text:
\section{Introduction}

A language $L$ is prefix-closed
if $w\in L$ implies that every prefix of $w$ is in $L$.
It is known that a regular language is prefix-closed
if and only if it is accepted by a nondeterministic finite automaton (NFA)
with all states final \cite{krs09}.
In the minimal incomplete deterministic finite automaton (DFA)
for a prefix-closed language, all the states are final as well.

The authors of \cite{krs09} examined several questions
concerning NFAs with all states final.
They proved that the inequivalence problem for NFAs with all states final
is PSPACE-complete in the binary case,
but polynomially solvable in the unary case.
Next, they showed that minimizing a binary NFA with all states final is
PSPACE-hard, and that deciding whether a given NFA accepts a language that is not prefix-closed is PSPACE-complete, while the same problem for DFAs
can be solved in polynomial time.
The NFA-to-DFA conversion and complementation of NFAs with all states final
 have been also considered in \cite{krs09},
and the tight bound $2^n$ for the first problem, and the lower bound $2^{n-1}$
for the second one have been obtained.

The quotient complexity of prefix-closed languages
has been studied  in \cite{bjz14}.
The quotient of a language $L$ by the string $w$
is the set $L_w=\{x\mid wx\in L\}$.
The quotient complexity of a language $L$, $\kappa(L)$,
is the number of distinct quotients of $L$.
Quotient complexity  is defined for any language,
and it is finite if and only if the language is regular.
The quotient automaton of a regular language $L$
is the DFA $(\{L_w\mid w\in\Sigma^*\},\Sigma,\cdot,L_\varepsilon,F)$,
where $L_w\cdot a = L_{wa}$, and a quotient $L_w$ is final
if it contains the empty string.
The quotient automaton of $L$ is a minimal complete DFA for $L$,
so quotient complexity is the same as the state complexity of $L$
which is defined as the number of states in the minimal DFA for $L$.
In \cite{bjz14}, the tight bounds on the quotient complexity
of basic regular operation have been obtained,
and to prove upper bounds, the properties of quotients have been used
rather than automata constructions.

Automata with all states final
represent  systems,  for example,  production lines,
and their intersection or parallel composition
represents the composition of these systems \cite{ma13}.
A question that arises here is,
whether the complexity of intersection
of automata with all states final
is the same as in the general case of arbitrary DFAs or NFAs.
At the first glance, it seems that this complexity could be smaller.
Our first result shows that this is not the case.
We show that both incomplete and nondeterministic state complexity
of intersection on prefix-closed languages is given by the function $m n$,
which is the same as in the general case of regular languages.

In the deterministic case, to have all the states final,
we have to consider incomplete deterministic automata
because otherwise, the complete automaton
with all states final would accept
the language consisting of all the strings over an input alphabet.
Notice that the model of incomplete deterministic automata
has been considered already by Maslov \cite{ma70}.
The same model has been used in the study of the complexity
of the shuffle operation \cite{csy02};
here, the complexity on complete DFAs is not known yet.

We next study the complexity of complement, union,
concatenation, square, star, and reversal on languages
represented by incomplete DFAs or NFAs with all states final.
We get  tight bounds in both nondeterministic
and incomplete deterministic cases.
In the nondeterministic case, all the bounds are the same
as in the general case of regular languages,
except for the bound for star that is $n$ instead of  $n+1$.
However, to prove the tightness of these bounds,
we usually  use  larger alphabets
than in the general case of regular languages where all the upper bounds
can be  met by binary languages \cite{hk03,ji05}.

To get lower bounds, we use a fooling-set lower-bound method
\cite{auy83,bi92,bi93,gs96,hr97}.
In the case of union and reversal,
the method does not work since it provides a lower bound
on the size of NFAs with multiple initial states.
Since the nondeterministic state complexity of a regular language
is defined using a model of NFAs with a single initial state \cite{hk03},
we have to use a modified fooling-set technique to get the
tight bounds $m+n+1$ and $n+1$ for union and reversal,
respectively.

In the case of incomplete deterministic finite automata,
the tight bounds for complement, union,
concatenation,  star, and reversal are
$n+1, mn+m+n, m\cdot2^{n-1}+2^n-1$,
$2^{n-1}$, and
$2^n-1$,
respectively.
To define worst-case examples,
we use a binary alphabet for union, star, and reversal,
and a ternary alphabet for concatenation.

The paper is organized as follows.
In the next section, we give some basic definitions
and preliminary results.
In Sections~\ref{se:co} and \ref{se:bo}, we  study boolean operations.
Concatenation is discussed in Section~\ref{se:cs},
and star and reversal in Section~\ref{se:sr}.
The last section contains some concluding remarks.

\section{Preliminaries}

In this section, we recall some basic definitions
and preliminary results.
For details and all unexplained notions,
the reader may refer to \cite{si97}.

A \emph{nondeterministic finite automaton} (NFA)
is a quintuple $A=(Q,\Sigma,\delta,I,F)$,
where $Q$ is a finite set of states,
$\Sigma$ is a finite alphabet,
$\delta\colon Q\times\Sigma\to 2^Q$ is the transition function
which is extended to the domain $2^Q\times\Sigma^*$
in the natural way,
$I\subseteq Q$ is the set of initial states, and
$F\subseteq Q$ is the set of final states.
The language accepted by $A$ is the set
$L(A)=\{w\in\Sigma^*\mid \delta(I,w)\cap F \neq \emptyset\}$.

The \emph{nondeterministic state complexity}
 of a regular language $L$, $\mathrm{nsc}(L)$,
is the smallest number of states in any NFA
with a \emph{single initial state}  recognizing $L$.

An NFA $A$ is \emph{incomplete deterministic} (DFA)
if $|I|=1$ and $|\delta(q,a)|\le1$
for each $q$ in $Q$ and each $a$ in $\Sigma$.
In such a case, we write  $\delta(q,a)=q'$
instead of $\delta(q,a)=\{q'\}$.
A non-final state $q$ of a DFA is called a \emph{dead} state
if $\delta(q,a)=q$ for each symbol $a$ in $\Sigma$.

The \emph{incomplete state complexity} of a regular language $L$,
$\mathrm{isc}(L)$,
is the smallest number of states in any incomplete DFA recognizing $L$.
An incomplete DFA is minimal (with respect to the number of states)
if it does not have any dead state,
all its states are reachable,
and no two distinct states are equivalent.

Every NFA $A=(Q,\Sigma,\delta,I,F)$ can be converted
to an equivalent DFA $A'=(2^Q,\Sigma,\cdot,I,F')$,
where $R\cdot a =\delta(R,a)$ and $F'=\{R\in 2^Q\mid R\cap F\neq\emptyset\}$.
The DFA $A'$ is called
the \emph{subset automaton} of the NFA~$A$.
The subset automaton need not be minimal
since some of its states may be unreachable or equivalent.
However, if for each state $q$ of an NFA $A$,
there exists a string $w_q$ that is accepted by $A$
only from the state~$q$,
then the subset automaton of the NFA $A$
does not have equivalent states
since if two  subsets of the subset automaton
differ in a state $q$,
then they are distinguishable by $w_q$.

To prove the minimality of NFAs,
we use a fooling set lower-bound technique,
see~\cite{auy83,bi92,bi93,gs96,hr97}.

\begin{definition}\label{def:fool}
 A set of pairs of strings $\{(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)\}$
 is called a \emph{fooling set} for a language $L$ if
 for all $i,j$ in $\{1,2,\ldots,n\}$, the following two conditions hold:\\
 \hglue10pt
 \textbf{\emph{(F1)}} $x_iy_i\in L$, and \\
 \hglue10pt
 \textbf{\emph{(F2)}} if $i\neq j$, then  $x_iy_j\notin L$ or $x_jy_i \notin L$.
\end{definition}

It is well known that the size of a fooling set for a regular language
provides a lower bound on the number of states
in any NFA (with multiple initial states)
for the language.
The argument is simple.
Fix the accepting computations of any NFA on strings $x_i y_i$ and $x_j y_j$.
Then, the states on these computations reached after reading $x_i$ and $x_j$
must be distinct, otherwise the NFA
accepts both $x_i y_j$ and $x_j y_i$ for two distinct pairs.
Hence we get the following observation.

\begin{lemma}[\cite{bi93,gs96,hr97}]
\label{le:fool}
  Let $\mathcal{F}$ be a fooling set for a language $L$.
 Then every NFA (with multiple initial states) for the language $L$
 has at least $|\mathcal{F}|$ states.
\qed
\end{lemma}

The next lemma  shows that sometimes,
if we insist on having a single initial state in an NFA,
one more state is necessary.
It can be used in the case of union, reversal, cyclic shift \cite{jo08},
and AFA-to-NFA conversion \cite{ji12}.
In each of these cases,  NFAs
with a single initial state
require one more state than NFAs
with multiple initial states.
For the sake of completeness,
we recall the proof of the lemma here.

\begin{lemma}[\cite{jm11}]\label{le:fool1}
 Let $\mathcal{A}$ and $\mathcal{B}$ be sets of pairs of strings
 and let $u$ and $v$ be two strings such that
 $\mathcal{A}\cup\mathcal{B}$,
 $\mathcal{A}\cup\{(\varepsilon,u)\}$, and
 $\mathcal{B}\cup\{(\varepsilon,v)\}$ are fooling sets for a  language $L$.
 Then every NFA with a single initial state for the language $L$
 has at least $|\mathcal{A}|+|\mathcal{B}|+1$ states.
\end{lemma}

\begin{proof}
 Consider an NFA for a language $L$, and let
 $\mathcal{A}=\{(x_i,y_i)\mid i=1,2,\ldots,m\}$ and
 $\mathcal{B}=\{(x_{m+j},y_{m+j})\mid j=1,2,\ldots,n\}$.
 Since the strings $x_k y_k$ are in $L$,
 we fix an accepting computation of the NFA  on each string $x_k y_k$.
 Let $p_k$ be the state on this computation
 that is reached after reading $x_k$.
 Since $\mathcal{A}\cup\mathcal{B}$ is a fooling set for $L$,
 the states $p_1$, $p_2$,~\ldots, $p_{m+n}$ are pairwise distinct.
 Since $\mathcal{A}\cup\{(\varepsilon,u)\}$ is a fooling set,
 the  initial state is distinct from all the states $p_1$, $p_2$, \ldots, $p_m$.
 Since $\mathcal{B}\cup\{(\varepsilon,v)\}$ is a fooling set,
 the (single) initial state is also distinct
 from all the states $p_{m+1}$, $p_{m+2}$, \ldots, $p_{m+n}$.
 Thus the NFA has at least $m+n+1$ states.
\end{proof}

\begin{example}\label{ex:fooling}
  Let $K=(a^3)^*$ and $L=(b^3)^*$.
 Then $\mathrm{nsc}(K)=3$ and $\mathrm{nsc}(L)=3$,
 and the language $K\cup L$ is accepted
 by a 6-state NFA with two initial states.
 Therefore, we cannot expect that we will be able to find a fooling set
 for $K\cup L$ of size $7$.
 However, every NFA with a \emph{single} initial state for the language $K\cup L$
 requires at least $7$ states since Lemma~\ref{le:fool1}
 is satisfied for the language $K\cup L$
 with
 \begin{align*}
   \mathcal{A} &=\{(a,a^2),(a^2,a),(a^3,a^3)\},\\
   \mathcal{B} &=\{(b,b^2),(b^2,b),(b^3,b^3)\},\\
   u                &=b^3, \text{ and} \\
   v                &=a^3.
 \end{align*}
\end{example}

If $w=uv$ for strings $u$ and $v$, then $u$ is a \emph{prefix} of $w$.
A language $L$ is \emph{prefix-closed}
if $w\in L$ implies that every prefix of $w$ is in $L$.
The following observations are easy to prove.

\begin{proposition}[\cite{krs09}]
 A regular language is prefix-closed if and only if
 it is accepted by some NFA with all states final.
\qed
\end{proposition}

\begin{proposition}
 Let $A$ be a minimal incomplete DFA for a language $L$.
 Then the language $L$ is prefix-closed
 if and only if all the states of the DFA $A$ are final.
\qed
\end{proposition}

\section{Complementation}
\label{se:co}

If $L$ is a language over an alphabet $\Sigma$,
then the complement of $L$ is the language $L^c=\Sigma^*\setminus L$.
If $L$ is accepted by a minimal complete DFA $A$,
then we can get a minimal DFA for $L^c$
from the DFA $A$ by interchanging the final and non-final states.
In the case of incomplete DFAs, we first have to add a dead state,
that is, a non-final state which goes to itself on each input,
and let all the undefined transitions go to the dead state.
After that, we can interchange the final and non-final states
to get a (complete)  DFA for the complement.
This gives the following result.

\begin{theorem}
 Let $n\ge1$.
 Let $L$ be a prefix-closed regular language over an alphabet $\Sigma$
 with $\mathrm{isc}(L)=n$.
 Then $\mathrm{isc}(L^c)\le n+1$, and the bound is tight if $|\Sigma|\ge1$.
\end{theorem}

\begin{proof}
 For tightness, we can consider the unary prefix-closed language
 $\{a^i\mid 0\le i\le n-1\}$.
\end{proof}

If a language $L$ is represented by an $n$-state NFA,
then we first construct  the corresponding subset automaton,
and then interchange the final and non-final states to get a DFA for the language $L^c$ of at most $2^n$ states.
This upper bound on the nondeterministic state complexity of complement
on regular languages
is know to be tight in the binary case \cite{ji05}.

