Specialized Grammars. Overview. Linear Grammars. Linear Grammars and Normal Forms. Yet Another Way to Specify Regular Languages

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1 Overview Specialized Grammars Linear Grammars and Normal Forms. Introduce right-linear and left-linear grammars and use these to prove that regular languages are a proper subset of context-free languages. Friday, November 5, 2 Reading: Sipser 2.; Stoughton 4.4, ; Kozen 2 S235 and Automata Department of omputer Science Wellesley ollege ontext-free Regular 2. Introduce homsky Normal Form and Greibach Normal Form and show how to convert any context-free grammars to these forms. Specialized Grammars 24-2 Yet Another Way to Specify Regular Linear Grammars Reg = Regular Deterministic Finite Automaton Nondeterministic Finite Automaton ** ()* Regular Expression Right-Linear Grammar **+()* FL = ontext-free Language ontext-free Grammar Nondeterministic Pushdown Automaton Dec = Recursive (Turing-Decidable) Language Turing Machine Unrestricted Grammar RE = Recursively Enumerable (Turing-Recognizable/Acceptable) Language Lan = All n n ww R n n 2 n ww A FG is linear iff every production has at most one variable in its RHS. A FG is right-linear iff every production has one of these two forms, where V and W are any variables and x is any string of terminals. A FG is left-linear iff every production has one of these two forms, where V and W are any variables and x is any string of terminals. V % V xw V % V Wx Via a simple transformation, we can also include productions with the form V y, where y is a string of terminals. How? Specialized Grammars 24-3 Specialized Grammars 24-4

2 Right-Linear FGs Generate All Regular () We ll show that every right-linear grammar can be converted to an FA, and so generates a regular language. S % T U T T U V U % S V V % V (2) We ll show that every FA (and thus every regular language) can be converted to a right-linear grammar. A B FA right-linear grammar ontext-free Regular Together, () and (2) imply that regular languages are a subset of the context-free languages. Non-regular FLs (like n n ) show that the subset relation is proper. Specialized Grammars 24-5 onverting Right-Linear Grammars to FAs S % T U T T U V U % S V V % V () The states of the FA are named by the variables of the FG. (2) The start state is the state named by the start variable. (3) A state Q is an accepting state iff there is a production of the form Q %. (4) There is a transition (P, x, Q) for each production P xq. S T U V Specialized Grammars 24-6 onverting FAs to Right-Linear Grammars A B A B B % A B % A () The variables of the FG are the states of the FG. (2) The nonterminals of the FG are the transition symbols in the FA. (3) The start variable of the FG is the FA s start state. (4a) There is a production Q % for each accepting state Q. What About Left-Linear Grammars? We have seen that a language is regular iff it can be expressed via a right-linear grammar. It turns out that the same holds for left-linear grammars: A language is regular iff it can be expressed via a left-linear grammar. P P Q R (You will work out the details on PS8.) Q % P B R R % Q R B A left-linear grammar FA (4b) There is a production P xq for each transition (P, x, Q). areful! Regularity is not guaranteed if right-linear and left-linear productions are mixed! E.g.: A % B B A Specialized Grammars 24-7 Specialized Grammars 24-8

3 homsky Normal Form (NF) Sometimes it s helpful to require a FG to be in a standard form. E.g., we ll see this next time in regards to a pumping lemma for FLs. One such form is homsky Normal Form (NF), in which all productions must have one of the following two forms: V UW V t Variable rewrites to two variables; U,W can t be start variable Variable rewrites to a single terminal t homsky Normal Form can generate any FL not containing %. In order to allow languages with %, we also allow the production: homsky Normal Form: Example S % AB A A B DB D AB A Here s a sample grammar in homsky Normal Form. Draw derivations for the 3 shortest strings in the language generated by this grammar. S % % production allowed only for start variable S Intuitions: Parse trees for NF have variables arranged in binary trees. Every step in a derivation from a NF grammar makes nontrivial progress toward the terminal string. an t have subtrees yielding % or long sequences of unit productions: A B Specialized Grammars 24-9 Specialized Grammars 24- FG to NF, Step : Add New Start Variable* Introduce a new start variable that rewrites to the given one. (Guarantees that the new start variable does not occur in a RHS.) Our running example (what language does it generate?): S S FG to NF, Step 2: Find Nullable Variables A variable is nullable iff it can yield %. To find all nullable variables, first color every %. Then repeat the following steps until there are no changes (an iterative fixed point technique) S S color a LHS variable if one of its RHSs is completely colored; color every occurrence of a colored LHS variable in a RHS. S S S S S S * Sipser does this, but Stoughton and Kozen do not (because they don t handle languages containing %). Specialized Grammars 24- S S S S So the nullable variables in our example are: {S, S, Q} Specialized Grammars 24-2

