text:
\section*{Introduction}

	\title[ On the intersection ideal graph of semigroups]{On the intersection ideal graph of semigroups}
	\author[Barkha Baloda, Jitender Kumar]{Barkha Baloda, $\text{Jitender Kumar}^{^*}$}
	\address{Department of Mathematics, Birla Institute of Technology and Science Pilani, Pilani, India}
	\email{barkha0026@gmail.com,jitenderarora09@gmail.com}

	\begin{abstract}
		The intersection ideal graph $\Gamma(S)$ of a semigroup $S$ is a simple undirected graph whose vertices are all nontrivial left ideals of $S$ and two distinct left ideals $I, J$ are adjacent if and only if their intersection is nontrivial. In this paper, we investigate the connectedness of $\Gamma(S)$. We show that if $\Gamma(S)$ is connected then $diam(\Gamma(S)) \leq 2$. Further we classify the semigroups such that the diameter of their intersection graph is two. Other graph invariants, namely perfectness, planarity, girth, dominance number, clique number, independence number etc. are also discussed. Finally, if $S$ is union of $n$ minimal left ideals then we obtain the automorphism group of $\Gamma(S)$.
		\end{abstract}

	\subjclass[2010]{05C25}

	\keywords{Semigroup, ideals, clique number, graph automorphism\\ *  Corresponding author}

	\maketitle
	\section{Introduction}
Literature is abound with numerous remarkable results concerning a number of constructions of  graphs from rings, semigroups or groups.
The  intersection graph of a  semigroup was introduced by Bos\'ak  \cite{i.Bosak} in $1964$.
 The \emph{intersection subsemigroup graph} $\Gamma(S)$  of $S$ is an undirected simple graph whose vertex set is the collection of proper subsemigroups of  $S$ and two distinct vertices $A, B$ are adjacent if and only if $A \cap B \neq \emptyset$.
In \cite{i.Bosak}, it was shown that if $S$ is a nondenumerable semigroup or a periodic semigroup with more than two elements, then the graph $\Gamma(S)$ is connected. Bos\'ak then  raised the following  open problem: Does there exists a semigroup with more than two elements whose graph is disconnected? Y. F. Lin \cite{lin1969}, answer the problem posed by Bos\'ak, in the negative manner and proved that every semigroup with more than two elements has a connected graph. Also, B. Pond\v{e}li\v{c}ek \cite{abc} proved that  the diameter of a semigroup with more than two  elements does not exceed three.

Inspired by the work of J. Bos\'ak , Cs\'ak\'any and Poll\'ak \cite{a.Cskany1969} studied the intersection graphs of groups and showed that there is an edge between two proper subgroups if they have at least two elements common. Further, Zelinka \cite{B.kaka}, continued the work for finite abelian groups. R. Shen \cite{R.shen}, characterized all finite groups whose intersection graphs are disconnected. This solves the problem posed in \cite{a.Cskany1969}.
The groups whose intersection graphs of normal subgroups are connected, complete, forests or bipartite  are classified in \cite{a.jafari}. Tamizh \emph{et al.} \cite{T.Chelvam2012},  continued the seminal paper of Cs\'ak\'any and Poll\'ak to introduce the subgroup intersection graph of a finite group $G$. Further, in \cite{X.Ma}, it was shown that the diameter of intersection graph of a finite non-abelian simple group has an upper bound $28$. Shahsavari \emph{et al.} \cite{a.Shahsavari2017} have studied the structure  of the automorphism group of this graph. The intersection graph on cyclic subgroups of a group has been studied in \cite{a.Haghi2017}. Further, Kayacan \emph{et al.} \cite{kayacan2015abelian} studied the conjecture given in \cite{B.kaka}, that two (noncyclic) finite abelian groups with isomorphic intersection graphs are isomorphic. In \cite{kayacan2018connectivity}, finite solvable groups whose intersection graphs are not 2-connected, finite nilpotent groups whose intersection graphs are not 3-connected is classified. Further, the dominating sets of the intersection graph of finite groups is investigated in \cite{a.kalyacan}.

Recently, Chakrabarty et al. \cite{a.sen2009}  introduced the notion of  intersection ideal graph of rings. The \emph{intersection ideal graph} $\Gamma(R)$  of a ring $R$ is an undirected simple graph whose vertex set is the collection of nontrivial left ideals of  $R$ and two distinct vertices $I, J$ are adjacent if and only if $I \cap J \neq \{0\}$. They characterized the rings $R$ for which the graph $\Gamma(R)$ is connected and obtain several necessary and sufficient conditions on a ring $R$ such that $\Gamma(R)$ is complete. Planarity of intersection graphs of ideals of ring with unity is described in \cite{MR2660547} and domination number in \cite{jafari2011dominion}. Akbari \emph{et al.} \cite{S.Akbari2013} classified all rings whose intersection graphs of ideals are not connected and also determined all rings whose clique number is finite. The intersection graphs of ideals of direct product of rings have been discussed in \cite{MR3310566}. Pucanovic \emph{et al.} \cite{MR3190084} classified all graphs of genus two that are intersection graphs of ideals of some commutative rings and obtain some lower bounds for the genus of the intersection graph of ideals of a non local commutative ring. In \cite{das2017}, Das  characterized the positive integer $n$ for which the intersection graph of ideals of $\mathbb{Z}_n$ is perfect. The Intersection graph for submodules of modules have been studied in \cite{akbari2012intersection,akbari2017some, yaraneri2013intersection}. The intersecton graph on algebraic structures have also been studied in \cite{ahmadi2016planarity,  akbari2015intersectiongroup, akbari2014some, jafari2011results,  laison2010subspace,  xu2020automorphism}.

It is pertinent as well as interesting to associate graphs to ideals of a semigroup as ideals gives a lot of information about the structure of semigroups.
Motivated with the work of \cite{S.Akbari2013, a.sen2009},
in this paper, we consider the intersection ideal graph associated with semigroups. The \emph{intersection ideal graph} $\Gamma(S)$ of a semigroup $S$ is an undirected simple graph whose vertex set is nontrivial left ideals of  $S$ and two distinct nontrivial left ideals $I, J$ are adjacent if and only if their intersection is nontrivial. The paper is arranged as follows. In Section 2, we state necessary fundamental notions  and recall some necessary results. Section 3 comprises the results concerning the connectedness of intersection ideal graph of an arbitrary semigroup. In Section 4, we study various graph invariants of  $\Gamma(S)$ viz. girth, dominance number, independence number and clique number etc.  Further, if $S$ is union of $n$ minimal left ideals then the automorphism group of $\Gamma(S)$ is obtained.

	\section{Preliminaries}
In this section, first we recall necessary definitions and results of semigroup theory from \cite{b.clifford61vol1}. A \emph{semigroup} $S$ is a non-empty set together with an associative binary operation on $S$. The Green's $\mathcal{L}$-relation on a semigroup $S$ defined as $x$         $\mathcal{L}$ $y \Longleftrightarrow S^{1}x = S^{1}y$ where $S^{1}x = Sx \cup    \{x\}$.
	 The $\mathcal{L}$-class of an element $a \in S$ is denoted by $L_a$. 	A non-empty subset $I$ of $S$ is said to be a \emph{left [right] ideal} if $SI \subseteq I  [IS \subseteq I]$ and an  \emph{ideal} of $S$ if $SIS \subseteq I$. Union of two left [right] ideals of $S$ is again a left [right] ideal of $S$. A left ideal $I$ is \emph{maximal} if it does not contained in  any nontrivial left ideal of $S$. If $S$ has a unique maximal left ideal then it contains every nontrivial left ideal of $S$.
	A left ideal $I$ of $S$ is \emph{minimal} if it does not properly contain any left ideal of $S$. It is well known that every non-zero element of a minimal left ideal of $S$ is in same $\mathcal{L}$-class. If $S$ has a minimal left ideal then every nontrivial left ideal contains at least one minimal left ideal. If $A$ is any other left ideal of $S$ other than $I$, then either $I \subset A$ or $I \cap A = \emptyset$. Thus we have the following remark.
	\begin{remark}\label{disjoint intersection minimal}
		Any two different minimal left ideals of a semigroup $S$ are disjoint.
	\end{remark}
	\begin{remark}\label{everynontrivial left ideal is union}
	Let $S$ be union of $n$ minimal left ideals. Then each nontrivial left ideal is union of these minimal left ideals.
	\end{remark}
The following lemma is useful in the sequel and we shall use this without referring to it explicitly.
	\begin{lemma}\label{S minus K is lclass}
A left ideal $K$ of $S$ is maximal if and only if $S \setminus K$ is an $\mathcal{L}$-class.
\end{lemma}
\begin{proof}
First suppose that $S \setminus K$ is an $\mathcal{L}-$class. Let if possible, $K$ is not maximal left ideal of $S$. Then there exists a nontrivial left ideal $K'$ of $S$ such that $K \subset K'$. There exists $a \in K'$ but $a \notin K$. Thus,  $L_a = S \setminus K$. Consequently, $L_a \subset K'$ gives $S = K'$, a contradiction.
Conversely, suppose that $K$ is a maximal left ideal of $S$. For each $a \in S \setminus K$, maximality of $K$ implies $K \cup S^1a = S$. Consequently, $a$  $\mathcal{L}$  $b$ for every $a, b \in S \setminus K$. Thus $S \setminus K$ is contained in some $\mathcal{L}-$class and this $\mathcal{L}-$class is disjoint from $K$. It follows that $S \setminus K$ is an $\mathcal{L}-$class.
\end{proof}

	We also require the following graph theoretic  notions \cite{westgraph}. A \emph{graph} $\Gamma$ is a pair  $\Gamma = (V, E)$, where $V = V(\Gamma)$ and $E = E(\Gamma)$ are the set of vertices and edges of $\Gamma$, respectively. We say that two different vertices $u, v$ are $\mathit{adjacent}$, denoted by $u \sim v$ or $(u,v)$, if there is an edge between $u$ and $v$. We write $u \nsim v$, if there is no edge between $u$ and $v$. The \emph{distance} between two vertices $u, v$ in $\Gamma$ is the number of edges in a shortest path connecting them and it is denoted by $d(u, v)$. If there is no path between $u$ and $v$, we say that the distance between  $u$ and $v$ is \emph{infinity} and we write as $d(u, v) = \infty$. The diameter $diam(\Gamma)$ of $\Gamma$ is the greatest distance between any pair of vertices. The \emph{degree} of the vertex $v$ in $\Gamma$ is the number of edges incident to $v$ and it is denoted by $deg(v)$.

