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\title{\large \bf  On the Block Coloring of Steiner Triple Systems}
%\thanks
%{{\it Key Words}: Steiner triple system, chromatic index}
%\thanks {2000{ \it Mathematics Subject Classification}: 05B07, 05C15.
 %}}

\author{{\normalsize{\sc R.
Manaviyat${}^{\mathsf{}}$}}\, \vspace{3mm}
\\{\footnotesize{${}^{\mathsf{}}$\it Department of Mathematics, Payame Noor University, 19395-4697 Tehran, Iran}}
\thanks{{\it E-mail addresses}: $\mathsf{ra\_manaviyat@yahoo.com}$.} }






\date{}

\begin{document}
\maketitle



\begin{abstract}
{\small \noindent A {\it Steiner triple system} of order $v$,
STS($v$), is an ordered pair $S=(V,B)$, where $V$ is a set of size
$v$ and $B$ is a collection of triples of $V$ such that every pair
of $V$ is contained in exactly one triple of $B$. A {\it $k$-block
coloring} is a partitioning of the set $B$ into $k$ color classes
such that every two blocks in one color class do not intersect. In this paper, it is proved that for every
$k$-block colorable STS($v$) and $l$-block colorable STS($w$), there
exists a $(k+lv)$-block colorable STS($vw$) obtained from an introduced construction. Moreover, it is shown
that
 for every $k$-block colorable   STS($v$), every
STS($2v+1$) obtained from the well-known construction is
$(k+v)$-block colorable.  }
\end{abstract}

{\it Keywords}: Steiner triple system, Chromatic index, Matching, Coloring.\par
2000{ \it Mathematics Subject Classification}: 05B07, 05C15.
 


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\section{Introduction}

Let $G$ be a graph. We denote the  vertex set and the edge set of
$G$ by  $V(G)$ and $E(G)$, respectively. A graph $G$ called {\it
strongly $(k,\lambda,\mu)$-regular} if there are parameters $k,\ \lambda$ and $\mu
$ such that $G$ is $k$-regular, every adjacent pair of vertices have
$\lambda$ common neighbors, and every nonadjacent pair of vertices
have $\mu$ common neighbors. A {\it proper vertex coloring} of $G$
is a function $c: V(G)\longrightarrow L$, with this property that if
$u,v\in V(G)$ are adjacent, then $c(u)$ and $c(v)$ are different. A
{\it vertex $k$-coloring} is a proper vertex coloring with $|L|=k$.
The {\it  chromatic number } of $G$, denoted by $\chi(G)$, is the
minimum number $k$ for which $G$ has a vertex $k$-coloring.

A celebrated theorem due to Brooks \cite{west} states that:


\begin{thm} \label{brooks}
{If $G$ is not an odd cycle or a complete graph, then $\chi(G)\leq
\Delta(G)$. }
\end{thm}


A {\it vertex $k$-coloring} is a proper vertex coloring with
$|L|=k$. A {\it (proper) $k$-edge coloring} of a graph $G$ is a
function $f: E(G)\longrightarrow L$, where $|L| = k$ and $f(e_1)\neq
f(e_2)$, for every two adjacent edges of $G$.  A {\it matching} in a
graph is a set of non-adjacent edges. A {\it perfect matching} of
$G$ is a matching that covers all vertices of $G$. Given an edge
coloring of a graph, a {\it rainbow matching} is a matching whose
edges have distinct colors.

An $n\times n$ matrix $L=(l_{ij})$ whose entries are taken from a set $S$ of $n$ symbols
 is called a  latin square of order $n$ on $S$ if each symbol appears precisely once in each
 row and in  each column of $L$.
A pair of latin squares $L=(l_{ij})$ and $L'=(l'_{ij})$ are called {\it orthogonal latin squares}
 if and only if the ordered pairs $(l_{ij},l'_{ij})$ are distinct for all $i$ and $j$.
 Here we say that $L$ is orthogonal to $L'$.
The following theorem states the condition for existence orthogonal latin squares of order $n$.


\begin{thm} \label{l}{\rm\cite{and}}
{For every natural number $n\neq 2,6$, there is a pair of orthogonal
latin squares of order $n$.}
\end{thm}


