Ìàòåìàòè÷íi
Ñòóäi¨. Ò.29, 1
Matematychni Studii. V.29, No.1
ÓÄÊ 519.713.2
A. S. Antonenko
ON TRANSITION FUNCTIONS OF MEALY AUTOMATA
OF FINITE GROWTH
A. S. Antonenko. On transition functions of Mealy automata of nite growth , Matematychni
Studii, 29 (2008) 317.
The transition functions of Mealy automata that generate nite automaton transformation
semigroups are considered in the paper. We formulate conditions on transition functions such
that any Mealy automaton with this transition generates a nite semigroup. Also we prove
that these requirements are maximal, because for any transition function that does not satises them there exists an output function such that the Mealy automaton denes an innite
semigroup.
À. C. Àíòîíåíêî. Î ôóíêöèÿõ ïåðåõîäîâ àâòîìàòîâ Ìèëè êîíå÷íîãî ðîñòà // Ìàòåìàòè÷íi Ñòóäi¨. 2008. Ò.29, 1. C.317.
 ñòàòüå ðàññìàòðèâàþòñÿ ôóíêöèè ïåðåõîäîâ àâòîìàòîâ Ìèëè, êîòîðûå ïîðîæäàþò
êîíå÷íûå ïîëóãðóïïû. Ïîëó÷åíû óñëîâèÿ íà ôóíêöèè ïåðåõîäîâ, ïðè êîòîðûõ âñå àâòîìàòû
Ìèëè ïîðîæäàþò êîíå÷íûå ïîëóãðóïïû. Äîêàçàíî, ÷òî ýòè óñëîâèÿ ÿâëÿþòñÿ ìàêñèìàëüíûìè,
òî åñòü äëÿ ôóíêöèè ïåðåõîäîâ, êîòîðàÿ íå óäîâëåòâîðÿåò óêàçàííûì âûøå óñëîâèÿì,
ñóùåñòâóåò ôóíêöèÿ âûõîäîâ òàêàÿ, ÷òî ñîîòâåòñòâóþùèé àâòîìàò Ìèëè ïîðîæäàåò áåñêîíå÷íóþ
ïîëóãðóïïó.
1. Introduction. The groups and semigroups of automaton transformations are actively
studied since 60ies of the last century [1, 2, 3]. In these papers close relation between Mealy
automata properties and properties of semigroups and groups of automaton transformations
that are generated by these automata is shown. The review of recent results in the area
of automata and groups of automaton transformations is represented in the paper [4]. For
a recent review of properties of automaton transformation semigroups see [5]. However, there
are a great number of open questions characterization of dierent properties of groups and
semigroups generated by automata in general case (see [4]).
One of the most interesting questions concerning semigroups dened by Mealy automata
is the question on niteness or innity of a semigroup. This property has a natural interpretation. Mealy automaton determines a particular transformation of words at each state,
and niteness of semigroup means that successive application of these transformations produce only a nite set of automaton transformations. In addition, any automaton generates
nite semigroup if and only if it has nite growth order. The problem of the determining
the niteness of the (semi)group generated by a Mealy automaton is mentioned in the list
of open problems [4]. All two-state Mealy automata over the two-symbol alphabet which
denes nite semigroups are investigated in [6] (see also [7]). However the problem has not
been solved in the general case.
2000 Mathematics Subject Classication: 20M35, 68Q70.
c A. S. Antonenko, 2008
°
4
A. S. ANTONENKO
For arbitrary transition function there exists an output function such that the automaton generates a nite semigroup. Dierent classes of Mealy automata that generate nite
(semi)groups for all output functions described by limitations on transition functions were
investigated in [8, 9] and in the verbal message of Reznykov and Sushchansky). The class
of invertible automata is considered in the paper [9], and niteness of groups of automata
transformations generated by such automata is proved.
The class of transition functions that dene automata of nite growth for any output
function is considered in the paper. This class of transition functions includes all classes
from the papers mentioned above. Moreover, we show maximality of this class of transition
functions. That is for any transition function that does not belong to the class there exists
an output function such that the corresponding automaton generates an innite semigroup.
This result allows to simplify the research of Mealy automata growth, because it covers a
wide class of automata of nite growth order.
In section Preliminaries we give preliminaries of Mealy automata and semigroups dened by them and set up notations. In section Automata without branches the automata
without branches [8] and the class of their transition functions are considered. Such automata dene only nite semigroups. We extend this class of transition functions to the class
of transition functions with limitary cycle in Section Mealy automata of nite growth.
Any a automaton with the latter functions also generates only nite semigroups. In Section
Transition functions without limitary cycle it is shown that this class is maximal.
The author wishes to express his gratitude to I. I. Reznykov for suggesting the problem,
permanent attention and help in preparation of the article, and also to Yu. G. Leonov,
P. D. Varbanets, and E. Berkovich for the useful remarks.
2. Preliminaries.
Denition 1 ([10, 11]). A nite Mealy automaton is an ordered quintuple A = (X, Y, Q, π,
λ), where X is the input alphabet, Y is the output alphabet, Q is the nite nonempty set of
states, π : X × Q → Q is the transition functions and λ : X × Q → X is the output function.
X and Y are nite nonempty sets.
We will consider only nite automata whose input and output alphabet coincide (X = Y ).
We denote such automata by quadruples A = (X, Q, π, λ). For convenience, we x a nite
alphabet X = {0, 1, . . . , m − 1}, where m ≥ 2 is some natural number.
Let TX = {f | f : X → X} be the semigroup of all transformation of the alphabet X (the
full symmetric semigroup), X ∗ be the set of all nite words over X , X ω be the set of all
ω -words (innite words) over X .
It is convenient to describe nite automata by the Moore diagrams. We will use the
following modication of them. The Moore diagram of an automaton A is an edge-labeled
and vertex-labeled directed multigraph DA with the set of vertices Q. Vertices qi and qj of the
graph DA are connected by the oriented edge in direction from qi to qj marked by the label
x, if π(x, qi ) = qj . Here x ∈ X, qi , qj ∈ Q. Every vertex q is labeled by the transformation
λq ∈ TX of the alphabet X that corresponds to the output function at the state q , i.e.
λq (x) = λ(x, q), where x ∈ X, q ∈ Q.
