. On the Size of the Universal Automaton of a Regular Language - PowerPoint PPT Presentation

. On the Size of the Universal Automaton of a Regular Language Sylvain Lombardy IGM - Universit´ e de Marne-la-Vall´ ee STACS’07 - February 22, 2007 – p. 1/17

Motivation a b Minimizing deterministic automata a − → Merging states with the same future b b (Nerode equivalence) a − → Hopcroft algorithm O ( n log n ) STACS’07 - February 22, 2007 – p. 2/17

Motivation a b Minimizing deterministic automata − → Merging states with the same future b b a (Nerode equivalence) a − → Hopcroft algorithm O ( n log n ) STACS’07 - February 22, 2007 – p. 2/17

Motivation a b Minimizing deterministic automata b − → Merging states with the same future (Nerode equivalence) a − → Hopcroft algorithm O ( n log n ) STACS’07 - February 22, 2007 – p. 2/17

Motivation a b Minimizing deterministic automata b − → Merging states with the same future (Nerode equivalence) a − → Hopcroft algorithm O ( n log n ) Minimizing non deterministic automata − → Much harder : PSPACE-complete (Jiang - Ravikumar, 93) STACS’07 - February 22, 2007 – p. 2/17

Motivation a b Minimizing deterministic automata b − → Merging states with the same future (Nerode equivalence) a − → Hopcroft algorithm O ( n log n ) Minimizing non deterministic automata − → Much harder : PSPACE-complete (Jiang - Ravikumar, 93) − → However merging states may reduce the size of NFAs STACS’07 - February 22, 2007 – p. 2/17

Motivation Problem 1 : How to guarantee that merging does not modify the language ? − → "Equivalences" (Ilie - Yu, 02) "Inequalities" (Champarnaud - Coulon, 04) STACS’07 - February 22, 2007 – p. 3/17

Motivation Problem 1 : How to guarantee that merging does not modify the language ? − → "Equivalences" (Ilie - Yu, 02) "Inequalities" (Champarnaud - Coulon, 04) Problem 2 : What is the difference between this reduce NFA and any minimal NFA for the language ? − → L fixed, ∃ E L ∀A s.t. L ( A ) = L �A� > E L = ⇒ A has merging states (Câmpeanu - Sântean - Yu, 04) STACS’07 - February 22, 2007 – p. 3/17

Motivation Problem 1 : How to guarantee that merging does not modify the language ? − → "Equivalences" (Ilie - Yu, 02) "Inequalities" (Champarnaud - Coulon, 04) Problem 2 : What is the difference between this reduce NFA and any minimal NFA for the language ? − → L fixed, ∃ E L ∀A s.t. L ( A ) = L �A� > E L = ⇒ A has merging states (Câmpeanu - Sântean - Yu, 04) → E L � 2 k ( k : size of the minimal DFA) − (Grunsky - Kurganskyy - Potapov, 06) STACS’07 - February 22, 2007 – p. 3/17

Motivation Problem 1 : How to guarantee that merging does not modify the language ? − → "Equivalences" (Ilie - Yu, 02) "Inequalities" (Champarnaud - Coulon, 04) Problem 2 : What is the difference between this reduce NFA and any minimal NFA for the language ? − → L fixed, ∃ E L ∀A s.t. L ( A ) = L �A� > E L = ⇒ A has merging states (Câmpeanu - Sântean - Yu, 04) → E L � 2 k ( k : size of the minimal DFA) − (Grunsky - Kurganskyy - Potapov, 06) − → E L � D ( n ) ( n : size of a minimal NFA) STACS’07 - February 22, 2007 – p. 3/17

Factorizations A accepts L p state of A w w Past ( p ) = { w |→ i − → p } Fut ( p ) = { w | p − → t →} − → Past ( p ) . Fut ( p ) ⊆ L The universal automaton is an automaton s.t. the pairs ( Past ( p ) , Fut ( p )) are maximal w.r.t. L − → Factorizations of the language (Conway,71) Example : L = a + b + : ( a + b + , b ∗ ) ( a + b ∗ , b + ) ( A ∗ , ∅ ) ( a ∗ , a + b + ) ( a + , a ∗ b + ) ( ∅ , A ∗ ) STACS’07 - February 22, 2007 – p. 4/17

Universal automaton of L a, b a, b a, b a b ∅ , A ∗ a, b a, b b a, b a + , a ∗ b + a + b + , b ∗ a b a b a, b a a ∗ , a + b + a + b ∗ , b + a, b a, b A ∗ , ∅ a, b a, b a b a, b SL - Sakarovitch, 2000 STACS’07 - February 22, 2007 – p. 5/17

Universal automaton of L a, b a, b a, b a b ∅ , A ∗ a, b a, b b a, b a + , a ∗ b + a + b + , b ∗ a b a b a, b a a ∗ , a + b + a + b ∗ , b + a, b a, b A ∗ , ∅ a, b a, b a b a, b SL - Sakarovitch, 2000 STACS’07 - February 22, 2007 – p. 5/17

Universal automaton and reduction Proposition : The universal automaton is reduced. STACS’07 - February 22, 2007 – p. 6/17

