Grammars (part II) Prof. Dan A. Simovici UMB

Transcription

1 rammars (part II) Prof. Dan A. Simovici UMB 1 / 1

2 Outline 2 / 1

3 Length-Increasing vs. Context-Sensitive rammars Theorem The class L 1 equals the class of length-increasing languages. 3 / 1

4 Length-Increasing vs. Context-Sensitive rammars Proof Clearly, every type-1 grammar is length-increasing. Therefore, as we observed earlier, L 1 is included in the class of length-increasing languages. To prove the converse inclusion, consider a length-increasing grammar = (A N, A T, S, P). We can assume that every production of P that contains a terminal symbol is of the form X a. 4 / 1

5 Length-Increasing vs. Context-Sensitive rammars Proof (cont d) Let π : α β P be a production such that α = X 0 X n 1, β = Y 0 Y m 1. If n = 1, then π is already a context-sensitive production. Therefore, suppose that 2 n m. By hypothesis, X 0,..., X n 1, Y 0,..., Y m 1 are nonterminals. Consider n new nonterminals Z π 0,..., Z π n 1 and the set of productions P π that consists of: X 0 X n 1 Z π 0 X 1 X n 1 Z π 0 X 1 X n 1 Z π 0 Z π 1 X 2 X n 1. Z π 0 Z π 1 Z π n 2 X n 1 Z π 0 Z π 1 Z π n 1 Y n Y m 1 Z π 0 Z π 1 Z π n 1 Y n Y m 1 Y 0 Z π 1 Z π n 1 Y n Y m 1 Y 0 Z π 1 Z π n 1 Y n Y m 1 Y 0 Y 1 Z π n 1 Y n Y m 1. Y 0 Y 1 Z π n 1 Y n Y m 1 Y 0 Y 1 Y n 1 Y n Y m 1. The set P π consists of context-sensitive productions. 5 / 1

6 Length-Increasing vs. Context-Sensitive rammars Proof (cont d) Let P 1 be the set of productions that consists of {P α β 2 α β } and the productions of the form X β or S λ (whenever such productions belong to P). Consider the context-sensitive grammar 1 = (A N A, A T, S, P 1 ), where A consists of all new nonterminal symbols introduced when the production sets P π were constructed. 6 / 1

7 Length-Increasing vs. Context-Sensitive rammars Proof (cont d) Note that a derivation step in that consists of the application of a production π : α β with 2 α β corresponds to the successive application of the productions of P π in 1 ; conversely, the productions of P π can be applied only in this order in a derivation in 1, and they simulate a step in a derivation in that makes use of the production π. A step that uses a production X β or S λ is the same in both and 1. Thus, S x if and only if S 1 x, so L() = L( 1 ). This shows that the class of length-increasing languages is included in L 1. 7 / 1

8 Length-Increasing vs. Context-Sensitive rammars Example Let = ({S, X, Y, X a, X b, X c }, {a, b, c}, S, P ) be the length-increasing previously constructed grammar. The set of productions P 1 consists of: The context-sensitive productions in P, namely: π 0 : S X a X b X c, π 1 : S X a XX b X c, π 7 : X a a, π 8 : X b b π 9 : X c c 8 / 1

9 Length-Increasing vs. Context-Sensitive rammars Example (cont d) Each remaining production in P generates the following productions: For π 2 : XX b X b X, include the productions: π 10 : XX b Z 0 X b π 11 : Z 0 X b Z 0 Z 1 π 12 : Z 0 Z 1 X b Z 1 π 13 : X b Z 1 X b X For π 3 : XX c YX b X c X c, include the productions: π 14 : XX c Z 2 X c π 15 : Z 2 X c Z 2 Z 3 X c X c π 16 : Z 2 Z 3 X c X c YZ 3 X c X c π 17 : YZ 3 X c X c YX b X c X c 9 / 1

