1. COSC 3340: Introduction to Theory of Computation
University of Houston
Dr. Verma
Lecture 16
Lecture 16
UofH - COSC Dr. Verma

2. Turing Machine (TM) . . . Bi-direction Read/Write Finite State control
Bi-direction
Read/Write
Finite
State
control
Lecture 16
UofH - COSC Dr. Verma

3. 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 (1936-7); correction ibid. 43, pp (1937).
Lecture 16
UofH - COSC Dr. Verma

4. 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.
Lecture 16
UofH - COSC Dr. Verma

5. 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!
Lecture 16
UofH - COSC Dr. Verma

6. Example of TM for {0n1n | n > 0}
English description of how the machine works:
Look for 0’s
If 0 found, change it to x and move right, else reject
Scan past 0’s and y’s until you reach 1
If 1 found, change it to y and move left, else reject.
Move left scanning past 0’s and y’s
If x found move right
If 0 found, loop back to step 2.
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
Lecture 16
UofH - COSC Dr. Verma

7. Example of TM for {0n1n | n > 0} contd.
Head is on the left or start of the string.
State
Symbol
Next state action q0
(q1, x, R) 1
halt/reject x y
(q3, y, R)
Lecture 16
UofH - COSC Dr. Verma

8. Example of TM for {0n1n | n > 0} contd.
Head is on the left or start of the string.
State
Symbol
Next state action q1
(q1, 0, R) 1
(q2, y, L) x
halt/reject y
(q1, y, R)
Lecture 16
UofH - COSC Dr. Verma

9. Example of TM for {0n1n | n > 0} contd.
Head is on the left or start of the string.
State
Symbol
Next state action q2
(q2, 0, L) 1
halt/reject x
(q0, x, R) y
(q2, y, L)
Lecture 16
UofH - COSC Dr. Verma

10. Example of TM for {0n1n | n > 0} contd.
Head is on the left or start of the string.
State
Symbol
Next state action q3
halt/reject 1 x y
(q3, y, R) □
(q4, □, R)
Lecture 16
UofH - COSC Dr. Verma

11. Example of TM for {0n1n | n > 0} contd.
Head is on the left or start of the string.
State
Symbol
Next state action q4
illegal i/p 1 x y □
halt/accept
Lecture 16
UofH - COSC Dr. Verma

12. Example of TM for {0n1n | n  0} contd.
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13. JFLAP SIMULATION
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14. JFLAP SIMULATION
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15. JFLAP SIMULATION
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16. JFLAP SIMULATION
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17. JFLAP SIMULATION
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18. JFLAP SIMULATION
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19. JFLAP SIMULATION
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20. JFLAP SIMULATION
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21. JFLAP SIMULATION
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22. JFLAP SIMULATION
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23. JFLAP SIMULATION
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24. JFLAP SIMULATION
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25. JFLAP SIMULATION
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26. JFLAP SIMULATION
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27. JFLAP SIMULATION
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28. JFLAP SIMULATION
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29. JFLAP SIMULATION
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30. 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.
Lecture 16
UofH - COSC Dr. Verma