The Accepting Power of Finite
Automata Over Groups
Victor Mitrana
Faculty of Mathematics University of Bucharest
Str Academiei Bucharest ROMANIA
Ralf STIEBE
Faculty of Computer Science University of Magdeburg
P O Box D Magdeburg GERMANY
Turku Centre for Computer Science
TUCS Technical Rep ort No
Novemb er
ISBN ISSN
Abstract
Some results from are generalized for nite automata over arbitrary
groups The accepting p ower is smaller when ab elian groups are considered
in comparison with the non ab elian groups We prove that this is due to
the commutativity Each language accepted by a nite automaton over an
ab elian group is actually a unordered vector language Finally deterministic
nite automata over groups are investigated
TUCS Research Group
Mathematical Structures of Computer Science
Intro duction
One of the oldest and most investigated machine in the automata theory is
the nite automaton Many fundamental prop erties have b een established
and many problems are still op en
Unfortunately the nite automata without any external control have a
very limited accepting p ower Di erent directions of research have b een con
sidered for overcoming this limitation The most known extension added to a
nite automata is the pushdown memory In this way a considerable increas
ing of the accepting capacity has b een achieved The pushdown automata
are able to recognize all context free languages
Another simple and natural extension related somehow to the pushdown
memory was considered in a series of pap ers namely to asso ciate to
each con guration an element of a given group but no information regarding
the asso ciated element is allowed This value is stored in a counter An input
string is accepted if and only if the automaton reaches a designated nal state
with its counter containing the neutral element of the group
Thus new characterizations of unordered vector languages or context
free languages have b een rep orted These results are in a certain sense
unexp ected since in such an automaton the same choice is available regardless
the content of its counter More precisely the next action is determined just
by the input symb ol currently scanned and the state of the machine
In this pap er we shall consider only acceptors with a one way input tap e
read from left to right and a counter able to store elements from a given
group The aforementioned pap ers deal with nite automata over very well
de ned groups e g the additive group of integers the multiplicative group
of non null rational numb ers or the free group The aim of this pap er is to
provide some general results regardless the asso ciated group
We shall prove that the addition of an ab elian group to a nite automaton
is less p owerful than the addition of the multiplicative group of rational
numb ers An interchange lemma p oints out the main reason of the p ower
decrease of nite automata over ab elian groups Characterizations of the
context free and recursively enumerable languages classes are set up in the
case of non ab elian groups
As far as the deterministic variants of nite automata over groups are
concerned we shall show their considerable lack of accepting p ower
Preliminaries
We assume the reader familiar with the basic concepts in automata and
formal language theory and in the group theory For further details we refer
to and resp ectively
For an alphab et we denote by the free monoid generated by
under the op eration of concatenation the empty string is denoted by and
+
the semigroup f g is denoted by The length of x is denoted
by jxj
Let K M e b e a group under the op eration denoted by with the
neutral element denoted by e An extended nite automaton EFA shortly
over the group K is a construct
A Z K q F
0
where Z q F have the same meaning as for a usual nite automaton
0
namely the set of states the input alphab et the initial state and the set of
nal states resp ectively and
Z f g P Z M
f
This sort of automaton can b e viewed as a nite automaton having a counter
