1. Today Chapter 1: RE = Regular Languages, nonregular languages RL pumping lemma Chapter 2: Context-Free Languages (CFLs)

2. Regular Expressions (Def. 1.26) Given an alphabet , R is a regular expression if: 1.R = a, with a  2.R =  3.R =  4.R = (R 1  R 2 ), with R 1 and R 2 regular expressions 5.R = (R 1  R 2 ), with R 1 and R 2 regular expressions 6.R = (R 1 *), with R 1 a regular expression

3. Thm 1.28: RL ~ RE We need to prove both ways: If a language is described by a regular expression, then it is regular (Lemma 1.29) (Last week we saw how we can convert a regular expression R into an NFA M such that L(R)=L(M)) Today we do the second part: If a language is regular, then it can be described by a regular expression (Lemma 1.32)

4. Generalized NFA Generalized nondeterministic finite automaton M=(Q, , , q start, q accept ) with Q finite set of states  the input alphabet q start the start state q accept the accept state  :(Q\{q accept })  (Q\{q start })  R the transition function ( R is the set of regular expressions over  )

5. Example GNFA qSqS qAqA 01* 0 0*  11 0110  

6. Characteristics of GNFA’s   :(Q\{q accept })  (Q\{q start })  R The interior Q\{q accept,q start } is fully connected by  From q start only ‘outgoing transitions’ To q accept only ‘ingoing transitions’ Impossible q i  q j transitions are “  (q i,q j ) =  ” qSqS qAqA RRRR Observation: This GNFA: recognizes the language L(R)

7. Proof Idea of Lemma 1.32 Proof idea (given a DFA M): Construct an equivalent GNFA M’ with k  2 states Reduce one-by-one the internal states until k=2 This GNFA will be of the form This regular expression R will be such that L(R) = L(M) qSqS qAqA R

8. DFA M  Equivalent GNFA M’ Let M have k states Q={q 1,…,q k } - Add two states q accept and q start qSqS q1q1  - Connect q start to earlier q 1 : qiqi  qjqj - Complete missing transitions by qAqA qjqj  - Connect old accepting states to q accept - Join multiple transitions: qiqi 0 qjqj 1 becomes qiqi 0101 qjqj

9. Remove Internal state of GNFA If the GNFA M has more than 2 states, ‘rip’ internal q rip to get equivalent GNFA M’ by: - Removing state q rip : Q’=Q\{q rip } - Changing the transition function  by  ’(q i,q j ) =  (q i,q j )  (  (q i,q rip )(  (q i,q j ))*  (q rip,q j )) for every q i  Q’\{q accept } and q j  Q’\{q start } qiqi R 4  (R 1 R 2 *R 3 ) qjqj qiqi R2R2 qjqj R4R4 q rip R1R1 R3R3 =

10. Proof Lemma 1.32 Let M be DFA with k states Create equivalent GNFA M’ with k+2 states Reduce in k steps M’ to M’’ with 2 states The resulting GNFA describes a single regular expressions R The regular language L(M) equals the language L(R) of the regular expression R

11. Recap Regular Languages = Regular Expressions Let R be a regular expression, then there exists an NFA M such that L(R) = L(M) The language L(M) of a DFA M is equivalent to a language L(M’) of a GNFA = M’, which can be converted to a two-state M’’ The transition q start  R  q accept of M’’ obeys L(R) = L(M’’) Hence: RE  NFA = DFA  GNFA  RE

12. Nonregular Languages §1.4 Which languages cannot be recognized by finite automata? Example: L={ 0 n 1 n | n  N } ‘Playing around’ with FA convinces you that the ‘finiteness’ of FA is problematic for “all n  N” The problem occurs between the 0 n and the 1 n Informal: the memory of a FA is limited by the the number of states |Q|

13. Repeating DFA Paths q1q1 qkqk qjqj Consider an accepting DFA M with size |Q| On a string of length p, p+1 states get visited For p  |Q|, there must be a j such that the computational path looks like: q 1,…,q j,…,q j,…,q k

14. Repeating DFA Paths q1q1 qkqk qjqj The action of the DFA in q j is always the same. If we repeat (or ignore) the q j,…,q j part, the new path will again be an accepting path