For prefix-closed languages, we get the same bound,
however, to prove tightness, we use a ternary alphabet.
Whether or not the bound $2^n$ can be met by a binary language
remains open.

\begin{theorem}
 Let $n\ge2$.
 Let $L$ be a prefix-closed regular language
 over an alphabet  $\Sigma$ with $\mathrm{nsc}(L)=n$.
 Then $\mathrm{nsc}(L^c)\le 2^n$,
 and the bound is tight if $|\Sigma|\ge3$.
\end{theorem}

\begin{proof}
 The upper bound is the same as
 in the general case of regular languages \cite{hk03}.
 To prove tightness,
 consider the language $L$
 accepted by the NFA $N$ shown in Figure~\ref{fig:complement},
 in which state $n$ goes to the empty set on both $a$ and $b$,
 and to $\{1\}$ on $c$.
 Each other state $i$ goes to $\{i+1\}$ on both $a$ and $c$, and
 to $\{1,i+1\}$ on~$b$.
 Our~aim is to describe a fooling set
 $\mathcal{F}=\{(x_S,y_S)\mid S \subseteq \{1,2,\ldots,n\}\}$
 of size $2^n$ for $L^c$.

 \begin{figure}[b]\label{-----fi1}
 \centerline{\includegraphics[scale=.40]{complement.eps}}
 \caption{The NFA $N$ of a prefix-closed language $L$ with $\mathrm{nsc}(L^c)=2^n$.}
 \label{fig:complement}
 \end{figure}

 First, let us show that each subset of $\{1,2,\ldots,n\}$
 is reachable in the subset automaton of the NFA $N$.
 The initial state is $\{1\}$,
 and each singleton set $\{i\}$ is reached from $\{1\}$
 by $a^{i-1}$.
 The empty set is reached from $\{n\}$ by $a$.
 The set $\{i_1,i_2,\ldots,i_k\}$ of size $k$,
 where  $2\le k\le n$ and $1\le i_1 < i_2 <\cdots < i_k\le n$,
 is reached from the set  $\{i_2-i_1,\ldots,i_k-i_1\}$ of size $k-1$
 by the string $b a^{i_1-1}$.
 This proves reachability by induction.
 Now, define $x_S$ as the string, by which the initial state $1$
 of the NFA $N$ goes to the set $S$.

 Next, for a subset $S$ of $\{1,2,\ldots,n\}$,  define the string $y_S$
 as the string $y_S=y_0y_1\cdots y_{n-1}$
 of length $n$, where
 $$
     y_i = \begin{cases}
             a, &\text{if $n-i\in S$,}\\
             c, &\text{if $n-i\notin S$.}
       \end{cases}
 $$
 We claim that the string $y_S$ is rejected by the NFA $N$
 from each state in $S$ and accepted from each state that is not in $S$.
 Indeed, if $i$ is a state in $S$,
 then $y_{n-i}=a$ and $y_S=u a v$ with $u=y_0y_1\cdots y_{n-i-1}$
 and $v=y_{n-i+1}y_{n-i+2}\cdots y_{n-1}$.
 Hence $|u|=n-i$,
 which means that the state $i$ goes to $\{n\}$ by $u$
 since both $a$ and $c$ move each state $q$ to state $q+1$.
 However, in state $n$ the NFA $N$ cannot read $a$,
 and therefore the string $y_S=uav$ is rejected from $i$.
 On the other hand,
 if $i\notin S$, then $y_{n-i}=c$,
 and the string $y_S=u c v$ with $|u|=n-i$ and $|v|=i-1$
 is accepted from $i$ through the computation
 $i\xrightarrow{u} n \xrightarrow{c} 1 \xrightarrow{v} i$.

 Now, we are ready to prove that the set of pairs of strings
 $ \mathcal{F} = \{(x_S,y_S) \mid S\subseteq\{1,2,\ldots,n\}\}$
 is a fooling set for the language $L^c$.

 (F1) By $x_S$, the initial state 1 goes to the set $S$.
 The string $y_S$ is rejected by $N$ from each state in $S$.
 It follows that the NFA $N$ rejects the string $x_S y_S$.
 Thus the string $x_S y_S$ is in $L^c$.

 (F2) Let $S\neq T$. Then without loss of generality,
 there is  a state $i$ such that $i\in S$ and $i\notin T$.
 By $x_S$, the initial state $1$ goes to $S$, so it also goes to the state $i$.
 Since $i\notin T$,
 the string $x_T$ is accepted by $N$ from $i$.
 Therefore, the NFA $N$ accepts the string $x_S y_T$,
 and so this string is not in $L^c$.

 Hence $\mathcal{F}$ is a fooling set for $L^c$ of size $2^n$.
 By Lemma~\ref{le:fool}, we have  $\mathrm{nsc}(L^c)\ge 2^n$.
\end{proof}

\section{Intersection and Union}
\label{se:bo}

In this section, we study the incomplete and nondeterministic state complexity
of intersection and union of prefix-closed languages.
If regular languages $K$ and $L$ are accepted
by $m$-state and $n$-state NFAs, respectively,
then the language $K\cap L$
is accepted by an NFA of at most $m n$ states,
and this bound is known to be tight in the binary case \cite{hk03}.
Our first result shows that the bound $m n$ can be met
by binary prefix-closed languages.
Then, using this result, we  get the same bound on
the incomplete state complexity of intersection on prefix-closed languages.

\begin{theorem}
\label{thm:int_nsc}
 Let $K$ and $L$ be prefix-closed languages
 over an alphabet $\Sigma$ with $\mathrm{nsc}(K)=m$ and \mbox{$\mathrm{nsc}(L)=~n$.}
 Then $\mathrm{nsc}(K\cap L)\le m n$,
 and the bound is tight if $|\Sigma|\ge2$.
\end{theorem}

\begin{proof}
 The upper bound is the same as for regular languages \cite{hk03}.
 For tightness, consider prefix-closed binary languages
 $K = \{w\in\{a,b\}^*\mid \#_a(w) \le m-1\}$  and
 $L = \{w\in\{a,b\}^*\mid \#_b(w) \le n-1\}$
 that are accepted by an $m$-state and an $n$-state
 incomplete DFAs $A$ and $B$, respectively, shown in Figure~\ref{fig:intersection}.

 \begin{figure}[b]\label{-----fi2}
 \centerline{\includegraphics[scale=.40]{intersection.eps}}
 \caption{The incomplete DFAs $A$ and $B$
              of prefix-closed languages $K$ and $L$ with $\mathrm{nsc}(K\cap L)=m n$.}
 \label{fig:intersection}
 \end{figure}

 Consider the set of pairs of strings
 $\mathcal{F}=\{(a^i b^j, a^{m-1-i} b^{n-1-j})
 \mid 0\le i \le m-1, 0 \le j \le n-1\}$
 of size $m n$.
 Let us show that $\mathcal{F}$
 is a fooling set for the language $K\cap L$.

 (F1) The string $a^i b^j \cdot a^{m-1-i} b^{n-1-j}$
 has exactly $m-1$ $a$'s and $n-1$ $b$'s.
 It follows that it is in $K\cap L$.

 (F2) Let $(i,j)\neq(k,\ell)$.
 If $i<k$, then the string $a^k b^{\ell}\cdot  a^{m-1-i} b^{n-1-j}$
 contains $m-1+(k-i)$ $a$'s,
 and therefore it is not in $K$.
 The case of $j<\ell$ is symmetric.

 Hence $\mathcal{F}$ is a fooling set for $K\cap L$,
 and the theorem follows.
\end{proof}

\begin{theorem}
 Let $K$ and $L$ be prefix-closed languages
 over an alphabet $\Sigma$ with $\mathrm{isc}(K)=m$ and \mbox{$\mathrm{isc}(L)=~n$.}
 Then $\mathrm{isc}(K\cap L)\le m n$,
 and the bound is tight if $|\Sigma|\ge2$.
\end{theorem}

\begin{proof}
 Let $A=(Q_A,\Sigma,\delta_A,s_A,Q_A)$ and
 $B=(Q_B,\Sigma,\delta_B,s_B,Q_B)$
 be incomplete DFAs for  $K$ and $L$, respectively.
 Define an incomplete product automaton
 $M=(Q_A\times Q_B, \Sigma, \delta, (s_A,s_B), Q_A\times Q_B)$,
 where
 $$
     \delta((p,q),a) = \begin{cases}
            (\delta_A(p,a),\delta_B(q,a)),
                 & \text{if both $\delta_A(p,a)$  and $\delta_B(q,a)$ are defined,}\\
          \text{undefined},  &\text{otherwise}.
   \end{cases}
 $$
 The  DFA $M$ accepts the language $K\cap L$.
 This gives the upper bound $m n$.
 For tightness, consider the same languages $K$ and $L$
 as in the proof of the previous theorem.
 Notice that $K$ and $L$ are accepted by  $m$-state and $n$-state
 incomplete DFAs, respectively.
 We have shown that nondeterministic state complexity of their intersection
 is $m n$. It follows that the incomplete state complexity is also at least $m n$.
\end{proof}

Our next result
on the incomplete state complexity
of union on prefix-closed languages
can be derived from the result on the quotient complexity of union in \cite{bjz14}.
For the sake of completeness, we restate it in terms of incomplete complexities,
and recall the proof.

\begin{theorem}
  Let $K$ and $L$ be prefix-closed languages
 over an alphabet $\Sigma$ with $\mathrm{isc}(K)=m$ and \mbox{$\mathrm{isc}(L)=~n$.}
 Then $\mathrm{isc}(K\cup L)\le m n + m +n$,
 and the bound is tight if $|\Sigma|\ge2$.
\end{theorem}

\begin{proof}
 Let $A=(\{0,1,\ldots,m-1\},\Sigma,\delta_A,0,F_A)$ and
 $B=(\{0,1,\ldots,n-1\},\Sigma,\delta_B,0,F_B)$
 be incomplete DFAs for the languages $K$ and $L$, respectively.
 To construct a DFA for the language $K\cup L$,
 we first add the dead states $m$ and $n$ to the DFAs $A$ and $B$,
 and let go all the undefined transitions to the dead states.
 Now we construct the classic product-automaton
 from the resulting complete DFAs
 with the state set
 $\{0,1,\ldots,m\}\times\{0,1,\ldots,n\}$.
 All its states are final, except for the state $(m,n)$
 that is dead, and we do not count it.
 Hence we get the upper bound $m n + m +n$
 on the incomplete state complexity of union.

 \begin{figure}[h!]\label{-----fi3}
  \centerline{\includegraphics[scale=.40]{union_product.eps}}
  \caption{The product automaton for incomplete DFAs $A$ and $B$
        from Figure~\ref{fig:intersection}; $m=3$ and $n=4$.}
  \label{fig:union_product}
  \end{figure}

 For tightness, we again consider the languages
 described in the proof of Theorem~\ref{thm:int_nsc}.
 We add the dead states $m$ and $n$ and construct the product automaton.
 The product automaton in the case of $m=3$ and $n=4$
 is shown in Figure~\ref{fig:union_product}.

 Each state $(i,j)$ of the product automaton is reached
 from the initial state $(0,0)$ by the string $a^ i b^j$.
 Let $(i,j)$ and $(k,\ell)$ be two distinct states of the product automaton.
 If $i<k$, then the string $a^{m-k} b^n$ is rejected from  $(k,\ell)$
 and accepted from $(i,j)$.
 If $j<\ell$, then the string $b^{n-\ell} a^m$
 is rejected from $(k,\ell)$
 and accepted from $(i,j)$.
 Thus all the states in the product-automaton
 are reachable and pairwise distinguishable,
 and the lower bound $mn+m+n$ follows.
\end{proof}

In the nondeterministic case,
the upper bound for union on regular language is $m+n+1$,
and it is tight in the binary case \cite{hk03}.
We get the same bound for union on prefix-closed languages,
however, to define witness languages,
we use a four-letter alphabet.

\begin{theorem}
 Let $K$ and $L$ be prefix-closed languages
 over an alphabet $\Sigma$ with $\mathrm{nsc}(K)=m$ and \mbox{$\mathrm{nsc}(L)=n$.}
 Then $\mathrm{nsc}(K\cup L)\le m+ n+1$,
 and the bound is tight if $|\Sigma|\ge4$.
\end{theorem}

\begin{proof}
 \begin{figure}[t]\label{-----fi4}
 \centerline{\includegraphics[scale=.40]{union.eps}}
 \caption{The NFAs $A$ and $B$ of prefix-closed languages $K$ and $L$ with
       $\mathrm{nsc}(K\cup L)=m+n+1$.}
 \label{fig:union}
 \end{figure}
 The upper bound is the same as for regular languages \cite{hk03}.
 To prove tightness, let $K$ and $L$ be the prefix-closed languages
 accepted by the NFAs $A$ and $B$, respectively,
 shown in Figure~\ref{fig:union}.
 Let
 \begin{align*}
   &\mathcal{A}= \{(a^i , a^{m-1-i} b)
              \mid i=1,2,\ldots,m-1\}\cup\{(a^{m-1}b,a)\},\\
   &  \mathcal{B}= \{(c^j , c^{n-1-j} d)
              \mid j=1,2,\ldots,n-1\}\cup\{(c^{n-1}d,c)\}.
 \end{align*}
 Let us show that $\mathcal{A} \cup \mathcal{B}$
 is a fooling set for the language $K\cup L$.