	A \emph{cycle} is a closed walk with distinct vertices except for the initial and end vertex, which are equal and a cycle of length $n$ is denoted by $C_n$. The \emph{girth} of $\Gamma$ is the length of its shortest cycle and is denoted by ${g(\Gamma)}$. A subset $X$ of $V(\Gamma)$ is said to be \emph{independent} if no two vertices of $X$ are adjacent. The \emph{independence number} of $\Gamma$  is the cardinality of the largest independent set and it is denoted by $\alpha(\Gamma)$. A graph $\Gamma$ is \emph{bipartite}  if $V(\Gamma)$ is the union of two disjoint independent  set. It is well known that a graph is bipartite if and only if it has no odd cycle {\cite[Theorem 1.2.18]{westgraph}}. A connected graph $\Gamma$ is Eulerian if and only if degree of every vertex is even {\cite[Theorem 1.2.26]{westgraph}}. A \emph{subgraph}  of $\Gamma$ is a graph $\Gamma'$ such that $V(\Gamma') \subseteq V(\Gamma)$ and $E(\Gamma') \subseteq E(\Gamma)$.  A subgraph $\Gamma'$ of  $\Gamma$ is called an \emph{induced subgraph} by the elements of  $V(\Gamma') \subseteq V(\Gamma)$ if for $u, v \in V(\Gamma')$, we have $u \sim v$   in $\Gamma'$ if and only if $u \sim v$ in $\Gamma$.  The \emph{chromatic number} of $\Gamma$, denoted by $\chi(\Gamma)$, is the smallest number of colors needed to color the vertices  of $\Gamma$ so that no two adjacent vertices share the same color. A \emph{clique} in $\Gamma$ is a set of pairwise adjacent vertices. The \emph{clique number} of $\Gamma$ is the size of maximum clique in $\Gamma$ and it is denoted by $\omega(\Gamma)$. It is well known that $\omega(\Gamma) \leq \chi(\Gamma)$ (see \cite{westgraph}). A graph $\Gamma$ is \emph{perfect} if $\omega(\Gamma') = \chi(\Gamma')$ for every induced subgraph $\Gamma'$ of $\Gamma$.
	Recall that the {\em complement} $\overline{\Gamma}$ of $\Gamma$ is a graph with same vertex set as $\Gamma$ and distinct vertices $u, v$ are adjacent in $\overline{\Gamma}$ if they are not adjacent in $\Gamma$. A subgraph $\Gamma'$ of $\Gamma$ is called \emph{hole} if $\Gamma'$  is a  cycle as an induced subgraph, and $\Gamma'$  is called an \emph{antihole} of   $\Gamma$ if   $\overline{\Gamma'}$ is a hole in $\overline{\Gamma}$.

	\begin{theorem}\label{strongperfecttheorem}\cite{strongperfectgraph}
		A finite graph $\Gamma$
		is perfect if and only if it does not contain hole or antihole of odd length at least $5$.
	\end{theorem}

	A subset $D$ of $V(\Gamma)$ is said to be a dominating set if any vertex in $V(\Gamma) \setminus D$ is adjacent to at least one vertex in $D$. If $D$ contains only one vertex then that vertex is called dominating vertex. The \emph{domination number} $\gamma(\Gamma)$ of $\Gamma$ is the minimum size of a dominating set in $\Gamma$. A graph $\Gamma$ is said to be planar if it can be drawn on a plane without any crossing of its edges. In $\Gamma$, a vertex $z$ resolves a pair of distinct vertices $x$ and $y$ if
	$d(x, z) \neq d(y, z)$. A resolving set of $\Gamma$ is a subset $R \subseteq V (\Gamma)$ such that every pair of distinct vertices of $\Gamma$ is resolved by some vertex in $R$. The metric dimension of $\Gamma$,
	denoted by $\beta(\Gamma)$, is the minimum cardinality of a resolving set of $\Gamma$. For  vertices $u$ and $v$ in a graph $\Gamma$, we say that $z$ \emph{strongly resolves} $u$ and $v$ if there exists a shortest path from $z$ to $u$ containing $v$, or a shortest path from $z$ to $v$ containing $u$. A subset $U$ of $V(\Gamma)$ is a \emph{strong resolving set} of $\Gamma$ if every pair of vertices of $\Gamma$ is strongly resolved by some vertex of $U$. The least cardinality of a strong resolving set of $\Gamma$ is called the \emph{strong metric dimension} of $\Gamma$ and is denoted by $\operatorname{sdim}(\Gamma)$. For  vertices $u$ and $v$ in a graph $\Gamma$, we write $u\equiv v$ if $N[u] = N[v]$. Notice that that $\equiv$ is an equivalence relation on $V(\Gamma)$.
We denote by $\widehat{v}$ the $\equiv$-class containing a vertex $v$ of $\Gamma$.
 Consider a graph $\widehat{\Gamma}$ whose vertex set is the set of all $\equiv$-classes, and vertices $\widehat{u}$ and  $\widehat{v}$ are adjacent if $u$ and $v$ are adjacent in $\Gamma$. This graph is well-defined because in $\Gamma$, $w \sim v$ for all $w \in \widehat{u}$ if and only if $u \sim v$.  We observe that $\widehat{\Gamma}$ is isomorphic to the subgraph $\mathcal{R}_{\Gamma}$ of $\Gamma$ induced by a set of vertices consisting of exactly one element from each $\equiv$-class. Subsequently, we have the following result of \cite{ma2018strong} with $\omega(\mathcal{R}_{\Gamma})$ replaced by $\omega(\widehat{\Gamma})$.

\begin{theorem}[{\cite[Theorem 2.2]{ma2018strong}}]\label{strong-metric-dim}
For any graph $\Gamma$ with diameter $2$,  $\operatorname{sdim}(\Gamma) = |V(\Gamma)| - \omega(\widehat{\Gamma})$.
\end{theorem

	\section{Connectivity of the Intersection graph $\Gamma(S)$}
	In this section, we investigate the connectedness of $\Gamma(S)$. We show that $diam(\Gamma(S)) \leq 2$ if it is connected. Also, we classify the semigroups, in terms of left ideals, such that the diameter of $\Gamma(S)$ is two.
	\begin{theorem}\label{disconnectedintersection}
		The intersection ideal graph $\Gamma(S)$ is disconnected if and only if $S$ contains at least two minimal left ideals and every nontrivial left ideal of $S$ is minimal as well as maximal.
	\end{theorem}
	\begin{proof}
	    First suppose that $\Gamma(S)$ is not connected. Then $S$ has at least two nontrivial left ideals, namely $I_1, I_2$. Without loss of generality, assume that $I_1 \in C_1$ and $I_2 \in C_2$, where $C_1$ and $C_2$ are distinct components of $\Gamma(S)$.  If $I_1$ is not minimal then there exists at least one nontrivial left ideal $I_k$ of $S$ such that $I_k \subset I_1$ so that their intersection is nontrivial. Therefore, $I_1 \sim I_k$. Now if the intersection of $I_2$ and $I_k$ is nontrivial then $I_1 \sim I_k \sim I_2$, a contradiction. Therefore the  intersection of $I_2$ and $I_k$ is trivial. If $I_2 \cup I_k \neq S$ then $I_1 \sim I_2 \cup I_k \sim I_2$, a contradiction. Thus, $I_k \cup I_2 = S$. It follows that $I_1 \sim I_2$, again a contradiction. Thus $I_1$ is minimal. Similarly, we get $I_2$ is minimal.

	    Further assume that $I_1$ is not maximal. Then  there exists a nontrivial left ideal $I_k$ of $S$ such that $I_1 \subset I_k$ so that $I_1 \sim I_k$. If $I_1 \cup I_2 \neq S$ then $I_1 \sim I_1 \cup I_2 \sim I_2$, a contradiction to the fact that $\Gamma(S)$ is disconnected. It follows that $I_1 \cup I_2 = S$ so that the intersection of $I_k$ and $I_2$ is nontrivial. Thus we have $I_1 \sim I_k \sim I_2$, a contradiction. Hence $I_1$ is maximal. Similarly, we observe that $I_2$ is maximal. The converse follows from the Remark \ref{disjoint intersection minimal}.
	\end{proof}
		\begin{corollary}\label{null graphintersection}
If the graph $\Gamma(S)$ is disconnected then it is a null graph (i.e. it has no edge).
\end{corollary}

\begin{theorem}\label{two minimalintersection}
		The graph $\Gamma(S)$ is disconnected if and only if $S$ is the union of exactly two minimal left ideals.
	\end{theorem}
\begin{proof}
		Suppose first that $\Gamma(S)$ is disconnected. Then by Theorem \ref{disconnectedintersection}, each nontrivial left ideal of $S$ is minimal. Suppose $S$ has at least three minimal left ideals, namely $I_1, I_2$
		and $I_3$. Then $I_1 \cup I_2$ is a nontrivial left ideal of $S$ which is not minimal. Consequently, by Theorem \ref{disconnectedintersection}, we get a contradiction of the fact that $\Gamma(S)$ is disconnected. Thus, $S$ has exactly two minimal left ideals. If $S \neq I_1 \cup I_2$, then $I_1 \cup I_2$ is a nontrivial left ideal which is not minimal, a contradiction ( cf. Theorem \ref{disconnectedintersection}). Thus, $S = I_1 \cup I_2$.

		Conversely, suppose $S = I_1 \cup I_2$ where $I_1$ and  $I_2$ are minimal left ideals of $S$. If there exists another nontrivial left ideal $I_k$ of $S$ then either $I_1 \subset I_k$ or $I_2 \subset I_k$. Without loss of generality, assume that $I_1 \subset I_k$, we have $I_1 \sim I_k$. Since $I_1 \cup I_2 = S$ we get $I_k \cup I_2 = S$. It follows that the intersection of $I_2$ and $I_k$ is nontrivial. By minimality of $I_2$, we can observe that $I_2 \subset I_k$. Consequently, $S \subseteq I_k$, a contradiction.  Thus, by Theorem \ref{disconnectedintersection},  $\Gamma(S)$ is disconnected.

\end{proof}
	\begin{theorem}\label{diameter2}
		If $\Gamma(S)$ is a connected graph then $diam(\Gamma(S))$ $\leq$ $2$.
	\end{theorem}
	\begin{proof}
		Let $I_1, I_2$ be two nontrivial left ideals of $S$. If $I_1 \sim I_2$ then $d(I_1, I_2)$ = 1. If $I_1 \nsim I_2$ i.e. $I_1 \cap I_2$ is trivial then in the following cases we show that $d(I_1, I_2) $$\leq 2$.

		\noindent\textbf{Case 1.} $I_1 \cup I_2 \neq S$. Then $I_1 \sim (I_1 \cup I_2) \sim I_2$ so that $d(I_1, I_2)$ = 2.

		\noindent\textbf{Case 2.} $I_1 \cup I_2 = S$. Since $\Gamma(S)$ is a connected graph, there exists a nontrivial left ideal $I_k$ of $S$ such that either $I_1 \cap I_k$ is nontrivial or $I_2 \cap I_k$ is nontrivial. Now we have the following subcases.

		\textbf{Subcase 1.} $I_1 \not \subset I_k$ and $I_k \not \subset I_1$. Since $I_1 \not \subset I_k$ it follows that there exists $x \in I_k$ but $x \notin I_1$ so that $x \in I_2$. Consequently, $I_2 \cap I_k$ is nontrivial. Therefore, we get a path $I_1 \sim I_k \sim I_2$ of length two. Thus, $d(I_1, I_2) = 2$.

		\textbf{Subcase 2.} $I_k \subset I_1$. There exists $x \in I_1$ but $x \notin I_k $. If $I_2 \cup I_k = S$ then $x \in I_2$. Thus, we  get $I_1 \cap I_2$ is nontrivial, a contradiction. Consequently, $I_2 \cup I_k \neq S$. Further, we get a path $I_1  \sim (I_2 \cup I_k) \sim I_2$ of length two. Thus, $d(I_1, I_2) = 2$.

		\textbf{Subcase 3.} $I_1 \subset I_k$. Since $I_1 \cup I_2 = S$ we get $I_k \cup I_2 = S$. Further, the intersection of  $I_k$ and $I_2$ is nontrivial. Consequently, $I_1 \sim I_k \sim I_2$ gives a path of length two between $I_1$ and $I_2$. Thus, $d(I_1, I_2) = 2$. Hence, $diam(\Gamma(S))$ $\leq$ $2$.

	\end{proof}
\begin{lemma}
  Let $S$ be a semigroup having minimal left ideals. Then $\Gamma(S)$ is complete if and only if $S$ has unique minimal left ideal.
\end{lemma}
\begin{proof}
    Suppose that $S$ contains a unique minimal left ideal $I_1$. Note that every nontrivial left ideal of $S$ contains at least one minimal left ideal. Since $I_1$ is unique then it must contained in every nontrivial left ideals of $S$. Thus, the graph $\Gamma(S)$ is complete.