Universal automaton and reduction Proposition : The universal automaton is reduced. Proposition : If L ( A ) = L , A can be reduced into a subautomaton of U L . STACS’07 - February 22, 2007 – p. 6/17

Universal automaton and reduction Proposition : The universal automaton is reduced. Proposition : If L ( A ) = L , A can be reduced into a subautomaton of U L . A − → U L X p = { u ∈ A ∗ | u Fut ( p ) ⊆ L } � �− → p Y p = { v ∈ A ∗ | X p v ⊆ L } STACS’07 - February 22, 2007 – p. 6/17

Construction of the universal automaton The co-determinized automaton of A is obtained from A by a “backward” determinization. Example : a b a b p q r a b a b p, q q, r r STACS’07 - February 22, 2007 – p. 7/17

Construction of the universal automaton Theorem : (ICALP 2002) The set of states of the universal automaton of L is obtained from the minimal automaton A of L by computing the intersection of states of the co-determinized automaton of A a b p, q, r a b p q r p, q q, r a b q r a b p, q q, r r ∅ STACS’07 - February 22, 2007 – p. 8/17

Construction of the universal automaton a b b q ↔ a + r ↔ a + b + a b a a, b b a q, r ↔ a + b ∗ p, q ↔ a ∗ a b a b a b p q r a b a b p, q q, r r STACS’07 - February 22, 2007 – p. 9/17

Bound w.r.t minimal DFA Proposition : For every n , there exists a deterministic automaton with n states that recognizes a language for which the universal automaton has 2 n states. b q a, b a p a r b STACS’07 - February 22, 2007 – p. 10/17

Bound w.r.t minimal DFA Proposition : For every n , there exists a deterministic automaton with n states that recognizes a language for which the universal automaton has 2 n states. b q p, q a b a a, b a p q, r p, q, r a b a a b p, r r b STACS’07 - February 22, 2007 – p. 10/17

Bound w.r.t minimal DFA Proposition : For every n , there exists a deterministic automaton with n states that recognizes a language for which the universal automaton has 2 n states. b b q p, q a a, b a a, b b b a p q, r p, q, r ∅ a a a b p, r r b STACS’07 - February 22, 2007 – p. 10/17

Unary alphabet q a a p a r STACS’07 - February 22, 2007 – p. 11/17

Unary alphabet q, r a a p, q a p, r STACS’07 - February 22, 2007 – p. 11/17

Unary alphabet q p, q a a a a a p q, r p, q, r ∅ a a a p, r r STACS’07 - February 22, 2007 – p. 11/17

From a NFA : unary alphabet Construction of the universal automaton : first determinization (at most k = G ( n ) states) then construction of the universal automaton ( 2 k states). → Upper bound : 2 G ( n ) − � G ( n ) = max { lcm ( n 1 , . . . , n r ) | n i = n } 1 1 1 a a a . . . �− → 0 0 0 a a a n 1 − 1 n r − 1 n − 1 − → The upper bound is tight. STACS’07 - February 22, 2007 – p. 12/17

From a NFA Construction of the universal automaton : first determinization (at most k = 2 n − 1 states) then construction of the universal automaton ( 2 k states). → Upper bound : 2 2 n − 1 − p b b a q r a STACS’07 - February 22, 2007 – p. 13/17

From a NFA Construction of the universal automaton : first determinization (at most k = 2 n − 1 states) then construction of the universal automaton ( 2 k states). → Upper bound : 2 2 n − 1 − p p, pr, pq b b b b a a q r r, pr q, pq ba a a a a b a pr pq p, q, pq, pr p, r, pq, pr b b STACS’07 - February 22, 2007 – p. 13/17

Characterization of the states Proposition : D determinized of A , C co-determinized of D . Let P ⊂ R be two states of D . If P belongs to a state X of C , so does R . − → States of C are upperset. Lemma : The uppersets are closed under intersection. Theorem : The states of the universal automaton are uppersets. Theorem : The universal automaton of a language recognized by a NFA with n states has at most D ( n ) states ( n -th Dedekind number). STACS’07 - February 22, 2007 – p. 14/17

D(n) ? 1 D ( n ) n 0 2 0.8 1 3 0.6 2 6 3 20 0.4 4 168 5 7581 0.2 6 7828354 7 2414682040998 0 0 5 10 15 20 25 8 56130437228687557907788 → log 2 D ( n ) n �− 2 n STACS’07 - February 22, 2007 – p. 15/17

Lower bound b 2 a 0 a a 1 b STACS’07 - February 22, 2007 – p. 16/17

Lower bound b b 01 1 a a, b a b a b b a 012 12 0 ∅ a a a b 02 2 b STACS’07 - February 22, 2007 – p. 16/17

Lower bound b 2 02 b a, b a b a a b b a 012 0 12 ∅ a a a b 1 01 b STACS’07 - February 22, 2007 – p. 16/17

Lower bound b 2 a a, b a, b 0 a ∅ a 1 b a a 2 , 01 02 , 12 a a 0 , 12 a a 012 01 , 02 0 , 1 , 2 a a b b 1 , 02 01 , 12 a 1 , 2 12 a a a 0 , 2 01 01 , 02 , 12 a a a 0 , 1 02 D (3) = 20 STACS’07 - February 22, 2007 – p. 16/17