finals

regular language

$$\begin{align*} \hat{\delta}(q, \epsilon) &= q \\\ \hat{\delta}(q, xa) &= \delta(\hat{\delta}(q, x), a) \end{align*}$$

All finite languages are regular, but not all regular languages are finite

Pumping Lemma

$$\text{L is regular} \implies (\exists \mid k \geq 0: (\forall x,y,z \in L \land |y| \geq k : (\exists u,v,w | y=uvw \land |v| > 1: (\forall i \mid i \geq 0: xuv^iwz \in L))))$$

• demon picks $k$
• you pick $x,y,z \leftarrow xyz \in L \land |y| \geq k$
• demon picks $u,v,w \leftarrow uvw = y \land |v| \geq 1$
• you pick an $i \geq 0$, and show $xuv^2wz \notin L$

context-free grammar

$$\begin{align*} \mathbb{G} = (N, \Sigma, P, S) &\quad N: \text{non-terminal symbols} \\\ &\quad \Sigma: \text{terminal symbols} \space s.t \space \Sigma \cap N = \emptyset \\\ &\quad P: \text{production rules} \space s.t \text{a finite subset of } N \times (N \cup \Sigma)^{*} \\\ &\quad S: \text{start symbol} \in N \end{align*}$$

Properties

• $\exists \text{ CFG} | L(G) = L \iff L \text{ is a context-free language}$
• L is regular $\implies$ L is context-free
• $L_{1}, L_{2} \text{ are context-free} \implies L_{1} \cup L_{2} \text{ are context-free}$
• context-free languages are not closed under complement, and not closed under intersection. ($L_1 \cap L_2, \sim{L_1} \text{ are not context-free}$)

We know that $\{a^nb^nc^n\mid n \geq 0\}$ is not CF

Pushdown Automata PDA

$$\begin{align*} \text{PDA} = (Q, \Sigma, \Gamma, \delta, s, \bot, F) &\quad Q: \text{Finite set of state} \\\ &\quad \Sigma: \text{Finite input alphabet} \\\ &\quad \Gamma: \text{Finite stack alphabet} \\\ &\quad \delta: \subset (Q \times (\Sigma \cup \{\epsilon\}) \times \Gamma) \times (Q \times \Gamma^{*}) \\\ &\quad s: \text{start state} \in Q \\\ &\quad \bot: \text{empty stack} \in \Gamma \\\ &\quad F: \text{final state} \in Q \end{align*}$$

Properties

$\mathcal{L}(M) = L \iff L \text{ is context-free}$

Turing machine

$$\begin{align*} \text{TM} = (Q, \Sigma, \Gamma, \delta, s, q_{\text{accept}}, q_{\text{reject}}, \square) &\quad Q: \text{Finite set of state} \\\ &\quad \Sigma: \text{Finite input alphabet} \\\ &\quad \Gamma: \text{Finite tape alphabet} \\\ &\quad \delta: (Q \times \Gamma) \rightarrow Q \times \Gamma \times \{L, R\} \\\ &\quad s: \text{start state} \in Q \\\ &\quad q_{\text{accept}}: \text{accept state} \in Q \\\ &\quad q_{\text{reject}}: \text{reject state} \in Q \\\ &\quad \square: \text{blank symbol} \in \Gamma \end{align*}$$

Transition function: $\delta(q, x) = (p, y, D)$: when in state $p$ scan symbol $a$, write $b$ on tape cell, move the head in direction $d$ and enter state $q$

transition to $q_{\text{accept}}$ or $q_{\text{reject}}$ is a halting state and accept/reject respectively.

Properties

• A TM is “total” iff it halts on all input
• $\mathcal{L}(M) = L \iff (\forall s \mid s \in L \iff M \text{ accepts s})$
• L is recognizable: $\iff \exists \text{ TM M s.t } \mathcal{L}(M)=L$
• L is decidable: $\iff \exists \text{ total TM M s.t } \mathcal{L}(M)=L \land \forall s \in \Sigma^{*} \text{ M halts on s}$
• $\text{L is decidable} \implies \text{L is recognizable}$

Church-Turing Thesis

Conjecture 1: All reasonable models of computation are equivalent:

• perfect memory
• finite amount of time

Conjecture 2: Anything a modern digital computer can do, a Turing machine can do.

Equivalence model

• TMs with multiple tapes.
• NTMs.
• PDA with two stacks.

