Turing Machines

Turing Machine

CS154

FINITE STATE q10 CONTROL

Turing Machines AI

N

P

U

T

INFINITE TAPE

read

write

move

read → , R

0 → 0, R

write

move → , R

0 → 0, R

qaccept

0 → 0, R → , R

0 → 0, R → , R

qreject Language = {0}

qaccept

0 → 0, R → , L

0 → 0, R → , L This Turing machine recognizes the language {0}

Deciding the language L = { w#w | w ∈ {0,1}* }

Turing Machines versus DFAs TM can both write to and read from the tape

STATE q1,FIND qGO LEFT q1,#FIND # q#, FIND q0, FIND q0,FIND

The head can move left and right The input doesn’t have to be read entirely, and the computation can continue further (even, forever) after all input has been read Accept and Reject take immediate effect

0

1 0 X

1 X

# 1

0 #

0 X 1

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1. If there’s no # on the tape, reject. 2. While there is a bit to the left of #, Replace the first bit with X, and check if the first bit to the right of the # is identical. (If not, reject.) Replace that bit with an X too. 3. If there’s a bit to the right of #, then reject else accept

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Definition: A Turing Machine is a 7-tuple T = (Q, Σ, Γ, δ, q0, qaccept, qreject), where:

Turing Machine Configurations

Q is a finite set of states

q7

Σ is the input alphabet, where ∉ Σ Γ is the tape alphabet, where ∈ Γ and Σ ⊆ Γ δ : Q × Γ → Q × Γ × {L, R} q0 ∈ Q is the start state

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qreject ∈ Q is the reject state, and qreject ≠ qaccept

Let C1 and C2 be configurations of M We say C1 yields C2 if, after running M in C1 for one step, M is then in configuration C2 Suppose δ(q1, b) = (q2, c, L) Then aaq1bb yields aq2acb Suppose δ(q1, a) = (q2, c, R) Then cabq1a yields cabcq2 Let w ∈ Σ* and M be a Turing machine M accepts w if there are configs C0, C1, ..., Ck, s.t. • C 0 = q0 w • Ci yields Ci+1 for i = 0, ..., k-1, and • Ck contains the accepting state qaccept

A language is called recognizable (recursively enumerable) if some TM recognizes it A language is called decidable (recursive) if some TM decides it

recognizable languages

decidable languages

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0

0

0

1

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0

corresponds to the configuration:

qaccept ∈ Q is the accept state

Defining Acceptance and Rejection

0

11010q700110

A TM M recognizes a language L iff M accepts all and only those strings in L A language L is called recognizable or recursively enumerable (r.e.) iff some TM recognizes L

A TM M decides a language L iff M accepts all strings in L and rejects all strings not in L A language L is called decidable or recursive iff some TM decides L

n

{ 02 | n ≥ 0 } PSEUDOCODE: 1. Sweep from left to right, cross out every other 0 2. If in stage 1, the tape had only one 0, accept 3. If in stage 1, the tape had an odd number of 0’s, reject 4. Move the head back to the first input symbol. 5. Go to stage 1. Why does this work? Idea: Every time we return to stage 1, the number of 0’s on the tape is halved.

2

x → x, L 0 → 0, L

n

{ 02 | n ≥ 0 }

C = {aibjck | k = i*j, and i, j, k ≥ 1}

q2 → , R

PSEUDOCODE: → , L

x → x, R

x → x, R

q0

q1

0 → , R x → x, R → , R

q3 0 → x, R

→ , R

qreject

0 → 0, R

0 → x, R

qaccept

q4 x → x, R → , R

C = {aibjck | k = i*j, and i, j, k ≥ 1}

aabbbcccccc xabbbcccccc xayyyzzzccc xabbbzzzccc xxyyyzzzzzz

1. If the input doesn’t match a*b*c*, reject. 2. Move the head back to the leftmost symbol. 3. Cross off an a, scan to the right until b. Sweep between b’s and c’s, crossing off one of each until all b’s are crossed off. If all c’s get crossed off while doing this, reject. 4. Uncross all the b’s. If there’s another a left, then repeat stage 3. If all a’s are crossed out, Check if all c’s are crossed off. If yes, then accept, else reject.

Multitape Turing Machines

FINITE STATE CONTROL

k

δ : Q × Γk → Q × Γk × {L,R}k

Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine


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Nondeterministic TMs Theorem: Every nondeterministic Turing machine N can be transformed into a single tape Turing Machine M that recognizes the same language. Proof Idea (more details in Sipser): M(w): For all strings C ∈ {Q ∪ Γ ∪ #}* in lex. order, Check if C = C0# < #Ck where C0, < ,Ck is some accepting computation history for N on w. If so, accept.


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We can encode a TM as a bit string. n states

start state

reject state

0n10m10k10s10t10r10u1 m tape symbols (first k are input symbols)

accept state

blank symbol

( (p, a), (q, b, L) ) = 0p10a10q10b10 ( (p, a), (q, b, R) ) = 0p10a10q10b11

Similarly, we can encode DFAs and NFAs as bit strings. So we can define the following languages: ADFA = { (B, w) | B is a DFA that accepts string w }

ADFA = { (D, w) | D is a DFA that accepts string w } Theorem: ADFA is decidable Proof Idea: Directly simulate D on w! ANFA = { (N, w) | N is an NFA that accepts string w } Theorem: ANFA is decidable. (Why?)

ANFA = { (B, w) | B is an NFA that accepts string w } ATM = { (C, w) | C is a TM that accepts string w } (Can define (x, y) := 0|x|1xy)

ATM = { (M, w) | M is a TM that accepts string w }

ATM is recognizable but not decidable

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Universality of Turing Machines The Church-Turing Thesis

Theorem: There is a universal Turing machine U that can take as input - the code of an arbitrary TM M - an input string w, such that U(M, w) accepts iff M(w) accepts.

Everyone’s Intuitive Notion = Turing Machines of Algorithms

This is a fundamental property: Turing machines can run their own code!

This thesis is tested every time you write a program that does something!

Note that DFAs/NFAs do not have this property. That is, ADFA is not regular.

A language is called recognizable or recursively enumerable, (or r.e.) if some TM recognizes it A TM decides a language if it accepts all strings in the language and rejects all strings not in the language A language is called decidable (or recursive) if some TM decides it

Recall: Given L ⊆ Σ *, define ¬L := Σ * - L Theorem: L is decidable iff both L and ¬L are recognizable Given: a TM M1 that recognizes A and a TM M2 that recognizes ¬A, we can build a new machine M that decides A How? Any ideas? M1 always accepts x, when x is in A M2 always accepts x, when x is not in A

w ∈ Σ*

w ∈ Σ*

A TM recognizes a language if it accepts all and only those strings in the language TM

TM

w∈L? yes

no

accept

reject

L is decidable (recursive)

w∈L? yes

no

accept reject or no output L is recognizable (recursively enumerable)

Theorem: L is decidable if both L and ¬L are recursively enumerable

There are languages over {0,1} that are not decidable Assuming the Church-Turing Thesis, this means there are problems that NO computing device can solve We can prove this using a counting argument. We will show there is no onto function from the set of all Turing Machines to the set of all languages over {0,1}. (But it works for any Σ) That is, every mapping from Turing machines to languages must “miss” some language.

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