From 19ca631705932d4e5e8cd856a694c327bb3c0e35 Mon Sep 17 00:00:00 2001 From: Janis Hutz Date: Tue, 4 Nov 2025 15:57:36 +0100 Subject: [PATCH] [TI] Compact: Add better explanation --- .../ti-compact/parts/02_finite-automata.tex | 6 +++--- .../ti-compact/parts/03_turing-machines.tex | 4 ++-- .../ti-compact/parts/04_computability.tex | 19 +++++++++--------- semester3/ti-compact/ti-compact.pdf | Bin 330522 -> 330810 bytes 4 files changed, 15 insertions(+), 14 deletions(-) diff --git a/semester3/ti-compact/parts/02_finite-automata.tex b/semester3/ti-compact/parts/02_finite-automata.tex index eb26576..237f9c9 100644 --- a/semester3/ti-compact/parts/02_finite-automata.tex +++ b/semester3/ti-compact/parts/02_finite-automata.tex @@ -20,7 +20,7 @@ We can note the automata using graphical notation similar to graphs or as a seri \item $\delta(q, a) = p$ transition from $q$ on reading $a$ to $p$ \item $q_0$ initial state \item $F \subseteq Q$ accepting states - \item $\mathcal{L}_{EA}$ regular languages (accepted by FA) + \item $\cL_{EA}$ regular languages (accepted by FA) \end{itemize} \end{multicols} @@ -144,7 +144,7 @@ Thus, all four words have to lay in pairwise distinct states and we thus need at \subsection{Non-determinism} -The most notable differences between deterministic and non-deterministic FA is that the transition function maps is different: $\delta: Q \times \Sigma \rightarrow \mathcal{P}(Q)$. +The most notable differences between deterministic and non-deterministic FA is that the transition function maps is different: $\delta: Q \times \Sigma \rightarrow \cP(Q)$. I.e., there can be any number of transitions for one symbol from $\Sigma$ from each state. This is (in graphical notation) represented by arrows that have the same label going to different nodes. @@ -162,5 +162,5 @@ States are no now sets of states of the NFA in which the NFA could be in after p For each state, the set of states $P = \hdelta(q_0, z)$ for $|z| = n$ represents all possible states that the NFA could be in after doing the first $n$ calculations. -Correspondingly, we add new states if there is no other state that is in the same branch of the calculation tree $\mathcal{B}_M(x)$. +Correspondingly, we add new states if there is no other state that is in the same branch of the calculation tree $\cB_M(x)$. So, in other words, we execute BFS on the calculation tree. diff --git a/semester3/ti-compact/parts/03_turing-machines.tex b/semester3/ti-compact/parts/03_turing-machines.tex index c708000..90fb2de 100644 --- a/semester3/ti-compact/parts/03_turing-machines.tex +++ b/semester3/ti-compact/parts/03_turing-machines.tex @@ -26,14 +26,14 @@ As with normal TMs, the Turing Machine $M$ accepts $w$ if and only if $M$ reache Church's Thesis states that the Turing Machines are a formalization of the term ``Algorithm''. It is the only axiom specific to Computer Science. -All the words that can be accepted by a Turing Machine are elements of $\mathcal{L}_{RE}$ and are called \bi{recursively enumerable}. +All the words that can be accepted by a Turing Machine are elements of $\cL_{RE}$ and are called \bi{recursively enumerable}. \subsection{Non-Deterministic Turing Machines} The same ideas as with NFA apply here. The transition function also maps into the power set: \rmvspace \begin{align*} - \delta : (Q - \{ \qacc, \qrej \}) \times \Gamma \rightarrow \mathcal{P}(Q \times \Gamma \times \{ L, R, N \}) + \delta : (Q - \{ \qacc, \qrej \}) \times \Gamma \rightarrow \cP(Q \times \Gamma \times \{ L, R, N \}) \end{align*} Again, when constructing a normal TM from a NTM (which is not required at the Midterm, or any other exam for that matter in this course), diff --git a/semester3/ti-compact/parts/04_computability.tex b/semester3/ti-compact/parts/04_computability.tex index fea1343..9273a2c 100644 --- a/semester3/ti-compact/parts/04_computability.tex +++ b/semester3/ti-compact/parts/04_computability.tex @@ -18,19 +18,18 @@ Below is a list of countable objects. They all have corresponding Lemmas in the \rmvspace \drmvspace -The following objects are uncountable: $[0, 1]$, $\R$, $\mathcal{P}(\wordbool)$ +The following objects are uncountable: $[0, 1]$, $\R$, $\cP(\wordbool)$ -\inlinecorollary $|\text{KodTM}| < |\mathcal{P}(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$ +\inlinecorollary $|\text{KodTM}| < |\cP(\wordbool)|$ and thus there exist infinitely many not recursively enumerable languages over $\alphabetbool$ \fhlc{Cyan}{Proof of $L$ (not) recursively enumerable} -Proving that a language \textit{is} recursively enumerable is as easy as providing a Turing Machine that accepts it. +Proving that a language \textit{is} recursively enumerable is as difficult as providing a Turing Machine that accepts it. -Proving that a language is \textit{not} recursively enumerable is a bit harder. For it, let $d_{ij} = 1 \Longleftrightarrow M_i$ accepts $w_j$. +Proving that a language is \textit{not} recursively enumerable is likely easier. For it, let $d_{ij} = 1 \Longleftrightarrow M_i$ accepts $w_j$. -As an example, we'll use the following language - -Assume towards contradiction that $L_\text{diag} \in \mathcal{L}_{RE}$. Let +\inlineex Assume towards contradiction that $L_\text{diag} \in \cL_{RE}$. Let +\rmvspace \begin{align*} 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 \} \\ & = \{ w \in \wordbool \divides w = w_i \text{ for an } i \in \N - \{ 0 \} \text{ and } d_{ii} = 0\} @@ -40,7 +39,9 @@ Since $M$ is a Turing Machine in the canonical ordering of all Turing Machines, This however leads to a contradiction, as $w_i \in L_\text{diag} \Longleftrightarrow d_{ii} = 0 \Longleftrightarrow w_i \notin L(M_i)$. -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 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 $L_\text{diag} \in \cL_{RE}$ + +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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