        Conversely, suppose that $\Gamma(S)$ is a complete graph. On  contrary if $S$ has at least two minimal left ideals, viz. $I_1, I_2$. By Remark \ref{disjoint intersection minimal}, $I_1 \nsim I_2$, a contradiction to the fact that $\Gamma(S)$ is complete. Thus $S$ has unique minimal left ideal.
\end{proof}

\begin{lemma}
 If graph $\Gamma(S)$ is a regular if and only if either $\Gamma(S)$ is null or a complete graph.
\end{lemma}
\begin{proof}
        First suppose that $\Gamma(S)$ is not a null graph. Let if possible, $S$ has at least two minimal left ideals, namely $I_1, I_2$. Since $\Gamma(S)$ is not a null graph then $I_1$ and $I_1 \cup I_2$ forms a nontrivial left ideals of $S$ and $I_1 \sim (I_1 \cup I_2)$. Suppose $J$ is any nontrivial left ideal of $S$ such that $J \sim I_1$ then $J \sim (I_1 \cup I_2)$. It follows that every nontrivial left ideal of $S$ which is adjacent with $I_1$ is also adjacent with $(I_1 \cup I_2)$ and $I_2 \sim I_1 \cup I_2$ but $I_2 \nsim I_1$ implies that $deg(I_1) < deg(I_1 \cup I_2)$, a contradiction. Therefore, $\Gamma(S)$ is a complete graph.
\end{proof}
Next we classify the semigroups such that the diameter of intersection ideal graph $\Gamma(S)$ is two.

\begin{theorem}\label{diameter2clasification}
 Let $S$ be a semigroup having minimal left ideals. Then for a connected graph $\Gamma(S)$, we have  $diam(\Gamma(S)) = 2$ if and only if $S$ has at least two minimal left ideals.
\end{theorem}
\begin{proof}
Suppose that $diam(\Gamma(S)) = 2$. Assume that $I_1$ is the only minimal left ideal of $S$. Since $I_1$ is unique minimal left ideal then it is contained in all other nontrivial left ideals of $S$. Therefore, for any nontrivial left ideals $J, K$, we have $I_1 \subset (J \cap K)$. Consequently, $d(J, K) = 1$ for any $J, K \in V(\Gamma(S))$. Therefore $S$ has at least two minimal left ideals. Conversely suppose that $S$ has at least two minimal left ideals, viz. $I_1, I_2$. Then by Remark \ref{disjoint intersection minimal}, we have $I_1 \nsim I_2$. Consequently, by Theorem \ref{diameter2}, $d(I_1, I_2) = 2$. Thus,  $diam(\Gamma(S)) = 2$.
\end{proof}

\section{Invariants of  $\Gamma(S)$}
In this section, first we obtain the girth of $\Gamma(S)$. Then we discuss planarity and perfectness of $\Gamma(S)$. Also we classify the semigroup $S$ such that  $\Gamma(S)$ is bipartite, star graph and tree, respectively. Further, we investigate other graph invariants viz. clique number, independence number and  strong metric dimension of $\Gamma(S)$.
\begin{theorem} \label{girthofintersection}
Let $S$ be a semigroup such that $\Gamma(S)$ contains a cycle. Then $g(\Gamma(S)) = 3$.
\end{theorem}
\begin{proof}
If $\Gamma(S)$ is disconnected or a tree, then clearly $g(\Gamma(S)) = \infty$. Suppose that the semigroup $S$ has $n$ minimal left ideals. Now we prove the result through following cases.

\noindent\textbf{Case 1.} $n = 0$. If $S$ has no nontrivial left ideals then there is nothing to prove. Otherwise, there exists a chain of nontrivial left ideals of $S$ such that $I_1 \supset I_2 \supset \cdots \supset I_k \supset \cdots$. Thus, $g(\Gamma(S)) = 3$.

\noindent\textbf{Case 2.} $n = 1$. Suppose that $I_1$ is the only minimal left ideal of $S$. Since $I_1$ is unique minimal left ideal then it is contained in all other nontrivial left ideals of $S$. Therefore, for any nontrivial left ideals $I, J$, we have $I_1 \subset I \cap J \neq \emptyset$.  If $S$ has at least three nontrivial left ideals then $g(\Gamma(S)) = 3$. Otherwise, $g(\Gamma(S)) = \infty$.

\noindent\textbf{Case 3.} $n = 2$. Let $I_1, I_2$ be two minimal left ideals of $S$. If $I_1 \cup I_2 = S$ then by Theorem \ref{two minimalintersection} and Corollary \ref{null graphintersection}, $g(\Gamma(S)) = \infty$. If $I_1 \cup I_2 \neq S$, then $J = I_1 \cup I_2$ is a nontrivial left ideal of $S$. If $S$ has only these three, namely  $I_1, I_2$ and $J$, left ideals then we obtain $I_1 \sim J \sim I_2$ so that $g(\Gamma(S)) = \infty$. Now suppose that  $S$ has a nontrivial left ideal $K$ other than $I_1, I_2$ and $J$. Since  $I_1, I_2$ are minimal left ideals of $S$ we have either $I_1 \subset K$ or $I_2 \subset K$. Without loss of generality, assume that $I_1 \subset K$, then we get a triangle $I_1 \sim K \sim J \sim I_1$. It follows that $g(\Gamma(S)) = 3$.

 \noindent\textbf{Case 4.} $n \geq 3$. Let $I_1, I_2, I_3$ be the minimal left ideals of $S$. Then we have a cycle $(I_1 \cup I_2) \sim (I_2 \cup I_3) \sim (I_1 \cup I_3) \sim (I_1 \cup I_2)$ of length 3. Thus, $g(\Gamma(S)) = 3$.
\end{proof}

	Let ${\rm Min}(S)$ (${\rm Max}(S)$) be the set of all minimal (maximal) left ideals of $S$. For a nontrivial left ideal $I_{{i_1}{i_2}\cdots{i_k}}$, we mean $I_{i_1} \cup I_{i_2} \cup \cdots \cup I_{i_k}$, where $I_{i_1}, I_{i_2}, \cdots, I_{i_k}$ $\in {\rm Min}(S)$
	\begin{theorem}
		For the graph $\Gamma(S)$, we have the following results:

		\begin{enumerate}
		 \item[{\rm(i)}] If $\Gamma(S)$ is  planar  then $| {\rm Min}(S) | \leq 3$.
			\item [{\rm(ii)}] For $S = I_{{i_1}{i_2}\cdots{i_n}}$, we have  $\Gamma(S)$ is planar if and only if $n \leq 3$.
	\end{enumerate}
	\end{theorem}
	\begin{proof}
		(i) Suppose that $| {\rm Min}(S) | =4$ with ${\rm Min}(S) = \{I_1, I_2, I_3, I_4\}$. Then note that the subgraph induced by the vertices $I_1, I_{12}, I_{123}, I_{14}$ and $I_{124}$ is isomorphic to $K_5$. Thus, $\Gamma(S)$ is nonplanar.

		(ii) The proof for $\Gamma(S)$ is nonplanar for $n \geq 4$ follows  from part (i).
		If $n = 2$ then by Corollary \ref{null graphintersection} and Theorem \ref{two minimalintersection}, $\Gamma(S)$ is planar. For $n= 3$, $\Gamma(S)$ is planar as shown in Figure 1.
		\begin{figure}[h!]
			\centering
			\includegraphics[width=0.2 \textwidth]{s123newone.pdf}
			\caption{Planar drawing of $\Gamma(S)$ for $S = I_{123}$}
	\end{figure}
		\end{proof}

\begin{theorem}
		For the graph $\Gamma(S)$, we have the following results:

		\begin{enumerate}
		 \item[{\rm(i)}] If
		 $\Gamma(S)$ is a perfect graph then $| {\rm Min}(S) | \leq 4$.
			\item [{\rm(ii)}] Let $S$ be the union of $n$ minimal left ideals. Then  $\Gamma(S)$ is perfect if and only if $n \leq 4$.
	\end{enumerate}
	\end{theorem}
	\begin{proof}
	(i) Suppose that $| {\rm Min}(S) | = 5$ with ${\rm Min}(S) = \{I_1, I_2, I_3, I_4, I_5 \}$. Note that $I_{12} \sim I_{23} \sim I_{34} \sim I_{45} \sim I_{15} \sim I_{12}$ induces a cycle of length 5. Then by Theorem \ref{strongperfecttheorem}, $\Gamma(S)$ is not  perfect.

	(ii) The proof for $\Gamma(S)$ is not a perfect graph for $n \geq 5$ follows  from part (i). If $n = 2$ then by Corollary \ref{null graphintersection} and Theorem \ref{disconnectedintersection}, $\Gamma(S)$ is disconnected. Thus, being a null graph, $\Gamma(S)$ is perfect. For $n \in \{ 3, 4\}$, we show that $\Gamma(S)$ does not contain a hole or an antihole of odd length at least five (cf. Theorem \ref{strongperfecttheorem}). If $n =3$, $\Gamma(S)$ is perfect as shown in Figure 1. If $n = 4$ then  from Figure 2 note that  $\Gamma(S)$ does not contain a hole or an antihole of odd length at least five.
	\begin{figure}[h!]
			\centering
			\includegraphics[width=0.5 \textwidth]{s1234.pdf}
			\caption{}
	\end{figure

	\end{proof}

	\begin{theorem}
	Let $S$ be a semigroup having minimal left ideals such that $V(\Gamma(S)) > 1$. Then the following conditions are equivalent:

	\begin{enumerate}
		 \item[{\rm(i)}] $\Gamma(S)$ is star graph.

		 \item[{\rm(ii)}] $\Gamma(S)$ is a tree.

		 \item[{\rm(iii)}] $\Gamma(S)$ is bipartite.

		  \item[{\rm(iv)}] Either $S$ has exactly three nontrivial left ideals $I_1$, $I_2$ and $I_{12}$ such that $I_1$ and $I_{2}$ are minimal or $S$ has two nontrivial left ideals $I_1, I_2$ such that $I_1 \subset I_2$.
		 \end{enumerate}
	\end{theorem}
	\begin{proof}

	  We prove (ii), (iii) $\Rightarrow$ (iv). The proof of remaining parts is straightforward. Suppose  $\Gamma(S)$ is a tree. Then clearly $| {\rm Min}(S) | \leq 2$. Otherwise, for minimal left ideals $I_1, I_2, I_3$ we have $I_{12} \sim I_{13} \sim I_{23} \sim I_{12}$ a cycle, a contradiction. Suppose that $| {\rm Min}(S) | = 1$. Let $I_1$ be the unique minimal left ideal of $S$. Consequently, $I_1$ is contained in all other nontrivial left ideals of $S$. If $S$ has at least three nontrivial left ideals then we get a cycle, a contradiction. Thus $|V(\Gamma(S))| = 2$. Now we assume that $| {\rm Min}(S) | = 2$. Let $I_1, I_2$ be two minimal left ideals of $S$. Let if possible, $S = I_{12}$. Then by Corollary \ref{null graphintersection} and Theorem \ref{two minimalintersection}, $\Gamma(S)$ is disconnected so is not a tree. Thus $S \neq I_{12}$.  Then $J = I_{12}$ is a nontrivial left ideal of $S$.
	  Now if $S$ has a nontrivial left ideal $K$ other than $I_1, I_2$ and $J$. Without loss of generality, assume that $I_1 \subset K$ then we get a cycle $I_1 \sim I_{12} \sim K \sim I_1$, a contradiction. Thus, for $S \neq I_{12}$, we have $V(\Gamma(S)) = \{I_1, I_2, I_{12}  \}$.

	   (iii) $\Rightarrow$ (iv). If $\Gamma(S)$ is bipartite then we have $| {\rm Min}(S) | \leq 2$. In the similar lines of the work discussed above, (iv) holds.
	\end{proof}

\begin{theorem}
Let $S$ be a semigroup with $n$ minimal left ideals. Then the following results hold:
\begin{enumerate}
		 \item[{\rm(i)}] If $S \neq I_{12 \cdots n}$ then $\gamma(\Gamma(S)) = 1$.

		\item[{\rm(ii)}] If $S = I_{12 \cdots n}$ then $\gamma(\Gamma(S)) = 2$.
\end{enumerate}
\end{theorem}
\begin{proof}
     (i) Suppose that $S \neq I_{12 \cdots n}$. It follows that $J = I_{12 \cdots n}$ is a nontrivial left ideal of $S$. It is well known that every nontrivial left ideal of $S$ contains at least one minimal left ideal. Consequently, for any nontrivial left ideal $K$ of $S$, we have $J \cap K$ is nontrivial. Thus, $J$ is a dominating vertex. Hence, $\gamma(\Gamma(S)) = 1$.

