CS 3240: Languages and Computation

1
CS 3240 Languages and Computation

• Turing Machine

2
Turing and Turing Machine

• Alan Turing (19121954), British
  mathematician/engineer.
• In 1936, Turing introduced his abstract model for
  computation in his article On Computable
  Numbers, with an application to the
  Entscheidungsproblem
• Near the same time, Alonzo Church published
  similar ideas and results.
• Turing machine is the ultimate model of
  computation.

3
Comparison of TM and FA

• TM has tape you can read from and write to
• Read-write head can be moved in either direction
• Tape is infinite
• Accept and reject states take effect immediately

4
Example I

• Language ww w ? 0,1
• How can we design a Turing machine to test its
  membership?
• Zig-zag across tape to corresponding positions on
  either side of to check whether they contain
  same symbol. If not or if no found, reject.
• Cross off symbols as they are checked
• When all symbols to the left of are crossed
  off, if anything remain to the right o , reject
  otherwise, accept.
• Example test on string 011001011001

5
Formal definition of TM

• Definition A Turing machine is a 7-tuple
  (Q,?,?,?,q0,qaccept,qreject), where Q, ?, and ?
  are finite sets and
• Q is the set of states,
• ? is the input alphabet not containing the
  special blank symbol
• ? is tape alphabet, where ????,
• ? Q???Q???L,R is the transition function,

6
Formal definition of a TM

• Definition A Turing machine is a 7-tuple
  (Q,?,?,?,q0,qaccept,qreject), where Q, ?, and ?
  are finite sets and
• q0?Q is the start state,
• qaccept?Q is the accept state, and
• qreject?Q is the reject state, where
  qreject?qaccept

7
More on Transition Function

• Q set of states, ? tape alphabet
• ? Q???Q???L,R
• Given
• The current internal state ? Q
• The symbol on the current tape cell
• Then ? tells us what the TM does
• Changes to new internal state ? Q
• Either writes new symbol ? ?
• Or moves one cell left or right

8
Computing with TM

• M receives input w w1w2wn?? on leftmost n
  squares of tape
• Rest of tape is blank (all symbols)?
• Head position begins at leftmost square of tape
• Computation follows rules of ?
• Head never moves left of leftmost square of the
  tape
• If ? says to move L, head stays put!

9
Completing computation

• Continue following ? transition rules until M
  reaches qaccept or qreject
• Halt at these states
• May never halt if machine never reaches one of
  these states!

10
Turing machine notation

• ?(qi,b)(qk,c,D), where D L or R, is
  represented by the following transition

• In the special case where ?(qi,b)(qk,b,D), i.e.,
  the tape is unchanged, the right-hand side will
  just display the direction

11
Revisiting Example I

• ?(q1, 0) (q2, x, R)
• ?(q2, 0) (q2, 0, R) ?(q2, 1) (q2, 1, R)
  ?(q2, ) (q4, , R)?
• ?(q4, x)(q4, x, R) ?(q4, 0)(q6, x, L)?
• ?(q1, 1) (q3, x, R)?
• ?(q3, 0) (q3, 0, R) ?(q3, 1) (q3, 1, R)
  ?(q3, ) (q5, , R)?
• ?(q5, x)(q5, x, R) ?(q5, 1)(q6, x, L)?
• ?(q6, a)(q6, a, L) for a0, 1, or x
• ?(q6, ) (q7, , L) ?(q7, 0) (q7, 0, L) ?(q7,
  1) (q7, 1, L) ?(q7, x) (q1, x, R)?
• ?(q1, ) (q8, , R) ?(q8, x) (q8, x, R)
  ?(q8, ) (qaccept, , R)?

12
TM state diagram
13
Example

• Ex1x2x3xn xi? 0,1 and xi?xj for i ?
  j
• Basic idea Compare each pair of (xi , xj) by
  zig-zagging. Use special marks to replace to
  indicate which pair of (xi , xj) is being
  compared
• Must take care special cases such as n0, n1,
  missing , etc.

14
Question

• How to detect whether we have reached left-most
  cell?
• Alternative 1 If leftmost cell marked by special
  symbol, then easy
• Alternative 2 Use special property that TM
  stays put at left-most cell by replacing cell
  with special symbol and then recover

15
TM configurations

• The configuration of a Turing machine is the
  current setting
• Current state
• Current tape contents
• Current tape location
• Notation uqv
• Current state q
• Current tape contents uv
• Only symbols after last symbol of v
• Current tape location first symbol of v

16
Configuration C1 yields C2

• C1 yields C2 if the TM can legally go from C1 to
  C2 in one step
• Assume a, b ? ? and u, v ? ?
• uaqibv yields uqkacv
• if ?(qi,b)(qk,c,L)?
• uaqibv yields uacqkv
• if ?(qi,b)(qk,c,R)?

17
Configuration C1 yields C2

• Special cases if head is at beginning or end of
  tape
• uaqibv yields uqjacv if ?(qi,b)(qk,c,L)?
• uaqibv yields uacqkv if ?(qi,b)(qk,c,R)?
• Special case
• qibv yields qkcv if ?(qi,b)(qk,c,L) and tape
  head is at beginning of tape

18
Special configurations

• Start configuration
• q0w
• Halting configurations
• Accepting configuration uqacceptv
• Rejecting configuration uqrejectv
• u, v ? ?

19
Strings accepted by a TM

• A Turing machine M accepts input sequence w if a
  sequence of configurations C1, C2, , Ck exist,
  where
• C1 is the start configuration of M on input w
• each Ci yields Ci1 for i 1, 2, , k-1
• Ck is an accepting configuration

20
Language of a TM

• The language of M, denoted L(M), is
• L(M) w M accepts w
• A language is called Turing-recognizable if some
  Turing machine recognizes it

21
Deciders

• A Turing machine is called a decider if every
  string in ? is either accepted or rejected
• A language is called Turing-decidable if some
  Turing machine decides it
• These languages are often just called decidable