The functions π and λ can be extended naturally to mappings of the set X ∗ × Q into the
sets Q and X ∗ by the following equalities [11]:
π(Λ, q) = q, π(wx, q) = π (x, π(w, q)) ,
λ(Λ, q) = Λ, λ(wx, q) = λ(w, q)λ (x, π(w, q)) ,
ON MEALY AUTOMATA OF FINITE GROWTH
5
where Λ ∈ X ∗ is the empty word, q ∈ Q, w ∈ X ∗ , x ∈ X . The function λ can also be
extended in a natural way to a mapping λ : X ω × Q → X ω (see for example, [11]).
Let π (X1 , Q1 ) = {π(x, q)|x ∈ X1 , q ∈ Q1 } and λ (X1 , Q1 ) = {λ(x, q)|x ∈ X1 , q ∈ Q1 } for
an arbitrary X1 ⊆ X , Q1 ⊆ Q.
Denition 2 ([11]). The transformation fq : X ∗ → X ∗ (fq : X ω → X ω ), dened by the
equality fq (u) = λ(u, q), where u ∈ X ∗ (u ∈ X ω ), is called an automaton transformation
dened by the automaton A = (X, Q, π, λ) at the state q .
Mealy
© automaton Aª= (X, Q, π, λ), where Q = {q0 , q1 , . . . , qn−1 }, determines the set
FA = fq0 , fq1 , . . . , fqn−1 of automaton transformations over X ∗ . The Mealy automaton A
is called invertible if all transformations from the set FA are bijections. It is easy to show
(see for example [4]) that A is invertible if and only if the transformation λq is a permutation
of X for each state q ∈ Q.
Denition 3 ([11]). Mealy automata Ai = (X, Qi , πi , λi ), for i = 1, 2, are called isomorphic
if there exist permutations ξ, ψ ∈ Sym(X) and θ : Q1 → Q2 such that
θπ1 (x, q) = π2 (ξx, θq), ψλ1 (x, q) = λ2 (ξx, θq)
for all x ∈ X and q ∈ Q1 .
Denition 4 ([11]). Mealy automata Ai for i = 1, 2, are called equivalent if FA1 = FA2 .
Proposition 1 ([11]). Each class of equivalent Mealy automata over an alphabet X contains,
up to isomorphism, a unique automaton that is minimal with respect to the number of states
(such an automaton is called reduced).
The minimal automaton can be found using the standard algorithm of minimization.
Denition 5 ([12]). For i = 1, 2 let Ai = (X, Qi , πi , λi ) be arbitrary Mealy automata. The
automaton A = (X, Q1 × Q2 , π, λ) such that its transition and output functions are dened
in the following way:
π (x, (q1 , q2 )) = (π1 (λ2 (x, q2 ) , q1 ) , π2 (x, q2 )), λ (x, (q1 , q2 )) = λ1 (λ2 (x, q2 ) , q1 ) ,
where x ∈ X and (q1 , q2 ) ∈ Q1 × Q2 , is called the product of the automata A1 and A2 .
Proposition 2 ([12]). For any states q1 ∈ Q1 and q2 ∈ Q2 and an arbitrary word u ∈ X ∗
the following equality holds:
f(q1 ,q2 ),A (u) = fq1 ,A1 (fq2 ,A2 (u)) .
The power An is dened for any automaton A and any positive integer n. Let us denote by
A(n) the minimal Mealy automaton equivalent to An . It follows from denition of the product
that |QA(n) | ≤ |QA |n .
Denition 6 ([13]). The function γA of a natural argument dened by γA (n) = |QA(n) | , n ∈
N, is called the growth function of a Mealy automaton A.
ª
©
Denition 7. A semigroup generated by the set FA = fq0 , fq1 , . . . , fqn−1 of transformations dened by a Mealy automaton in all its states A is called the semigroup generated by
automaton A. In the case of invertible automaton A the group generated by FA is called the
group generated by the automaton A.
Denition 8. An automaton A is called an automaton of nite growth (order) if its growth
function is bounded.
6
A. S. ANTONENKO
It is easy to show that an automaton A is an automaton of nite growth if and only if
the semigroup generated by it is nite.
Denition 9 ([14]). The wreath product by innite sequence (H1 , M1 ), (H2 , M2 ), . . . (by nite
sequence (H1 , M1 ), (H2 , M2 ), . . . , (Hk , Mk )Q
) of transformation
Qk semigroups is the semigroup
of all transformations h of the set M = ∞
M
(
M
=
i
i=1
i=1 Mi ) satisfying the following
conditions:
1. if (y1 , y2 , . . .) = (x1 , x2 , . . .)h then yi depends only on i rst coordinates x1 , x2 , . . . for
i = 1, 2, . . .;
¡
¢
2. if x01 , . . . , x0i−1 is xed then the transformation hi x01 , . . . , x0i−1 = gi , hi : xi → yi induced
by h is a transformation from the semigroup Hi .
We denote the dened wreath product of transformation semigroups by H = o∞
i=1 (Hi , Mi ) =
∞
k
k
oi=1 Hi (in the nite sequence case H = oi=1 (Hi , Mi ) = oi=1 Hi = H1 o H2 o · · · o Hk ).
3. Automata without branches. In this section we consider a special set of Mealy au-
tomata that dene nite (semi)groups. We will use the following classication of automaton
states [8].
Let Q be a nonempty nite set of states and π : X ×Q → Q be a xed transition function.
Let q ∈ Q be an arbitrary state.
Denition 10 ([8]). We say that q is a rest state if the automaton stays in this state under
the action of any input symbol, i.e.
∀x ∈ X : π (x, q) = q.
Denition 11 ([8]). We say that q is an unconditional jump state if there exists the state
q 0 ∈ Q, q 0 6= q , such that the automaton moves from the state q to the state q 0 under the
action of any input symbol, i.e.
∃q 0 ∈ Q, q 0 6= q : ∀x ∈ X, π (x, q) = q 0 .
Denition 12. We say that q is a state without branches if it is either a rest state or an
unconditional jump one, otherwise we say that q is a branch state.
A state q ∈ Q is a state without branches if and only if for all x1 , x2 ∈ X the equality
π (x1 , q) = π (x2 , q) holds.
Examples of a rest state, an unconditional jump state and a branch state for the case
X = {0, 1} are shown, respectively, in Figures 1 a, b and c.
Denition 13. We say that a Mealy automaton A = (X, Q, π, λ) has no branches (and,
correspondingly, its transition function π has no branches) if any state q ∈ Q is a state
without branches (i.e. either a rest state or an unconditional jump state).
Note that an automaton would be an automaton without branches if and only if its
transition function depends only on the state and does not depend on the input symbol.
