. His death was confirmed by Variety.
Miller is best known for his 2015 album GO:OD AM and songs like "Weekend" and "Self Care." He was also featured on Ariana Grande's "The Way," which the two collaborated on in 2013. He and Grande dated from 2016 to 2018.
Miller's new album Swimming was just released on August 3, and he was set to go on tour in October. The last message posted to Miller's official Twitter account was a tweet on Thursday night that expressed his excitement about kicking off his concert series:
Go get tickets for tour.
J.I.D.
I'm bringing a band.
The show is going to be special every night.
I wish it started tomorrow.
It starts October 27th.https://t.co/RGYZC5DTza
— Mac (@MacMiller) September 7, 2018
Miller never flinched in writing music about his personal battles with rejection and the music industry, or about his struggles with drug and alcohol addiction. That put a spotlight on the life he led in the public eye. In May, Miller was charged with a DUI after crashing his Mercedes into a power pole.
"I used to rap super openly about really dark shit," Miller told Vulture in an interview this month, when discussing his early music and its allusions to hard drugs and the theme of dying young. "Because that's what I was experiencing at the time. That's fine, that's good, that's life. It should be all the emotions."
Next Up In Culture


text:
Q: Is my PCA code written in Python correct? I'm doing a research that demands a PCA plot, so I can see the agglomeration patterns of the plots thanks to the dimension reduction that PCA provides. So, I found a code and wanted to be sure that it does what it is supposed to do: a PCA plot. The thing is, I don't know if the code is 100% correct, because I think that for PCA to work, we first need no standardize the data so that the mean of the values for each dimension is 0, and I couldn't identificate that in the code. I'm writing this in hope to find someone that have experience doing PCA to enlighten me. Thanks, here's the code:
import numpy as np
import matplotlib.pyplot as plt

with open('B0.txt', 'r') as simulation_data:
    simulation_data = [x.strip() for x in simulation_data if x.strip()]

    data = [tuple(map(float, x.split())) for x in simulation_data[2:]]

    x = [x[1] for x in data]
    y = [x[2] for x in data]
    z = [x[3] for x in data]
    mx = [x[4] for x in data]
    my = [x[5] for x in data]
    mz = [x[6] for x in data]

dados = np.array([x, y, z, mx, my, mz])

matriz_cov = np.cov(dados)

v, w = np.linalg.eig(matriz_cov)

ordem = v.argsort()

maiorautovetor = w[:, ordem[-1]]
segundomaiorautovetor = w[:, ordem[-2]]

Wpca = np.array([maiorautovetor, segundomaiorautovetor])

novos_dados = np.dot(Wpca, dados)

plt.figure(1)
plt.scatter(novos_dados[0, :], novos_dados[1, :])
plt.show()

print(maiorautovetor)
print(segundomaiorautovetor)


text:
@interface HomeWebViewController ()

@end

@implementation HomeWebViewController

- (void)viewDidLoad {
    [super viewDidLoad];
    // Do any additional setup after loading the view.

    self.view.backgroundColor = kGlobalBg;
}

- (void)didReceiveMemoryWarning {
    [super didReceiveMemoryWarning];
    // Dispose of any resources that can be recreated.
}

@end

redpajama_set_name: RedPajamaGithub


I found my pictures! They were right where they were supposed to be – I was just having a little trouble figuring out where "supposed to be" was. But, now that I've found them, I can tell that I have several posts worth of pictures. Yay!
Still a cute couple, even at 90!
Two of their three kids – we'll have to take another one this fall when Dad gets home.
Abigail on her daddy perch waiting patiently for cake. She was in SUCH a fun mood that day!
Visiting.(My uncle Glen in the rocker, my two brothers Ben & Tim and my aunt Polly Ann on the couch. Ben's wife Ashley in the chair. And, of course, precious Abigail visiting with Aunt Polly Ann. We missed visiting with Tim's wife DeAnn who had to work that day.
But, there's no doubt as to who is the most popular one with all four cousins. Yep, Aunt Alli! It took about five seconds for them all to descend on her when we walked in the door.
Although, Abigail did make her little beeline toward Olivia pretty quickly! Aren't they cute?
Grandma needed some great-granddaughter loving, too.
It was a whirlwind day for all of us – none of us could stay very long. But, it was a good visit. We are so blessed to still have my grandparents. I didn't know any of my great-grandparents, and even my mom's parents passed away when I was pretty young. So it's pretty special to me that my kids have both sets of grandparents and a set of great-grandparents to know.
Happy birthday, Grandpa! We're thankful for every single year God allows you to remain with us!


