CS 3240 – Chapter 11

They may not halt on every possible input! And not just because the creator of a specific

TM was a doofus

This is related to the major mathematical/computational discovery of the 20th century! There are propositions that cannot be decided

(“proven”)

A question is decidable if there is a TM that always halts and answers “yes” or “no” for each possible input The TM therefore constitutes an algorithm

A language is decidable if there is a TM that always halts and answers “accept” or “reject” whenever an input string is in the language or not aka “Turing acceptable”

A function is computable if there is a TM that always halts with the appropriate output for each possible input in the function domain aka “Total function”

Let g(x,y) be some computable function Let f(x) =

the smallest p where g(x, p) = 1, or0, if such a p does not exist

“Pseudo-algorithm” for f(x):m = 0;while (g(x, m) != 1)

++m;cout << m;

If there is no m for a given g and x, then we hang!

3 possibilities in general when a TM processes an input string: Accepts (goes to an accepting halt state) Rejects (e.g., crashes, or gives a “no”

answer) Hangs (infinite loop)

A language for which there is a TM that always halts for and accepts strings in the language is recognizable “It knows one when it sees one” :-) But it may hang on strings not in the

language

A function that is defined for only some of its domain elements is a partial function It may hang (or “blow up”) on some inputs

e.g., divide by zero This is the computational analogue to a

recognizable language you’ll always get an answer with a valid input

it may hang on invalid input

Language Machine Grammar

Regular Finite Automaton Regular Expression,Regular Grammar

Context-Free Pushdown Automaton

Context-Free Grammar

Recursively Enumerable

Turing Machine Unrestricted Phrase-Structure Grammar

9CS 3240 - Introduction

A language L over the alphabet is called recursively enumerable (aka “recognizable”) if there is a TM T that accepts every word in L and either rejects (crashes) or loops forever for every word in the language L', the complement of L:

accept(T) = L reject(T) + loop(T) = L'

recursively enumerable (r.e.) = recognizable

We just saw one (slide 5) Let g(x,y) be some computable function Let f(x) =

the smallest p where g(x, p) = 1, or0, if such a p does not exist

Let Lg be the set of functions, f, corresponding to all computable functions g, as explained above Such functions can be encoded as strings, and are

therefore countable

Lg can then be seen as a language (a set of strings)

A language L over the alphabet is called recursive (aka “decidable”) if there is a TM T that accepts every word in L and rejects every word in L'; that is,

accept(T) = L

reject(T) = L'

A TM that accepts a recursive set is a model of an algorithm. It always halts.

recursive = decidable

There exist languages that are r.e. but not recursive We just saw one

There exist languages that aren't even r.e.! (You'll see one soon)

All are “contrived” Languages generated by grammars are

r.e. or “better”

All r.e. languages are closed under union, intersection, concatenation, and Kleene* Everything but complement!

Recursive languages are also closed under complement

Also: If L and L' are r.e., then L is recursive

Here come the Proofs…

Let M be a machine that decides a recursive language, L

Form the machine M' by inverting the acceptability output of M Goes to a reject state instead

Then M' decides L' So L' is recursive

Suppose L and L' are both r.e. Let M recognize L, and M' recognize L'

M may hang on elements of L', but M' doesn't Form a new machine, M* that calls M and

M' in parallel (non-deterministically) If M accepts w, so does M*

If M' accepts w, reject w There are no other possibilities! (No hanging) Therefore, L is decidable/recursive, by

definition

TMs can recognize/accept strings from certain languages and/or compute functions

If there is a TM, M, that accepts a language, L, and M always halts, then L is recursive

If there is no such M for L, but there is instead a machine M that accepts every string in L, but M may hang on strings not in L, then L is recursively enumerable

The complement of a recursive language is recursive In fact, recursive languages are closed

under all operations, like regular languages are

r.e. languages are closed under intersection

The complement of a r.e. language may not be r.e. But if it is, then both languages are

actually recursive!

A Timely Interlude

The real numbers in (0,1) are uncountable They cannot be mapped in a 1-to-1

fashion to the counting numbers Proof:

Assume they can be: r0, r1, r2, … Arrange their digits in a table a[ ][ ]▪ each row, a[i], contains the digits or ri

The diagonal sequence (a[n][n]+1) mod 10, representing a valid real number, is not in the table! Contradiction!