Hence, we will denote the transition function of an automaton without branches by the
symbol s(q) = π (x, q), for all q ∈ Q, x ∈ X .
Let g = g1 g2 . . . ∈ (TX )ω be an arbitrary innite word. Let Fg : X ω → X ω be the
transformation over the set of innite words dened by the equality
Fg (x1 x2 . . .) = g1 (x1 )g2 (x2 ) . . . .
7
ON MEALY AUTOMATA OF FINITE GROWTH
a
q
b
0,1
q
0,1
c
q'
0
q'
1
q''
q
Fig. 1: A rest state, an unconditional jump state and a branch state
In the sequel, we will call such transformations as symbol-by-symbol transformations. It
is clear that Fu ◦ Fv = Fu◦v , where u = u1 u2 . . . ∈ (TX )ω , v = v1 v2 · · · ∈ (TX )ω , u ◦ v =
(u1 ◦ v1 ) (u2 ◦ v2 ) . . . , is an elementwise composition.
Lemma 1 ([8]). Let q ∈ Q be an arbitrary state of an automaton without branches A =
(X, Q, π, λ). Then the transformation dened by the automaton A at the state q is a symbolby-symbol one.
Proof. Let fq be an automaton transformation generated by an automaton without branches
at the state q and let v be an input word. The automaton goes to the xed (independent of
the rst symbol) state s (q) under the action of the rst symbol. Similarly, it goes to the state
s (s (q)) = s2 (q) after reading two rst symbols, and so on. Therefore the equality fq = Fu
holds, where u = λq λs(q) λs2 (q) · · · λsk (q) · · · .
Let Pu,v = Fuv∗ , where u, v ∈ TX∗ , v ∗ = vv · · · For any u1 , u2 , v1 , v2 ∈ TX∗ such that
|u1 | = |u2 |, |v1 | = |v2 | the equality Pu1 ,v1 ◦ Pu2 ,v2 = Pu1 ◦u2 ,v1 ◦v2 holds. Here |u| denotes the
length of a word u.
Lemma 2. Let A = (X, Q, π, λ) be an automaton without branches, |Q| = n. Then there
exist integers l, k ∈ N, l < n, such that the transformation generated by the automaton A
at any state q ∈ Q has the form of Pu,v , where |u| = l, |v| = k .
Proof. The proof is similar to the proof of the pumping lemma [15]. Let q ∈ Q. Since
|Q| = n, there are two equal states among states q, s(q), s2 (q), . . . , sn (q). Let such rst states
be sa (q) = sb (q), a < b and let lq = a − 1, kq = b − a. It is easily seen that the sequence
{si (q)}∞
i=1 is an ultimately periodic
¡ period
¢ kq . Indeed for all natural
¡ sequence ¢with the least
p > lq we have sp+kq (q) = sp−lq −1 slq +1+kq (q) = sp−lq −1 slq +1 (q) = sp (q).
Let l = maxq∈Q (lq ), and let k be the least common multiple of all kq where q ∈ Q.
It is clear that l < n and k is not more than the least common multiple of all integers
between 2 and n. Then all sequences {si (q)}∞
i=1 , where q ∈ Q, are ultimately periodic with
period k and we can regard that the periodic parts start at the position l. By Lemma 1 the
automaton A at the state q denes the transformation Fλq λs(q) λs2 (q) ...λsk (q) ... = Puq ,vq , where
uq = λq λs(q) λs2 (q) · · · λsl−1 (q) ,vq = λsl (q) λsl+1 (q) · · · λsl+k−1 (q) , |uq | = l,|vq | = k .
Theorem 1. A semigroup S is isomorphic to some semigroup generated by a nite automa-
ton without branches over an alphabet X (|X| = m), if and only if S is isomorphic to a
8
A. S. ANTONENKO
semigroup S 0 ≤ (TX )k for some natural k such that there exists a natural number l such
that 1 ≤ l ≤ k and
v1 v2 · · · vk ∈ S 0 ⇒ v2 · · · vk vl ∈ S 0
(1)
where vi ∈ TX for 1 ≤ i ≤ k .
Proof. The proof falls naturally into two parts. First we will construct a semigroup S 0 ≤
(TX )k with property (1) such that it is isomorphic to a semigroup S generated by an automaton without branches. Then for some semigroup S 0 ≤ (TX )k with property (1) we will
construct an automaton without branches such that semigroup generated by it is isomorphic
to S 0 .
Let S be a semigroup generated by the automaton without branches A = (X, Q, π, λ).
By Lemma 2 there exist integers l0 , k 0 ∈ N such that the transformation generated by the
0
0
automaton A at any state q ∈ Q has the form of Pu,v , where |u| = l0 , |v| = k 0 , u ∈ TXl , v ∈ TXk .
Let k = l0 + k 0 . The desired isomorphism is F1 : S → S 0 ≤ (TX )k , where F1 (z) = uv ,
z = Pu,v ∈ S . The isomorphism F1 is well dened because uv ∈ (TX )k and the words u
and v are uniquely dened by z ∈ S . It is clear that F1 (Pu1 ,v1 ◦ Pu2 ,v2 ) = F1 (Pu1 ◦u2 ,v1 ◦v2 ) =
(u1 ◦u2 )(v1 ◦v2 ) = u1 v1 ◦u2 v2 = F1 (Pu1 ,v1 )◦F1 (Pu2 ,v2 ). Let us prove property (1). Let l = l0 +1.
Suppose v = v1 v2 . . . vk ∈ S 0 . Then v 0 = F1−1 (v) = F1−1 (v1 v2 . . . vk ) = Pv1 v2 ...vl0 ,vl0 +1 ...vk ∈ S
and let v 0 = fq1 ◦ fq2 ◦ · · · ◦ fqr , where q1 , . . . , qr ∈ Q are not necessarily dierent states, r ∈ N.
Let v 00 = fs(q1 ) ◦ fs(q2 ) ◦ · · · ◦ fs(qr ) . It is clear that v 00 ∈ S and v 00 = Pv2 ...vl0 vl0 +1 ,vl0 +2 ...vk vl0 +1 . So
F1 (v 00 ) = v2 . . . vl0 vl0 +1 vl0 +2 . . . vk vl0 +1 = v2 . . . vk vl ∈ S 0 .
