Turing machines COMS Ashley Montanaro 21 March Department of Computer Science, University of Bristol Bristol, UK

Transcription

1 COMS11700 Turing machines Department of Computer Science, University of Bristol Bristol, UK 21 March 2014 COMS11700: Turing machines Slide 1/15

2 Introduction We have seen two models of computation: finite automata and pushdown automata. We now discuss a model which is much more powerful: the Turing machine. COMS11700: Turing machines Slide 2/15

3 Introduction We have seen two models of computation: finite automata and pushdown automata. We now discuss a model which is much more powerful: the Turing machine. A Turing machine is like a finite automaton, with three major differences: It can write to its tape; It can move both left and right; The tape is infinite in one direction. Control a b b a a... Initially, the input is provided on the left-hand end of the tape, and followed by an infinite sequence of blank spaces ( ). COMS11700: Turing machines Slide 2/15

4 Example q 0 a b b a a... COMS11700: Turing machines Slide 3/15

5 Example q 1 c b b a a... COMS11700: Turing machines Slide 3/15

6 Example q 1 c a b a a... COMS11700: Turing machines Slide 3/15

7 Example q 3 c a b a a... COMS11700: Turing machines Slide 3/15

8 Example q 2 c a b a a... COMS11700: Turing machines Slide 3/15

9 Example q 2 c a b a a... COMS11700: Turing machines Slide 3/15

10 Example q 2 c a b a a... COMS11700: Turing machines Slide 3/15

11 Example q 2 c a b a a... COMS11700: Turing machines Slide 3/15

12 Example q 1 c a b a a c... COMS11700: Turing machines Slide 3/15

13 Alan Turing ( ) 1936: Invented the Turing machine and the concept of computability : Worked at Bletchley Park on cracking the Enigma cryptosystem and others : Work on practical computers, AI, mathematical biology, : Convicted of indecency. Died of cyanide poisoning in : Received a royal pardon. Pic: Wikipedia/Bletchley Park COMS11700: Turing machines Slide 4/15

14 Turing machines Turing machines have two special states: an accept state and a reject state. COMS11700: Turing machines Slide 5/15

15 Turing machines Turing machines have two special states: an accept state and a reject state. If the machine enters the accept or reject state, it halts (stops). COMS11700: Turing machines Slide 5/15

16 Turing machines Turing machines have two special states: an accept state and a reject state. If the machine enters the accept or reject state, it halts (stops). If it doesn t ever enter either of these states, it never halts (i.e. it runs forever). COMS11700: Turing machines Slide 5/15

17 Turing machines Turing machines have two special states: an accept state and a reject state. If the machine enters the accept or reject state, it halts (stops). If it doesn t ever enter either of these states, it never halts (i.e. it runs forever). The language L(M) recognised by a Turing machine M is the set {x M halts in the accept state on input x} COMS11700: Turing machines Slide 5/15

18 Turing machines Turing machines have two special states: an accept state and a reject state. If the machine enters the accept or reject state, it halts (stops). If it doesn t ever enter either of these states, it never halts (i.e. it runs forever). The language L(M) recognised by a Turing machine M is the set {x M halts in the accept state on input x} For some language L, if there exists a Turing machine M such that L = L(M), we say that L is Turing-recognisable. (These languages are also sometimes called recursively enumerable.) COMS11700: Turing machines Slide 5/15

19 Describing Turing machines We can describe a Turing machine by its state diagram. As with DFAs and PDAs, we have a graph whose vertices are labelled by states of the machine, and whose edges are labelled by possible transitions. a b, R q 1 q 2 COMS11700: Turing machines Slide 6/15

20 Describing Turing machines We can describe a Turing machine by its state diagram. As with DFAs and PDAs, we have a graph whose vertices are labelled by states of the machine, and whose edges are labelled by possible transitions. A label of the form a b, R q 1 q 2 a b, R means that on reading tape symbol a, the machine writes b to the tape and then moves right. COMS11700: Turing machines Slide 6/15

21 Describing Turing machines We can describe a Turing machine by its state diagram. As with DFAs and PDAs, we have a graph whose vertices are labelled by states of the machine, and whose edges are labelled by possible transitions. A label of the form a b, R q 1 q 2 a b, R means that on reading tape symbol a, the machine writes b to the tape and then moves right. Another example: a label a, b L means that on reading either tape symbol a or b, the machine doesn t write anything to the tape and then moves left. COMS11700: Turing machines Slide 6/15

22 Turing machines Imagine we want to test membership in the language L EQ = {w#w w {0, 1} }. x L EQ if it is made up of two equal bit-strings, separated by a # symbol. COMS11700: Turing machines Slide 7/15

23 Turing machines Imagine we want to test membership in the language L EQ = {w#w w {0, 1} }. x L EQ if it is made up of two equal bit-strings, separated by a # symbol. Idea for recognising this language 1. Our algorithm will scan forwards and backwards, testing each corresponding pair of bits either side of the # for equality in turn. 2. We can overwrite each bit with an x symbol after checking it so we don t check the same bits twice. COMS11700: Turing machines Slide 7/15

24 State diagram (Sipser, Figure 3.10) q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept q # L 6 q 7 1 x, R q 3 q 5 1 x, L COMS11700: Turing machines Slide 8/15

25 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q # COMS11700: Turing machines Slide 9/15

26 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 2 x 1 # COMS11700: Turing machines Slide 9/15

