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 LC? Deterministic Automata Trivial: it suffices to complement the set of final states. Nondeterministic Automata Upper bound: 2n (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 LC 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 eΘ

√

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 having size at most

• n, O
• n2

State complexity of a regular language: sc (L) ≡ number of states

• f the smallest deterministic

automaton accepting it. Nondeterministic state complexity: nsc (L) ≡ number of states

• f a minimal nondeterministic

automaton accepting it. Theorem [Jiang, McDowell, Ravikumar 91] If λ is the ultimate period of L and λ factorizes as p1k1 · p2k2 · · · · · psks, then we have nsc (L) ≥ p1k1 + p2k2 + · · · + psks.

Main result:

If a unary language L with ultimate pe- riod λ = p1k1 · p2k2 · · · · · psks is accepted by an nfa A with p1k1 + p2k2 + · · · + psks states in its cycles, then LC 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 LC

Proof(sketch) W.l.o.g. A is in Chrobak Normal Form with cy- cles of lengths p1k1, . . . , psks Choose mi and m such that:

• ami is accepted in the i-th cycle,
• ami+piki−1 /

∈ L, and

• ∀i m ≡ mi + piki−1 (mod pki

i )

(Chinese Remainder Theorem). Then:

• ami+pikix ∈ L, for i = 1, . . . , s, x ≥ 0,
• am /

∈ L.

Let p be the length of a cycle of the automa- ton A′ accepting LC visited during an accepting computation on am. Then:

• A′ accepts each am+py, with y ≥ 0,
• A′ must reject each ami+pki

i x, i = 1, . . . , s,

x ≥ 0. Hence, for i = 1, . . . , s, there are no integers x, y ≥ 0 such that m + py = mi + pikix, But m = mi + hpiki + piki−1, for some integer h. Hence, there are no integers x, y ≥ 0 such that hpiki + piki−1 = pikix − py This implies that piki | p, for i = 1, . . . , s.

Nonunary alphabets Main Theorem cannot be extended to nonunary languages: Theorem A sequence Ln of languages can be exhibited such that:

• nsc (Ln) = n
• sc (Ln) = 2n

(Ln is thus a witness of the gap between nondeterministic and deterministic automata) and

• nsc
• LnC

≤ 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 LC has a “small” nfa? The answer is negative: Theorem Let p1k1 · p2k2 · · · · · psks be the prime fac- torization for an arbitrary integer λ and consider Lλ = {am | #{i | pki

i

divides m} is even}. We have both nsc (Lλ) = λ and nsc

• LλC

= λ. The smallest nfa accepting Lλ (or LλC) is actu- ally a dfa made of a single cycle of length λ.

Sketch of the proof in the case of λ = p1k1 · · · · · psks, with s even. We show that each nfa A accepting the language Lλ = {am | #{i | pki

i

divides m} is even} 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 aλH, for a sufficiently large H. Let mj = Hλ + ℓ λ

p

kj j

, for j = 1, . . . , s. By “pumping” the computation C with the cycle

• f length ℓ, we get that amj ∈ Lλ.

Since each pki

i , with i = j, divides mj, this im-

plies that even pkj

j

must divide mj. Hence, we can easily conclude that each pkj

j

di- 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-

• usly apply to nsc (L) and nsc
• LC

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.