Let S 0 ≤ (TX )k be a semigroup such that property (1) holds. Let us construct the
automaton A = (X, S 0 , π, λ), where π(x, v1 v2 . . . vk ) = v2 . . . vk vl and λ(x, v1 v2 . . . vk ) = v1 (x),
where v1 v2 . . . vk ∈ S 0 . The automaton is well dened because v2 . . . vk vl ∈ S 0 by property
(1) and v1 (x) ∈ X , because v1 ∈ TX . It is clear that A is an automaton without branches.
Moreover, the automaton transformation generated by state w = v1 v2 . . . vk has the form
Fww2 w2 ···w2 ··· = Pw1 ,w2 . Here w1 = v1 v2 · · · vl−1 , w2 = vl vl+1 vl+2 · · · vk . So we construct the
isomorphism F1 : S → S 0 ≤ (TX )k as above, where A generates the semigroup S .
Thus, it follows from Theorem 1 that automata without branches always generate nite
semigroups independently of their output function.
4. Mealy automata of nite growth. Let Q be a nonempty nite set of states. We say
that a transition function π : X × Q → Q generates an innite semigroup (group) of nite
automaton transformation if there exists an output function λ : X × Q → Q such that the
noninitial automaton A = (X, Q, π, λ) generates innite semigroup (group) of automaton
transformations. Otherwise we say that transition function generates only nite semigroups
(groups) of nite automaton transformations.
Transition functions without branches generate only nite semigroups and groups. However, there exists an other transition function and, therefore, other automata classes which
generates only nite semigroups. A more general (but not maximal) class was given in
Reznykov's verbal message by the following theorem.
Proposition 3. Let A = (Xm , Qn , π, λ) be a Mealy automaton such that the transition
function satises the following condition. Let there exists the number k ≥ 1 such that for
∗
of length k the equality π (u1 , q) = π (u2 , q)
any q ∈ Qn and for arbitrary words u1 , u2 ∈ Xm
holds. Thus, the automaton A generates the nite semigroup and the growth function γA is
almost constant.
9
ON MEALY AUTOMATA OF FINITE GROWTH
We will formulate a maximal class of transition functions which generates nite semigroups. Similar class for groups is considered in [9].
Recall that a state q ∈ Q is a state without branches if for all x1 , x2 ∈ X the equality
π(q, x1 ) = π(q, x2 ) holds, otherwise the state q ∈ Q is called a branch state. A state q ∈ Q
is called cyclic if there exists a nonempty word w ∈ X ∗ such that π (w, q) = q .
Let Q1 ⊆ Q, write Π (Q1 ) = π (X, Q1 ) = {π (x, q) |q ∈ Q1 , x ∈ X}. Set Π0 (Q1 ) = Q1 . Let
us prove by induction that Πk (Q1 ) = Π (· · · Π (Q1 ) · · ·) = {π (v, q) |q ∈ Q1 , v ∈ X ∗ , |v| = k }.
For k = 0 we have Π0 (Q1 ) = Q1 = {π (Λ, q) |q ∈ Q1 }, where Λ is the empty word. Assuming
Πl (Q1 ) = {π (v, q) |q ∈ Q1 , v ∈ X ∗ , |v| = l } for k = l, we have for k = l + 1
¡
¢
Πl+1 (Q1 ) =Π Πl (Q1 ) = Π ({π (v, q) |q ∈ Q1 , v ∈ X ∗ , |v| = l })
={π (x, π (v, q)) |q ∈ Q1 , v ∈ X ∗ , |v| = l , x ∈ X}
={π (vx, q) |q ∈ Q1 , v ∈ X ∗ , |v| = l , x ∈ X}
={π (v 0 , q) |q ∈ Q1 , v 0 ∈ X ∗ , |v 0 | = l + 1}
Let Q1 be a nonempty set such that Π (Q1 ) ⊆ Q1 (in particular, one can set Q1 = Q
since Π (Q) ⊆ Q). Let k ∈ N. If q ∈ Πk (Q1 ) then there exists a word v = v1 v2 . . . vk ∈ X ∗
and a state q 0 ∈ Q1 such that π (v, q 0 ) = q and, consequently, q = π (v2 . . . vk , π (v1 , q 0 )),
π (v1 , q 0 ) ∈ Π (Q1 ). It follows that π (v1 , q 0 ) ∈ Q1 , |v2 . . . vk | = k − 1, whence q ∈ Πk−1 (Q1 ).
Therefore if Π (Q1 ) ⊆ Q1 then the equality Πk (Q1 ) ⊆ Πk−1 (Q1 ) holds for T
any integer k ≥ 1.
k
∞
k
Furthermore, for any natural k we have Π (Q1 ) 6= ∅. Let Π (Q1 ) = ∞
k=1 Π (Q1 ).
k
k+1
k
k+1
k+2
If Π (Q1 ) = Π
(Q1 ) for some k ∈ N, then Π (Q1 ) = Π
(Q1 ) = Π
(Q1 ) =
· · · = Π∞ (Q1 ). Since the set Q1 is nite, there always exists k ∈ N, k ≤ |Q| such that
Πk (Q1 ) = Πk+1 (Q1 ) = Π∞ (Q1 ). Moreover Π∞ (Q) = Πk (Q) 6= ∅. Thus, we have
Q1 ⊇ Π (Q1 ) ⊇ Π2 (Q1 ) ⊇ · · · ⊇ Πk (Q1 ) = Πk+1 (Q1 ) = · · · = Π∞ (Q1 ) .
(2)
Summarizing the properties of Π, we have the following lemma.
Lemma 3. Let Q1 be a nonempty set of states such that Π (Q1 ) ⊆ Q1 . The sets Πi (Q1 ),
i ≥ 0, satisfy the following equalities:
1. Π(Π∞ (Q1 )) = Π∞ (Q1 );
2. there exists k ≤ |Q1 | such that the equality Π∞ (Q1 ) = Πk (Q1 ) holds;
3. for all l ≥ k the equality Πl (Q1 ) = Πk (Q1 ) holds.
Recall that the above statements are valid in the case Q1 = Q, too. In this case we have
the following characterization of the set Π∞ (Q).
Lemma 4. The set Π∞ (Q) consists of all states that are accessible from the cyclic ones.
Proof. Let us denote by S the set of all states that are accessible from the cyclic ones. Let q
be a cyclic state, i.e. for some nonempty word w ∈ X ∗ we have π (w, q) = q . Let l = |w|, then
l > 0. We have q = π (w, q) ∈ Πl (Q), q = π (ww, q) ∈ Π2l (Q), . . . , q ∈ Πil (Q), . . . , i ≥ 1.
