Recursively Enumerable and Recursive Languages

Definition: A language is recursively enumerable if some Turing machine accepts it

For string : Let be a recursively enumerable language and the Turing Machine that accepts it ifthen halts in a final state ifthen halts in a non-final state or loops forever

Definition: A language is recursive if some Turing machine accepts it and halts on any input string In other words: A language is recursive if there is a membership algorithm for it

For string : Let be a recursive language and the Turing Machine that accepts it ifthen halts in a final state ifthen halts in a non-final state

We will prove: 1. There is a specific language which is not recursively enumerable (not accepted by any Turing Machine) 2. There is a specific language which is recursively enumerable but not recursive

Recursive Recursively Enumerable Non Recursively Enumerable

A Language which is not Recursively Enumerable

We want to find a language that is not Recursively Enumerable This language is not accepted by any Turing Machine

Consider alphabet Strings:

Consider Turing Machines that accept languages over alphabet They are countable:

Example language accepted by Alternative representation

Consider the language consists from the 1’s in the diagonal

Consider the language consists of the 0’s in the diagonal

Theorem: Language is not recursively enumerable

Proof: is recursively enumerable Assume for contradiction that There must exist some machine that accepts

Question:

Answer:

Question:

Answer:

Question:

Answer:

Similarly: for any Because either: or

Therefore, the machine cannot exist Therefore, the language is not recursively enumerable End of Proof

Observation: There is no algorithm that describes (otherwise would be accepted by some Turing Machine)

Recursive Recursively Enumerable Non Recursively Enumerable

A Language which is Recursively Enumerable and not Recursive

We want to find a language which There is a Turing Machine that accepts the language The machine doesn’t halt on some input Is recursively enumerable But not recursive

We will prove that the language Is recursively enumerable but not recursive

The language Theorem: is recursively enumerable

Proof: We will give a Turing Machine that accepts

Turing Machine that accepts For any input string Compute, for which Find Turing machine (using an enumeration procedure for Turing Machines) Simulate on input If accepts, then accept End of Proof

Observation: Recursively enumerable Not recursively enumerable (Thus, also not recursive)

Theorem: The language is not recursive

Proof: Assume for contradiction that is recursive Then is recursive: Take the Turing Machine that accepts halts on any input: If accepts then reject If rejects then accept

Therefore: is recursive But we know: is not recursively enumerable thus, not recursive CONTRADICTION!!!!

Therefore, is not recursive End of Proof

Recursive Recursively Enumerable Non Recursively Enumerable

Turing acceptable languages and Enumeration Procedures

We will prove: If a language is recursive then there is an enumeration procedure for it A language is recursively enumerable if and only if there is an enumeration procedure for it (weak result) (strong result)

Theorem: if a language is recursive then there is an enumeration procedure for it

Proof: Enumeration Machine Accepts Enumerates all strings of input alphabet

If the alphabet is then can enumerate strings as follows:

Enumeration procedure Repeat: generates a string checks if YES: print to output NO: ignore End of Proof

Example: Enumeration Output

Theorem: if language is recursively enumerable then there is an enumeration procedure for it

Proof: Enumeration Machine Accepts Enumerates all strings of input alphabet

If the alphabet is then can enumerate strings as follows:

Enumeration procedure Repeat:generates a string checks if YES: print to output NO: ignore NAIVE APPROACH Problem: If machine may loop forever

executes first step on BETTER APPROACH Generates second string executes first step on second step on Generates first string

Generates third string executes first step on second step on third step on And so on

1 Step in string

If for any string machine halts in a final state then it prints on the output End of Proof

Theorem: If for language there is an enumeration procedure then is recursively enumerable

Proof: Input Tape Enumerator for Compare Machine that accepts

Turing machine that accepts Repeat: Using the enumerator, generate the next string of For input string Compare generated string with If same, accept and exit loop End of Proof

We have proven: A language is recursively enumerable if and only if there is an enumeration procedure for it

The Chomsky Hierarchy

Non-recursively enumerable Recursively-enumerable Recursive Context-sensitive Context-free Regular The Chomsky Hierarchy

Decidability

Consider problems with answer YES or NO Examples: Does Machine have three states ? Is string a binary number? Does DFA accept any input?

A problem is decidable if some Turing machine decides (solves) the problem Decidable problems: Does Machine have three states ? Is string a binary number? Does DFA accept any input?

Turing Machine Input problem instance YES NO The Turing machine that decides (solves) a problem answers YES or NO for each instance of the problem

The machine that decides (solves) a problem: If the answer is YES then halts in a yes state If the answer is NO then halts in a no state These states may not be final states

YES states NO states Turing Machine that decides a problem YES and NO states are halting states

Difference between Recursive Languages and Decidable problems The YES states may not be final states For decidable problems:

Some problems are undecidable: which means: there is no Turing Machine that solves all instances of the problem A simple undecidable problem: The membership problem

The Membership Problem Input:Turing Machine String Question: Does accept ?

