Pumping Lemma Revisited

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Presentation transcript:

Pumping Lemma Revisited Is { w | w contains and equal number of As and Bs } regular? What MUST you do to answer this question?

Turing Machines Section 3.1 November 3, 2008 CS310 Turing Machines Section 3.1 November 3, 2008

Turing Machines Similar to Finite Automata unlimited and unrestricted memory random access more accurate model of modern computer Problems that cannot be solved by a Turing Machine cannot be solved by a “real” digital computer theoretical limits of computation What are the fundamental capabilities and limitations of computers? Computer Science is really the science of computation, not of computers.

Turing Machine State Transitions plus infinite “data tape” a b d c e f read tape write tape move around on tape Read/Write Head Data Tape (containing input string) State Transitions a b d c e f

Differences with FA TM can read and write from tape FA can only read Read/Write head can move left or right FA can only move one direction TM tape is infinite TM accept and reject states take effect immediately

Church-Turing Thesis Turing Model is and always will be the most powerful model it can simulate other models: D/NFA, PDA variations do no improve it extra tape nondeterminism extra read/write heads

Formal Definition (7 Tuple) {Q, Σ, ,  ,q0, qaccept, qreject} Q: set of states Σ: input alphabet, not containing the blank character:  : tape alphabet,    and Σ   : Q x   Q x  x {L, R} : transition function q0  Q : start state qaccept  Q : accept state qreject  Q : reject state, qaccept  qreject

Operation Start configuration of M on input w is:q0w Accepting configuration: qaccept Rejecting configuration: qreject Yield: uaqibv yields uqnacv if (qi, b) = (qn, c, L) Accepting and Rejecting configurations are called halting configurations the TM stops operating otherwise, loops forever

Definition of Computing A TM, M, accepts a string, w, if there exists a sequence of configurations, c0,c1,…,cn, such that: c0 is the start configuration ci yields ci+1 for all i cn is an accept configuration The set of strings M accepts is L(M) language of M

Notes Deterministic May make multiple passes over input Reject string by entering reject configuration or looping forever hard to tell if a machine will loop forever halting problem

Example L = { w#w | w  { 0,1} * } Conceptually, we want to do what? input string: Configuration of the TM: u qn v u, v  * and qn  Q the read/write head is on the first character of v and the TM is in state qn 1 # 

Definitions Turing recognizable a language is Turing Recognizable if some TM recognizes it Turing decidable a language is Turing decidable if some TM decides it halts on rejected strings rather than looping forever hard to tell if a looping machine is really going to reject the string

Example L = {ΣΣ0 | Σ = {0 ,1} }

Example L = {anbn | n>=0 }

Example TM L = { w | |w| is even }, Σ = {a, b}

Transducer A machine that produces output A function F with domain D is Turing- Computable if there exists a TM, M, such that the configuration q0w yields qaccept, F(w) for all w D. x = number in base 1, F(x) = 2x x = 111 2x = 111111

Transducer x, y positive integers in base 1 design TM that computes x+y

Transducer x, y positive integers in base 1, x > y design TM that computes x-y