[TI] Start compact summary

README.md

@@ -16,6 +16,7 @@ If you find mistakes, please open an issue or fix them yourself and open a PR.
 
 ### Semester 3
 - [Theoretical Computer Science Summary (DE)](./semester3/ti/ti-summary.pdf) Author: Janis Hutz
+- [Theoretical Computer Science Compact Summary (EN)](./semester3/ti-compact/ti-compact.pdf) Author: Janis Hutz
 - [Numerical Methods for Computer Science Summary (DE)](./semester3/numcs/numcs-summary.pdf) Authors: Robin Bacher, Janis Hutz
 - [Systems Programming and Computer Architecture Summary (EN)](./semester3/spca/spca-summary.pdf) Author: Janis Hutz (*NOTE: has not been started yet, may or may not happen*)
 - [Analysis II Cheat-Sheet (EN)](./semester3/analysis-ii/analysis-ii-cheat-sheet.pdf) Author: Janis Hutz

semester3/ti-compact/parts/01_words-alphabets.tex

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+\newsection
+\section{Alphabets, Words, etc}
+\stepcounter{subsection}
+\subsection{Alphabets, Words, Languages}
+\fancydef{Alphabet} Set $\Sigma$.
+Important alphabets: $\alphabetbool$, $\alphabets{lat}$ (all latin chars), $\alphabets{Keyboard}$ (all chars on keyboard), $\Sigma_m$ ($m$-ary numbers)
+
+\fancydef{Word} Possibly empty (denoted $\lambda$) sequences of characters from $\Sigma$. $|w|$ is the length, 
+$\Sigma^*$ is the set of all words and $\Sigma^+ = \Sigma^* - \{ \lambda \}$
+
+\fancydef{Konkatenation} $\text{Kon}(x, y) = xy$, (so like string concat). $(xy)^n$ is $n$-times repeated concat.
+
+\fancydef{Reversal} $a^R$, simply read the word backwards.
+
+\stepLabelNumber{definition}
+\fancydef{Prefix, Suffix, Subword} $v$ in $w = vy$; $s$ in $w = sx$; Subword is $u$ in $w = xuy$
+
+\fancydef{Appearance} $|x|_a$ is the number of times $a \in \Sigma$ appears in $x$
+
+\fancydef{Canonical ordering} Ordered by length and then by first non-common letter:
+\rmvspace
+\begin{align*}
+    u < v \Longleftrightarrow |u| < |v| \lor (|u| = |v| \land u = x \cdot s_i \cdot u' \land v = x \cdot s_j \cdot v') \text{ for any } x, u', v' \in \word \text{ and } i < j
+\end{align*}
+
+\drmvspace
+\fancydef{Language} $L \subseteq \word$, and we define $L^C = \word - L$ as the complement, with $L_{\emptyset}$ being the empty language, 
+whereas $L_\lambda$ is the language with just the empty word in it.
+
+\bi{Concatenation}: $L_1 \cdot L_2 = \{ vw | v \in L_1 \land w \in L_2 \}$ and $L^{i + 1} = L^i \cdot L \ \forall i \in \N$.
+
+\bi{Cleen Star}: $L^* = \bigcup_{i \in \N} L^i$ and $L^+ = L \cdot L^*$
+
+
+\inlinelemma $L_1L_2 \cup L_1 L_2 = L_1(L_2 \cup L_3)$
+\inlinelemma $L_1(K_2 \cap L_3) \subseteq L_1 L_2 \cap L_1 L_3$
+
+
+\stepcounter{subsection}
+\subsection{Kolmogorov-Complexity}
+\setLabelNumber{definition}{17}
+\fancydef{Kolmogorov-Complexity} $K(x)$ for $x \in \wordbool$ is the minimum of all binary lengths of Pascal programs that output $x$, 
+where the Program doesn't have to compile, i.e. we can describe processes informally
+
+\stepLabelNumber{lemma}
+\inlinelemma For each word $x$ exists constant $d$ s.t. $K(x) \leq |x| + d$, for which we can use a program that simply includes a \texttt{write(x)} command
+
+\fancydef{Of natural number} $K(n) = K(\text{Bin}(x))$ with $|\text{Bin}(x)| = \ceil{\log_2(x + 1)}$
+
+\inlinelemma For each $n \in \N \exists w_n \in (\alphabetbool)^n$ s.t. $K(w_n) \geq |w_n| = n$, i.e. exists a non-compressible word.
+
+\inlinetheorem Kolmogorov-Complexity doesn't depend on programming language. It only differs in constant
+
+\fancydef{Randomness} $x$ random if $K(x) \geq |x|$, thus for $n$, $K(n) \geq \ceil{\log_2(n + 1)} - 1$
+
+\stepLabelNumber{theorem}
+\fancytheorem{Prime number} $\displaystyle \limni \frac{\text{Prime}(n)}{\frac{n}{\ln(n)}}$

