Lecture 6 Regular Grammars
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1 Lecture 6 Regulr Grmmrs COT 4420 Theory of Computtion Section 3.3
2 Grmmr A grmmr G is defined s qudruple G = (V, T, S, P) V is finite set of vribles T is finite set of terminl symbols S V is specil vrible clled strt symbol P is finite set of production rules of the form x y where x (V T) +, y (V T) *
3 Liner Grmmrs Grmmrs with t most one vrible t the right side of the production. Exmple: S Sb λ Exmple: S Ab A Ab λ
4 Right-liner Grmmr A grmmr G=(V, T, S, P) is sid to be rightliner if ll productions re of the form: A xb or A x x T*, A,B V string of terminls
5 Left-liner Grmmr A grmmr G=(V, T, S, P) is sid to be leftliner if ll productions re of the form: A Bx or A x x T*, A,B V string of terminls
6 Liner Grmmrs Exmple Right-Liner Grmmr S bs Left-Liner Grmmr S Ab A Ab B B
7 Regulr Grmmrs A regulr grmmr is ny right-liner or leftliner grmmr. G 1 : G 2 : S bs S Ab A Ab B B
8 Regulr Grmmrs Wht bout this Grmmr? S A A B λ B Ab Is this grmmr liner? Is this grmmr regulr? This grmmr is neither right-liner nor left-liner it is not regulr.
9 Regulr grmmrs nd Regulr lnguges Regulr grmmrs generte regulr lnguges. Exmple: G 1 : G 2 : S bs S Ab A Ab B B L(G 1 ) = (b)* L(G 2 ) = b(b)*
10 Theorem Lnguges generted by Regulr Grmmrs = Regulr lnguges Prt 1) Any regulr grmmr genertes regulr lnguge Prt2) Any regulr lnguge is generted by regulr grmmr
11 Proof - Prt 1 Theorem: Let G = (V, T, S, P) be right-liner grmmr. Then L(G) is regulr lnguge. In order to prove this we construct n NFA M such tht L(M) = L(G).
12 Proof - Prt 1 Right-liner grmmr to NFA 1. For every Nonterminl there is stte in our NFA (The strt symbol is the strting stte) 2. For production rule of the form V i 1 2 m V j the utomton will hve trnsitions to connect V i nd V j such tht δ*(v i, 1 2 m ) = V j V i 1 2 m... V j
13 Proof - Prt 1 Right-liner grmmr to NFA 3. For ech production V i 1 2. m the corresponding trnsition will be δ*(v i, 1 2 m ) = V f V f : finl stte V 1 2 m i... V f
14 Right-liner grmmr to NFA Exmple Grmmr G is right-liner: S A S B A B B bb S B A V f B
15 Right-liner grmmr to NFA Exmple Grmmr G is right-liner: S A S B A B B bb S B A V f B
16 Right-liner grmmr to NFA Exmple Grmmr G is right-liner: S A S B A B B bb S B λ A B V f
17 Right-liner grmmr to NFA Exmple Grmmr G is right-liner: S A S B A B B bb S B λ A B V f
18 Right-liner grmmr to NFA Exmple Grmmr G is right-liner: S A S B A B B bb S B λ A V f B b
19 Right-liner grmmr to NFA Exmple Grmmr G is right-liner: S A S B A B B bb S A V f B λ B b
20 S A Right-liner grmmr to NFA Exmple B S => A => B => bb => b A B B bb A S V f λ B b
21 Proof - Prt 1 In the cse of Left-liner grmmr Theorem: Let G = (V, T, S, P) be left-liner grmmr. Then L(G) is regulr lnguge. Proof ide: We construct right-liner grmmr G such tht L(G) = L(G ) R
22 Proof - Prt 1 In the cse of Left-liner grmmr G is left-liner grmmr of the form: A Bv 1 v 2 v k or A v 1 v 2 v k Construct right-liner grmmr G : A v k.v 2 v 1 B A v k.v 2 v 1
23 Proof - Prt 1 In the cse of Left-liner grmmr It is esy to see tht L(G) = L(G ) R Since G is right-liner, L(G ) is regulr lnguge. Therefore, L(G ) R is lso regulr lnguge L(G) is regulr lnguge.
24 Proof Prt 2 Theorem: If L is regulr lnguge on the lphbet Σ, then there exists right-liner grmmr G = (V, Σ, S, P) such tht L = L(G). Let M be the NFA with L = L(M), we construct regulr grmmr G such tht L(M) = L(G)
25 Proof Prt 2 NFA to right-liner grmmr For ny trnsition Add production: q p Vrible Vrible Terminl
26 Proof Prt 2 NFA to right-liner grmmr The strting stte is your strt symbol For ny finl stte Add production: q f λ
27 NFA to right-liner grmmr Exmple q 0 q 1
28 NFA to right-liner grmmr Exmple q 0 q 1 q 1 q 2 q 1 bq 1
29 NFA to right-liner grmmr Exmple q 0 q 1 q 1 q 2 q 1 bq 1 q 2 bq 3
30 NFA to right-liner grmmr Exmple q 0 q 1 q 1 q 2 q 1 bq 1 q 2 bq 3 q 3 q 1
31 NFA to right-liner grmmr Exmple q 0 q 1 q 1 q 2 q 1 bq 1 q 2 bq 3 q 3 q 1 q 3 λ
32 Summry Regulr Lnguges Regulr expressions NFA or DFA Regulr Grmmrs
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