15. Line of Reasoning Proof by contradiction: Assume that L is regular Hence, there is a DFA M that recognizes L For strings of length  |Q| the DFA M has to ‘repeat itself’ Show that M will accept strings outside L Conclude that the assumption was wrong Note that we use the simple DFA, not the more elaborate (but equivalent) NFA or GNFA

16. Pumping Lemma (Thm 1.37) For every regular language L, there is a pumping length p, such that for any string s  L and |s|  p, we can write s=xyz with 1) x y i z  L for every i  {0,1,2,…} 2) |y|  1 3) |xy|  p Note that 1) implies that xz  L 2) says that y cannot be the empty string  Condition 3) is not always used

17. Formal Proof of Pumping Lemma Let M = (Q, , ,q 1,F) with Q = {q 1,…,q p } Let s = s 1 …s n  L(M) with |s| = n  p Computational path of M on s is the sequence r 1,…,r n+1  Q n+1 with r 1 = q 1, r n+1  F and r t+1 =  (r t,s t ) for 1  t  n Because n+1  p+1, there are two states such that r j = r k (with j<k and k  p+1) Let x = s 1 …s j–1, y = s j …s k–1, and z = s k …s n+1 x takes M from q 1 =r 1 to r j, y takes M from r j to r j, and z takes M from r j to r n+1  F As a result: xy i z takes M from q 1 to r n+1  F (i  0)

18. Formal Proof of Pumping Lemma Let M = (Q, , ,q 1,F) with Q = {q 1,…,q p } Let s = s 1 …s n  L(M) with |s| = n  p Computational path of M on s is the sequence r 1,…,r n+1  Q n+1 with r 1 = q 1, r n+1  F and r t+1 =  (r t,s t ) for 1  t  n Because n+1  p+1, there are two terms such that r j = r k (with j<k and k  p+1) Let x = s 1 …s j–1, y = s j …s k–1, and z = s k …s n+1 x takes M from q 1 =r 1 to r j, y takes M from r j to r j, and z takes M from r j to r n+1  F As a result: xy i z takes M from q 1 to r n+1  F (i  0) |y|  1 and |xy|  p x y i z  L(M) for every i  {0,1,2,…}

19. Pumping 0 n 1 n (Ex. 1.38) Assume that B = {0 n 1 n | n  0} is regular Let p be the pumping length, and s = 0 p 1 p  B P.L.: s = xyz = 0 p 1 p, with xy i z  B for all i  0 Three options for y: 1) y=0 k, hence xyyz = 0 p+k 1 p  B 2) y=1 k, hence xyyz = 0 p 1 k+p  B 3) y=0 k 1 l, hence xyyz = 0 p 1 l 0 k 1 p  B Conclusion: The pumping lemma does not hold, the language B is not regular.

20. F = { ww | w  {0,1}* } (Ex. 1.40) Let p be the pumping length, and take s = 0 p 10 p 1 Let s = xyz = 0 p 10 p 1 with condition 3) |xy|  p Only one option: y=0 k, with xyyz = 0 p+k 10 p 1  F Without 3) this would have been a pain.

21. Intersecting Regular Languages Let C = { w | # of 0s in w equals # of 1s in w} Problem: If xyz  C with y  C, then xy i z  C Idea: If C is regular and F is regular, then the intersection C  F has to be regular as well Solution: Assume that C is regular Take the regular F = { 0 n 1 m | n,m  N}, then for the intersection: C  F = { 0 n 1 n | n  N } But we know that C  F is not regular Conclusion: C is not regular

22. Pumping Down E = { 0 i 1 j | i  j } Problem: ‘pumping up’ s=0 p 1 p with y=0 k gives xyyz = 0 p+k 1 p, xy 3 z = 0 p+2k 1 p, which are all in E (hence do not give contradictions) Solution: pump down to xz = 0 p–k 1 p. Overall for s = xyz = 0 p 1 p (with |xy|  p): y=0 k, hence xz = 0 p–k 1 p  E Contradiction: E is not regular End Chapter One