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Lecture 16 Oct 18

Lecture 16 Oct 18. Context-Free Languages (CFL) - basic definitions Examples. Context-Free Languages (Ch. 2). Context-free languages allow us to describe non-regular languages like { 0 n 1 n | n  0}.

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Lecture 16 Oct 18

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  1. Lecture 16 Oct 18 • Context-Free Languages (CFL) - basic definitions • Examples

  2. Context-Free Languages (Ch. 2) Context-free languages allow us to describe non-regular languages like { 0n1n | n0} General idea: CFL’s are languages that can be recognized by automata that have one stack: { 0n1n | n0} is a CFL { 0n1n0n | n0} is not a CFL

  3. Context-Free Grammars Start symbol S with rewrite rules: 1) S  0S1 2) S  e S yields 0n1n : S  0S1  00S11  …  0nS1n  0n1n

  4. Context-Free Grammars (Def.) • A context free grammar G=(V,,R,S) is defined by • V: a finite set variables (non-terminals) • : finite set terminals (with V=) • R: finite set of substitution rules V  (V)* • S: start symbol V • The language of grammar G is denoted by L(G): • L(G) = { w* | S * w }

  5. Derivation * A single step derivation “” consist of the substitution of a variable by a string accordingto a substitution rule. Example: with the rule “ABB”, we can havethe derivation “01AB0  01BBB0”. A sequence of several derivations (or none)is indicated by “ * ”Same example: “0AA * 0BBBB”

  6. Some Remarks The language L(G) = { w* | S * w } contains only strings of terminals, not variables. Notation: we summarize several rules, like A  B A  01 by A  B | 01 | AA A  AA Unless stated otherwise: topmost rule has the start variable on the left side.

  7. Context-Free Grammars (Ex.) Consider the CFG G=(V,,R,S) with V = {S}  = {0,1} R: S  0S1 | 0Z1 Z 0Z |  Then L(G) = {0i1j | ij and j > 0} S yields0j+k1j according to:S  0S1  …  0jS1j  0jZ1j  0j0Z1j …  0j+kZ1j  0j+k1j = 0j+k1j

  8. Importance of CFL • Model for natural languages (Chomsky) • Specification of programming languages: “parsing of a computer program” • parser for HTML (and some special cases of SGML) • Describes mathematical structures. • Intermediate between regular languages and other language families of Chomsky hierarchy

  9. Set of boolean expressions is a CFL Consider the CFG G=(V,,R,S) with V = {S}  = {0,1,(,),,,} R: S  0 | 1 | (S) | (S)(S) | (S)(S) Some elements of L(G): 0 (((0))(1)) (1)((0)(0)) Note: Parentheses prevent “100” confusion. This language requires full-parenthesizing.

  10. A very small subset of English Rules:<SENTENCE>  <NOUN-PHRASE><VERB-PHRASE> <NOUN-PHRASE>  <CMPLX-NOUN> | <CMPLX-NOUN><PREP-PHRASE><VERB-PHRASE>  <CMPLX-VERB> | <CMPLX-VERB><PREP-PHRASE> <CMPLX-NOUN>  <ARTICLE><NOUN> <CMPLX-VERB>  <VERB> | <VERB><NOUN-PHRASE> … <ARTICLE>  a | the<NOUN>  boy | girl | house<VERB>  sees | ignores A string that can be generated by this grammar: the boy sees the girl

  11. ( )  S ( ) S 0 S ( ) ( ) S 0 1 Parse Trees The parse tree of (0)((0)(1)) via rule S  0 | 1 | (S) | (S)(S) | (S)(S): S

  12. Ambiguity A grammar is ambiguous if some strings are derived ambiguously. A string is derived ambiguously if it has more than one leftmost derivations or more than one parse tree. Typical example: rule S  0 | 1 | S+S | SS S  S+S  SS+S  0S+S  01+S  01+1 versus S  SS  0S  0S+S  01+S  01+1

  13. S S S  S S + S 0 1 S S  + S S 0 1 1 1 Ambiguity and Parse Trees The ambiguity of 01+1 is shown by the two different parse trees:

  14. More on Ambiguity The two different derivations: S  S+S  0+S  0+1and S  S+S  S+1  0+1 do not constitute an ambiguous string 0+1 (they will have the same parse tree) Languages that can only be generated by ambiguous grammars are “inherently ambiguous”

  15. Context-Free Languages Any language that can be generated by a contextfree grammar is a context-free language (CFL). The CFL { 0n1n | n0 } shows us that certain CFLs are nonregular languages. Q1: Are all regular languages context free?Q2: Which languages are outside the class CFL?

  16. Example. A context-free grammar for the set of strings over {0,1} with an equal number of 0’s and 1’s. We need a grammar for the language L = { w | w has an equal number of 0’s and 1’s} Consider the grammar: S  0 S 1 | 1 S 0 | e This does not cover all the strings. Exhibit a string in L that is not generated by G.

  17. We need to add the rule S  SS The complete grammar is: S 0S1 | 1S0 | e | SS How do we get a string like 011010? S  SS  S S S  0S1SS  01SS 011S0S  0110S  01101S0  011010 How do we show that L = L(G)? We need the following Lemma: Let x be a string in L. Then, (exactly) one of the following holds: (a) x = 0 y 1 for some y in L, (b) x = 1 y 0 for some y in L, or (c) x = yz for some y and z (both of them non-null) such that both y and z are in L.

  18. Chomsky Normal Form A context-free grammar G = (V,,R,S) is in Chomsky normal form if every rule is of the form A  BC or A  x with variables AV and B,CV \{S}, and x  For the start variable S we also allow the rule S   Advantage: Grammars in this form are far easier to analyze.

  19. Theorem 2.6 Every context-free language can be describedby a grammar in Chomsky normal form. Outline of Proof:We rewrite every CFG in Chomsky normal form.We do this by replacing, one-by-one, every rulethat is not in Chomsky Normal Form (CNF). We have to take care of: Starting Symbol,  symbol, all other violating rules.

  20. Proof Theorem 2.6 Given a context-free grammar G = (V,,R,S),rewrite it to Chomsky Normal Form by 1) New start symbol S0 (and add rule S0S) 2) Remove A rules (from the tail): before: BxAy and A, after: B xAy | xy 3) Remove unit rules AB (by the head): “AB” and “BxCy”, becomes “AxCy” and “BxCy” 4) Shorten all rules to two: before: “AB1B2…Bk”, after: AB1A1, A1B2A2,…, Ak-2Bk-1Bk 5) Replace ill-placed terminals “a” by Ta with Taa

  21. Careful Removing of Rules Do not introduce new rules that you removedearlier. Example: AA simply disappears When removing A rules, insert all newreplacements: BAaA becomes B AaA | aA | Aa | a

  22. Example of CNF Initial grammar: S aSb |  In Chomsky normal form: S0   | TaTb | TaX X  STbS  TaTb | TaX Ta  a Tb  b

  23. RL  CFL Every regular language can be expressed by a context-free grammar. Proof Idea:Given a DFA M = (Q,,,q0,F), we construct a corresponding CF grammar GM = (V,,R,S)with V = Q and S = q0 Rules of GM: qi x(qi,x) for all qiV and all xqi  for all qiF

  24. 0 1 1 0 q1 q2 q3 0,1 Example RL  CFL The DFA leads to thecontext-free grammarGM = (Q,,R,q1) with the rules q1 0q1 | 1q2 q2 0q3 | 1q2 |  q3 0q2 | 1q2

  25. Picture Thus Far ?? context-free languages Regularlanguages { 0n1n }

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