CS 275 Automata and Formal Language Theory. Proof of Lemma II Lemma (II )

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1 CS 275 Automata and Formal Language Theory Course Notes Part II: The Recognition Problem (II) Additional Material Sect II.2.: Basics of Regular Languages and Expressions Anton Setzer (Based on a book draft by J. V. Tucker and K. Stephenson) Dept. of Computer Science, Swansea University csetzer/lectures/ automataformallanguage/current/index.html March 18, 2015 CS 275 Sect. II.2. (Additional Material) 1/ 26 CS 275 Sect. II.2. (Additional Material) 2/ 26 Proof of Lemma II Lemma (II ) 1. Assume a grammar G which has only productions of the form A Bw or A w for some w T, A, B N. Then L(G) = L(G ) for some left-linear grammar G, which can be computed from G. 2. Assume a grammar G which has only productions of the form A wb or A w for some w T, A, B N. Then L(G) = L(G ) for some right-linear grammar G, which can be computed from G. CS 275 Sect. II / 26 CS 275 Sect. II / 26

2 Proof of Lemma II Proof In a first step we omit all transitions A B for A, B N: Let G = (N, T, S, P) be a grammar having such transitions. We form a grammar G having no such transitions as follows: grammar G terminals nonterminals start symbol productions N T S A w if A G A w for some A, A N, w T A wb if A G A wb for some A, A, B N, w T So in G we just jump over all silent transitions A B in G. We can in fact decide whether A A, since such a derivation must have the form A = A or A = A 1 A 2 A n = A for some A i N. And if such derivation exists then a derivation exists in which all A i are distinct (omit loops). Therefore n can be resricted to the number of elements in N, and therefore there are only finitely many possible derivations, which we can enumerate. For each of them we can check whether it is in fact a derivation, and therefore determine all possible derivaitons A A. Proof CS 275 Sect. II / 26 CS 275 Sect. II / 26 End of Proof of II Now one can easily see that for w T S G w iff S G w We have now obtained a grammar which doesn t contain productions of the form A B for nonterminals A, B. silent The following lemma shows that such languages are definable by left-linear or right-linear grammars. CS 275 Sect. II / 26 CS 275 Sect. II / 26

3 Proof of Lemma II Proof of Lemma II Lemma (II ) 1. Assume a grammar G which has only productions of the form A Bw or A w for some w T +, w T, A, B N. Then L(G) = L(G ) for some left-linear grammar G, and G can effectivly computed from G. 2. Assume a grammar G which has only productions of the form In (2) replace Productions A a 1 a 2 a n B with n 2 by A a 1 A 1, A 1 a 2 A 2,..., A n 1 a n B for some new nonterminals A i. Productions A a1 a 2 a n with n 2 by A a 1 A 1, A 1 a 2 A 2,..., A n 1 a n for some new nonterminals A i. (1) is proved similarly. A wb or A w for some w T +, w T, A, B N. Then L(G) = L(G ) for some right-linear grammar G, and G can effectivly computed from G. CS 275 Sect. II / 26 Derivations in Regular Grammars Theorem (a) Let G = (N, T, S, P) be a left-linear grammar, A N, w (N T ), A w. Then the derivation of A w is CS 275 Sect. II / 26 Derivations in Regular Grammars Theorem (b) Let G = (N, T, S, P) be a right-linear grammar, A N, w (N T ), A w. Then the derivation of A w is A A 1 a 1 A 2 a 2 a 1 A n a n a 2 a 1 = w (1) or A A 1 a 1 A 2 a 2 a 1 A n a n a 2 a 1 (2) a n+1 a n a 2 a 1 = w or A A 1 a 1 A 2 a 2 a 1 A n a n a 2 a 1 a n a 2 a 1 = w (3) or or A a 1 A 1 a 1 a 2 A 2 a 1 a 2 a n A n = w (1) A a 1 A 1 a 1 a 2 A 2 a 1 a 2 a n A n (2) a 1 a 2 a n a n+1 = w A a 1 A 1 a 1 a 2 A 2 a 1 a 2 a n A n (3) a 1 a 2 a n = w for productions A i A i+1 a i+1 (in (1) (3)), A n a n+1 (in (2)) A n ɛ (in (3)) for productions A i a i+1 A i+1 (in (1) - (3)) A n a n+1 (in (2)) A n ɛ (in (3)). CS 275 Sect. II / 26 CS 275 Sect. II / 26

