Applied Computer Science II Chapter 3 : Turing Machines

Transcription

1 Applied Computer Science II Chapter 3 : Turing Machines Prof. Dr. Luc De Raedt Institut für Informatik Albert-Ludwigs Universität Freiburg Germany

2 Overview Turing machines Variants of Turing machines Multi-tape Non-deterministic The definition of algorithm The Church-Turing Thesis

3 Turing Machine Infinite tape Both read and write from tape Move left and right Special accept and reject state take immediate effect Machine can accept, reject or loop

4 F = { w# w w {0,1}} *

5 F = { w# w w {0,1}} *

6 Turing Machines

7 Configurations ua q bv yields u q acv if δ ( q, b) = ( q, c, L) i j i j ua q bv yields uac q v if δ ( q, b) = ( q, c, R) i j i j cannot go beyond left border! start configuration accepting configuration - state is rejecting configuration - state is 1 qw 1 0 accept reject A Turing Machine accepts input w if a sequence of configurations C,..., C exists where 1. C is start configuration 2. Each Cyields C i i C is an accepting state k k q q

8 Languages The collection of strings that M accepts is the language of M, L(M) A language is Turing-recognizable (recursively enumerable) if some Turing machine accepts it Deciders halt on every input (i.e. they do not loop) A language is Turing-decidable (recursive) if some Turing machine decides it

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11 F = w w w * { # {0,1}}

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14 Variants of Turing Machines Most of them turn out to be equivalent to original model E.g. consider movements of head on tape {L,R,S} where S denotes same Equivalent to original model (represent S transition by first R and then L, or vice versa)

15 Multi-tape Turing Machines The input appears on Tape 1; others start off blank Transition function becomes δ : Q Γ Q Γ { L, R} k k k δ ( q, a,..., a ) = ( q, b,..., b, L, R,..., R) i 1 k j 1 k

16 Theorem Every multitape Turing machine has an equivalent single tape Turing machine. Proof idea See p. 137

17 Corollary A language is Turing recognizable if and only if some multitape TM recognizes it.

18 Non-deterministic TMs Transition function becomes δ : Q Γ Ρ ( Q Γ { L, R}) δ ( qa, ) = {( q, b, L),...,( q, b, R)} 1 1 Same idea/method as for NFAs k k Theorem Every non-deterministc Turing machine has an equivalent deterministic Turing machine. Proof idea Numbering the computation. Work with three tapes : 1.input tape (unchanged) 2.simulator tape 3.index for computation path in the tree - alphabet Σ = {1,..., b} b

19 Insert p 139

20 Theorem A language is Turing-recognizable if and only if some non-deterministic TM recognizes it. Corollary A language is decidable if and only if some non-deterministic TM decides it.

21 Enumerators Turing recognizable = Recursively enumerable Therefore, alternative model of TM, enumerator Works with input tape (initially empty) and output tape

22 Insert theorem 3.13

23 Equivalence with other models Many variants of TMs (and related constructs) exist. All of them turn out to be equivalent in power (under reasonable assumptions, such as finite amount of work in single step) Programming languages : Lisp, Haskell, Pascal, Java, C, The class of algorithms described is natural and identical for all these constructs. For a given task, one type of construct may be more elegant.

24 The definition of an David Hilbert algorithm Paris, 1900, Intern. Congress of Maths. 23 mathematical problems formulated 10 th problem to devise an algorithm that tests whether a polynomial has an integral root Algorithm = a process according to which it can be determined by a finite number of operations

25 Integral roots of xyz+ 3xy x 10 polynomials root = assignment of values to variables so that value of polynomial equals 0 integral root = all values in assignment are integers Church Turing Thesis There is no algorithm that solves this task. A formal Intuitive notion notion of algorithm of algorithm is necessary. Alonso Church : λ-calculus (cf. functional programming) = Allen Turing : Turing machines Turing machine algorithms

26 1 1 Integral roots of polynomials D= { p pis a polynomial with an integral root} Hilbert's 10th problem : is D decidable? D is not decidable, but Turing recognizable Consider D = { p pis a polynomial over x with an integral root} Define M : "the input is a polynomial over x 1. Evaluate p wrt xset to 0,1,-1,2,-2,... There If at any is point no p evaluates algorithm to 0, accept" that solves this task. A formal notion D of algorithm is necessary. 1 but not a decider Alonso Church : λ-calculus (cf. functional cmax can be converted into a decider using the bounds ± k for x programming) c1 Allen Turing 1 : Turing machines max This is a recognizer for M 1 k: number of terms; c : coefficient highest order term; c : largest absol. value coef Extension of M exist to D but remains a recognizer 1

27 Turing machines Three levels of description Formal description Implementation level High-level description The algorithm is described From now on, we use this level of description O : describes object O O,..., O : describes objects O,..., O 1 k 1 Encodings can be done in multiple manners; often not relevant because one encoding (and therefore TM can be transformed into another one) k

28 Connected graphs A= { G Gis a connected undirected graph} connected = every node can be reached from every other node

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30 Overview Turing machines Variants of Turing machines Multi-tape Non-deterministic The definition of algorithm The Church-Turing Thesis