计算理论笔记（三）TM Decidablity

Turing Machine

• one-way infinite
• its head can move left and right
• can read and write

Turing Machine is a 5-tuple $P=(K,\Sigma ,\delta ,s,H)$

• $K$ is a finite set of states
• $\Sigma$ is a alphabet containing left end symbol $⊳$ and blank symbol $\bigsqcup$
• $s\in K$ is the initial state
• $H\subseteq K$ is the set of halting states
• transition function δ:((K−H)×Σ→K×(Σ∪{←,→})\delta:((K-H)\times\Sigma\to K\times(\Sigma\cup\{\leftarrow,\rightarrow\})δ:((K−H)×Σ→K×(Σ∪{←,→})
  • $\forall q\in K-H$, if $\delta (q,⊳)=(p,b)$, then $b=\to$
  • $\forall q\in K-H,\forall a\in \Sigma$, if $\delta (q,a)=(p,b)$, then $b\ne ⊳$

A configuration is a member of K×⊳(Σ−{⊳})∗×((Σ−{⊳})∗(Σ−{⊳,⊔})∪{e})K\times\rhd(\Sigma-\{\rhd\})^*\times((\Sigma-\{\rhd\})^*(\Sigma-\{\rhd,\sqcup\})\cup\{e\})K×⊳(Σ−{⊳})∗×((Σ−{⊳})∗(Σ−{⊳,⊔})∪{e})

• $(q,⊳b\bigsqcup a,a\bigsqcup \bigsqcup b)\equiv (q,⊳b\bigsqcup \underline{a}a\bigsqcup \bigsqcup b)$
• halting configuration if $q\in H$

Yields in one step: $(q_1,⊳w_1{a}_1{u}_1){⊢}_M(q_2,⊳w_2{a}_2{u}_2)$ if

• (writing) w1=w2,u1=u2,a2∈Σ−{⊳}w_1=w_2,u_1=u_2,a_2\in\Sigma-\{\rhd\}w1​=w2​,u1​=u2​,a2​∈Σ−{⊳} and $\delta (q_1,a_1)=(q_2,a_2)$
• (moving left) $\delta (q_1,a_1)=(q_2,\leftarrow ),w_1=w_2{a}_2$ and either
  • $u_2=e$ if $u_1=e$ and $a_1=\bigsqcup$
  • $u_2=a_1{u}_1$, otherwise
• (moving right) $\delta (q_1,a_1)=(q_2,\to ),w_2=w_1{a}_1$ and
  • $u_1=e$ if $u_2=e$ and $a_2=\bigsqcup$
  • $u_1=a_2{u}_2$, otherwise

We say two configurations $C{⊢}_M^{\ast }{C}^{\prime }$ iff

• $C=C^{\prime }$ or
• $\exists C_0,C_1,\dots ,C_k(k\ge 1)$ s.t. $C=C_0{⊢}_M{C}_1{⊢}_M\cdots {⊢}_M{C}_k=C^{\prime }$

$M$ halts whenever it reaches a halting configuration

Exercise

• for any a∈Σ−{⊳}a\in\Sigma-\{\rhd\}a∈Σ−{⊳}, symbol-writing machine Ma=({s,h},Σ,δ,s,{h})M_a=(\{s,h\},\Sigma,\delta,s,\{h\})Ma​=({s,h},Σ,δ,s,{h})
  • $\delta (s,⊳)=(s,\to )$
  • $\delta (s,b)=(h,a)$ for each b∈Σ−{⊳}b\in\Sigma-\{\rhd\}b∈Σ−{⊳}
• left-head-moving machine $M_{\leftarrow }$
  • $\delta (s,⊳)=(s,\to )$
  • $\delta (s,b)=(h,\leftarrow )$ for each b∈Σ−{⊳}b\in\Sigma-\{\rhd\}b∈Σ−{⊳}
• right-head-moving machine $M_{\to }$
  • $\delta (s,⊳)=(h,\to )$
  • $\delta (s,b)=(h,\to )$ for each b∈Σ−{⊳}b\in\Sigma-\{\rhd\}b∈Σ−{⊳}