     (ii) Suppose that $S = I_{12 \cdots n}$. Note that  there is no dominating vertex in $\Gamma(S)$ so that $\gamma(\Gamma(S)) \geq 2$. Now we show that $D = \{I_1, I_{23 \cdots n}\}$ is a dominating set. Since $S$ is the union of $n$ minimal left ideals so any nontrivial left ideal of $S$ is union of these minimal left ideals (cf. Remark \ref{everynontrivial left ideal is union}). Let $J \in V(\Gamma(S)) \setminus D$ be any nontrivial left ideal of $S$. Then $J$ is union of $k$ minimal left ideals of $S$, where $1 \leq k \leq n-1$. If $I_1 \subset J$, then we are done. If $I_1 \not\subset J$ then $J$ must be union of $I_2, I_3, \ldots, I_n$. It follows that intersection of $J$ and $I_{23 \cdots n}$ is nontrivial. Consequently, $J \sim I_{23 \cdots n}$. Thus $D$ is a dominating set. This completes the proof.
\end{proof}
\begin{theorem}
    Let $S$ be a semigroup with $n$ minimal left ideals. Then $\alpha(\Gamma(S)) = n$.
\end{theorem}
\begin{proof}
  Let ${\rm Min}(S) = \{{I_{i_1} : i_1 \in [n]}\}$ be the set of all minimal left ideals of S. Then, by Remark \ref{disjoint intersection minimal}, ${\rm Min}(S)$ is an independent set of $\Gamma(S)$. It follows that $\alpha(\Gamma(S)) \geq n$. Now we prove that for any arbitrary independent set $U$, we have $|U| \leq n$. Assume that $I \in V(\Gamma(S))$ such that $I \in U$. Since every nontrivial left ideal contains at least one minimal left ideal. Without loss of generality, assume that  $I_{{i_1}{i_2}\cdots{i_k}} \subseteq I$ for some $i_1,i_2, \cdots ,i_k
 \in [n]$. Then note that $|U| \leq n-k+1$. Otherwise, there exist at least two nontrivial left ideals which are adjacent, a contracdiction. Consequently, we have $|U| \leq n$. Thus, $\alpha(\Gamma(S)) = n$.
\end{proof}
\begin{lemma}
    Let $S$ be a semigroup with $n~ (\geq 3)$ minimal left ideals. Then there exists a clique in $\Gamma(S)$ of size $n$.
\end{lemma}
\begin{proof}
Let $I_1, I_2, \ldots, I_n$ be $n$ minimal left ideals. Consider $\mathcal{C} = \{I_{{i_1}{i_2} \cdots {i_{n-1}}} : i_1, i_2, \ldots, i_{n-1} \in [n]\}$. Clearly, $|\mathcal{C}| = n$. Notice that for any $J , K \in \mathcal{C}$, we have  $J \cap K$ is nontrivial so that $J \sim K$. Thus, $\mathcal{C}$ becomes a clique of size $n$.
\end{proof}

\begin{theorem}
Let $S$ be a semigroup with $n(>1)$ minimal left ideals. Then $\omega(\Gamma(S)) = n$ if and only if one of the following holds:
\begin{enumerate}
		 \item[{\rm(i)}] $S = I_{123}$.
		 \item[{\rm(ii)}] $S$ has only two minimal left ideals $I_1$ and $I_2$ and a unique maximal left ideal $I_{12}$.
		 \end{enumerate}
\end{theorem}
\begin{proof}
First suppose that $\omega(\Gamma(S)) = n$. Assume that $S$ has $n (\geq 4)$ minimal left ideals, namely $I_1, I_2, \ldots, I_n$. Then $\mathcal{C} = \{I_{{i_1}{i_2}\cdots{i_{n-1}}}, I_{{i_1}{i_2}} : i_1, i_2, \ldots, i_n \in [n] \}$ forms a clique of size greater than $n$ of $\Gamma(S)$. It follows that $\omega(\Gamma(S)) > n$. If $n =3$, assume  that $S \neq I_{123}$. Then $\mathcal{C} = \{I_{12}, I_{13}, I_{23}, I_{123}\}$ forms a clique of size four of $\Gamma(S)$. It follows that $S = I_{123}$.
For $n =2$, we have either $S = I_{12}$ or $S \neq I_{12}$. For $S = I_{12}$, by Corollary \ref{null graphintersection} and by Theorem \ref{two minimalintersection}, $\Gamma(S)$ is disconnected. Thus, $\omega(\Gamma(S)) < n$. Thus $S \neq I_{12}$. If $S$ has a nontrivial left ideal $K \notin \{I_1, I_2, I_{12}\}$ then we get a clique of size three. Therefore, $I_{12}$ is a unique maximal left ideal. Converse follows trivially
\end{proof}

\begin{lemma}\label{maximal ideal intersection nonempty}
 If $\Gamma(S)$ is connected then ${\rm Max}(S)$ forms a clique of $\Gamma(S)$.
\end{lemma}
\begin{proof}
We prove the result by showing that if $J, K \in {\rm Max}(S)$ then $J \sim K$.
Let if possible, $J \nsim K$. The maximality of $J$ and $K$ follows that $J \cup K = S$. By Lemma \ref{S minus K is lclass}, $S \setminus J$ and $S \setminus K$ are $\mathcal{L}-$classes of $S$. It follows that $J$ and $K$ are  only nontrivial left ideals of $S$. Thus, being a null graph $\Gamma(S)$ is disconnected, a contradiction.
\end{proof}

\begin{theorem}
If $K$ is a maximal left ideal of $S$ such that $deg(K)$ is finite, then $\chi(\Gamma(S)) < \infty$.
\end{theorem}
\begin{proof}
 Let $J$ be an arbitrary nontrivial left ideal of $S$ such that $J \nsim K$. Note that $J$ is minimal left ideal of $S$. On contrary, suppose that $J$ is not a minimal left ideal of $S$. Then there exists a nontrivial left ideal $J'$ of $S$ such that $J' \subset J$. Since $K$ is maximal left ideal of $S$. Consequently, $J' \cup K = S$. It follows that  intersection of $J$ and $K$ is nontrivial, a contradiction. By Remark \ref{disjoint intersection minimal}, we can color all the vertices which are not adjacent with $K$ with one color. Since $deg(K)$ is finite, we have $\chi(\Gamma(S)) < \infty$.
\end{proof}

\begin{lemma}\label{chromaticnumberintersection}
        For $S = I_{{i_1}{i_2}\cdots{i_n}}$, we have $\omega(\Gamma(S)) = \chi(\Gamma(S)) = 2^{n-1}-1$.
\end{lemma}
\begin{proof}
       First note that $S$ has $2^n-2$ nontrivial left ideals and every nontrivial left ideal of $S$ is of the form $I_{{i_1}{i_2}\cdots{i_k}}$ and $1 \leq k \leq n-1$ (cf. Remark \ref{everynontrivial left ideal is union}). If $n$ is odd then consider $\mathcal{C} = \{I_{{j_1}{j_2}\cdots{j_t}} : \lceil \frac{n}{2} \rceil \leq t\leq n-1\}$. Note that $\mathcal{C}$ forms a clique of size $2^{n-1}-1$. We may now suppose that $n$ is even. Consider $\mathcal{C}_1 = \{I_{{j_1}{j_2}\cdots{j_t}} :
 \frac{n}{2} + 1 \leq t\leq n-1\}$. Notice that $\mathcal{C}_1$ forms a clique. Further, observe that $\mathcal{C}^{'} = \{I_{{i_1}{i_2}\cdots{i_{\frac{n}{2}}}} :  i_1, i_2, \ldots, i_{\frac{n}{2}} \in [n]\}$ do not form a clique because for $j_1, j_2, \ldots, j_{\frac{n}{2}} \in [n] \setminus \{i_1, i_2, \ldots, i_{\frac{n}{2}}\}$, $I_{{i_1}{i_2}\cdots{i_{\frac{n}{2}}}} \nsim I_{{j_1}{j_2}\cdots{j_{\frac{n}{2}}}}$. However, $\mathcal{C}^{''} = \{I_{{i_1}{i_2}\cdots{i_{\frac{n}{2}}}} \in \mathcal{C}^{'} \setminus \{I_{{j_1}{j_2}\cdots{j_{\frac{n}{2}}}}\} : j_1, j_2, \ldots, j_{\frac{n}{2}} \notin \{i_1, i_2, \ldots, i_{\frac{n}{2}}\} \}$ forms a clique of size $\frac{|\mathcal{C}^{'}|}{2}$. Further note that the set $\mathcal{C}_1 \cup \mathcal{C}^{''}$ also forms a clique of size $2^{n-1}-1$. Thus, $\omega(\Gamma(S)) \geq 2^{n-1}-1$. To complete the proof, we show that  $\chi(\Gamma(S)) \leq 2^{n-1}-1$. For $I = I_{{i_1}{i_2}\cdots{i_k}}$ and $J = I_{{j_1}{j_2}\cdots{j_{n-k}}}$, where $i_1, i_2, \ldots, i_k \in [n] \setminus \{j_1, j_2, \ldots, j_{n-k}\}$ we have $I \cap J$ is trivial. Consequently, we can color these vertices with same color so that we can cover all the vertices with $2^{n-1}-1$ colors. Thus $\chi(\Gamma(S)) \leq 2^{n-1}-1$. Hence $\omega(\Gamma(S)) = \chi(\Gamma(S)) = 2^{n-1}-1$.
\end{proof}
\begin{corollary}
	If $S = I_{{i_1}{i_2}\cdots{i_n}}$  then $\Gamma(S)$ is a weakly perfect graph.
	\end{corollary}

In order to find the upper bound of the chromatic number of $\Gamma(S)$, where $S$ is an arbitrary semigroup, first we define
\begin{align*}
    X_1 & = \{I \in V(\Gamma(S)) :  I_{{i_1}{i_2}\cdots{i_n}} \subseteq I \}\\
        X_2 & = \{I \in V(\Gamma(S)) :  I \subset I_{{i_1}{i_2}\cdots{i_n}} ~\text{and}~ I \neq I_{{i_1}{i_2}\cdots{i_n}} \}\\
X_3 & = V(\Gamma(S)) \setminus (X_1 \cup X_2).
\end{align*}

Let $\widetilde{{\text{Min}(I)}}$ be the set of all minimal left ideals contained in $I$
. Further define a relation $\rho$ on $X_3$ such that \begin{center}
    $J ~~ \rho ~~K \Longleftrightarrow \widetilde{{\text{Min}(J)}} = \widetilde{{\text{Min}(K)}}$
\end{center}
Note that $\rho$ is an equivalence relation.
\begin{theorem}
    Let $S$ be a semigroup with $n$ minimal left ideals and $\chi(\Gamma(S)) < \infty$. Then
    \begin{center}$\chi(\Gamma(S)) \leq |X_1| + (2^{n-1}-1) + (2^{n-1}-1)m$,
    \end{center}
    where $m = {\rm{max}} \{|C(I)| : C(I) \
    \text{is an equivalence class of} \; \rho \}$.
\end{theorem}
\begin{proof}

       Note that for any $I, J \in X_1$, we have $I \sim J$. Since every nontrivial left ideal contains at least one minimal left ideal, consequently each element of $X_1$ is a dominating vertex of $\Gamma(S)$. Therefore, we need at least $|X_1|$ colors in any coloring of $\Gamma(S)$.
       By proof of Lemma \ref{chromaticnumberintersection}, we can color all the vertices of $X_2$ with at least $2^{n-1}-1$ colors so that we need at least $2^{n-1}-1 + |X_1|$ colors to color $X_1 \cup X_2$.

     To prove our result we need to show that the vertices of $X_3$ can be colored by using $(2^{n-1}-1)m$ colors. Now let $J, K \in X_3$ such that $I_{{i_1}{i_2}\cdots{i_k}} \subset J$ and $I_{{j_1}{j_2}\cdots{j_t}} \subset K$. Note that $J \cap K$ is nontrivial if and only if $I_{{i_1}{i_2}\cdots{i_k}} \cap I_{{j_1}{j_2}\cdots{j_t}}$ is nontrivial. It follows that $J \sim K$ in $\Gamma(S)$ if and only if either $I_{{i_1}{i_2}\cdots{i_k}} = I_{{j_1}{j_2}\cdots{j_t}}$ or $I_{{i_1}{i_2}\cdots{i_k}} \sim I_{{j_1}{j_2}\cdots{j_t}}$.