in which any element of M can b e stored The relation q m s a q s
Z a f g m M means that the automaton A changes its current
state s into q by reading the symb ol a on the input tap e and writes in the
register x m where x is the old content of the register The initial value
registered is e
We shall use the notation
q aw m j s w m r i s r q a
A
for all s q Z a f g m r M The re exive and transitive closure
Sometimes the subscript identifying of the relation j is denoted by j
A
A
the automaton will b e omitted when it is self understo o d
The word x is accepted by the automaton A if and only if there
is a nal state q such that q x e j q e In other words a string is
0
accepted if the automaton completely reads it and reaches a nal state when
the content of the register is the neutral element of M
The language accepted by an extended nite automaton over a group A
as ab ove is
L A fx j q x e j q e for some q F g
0
A
For two groups K M e and K M e we de ne the triple
1 1 1 1 2 2 2 2
K K M M e e with m m n n m n m n
1 2 1 2 1 2 1 2 1 2 1 1 1 2 2 2
It is well known that K K is also a group
1 2
We are going to provide some results that will b e useful in what follows
The notation L RE G identi es the class of regular languages
Theorem For any group K L EFA K L RE G i al l nitely gen
erated subgroups of K are nite
Proof Let K b e a group such that any nitely generated subgroup of K is
nite Let A Z K z F b e an EFA over K M e We denote
0
by X the nite subset of M
X fm M j z m z a for some z z Z a f gg
Let H hX i e b e the subgroup generated by X
We construct the nite automaton with moves B Z hX i z e
0
F feg with z m a f z m n j z n z a for all z Z
z m i m M a f g One can easily prove that z w e j
0
A
z e w j z m which implies L A L B
0
B
It remains to prove that for any in nite group K nitely generated exists
an EFA over K accepting a non regular language Let K hX i e b e such
a group with the nite set of generators X Consider the deterministic EFA
1
A fz g Y K z fz g with Y X fx j x X g and z a z a
for all a Y The following facts ab out L A are obvious
z m For any m hX i exist a word v Y such that z v e j
A
For any v Y exists a word w Y such that v w L A
For any k the set
z m g X fm hX i j v Y jv j k z v e j
k
A
is nite
As a consequence of these facts and of the in niteness of hX i we obtain For
all k there is a word v such that v v L A for all v Y jv j k
k k
But one can easily prove that for any regular language L Y exists
k such that for all v w L there is w Y jw j k with v w L
Hence L A cannot b e regular 2
A nitely generated ab elian group is nite if all its elements are of nite
order Hence for an ab elian group K L EFA K L RE G i all ele
ments of K have nite order This is not necessarily true for non ab elian
groups We can however prove a pumping lemma which is very similar to
the pumping lemma for regular languages
Lemma Let K be some group without elements of in nite order For any
language L L EFA K there is a constant n such that for al l x L
jxj n there exist a decomposition x uv w and a natural number q
iq +1
with juv j n jv j uv w L for al l i
Moreover if K has the nite exponent p then q can uniformly be chosen
as q p
Proof Let A Z K z F b e an EFA over K We cho ose n jZ j
0
Now consider a word x with jxj n Similar to the pro of of the pumping
lemma for regular languages it can b e shown that there is a decomp osition
x uv w juv j n jv j such that
z uv w e j z v w m j z w m m j f e z Z f F
0 1 1 2
A A A
q
iq +1
Now cho ose q such that m e Obviously any word uv w is accepted
2
by A 2
As a consequence of the ab ove pumping lemma we obtain
Theorem For any group K L EFA K contains the language L
n n
fa b j n g i at least one element of K has an in nite order
Proof Let K M e If M contains an element m of in nite order then
the nitely generated subgroup hmi e is isomorphic to ZZ hence
L is in L EFA K
If all elements of M have nite order a simple application of the ab ove
