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Introduction to Turing Machines

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Title: Introduction to Turing Machines


1
Introduction to Turing Machines
  • Juan Carlos Guzmán
  • CS 6413 Theory of Computation
  • Southern Polytechnic State University

2
What Are Programs?
  • Programs can be thought of as implementation of
    languages (or properties)
  • P(x) yes, if x?LP
  • P(x) no, otherwise
  • How do we describe programs
  • Programs are themselves finite strings

3
On Counting Languages
  • Consider S 0,1
  • S e,0,1,00,01,10,11,000,
  • If L is a language over S
  • then L ? S
  • If LS is the set of all languages over S
  • LS P (S) (or 2S) (parts of S)

4
On Counting Languages
  • It is a well-known mathematical fact that, for
    any set A
  • A lt 2A
  • In our case
  • S lt LS
  • However
  • Programs are strings
  • They solve problems on strings (languages)

5
On Counting Languages
  • There must be languages that cannot be described
  • In fact, for any language described, there are
    infinitely many that are not described

6
Decidable Problem
  • A problem is decidable if there is an algorithm
    that correctly answers yes or no to each
    instance of the problem
  • Otherwise, the problem is undecidable

7
The Halting Problem
  • Determining whether a program terminates for a
    given input is undecidable

8
The Halting Problem
  • Lets assume there is such a program, called
    halting
  • halting Program Input ? yes,no
  • There is another program looping, that does not
    terminate
  • looping Input ? __

9
The Halting Problem
  • H1 ? P . halting(P,P )
  • H2 ? P . if halting(P,P )
  • then looping(P )
  • else yes
  • What is the answer of H2(H2)

10
Paradox
  • If H2 really halts on H2, then it will loop
  • If H2 loops, then it really should halt
  • The paradox arises from assuming the existence of
    the halting program

11
Problem Reduction
  • Suppose P1 is undecidable
  • We want to prove that P2 is also undecidable
  • We need to find a transformation that would take
    any instance of P1 into an instance of P2
  • We have thus proven that P2 is undecidable

12
Church-Turing Thesis
  • All general-purpose models of computation have
    the same power
  • they compute exactly the same functions
  • partial recursive functions

13
Turing Machine
  • Finite set of states
  • Unbounded tape
  • blank characters, except
  • a finite string (input) in the tape
  • Tape head
  • always scanning one cell
  • initially scanning leftmost character of input
  • Movement
  • moves left or right
  • reads the current character on the tape
  • writes a character symbol

14
Turing Machine
  • M (Q,S,G,d,q0,B,F ), where
  • Q is the set of states
  • S is the set of input symbols
  • G is the set of tape symbols (S ? G)
  • d is the transition function
  • d (Q G) ? (Q G left,right)
  • q0 is the starting state
  • B is the blank symbol (B ? G - S)
  • F is the set of final states

15
Instantaneous Description
  • An instantaneous description is a sequence
    X1X2Xi-1qXiXn, where
  • Xk in G
  • q in Q
  • All Xis are nonblank, except when q is at one of
    the extremes, when there may be a sequence of
    blanks between q and the string
  • The elements not represented in the tape are all
    blank
  • In this description, q is scanning Xi, so, if we
    were more sophisticated, we could write

16
Movement Relation ()
  • If the current ID of the machine is
    X1X2Xi-1qXiXn
  • If d(q,Xi) (p,Y,left )
  • X1X2Xi-1qXiXn X1X2Xi-2pXi-1YXn
  • If d(q,Xi) (p,Y,right )
  • X1X2Xi-1qXiXn X1X2Xi-1YpXi1Xn
  • We will not notate blanks to the end of the ID,
    so

17
Movement Relation(Special Cases)
18
Example (0n1n)
Y/Y? 0/0?
Y/Y ? 0/0 ?
start
0/X?
1/Y?
Y/Y?
B/B?
q0
q4
q1
q2
q3
X/X?
Y/Y?
19
Language of a Turing Machine
  • Acceptance by final state
  • q0w ?p? with p ? F
  • Acceptance by halting

20
Programming Techniques
  • Storage in the State
  • Multiple tracks
  • Subroutines

21
Storage in the State
  • Store information along with the state
  • Otherwise, the TM really can only store info in
    the tape
  • ?q,A,B,C ?

