Homework 4 Solutions. 2. Find context-free grammars for the language L = {a n b m c k : k n + m}. (with n 0,

Transcription

1 Introduction to Formal Language, Fall 2016 Due: 21-Apr-2016 (Thursday) Instructor: Prof. Wen-Guey Tzeng Homework 4 Solutions Scribe: Yi-Ruei Chen 1. Find context-free grammars for the language L = {a n b m : n 2m} with n 0, m 0. [Solution 1] Parse L as L = L 1 L 2, where L 1 = {a n b m : n > 2m} and L 2 = {a n b m : n < 2m}. Then construct productions for L 1 and L 2, respectively. A context-free grammar for L is G = ({S, S 1, S 2, A, B}, {a, b}, S, P ) with the productions S S 1 S 2, S 1 aas 1 b A, A a aa, S 2 aas 2 b B, B b Bb ab. [Solution 2] Produce L = {a n b m : n = 2m} then add extra a s or b s. A context-free grammar for L is G = ({S, A, B}, {a, b}, S, P ) with the productions S aasb A B, A a aa, B b Bb ab. 2. Find context-free grammars for the language L = {a n b m c k : k n + m}. (with n 0, m 0, k 0) Parse L as L = L 1 L 2, where L 1 = {a n b m c k : k > n + m} and L 2 = {a n b m c k : k < n + m}. Then construct productions for L 1 and L 2, respectively. A context-free grammar for L is G = ({S, S 1, S 2, T 1, T 2, A, B, C}, {a, b, c}, S, P ) with the productions S S 1 S 2, S 1 as 1 c T 1, T 1 bt 1 c C, C cc c, S 2 as 2 c T 2 AB A B, T 2 bt 2 c B, A aa a, B bb b. 3. Show that L = {w {a, b, c} : w = 3n a (w)} is a context-free language. A context-free grammar for L is G = ({S, T }, {a, b, c}, S, P ) with the productions S SaST ST S ST SaST S ST ST SaS λ, T b c. 1

2 4. Let L = {a n b n : n 0}. Show that L and L are context-free. (1) L = {a m b k : m k} {(a + b) ba(a + b) }. A context-free grammar for L is G = ({S, S 1, S 2, A, B, T }, {a, b}, S, P ) with the productions S S 1 S 2, S 1 as 1 b A B, A a aa, B b bb, S 2 T bat, T at bt λ. (2) L = {(a n b n ) m : n, m 0}. A context-free grammar for L is G = ({S, S 1, S 2, A, B, T }, {a, b}, S, P ) with the productions S SS 1 λ, S 1 as 1 b λ. 5. Show a derivation tree for the string aabbbb with the grammar S AB λ, A ab, B Sb. 2

3 6. Define what one might mean by properly nested parenthesis structures involving two kinds of parentheses, say ( ) and [ ]. Intuitively, properly nested strings in this situation are ([ ]), ([[ ]])[( )], but not ([ )] or (( ]]. Using your definition, give a context-free grammar for generating all properly nested parentheses. A context-free grammar for generating all properly nested parentheses is G = ({S}, {(, ), [, ]}, S, P ) with production S [S] (S) λ. 7. Find an s-grammar for L = {a n b n+1 : n 2}. An s-grammar for L is G = ({S, S 1, S 2, B}, {a, b}, S, P ) with the productions S as 1 B, S 1 as 2 B, S 2 as 2 B b, B b. 8. Construct an unambiguous grammar equivalent to the following grammar. S AB aab, A a Aa, B b. This grammar generates the strings a + b. A desirable grammar is G = ({S, A}, {a, b}, S, P ) with the productions S as b, A aa a. 9. Show that the following grammar is ambiguous. S asbs bsas λ. The string w = abab has the following two derivation trees: 3

4 10. Eliminate useless productions from S a aa B C, A ab λ, B Aa, C ccd, D ddd. There are two cases for useless variables Case 1: Variables that cannot generate strings in T. V 1 = {}, (T V 1 ) = {a, b, c, d} ; Since S a, A λ, and D ddd, add S, A, and D to V 1 ; V 1 = {S, A, D}, (T V 1 ) = ({a, b, c, d, S, A, D}) ; Since S aa and B Aa, add S and B to V 1 ; V 1 = {S, A, B, D}, (T V 1 ) = ({a, b, c, d, S, A, B, D}) ; Since S B and A ab, B Aa, the algorithm stops since there is no new rules can be added to V 1 ; Thus, we have V 1 = {S, A, B, D}. After removing the related useless productions, we have: S a aa B, A ab λ, B Aa, D ddd. Case 2: Variables that cannot be reached from S. The dependency graph of the result grammar in Case 1 is as follows. 4