semester3/ti-compact/parts/02_finite-automata.tex

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+\newsection
+\section{Finite Automata}
+\stepcounter{subsection}
+\subsection{Representation}
+We can note the automata using graphical notation similar to graphs or as a series of instructions like this:
+\rmvspace
+\begin{align*}
+    \texttt{select } input & = a_1 \texttt{ goto } i_1 \\[-0.2cm]
+    \vdots                                             \\
+    input                  & = a_k \texttt{ goto } i_k
+\end{align*}
+
+\drmvspace
+\fancydef{Finite Automaton}
+\drmvspace
+\begin{multicols}{2}
+    \begin{itemize}
+        \item $Q$ set of states
+        \item $\Sigma$ input alphabet
+        \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)
+    \end{itemize}
+\end{multicols}
+
+\drmvspace
+$\hat{\delta}(q_0, w) = p$ is the end state reached when we process word $w$ from state $q_0$, and $(q, w) \bigvdash{M}{*} (p, \lambda)$ is the formal definition,
+with $\bigvdash{M}{*}$ representing any number of steps $\bigvdash{M}{}$ executed.
+
+The class $\class[q_i]$ represents all possible words for which the FA is in this state.
+Be cautious when defining them, make sure that no extra words from other classes could appear in the current class, if this is not intended.
+
+Sometimes, we need to combine two (or more) FA to form one larger one.
+We can do this easily with product automata. To create one from two automata $M_1$ (states $q_i$) and $M_2$ (states $p_j$) we do the following steps:
+\rmvspace
+\begin{enumerate}[noitemsep]
+    \item Write down the states as tuples of the form $(q_i, p_j)$ (i.e. form a grid by writing down one of the automata vertically and the other horizontally)
+    \item From each state, the automata on the horizontal axis decides for the input symbol if we move left or right,
+          whereas the automata on the vertical axis decides if we move up or down.
+\end{enumerate}
+
+\input{parts/02a_example-automata.tex}
+
+
+\stepcounter{subsection}
+\subsection{Proofs of nonexistence}
+We have three approaches to prove non-regularity of words.
+Below is an informal guide as to how to do proofs using each of the methods and possible pitfalls.
+
+For all of them start by assuming that $L$ is regular.
+
+\fhlc{Cyan}{Lemma 3.3}
+\rmvspace
+\begin{enumerate}[noitemsep]
+    \item Pick a FA $A$ over $\Sigma$ and say that $L(A) = L$
+    \item Pick $|Q| + 1$ words $x$ such that $xy = w \in L$ with $|y| > 0$.
+    \item State that via pigeonhole principle there exists w.l.o.g $i < j \in \{ 1, \ldots, |Q| + 1 \}$, s.t. $\hdelta_A(q_0, x_i) = \hdelta_A(q_0, x_j)$.
+    \item Build contradiction by picking $z$ such that $x_i z \in L$.
+    \item Then, if $z$ was picked properly, since $i < j$, we have that $x_j z \notin L$, since the lengths do not match
+\end{enumerate}
+
+\rmvspace
+That is a contradiction, which concludes our proof
+
+
+\fhlc{Cyan}{Pumping Lemma}
+\rmvspace
+\begin{enumerate}[noitemsep]
+    \item State that according to Lemma 3.4 there exists a constant $n_0$ such that $|w| \geq n_0$.
+    \item Choose a word $w \in L$ that is sufficiently long to enable a sensible decomposition for the next step.