10 Length-Increasing vs. Context-Sensitive rammars Example (cont d) For π 4 : X by YX b, include the productions: π 18 : X b Y Z 4 Y π 19 : Z 4 Y Z 4 Z 5 π 20 : Z 4 Z 5 YZ 5 π 21 : YZ 5 YX b For π 5 : X ay X a X a X, include the productions: π 22 : X a Y Z 6 Y π 23 : Z 6 Y Z 6 Z 7 X π 24 : Z 6 Z 7 X X a Z 7 X π 25 : X a Z 7 X X a X a X 10 / 1

11 Length-Increasing vs. Context-Sensitive rammars Finally, for π 6 : X ay X a X a, include the productions: π 26 : X a Y Z 8 Y π 27 : Z 8 Y Z 8 Z 9 π 28 : Z 8 Z 9 X a Z 9 π 29 : X a Z 9 X a X a This gives the context-sensitive grammar: 1 = ({S, X, Y, X a, X b, X c } {Z i 0 i 10}, {a, b, c}, S, {π 0, π 1, π 7, π 8, π 9} {π j 10 j 29}. that generates the language L = {a n b n c n n N}. This implies that L is a context-sensitive language. 11 / 1

12 Length-Increasing vs. Context-Sensitive rammars Theorem The classes L 0, L 1, L 2 are closed with respect to the reversal operation; in other words, if L L i, then L R L i for i {0, 1, 2}. 12 / 1

13 Length-Increasing vs. Context-Sensitive rammars Proof Let = (A N, A T, S, P) be a grammar, and let P R = {α R β R α β P}. Define R = (A N, A T, S, P R ). If is of type i, then so is R for i {0, 1, 2}. Further, we have γ γ if and only if γ R R γ R, as can be shown by induction on the length of the two derivations. Thus, S w if and only if S = S R R w R. This shows that L( R ) = L() R, so L() R has the same type as the language L(). 13 / 1

14 Closure Properties of Classes of Languages Theorem Each of the classes L i of Chomsky s hierarchy contains the class of finite languages, for i {0, 1, 2}. Proof. Let L = {u 0,..., u n 1 } be a finite, nonempty language over an alphabet A. The grammar = ({S}, A, S, {S u 0,..., S u n 1 }) is of type 3 and, therefore, of type 2, 1, and 0. If L =, then L is generated by the grammar = ({S}, A, S, {S S}) that is, again, of type / 1

15 Closure Properties of Classes of Languages Theorem If L is a language of type i, where i {0, 1}, then so is the language K, where K = L {λ}. Proof. If L is a type-1 language, then there exists a type-1 grammar = (A N, A T, S, P) such that L() = L. If λ L, the production S λ belongs to P, and S does not occur in the right member of any production. The language K is generated by the type-1 grammar = (A N, A T, S, P {S λ}). If λ L, we have K = L. If L is of type 0 and λ L, let P be the set of productions obtained from P by replacing all erasure productions α λ with Y α Y and αy Y for every Y A N A T. The grammar (A N, A T, S, P ) generates the language K, so K is a type-0 language. 15 / 1

16 Closure Properties of Classes of Languages Theorem L i is closed with respect to union, for i {0, 1, 2, 3}. Proof. Suppose that L, L are two languages of type i that are generated by the grammars = (A N, A T, S, P) and = (A N, A T, S, P ), respectively, where A N A N =. Consider the grammar = (A N A N {S 0}, A T, S, P P {S 0 S, S 0 S }), where S 0 is a new nonterminal symbol such that S 0 A N A N. Note that the grammar is of the same type i as the grammars and. To complete the proof, we need to show that L L = L( ). 16 / 1

17 Closure Properties of Classes of Languages Proof (cont d) Consider the grammar = (A N A N {S 0}, A T, S, P P {S 0 S, S 0 S }), where S 0 is a new nonterminal symbol such that S 0 A N A N. Note that the grammar is of the same type i as the grammars and. To complete the proof, we need to show that L L = L( ). Let x L L. If x L, then S x, so S 0 S x which shows that x L( ). The case when x L is entirely similar and is left to the reader. Thus, L L L( ). 17 / 1

