Computability Construct TMs. Decidability. Preview: next class: diagonalization and Halting theorem.

Homework/Classwork

• • Last Thursday, I showed a design for a TM that accepts {ww | w a string in {0,1}*}. It was decidable (always halts).

What about – {a n b n c n | n>=0}

Homework

• Video: reactions to video including reading from one of the original Turing papers?

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Enumerator

Recall: this was a type machine that takes no input but prints out members of a language L.

– If a language is Turing Recognizable, then there exists (we can design) an enumerator that prints out members of the language. If there exists an enumerator for a language L, then there is a TM that recognizes it.

– IF a language is decidable, we can build an enumerator that prints out the strings in lexical order. If an enumerator prints out the machines in lexical order, then it is Turing decidable.

• Note: we aren't saying that if a given enumerator doesn't print out the strings in lexical order, there couldn't be another one that does….

Enumerator Homework/classwork

• Provide formal definition of enumerator

Decidable vs Turing Recognizable

• • • A language is decidable if there is a TM that accepts the language and always halts.

A language is Turing-recognizable if there is a TM that accepts the language and may or may not halt on some of the strings NOT in the language.

All decidable languages are TM recognizable.

Hierarchy Bull's eye

Each is contained in and strictly smaller than next: FSM  NDFSM  regular expressions (deterministic push down automaton) Push Down Automation  Context Free Grammars Turing decidable  nondeterministic TM decidable Turing recognizable  nondeterministic TM recognizable

Decidability question

• • The language { | B is a FSM that accepts w} is decidable. That is, there exists a TM that says yes for a when B accepts w, and says no (stops and says no) if B does not accept w.

Informal proof: – Assume some encoding of B that a TM can check. If the first part of input doesn't fit a valid encoding for a FSM, reject. First states of TM check the encoding – Add to TM the simulation for B. Also record on the tape the state and the position scanning w. Do simulation. When done (known by position), check if state is accepting state.

More…

• The language { | E is a regular expression that generates w} is decidable – convert E to FSM – Build TM to simulate the FSM. Try on w.

– Accept or reject

• • •

The language { | A is a FSM and L(A) is empty} is decidable

Design encoding for FSM: list states and list triples representing the arcs: pairs of states and the letter. Build TM that checks for valid encoding. Note: alphabet includes symbol for each state.

Grow alphabet to include marked state for every state symbol.

Add to the TM coding that does following marking: mark start state. Scanning the triples, mark any state that has a transition from a marked state. Repeat for new round. Stop when no new state has been marked in the round. So this will take at most N rounds when N is number of states. Are any accept states marked? If yes, accept; otherwise reject.

Context Free Grammar

• • { | B is a CFG and generates w} is decidable.

Informal proof: – Generate the CFG in Chomsky normal form that is equivalent to B. Claim: If length of w is n, there is a derivation of w that is not greater than 2n + 1 steps. – Given the limit on the size of the parse tree/number of derivations, CLAIM we can construct a TM that tries each one.

Parse tree

S A B A1 B1 A2 B2 …..

w1 w2 w3 … wn 2*n -1 derivations

CFG case

• The fact that the number of derivations is bounding means that the TM knows when it has checked enough possibilities and can reject if there isn't a parse tree!!!

Preview: Halting problem

• Take two inputs: encoding of a TM T and input w for that TM and simulate / run T on w.

• Will show that {|M is a TM and w is a string and M accepts w} is – Turing-recognizable but not – Turing-decidable