Lecture 16 Deterministic Turing Machine (DTM)

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Lecture 16Deterministic Turing Machine (DTM)
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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 G. There exists
  a special symbol B which represents the empty
  cell.

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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).

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• 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
• d Q x G ? Q x G x R, L.

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b
a
q
p

• d(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.

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a
b
p
q

• d(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.

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s

• 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.

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x
h

• 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.

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• The DTM can be represented by
• M (Q, S, G, d, s)
• where S 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.

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• The transition diagram of a DTM is an
  alternative way to represent the DTM.
• For M (Q, S, G, d, s), the transition diagram
  of M is a symbol-labeled digraph G(V, E)
  satisfying the following
• V Q (s , h )
• E p q d(p, a) (q, b, D).

a/b,D
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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, S, G, d, s) where Q s, p, q, h,
• S 0, 1, ? 0, 1, B.
• d 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) (01)00(01).

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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.

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Review on Regular Set

• Regular sets on an alphabet S is defined
  recursively as follows
• (1) The empty set F is a regular set.
• (2) For every symbol a in S, a is a
• regular set.
• (3) If A and B are regular sets, then
• A n B, A U B, and A are all regular
• sets.

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Review on DFA
tape
a
b
c
d
e
f
head
finite control
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• The tape is divided into cells. Each cell
  contains a symbol in an alphabet S.
• The head scans at a cell on the tape and can read
  symbol from the cell. It can also move from left
  to right, one cell per move.

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• 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 function
  d Q x S ? Q. If the head reads symbol a and the
  finite control is in the state q, then the next
  state is p d(q, a).

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• There are some special states an initial state s
  and some final states which form a set F.
• The DFA can be represented by M
  (Q, S, d, s, F).

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• 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.

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Simulate DFA by DTM

• Given a DFA M (Q, S, d, s, F), we can construct
  a DTM M (Q, ?, S, d, s) to simulate M as
  follows
• Q Q U h, G S U B,
• If d(q, a) p, then d(q, a) (p, a, R).
  d(q, B) (h, B, R) for q in F.

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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
1
0, 1
0
0
s
p
q
1

• L(M) (01)00(01).

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Turing acceptable languages
Regular languages
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Why DTM can accept more languages than DFA?

• 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?

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Lecture 17Examples of DTM
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• (s, 0110B)
• (q0, B110B)
• (q1, B010B)
• (q1, B010B)
• (q0, B011B)
• (h, B0110B)

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Remark
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Turing-acceptable

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• ww w e (01) 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) ?

R
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R

• L ww w in (01)

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Turing-Computable Functions

• A total function f S ? S is Turing-computable
  if there exists a DTM M such that for every x in
  S,
• (s, BxB) (h, Bf(x)B).
• A partial f O ? S is Turing-computable if
  there exists a DTM M such that L(M)O and for
  every x in O,
• (s, BxB) (h, Bf(x)B).

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• 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, B011B)
• (ok, B011B) (q, B011B) (q1, B01BB)
• (q1, B01B) (p1, B01B) (ok, B01B)
• (q, B01B) (r, B01B) (r, B01B)
• (o, B01B) (o, B01B) (k, B01BB)
• (h, B01B)

R
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R

• f(x) w if x ww ?, otherwise

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Turing-decidable

• A language A is Turing-decidable if its
  characteristic function is Turing-computable.
• ? (x)

1, if x e A
0, otherwise
A
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• ww w e (01) is Turing-decidable.
• (s, B0110B) (q, B0110B) (q0, B011BB)
• (q0, B011B) (p0, B011B) (ok, B11B)
• (ok, B11B) (q, B11B) (q1, B1BB)
• (q1, B1B) (p1, B1B) (ok, BB)
• (q, BB) (r, BBB) (r, BBB)
• (o, BBB) (h, B1B)

R
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Theorem

• A language A is Turing-acceptable iff there
  exists a total Turing-computable function f such
  that A f(x) x e (01).
• 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.).

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Theorem

• A language A is Turing-decidable iff A and its
  complement A are Turing-acceptable.
• Proof. Suppose L(M) A and L(M) A.
• Construct M to simulate M and M
  simultaneously

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Turing-decidable (recursive)
Turing-acceptable (r.e.)
regular