 (F1) We have $a^i \cdot a^{m-1-i} b = a^{m-1}b$
 and $c^j \cdot c^{n-1-j} d = c^{n-1} d$. Both these strings are in $K\cup L$.
 The strings $a^{m-1} b\cdot a$
 and $c^{n-1} d \cdot c$ are in $K\cup L$ as well.

 (F2) If $1\le i < i' \le m-1$, then the string $a^i \cdot a^{m-1-i'} b$
 is not in $K$ since $ m-1-(i'-i)<m-1$.
 Next, if $1 \le i\le m-1$, then $a^{m-1} b\cdot a^{m-1-i} b$  is not in $K$.
 The argumentation for two pairs from $\mathcal{B}$
 is similar.
 If~we concatenate the first part of a pair in  $\mathcal{A}$
 with the second part of a pair in $\mathcal{B}$,
 then we get a string that either contains all three symbols $a,c,d$,
 or contains both symbols $a$ and $d$.
 No such string is in $K\cup L$.

 Thus $\mathcal{A} \cup \mathcal{B}$
 is a fooling set for the language $K\cup L$.
 Moreover, the sets $\mathcal{A}\cup \{(\varepsilon,c)\}$
 and  $\mathcal{B}\cup \{(\varepsilon,a)\}$
 are fooling sets for $K\cup L$ as well.
 By Lemma~\ref{le:fool1},
 we have $\mathrm{nsc}(K\cup L)\ge m+n+1$.
\end{proof}

\section{Concatenation}
\label{se:cs}

In this section, we deal with the concatenation operation
on prefix-closed languages.
We start with incomplete state complexity.
We use a slightly different ternary witness language than in \cite{bjz14},
and prove the upper bound using automata constructions.

\begin{theorem}
 Let $m,n\ge3$.
 Let $K$ and $L$ be  prefix-closed languages
 over an alphabet $\Sigma$ with $\mathrm{isc}(K)=m$ and $\mathrm{isc}(L)=n$.
 Then $\mathrm{isc}(KL)\le m\cdot 2^{n-1}+2^n-1$,
 and the bound is tight if $|\Sigma|\ge3$.
\end{theorem}

\begin{proof}
 Let
 $A=(Q_A,\Sigma,\delta_A,s_A,Q_A)$ and
 $B=(Q_B,\Sigma,\delta_B,s_B,Q_B)$ be incomplete DFAs
 with all states final
 accepting the languages $K$ and $L$,
 respectively.
 Construct an NFA $N$ for the language $KL$
 from the DFAs $A$ and $B$
 by adding the transition on a symbol $a$
 from a state $q$ in $Q_A$ to the initial state $s_B$ of $B$
 whenever the transition on $a$ in state $q$ is defined in $A$.
 The initial states of the NFA $N$ are $s_A$ and $s_B$,
 and the set of final states is $Q_B$.
 Each reachable subset of the subset automaton
 of the NFA $N$ contains at most one state of $Q_A$,
 and several states of $Q_B$.
 Moreover, if a state of $Q_A$ is in a reachable subset $S$,
 then $S$ must contain the state $s_B$.
 This gives the upper bound $m\cdot 2^{n-1}+2^n-1$
 on $\mathrm{isc}(KL)$ since the empty set is not counted.

 For tightness,
 consider the prefix-closed languages $K$ and $L$
 accepted by incomplete DFAs  $A$ and $B$, respectively,
 shown in Figure~\ref{fig:concatenation},
 in which the transitions are as follows:

 on $a$, state $q_0$ goes to itself,
 and each state $j$ goes to $(j+1)\bmod n$;

 on $b$, each state $q_i$ goes to state $q_0$, state $0$ goes to itself,
 and state $j$ with $1\le j\le n-2$ goes to $j+1$;

 on $c$, each state $q_i$ with $0 \le i \le m-2$ goes to $q_{i+1}$,
 and each state $j$ goes to itself;\\
\noindent
 and all the remaining transitions are undefined.

 Construct an NFA $N$ for the language $KL$ as described above.
 Let us show that the subset automaton of the NFA $N$
 has $m\cdot 2^{n-1}+2^n-1$
 reachable and pairwise distinguishable non-empty subsets.

 \begin{figure}[t]\label{-----fi5}
  \centerline{\includegraphics[scale=.40]{concatenation.eps}}
  \caption{The incomplete DFAs $A$ and $B$ of languages $K$ and $L$
      with $\mathrm{isc}(KL)=m\cdot2^{n-1}+2^n-1$.}
  \label{fig:concatenation}
  \end{figure}

 (1) First, let us show that each set $\{q_0\}\cup S$ is reachable, where
 $S\subseteq\{0,1,\ldots,n-1\}$ and $0\in S$.
 The proof is by induction on the size of subsets.
 The set $\{q_0,0\}$ is the initial subset.
 The set $\{q_0,0, j_1,j_2,\ldots,j_k\}$ with
 $1\le j_1<j_2<\cdots  <j_k\le n-1$ is reached from the set
 $\{q_0,0,j_2-j_1,\ldots,j_k-j_1\}$ by the string $a b^{j_1-1}$,
 and the latter set is reachable by induction.

 (2) Now, let us show that each set $\{q_i\}\cup S$,
 is reachable, where
 $1\le i \le m-1$, $S\subseteq\{0,1,\ldots,n-1\}$ and $0\in S$.
 The set $\{q_i\}\cup S$ is reached from $\{q_0\}\cup S$ by $c^i$,
 and the latter set is reachable as shown in  (1).

 (3) Next, we show that each set $S$ with
 $S\subseteq\{0,1,\ldots,n-1\}$ and $0\in S$ is reachable.
 The set $S$ is reached from $\{q_{m-1}\}\cup S$ by $c$,
 and the latter set is reachable as shown in case (2).

 (4) Finally, we show that each non-empty set $S$ with
 $S\subseteq\{0,1,\ldots,n-1\}$ and $0\notin S$ is reachable.
 If $S=\{j_1,j_2,\ldots,j_k\}$ with $j_1\ge1$,
 then $S$ is reached from the set $\{0,j_2-j_1,\ldots,j_k-j_1\}$ by $a^{j_1}$,
 and the latter set is reachable as shown in case (3).

 This proves the  reachability of $m\cdot 2^{n-1}+2^n-1$ non-empty subsets.

 To prove distinguishability,
 notice that the string $b^n$ is accepted by the DFA $B$
 only from the state 0,
 and the string $a^{n-1-i}a b^n$ is accepted only from the state $i$
 ($1\le i\le n-1$).
 If $S$ and $T$ are two distinct subsets of $\{0,1,\ldots,n-1\}$,
 then $S$ and $T$
 differ in a state $i$.
 If $i=0$, then $b^n$ distinguishes $S$ and $T$,
 and if $i\ge 1$, then $a^{n-i}b^n$ distinguishes $S$ and $T$.

 Next, the sets $\{q_i\}\cup S$ and $\{q_i\}\cup T$,
 where $S$ and $T$ are distinct subsets of $\{0,1,\ldots,n-1\}$,
 go to $S$ and $T$, respectively, by $c^m$.
 Since  $S$ and $T$ are distinguishable,
 the sets $\{q_i\}\cup S$ and $\{q_i\}\cup T$
 are distinguishable as well.

 Finally, notice that the string $b^n a b^n$
 is accepted by the NFA $N$ from each state $q_i$,
 but rejected from each state $i$ in $\{0,1,\ldots,n-1\}$.
 Hence the sets $\{q_i\}\cup S$ and $T$,
 where $S$ and $T$ are subsets of $\{0,\ldots,n-1\}$,
 are distinguishable.
 Now let $0\le i <j\le m-1$.
 Then $\{q_i\}\cup S$ and $\{q_j\}\cup T$
 go  to
 $\{q_{i+m-j}\}\cup S$ and $T$, respectively,  by $c^{m-j}$.
 Since  $\{q_{i+m-j}\}\cup S$ and $T$ are distinguishable,
 the sets $\{q_i\}\cup S$ and $\{q_j\}\cup T$
 are distinguishable as well.
 This proves the distinguishability of all the reachable subsets,
 and completes the proof.
\end{proof}

In the next theorem, we consider the nondeterministic case.
For regular languages, the upper bound
on the nondeterministic state complexity of concatenation
is $m+n$, and it is tight in the binary case \cite{hk03}.
For prefix-closed languages,
we get the same bound for concatenation.
However, we define witness languages over a ternary alphabet.

\begin{theorem}
 Let $m,n\ge3$.
 Let $K$ and $L$ be  prefix-closed languages
 over an alphabet $\Sigma$ with $\mathrm{nsc}(K)=m$ and $\mathrm{nsc}(L)=n$.
 Then  $\mathrm{nsc}(KL)\le m+n$,
 and the bound is tight if $|\Sigma|\ge3$.
\end{theorem}

\begin{proof}
 The upper bound is the same as for regular languages \cite{hk03}.
 For tightness, consider the ternary prefix-closed languages $K$ and $L$
 accepted by incomplete DFAs $A$ and $B$, respectively,
 shown in  Figure~\ref{fig:productNSC}.
 Notice that if a  string $w$ is  in $KL$, then $w$ is in the language
 $b^* a^* c^* b^* a^* c^*$, and the number of $a$'s in $w$
 is at most $(n+m-2)$.

 \begin{figure}[t]\label{-----fi6}
 \centerline{\includegraphics[scale=.40]{product_nsc.eps}}
 \caption{The incomplete DFAs of prefix-closed languages $K$ and $L$
              with $\mathrm{nsc}(KL)=m+n$.}
 \label{fig:productNSC}
 \end{figure}

 For $i=0,1,\ldots,m+n-1$, define the pair $(x_i, y_i)$ as follows:
 \begin{align*}
      (x_i, y_i) &= ( a^i, a^{m-1-i} c b a^{n-1} ),
     \quad\text{for $i=0,1,\ldots m-1$,}\\
    (x_{m+j}, y_{m+j} ) &= ( a^{m-1} c b a^j , a^{n-1-j} ),
                      \quad \text{for $j=0,1,\ldots n-1$.}
 \end{align*}
 Let us show that the set of pairs
 $\mathcal{F} = \{ (x_i, y_i)\mid i=0,1,\ldots,m+n-1\}$
 is a fooling set for the language~$KL$.

 (F1) For each $i$, we have $x_i y_i = a^{m-1} c b a^{n-1}$.
 Thus $x_i y_i$ is in $KL$ since $a^{m-1} c $ is in $K$ and $b a^{n-1}$ is in $L$.

 (F2) Let $i<j$ and $(i,j)\neq(m-1,m)$.
 Then the number of $a$'s in the string $x_j y_i$ is greater than  $m+n-2$,
 and therefore the string $x_j y_i$ is not in $KL$.
 If  $(i,j)=(m-1,m)$, then $x_m y_{m-1} = a^{m-1} c b c b a^{n-1}$.
 Thus $x_m y_{m-1}$ is not in $b^* a^* c^* b^* a^* c^*$,
 and therefore it is not in $KL$.

 Hence the set $\mathcal{F}$ is a fooling set for the language $KL$,
 so $\mathrm{nsc}(KL)\ge m+n$.
\end{proof}

\section{Star and Reversal}
\label{se:sr}

We conclude our paper with the star and reversal operation
on prefix-closed languages.
The star of a language $L$ is the language $L^*=\bigcup_{i\ge0} L^i$,
where $L^0=\{\varepsilon\}$ and $L^{i+1}=L^i\cdot L$.

If a regular language  $L$ is accepted by a complete $n$-state DFA,
then the language $L^*$ is accepted by a DFA of at most
$3/4\cdot 2^n$ states, and the bound is tight in the binary case
\cite{ma70,yzs94}.

For prefix-closed languages,
the upper bound on the quotient complexity for star is $2^{n-2}+1$,
and it has been shown to be tight in the ternary case \cite{bjz14}.
In the case of incomplete state complexity,
we get the bound $2^{n-1}$.
For the sake of completeness, we give a simple proof of the upper bound
using automata constructions.
Moreover, we are able to define a witness language over a binary alphabet.

\begin{theorem}
 Let $n\ge4$.
 Let $L$ be a prefix-closed regular language
 over an alphabet $\Sigma$ with $\mathrm{isc}(L)=n$.
 Then $\mathrm{isc}(L^*)\le2^{n-1}$,
 and the bound is tight if $|\Sigma|\ge2$.
\end{theorem}

\begin{proof}
  Let $A=(Q,\Sigma,\cdot,s,Q)$
 be an incomplete DFA for $L$.
 Construct an NFA $A^*$ for $L^*$ from the DFA $A$
 by adding the transition on a symbol $a$
 from a state $q$ to the initial state $s$
 whenever the transition $q\cdot a$ is defined.
 In the subset automaton of the NFA $A^*$,
 each reachable set is either empty,
 or it contains the initial state $s$.
 It follows that $\mathrm{isc}(L^*)\le2^{n-1}$.