The power set of the natural numbers, N the set of all subsets (2N)

Suppose it is countable The we can enumerate the sets: p0, p1, p2, …

Now consider the set T = {i | i ∉ f(i) = pi } Certainly T is a set of integers, so T ∈ P(N)▪ Call it pk

Question: Is k in T = pk?

The power set of any countably infinite set is uncountable

Much bigger than a countable set! The number of countable sets is

negligible compared to the number of uncountable sets

Well?

A language over an alphabet Σ is a subset of Σ*

the latter being an infinite setThe set of all languages over Σ is

therefore the power set of Σ* (2∑*) which we just showed is uncountable

So… the number of languages over any finite alphabet is uncountable

The # of TMs are countable, but the # of languages is not

Therefore, some languages cannot be recognized by a TM There aren't enough TMs to go around! Just like there are more reals than

integersSo, non-r.e. languages must exist!

Just take the complement of any r.e. language that is not recursive

Example (page 279): Consider all TMs, Mi with alphabet Σ={a}.

Let X = {ai: ai ∈ L(Mi)}▪ This is r.e., but not recursive▪ Because we can construct a TM that carries out

the computation, but it may not halt when ai ∉ L(Mi)

Then X’ must be non-r.e.!

Suppose X ’ is r.e. Then there is a TM, Mk, that recognizes it

Now ask the question, is ak ∈ X’ ? If it is, this means ak ∉ L(Mk) = X’, by

definition Contradiction!

If it isn’t, then this means that ak ∈ L(Mk) = X’ Again, by definition Again, a contradiction!

⇒ X ’ is not r.e.

What we have just described is the membership problem: “Given a r.e. language, L, and a string,

w, is w in L?”We have just shown that the

membership problem is, in general, undecidable

Language Machine Grammar

Regular Finite Automaton Regular Expression,Regular Grammar

Context-Free Pushdown Automaton

Context-Free Grammar

Recursively Enumerable

Turing Machine Unrestricted Grammar

29CS 3240 - Introduction

Left-hand side of the rule is a concatenation of one or more symbols There must be at least one variable on

the left▪ λ is not allowed on the left

Any string is allowed on the right, including λ

aka “Type 0” Grammar

S → aAbc | abc | λAb → bAAc → BbccbB → BbaB → aa | aaA

S ⇒aAbc ⇒abAc ⇒abBbcc ⇒aBbbcc ⇒aabbcc

An A is created as the left-most variable. It travels to the right until it finds a 'c', then creates a new b and c, and becomes a B. The B moves back to create the extra needed a. The last rule allows the option to do it all over again (by introducing another A). This is similar to what the Turing machine for the same language does, except the TM marks instead of generates the letters.

We have already seen CFGs for this

This unrestricted grammar: Introduces X’s and

Y’s together Allows moving X’s

and Y’s arbitrarily by swapping them

S → XYS | λXY → YXYX → XYX → aY → b

Cannot use the decidability algorithms for CFGs e.g., CYK algorithm does not apply

No “Normal Form”Non-null productions may create

shorter stringsTerminals can disappear!

For every r.e. language, there is an unrestricted grammar that generates it

For every unrestricted grammar, there is a TM that recognizes its language It may or may not decide it

We will not prove this but the book does

Like Unrestricted Grammars except: Right-hand side must be no shorter than left-

hand side▪ So strings never shrink

Hence the name “non-contracting” or “monotonic” Cannot contain λ anywhere

Context-sensitive languages don't need unlimited memory Since intermediate sentential forms never shrink,

the largest memory requirement is proportional to |w|

Accepting machine: Linear Bounded Automaton

S → aAbc | abcAb → bAAc → BbccbB → BbaB → aa | aaA

Every context-sensitive language is recursive there is a TM that accepts it (i.e., always

halts)But not all recursive languages are

context-sensitiveSo Context Sensitive languages form a

proper subset of Recursive Languages Just like Recursive languages are a proper

subset of Recursively enumerable languages

Type 3 grammar

Type 2 grammar

Type 1 grammar

Type 0 grammar

Language Grammar Machine

Recursively Enumerable(“Turing Recognizable”)

Unrestricted Phrase Structure (Type 0)

TM (that may not halt)

Recursive(“Turing Decidable”)

" TM (that always halts)

Context-Sensitive Context Sensitive (Type 1)(monotonic/non-contracting)

Linear-bounded automata(bounded TM)

Context Free Context Free (Type 2) PDA

Deterministic Context Free

" DPDA

Regular Regular (Type 3)(Right-linear or Left-linear)

FA