CS 154, Lecture 3: DFA  NFA, Regular Expressions

1. CS 154, Lecture 3:DFANFA, Regular Expressions

2. Homework 1 is coming out …

3. Deterministic Finite Automata Computation with finite memory

4. Non-Deterministic Finite Automata Computation with finite memoryand “verified guessing”

5. From NFAs to DFAs Input: NFA N = (Q, Σ,, Q0, F) Output: DFA M = (Q, Σ, , q0, F) To learn if an NFA accepts, we could do the computation in parallel, maintaining the set of all possible states that can be reached Idea: Set Q = 2Q

6. (R,) =  ε( (r,) ) rR From NFAs to DFAs: Subset Construction Input: NFA N = (Q, Σ, , Q0, F) Output: DFA M = (Q, Σ, , q0, F) Q = 2Q  : Q Σ→ Q * q0 = ε(Q0) F = { R  Q | f  R for some f  F } * For S  Q, the ε-closure of S is ε(S) = {q | q reachable from some s  S by taking 0 or more ε transitions}

7. Example of the ε-closure {q0 , q1, q2} ε({q0}) = {q1, q2} ε({q1}) = ε({q2}) = {q2}

8. Given: NFA N = ( {1,2,3}, {a,b},  , {1}, {1} ) Construct: Equivalent DFA M = (2{1,2,3}, {a,b}, , {1,3}, …) M N a a b 1 {1,3} {3}  a b b a b b a,b ε a , b {2} {2,3} a 2 3 a b a {1}, {1,2} ? ε({1}) = {1,3} {1,2,3}

9. Reverse Theorem for Regular Languages Theorem: The reverse of a regular language is also a regular language Ifa language can be recognized by a DFA that reads strings from right to left, thenthere is an “normal” DFA that accepts the same language Proof? Given a DFA for a language L, “reverse” its arrows and flip its start and accept states, getting an NFA. Convert that NFA back to a DFA!

10. Using NFAs in place of DFAs can make proofs about regular languages much easier! Remember this on homework/exams!

11. Union Theorem using NFAs?

12. 1 1 0 0 0,1 0,1 1 1 0 0 0 0 1 1 Regular Languages are closed under concatenation Concatenation: A  B = { vw | v  A and w  B } 1 Given DFAs M1 and M2, connect the accept states of M1 to the start states of M2 0 0,1 1 ε 0 0 1 ε L(N) = L(M1)  L(M2)

13. 1 0 0,1 1 0 0 1 Regular Languages are closed under star A* = { s1 … sk | k ≥ 0 and each si A } Let M be a DFA, and let L = L(M) We can construct an NFA N that recognizes L* ε ε ε

14. Formally, the construction is: Input: DFA M = (Q, Σ, , q1, F) Output: NFA N = (Q, Σ, , q0, F) Q = Q  {q0} F = F  {q0} {(q,a)} if q  Q and a ≠ ε if q  F and a = ε {q1} if q = q0 and a = ε {q1} (q,a) = if q = q0 and a ≠ ε  else 

15. Regular Languages are Closed Under Star • How would we prove that this NFA construction works? • Want to show: L(N) = L* • L(N) L* • L(N) L*

16. 1. L(N) L* Assume w = w1…wk is in L* where w1,…,wk L We show N accepts w by induction on k Base Cases: (w = ε) k = 0   (w L) k = 1 Inductive Step: Assume N accepts all strings v = v1…vk L*, vi L Let u = u1…ukuk+1  L* , uj L Since N accepts u1…uk(by induction) and M accepts uk+1, N also accepts u (by construction)

17. 2. L(N) L* Assume w is accepted by N; we want to show w  L* If w = ε, then w  L* ε I.H. N accepts u and takes at most k ε-transitionsu  L* uL(N), so By I.H. uL* u • Let w be accepted by N with k+1 ε-transitions. • Write w as w=uv, • where v is the substring read after the lastε-transition ε v vL w = uv L* accept

18. Closure Properties for Regular Languages Union: A  B = { w | w  A or w  B }  Intersection: A  B = { w | w  A and w  B }  Complement:A = { w  Σ*| w  A }  Reverse: AR = { w1 …wk | wk …w1 A, wi Σ}  Concatenation: A  B = { vw | v  A and w  B }  Star: A* = { s1 … sk | k ≥ 0 and each si A }  Theorem: if A and B are regular then so are: A  B, A  B, A, AR, A  B, and A*

19. Regular Expressions Computation as simple, logical description A totally different way of thinking about computation: What is the complexity of describing the strings in the language?

