Turing Machine UNIT V. Prepared by Prof. T.H.Gurav,SKNCOE, Pune

Transcription

1 Turing Machine UNIT V

2 Devices of Increasing Computational Power So far: Finite Automata good for devices with small amounts of memory, relatively simple control Pushdown Automata stack-based automata Limitations: 1. Amount of Memory(Limited) 2. The way of accessing the memory (PDA can access only top of the stack ) 3. Input Tape (Finite) 4. Tape head is read only.

3 Who was Alan Turing? Founder of computer science, mathematician, philosopher, code breaker. Introduced TM in 1936!

4 Turing machine! - models a human computer (human writes/rewrites symbols on a sheet of paper, the human s state of mind changes based on what s/he has seen) - The Turing machine is a mathematical model not of computers, but of computation. His computer model the Turing Machine was inspiration of the electronic computer that came two decades later

5 Turing Machines TM s described in 1936 by Alen Turing Abstract model for today s computers Components of TM : Finite Control : A TM consists of a finite control (i.e. a finite state automaton). Infinite tape : FC is connected to an infinite tape.the tape consists of cells where each cell holds a symbol from the tape alphabet. Initially each cell has blanks. Tape Head : Reads one cell at a time and moves to L/ R / N States : Initial State, Final State

6 TM

7 States & Transitions Read Write Move Left q q2 1 a / b / L Move Right q q2 1 a / b / R

8 Turing Machine Details In one move the TM will: 1. Change state, which may be the same as the current state 2. Write a tape symbol in the current cell, which may be the same as the current symbol 3. Move the tape head left or right one cell 4. The special states for rejecting and accepting take effect immediately

9 Formal definition of TM Formally, the TM is denoted by the 7-tuple: M = (Q,, Γ, δ, q 0, B, q a ) Q = finite states of the control = finite set of input symbols, which is a subset of Γ below Γ = finite set of tape symbols δ = transition function. δ(q,x)=(p,y,d) p,q ϵ Q and X,Y ϵ Γ and D =,L/R/N q 0 = start state for finite control B = blank symbol. This symbol is in Γ but not in. q accept = set of final or accepting states of Q.

10 Turing Machine: States Input alphabet Tape alphabet M ( Q,,,, q0, B, F) Transition function Final states Initial state blank

11 Transition function q X/Y/D p

12 Instantaneous description processing of a string by TM can be shown using the instantaneous description. The ID of TM includes : 1. All non-blank cells in the tape 2. Position of head. 3. Current state of machine

13 Instantaneous Description (ID)

14 Turing Machines are deterministic Allowed Not Allowed a b, R q 2 a b, R q 2 q 1 q 1 b d, L q 3 a d, L q 3 No empty move is allowed

15 Halting and Acceptance The machine halts if there are no possible transitions to follow. A i/p string is accepted by TM if scanning of the string is completed from left to right and machine is in the final State.

16 Transition Table State I/P a B q0 (q0,a,r) (q1,b,n) q1 - -

17 Handrun Let us trace aa 1. a a B q0 2. a a B 3. a a B q0 q1 Accept

18 TM for calculations TMs can also be used for calculating values Like arithmetic computations Eg., addition, subtraction, multiplication, etc.

19 Equivalence of TM s and Computers In one sense, a real computer has a finite amount of memory, and thus is weaker than a TM. But, we can postulate an infinite supply of tapes, disks, or some peripheral storage device to simulate an infinite TM tape. Additionally, we can assume there is a human operator to mount disks, keep them stacked neatly on the sides of the computer, etc. Need to show both directions, a TM can simulate a computer and that a computer can simulate a TM

20 Computer Simulate a TM This direction is fairly easy - Given a computer with a modern programming language, certainly, we can write a computer program that emulates the finite control of the TM. The only issue remains the infinite tape. Our program must map cells in the tape to storage locations in a disk. When the disk becomes full, we must be able to map to a different disk in the stack of disks mounted by the human operator.

21 TM Simulate a Computer In this exercise the simulation is performed at the level of stored instructions and accessing words of main memory. TM has one tape that holds all the used memory locations and their contents. Other TM tapes hold the instruction counter, memory address, computer input file, and scratch data. The computer s instruction cycle is simulated by: 1. Find the word indicated by the instruction counter on the memory tape. 2. Examine the instruction code (a finite set of options), and get the contents of any memory words mentioned in the instruction, using the scratch tape. 3. Perform the instruction, changing any words' values as needed, and adding new address-value pairs to the memory tape, if needed.

22 Turing Machine Variants There are many variations we can make to the basic TM Extensions can make it useful to prove a theorem or perform some task However, these extensions do not add anything extra the basic TM.

23 TM variants 1. Multitape TM 2. Multitrack TM 3. Non deterministic TM

24 Multitape Turing Machines A multitape Turing machine has several tapes instead of one tape and each tape with a separate head. Each head can move independently of the others The transition function is changed to allow for reading, writing, and moving the heads on all the tapes simultaneously. This means we could read on multiples tape and move in different directions on each tape as well as write a different symbol on each tape, all in one move.

25 Continue..

26 Multitape Machine B M a a a B b a B Equivalent Single Tape Machine:

27 How a multitape TM operates? Initially: A move: The input is in tape #1 surrounded by blanks All other tapes contain only blanks The tape head for tape #1 points to the 1 st symbol of the input The heads for all other tapes point at an arbitrary cell (doesn t matter because they are all blanks anyway) Is a function (current state, the symbols pointed by all the heads) After each move, each tape head can move independently (left or right) of one another

28 Multitrack TM

29 continue

30 Deterministic and Nondeterministic TM Deterministic : when the transition rules are represented as five-tuples (s, x, s, x, d), where s, x are the current state and tape symbol, s, x are the next state and tape symbol, and d is the move direction, then no two transition rules begin with the same pair (s, x), i.e., the mapping (s,x) (s, x, d) is a function.