Applying (2), for each natural j we obtain q ∈ Πj (Q), i.e. q ∈ Π∞ (Q). Let q 0 be accessible
from q , i.e. for some word v ∈ X ∗ (|v| = t) we have q 0 = π (v, q). Since q ∈ Π∞ (Q), we have
q 0 ∈ Πt (Π∞ (Q)). By Lemma 3, Πt (Π∞ (Q)) = Π∞ (Q). Therefore, q 0 ∈ Π∞ (Q). Hence, all
states accessible from the cyclic ones belong to Π∞ (Q), i.e. S ⊆ Π∞ (Q).
10
A. S. ANTONENKO
Let us prove that Π∞ (Q) ⊆ S . Let q ∈ Π∞ (Q) and let |Q| = n, then q ∈ Πn (Q) by
Lemma 3, i.e. there exists q0 such that q = π (w, q0 ), where w = w1 w2 · · · wn−1 wn ∈ X ∗ , q0 ∈
Q. Set qi = π (w1 w2 · · · wi−1 wi , q0 ) = π (wi , qi−1 ), where 0 < i 6 n. Since the number of
states |Q| = n, there are two equal states among the (n + 1) states q0 , q1 , . . . , qn−1 , qn . Let
qj = qj+r , r > 0, 0 6 j < j + r 6 n. Since qj = qj+r = π (wj+1 wj+2 . . . wj+r , qj ), qj is a cyclic
state. Moreover, q = qn+1 = π (wj+1 wj+2 · · · wn wn+1 , qj ) is accessible from the cyclic state qj ,
i.e. q ∈ S . We obtain Π∞ (Q) ⊆ S , thus Π∞ (Q) = S , which proves the lemma.
Similarly to the proof of Lemma 4, if Π (Q1 ) ⊆ Q1 ⊂ Q, then the set Π∞ (Q1 ) consists of
all states that are accessible from the cyclic states that belong to Q1 .
Denition 14. Let Q be a nite set of states. A transition function π : X × Q → Q is called
a transition function with limitary cycle if the set Π∞ (Q) consists of states without branches,
otherwise it is called a transition function without limitary cycle.
Theorem 2. Let π : X × Q → Q be a transition function with limitary cycle and A =
(X, Q, π, λ) be an arbitrary automaton with this transition function. Then A generates a
nite automaton transformation semigroup.
Proof. Let SA denote the semigroup generated by the automaton A. Let Π∞ (Q) contains
only states without branches, and let Π∞ (Q) = Πk (Q) (such a number k always exists by
Lemma 3). Let us denote by Ak the automaton with the set of states Πk (Q), the same
alphabet X and the corresponding restrictions of the transition
function
π and the output
¡
¢
function λ. This automaton is well dened, because π X, Πk (Q) = Πk+1 (Q) = Πk (Q) by
the denition of k . Note that the automaton Ak is an automaton without branches. Therefore
it denes a nite semigroup of automaton transformations SAk .
An automaton transformation dened by the automaton A at the state q has the form
fq (wk w) = fq (wk ) fπ(wk ,q) (w), where wk ∈ X ∗ , |wk | = k, w ∈ X ω , π (wk , q) ∈ Πk (Q),
fπ(wk ,q) ∈ SAk . Thus every f ∈ SA has the form f (wk w) = f 0 (wk )fw00k (w), where f 0 : X k →
X k , wk ∈ X k , fw00k ∈ SAk . The transformation f ∈ SA is uniquely determined by the function
f 0 : X k → X k and the function f 000 : X k → SAk , where f 000 (wk ) = fw00k , wk ∈ X k . Since there
are only a nite number of functions f 0 : X k → X k and f 000 : X k → SAk , the number of
elements of SA is nite. Moreover, SA ∼
= S ≤ TX o · · · o TX o SAk .
{z
}
|
k
This theorem gives a sucient condition for transition functions that determines only
nite semigroups. In the following section we will show that this condition is necessary.
5. Transition functions without limitary cycle. Now we can formulate our main result.
A transition function of Mealy automaton determines only nite semigroups if and only if it
is a transition functions with limitary cycle. In other words, a transition function determines
only nite semigroups (i.e. any automaton with such transition function determines a nite
semigroup independently of an output function) if and only if there exists an integer k such
that the automaton goes to a state without branches from any initial state under action of
any word of length greater than or equal to k . Formally, we have the following theorem.
Theorem 3. Let Q be a nite set of states. Let π : X × Q → Q be a transition function.
Then the following three conditions are equivalent:
1. the transition function π denes an innite semigroup;
ON MEALY AUTOMATA OF FINITE GROWTH
11
2. there exists a cyclic branch state for π ;
3. the transition function π is transition function without limitary cycle (in other words,
Π∞ (Q) contains a branch state).
In order to prove this theorem we will analyze transition functions without limitary cycle.
We will consider the three types of such transition functions. For an arbitrary transition
function π of each type we will construct a output function λ : X × Q → Q such that
the noninitial automata A = (X, Q, π, λ) generates an innite automaton transformation
semigroup SA . To this end we will select some transformation of the alphabet λq (x) =
λ (x, q) ∈ TX for each state. Let us x several elements of the set TX : α (x) ≡ 0, β (x) ≡
1, ε (x) ≡ x, α, β, ε ∈ TX , x ∈ X . We will denote by fq the automaton transformation
determined by the automaton A at the state q ∈ Q. In order to prove that the semigroup SA
is innite we will show innite order of some element f ∈ SA , namely we will select a word
¡
¢
2
i
w ∈ X ω such that the words f (w) , f (w) = f f (w) , . . . , f (w) , . . . are pairwise dierent.
2
i
This condition provides pairwise dierence of f , f , . . . , f , . . . as a transformation over X ω ,
whence the semigroup SA is innite.
Let π be a transition function such that Π∞ (Q) contains a branch state. Let us consider
all possible sequences of states of the form q0 , q1 = π (0, q0 ) , . . . , qi+1 = π (0, qi ) , . . .. Each
of them is ultimately periodic. So we can select from them their periodical parts (simple
cycles, marked by 0 in the Moore diagram) of the form C = {qz , qz+1 , . . . , qz+k = qz }, where
z > 0, k > 1. Only one of the following three cases is possible.
a) All these cycles consist of states without branches.
b) There is a cycle C , marked by 0, such that there exists a state q ∈ C and a symbol
x ∈ X such that π(x, q) 6∈ C (i.e. there exists a cycle which we can leave).
c) There exists a cycle with a branch state but there is no cycle described in case b).