27 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 2 x 1 # COMS11700: Turing machines Slide 9/15

28 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 4 x 1 # COMS11700: Turing machines Slide 9/15

29 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 6 x 1 # x 1... COMS11700: Turing machines Slide 9/15

30 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 7 x 1 # x 1... COMS11700: Turing machines Slide 9/15

31 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 7 x 1 # x 1... COMS11700: Turing machines Slide 9/15

32 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 1 x 1 # x 1... COMS11700: Turing machines Slide 9/15

33 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 3 x x # x 1... COMS11700: Turing machines Slide 9/15

34 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 5 x x # x 1... COMS11700: Turing machines Slide 9/15

35 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 5 x x # x 1... COMS11700: Turing machines Slide 9/15

36 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 6 x x # x x... COMS11700: Turing machines Slide 9/15

37 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 6 x x # x x... COMS11700: Turing machines Slide 9/15

38 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 7 x x # x x... COMS11700: Turing machines Slide 9/15

39 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 1 x x # x x... COMS11700: Turing machines Slide 9/15

40 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 8 x x # x x... COMS11700: Turing machines Slide 9/15

41 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 8 x x # x x... COMS11700: Turing machines Slide 9/15

42 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q 8 x x # x x... COMS11700: Turing machines Slide 9/15

43 Example: testing whether 01#01 L EQ q 2 q 4 0 x, R 0 x, L 0, 1, x L 0, 1 L q 1 q 8 R q accept # L q 6 q 7 1 x, R q 3 q 5 1 x, L q accept x x # x x... COMS11700: Turing machines Slide 9/15

44 Turing machines: formal definition Definition A Turing machine is described by a 7-tuple (Q, Σ, Γ, δ, q 0, q accept, q reject ), where: 1. Q is the set of states, 2. Σ is the input alphabet (which must not contain ), 3. Γ is the tape alphabet, where Γ and Σ Γ, 4. δ : Q Γ Q Γ {L, R} is the transition function, 5. q 0 Q is the start state, 6. q accept is the accept state, 7. q reject is the reject state, where q reject q accept. COMS11700: Turing machines Slide 10/15

45 Example The Turing machine we just saw is described by ({q 1,..., q 8, q accept, q reject }, {0, 1, #}, {0, 1, #, x, }, δ, q 1, q accept, q reject ) where the transition function δ is defined by the table 0 1 # x q 1 q 2, x, R q 3, x, R q 8, R q 2 q 2, R q 2, R q 4, R q 3 q 3, R q 3, R q 5, R q 4 q 6, x, L q 4, R q 5 q 6, x, L q 5, R q 6 q 6, L q 6, L q 7, L q 6, L q 7 q 7, L q 7, L q 1, R q 8 q 8, R q accept, R Blank entries in the table correspond to transitions where the machine rejects. COMS11700: Turing machines Slide 11/15

46 Turing machines: formal definition (2) The full description of what a Turing machine M is doing at any point in time is called its configuration. We write uqv for the configuration where: the tape to the left of the current position contains u; the tape to the right of the current position (and including the current position) contains v (followed by an infinite number of s); the machine is in state q. COMS11700: Turing machines Slide 12/15

47 Turing machines: formal definition (2) The full description of what a Turing machine M is doing at any point in time is called its configuration. We write uqv for the configuration where: the tape to the left of the current position contains u; the tape to the right of the current position (and including the current position) contains v (followed by an infinite number of s); the machine is in state q. For example, 01q describes the following situation: q COMS11700: Turing machines Slide 12/15

48 Turing machines: formal definition (3) Configuration C 1 yields C 2 if M can go from C 1 to C 2 in one step. So: uaqj bv yields uq k acv if δ(q j, b) = (q k, c, L); uaqj bv yields uacq k v if δ(q j, b) = (q k, c, R). COMS11700: Turing machines Slide 13/15

49 Turing machines: formal definition (3) Configuration C 1 yields C 2 if M can go from C 1 to C 2 in one step. So: uaqj bv yields uq k acv if δ(q j, b) = (q k, c, L); uaqj bv yields uacq k v if δ(q j, b) = (q k, c, R). If M is at the left-hand end of the tape, it cannot move any further to the left; but M can move arbitrarily far to the right (the tape is one-way infinite). COMS11700: Turing machines Slide 13/15

50 Turing machines: formal definition (3) Configuration C 1 yields C 2 if M can go from C 1 to C 2 in one step. So: uaqj bv yields uq k acv if δ(q j, b) = (q k, c, L); uaqj bv yields uacq k v if δ(q j, b) = (q k, c, R). If M is at the left-hand end of the tape, it cannot move any further to the left; but M can move arbitrarily far to the right (the tape is one-way infinite). M accepts input x if there is a sequence of configurations C 1,..., C k such that: 1. C 1 is the start configuration of M on input x; 2. For all 1 i k 1, C i yields C i+1 ; 3. C k is an accepting configuration (M is in state q accept). COMS11700: Turing machines Slide 13/15

51 Real-world implementations As well as being a mathematical tool, a Turing machine is a real machine that we can build COMS11700: Turing machines Slide 14/15

52 Summary and further reading A Turing machine is a generalisation of a finite automaton which has access to an infinite tape which it can read from and write to. Turing machines can perform complicated computations (although writing these down formally can be a painful process). Further reading: Sipser 3.1. COMS11700: Turing machines Slide 15/15