 For tightness, consider
 the binary incomplete DFA
 with the state set $\{1,2,\ldots,n\}$,
 the initial state $1$ and with all states final.
 The transitions are as follows.
 By $a$, the transitions in states 1 and 2 are undefined,
 each odd state $i$ with $3\le i \le n-1$ goes to $i+1$, and
 each even state $i$ with $3\le i \le n-1$ goes to $i-1$.
 By $b$, there is a cycle $(1,2,3)$,
 each odd state $i$ with $4\le i \le n-1$ goes to $i-1$, and
 each even state $i$ with $4\le i \le n-1$ goes to $i+1$.
 If $n$ is odd, then $n$ goes to itself by $a$,
 otherwise it goes to itself by~$b$.
 The DFA for $n=6$ is shown in Figure~\ref{fig:star}.

 \begin{figure}[t]\label{-----fi7}
  \centerline{\includegraphics[scale=.40]{star.eps}}
  \caption{The incomplete DFA $A$ of a prefix-closed language $L$
           with $\mathrm{isc}(L^*)=2^{n-1}$; $n=6$.}
  \label{fig:star}
  \end{figure}

 Notice that each state $i$ with $3\le i \le n$
 has exactly one in-transition on $a$ and on $b$.
 Denote by $a^{-1}(i)$ the state that goes to $i$ on $a$,
 and by $b^{-1}(i)$ the state that goes to $i$ on $b$.

 Construct an NFA $A^*$ as described above.
 Let us show that in the subset automaton of the NFA $A^*$,
 all subsets of $\{1,2,\ldots,n\}$ containing state $1$
 are reachable and pairwise distinguishable.

 We prove reachability by induction on the size of subsets.
 The basis is $|S|=1$,
 and the set $\{1\}$ is reachable since it is the initial state of
 the subset automaton.
Assume that every set $S$ containing $1$
with $|S| = k,$ where  $1\leqslant{k}\leqslant{n-1},$ is reachable.
Let $S=\{1,i_1,i_2,i_3,\ldots,i_k\},$ where $2\leqslant{i_1}<{i_2}<{\cdots}<{i_k}\leqslant{n},$
be a set of size $k+1$.
Consider three cases:
\begin{enumerate}
     \item [(\textit{i})]$ i_{1}= 2.$
Take $S^{'} = \{1,b^{-1}(i_2),b^{-1}(i_3),\ldots,b^{-1}(i_k)\} $.
Then $|S^{'}| = k $, and therefore $S^{'}$ is reachable by the induction hypothesis.
Since we have
 $
          S'\xrightarrow{b}\{1,2, i_2,\ldots,i_k\} = S,
$
 the set $S$ is reachable.

 \item[(\textit{ii})]$ i_{1}= 3.$
Take $S'=\{1,2,b^{-1}(i_2),b^{-1}(i_3),\ldots,b^{-1}(i_k)\}$.
Then $|S'|=k+1$ and $S'$ contains states $1$ and $2$. Therefore, the set $S'$ is reachable as shown in case $(i)$.
Since we have
$
     S^{'}\xrightarrow{b}\{1,2,3,i_2,i_3,\ldots,i_k\}
              \xrightarrow{aa}\{1,3,i_2,i_3,\ldots,i_k\}=S,
$
the set $S$ is reachable.

 \item[(\textit{iii})] Let $i_1=j\ge3$,
 and assume that each set $\{1,j,i_2,\ldots,i_k\}$ is reachable.
 Let us show that then also each set $\{1,j+1,i_2,\ldots,i_k\}$
 is reachable.
 If $j$ is odd, then the set
 $\{1,j+1,i_2,\ldots,i_k\}$ is reached from the set
 $\{1,j,a^{-1}(i_2),a^{-1}(i_3),\ldots,a^{-1}(i_k)\} $ by $a$.
 If $j$ is even, then  the set
 $\{1,j+1,i_2,\ldots,i_k\}$ is reached from the set
 $\{1,j,b^{-1}(i_2),b^{-1}(i_3),\ldots,b^{-1}(i_k)\} $ by $b a a$.
\end{enumerate}

This proves reachability.
To prove distinguishability,
 notice that the string $(ab)^{n-2}$
 is accepted by the NFA $A^*$ from state $3$ since
 state $3$ goes to the initial state $1$ by $(ab)^{n-2}$
 through the computation
  $$3\xrightarrow{ab}5\xrightarrow{ab}7\xrightarrow{ab}
   \cdots\xrightarrow{ab} n \xrightarrow{a} n\xrightarrow{b} n-1
\xrightarrow{ab} n-3\xrightarrow{ab}
   \cdots\xrightarrow{ab} 4\xrightarrow{ab}1 $$
if $n$ is odd, and through a similar computation if $n$ is even.
 On the other hand, the  string $(ab)^{n-2}$
cannot be read from any  other state $2i$ with $2\le i\le n/2$
 since we have
$$
    2i \xrightarrow{ab} \{2i-2,1,2\}
        \xrightarrow{ab} \{2i-4,1,2\}
 \xrightarrow{ab} \cdots  \xrightarrow{ab}\{4,1,2\} \xrightarrow{a}\{3,1\}
 \xrightarrow{b}\{1,2\} \xrightarrow{ab} \emptyset,
$$
 thus $2i$ goes to the empty set by $(ab)^i$, so also by $(ab)^{n-2}$.
 If $n$ is odd, then we have
$$
      2i+1  \xrightarrow{ab} \{2i+3,1,2\}
 \xrightarrow{ab} \{2i+5,1,2\}
 \xrightarrow{ab} \cdots  \xrightarrow{ab} \{n,1,2\}
\xrightarrow{a} \{n,1\} \xrightarrow{b}\{n-1,1,2\}
\xrightarrow{ab}$$ $$  \{n-3,1,2\}\xrightarrow{ab} \cdots \xrightarrow{ab}
\{2i,1,2\} \xrightarrow{(ab)^i} \emptyset,
$$
thus $2i+1$ goes to the empty set by $(ab)^{n-i}$, $i\ge2$,
and so also by $(ab)^{n-2}$.
For $n$ even, the argument is similar.
The string $(ab)^{n-2}$ is not accepted from states 1 and 2.
Hence the NFA $A^*$ accepts the string $(ab)^{n-2}$
only from the state 3.
 Since there is exactly one in-transition on $b$ in state $3$,
 and it goes from state $2$,
 the string $b (ab)^{n-2}$
 is accepted by $A^*$ only from state $2$.
 Similarly, the string $ b b (ab)^{n-2}$
 is accepted
 by $A^*$ only from state $1$.
 Next, for similar reasons,
 the string $a (ab)^{n-2}$ is accepted only from $4$,
 the string $ b a (ab)^{n-2}$ is accepted only from $5$,
 and in the general case, the string
 $(ab)^i a (ab)^{n-2}$ is accepted only from $4+2i$  ($i\ge0$),
 and the string
 $(ba)^i (ab)^{n-2}$ is accepted only from $3+2i$  ($i\ge1$).
 Hence for each state $q$ of the NFA $A^*$,
 there exists a string $w_q$ that is accepted by $A^*$
 only from the state $q$.
 It follows that all the subsets of the
 subset automaton of the NFA $A^*$
 are pairwise distinguishable
 since two distinct subsets differ in a state $q$,
 and the string $w_q$ distinguishes the two subsets.
 This completes the proof.
\end{proof}

We did some computations in the binary case.
Having the files of $n$-state minimal binary pairwise non-isomorphic complete DFAs with a dead state and all the remaining states final,
we computed the state complexity of the star of languages accepted
by DFAs on the lists;
here the state complexity of a regular language $L$, $\sc(L)$,
is defined as the smallest number of states
in any \emph{complete} DFA for the language~$L$.
We computed the frequencies of the resulting complexities,
and the average complexity of star.
Our results are summarized in Table~\ref{ta}.
Notice that for $n=3,4,5$,
there is just one language with $\sc(L)=n$ and $\sc(L^*)=2$.
Let us show that this holds for every $n$ with $n\ge3$.

\begin{table}[b]
\begin{center}
\setlength{\extrarowheight}{4pt}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n \backslash \sc(L^*) $      & \ \ \ 1\  \ \ &\ \ \  2\  \ \   &\  \ \   3 \ \ \
 &\ \ \    4 \ \ \   &  \ \ \  5  \ \ \  &  \ \ \  6  \ \ \  & \ \ \   7  \ \ \
&\ \ \ 8 \ \ \ &\ \ \ 9 \ \ \ & average\\
\hline
2       &-  &2 & - & -  & - & -  & -  &-&-&  2\\
\hline
3         & 8  & 1  & 6 & -  & - & -  & -  &  -&-&  1.866\\
\hline
4        &161 &1 &48& 30& 6& -  & -  &  -&-&  1.857\\
\hline
5        &4177 &1&771& 275& 350&84&84& -&26&   1.849\\
\hline
\end{tabular}
\vskip10pt
\caption{The frequencies of the complexities
     and the average complexity of star
 on prefix-closed languages in the binary case; $n=2,3,4,5.$}
\label{ta}
\end{center}
\end{table}

\begin{proposition}
 Let $n\ge3$.
 There exists exactly one (up to renaming of alphabet symbols)
 binary prefix-closed regular language $L$
 with $\sc(L)=~n$ and $\sc(L^*)=2$.
\end{proposition}

\begin{proof}
 Let $A=(\{0,1\}, \{a,b\}, \delta, 0, F)$
 be a minimal two-state DFA for the language $L^*$.
 Since $L$ is prefix-closed,
 the language $L^*$ is prefix-closed as well.
 It follows that state 0 is final, and state 1 is dead,
 thus $F=\{0\}$ and $\delta(1,a)=\delta(1,b)=1$.

 Without loss of generality,
 state 1 is reached from the initial state 0 by $a$,
 thus $\delta(0,a) =1$.

 Since $n\ge3$, the language $L$ contains a non-empty string.
 This means that the language $L^*$  contains a non-empty string as well.
 Therefore, we must have $\delta(0,b)=0$,
 and so $L^*=b^*$.

 \begin{figure}[t]\label{-----fi8}
 \centerline{\includegraphics[scale=.40]{allFinal2.eps}}
 \caption{The only binary $n$-state complete DFA of
              a prefix-closed language $L$ with $\sc(L^*)=2$.}
 \label{fig:allFinal2}
 \end{figure}

 Now let $B$ be the minimal $n$-state  DFA for $L$.
 Then all the states of $B$ are final, except for  the dead state.
 Since $L^*=b^*$,
 no $a$ may occur in any string of $L$.
 Hence each non-dead state of $B$
 must go to the dead state on $a$.
 Since all states must be reachable,
 we must have a path labeled by $b^{n-2}$ and
 going through all the final states.
 The last final state must go to the dead state on $b$
 because otherwise all  final states
 would be equivalent.
 The resulting $n$-state DFA $B$
 is shown in Figure~\ref{fig:allFinal2}.
\end{proof}

The reverse $w^R$ of a string $w$ is defined by $\varepsilon^R=\varepsilon$,
and $(w a)^R= a w^R$ for $a$ in $\Sigma$ and $w$ in $\Sigma^*$.
The reverse of a language $L$ is the language
$L^R=\{w^R\mid w\in L\}$.
If a regular language $L$ is accepted by a complete $n$-state DFA,
then the language $L^R$ is accepted by  a complete DFA of at most $2^n$
states \cite{mi66,yzs94},
and the bound is tight in the binary case \cite{js12,le81}.

For prefix-closed languages, the quotient complexity of reversal
is $2^{n-1}$  \cite{bjz14}, and it
follows from the results on ideal languages \cite{bjl13}
since reversal commutes with complementation,
and the complement of a prefix-closed language is a right ideal;
here a language $L$ is a right ideal if $L=L\cdot \Sigma^*$.

We restate the result for reversal in terms of incomplete state complexity,
and prove tightness using a slightly different witness language.

\begin{theorem}
 Let $n\ge2$.
 Let $L$ be a prefix-closed regular language
 over an alphabet $\Sigma$ with $\mathrm{isc}(L)=n$.
 Then $\mathrm{isc}(L^R)\le2^{n}-1$,
 and the bound is tight if $|\Sigma|\ge2$.
\end{theorem}

\begin{proof}
 Let $A$ be an incomplete DFA for $L$.
 Construct an NFA $A^R$ for the language $L^R$
 from the DFA $A$
 by swapping the role of the initial and final states,
 and by reversing all the transitions.
 The subset automaton of the NFA $A^R$
 has at most $2^n-1$ non-empty reachable states,
 and the upper bound follows.

 For tightness,
 consider the incomplete DFA $A$ with all states final,
 shown in Figure~\ref{fig:reversal}.
 Construct an NFA $A^R$ as described above.
 In the subset automaton of the NFA $A^R$,
 the initial state is $\{1,2,\ldots,n\}$.
 If $S$ is a subset and if $i\in S$,
 then the subset $S\setminus\{i\}$
 is reached from $S$ by $a^ i b a^{n-i}$.
 This proves the reachability of all non-empty
 subset by odd induction.
 Since the states of the subset automaton of
 any reversed DFA are pairwise distinguishable \cite{ckp,js12,mi66},
 the theorem follows.

 \begin{figure}[t]\label{-----fi9}
 \centerline{\includegraphics[scale=.35]{reversal.eps}}
 \caption{The incomplete DFA $A$ of a  language $L$
               with $\mathrm{isc}(L^R)=2^n-1$, and the NFA $A^R$. }
 \label{fig:reversal}
 \end{figure}
 \end{proof}

Now, let us turn to the nondeterministic case.
For regular languages, the tight bound for both star and reversal is
$n+1$. It is met by a unary language for star \cite{hk03},
and by a binary language for reversal \cite{ji05}.