X1
X2
Xi
Xn
B
B




q
A
C
D
22
Multiple Tracks
  • Each symbol is actually a tuple of several
    symbols
  • Initially, only the upper track contains
    information

23
Subroutines
  • TMs do not have any abstraction mechanism no
    call/return pair
  • We can reason about code with some level of
    abstraction (e.g. swap), and splice the operation
    every time it is used

24
Extensions to Turing Machines
  • Multi-tape Turing Machine
  • Nondeterministic Turing Machine

25
Multi-Tape Turing Machines
  • Several tapes
  • Ability to move each of them independently
  • The initial string is in the first tape
  • All other tapes are initially empty
  • There is one head per tape each tape
  • Scans an input symbol
  • Upon transition
  • Writes a new symbol
  • moves left, right, or remains stationary

26
Multi-Tape Turing Machines
Z1
Z2
Zk
Zn
B
B




q
27
Multi-Tape Turing Machines
  • Are many tapes better than one?
  • Does a multi-tape TM have more power than a
    (single tape) TM?
  • Can it compute more functions?
  • See Turing-Church Thesis
  • A multi-tape TM computes exactly the same
    functions as a (single tape) TM

28
Multi-Tape Turing Machines
  • A (single tape) TM is already a multi-tape TM,
    with only one tape
  • A k-tape TM can be simulated by a (single tape)
    Turing Machine
  • with 2k tracks
  • k tracks to store the k tapes
  • k corresponding tracks to mark the positions of
    the heads of the multi-tape TM
  • a stored value meaning the number of heads are
    to the left of the current position

29
Nondeterministic Turing Machines
  • Delta(q,X) can be a set of triplets, rather than
    a single triplet
  • Therefore, the machine may have a choice of
    movements on certain configurations

30
Nondeterministic Turing Machines
  • Nondeterminstic Turing Machines (NTM) accept the
    same class of languages as the deterministic
    Turing Machines (DTM)
  • Use a two-tape deterministic TM to simulate a NTM
  • Use one tape to keep a queue of IDs
  • Use the other as scratch to simulate executions
    of the NTM
  • Separate the IDs and mark current one
  • Each NTM is emulated by a different DTM

x
ID1 ID2 ID3 ID4

q
31
Nondeterministic Turing Machines
  • To execute current ID find q and a
  • Encoded in the machine is knowledge on how many
    different movements it can do on such input
  • For each choice
  • Copy the ID into the scratch tape
  • Do the corresponding movement
  • Copy the resulting ID to the end of the first
    tape
  • Advance current ID marker

x
ID1 ID2 ID3 ID4 ID5 ID6 ID7

q
32
Running Time of Simulated Machines
  • Machines run in simulation take more steps to
    accept a string than if they were running
    natively
  • Multi-tape by One-tape TM
  • O (n2)
  • Nondeterministic to Deterministic TM
  • exponential

33
Restricted Turing Machines
  • Turing Machines with semi-infinite tapes
  • Multi-stack machines
  • Counter machines

34
Turing Machines with Semi-Infinite Tapes
  • There is the concept of the beginning of the
    tape
  • Simulate the two-way infiniteness by having a
    two-track tape

X1
X2
Xi
X0


X-1
X-2
X-i



q
35
Multi-Stack Machines
  • These are PDAs with multiple stacks
  • Each stack is handled independently
  • An symbol is popped
  • A group of symbols is pushed
  • Two-stack machines recognize the same class of
    languages as Turing Machines

36
Counter Machines
  • Similar to stack machines, but more restricted
  • Instead of stacks, the machines have counters,
    that can hold any positive value, or zero, and
    can be incremented or decremented independently

37
Turing Machines and Computers
  • A computer can simulate a Turing Machine
    (although with some difficulty)
  • A Turing Machine can simulate a computer (in
    polynomial time)

38
Computer Simulation of a Turing Machine
  • No computer with a finite amount of memory is
    able to simulate a Turing Machine
  • Consider that the hard drive is the tape you
    can buy very large ones, but they are finite
  • A TM can solve problems that require more memory
    than the memory available to the computer
  • Consider a string so awfully large that is larger
    than the computers memory

39
Computer Simulation of a Turing Machine
  • Assume
  • Unbounded amounts of removable media
  • Each disk represents a segment of the tape
  • The computer can have one segment at any given
    time

Computer
Tape to the left
Tape to the right
40
Turing Machine Simulation of a Computer
  • A computer can be simulated by a multi-tape
    Turing Machine

Memory
Instruction Counter
Memory Address
Input File
Scratch

q
41
Turing Machine Simulation of a Computer
  • The good news
  • The simulation only takes polynomial time!

42
Languages Recognized by Machines
  • Some languages cannot be not recognized
  • Some languages are recognized, but not reasonably
    fast
  • Polynomial time, vs.
  • Exponential time

All languages
TM
NPDA
DPDA
FA
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