Regular Languages’ Closure Properties, Take 2 CS
Description: Regular Languages Closure Properties, Take 2 CS 154, Omer Reingold Reverse Theorem for Regular Languages Theorem: The reverse of a regular language is also a regular language If a language can be recognized by a DFA that reads strings from
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slide1. Regular Languages’
Closure Properties, Take 2 CS 154, Omer Reingold<br>
slide2. Reverse Theorem for Regular Languages Theorem: The reverse of a regular language is also a regular language If a language can be recognized by a DFA that reads strings from right to left,
then there 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!<br>
slide3. Using NFAs in place of DFAs can make proofs about regular languages much easier! Remember this on homework/exams!<br>
slide4. Union Theorem using NFAs?<br>
slide5. Regular Languages are closed under concatenation Given DFAs M1 and M2, connect the accept states of M1 to the start states of M2 Concatenation: A B = { vw | v A and w B } ε ε L(N) = L(M1) L(M2)<br>
slide6. Regular Languages are closed under star Let M be a DFA, and let L = L(M) We can construct an NFA N that recognizes L* A* = { s1 … sk | k ≥ 0 and each si A } ε ε ε<br>
slide7. 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) = {(q,a)} {q1} {q1} if q Q and a ≠ ε if q F and a = ε if q = q0 and a = ε if q = q0 and a ≠ ε else<br>
slide8. How would we prove that this NFA construction works?
Want to show: L(N) = L*
L(N) L*
L(N) L* Regular Languages are Closed Under Star<br>
slide9. 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: k = 0 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) (w = ε) (w L)<br>
slide10. Assume w is accepted by N; we want to show w L* If w = ε, then w L* I.H. N accepts u and takes at most k ε-transitions u L* uL* vL 2. L(N) L* By I.H. w = uv L* uL(N), so Let w be accepted by N with k+1 ε-transitions.
Write w as w=uv,
where v is the substring read after the last ε-transition accept ε ε u v<br>
slide11. 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* <br>
Closure Properties, Take 2 CS 154, Omer Reingold<br>
slide2. Reverse Theorem for Regular Languages Theorem: The reverse of a regular language is also a regular language If a language can be recognized by a DFA that reads strings from right to left,
then there 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!<br>
slide3. Using NFAs in place of DFAs can make proofs about regular languages much easier! Remember this on homework/exams!<br>
slide4. Union Theorem using NFAs?<br>
slide5. Regular Languages are closed under concatenation Given DFAs M1 and M2, connect the accept states of M1 to the start states of M2 Concatenation: A B = { vw | v A and w B } ε ε L(N) = L(M1) L(M2)<br>
slide6. Regular Languages are closed under star Let M be a DFA, and let L = L(M) We can construct an NFA N that recognizes L* A* = { s1 … sk | k ≥ 0 and each si A } ε ε ε<br>
slide7. 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) = {(q,a)} {q1} {q1} if q Q and a ≠ ε if q F and a = ε if q = q0 and a = ε if q = q0 and a ≠ ε else<br>
slide8. How would we prove that this NFA construction works?
Want to show: L(N) = L*
L(N) L*
L(N) L* Regular Languages are Closed Under Star<br>
slide9. 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: k = 0 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) (w = ε) (w L)<br>
slide10. Assume w is accepted by N; we want to show w L* If w = ε, then w L* I.H. N accepts u and takes at most k ε-transitions u L* uL* vL 2. L(N) L* By I.H. w = uv L* uL(N), so Let w be accepted by N with k+1 ε-transitions.
Write w as w=uv,
where v is the substring read after the last ε-transition accept ε ε u v<br>
slide11. 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* <br>