Understanding Inherent Limitations of Computer Programs through Regular Operations

1. CSci 4011 INHERENT LIMITATIONS OF COMPUTER PROGRAMS

2. THE REGULAR OPERATIONS Negation: A = { w | w  A } Union: A  B = { w | w  A or w  B } Intersection: A  B = { w | w  A and w  B } Reverse: AR = { w1 …wk | wk …w1 A } Concatenation: A  B = { vw | v  A and w  B } Star: A* = { w1 …wk | k ≥ 0 and each wi A }

3. 0,1 1 0 0 0 1 1

4. NON-DETERMINISM 0,1 1 0 0 0 1 1 An NFA accepts if there is a series of choices that take it to an accept state

5. Deterministic Computation Non-Deterministic Computation reject accept or reject accept

6. Theorem: Every NFA has an equivalent DFA Corollary: A language is regular iff it is recognized by an NFA

7. FROM NFA TO DFA Input: N = (Q, Σ, , Q0, F) Output: M = (Q, Σ, , q0, F) To learn if NFA accepts, we could do the computation in parallel, maintaining the set of states where all threads are reject Idea: Q = 𝒫(Q) accept

8. FROM NFA TO DFA Input: N = (Q, Σ, , Q0, F) Output: M = (Q, Σ, , q0, F) Q = 𝒫(Q)  : Q Σ→ Q (R,) =  ε( (r,) ) (R,) =  ε( (r,) ) (R,) =  ε( (r,) ) rR q0 = ε(Q0) F = { R  Q | f  R for some f  F }

9. EXAMPLES: NFA TO DFA 1 a b 0,1 ε 0 3 1 2 1

10. REGULAR LANGUAGES CLOSED UNDER CONCATENATION Given DFAs M1 and M2, construct NFA by connecting all accept states in M1 to start states in M2 ε L(M1)=A L(M2)=B ε

11. REGULAR LANGUAGES ARE CLOSED UNDER REGULAR OPERATIONS Union: A  B = { w | w  A or w  B } Intersection: A  B = { w | w  A and w  B } Reverse: AR = { w1 …wk | wk …w1 A } Negation: A = { w | w  A } Concatenation: A  B = { vw | v  A and w  B } Star: A* = { w1 …wk | k ≥ 0 and each wi A }

12. 1 0 0,1 1 0 0 1 REGULAR LANGUAGES CLOSED UNDER STAR Let L be a regular language and M be a DFA for L We construct an NFA N that recognizes L* ε ε ε

13. Formally: Input: M = (Q, Σ, , q1, F) Output: 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 

14. 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: k = 0  k = 1  Inductive Step: Assume N accepts all strings v = v1…vk L*, and let u = u1…ukuk+1  L* Since N accepts u1…uk and M accepts uk+1, N must accept u

15. Assume w is accepted by N, we show w  L* If w = ε, then w  L* If w ≠ε ε ε accept

16. REGULAR LANGUAGES ARE CLOSED UNDER REGULAR OPERATIONS Union: A  B = { w | w  A or w  B } Intersection: A  B = { w | w  A and w  B } Reverse: AR = { w1 …wk | wk …w1 A } Negation: A = { w | w  A } Concatenation: A  B = { vw | v  A and w  B } Star: A* = { w1 …wk | k ≥ 0 and each wi A }

17. REGULAR EXPRESSIONS

18. REGULAR EXPRESSIONS • is a regular expression representing {} ε is a regular expression representing {ε}  is a regular expression representing  If R1 and R2 are regular expressions representing L1 and L2 then: (R1R2) represents L1L2 (R1  R2) represents L1 L2 (R1)* represents L1*

19. PRECEDENCE * 

20. EXAMPLE ( ( R1* ) R2 )  R3 R1*R2 R3 =

21. { w | w has exactly a single 1 } 0*10*

22. { w | w has length ≥ 3 and its 3rd symbol is 0 } 000(01)*  010(01)*  100(01)*  110(01)*

23. { w | every odd position of w is a 1 }