      Note that the equivalence class of $I \in X_3$ under $\rho$ is $C(I) = \{J \in X_3 : \widetilde{{\text{Min}(I)}} = \widetilde{{\text{Min}(J)}} \}$. Since $\chi(\Gamma(S)) < \infty$ we get $|C(I)| < \infty$. Consequently, $|C(I)| \leq m$. Observe that $C(I)$ forms a clique, we require maximum $m$ colors to color each class under $\rho$. Note that $J \in C(J)$ and $K \in C(K)$ such that $J \sim K$ if and only if $I_{{i_1}{i_2}\cdots{i_k}} \sim I_{{j_1}{j_2}\cdots{j_t}}$ in $\Gamma(S)$. Consequently, we can color the vertices in $X_3$ by using $(2^{n-1}-1)m$ colors.
       \end{proof}

       \begin{theorem}
    Let $S$ be a semigroup with $n$ minimal left ideals. Then
 \[\operatorname{sdim}(\Gamma(S)) =  \begin{cases}
|X_1| + |X_3| + 2^{n-1}-2 &  \text{\rm if} ~ S \neq I_{{i_1}{i_2}\cdots{i_n}} \\
2^{n-1}-1 &  \text{\rm if}~ S = I_{{i_1}{i_2}\cdots{i_n}}
\end{cases}\]
\end{theorem}
\begin{proof}
        Let $I, J \in V(\Gamma(S))$ such that $I_{{i_1}{i_2}\cdots{i_k}} \subseteq I$ and $I_{{j_1}{j_2}\cdots{j_t}} \subseteq J$. Then $I \sim J$ if and only if either $I_{{i_1}{i_2}\cdots{i_k}} = I_{{j_1}{j_2}\cdots{j_t}}$ or $I_{{i_1}{i_2}\cdots{i_k}} \sim I_{{j_1}{j_2}\cdots{j_t}}$. Define a relation $\rho_1$ on $V(\Gamma(S))$ such that $I$  $\rho_1$  $J$ if and only if $\widetilde{{\text{Min}(I)}} = \widetilde{{\text{Min}(J)}}$. Clearly, $\rho_1$ is an equivalence relation on $V(\Gamma(S))$. Let $N[I_{{i_1}{i_2}\cdots{i_k}}] = \{I \in V(\Gamma(S)) : \widetilde{{\text{Min}(I)}} = I_{{i_1}{i_2}\cdots{i_k}}\}$  be equivalence class containing $I_{{i_1}{i_2}\cdots{i_k}}$. If $S \neq  I_{{i_1}{i_2}\cdots{i_n}}$, then by Theorem \ref{strong-metric-dim}, we have $\mathcal{R}_{\Gamma(S)}$ whose vertex set $V(\mathcal{R}_{\Gamma(S)}) = \{I_{{i_1}{i_2}\cdots{i_k}} : i_1, i_2, \cdots, i_k \in [n] ~  \text{and} ~ 1 \leq k \leq n\}$. By using Lemma \ref{chromaticnumberintersection}, note that $\omega (\mathcal{R}_{\Gamma(S)}) = 2^{n-1}$. Then $\operatorname{sdim}(\Gamma(S)) = |X_1| + |X_3| + 2^{n-1}-2$. Next, if $S = I_{{i_1}{i_2}\cdots{i_n}}$, then $V(\mathcal{R}_{\Gamma(S)}) = \{I_{{i_1}{i_2}\cdots{i_k}} : i_1, i_2, \cdots, i_k \in [n] ~  \text{and} ~ 1 \leq k \leq n-1\}$. By using Lemma \ref{chromaticnumberintersection}, note that $\omega (\mathcal{R}_{\Gamma(S)}) = 2^{n-1}-1$. Then $\operatorname{sdim}(\Gamma(S)) =  2^{n-1}-1$.
\end{proof}
Now in the remaining section, we consider a class of those semigroups which are union of $n$ minimal left ideals. In particular, completely simple semigroups belongs to this class. In what follows, the semigroup $S$ is assumed to be the union of $n$ minimal left ideals $I_{i_1}, I_{i_2}, \ldots, I_{i_n}$ i.e. $S = I_{{i_1}{i_2}\cdots{i_n}}$. The following lemma gives the lower bound of the metric dimension of $\Gamma(S)$.
\begin{lemma}[{\cite[Theorem 1]{chartrand2000resolvability}}]\label{metric dimension theorem}
        For positive integers $d$ and $m$ with $d < m$, define $f(m, d)$ as the least positive integer $k$ such that $k + d^k \geq m$. Then for a connected graph $\Gamma$ of order $m \geq 2$ and diameter $d$, the metric dimension $\beta(\Gamma) \geq f(m, d)$.
\end{lemma}

\begin{theorem}
    If $S = I_{{i_1}{i_2}\cdots{i_n}}$ then  the metric dimension of $\Gamma(S)$ is given below:
\[\beta(\Gamma(S)) = \begin{cases}
2 &  \text{\rm if} ~ n = 3 \\
n &  \text{\rm if}~ n \geq 4
\end{cases}\]
\end{theorem}
\begin{proof}
For $n =3$, it is easy to observe that $\{I_{i_1}, I_{i_2}\}$ forms a minimum resolving set. If $n \geq 4$ then by Remark \ref{everynontrivial left ideal is union}, we have $|V(\Gamma(S))| = 2^n-2$. In view of Lemma \ref{metric dimension theorem}, we get \begin{center}
 $n = f(2^n - 2, 2) \leq  \beta(\Gamma(S))$.
\end{center}
It is easy to observe that for $k = n-1$, $2^k + k \not\geq 2^n - 2$. Therefore, the least positive integer $k$ is $n$ for which $k + 2^k \geq 2^n-2$. Thus $n \leq \beta(\Gamma(S))$.
To obtain upper bound of $\beta(\Gamma(S))$, let $J = I_{{i_1}{i_2}\cdots{i_k}}$ and $K = I_{{j_1}{j_2}\cdots{j_t}}$ be distinct arbitrary vertices $\Gamma(S)$. Since $J \neq K$, there exists at least $I_{i_s} \in {\rm Min}(S)$ such that $I_{i_s} \sim J$ and $I_{i_s} \nsim K$. It follows that $d(J, I_{i_s}) \neq d(K, I_{i_s})$. Thus ${\rm Min}(S) = \{I_{i_1} : i_1 \in [n]\}$ forms a resolving set for $\Gamma(S)$ of size $n$. It follows that $\beta(\Gamma(S)) \leq n$. This completes our proof.
\end{proof}

An automorphism of a graph $\Gamma$  is a permutation $f$ on $V (\Gamma)$ with the property that, for any vertices $u$ and $v$, we
	have $uf \sim vf$ if and only if $u \sim v$. The set $Aut(\Gamma)$ of all graph automorphisms of a graph $\Gamma$ forms a group with
	respect to composition of mappings. The symmetric group of degree $n$ is denoted by $S_n$. Now we obtain the automorphism group of $\Gamma(S)$, when $S$ is union of $n$ minimal left ideal.
	\begin{lemma}\label{degree k}
	        Let $S = I_{{i_1}{i_2} \cdots {i_{n}}}$ and let $K = I_{{i_1}{i_2} \cdots {i_{k}}}$ be a nontrivial left ideal of $S$. Then $deg(K) = (2^k-2) + (2^{n-k}-2) + (2^{n-k}-1)(2^{k}-2)$.
	\end{lemma}
	\begin{proof}
	   Let $J$ be a nontrivial left ideal of $S$  such that $J \sim K$. Clearly $J \cap K$ is a nontrivial left ideal. Now we discuss the following cases:

		\noindent\textbf{Case 1.}  $J \not\subset K$ and $K \not\subset J$. Since $J \sim K$ and $K = I_{{i_1}{i_2} \cdots {i_{k}}}$ then note that the number of nontrivial left ideals such that $J \not\subset K$ and $K \not\subset J$ is

\begin{align*}
 &= \left(\sum_{i=1}^{n-k} \binom{n-k}{i}\right) \left(\sum_{i=1}^{k-1} \binom{k}{i}\right) = (2^{n-k}-1)(2^k-2)
 \end{align*}

\noindent\textbf{Case 2.} $J \subset K$. The number of nontrivial left ideals of $S$ which are properly contained in $K$ are $2^k-2$.

\noindent\textbf{Case 3.} $K \subset J$. The number of nontrivial left ideals of $S$ properly containing $K$ are $2^{n-k}-2$.
Thus, from  the above cases we have the result.
	\end{proof}

	\begin{corollary}
	If $S = I_{{i_1}{i_2}\cdots{i_n}}$  then the graph $\Gamma(S)$ is Eulerian for $n \geq 3$.
	\end{corollary}

\begin{lemma}\label{symmetric group}
		For $\sigma \in S_n$, let $\phi_{\sigma} : V(\Gamma(S)) \rightarrow V(\Gamma(S))$ defined by $\phi_{\sigma}(I_{{i_1}{i_2}\cdots {i_k}}) = I_{\sigma({i_1})\sigma({i_2})\cdots \sigma({i_k})}$. Then $\phi_{\sigma} \in Aut(\Gamma(S))$.
	\end{lemma}
	\begin{proof}
	It is easy to verify that $\phi_{\sigma}$ is a permutation on $V(\Gamma(S))$. Now we show that $\phi_{\sigma}$ preserves adjacency. Let $I_{{i_1}{i_2}\cdots {i_t}}$ and $I_{{j_1}{j_2}\cdots {j_k}}$ be arbitrary vertices of $\Gamma(S)$ such that $I_{{i_1}{i_2}\cdots {i_t}} \sim I_{{j_1}{j_2}\cdots {j_k}}$. Then  $I_{{i_1}{i_2}\cdots {i_t}} \cap I_{{j_1}{j_2}\cdots {j_k}} \neq \emptyset$.
Now
\begin{align*}
I_{{i_1}{i_2}\cdots {i_t}} \sim I_{{j_1}{j_2}\cdots {j_k}}
&\Longleftrightarrow I_{\sigma({i_1})\sigma({i_2})\cdots \sigma({i_t})} \sim I_{\sigma({j_1})\sigma({j_2})\cdots \sigma({j_k})}\\
& \Longleftrightarrow \phi_{\sigma}(I_{{i_1}{i_2}\cdots {i_t}}) \sim \phi_{\sigma}(I_{{j_1}{j_2}\cdots {j_k}}).
\end{align*}
Thus, $\phi_{\sigma} \in Aut(\Gamma(S))$.
\end{proof}

	\begin{proposition}\label{phisigma}
		For each $f \in  Aut(\Gamma(S))$, we have  $f = \phi_{\sigma}$ for some $\sigma \in S_n$.
	\end{proposition}
	\begin{proof}
	 In view of Lemma \ref{degree k} and Lemma \ref{symmetric group}, suppose that
	 $f(I_{i_1}) = I_{j_1}$, $f(I_{i_2}) = I_{j_2}$, $\ldots$, $f(I_{i_n}) = I_{j_n}$. Consider $\sigma \in S_n$ such that $\sigma(i_1) = j_1, \sigma(i_2) = j_2, \ldots, \sigma(i_n) = j_n$. Then $\phi_{\sigma}(I_{{i_1}{i_2}\cdots{i_k}}) = I_{\sigma({i_1})\sigma({i_2})\cdots \sigma({i_k})} = I_{{j_1}{j_2}\cdots{j_k}}$ (cf. Lemma \ref{symmetric group}). Clearly, $I_{i_1} \sim I_{{i_1}{i_2}\cdots{i_k}}$, $I_{i_2} \sim I_{{i_1}{i_2}\cdots{i_k}}$, $\ldots$, $I_{i_k} \sim I_{{i_1}{i_2}\cdots{i_k}}$. Also note that    $I_{i_t} \cap I_{{i_1}{i_2}\cdots{i_k}}$ is trivial for ${i_t} \in \{i_{k+1}, i_{k+2}, \ldots, i_{n}\}$  where $i_{k+1}, i_{k+2}, \ldots, i_{n}\in [n] \setminus \{i_1, i_2, \ldots, i_k\}$. It follows that $I_{i_{k+1}} \nsim I_{{i_1}{i_2}\cdots{i_k}}$, $I_{i_{k+2}} \nsim I_{{i_1}{i_2}\cdots{i_k}}$, $\ldots$, $I_{i_{n}} \nsim I_{{i_1}{i_2}\cdots{i_k}}$. Thus, $f(I_{i_1}) \sim f(I_{{i_1}{i_2}\cdots{i_k}})$, $f(I_{i_2}) \sim f(I_{{i_1}{i_2}\cdots{i_k}})$, $\ldots$, $f(I_{i_k}) \sim f(I_{{i_1}{i_2}\cdots{i_k}})$ and $f(I_{i_{k+1}}) \nsim f(I_{{i_1}{i_2}\cdots{i_k}})$, $f(I_{i_{k+2}}) \nsim f(I_{{i_1}{i_2}\cdots{i_k}})$, $\ldots$, $f(I_{i_{n}}) \nsim f(I_{{i_1}{i_2}\cdots{i_k}})$. Consequently, $I_{j_1} \subset f(I_{{i_1}{i_2}\cdots{i_k}})$, $I_{j_2} \subset f(I_{{i_1}{i_2}\cdots{i_k}})$, $\ldots$, $I_{j_k} \subset f(I_{{i_1}{i_2}\cdots{i_k}})$ and $I_{j_{k+1}} \not \subset f(I_{{i_1}{i_2}\cdots{i_k}})$, $I_{j_{k+2}} \not \subset f(I_{{i_1}{i_2}\cdots{i_k}})$, $\ldots$, $I_{j_n} \not \subset f(I_{{i_1}{i_2}\cdots{i_k}})$. It follows that $f(I_{{i_1}{i_2}\cdots{i_k}}) = I_{{j_1}{j_2}\cdots{j_k}} = \phi_{\sigma}(I_{{i_1}{i_2}\cdots{i_k}})$. Thus, $f = \phi_{\sigma}$.
	\end{proof}