A {\it Steiner triple system} of order $v$, STS($v$), is an ordered
pair $S=(V,B)$, where $V$ is a  set of size $v$ and $B$ is a set of
size $b$ which is a collection of triples of $V$ such that every
pair of $V$ is contained in exactly one triple of $B$. Every
triple of STS($v$) called a block. The number of times that each $v
\in V$ appears in the blocks is denoted by $r$. One can easily see
that for every STS($v$), $r=\frac{v-1}{2}$. It is not
hard to see that a STS($v$) exists if and only if the edges of the
complete graph $K_v$ partitions into triangles. It is well known
that a necessary and sufficient condition for a STS($v$), $v\geq 3$
to exist is that $v\equiv1$ or $3$ (mod $6$). Such a $v$ is said to
be {\it admissible}.
 A Steiner triple system $(V,B)$ is called {\it resolvable}
 if the triples of $B$ can be partitioned into $\frac{b}{r}$ classes, where each class is a partition of $V$.
 By Lemma $9.1.1$ of \cite{and}, a resolvable STS($v$) can exist
 only if $v \equiv 3$, (mod $6$).

Let $S=(V,B)$ be a {\it Steiner triple system}. A {\it color class}
is a system of pairwise disjoint triples. A {\it $k$-block coloring}
is a partitioning of the set $B$ into $k$ color classes. Here we say
that $(V,B)$ is $k$-block colorable.
 The chromatic index, $\chi'(S)$, of a Steiner triple system
$S$ is the least $k$ for which a $k$-block coloring exists. We say
two blocks are adjacent if they have an element of $S$ in common. A
{\it block intersection graph} of a Steiner triple system $S =
(V,B)$, denoted by $G_S$, is a graph with the vertex set $B$; the
vertices are adjacent if and only if the respective blocks are
adjacent. Moreover, it is not hard to see that $G_S$ is a strongly
$(3r-3,r+2,9)$-regular graph. So,  by  Theorem \ref{brooks},
$\chi'(S)\leq 3r-3$ for $v>7$. Also, since the the clique number of
$G_S$ is $r$, $\chi'(S) \geq r$ if $v \equiv 3$ (mod $6$).  The following well
known theorem states that in what conditions  $\chi'(S)=r$.


\begin{thm} \label{resolvable}
{Let $S$ be a STS($v$). Then $\chi'(S)=r$ if and only if $S$ is
resolvable.}
\end{thm}

Now, By Theorem \ref{brooks} and  Theorem \ref{resolvable}, we conclude that
$r\leq \chi'(S) \leq 3r-3$ if $v \equiv 3$ (mod $6$) and $r+1 \leq
\chi'(S) \leq 3r-3$ if $v \equiv 1$ (mod $6$).

The upper bound $\chi'(S) \leq  3r-3$ seems to be weak in general.
In fact, using probabilistic methods Pippenger and Spencer in
\cite{8'} proved that $\chi'$(STS($v$)) is asymptotic to $\frac{v}{2}$. For more information on the chromatic
index of Steiner triple systems the reader is referred to Chapter
$18$ of \cite{6'}. For some classes of STS($v$) the upper bound was
improved. In particular, Colbourn in \cite{6} improved it for cyclic
STS($v$) by proving $\chi'$(STS($v$))$\leq v$.




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\section{ Block Coloring of Steiner
Triple Systems of Order $vw$}


In this section, we introduce a block coloring of a Steiner
triple system STS($vw$)  obtained from two Steiner triple systems
STS($v$) and STS($w$). For this purpose, first we establish the following Construction.
In particular, we introduce some resolvable STS($vw$) of STS($v$) and STS($w$).



\begin{construction} \label{cvw}
 {\rm  Let $(V,B)$ be a STS($v$) on the set $V:=\{x_1,\ldots,x_v\}$ and $(W,B')$ be a STS($w$) on the
  set $W:=\{y_1,\ldots,y_w\}$. Then define $(Z,S)$ as a STS($vw$) on the set $Z := \{z_{ij} , 1\leq i \leq v, 1\leq j \leq w\}$ with two types of blocks as follows:

For every $j$, $1\leq j\leq w$, consider a copy of the complete
graph $K_v$, $K_{v}^{j}$ , with the vertex set $\{z_{1j} ,\ldots ,
z_{vj}\}$.
 Using $(V, B)$ one can partition the edges of
each $K_{v}^j$, for every $1\leq j \leq w$, into triangles. Call the
blocks made by these triangles, Type $1$.
  Now, consider the complete graph $K_w$ with
the vertex set $\{K_{v}^j , 1 \leq j \leq w\}$. Using $(W,B')$ one
can partition the edges of  $K_w$ into triangles. Let us call this
partition by $F$. For every $i,j$, $1\leq i,j\leq w$, $i\neq j$,
join every vertex of $K_{v}^i$ to every vertex of $K_{v}^j$. So,
every triangle in the $K_w$ is corresponding to $3v^2$ edges. Now,
for each triangle of $F$ such as $\{K_{v}^p,K_{v}^s,K_{v}^t\}$,
$1\leq p,s,t\leq w$, consider a latin
 square $L$ of order $v$ on the set $\{z_{1t},\ldots,z_{vt}\}$
 such that the rows and the columns are indexed
 by  $\{z_{1p},\ldots,z_{vp}\}$ and  $\{z_{1s},\ldots,z_{vs}\}$,
 respectively. For every $z_{ip}$ and $z_{js}$,
 $1\leq i,j \leq v$,  $\{z_{ip},z_{js},L_{z_{ip} z_{js}}\}$ is
 considered as a block of Type $2$.
 It is not hard to see that all blocks of Type $1$ and Type $2$
  form a STS$(vw)$. Call a Steiner triple
 Systems obtained from  this Construction such that every used latin square
  has an orthogonal latin square, by OLS$(vw)$. Note that by Theorem \ref{l}, OLS$(vw)\neq \emptyset$ for each admissible
 $v$ and $w$.}
\end{construction}

\begin{thm} \label{tvw}
{ For every  $k$-block colorable STS{\rm (}$v${\rm )} and $l$-block
colorable STS{\rm (}$w${\rm )}, there exists a $(k+lv)$-block
colorable STS{\rm (}$vw${\rm )} OLS{\rm (}$vw${\rm )}.}
\end{thm}

\begin{proof}
{Let $(V,B)$ be a $k$-block colorable STS($v$) and
$f:B\longrightarrow\{1,\ldots,k\}$ be a such coloring and
$(W,B')$ be a $l$-block colorable STS($w$) with the function
$f':B'\longrightarrow\{1,\ldots,l\}$. Moreover, let
$(Z,S)$ be an OLS($vw$).
 Now, define $c:S\longrightarrow\{1,\ldots,
lv+k\}$ as follows. First, we color the blocks of Type $1$. For all
$1\leq j \leq w$ and  $\{x_{m},x_{n},x_{p}\}\in B$, let
$c(\{z_{mj},z_{nj},z_{pj}\})=f(\{x_{m},x_{n},x_{p}\})$.  Next, to
color the blocks of Type $2$, consider a triangle
$t=\{K_{v}^p,K_{v}^s,K_{v}^t\}$ of partition $F$ in Construction
\ref{cvw}. Note that $t$ is corresponding to a block of
$(W,B')$, say $b$. Let $L$ be a latin square
used to partition the edges of $t$ and $L'$ be a latin square
on the set $\{k+(f'(b)-1)v+1,\ldots,k+f'(b)v\}$ orthogonal to $L$.
Then, let $c(\{z_{ip},z_{js},L_{z_{ip} z_{js}}\})=L_{z_{ip}
z_{js}}'$
 and repeat this procedure for every triangle of $F$. We show that $c$ is
 a $(k+lv)$-block coloring of $(Z,S)$.
First, note that if two adjacent blocks call $b_1$ and $b_2$ have
the same color, since the set of colors used to color the blocks of
Type $1$ and Type $2$ have no color in common, then $b_1$ and $b_2$
belong to the same type. First,
suppose that $b_1$ and $b_2$ are the blocks of Type $1$. Since $f$
is a $k$-block coloring of $(V,B)$, $c(b_1)\neq c(b_2)$.  Now,
suppose that $b_1$ and $b_2$ are the blocks of Type $2$. Two cases
may be assumed. Suppose that $b_1=\{m,n,L_{mn}\}$ and
$b_2=\{p,q,L_{pq}\}$ are obtained from the same
triangle of partition $F$. If $m=p$ or $n=q$, since $L'$ is a latin
square, then $c(b_1)\neq c(b_2)$. If $L_{mn}=L_{pq}$, since $L$ and
$L'$ are orthogonal latin squares, then $L'_{mn}\neq L'_{pq}$. So,
in this case $c(b_1)\neq c(b_2)$. Now, assume that $b_1$ and $b_2$ belong to different triangles, call $t_1$ and $t_2$.
The adjacency of $b_1$ and $b_2$ concludes the adjacency of $t_1$ and $t_2$. So, $t_1$ and $t_2$ are corresponding to two adjacent blocks in $B'$. Since $f'$
is a block coloring of $B'$, the sets of colors used to color the blocks obtained from $t_1$ and $t_2$ have no color in common. Thus
 $c(b_1)\neq c(b_2)$ and the proof is complete.
  }
\end{proof}




