Synchronized automata (1)

Synchronized automataNondeterministic automata

Unambiguous automata

Synchronized automata (1)

Marie-Pierre Beal, Dominique Perrin

May 24, 2012

Marie-Pierre Beal, Dominique Perrin Synchronized automata (1)



Outline

Cerny conjecture (DP)

Synchronized automataNondeterministic automataUnambiguous automataAperiodic automataLinear representations

Road coloring (MPB)




Synchronizing words

We first consider deterministic automata.A synchronizing word is a word w such that there are paths labeledby w and all have the same terminal state. An automaton whichhas a synchronizing word is called synchronized.Elementary arguments give the cubic bound 1

2n(n − 1)2 for the

length of a synchronizing word for an n-state synchronizedautomaton. Computing the minimal length of a synchronizingword is NP-complete (Eppstein, 1990).




An example

A deterministic automaton with four vertices and two edges goingout of each vertex labeled B or R .

1 2

34

BRB is a synchronizing sequence: starting anywhere, it leads to 1.




Another example

With a different coloring:

1 2

34

there is no synchronizing sequence: the sets {1, 3} and {2, 4}cannot be collapsed further:

1, 3 2, 4




Cubic bounds

Let A be a synchronized automaton with n states. For each set P

of states there is a word w of length at most n(n − 1)/2 such thatCard(P · w) < Card(P). By iteration, this proves the existence of asynchronizing word of length at most 1

2n(n − 1)2.

Refinement: a sequence P1,P2, . . . of k element subsets of an n

element set is called 2-renewing if each Pi contains a pair Ri suchthat Ri is not contained in Pj for j < i .

Proposition (Frankl, 1982)

A 2-renewing sequence has at most(

n−k+2

2

)

elements.

Consequence: existence of a synchronizing word of length at most(n3 − n)/6.




Complexity

Proposition (Eppstein,1990)

The existence of a synchronizing word of length ℓ is NP-complete.

Proof: polynomial reduction from the satisfiability of booleanformulas. Given a formula ψ = c1 ∧ c2 ∧ . . . ∧ cm with each ci adisjuction of x1, x2, . . . , xn or their negations, one builds anautomaton A(ψ) which has a synchronizing word on length n ifand only if ψ is satisfiable.As usual a binary word of length n is interpreted as a truthassignignement for the n variables.




Cerny conjecture

Conjecture (Cerny, 1964)

Any complete n-state synchronized automaton has a synchronizing

word of length at most (n − 1)2.

The hypothesis that the automaton is complete is perhaps notnecessary?




Cerny automata

0

1

2

3

a, b

aa

a

b

b

b

0

1

23

4

a, b

a

a

a

ab

bb

b0

1

2

3

4

5

a, b

a

aa

a

ab

b

b

b

b

Figure: The automata C4, C5, C6

The word (ban−1)n−2b is a synchronizing word of length (n − 1)2.There is no shorter one.




A nice proof (Ananisev, Gusev, Volkov)

0

1

2n − 2

n − 1

a, b

aa

ab

bb

b 0

1

2n − 2

n − 1

a

c a, ca, c

a, c

Figure: The automata Cn and Dn with c = ba

The cycles of Dn are nonnegative integer combinations of n andn − 1. The following lemma is well-kown.

Lemma

If k, ℓ ≥ 1 are relatively prime kℓ− k − ℓ is the largest integer not

expressible as a nonnegative integer combination of k, ℓ.




Proof (Cont’d)

Let w be a synchronizing word of minimal length for Cn. Setw = w ′b. Then Q · w ′ = {0, 1}. Let v be obtained from w ′

replacing each ba by c (there are no consecutive b in w).Then vc is synchronizing for Dn and thus Dn has cycles of alllengths ≥ |vc |. This forces

|vc | > n(n − 1) − n − (n − 1) = n2 − 3n + 1.

We cannot have |vc | = n2 − 3n + 2 because 1 · vc = 2 would implythat vc minus its first letter sends 2 to 2 although there is no cycleof length n2 − 3n + 1 in Dn. Thus |vc | ≥ n2 − 3n + 3 and|w ′| = |v | − (n − 2) ≥ n2 − 2n and finally |w | ≥ n2 − 2n + 1.




The generalized conjecture

The rank of a word w in a deterministic automaton A = (Q,E ) isthe size of its image Q · w . A synchronizing word has rank 1.It has been conjectured by Pin that if an automaton admits a wordof rank at most k, then there exists such a word of length at most(n − k)2. This generalizes the Cerny conjecture. Pin’s conjecturewas proved to be false (Kari, 2001). The counterexamplecorresponds to n = 6 and k = 2.A new conjecture proposed by Volkov states that an automaton ofminimal rank k admits a word of rank k of length at most (n− k)2.




