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tape head Finite Control Lecture 16 Deterministic Turing Machine (DTM) e p h B a l a The tape has the left end but infinite to the right. It is divided into cells. Each cell contains a symbol in an alphabet Γ . There exists a special symbol B which represents the empty cell. a

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Lecture 16 deterministic turing machine dtm l.jpg

tape

head

Finite Control

Lecture 16Deterministic Turing Machine (DTM)


Slide2 l.jpg

e

p

h

B

a

l

a

The tape has the left end but infinite to the right. It is divided into cells. Each cell contains a symbol in an alphabet Γ. There exists a special symbol B which represents the empty cell.


Slide3 l.jpg

a

  • The head scans at a cell on the tape and can read, erase, and write a symbol on the cell. In each move, the head can move to the right cell or to the left cell (or stay in the same cell).


Slide4 l.jpg

  • The finite control has finitely many states which form a set Q. For each move, the state is changed according to the evaluation of a transition function

    δ : Q x Γ → Q x Γ x {R, L}.


Slide5 l.jpg

b

a

  • δ(q, a) = (p, b, L) means that if the head reads symbol a and the finite control is in the state q, then the next state should be p, the symbol a should be changed to b, and the head moves one cell to the left.

q

p


Slide6 l.jpg

a

b

  • δ(q, a) = (p, b, R) means that if the head reads symbol a and the finite control is in the state q, then the next state should be p, the symbol a should be changed to b, and the head moves one cell to the right.

q

p


Slide7 l.jpg

  • There are some special states: an initial state s and an final states h.

  • Initially, the DTM is in the initial state and the head scans the leftmost cell. The tape holds an input string.

s


Slide8 l.jpg

x

  • When the DTM is in the final state, the DTM stops. An input string x is accepted by the DTM if the DTM reaches the final state h.

  • Otherwise, the input string is rejected.

h


Slide9 l.jpg

  • The DTM can be represented by

    M = (Q, Σ, Γ, δ, s)

    where Σ is the alphabet of input symbols.

  • The set of all strings accepted by a DTM M is denoted by L(M). We also say that the language L(M) is accepted by M.


Slide10 l.jpg

  • The transition diagram of a DTM is an alternative way to represent the DTM.

  • For M = (Q, Σ, Γ, δ, s), the transition diagram of M is a symbol-labeled digraphG=(V, E) satisfying the following:

    V = Q (s = , h = )

    E = { p q | δ(p, a) = (q, b, D)}.

a/b,D


Slide11 l.jpg

1/1,R

0/0,R; 1/1,R

0/0,R

0/0,R

B/B,R

s

p

q

h

1/1,R

M=(Q, Σ, Γ, δ, s) where Q = {s, p, q, h},

Σ = {0, 1}, Г = {0, 1, B}.

δ 0 1 B

s (p, 0, R) (s, 1, R) -

p (q, 0, R) (s, 1, R) -

q (q, 0, R) (q, 1, R) (h, B, R)

L(M) = (0+1)*00(0+1)*.


Theorem every regular set can be accepted by a dtm l.jpg
Theorem. Every regular set can be accepted by a DTM.

Proof.

  • Every regular set can be accepted a deterministic finite automata (DFA).

  • Every DFA can be simulated by a DTM.


Review on regular set l.jpg
Review on Regular Set

  • Regular sets on an alphabet Σis defined recursively as follows

    (1) The empty set Φ is a regular set.

    (2) For every symbol a in Σ, {a} is a

    regular set.

    (3) If A and B are regular sets, then

    A ∩ B, A U B, and A* are all regular

    sets.


Review on dfa l.jpg
Review on DFA

tape

a

b

c

d

e

f

head

finite control


Slide15 l.jpg


Slide16 l.jpg


Slide17 l.jpg


Slide18 l.jpg

  • Initially, the DFA is in the initial state and the head scans the leftmost cell. The tape holds an input string x. When the head moves off the tape, the DFA stops. An input string is accepted by the DFA if the DFA stops in a final state. Otherwise, the input string is rejected.


Simulate dfa by dtm l.jpg
Simulate DFA by DTM scans the leftmost cell. The tape holds an input string x.

  • Given a DFA M = (Q, Σ, δ, s, F), we can construct a DTM M’ = (Q’, Г, Σ, δ’, s) to simulate M as follows:

  • Q’ = Q U {h}, Γ = Σ U {B},

  • If δ(q, a) = p, then δ’(q, a) = (p, a, R). δ’(q, B) = (h, B, R) for q in F.


Slide20 l.jpg

1/1,R scans the leftmost cell. The tape holds an input string x.

0/0,R; 1/1,R

0/0,R

0/0,R

B/B,R

s

p

q

h

1/1,R

L(M) = (0+1)*00(0+1)*.

1

0, 1

0

0

s

p

q

1


Slide21 l.jpg

Turing acceptable languages scans the leftmost cell. The tape holds an input string x.

