Concatenation Hierarchies and Forbidden Patterns Christian Glaßer Heinz Schmitz Theoretische Informatik

Transcription

Concatenation Hierarchies and Forbidden Patterns
Christian Glaßer
Heinz Schmitz
Theoretische Informatik
Julius–Maximilians–Universität Würzburg
97074 Würzburg, Germany
glasser, schmitz@informatik.uni-wuerzburg.de
Abstract
We make the following progress on the dot–depth problem:
(1) We introduce classes and of starfree languages defined via forbidden patterns in finite
automata. It is shown for all that ( ) contains level of the dot–depth hierarchy
(Straubing–Thérien hierarchy, resp.). Since we prove that and have decidable membership
problems, this yields a lower bound algorithm for the dot–depth of a given language.
(2) We prove many structural similarities between our hierarchies and , and the mentioned concatenation hierarchies. Both show the same inclusion structure and can be separated by
the same languages. Moreover, we see that our pattern classes are not too large, since does not
capture level of the dot–depth hierarchy. We establish an effective conjecture for the dot–
depth problem, namely that and , and the respective levels of concatenation hierarchies in fact
coincide. It is known from literature that this is true for .
(3) We prove the decidability of level of the Straubing–Thérien hierarchy in case of a two–
letter alphabet — in exactly the way as predicted by our conjecture. To our knowledge, no results
concerning the decidability of this class (and of any level of these concatenation hierarchy)
were previously known. The result is based on a more general upward translation of decidability,
connecting concatenation hierarchies and our pattern hierarchies.
Keywords: Dot–depth problem, concatenation hierarchies, finite automata, decidability
Supported by the Studienstiftung des Deutschen Volkes.
Supported by the Deutsche Forschungsgemeinschaft (DFG), grant Wa 847/4-1.
1
Contents
1
Introduction
2
Preliminaries
2.1 Basic Definitions . . . . . . . . . . . .
2.2 Definition of Concatenation Hierarchies
2.3 Alternative Definitions . . . . . . . . .
2.3.1 Polynomial Closure . . . . . . .
2.3.2 Boolean Closure . . . . . . . .
2.3.3 Comparing the Lower Levels . .
2.3.4 Comparing the Hierarchies . . .
2.4 Normalforms and Closure Properties . .
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A Theory of Forbidden Patterns
3.1 Definition of Iterated Patterns . . . . . . .
3.2 Auxiliary Results for Iterated Patterns . .
3.3 Pattern Iterator versus Polynomial Closure
3.4 Inclusion Structure of the classes . . .
3.5 Pattern Iteration remains Starfree . . . . .
3.6 Decidability of the Pattern Classes . . . .
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18
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Consequences for Concatenation Hierarchies
4.1 Some Auxiliary Results . . . . . . . . . . . . . . . . . . . . .
4.2 Inclusion Relations and Lower Levels . . . . . . . . . . . . .
4.3 Pattern Classes are Starfree . . . . . . . . . . . . . . . . . . .
4.4 The Hierarchy of Pattern Classes is Strict . . . . . . . . . . .
4.5 Decidability of the Pattern Classes and . . . . . . . . .
4.6 Lower Bounds and a Conjecture for Concatenation Hierarchies
5.1
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is decidable for two–letter alphabets
Changing the Alphabet . . . .
Automata Construction . . . .
Transformation of Patterns . .
Transformation of Expressions
Connecting Hierarchies . . . .
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2
List of Figures
1
2
3
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9
10
Example for an appearance of a pattern at some state . . . . . . . . . . . . . . . . . . .
Example for a connection of two states via a pattern . . . . . . . . . . . . . . . . . . . .
Concatenation Hierarchies and Forbidden Pattern Classes . . . . . . . . . . . . . . . . .
Forbidden Pattern for . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Forbidden Pattern for . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Forbidden Pattern for . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Forbidden Pattern for from [GS00] . . . . . . . . . . . . . . . . . . . . . . . . . .
3
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65
1 Introduction
We contribute to the study of concatenation hierarchies and focus on starfree regular languages, which
are constructed from alphabet letters using Boolean operations together with concatenation. Alternating
these two kinds of operations leads to the definition of hierarchies that further classify starfree languages.
Here we deal with the dot–depth hierarchy [CB71] and the closely related Straubing–Thérien hierarchy
[Str81, Thé81, Str85]. Their investigation is of major interest in many research areas since surprisingly
close connections have been exposed, e.g., to finite model theory, theory of finite semigroups, complexity
theory and others (for an overview, see [Pin96]).
Let be some finite alphabet with . For a class of languages let be its polynomial
closure, i.e., the closure under finite union and concatenation. Denote by its Boolean closure
(taking complements with respect to ). Then the classes of the dot–depth hierarchy and the
classes of the Straubing–Thérien hierarchy can be defined as follows.
for for By definition, all these classes are closed under union and it is known, that they are also closed under
intersection [Arf91].
The question whether there exists an algorithm solving the membership problem for the levels of
these hierarchies is known as the dot–depth problem, recently considered as one of the most important
open questions on regular languages [Pin98]. Although many researchers believe the answer should be
positive, some suspect the contrary. Up to now, levels , and of both hierarchies are known to be
decidable [Sim75, Kna83, Arf87, PW97, GS00], while the question is open for any other level. Partial
results are known only for level of the Straubing–Thérien hierarchy. It is decidable if a two–letter
alphabet is considered [Str88].