\begin{cor}
{ If there exists a resolvable  STS{\rm (}$v${\rm )} and  a
resolvable  STS{\rm (}$w${\rm )}, then there exists a resolvable
STS{\rm (}$vw${\rm )}.}
\end{cor}

\begin{proof}
{ Let $(V,B)$ and $(W,B')$ be a resolvable STS($v$) and a resolvable
STS($w$). Moreover, let $(Z,S)$ be an OLS$(vw)$. By  Theorem \ref{tvw},
$(Z,S)$ is $(\frac{vw-1}{2})$-block colorable. Since the chromatic
index of $(Z,S)$ is at least $\frac{vw-1}{2}$,  by Theorem \ref{resolvable} we are done.}
\end{proof}






\begin{thm} \label{r3v}
{ If $(Z,S)$ is a resolvable STS{\rm (}$vw${\rm )} obtained from  Construction \ref{cvw},
 then $(Z,S)$ is an OLS{\rm (}$vw${\rm )}. }
\end{thm}

\begin{proof}
{Let $f:Z\longrightarrow \{1,\ldots,\frac{vw-1}{2}\}$ be a
$(\frac{vw-1}{2})$-block coloring of $(Z,S)$. First we claim that exactly $v$ colors are appeared
in the coloring of the blocks obtained from each used latin square. Note that for every $x\in Z$, $x$
appears in $\frac{v-1}{2}$ blocks of Type $1$ and $\frac{v(w-1)}{2}$
blocks of Type $2$.
Call $q=\frac{w-1}{2}$ latin square used in Construction \ref{cvw} by $L_1,\ldots,L_q$. Note that $x$ appears in $v$ blocks obtained from $L_i$
for eah $1\leq i \leq q$. Thus for every $1\leq i,j\leq q $, $i\neq j$, there are $v$ colors appeared in the blocks obtained from $L_i$ not in the blocks
obtained from $L_j$. Moreover, since $f$ is a
$(\frac{vw-1}{2})$-block coloring, for each latin square exactly $v$ colors are used in $f$. Now, for each latin square $L$ define $L'$
be  a square of size $v$ such that $L'_{ij}=f(\{i,j,L_{ij}\})$. The
properties of block coloring conclude that $L'$ is orthogonal to $L$
and the proof is complete.}
\end{proof}






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\section{ Block Coloring of Steiner
Triple Systems of Order $2v+1$}



\begin{defi}\label{rooms} {\rm {\cite[p.584] {and}}}
{ 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:\\
\noindent$(1)$ every cell of $F$ either is empty or contains an
unordered pair
of symbols from $S$.\\
$(2)$ Each symbol of $S$ occurs once in each row and column of $F$.\\
$(3)$ Every unordered pair of symbols occurs in precisely one cell
of $F$.}
\end{defi}






\begin{thm}\label{room}{\rm {\cite[p.585] {and}}}
A Room square of side $n$ exists if and only if $n$ is odd and
$n\neq 3,5$.
\end{thm}


