Complementing Unary Nondeterministic Automata Filippo Mera and - PDF document

Complementing Unary Nondeterministic Automata Filippo Mera and Giovanni Pighizzini Dipartimento di Informatica e Comunicazione Universit` a degli Studi di Milano, Italy

PROBLEM Comparing the number of states needed to ac- cept a language and its complement, i.e., Given an n -state automaton accepting L , how many states are necessary and sufficient to accept L C ? Deterministic Automata Trivial: it suffices to complement the set of final states. Nondeterministic Automata Upper bound: 2 n (just convert to a deterministic automaton). This bound cannot be improved [Next talk].

Present paper Investigation of this problem for the unary case. Main result: L is accepted by a “small” nfa ⇓ each nfa accepting L C must be “large”

What do “small” and “large” mean? Small: The automaton and the language it ac- cepts are witnesses of the gap between nfa’s and dfa’s, i.e., the nfa has n -states and each dfa accepting the same language has at least � √ � e Θ n ln n states. Large: The automaton has at least as many states as a deterministic automaton (non- determinism is thus useless.)

Unary Deterministic Automata ✓✏ ✛ ✒✑ A ✓✏ ✓✏ ✓✏ ✓✏ ✎ ✲ ✲ ✲ ✲ ✓✏ ✒✑ ✒✑ ✒✑ ✒✑ ✒✑ � �� � ✓✏ µ states ✒ ❥ ✒✑ � �� � λ states Size of an automaton A ≡ ( λ, µ ) Any unary regular language is ultimately cyclic. It is cyclic for words of length ≥ µ , being ( λ, µ ) the size of a dfa accepting it. If A is minimum then λ is the ultimate period of the language considered.

Unary Nondeterministic Automata The Chrobak Normal Form ✗ ✔ ✗ ✔ ✓ ✏ a ✲ ✛ a ✒ ✑ ✖ ✕ ✖ ✕ ✒ � � a � ✗ ✔ � ✗ ✔ ✗ ✔ ✗ ✔ ✓ ✏ � a ✲ a ✲ ✖ ✕ ✲ � ✒ ✒ ✑ ✖ ✕ ✖ ✕ ✖ ✕ � ❅ � a a ❅ � a ❅ � ✗ ✔ ✗ ✔ ❅ ✓ ✏ ❄ � ❅ ❘ ✛ a ✒ ✑ ✖ ✕ ✖ ✕ Theorem [Chr86]: Any nfa with n states can be simulated with an nfa in Chrobak Normal Form � � n 2 �� having size at most n, O

State complexity of a regular language: sc ( L ) ≡ number of states of the smallest deterministic automaton accepting it. Nondeterministic state complexity: nsc ( L ) ≡ number of states of a minimal nondeterministic automaton accepting it. Theorem [Jiang, McDowell, Ravikumar 91] If λ is the ultimate period of L and λ factorizes as p 1 k 1 · p 2 k 2 · · · · · p sk s , then we have nsc ( L ) ≥ p 1 k 1 + p 2 k 2 + · · · + p sk s .

Main result: If a unary language L with ultimate pe- riod λ = p 1 k 1 · p 2 k 2 · · · · · p sk s is accepted by an nfa A with p 1 k 1 + p 2 k 2 + · · · + p sk s states in its cycles, then L C requires at least λ cyclic states. Notes: • A has the smallest possible number of states with respect to the ultimate period of L • Nondeterminism does not allow to reduce the number of states for automata accepting L C

Proof (sketch) W.l.o.g. A is in Chrobak Normal Form with cy- cles of lengths p 1 k 1 , . . . , p sk s Choose m i and m such that: • a m i is accepted in the i -th cycle, • a m i + p iki − 1 / ∈ L , and • ∀ i m ≡ m i + p ik i − 1 (mod p k i i ) (Chinese Remainder Theorem). Then: • a m i + p iki x ∈ L , for i = 1 , . . . , s , x ≥ 0, • a m / ∈ L .

Let p be the length of a cycle of the automa- ton A ′ accepting L C visited during an accepting computation on a m . Then: • A ′ accepts each a m + py , with y ≥ 0, • A ′ must reject each a m i + p ki i x , i = 1 , . . . , s , x ≥ 0. Hence, for i = 1 , . . . , s , there are no integers x, y ≥ 0 such that m + py = m i + p ik i x, But m = m i + hp ik i + p ik i − 1 , for some integer h . Hence, there are no integers x, y ≥ 0 such that hp ik i + p ik i − 1 = p ik i x − py This implies that p ik i | p , for i = 1 , . . . , s .

Nonunary alphabets Main Theorem cannot be extended to nonunary languages: Theorem A sequence L n of languages can be exhibited such that: • nsc ( L n ) = n • sc ( L n ) = 2 n ( L n is thus a witness of the gap between nondeterministic and deterministic automata) and � L nC � • nsc ≤ n + 1

Question What can be said about the converse of Main Theorem? i.e., does the fact that each nfa accepting L is “large” imply that L C has a “small” nfa? The answer is negative: Theorem Let p 1 k 1 · p 2 k 2 · · · · · p sk s be the prime fac- torization for an arbitrary integer λ and consider L λ = { a m | # { i | p k i divides m } is even } . i � L λC � We have both nsc ( L λ ) = λ and nsc = λ . The smallest nfa accepting L λ (or L λC ) is actu- ally a dfa made of a single cycle of length λ .

Sketch of the proof in the case of λ = p 1 k 1 · · · · · p sk s , with s even . We show that each nfa A accepting the language L λ = { a m | # { i | p k i divides m } is even } i contains one simple cycle of at least λ states. Consider the length ℓ of a simple cycle crossed in an accepting computation C on an input a λH , for a sufficiently large H . Let m j = Hλ + ℓ λ , for j = 1 , . . . , s . kj p j By “pumping” the computation C with the cycle of length ℓ , we get that a m j ∈ L λ . Since each p k i i , with i � = j , divides m j , this im- plies that even p k j must divide m j . j Hence, we can easily conclude that each p k j di- j vides ℓ and, finally, that λ divides ℓ .

Conclusions and open problems We have been studying the problem in its “ex- treme” case, investigating languages that wit- ness the gap considered. Problem: Bounds should be found, that simultane- � L C � ously apply to nsc ( L ) and nsc for unary languages. In other words, it is of some interest to investi- gate the trade-off between the nondeterministic complexity of unary languages and that of their complements.