+    \item Choose a decomposition, such that $|yx| = n_0$ (makes it quite easy later). Specify $y$ and $x$ in such a way that for $|y| = l$ and $|x| = m$ we have $l + m \leq n_0$
+    \item According to Lemma 3.4 (ii), $m \geq 1$ and thus $|x| \geq 1$. Fix $z$ to be the suffix of $w = yxz$
+    \item Then according to Lemma 3.4 (iii), fill in for $X = \{ yx^k z \divides k \in \N \}$ we have $X \subseteq L$.
+    \item This will lead to a contradiction commonly when setting $k = 0$, as for a language like $0^n1^n$, we have $0^{(n_0 - m) + km}1^{n_0}$ as the word (with $n_0 - m = l$),
+          which for $k = 0$ is $u= 0^{n_0 - m} 1^{n_0}$ and since $m \geq 1$, $u \notin L$ and thus by Lemma 3.4, $X \cap L = \emptyset$
+\end{enumerate}
+
+
+\fhlc{Cyan}{Kolmogorov Complexity}
+\begin{enumerate}[noitemsep]
+    \item We first need to choose an $x$ such that $L_x = \{ y | xy \in L \}$.
+          If not immediately apparent, choosing $x = a^{\alpha + 1}$ for $a \in \Sigma$
+          and $\alpha$ being the exponent of the exponent of the words in the language after a variable rename.
+          For example, for $\{ 0^{n^2 + 2n} \divides n \in \N \}$, $\alpha(m) = m^2 + 2m$.
+          Another common way to do this is for languages of the form $\{ a^n b^n \divides n \in \N \}$ to use $x = a^m$ and
+          $L_{0^m} = \{ y | 0^m y \in L \} = \{ 0^j 1^{m + j} | j \in \N \}$.
+    \item Find the first word $y_1 \in L_x$. In the first example, this word would be $y_1 = 0^{(m + 1)^2 \cdot 2(m + 1) - m^2 \cdot 2m + 1}$,
+          or in general $a^{\alpha(m + 1) - \alpha(m) + 1}$.
+          For the second example, the word would be $y_1 = 1^m$, i.e. with $j = 0$
+    \item According to Theorem 3.1, there exists constant $c$ such that $K(y_k) \leq \ceil{\log_2(k + 1)} + c$. We often choose $k = 1$,
+          so we have $K(y_1) \leq \ceil{\log_2(1 + 1)} + c = 1 + c$ and with $d = 1 + c$, $K(y_1) \leq d$
+    \item This however leads to a contradiction, since the number of programs with length $\leq d$ is at most $2^d$ and thus finite and our set $L_x$ is infinite.
+\end{enumerate}
+
+
+\subsection{Non-determinism}
+The most notable differences between deterministic and non-deterministic FA is that the transition function maps is different: $\delta: Q \times \Sigma \rightarrow \mathcal{P}(Q)$.
+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.
+
+It is also possible for there to not be a transition function for a certain element of the input alphabet.
+In that case, regardless of state, the NFA rejects, as it ``gets stuck'' in a state and can't finish processing.
+
+Additionally, the NFA accepts $x$ if it has at least one accepting calculation on $x$.
+
+\stepLabelNumber{theorem}
+\inlinetheorem For every NFA $M$ there exists a FA $A$ such that $L(M) = L(A)$. They are then called \bi{equivalent}
+
+
+\fhlc{Cyan}{Potenzmengenkonstruktion}
+States are no now sets of states of the NFA in which the NFA could be in after processing the preceding input elements and we have a special state called $q_{\text{trash}}$.
+
+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 $\mathcal{B}_M(x)$.
+So, in other words, we execute BFS on the calculation tree.
+
+% TODO: Insert worked example for minimum number of states