We take up the discussion started in [GS00] and continue known characterizations of lower levels
in a natural way to a more general approach. More precisely, based on recent forbidden pattern characterizations of , and in [PW97], and of in [GS00], we introduce hierarchies defined
by forbidden patterns in deterministic finite automata (dfa, for short). We obtain strict and decidable
hierarchies of classes and , which exhaust the class of starfree languages and which are closely
related to the dot–depth hierarchy and the Straubing–Thérien hierarchy. We prove that
for .
These inclusions imply a lower bound algorithm for the dot–depth of a given language . Just determine
the class or for minimal to which belongs. It follows that has at least dot–depth .
For another result of this type see [Wei93].
We obtain more structural similarities. The pattern hierarchies show the same inclusion structure as
known for concatenation hierarchies. Typical languages separating the levels of the dot–depth hierarchy
witness also the strictness of our pattern hierarchies, and it holds that does not capture . We
take all this as reasons to conjecture that in fact and for all . This
conjecture agrees with the known decidability results, and we uniformly iterate with the definition of
the pattern classes, what we have learned from these levels. If an effective characterization in terms of
forbidden patterns in automata is possible at all, we believe that our conjecture provides a step in this
direction. Other conjectures for the dot–depth problem can be found in the literature, e.g., the one in
[PW97], note that we would have as a consequence an effective characterization.
4
Our results are applications of more general ones. For a so–called initial pattern (fulfilling some
reasonable weak prerequisites) we define to be the class of languages defined by in a forbidden
pattern manner, and we recursively iterate to obtain . Then we prove that and that hold. If one could also show then our conjecture
would follow.
Finally, we give another argument in favour of the conjecture, i.e., we show that it is true in one more
case, namely that holds if we consider a two–letter alphabet. This implies the decidability of
for . To our knowledge, nothing was previously known concerning the decidability of (nor for any other level ). It has immediate consequences in first–order logic. The connection of
the Straubing–Thérien hierarchy to this logic goes back to the work of McNaughton and Papert [MP71].
It is related to the first–order logic having unary relations for the alphabet symbols from and
the binary relation . Let be the subclass of this logic which is defined by at most quantifier
alternations, starting with an existential quantifier. It has been proved in [Tho82, PP86] that –formulas
describe just the languages. Due to this characterization we can conclude the decidability of the
class of languages over definable by –formulas of the logic .
The papers starts in section 2 with a discussion of various notions of polynomial closure and of
dot–depth. It is shown in detail that if we define all hierarchy classes as classes of languages of , as
we did here, then the languages on each level differ from other definitions in the literature at most by
the empty word (see Theorem 2.20). This is easy to handle when dealing with automata. The benefits
we obtain hereby are on one hand the levelwise comparability by inclusion between the concatenation
hierarchies (see Fig. 3), on the other hand we see that we can use the same operations to build up the
dot–depth hierarchy and the Straubing–Thérien hierarchy. This is reflected on the pattern side by the
uniform iteration rule in both cases. We just start with two different initial patterns.
5
2 Preliminaries
2.1 Basic Definitions
We fix some arbitrary finite alphabet with . All definitions of language classes will be made
with respect to and it will always be clear from the context if we deal with a particular alphabet. The
and the set of all non–empty
empty word is denoted by , the set of all words over is denoted by
words over is denoted by . We take complements with respect to
or , depending
on what
denote domain we consider. So for a class of languages of
the set
of complements w.r.t.
, and for a class of languages of
let be set
of complements w.r.t. . Since we focus on we write instead of .
For a word denote by its number of letters, and by the set of letters occurring
in .
and we
left and right residuals of as For define the
. We write for the powerset of an arbitrary set .
and Regular languages are build up from the empty set and the singletons for using Boolean
operations, concatenation and iteration. Of particular interest for us is the subclass of starfree languages,
denoted as . Here the iteration operation is not allowed and we look at languages of (taking
complements with respect to ).
is its input
A deterministic finite automaton (dfa) is given by Æ , where
alphabet, is its set of states, Æ is its total transition function, is the starting state
and is the set of accepting states. For a dfa we denote by the language accepted by .
As usual, we extend transition functions to input words, and we denote by the number of states of
. In this paper we will look only at dfa’s where the starting state is not accepting. We say that a state
has a loop (has a -loop, for short) if and only if Æ . Every induces a
total mapping Æ with Æ Æ . We define that a total mapping Æ leads to
leads to a
a -loop if and only if Æ has a -loop for all . We may also say for short that -loop if Æ leads to a -loop. Moreover, for we write if and only if Æ Æ .
Recall the following theorem concerning starfree languages.
Theorem 2.1 [Sch65, MP71]. Let Æ be a minimal dfa. Then is starfree if and
only if there is some such that for all and for all it holds that Æ Æ .
We call a minimal dfa permutationfree if it has the above property. The decision of this property for
given is known to be PSPACE–complete [CH91]. For later use we restate the previous theorem as
follows.
Proposition 2.2. Let Æ be a minimal dfa. is not starfree if and only if there exist
, some and distinct states such that Æ for (with ).
An obvious property of permutationfree dfa’s is that they run into a -loop after a small number of
successive ’s in the input.