4 FG to NF, Step 3: Remove Nullable Variables For each nullable V and production W!V" (where at least one of! or " is nonempty), add the production W!". If S is nullable, add the production S %. (c) Remove all productions of the form V %, where V # S. Nullable variables = {S, S, Q} S S S S ba Q S b, (c) S % S ba Q QS Q S b Specialized Grammars 24-3 FG to NF, Step 4: Remove Unit Productions A unit production is one of the form V W. For each rewrite sequence V $* V n $!, where! isn t a single variable, add a production V! (if it doesn t already exist). Remove all unit productions. S % S ba Q QS Q S b S % S ar QS bsa ba ba ar QS Q QS Q S ar bsa ba b S % ar QS bsa ba S ar QS bsa ba Q ar QS bsa ba b Specialized Grammars 24-4 FG to NF, Step 5: ompleting NF Every production now has one of the following three forms: () V t (2) V!, where! has at least two variables and/or terminals. (3) S % Introduce new variables and productions to replace all productions of form (2) not in NF by NF productions. S % ar QS bsa ba S ar QS bsa ba Q ar QS bsa ba b In this example, we cleverly added only three new variables, but a straightforward implementation would add many more. S % TR QS UV UT P TR Specialized Grammars 24-5 Simplification A variable in a FG is useless if it can never appear in a parse tree rooted at the start symbol. It is always safe to simplify a FG by removing all productions that mention a useless variable. S % TR QS UV UT P TR P is useless See Stoughton 4.4 for details on simplification. S % TR QS UV UT Specialized Grammars 24-6

5 NF in Forlan - val Lgram = Gram.input "L.gram"; val Lgram = - : gram - Gram.output ("", Lgram); A, B, S S A -> % A; B -> % B; S -> AB - val LgramNF = Gram.chomskyNormalForm Lgram; val LgramNF = - : gram - Gram.output ("", LgramNF); <,A>, <,B>, <,S>, <2,>, <2,>, <3,A>, <3,B> <,S> <,A> -> <2,><2,> <2,><3,A>; <,B> -> <2,><2,> <2,><3,B>; <,S> -> <,A><,B> <2,><2,> <2,><3,A> <2,><2,> <2,><3,B>; <2,> -> ; <2,> -> ; <3,A> -> <,A><2,>; <3,B> -> <,B><2,> Forlan Grammar anonicalization - Gram.output ("", (Gram.renameVariablesanonically LgramNF)); A, B,, D, E, F, G A -> DE DF; B -> ED EG; -> AB DE DF ED EG; D -> ; E -> ; F -> AE; G -> BD (* Note: Forlan s NF *cannot* generate % even though it was in the original language! *) Using the above NF grammar, what is the parse tree for? Specialized Grammars 24-7 Specialized Grammars 24-8 Greibach Normal Form (GNF) Another normal form for FGs is Greibach Normal Form (GNF), in which every production (except for S %) has the form: V tv V 2 V n, n! (where t is a single terminal) GNF has the nice property that every rewrite (except S %) makes progress toward the final string by adding a terminal. Not-quite-right idea: To obtain GNF, start with NF, and create a GNF production of the above form for every collection of NF productions with the following form: V W n V n W n W n- V n- W 2 W V W t V tv V 2 V n W n tv V 2 V n- W 2 tv W t In fact, it s more complicated than this. See Kozen Lecture 2 for details. Specialized Grammars 24-9 Greibach Normal Form: Example S % TR QS UV UT S % ar ars bvs bts S ar ars bvs bts R aru arsu bvsu btsu bvu btu b V art arst bvst btst bvt btt expand initial T, U expand initial S and remove Q (now useless) * This step is glossing over some details. S % ar QS S ar QS Q ar QS expand initial Q * S % ar ars bvs bts S ar ars bvs bts Q ar ars bvs bts Specialized Grammars 24-2

Languages and Automata

Languages and Automata What are the Big Ideas? Tuesday, August 30, 2011 Reading: Sipser 0.1 CS235 Languages and Automata Department of Computer Science Wellesley College Why Take CS235? 1. It s required