From: Nathaniel Wesley Filardo Date: Sun, 15 May 2016 02:39:07 +0000 (-0400) Subject: Accumulated small tweaks X-Git-Url: https://hydra-www.ietfng.org/gitweb/?a=commitdiff_plain;p=autzoo Accumulated small tweaks --- diff --git a/strautintro.tex b/strautintro.tex index f465637..eb33f2e 100644 --- a/strautintro.tex +++ b/strautintro.tex @@ -49,7 +49,7 @@ transition process.} Now a run of our nondeterministic machine looks like $C_0 s_1 C_1 \ldots s_n C_n$ where each $C_i \subseteq \config$ and $c' \in C_{i+1}$ iff $\exists_{c \in C_i} . c' \in \delta\paren{c,s_{i+1}}$. We consider the string -$\vec{s}$ accepted by the automaton if $c \in C_n \cap \config_F$; +$\vec{s}$ accepted by the automaton if $C_n \cap \config_F \ne \emptyset$; equivalently, we may say that a string is accepted if {\em there exists} a deterministic run, as defined earlier, where each $c_i \in C_i$, and $c_n \in \config_F$. @@ -130,8 +130,8 @@ Here, we use the common table to describe itself. Given a string $s \in compl={Find $B$ s.t. $\alang{B} = \alphabet^* \setminus \alang{A}$}, relcompl={Find $B$ s.t. $\alang{B} = \alang{A} \setminus \alang{A'}$}, % - intersect={Find $B$ s.t. $\alang{B} = \alang{A} \cap \alang{A'}$?}, - union={Find $B$ s.t. $\alang{B} = \alang{A} \cup \alang{A'}$?}, + intersect={Find $B$ s.t. $\alang{B} = \alang{A} \cap \alang{A'}$}, + union={Find $B$ s.t. $\alang{B} = \alang{A} \cup \alang{A'}$}, % hom={Find $B$ for hom. $h$ s.t. $\alang{B} = h\paren{\alang{A}}$}, invhom={Find $B$ for hom. $h$ s.t. $\alang{A} = h\paren{\alang{B}}$}, @@ -153,11 +153,16 @@ Here, we use the common table to describe itself. Given a string $s \in Of course, not every row in the table is independent of the others. The following implications (at least) hold in all cases: \begin{itemize} - \item Subset testing implies equivalence testing. - \item Relative complement closure implies general complement closure. - \item Intersection and general complement closures implies - relative complement closure. If, additionally, a class has - a decidable emptiness test, then it has a decidable subset test. + \item Subset testing implies equivalence testing, of which universality + and emptiness testing are often special cases (some automata + families may not be able to represent $\emptyset$ or + $\alphabet^*$). + \item Relative complement closure implies general complement closure, assuming $\alphabet^*$ is expessible. + \item Intersection and general complement closures imply + relative complement closure. + \item If a class has relative complement closure and a decidable emptiness test, + then it has a decidable subset test: + $S \subseteq T \Leftrightarrow S \setminus T = \emptyset$. \item Closure under arbitrary homomorphism implies closure under $\epsilon$-free homomorphisms. \end{itemize} diff --git a/zoo-tree/ra.tex b/zoo-tree/ra.tex index bef0a55..942cdd9 100644 --- a/zoo-tree/ra.tex +++ b/zoo-tree/ra.tex @@ -22,11 +22,10 @@ C]{jacquemard:tamodeq} and in contradiction to the statement made in {\em deterministic} and {\em complete} RAs, carries over to the non-deterministic case as well.} % -that emptiness is not decidable for the class of nondeterminstic RAs. (This -proof generalizes to a large class of tree automata; see -\autoref{sec:tree-sepex:twocm}.) We feel that a few words said about this +that emptiness is not decidable for the class of nondeterminstic RAs. We feel that a few words said about this proof here may help to de-mystify it. A decomposition if this machine into -simpler automata whose intersection is this machine may be found in +simpler automata whose intersection is this machine, which demonstrates its +applicability to a large collection of automata classes, may be found in \ref{sec:tree-sepex:twocm}. \begin{wrapfigure}{r}{3.5in}\centering\begin{tikzpicture} @@ -63,9 +62,9 @@ of the machine's execution, denoted $C_i$, and h$_{i-1}$. The equality constraints are introduced only at the top and at ``g'' nodes in the tree; the state labels are carefully constructed so that only three ranks are required to satisfy the Reduction metaconstraint: a base rank, which -includes all labels in the $C_i$ trees is $\le$ a ``g''-state rank which +includes all labels in the $C_i$ trees, is $\le$ a ``g''-state rank, which includes the labels of all ``g'' and ``h'' nodes in the tree except the -apex; the accepting state at the root is the sole member of the third rank. -There will be $2^{n-i}$ copies of $h_i$ (or $C_i$) in a tree encoding $n$ -steps. This construction is {\em deeply} dependent on the fact that +apex, and the accepting state at the root is the sole member of the third +rank. There will be $2^{n-i}$ copies of $h_i$ (or $C_i$) in a tree encoding +$n$ steps. This construction is {\em deeply} dependent on the fact that non-determinism allows equal trees to have several distinct runs.