We will review all this cases sequentially and construct automata which generate innite
semigroups for each case.
Lemma 5. Let π be a transition function such that all cycles, marked by 0, consists of
states without branches but the set Π∞ (Q) contains a branch state. Then there exists the
output function λ such that A = (X, Q, π, λ) generates an innite automaton transformation
semigroup.
Proof. Let us consider the following sets.
• the set Q1 of all states q ∈ Q such that there exists a nonnegative integer number
k = k (q) such that for each word w ∈ X ∗ of length greater than or equal to k (|w| > k)
the state π (w, q) will be a state without branches,
• the set Q2 = {q ∈ Q\Q1 |∃x ∈ X : π (x, q) ∈ Q1 } of all states, from which the automaton
can get to a state from Q1 under action of one input symbol.
• Q3 = Q\ (Q1 ∪ Q2 ) is the set of all other states.
Let us consider an example of an automaton which satises the conditions of Lemma 5.
Let X = {0, 1} and let the Moore diagram of the automaton be shown in Figure 2. We have
Q = {q0 , q1 , q2 , q3 , q4 , q5 }, Q1 = {q3 , q4 , q5 }, Π∞ (Q1 ) = {q4 , q5 }, Q2 = {q1 , q2 }, Q3 = {q0 }.
12
A. S. ANTONENKO
ε
q0
α
q1
0
1
0
1
1
β
q2
ε
q3
ε
q4
0
1
0,1
0,1
ε
q5
0
Fig. 2: Example of an automaton for Lemma 5
Let us prove that Π (Q1 ) ⊆ Q1 , and all states q ∈ Π∞ (Q1 ) are states without branches.
Indeed, let q ∈ Π (Q1 ), i.e. q = π (x, q1 ) , x ∈ X, q1 ∈ Q1 and also for q1 there exists k1 = k (q1 )
such that for each word w1 ∈ X ∗ of length greater then or equal to k1 (|w1 | > k1 ) all states
of the form π (w1 , q1 ) will be states without branches. Set k = k (q) = k1 − 1 (if k1 = 0, than
set k = 0). Then for each word w ∈ X ∗ of length greater than or equal to k (|w| > k) we
have π (w, q) = π (w, π (x, q1 )) = π (wx, q1 ).
Since the length of the word is greater than or equal to (k + 1), i.e. this length is greater
than or equal to k1 , the state π (w, q) is a state without branches. Therefore, it follows from
the denition of Q1 that q ∈ Q1 . Since q is arbitrary state from the set Π (Q1 ), we conclude
that Π (Q1 ) ⊆ Q1 . Since the set Q1 is nite, there exists kmax = max{k (q)} and Πkmax (Q1 )
q∈Q1
contains only states without branches, hence all states in the set Π∞ (Q1 ) are states without
branches. By lemma condition, Q 6= Q1 , hence the sets Q1 and Q2 are not empty, but the
set Q3 can be empty.
We will assign the transformation ε to each state from Q1 and Q3 . For each state q ∈ Q2
the set Xq = {x ∈ X|π (x, q) ∈ Q2 ∪ Q3 } is not empty (otherwise q ∈ Q1 ), but it does not
coincide with the alphabet X . For each q ∈ Q2 we choose an arbitrary element xq ∈ Xq , and
we assign to the state q the transformation
½
x, x ∈ Xq ;
λq (x) =
xq , x ∈
/ Xq .
Fix an arbitrary state q ∈ Q3 ∪ Q2 , the word w = 0∗ and the transformation f = fq .
The set Π∞ (Q1 ) is the union of all disjoint cycles which consist of states without branches,
Q1 is the set of all states, from which the automaton always moves to a state from Π∞ (Q1 )
after a while. Since Π (Q1 ) ⊆ Q1 , the property (2) holds. Thus for all q ∈ Q1 and v ∈ X ∗
the equality π (v, q) ∈ Q1 holds, i.e. the automaton moves from state which belongs Q1 only
to a state from Q1 .
Consider an action of an automaton transformation fq on arbitrary innite word of the
form
v = v 0 0∗ ∈ X ω , v 0 ∈ X ∗
(3)
ON MEALY AUTOMATA OF FINITE GROWTH
13
which contains only nite number of nonzero symbols. Let us prove that the automaton
starting with the state q under action of this word goes to a state from the set Q1 . Indeed,
either π (v 0 , q) ∈ Q1 , or if π (v 0 , q) ∈ Q2 ∪ Q3 , then the automaton starts to work at the state
π (v 0 , q), and processes the input sequence that consists of zeroes. Since there is no 0-cycles in
(n)
Q2 ∪ Q3 , the automaton goes to a state from
the set
¢ Q1 . We will denote by v = v1 v2 . . . vn
¡ (n−1)
(n)
the initial segment of v and by q = π v
, q the state to which the automaton goes
under action of rst n − 1 symbols of word v . Let c = c (v) be the minimal number such that
q (c+1) ∈ Q1 . It is obvious, that c > 1, q (c) ∈ Q2 . For all r < c − 1, we have q (r) ∈ Q2 ∪ Q3 ,
0
and for all r0 > c, we have q (r ) ∈ Q1 .
It follows from construction of the output function that the automaton changes symbol
of an input word if and only if it goes from a state from the set Q2 to a state from the
set Q1 . Therefore fq (v) diers from v only at the c-th position. The word fq (v) also has
form (3), so we can apply similar arguments to it, i.e. fq (fq (v)) diers from fq (v) only in
the position c (fq (v)). Moreover, c (fq (v)) > c (v), since the automaton xes symbol at
position c (v), after which the automaton goes to state from Q1 . Processing the input word
fq (v), the automaton works in the same way¡ until it reaches
position c (v) − 1 , i.e.
¢ the(c(v)+1)
(r)
(r)
(c(v))
q (fq (v)) = q (v) , r 6 c (v). But since π vc(v) , q
(v) = q
(v) ∈ Q1 and the
symbol xq(c(v)) (v) ∈ Xq(c(v)) (v) is located in c (v)-th position of the word fq (v), we see that
¡
¢
¡
¢
/ Q1 .
q (c(v)+1) (fq (v)) = π xq(c(v)) (v) , q (c(v)) (fq (v)) = π xq(c(v)) (v) , q (c(v)) (v) ∈
Therefore the automaton moves to a state that belongs to Q1 and changes the symbol later,
i.e. c (fq (v)) > c (v).
