Non-Regular Languages - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

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Chapter Eleven:

Non-Regular Languages

We have now encountered regular languages in several different places. They are the languages that can be recognized by a DFA. They are the languages that can be recognized by an NFA. They are the languages that can be denoted by a regular expression. They are the languages that can be generated by a right-linear grammar. You might begin to wonder: are there any languages that are not regular? In this chapter, we will see that there are. There is a proof tool that is often used to prove languages non-regular. It is called the pumping lemma, and it describes an important property that all regular languages have. If you can show that a given language does not have this property, you can conclude that it is not a regular language.

S → aSb | ε

The Language { a n^ b n }

• Any number of a s followed by the same number of b s
• Easy to give a grammar for this language:
• All derivations of a fully terminal string use the first production n =0 or more times, then the last production once: a nbn
• Is it a regular language? For example, is there an NFA for it?

Trying To Build An NFA

• We'll try working up to it
• The subset { a n^ b n^ | n ≤ 0}:
• The subset { a n^ b n^ | n ≤ 1}:

a

b

The Subset { a n^ b n^ | n ≤ 3}

a

b

b

a a

b

A Futile Effort

• For each larger value of n we added two more states
• We're using the states to count the a s, then to check that the same number of b s follow
• That's not going to be a successful pattern on which to build an NFA for all of { a nbn } - NFA needs a fixed, finite number of states - No fixed, finite number will be enough to count the unbounded n in { a nb n }
• This is not a proof that no NFA can be constructed
• But it does contain the germ of an idea for a proof…

A Word About That Proof

• Nothing was assumed about the DFA M ,

except its alphabet { a , b }

• In spite of that, we were able to infer quite a

lot about its behavior

• The basic insight: with a sufficiently long

string we can force any DFA to repeat a state

• That's the basis of a wide variety of non-

regularity proofs

Outline

• 11.1 The Language { a n^ b n }
• 11.2 The Languages { xx R }
• 11.3 Pumping
• 11.4 Pumping-Lemma Proofs
• 11.5 Strategies
• 11.6 Pumping And Finite Languages

S → aSa | bSb | ε

A Grammar For { xx R^ | x ∈ { a , b }*}

• A derivation for abba :
  • S ⇒ aSa ⇒ abSba ⇒ abba
• A derivation for abaaba :
  • S ⇒ aSa ⇒ abSba ⇒ abaSaba ⇒ abaaba
• Every time you use one of the first two productions, you add a symbol to the end of the first half, and the same symbol to the start of the second half
• So the second half is always the reverse of the first half: L ( G ) = { xxR^ | x ∈ { a , b }*}
• But is this language regular?

Intuition

• After seeing the first example, you may

already have the feeling this can't be regular

• A finite state machine would have to use states to keep track of x , then check that it is followed by a matching x R
• But there is no bound on the length of x , so no fixed, finite number of states will suffice
• The formal proof is very similar to the one we

used for { a n^ b n }…

Outline

• 11.1 The Language { a n^ b n }
• 11.2 The Languages { xx R }
• 11.3 Pumping
• 11.4 Pumping-Lemma Proofs
• 11.5 Strategies
• 11.6 Pumping And Finite Languages

Review

• We've shown two languages non-regular:

{ a n^ b n } and { xx R }

• In both cases, the key idea was to choose a

string long enough to make any given DFA

repeat a state

• For both those proofs we just used strings of

a s, and showed that ∃ i and j with i < j such

that δ( q 0 , a i ) = δ( q 0 , a j )

Pumping

• We say that the substring a ( j-i)^ can be pumped

any number of times, and the DFA always

ends up in the same state

• All regular languages have an important

property involving pumping

• Any sufficiently long string in a regular

language must contain a pumpable substring

• Formally, the pumping lemma…

Lemma 11.3: The Pumping

Lemma for Regular Languages

• Let M = ( Q , Σ, δ, q 0 , F ) be any DFA with L ( M ) = L
• Choose k = | Q|
• Consider any x , y , and z with xyz ∈ L and | y | ≥ k
• Let r be a state that repeats during the y part of xyz
  • We know such a state exists because we have | y | ≥ |Q|…

For all regular languages L there exists some integer k such that for all xyz ∈ L with | y | ≥ k , there exist uvw = y with | v | >0, such that for all i ≥ 0, xuviwz ∈ L.

x y z

In state r here And again here