Turing Decidable and Recognizable Languages
TMs and Infinite Loops
- Question: If a TM is given a string, what can happen to the computation?
- Answer: The machine can either
- Accept the string (ie enter the Accept state and halt the computation)
- Reject the string (ie enter the Reject state and halt the computation)
- Enter an Infinite Loop (ie the computation never ends)
Looping on a String
- We say that a machine loops on a string if the string puts
the machine into an infinite loop
- When a machine loops, it never enters the Accept state or the Reject state
- When a machine loops, it transitions forever among states that are
neither the Accept state nor the Reject state
Looping and DFA, NFA, PDA
- In the context of TMs and looping, it's useful to think about looping for all of our machines.
- Can a DFA M loop?
- No: a computation halts when the end of the input is reached
- Can an NFA loop?
- Yes: a computation can make epsilon transitions forever
- But, every NFA has an equivalent DFA that doesn't loop
- Can a PDA loop?
- Yes: a computation can make epsilon transitions forever
- But, every PDA has an equivalent PDA that halts on all input
- Create a grammar that accepts the same language as the PDA
- Create a PDA for that grammar
- This PDA will halt on all strings [true, but not proved]
- Can an TM loop?
- Yes: a computation can make epsilon transitions forever
- The TM is a recognizer and not a decder
- If TM M accepts L and M loops on some w ∈ L^{C},
is it possible to make a TM M' accepts L and that
halts on all w ∈ L^{C}?
- Sometimes, but not always.
- That is, sometimes you can create a decider, sometimes not
Languages Accepted by DFA, NFA, PDA
- In the context of TMs and looping, it's useful to think about the language
accepted (and accepting the complement) for all of our machines.
- DFA M:
- L = L(M) = {w| M's unique computation on w reaches accept state}
- L^{C} = {w| M's unique computation on w reaches non-accept state}
- NFA N:
- L = L(N) = {w| Some computation on w reaches an accept state}
- L^{C} = {w| NO computation on w reaches an accept state}
- Can determine if w ∈ L^{C} using the equivalent DFA
- PDA P:
- L = L(P) = {w| Some computation on w reaches an accept state}
- L^{C} = {w| NO computation on w reaches an accept state}
- Can we determine if w ∈ L^{C} using the equivalent PDA that
halts on all inputs? No.
- Can't simply swap states - why?
- CFL not closed under complement.
- TM:
- L = L(M) = {w| Some computation on w reaches the Accept state}
- L^{C} = {w| NO computation on w reaches an accept state}
- A computation that doesn't reach the Accept state can either
- Reach the Reject state
- Loop
- Can we tell if w ∈ L^{C}:
- If it's possible to build a decider, then yes.
- If not, then not always.
- For some TM, it's impossible to build a decider for the machines
language.
Looping and Defining Languages
- The possibility of looping affects how we define the relationship
between a machine and the language it defines
- TM M recognizes language L iff
- the strings L put M into the Accept state
- the strings NOT IN L
- EITHER put M into the Reject state
- OR cause M to loop
- TM M decides language L iff
- the strings L put M into the Accept state
- the strings NOT IN L put M into the Reject state
Recognizers and Deciders
- A recognizer of a language is a machine that recognizes that language
- A decider of a language is a machine that decides that language
- Both types of machine halt in the Accept state on strings that are in the language
- A Decider also halts if the string is not in the language
- A Recogizer MAY or MAY NOT halt on strings that are not in the language
- On all input:
- A Decider MUST halt (in Accept or Reject state)
- A Recogizer MAY or MAY NOT halt on some strings (Q: Which ones?)
- Is every Recognizer a Decider, or is every Decider a Recognizer?
- Finish this statement: TM M decides language L if M
recognizes L and it also ...
- It's useful to look at two subset diagrams:
- All string > strings L^{C}, string L=L(M) for recognizer M
- All string > strings L^{C}, string L=L(M) for decider M
Examples
- Think about the Turing machines for these languages:
- Are they deciders or recognizers?
Recognizers
- Examples of recognizers that aren't deciders?
- If L has a Decider D, can we find a machine R
that recognizes L but that does not decide it?
- It's easy
- How can we modify the previous examples.
- Another question: Can we find L that has a recognizer but not a decider?
- Yes, as we will see later
- First, define some more terms: TR and TD languages
Definition: Turing Decidable Language
- A language is Turing-decidable (or decidable) if some Turing
machine decides it
- Aka Recursive Language
Review: Turing Recognizable Language
- A language is Turing-recognizable if some Turing
machine recognizes it
- Aka Recursively Enumerable Language
Showing that a Language is Decidable/Recognizable
- How do we show that a language is TD: Build a decider for it
- How do we show that a language is TR: Build a decider or recognizer for it
- How do we show that language L is TR and not TD:
- Build a Recognizer for L
- PROVE that there is no Decider for l
- We will see examples soon
Relationship among Language Classes
- How do these language classes fit togther
- Regular
- Context Free
- Turing Decidable
- Turing Recognizable
- (We haven't yet proved the relation between CF and TD langs)
- What are some examples in each class (and not in smaller class)?
- First three are easy: Regular, CF, decidable
- Are there languages that are recognizable but NOT decidable?
- Are the languages that are NOT recognizable?