Applying sequentially the automaton transformation f = fq to the word w = 0∗ (this
(i)
word has the form (3)), we obtain the words of the form (3) fq (w) , fq (fq (w)) , . . . , fq (w) , . . ..
Each of these words diers from previous word only by one symbol (zero has been changed
to nonzero symbol). Moreover, the position of this symbol is located farther and farther from
the beginning of word. This proves innite order of fq as well as inniteness of semigroup
generated by the automaton.
The output function constructed by the rules described in Lemma 5 for the example given
above is shown in Figure 2. The automaton transformations fq0 , fq1 , fq2 have innite order,
so the semigroup dened by the automaton also has innite order. Indeed, fq2 (0∗ ) = 10∗ ,
fq22 (0∗ ) = fq2 (10∗ ) = 1010∗ and so on.
Lemma 6. Let π be a transition function such that there exist a cycle q0 , q1 , . . . , qk−1 ,
marked by 0, a state q ∈ {q0 , q1 , . . . , qk−1 } and a symbol x ∈ X \ {0} such that π (x, q) =
q0 ∈
/ {q0 , q1 , . . . , qk−1 }. Then there exists the output function λ such that A = (X, Q, π, λ)
generates an innite automaton transformation semigroup.
/
Proof. Without loss of generality we can assume q = q0 , x = 1, i.e. π(q0 , 1) = q 0 ∈
{q0 , q1 , . . . , qk−1 }. We will assign the transformation α to the states q0 , q1 , . . . , qk−1 , and the
transformation β to the state q 0 . See Figure 3 for an example in the case k = 4 (only
important transitions are shown). Set w = 1∗ .
Consider the action of automaton transformation fqj (0i 1v), where v ∈ X w , i > 0, 0 6
j < k −1, j +i ≡ 0 (modk). From the last congruence it follows that π (0i , qj ) = q0 . Therefore
under action of i zeroes the automaton goes from state qj to state q0 and outputs i zeroes.
14
A. S. ANTONENKO
β
q'
1
1
α
q0
0
α
q1
1
0
0
0
1
q3
α
0
q2
α
1
Fig. 3: Example of an automaton for Lemma 6
Then under the symbol 1, the automaton goes from state q0 to state q 0 and outputs the
symbol 0. After that the automaton emits the symbol 1. Thus
¡
¢
¡ ¢
fqj 0i 1v = fqj 0i fq0 (1) fq0 (v) = 0i 0fq0 (v) = 0i+1 1v 0
for some v 0 ∈ X w .
¡
¢
Let f = fq1 ◦ fq2 ◦ · · · ◦ fqk−1 ◦ fq0 and consider f 0ik 1v , where i is a nonnegative integer.
We obtain
¡
¢
fq0 0ik 1v = 0ik+1 1v1 since 0 + ik ≡ 0 (modk)
¡
¢
fqk−1 0ik+1 1v1 = 0ik+2 1v2 since k − 1 + ik + 1 ≡ 0 (modk))
···
¡ ik+k−1 ¢
fq1 0
1v1 = 0ik+k 1vk
¡
¢
¡ ¡
¡ ¡
¢¢¢ ¢
where v1 , . . . , vk ∈ X ω . Therefore f 0ik 1v = fq1 · · · fqk−1 fq0 0ik 1v
· · · = 0(i+1)k 1v 0
0
ω
for some word v ∈ X .
i
We have f (w) = f (1∗ ) = 0k 1v1 , . . . . By induction, f (w) = 0ik 1vi , vi ∈ X ω , which
proves innite order of f ∈ SA . Therefore semigroup SA dened by the automaton A is
innite.
Lemma 7. Let π be a transition function such that there is no a cycle C , marked by 0, a
state q ∈ C and a symbol x ∈ X \ {0} such that π (x, q) ∈
/ C . Assume that there exists a
0-cycle q0 , q1 , . . . , qk−1 which contains a branch state. Then there exists an output function
λ such that A = (X, Q, π, λ) generates an innite automaton transformation semigroup.
Proof. Without loss of generality we can assume that q0 is a branch state. Thus there exists
x ∈ X\{0} such that π (x, q0 ) 6= π (0, q0 ) = q1 . We can assume x = 1. Consider cyclic part
0
= π (1, qi0 ) , . . . of transitions under symbol 1
of sequence q00 = q0 , q10 = π (1, q00 ) , . . . , qi+1
0
0 0
= qt0 , where t > 0, l > 1. The
from the state q0 . This part has the form qt , qt+1 , . . . , qt+l
case where this cyclic part does not contain all states q0 , q1 , . . . , qk−1 is similar to Lemma 6
ON MEALY AUTOMATA OF FINITE GROWTH
ε
q0
0
0
1
1 1
q3
α
15
α
q1
1 0
0
q2
α
Fig. 4: Example of an automaton for Lemma 7
with interchanging of symbols 0 and 1 (i.e. there exists cycle marked by 1 with exit by 0 at
e = {q0 , q1 , . . . , qk−1 }, the
least in one state). In the other case, we consider the set of states Q
e = {0, 1} and the corresponding restriction of the transition function. They dene
alphabet X
group automaton with two dierent permutations ς0 è ς1 of states under actions of symbols
e q ∈ Q
e and ς1 is an permutation because it is
of alphabet. Here ςx (q) = π (x, q) , x ∈ X,
surjective transformation of a nite set on itself. These two permutations are cycles of length
k , so their orders are k . Moreover, ς0 is a cyclic permutation, that is ς0 (qi ) = q(i+1) mod k . We
have ς0 (q0 ) 6= ς1 (q0 ) because q0 is a branch state.
We will assign the transformation ε to the state q0 , transformation α to the states
q1 , . . . , qk−1 and the transformation
³ ε to other
´ states. Thus we have dened the two aue , where π
e: X
e = X,
e Q,
e π
e ×Q
e→Q
e and λ
e ×Q
e→X
e
tomata A = (X, Q, π, λ) and A
e, λ
e: X
are
restrictions
³ corresponding
´
³
´of functions π and λ (these function are well dened because
e Q
e ⊂ Q
e and λ X,
e Q
e ⊂ X)
e . See Figure 4 for an example of A
e. The semigroup
π X,
e is a subsemigroup of the factor semigroup of the semigroup generated by
generated by A
e, whence the innity of semigroup
A. Let us prove the innity of semigroup generated by A
∗
generated by A concludes. To this end
we ¢set w = 1 , f = fq0 .