18 Closure Properties of Classes of Languages Proof (cont d) Conversely, suppose that x L( ). We have S 0 x. If the first production applied in this derivation is S 0 S, then the derivation can be written as S 0 S x. The last part of this derivation S x uses only productions from P since A N A N = implies P P =. Therefore, we have S x, so x L(). Similarly, if the first production applied is S 0 S, then x L( ). Therefore, L( ) L() L( ), hence L( ) = L() L( ). 18 / 1

19 Closure Properties of Classes of Languages Closure with respect to Lemma The classes L 0 and L 1 are closed with respect to the operation. Proof. We must prove that if L L i, then L L i for i {0, 1}. Let us assume initially that λ L. Let = (A N, A T, S, P) be a grammar of type 0 or type 1 that generates the language L. We can assume that terminal symbols do not occur in the left member of any production of P. Let S 0, S 1 be new symbols such that S 0, S 1 A N, and let = (A N {S 0, S 1 }, A T, S 0, P ) be the grammar whose set of productions is P = P {S 0 λ, S 0 S, S 0 S 1 S} {S 1 a S 1 Sa a A T } {S 1 a Sa a A T }. 19 / 1

20 Closure Properties of Classes of Languages Proof (cont d) If is of type 1, then so is. We have L n L( ) for n N. Let x L n. If n = 0 we have x = λ, and x L( ) because S 0 λ is in P. Suppose now that x = x 0 x 1... x n 1, where x 0,..., x n 1 L for n 1. We have S x j for 0 j n 1, and the same derivations are valid in because P P. Thus, we can put together the following derivation in the grammar : S 0 S 1 S S 1 x n 1 S 1 Sx n 1 S 1 x n 2 x n 1. S 1 Sx 2 x n 1 S 1 x 1 x n 1 Sx 1 x n 1 x 0 x 1 x n 1 = x. 20 / 1

21 Closure Properties of Classes of Languages Proof (cont d) This proves that x L( ). Thus, L = n N Ln L( ). Observe that the regeneration of the symbol S is made possible in the above derivation by the fact that the words x j are not null (which puts S 1 adjacent with terminal symbols, thereby allowing the application of the productions S 1 a S 1 Sa). 21 / 1

22 Closure Properties of Classes of Languages Proof (cont d) Conversely, let x L( ). There exists a derivation S 0 x. If we write this derivation explicitly, S 0 α 1 α n 1 = x, we have to consider three cases: Case 1: If the first production applied in this derivation is S 0 λ, then x = λ, so x L. Case 2: If the first production applied is S 0 S, then we have S x, and the same derivation is valid in, which shows that x L() = L. 22 / 1

23 Closure Properties of Classes of Languages Proof (cont d) Case 3: If the first production applied is S 0 S 1 S, then each word α l in the above derivation falls into one of the following two cases: Case 3a: α l is of the form S 1 β 1 β k, where k 1, S β q for 1 q k, and the first symbol of each of the words β 2,..., β k is a terminal. Case 3b: α l is of the form β 0 β 1 β k, where k 1, the first symbol of each β p is a terminal for p 1, and S β p for 0 p k. 23 / 1

24 Closure Properties of Classes of Languages Proof (cont d) This argument is by induction on l 1. It is clear that α 1 falls in Case 3a. Suppose that α h satisfies the conditions of either Case 3a or Case 3b. Since the productions of P do not have terminal symbols in their left members, their application may involve only one word β i. This guarantees that if α h+1 was obtained through the application of a production in P, then α h+1 satisfies one of the conditions of the previous cases. The same conclusion can be reached if α h+1 was obtained by applying a production of the form S 1 a S 1 Sa or S 1 a Sa. 24 / 1

25 Closure Properties of Classes of Languages Proof (cont d) The existence of the derivation S 0 x A T implies that x is a word of the form 3b. Therefore x can be written as a product, x = β 0 β k, where k 1, and S β p for 0 p k. Since every word β 0, β 1,..., β k A T it follows that β 0, β 1,..., β k L(). Thus, x (L()) k+1 L() which implies L( ) L(). If λ L, then consider the language K = L {λ}. The language K is of the same type as L and, by the above argument, so is K. Since K = L we obtain the desired closure property. 25 / 1

