Direct automaton construction for Parikhs Theorem Abhishek Mukati, - PowerPoint PPT Presentation

Direct automaton construction for Parikh’s Theorem Abhishek Mukati, Rishi Tulsyan, Shubham Waghmare November 23, 2018 Abhshek, Rishi, Shubham Automata Construction November 23, 2018 1 / 61

Parikh’s Theorem Theorem Every CFL has the same Parikh image as some regular language. Parikh’s proof of theorem Given a CFG G, the proof produces (implicitly) an automaton or regular expression whose language has the same Parikh image as G. Constructed automata is of size O(n n ), where n - number of variables in the CNF of G. Enhanced automaton construction Given a CFG G, construct an automaton explicitly. Instead of O(n n ), only O(4 n ) states. Abhshek, Rishi, Shubham Automata Construction November 23, 2018 2 / 61

Terms and Notations Context Free Grammar G=( V , T , P , S ) V is set of variables, denoted as A 1 , A 2 , A 3 ,...... T is a set of terminal symbols, denoted as a, b, c..... S is the start variable. → α where α ∈ (V U T) ∗ P is a set of productions of the form A i − If w is a word over some T, we denote by Π T (w) the Parikh image of w over alphabet T. Π T (w)maps a character in T to its number of occurrences in w. Abhshek, Rishi, Shubham Automata Construction November 23, 2018 3 / 61

Terms and Notations Cont. Context Free Grammar G=( V , T , P , S ) The Parikh image of a language L over T is { Π T ( w ) | w ∈ L } . It is denoted by Π T (L). Examples :- Π { a , b , c } (bccba) = (1, 2, 2) where (1, 2, 2) stands for { ( a , 1) , ( b , 2) , ( c , 2) } Π { a , b , c } (cabaaabb) = (4, 3, 1) α / V means α projected on V α / T means α projected on T A pair ( α, β ) is a step , denoted by α ⇒ β , if there exists a production → γ and α 1 , α 2 ∈ ( V ∪ T) ∗ such that: α = α 1 A α 2 and β = α 1 γα 2 . A − Transition t( α ⇒ β ) = (Π V ( α ), γ / T , Π V ( β )) where, α , β ∈ ( V ∪ T ) ∗ Abhshek, Rishi, Shubham Automata Construction November 23, 2018 4 / 61

k-Parikh automaton of G The k-Parikh automaton of G is the NFA M k G = (Q, T ∗ , δ , q 0 , q f ) n= | V | Q= { (x 1 .... x n ) ∈ N n | � n i =1 x i ≤ k } δ = { t( α ⇒ β ) | α ⇒ β is a step and Π V ( α ), Π V ( β ) ∈ Q } q f = Π V ( ǫ ) = (0, ..... , 0) q 0 = Π V (S); Abhshek, Rishi, Shubham Automata Construction November 23, 2018 5 / 61

3-Parikh Automaton Example A 1 − → A 1 A 2 | a A 2 → bA 2 aA 2 | cA 1 Transition example: (0, 2) c − → (1, 1) A 2 A 2 → cA 1 A 2 Abhshek, Rishi, Shubham Automata Construction November 23, 2018 6 / 61

Example Abhshek, Rishi, Shubham Automata Construction November 23, 2018 7 / 61

Theorem If G is a context-free grammar with n variables and degree m, then L(G) and L(M nm +1 ) have same Parikh image. G the degree of G denoted by m := -1 + max {| γ / V | ( A → γ ) ∈ P } ;i.e., m+1 is the maximal number of variables on the right hand sides. For the above grammar G, n=2 and m=1, which means L(G) = Π L(M 3 G ). Abhshek, Rishi, Shubham Automata Construction November 23, 2018 8 / 61

Proof of L ( M k G ) ⊆ Π L ( G ) Claim σ → q be a run of M k G on the word σ ∈ T ∗ , then there exists If k ≥ 1, let q 0 − a step sequence S ⇒ ∗ α satisfying Π V ( α ) = q Π T ( α ) = Π T ( σ ) Proof of claim (by induction on length of path): If l = 0 then σ = ǫ , so α = S and Π V ( S ) = q 0 and Π T ( S ) = Π T ( ǫ ) → q ′ γ σ ′ If l > 0, let σ = σ ′ γ and q 0 − − → q , then according to I.H.: S ⇒ ∗ α ′ where Π V ( α ′ ) = q ′ and Π T ( α ′ ) = Π T ( σ ′ ) Since q ′ γ → q is a transition there exists a step α 1 A α 2 ⇒ α 1 γ ′ α 2 and − a production A → γ ′ where Π V ( α 1 A α 2 ) = q ′ , Π V ( α 1 γ ′ α 2 ) = q and γ ′ / T = γ Abhshek, Rishi, Shubham Automata Construction November 23, 2018 9 / 61

Claim’s proof cont. Since Π V ( α ′ ) = q ′ = Π V ( α 1 A α 2 ), α ′ = α ′ 1 A α ′ 2 for some α ′ 1 , α ′ 2 2 so that S ⇒ ∗ α ′ ⇒ α and Let α = α ′ 1 γ ′ α ′ = Π V ( α ′ 1 γ ′ α ′ Π V ( α ) 2 ) = Π V ( α ′ 1 A α ′ 2 ) − Π V ( A ) + Π V ( γ ′ ) = Π V ( α ′ ) − Π V ( A ) + Π V ( γ ′ ) = Π V ( α 1 A α 2 ) − Π V ( A ) + Π V ( γ ′ ) = Π V ( α 1 γ ′ α 2 ) = q = Π T ( α ′ 1 γ ′ α ′ Π T ( α ) 2 ) = Π T ( α ′ 1 A α ′ 2 ) + Π T ( γ ′ ) = Π T ( α ′ ) + Π T ( γ ′ ) = Π T ( α ′ ) + Π T ( γ ) = Π T ( σ ′ ) + Π T ( γ ) = Π T ( σ ) Abhshek, Rishi, Shubham Automata Construction November 23, 2018 10 / 61