BE QUICK - MUST BE SOLD!!
Highly successful ladies fashion and styling business. Town and Country Boutique has had the same owner for the past 21 years with well established clientele. This opportunity rarely comes along, don't delay your inspection.
Excellent facility located on the Armidale Road Nemingha offering high visibility and exposure. Could easily be run as 2 businesses/offices with 2 separate kitchens and bathrooms plus several different office spaces. Previously housed a successful tarp repair business for over 40 years.
The perfect combination of retail & auxiliary floor space complimented with a very stylish first floor 150sqm residence complete with car accommodation. This is ideal for an investor or a business retailer looking for the perfect site. The site is in excellent presentation including new roofing, renovations and many features. A must to look at!
That's right, this highly exposed trading centre has just been offered for sale. It boasts 4 strong tenants, strong leases together with excellent exposure in the CBD. It ticks every box. These type of sites rarely become available and this site is returning a net annual income of $169,000 along with annual reviews with long standing tenants. Very hard to beat. Call for more details.
This strategically positioned property on Gunnedah Road has been totally renewed inside and out. It offers a one off opportunity for an astute business owner to occupy or investor to benefit from capital appreciation and or rental/lease return. Current configuration includes reception and consultation area, boardroom, large open office, two private offices, storeroom with built in shelving, bathroom and kitchen facilities and an outdoor courtyard.
This large 25,000 sqm site sits in the Longyard precinct with the potential to accommodate a wide range of commercial / business uses.
It has perfect location with high exposure to the New England Highway. The site is nicely shaped to build larger structures, suitable for bulky goods, accommodation and a range of other options. Full details are available on request.
This is a unique opportunity in a very progressive area of Tamworth.
This East Tamworth residence is a rare opportunity to secure an investment together with a strong capitol growth. As well as a residence it also includes a well established restaurant. A strong lease is in place and all in one of Tamworth's prime locations. Owner wants to sell and sell now!!
An opportunity to secure a high yielding property!
The iconic Tamworth landmark of Heemskerks Nursery is up for sale after 50 years of solid trading all within the one family name.


This is our uniquely exclusive Pacific-Island Themed Name-Art Illustration in a Hafa Adai motif.
We are currently offering this product as a 9 3/8 x 23 3/4 inch, 10 mil, museum-print French Giclee print. This means that we use a process that ensures a vivid presentation of the image without fading, cracking, or peeling for decades.
This print fits a 10x24 inch frame that can be purchased seperately at any Michael's Art Store across the mainland United States.
This illustration is rolled and wrapped in a clear protective sleeve and carefully packaged in a 15" mailing tube - with an exterior label appropriate for gift-giving. The label reads "A Gift of the Islands" (see below).
This is how the print will look (copyright watermark is not printed on purchased products).
This is a sample of how the print looks framed with a border mat.
If you plan to order this product as a Christmas gift, we suggest that you place weeks before the holidays, because of production lead times and shipping timelines. Placing orders after this date, will subject your shipment to potential production and USPS backlogs.
This product comes standard with no modifications allowed.


text:
1001 Songs Challenge #397: Evie (1974)
Stevie Wright – Evie (1974)
We're leaving the US behind today, dear reader, and making our way Down Under to Australia. Our artist today was originally born in the UK in Leeds which isn't a million miles from where I live, but he and his family moved to Australia when he was only nine years old. I could think of worse places. Stevie Wright was the singer in the group The Easybeats, who previously appeared on our list with Friday on My Mind. The group disbanded in 1969 and we now join Stevie Wright in 1974 with his solo career in full flow. From his setlist at this time, 1001 Songs have gone with the track – Evie.
Evie is essentially three songs for the price of one but they are bound by a common theme and person so we'll allow this minor
discretion on our prestigious list. Part 1 is a combination of hard rock and blues and sees Wright's narrator singing about the Evie of the title who is seventeen and sounds shy and quite reserved. The narrator thinks she is beautiful though and beseeches her to let her hair down and enjoy the music of a band he has gone to see. He very much wants to woo her at this stage. Part 2 is a gentler melody, more soft rock with a piano accompaniment. In this section the narrator and Evie are together and they are happy, immersed in the waves of love. Things are going very well by the sound of things and the narrator simply hasn't the words to convey how much he feels for Evie. He is also grateful to her for she is pregnant and their future looks bright. Part 3 is heavier rock and now the song takes a dark turn. Evie does indeed have a baby but something goes wrong and she sadly does not survive. The narrator is left alone with their child and laments the injustice of this tragedy and the end of their wonderful union.
Evie is an intricately crafted and moving song with the three parts that make up the suite having a different sound and feel to them but all are equally absorbing. All the ingredients are there for this romance to go the distance and it does but Part 3 is a reminder that sometimes things happen that are beyond our control. I love that Wright mixes up the sound between the parts and how none of the three segments suffer for it. This is a truly accomplished piece.
The Kinks – Days (1968)
Derek & The Dominos – Layla (1970)
David Bowie – Life on Mars? (1971)
Stevie Wonder – Living for the City (1973)
Patti Smith Group – Piss Factory (1974)
Pink Floyd Pilgrimage #6: Meddle (1971)
1001 Songs Challenge #41: Rocket 88 (1951)
1001 Songs Challenge #213: I'm a Believer (1966)
1001 Songs Challenge #396: Piss Factory (1974)
#396 of the 1001 Songs Challenge is Piss Factory by Patti Smith Group ...
1001 Songs Challenge #395: Sweet Home Alabama (1974)
1001 Songs Challenge #394: (Looking for) The Heart of Saturday Night (1974)
1001 Songs Challenge #398: Free Man in Paris (1974)
#398 of the 1001 Songs Challenge is Free Man in Paris by Joni Mitchell ...
1001 Songs Challenge #399: I Will Always Love You (1974)
1001 Songs Challenge #400: The Grand Tour (1974)


Year: 1957 Item #: 384 Views: 4,695 Comments: No Comments.
Before you buy for 1957 you must not fail to see the new BLUEBIRD THUNDERBOLT. There's never been a caravan like this before! It's brilliantly spectacular, it's gloriously colourful, it's years ahead of its time!