31 NTM In a nondeterministic Turing machine there maybe more that one transition rule beginning with the same pair eg (q, X). A non-deterministic Turing Machine allows more than one possible action per given state-tape symbol pair. A string w is accepted by TM if after being put on the tape and letting TM run, TM eventually enters q acc on some computation branch.

32 A Universal Turing Machine Prepared by Prof. T.H.Gurav

33 A limitation of Turing Machines: Turing Machines are hardwired they execute only one program Real Computers are re-programmable Prepared by Prof. T.H.Gurav

34 Solution: Universal Turing Machine A Universal Turing Machine is a TM that can simulate the behavior of any other Turing Machine Prepared by Prof. T.H.Gurav

35 Continue A universal Turing machine, a machine that can do any computation if the appropriate program is provided, was the first description of a modern computer. It can be proved that a very powerful computer and a universal Turing machine can compute the same thing. We need only provide the data and the program the description of how to do the computation to either machine. In fact, a universal Turing machine is capable of computing anything that is computable. Prepared by Prof. T.H.Gurav

36 Input string of M Prepared by Prof. T.H.Gurav Universal Turing Machine simulates any Turing Machine M Input of Universal Turing Machine: Description of transitions of M

37 Three tapes Tape 1 Description of M Universal Turing Machine Tape 2 Tape Contents of M Tape 3 Prepared by Prof. T.H.Gurav State of M

38 Tape 1 Description of M We describe Turing machine as a string of symbols: M We encode M as a string of symbols Prepared by Prof. T.H.Gurav

39 Alphabet Encoding Symbols: a b c d Encoding: Prepared by Prof. T.H.Gurav

40 Encoding: 1 11 Prepared by Prof. T.H.Gurav State Encoding States: q 1 q2 q3 q4 Encoding: Head Move Encoding Move: L R

41 Transition Encoding Transition: ( q1, a) ( q2, b, L) Encoding: separator Prepared by Prof. T.H.Gurav

42 separator Prepared by Prof. T.H.Gurav Turing Machine Encoding Transitions: ( q1, a) ( q2, b, L) ( q2, b) ( q3, c, R) Encoding:

43 Tape 1 contents of Universal Turing Machine: binary encoding of the simulated machine M Tape Prepared by Prof. T.H.Gurav

44 A Turing Machine is described with a binary string of 0 s and 1 s Therefore: The set of Turing machines forms a language: each string of this language is the binary encoding of a Turing Machine Prepared by Prof. T.H.Gurav

45 Language of Turing Machines L = { , , , (Turing Machine 1) (Turing Machine 2) } Prepared by Prof. T.H.Gurav

46 Halting Problem Given a program and an input to the program, determine if the program will eventually stop when it is given that input. Prepared by Prof. T.H.Gurav

47 proof that the Halting Problem is unsolvable This proof was devised by Alan Turing, Suppose you have a solution to the halting problem called H. H takes two inputs: 1. a program P and 2. an input I for the program P. H generates an output "halt" if H determines that P stops on input I or it outputs "loop" otherwise. Prepared by Prof. T.H.Gurav

48 Continue.. So now H can be revised to take P as both inputs (the program and its input) and H should be able to determine if P will halt on P as its input. Let us construct a new, simple algorithm K that takes H's output as its input and does the following : 1. if H outputs "loop" then K halts, 2. otherwise H's output of "halt" causes K to loop forever. That is, K will do the opposite of H's output. Prepared by Prof. T.H.Gurav

49 Continue.. function K() { if (H()=="loop"){ return; } else { while(true); //loop forever } } Since K is a program, let us use K as the input to H. Prepared by Prof. T.H.Gurav

50 Continue.. If H says that K halts then K itself would loop (that's how we constructed it). If H says that K loops then K will halt. In either case H gives the wrong answer for K. Thus H cannot work in all cases. We've shown that it is possible to construct an input that causes any solution H to fail. Prepared by Prof. T.H.Gurav

51 Turing Languages Three classes of languages: Turing-decidable or recursive: TM can accept the strings in the language and tell if a string is not in the language. Sometimes these are called decidable problems. Turing-recognizable or recursively enumerable : TM can accept the strings in the language but cannot tell for certain that a string is not in the language. Sometimes these are called partially-decidable. Undecidable : no TM can even recognize ALL members of the language.

52 Recursively Enumerable Languages A TM accepts a string w if the TM halts in a final state. A TM rejects a string w if the TM halts in a non final state or the TM never halts. A language L is recursively enumerable if some TM accepts it. Hence they are also called as Turing Acceptable L. Recursively Enumerable Languages are also called Recognizable

53 Recursive Language Recursive Language : A language L is recursive if some TM accepts it and halts on every input. Recursive languages are also called Decidable Languages because a Turing Machine can decide membership in those languages (it can either accept or reject a string).

54 Undecidable Languages undecidable language = not decidable language If there is no Turing Machine which accepts the language and makes a decision (halts) for every input string. Note : (machine may make decision for some input strings) For an undecidable language, the corresponding problem is undecidable (unsolvable):

55 Comparison with Previous Models Device Separate Input? Read/Write Data Structure Deterministic by default? FA Yes None Yes PDA Yes LIFO Stack No TM No 1-way infinite tape. 1 cell access per step. Yes (but will also allow crashes)