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57d96ac156 [TI] Compact: Improve section on Kolmogorov complexity proofs 2025-11-09 07:26:49 +01:00
cc0ff6d3cd [TI] Compact: Fix spacing 2025-11-09 07:14:53 +01:00
7f0392be0b [TI] Compact: More notes 2025-11-09 07:12:11 +01:00
4 changed files with 9 additions and 5 deletions

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@@ -62,11 +62,13 @@ where the Program doesn't have to compile, i.e. we can describe processes inform
\stepLabelNumber{theorem}
\fancytheorem{Prime number} $\displaystyle \limni \frac{\text{Prime}(n)}{\frac{n}{\ln(n)}}$
\fhlc{Cyan}{Proofs} Most of the proofs start with defining the number of words of exactly the required length and we can then usually deduce some kind of indirect proof
(using the fact that there are at most $2^k - 1$ words $x$ with $K(x) < k$).
\fhlc{Cyan}{Proofs} Proofs in which we need to show a lower bound for Kolmogorov-Complexity (almost) always work as follows:
Assume for contradiction that there are no words with $K(w) > f$ for all $w \in W$.
We count the number $m$ of words in $W$ and the number $n$ of programs of length $\leq f$ ($f$ being the given, lower bound).
We will have $m - n > 0$, which means, there are more different words than there are Programs with Kolmogorov-Complexity $\leq f$,
which is a contradiction to our assumption.
It is useful to remember the laws of logarithm and the fact that there are $\floor{\frac{n}{k}} + 1$ numbers divisible by $k$ in the set $\{ 0, 1, \ldots, n \}$.
Additionally, the pigeonhole principle can come in very handy
There are $\floor{\frac{n}{k}} + 1$ numbers divisible by $k$ in the set $\{ 0, 1, \ldots, n \}$.
\shade{Orange}{Laws of logarithm}
\drmvspace

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@@ -119,7 +119,8 @@ That is a contradiction, which concludes our proof
To show that a language needs \textit{at least} $n$ states, use Lemma 3.3 and $n$ words. We thus again do a proof by contradiction:
\begin{enumerate}
\item Assume that there exists FA with $|Q| < n$. We now choose $n$ words (as short as possible), as we would for non-regularity proofs using Lemma 3.3 (i.e. find some prefixes).
It is usually beneficial to choose prefixes with $|w|$ small (consider just one letter, $\lambda$, then two and more letter words)
It is usually beneficial to choose prefixes with $|w|$ small (consider just one letter, $\lambda$, then two and more letter words).
An ``easy'' way to find the prefixes is to construct a finite automaton and then picking a prefix from each class
\item Construct a table for the suffixes using the $n$ chosen words such that one of the words at entry $x_{ij}$ is in the language and the other is not. ($n \times n$ matrix, see below in example)
\item Conclude that we have reached a contradiction as every field $x_{ij}$ contains a suffix such that one of the two words is in the language and the other one is not.
\end{enumerate}

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@@ -36,6 +36,7 @@ The same ideas as with NFA apply here. The transition function also maps into th
\delta : (Q - \{ \qacc, \qrej \}) \times \Gamma \rightarrow \cP(Q \times \Gamma \times \{ L, R, N \})
\end{align*}
\drmvspace
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.

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