	\begin{theorem}\label{automorphism group}
		Let $S$ be the union of $n$ minimal left ideals. Then for $n \geq 2$, we have $ Aut(\Gamma(S)) \cong S_n$. Moreover, $|Aut(\Gamma(S))| = n!$.
	\end{theorem}

	\begin{proof}
			In view of Lemma \ref{symmetric group} and by Proposition \ref{phisigma},
		note that the underlying set of the automorphism group of $\Gamma(S)$ is
		$Aut(\Gamma(S)) = \{\phi_{\sigma} \; : \; \sigma \in S_n \}$, where $S_n$ is a symmetric group of degree $n$. The groups $Aut(\Gamma(S))$ and $S_n$ are isomorphic under the assignment $\phi_{\sigma} \mapsto \sigma$.  Since all the elements in $Aut(\Gamma(S))$ are distinct, we have $|Aut(\Gamma(S))| = n!$.
	\end{proof}

\section{Acknowledgement}
The first author gratefully acknowledge for providing financial support to CSIR  (09/719(0093)/2019-EMR-I) government of India. The second author wishes to acknowledge the support of MATRICS Grant  (MTR/2018/000779) funded by SERB, India.

redpajama_set_name: RedPajamaArXiv


text:
screaming tribesmen - The I-94 Bar
2019 Barfly Top Tens: Mick Medew
February 2 at The Lyrebyrd Lounge, Ripponlea: The Lyrebyrd Lounge was the first of seven shows for me in Melbourne this year. It felt like coming home; what a marvellous club - run by my buddy Leon Storch. This show also featured Penny Ikinger and would turn out to be the start of something good.
Late January marks the birth of my special new friend '' Arthur Robert Collie''
Stephen Cummings at The Triffid on April 4:
An unlikely choice for my Top Ten but when you are hot you are hot. This was the gig of the year, no doubt about it. Clare Moore on drums and Sam Lemann on guitar.
A Date With Some Vampires or a Dark Surprise? Masuak joins Phantom Fifth for US mini-tour
Written by The Barman on 08 June 2016 .
New York City-based band Phantom Fifth (pictured right) has announced that Chris "Klondike" Masuak will be joining them for shows in California and New York in July.
Phantom Fifth is from New York City and is comprised of expatriate Australian bassist Greg Delves (ex-Inner Sleeves, Radio Luxembourg), Irishman Colin McCarthy (drums) and US-born Paul Stinson (guitar, vocals.)
Formed in 2015, the band bio says it "embraces the sonic sound and dark energy of Nick Cave, Joy Division, The Strokes and The Stooges. The Phantoms' sound is built around dark melodies, shimmering guitars, and propulsive bass lines, with lyrics about wolves, ghosts, murderous paramours, Italian discos, and other manifestations of love and madness".
Originally from California, where he co-founded the "punk Americana" band The Stripminers with members of X and The Donnas, Paul Stinson moved to Brooklyn in 2014 and quickly met up with power pop guru Delves and McCarthy (formerly of Amsterdam-based Belsonic Sound).
The trio immediately began writing, recording and performing in and around NYC and have plans in the works for tours of both the UK and Australia. The band's debut EP is due out in the US Summer of 2016.
July 20 – Brick & Mortar Music Hall, San Francisco, CA
July 22 – The Uptown Nightclub, Oakland, CA
July 24 – The Viper Room, Hollywood, CA
July 29 – The Bowery Electric, NYC
Phantom Fifth on the Web
Chris Masuak on the Web
A two-night Date With Some Mesmerisers
Written by The Barman on 26 February 2018 .
Brian Mann and Mick Medew trading licks in Thirroul. Shona Ross photo.
Gonna break that rule about not reviewing I-94 Bar gigs, as the second of this two-night engagement was badged thus. If you can't write about things you like, what's the point? It's the ethos of why we do this e-zine thing.
So let's be up front and say that Mick Medew is a good mate and his band, The Mesmerisers, are lovely people. In customary evangelical spirit, I'm unashamedly going to tell you that if you love rock and roll then you have to see them - in their native Brisbane or on one of their few forays outside of Queensland.
The first leg of this two-night weekend stand was a support to the mighty Sunnyboys, opening a bill shared with the mercurial Ups & Downs. The venue was the magic Anita's Theatre in Thirroul, an hour-and-a-bit south of Sydney. The second was a Sydney show at the increasingly familiar stamping ground of Marrickville Bowling Club. The Mesmerisers are making the road trip a family affair with partners and two offspring in tow.
BARFLY TOP TEN: Mick Medew of Mick Medew & The Mesmerisers
In no special order:
1. The Damned at The Triffid, Brisbane, March 15
A school night: Wednesday. The Mesmerisers go on at 7.30pm to a packed house. We carve, the crowd makes us feel like they are there to see us. The Damned 's tour manager remarks that he has never encountered a support band being granted a bottle of Gordon's gin as part of their drink rider: another milestone ticked.
The Damned play for two-and-a-half hours - brilliantly. They are a big hit with the audience - and with Captain Sensible back in the band, they could hardly miss.
2. Perfect Match
Now I do know where she comes from: Banyo. I've got a Date with a Banyo girl, tonight.
3. Died Pretty, Radio Birdman and The Mesmerisers at The Tivoli Theatre, Brisbane, June 23
The crowd have all turned up early and turn on to us straight away. Died Pretty get better every time I see them. They always were a fantastic band.
4. Perfect Purchase
My Zoom H5 portable recording device does everything i wanted it to.
5. Gap Farmers Markets, Brisbane, June 25
Andrew Ross and Co sure know how to put on a festival . We go on before dusk. I wear my sunglasses for half the set. Michael again chooses the right shirt for the occasion.
BARFLY TOP TEN: Viveiro Wave Riders and ex-Radio Birdman and Hitmen guitarist Chris Klondike Masuak
Hmmm. What wondrous things have defined my year?
I suppose my prevailing sentiment has been one of simmering frustration and anger at a world more out of control and stupid than I can remember. Perhaps it's always been this way and I'm only noticing it in my dotage.
Whatever the reason, I'm noticing that personal selfishness and cultural callousness are the new black and that I just haven't been able to continue being my usual conciliatory self.
I Want, Need, Love You: Garage-Beat Nuggets From The Festival Vaults - Various Artists (Playback Records)
A vinyl fanatic you may be, but owning most of these 26 tracks in their original format would set you back almost as much as a down-payment on a modest piece of real estate in the over-heated Sydney property market. So put aside your collector scum aspirations and focus on the merits of this formidable collection of '60s gems on CD instead.
Australia's garage and beat history is under-appreciated - despite the best efforts of labels like Raven and others of dubious legal standing - so new imprint Playback is a welcome market entrant. "I Want, Need, Love You" focuses on the output of The 5, Toni McCann, The Black Diamonds and The Pogs, spanning the period 1965-68, with about half the tracks new to CD. As many as possible have been sourced from the original masters.
It's nearly Open Season
Our own loss-leading I-94 Bar Records is proud to announce the imminent release of "OPen Season", the new album by Mick Medew and The Mesmeriers.
Led by Brisbane legend Mick Medew, the band includes members of his Screaming Tribesmen, the Lipstick Killers, The 31st and Shy Impostors. "Open Season" is available for pre-order from our Bandcamp now and will be out on CD and digitally from June 4. Enjoy this teaser and the ordering link is here.
Mick Medew returns to Sydney for first show in five years
Written by The Barman on 25 January 2018 .
Long regarded as one of Queensland rock and roll's most venerable singer-songwriters, Mick Medew is returning to a Sydney stage after a five-year absence with his band The Mesmerisers.
With one album ("The Mesmerisers") under their belt and another underway, Mick Medew and The Mesmerisers will play Marrickville Bowling Club on Saturday, February 24, with support from Loose Pills and The Dark Clouds. Tickets are on sale here.
The Sydney show has been added to follow a support to The Sunnyboys in Thirroul on February 23.
Vocalist-guitarist Medew is co-founder of the Screaming Tribesmen and more recently leader of Mick Medew and The Rumours, and he's been making a mark on Brisbane (and Australian) music for more than 30 years.
Originally a member of seminal inner-city Brisbane band The 31st, whose ranks included future members of the Hoodoo Gurus, the Hitmen and Died Pretty, Medew went on to front the Screaming Tribesmen, a band schooled in tough guitar rock overlaid with alternately hooky and plaintive melodies.
The Tribesmen outgrew their home town and moved to Sydney in the 1980s, becoming Australian independent music chart toppers with the classic single "Igloo".
The band lasted 16 years, signed to a US label, released three albums and toured North America and Europe (twice) before Mick moved home to Queensland and put them on hiatus in 1993. Reunions for tours followed in 2011 (Australia) and 2012 (Europe).
The 31st - The 31st (LCMR)
Copies of these four songs have been circulating for years and two have surfaced on compilations. The balance were re-recorded by members' subsequent bands. But don't kid yourself that you don't need this vinyl only 12-inch EP.
The 31st started when future members of Died Pretty (Ron Peno and Chris Welsh), the Screaming Tribesmen (MIck Medew) and the Hitmen/New Christs/Screaming Tribesmen (Tony Robertson) started playing shows in a strip club and anywhere else that would have them. Evidently, they played no one style of music - which must have been confusing for the Brisbane punks, boogie-heads and blues fans to pin a tail on.
The 31st were a future supergroup before those things were called that in Australia. They kicked around the undergrowth of Brisbane's downtrodden music scene in the early 1980s, and fell to pieces before anyone outside of it saw or heard them.
Future Hoodoo Guru Brad Shepherd was to briefly become a member although he's not on these recordings.
The Glory Days of Aussie Pub Rock Vol 1 - Various Artists (Festival)
Written by The Barman on 01 March 2017 .
"What's he doing reviewing THAT?"
Only people of a certain age will "get" this review. The term "Guilty Pleasure" will not be used at any point.
Admit it, punk. If you grew up in Australia in the 1970s and '80s (OK, you were might have been underage and still growing up, but you could sneak into licensed premises) and lived anywhere outside of Melbourne and Sydney's inner-city regions, a dose of Pub Rock was unavoidable. A way of life, even.
Welcome Aboard - The On and Ons (Citadel)
Great guitar pop is timeless and that's what Sydney's The On and Ons have delivered (again) on their second long-player.
Well established on the strength of their 2015 debut, "It's The On And Ons Calling", Morris and Co have doubled down on the pop factor on "Welcome Aboard". The rock is turned down just a tad and (to these ears at least) it takes a few more listens for the songs to take hold.
Truth-be-told, I almost marked it down half-a-beer for not rocking as much as the debut - but the pop smarts won out.
Who Is Innocent? A Singles Collection 1983-87 - Fixed Up (Nineteen Something)
Their legacy was just two LPs and a stack of singles but Fixed Up's punky and soulful garage rock touched people in their native France and all the way around to the other side of the world in Australia.
A lot's been made about the Sydney-Detroit connection, mainly through Radio Birdman and its now fading local musical legacy. The irrefutable fact was that Birdman and its associated influences ruled the Sydney roost in the early 1980s. As true as that was, you can make a strong case for the affinity between Australia and France being almost as important, once the Sydney underground scene started to diversify and expand.
The Franco-Ausstralian link was made when John Needham, chief of seminal Sydney label Citadel Records, started dealing with the likes of Sonics Records in France. Suddenly, there was a pipeline for Australian bands to have their music heard on the Continent - meaning outside the UK where the perpetually jaded music press briefly adopted Aussie arty pop, junkie rock and the swamp sound for a time.