Proposition 2.3. Let Æ be a permutationfree dfa and
Æ for all and .
. Then Æ Proof. Since Æ there must be with Æ Æ . Then Æ for since otherwise is not permutationfree. It follows that
Æ for and in particular Æ Æ because .
❑
6
Proposition 2.4. Let and , then leads to a -loop in every dfa with .
Proof. Observe that leads to a -loop for some such -loop can be considered as a -loop.
7
. The proposition follows since every
❑
2.2 Definition of Concatenation Hierarchies
We study in this paper two well–known concatenation hierarchies, namely the dot–depth hierarchy
(DDH, for short) and the Straubing–Thérien hierarchy (STH, for short). For a class of languages
we denote its closure under finite (possibly empty) union by . Set
and as the polynomial closure of . Note that is exactly the closure of under finite (possibly empty)
union and finite (non–empty) concatenation. Furthermore, is a subset of the polynomial closure of .
For a second closure operation we consider Boolean operations. We take as our universe and denote
the Boolean closure of a class of languages of by (taking complements with respect to ).
Definition 2.5 (DDH). We define the levels of the dot–depth hierarchy as follows:
for for Definition 2.6 (STH). We define the levels of the Straubing–Thérien hierarchy as follows:
for for The discussion in the forthcoming Section 2.3 relates our definitions to the ones known from literature.
The following inclusion relations in each hierarchy are easy to see from the definitions.
Proposition 2.7. For the following holds.
1. 2. We can also compare the hierarchies in both directions by inclusion.
Proposition 2.8. For the following holds.
1. 2. 3. Proof. Since for all it holds that . Moreover, it holds that , and for for letters and we obtain
we obtain
with In particular and for all from which we get . So we
have seen and the proposition follows from the monotony of Pol and BC, and
complementation.
❑
It is shown in [Eil76] that
Proposition 2.9.
. Together with Proposition 2.8 we get the following.
.
8
2.3 Alternative Definitions
The dot–depth hierarchy and the Straubing–Thérien hierarchy have gained much attention due to the
still pending dot–depth problem. The purpose of this section is to make our work comparable to other
investigations, so we discuss alternative definitions. There are two points to look at: First, one finds other
versions of the polynomial closure operation in the literature. Let be a class of languages. Here are the
definitions of polynomial closure as chosen, e.g., with an algebraic approach recently in [PW97].
and and if then A second point is that languages may be defined in a way such that they contain the empty word. So
we also want to see if that makes any difference. It is pointed out, e.g., in [Pin95] that this is a crucial
point in the theory of varieties of finite semigroups. Since most results in the field were obtained via this
theory, the following definitions of concatenation hierarchies are widely used. We denote the Boolean
closure of a class of languages of
by (taking complements with respect to ).
Definition 2.10 (DDH due to [Pin96]). Let be the class of all languages of which can be written as finite unions of languages of the form where and . For let and .
which can
Definition 2.11 (STH due to [Str81, Thé81]). Let be the class of all languages of
where and . For
be written as finite unions of languages of the form let and .
These definitions are local to the remainder of Section 2. In this paper, we follow an automata–theoretic
approach, and we find it suitable to have the inclusion relations from Proposition 2.8 at hand. Theorem 2.20 below shows that the classes from Definition 2.10 and our classes coincide, and
that the languages in are up to the empty word the languages in . Let us recall known closure
properties of the just defined classes.
Lemma 2.12 [Arf91, PW97, Gla98]. Let and .
1. The classes and are closed under finite union and intersection.
2. Let
be one of the classes or . Then 3. Let
be one of the classes or . Then 2.3.1
for .
for .
Polynomial Closure
We investigate the relationships between the different notions of polynomial closure, and identify a
condition for under which these notions coincide.
Theorem 2.13. for a class of languages
(a) (b)
for every word and for every and The proof is an easy consequence of the following lemma.
Lemma 2.14. Let
be a class of languages. Then the following holds.
1. 9
satisfying the conditions:
2. 3. If for all 4. If 5. If ,
for all then .
for all ,
, , then .
then .
Proof. The statements 1, 2 and 5 can be easily verified. If we are given a language of the form
with , languages and words such that ,
then we can first take out every from this representation without changing the language. If
for all , then we can replace all remaining by languages from . We obtain
an equivalent expression of the form with and (note that if , then
). This shows statement 3.
Let us turn to statement 4. Here we have for all words with length or ,
and languages . It suffices to show that
and for letters for and . We will prove this by induction on . For (induction base) this is trivial. We assume that we have proven statement 4 for and we want to
show that for and . Since by induction
hypothesis, it suffices to show that for , and
. With we obtain
By assumption we have if . It follows that
Since also , it remains to show that for letters . Now consider the following case study of .
By assumption, following holds.
otherwise.
for letters
and for for 10
then define
if and . If if and if and if and . Hence for
for the
Together with the case study this implies
proof of statement 4.
. This completes both the induction and the
❑
If the languages of two classes and differ only by the empty word, we show that this property
is preserved by the polynomial closure operation. In other words, we obtain that also is equal to
up to the empty word.
Lemma 2.15. Let be a class of languages of and be a class of languages of
then the following holds.