For prefix-closed languages, we get the same bound for reversal.
However, for star, the upper bound is $n$
since every prefix-closed language contains the empty string,
and there is no need to add a new initial state in the
 construction of an NFA for star.
In the following theorem,
we show that both these bounds are tight in the binary case.

\begin{theorem}
 Let $n\ge2$.
 Let $L$ be a prefix-closed language
 over an alphabet $\Sigma$ with $\mathrm{nsc}(L)=n$.
 Then \\
 \hglue10pt (1) $\mathrm{nsc}(L^*)\le n$,\\
 \hglue10pt (2) $\mathrm{nsc}(L^R)\le n+1$,\\
 and both bounds are tight if $|\Sigma|\ge2$.
\end{theorem}

\begin{proof}
 (1) Let $N=(Q,\Sigma,\delta,s, F)$ be an $n$-state NFA for $L$.
 Since $L$ is prefix-closed, the empty string is in~$L$.
 Therefore, we can get an $n$-state NFA for the language $L^*$
 from the NFA $N$  as follows:
 for each state $q$ and each symbol $a$ such that
 $\delta(q,a)\cap F\neq \emptyset$,
 we add a transition on $a$ from $q$ to the initial state $s$.
 Thus the upper bound is $n$.

 For tightness, consider the prefix-closed language $L$
 accepted by the NFA $A$ shown in Figure~\ref{fig:star_nsc}.
 Consider the set of pair of strings
 $\mathcal{F}=\{ (a^i, a^{n-1-i}b) \mid i=0,1,\ldots,n-1\}$
 of size $n$.
 Let us show that $\mathcal{F}$
 is a fooling set for the language~$L^*$.

 \begin{figure}[b]\label{-----fi10}
 \centerline{\includegraphics[scale=.40]{star_nsc.eps}}
 \caption{The NFA of a prefix-closed language $L$ with $\mathrm{nsc}(L^*)=n$
              and $\mathrm{nsc}(L^R)=n+1$.}
 \label{fig:star_nsc}
 \end{figure}

 (F1) We have $a^i  a^{n-1-i}b=a^{n-1}b$.
 Since the string $a^{n-1}b$ is in $L$,
 it also is in $L^*$.

 (F2) Let $i<j$. Then $a^i a^{n-1-j}b = a^{n-1-(j-i)} b$.
 Since no string $a^\ell b$ with $\ell<n-1$ is in $L$,
 the string $ a^{n-1-(j-i)} b$ is not in $L^*$.

 Hence the set $\mathcal{F}$
 is a fooling set for the language $L^*$,
 and the lower bound follows.

\smallskip
 (2) The upper bound is the same as
 for regular languages \cite{hk03}.
 It is shown in \cite[Theorem~2]{ji05}
 that this bound is met by the binary prefix-closed language $L$
 accepted by the NFA   shown in Figure~\ref{fig:star_nsc}.
 The~proof~in \cite{ji05} is by a counting argument.
 Notice that Lemma~\ref{le:fool1} is satisfied for the language $L^R$ with
 $
 \mathcal{A}=\{(b a^i, a^{n-1-i})\mid i=0,1,\ldots,n-2\},
 \mathcal{B}=\{(b a^{n-1}, b a^{n-1})\},
 u = b a^{n-1}$, and
$ v = a$. This gives $\mathrm{nsc}(L^R)\ge n+1$ immediately.
\end{proof}

\section{Conclusions}

In this paper we considered operations
on languages recognized by incomplete deterministic
or nondeterministic finite automata with all states final.
Our results are summarized in Tables~\ref{ta} and~\ref{ta2}.
The results on quotient (state) complexity
on prefix-closed languages are from \cite{bjz14},
and the results for regular languages are from \cite{hk03,ji05,ma70,yzs94}.
Notice that in the nondeterministic case,
our results are the same as in the general case of regular languages,
except for the star operation.
However, to prove tightness,
we usually  used larger alphabets than in the general case.
Whether or not these bounds are tight also for smaller alphabets remains open.

\begin{table}[t]
\begin{center}
\begin{tabular}{|ll|cc|cc|cc|}
\hline
&& complement &$|\Sigma|$&intersection&$|\Sigma|$& union & $|\Sigma|$    \\
\hline
$\mathrm{isc}$     &on prefix-closed
& $n+1$ & 1 & $m n$ & 2& $m n+m+n$ &2  \\
\hline
$\sc$ & on prefix-closed \cite{bjz14}
& $n$ & 1 & $m n-m-n+2$ & 2& $m n$ &2  \\
\hline
$\sc$ & on regular \cite{ma70,yzs94}
& $n$ & 1 & $m n$ & 2& $m n$ &2  \\
\hline
\hline
$\mathrm{nsc}$ &on prefix-closed
& $2^n$ & 3  & $m n$ &2& $m+n+1$ & 4\\
\hline
$\mathrm{nsc}$ &on regular \cite{hk03,ji05}
& $2^n$ & 2  & $m n$ &2& $m+n+1$ & 2\\
\hline
\end{tabular}
\end{center}
\caption{The complexity
    of boolean operations on prefix-closed and regular languages.}
\label{ta}
\end{table}

\begin{table}[t]
\begin{center}
\begin{tabular}{|ll|cc|cc|cc|}
\hline
&& concatenation &$|\Sigma|$&star&$|\Sigma|$& reversal & $|\Sigma|$ \\
\hline
$\mathrm{isc}$ &on prefix-closed
 & $m2^{n-1}+2^n-1$ & 3& $2^{n-1}$ & 2& $2^n-1$ &2\\
\hline
$\sc$ & on prefix-closed \cite{bjz14}
  & $(m+1)2^{n-2}$ & 3 & $2^{n-2}+1$ &3& $2^{n-1}$ &2  \\
\hline
$\sc$ & on regular \cite{ma70,yzs94}
  & $m2^{n}-2^{n-1}$ &2 & $2^{n-1}+2^{n-2}$ & 2& $2^n$ &2  \\
\hline
\hline
$\mathrm{nsc}$  &on prefix-closed
 & $m+n$ &3 &  $n$ &2 & $n+1$ &2\\
\hline
$\mathrm{nsc}$ &on regular \cite{hk03,ji05}
 & $m+n$ & 2  & $n+1$ &2& $n+1$ & 2\\
\hline
\end{tabular}
\end{center}
\caption{The complexity of concatenation, star, and reversal on prefix-closed
    and regular languages.}
\label{ta2}
\end{table}

 \nocite{*}
 \bibliographystyle{eptcs}

redpajama_set_name: RedPajamaArXiv


text:
Caldwell County, Caldwell County History, Korean War, Veterans Stories, Vietnam
COLONEL STAN REECE – AN INCREDIBLE LIFE OF SERVICE TO HIS COUNTRY AND COMMUNITY
October 24, 2017 tripodtab 3 Comments
COLONEL STAN REECE
AN INCREDIBLE LIFE OF SERVICE TO HIS COUNTRY AND COMMUNITY
Todd Blomerth
It was early 1953. The Allies were locked in vicious combat with communist forces along a line across the Korean peninsula. Ten miles inside enemy lines, Captain Stan Reece was in the back seat on an AT6 "Texan" artillery spotter aircraft. After flying over thirty missions, the 6147th Group commander had made Captain Reece the group's operations officer. The unit's casualty rate was too high, and as operations officer, Captain Reece was told to check out each pilot, to ensure all were qualified and performing at the high standard needed. Stan sat in the rear seat of the two-seater aircraft, where an Army forward observer usually was placed. He was evaluating the skill and proficiency (we'll call him Jones), trying to ensure Jones would not be another casualty while providing the US Army with an aerial platform for forward observers.
Very concerned about the Jones's too-casual attitude about the enemy's willingness to shoot down "low and slow" unarmed intruders, Captain Reece instructed him to make corrections. Before that could happen, the AT6 took 37mm antiaircraft rounds that damaged its right wing and flipped the aircraft onto its back. Smoke poured into the cockpit Jones failed to respond to Reece's intercom, and Captain Reece was forced to take over piloting the wounded airplane. Flinging open the canopy to clear blinding smoke, Captain Reece fought to regain control of the damaged aircraft,. He prayed the wounded bird would get the two back across the front lines before it came apart. Fortunately, it did.
"Where were you when I needed you to flying the airplane?" he frantically demanded of Jones. Finally responsive, Jones meekly replied, "I was scared.". "Well I was too," responded Captain Reece. "But you still have to fly the plane!"
As Captain Reece landed, the ground crew furiously waved him away from revetments and parked aircraft. It turned out that the anti-aircraft rounds had ignited white phosporus rockets under the aircraft's wings, used to spot enemy positions. The ground crew's well founded fear of an horrific explosion and fire was not realized. Fire crews extinguished the flames, as a forward observer pilot and his evaluator leapt from the plane. Captain Reece made sure Jones never flew with the unit again. It was a tough call, as Jones was a friend, but the only one to make.
Sixty-four years later, Colonel Stan Reece sat in his living room, telling me of his life. Now ninety six years old, he expresses some doubt as to why he is being interviewed. "I don't think my career was all that interesting," he tells me.
This is his story. You decide for yourselves whether his belief is correct.
Horace Stanley "Stan" Reece was born in the community of Martin's Mill, Van Zandt County, Texas on September 21, 1921. He was the oldest of five children of Flake Elijah Reece and Nettie (Lancaster) Reece. Flake and Nettie taught school for a period of time. Flake was also a farmer, but later moved the family to Athens, Texas where he owned a garage. Stan and his younger siblings, Charles, Billy, Bonetia and Mary Jo all graduated from Athens High School. Like all children of the Depression, Stan knew about hard work. When he wasn't in school, he worked for his dad as a mechanic. He graduated in 1940, and then worked construction in Ft. Worth and at Camp Wolters.
Flight Cadet Reece – Eagle Pass, Texas
On June 30, 1942, Stan was sworn into the United States Army Air Forces as an aviation cadet. Because of training facility shortages, he was told to go home. "They said they'd call me in a month," he recalls. He worked in his dad's garage for almost six months before he received orders to report to the San Antonio Aviation Cadet Center (later Lackland Air Force Base). Stan took to the air naturally. Without bragging, he calmly states, "I was a very good pilot." After Basic at Brady, Texas and Primary at Waco Airfield, he finished second in his class at Advanced training at Eagle Pass Army Airfield. With over twenty hours already in a high-powered P-40, he hoped for a combat assignment. Enamored with the monstrous P-47 Thunderbolt, he tried for an assignment flying that fighter. No luck. Then he applied for a unit flying the magnificent P-51 Mustang. Again, no luck. The Army Air Forces had other ideas. 2nd Lieutenant Stan Reece was assigned as an instructor pilot at Randolph Army Airfield, outside of San Antonio. It was a bitter disappointment, but just one of the breaks. He loved flying, and was a natural teacher.
Edith Clark Reece
The stateside duty proved a blessing in disguise. In mid-1944 he was diagnosed with osteomylitis. Prior to the advent of strong antibiotics, the best someone with this dangerous bone infection could hope for was to survive with an amputation. Stan was lucky, as penicillin was coming available. It still took fifteen months in the hospital to be cured. "They didn't know how long I had had the infection in my leg before it was discovered," he recalls. By the time he was released in October of 1945, the war was over. The interminable time recovering was not all a loss. Stan met and married a beautiful Army officer (and nurse) – Edith Clark, from Prairie Lea. Their happy marriage lasted for sixty-six years. Edith died in 2011. They have two children. Ralph was born in 1946. Jacqueline was born in 1948.
Because of his skills, Stan was one of the small minority of pilots who were allowed to stay in the Army Air Force after World War II, as the services drastically reduced manpower. The newly married pilot, soon to be a father, was assigned to post-war Germany, first in Wiesbaden, and then in Kassal. He had several assignments.
Then the Soviet Union and its East German lackeys began closing road and canal access to areas of Berlin controlled by the French, British, and Americans. With one and one half million Soviet troops surrounding the former capital of Nazi Germany, Stalin felt sure that he could force the Western Allies out of Berlin, subjecting its citizens to the communist rule beling inflicted on East Germany. The Americans and British undertook to fly into the beleaguered city over 6000 tons daily, of food, supplies and coal. In was considered an impossible task.
Beginning in late June, 1948, with C-47 Dakotas, and later with C-54s and other, heavier transports, pilots flew narrow air corridors at three minute intervals, day and night, for almost one year. Airplanes got one shot at landing. Upon landing in West Berlin, engines were kept running, air crews remained in their transports, as Berliners unloaded the precious food and supplies. Then it was an immediate take-off and return to friendly territory. A missed approach meant returning to West Germany, fully loaded. There was no room for error.
Despite Soviet harrassment, eventually the Airlift was moving over 11,000 tons a day! In April of 1949, the Soviets gave up. Berliners actually had a surplus of food and supplies. Stan Reece flew over twenty flights in and out of Berlin. His younger brother Charles, also an aviator, was part of the all-out effort. The Airlift is one of the proudest moments in our nation's history.
Captain Reece was then assigned to Tinker Air Force Base as an instructor pilot. He had flown and mastered all sorts of aircraft. A small sampling: P-40, P-47, and P-51 fighters; A-26, B-25 bombers; C-47, C-54 transports, AT-6 trainer. Because of the rush to get pilots into aircraft in World War II, and the budget and manpower cuts afterward, training had suffered grievously. Stan recalls that "many of these pilots, including some who'd flown in World War II, weren't trained properly." What better man to help fix the problem than someone of Stan's stature and reputation?
Then, war in an unexpected place threw the United States in a frenzied effort to save South Korea from subjugation by North Korean, and later, Chinese communists. Bidding farewell to his wife and kids, Captain Reece arrived in the war-torn Korean peninsula in 1952. He was assigned to the 6147th Tactical Control Group. He flew unarmed AT-6 "Texans" over enemy lines, with an Army artillery forward observer onboard. In over twenty-five missions, he was shot at "plenty of times" but never hit. That is, until he was a check pilot for "Jones," and was almost shot out of the sky over enemy lines.
In September of 1953, Captain Reece was back in the States. The 1950s were a transition period for the United States Air Force. Propeller driven aircraft were phased out, and jets of all types were tested. There were fewer straight winged aircraft, as jets' speeds required swept wings. Thousands of aircraft were moved back to the continental United States from overseas. Some were scrapped, some were sold to friendly countries, some were given to Air National Guard units, and some were retrofitted. Stan was part of this complicated and often dangerous task. Between 1953 and 1962, Stan took on hundreds of ferrying operations.
Assigned to the Military Air Transport Command (MATS) and its 1708th Ferrying Wing, he coordinated and flew transports to and from overseas air force units. One trip nearly got him killed. He and his crew flew a new C-119 to Japan. They exchanged it for one that was on its last legs. Taking off, the overloaded aircraft needed all the runway and overrun to make it into the air. Then, the crew discovered that the extra 900 gallon tank in the fuselage was leaking fuel. Stan was piloting a potential fiery bomb. It turned out that someone had installed a fuel vent pipe the wrong way. Remaining cool, Stan landed the plane without incident. The crew fixed the problem, and made it to Hawaii before the tired bird gave up the ghost. He and his crew caught another flight home.
After an assignment at Andrews AFB, Major Stan Reece and family spent three years in Japan, where he commanded the 531st Fighter Squadron flying F-100s. Later, as the Wing operations officer. He oversaw pilots flying RF-101s, and AD-102s. As he recalls, most of his pilots were "a bunch of kids." They were lucky to have a proficient and demanding boss, who expected nothing less than the best, and made sure they achieved it. The Wing was equipped with nuclear weapons.
Stan knew that to move up the promotion ladder, it was expected that he be proficient in all areas of aviation. Often that meant accepting assignments to non-glamorous jobs. Because of his willingness and skill, he was often picked to 'clean up' situations, particularly those where pilots' lives were in danger because of inefficiency or poor training standards. The 1960s found him and the family at Eglin AFB in Florida, in another of those roles.
His next assignment put Lieutenant Colonel Reece back in Germany. He was the first commander of the 7th Air Special Operations Squadron, tasked with supporting the 10th Special Forces, one of the first Green Beret units. Comprised of a hodgepodge of propeller-driven aircraft, and designed for special operations, the squadron's pilots and crews helped the SF forces conduct airborne training operations. "A large percentage of the Special Forces unit were first generation Americans from Eastern Europe. They spoke with thick accents. They hated the communists," recalls Stan. Maximum flexibility best describes the operations of the Squadron. It was a heady time, as the United States continued its experiment with small and specialized forces. Stan spent two and a half years as squadron commander. His Army counterp art was "an old OSS [the World War II predecessor to the CIA] colonel. We got along famously." Practice for insertions and extractions behind the Iron Curtain, providing air support for British Special Air Services commandos, conducting parachute drops, and retrofitting aircraft for unorthodox warfare made life, in Stan's word, "interesting." That has to be a understatement
Military spouses have unenviable roles to play. Stan was transferred many times during his career. Even when assigned to one base, duties often took him away from his family. A full-time mom who often had to parent alone, Edith was the rock that kept the Reece family together. Edith's duties expanded when Stan was a unit's commanding officer. She then became a mentor, advisor, and, occasionally, a shoulder to cry on. It was tough duty, and deeply appreciated. Stan is unstinting in his praise of Edith. "Edith was a stabilizing influence on many young wives," Stan recalls.
Between 1966 and 1968, Lieutenant Colonelr Stan Reece commanded the 492nd Tactical Fighter squadron, based in England. Flying F100s, the squadron participated in numerous NATO exercises. These included deployments to Bodo, Norway, Aviano, Italy, and Ishmir, Turkey. Joint exercises, intended as training for possible response to Soviet aggression, came at the heighth of the Cold War. The training was arduous, and demanding. If war came, the Soviets had a numerical superiority that would test the very fiber of the North Atlantic Treaty Organization's ability to defend the free countries in western Europe, Turkey, and Scandinavia.
It was inevitable that an aviator of Stan Reece's ability would end up in Vietnam. Saying goodbye to Edith, he reported for duty at the Phu Cat Air Base. A new airfield, capable of jet powered aircraft, it was completed in late 1967. Lt. Colonel Reece was assigned as the 37th Tactical Fighter Wing's operations officer. Part of the unit was the Iowa National Guard's 174th Tactical Fighter Squadron. Equipped with the now aging and quirky F100s, the Wing flew interdiction and combat support missions. In just six months, Stan flew 135 combat missions. Most were in South Vietnam, but over thirty were in Laos, and there were a few into North Vietnam. Thankfully, he came through that experience in one piece. His final six months in Vietnam were at 7th Air Force Headquarters in Saigon.
Returning to the United States, newly promoted Colonel Stan Reece finished out a most colorful and exciting career in St. Louis, Missouri as Wing Advisor to the Air National Guard. He retired, after 30 years of service, in 1974.
Edith longed to return to her Texas and Caldwell County roots. In 1975, the Reeces purchased forty-nine acres of land outside of Luling. Anyone who knew him realized that Stan Reece wasn't someone to sit back and let the world pass him by. He raised cattle. He served on the Luling Independent School District Board of Trustees for twenty-four years, of which nineteen he was the Board's president. He joined the Luling Kiwanis Club in 1975 and remains active today. He was an officer on the Caldwell County Farm Bureau, and served on the Caldwell County Agent supervisory board for many years. Trying to track him a couple of weeks ago, I found him awash with plans for Night In Old Luling.
Stan will tell you his has been a blessed life. In 2014 at the young age of ninety three, he tied the knot for the second time when Madeline Manford, a dear family friend of many years and a widow, agreed to be his bride.
So this is a short-hand version of Stan's life (so far). After reading it, I know you will agree that Colonel Stan Reece has indeed led a most incredibly interesting and fruitful life. He truly is a shining example of the best of the Greatest Generation. If you can catch him, please be sure to thank him for his service to his country and community.

