Grammars and the Pumping Lemma 10/2/19 (Using slides adapted from - PowerPoint PPT Presentation

Grammars and the Pumping Lemma 10/2/19 (Using slides adapted from the book)

Administrivia • No class Friday

A Little English • An article can be the word a or the : A → a A → the • A noun can be the word dog , cat or rat : N → dog N → cat N → rat A noun phrase is an article followed by a noun: P → AN

A Little English • An verb can be the word loves, hates or eats : V → loves V → hates V → eats A sentence can be a noun phrase, followed by a verb, followed by another noun phrase: S → PVP

The Little English Grammar • Taken all together, a grammar G 1 for a small subset of unpunctuated English: S → PVP A → a P → AN A → the V → loves N → dog V → hates N → cat V → eats N → rat • Each production says how to modify strings by substitution • x → y says, substring x may be replaced by y

S → PVP A → a P → AN A → the V → loves N → dog V → hates N → cat V → eats N → rat • Often there is more than one place in a string where a production could be applied • For example, P loves P : – P loves P ⇒ AN loves P – P loves P ⇒ P loves AN • The derivations on the previous slide chose the leftmost substitution at every step, but that is not a requirement • The language defined by a grammar is the set of lowercase strings that have at least one derivation from the start symbol S

S → PVP P → AN V → loves | hates | eats A → a | the N → dog | cat | rat • Often, a grammar contains more than one production with the same left- hand side • Those productions can be written in a compressed form • The grammar is not changed by this • This example still has ten productions

Complaint letter generator https://www.pakin.org/complaint/

4-Tuple Definition • A grammar G is a 4-tuple G = ( V , Σ , S , P ), where: – V is an alphabet, the nonterminal alphabet – Σ is another alphabet, the terminal alphabet , disjoint from V – S ∈ V is the start symbol – P is a finite set of productions, each of the form x → y , where x and y are strings over Σ ∪ V and x ≠ ε

Chapter 11: Non-regular languages 11.1 The Language { a n b n } • • 11.2 The Languages { xx R } • 11.3 Pumping • 11.4 Pumping-Lemma Proofs • 11.5 Strategies • 11.6 Pumping And Finite Languages

Lemma 11.3: The Pumping Lemma for Regular Languages For all regular languages L there exists some integer k such that for all xyz ∈ L with | y | ≥ k , there exist uvw = y with | v | >0, such that for all i ≥ 0, xuv i wz ∈ L. • Let M = ( Q , Σ , δ , q 0 , F ) be any DFA with L ( M ) = L • Choose k = | Q| • Consider any x , y , and z with xyz ∈ L and | y | ≥ k • Let r be a state that repeats during the y part of xyz – We know such a state exists because we have | y | ≥ |Q|… In state r here And again here x y z

Lemma 11.3: The Pumping Lemma for Regular Languages For all regular languages L there exists some integer k such that for all xyz ∈ L with | y | ≥ k , there exist uvw = y with | v | >0, such that for all i ≥ 0, xuv i wz ∈ L. • Let M = ( Q , Σ , δ , q 0 , F ) be any DFA with L ( M ) = L • Choose k = | Q| • Consider any x , y , and z with xyz ∈ L and | y | ≥ k • Let r be a state that repeats during the y part of xyz • Choose uvw = y so that δ *( q 0 , xu ) = δ *( q 0 , xuv ) = r • Now v is pumpable: for all i ≥ 0, δ *( q 0 , xuv i ) = r… In state r here And again here x u v w z

Lemma 11.3: The Pumping Lemma for Regular Languages For all regular languages L there exists some integer k such that for all xyz ∈ L with | y | ≥ k , there exist uvw = y with | v | >0, such that for all i ≥ 0, xuv i wz ∈ L. • Let M = ( Q , Σ , δ , q 0 , F ) be any DFA with L ( M ) = L • Choose k = | Q| • Consider any x , y , and z with xyz ∈ L and | y | ≥ k • Let r be a state that repeats during the y part of xyz • Choose uvw = y so that δ *( q 0 , xu ) = δ *( q 0 , xuv ) = r • Now v is pumpable: for all i ≥ 0, δ *( q 0 , xuv i ) = r • Then for all i ≥ 0, δ *( q 0 , xuv i wz ) = δ *( q 0 , xuvwz ) = δ *( q 0 , xyz ) ∈ F • Therefore, for all i ≥ 0, xuv i wz ∈ L … x u v v v w z

Pumping Lemma Structure For all regular languages L there exists some integer k such that for all xyz ∈ L with | y | ≥ k , there exist uvw = y with | v | >0, such that for all i ≥ 0, xuv i wz ∈ L. • Notice the alternating "for all" and "there exist" clauses: 1. ∀ L … 2. ∃ k … 3. ∀ xyz … 4. ∃ uvw … 5. ∀ i … Our proof showed how to construct the ∃ parts • • But that isn't part of the lemma: it's a black box • The lemma says only that k and uvw exist

Pumping-Lemma Proofs • The pumping lemma is very useful for proving that languages are not regular • For example, { a n b n }…

{ a n b n } Is Not Regular 1. Proof is by contradiction using the pumping lemma for regular languages. Assume that L = { a n b n } is regular, so the pumping lemma holds for L . Let k be as given by the pumping lemma. 2. Choose x , y, and z as follows: x = a k y = b k z = ε Now xyz = a k b k ∈ L and | y | ≥ k as required. 3 Let u , v , and w be as given by the pumping lemma, so that uvw = y , | v | > 0, and for all i ≥ 0, xuv i wz ∈ L. 4 Choose i = 2. Since v contains at least one b and nothing but b s, uv 2 w has more b s than uvw . So xuv 2 wz has more b s than a s, and so xuv 2 wz ∉ L . 5 By contradiction, L = { a n b n } is not regular.