Positive Properties are closed under: Union of Context-Free - PDF document

Review • Languages and Grammars – Alphabets, strings, languages • Regular Languages CS 301 - Lecture 18 – Deterministic Finite and Nondeterministic Automata – Equivalence of NFA and DFA – Regular Expressions Properties of Context Free – Regular Grammars – Properties of Regular Languages – Languages that are not regular and the pumping lemma Grammars • Context Free Languages – Context Free Grammars Fall 2008 – Derivations: leftmost, rightmost and derivation trees – Parsing and ambiguity – Simplifications and Normal Forms – Nondeterministic Pushdown Automata – Pushdown Automata and Context Free Grammars – Deterministic Pushdown Automata – Pumping Lemma for Context Free Grammars • Today: – Closure and Decidable Properties for Context Free Languages Union Context-free languages Positive Properties are closed under: Union of Context-Free languages L is context free 1 L ∪ L 1 2 L is context free is context-free 2 1

Example In general: L 1 , L Language Grammar For context-free languages 2 with context-free grammars G 1 , G 2 n b n L = { a } S → aS b | and start variables λ 1 S 1 , S 1 1 2 R L = { ww } S → aS a | bS b | λ 2 2 2 2 L ∪ L The grammar of the union 1 2 Union S has new start variable and additional production S → S 1 | S n n R 2 S → S 1 | S L { a b } { ww } = ∪ 2 Example Concatenation Language Grammar Context-free languages n b n Concatenation are closed under: L = { a } S → aS b | λ 1 1 1 R L = { ww } S → aS a | bS b | λ 2 L is context free 2 2 2 1 L 1 L 2 Concatenation L is context free is context-free 2 n n R S → S 1 S L = { a b }{ ww } 2 2

In general: Star Operation L 1 , L For context-free languages 2 with context-free grammars G 1 , G 2 and start variables S 1 , S Context-free languages 2 are closed under: Star-operation L 1 L The grammar of the concatenation 2 S has new start variable * L is context free L is context-free and additional production S → S 1 S 2 Example In general: For context-free language L Language Grammar with context-free grammar G and start variable n b n S L = { a } S → aSb | λ Star Operation The grammar of the star operation L * S has new start variable 1 and additional production S → SS | n b n λ L = { a } * S → SS | 1 1 λ 1 1 3

Intersection Negative Properties Context-free languages are not closed under: intersection of Context-Free Languages L is context free 1 L ∩ L 1 2 L is context free not necessarily 2 context-free Example Complement n n m n m m L = { a b c } L = { a b c } 1 2 Context-free: Context-free: Context-free languages S AC S AB are not closed under: complement → → A aAb | A aA | → λ → λ C cC | B bBc | → λ → λ is context free L not necessarily L context-free Intersection n n n L L { a b c } NOT context-free ∩ = 1 2 4

Example n n m n m m L = { a b c } L = { a b c } Intersection 1 2 Context-free: Context-free: of S AC S AB Context-free languages → → A aAb | A aA | and → λ → λ C cC | B bBc | Regular Languages → λ → λ Complement n n n L L L L { a b c } ∪ = ∩ = 1 2 1 2 NOT context-free The intersection of Machine M Machine M 1 2 a context-free language and DFA for L NPDA for L 2 a regular language 1 regular is a context-free language context-free Construct a new NPDA machine M L context free that accepts L ∩ L 1 1 2 L ∩ L 1 2 regular L M M context-free M simulates in parallel and 2 1 2 5

NPDA M DFA M NPDA M DFA M 1 2 1 2 a , b c , b → c a → λ q q q q p p p 1 2 1 2 1 2 1 transition transition transition NPDA NPDA M M a , b c , b → c → λ q 1 , p q 1 , p q 2 , p q 2 , p 1 1 2 1 transition transition NPDA M DFA M NPDA M DFA M 1 2 1 2 q p q p p 1 1 0 2 0 initial state final state initial state final states NPDA NPDA M M q 0 , p q 1 , p q 1 , p 0 1 2 Initial state final states 6