pumping lemma yields L L EFA K 2
For a group K let F K denote the family of all nitely generated sub
groups of K
Theorem For any group K
L EFA K L EFA H
H F (K)
Proof Let K M e b e a group The inclusion
L EFA K L EFA H
H F (K)
holds since L EFA K L EFA H for any subgroup H of K
On the other hand let A Z K z F b e an EFA over K The
0
group H hX i e where X fm M j q m z a for some q z
Z a f gg is a nitely generated subgroup of K Obviously during any
computation in the counter of A app ear only elements of hX i Therefore
the automaton A can b e viewed as an automaton over H More precisely
A Z H z F accepts the same language as A do es This proves the
0
second inclusion and thus the theorem 2
EFA over ab elian groups
Valence grammars and EFA have initially b een intro duced for the groups
k
Z ZZ k and Q Q f g In what follows we shall
k
show that the accepting capacity of EFA do es not increase if we consider
arbitrary ab elian groups instead of Q Thus every language accepted by an
EFA over an ab elian group is a unordered vector language The deep er
reason of this fact is the following fundamental result in the group theory
Theorem A nitely generated abelian group is the direct product of a nite
number of cyclic groups
As a consequence a nitely generated ab elian group is either nite or
isomorphic to a group Z H where k is a p ositive integer and H is a nite
k
ab elian group
Theorem For a group K and a nite group H
L EFA K H L EFA K
Proof Let K and H b e given by K M e and H M e
1 1 1 2 2 2
and let A Z K H z F b e an EFA over K H We construct
0
z e F with Z Z M z the EFA over K A Z K z
0 2 2
0 0
F F fe g and
2
z n a f z n m m j z m m z a g
2 2 2 2 1 1 2
z z Z a f g m M m n M
1 1 2 2 2
By induction on the numb er of steps one can show that z e w e j
0
0 2 1
A
z w m m hence L A L A z m w m i z w e e j
1 2 2 1 0 1 2
A
2
We are now ready to prove the main result of this section
Theorem For an abelian group K one of the fol lowing relations hold
L EFA K L RE G
L EFA K L EFA Z for some k
k
L EFA K L EFA Q
S
Proof As it was shown in Theorem L EFA K L EFA H
H F (K)
Every H F K is either nite or isomorphic to a group Z H where k
k
and H is a nite group Hence for all H F K either L EFA H
L RE G or L EFA H L EFA Z H L EFA Z for some
k k
k
If all nitely generated subgroups of K are nite then L EFA K
L RE G holds due to Theorem
Otherwise let N K b e the set of all k such that L EFA H L EFA Z
k
for some H F K If N K is nite then L EFA K L EFA Z
k
where k max N K If N K is in nite then L EFA K L EFA Q
2
n n R
It is known that languages as L fa b j n g or L fw cw j
1 2
+
w fa bg g are not in L EFA Q Therefore L L L EFA K for
1 2
any ab elian group K In it was conjectured that the commutativity of
the multiplication of rational numb ers is resp onsible for this fact We shall
formally prove this conjecture by help of the following interchange lemma
Lemma Let K M e be some abelian group and let L be a language
in L EFA K There is a constant k such that for any x L jxj k and
any decomposition x v w v w v w v jw j exist two integers
1 1 2 2 k k k +1 i
w v with w v w v w r s k such that the word x v w
s k +1 k 2 1
r k 2 1
w for i fr sg is in L w w w
i r
i s
Proof Let A Z K z F b e an EFA over K M e We cho ose
0
2
k jZ j For a word x L A jxj k let b e given a decomp osition
x v w v w v w v jw j There are the states y z Z i
1 1 2 2 k k k +1 i i i
k q F with
z v e j y m i k
i 1 i i i
y w e j z n i k
i i i i
z v e j q m
k k +1 k +1
and m n m n m n m e
1 1 2 2 k k k +1
By the pigeon hole principle there are two numb ers r s k with
y z y z
r r s s
v w v w v with w w w w w Now consider x v w
2 k k +1 s r 1
1 2 k r s i
i k the w for i fr sg For the words v i k and w
i i
i
relations
z v e j y m i k
i 1 i i i
i k e j z n y w
i i
i i
z v e j q m
k k +1 k +1
hold with n n n n n n for i fr sg By the commutativity of
s r i
r s i
K it follows
m n m n m n m
1 2 k k +1
1 2 k
m n m n m n m
1 1 2 2 k k k +1
e
implying that x is accepted by A 2
n n
