Do any six problems. Write your answers clearly and neatly. Use the back if necessary.

Transcription

1 CS5003 Foundations of C.S. Spring, 2016 Final Exam PRINT NAME: SIGN : Do any six problems. Write your answers clearly and neatly. Use the back if necessary. 1. Let L {a, b, c} be the language with definition L = {a 2n b m c 3k {a, b, c} 0 n, 0 m 2, 0 k}. a) Give a recursive definition of L. b) Compute L 3 according to your definition. c) Construct a context free grammar G which realizes L. d) Prove that L = L(G). BASIS: λ, b, b 2 L RECURSIVE STEP: u L a 2 u, uc 3 L. CLOSURE: All elements in the language can be obtained from the basis with a finite number of applications of the recursive step. With this definition: L 3 = {λ, b, b 2, a 2, a 2 b, a 2 b 2, a 4, a 4 b, a 4 b 2, a 6, a 6 b, a 6 b 2, c 3, bc 3, b 2 c 3, a 2 c 3, a 2 bc 3, a 2 b 2 c 3, a 4 c 3, a 4 bc 3, a 4 b 2 c 3, c 6, bc 6, b 2 c 6, a 2 c 6, a 2 bc 6, a 2 b 2 c 6, c 9, bc 9, b 2 c 9 }. G : S a 2 S Sc 3 a 2 a λ Proof: L L(G): Let a 2n b m c 3k L. A derivation sequence is S n a 2n S k a 2n Sc 3k a 2n b m c 3k L(G) L: We show that the set of sentential forms is contained in X = {a 2n Sc 3k, a 2n b m c 3k } with n, k 0 and m {0, 1, 2}. by induction on the number of rules applied. Base case: No rules applied. Sentential form is S X with n = k = 0, so base case is satisfied. Inductive step: Suppose after j rules the a sentential form is of form a 2n Sc 3k or a 2n b m c 3k. For the second, no rule applies so there is nothing to show. For the first, we get a 2(n+1) Sc 3k, a 2n Sc 3(k+1) a 2n c 3k a 2n bc 3k or a 2n b 2 c 3k depending on which S rules we use, and all are in X, as required. Since X contains all sentential forms, L(G) L. 1 of 7

2 2. Give regular expressions for each of the following subsets of {a, b, c}. a) Strings with an even number of b s. : Looking from left to right, the b s come in pairs, with perhaps a s in between. (b(a c) b) (a c)) b) strings which either start with a or have an even number of letters. : Simple with the union [a(a b c) ] [((a b c) 2 ) ] c) All strings not containing aaa. : More than two a s must be separated by non-a s. First, such strings that start and end with a. (a a 2 )((b c) + (a a 2 )) Then we add the possibility that it starts or ends with b, or has no a s at all. [(b c) (a a 2 )((b c) + (a a 2 )) (b c) ] (b c) 2 of 7

3 3. Let L {a, b, c} be given by BASIS: λ L RECURSIVE STEP: If u L then aub L, a 2 uc 2 L, and a 2 uc 3 L 2. CLOSURE: A string is in L if if can be obtained from the basis by a finite number of applications of the recursive step. Prove by induction that for all w L, n a (w) n b (w) + n c (w) 2n a (w). : Proof. We will prove the statement by induction on the number of steps in the recursive definition of w L. Base Case: Suppose w L 0. Then w = λ and n a (w) = n b (w) = n b (w) = 0, and the statement is true. Inductive step. Suppose for all w L k and n a (w) n b (w) + n c (w) 2n a (w). The element in L k+1 are of the form awb, a 2 wc 2, or a 2 wc 3 We have n a (awb) = n a (w) + 1, n b (awb) = n b (w) + 1, n c (awb) = n c (w), so n a (awb) = n a (w) + 1 n b (w) + n c (w) + 1 = (n b (w) + 1) + n c (w) = n b (awb) + n c (awb) and n b (awb) + n c (awb) = n b (w) + n c (w) + 1 2n a (w) + 1 2n a (w) + 2 = 2n a (awb) as required. Next n a (a 2 wc 2 ) = n a (w) + 2, n b (a 2 wc 2 ) = n b (w), n c (a 2 wc 2 ) = n c (w) + 2, so and n a (a 2 wc 2 ) = n a (w) + 2 n b (w) + n c (w) + 2 = n b (a 2 wc 2 ) + n c (a 2 wc 2 ) n b (a 2 wb 2 ) + n c (a 2 wb 2 ) = n b (w) + n c (w) + 2 2n a (w) + 2 2n a (w) + 4 = 2n a (a 2 wc 2 ) as required. Last n a (a 2 wc 3 ) = n a (w) + 2, n b (a 2 wc 2 ) = n b (w), n c (a 2 wc 3 ) = n c (w) + 3, so n a (a 2 wc 3 ) = n a (w) + 2 n b (w) + n c (w) + 2 n b (w) + n c (w) + 3 = n b (a 2 wc 3 ) + n c (a 2 wc 3 ) and n b (a 2 wc 3 )+n c (awb) = n b (w)+n c (w)+3 2n a (w)+3 2n a (w)+3 2n a (w)+4 = 2n a (a 2 wc 3 ) as required. So the inequality is satisfied in every case, and so is true for all w L by induction. 3 of 7

