Variants of Turing machines

1
Variants of Turing machines
CSC 4170 Theory of Computation
Section 3.2
2
Turing machines with the stay put option
3.2.a
A transition of this type of machine may have
either L (move left), Or R (move right), or S
(stay put).
a?b,S
q1
q2
Replace a with b and go to state q2 without
moving the head
This does not increase the power of the machine,
because the above transition can be simulated
with the ordinary TM as follows



3
Multitape Turing machines
3.2.b
A multitape TM has k tapes, each with its own
read/write had. Initially, the input is written
on the first tape, and all the other tapes are
blank, with each head at the beginning of the
corresponding tape. For a 3-tape TM, a
transition will look like
0,1,1?1,0,1,R,R,L
q1
q2

• If you are in state q1 and see 0 on Tape1, 1 on
  Tape2 and 1 on Tape3,
• type 1 on Tape1, 0 on Tape2 and 1 on Tape3
• move Head1 right, Head2 right and Head3 left
• go to state q2.

4
Multitape Turing machines Example
3.2.c
Design a fragment of a 2-tape TM that swaps the
contents of the tapes, from the position where
the heads are at the beginning and there are no
blanks followed by non-blank symbols. Tape
alphabet 0,1,-
0,0?0,0,R,R 0,1?1,0,R,R 0,- ? -,0,R,R 1,0?0,1,R,R
1,1?1,1,R,R 1,- ? -,1,R,R -,0?0,-,R,R -,1?1,-,R,R
-,- ?- ,-,L,L
Done
Swap
Tape1 Tape2
1
1
0
-
0
0
-
-



5
Simulating multitape TM with ordinary TM
3.2.d
Theorem 3.13 Every multitape TM M has an
equivalent single-tape TM S.
Proof idea The tape contents and the head
positions of M can be represented on the single
tape of S and correspondingly updated as shown
on the following example for a 3-tape M S
follows the steps of M and accepts iff M
accepts.
Tape1 M Tape2
Tape3 S
0
0
1
-
q0
-
-
-
-
-
-
-
-
.
.
.

0
0
1
-

-


-
-
-
-
-
-




6
Nondeterministic Turing machines
3.2.e
A nondeterministic TM is allowed to have more
than 1 transition for a given tape
symbol A string is accepted, if one of the
branches of computation takes us to the accept
state.
q2
a?b,R
q1
a?c,L
q3
Theorem 3.16 Every nondeterministic TM has an
equivalent deterministic TM.
Proof omitted. Idea Simulate every possible
branch of computation in a breadth-first manner.



7
Enumerators
3.2.f
An enumerator is a TM with two tapes the work
tape and the printer.

• Initially both tapes are blank.
• On the work tape it can read, write and move in
  either direction just as an ordinary TM.
• On the printer, it can only print one of the
  symbols of the language alphabet, or ,
• serving as the end of string symbol, which
  is not in the language alphabet.
• So that transitions look like
• meaning if, in state q1, you see an a on the
  work tape, replace it with b, move in the
• direction D (DL or DR), go to state q2 and
  print c on the printer.
• Every time a symbol is printed on the printer,
  the printer head moves right.
• If c is absent, nothing happens on the
  printer.
• There is a start state, but there are no accept
  or reject states. Entering a configuration
• from which there is no transition causes the
  enumerator to halt.
• The language enumerated by an enumerator is the
  set of strings (separated with )
• that it prints on the printer.
• It is OK if some of the strings are printed
  more than once.

c
a ? b,D
q1
q2


8
An example of an enumerator
3.2.g

- ? 0,L
? R
- ? ,L
start
go to the beginning
print
0
0 ? L
0 ? R
-
-
-
-
-
-
-
-
-
-
Work tape
-
-
-
-
-
-
-
-
-
-
Printer


9
Enumerability vs Turing recognizability
3.2.h
Definition A language is said to be
(recursively) enumerable iff some enumerator
enumerates it.
Theorem 3.21 A language is Turing recognizable
iff it is enumerable.
Proof sketch. (?) Suppose E enumerates L.
Construct a TM M that works as follows
M On input w 1. Simulate
E. Every time E prints a new string, compare it
with w. 2. If w is ever
printed, accept.
(?) Suppose M recognizes L. Let s1,s2,s3, be
the lexicographic list of all strings over the
alphabet of L. Construct an enumerator E that
works as follows E 1. Repeat the
following for i1,2,3, 2.
Simulate M for i steps on each of the inputs
s1,s2,,si. 3. If any
computations accept, print out the corresponding
sj.


10
Turing machines with an output(not in the
textbook!)
3.2.i
The only difference with ordinary TM is
that a TM with an output (TMO) has a state halt
instead of accept and reject if and when such a
machine reaches the halt state, the contents of
the tape (up to the first blank cell) will be
considered its output.
Example Design a machine that, for every input
w, returns w0.



11
Computable functions
3.2.j
A function g ?? ? is said to be
computable, iff there is a TMO M such that for
every input w?? M returns the output u with
ug(w). In this case we say that M computes g.
Example Let f 0,1 ?0,1 be the function
defined by
f(w)w0, so that f(?)0, f(0)00, f(1)10,
f(00)000, f(01)010, etc. Then f is
computable as we saw on the previous slide.
The graph of such a function is the language
(w,u) w??, ug(w) E.g.,
the graph of the above function f is
(?,0), (0,00), (1,10), (00,000), (01,010), ,
i.e.
(w,u) w?0,1, uw0


12
Computability vs decidability
3.2.k
Theorem A function is computable iff its graph
is decidable.
Proof sketch. Let g ?? ? be a function. (?)
Suppose C is a TMO that computes g. Construct a
TM D that works as follows D On
input t 1. If t does not have
the form (w,u), where w,u??, then reject.
Otherwise, 2. Simulate C for
input w. If it returns u, accept otherwise
reject.
(?) Suppose a TM D decides the graph of g. Let
s1,s2,s3, be the lexicographic list of all
strings over the alphabet ?. Construct a TMO C
that works as follows C On input t
Simulate D for each of the
inputs (t,s1), (t,s2),(t,s3), until you find si
such that (t,si) is
accepted, and return this si