1. is a class of languages of
. If and
and is a class of languages of
2. 3. Proof. The statements 1 and 2 follow immediately from the definition of Pol. Now we want to show
. Since the right hand side is closed under finite union, it
suffices to show the following claim.
Claim: for and .
We will show this by induction on . Observe that for (induction base) the claim holds. Now
assume that we have already proven the claim for the case and we want to show it for .
Let with . By induction hypothesis there exists an , such that
or . Since , there exists an such that
or . It follows that
or . This proves the
such that This shows that in any case there exists an claim.
Finally we have to show the inclusion . Since ,
we have . Furthermore, we have for , since ❑
and . This proves the lemma.
2.3.2
Boolean Closure
We show a counterpart of Lemma 2.15, this time for Boolean closures.
Lemma 2.16. Let be a class of languages of and be a class of languages of
then the following holds.
and is a class of languages of
2. 3. 11
. If and
Proof. The statements 1 and 2 follow from the definition of BC and BC . Now we want to show the
inclusion . Note that elements from can be written
as finite unions of finite intersections of literals which in turn are of the form (positive literal) or (negative literal) for some . Observe that the right hand side of the equation in statement 3 is
closed under finite union and intersection. Thus it suffices to show that all literals are elements of the
right hand side. Since , all positive
literals are elements of the right hand side, and it remains to show the following claim.
for all .
Claim:
Let then there exists an such that or . Since , we
obtain
in the first case, and
in the second case. Since
, we conclude that
. This proves our claim,
and it follows that .
Let us turn to the reverse inclusion. From , and for , it follows that and . This shows the
❑
inclusion .
2.3.3
Comparing the Lower Levels
To apply Lemma 2.15 and 2.16 for the general case, we need to compare the hierarchies on their lower
levels. This will be the induction base for the proof of Theorem 2.20 below.
Proposition 2.17. The following holds.
1. 2. equals the class of languages of which can be written as finite unions of languages of the
form where and .
Proof. Observe that elements of are languages of which can be written as finite unions of
(add some if necessary).
languages of the form where and having the form It follows that . On the other hand, languages of
with and , can be written as concatenations of and non–empty words (just drop all
). This shows and statement 1 follows.
Now we can exploit statement 1 for the proof of statement 2. Therefore, it suffices to show
that for each language the following holds: is a finite union of languages of the form
with , if and only if is a finite union of languages of the form
with , . This is easy to see since we can replace
and due to
and
. This shows statement 2.
❑
Proposition 2.18. The following holds.
1. equals the class of languages of which can be written as finite unions of languages of the
where and .
form 2. is a class of languages of
and .
and .
Proof. The first parts of all statements are by definition. To see the remaining parts of statement 1 it
suffices to mention that (i) the case can not occur since we speak about languages of and (ii)
. The second part of statement 2 follows immediately from statement 1 and the definition of
.
12
For the second part of statement 3, observe that and are classes being
closed under finite union and finite intersection. Furthermore, we have
Thus it suffices to show that (i) for and (ii) for . Since by the second statement of the lemma we have , and since
it follows that and . Hence it remains to show the following.
for 1.
2.
for First of all let . If then and we obtain . Otherwise we
can write
.
Now let . Since and , we obtain .
This proves the proposition.
❑
2.3.4
Comparing the Hierarchies
We apply our Theorem 2.13.
Proposition 2.19.
1. For , let
2. For , let
be one of the classes or . Then .
Proof. To show this we need to prove that the mentioned classes fulfil the condition from Theorem 2.13.
(a) for every word and for every and So let . Then we have by definition, and we see that
(b)
This shows (a) for statement 1 due to the known inclusions. If
, we obtain (a) for statement 2 with
for
and letters
Condition (b) is fulfilled for statement 1 by Lemma 2.12. We use the same lemma for statement 2
❑
together with the closure under intersection. So it suffices to note that .
13
Theorem 2.20. The following holds.
1. for 2. 3. for Proof. We show statement 1 by induction on . By Proposition 2.17, the assertion holds for (induction base). We first assume that it holds for with , and we want to
prove it for . Then
(1)
where we use the induction hypothesis to obtain equation (1). This shows in particular
statement 1 for . So now we assume that it holds for with , and we
want to prove it for . Then we have , and from Proposition 2.19 we get
. The induction hypothesis provides .
Statement 2 is given in Proposition 2.18.
We prove statement 3 also by induction on . For , the induction base is given in Proposition 2.18. We first assume that it holds for with , and we want to prove
it for . Then we have and . It holds that
is a class of languages of , , and from the induction hypothesis we obtain
. This shows that the classes and satisfy the assumptions of Lemma 2.15 and we obtain
(2)
Because we get from Proposition 2.19 that . This shows
in particular statement 3 for . So now we assume that it holds for with ,
and we want to prove it for . Then we have and .
By definition, is a class of languages of , and still . Furthermore, from the
induction hypothesis we obtain . Thus the classes and satisfy the assumptions of Lemma 2.16 and we obtain
This shows
❑
Let us carry over Theorem 2.20 to the classes of complements.
Corollary 2.21. The following holds.
1. 2. for 14
for Proof. Statement 1 is an immediate consequence of Theorem 2.20, from which we also get for :
❑
This shows the second statement.
Note that and in Theorem 2.20 are some kind of exception since all classes with
have the following property: for all languages .
is the only language in which contains the empty word.