26 Closure Properties of Classes of Languages Lemma The class L 2 is closed with respect to the operation. Proof. Let L be a context-free language generated by the type-2 grammar = (A N, A T, S, P). Suppose that S 0 is a new nonterminal symbol and consider the type-2 grammar = (A N {S 0 }, A T, S 0, P {S 0 λ, S 0 S 0 S}). It is easy to verify that L( ) = L, so L L / 1

27 Closure Properties of Classes of Languages Lemma The class L 3 is closed with respect to the operation. Proof. Let L L 3 such that L = L(), where = (A N, A T, S, P) is a type-3 grammar. Define the set of productions P 1 = {X us X u P}. Consider the type-3 grammar = (A N {S 0 }, A T, S 0, P P 1 {S 0 λ, S 0 S}), where S 0 be a new nonterminal symbol, S 0 A N. It is easy to verify that L( ) = L, so L L / 1

28 Closure Properties of Classes of Languages Theorem The classes L i are closed with respect to the operation for i {0, 1, 2, 3}. Proof. This follows from the previous lemmas. 28 / 1

29 Closure Properties of Classes of Languages Corollary The classes L i are closed with respect to the + operation for i {0, 1, 2, 3}. Proof. Let L i be one of the classes of Chomsky s hierarchy, and let L L i. Note that { L + L if λ L = L {λ} if λ L. In all cases L + L i. 29 / 1

30 Closure Properties of Classes of Languages Lemma The classes L 0 and L 2 are closed with respect to the product operation. Proof. Let L, L be two languages of type i, and let = (A N, A T, S, P), = (A N, A T, S, P ) be two grammars of type i such that L() = L and L( ) = L, where i {0, 2}. Without any loss of generality, we can assume that A N A N =. If S 0 is a new symbol, S 0 A N A N, then the grammar p = (A N A N {S 0}, A T, S 0, P P {S 0 SS }) is also of type i. We claim that L( p ) = LL. 30 / 1

31 Closure Properties of Classes of Languages Proof (cont d) Let x LL. We can write x = uv for some u L and v L. By hypothesis, S u and S v, so S 0 p SS p us p uv = x. Thus, x L( p ). 31 / 1

32 Closure Properties of Classes of Languages Conversely, suppose that x L p. There is a derivation S 0 p SS p x. Since A N and A N are disjoint sets, the sets of productions P and P are disjoint. Therefore, the productions of p used to transform S into a word over A T belong to P, while the ones used to rewrite S belong to P. Thus, we can write x = uv, where S u and S v, which implies x LL. 32 / 1

33 Closure Properties of Classes of Languages Lemma The class L 1 is closed with respect to the product operation. Proof. Let L, L be two languages in L 1. If neither L nor L contains the null word, we may assume that both languages are generated by type-1 grammars that have no erasure rules. It is easy to see that, in this case, the construction of the grammar p given in the proof of previous Lemma yields a type-1 grammar, so LL belongs to L / 1

34 Closure Properties of Classes of Languages Proof cont d Suppose now that λ L or λ L. By The languages L 1 = L {λ} and L 1 = L {λ} also belong to L 1 and, by the previous argument L 1 L 1 L 1. We need to consider the cases summarized below. Case λ L λ L L L LL 1 yes no L 1 {λ} L 1 L 1 L 1 L 1 2 no yes L 1 L 1 {λ} L 1L 1 L 1 3 yes yes L 1 {λ} L 1 {λ} L 1L 1 L 1 L 1 {λ} In each case, we have LL L / 1

35 Closure Properties of Classes of Languages Lemma The class L 3 is closed with respect to the product operation. Proof. Let L, L be two languages in L 3 and assume that L, L are generated by the grammars = (A N, A T, S, P) and = (A N, A T, S, P ), respectively. Without loss of generality, we may assume that A N A N = ; this also implies P P =. Consider the set of productions P 1 = {X uy X uy P} {X us X u P}, which is obtained from P by replacing every production X u by a production X us. 35 / 1