Combining of basic machines

• $M_i=(K_i,\Sigma ,{\delta }_i,s_i,H_i),\forall i=1,2,3$
• Construct $M=(K,\Sigma ,\delta ,s,H)$
  • K=K1∪K2∪K3∪{h}K=K_1\cup K_2\cup K_3\cup\{h\}K=K1​∪K2​∪K3​∪{h}
  • $s=s_1$
  • H=H2∪H3∪{h}H=H_2\cup H_3\cup\{h\}H=H2​∪H3​∪{h}
  • $\forall q\in K-H,\forall a\in \Sigma$
    • If $q\in K_1-H_1$
      • $\delta (q,a)={\delta }_1(q,a)$
    • If $q\in H_1$
      • $\delta (q,0)=(s_2,0)$
      • $\delta (q,1)=(s_3,1)$
      • δ(q,a)=(h,→),∀a∈Σ−{0,1}\delta(q,a)=(h,\rightarrow),\forall a\in\Sigma-\{0,1\}δ(q,a)=(h,→),∀a∈Σ−{0,1}
    • If $q\in K_2-H_2$
      • $\delta (q,a)={\delta }_2(q,a)$
    • If $q\in K_3-H_3$
      • $\delta (q,a)={\delta }_3(q,a)$

left-shifting machine $S_{\leftarrow }$

• ∀w∈(Σ−{⊳,⊔})∗:⊳⊔⊔w⊔‾→⊳⊔w⊔‾\forall w\in(\Sigma-\{\rhd,\sqcup\})^*: \rhd\sqcup\sqcup w\underline{\sqcup}\to\rhd\sqcup w\underline{\sqcup}∀w∈(Σ−{⊳,⊔})∗:⊳⊔⊔w⊔​→⊳⊔w⊔​
  

copy machine

• ∀w∈(Σ−{⊳,⊔})∗:⊳⊔‾w⊔→⊳⊔‾w⊔w⊔\forall w\in(\Sigma-\{\rhd,\sqcup\})^*: \rhd\underline{\sqcup}w\sqcup\to\rhd\underline{\sqcup}w\sqcup w\sqcup∀w∈(Σ−{⊳,⊔})∗:⊳⊔​w⊔→⊳⊔​w⊔w⊔
  

Recognize language

• input alphabet: Σ0=Σ−{⊳,⊔}\Sigma_0=\Sigma-\{\rhd,\sqcup\}Σ0​=Σ−{⊳,⊔}
• $M=(K,{\Sigma }_0,\Sigma ,\delta ,S,H)$
• initial configuration: given $w\in {\Sigma }^{\ast }$, $M$ starts with $(s,⊳\underline{\bigsqcup }w)$
• L(M)={w∈Σ0∗L(M)=\{w\in\Sigma_0^*L(M)={w∈Σ0∗​: $M$ halts on $w\}$, $M$ semidecides $L(M)$
• Example: Fix some input alphabet ${\Sigma }_0^{\ast }$
  • Let $a\in {\Sigma }_0$, $R_a$ semidecides L={w∈Σ0∗L=\{w\in\Sigma_0^*L={w∈Σ0∗​: $w$ contains $a\}$

Let M=(K,Σ,δ,S,{y,n})M=(K,\Sigma,\delta,S,\{y,n\})M=(K,Σ,δ,S,{y,n}) be a TM with input alphabet ${\Sigma }_0$, we say $M$ decides a language $L\subseteq {\Sigma }_0^{\ast }$ if

• For every $w\in L$, $(s,⊳\underline{\bigsqcup }w){⊢}_M^{\ast }(y,⊳u\underline{a}v)$ for some $u,v\in {\Sigma }^{\ast }$ and $a\in \Sigma$
  • $M$ accepts $w$
• For every $w\in L$, $(s,⊳\underline{\bigsqcup }w){⊢}_M^{\ast }(n,⊳u\underline{a}v)$ for some $u,v\in {\Sigma }^{\ast }$ and $a\in \Sigma$
  • $M$ rejects $w$
• A language is recursive/decidable if it is decided by a TM