Theorem: The membership problem is undecidable Proof:Assume for contradiction that the membership problem is decidable (there are and for which we cannot decide whether )

Thus, there exists a Turing Machine that solves the membership problem YES accepts NO rejects

Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input

accepts ? NO YES accept Turing Machine that accepts and halts on any input reject

Therefore,is recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Since is chosen arbitrarily, every recursively enumerable language is also recursive

Therefore, the membership problem is undecidable END OF PROOF

Another famous undecidable problem: The halting problem

The Halting Problem Input:Turing Machine String Question: Does halt on input ?

Theorem: The halting problem is undecidable Proof:Assume for contradiction that the halting problem is decidable (there are and for which we cannot decide whether halts on input )

Thus, there exists Turing Machine that solves the halting problem YEShalts on doesn’t halt on NO

Input: initial tape contents Encoding of String YES NO Construction of

Construct machine : If returns YES then loop forever If returns NO then halt

NO Loop forever YES

Construct machine : Input: If halts on input Then loop forever Else halt (machine )

copy

Run machine with input itself: Input: If halts on input Then loop forever Else halt (machine )

on input If halts then loops forever If doesn’t halt then it halts : NONSENSE !!!!!

Therefore, we have contradiction The halting problem is undecidable END OF PROOF

Another proof of the same theorem: If the halting problem was decidable then every recursively enumerable language would be recursive

Theorem: The halting problem is undecidable Proof:Assume for contradiction that the halting problem is decidable

There exists Turing Machine that solves the halting problem YEShalts on doesn’t halt on NO

Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input

halts on ? YES NO Run with input reject accept reject Turing Machine that accepts and halts on any input Halts on final state Halts on non-final state

Thereforeis recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Since is chosen arbitrarily, every recursively enumerable language is also recursive

Therefore, the halting problem is undecidable END OF PROOF

Reducibility

Problem is reduced to problem If we can solve problem then we can solve problem

If is undecidable then is undecidable If is decidable then is decidable Problem is reduced to problem

Example: the halting problem is reduced to the state-entry problem

The state-entry problem Inputs: Turing Machine State Question: Does String enter state on input ?

Theorem: The state-entry problem is undecidable Proof: Reduce the halting problem to the state-entry problem

state-entry problem decider YES NO enters doesn’t enter Suppose we have a Decider for the state-entry algorithm:

Halting problem decider YES NO halts on doesn’t halt on We want to build a decider for the halting problem:

State-entry problem decider Halting problem decider YES NO YES NO We want to reduce the halting problem to the state-entry problem:

Halting problem decider YES NO YES NO Convert Inputs ? We need to convert one problem instance to the other problem instance State-entry problem decider

Convert to : Add new state From any halting state of add transitions to halting states Single halt state

halts on input halts on state on input if and only if

Generate Halting problem decider YES NO YES NO State-entry problem decider

Since the halting problem is undecidable, the state-entry problem is undecidable END OF PROOF We reduced the halting problem to the state-entry problem

Another example: the halting problem is reduced to the blank-tape halting problem

The blank-tape halting problem Input:Turing Machine Question:Doeshalt when started with a blank tape?

Theorem: Proof: Reduce the halting problem to the blank-tape halting problem The blank-tape halting problem is undecidable

blank-tape halting problem decider YES NO halts on blank tape doesn’t halt on blank tape Suppose we have a decider for the blank-tape halting problem:

halting problem decider YES NO halts on doesn’t halt on We want to build a decider for the halting problem:

Blank-tape halting problem decider Halting problem decider YES NO YES NO We want to reduce the halting problem to the blank-tape halting problem:

Blank-tape halting problem decider Halting problem decider YES NO YES NO We need to convert one problem instance to the other problem instance Convert Inputs ?

Construct a new machine When started on blank tape, writes Then continues execution like then write step 1step2 if blank tapeexecute with input

halts on input string halts when started with blank tape if and only if

Generate blank-tape halting problem decider Halting problem decider YES NO YES NO

Since the halting problem is undecidable, the blank-tape halting problem is undecidable END OF PROOF We reduced the halting problem to the blank-tape halting problem

Does machine halt on input ? Summary of Undecidable Problems Halting Problem: Membership problem: Does machine accept string ?