This does not hold for because
However, we have the following uniform statement of this relation.
Corollary 2.22. For we have .
Proof. For this follows from Theorem 2.20. By definition, is a class of languages of
and if we intersect both sides of with we get .
,
❑
2.4 Normalforms and Closure Properties
Finally, we give in this preliminary section some normalforms and closure properties for the hierarchy
classes and . We first mention the normalform for from [Arf87].
Proposition 2.23.
1. is equal to the class of languages of which can be written as finite
unions of languages of the form where , and .
2. is equal to the class of languages of which can be written as finite unions of languages of
and .
the form where , which can be written
Proof. In [Arf87] it is shown that is equal to the class of languages of
as finite unions of languages of the form where , and . By
Corollary 2.22 we have
which shows statement 1. Observe that , and for and . So it is easy to see by mutual substitution that the following statements are equivalent
for every language .
is a finite union of languages of the form with , is a finite union of languages of the form with , is a finite union of languages of the form with , In [Gla98] normalforms for the definition of levels
Straubing–Thérien hierarchy are proven.
Lemma 2.24. For the following holds.
2. .
, .
, .
❑
This shows statement 2.
1. and 15
of the dot–depth hierarchy and the
Proof. By definition, and for . Thus we have and for . It remains to show the other inclusions.
For this end, we recall from [Gla98] that for it holds that
(3)
(4)
First of all we consider statement 1 for . By Proposition 2.23, languages from can be written
and .
as finite unions of languages of the form where , Note that if then , since languages of do not contain the empty word. Hence it
suffices to show that for and all . This can be seen as follows.
This shows . Now we consider statement 1 for some . Here we have
by Theorem 2.20. Since , we can apply Lemma 2.15 as before and we obtain
From Proposition 2.19 we see that rewrite (5) as
(5)
. So together with (3) we can
(6)
We can compare this to Theorem 2.20 where we have
(7)
Because the unions in (6) and (7) are disjoint, we see that and statement 1 is
proven.
Let us consider statement 2 for . From (4) and Theorem 2.20 we obtain . Together with Proposition 2.19 this yields . With Corol❑
lary 2.21 we get .
Now we translate the closure properties from Lemma 2.12 to our definitions.
Lemma 2.25. Let and .
1. The classes and are closed under finite union and intersection.
2. Let
16
for
Proof. For the classes and the lemma follows from Theorem 2.20 and Lemma 2.12.
The closure of under finite union and intersection for is immediate from Lemma 2.12
and Corollary 2.22. This carries over to .
Now let , and . By Theorem 2.20 we have . Thus and
we obtain by Lemma 2.12. Since it follows from the closure
under intersection and from Theorem 2.20 that . Analogously this can
be shown for , and using Corollary 2.21 and Lemma 2.12.
Finally let with for some . Then we have
Analogously one shows .
❑
17
3 A Theory of Forbidden Patterns
The dot–depth hierarchy and the Straubing–Thérien hierarchy are hierarchies of classes of regular languages, and it is known from literature that their lower levels have characterizations in terms of forbidden
patterns. In this section we take up this idea and develop a method for a uniform definition of hierarchies
via iterated forbidden patterns. Such a definition starts with a so–called initial pattern which determines
the first level of the corresponding hierarchy. Using an iteration rule we obtain more complicated patterns
which in turn define the higher levels.
In subsection 3.1 we introduce the notion of iterated patterns, and in subsection 3.2 we prove some
auxiliary results which help to handle these patterns in a better way. In subsection 3.3 we prove an interesting relation between the polynomial closure operation on the one hand and our iteration rule on the
other hand. More precisely we show that a complementation followed by a polynomial closure operation
on the language side is captured by our iteration rule on the forbidden pattern side. In subsection 3.4 we
investigate the general inclusion structure between hierarchies which are defined via iterated forbidden
patterns. In subsection 3.5 we study in how far the pattern iteration rule preserves starfreeness, and in
subsection 3.6 we treat the decidability of our pattern hierarchies.
3.1 Definition of Iterated Patterns
Starting with a so–called initial pattern (fulfilling some reasonable weak prerequisites) and using a certain
iteration rule, we define hierarchies of patterns in dfa’s. In later subsections we will use these hierarchies
of patterns for the definition of hierarchies of languages which share some interesting properties with
concatenation hierarchies.
Definition 3.1. We define an initial pattern to be a subset of
it holds that .
Definition 3.2. For every set we define such that for all
and
.
Definition 3.3. For an initial pattern we define and for .
Definition 3.4. Let be an initial pattern, Æ a dfa and .
1. For we define:
and Æ (b) appears at state Æ (c) the states are connected via appears at and , and Æ For and we define the following:
(a) and Æ (b) appears at state there exist states with such that Æ , , Æ for and the states are connected via
for .
(c) the states are connected via appears at state , appears at state and
there exist states with such that Æ , , Æ for and appears at state for (a)
2.
18
We give the following interpretation of the definitions made so far. In the known forbidden patterns
for concatenation hierarchies there usually appear two states , both having a loop with the same
structure in the dfa (loop–structure for short). This structure in turn determines the subgraph we need
to find between and (bridge–structure). Each pattern class with can be understood as
a set of structures in dfa’s, and each such structure determines a certain loop–structure and a certain
bridge–structure. If a appears at some state , then this means that we can find the loop–structure
described by at . If two states , are connected via , then (i) appears at both states and (ii) we
find the bridge–structure described by between them. To handle these patterns, we define for a
word Æ obtained from the loop–structure described by (call this the loop–word), and a word which
is derived from the bridge–structure of (bridge–word). It will turn out that if appears at some state ,
then this state has a Æ -loop. If and are connected via then the bridge–word leads directly from
to .
