Homework Homework #5 Turing Machines Returned today Homework - PDF document

Homework • Homework #5 Turing Machines – Returned today • Homework #7 (due Thursday) – Help after lecture And the languages they handle Before We Start Languages • Any questions? • Is that your final answer? – What is a language? – What is a class of languages? Now our picture looks like Turing Machine • A Machine consists of: Context Free Languages – A state machine Deterministic Context Free Languages – An input tape – A movable r/w tape head Regular Languages • A move of a Turning Machine – Read the character on the tape at the current position of Finite the tape head Languages – Change the character on the tape at the current position of the tape head – Move the tape head We’re going to start to look at – Change the state of the machine based on current state languages out here and character read 1

Turing Machine Turing Machine • In the original Turing Machine • Tape that holds Input tape (input/memory) character string – The read tape was infinite in both directions ... • Movable tape head that – The text describes a semi-infinite TM where Read head reads and writes • The tape is bounded on the left and infinite on the State right character Machine • It can be shown that the semi-infinite TM is • Machine that changes equivalent to the basic TM. (program) state based on what is – As not to confuse, I’ll follow the conventions in read in the book…(as much as that bothers me!) Turing Machines Turing Machines • About the input tape • About the machine states – Bounded to the left, infinite to the right – A Turing Machine does not have a set of accepting states – All cells on the tape are originally filled with a special “blank” character ∆ – Instead, each TM has a special halting state, h. – Tape head is read/write • Once in the halting state, the machine halts. – Unlike PDAs, the basic TM is deterministic! – Tape head can not move to the left of the start of the tape • If it tries, the machine “crashes” Turing Machines Turing Machines • Transition function: • Let’s formalize this – δ : Q x ( Γ ∪ { ∆ }) → (Q ∪ {h}) x ( Γ ∪ { ∆ }) x {R, L, – A Turing Machine M is a 5-tuple: S} – M = (Q, Σ , Γ , q 0 , δ ) where • Q = a finite set of states (assumed not to contain the halting – Input: state h) • Current state • Σ = input alphabet (strings to be used as input) • Tape symbol read at current position of tape head • Γ = tape alphabet (chars that can be written onto the tape. – Output: Includes symbols from Σ ) • State in which to move the machine (can be the halting state) • Both Σ and Γ are assumed not to contain the “blank” symbol • Tape symbol to write at the current position of the tape head • δ = transition function (can be the “blank” symbol) • Direction in which to move the tape head (R = right, L = left, S= stationary) 2

Turing Machine Turing Machine • Accepting a string • Running a Turing Machine – A string x is accepted by a TM, if – The execution of a TM can result in 4 possible cases: • Starting in the initial configuration • With x on the input tape • The machine “halts” (ACCEPT) • The machine has nowhere to go (REJECT) • The machine eventually ends up in the halting state. • The machine “crashes” (REJECT) – I.e. • The machine goes into an “infinite loop” (REJECT • (q 0 , ∆ x) a * (h, xay) but keeps us guessing!) – for x,y ∈ ( Γ ∪ { ∆ }) * , a ∈ ( Γ ∪ { ∆ }) Turing Machine Computation with Turing Machines • Language accepted by a TM • Computing functions with TMs – The language accepted by a TM is the set of all – The result of the function applied to an input input strings x on which the machine halts. string x, will be left on the tape when the machine halts. – Functions with multiple arguments can be placed on the input tape with arguments separated by blanks. • Questions? Computation with Turing Machines Computation with Turing Machines • Computing functions with TMs • Computing functions with TMs – Formally, – Formally (with multiple arguments) • Let T = (Q, Σ , Γ , q 0 , δ ) be a TM and let f be a • Let T = (Q, Σ , Γ , q 0 , δ ) be a TM and let f be a partial function on Σ * . We say T computes f if for partial function on ( Σ *) k . We say T computes f if every x ∈ Σ * where f is defined: for every (x 1 , x 2 . …, x k ) where f is defined: – (q 0 , ∆ x) a * (h, ∆ f(x)) – (q 0 , ∆ x 1 ∆ x 2 … ∆ x k ) a * (h, ∆ f(x 1 , x 2 . …, x k )) • And for every other x, T fails to halt on input x. • And for every other k-tuple, T fails to halt on input x. 3