Kari’s automaton

1

2

3 4

5

6

a a

a

a a

a

b

b

b

b b

bShortest words of rank 2: x = baabababaabbabaab andy = baababaabaababaab of length 17. Shortest synchronizing wordxaababaab of length 24 = 52 − 1.




A funny conjecture

Conjecture (Trahtman,2006)

The set of n-state complete deterministic automata with minimal

synchronizing word of length (n − 1)2 consists of the sequence of

Cerny automata and eight automata of size at most 6.




The eight exceptional automata

1

2

3 4

5

6a aa a aa

b

b

b

b b

b

1 2 3

4

b

a b

a

b

b

a

a

1 2 3

b

a

c

c

a, c b a, b

1

2

3

c

c

a, b

a, ba

b c




1

2

3

4

5a, b

cc

bb

a

a

a

b

c

c

a, b

c

1

2

3

aa, b

b

bb

1 2 3

4

a

cc a, bb

a, c

c

a

bb 1

2

3

4

b

b

a

a

c c

c b, c

a, b

a




Nondeterministic automata

A word w is synchronizing for a nondeterministic automaton ifthere is at least one path labeled by w and, for any states p, q, r , s,

pw−→ q, r

w−→ s implies p

w−→ s and r

w−→ q.

The automaton is synchronized if there is a synchronizing word.This is consistent with the previous definition. Indeed, if theautomaton is deterministic, the condition implies that q = s.




Rank

Associate to a word w the relation ϕA(w) = {(p, q)|pw→ q}.

The rank of a relation m on a set Q is the minimal cardinality of aset R such that m = uv with u a Q × R relation and v an R × Q

relation.A word w is synchronizing for a nondeterministic automaton A ifand only if the relation ϕA(w) has rank one. Indeed, if w issynchronizing, then ϕA(w) = uv where u is the column Q-vector v

is the row Q-vector defined as follows. One has up = 1 if and onlyif the is a path labeled w starting from p and vq = 1 if and only ifthere is a path labeled w ending in q.




Example

The automaton below is synchronized. Indeed, the word ab is suchthat

ϕA(ab) =

0 0 00 0 10 0 1

=

011

[

0 0 1]

and thus is synchronizing.

1

2

3

a

ab

a, b

a, b

Figure: A synchronized nondeterministic automaton.





A nondeterministic automaton is unambiguous if, for any pair p, qof states, and any word w , there is at most one path labeled by w

going from p to q. A deterministic automaton is unambiguous.The converse is not true.One may check that an automaton is unambiguous by computingits square. The square of A is the automaton on Q ×Q with edges(p, q)

a→ (r , s) if p

a→ r and q

a→ s are edges of A. The automaton

A is unambiguous if and only if there is no path in its square ofthe form (p, p)

u→ (r , s)

v→ (q, q) with r 6= s.




Example

Let A be the automaton represented on the left.

1

2

3

a

ab

a, b

a, b

1, 1

2, 2

3, 3

1, 2

2, 1

a

ab

a, b

a, b

a, b a, b

a, b

a, b

Figure: An unambiguous automaton and part of its square.

This automaton is unambiguous as one may check by computingthe square of the automaton A represented on the right (with onlythe states accessible from the states (p, p) represented).




Synchronizing words and recognizability of morphisms

Let f : A∗ → A∗ be a morphism with f (a) ∈ aA∗. Let x = f ω(a).Let

I = {|f (x0 · · · xn)| | n ≥ 0}

The morphism f is n-recognizable if for i ∈ I ,xi−n · · · xi+n = xj−n · · · xj+n implies j ∈ I .Example: The Fibonacci morphism f (a) = ab, f (b) = a is2-recognizable.

a b a a b a b . . .

Martin’s theorem: any primitive morphism such that x is notperiodic is recognizable.




Using the fact that

(i) for any code X there is an unambiguous automatonA = (Q, 1, 1) recognizing X ∗

(ii) The fixpoint of a primitive morphism is uniformly recurrent

one obtains

Proposition

Assume that f is injective and primitive and let A = (Q, 1, 1) be

an unambiguous automaton recognizing f (A)∗. If there is a

synchronizing word w ∈ F (x), then

(i) There is an n ≥ 1 such that any word of F of length n is

synchronizing.

(ii) The morphism f is n-recognizable.