Let us make this more precise, first for an initial pattern. Let be an element of an
initial pattern . Then we interpret as a loop (the loop–structure of ) and as a certain path (the
bridge–structure of ) in a dfa. Therefore, the loop–word Æ is defined as and the bridge–word is
defined as . So appears at state if and only if this state has a -loop. Two states , are connected
via if and only if (i) both states have a -loop and (ii) leads from to , i.e., Æ .
Now let be an element of for . This means that for words and elements . We interpret the loop–structure described by as a loop which can be divided
in words in this ordering, and between each , we find (i.e., the states at the
end of each and at the beginning of each are connected via ). Here we see that elements of
appear in the loop–structure of elements of . This is a phenomenon which can also be observed
in the known forbidden pattern characterizations for the levels and of the dot–depth hierarchy
and the Straubing–Thérien hierarchy (cf. Figures 4, 5, 7, 8). The bridge–structure of can be interpreted
as a path in a dfa labeled with and going from state to such that (i) we find the
loop–structure of at both states and (ii) after the word we reach a state where we find the
loop–structure of .
The known forbidden pattern characterizations for levels and of concatenation hierarchies
describe patterns which consist of reachable states , having certain structural properties such that
some word is accepted when starting in and rejected when starting in . The following definition
takes this into account.
Definition 3.5. Let Æ be a dfa, an initial pattern and . We say that has
pattern if and only if there exist , , such that Æ , Æ ,
Æ and the states are connected via .
3.2 Auxiliary Results for Iterated Patterns
In order to establish a relation between the polynomial closure operation and the iteration rule for the
patterns classes (cf. Theorem 3.15), we isolate the main argument of this proof in Lemma 3.11 below,
for which the following two constructions are needed. First, for some can be defined
such that if appears at some state then , are connected via (cf. Definition 3.7 and Lemma 3.8).
Secondly, in Definition 3.9 and Lemma 3.10 we pump up the loop–structure of to construct for given
some such that
(i) if two states are connected via , then they are also connected via and
(ii) in every dfa with the words and lead to states where appears.
Æ
First of all let us states some basic properties of iterated patterns.
Lemma 3.6. Let be an initial pattern, , , 19
Æ a dfa and .
1. If 2.
3.
4.
5.
are connected via then appears at and at .
If are connected via then Æ .
If appears at state then Æ Æ .
If then Æ .
Let and for suitable , appears at state then also appears at state .
and . If
Proof. We only have to consider statement 3, since the statements 1, 2, 4 and 5 are easy consequences
from the definition. Let for some such that appears at state . If then we
have Æ Æ Æ . If then there exist states with such that
Æ , , Æ for and the states are connected via for
. From statement 2 it follows that Æ for . This implies:
Æ Æ Æ ..
.
Æ Æ Æ Æ Æ ❑
Let be an initial pattern. Given some , a dfa and states which are connected
via some , we want to construct a such that (i) the occurrence of is implied by the
Æ
occurrence of and (ii) and lead to a state where appears in every dfa with . First of
all we will see that a appearing at state can be considered as a connection between and . For
this we have to carry out a slight modification of . For and we will construct a similar
pattern such that the following holds for every state of some dfa : If appears at , then
are connected via .
Definition 3.7. Let be an initial pattern. For we define . Now let
such that for some , and we have . We
define Æ .
Lemma 3.8. Let be an initial pattern, and , then . Furthermore, let
Æ be a dfa and such that appears at state . Then the states are
connected via .
Proof. We prove the lemma by induction on . For we have . It follows
from Definition 3.1 that . Since appears at state , we have Æ .
Therefore, the states are connected via by Definition 3.4.
We assume that the lemma has been shown for all , we want to prove it for . Let
such that for some , and we have . By
Lemma 3.6.4 we have Æ , and from induction hypothesis we know that . From the
Definitions 3.2 and 3.3 it follows that Æ .
It remains to show that the states are connected via Æ in dfa . By
Lemma 3.6.5, appears at state . From the induction hypothesis it follows that are connected
via . Since Æ Æ (Lemma 3.6.3), we obtain that appears at state . Now let ,
and . Since appears at state , it follows from Lemma 3.6.3 that Æ Æ.
We have already seen that are connected via , particularly appears at state (Lemma 3.6.1). This shows that are connected via .
❑
20
Definition 3.9. Let be an initial pattern. For and we define . Now let and such that for some , and we have
. For we define the following.
Æ Æ Æ
Æ
times “ ”
Lemma 3.10. Let be an initial pattern, , , and .
2. If appears at state of some dfa , then also appears at this state.
3. If the states of some dfa are connected via , then these states are also connected via .
Æ
4. If is a dfa with , then and lead to a state in where appears.
Æ
Æ
5. If is a dfa with , then lead to states in which are connected via .
6. If is a dfa with , then lead to states in which are connected via .
Proof. Let be an initial pattern and . We will prove the lemma by induction on .