Finite Automata from Church-Turing Thesis

Finite automata can be encoded as a string:

Let $0^n10^m10^j0^{k_1}\ldots 10^{k_n}$ be a DFA with $n$ states, $m$ input characters, $j$ final states, $k_1\ldots k_n$ transitions

$$\begin{align*} A_{\text{DFA}} &= \{M\#w \mid M \text{ is a DFA which accepts } w\} (1) \\\ A_{\text{TM}} &= \{M\#w \mid M \text{ is a TM which accepts } w\} (2) \end{align*}$$

M is a “recognizer” $\implies M(x) = \begin{cases} \text{accept} & \text{if } x \in L \\\ \text{reject or loop} & \text{if } x \notin L \end{cases}$

M is a “decider” $\implies M(x) = \begin{cases} \text{accept} & \text{if } x \in L \\\ \text{reject} & \text{if } x \notin L \end{cases}$

Decidability and Recognizability

(1) is deciable: Create a TM $M^{'}$ such that $M^{'}(M\#w)$ runs $M$ on $w$, therefore $M'$ is total, or $\mathcal{L}(M) = A_{\text{DFA}}$ > $M\#w \in \mathcal{L}(M^{'}) \iff M \text{ accepts } w \iff M\#w \in A_{\text{DFA}}$

(2) is recognizable: Create a TM $M^{'}$ such that $M^{'}(M\#w)$ runs $M$ on $w$ > $M\#w \in \mathcal{L}(M^{'}) \iff M \text{ accepts } w \iff M\#w \in A_{\text{TM}} \implies \mathcal{L}(M^{'}) = A_{\text{TM}}$

Note that all regular language are deciable language

Proof for $A_{\text{TM}}$ is undeciable:

Assume $A_{\text{TM}}$ is decidable

$\exists$ a decider for $A_{\text{TM}}$, $D$.

Let $P$ another TM such that $P(M)$: Call $D$ on $M\#M$

Paradox machine: P never loops: $P(M) = \begin{cases} \text{accept} & \text{if P reject M} \\\ \text{reject} & \text{if P accepts M} \end{cases}$

Countability

• A set $S$ is countable infinite if $\exists$ a monotonic function $f: S \rightarrow \mathbb{N}$ (isomorphism)
• A set $S$ is uncountable if there is NO injection from $S$

Theorem:

• The set of all PDAs is countably infinite
• $\Sigma^{*}$ is countably infinite (list out all string n in finite time)
• The set of all TMs is countably infinite ($\Sigma = \{0,1\} \mid \text{set of all TMs that } S \subseteq \Sigma^{*}$, so does REC, DEC, CF, REG
• The set of all languages is uncountable.

Diagonalization and Problems

The set of unrecognizable languages is uncountable. The set of all languages is uncountable.

Proof: I can encode a language with a infinite string. $\Sigma = \{0,1\}$ Consider a machine $N$ that on input $x \in \{0,1\}^{*}$ such that $L^{*}(i)$ is undeciable from the diagonalization. Make sure to use the negation of the dia

Theorem

• L is deciable $\iff$ L and $\sim L$ are both recognizable

Proof: $L$ is deciable $\iff \sim L$ is deciable. $L$ is deciable $\implies$ L is recognizable

Let $R_L, R_{\sim L}$ be recognizer. Create TM $M$ that runs $R_L$ and $R_{\sim L}$ on $x$ concurrently. if $R_L$ accepts $\implies$ accept, $R_{\sim L}$ accepts $\implies$ reject.

If M never halts, M decides L. If $x \in L \implies R_L(x) \text{ halts}$, and $x \notin L \implies R_{\sim L}(x) \text{ halts}$.

Reduction on universal TMs

$\sim A_{\text{TM}} = \{M\#w \mid M \text{ does not accept w}\}$. Which implies $\sim A_{\text{TM}}$ is unrecognizable

HP is undeciable, and recognizable.

$$\text{Halting problem} = \{M\#w \mid M \text{ halts on w} \}$$

Proof: Assume HP is deciable. $\exists D_{MP}(M\#w) = \begin{cases} \text{accept} & \text{if M halts on w} \\\ \text{reject} & \text{if M loops on w} \end{cases}$

Build a TM $M^{'}$ where $M^{'}(M\#v)$:

calls $D_{MP}$ on $M\#v$:
accepts:
  - run $M$ on $v$
    - accept -> accept
    - reject -> reject
reject: reject

Therefore $M^{'}$ is total. Since $M\#w \in \mathcal{L(M^{'})} \iff \text{M accepts w} \iff M\#w \in A_{\text{TM}}$. Therefore $\mathcal{L}(M^{'}) = A_{\text{TM}}$. Which means $M^{'}$ is a decider for $A_{\text{TM}}$ (which is a paradox) $\square$