Nathan Brunelle

Department of Computer Science

University of Virginia

www.cs.virginia.edu/~njb2b/theory

Theory of ComputationCS3102 – Spring 2014

A tale of computers, math, problem solving, life, love and tragic death

MidtermTake home? Vote now.

Date:1. During the week before Spring break (out March 3, due March 6)?• Don’t have to worry over the break

2. After Spring break (4 days within March 16-23)?• 1 more week of content but 2 more weeks to solve problems• I will hold Skype/Google Hangout office hours over the break

Think about it, vote on Thursday.

Pumping LemmaGoal: Give a sufficient condition for showing non-regularity

Consider that L is an infinite regular languageLet M be a DFA, L(M)=L

Let’s say M has p statesThere must be someString w in L s.t. |w|>p

By the pigeon-hole principle:Some state was visited at least twice!Taking that loop another time must giveanother string in the language.

M

1

8

12

20

p

3x2

Pumping LemmaIf L is a regular language then there is some number p (called pumping length) where if w is a string in L s.t. |w|>p then w can be divided into 3 pieces: w=xyz which satisfy:

1. For each i≥0, L2. |y|>03. |xy|≤p

Example: Consider:By condition 3 we know:Thus for i=2 we have more a’s than b’s,so L cannot be regular

M

1

8

12

20

p

3

zxy i x

y

z

ℕnba nn |{ }ppba

ay

Pumping Lemma1. For each i≥0, L2. |y|>03. |xy|≤p

Show: is not regular

Consider By condition 3 we know y , this leaves 4 cases:4. y5. y6. y7. y

zxy i

}|{ *wwwR

pp baab )()( )(ba

)(ab*)(abbaab *)(aabb )(

Problem: in all of these I only change the first half!

I needed to “remember” the first half of the string.

Pumping Lemma1. For each i≥0, L2. |y|>03. |xy|≤p

Show: is not regular

Consider where q is a prime greater than py may have only a’s, let y= where m≤pFor i=p+1 we have Clearly, p(m+1) is not prime

zxy i

qa

}prime is |{ nan

ma)1( mpi azxy

For let then

So

7a3ay 2417 ay

)13(728424 aaaaxyz

Pumping Lemma1. For each i≥0, L2. |y|>03. |xy|≤p

Consider: This language is pumpable but not regular3 cases:4. If then let x=ε, y=a5. If then let x=ε, y=c6. If then let x=ε, y=cThus every string is pumpable

Nonregular: Then use pumping Lemma!

zxy i

}1|{ kjibac kji

kjbajjbca

1, ibac jji

}{}1|{ ** nnkji bcabcakjibac

Myhill-Nerode TheoremGives a necessary and sufficient condition for regularity!

Idea: If two strings terminate in the same state in a DFA then their membership must be equivalent for any suffix.

If and meet at state q then for any string , ends in the samestate as

We say and haveno distinguishing extensions

M1w

2w

qx

1w 2w

x xw1

xw2

2w1w

Myhill-Nerode TheoremRecall Equivalence Relation:A relation ~ is called an equivalence relation if:Reflexive:Symmetric:Transitive:

The relation if and have no distinguishing extensions in language L forms an equivalence relation.

Recall Equivalence Class:For equivalence relation ~, equivalence class

Myhill Nerode Theorem: L is regular iff has a finite number of equivalence classes under relation

xx ~

xyyx ~~ zyzyyx ~~~

21 ~ ww L 1w 2w

}~|{][ baba

L~

*

Myhill-Nerode Theorem

Myhill Nerode Theorem: L is regular iff has a finite number of equivalence classes under relation

Given a finite set of equivalence classes, construct a DFA:

Each equivalence class becomes a state:

If character takes a string from equivalence class to then add a transition from the state for to the state for .

Accept states are those for equivalence classes of strings in the language.

DE~

*

][0 q

][ 1w ][ 2w

][ 1w][ 2w

Myhill-Nerode Theorem

Extra Credit: Use the Myhill-Nerode Theorem to show the following languages are non-regular:

}|{ *wwwR }prime is |{ nan ℕnba nn |{ }

Language ReversalTheorem: The regular languages are closed under reversal.Proof: Construction.Given a regular language L, show that the language is also regular.Let M be a DFA for L, construct M’ to be a NFA for L:

}|{ LwwL RR

M M’

q0

ε

ε

HalfLetShow that HALF preserves regularity.e.g. if “RingoStarr” is in L, then “Ringo” is in HALF(L)

Let M be a DFA for language L

Intuition: follow the transitions of M for string v. “Check” that there is a path from v to an accept state that consumes |v| characters.

How do we do this “check”?

}|||| s.t. ,|{)( * LvwwvwvvLHALF

M

v

w

HalfLetShow that HALF preserves regularity.

Intuition: follow the transitions of M for string v. “Check” that there is a path from v to an accept state that consumes |v| characters.

How do we do this “check”? Use the machine for LR

}|||| s.t. ,|{)( * LvwwvwvvLHALF

M

v

w

M’

ε

ε

Σ

ΣΣ

ΣΣ

F={(q,q)}

DoubleLetShow that DOUBLE preserves regularity.e.g. if “BamBam” is in L, then “Bam” is in DOUBLE(L)

Let M be a DFA for language L

Intuition: Run L on w, in parallel non-deterministically “guess” the end state in machine M on w and check if starting from that guess puts the machine in an accept state

}|{)( LwwwLDOUBLE

M

w

M

w

“Guess”

DoubleLetIntuition: Run L on w, in parallel non-deterministically “guess”

the end state in machine M on w and check if starting from that guess puts the machine in an accept state

}|{)( LwwwLDOUBLE

M

…

Accept if:•M ends in state •M ends in an accept state

} '|)', {( FqqF

M M

M2 MM1

εε

ε

||Q