redpajama_set_name: RedPajamaCommonCrawl


Almost two decades ago we made a decision to raise a family in this picture-perfect community, and rarely have I regretted that decision. Here the community parades are composed of moms and dads leading the pack with little ones trailing behind on decked out bikes; brisk Saturday- mornings throughout the fall are filled with squeals of delight as not -much- bigger- than- toddler sized competitors swarm the soccer ball relentlessly vying for the next toe tap in the direction of the goal, and blue- ribbon awarded elementary school classrooms boast as many parent volunteers as students. It is picturesque, indeed. However, raising children here has proven to be a double-edged sword though as parents are ever ready to slice clear through any impropriety.
One of the beauties of a small community is everyone knows everyone else. What a sense of peace to send your child home with friends or neighbors you have known since birth. Throughout elementary school we were blessed with friendships and families whose values reflected our own. The picturesqueness continued.
Once we ventured from the safe, secure world of elementary school into the tumultuous world of middle school, I caught my first glimpse of what could be considered as the flaw of our community. The same close knit neighbors who shared in each step of our childhood adventures continued to look out for our children.....but somewhere along the line the collaborative approach to raising a community of well- rounded individuals had been replaced with an air of righteousness, superiority, and condemnation. The complex situation we were placing adolescents in became clear. It was like our teens were being placed in a fishbowl where all of their actions were magnified. To make matters worse, these eyes belonged to parents who somehow felt other adolescents mistakes somehow validated their superiority as a parent....so everyone patted themselves on the back by focusing on the indiscretions of others. I have never really understood how discussing the length of Mary Sue's shorts made someone feel like a better parent?


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text:
Bryant, Edward
Among the Dead and Other Events Leading up to the Apocalypse (New York: The Macmillan Company, 1973)
Among the Dead and Other Events Leading up to the Apocalypse (New York: Collier Books, 1974)
Phoenix without Ashes (Greenwich, Connecticut: Fawcett Gold Medal, 1975) with Harlan Ellison
Cinnabar (New York: Macmillan Publishing Co, 1976)
2076: The American Tricentennial (New York: Pyramid, 1977)
Wyoming Sun (Laramie, Wyoming: Jelm Mountain Press, 1980)
Particle Theory (New York: Pocket Books/Timescape, 1981)
Trilobyte (Eugene, Oregon: Axolotl Press, 1987)
Neon Twilight (Eugene, Oregon: Pulphouse Publishing, 1990)
The Man of the Future (Arvada, Colorado: Roadkill Press, 1990)
The Cutter (Eugene, Oregon: Pulphouse Publishing, 1991)
Fetish (Eugene, Oregon: Axolotl Press, 1991)
The Thermals of August (Eugene, Oregon: Pulphouse Publishing, 1992)
Aqua Sancta (Arvada, Colorado: Roadkill Press, 1993)
Darker Passions (Arvada, Colorado: Roadkill Press, 1993)
The Baku: Tales of the Nuclear Age (Burton, Michigan: Subterranean Press, 2001)
A Sad Last Love at the Diner of the Damned (Fort Wayne, Indiana: Wormhole Books, 2001)
While She Was Out (Denver, Colorado: Wormhole Books, 2001)
Predators and Other Stories (Golden, Colorado: ReAnimus Press, 2014)


text:
Q: Mail won't send I have this script:
$to      = $_GET["to"];
$subject = $_GET["sub"];
$message = $_GET["message"];
$from    = "admin@chipperyman573.com";
$headers = "From:" . $_GET["from"];
mail($to, $subject, $message, $headers);

I go to:
http://chipperyman573.com/THEURL?to=myemail@gmail.com&sub=Subject&message=Hi&from=admin@chipperyman573.com

I want to always send it from admin@chipperyman573.com, however if I want to change the first part (say, to personal@chipperyman573.com) it won't send. I want to use admin@chipperyman573.com.
(admin@chipperyman573.com is the only one that exists).

redpajama_set_name: RedPajamaStackExchange


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text:
"I could just go to a gun show, and do it a couple of days, or once a year if I felt like it," he said. "I have 10, 15 firearms to take and display. It's not that big a deal. Believe me, if I ever knew it was going to create havoc like this, I'd have thought twice about it."
In this case, the problem got solved without really resolving the issue. At a public hearing last week, Kosienski announced he was withdrawing his request for a zoning variance because he is moving out of Ocean Pines. He said it had nothing to do with the controversy his zoning variance request stirred up.
So, this particular case notwithstanding, we ask: Where do you stand on gun rights? More importantly, how would you feel if a neighbor was trying to do what Kosienski attempted to do? Would neighborly proximity make you think twice about the right to bear arms or do you trust the system to keep you safe?
This is truly a case of NIMBY — not in my backyard, please.
One neighbor, expressing his objections and concerns, asked if the gun storage provision would attract what he referred to as "a bad element" — people who might try to steal the guns. More pertinent to him, he wondered aloud whether those would-be thieves might break into the wrong house — his home, perhaps? — because "burglars aren't the smartest people in the world."
On the other hand, Kosienski did receive strong support from other neighbors, some of whom he had never met, who called to wish him luck or told him not to be concerned about the opposition.
That's not to say neighbors' concerns were not valid. Some of the issues seemed legitimate, especially in light of well-publicized mass shootings like the Newtown, Connecticut, tragedy a few years ago.
Legal ownership of firearms does not guarantee safety or responsible use of them.
The problem mostly seems to be that fear and speculation have taken over, helped along by some less-than-accurate and mean-spirited social media postings.
Kosienski jumped through all the hoops, was thoroughly vetted by the federal Bureau of Alcohol, Tobacco, Firearms and Explosives and talked with as many neighbors as he could. He followed all the rules. He did not hide anything, he simply wanted to pursue a hobby in retirement. It happened to be a hobby that raised controversy, is tightly regulated and evokes fear in the hearts of many people.
It might be beneficial for everyone to step back, take a deep breath and, in the absence of the pressure of a neighbor asking for a variance, really consider the answer to that question: Do you support the right to bear arms strongly enough to be OK with a neighbor storing a cache or inventory of firearms on the property next door or down the street?
If you truly believe you support the right of the people to bear arms, the objections raised in this case can be difficult to justify.
WHERE DO YOU STAND?
Would you support or oppose a request to store legal firearms on private property in your own neighborhood? Share your thoughts in a letter to the editor. Email your comments for publication to opinions@delmarvanow.com.


Gentlemen, I hereby propose a challenge to all.
It may serve as an "opinion" for a possible approach to the next year in the course of simulation in edX. Basically the idea I had is to assemble in a "solid" and complete, step by step the recreation of the event of the bridge Tacoma Narrows. For those who do not know, a text about this is available. The idea is to assemble the geometry and mesh, to simulate an external flow for the purpose of modeling vortex formation, since the collapse is explained by the induction of vibration through vortices, according to Billah and Scanian (1991). And from the obtained data, starting for structural analysis, thus composing a multiphysical simulation and at the end proposing solutions to the problem, since it is a typical problem where many areas of engineering can contribute, like my area, mechanical engineering. I would like it to be something done successively, with the help of everyone. Some are enabled?
Billah, K. Yusuf, Scanian Robert H., "Resonance, Tacoma Narrows bridge failure, and undergraduate physics textbooks" – Am. Journal Physics, Vol. 59, No 2. Feb. 1991.
That is a bold challenge José!
Shell elements may be required to fit a model into the Structural Student license limit of 32k nodes+elements.