semester3/ti-compact/parts/02a_example-automata.tex

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+\begin{center}
+    \begin{tikzpicture}[node distance = 1cm and 2cm, >={Stealth[round]}]
+        \node[state, initial left, accepting] (q0p0) {$q_0, p_0$};
+        \node[state] (q0p1) [right=of q0p0] {$q_0, p_1$};
+        \node[state] (q0p2) [right=of q0p1] {$q_0, p_2$};
+        \node[state, accepting] (q1p0) [below=of q0p0] {$q_1, p_0$};
+        \node[state] (q1p1) [right=of q1p0] {$q_1, p_1$};
+        \node[state] (q1p2) [right=of q1p1] {$q_1, p_2$};
+        \node[state, accepting] (q2p0) [below=of q1p0] {$q_2, p_0$};
+        \node[state, accepting] (q2p1) [right=of q2p0] {$q_2, p_1$};
+        \node[state, accepting] (q2p2) [right=of q2p1] {$q_2, p_2$};
+
+        \path[->]
+        % Level 0
+        (q0p0) edge node [above] {a} (q0p1)
+        (q0p1) edge node [above] {a} (q0p2)
+        (q0p2) edge [bend right] node [above] {a} (q0p0)
+        % Level 0 to level 1
+        (q0p0) edge node [right] {b} (q1p0)
+        (q0p1) edge node [right] {b} (q1p1)
+        (q0p2) edge node [right] {b} (q1p2)
+        % Level 1 to level 2
+        (q1p0) edge node [above] {a} (q2p1)
+        (q1p1) edge node [above] {a} (q2p2)
+        (q1p2) edge node [right, xshift=0.3cm] {a} (q2p0)
+        % Level 2 to level 1
+        (q2p0) edge node [right] {b} (q1p0)
+        (q2p1) edge node [above left, yshift=0.1cm] {b} (q1p1)
+        (q2p2) edge node [right] {b} (q1p2)
+        % Level 2
+        (q2p0) edge node [above] {a} (q2p1)
+        (q2p1) edge node [above] {a} (q2p2)
+        (q2p2) edge [bend left] node [below] {a} (q2p0)
+        % ────────────────────────────────────────────────────────────────────
+        % Loops on level 1
+        (q1p0) edge [loop left] node {b} ()
+        (q1p1) edge [loop left] node {b} ()
+        (q1p2) edge [loop left] node {b} ();
+    \end{tikzpicture}
+\end{center}
+
+For the automata
+\begin{figure}[h!]
+    \begin{subfigure}{0.49\textwidth}
+        \begin{center}
+            \begin{tikzpicture}[node distance = 1cm, >={Stealth[round]}]
+                \node[state, initial left, accepting] (p_0) {$p_0$};
+                \node[state] (p_1) [right=of p_0] {$p_1$};
+                \node[state] (p_2) [right=of p_1] {$p_2$};
+
+                \path[->]
+                (p_0) edge node [above] {a} (p_1)
+                (p_1) edge node [above] {a} (p_2)
+                (p_2) edge [bend left] node [below] {a} (p_0)
+                (p_0) edge [loop above] node {b} ()
+                (p_1) edge [loop above] node {b} ()
+                (p_2) edge [loop above] node {b} ();
+            \end{tikzpicture}
+        \end{center}
+        \caption{Module to compute $|w|_b \equiv |w| (\text{mod } 3$). States $q \in Q_a$}
+    \end{subfigure}
+    \begin{subfigure}{0.49\textwidth}
+        \begin{center}
+            \begin{tikzpicture}[node distance = 1cm, >={Stealth[round]}]
+                \node[state, initial left] (q_0) {$q_0$};
+                \node[state] (q_1) [right=of q_0] {$q_1$};
+                \node[state, accepting] (q_2) [right=of q_1] {$q_2$};
+
+                \path[->]
+                (q_0) edge node [above] {b} (q_1)
+                (q_1) edge [bend left] node [above] {a} (q_2)
+                (q_2) edge [bend left] node [below] {b} (q_1)
+                (q_0) edge [loop above] node {a} ()
+                (q_1) edge [loop above] node {b} ()
+                (q_2) edge [loop above] node {a} ();
+            \end{tikzpicture}
+        \end{center}
+        \caption{Module to compute $w$ contains sub. $ba$ and ends in $a$. States $p \in Q_b$}
+    \end{subfigure}
+    \caption{Graphical representation of the Finite Automaton of Task 9 in 2025}
+\end{figure}