Example

Set f (a) = ab and f (b) = a. Then x = abaababa · · · is theFibonacci word. The automaton A is represented below.

1 2a

a

b

The word a is synchronizing. Thus aa, ab and ba are synchronizingand f is 2-recognizable.




Example

Set f (a) = ab and f (b) = ba. Then x = abbabaab · · · is theThue-Morse word. The automaton A is represented below.

2 1 3

a

b

a

b

The words aa and bb are synchronizing. Thus f is 5-recognizable.In this case the automaton is deterministic and one could use aone-sided notion of recognizability from left to right (in the sameway for the previous case from right to left).




A nondeterministic automaton is complete if, for any word w ,there exists a pair p, q of states such that p

w−→ q is a path.

The following result gives a cubic upper bound for the length ofsynchronizing words.

Theorem (Carpi, 1988)

Let A be an n-state strongly connected complete unambiguous

automaton. If A is synchronized, it has a synchronizing word of

length at most (n2 − n + 2)(n − 1)/2.




We introduce some terminology concerning monoids of relations ona set. For a relation m on a set Q, we denote indifferently(p, q) ∈ m or mpq = 1, considering the relation either as a subsetof Q × Q or as a boolean Q × Q matrix. We denote by mp∗ therow of of index p of the matrix m, which is the characteristicvector of the set {q ∈ Q | (p, q) ∈ m}.A monoid of relations M on a set Q is unambiguous if for anym, n ∈ M and p, q ∈ Q there exists an most one element r of Q

such that (p, r) ∈ m and (r , q) ∈ n. The monoid is transitive if forany p, q ∈ M there is an m ∈ M such that (p, q) ∈ m.An automaton is unambiguous if and only if its transition monoidis unambiguous. The automaton is strongly connected if and onlyits transition monoid is transitive.




We use the following lemmas.A row of an element of M which is maximal among the rows of theelements of M is called a maximal row.

Lemma

Let M be a transitive unambiguous monoid of relations not

containing zero. If v is a maximal row, then vm is a maximal row

for any m ∈ M.

Lemma

Let M be a transitive unambiguous monoid of relations not

containing zero. For two elements m,m′ of M, if m ≤ m′ then

m = m′.




The following lemma is the key of Carpi’s theorem.

Lemma

For a state p ∈ Q and a word u ∈ A∗, if ϕ(u)p∗ is not a maximal

row, there is a state q and a word v of length at most n(n − 1)/2such that ϕ(u)p∗ < ϕ(vu)q∗.




Proof of Carpi’s theorem.By the last Lemma and its symmetric form, there exist pairs(p1, u1), (p2, u2), . . . , (ps , us) in Q × A∗ and(v1, q1), (v2, q2), . . . , (vt , qt) in A∗ × Q such that, withxi = ϕ(ui · · · u1)pi∗ and yj = ϕ(v1 · · · vj)∗qi

,

(i) u1 = v1 = 1 and p1 = q1.

(ii) for 2 ≤ i ≤ s, the word ui has length at most n(n − 1)/2 andxi > xi−1.

(iii) for 2 ≤ j ≤ t, the word vj has length at most n(n − 1)/2 andyj > yj−1.

(iv) xs is a maximal row and yt is a maximal column.

Let u = us · · · u1 and v = v1 · · · vt . We have|u| ≤ (s − 1)n(n − 1)/2 and |v | ≤ (t − 1)n(n − 1)/2. Thus|uv | ≤ (s + t − 2)n(n − 1)/2. Since A is unambiguous, we havexsyt = 1.




Thus s + t ≤∑

q∈Q(xs)q +∑

q∈Q(yt)q ≤ n + 1. Let finally z ∈ A∗

be such that qtz→ ps with |z | ≤ n − 1. Then w = vzu is such that

ytxs ≤ ϕ(w). By the second Lemma, this implies ϕ(w) = ytxs ,whence the conclusion.




Example

We illustrate the proof on the automaton of the previous example.We start with u1 = v1 = ε and p1 = q1 = 1. The rowϕ(ǫ)1∗ =

[

1 0 0]

is not maximal butϕ(ǫ)1∗ < ϕ(a)3∗ =

[

1 1 0]

. Thus we choose u2 = a and p2 = 3.

Symmetrically, the column ϕ(ǫ)∗1 =[

1 0 0]t

is not maximal but

ϕ(ǫ)∗1 < ϕ(b)∗3 =[

1 1 0]t

. Thus we choose v2 = b andq2 = 3. Then ab is a synchronizing word.




Question: is Cerny conjecture true for unambiguous automata?


Synchronized automata (1)

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