1.
Induction Base:
For we have and . From Definition 3.1 it follows that
. Let Æ be dfa and . If appears at state , then we have
Æ . Hence Æ and it follows that appears at state . If are connected via ,
then Æ and Æ Æ . It follows that Æ and Æ Æ . Thus are also connected
via . By Proposition 2.4, leads to a -loop
Æ
in every dfa with . Since and , in every dfa with both words
Æ and lead to states which have a -loop. It follows that
appears at these states. Furthermore, it
Æ Æ is easy to see that in every dfa with the words lead to states (resp.), such that
Æ and Æ Æ . Hence are connected via . Analogously we
see that lead to states which are connected via . This shows the induction base.
Induction Step:
Suppose now that we have proven the lemma for and we want to show it for .
Let , and choose suitable , , such that Æ
Æ
. As in Definition 3.9 let and .
Moreover let Æ be a dfa and such that (i) appears at state and (ii) the
states are connected via .
First of all let us prove that Æ for every state where appears. If appears at ,
then there exist states with such that Æ , , Æ for and the states are connected via for . By induction hypothesis,
are also connected via for . From Lemma 3.6.2 it follows that Æ for
. Hence Æ and Æ Æ
for . By the Lemmas 3.6.1 and 3.6.3, the state has a -loop for . It follows that
Æ Æ Æ for . This shows Æ .
Statement 1: By induction hypothesis we have . By Lemma 3.8, .
Æ
Æ
Since and , it follows that .
Statement 2: Since appears at state , there exist states with such that
Æ , , Æ for and the states are connected via for
21
. Using the additional states and for we will
show that also appears at state . With it suffices to show the following:
2a. Æ Æ
2b. 2c. Æ for Æ
2d. Æ for 2e. the states are connected via for 2f. the states are connected via for We already know that Æ and that the states are connected via . From Lemma 3.6.1
it follows that appears at state . By induction hypothesis appears at state . From Lemma 3.6.3
we conclude Æ Æ . This shows 2a. Statement 2b follows from the definition of . Now
and the states choose some with . By assumption we have Æ are connected via . Thus appears at state (Lemma 3.6.1). From the induction hypothesis
Æ (Lemma 3.6.3), this shows
it follows that appears at state . We obtain Æ statement 2c. Note that for and for . Therefore, statement 2d
follows, since we have already seen that Æ . By assumption, the states are connected via
for . From the induction hypothesis it follows that are also connected via for
. This shows statement 2e. From Lemma 3.6.5 it follows that appears at state . From
induction hypothesis it follows that appears at state . Thus the states are connected via by Lemma 3.8. This shows statement 2f, and we have finished the induction step for statement 2 of the
lemma.
Statement 3: By assumption, the states are connected via . It follows that appears at
and at , and there exist states with such that Æ , ,
Æ for and appears at state for . From statement 2 it follows
that appears at and at . Let and for . The
states with will witness that are connected via . To this end it suffices to show the
following.
3a. Æ Æ 3b. 3c. Æ for Æ
3d. Æ for 3e. appears at state for 3f. appears at state for By assumption, Æ and appears at . From induction hypothesis it follows that also appears at . With Lemma 3.6.3 we obtain Æ Æ . This shows statement 3a. Note that 3b is by
by assumption. SinceÆ appears at , it
definition. For we have Æ (Lemma 3.6.3).
follows that also appears at (induction hypothesis) and Æ This implies statement 3c. At the beginning of the induction step we have already seen that Æ for all where appears. It follows that Æ . This shows statement 3d, since for
. We know that appears at for . Thus appears at (induction hypothesis)
22
and statement 3e follows. Particularly, appears at for . Together with the
Lemmas 3.8 and 3.6.1 this shows statement 3f. Thus we have proven statement 3.
Statement 4: Let be a dfa with . We will show that leads to states in where appears. Assume for the moment that we have already shown this, then we argue as follows. First of all
we observe that the following holds.
Æ
Æ Æ Æ Æ Hence leads to states in where appears. Note that is a suffix of . From induction hypothesis
it follows that leads to states in where appears. From the Lemmas 3.8 and 3.6.2 we obtain that
leads to a Æ -loop in . Hence and Æ Æ Æ .
It follows that leads to states in where appears. This implies statement 4 and it remains to show
that leads to states in where appears.
By Proposition 2.4, leads to a -loop in . Thus it suffices to show that appears at for every state with Æ . For this purpose let and
define the following states.
Æ Æ Æ Æ Æ for Æ for Æ for for We want to show that these are the states witnessing that appears at state . It is easy to see that
Æ ÆÆ Æ and Æ Æ Æ for . Since
Æ , we have Æ and Æ Æ for . It follows that Æ . To see that appears at , it remains to show
the following.
the states are connected via for the states are connected via for The first item follows from the induction hypothesis, since lead to states which are connected
via (by definition are such states). Note that is a suffix of . From induction hypothesis it
follows that leads to states in where appears. Therefore, appears at for .
From Lemma 3.8 it follows that are connected via . This shows the second item. Hence we
have proven statement 4 of the lemma.
Statement 5: Let be some dfa with and let be an arbitrary state of . For Æ Æ and Æ Æ we want to show that are connected via . For this let and define the following witnessing states.