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Published on: August 31, 2019 August 31, 2019 by RRadmin7 Category:Rapture Ready End Times News
Hong Kong protests: Demonstrators defy ban on march
Hong Kong police have used tear gas and water cannon to disperse crowds of protesters who defied a police ban and marched through the city. Demonstrators lit fires and attacked the parliament building as thousands joined the march. An event to mark five years since Beijing ruled out fully democratic elections was banned by officials and called off by organisers.
Assam NRC: What next for 1.9 million 'stateless' Indians?
India has published the final version of a list which effectively strips about 1.9 million people in the north-eastern state of Assam of their citizenship. The National Register of Citizens (NRC) is a list of people who can prove they came to the state by 24 March 1971, the day before neighbouring Bangladesh declared independence from Pakistan. People left off the list will have 120 days to appeal against their exclusion.
Iran tanker row: US blacklists vessel 'bound for Syria'
The US Treasury has blacklisted an Iranian tanker it says is being used to ship oil to Syria. The Adrian Darya 1, previously known as Grace 1, was detained by British authorities in Gibraltar in July after it was suspected of moving oil to Syria in breach of EU sanctions. It was released earlier this month after Iran gave assurances it would not discharge its cargo in Syria.
Syria war: Russia announces ceasefire in Idlib rebel stronghold
A ceasefire by Syrian government forces will come into force on Saturday morning in the rebel-held province of Idlib, Russia has announced. The unilateral agreement will begin at 06:00 (03:00 GMT), Russia's defence ministry said. Syrian forces intensified their efforts to recapture Idlib in April.
UN chief Guterres fears eruption of 'full civil war' in Liby
The United Nations Secretary-General, Antonio Guterres, has said he fears Libya will end up in a "full civil war" unless moves are soon made to end the conflict. "Unless action is taken in the near term, it is highly likely that the current conflict will escalate into full civil war," Guterres said on Thursday in his latest report on the UN Support Mission in Libya.
Lebanon's Hezbollah pledges to retaliate against Israel
Lebanon's Hezbollah has promised to retaliate against Israel after accusing it of being behind what it called a suicide attack by self-destructing drones last weekend. Domestic rivals have long accused the group of making security and military decisions on its own, dragging the country to war. But this time Lebanese officials are rallying behind Hezbollah.
Erdogan suggests Turkey could look to Russia for jets
President Recep Tayyip Erdogan on Friday suggested Turkey could look to Russia for an alternative after the US excluded Ankara from its F-35 fighter jet program. Following Turkey's controversial purchase of the Russian S-400 missile defense system, Washington discontinued Turkey's involvement in the F-35 program.
Arab leader seeks to shake up Israeli election
The leader of the main Arab faction in parliament has shaken up Israel's election campaign by offering to sit in a moderate coalition government — …that would end decades of Arab political marginalization and could potentially bring down Prime Minister Benjamin Netanyahu. Ayman Odeh's offer to back…Benny Gantz…reflects the growing desire of Israel's large Arab minority to take a more active role in shaping the country.
Muslims Demand That 'Offensive' Crosses Be Removed…From CATHOLIC School
A complaint that says crosses at Catholic school are "offensive", and prevent Muslim prayers, has been filed against Catholic University in Washington, D.C. It is only the latest in an endless series of demands for Sharia Law adherence by Muslims in America.
Buddhist Monk says all races and religions could live together if it weren't for Muslims
In Myanmar, Muslims are killing Buddhists and trying to destroy Buddhism in order to create an Islamic State.
World's 2nd-largest Ebola outbreak surpasses 3,000 cases
The second-largest, second-deadliest Ebola outbreak in history has exceeded 3,000 cases and 2,000 deaths as the yearlong epidemic continues despite access to an experimental vaccine and developmental treatments.
Toxic Algae Kills Dogs Across the Country
A deadly variety of algae has caused a recent spate of dog deaths in the Southern United States, causing concern among canine owners nationwide.
DNC Resolution Celebrates Democrats' 'Largest Religious Group' — And It's Not Christians
The Democratic National Committee (DNC) passed a resolution at its summer meeting championing "religiously unaffiliated" voters, who the resolution noted are the "largest religious group" within the party.
Florida preps for 'absolute monster' of a storm: Hurricane Dorian
Unsure where Hurricane Dorian is going to land over Labor Day weekend, many Florida residents faced a sense of helplessness as they prepared for what US President Donald Trump said could be an "absolute monster" of a storm.
Bolivia loses 1.2 million ha (3.2 million acres) of forest land this year as wildfires continue
Forest fires continue raging across Bolivia's forests, damaging almost 1.2 million hectares (3.2 million acres) of land this year.
Liberals REALLY hate God, and the REASON will surprise you. Patriots, pay attention
…It turns out it never was about sex, or drugs, or even Rock and Roll—that was just the Trojan Horse; it was about that damned Constitution. Not all of it, just those pesky "inalienable rights." You remember them…Life, Liberty, and the Pursuit of Happiness.
Report: Google to Pay up to $200 Million for Child Privacy Violations
Google will reportedly pay between $150 and $200 million to resolve a Federal Trade Commission (FTC) investigation into YouTube regarding its potential violation of a children's privacy law, according to a report.
Why Does Darpa Need a Huge Underground Facility by Friday?
…Darpa sent out a tweet today that's making people a little nervous. The agency wants a large underground facility for "testing" of some sort. And they need it by Friday at 5pm ET.
The Nazi-Like Boycott of Jews Is a Global Menace: Why BDS Is All About Eradicating the State of Israel
Since its establishment in 2001 by the major Palestinian terrorist organizations, the Boycott, Divestment, and Sanctions (BDS) movement has masqueraded as a nonviolent grassroots human rights organization that aims to "improve" the well-being of Palestinians. Instead of protecting the Palestinians, the movement is laser-focused on economically, culturally, and politically isolating and eradicating the State of Israel, using the model that was applied previously to the apartheid regime of South Africa.
Illinois Passes Law Requiring LGBTQ Classes in Middle School
Illinois Gov. J.B. Pritzker has signed a new bill into law requiring the state's schools to teach "LGBTQ history" to students by the eighth grade.
Democrats Attack Religious Liberty in New Resolution, Praise Nonreligious Voters
In a new resolution, the Democratic Party praises nonreligious voters and "religiously unaffiliated" Americans while attacking "religious liberty."
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Embedding Change at West Midlands Police
Discover how West Midlands Police ran a 30-Day Challenge that has resulted in officers thinking digital first.
Celebrating 10 years of changing hearts, minds, and habits at scale.
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Read about how West Midlands Police ran a 30-Day Challenge on the Cognician platform, which contributed to digital transformation and a change in culture that has resulted in officers thinking digital first. Not only did the program positively influence the attitudes of more than 1,000 officers to the force's digital capabilities, it also doubled the number of crimes reported through digital channels – all in 30 days. As a mechanism for behavior change, the 30 Day Challenge consists of a series of daily actions that take less than 10 minutes to complete over 30 days. Accenture worked with West Midlands Police to create the challenges, and Cognician adapted them for digital delivery. But the success of a program like this depends on more than good design and development. This case study shares detail on critical success factors in the implementation of a behavior change program, which include:
Stakeholder engagement through a series of workshops
Continual action and reflection
Repetition with variation
Using technology and data to monitor and drive engagement
Support and encouragement from managers
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30-day hosting money-back guarantee. No-questions-asked refund.
How many times have you come across such marketing and sales gimmicks when buying a product online?
Chances are most of us don't ever usually get to that point, but at those rare times when we have to, the experience is not always pleasant, to say the least.
When I recommend a product, I ensure I do all kinds of testing before recommending it to millions of ShoutMeLoud users. Bluehost is one such web-hosting which is on our list of recommended WordPress hosting. I like it because it has many unique features to offer. This article, however, is not about what Bluehost hosting offers but its marketing gimmick and how reliable it actually is.
I thought of testing out how effective Bluehost money-back guarantee is. I'm sharing the exact steps and my overall experience with canceling a brand new Bluehost hosting account and getting the refund.
3 weeks ago I purchased a new hosting package from Bluehost to test out their new features. I purchased a new hosting package and have been playing around with it. Only yesterday did I realize that it had been over 25 days since I purchased it and that is when the idea of testing their refund policy came into my notorious mind.
Usually, companies make it really difficult for users to cancel their accounts, and not everyone has a blog to share their experience.
Therefore, I took it upon myself to find out what its like to cancel the Bluehost hosting account and whether or not I will get a complete refund.
I was hoping to find an account cancellation form. However, Bluehost offers account cancellation over the telephone. Their cancellation department is based out of USA and I had to dial the USA number from my country.
And we all know how much time tech support usually takes.
I used Google voice to call Bluehost account cancellation department and for the extra info, you can make free calls to the USA using Google voice.
I called 844-853-6296 and the operator on the other side was Daniel.
Operator: What's your domain name?
Operator: The last 4 digits of your password?
Operator: I have sent you an email with Validation Token. Kindly provide me with the information.
Operator: Is there anything I could do to change your mind?
Me: Thanks but I would like to go ahead with my account cancellation and get the full refund.
Operator: Alright! Would you like to keep the domain name or cancel that as well?
Me: If I cancel the domain, will I get a refund for that as well?
Operator: Nope. Domain money can't be refunded.
Me: Alright. I would like to keep the domain name.
Operator: I have canceled your account. You will get the full refund within 3-5 working days.
I acknowledged that and hung up.
Was that for real, I asked myself! It took me a few minutes to soak that up.
In no more than 8-10 minute I was able to cancel my account without any questions being asked. Usually, account canceling experiences are never one worth talking about.
Another thing is, depending on your payment method, you can get the refund within 24 hours as well.
PayPal: Refund will be processed within 24 hours of account cancellation.
Credit card: Refund can take up to 10 days, depending on your card issuer.
Overall, I was impressed by the whole process of Bluehost account cancellation. It was smooth and actually a no-questions-asked refund. The only thing not likable in this process, also a feedback to the Bluehost team, is the lack of a confirmation email as soon as the account is canceled. However, you can verify your account status by visiting the billing section of your Bluehost account dashboard.
Overall, Bluehost scored well in this experiment and they definitely deserve a big round of applause for being customer-centric.
Have you ever tried claiming the full refund with Bluehost or any other online services? How was your experience? Let us know in the comments below.
Harsh, thanks for telling the full procedure of cancelling bluehost hosting in case of problem. The only thing to note is that we have to make phone call to there US call center which is quite expensive. Other that that everything looks good.
2 years back I remember cancelling my '1and1' hosting, there too I had to call their US call center and had to spend some 120 rupees in that call.
Harsh, I will try Bluehost hosting. And thanks for telling this procedure.
For telling us the cancellation process of bluehost hosting. I'm looking for a hosting site. It will surely help me and others if someone is unsatisfied with bluehost's hosting.
Very nice article…I will try this month bluehost.
I've been using blue host for quite awhile and I'm quite happy with the their service for WordPress hosting. I'd still recommend WP Engine if you have the budget but for a low cost option their features and support are pretty good.
Thanks for your comment and adding more value. If you like WPEngine, then you should also check out Kinsta. I have used WPEngine and later on shifted to Kinsta hosting which is where currently ShoutMeLoud is hosted. Better service and pricing.
Many time I chatted with them for technical support and they always help up to their level. Till now I am satisfied with their services.
3. Difficult to use the cpanel.
So sorry to hear about your issues. Are you talking about Bluehost USA or Bluehost India?
I see you have a .in domain name.
Do remember, Bluehost India is not the same as Bluehost USA. I have highlighted this earlier.
Looking forward to hearing your comment on this.
I have tried both of them the indian version is better but the US version sticks.
Really informative content. There are so many hosting sites in the market and to decide which could be relevant to your application is really a challenging task.
I wanted to know that while choosing a hosting site, what are the other factors that I should have to consider apart from their backend support? How to test their servers? Is there any method to check? What you recommend?
Brand value (This way you are certain you working with a reputed company)..
Read a lot of reviews which is based on experience and then pick one.
I usually pick random hosting and test them. It's something I like doing and it help me to compare different hosting to see which is better than other.
I'm trying to cancel my account right now and they're telling me I'm not allowed. Giving me a hard time saying they don't even have a 30 day money back guarantee even though it is right on their website.
A perfect article for beginners. Hey! Harsh just after reading this article I feel blue host is far much reliable and better hosting provider I should give a try at. Some of my friends are using it already, I will give give it a try as well. Thanks for such a detailed article.
i buy hosting from godaddy already. can i convert it in bluehost? and is there any coupon for bluehost that after using that i can get discount?
You can migrate your blog from godaddy to bluehost. Since, you are buying bluehost for the first time, you will get a discount. I recommend you to buy for 3 years as it is the most cost-effective. After 3 years, there will be no discount on renewal.
This month I have experience with Bluehost cancellation.
I purchased Bluehost with full of hope. But that not perform well compare to my old hosting. I got frequent emails Jetpack about site down.
I contacted customer care and they told avoid it.
Finally I decided to cancel and refund the amount. I purchased another hosting and migrated my website.
I just used chat option and asked for cancellation. They asked 5 questions and I replied it. I paid $96 for 3 years. They deducted $15 domain fee and refunded $81. I received money within 2 days. Service disappointed but refund policy nice.
It took me about 10 minutes using live chat. Go to "support" and click a few boxes, they'll make contact, you'll need your password and credit card (last 4 digits) and that's it, I got a full refund. Very impressed.
Today, i am cancel my Bluehost account with the help of this post. But they did not send me any Email Validation Token. I think the change the process. They simply asking my last four characters of account password, and cancel my account.
I also learning the method of free calls to USA with this post.


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Off-Duty Racine Officer Fatally Shot While Trying to Intervene in Bar Robbery
Officer John Hetland, who was off duty at the time, was fatally shot at around 9:40 p.m. at Teezers Tavern
Published Jun 18, 2019 at 6:57 AM | Updated at 5:20 PM CDT on Jun 18, 2019
//www.nbcchicago.com/on-air/as-seen-on/racine-officer-killed-while-trying-to-intervene-in-bar-robbery--511481872.html
Police in Racine, Wisconsin, say one of their officers was killed while trying to stop an armed robbery at a local bar on Monday night. NBC 5's Christian Farr reports.
(Published Tuesday, June 18, 2019)
Police in Racine, Wisconsin, say one of their officers was killed while trying to stop an armed robbery at a local bar on Monday night.
Officer John Hetland, who was off duty at the time, was fatally shot at around 9:40 p.m. at Teezers Tavern, according to a statement from Racine Police Chief Arthel Howell.
Authorities said Hetland witnessed an armed robbery in progress at the bar, located at 1936 Lathrop Ave., and attempted to intervene when he was shot.
Multiple law enforcement agencies are searching for the gunman, who was masked when he entered the bar. One of the bar's owners posted on Facebook that the suspect fired his gun and fled out the back door after a struggle.
Top News: PR Governor Faces Protests Over Scandal, and More
Carlos Giusti/AP
Police officers could be seen embracing at the scene of the shooting. Hetland was a 24-year veteran of the Racine Police Department, Howell said, calling him a "trusted and highly respected member of the department" who had served in several positions, including as a field training officer and a member of the Greater Racine FBI Gang Task Force.
A memorial for the father of two continues to grow outside of the Racine Police Department, which had not experienced a loss of a police officer in the line of duty for more than four decades.
Frank Pistulka says he's known the fallen officer since he was a teenager and that becoming a cop was a life ambition of hits, always concerned about the community.
No one is in custody in connection with the shooting, according to police, who said the investigation remained ongoing.
Anyone with information on the shooting is asked to contact the Kenosha County Sheriff's Department at (262) 656-1234. Those who wish to stay anonymous can submit tips via Crime Stoppers at (262) 636-9330 or by texting a message beginning with "RACS" to 274637.
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