TMs and languages Computation with Turing Machines • Characteristic function of a set: • Two classes of languages that involve TMs: – For any language L, the characteristic function is – Languages, L accepted by TM defined as: • Halts on x ∈ L, “rejects” on x ∉ L • 1 if x ∈ L • 0 otherwise – Languages, L recognized by a TM • TM computes the characteristic function of L – If a TM computes the characteristic function of a language, it will always halt with either a 1 or 0 left on • Halts on all x the tape • Give yes/no answer as to whether x ∈ L • This is stronger than saying that x is accepted by the TM since if x ∉ L, for a TM that accepts L, there is no guarantee that the the machine will halt. TMs and languages TMs and languages • Two classes of languages that involve TMs: • First observation: – A language is recursively enumerable if there is – Every recursive language is also recursively a TM that accepts it enumerable – A language is recursive if there is a TM that • Modify the TM that recognizes the language so that it goes into a “nowhere state” just before placing the recognizes it. 0 on the tape indicating that a string is not in the language. TMs and languages TMs and languages • Game plan Is there 1. Show that there exists a language that is not something out recursively enumerable here Recursively 2. Show that there exists a language that is Enumerable recursively enumerable but not recursive. Recursive Is there something in here 4

Universal Turing Machine A language that is not recursively enumerable • Universal Turing Machine • Sorry, no pumping Lemma for TMs! – A TM T u that takes as input, an encoded version of another TM M’ and an input to M’. • Instead we introduce the notion of a – M will simulate the processing of M’ Universal Turning Machine • If M’ halts on input x, T u will halt • If M’ rejects on input x,T u will reject. – T u is a general purpose TM • M’ is a program run on the general purpose TM Universal Turing Machine Universal Turing Machine • Encoding scheme • Encoding scheme – We need a way to encode a TM so that it can be – Associate states, symbols, and directions with provided as input to T u binary numbers: – Define sets: • States: • Q = {q 1 , q 2 , … } = set of all possible states that may appear in – s(h) → 0 any TM – s(q i ) → 0 i+1 • S = {a 1 , a 2 , a 3 , …} = set of all possible tapes symbols that can • Symbols be written on any TM. – s( ∆ ) → 0 – So for any TM • Q ⊆ Q and Γ ⊆ S – s(a i ) → 0 i+1 Universal Turing Machine Universal Turing Machine • Encoding scheme • Encoding scheme – Associate states, symbols, and directions with – Encoding transitions: binary numbers: • Encode each state, symbol, direction • Directions: • Separate by 1’a – s(S) → 0 – The transition δ (p,a) = (q, b, D) can be – s(L) → 00 encoded as: – s(R) → 000 • s(p)1s(a)1s(q)1s(b)1s(D) 5

Universal Turing Machine Universal Turing Machine • Encoding • The input to T u will be – Encoding for a TM – an encoded TM and • Encode the start state – a string x to run on that TM • Encode each of it’s moves – We encode x by individually encoding the • Separate by 1’s characters of x (separated by 1’s) – e(T) = s(q)1e(m 1 )1e(m 2 )1…1e(m k )1 • x = z 1 z 2 …z n • e(x) = s(z 1 )1s(z 2 )1…1s(z n )1 Universal Turing Machine Universal Turing Machine • Encoding Scheme • Encoding example – States – Finally, the description of the TM and the • q 0 – state 1 encoding of the string to run on the TM will be • p – state 2 separated by 2 1’s • s – state 3 • Input to T u – Symbols – e(M)11e(x) • a – symbol 1 • b – symbol 2 Universal Turing Machine Universal Turing Machine • Encoding example • Encoding scheme – States – The actual scheme isn’t really important (in • s (h) = 0 • s(q 0 ) = 00 fact, there are many) • s(p) = 000 – What is important is that we can encode a TM • s(r) = 0000 – Symbols and it’s input as a binary string. • s( ∆ ) = 0 • s(a) = 00 00 1 00 1 0 1 000 1 0 1 000 • s(b) = 000 – Directions • s(S) → 0 • s(L) → 00 • s(R) → 000 6