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177 lines
9.9 KiB
TeX
177 lines
9.9 KiB
TeX
\newsection
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\section{Computability}
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\stepcounter{subsection}
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\subsection{Diagonalization}
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The \bi{set of binary encodings of all TMs} is denoted $\text{KodTM}$ and $\text{KodTM} \subseteq \wordbool$ and the upper bound of the cardinality is $|\wordbool|$,
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as there are infinitely many TMs.
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Below is a list of countable objects. They all have corresponding Lemmas in the script, but omitted here:
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\drmvspace
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\begin{multicols}{4}
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\begin{itemize}
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\item $\word$ for any $\Sigma$
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\item $\text{KodTM}$
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\item $\N \times \N$
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\item $\Q^+$
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\end{itemize}
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\end{multicols}
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\rmvspace
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\drmvspace
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The following objects are uncountable: $[0, 1]$, $\R$, $\cP(\wordbool)$
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\inlinecorollary $|\text{KodTM}| < |\cP(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$
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\fhlc{Cyan}{Proof of $L$ (not) recursively enumerable}
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Proving that a language \textit{is} recursively enumerable is as difficult as providing a Turing Machine that accepts it.
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Proving that a language is \textit{not} recursively enumerable is likely easier. For it, let $d_{ij} = 1 \Longleftrightarrow M_i$ accepts $w_j$.
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\inlineex Assume towards contradiction that $L_\text{diag} \in \cL_{RE}$. Let
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\rmvspace
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\begin{align*}
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L_{\text{diag}} & = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and $M_i$ does not accept } w_i \} \\
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& = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and } d_{ii} = 0\}
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\end{align*}
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Thus assume that, $L_\text{diag} = L(M)$ for a Turing Machine $M$.
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Since $M$ is a Turing Machine in the canonical ordering of all Turing Machines, so there exists an $i \in \N - \{ 0 \}$, such that $M = M_i$.
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This however leads to a contradiction, as $w_i \in L_\text{diag} \Longleftrightarrow d_{ii} = 0 \Longleftrightarrow w_i \notin L(M_i)$.
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In other words, $w_i$ is in $L_\text{diag}$ if and only if $w_i$ is not in $L(M_i)$, which contradicts our statement above, in which we assumed that $L_\text{diag} \in \cL_{RE}$.
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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 $w_i \notin L(M_i)$.
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\setLabelNumber{theorem}{3}
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\inlinetheorem $L_\text{diag} \notin \cL_{RE}$
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% ────────────────────────────────────────────────────────────────────
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\subsection{Reductions}
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\label{sec:reductions}
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This is the start of the topics that are explicitly part of the endterm.
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For a language to be in $\cL_R$, in contrast to $L \in \cL_{RE}$, the TM has to halt also for \texttt{no} instances, i.e. it has to be an algorithm.
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In other words: A TM $A$ can enumerate all valid strings of a \textit{recursively enumerable language} ($L \in \cL_{RE}$),
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where for \textit{recursive languages}, it has to be able to difinitively answer for both \texttt{yes} and \texttt{no} and thus halt in finite time for both.
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First off, a list of important languages for this and the next section:
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\begin{itemize}
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\item $L_U = \{ \text{Kod}(M)\# w \divides w \in \wordbool \text{ and TM $M$ accepts } w \}$ ($\in \cL_{RE}$, but $\notin \cL_R$)
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\item $L_H = \{ \text{Kod}(M)\# x \divides x \in \wordbool \text{ and TM $M$ halts on } x \}$ ($\in \cL_{RE}$, but $\notin \cL_R$)
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\item $L_{\text{diag}} = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and $M_i$ does not accept } w_i \}$ ($\notin \cL_{RE}$ and thus $\notin \cL_R$)
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\item $(L_{\text{diag}})^C$ ($\in \cL_{RE}$, but $\notin \cL_R$)
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\item $L_{EQ} = \{ \text{Kod}(M)\# \text{Kod}(\overline{M}) \divides L(M) = L(\overline{M}) \}$ ($\in \cL_{RE}$, but $\notin \cL_R$)
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\item $\lempty = \{ \text{Kod}(M) \divides L(M) = \emptyset \}$ ($\in \cL_{RE}$, but $\notin \cL_R$)
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\item $(\lempty)^C = \{ x \in \wordbool \divides x \notin \text{Kod}(\overline{M}) \forall \text{ TM } \overline{M} \text{ or } x = \text{Kod}(M) \text{ and } L(M) \neq \emptyset \}$
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($\in \cL_{RE}$, but $\notin \cL_R$)
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\item $L_{H, \lambda} = \{ \text{Kod}(M) \divides M \text{ halts on } \lambda \}$ ($\in \cL_{RE}$, but $\notin \cL_R$)
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\end{itemize}
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\newpage
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\setLabelNumber{definition}{3}
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\fancydef{Recursively reducible languages} $L_1 \leq_R L_2$ ($L_1$ reducible into $L_2$), if $L_2 \in \cL_R \Rightarrow L_1 \in \cL_R$
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\fancydef{$EE$-Reductions} $L_1 \leq_{EE} L_2$ if there exists a TM $M$ that implements image $f_M : \word_1 \rightarrow \word_2$,
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for which we have $x \in L_1 \Leftrightarrow f_M(x) \in L_2$ for all $x \in \wordbool_1$
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\setLabelNumber{lemma}{3}
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\inlinelemma If $L_1 \leq_{EE} L_2$ then also $L_1 \leq_R L_2$
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\inlinelemma For each language $L \subseteq \word$ we have: $L \leq_R L^C$ and $L^C \leq_R L$
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\setLabelNumber{theorem}{6}
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\fancytheorem{Universal TM} A TM $U$, such that $L(U) = L_U$