Example: The following TM decides {anbncn:n≥0}\{a^nb^nc^n:n\ge0\}{anbncn:n≥0}

Theorem: If $L$ is recursive, it must be recursively enumerable

• $L$ is recursive ⇒∃M=(K,Σ,δ,S,{y,n})\Rightarrow \exists M=(K,\Sigma,\delta,S,\{y,n\})⇒∃M=(K,Σ,δ,S,{y,n}) decides $L$
  • $\delta (n,a)=(n,a)$ for any a∈Σ−{⊳}a\in\Sigma-\{\rhd\}a∈Σ−{⊳}
  • $\delta (n,⊳)=(n,\to )$

Theorem: If $L$ is recursive, so is $\overline{L}$

• Exchange the role of $y$ and $n$
• False generalization: If $L$ is recursive enumerable, so is $\overline{L}$

Compute function

• Let $M=(K,\Sigma ,\delta ,S,H)$ be a TM with input alphabet ${\Sigma }_0$, let $w\in {\Sigma }_0^{\ast }$, if $(s,⊳\underline{\bigsqcup }w){⊢}_M^{\ast }(h,⊳\underline{\bigsqcup }y)$ for some $h\in H$ and $y\in {\Sigma }_0^{\ast }$
  • $y$: output of $M$ on $w$
• Let $f:{\Sigma }_0^{\ast }\to {\Sigma }_0^{\ast }$, we say $M$ computes $f$ if for any $w\in {\Sigma }_0^{\ast }$ $M(w)=f(w)$, i.e. $M$ halts with $⊳\underline{\bigsqcup }f(w)$

Example: $f(w)=ww$ is computed by the following TM

Enumrator

• We say a TM $M$ enumerates a language $L$ if for some state $q$, L={w:(s,⊳⊔‾)⊢M∗(q,⊳⊔‾w)}L=\{w:(s,\rhd\underline{\sqcup})\vdash^*_M(q,\rhd\underline{\sqcup}w)\}L={w:(s,⊳⊔​)⊢M∗​(q,⊳⊔​w)}
  • $q$ is called output state
• A language is Turing enumerable if some TM enumerates it

Extensions of TM

• K-tape TM
  • δ:(K−H)×Σk→K×(Σ∪{←,→})k\delta:(K-H)\times\Sigma^k\to K\times(\Sigma\cup\{\leftarrow,\rightarrow\})^kδ:(K−H)×Σk→K×(Σ∪{←,→})k
  • Theorem: Any k-tape TM can be simulated by a standard TM
  • K-track(conceptually), Σ′=(Σ∪{a‾:a∈Σ})k\Sigma'=(\Sigma\cup\{\underline{a}:a\in\Sigma\})^kΣ′=(Σ∪{a​:a∈Σ})k
• Two-way Infinite Tape
  • Simulated by 2-tape TM, split indices to ${\mathbb{N}}^{+}$ and $\mathbb{Z}-{\mathbb{N}}^{+}$
• K-head TM
• Two-dimensional Tape
  • countable, because $|{\mathbb{N}}^{+}\times {\mathbb{N}}^{+}|=|{\mathbb{N}}^{+}|$
• Random Access
• Non-determinism

A non-deterministic TM(NTM) is 5-tuple $M=(K,\Sigma ,\Delta ,S,H)$

• Δ⊆((K−H)×Σ)×(K×(Σ∪{←,→}))\Delta\subseteq((K-H)\times\Sigma)\times(K\times(\Sigma\cup\{\leftarrow,\rightarrow\}))Δ⊆((K−H)×Σ)×(K×(Σ∪{←,→}))

A NTM $N$ with input alphabet ${\Sigma }_0$ semidecides a language $L\subseteq {\Sigma }_0^{\ast }$ if for any $w\in {\Sigma }_0^{\ast }$, the following is true:

• $w\in L$ iff $(s,⊳\underline{\bigsqcup }w){⊢}_M^{\ast }(h,⊳u\underline{\bigsqcup }v)$ for some $h$

Let M=(K,Σ,Δ,S,{y,n})M=(K,\Sigma,\Delta,S,\{y,n\})M=(K,Σ,Δ,S,{y,n}) be a NTM with input alphabet ${\Sigma }_0$, we say $M$ decides a language $L\subseteq {\Sigma }_0^{\ast }$ if

• There is a natural number $N$, depending on $M$ and $w$, such that there is no configuration $C$ satisfying $(s,⊳\underline{\bigsqcup }w){⊢}_M^{N}C$
• $w\in L$ iff $(s,⊳\underline{\bigsqcup }w){⊢}_M^{\ast }(y,⊳u\underline{\bigsqcup }v)$

Example: Let $C$ be the set of binary strings representing composite number, design a NTM decides $C$

• $⊳\bigsqcup w\bigsqcup p\bigsqcup q$: non-deterministicly generates a binary string $p$ with $2\le \overline{p}\le \overline{w}$ and then non-deterministicly generates a binary string $q$ with $2\le \overline{q}\le \overline{w}$
• Multiply $\overline{p}$ with $\overline{q}$, compare the product with $\overline{w}$
• Halt with $y$ if match, $n$ if not

Theorem: A NTM $N$ can be simulated by a DTM $M$

• Choices for every step of NTM is finite (fanout): ∣K×(Σ∪{←,→}∣=∣K∣×(∣Σ∣+2)=Δr|K\times(\Sigma\cup\{\leftarrow,\rightarrow\}|=|K|\times(|\Sigma|+2)\xlongequal{\Delta}r∣K×(Σ∪{←,→}∣=∣K∣×(∣Σ∣+2)Δr
• Construct 3-tap DTM to simulate NTM, by BFS
  • 1st-tape: input
  • 2nd-tape: simulate a branch of NTM on $w$
  • 3rd-tape: enumerate strings over {1,2,…,r}\{1,2,\dots,r\}{1,2,…,r}, addressing one node to access

Church-Turing Thesis

• Intuitive notation of algorithms equals to Turing Machine that halt on every input

Description of TM

• Formal description: 5-tuple
• Implement-level description: diagram
• High-level description: pseudo code

Fact: Every object $O$ can be encoded into a string “$O$”

Example: Every TM can be encoded into a string

• Encoding states
  • i=min⁡j{j:2j≥∣K∣}i=\displaystyle\min_j\{j:2^j\ge|K|\}i=jmin​{j:2j≥∣K∣}
  • “q{0,1}iq\{0,1\}^iq{0,1}i”
  • $s:q{0}^i$
• Encoding alphabet
  • t=min⁡j{j:2j≥∣Σ∣+2}t=\displaystyle\min_j\{j:2^j\ge|\Sigma|+2\}t=jmin​{j:2j≥∣Σ∣+2}
  • “a{0,1}ta\{0,1\}^ta{0,1}t”
  • $\bigsqcup :a{0}^t$
  • $⊳:a{0}^{t-1}1$
  • $\leftarrow :a{0}^{t-2}10$
  • $\to :a{0}^{t-2}11$
• Encoding transition
  • δ:(K−H)×Σ→K×(Σ∪{←,→})\delta: (K-H)\times\Sigma\to K\times(\Sigma\cup\{\leftarrow,\rightarrow\})δ:(K−H)×Σ→K×(Σ∪{←,→})
  • in 4-tuple like $(q00,a100,q01,a00)$

Fact: Given a valid encoding object, a TM is always able to decode it

Example

• Problem: Given a graph $G$, is $G$ connected?
• Language: L={L=\{L={“$G$”: $G$ is a connected graph$\}$
• $M=$ on input “$G$”
  1. select a node of $G$ and mark it
  2. repeat the following until no new nodes are marked: for each marked node, mark all its neighbors
  3. if all nodes of $G$ are marked, accept “$G$”
  4. else reject