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

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+\newsection
+\section{Turing Machines}
+\setcounter{subsection}{2}
+\subsection{Representation}
+Turing machines are much more capable than FA and NFA. A full definition of them can be found in the book on pages 96 - 98 (= pages 110 - 112 in the PDF).
+
+For example, to detect a recursive language like $\{ 0^n 1^n \divides n \in \N \}$ we simply replace the left and rightmost symbol with a different one
+and repeat until we only have the new symbol, at which point we accept, or there are no more $0$s or $1$s, at which point we reject.
+
+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.
+
+
+\subsection{Multi-tape TM and Church's Thesis}
+$k$-Tape Turing machines have $k$ extra tapes that can be written to and read from, called memory tapes. They \textit{cannot} write to the input tape.
+Initially the memory tapes are empty and we are in state $q_0$.
+All read/write-heads of the memory tapes can move in either direction, granted they have not reached the far left end, marked with $\cent$.
+
+As with normal TMs, the Turing Machine $M$ accepts $w$ if and only if $M$ reaches the state $\qacc$ and rejects if it does not terminate or reaches the state $\qrej$
+
+\inlinelemma There exists and equivalent $1$-Tape-TM for every TM.
+
+\inlinelemma There exists an equivalent TM for each Multi-tape TM.
+
+
+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}.
+
+
+\subsection{Non-Deterministic Turin 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 \})
+\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),
+we again apply BFS to the NTM's calculation tree.
+
+\stepLabelNumber{theorem}
+\inlinetheorem For an NTM $M$ exists a TM $A$ s.t. $L(M) = L(A)$ and if $M$ doesn't contain infinite calculations on words of $(L(M))^C$, then $A$ always stops.

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

@@ -0,0 +1,43 @@
+\newsection
+\section{Computability}
+\stepcounter{subsection}
+\subsection{Diagonalization}
+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 $|\wordbool|$,
+as there are infinitely many TMs.
+
+Below is a list of countable objects. They all have corresponding Lemmas in the script, but omitted here:
+\drmvspace
+\begin{multicols}{4}
+    \begin{itemize}
+        \item $\word$ for any $\Sigma$
+        \item $\text{KodTM}$
+        \item $\N \times \N$
+        \item $\Q^+$
+    \end{itemize}
+\end{multicols}
+
+\rmvspace
+\drmvspace
+The following objects are uncountable: $[0, 1]$, $\R$, $\mathcal{P}(\wordbool)$
+
+\inlinecorollary $|\text{KodTM}| < |\mathcal{P}(\wordbool)|$ 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 is \textit{not} recursively enumerable is a bit harder. 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 
+\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\}
+\end{align*}
+Thus assume that, $L_\text{diag} = L(M)$ for a Turing Machine $M$.
+Since $M$ is a Turing Machine in the canonical ordering of all Turing Machines, so there exists an $i \in \N - \{ 0 \}$, such that $M = M_i$.
+
+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.

semester3/ti-compact/ti-compact.tex

@@ -0,0 +1,34 @@
+\documentclass{article}
+
+\newcommand{\dir}{~/projects/latex} % IMPORTANT: No trailing slashes!
+\input{\dir/include.tex}
+\load{recommended}
+\usetikzlibrary{automata, positioning, arrows.meta}
+\newcommand{\hdelta}{\hat{\delta}}
+\newcommand{\qacc}{q_{\text{accept}}}
+\newcommand{\qrej}{q_{\text{reject}}}
+
+\setup{Theoretical Computer Science - Compact}
+
+\begin{document}
+\startDocument
+\usetcolorboxes
+
+
+\section{Introduction}
+This summary aims to provide a simple, easy to understand and short overview over the topics covered, with approaches for proofs, important theorems and lemmas,
+as well as definitions.
+
+It does not aim to serve as a full replacement for the book or my main summary, but as a supplement to both of them.
+
+It also lacks some formalism and is only intended to give some intuition
+
+
+\input{parts/01_words-alphabets.tex}
+\input{parts/02_finite-automata.tex}
+\input{parts/03_turing-machines.tex}
+\input{parts/04_computability.tex}
+
+
+
+\end{document}