5 D is unreachable from S; Thus, after removing the related useless productions, we have: S a aa B, A ab λ, B Aa. 11. Eliminate all λ-productions from S AaB abb, A λ, B bba λ. A procedure of removing all λ-productions is as follows. Find the nullable variable set V N = {A, B}; The λ-production A λ can be removed after adding new productions obtained by substituting λ for A where it occurs on the right: S AaB abb ab, B bba λ bb; The λ-production B λ can be removed after adding new productions obtained by substituting λ for B where it occurs on the right: S AaB Aa abb ab a, B bba bb; 5

6 12. Eliminate all unit-productions in Question 10. From the dependency graph of the grammar, we add the new rules S Aa ccd to the non-unit productions S a aa Aa ccd, A ab λ, B Aa, C ccd, D ddd. 13. Transform the grammar with production into Chomsky normal form. The transform procedure is as follows. Removing λ-productions: S abab, A baa λ, B BAa A λ Removing A λ: S abab abb, A baa ba, B BAa A λ Ba. Removing B λ: S abab abb aba ab, A baa ba, B BAa A Ba Aa a. Removing unit-production B A: S abab abb aba ab, A baa ba, B BAa baa ba Ba Aa a. Removing useless productions: No useless productions. Convert the grammar into Chomsky normal form: Introduce new variables S x for each x T : S S a S b AB S a S b B S a S b A S a S b, A S b AS a S b S a, B BAS a S b AS a BS a AS a a, S a a, S b b. Introduce additional variables to get the first two productions into normal 6

7 form and we get the final result S S a U S a X S a Y S a S b, A S b W S b S a, B BZ S b V S b B S b A S b BS a AS a a, U S b V, V AB, X S b B, Y S b A, W AS a, Z AS a, S a a, S b b. 14. Convert the grammar with production into Greibach normal form. We introduce new variables X and Y : S ABb a, A aaa B, B bab λ S ABX Y, A Y Y A B, B XAX λ, X b, Y a Then, by using the substitution, we immediately get the equivalent grammar S ay ABX baxbx baxx b a, A ay A B, B bax λ, X b, Y a 7

8 15. Use the CYK algorithm to determine whether the strings aabb, aabba, and abbbb are in the language generated by the grammar in Example (1) For the string aabb: Firstly, we have V 1,1 = {A}, V 2,2 = {A}, V 3,3 = {B}, V 4,4 = {B}. 8

9 Then, by using the equation V ij = {A BC, with B V ik, C V k+1,j }, k {i,i+1,...,j 1} we have V 1,2 = {A BC, B V 1,1, C V 2,2 } = {}, V 2,3 = {A BC, B V 2,2, C V 3,3 } = {S, B}, V 3,4 = {A BC, B V 3,3, C V 4,4 } = {A}, V 1,3 = {A BC, B V 1,1, C V 2,3 } {A BC, B V 1,2, C V 3,3 } = {S, B}, V 2,4 = {A BC, B V 2,2, C V 3,4 } {A BC, B V 2,3, C V 4,4 } = {A}, V 1,4 = {A BC, B V 1,1, C V 2,4 } {A BC, B V 1,2, C V 3,4 } Thus, we have {A BC, B V 1,3, C V 4,4 } = {A}. 9

10 V 1,4 = {A} V 1,3 = {S, B} V 2,4 = {A} V 1,2 = {} V 2,3 = {S, B} V 3,4 = {A} V 1,1 = {A} V 2,2 = {A} V 3,3 = {B} V 4,4 = {B} Because V 1,4 = {A}, S V 1,4, we conclude that aabb is not in the language generated by the given grammar, i.e., aabb L(G), by using the CYK algorithm. (2) For the string aabba: Similarly, we have 10

11 V 1,5 = {} V 1,4 = {A} V 2,5 = {} V 1,3 = {S, B} V 2,4 = {A} V 3,5 = {} V 1,2 = {} V 2,3 = {S, B} V 3,4 = {A} V 4,5 = {} V 1,1 = {A} V 2,2 = {A} V 3,3 = {B} V 4,4 = {B} V 5,5 = {A} Because V 1,5 = {}, we conclude that aabba is not in the language generated by the given grammar, i.e., aabba / L(G), by using the CYK algorithm. (3) For the string abbbb: Similarly, we have V 1,5 = {A} V 1,4 = {S, B} V 2,5 = {A} V 1,3 = {A} V 2,4 = {S, B} V 3,5 = {S, B} V 1,2 = {S, B} V 2,3 = {A} V 3,4 = {A} V 4,5 = {A} V 1,1 = {A} V 2,2 = {B} V 3,3 = {B} V 4,4 = {B} V 5,5 = {B} Because V 1,5 = {A}, S V 1,5, we conclude that abbbb is not in the language generated by the given grammar, i.e., abbbb / L(G), by using the CYK algorithm. 11