Proof of L ( M k G ) ⊆ Π L ( G ) from the claim σ If σ ∈ L ( M k G ), then there is a run q 0 − → Π V ( ǫ ) From the claim, there exists a step sequence S ⇒ ∗ α satisfying Π V ( α ) = Π V ( ǫ ) = (0 , 0 , ..., 0) and Π T ( α ) = Π T ( σ ) So α ∈ T ∗ and hence α ∈ L ( G ) Since Π T ( α ) = Π T ( σ ), we have α = Π σ Abhshek, Rishi, Shubham Automata Construction November 23, 2018 11 / 61

Notations and Definitions Yield of a parse tree t is denoted by Y ( t ) Set of yields of a set T of trees by Y ( T ) A child of t is a subtree of t whose root is a child of the root of t . A child is proper if its root is not a leaf. The dimension d ( t ) of a parse tree is defined as follows : If t has no proper children then d ( t ) = 0. Otherwise let t 1 , t 2 , ... t r be the proper children of t sorted s . t . d ( t 1 ) ≥ d ( t 2 ) ≥ ... ≥ d ( t r ). Then � d ( t 1 ) if r = 1 or d ( t 1 ) > d ( t 2 ) d ( t ) = d ( t 1 ) + 1 if d ( t 1 ) = d ( t 2 ) The set of parse trees with dimension k is denoted by T ( k ) Fact 2.1 : The height of a tree, h ( t ) > d ( t ). Abhshek, Rishi, Shubham Automata Construction November 23, 2018 12 / 61

Proof idea of L ( G ) ⊆ Π L ( M nm +1 ) G Definition A derivation S ⇒ α 0 ⇒ ... ⇒ α l has index k if ∀ i ∈ { 0 , ..., l } , the word ( α i ) / V has length atmost k . The set of words derivable through derivations of index k is denoted by L k ( G ). Clearly L 1 ( G ) ⊆ L 2 ( G ) ⊆ L 3 ( G ) ⊆ ... and L ( G ) = � k ≥ 1 L k ( G ) We can prove L ( G ) ⊆ Π L ( M nm +1 ) by proving G L ( G ) ⊆ Π L nm +1 ( G ) (Collapse Lemma) -proved using � n i =0 Y ( T ( i ) ) Y ( T ) ⊆ Π Y ( T ( k ) ) ⊆ L km +1 ( G ) L k ( G ) ⊆ Π L ( M k G ) Abhshek, Rishi, Shubham Automata Construction November 23, 2018 13 / 61

Lemma 2.1 Lemma 2.1 Y ( T ) ⊆ Π ∪ n i =0 Y ( T ( i ) ) Π (Yields of all the parse trees) ⊆ Π ( Union of Yields of all the parse trees of dimension i, where (i=0 to n)) Proof Idea For every parse tree t ∈ T , we can find a parse tree t’ ∈ T ( i ) , such that: Π ( Y ( t )) = Π ( Y ( t’ )) d ( t ′ ) = i ≤ n . (n = | V | ) Abhshek, Rishi, Shubham Automata Construction November 23, 2018 14 / 61

Lemma 2.1: Proof - Preliminaries In this proof, we write t = t 1 t 2 to denote that t 1 is a parse tree except that exactly one leaf is labelled by a variable, say A 2 , instead of a terminal. t 2 is a parse tree with root A 2 . t is obtained from t 1 and t 2 by replacing the leaf A 2 of t 1 by the tree t 2 . Figure: Tree t. Figure: Tree t1 and t2. Abhshek, Rishi, Shubham Automata Construction November 23, 2018 15 / 61

Lemma 2.1: Proof - Preliminaries Ω − equivalence Two trees t, t’ are Ω -equivalent if: They have same number of nodes. The sets of variables in t and t ′ coincide. Y ( t ) = Π Y ( t’ ) Compact tree A tree t is compact if d ( t ) ≤ K ( t ). where K ( t ) denotes the number of variables that appear in t . K ( t ) ≤ n . Why? Proof Idea For every parse tree t ∈ T , we can find a parse tree t’ ∈ T ( i ) , such that: Π ( Y ( t )) = Π ( Y ( t’ )) d ( t ′ ) = i ≤ n . (n = | V | ) Abhshek, Rishi, Shubham Automata Construction November 23, 2018 16 / 61

Compactify(t) Compactify(t) transforms a tree t into an Ω − equivalent compact tree. Steps : 1 If t is compact then return t and terminate. 2 If t is not compact then: 2.1 Let t 1 , ..... t r be the proper children of t , r ≥ 1. 2.2 For every 1 ≤ i ≤ r : t i := Compactify( t i ). (i.e., replace in t the subtree t i , by the result of compactifying t i . 2.3 Let x be the smallest index 1 ≤ x ≤ r such that K ( t x ) = max i K ( t i ). 2.4 Choose an index y � = x such that d ( t y ) = max i d ( t i ). 2.5 Choose subtrees t a x , t b x of t x and subtrees t a y , t b y , t c y of t y such that: i t x = t a x . t b x and t y = t a y . ( t b y . t c y ) ; and ii the roots of t b x , t b y and t c y are labelled by the same variable. 2.6 Remove t b y from t y and insert it into t x . t x = t a x .( t b y . t b x ) ; t y = t a y . t c y . 2.7 Goto (1). Abhshek, Rishi, Shubham Automata Construction November 23, 2018 17 / 61

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