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\fhlc{Cyan}{Showing reductions} First, a general guide to reductions and below what else we need to keep in mind for specific reductions:
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\begin{enumerate}
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\item We construct a TM $A$ that:
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\begin{enumerate}
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\item Checks if the input has the right form and if it does not, returns some output that is $\notin L_2$
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\item Applies the transformation to all remaining input
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\end{enumerate}
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\item We show $x \in L_1 \Leftrightarrow A(x) \in L_2$ by showing the implications:
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\begin{enumerate}
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\item For $\Rightarrow$, we show it directly, by assuming that $x \in L_1$ (obviously) and we can ignore the invalid input (as that $\notin L_1$ anyway)
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\item For $\Leftarrow$, we have two options (mention what happens to invalid input here):
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\begin{itemize}
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\item We show $A(x) \in L_2 \Rightarrow x \in L_1$ directly (usually harder)
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\item We show $x \notin L_1 \Rightarrow A(x) \notin L_2$ (contraposition)
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\end{itemize}
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\end{enumerate}
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\item Show that the TM always halts (for $P$, $EE$ and $R$ reductions at least)
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\end{enumerate}
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\shade{Cyan}{$EE$-reductions} They follow the above scheme exactly
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\shade{Cyan}{$R$-reductions} It is usually a good idea to draw the setup here. We have a TM $C$ that basically executes an $EE$-reduction and we have a TM $A$ that can check $L_2$.
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Then, we have a TM $B$ that wraps the whole thing: It first executes TM $C$, which will either output an transformation of $L_1$ for $L_2$ (i.e. execute an $EE$-reduction) or
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output some encoding for \textit{invalid word}.
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If it outputs the encoding for \textit{invalid word}, $B$ will output $x \notin L_1$.
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If $C$ does not output an encoding for \textit{invalid word}, then $B$ will execute $A$ on the output of $C$ and then use the output of $A$ (either accepting or rejecting)
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to output the same (i.e. if $A$ accepts, then $B$ will output $x \in L_1$ and if $A$ rejects, $B$ outputs $x \notin L_1$)
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\inlineintuition In $R$-reductions, we construct a full verifier for $L_1$ using the verifier for $L_2$, i.e. we can use TM $B$ directly to check if a given word is in $L_1$
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given that the transformed word is also in $L_2$.
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\shade{Cyan}{$P$-reductions} (Used in Chapter \ref{sec:complexity}). We need to also show that $A$ terminates in polynomial time.
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\shade{orange}{Tips \& Tricks:}
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\begin{itemize}
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\item The TM $A$ has to terminate always
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\item Check the input for the correct form first
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\item For the correctness, show $x \in L_1 \Leftrightarrow A(x) \in L_2$
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\item The following tricks can be useful:
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\begin{itemize}
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\item Transitions into $\qacc$ and $\qrej$ can be redirected to $\qacc / \qrej$ or into an infinite loop
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\item Construct TM $M'$ that ignores input and does the same, regardless of input
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\end{itemize}
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\item Generate encoding of a $TM$ with special properties (e.g. accepts all input, never halts, \dots)
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\end{itemize}
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% ────────────────────────────────────────────────────────────────────
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\newpage
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\subsection{Rice's Theorem}
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\setLabelNumber{definition}{7}
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\inlinedef $L$ is called a \bi{semantically non-trivial decision problem}, if these conditions apply:
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\begin{enumerate}[label=\textit{(\roman*)}]
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\item There exists a TM $M_1$, such that $\text{Kod}(M_1) \in L$ (i.e. $L \neq \emptyset$)
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\item There exists a TM $M_2$, such that $\text{Kod}(M_2) \notin L$ (not all encodings are in $L$)
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\item For two TM $A$ and $B$: $L(A) = L(B) \Rightarrow \text{Kod}(A) \in L \Rightarrow \text{Kod}(B) \in L$
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\end{enumerate}
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\setLabelNumber{theorem}{9}
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\fancytheorem{Rice's Theorem} Every semantically non-trivial decision problem over TMs is undecidable
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\fhlc{Cyan}{Using Rice's Theorem} We only need to show that a language is semantically non-trivial, which we do by checking the above conditions.
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For the third condition, intuitively, we only need to check if in the definition of $L$ only $L(M)$ appears and nowhere $M$ directly (except of course, to say that $M$ has to be a TM),
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or the condition can be restated such that only $L(M)$ is described by it.
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For a more formal proof of that condition, simply show that the implication holds
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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:
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\rmvspace
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\begin{align*}
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L, L^C \in \cL_{RE} \Longleftrightarrow L \in \cL_R
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\end{align*}
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\stepcounter{subsection}
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\subsection{The method of the Kolmogorov-Complexity}
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\inlinetheorem The problem of computing the Kolmogorov-Complexity $K(x)$ for each $x$ is algorithmically unsolvable.
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\inlinelemma If $L_H \in \cL_R$, then there exists an algorithm to compute the Kolmogorov-Complexity $K(x)$ for each $x \in \wordbool$
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As of HS2025, chapters 5.5 and 5.7 are not relevant for the Endterm or Session exam, so they are omitted here
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