redpajama_set_name: RedPajamaCommonCrawl


text:
BR Home Page > Major League Players > E Listing > Darrell Evans Statistics and History
Positions: Third Baseman and First Baseman
Bats: Left • Throws: Right
Born: May 26, 1947 in Pasadena, CA us
Draft: Drafted by the Chicago Cubs in the 13th round of the 1965 MLB June Amateur Draft from Muir HS (Altadena, CA), the New York Yankees in the 2nd round of the 1966 MLB January Draft-Secondary Phase from Pasadena City College (Pasadena, CA), the Detroit Tigers in the 5th round of the 1966 MLB June Draft-Secondary Phase from Pasadena City College (Pasadena, CA), the Philadelphia Phillies in the 3rd round of the 1967 MLB January Draft-Secondary Phase from Pasadena City College (Pasadena, CA) and the Kansas City Athletics in the 7th round of the 1967 MLB June Draft-Secondary Phase from Pasadena City College (Pasadena, CA).
High Schools: John Muir HS (Pasadena, CA), Muir HS (Altadena, CA)
School: Pasadena City College (Pasadena, CA)
Debut: April 20, 1969 (Age 21-329d, 12,821st in major league history)
vs. CIN 1 AB, 0 H, 0 HR, 0 RBI, 0 SB
vs. LAD 1 AB, 0 H, 0 HR, 0 RBI, 0 SB
Agents: Jerry Kapstein
Full Name: Darrell Wayne Evans
Nicknames: Howdy Doody
View Player Bio from the SABR BioProject
2x All-Star
1984 World Series
11 2 11 11 11 41 41 41 11
Darrell Evans Overview
More Evans Pages
Minor, Winter & Independent Lg Stats
vs. Pitcher
Postseason Event Finder
More Darrell Evans Pages at Baseball Reference
Darrell Evans page at the Bullpen Wiki
Advanced Batting
Postseason Batting
Hall of Fame Statistics
Name + "Statistics" Translations
20 KCA-min Rk,A 61 248 215 35 70 8 6 2 29 7 1 28 20 .326 .402 .447 .849 96 1 2 2 LEE,ALT,PNS · FLOR,GULF,CARL
21 OAK-min AA 56 205 187 18 45 6 3 3 25 1 0 16 36 .241 .299 .353 .652 66 0 1 1 BIR · SOUL
22 ATL-min AAA,WRk,AA 109 459 393 72 126 21 11 10 78 6 6 54 77 .321 .401 .506 .908 199 3 3 6 RMD,BRV,SHV · IL,AZIL,TL
22 ATL NL 12 28 26 3 6 0 0 0 1 0 0 1 8 .231 .250 .231 .481 35 6 0 0 0 1 0 /5H
23 ATL-min AAA 120 535 447 92 134 20 7 20 83 1 4 78 80 .300 .403 .510 .913 228 3 2 5 RMD · IL
23 ATL NL 12 52 44 4 14 1 1 0 9 0 0 7 5 .318 .423 .386 .809 115 17 2 1 0 0 0 5
24 ATL-min AAA 31 126 101 20 31 2 2 6 30 0 1 23 17 .307 .437 .545 .981 55 1 0 1 RMD · IL
24 ATL NL 89 305 260 42 63 11 1 12 38 2 3 39 54 .242 .338 .431 .768 112 112 4 1 0 5 4 5H/7
25 ATL NL 125 521 418 67 106 12 0 19 71 4 2 90 58 .254 .384 .419 .803 121 175 6 4 0 9 4 *5/H
26 ATL NL 161 733 595 114 167 25 8 41 104 6 3 124 104 .281 .403 .556 .959 156 331 9 3 3 8 8 *53/H AS,MVP-18
27 ATL NL 160 710 571 99 137 21 3 25 79 4 2 126 88 .240 .381 .419 .800 121 239 11 6 4 3 9 *5/H
28 ATL NL 156 681 567 82 138 22 2 22 73 12 3 105 106 .243 .361 .406 .766 111 230 10 2 2 5 5 *5/3H
29 TOT NL 136 471 396 53 81 9 1 11 46 9 1 72 71 .205 .326 .316 .641 81 125 5 0 1 2 4 *3H5
29 ATL NL 44 169 139 11 24 0 0 1 10 3 0 30 33 .173 .320 .194 .514 45 27 3 0 0 0 0 3/5H
29 SFG NL 92 302 257 42 57 9 1 10 36 6 1 42 38 .222 .329 .381 .710 100 98 2 0 1 2 4 3H/5
30 SFG NL 144 539 461 64 117 18 3 17 72 9 6 69 50 .254 .351 .416 .768 107 192 8 3 1 5 3 735H
31 SFG NL 159 668 547 82 133 24 2 20 78 4 5 105 64 .243 .360 .404 .764 118 221 5 0 6 10 12 *5/H
32 SFG NL 160 661 562 68 142 23 2 17 70 6 7 91 80 .253 .356 .391 .748 111 220 6 2 1 5 14 *5/H
33 SFG NL 154 653 556 69 147 23 0 20 78 17 5 83 65 .264 .359 .414 .772 119 230 12 2 6 6 6 *53/H
34 SFG NL 102 419 357 51 92 13 4 12 48 2 3 54 33 .258 .356 .417 .773 121 149 4 2 3 3 8 *53/H
35 SFG NL 141 553 465 64 119 20 4 16 61 5 4 77 64 .256 .360 .419 .779 119 195 8 2 3 6 7 53H6
36 SFG NL 142 611 523 94 145 29 3 30 82 6 6 84 81 .277 .378 .516 .894 150 270 8 2 0 2 12 *35/6H AS,MVP-14
37 DET AL 131 483 401 60 93 11 1 16 63 2 2 77 70 .232 .353 .384 .737 106 154 7 0 1 4 10 D3H5
38 DET AL 151 594 505 81 125 17 0 40 94 0 4 85 85 .248 .356 .519 .875 138 262 5 1 1 2 12 *3DH/5 MVP-14
39 DET AL 151 601 507 78 122 15 0 29 85 3 2 91 105 .241 .356 .442 .798 116 224 6 1 0 2 5 3DH/5
40 DET AL 150 609 499 90 128 20 0 34 99 6 5 100 84 .257 .379 .501 .880 135 250 2 2 2 6 8 3DH/5 MVP-12
41 DET AL 144 522 437 48 91 9 0 22 64 1 4 84 89 .208 .337 .380 .717 105 166 14 1 0 0 4 D3H
42 ATL NL 107 323 276 31 57 6 1 11 39 0 1 41 46 .207 .303 .355 .658 87 98 1 0 0 6 6 3H5
2687 10737 8973 1344 2223 329 36 414 1354 98 68 1605 1410 .248 .361 .431 .792 119 3866 133 35 34 90 141
162 647 541 81 134 20 2 25 82 6 4 97 85 .248 .361 .431 .792 119 233 8 2 2 5 9
ATL (9 yrs)
866 3522 2896 453 712 98 16 131 424 31 14 563 502 .246 .368 .426 .794 118 1235 46 17 9 37 36
SFG (8 yrs)
1094 4406 3728 534 952 159 19 142 525 55 37 605 475 .255 .358 .422 .781 119 1575 53 13 21 39 66
727 2809 2349 357 559 72 1 141 405 12 17 437 433 .238 .357 .450 .806 121 1056 34 5 4 14 39
NL (16 yrs)
1960 7928 6624 987 1664 257 35 273 949 86 51 1168 977 .251 .362 .424 .787 119 2810 99 30 30 76 102
AL (5 yrs)
Bold season totals indicate player led league. Italic season totals indicate player led all major leagues.
See More Advanced Batting
Compare Darrell Evans to
Ex: Graig Nettles, Dale Murphy
Postseason Game Log
Convert to YYYY $'s Back to Actual 1871 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010 2021
Salaries may not be complete (especially pre-1985) and may not include some earned bonuses
Note, this is done in an automated way, so we apologize for any errors, & please report any suggested corrections. "Statistics" is included to allow non-English speakers to find our pages.
You are here: BR Home Page > Major League Players > E Listing > Darrell Evans Statistics and History