36 Closure Properties of Classes of Languages Proof cont d The type-3 grammar 1 = (A N A N, A T A T, S, P 1 P ) generates the language LL. Indeed, if x L() and y L( ), then S x and S y. Since P was replaced by P 1 in 1, we have S xs. Note that 1 we also have S 1 y. Combining the last two derivations we can write S xy, so LL L( 1 ) / 1

37 Closure Properties of Classes of Languages Proof cont d Conversely, let z L( 1 ). We have S 1 z. This derivation begins with a symbol from A N and must eventually use a production from P since, otherwise, nonterminal symbols cannot be erased. Therefore, the last derivation can be written as S 1 us 1 uv = z, where u A T and v A T. This implies the existence of the derivations S 1 us and S 1 v. Note that the first derivation corresponds to S u; the second corresponds to S v. Thus, u L(), v L( ), and this implies z = uv LL. Therefore, L( 1 ) LL. 37 / 1

38 Closure Properties of Classes of Languages Theorem Each of the classes L i is closed with respect to the product operation. Proof. Follows from previous lemmas. 38 / 1

39 Properties of Type-2 rammars Theorem Let = (A N, A T, S, P) be a context-free grammar. If n X 0 X k 1 α, where X 0,..., X k 1 A N A T and α (A N A T ), then we can write α = α 0 α k 1, where X i n i α i for 0 i k 1 and 0 i k 1 n i = n. Proof. We use an argument by induction on n, n 0. For n = 0, we have α i = X i for 0 i k 1, and the statement is obviously true; in this case, n 0 = = n k 1 = / 1

40 Properties of Type-2 rammars Proof cont d Assume that the statement is true for derivations of length n, and let X 0 X k 1 n+1 α. If X 0 X k 1 n γ α, by the inductive hypothesis, we have n i γ = γ 0 γ k 1, where X i {ni 0 i k 1} = n. γ i for 0 i k 1 and 40 / 1

41 Properties of Type-2 rammars Proof cont d Let Y β be the production applied in the last step γ α. Y occurs in one of the words γ 0,..., γ k 1, say, γ j. In this case, we can write γ j = γ j Y γ j and α can be written as α = α 0 α k 1, where α i = γ i for n j 0 i j 1, and j + 1 i k 1, X j γ j γ j βγ j = α j, which proves the statement. 41 / 1

42 Properties of Type-2 rammars Definition A derivation γ 0 γ 1 γ n in a context-free grammar = (A N, A T, S, P) is complete if γ n A T. Note that if X 0 X k 1 α is a complete derivation in, then every derivation that results from splitting this derivation is also complete. 42 / 1

43 Properties of Type-2 rammars Example Let = (A N, A T, S 0, P) be a context-free grammar, where A N = {S 0, S 1, S 2 }, A T = {a, b}, and P contains the following productions: S 0 as 2, S 0 bs 1, S 1 a, S 1 as 0, S 1 bs 1 S 1, S 2 b, S 2 bs 0, S 2 as 2 S / 1

44 Properties of Type-2 rammars Example cont d We prove that L() consists of all nonnull words over {a, b} that contain an equal number of a s and b s. Recall that n X (α) is the number of occurrences of symbol X in the word α. We will show by strong induction on p, p 1, that 1 if n a (u) = n b (u) = p, then S 0 u; 2 if n a (u) = n b (u) + 1 = p, then S 1 u; 3 if n b (u) = n a (u) + 1 = p, then S 2 u. 44 / 1

45 Properties of Type-2 rammars Example cont d In the first case, for p = 1, we have either u = ab or u = ba; hence, we have either S 0 as 2 ab or S 0 bs 1 ba. For the second case, u = a, and we have S 1 u = b, is similar. a; the third case, for 45 / 1