\begin{cor}\label{rainbow}
Let $n$ be an even integer where $n\neq 4,6$. Then there exists a
$(n-1)$-edge coloring of $K_n$ such that the edges partition into
$n-1$ rainbow perfect matchings.
\end{cor}



\begin{proof}
{Let $S=V(K_n)=\{v_1,\ldots,v_n\}$. Since $n\neq 4,6$ is an even
integer, by Theorem \ref{room}, there exists a room square $F$ of
side $n-1$ on $S$. Note that each unordered pair appeared in each cell
of $F$ is corresponding to an edge of $K_n$. Moreover, by Part $(2)$ of
Definition \ref{rooms},  the union of edges appeared in each row or
column is a perfect matching of $K_n$. Now,  assign color $i$ to all
edges appeared in cells of row $i$ in $F$, for every $i$, $1\leq i
\leq n-1$. Note that since every unordered  pair of symbols occur in
precisely one cell of $F$, we obtain a $(n-1)$-edge coloring of
$K_n$. Now, the columns of $F$ partition the edges of $K_n$ to
$(n-1)$ rainbow perfect matchings.}
\end{proof}

\begin{construction}\label{2}
{Let $(V,B)$ be a STS$(v)$. Define a STS$(2v+1)$, $(W,B')$, on the
set $W=\{x_1,\ldots,x_{2v+1}\}$ with two types of blocks as follows.
The blocks of Type $1$ are the blocks of $B$ on
$\{x_1,\ldots,x_v\}$. Now, consider the complete graph $K_{v+1}$
with the vertex set $\{x_{v+1},\ldots,x_{2v+1}\}$. Since $v$ is odd,
the edges of $K_{v+1}$ can be partitioned to $v$ perfect matchings
$F_1,\ldots,F_v$. Now, every triangle obtained from  vertex $x_i$,
$1\leq i \leq v$, and two vertices of every edge of  $F_i$ introduce
a block. These blocks are blocks of Type $2$.}
\end{construction}

\begin{thm}
{ For every $k$-block colorable   STS{\rm (}$v${\rm )}, every
STS{\rm (}$2v+1${\rm )} obtained from Construction {\rm\ref{2}} is
$(k+v)$-block colorable.}
\end{thm}


\begin{proof}
{Let $(V,B)$ be a $k$-block colorable STS($v$) and
$f:B\rightarrow\{1,\ldots,k\}$ be a such coloring. Moreover, let
$(W,B')$ be a STS$(2v+1)$ obtained from Construction \ref{2}. Define
$c:B'\rightarrow\{1,\ldots,{k+v}\}$ as follows.   For $1\leq
i,j,t\leq v$, let $c(\{x_i,x_j,x_t\})=f(\{x_i,x_j,x_t\})$. Since
$v+1$ is even, by Theorem \ref{rainbow},there exists a $v$-edge
coloring $\phi$ of $K_{v+1}$ on the vertices
$\{x_{v+1},\ldots,x_{2v+1}\}$ such that all edges partitions into
$v$ rainbow perfect matchings $F_1,\ldots,F_v$. Now, for every block
$\{x_i,x_j,x_t\}$, $1\leq i \leq v$ and $v+1\leq j,t \leq 2v+1$, let
$c(\{x_i,x_j,x_t\})=\phi(x_j x_t)$. We claim that $c$ is a
$(k+v)$-block coloring. Notice that if two adjacent blocks call
$b_1$ and $b_2$ have the same color, since the set of colors used to
color the blocks of Type $1$ and Type $2$ have no color in common,
then $b_1$ and $b_2$ belong to the same type. So, two cases may be
considered. First, suppose that $b_1$ and $b_2$ are the blocks of
Type $1$. Since $f$ is a $k$-block coloring of $(V,B)$, $c(b_1)\neq
c(b_2)$.
 Otherwise, two cases
may be considered. First assume that $b_1\cap b_2=\{x_i\}$ such that
$1\leq i \leq v$. Since every perfect matching $F_p$, $1\leq p \leq
v$ is rainbow, $c(b_1)\neq c(b_2)$. Now, suppose that $b_1\cap
b_2=\{x_i\}$, $v+1\leq i \leq 2v+1$. Since $\phi$ is a proper edge
coloring, we are done and the proof is complete.}
\end{proof}



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