Theorem For any abelian group K the languages L fa b j n g
1
R +
and L fw cw j w fa bg g are not in L EFA K
2
Proof Assume that L L EFA K for some K and let k b e the constant
1
2 2 k k
from the interchange lemma Now consider the word x aba b a b
i i
and the decomp osition v a w b i k v There are
i i k +1
2 k s r
r s k such that x aw a w a w with w b w b
1 2 k r s
i
b for i fr sg is in L contradiction w
1
i
A similar reasoning for the relation L L EFA K is left to the reader
2
2
EFA over non ab elian groups
In this section we restrict our investigation to the free groups since for any
non ab elian group K there is a homomorphism from a free group to K
In this way we get a characterization of the context free languages class in
terms of languages accepted by extended nite automata over the free group
with just two generators The free group with n generators is denoted by
F
n
Recall from
Theorem The family of context free languages equals L EFA F
2
Lemma Let K and K be two groups For two languages L L EFA K
1 2 i i
i the languages L L and L L are in L EFA K K
1 2 1 2 1 2
0
Proof Let A Z K z F i b e two EFA over K resp ec
i i i i i i i
i
tively We assume that Z Z is empty
1 2
We have L B L L and L C L L where B Z Z
1 2 1 2 1 2 1
0 0
K K z z F F with
2 1 2 1 2
1 2
z z a f z z m m j z m z a z m z a g
1 2 1 2 1 1 1 2 2 2
1 2 1 2
z Z z Z a
1 1 2 2 1 2
z z f z z m e j z m z g
1 2 2 1 2 1 1 1
1 1
f z z e m j z m z g
1 1 2 2 2 2
2 2
z Z z Z
1 1 2 2
0
and C Z Z K K z F with
1 2 1 2 1 2 2
1
z a f z m e j z m z a g z Z a
2 1 1 1
f z m e j z m z g z Z n F
2 1 1 1
z
0
f z m e j z m z g f z e e g z F
2 1 1 2
2
z a f z e m j z m z a g z Z a f g
1 2 2 2
which concludes the pro of 2
It is well known that every recursively enumerable language can b e ex
pressed as the homomorphical image of the intersection of two linear lan
guages It is obvious that each family L EFA K is closed under homomor
phims In conclusion due to the previous lemma as well as to Theorem
we have just proved
Theorem L EFA F F equals the family of recursively enumerable
2 2
languages
Deterministic EFA
In the case of EFA the determinism signi cantly decreases the accepting
capacity Denote by L DEFA K the family of languages recognized by
deterministic EFA over the group K
n n n
Lemma For any group K the languages L fa j n g fa b j
1
m n n n
n g L fa b j m ng and L fa bg n fa b j n g are not in
2 3
L DEFA K
Proof Let K M e and let A Z fa bg K z F b e a DEFA
0
n
over K such that L L A Since a L A for all n there are
1
r
q e and the integers r s jF j such that z a e j
0
A
n n s
q e for some q Z Since a b L A for all n z a e j
0
A
r r r
q b e j q e for some q F Hence we have also z a b e j
0
A A
r s r s r
q b e j q e implying a b L A contradiction z a b e j
0
A A
In conclusion L L A
1
By similar arguments it can b e shown that L and L are not in L DEFA K
2 3
2
On the other hand L DEFA K generally contains languages with un
decidable memb ership problem
Theorem There is a nitely generated group K such that the fol lowing
question is undecidable Given a DEFA A over K and a word w over the
input alphabet of A is w L A
Proof For a nitely generated group K M e with the set of generators
X the word problem is the following question Is a given term x x x
1 2 n
1 1 1
x X X i n X fx j x X g equal to the neutral
i
element e It is a well known result from group theory that there is a nitely
generated group K with undecidable word problem
Let K M e with the nite set of generators X b e such a group
1
We construct the DEFA A fz g X X K z fz g with the transition
1
function z x z x for all x X X Obviously A accepts a word
x x x i x x x e The undecidability of the memb ership
1 2 n 1 2 n
problem for A follows directly from the undecidability of the word problem
for K 2
Theorem For every group K having at least one element of in nite or
der we have L DEFA K L EFA K strict inclusion
Proof Let K b e an arbitrary given group and A Z K z F b e a
0
deterministic EFA over K
