1. INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011

2. Q = {q 0, q 1, q 2, q 3 } Σ = {0,1}  : Q  Σ → Q transition function * q 0  Q is start state F = {q 1, q 2 }  Q accept states M = (Q, Σ, , q 0, F) where  01 q0q0 q0q0 q1q1 q1q1 q2q2 q2q2 q2q2 q3q3 q2q2 q3q3 q0q0 q2q2 * q2q2 0 0,1 0 0 1 1 1 q0q0 q1q1 q3q3 M

3. THE REGULAR OPERATIONS Union: A  B = { w | w  A or w  B } Intersection: A  B = { w | w  A and w  B } Negation:  A = { w | w  A } Reverse: A R = { w 1 …w k | w k …w 1  A } Concatenation: A  B = { vw | v  A and w  B } Star: A* = { w 1 …w k | k ≥ 0 and each w i  A }

4. REGULAR EXPRESSIONS

5.  is a regular expression representing {  } ε is a regular expression representing {ε}  is a regular expression representing  If R 1 and R 2 are regular expressions representing L 1 and L 2 then: (R 1 R 2 ) represents L 1  L 2 (R 1  R 2 ) represents L 1  L 2 (R 1 )* represents L 1 *

6. REGULAR EXPRESSIONS (+ [ - [  ) {0…9}* (. [  ) {0…9} +  * NICK STINKS  *  * NICK RULES  * CSCI {1...4}{0...9}{0...9}{0…9}

7. PRECEDENCE *   EXAMPLE R 1 R 2 *  R 3 = (R 1 (R 2 *)) ∪ R 3

8. { w | w has exactly a single 1 } 0*10* { w | | w | ≥ 3 and its 3rd symbol is 0 } { w | every odd position of w is a 1 } ΣΣ0Σ*ΣΣ0Σ* (1Σ)*(1 ∪ ε)

9. EQUIVALENCE L can be represented by a regexp  L is a regular language

10. L can be represented by a regexp  L is a regular language EQUIVALENCE Given regular expression R, we show there exists NFA N such that R represents L(N) Induction on the length of R:

11. Given regular expression R, we show there exists NFA N such that R represents L(N) Induction on the length of R: L can be represented by a regexp  L is a regular language Base Cases (R has length 1): R =   R = ε R = 

12. Inductive Step: Assume R has length k > 1 and that any regular expression of length < k represents a language that can be recognized by an NFA Three possibilities for R: R = R 1  R 2 R = R 1 R 2 R = (R 1 )*

13. L can be represented by a regexp  L is a regular language

14. Transform (1(0  1))* to an NFA 1ε1,0 ε

15. a*(b ∪ ε)a* (1  )* ∪ 1(  1)* (lol ∪ zomg)* CONVERT TO NFAs

16. L can be represented by a regexp  L is a regular language  Proof idea: Transform an NFA for L into a regular expression by removing states and re- labeling the arrows with regular expressions 0 1 0 01*0

17. GENERALIZED NFAs q2q2 a∪ba∪b a a*b qsqs qaqa R(q s,q 2 ) = a*b R(q a,q) = Ø R(q,q s ) = Ø G accepts w if it finds q 0 q 1 …q k, w 1 …w k : w i ∈ R(q i,q i+1 ) w = w 1 w 2 …w k q 0 =q s, q k =q a

18. NFA ε ε ε ε ε Add unique and distinct start and accept states NFA TO GNFA

19. NFA ε ε ε ε ε While machine has more than 2 states: Pick an internal state, rip it out and re- label the arrows with regular expressions to account for the missing state GNFA TO REGEXP

20. q1q1 b a ε q2q2 a,b ε a*b (a*b)(a  b)* q0q0 q3q3 R(q 0,q 3 ) = (a*b)(a  b)*

21. q3q3 q2q2 b a b q1q1 b a a ε ε ε bb

22. q2q2 b a b q1q1 a a ε ε ε a  ba b

23. bb  (a  ba)b*a a  ba q2q2 b q1q1 a ε ε bb b b  (a  ba)b* = R(q 1,q 1 )

24. Convert the NFA to a regular expression q3q3 q2q2 b b q1q1 a a, b ε ε ε b (a  b)b*b bb*b (a  b)b*b(bb*b)* (a  b)b*b(bb*b)*a

25. ((a  b)b*b(bb*b)*a)*  ((a  b)b*b(bb*b)*a)*(a  b)b*b(bb*b)*

26. Formally: Run CONVERT(G): If #states = 2 return the expression on the arrow going from q start to q accept Add q start and q accept to create G If #states > 2

27. Formally: Run CONVERT(G): Add q start and q accept to create G If #states > 2 select q rip  Q different from q start and q accept define Q = Q – {q rip } define R as: R(q i,q j ) = R(q i,q rip )R(q rip,q rip )*R(q rip,q j )  R(q i,q j ) return CONVERT(G)

28. DFA NFA Regular Language Regular Expression DEF