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