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Languages & Machines/Introduction.md
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55
Languages & Machines/Introduction.md
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---
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type: theoretical
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ty:
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---
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## The foundations of computation
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Looking for answers for basic questions like:
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- Computability?
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- Power $\leftrightarrow$ Programming constructs?
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Which leads us to fundamental concepts like:
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- State
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- Transitions
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- Non-determinism
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- Undecideability
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## Models
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### Finite memory
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Finite automata, regexp
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### Finite memory with stack
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Push down automata
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### Unrestricted
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Turing machines
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## Grammars
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On a higher level, it seems like grammars and machines are very different, but parsing a language (a set of strings) is quite similar to computation.
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## State-based systems and glossary
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An FSM can be a specification for OOP.
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- States ($q_0,\ldots, q_n$)
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- Transitions ($a,b,c,\ldots,z$)
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- We can interpret the transitions as class methods and specify the sequences of allowed invocations - **typestate**
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## Notation
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- $x \in X, X\subseteq Y$
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- $\forall x \in X: P(x), \exists x \in X: P(x)$
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- $R \subseteq X \times Y$ is a relation between $X$ and $Y$
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- $xRy \equiv (x,y) \in R$
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- $G = (V, E)$, where $E \subseteq V\times V$ is a directed graph
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Part of [Relations and Digraphs](Relations%20and%20Digraphs.md)
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Languages & Machines/Regular languages.md
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36
Languages & Machines/Regular languages.md
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---
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type: math
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---
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## Induction
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Similar (if not the same) to:
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- [Mathematical Proofs (Induction)](Mathematical%20Proofs%20(Induction).md)
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- [Structural Proofs](Proofs.md)
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- Base case $0\in \mathbb{N}$
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- Inductive step - if $n\in \mathbb{N} \implies n+1\in \mathbb{N}$
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- We allow a finite number of steps
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I.e.
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Given $f (n) = n(n + 1)$ for all $n\in N$, then $f (n)$ is even.
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**Base case:** $f(0) = 0\times 1 = 0$, which is even
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**I.S.:**
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$$
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f(n+1) = (n+1)(n+2)= n(n+1)+2(n+1) = f(n) + 2(n+1) \blacksquare
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$$
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## Strings and Languages
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Literally the same as [Mathematical Data Structures](Mathematical%20Data%20Structures.md), but on strings
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How to define the reversal of a string, inductively?
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Let $w$ be a finite string. We define $w^R$ by induction on $|w|$:
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**B.C.:**
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$|w| = 0$, then, trivially, $w = \epsilon \therefore w^R = \epsilon$
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**I.S.:**
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$|w| = n \geq 1$, so $w = u a$ with $|u| = n-1$,
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