Decidable problem/recursive language

• About DFA/NFA/regular languages

R1 Problem

• Given a DFA $B$ and a string $w$, does $B$ accept $w$?
• Language: ADFA={A_\mathrm{DFA}=\{ADFA​={“$B$”“$w$”: $B$ is a DFA that accepts $w\}$
• $M_{\mathrm{R}1}=$ on input “$B$”“$w$”
  1. simulate $B$ on $w$
  2. if $B$ accepts $w$, accept “$B$”“$w$”
  3. else reject

R1* Problem, for a fixed DFA $B$

• Given a string $w$, does $B$ accept $w$?
• Language: $L(B)$
• $M_{\mathrm{R}{1}^{\ast }}=$ on input $w$
  1. run $M_{\mathrm{R}1}$ on “$B$”“$w$”
  2. if $M_{\mathrm{R}1}$ accepts “$B$”“$w$”, accept $w$
  3. else reject

R2 Problem

• Given a NFA $B$ and a string $w$, does $B$ accept $w$?
• Language: ANFA={A_\mathrm{NFA}=\{ANFA​={“$B$”“$w$”: $B$ is a NFA that accepts $w\}$
• $M_{\mathrm{R}2}=$ on input “$B$”“$w$”
  1. convert $B$ into an equivalent DFA $D$
  2. run $M_{\mathrm{R}1}$ on “$D$”“$w$”
  3. if $M_{\mathrm{R}1}$ accepts “$D$”“$w$”, accept “$B$”“$w$”
  4. else reject

R3 Problem

• Language: AREX={A_\mathrm{REX}=\{AREX​={“$R$”“$w$”: $R$ is a regex and $w\in L(R)\}$
• $M_{\mathrm{R}3}=$ on input “$R$”“$w$”
  1. convert $R$ into an equivalent NFA $B$
  2. run $M_{\mathrm{R}2}$ on “$B$”“$w$”
  3. if $M_{\mathrm{R}2}$ accepts “$B$”“$w$”, accept “$R$”“$w$”
  4. else reject

R4 Problem

• Given a DFA $B$ and a string $w$, is $L(B)=\varnothing$?
• Language: EDFA={E_\mathrm{DFA}=\{EDFA​={“$B$”: $B$ is a DFA and $L(B)=\varnothing \}$
• $M_{\mathrm{R}4}=$ on input “$B$”
  1. mark the initial state of $B$
  2. repeat the following until no new states are marked: for each marked state, mark all its successors by transition function
  3. if no final states are marked, accept “$B$”
  4. else reject

Symmetric difference

• L(A)⊕L(B)={w:w∈L(A)∪L(B)∧w∉L(A)∩L(B)}L(A)\oplus L(B)=\{w:w\in L(A)\cup L(B)\land w\notin L(A)\cap L(B)\}L(A)⊕L(B)={w:w∈L(A)∪L(B)∧w∈/​L(A)∩L(B)}
• $L(A)\oplus L(B)=\left(L(A)\cap \overline{L(B)}\right)\cup \left(\overline{L(A)}\cap L(B)\right)$
• $L(A)=L(B)⟺L(A)\oplus L(B)=\varnothing$

R5 Problem

• Language: EQDFA={EQ_\mathrm{DFA}=\{EQDFA​={“$A$”“$B$”: $A$ and $B$ are two DFAs and $L(A)=L(B)\}$
• $M_{\mathrm{R}5}=$ on input “$A$”“$B$”
  1. construct a DFA $C$ s.t. $L(C)=L(A)\oplus L(B)$
  2. run $M_{\mathrm{R}4}$ on “$C$”
  3. if $M_{\mathrm{R}4}$ accepts “$C$”, accept “$A$”“$B$”
  4. else reject

CNF

• A CFG $G=(V,\Sigma ,S,R)$ is in Chomsky Normal Form(CNF) if every of its rules is in the following forms:
  • $A\to BC$, for $A,B,C\in (V-\Sigma )$
  • $A\to a$, for $A\in (V-\Sigma )$ and $a\in \Sigma$
  • $S\to e$
  • $S$ does no appear in the RHS of a rule