Example: context-free regular * * * L { w w : | w | | w |, w { a , b } , w { c , d } } = = ∈ ∈ L = { a , c } 1 1 2 1 2 1 2 2 NPDA M M 1 DFA 2 a , 1 c , 1 a , c λ → → λ b , 1 d , 1 λ → → λ p , , , 0 λ λ → λ λ λ → λ λ λ → λ q q q q 0 1 2 3 context-free In General: n n Automaton for: L L { a c : n 0 } ∩ = ≥ 1 2 M M M simulates in parallel and 1 2 NPDA M M accepts string if and only if w a , 1 c , 1 λ → → λ M accepts string and w 1 accepts string M w , , , λ λ → λ λ λ → λ λ λ → λ 2 q 0 , p q 1 , p q 2 , p q 3 , p 0 0 0 0 L ( M ) L ( M ) L ( M ) = ∩ 1 2 7

Therefore: is NPDA M Applications of L ( M ) L ( M ) is context-free ∩ Regular Closure 1 2 L ∩ L is context-free 1 2 The intersection of An Application of Regular a context-free language and Closure a regular language is a context-free language n n Prove that: L { a b : n 100 , n 0 } = ≠ ≥ Regular Closure is context-free L context free 1 L ∩ L 1 2 regular L context-free 2 8

We also know: We know: n n 100 100 { a b : n 0 } is context-free ≥ L = { a b } is regular 1 * 100 100 is regular L {( a b ) } { a b } = + − 1 n b n * 100 100 { a } L {( a b ) } { a b } = + − 1 Another Application of Regular context-free regular Closure L { w : n n n } Prove that: = = = a b c n n { a b } L (regular closure) context-free ∩ 1 is not context-free n n n n { a b } L { a b : n 100 , n 0 } L ∩ = ≠ ≥ = 1 is context-free 9

L { w : n n n } is context-free If = = = a b c (regular closure) Decidable Properties n n n Then L { a * b * c *} { a b c } ∩ = of Context-Free Languages context-free regular context-free Impossible!!! Therefore, is not context free L Empty Language Question: Membership Question: for context-free grammar G for context-free grammar G find if string w ∈ L ( G ) find if L ( G ) = ∅ Membership Algorithms: Parsers Algorithm: 1. Remove useless variables • Exhaustive search parser • CYK parsing algorithm S 2. Check if start variable is useless 10

Infinite Language Question: Example: S AB → G for context-free grammar A aCb | a → L ( G ) find if is infinite B bB | bb → Algorithm: C cBS → 1. Remove useless variables Infinite language Dependency graph 2. Remove unit and productions λ 3. Create dependency graph for variables A C 4. If there is a loop in the dependency graph S then the language is infinite B S AB → What’s Next A aCb | a → • Read B bB | bb – Linz Chapter 1,2.1, 2.2, 2.3, (skip 2.4), 3, 4, 5, 6.1, 6.2, (skip 6.3), 7.1, 7.2, 7.3, (skip → 7.4), 8, and 9.1 – JFLAP Chapter 1, 2.1, (skip 2.2), 3, 4, 5, 6, 7, (skip 8), 9.1, 9.2 C cBS → • Next Lecture Topics From 9.1 – Turing Machines • Quiz 3 in Recitation on Wednesday 11/12 S AB aCbB acBSbB acbbSbbb – Covers Linz 7.1, 7.2, 7.3, (skip 7.4), 8, and JFLAP 5,6,7 ⇒ ⇒ ⇒ ⇒ – Closed book, but you may bring one sheet of 8.5 x 11 inch paper with any notes you like. – Quiz will take the full hour ∗ ∗ 2 2 Homework • S acbbSbbb ( acbb ) S ( bbb ) ⇒ ⇒ – Homework Due Today – New Homework Available Friday – New Homework Due Next Thursday ∗ i i ( acbb ) S ( bbb ) ⇒ 11