Derivations of a CFG. MACM 300 Formal Languages and Automata. Contextfree Grammars. Derivations and parse trees


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1 Derivations of a CFG MACM 300 Formal Languages and Automata Anoop Sarkar strings grow on trees strings grow on Noun strings grow Object strings Verb Object Noun Verb Object Sentence 3 Contextfree Grammars Derivations and parse trees Set of rules by which val sentences can be constructed. xample: Sentence! Noun Verb Object Noun! trees strings Verb! are grow Object! on Noun Adjective Adjective! slowly interesting What strings can Sentence derive? Syntax only no semantic checking 2 Sentence Noun Verb Object Noun strings grow on trees 4
2 CFG Notation Normal CFG notation! *! + Backus Naur notation ::= * + (an orlist of right hand ses) 5 CFGs and Languages For the regular expression (01)* we know that it corresponds to a language L = {', 01, 0101, , } We know this because we know the definition of the Kleene closure operator * Similarly we define a relation between CFGs and a language as follows: 1. Write down the start variable 2. Find a variable written down so far and a rule that has that variable on the lefthand se of a rule in the CFG. Replace the variable with the righthand se of the rule 3. Repeat step 2 until no variable remains 7 CFG: Formal Definition CFGs and Languages A CFG is a 4tuple (V, ", R, S), where: V is a finite nonempty set of symbols (called variables, or nonterminals) " is a finite set of symbols (called set of terminal symbols, or the alphabet) We generally assume that V # " = $ R is a finite nonempty set of rules, where each rule is of the form: A! w, w % (V & ")* S % V is the start variable (or start nonterminal) We can formally specify the 3 rules for deriving a language from a CFG as follows: If u, v, w are strings from (V & ")* Then if we have a string uav written down And if we have a rule in the CFG, A! w, Then we can replace A with w, written as: uav ( wwv (called a derivation step) 6 8
3 CFGs and Languages Leftmost derivations for + * Conser CFG: A! 0A1 ', here are some strings derived from this grammar: A ( ' A ( 0A1 ( 0'1 = 01 (derived in 0 or more steps: A (* 01) A ( 0A1 ( 00A11 ( 00'11 = 0011 A ( 0A1 ( 00A11 ( 000A111 ( 000'111 = and so on The set of all strings that can be derived is the language of the CFG In this case the language is { 0 n 1 n n ) 0 } 9 ( + ( + ( + * ( + * ( + * (* lm + * + * 11! +! *! ( )! ! Arithmetic xpressions ( * ( + * Leftmost derivations for + * ( + * ( + * ( + * *
4 ! +! *! ( )! ! (* rm + * Rightmost derivation for + * ( * ( * ( + * ( + * ( + * Note that the parse tree is the same as one of the leftmost derivations. * 13 + A grammar G is ambiguous iff there are two parse trees or two distinct leftmost or two distinct rightmost derivations for any string in L(G) Ambiguity Alternatives Massage grammar to make it unambiguous Rely on default parser behavior Augment parser Conser the original ambiguous grammar:! +! *! ( )! ! How can we change the grammar to get only one tree for the input + * 15 Ambiguity Ambiguity Grammar is ambiguous if more than one parse tree is possible for some sentences xamples in nglish: Two sisters reunited after 18 years in checkout counter Ambiguity is not acceptable in PL Unfortunately, it s undecable to check whether a grammar is ambiguous 14 Original ambiguous grammar:! +! *! ( )! ! Unambiguous grammar:! + T T! T * F! T T! F F! ( ) F!  F! Input: + * + T T T * F F F 16
5 Other Ambiguous Grammars Conser the grammar R! R & R R R R * ( R ) a b What does this grammar generate? What s the parse tree for a&b*a Is this grammar ambiguous? First step, add S 0 and then remove epsilon rules S! ASA ab A! B S ' After adding S 0 and then 'removal (remove *): S 0! S S! ASA ab a SA AS S A! B S '* '* Grammar Transformations G is converted to G s.t. L(G ) = L(G) Left Factoring Removing cycles: A ( + A Removing 'rules of the form A! ' liminating left recursion Conversion to normal forms:, A! B C and A! a Greibach Normal Form, A! a * Second step, remove unit rules or chain rules S 0! S S! ASA ab a SA AS S A! B S After removal of unit rules S 0! S and S! S: S 0! ASA ab a SA AS S! ASA ab a SA AS A! B S 18 20
6 After removal of unit rules S 0! S and S! S: S 0! ASA ab a SA AS S! ASA ab a SA AS A! B S After removal of unit rules A! B and A! S: S 0! ASA ab a SA AS S! ASA ab a SA AS A! b ASA ab a SA AS Converting the rhs of each rule to have two nonterminals A! B N 1 C N 2 N 1! a N 2! d After converting to binary form: A! B N 3 N 1! a N 3! N 1 N 4 N 2! d N 4! C N Conser some simpler examples for next two steps Removing terminals from the rhs of rules A! B a C d After removal of terminals from the rhs: A! B N 1 C N 2 N 1! a N 2! d 22 Back to original example After removal of unit rules A! B and A! S: S 0! ASA ab a SA AS S! ASA ab a SA AS A! b ASA ab a SA AS After adding variables: S 0! AA 1 UB a SA AS S! AA 1 UB a SA AS A! b AA 1 UB a SA AS A 1! SA U! a 24
7 NonCF Languages CF Languages The pumping lemma for CFLs [BarHillel] is similar to the pumping lemma for RLs For a string wuxvy in a CFL for u,v! " and the string is long enough then wu n xv n y is also in the CFL for n # 0 Not strong enough to work for every non CF language (cf. Ogden s Lemma) NonCF Languages Contextfree languages and Pushdown Automata Recall that for each regular language there was an equivalent finitestate automaton The FSA was used as a recognizer of the regular language For each contextfree language there is also an automaton that recognizes it: called a pushdown automaton (pda) 26 28
8 Contextfree languages and Pushdown Automata Similar to FSAs there are nondeterministic pda and deterministic pda Unlike in the case of FSAs we cannot always convert a npda to a dpda Similar to the FSA case, a DFA construction proves us with the algorithm for lexical analysis, In this case the construction of a dpda will prove us with the algorithm for parsing (take in strings and prove the parse tree) Summary CFGs can be used describe PL Derivations correspond to parse trees Parse trees represent structure of programs Ambiguous CFGs exist CF languages can be recognized using Pushdown Automata Pushdown Automata PDA has an alphabet (terminals) and stack symbols (like nonterminals), a finitestate automaton, and stack ', '! $ 1, A! ' 0, '! A e.g. PDA for language L = { 0 n 1 n : n >= 0 }! implies a push/pop of stack symbol(s) push stack symbol A ', $! ' 1, A! ' pop stack symbol A check that stack is empty 30
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