Theroem: Every CFG has an equivalent CFG in CNF

• Given a CFG $G=(V,\Sigma ,S,R)$, convert it into CNF
• add a new start symbol $S_0$ and a new rule $S_0\to S$
• fix $e$-rules: $A\to e$ with $A\ne S_0$
• fix short rules: $A\to B$ with $A,B\in (V-\Sigma )$
• fix long rules: $A\to u$ with $u\in V^{\ast }$ and $|u|>2$
• fix $A\to u_1{u}_2$ with $u_1$ or $u_2$ as terminal

Observation: If CFG in CNF generates a string length $n(n\ge 1)$, it must take exactly $2n-1$ steps

• $A\to BC$ takes $n-1$ steps
• $A\to a$ takes $n$ steps

C1 Problem

• Language: ACFG={A_\mathrm{CFG}=\{ACFG​={“$G$”“$w$”: $G$ is a CFG that generates $w\}$
• $M_{\mathrm{C}1}=$ on input “$G$”“$w$”
  1. convert $G$ into an equivalent CFG $G^{\prime }$ in CNF
  2. if $|w|=0$
  • if $G^{\prime }$ has rule $S\to e$, accept “$G$”“$w$”
  • else reject
  3. if $|w|\ge 1$, list all the derivations(finite, $\le |R{|}^{2|w|-1}$) of length $2|w|-1$
  • if any listed derivation generates $w$, accept “$G$”“$w$”
  • else reject

C2 Problem

• Language: APDA={A_\mathrm{PDA}=\{APDA​={“$P$”“$w$”: $P$ is a PDA that accepts $w\}$
• $M_{\mathrm{C}2}=$ on input “$P$”“$w$”
  1. convert $P$ into an equivalent CFG $G$
  2. run $M_{\mathrm{C}1}$ on “$G$”“$w$”
  3. if $M_{\mathrm{C}1}$ accepts “$G$”“$w$”, accept “$P$”“$w$”
  4. else reject

C3 Problem

• Language: ECFG={E_\mathrm{CFG}=\{ECFG​={“$G$”: $G$ is a CFG and $L(G)=\varnothing \}$
• $M_{\mathrm{C}3}=$ on input “$G$”
  1. mark all the terminals and $e$
  2. repeat the following until no new symbols are marked: for each LHS symbol of a rule, mark it if all the RHS symbols are marked
  3. if $S$ is not marked, accept “$G$”
  4. else reject

Let $f:A\to B$ be a function

• $f$ is injective if $f(a)\ne f(a^{\prime })$ for any $a\ne a^{\prime }$
• $f$ is surjective if for any $b\in B$, there exists $a\in A$ satisfying $f(a)=b$
• $f$ is bijective if $f$ is injective and surjective

Equinumerous

• Two sets $A$ and $B$ are equinumerous if there exists a bijection $f:A\to B$
• A set is countable if it is finite or it is equinumerous with $\aleph$, and uncountable otherwise
• Collary: any subset of a countable set is countable

Theorem: Let $A$ be a set, the following statements are equivalent:

• a) $A$ is countable
• b) there exists an injection $f:A\to \aleph$
• c) there exists some way to enumerate the elements of $A$ such that each element is listed within finite steps
• Proof:
  • a) => c): there exists a bijection $f:A\to \aleph$, list the elements of $A$ in increasing order of $f(a)$, any element $a\in A$ can be listed in $f(a)$ steps
  • c) => b): for each $a\in A$, let $f(a)$ be # of steps used to list $a$ for the first time
  • b) => a): there exists a injection $f:A\to \aleph$, sort elements of $A$ in increasing order of $f(a)$, let $g(a)$ be the rank of $a$, which is bijective from $A$ to $\aleph$