20. Inductive Definition of Regexp Let Σ be an alphabet. We define the regular expressions overΣ inductively: For all ∊Σ, is a regexp ε is a regexp  is a regexp If R1 and R2 are both regexps, then (R1R2), (R1 + R2), and (R1)* are regexps

21. Precedence Order: * thenthen+ Example: R1*R2+ R3 = ( ( R1* ) ) + R3 · R2

22. Definition: Regexps Represent Languages The regexp∊Σrepresents the language {} The regexpε represents {ε} The regexp represents  If R1 and R2 are regular expressions representing L1 and L2 then: (R1R2) represents L1  L2 (R1 + R2) represents L1  L2 (R1)* represents L1*

23. Regexps Represent Languages For every regexpR, define L(R) to be the language that R represents A string w ∊Σ*is acceptedbyR(or, wmatchesR)if w ∊ L(R) Examples: 0, 010, and 01010 match (01)*0 110101110100100 matches (0+1)*0

24. Assume Σ = {0,1} { w | w has exactly a single 1 } 0*10* { w | w contains 001 } (0+1)*001(0+1)*

25. Assume Σ = {0,1} What language does the regexp* represent? {ε}

26. Assume Σ = {0,1} { w | w has length ≥ 3 and its 3rd symbol is 0 } (0+1)(0+1)0(0+1)*

27. Assume Σ = {0,1} { w | every odd position in w is a 1 } (1(0 + 1))*(1 + ε)

28. Assume Σ = {0,1} { w | w has equal number of occurrences of 01 and 10} = { w | w = 1, w = 0, or w = ε, or w starts with a 0 and ends with a 0, or w starts with a 1 and ends with a 1 } Claim: A string w has equal occurrences of 01 and 10 w starts and ends with the same bit. 1 + 0 + ε +0(0+1)*0 + 1(0+1)*1

29. DFAs NFAs Regular Expressions! L can be represented by some regexp  L is regular

30. L can be represented by some regexp L is regular

31. Given any regexp R, we will construct an NFA N s.t.N accepts exactly the strings accepted by R Proof by induction on the length of the regexp R L can be represented by some regexp L is regular Base Cases (R has length 1): R =  R = ε R = 

32. Induction Step: Suppose every regexp of length < k represents some regular language. Consider a regexpR of length k > 1 Three possibilities for R: R = R1+ R2 R = R1 R2 R = (R1)*

33. Induction Step: Suppose every regexp of length < k represents some regular language. Consider a regexp R of length k > 1 Three possibilities for R: R = R1+ R2 By induction, R1and R2 representsome regular languages, L1 and L2 R = R1 R2 But L(R) = L(R1+ R2) = L1L2so L(R) is regular, by the union theorem! R = (R1)*

34. Induction Step: Suppose every regexp of length < k represents some regular language. Consider a regexp R of length k > 1 Three possibilities for R: R = R1+ R2 By induction, R1and R2 representsome regular languages, L1 and L2 R = R1 R2 But L(R) = L(R1·R2) =L1·L2so L(R) is regular by the concatenation theorem R = (R1)*

35. Induction Step: Suppose every regexp of length < k represents some regular language. Consider a regexp R of length k > 1 Three possibilities for R: R = R1+ R2 By induction, R1and R2 representsome regular languages, L1 and L2 R = R1 R2 But L(R) = L(R1*) =L1*so L(R) is regular, by the star theorem R = (R1)*

36. Induction Step: Suppose every regexp of length < k represents some regular language. Consider a regexp R of length k > 1 Three possibilities for R: R = R1+ R2 By induction, R1and R2 representsome regular languages, L1 and L2 R = R1 R2 But L(R) = L(R1*) =L1*so L(R) is regular, by the star theorem R = (R1)* Therefore: If L is represented by a regexp, then L is regular

37. Give an NFA that accepts the language represented by (1(0 + 1))* ε 1 0,1 ε ( )* 1 (0+1) Regular expression:

38. Generalized NFAs (GNFA) L can be represented by a regexp  L is a regular language  Idea: Transform an NFA for L into a regular expression by removing states and re-labeling the arcs with regular expressions Rather than reading in just letters from the string on a step, we can read in entire substrings

39. Generalized NFA (GNFA) Is aaabcbcba accepted or rejected? Is bba accepted or rejected? Is bcba accepted or rejected? This GNFA recognizes L(a*b(cb)*a)

40. NFA Add unique start and accept states

41. 0 0 01*0 1 NFA While the machine has more than 2 states: Pick an internal state, rip it out and re-label the arrowswith regexps, to account for paths through the missing state

42. R(q1,q2)R(q2,q2)*R(q2,q3) + R(q1,q3) R(q1,q2) R(q2,q3) R(q2,q2) NFA G While the machine has more than 2 states: In general: R(q1,q3) q2 q3 q1

43. a,b a (a*b)(a+b)* ε b ε a*b q1 q2 q3 q0 R(q0,q3) = (a*b)(a+b)* represents L(N)

44. NFAs DFAs DEFINITION Regular Languages Regular Expressions

45. Parting thoughts: Regular Languages can be defined by their closure properties NFA=DFA, does it mean that non-determinism is free for Finite Automata? Questions?