4 4. Let L be the language consisting of those strings in {a, b, c} which begin with an a, contain exactly two b s and end with cc. Construct a Deterministic Finite Automaton which recognizes this language. 5. Let L be the language on Σ = {a, b, c, d} of all strings which either contain the substring bad or the substring baaad. Construct an Automaton, deterministic or not, which recognizes this language. Non-deterministic is much easier. 4 of 7

5 6. Let G be the grammar given by G : S S AB λ A aaabc ABC a B BCA b C bbcc λ Convert this grammar to Chomsky Normal Form. First note that the rule S S is useless and can be omitted. Second, the fact that C is nullable must be rectified: G : S AB λ A aaabc aaab ABC AB a B BCA BA b C bbcc bbc Lastly we perform our splitting moves: note that the duplicated rules from the nullity moves don t need to be done completely separated. G : S AB λ A U a A 1 AA 3 AB a A 1 AA 2 A 2 U a A 3 UaB A 3 BC B BB 1 BA b B 1 CA C U b C 1 C 1 BC 2 BU c C 2 U c C U a a U b b U c c 5 of 7

6 7. a) Let G be the grammar given by G : S A B A aa B C B bbb C D C cccc D a D λ ABC Construct an equivalent grammar which does not contain chain rules. Clearly CHAIN(S) = {S, A, B, C, D}, CHAIN(A) = {A, B, C, D}, CHAIN(B) = {B, C, D}, CHAIN(C) = {C, D}, CHAIN(D) = {D}, b) For the grammar G : S aa bbb cccc a λ ABC A aa bbb cccc a λ ABC B bbb cccc a λ ABC C cccc a λ ABC D λ ABC G : S ABE ADE EA A aa ae B BDB BBDD C BD AD b D BDD BB EE CC E BD AD b Construct REACH and TERM in the correct order are remove the useless symbols T ERM 0 = {C, E}, T ERM 1 = {A, C, D, E}, T ERM = T ERM 2 = {S, A, C, D, E}, so S T ERM and the language is non-empty. Also B is useless and can be removed: G : S ADE EA A aa ae C AD b D EE CC E AD b Now REACH 0 = {S}, REACH 1 = {S, A, D, E} and REACH = REACH 2 = {S, A, C, D, E} so all symbols are now reachable and terminable. Although this was not asked, and is not specified in the definition of useless, C and E are now exactly the same, so one of them is useless and can be replaced with the other. Also now D must be replace by CC, so is also useless. G : S ACCC CA A aa ac C ACC b 6 of 7

7 8. For the grammar G : S AB A AB DA a B BC DB b C CA DC c D DA DD a b c a) Trace the CYK algorithm to decide if cabbc is in the language. (Obviously the longest problem, by far. Most students chose it anyway. Why?) S, A, B, CD S, A, B, C, D S, A, B, C, D S, A, B, C, D S, A, D B, C, D A, C, D S, A, B, D B, D BCD C, D A, D B, D B, D C, D This was easy to fill out for the opposite reason than the quiz example, instead of many s, the are many {S, A, B, C, D} s. b) Convert to Greibach Normal Form. (also use the back of this page.) The grammar is already in Chomsky Normal Form and is already ordered for S < A < B < C < D. We can remove left recursion on D, put it in compliance, and remove the initial D s right away. G : S AB A AB aa ba ca ar d A br d A cr d A a B BC ab bb cb ar d B br d B cr d B b C CA ac bc cc ar d C br d C cr d C c D a b c ar d br d cr d R d A D AR d DR d G : S aab bab cab ar d AB br d AB cr d AB ab aar a B bar a B car a B ar d AR a B br d AR a B cr d AR a B ar a B A aa ba ca ar d A br d A cr d A a aar a bar a car a ar d AR a br d AR a cr d AR a ar a R a B BR a B ab bb cb ar d B br d B cr d B b abr b bbr b cbr b ar d BR b br d BR b cr d BR b br b R b C CR b C ac bc cc ar d C br d C cr d C c R c AR c A acr c bcr c ccr c ar d CR c br d CR c cr d CR c cr c D a b c ar d br d cr d R d A D AR d DR d 7 of 7

8 Lastly we fix the added variables: R a ab bb cb ar d B br d B cr d B b abr b bbr b cbr b ar d BR b br d BR b cr d BR b br b abr a bbr a cbr a ar d BR a br d BR a cr d BR a br a abr b R a bbr b R a cbr b R a ar d BR b R a br d BR b R a cr d BR b R a br b R a R b ac bc cc ar d C br d C cr d C c acr b bcr b ccr b ar d CR b br d CR b cr d CR b cr b acr b bcr b ccr b ar d CR b br d CR b cr d CR b cr b acr b R b bcr b R b ccr b R b ar d CR b R b br d CR b R b cr d CR b R b cr b R b R c aa ba ca ar d A br d A cr d A a aar a bar a car a ar d AR a br d AR a cr d AR a ar a aar c bar c car c ar d AR c br d AR c cr d AR c ar c aar a R c bar a R c car a R c ar d AR a R c br d AR a R c cr d AR a R c ar a R c R d aa ba ca ar d A br d A cr d A a Whew! aar a bar a car a ar d AR a br d AR a cr d AR a ar a a b c ar d br d cr d aar d bar d car d ar d AR d br d AR d cr d AR d ar d aar a R d bar a R d car a R d ar d AR a R d br d AR a R d cr d AR a R d ar a R d ar d br d cr d ar d R d br d R d cr d R d 8 of 7