4 Proof The above are the only derivations possible. CS 275 Sect. II / 26 Proof of Lemma II CS 275 Sect. II / 26 Proof of Lemma II Lemma (II ) Let G, G be both left-linear grammars or both right-linear grammars. Then we can define a left-linear or right-linear grammars G i s.t. 1. L(G 1 ) = L(G) L(G ), 2. L(G 2 ) = L(G).L(G ), 3. L(G 3 ) = L(G). These grammars can be computed from G and G. Assume in 1./2./3. G = (T, N, S, P), G = (T, N, S, P ). After renaming of nonterminals we can assume N N =. Let S be a new symbol not in N N T T. We define multi-step left/right-linear grammars with those properties, from which one can construct ordinary (one-step) left/right-linear grammars with those properties. We only carry out the proof for right-linear grammars. CS 275 Sect. II / 26 CS 275 Sect. II / 26

5 Proof of 1. Proof of 1. We define G 1 as follows: grammar G 1 terminals T T nonterminals N N {S } start symbol S productions Proof of 2. S S S S P P CS 275 Sect. II / 26 So G 1 has the productions from G and G plus Derivations in G 1 have the form and for derivations and S S and S S. S S w S S w S G w S G w So for w (T T ) we have S G 1 w iff S G w or S G w, so L(G ) = L(G) L(G ). Proof of 2. CS 275 Sect. II / 26 We define G 2 as follows: grammar G 2 terminals T T nonterminals N N start symbol S productions A aa for A aa P (A, A N, a T ) A as for A a P (A N, a T ) P So G 2 has the productions from G, the productions of the form A aa from G and productions A as, if A a is a production from G. A derivation in G 2 starts with a derivation S a 1 A 1 a 1 a 2 A 2 a 1 a 2 a 3 A 3 a 1 a 2 a n 1 A n 1 a 1 a 2 a n S for derivations in G of the form S a 1 A 1 a 1 a 2 A 2 a 1 a 2 a 3 A 3 a 1 a 2 a n 1 A n 1 a 1 a 2 a n. CS 275 Sect. II / 26 CS 275 Sect. II / 26

6 Proof of 2. Proof of 3. Then this is followed by a derivation a 1 a 2 a n S a 1 a 2 a n b 1 B 1 a 1 a 2 a n b 1 b 2 B 2 a 1 a 2 a n b 1 b 2 b m 1 B m 1 a 1 a 2 a n b 1 b 2 b m, We define G 3 as follows: grammar G 3 terminals T for a derivation in G of the form S b 1 B 1 b 1 b 2 B 2 b 1 b 2 b m 1 B m 1 b 1 b 2 b m nonterminals start symbol N S Therefore S G 2 w for some w (T T ) if and only if S G 1 w and S G 2 w for some w, w s.t. w = ww. So L(G 2 ) = L(G).L(G ). productions S ɛ, A aa for A aa P (A, A N, a T ) A as for A a P (A N, a T ) Proof of 3. CS 275 Sect. II / 26 Derivations in G 3 are S ɛ or they start similarly as for concatenation with S ws for a derivation in G S w and w N +. In the latter case it can continue either (using S ɛ) with ws w or with ws ww S for a derivation in G S w Again in the latter case we can continue (using S ɛ) with ww S ww or with ww S ww w S for a derivation in G S w CS 275 Sect. II / 26 Proof of 3. CS 275 Sect. II / 26 We obtain that in G 3 we have if there exist derivations in G of S w 1 S w 2 S w n s.t. w = w 1 w 2 w n. So we get S w L(G 3 ) = {w 1 w 2 w n n 0, w 1,..., w n L(G)} = L(G) CS 275 Sect. II / 26

7 Proof of Lemma II Proof of Lemma II Lemma (II ) Let E be a regular Expression. Then there exist both left-linear and right-linear grammars G, G s.t. G and G can be computed from L. L(E) = L(G) = L(G ) Proof: By Lemma II.2.2.1, and the fact that the finite languages, {ɛ} and {a} are regular. Induction on the definition of regular expressions. Case 1: L =, ɛ, a (where a T ). Then L is finite, therefore definable by a left/right-linear grammar. Case 2: L = (L 1 ) (L 2 ) or L = (L 1 )(L 2 ) or L = (L 1 ). By IH L i are defined by left/right-linear grammars G i. By Lemma II it follows that L can be defioned by a left/right-linear grammar. CS 275 Sect. II / 26 CS 275 Sect. II / 26