CSCI 2670Introduction to Theory of

Computing

September 15, 2004

Agenda

• Yesterday– Pumping lemma review

• Today– Prove the pumping lemma– Introduce context-free grammars

Announcement

• Quiz tomorrow– CGF parse trees & interpretation– Pumping lemma

Nonregular languages

• There are many nonregular languages– {0n1n | n 0}– {101,101001,1010010001,…}– {w | w has the same number of 0s and

1s}• All regular languages can be

generated by finite automata• States must be reused if the length of

a string is greater than the number of states– If states are reused, there will be

repetition

The pumping lemma

Theorem: If A is a regular language, then there is a number p where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions

1. for each i 0, xyiz is in A2. |y| > 0, and3. |xy| p

p is called the pumping length

Proof idea

• Pumping length is equal to the number of states in the DFA whose language is A– p = |Q|

• If A accepts a word w with |w| > p, then some state must be entered twice while processing w– Pigeonhole principle

Proof idea

1. for each i 0, xyiz is in A2. |y| > 0, and3. |xy| p

x

y

z

Proof

• Let A be any regular language

• Find DFA M=(Q,,,q0,F) with L(M)=A

• Let p=|Q|

• Let s=s1s2s3…sn be any string in A with |s| = n p– What if no such s is in A?

Proof (cont.)

• Let r1, r2, r3, …, rn+1 be the sequence of states entered while processing s– r1 = q0

– rn+1 F

– ri+1 = (ri, si)

Proof (cont.)

• Consider the first p+1 elements of this sequence– p+1 states must contain a repeated

state

• Let rk be the first state to be repeated and let rt be the second occurance of this state– t p+1

Proof (cont.)

• Let x=s1s2…sk-1, y=sksk+1…st-1, z=stst+1…sn

• There is an error in the proof on page 79 of Sipser (y is incorrectly defined)

– x takes M from r1 to rk

– y takes M from rk to rt

– z takes M from rt to rn+1, which is an accept state

• Since rk and rt are the same state, M must accept xyiz for any i=0, 1, 2, …

Proof (cont.)

• Have we satisfied the conditions of the theorem?

1. for each i 0, xyiz is in A

2. |y| > 0, and

3. |xy| p

• Have we satisfied the conditions of the theorem?

1. for each i 0, xyiz is in A Yes

2. |y| > 0, and Yes since t > k and y=sksk+1…st-1

3. |xy| p Yes since t p+1 and xy = s1s2…st-1

Regular languages -- Summary

• Let R be any language. The following are equivalent

1. R is a regular language2. R = L(M) for some finite automata

M, where M is a DFA, an NFA, or a GNFA

3. R is describe by some regular expression

4. R has a pumping length of p• If R can be shown not to have a

finite pumping length, then R is not regular

Context-free grammars

• The shortcoming of finite automata is that each state has very limited meaning– You have no memory of where you’ve

been – only knowledge of where you are

• Context-free grammars are a more powerful method of describing languages– Example: {0n1n | n 0} is a CFG

Example CFG

• Context-free grammars use substitution to maintain knowledge

S (S)S SSS ()

• All possible legal parenthesis pairings can be expressed by consecutive applications of these rules

• Is this a regular language?

Example

S (S) | SS | ()• (()())(())• S SS

(S)(S) (SS)(()) (()())(())

• This sequence of substitutions is called a derivation

Parse tree

S (S) | SS | ()S

SS

SS

()()()

S

)(

S

( )

Example 2

S Sb | BbB aBb | aCbC ε

• Derivation for aaabbbbbSSb

BbbaaBbbbbaaaεbbbbb

aBbbbaaaCbbbbb

Example 2 parse tree

S

ε

S

b

B

b

B

a b

B

a bba

C

Example 2

S Sb | BbB aBb | aCbC ε

Question 1: What language does this grammar accept?

Answer: {anbm | m > n > 0}Question 2: Can this CFG be simplified?Answer: yes.

Replace BaCb with Bab and remove Cε

Other languages described by CFG’s

• The foxtrot– http://linus.socs.uts.edu.au/~don/pubs/foxtrot.html

• A type of Tamil poetry called Venpa

• Ancient language of Sanscrit– http://en.wikipedia.org/wiki/Context-free_grammar

Tomorrow

• Formal definition for CFG• Designing CFG’s