Just wondering, does anyone know what the biggest (HP) outboard engine is?
Codders knows, it was something like a 400hp Evinrude from the 70's or 80's.
And it only weighed 225kgs cos it was a 2 stroke.
The Yamaha V8 350hp is 375kgs!!!
MErcury Verado 350 SCi 300kg.
The old Evinrude V8 2 stroke was often taken out to about 450hp. Where's the progress eh?
Yes the Verado 350 weighs about 300kg but the Mercury 300xs weighs only 225kg - I love 2 strokes!!!
it still amazes some people use tiller steering for massive outboards. I think mercury do it for even the big Verados!!!


text:
The 2017–18 College of Charleston Cougars men's basketball team represented the College of Charleston during the 2017–18 NCAA Division I men's basketball season. The Cougars, led by fourth-year head coach Earl Grant, played their home games at the TD Arena in Charleston, South Carolina as members of the Colonial Athletic Association. They finished the season 26–8, 14–4 in CAA play to share the regular season title with Northeastern. At the CAA tournament they defeated Drexel, William & Mary, and Northeastern to become CAA Tournament champions. They earned the CAA's automatic bid to the NCAA tournament where they lost in the first round to Auburn.

Previous season
The Cougars finished the 2016–17 season 25–10, 14–4 in CAA play to finish in second place. In the CAA tournament, they defeated James Madison in the quarterfinals and Towson in the semifinals, before losing to UNC Wilmington in the championship game. They were then invited to participate in the NIT, where they lost to Colorado State in the first round.

Offseason

Departures

2017 recruiting class

Preseason
In a poll of league coaches, media relations directors, and media members at the CAA's media day, the Cougars were picked to finish atop the CAA. Junior forward Jarrell Brantley and senior guard Joe Chealey were named to the preseason All-CAA first team, while sophomore guard Grant Riller was named to the preseason All-CAA second team. r

Roster

Schedule and results

|-
!colspan=12 style=| Exhibition

|-
!colspan=12 style=| Non-conference regular season
|-

|-
!colspan=12 style=| CAA regular season

|-
!colspan=12 style=| CAA tournament

|-
!colspan=12 style=| NCAA tournament

|-

Source

See also
2017–18 College of Charleston Cougars women's basketball team

References

College of Charleston Cougars men's basketball seasons
College of Charleston
College of Charleston

redpajama_set_name: RedPajamaWikipedia


text:
Harry Potter and Game of Thrones fans will love this magical event
Get your wands, cloaks and cauldrons at the ready
Nick Reid
Players will embark on a mysterious and magical journey into a world of witchcraft and wizardry (Image: CluedUpp/Destination Chesterfield)
Sign up to get the latest Chesterfield news in daily email bulletins
Wannabe witches and wizards, as well as fans of Harry Potter and Game of Thrones, are being called on to take part in a magical event in a Derbyshire town.
People are being urged to grab their wands, cloaks and cauldrons as the wizarding world is coming to Chesterfield in an all-new spell-binding witchcraft and wizardry adventure event.
Players will embark on a mysterious and magical journey into a world of witchcraft and wizardry plagued by dark magic, where teams of six muggles will compete to locate virtual witnesses, crack clues and defeat the darkest entities in the realm.
The event is due to take place in September.
Tref Griffiths, CluedUpp's founder, which has organised the event, said: "Due to the success of our previous games, we wanted to create something completely different and otherworldly.
"We're big fans of all things fantasy here at CluedUpp, take the biggest names in fantasy such as Harry Potter and Game of Thrones.
"We thought it was about time we created our own magical mystery within this genre."
Tickets for the event are on sale now. For more information and how to book, click here.
CluedUpp Detective Days are like giant, outdoor escape games.
How to get the latest Chesterfield news
Did you know you can now get the latest Chesterfield news, straight to your inbox?
To sign up, all you have to do is click here and select Chesterfield Live.
And if you later decide that you no longer wish to receive the emails, you can just follow the unsubscribe link in the newsletters - though we hope this won't ever be necessary.
This spell-binding outdoor adventure game will take your team on a mysterious and magical journey across the town.
Work with your team-mates to crack the clues, solve the puzzles and save the realm.
Taking place entirely via the CluedUpp smartphone app, all you need in order to take part is a phone and a team of six fellow witches or wizards to compete in the race to unravel the mystery on September, 19.


Doubt has a good public relations manager these days. The world seems awash with books, articles, sermons, even a few TED talks, praising its beneficial goodness. I too have tried to redeem the sullied reputation of doubt in the church with my preaching and writing. Over the past year I've started to wonder if the pendulum has swung too far though. Have we naively overestimated and championed the virtue of doubt without fully appreciating its destructive power?
The book of James (1:6) names one thing that will prevent us from receiving divine wisdom. Doubt.
So who is right? James or TED? Is doubt good or bad?
There are two primary modes of doubting. The first mode utilizes doubt as a tool to seek understanding. I call this "sincere skepticism." This mode analyzes ideas, beliefs, stories and arguments with logic and critical thinking, testing for validity and reliability. This mode of doubting is healthy and essential to understanding.
I call the second mode of doubting "toxic skepticism." Doubt, like most tools, can become deadly if used inappropriately. There are many genres of this deadly "toxic skepticism," but one of the ways doubt turns toxic is when it becomes our primary tool for life. For instance, if you wake up doubting gravity every morning, wondering if it's safe to get out of bed without putting on the NASA space boots you picked up on eBay, your doubt is no longer productive and healthy. It's pathological.
Many Christians are being conditioned to approach their faith with pathological doubt. Certain cultural influences equate intelligence with doubting the existence of God and Christian beliefs. However, doubting the existence of God is only rational if it is seeking to discover and understand. As soon as doubt isn't seeking to discover and understand, it's pathological, and pathological doubt is never rational. It's lazy. It just wants to doubt.
Healthy doubt wants to discover a new or better understanding of reality and adjust to what it discovers. In order to do that it can't always be doubting.
Sincere skepticism requires us to be good stewards of doubt. Too much of anything, even good things, can be harmful. For example, plants need water to grow but overwatering plants can kill them. Like water, doubt is good but too much is deadly.
Our relationship to "self" illustrates the need for balanced doses of doubt. A certain degree of self-doubt is necessary to have healthy relationships with other people and reality, because all of us are wrong about a great many things. If I'm unable to doubt my opinions, abilities, beliefs, knowledge, then I'm unable to adjust to reality. If I resist everything that contradicts my view of myself and the world, I'm not able to learn, grow and live in truth.
On the flip side, if self-doubt becomes excessive, it destroys my confidence, purpose, meaning and passion. It also kills my ability to take risks. We can all find "logical" reasons to doubt ourselves, causing us to avoid taking risks like starting up a business, confronting a bully or trying something new. But at some point we need to defy our doubts and take the necessary risks required to develop confidence and discover what we're really made of and what we're capable of. If not, our self-doubt becomes pathological and we lose the desire and ability to discover more about ourselves.
While it's true that doubt is beneficial to our lives, taking risks that defy our doubts is even more beneficial. You will never have enough answers to eradicate all your doubts about yourself. You can never be certain you won't fail if you try out for a sports team, enrol in university, ask someone out on a date, submit your name for a job promotion, become a parent, or try to lose weight.
There is no certainty in life. So it makes no sense to expect certainty with our faith.
Doubt can help us discover the truth about ourselves, reality and God, but it can also become the very thing that holds us back from discovery. Doubt, as a tool to discover, renews life. Doubt, as a way of life, is death.
We need to take risks of faith to discover who we are and who God is. In order to do this we need to remember to doubt our doubts sometimes.


"Let us pray that there can be an end to such terror attacks and hope for a future in which peace and security will prevail over violence & loss of human life."
Church inquiry concludes that Bishop Terbartz-van Elst should not continue in office.
Council is now operational and its first meeting is scheduled for May.


Linda Lonsdorf , born in Akron, Ohio, is a retired high school English and Speech instructor. She resides in Green, Ohio and Florida with her husband, Dr. W. Kevin Lonsdorf. Her lifelong passion for reading as well as her love for suspense and drama led her to write her first novel, Family Threat and now its sequel, Evil Injustice.


text:
#include "pcidsk_config.h"
#include "pcidsk_types.h"
#include "core/pcidsk_utils.h"
#include "pcidsk_exception.h"
#include <cassert>
#include <cstdio>

using namespace PCIDSK;

#if defined(HAVE_LIBJPEG)

extern "C" {
#include "jpeglib.h"
}

static void _DummyMgrMethod( j_compress_ptr /*pUnused*/ ) {}
static void _DummySrcMgrMethod( j_decompress_ptr /*pUnused*/ ) {}

/************************************************************************/
/*                             JpegError()                              */
/*                                                                      */
/*      Handle errors generated by the IJG library.  We treat all       */
/*      errors as fatal at this point.  Future handling may be          */
/*      improved by overriding other methods.                           */
/************************************************************************/

static void JpegError(j_common_ptr cinfo)
{
    char buf[256];

    cinfo->err->format_message(cinfo, buf);
    ThrowPCIDSKException( "%s", buf );
}

/************************************************************************/
/*                      LibJPEG_DecompressBlock()                       */
/************************************************************************/

void PCIDSK::LibJPEG_DecompressBlock(
    uint8 *src_data, int src_bytes, uint8 *dst_data, int dst_bytes,
    int xsize, int ysize, eChanType pixel_type )

{
    struct jpeg_decompress_struct sJCompInfo;
    struct jpeg_source_mgr	sSrcMgr;
    struct jpeg_error_mgr	sErrMgr;

    int				i;

/* -------------------------------------------------------------------- */
/*      Setup the buffer we will compress into.  We make it pretty      */
/*      big to ensure there is space.  The calling function will        */
/*      free it as soon as it is done so this shouldn't hurt much.      */
/* -------------------------------------------------------------------- */
    sSrcMgr.init_source = _DummySrcMgrMethod;
    sSrcMgr.fill_input_buffer = (boolean (*)(j_decompress_ptr))
        						_DummyMgrMethod;
    sSrcMgr.skip_input_data = (void (*)(j_decompress_ptr, long))
        						_DummyMgrMethod;
    sSrcMgr.resync_to_restart = jpeg_resync_to_restart;
    sSrcMgr.term_source = _DummySrcMgrMethod;

    sSrcMgr.next_input_byte = src_data;
    sSrcMgr.bytes_in_buffer = src_bytes;

/* -------------------------------------------------------------------- */
/*      Setup JPEG Decompression                                        */
/* -------------------------------------------------------------------- */
    jpeg_create_decompress(&sJCompInfo);

    sJCompInfo.src = &sSrcMgr;
    sJCompInfo.err = jpeg_std_error(&sErrMgr);
    sJCompInfo.err->output_message = JpegError;

/* -------------------------------------------------------------------- */
/*      Read the header.                                                */
/* -------------------------------------------------------------------- */
    jpeg_read_header( &sJCompInfo, TRUE );
    if (sJCompInfo.image_width != (unsigned int)xsize ||
        sJCompInfo.image_height != (unsigned int)ysize)
    {
        ThrowPCIDSKException("Tile Size wrong in LibJPEG_DecompressTile(), got %dx%d, expected %dx%d.",
                             sJCompInfo.image_width,
                             sJCompInfo.image_height,
                             xsize, ysize );
    }

    sJCompInfo.out_color_space = JCS_GRAYSCALE;

    jpeg_start_decompress(&sJCompInfo);

/* -------------------------------------------------------------------- */
/*	Read each of the scanlines.					*/
/* -------------------------------------------------------------------- */
    for( i = 0; i < ysize; i++ )
    {
        uint8	*line_data = dst_data + i*xsize;

        jpeg_read_scanlines( &sJCompInfo, (JSAMPARRAY) &line_data, 1 );
    }

/* -------------------------------------------------------------------- */
/*      Cleanup.                                                        */
/* -------------------------------------------------------------------- */
    jpeg_finish_decompress( &sJCompInfo );
    jpeg_destroy_decompress( &sJCompInfo );
}

/************************************************************************/
/*                      LibJPEG_CompressBlock()                         */
/************************************************************************/

void PCIDSK::LibJPEG_CompressBlock(
    uint8 *src_data, int src_bytes, uint8 *dst_data, int &dst_bytes,
    int xsize, int ysize, eChanType pixel_type, int quality )