46 Properties of Type-2 rammars Example cont d Suppose that the statement holds for p n. Again, we consider three cases for the word u: 1 n a (u) = n b (u) = n + 1; 2 if n a (u) = n b (u) + 1 = n + 1; 3 if n b (u) = n a (u) + 1 = n + 1. In the first case, we may have four situations: 1 1. u = abt, where t {a, b} and n a (t) = n b (t) = n, 1 2. u = bat, where t {a, b} and n a (t) = n b (t) = n, 1 3. u = aav with n b (v) = n + 1 and n a (v) = n 1, or 1 4. u = bbw with n a (w) = n + 1 and n b (w) = n / 1

47 Properties of Type-2 rammars Example cont d By the inductive hypothesis, we have S 0 t, and therefore, we obtain one of the following derivations: S 0 as 2 abs 0 abt = u, S 0 bs 1 bas 0 bat = u, for the cases (1 1 ) and (1 2 ), respectively. 47 / 1

48 Properties of Type-2 rammars Example cont d On the other hand, if u = aav, we can write v = v v, where v is the shortest prefix of v, where the number of bs exceeds the number of as. Clearly, we must have n b (v ) = n a (v ) + 1 = n, and therefore, n b (v ) = n a (v ) + 1 = n, where n + n = n + 1. By the inductive hypothesis, we have S 2 v, S 2 v ; hence, S 0 as 2 aas 2 S 2 aav v = u, which concludes the argument for (1 3 ). We leave to the reader the similar arguments for the remaining cases. This allows us to conclude that every word that contains an equal number of a s and b s belongs to L(). 48 / 1

49 Properties of Type-2 rammars Example cont d To prove the reverse inclusion, we justify the following implications: 1 If S 0 n α, then n a (α) + n S1 (α) = n b (α) + n S2 (α). 2 If S 1 n α, then n a (α) + n S1 (α) = n b (α) + n S2 (α) If S 2 n α, then n a (α) + n S1 (α) + 1 = n b (α) + n S2 (α). The proof is by strong induction on n, where n 1. For n = 1, the verification is immediate. For instance, if S 1 α, we have α = a, α = as 0, or α = bs 1 S 1 ; in every case, the equality is satisfied. 49 / 1

50 Properties of Type-2 rammars Example cont d Suppose that the implications hold for derivations no longer than n. n+1 If S 0 α, the first production applied in the derivation is S 0 as 2 or S 0 bs 1. In the first case, we have α = aβ, where S 2 n β, and by the inductive hypothesis, we have n a (β) + n S1 (β) + 1 = n b (β) + n S2 (β), so n a (α) + n S1 (α) = n a (β) n S1 (β) = n b (β) + n S2 (β) = n b (α) + n S2 (α). The second case has a similar treatment. 50 / 1

51 Properties of Type-2 rammars Example cont d n+1 If S 1 α, we have three possibilities. (a) If the first production of the derivation is S 1 a, then α = a and the equality corresponding to this case is obviously satisfied. (b) If the first production is S 1 as 0, we can write α = aβ, where S 0 n β; hence, n a (β) + n S1 (β) = n b (β) + n S2 (β), so n a (α) + n S1 (α) = n a (β) n S1 (β) = n b (β) + n S2 (β) + 1 = n b (α) + n S2 (α) / 1

52 Properties of Type-2 rammars Example cont d (c) If the derivation begins with S 1 bs 1 S 1 we can write α = bβγ, where S 1 p β and S 1 q γ, where p, q n. By the inductive hypothesis, n a (β) + n S1 (β) = n b (β) + n S2 (β) + 1, and n a (γ) + n S1 (γ) = n b (γ) + n S2 (γ) + 1. Consequently, n a (α) + n S1 (α) = n a (β) + n a (γ) + n S1 (β) + n S1 (γ) = n b (β) + n S2 (β) n b (γ) + n S2 (γ) + 1 = n b (α) + n S2 (α) + 1. The case of the derivation S 2 α can be treated in a similar manner. 52 / 1

53 Properties of Type-2 rammars Example cont d Let u L(). From the existence of the derivation S 0 u we obtain n a (u) = n b (u), which shows that L() {x {a, b} n a (x) = n b (x)}. 53 / 1