Regular

languages


Why dtm can accept more languages than dfa l.jpg
Why DTM can accept more languages than DFA? scans the leftmost cell. The tape holds an input string x.

Because

  • The head can move in two directions. (No!)

  • The head can erase. (No!)

  • The head can erase, write and move in two directions. (Yes!)

  • What would happen if the head can read, erase and write, but move in one direction?


Lecture 17 examples of dtm l.jpg
Lecture 17 scans the leftmost cell. The tape holds an input string x. Examples of DTM


Slide24 l.jpg

scans the leftmost cell. The tape holds an input string x.


Slide26 l.jpg

(s, scans the leftmost cell. The tape holds an input string x. 0110B)

├ (q0, B110B)

├ (q1, B010B)

├ (q1, B010B)

├ (q0, B011B)

├ (h, B0110B)


Remark l.jpg
Remark scans the leftmost cell. The tape holds an input string x.


Turing acceptable l.jpg
Turing-acceptable scans the leftmost cell. The tape holds an input string x.


Slide29 l.jpg

R scans the leftmost cell. The tape holds an input string x.

{ww | w ε (0+1)*} is Turing-acceptable.

(s, B0110B)├ (q, B0110B)├ (q0, B011BB)

├* (q0, B011B) ├ (p0, B011B)├ (ok, BB11B)

├* (ok, B11B) ├ (q, B11B) ├ (q1, B1BB)

├ (q1, B1B) ├ (p1, B1B) ├ (ok, BBB)

├ (q, BBB) ├ (h, BBB)

?? Could we do (ok, BBB) ├ (h, BBB) ?


Slide30 l.jpg

R scans the leftmost cell. The tape holds an input string x.

L= {ww | w in (0+1)*}


Turing computable functions l.jpg
Turing-Computable Functions scans the leftmost cell. The tape holds an input string x.

  • A total function f: Σ* → Σ* is Turing-computable if there exists a DTM M such that for every x in Σ*,

    (s, BxB) ├* (h, Bf(x)B).

  • A partialf: Ω→ Σ* is Turing-computable if there exists a DTM M such that L(M)=Ω and for every x in Ω,

    (s, BxB) ├* (h, Bf(x)B).


Slide32 l.jpg

R scans the leftmost cell. The tape holds an input string x.

The following function f is Turing-computable:

f(x) =w, if x = ww ;

↑, otherwise

(s, B0110B)├ (q, B0110B)├ (q0, B011BB)

├* (q0, B011B) ├ (p0, B011B)├ (ok, B0’11B)

├* (ok, B0’11B) ├ (q, B0’11B) ├ (q1, B0’1BB)

├ (q1, B0’1B) ├ (p1, B0’1B) ├ (ok, B0’1’B)

├ (q, B0’1’B) ├ (r, B0’1B) ├* (r, B01B)

├ (o, B01B) ├* (o, B01B)├ (k, B01BB)

├ (h, B01B)


Slide33 l.jpg

R scans the leftmost cell. The tape holds an input string x.

  • f(x) = w if x = ww ; ↑, otherwise


Turing decidable l.jpg
Turing-decidable scans the leftmost cell. The tape holds an input string x.

  • A language A is Turing-decidable if its characteristic function is Turing-computable.

    Χ (x) = {

1, if x ε A

0, otherwise

A


Slide35 l.jpg

R scans the leftmost cell. The tape holds an input string x.

  • { ww | w ε (0+1)*} is Turing-decidable.

    (s, B0110B)├ (q, B0110B)├ (q0, B011BB)

    ├* (q0, B011B) ├ (p0, B011B)├ (ok, B$11B)

    ├* (ok, B$11B) ├ (q, B$11B) ├ (q1, B$1BB)

    ├ (q1, B$1B) ├ (p1, B$1B) ├ (ok, B$$B)

    ├ (q, B$$B) ├ (r, B$BB) ├* (r, BBB)

    ├ (o, BBB) ├* (h, B1B)


Theorem l.jpg
Theorem scans the leftmost cell. The tape holds an input string x.

  • A language A is Turing-acceptable iff there exists a total Turing-computable function f such that A = { f(x) | x ε (0+1)*}.

  • If we look at each natural number as a 0-1 string, then f can be also a total Turing-computable function defined on N, the set of all natural numbers, that is,

    A = { f(1), f(2), f(3), …}

  • Therefore, “Turing-acceptable” is also called “recursively enumerable” (r.e.).


Theorem38 l.jpg
Theorem scans the leftmost cell. The tape holds an input string x.

  • A language A is Turing-decidable iff A and its complement Ā are Turing-acceptable.

    Proof. Suppose L(M) = A and L(M’) = Ā.

    Construct M* to simulate M and M’ simultaneously


Slide39 l.jpg

Turing-decidable scans the leftmost cell. The tape holds an input string x.

(recursive)

Turing-acceptable

(r.e.)

regular