¡ k−1
∗
∗
Then we have f (w) = fq0 (1 ) = 10
since the order of ς1 equals k , i.e. the automaton
e ∗,
will return to the³state q0 after
reading k symbols.³Let b0 =´0k , b1 = 10k−1 , b0 , b1 ∈ X
´
eω ∪ X
e ∗ , where u = u1 u2 . . . ∈ X
eω ∪ X
e ∗ is a nite or innite word.
Bu = bu1 bu2 . . . ∈ X
We have
π (b0 , qi ) = ς0k (qi ) = q(i+k) mod k = qi , λ (b0 , qi ) = b0 , and π (b1 , qi ) = ς0k−1 (ς1 (qi ))
for each integer i such that 0 ≤ i ≤ k − 1. Also λ (b1 , q0 ) = b1 , λ (b1 , qi ) = b0 for each integer
i such that 1 ≤ i ≤ k − 1. Note that g = ς0k−1 ◦ ς1 = ς0−1 ◦ ς1 is not the identity permutation.
Moreover, g (q0 ) = ς0−1 (ς1 (q0 )) 6= q0 , since otherwise ς1 (q0 ) = ς0 (g (q0 )) = ς0 (q0 ), that
contradicts the assumption that ς0 (q0 ) 6= ς1 (q0 ). Let the order of the permutation g be
e ∗ if and only if the
equal to r (r > 1). In this case, π (Bu , q0 ) = q0 for a nite word u ∈ X
quantity of ones in the word u is divisible by r, since each block of the form b1 acts on
16
A. S. ANTONENKO
state as permutation g , and each block of the form b0 acts on state as identity permutation.
e ω , u0 = u0 u0 . . . ∈ X
e ω,
Therefore fq0 (Bu ) = Bu0 , where
u = u1 u2 . . . ∈ X
1 2 ¢
¡
½
1, if ui = 1 and π Bu1 u2 ...ui−1 , q0 = q0
u0i =
0, otherwise,
i.e.
½
1, if ui = 1 and the count of ones in the word u1 u2 . . . ui−1 is divisible by r,
0
ui =
0, otherwise.
Thus in the word u only the rst, r + 1-th, 2r + 1-th and so on letters remain, and all
other letters are changed to 0.
³
´∗
j−1
Let us prove by induction that for all j > 1, fqj0 (1∗ ) = Bu(j) , where u(j) = 10(r −1) .
³
´∗
1−1
∗
For j = 1 we have u(1) = 10(1 −1) = (100 ) = 1∗ , fq0 (1∗ ) = B1∗ = Bu(1) . Assume that
³
´∗
rt−1 −1)
t
∗
(t)
(
for j = t the equality fq0 (1 ) = Bu(t) , u = 10
holds. Then for j = t + 1 we have
u(t)
¡ t ∗ ¢
∗
fqt+1
(1
)
=
f
fq0 (1 ) = fq0 (Bu(t) ) = Bu0 ,
q
0
0
Ã
!∗
³
´∗
t−1
t−1
t−1
t−1
(r −1) · 10(r −1) · · · · · 10(r −1) .
= 10(r −1) = 10
|
{z
}
r
Acting by transformation fq0 we obtain
!∗
Ã
u0 =
(
|10
) · 00(
rt−1 −1
) · · · · · 00(
rt−1 −1
{z
)
}
rt−1 −1
³
´∗
t
= 10(r −1) = u(t+1) .
r
Hence the words fqj0 (1∗ ) , j = 1, 2, . . . are pairwise dierent. It follows that the order of
e is innite. Thus the semigroup dened
fq0 is innite. Therefore the semigroup dened by A
by A is also innite.
Proof of Theorem 3. Let condition 2 hold, i.e. there exists a cyclic branch state for π . By
Lemma 4, this state belongs to Π∞ (Q), so condition 3 holds.
Let condition 3 hold, i.e. Π∞ (Q) contains a branch state. Let us prove that condition 2 holds. On the contrary, suppose that all cyclic states are states without branches.
Let q ∈ Q be an arbitrary cyclic state, i.e. for some nonempty word w = w1 . . . wk ∈ X ∗ the
equality π (w, q) = q holds. Then q 0 = π (w1 , q) is a cyclic state, since π (w2 . . . wk w1 , q 0 ) =
π (w2 . . . wk w1 , π (w1 , q)) = π (w1 w2 . . . wk w1 , q) = π (w1 , q) = q 0 . Since q by assumption is
a state without branches, for all x ∈ X we have π (x, q) = π (w1 , q) = q 0 , i.e. the automaton
being in a cyclic state goes only to cyclic states. Thus the set Π∞ (Q), being equal to the set
of states accessible from the cyclic ones, is equal to the set of all cyclic states. Since there
is no branch cyclic state, it follows that there is no branch state which belongs to Π∞ (Q).
This contradicts to condition 3.
Let us prove by contradiction that condition 3 follows from condition 1. Suppose that
a transition function π : X × Q → Q denes an innite semigroup but Π∞ (Q) consists only
of states without branches. By Theorem 2 an arbitrary automaton A = (X, Q, π, λ) with the
transition function π denes nite semigroup. This contradicts to our assumption.
It suces to prove that condition 1 follows from condition 3. Let condition 3 hold, i.e.
Π∞ (Q) contains a branch state. By the above, the transition function π satises conditions of
one of Lemmas 5, 6, 7. Therefore, there exists an output function λ such that A = (X, Q, π, λ)
ON MEALY AUTOMATA OF FINITE GROWTH
17
denes an innite automaton transformation semigroup. Thus the transition π generates an
innite automaton transformation semigroup, which proves the theorem.
6. Conclusion. A large number of automata of nite growth are described in this paper,
namely automata with limitary cycle. It is possible to make a conclusion that they have nite
growth judging only by transition function. For all other transition functions automata that
generate innite semigroups are constructed.
There are three interesting ways to extend the results of the paper:
1. for a transition function without limitary cycle to build an invertible automaton which
denes innite group;
2. to characterize all automata of nite growth;
3. to construct a criterion similar to Theorem 3 for other classes of growth: polynomial,
intermediate, exponential.
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Department of Computer Algebra and Discrete Mathematics
Institute of Mathematics, Economics and Mechanics
Odessa I.I. Mechnikov National University
Dvoryanskaya st.,2 65026 Odessa, Ukraine
[email protected]
Received 14.12.2006