cosc 3340: introduction to theory of computation

Lecture 16 UofH - COSC 3340 - Dr. Verma1

COSC 3340: Introduction to Theory of Computation

University of Houston

Dr. Verma

Lecture 16


Turing Machine (TM)

. . .

Finite State

control

Bi-directionRead/Write


Historical Note

Proposed by Alan Turing in 1936 in:

On Computable Numbers, with an application to the Entscheidungsproblem, Proc. Lond.

Math. Soc. (2) 42 pp 230-265 (1936-7); correction ibid. 43, pp 544-546 (1937).


Turing Machine (contd.)

Based on (q, ), q – current state, – symbol scanned by head, in one move, the TM can:

(i) change state

(ii) write a symbol in the scanned cell

(iii) move the head one cell to the left or right

Some (q, ) combinations may not have any moves. In this case the machine halts.


Turing Machine (contd.)

We can design TM’s for computing functions from strings to strings

We can design TM’s to decide languages– using special states accept/reject or by writing Y/N on tape.

We can design TM’s to accept languages.– if TM halts string is accepted

Note: there is a big difference between language decision and acceptance!


Example of TM for {0n1n | n > 0}

English description of how the machine works:1. Look for 0’s2. If 0 found, change it to x and move right, else reject3. Scan past 0’s and y’s until you reach 14. If 1 found, change it to y and move left, else reject.5. Move left scanning past 0’s and y’s6. If x found move right7. If 0 found, loop back to step 2.8. If 0 not found, scan past y’s and accept.

Head is on the left or start of the string.

x and y are just variables to keep track of equality


Example of TM for {0n1n | n > 0} contd.

State Symbol Next state action

q0 0 (q1, x, R)

q0 1 halt/reject

q0 x halt/reject

q0 y (q3, y, R)





q1 0 (q1, 0, R)

q1 1 (q2, y, L)

q1 x halt/reject

q1 y (q1, y, R)





q2 0 (q2, 0, L)

q2 1 halt/reject

q2 x (q0, x, R)

q2 y (q2, y, L)





q3 0 halt/reject

q3 1 halt/reject

q3 x halt/reject

q3 y (q3, y, R)

q3 □ (q4, □, R)





q4 0 illegal i/p

q4 1 illegal i/p

q4 x illegal i/p

q4 y illegal i/p

q4 □ halt/accept



Example of TM for {0n1n | n 0} contd.


JFLAP SIMULATION


Formal Definition of TM

Formally a TM M = (Q, , , , s) where, – Q – a finite set of states – input alphabet not containing the blank symbol # – the tape alphabet of M– s in Q is the start state : Q X Q X X {L, R} is the (partial) transition function.

Note: (i) We leave out special states. (ii) The model is deterministic but we just say TM instead of

DTM.

cosc 3340: introduction to theory of computation

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