[TI] Compact: Fixes

semester3/ti-compact/parts/03_turing-machines.tex

@@ -9,6 +9,11 @@ and repeat until we only have the new symbol, at which point we accept, or there
 
 The Turing Machines have an accepting $\qacc$ and a rejecting state $\qrej$ and a configuration is an element of
 $\{ \{ \cent \}\cdot \Gamma^* \cdot Q \cdot \Gamma^+ \cup Q \cdot \{ \cent \} \cdot \Gamma^+ \}$ with $\cdot$ being the concatenation and $\cent$ the marker of the start of the band.
+\rmvspace
+\begin{align*}
+    \cL_{RE} & = \{ L(M) \divides M \text{ is a TM} \}\\
+    \cL_R    & = \{ L(M) \divides M \text{ is a TM and it always halts} \}
+\end{align*}
 
 
 \subsection{Multi-tape TM and Church's Thesis}

semester3/ti-compact/parts/04_computability.tex

@@ -22,6 +22,9 @@ The following objects are uncountable: $[0, 1]$, $\R$, $\cP(\wordbool)$
 
 \inlinecorollary $|\text{KodTM}| < |\cP(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$
 
+\setLabelNumber{theorem}{3}
+\inlinetheorem $L_\text{diag} \notin \cL_{RE}$
+
 \fhlc{Cyan}{Proof of $L$ (not) recursively enumerable}
 
 Proving that a language \textit{is} recursively enumerable is as difficult as providing a Turing Machine that accepts it.
@@ -43,12 +46,26 @@ In other words, $w_i$ is in $L_\text{diag}$ if and only if $w_i$ is not in $L(M_
 
 In other, more different, words, $w_i$ being in $L_\text{diag}$ implies (from the definition) that $d_{ii} = 0$, which from its definition implies that $w_i \notin L(M_i)$.
 
-\setLabelNumber{theorem}{3}
-\inlinetheorem $L_\text{diag} \notin \cL_{RE}$
+\drmvspace
+\proven
+
+
+Another result (not formally proven in the script, but there is a proof by intimidation) that can come in useful, especially when trying to show $L \notin \cL_{RE}$ is:
+\rmvspace
+\begin{align*}
+    L, L^C \in \cL_{RE} \Longleftrightarrow L \in \cL_R
+\end{align*}
+
+\drmvspace
+Additionally, as a reminder, $\cL_{RE} = \{ L(M) \divides M \text{ is a TM} \}$, so to prove that a language $L \notin \cL_{RE}$,
+we only need to show that there exists no TM $M$, for which $L(M) \in \cL_{RE}$.
+
+
 
 % ────────────────────────────────────────────────────────────────────
 
 
+\newpage
 \subsection{Reductions}
 \label{sec:reductions}
 This is the start of the topics that are explicitly part of the endterm.
@@ -70,7 +87,7 @@ First off, a list of important languages for this and the next section:
     \item $L_{H, \lambda} = \{ \text{Kod}(M) \divides M \text{ halts on } \lambda \}$ ($\in \cL_{RE}$, but $\notin \cL_R$)
 \end{itemize}
 
-\newpage
+
 \setLabelNumber{definition}{3}
 \fancydef{Recursively reducible languages} $L_1 \leq_R L_2$ ($L_1$ reducible into $L_2$), if $L_2 \in \cL_R \Rightarrow L_1 \in \cL_R$
 
@@ -138,7 +155,6 @@ given that the transformed word is also in $L_2$.
 % ────────────────────────────────────────────────────────────────────
 
 
-\newpage
 \subsection{Rice's Theorem}
 \setLabelNumber{definition}{7}
 \inlinedef $L$ is called a \bi{semantically non-trivial decision problem}, if these conditions apply:
@@ -159,13 +175,6 @@ or the condition can be restated such that only $L(M)$ is described by it.
 For a more formal proof of that condition, simply show that the implication holds
 
 
-Another result (not formally proven in the script, but there is a proof by intimidation) that can come in useful, especially when trying to show $L \notin \cL_{RE}$ is:
-\rmvspace
-\begin{align*}
-    L, L^C \in \cL_{RE} \Longleftrightarrow L \in \cL_R
-\end{align*}
-
-
 \stepcounter{subsection}
 \subsection{The method of the Kolmogorov-Complexity}
 \inlinetheorem The problem of computing the Kolmogorov-Complexity $K(x)$ for each $x$ is algorithmically unsolvable.

semester3/ti-compact/parts/05_complexity.tex

@@ -175,14 +175,19 @@ $A$ is called a polynomial reduction of $L_1$ into $L_2$
 A few languages commonly used to show $NP$-completeness:
 \begin{itemize}
     \item $SAT = \{ \Phi \divides \Phi \text{ is a satisfiable formula in CNF} \}$
+    \item $3SAT = \{ \Phi \divides \Phi \text{ is a satisfiable formula in CNF with all clauses containing \textit{at most} three literals} \}$
     \item $\text{CLIQUE} = \{ (G, k) \divides G \text{ is an undirected graph that contains a $k$-clique } \}$
     \item $VC = \{ (G, k) \divides G \text{ is an undirected graph with a vertex cover of size $\leq k$ } \}$
-    \item $3SAT = \{ \Phi \divides \Phi \text{ is a satisfiable formula in CNF with all clauses containing \textit{at most} three literals} \}$
+    \item $SCP = \{ (X, \cS, k) \divides X \text{ has a set cover $\cC \subseteq \cS$ such that  $|\cC| \leq k$ } \}$
+    \item $DS = \{ (G, k) \divides G \text{ has a dominating set $D$ such that } |D| \leq k \}$
 \end{itemize}
-where a $k$-clique is a complete subgraph consisting of $k$ vertices in $G$, with $k \leq |V|$.
-and a vertex cover is any set $U \subseteq V$ where all edges $\{ u, v \} \in E$ have at least one endpoint $u, v \in U$
+where a $k$-clique is a complete subgraph consisting of $k$ vertices in $G$, with $k \leq |V|$;
+where a subset $\cC \subseteq \cS$ is a \textit{set cover} of $X$ if $X = \bigcup_{S \in \cC}$;
+where a \textit{dominating} set is is a set $D \subseteq V$ such that for every vertex $v \in V$, $v \in D$ or exists $w \in D$ such that $\{ v, w \} \in E$
+and where a vertex cover is any set $U \subseteq V$ where all edges $\{ u, v \} \in E$ have at least one endpoint $u, v \in U$
 
-We have $SAT \leq_p \text{CLIQUE}$, $SAT \leq_p 3SAT$, $\text{CLIQUE} \leq_p VC$
+We have $SAT \leq_p \text{CLIQUE}$, $SAT \leq_p 3SAT$, $\text{CLIQUE} \leq_p VC$, $VC \leq_p SCP$ and $SCP \leq_p DS$.
+Logically, we also have $SAT \leq_p DS$, etc, since $\leq_p$ is transitive (in fact, all reductions are transitive)
 
 Additionally, $\text{MAX-SAT}$ and $\text{MAX-CL}$, the problem to determine the maximum number of fulfillable clauses in a formula $\Phi$
 and the problem to determine the maximum clique, respectively, are $NP$-hard