{
    struct jpeg_compress_struct	sJCompInfo;
    struct jpeg_destination_mgr	sDstMgr;
    struct jpeg_error_mgr	sErrMgr;

    int		i;

/* -------------------------------------------------------------------- */
/*      Setup the buffer we will compress into.                         */
/* -------------------------------------------------------------------- */
    sDstMgr.next_output_byte = dst_data;
    sDstMgr.free_in_buffer = dst_bytes;
    sDstMgr.init_destination = _DummyMgrMethod;
    sDstMgr.empty_output_buffer = (boolean (*)(j_compress_ptr))
        						_DummyMgrMethod;
    sDstMgr.term_destination = _DummyMgrMethod;

/* -------------------------------------------------------------------- */
/*      Setup JPEG Compression                                          */
/* -------------------------------------------------------------------- */
    jpeg_create_compress(&sJCompInfo);

    sJCompInfo.dest = &sDstMgr;
    sJCompInfo.err = jpeg_std_error(&sErrMgr);
    sJCompInfo.err->output_message = JpegError;

    sJCompInfo.image_width = xsize;
    sJCompInfo.image_height = ysize;
    sJCompInfo.input_components = 1;
    sJCompInfo.in_color_space = JCS_GRAYSCALE;

    jpeg_set_defaults(&sJCompInfo);
    jpeg_set_quality(&sJCompInfo, quality, TRUE );
    jpeg_start_compress(&sJCompInfo, TRUE );

/* -------------------------------------------------------------------- */
/*	Write all the scanlines at once.				*/
/* -------------------------------------------------------------------- */
    for( i = 0; i < ysize; i++ )
    {
        uint8	*pabyLine = src_data + i*xsize;

        jpeg_write_scanlines( &sJCompInfo, (JSAMPARRAY)&pabyLine, 1 );
    }

/* -------------------------------------------------------------------- */
/*	Cleanup.							*/
/* -------------------------------------------------------------------- */
    jpeg_finish_compress( &sJCompInfo );

    dst_bytes = dst_bytes - sDstMgr.free_in_buffer;

    jpeg_destroy_compress( &sJCompInfo );
}

#endif /* defined(HAVE_LIBJPEG) */

redpajama_set_name: RedPajamaGithub


If you are missing one or more teeth, then you need to consider the dental restoration options that are available. There is no reason for you to leave the space empty after a tooth extraction! Instead, dental implants can be used to fill in the gap and improve the appearance of your smile.
When it is time to choose a dental restoration, you need to talk to an experienced dentist for personal recommendations. An examination will help you see if you are a good candidate for the type of restoration that you would like to use.
There are pros and cons to all of the restoration treatments that are offered. Look at the costs, the function of the restorations, and the long-term benefits that you will gain from each treatment.
Dental implants are usually the top recommendation if a patient has a jawbone that can hold the implant. This treatment offers the best solution to create a restoration that mimics the look and feel of a natural tooth. After the implant is done, you won't be able to tell that there is a fake tooth in place of the missing tooth.
Not only will you love the appearance of your smile, but you will be able to eat and talk normally. This treatment allows you to enjoy the full functionality as if the natural tooth was still in your mouth.
Can You Afford an Implant?
The biggest concern for many patients is the cost of dental implants. This treatment is one of the more expensive tooth restoration options. But, the function and long-term benefits are worth the cost. Investing in the quality of your mouth is essential to ensure that you have a smile that you love.
Not only will your smile impact your confidence and self-esteem, but the implant can also help you avoid other dental complications in the future. If you are a good candidate for the treatment, then you should put dental implants at the top of your priority list.
How do you know if dental implants are right for you? If you are missing one or more teeth, then you should schedule an exam and consultation with our team. We will evaluate your situation and help you choose the right treatments for your smile. For more information, call us at Simply Teeth to schedule an appointment.


text:
2019 MLB Draft
2019 MLB draft: Who's No. 2?
We present four great options for the make-or-break pick of the draft (no, really) — the White Sox's No. 45
By James Osborn May 31, 2019, 8:26am CDT
Share All sharing options for: 2019 MLB draft: Who's No. 2?
Second coming of Crede: Gunnar Henderson shouldn't be available at No. 45, but if he is, Lil Jimmy is chipping in for his signing bonus.
At top of this year's draft, there are six players.
Following them, there are about 15-18 players.
The next group has about 30 players.
After that, it's chicken salad or chicken s—-. The point is, we have a chance at a player from the last bunch with our second round pick, but only about a dozen will remain. For the last three years, Nick Hostetler has been making those picks: Alec Hansen at No. 49, Gavin Sheets at No. 49 and Steele Walker at No. 46.
These were solid picks, based on where they were chosen and who was available. There are several thoughts that Nick has shared this spring that may apply to this year's second pick. First, he indicated that because pitching is light this year, the first two picks would likely be position players. He also noted that when he took over a few year ago the "cupboard was bare" in the system, and he drafted college heavy. But after three of his own drafts, the system is replenished, so that may change.
Let's take a look at what players may make up the "group of 44" that will have been taken off the board.
Last year, 21 pitchers were gone by the No. 45 pick. That was in a good year for pitching. Seems about right: 22 being half of all players picked, and pitching being half the game. As most folks know, pitching this year is "not-a so good." This year, it looks like 16 pitchers will be selected in the first 44 picks.
On the other hand, the college hitting talent pool is considered above-average this year. Usually it's miserable. This is the best year I can recall. That said, I'd say 13 college hitters will be gone in the first 44, likewise prep hitters. That's 26 players.
Those picks all add up to 42 players in al. leaving two players' worth of wiggle room.
In some spots in the draft, you take a flier. Sometimes you fill out organizational needs. Sometimes you draft the GM's daughter. With the No. 45 pick you have to say, "can this player make it to Chicago?"
On draft day, there will be players available at No. 45 who will have solid careers. After this pick, drafting gets tougher. So, you know, don't screw up.
For almost a year, I had my eye on Gunnar Henderson, A third baseman from Selma, Ala. He reminds me of Joe Crede. People have gotten a good look at him by now, and now he's first-round bound. If somehow he's still there, pretty please Nick, I'll throw in money for the bonus.
Anyway, on draft night, we will all be watching closely to see who the White Sox choose at No. 3. When you and I wake up in the morning, one other player will be in the fold. I have looked long, and hard to find the player(s) that "can make it to Chicago" who might be, should be, could be, available when the Sox pick at No. 45.
Will they take a college pitcher, a college position player, a high school pitcher, or a high school position player? Here's a look at one of each.
College Pitcher
Mic Smith - Tim Cowie Photograph/Ball State University
Drey Jameson
Drey is a right-handed starting pitcher from Ball State. I featured him in my Players from Hoosierville article:
FanGraphs has Jameson moving up on their board, at No. 43 with a bullet. That would make Jameson our second-round pick. They write that "Jameson has flashed three plus pitches at times, so clubs can look past his smaller frame and command that comes and goes."
That command must be "coming" more than "going" to be this high on their board. MLB Pipeline has Jameson at No. 57 [Ed. Note: now No. 49] with a fastball that touches 98 mph. He's 94-95 all day. The curve and slider are better than average. He's thrown 42 innings, with 69 strikeouts and a 1.06 WHIP. Hmm ... command that comes and goes but a 1.06 WHIP. Wild in the zone: huh?"
If Jameson was 6´3´´, he'd be long gone by No. 45. That's just a fact. With three plus pitches, Drey has a starter's makeup. Finding starters this deep in the draft isn't easy. There won't be a better college pitcher on the board when the White Sox pick.
This kid has moxie. Nathan Detroit told me so.
College Hitter
Scott Chandler/UCLA
Mike Toglia
Hopefully the White Sox are foregoing the height-challenged, right-handed hitting, first baseman from Cal, Andrew Vaughn, at No. 3.
Instead, let's look at the middle of the lineup "Thumper" at UCLA, the best team in college ball. Where Gavin Sheets is 6´4´´ and 230 pounds, Toglia (the g is silent) is 6´5´´, 205. The terms "rangy" or "wiry" would apply. As a switch-hitter, Mike is equally at ease from both sides of the plate. He is from the state of Washington, where he was a two-way player, so he has a strong arm.
Mike is very athletic. Also a basketball star in high school, he thought if he was better handling the ball, his game would improve. The lefty went out to the garage, and practiced dribbling two balls at once. Once he was doing that easily, Mike talked his dad into coming out and trying to knock the balls away. That season, Toglia was one of the best players in the state, averaging 20 points plus per game.
Toglia is much better at baseball. He could be a Gold Glove-level first baseman, and is athletic enough to play a corner outfield position. He had a slow start this year, which kept his draft stock from rising early. He led UCLA last year with 58 RBIs, and has 56 this year. He's an extra-base-hitting machine.
This year and last, over 112 games, he's had 40 doubles, four triples and 24 home runs. Sixty-eight extra base hits in a season is hard to do.
This year, in Toglia's first 20 games, he hit .245. But 36 games later, he'd finished at .315.
The college playoffs start today. (Sorry Mike, I hope you stink, so you're still around for us next week.) Toglia only turns 21 in mid-August, so he's young for his class. Young, ambidextrous, with a ton of upside. Me likey. Gimme, gimme!
High School Pitcher
@SpencerJnes
Spencer Jones
Spencer was a lock for the first round, but while throwing a pitch this spring, the southpaw felt something that turned out to be a stress fracture of the ulna. He actually came back to hit towards the end of his team's season. A full recovery is expected.
The word is Jones would like attend college. The 6´7´´giant turned 18 this month, so he's pretty young. Draft him, pay him over slot, Tell him go to school: "Here's some money. Come see us next spring in Arizona. When you're ready, we'll send you to short season in Great Falls, we'll shut you down when school starts."
Seems like a lot of hassle for a kid with a fractured ulna? Yeah, well, Jones is really, really good at everything. He's worth the hassle.
I actually fractured my left ulna and it improved the spin on my tennis serve. Maybe it will improve Jones's curve ... it could happen!
Jones is a plus runner and can hit the ball clear out of sight. He's also smart and seems to have a good head on his shoulders.
High School Hitter
Nasim Nuñez
You know how in sports, nobody can do a "triple, double, half-twist, somersault." Then one kid just says, "I think I can do that," works at it, and in time, masters it? Nasim is that kid. Super athletic. Super quick, agile, and a sponge for anything baseball-related.
He hits well from both sides of the plate, but at 5´9´´, power will likely never be part of his game. Think Leury García, but faster and better in every way. Niñez is lock to stick at short. If the White Sox draft C.J. Abrams, sure, they'll look elsewhere when it comes to the No. 45. One thing I know for sure: Nasim is fun to watch.
So, to sum up, the White Sox need to hit on this pick. Any of these players could be fixtures in the 2025 lineup.
Mike Taglia could be a top-notch, Gold Glove first baseman.
Drey Jameson could be a South Side version of Marcus Stroman.
Spencer Jones could be a front line, left-handed monster who pounds the zone with upper 90's heat.
Nasim Nuñez could be the second coming of Ozzie Smith.
I'd gladly take them all, but we only get one. I'll let Nick make that choice.


Upload forms to us directly using this secure Health Information Exchange link.
Or you can fax forms to (850) 877-1338.
If you would rather mail or hand deliver forms to our office, the address is 1401 Centerville Road, Suite 202, Tallahassee, FL 32308.


text:
Home > The Area > Curious about our past?
City of Carrabelle-Chartered in 1893 by the Florida Legislature
The history of Dog Island and Carrabelle, Florida, includes a wonderful mix of Indians, shipping, bootlegging, logging and battles. This charming town was a fishing village prior to the Civil War, isolated from the rest of Franklin County by the lack of bridges and paved roads. After the war, it became an internationally known lumber town, shipping wood from the hundreds of thousands of acres of surrounding virgin forest. Turpentine, distilled from pine sap, was shipped via schooners from the natural deep water port on the Carrabelle River.
To read more, click here
Camp Gordon Johnston World War II Museum
New 4,000 square foot WW II Museum opened 5/4/18 across from Carrabelle Beach! Tuesday-Saturday 11 am - 5 pm Free admission Pa...
Carrabelle History Museum
Museum - Gift Shop Hours: Wednesday 12-5 pm Thursday, Friday, - Saturday 10 am-5 pm Sunday 12-5 pm The Carrabelle History Mus...
Museum and Gift Shop Hours: Wednesday -Thursday - Friday - 11 am - 5 pm Saturday 9 am - 5 pm Sunday 1 -5 pm Join the Carrabel...
Dog Island Lighthouse-Replaced by Crooked River
The Dog Island Lighthouse was the first lighthouse, a 50-foot brick tower and was completed in 1839. A storm in 1842 destroyed th...
Friends of the Governor Stone, Inc.
The National Historic Landmark Schooner GOVERNOR STONE is the last know working shallow draft Gulf Coast schooner. Our mission i...
World's Smallest Police Station
In the early 1960's, Carrabelle's police phone was located in a call box that was bolted to a building at the corner of U.S. High...
Visiting Month Not Sure January February March April May June July August September October November December
Number & Type of Guests 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Couple(s) Family Group with Children Single Adults
Number & Type of Pet(s)
Type (ctrl+click to select multiple) Accommodations Carrabelle Boat Parade of Lights Community Information Fishing Real Estate RV Park