Theorem: Let $\Sigma$ be an alphabet, ${\Sigma }^{\ast }$ is countable

• If $\Sigma =\varnothing$, Σ∗={e}\Sigma^*=\{e\}Σ∗={e} is countable
• Suppose $\Sigma \ne \varnothing$, enumerate all strings in ${\Sigma }^{\ast }$ by increasing order of length
• A string of length $i$ can be listed within $|\Sigma {|}^0+|\Sigma {|}^1+|\Sigma {|}^2+\cdots +|\Sigma {|}^i$ steps
  • $|\Sigma {|}^k$ denotes # of strings with length $k$
• Therefore ${\Sigma }^k$ is countable
• Collary: the set of all Turing machines is countable

Theorem: Let $\Sigma$ be some non-empty alphabet and $\mathcal{L}$ be the set of all languages over $\Sigma$, $\mathcal{L}$ is uncountable

• Supposed $\mathcal{L}$ is countable, the languages in $\mathcal{L}$ can be listed as $L_1,L_2,L_3,\dots$
• Since ${\Sigma }^{\ast }$ is countable, the strings in ${\Sigma }^{\ast }$ can be listed as $s_1,s_2,s_3,\dots$
• Diagonalization: D={si:si∉Li}D=\{s_i:s_i\notin L_i\}D={si​:si​∈/​Li​} is a language over $\Sigma$
  • for any $s_i\in D$ iff $s_i\notin L_i$, therefore $D\ne L_i$, therefore $D\notin \mathcal{L}$

Recursive $⊊$ R.E.

Theorem: Let $\Sigma$ be an alphabet, ${\Sigma }^{\ast }$ is countable

• If $\Sigma =\varnothing$, Σ∗={e}\Sigma^*=\{e\}Σ∗={e} is countable
• Suppose $\Sigma \ne \varnothing$, enumerate all strings in ${\Sigma }^{\ast }$ by increasing order of length
• A string of length $i$ can be listed within $|\Sigma {|}^0+|\Sigma {|}^1+|\Sigma {|}^2+\cdots +|\Sigma {|}^i$ steps
  • $|\Sigma {|}^k$ denotes # of strings with length $k$
• Therefore ${\Sigma }^k$ is countable
• Collary: the set of all Turing machines is countable

Theorem: Let $\Sigma$ be some non-empty alphabet and $\mathcal{L}$ be the set of all languages over $\Sigma$, $\mathcal{L}$ is uncountable

• Supposed $\mathcal{L}$ is countable, the languages in $\mathcal{L}$ can be listed as $L_1,L_2,L_3,\dots$
• Since ${\Sigma }^{\ast }$ is countable, the strings in ${\Sigma }^{\ast }$ can be listed as $s_1,s_2,s_3,\dots$
• Diagonalization: D={si:si∉Li}D=\{s_i:s_i\notin L_i\}D={si​:si​∈/​Li​} is a language over $\Sigma$
  • for any $s_i\in D$ iff $s_i\notin L_i$, therefore $D\ne L_i$, therefore $D\notin \mathcal{L}$

If $A$ and $B$ are recursive, so is $A\cup B$

• $M_{\cup }=$ on input “$w$”
  1. run $M_{\mathrm{A}}$ and $M_{\mathrm{B}}$ on “$w$”
  2. if at least one of them accepts $w$, accept “$w$”
  3. else reject

If $A$ and $B$ are recursive, so is $A\circ B$

• $M_{\circ }=$ on input “$w$”
  1. enumerate all possible division $(u,v)$ with $uv=w$
  2. run $M_{\mathrm{A}}$ on $u$ and $M_{\mathrm{B}}$ on $v$
  3. if both of them accepts $u$ and $v$, accept “$w$”
  4. else reject

If $A$ and $B$ are R.E., so is $A\cup B$

• $M_{\cup }=$ on input “$w$”
  1. run $M_{\mathrm{A}}$ and $M_{\mathrm{B}}$ on “$w$”
  2. if one of them halts, halt

If $A$ and $B$ are R.E., so is $A\circ B$

• $M_{\circ }=$ on input “$w$”
  1. non-deterministically divide $w$ into $uv$
  2. run $M_{\mathrm{A}}$ on $u$ and $M_{\mathrm{B}}$ on $v$
  3. if both of them halt, halt