Does machine enter state on input ? Does machine halt when starting on blank tape? Blank-tape halting problem: State-entry Problem:

Uncomputable Functions

A function is uncomputable if it cannot be computed for all of its domain Domain Values region

An uncomputable function: maximum number of moves until any Turing machine with states halts when started with the blank tape

Theorem:Function is uncomputable Proof: Then the blank-tape halting problem is decidable Assume for contradiction that is computable

Decider for blank-tape halting problem: Input: machine 1. Count states of : 2. Compute 3. Simulate for steps starting with empty tape If halts then return YES otherwise return NO

Therefore, the blank-tape halting problem is decidable However, the blank-tape halting problem is undecidable Contradiction!!!

Therefore, function in uncomputable END OF PROOF

Undecidable Problems for Recursively Enumerable Languages

is empty? is finite? contains two different strings of the same length? Take a recursively enumerable language Decision problems: All these problems are undecidable

Theorem: For any recursively enumerable language it is undecidable to determine whether is empty Proof: We will reduce the membership problem to this problem

empty language problem decider YES NO Suppose we have a decider for the empty language problem: Let be the TM with empty not empty

membership problem decider YES NO accepts rejects We will build the decider for the membership problem:

YES NO YES Membership problem decider empty language problem decider We want to reduce the membership problem to the empty language problem:

YES NO YES Membership problem decider empty language problem decider We need to convert one problem instance to the other problem instance Convert inputs ?

Construct machine : When enters a final state, compare with Accept only if executes the same as with On arbitrary input string

is not empty if and only if

construct YES NO YES Membership problem decider END OF PROOF empty language problem decider

Undecidable problems for Recursively enumerable languages continued…

is empty? is finite? contains two different strings of the same length? Take a recursively enumerable language Decision problems: All these problems are undecidable

Theorem: For a recursively enumerable language it is undecidable to determine whether is finite Proof: We will reduce the halting problem to this problem

finite language problem decider YES NO Suppose we have a decider for the finite language problem: Let be the TM with finite not finite

Halting problem decider YES NO halts on We will build a decider for the halting problem: doesn’t halt on

YES NO YES Halting problem decider finite language problem decider We want to reduce the halting problem to the finite language problem

YES NO YES Halting problem decider finite language problem decider We need to convert one problem instance to the other problem instance convert input ?

Construct machine : If enters a halt state, accept ( inifinite language) Initially, simulates on input Otherwise, reject ( finite language) On arbitrary input string

halts on is infinite if and only if

construct YES NO YES halting problem decider finite language problem decider

is empty? is finite? contains two different strings of the same length? Take a recursively enumerable language Decision problems: All these problems are undecidable

Theorem: For a recursively enumerable language it is undecidable to determine whether contains two different strings of same length Proof: We will reduce the halting problem to this problem

Two-strings problem decider YES NO Suppose we have the decider for the two-strings problem: Let be the TM with contains Doesn’t contain two equal length strings

Halting problem decider YES NO halts on We will build a decider for the halting problem: doesn’t halt on

YES NO YES NO Halting problem decider Two-strings problem decider We want to reduce the halting problem to the empty language problem

YES NO YES NO Halting problem decider Two-strings problem decider We need to convert one problem instance to the other problem instance convert inputs ?

Construct machine : When enters a halt state, accept if or Initially, simulate on input (two equal length strings ) On arbitrary input string Otherwise, reject ( )

halts on if and only if accepts two equal length strings accepts and

construct YES NO YES NO Halting problem decider Two-strings problem decider

Rice’s Theorem

Non-trivial properties of recursively enumerable languages: any property possessed by some (not all) recursively enumerable languages Definition:

Some non-trivial properties of recursively enumerable languages: is empty is finite contains two different strings of the same length

Rice’s Theorem: Any non-trivial property of a recursively enumerable language is undecidable

The Post Correspondence Problem

Some undecidable problems for context-free languages: Is context-free grammar ambiguous? Is ? are context-free grammars

We need a tool to prove that the previous problems for context-free languages are undecidable: The Post Correspondence Problem

Input: Two sequences of strings

There is a Post Correspondence Solution if there is a sequence such that: PC-solution: Indices may be repeated or omitted

Example: PC-solution:

Example: There is no solution Because total length of strings from is smaller than total length of strings from

1. The MPC problem is undecidable 2. The PC problem is undecidable (by reducing MPC to PC) (by reducing the membership to MPC) We will show:

Theorem: The PC problem is undecidable Proof: We will reduce the MPC problem to the PC problem

Some undecidable problems for context-free languages: Is context-free grammar ambiguous? Is ? are context-free grammars We reduce the PC problem to these problems

Theorem: Proof: Let be context-free grammars. It is undecidable to determine if Rdeduce the PC problem to this problem

Suppose we have a decider for the empty-intersection problem Empty- interection problem decider YES NO Context-free grammars

For a context-free grammar, Theorem: it is undecidable to determine if G is ambiguous Proof: Reduce the PC problem to this problem