n
If there is an element of M of in nite order then the language fa jn
n n
g fa b jn g L EFA K since L EFA K is closed under union
In conclusion L DEFA K L FEA K 2
Deterministic EFA over ab elian groups
The next result is a consequence of Theorem
Theorem For every abelian group K we have either L DEFA K
L RE G or L DEFA K L EFA K strict inclusion
Obviously the statement of Theorem is also valid for the deterministic
EFA over ab elian groups
As we have seen neither L EFA Q nor L DEFA Q are closed
under complement In the end of this section we will show that the com
plement of a language from L DEFA Q is in L EFA Q In order to
handle the di culties owing to the existence of steps we intro duce some
notations
Let A S Z s F b e a DEFA over Z for some k For all
k 0 k
s S we de ne the sets
k
N fr ZZ j s j q r for some q F g
s
Lemma Let A be a DEFA as above For al l s S there is an EFA A
s
Y Z s fq g such that s a for al l a and s j
s k s s s
k
q r i r ZZ n N
s s
k
Proof In what follows let e i k denote the i th unit vector of ZZ
i
For any s S we construct the EFA A s S fq g T Z s fq g with
1 k 1
q S T fa a g fb b g and the transition relation de ned
1 k 1 k 1
as
p a for p S a T
1
p p for p S n F
1
p p f q g for p F
1
q a f q e for i k
1 i i
q b f q e for i k
1 i i
q
1
Obviously the language accepted by A s is
1
jw j jw j N g jw j L fw T j jw j
a b s b s a
1 1
k k
2k
jw j Let T IN b e the Parikh mapping with w jw j
b a
1 1
jw j jw j for all w T The Parikh set L is semilinear and its
a b s
k k
complement L is semilinear to o Therefore there is a nite automaton
s
L such that the Parikh set of L A s is A s Y T s fq g
s 2 2 s s
s
From A s we can construct A as follows A Y Z s fq g with
2 s s s k s s
p a for p Y a
s s
p f p e j p p a i k g
s i i
s
p b i k g for p Y f p e j p
i s i
s
Obviously s j q r i L A s contains a word w with r
s 2
A
s
jw j jw j jw j jw j i e i L contains no word v with r
a b a b s
1 1
k k
jv j jv j jv j jv j hence i r N 2
a b a b z
1 1
k k
Theorem For al l L L DEFA Q the complement of L is contained
in L EFA Q
Proof Let A S Z s F b e a DEFA over Z The set of states can
k 0 k
b e partitioned into the set R consisting of all states s S such that A can
p erform only steps if s is reached and its complement S n R
The complement of L A consists of two sets rst the set of all words
k
+
w w w with w and s w j p w r for some r ZZ
1 2 2 0 2
A
+
and some p R second the set of all words w w a with a and
1
s w j p a r j s r for some p s S and r t for all
0 A
A
k
t N The last condition is equivalent to r t for some t ZZ n N
s s
Now let for all s S A Y Z s fq g b e the DEFA constructed
s s k s s
in the last lemma Without loss of generality we may assume that Y and
s
0
are disjoint for s s and that Y and S are disjoint for all s S Y
s s
Now we construct B S Z s F with the set of states S S
k 0
S
Y ff g the set of nal states F fq j s S g ff g and the following
s s
s S
transition mapping
s if s S fs g
0
s s f f g if s s
0
s if s Y for some x S
x x
0
r j s r s a g if s S R s a f y
s
s a
0
r j s r s a g f f g if s R s a f y
s
f a f f g
f f f m e g m f g i k
i
for all a
It is easy to see that B accepts all words not in L A On the other hand
no word from L A is accepted by B 2
An imp ortant consequence of the last theorem is the decidability of the
inclusion problem and of the equivalence problem to o for DEFA over Q
which is an interesting contrast to the undecidability of the universe problem
for nondeterministic one turn counter automata a prop er sub class of EFA
over ZZ
Theorem Let A and B be DEFA over Q It is decidable whether or not
L A L B
Proof Given B one can construct a nondeterministic EFA B over Q with
L B L B Clearly L A L B i L B L A is empty Now an EFA
C over Q can b e constructed such that L C L B L A Hence the
inclusion problem for DEFA over Q is reduced to the emptiness problem for
EFA over the same group which is decidable 2
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