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∞ 18 Turing Machines
- Turing Machines
- Nondeterministic Turing Machines
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∞ 18.1 Turing Machines
- Introduced by Alan Turing
- Abstract machine with many features of computer
- Designed before the stored program computer!
- No memory limit!
- An infinite tape, and a tape reader, that reads one symbol at a time.
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∞ Turing Machines (2)
0 1 2 3 4 5 6 ...
...
A quintuple M = (Q, Σ, Γ, δ, q0)
- A finite set of states Q.
- A finite set Γ, the tape alphabet, which includes blank B
- A subset Σ ⊆ Γ − {B} called the input alphabet
- A partial transition function δ : Q × Γ → Q × Γ × {L, R}
- A start state q 0 ∈ Q
∞ 18.1.1 Turing Machines: Initial
0 1 2 n
...
q
c1 c2 ... cn B B ...
...
B
Computation begins with
- An input string s = c 1 · · · cn in Σ∗^ on the tape from position 1 to n
- The remaining tape blank (filled with B)
- Initial state q 0
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∞ 18.1.2 Turing Machines: Transition
... ...
q
x ... y ...
q’
- Machine in state q reading tape symbol x ∈ Γ
- δ(q, x) = [q′, y, d]
- Transitions have three parts:
- Change the state to q′
- Write symbol y on square scanned by tape head
- Move head left or right (depending on direction d)
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∞ 18.1.3 Turing Machines: Final
... ...
q
x
- Machine halts if in state q reading tape symbol x ∈ Γ there is no δ(q, x) = [q′, y, d]
- Remember δ is a partial function.
- Output is the result remaining on the tape.
- Machine halts abnormally if the head moves off the left end of the tape.
∞ 18.1.7 Turing Machines: Another Example
Machine to add 1 to a binary number
q0 B/B R q1 B/B L q
0/0 R 1/1 R
0/1 R q
1/0 L
B/1 R
q 0 B 001 B q 0 B 11 B Bq 1001 B Bq 111 B B 0 q 101 B B 1 q 11 B B 00 q 11 B B 11 q 1 B B 001 q 1 B B 1 q 21 B B 00 q 21 B Bq 210 B B 0 q 200 B q 2 B 00 B B 01 q 30 B 1 q 300 B
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∞ 18.1.8 Turing Machines as Acceptors
0 1 2 3 4 5 6 ...
...
A sextuple M = (Q, Σ, Γ, δ, q 0 , F )
- F is the set of final states.
- If M halts in a state q ∈ F then the input is accepted
- If M halts in a state q 6 ∈ F the input is rejected.
- L(M ) is the set of strings accepted by M.
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∞ Turing Machines as Acceptors (2)
q
0/0 R 1/1 R
B/B R (^) q1 0/B R
1/B R
0/0 R 1/1 R
q
q
q
q
B/B L
B/B L
0/B L (^) q
0/0 L 1/1 L
1/B L
B/B R
A machine that accepts a string which is an even length palindrome of 0’s and 1’s. e.g 00 , 0110 , 0001001000 ∈ L(M )
∞ Exercise
Modify the even length palendrome machine to accept odd length palindromes too e.g. 101
q
0/0 R 1/1 R
B/B R (^) q1 0/B R
1/B R
0/0 R 1/1 R
q
q
q
q
B/B L
B/B L
0/B L (^) q
0/0 L 1/1 L
1/B L
B/B R
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∞ 18.1.9 Turing Machines in Prolog
- given delta(State,Sym,NewState,NewSym,LR)
- (maybe) given accepting(State) for accepting states
- We can write a Turing machine simulator in Prolog
- So Prolog can do anything a Turing machine can delta(q0, ’B’, q1, ’B’, right). delta(q1, ’B’, q2, ’B’, left). delta(q1, a, q1, b, right). delta(q1, b, q1, a, right). delta(q2, a, q2, a, left). delta(q2, b, q2, b, left).
initial(q0).
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∞ Main Code
run(Input, Output, State) :- initial(State0), tape_list_position(Tape0, Input, 0), run(State0, Tape0, State, Tape), tape_list_position(Tape, Output, _).
recognize(Input) :- run(Input, _, State), accepting(State).
∞ Example
q0 B/B R^ q1 c/c R
c/c L
q
q
q
q
a/a R
b/b L
b/b R (^) q
a/a L
a/a R b/b R c/c R
A machine which accepts strings of a’s, b’s and c’s where there is a c followed by or preceded by ab.
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∞ Example (2)
q 0 BacabB 1 q 0 BacabB 1 q 0 BacabB 2 Bq 1 acabB 1 Bq 1 acabB 1 Bq 1 acabB 2 Baq 1 cabB 1 Baq 1 cabB 1 Baq 1 cabB 3 Bacq 1 abB 1 Bacq 2 abB 2 Bq 3 acabB Bacaq 1 bB 1 Bacaq 4 bB 1 Bacabq 1 B ` Bacabq 6 B
First and third do not accept, but second does That’s good enough: the machine accepts
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