Æ
Æ
Æ Æ
(8)
Æ
Æ for (9)
Æ for (10)
By statement 4, appears at and at . Furthermore, it is easy to see that Æ .
It remains to show that (i) appears at state for and (ii) appears at state for
23
. By induction hypothesis, for it holds that Æ leads to states in where
appears. From the equations (8) and (9) it follows that appears at state for . From
induction hypothesis it follows that leads to states in where appears. Hence also leads to
states in where appears (note that is a suffix of ). Thus from equation (10) we obtain that appears at state for . From Lemma 3.8 it follows that are connected via for . Particularly, appears at state for (Lemma 3.6.1). This
shows statement 5.
Statement 6: This can be seen analogously to the proof of statement 5. There we only have to
Æ
❑
replace the terms “ ” by “ ”.
The following lemma isolates the main argument of the proof of Theorem 3.15 which establishes
a relation between the polynomial closure operation and the iteration rule for the patterns classes. The
lemma says that under certain assumptions we can replace bridge–words by their respective loop–words
without leaving the language of some dfa.
Lemma 3.11. Let be an initial pattern, , , and . Furthermore,
let Æ with be a dfa which does not have pattern . Then for all we have
Æ Proof. We choose suitable , and such that Æ
Æ
. For let and .
Æ
Æ
From Definition 3.9 it follows that where
the term “ ” is repeated times.
such that . Thus we have
Now let Æ
Æ
Æ
Æ We want to show . From Lemma 3.10.5 it follows that the states Æ Æ
Æ
and Æ are connected via . Note that by Lemma 3.10.1. If Æ
then we have Æ , Æ , Æ and the states are connected
via . It follows that has pattern . This is a contradiction to the assumption. Thus starting
from
Æ Æ Æ Æ
Æ
we have shown
Æ
Æ Æ Æ Æ
Analogously we obtain:
Æ Æ Æ Æ Æ
Æ Æ Æ Æ Æ
Æ Æ Æ Æ Æ
Æ Æ Æ Æ Æ
..
.
Æ
Æ
Æ
Æ
Æ Æ Æ Æ Æ Æ (11)
By definition, is a suffix of . From Lemma 3.10.4 it follows that leads to states in where
appears. From Lemma 3.8 we obtain that for all with Æ it holds that 24
are connected via . Now from Lemma 3.6.2 it follows that
-loop. Thus from equation (11) we obtain
leads to states in which have a
Æ Æ Æ Æ Æ Æ We can repeat this argument for the remaining occurrences of to see that
Æ Æ Æ Æ Æ Æ Æ
where the term “ ” is repeated times. This shows Æ
.
❑
In forthcoming proofs we make use of the fact that the structure of arbitrary patterns can be found at
each sink of some dfa. More precisely we can state the following proposition.
Proposition 3.12. Let Æ be a dfa and suppose there is some such that Æ for all . Let be an initial pattern, let and . Then appears at , and are
connected via .
Proof. The proof is by induction on . The induction base for is trivial, just consider Definition 3.4. Now suppose the proposition holds for some and we want to show it for . So let
for some with and . By hypothesis, we
know for that appears at , and and are connected via . If we set the required states
in Definition 3.4 all to it is easy to verify with help of the hypothesis that appears at , and and are connected via . Just observe that Æ for .
❑
3.3 Pattern Iterator versus Polynomial Closure
The main result in this section is stated in Theorem 3.15: A complementation followed by a polynomial
closure operation on the language side is captured by our iteration rule on the forbidden pattern side. To
prove this we make use of the auxiliary results given in subsection 3.2 where Lemma 3.11 contains the
main argument.
First of all we use our hierarchies of pattern classes to define hierarchies of language classes via a
forbidden pattern approach. So each initial pattern induces a hierarchy of language classes. We treat the
inclusion structure and the decidability question of these hierarchies in the subsections 3.4 and 3.6.
Definition 3.13. Let be an initial pattern. For we define the class of languages corresponding
to as is accepted by some dfa which does not have pattern .
The following proposition shows that the classes from Definition 3.13 are well defined.
Proposition 3.14. For , an initial pattern and dfa’s with we have:
has pattern Proof. Suppose has pattern . Then there are states in which are connected via some
such that Æ , Æ , Æ for suitable , and for the starting
state of and the set of accepting states of . Define and . We
Æ
obtain from Lemma 3.10.3 that in are also connected via . So Æ
(Lemma 3.6.3) and (Lemma 3.6.2). Now define and to be the states in
that can be reached from its starting state by Æ and Æ, respectively. By Lemma 3.10.5 we get
that and are connected via in . Since we reach from ( ) with an accepting state (rejecting
state, respectively) of , this shows that also has pattern .
❑
If we change the roles of and the same arguments also show the other implication.
has pattern 25
Theorem 3.15. Let be an initial pattern and , then .
Proof. Let be an initial pattern and . We assume that there exists an which is no
element of , this will lead to a contradiction. From it follows that
for some , and . Let Æ be a dfa with .
be a dfa with
For and let be a dfa with and let . Furthermore, we define
! The dfa has pattern , since and (cf. Definition 3.13 and Proposition 3.14). So
and the
there exist , , such that Æ , Æ , Æ states are connected via . It follows that and . By Lemma 3.10.2, the states are
Æ also connected via pattern . From Lemma 3.6.3 it follows that for all .
Thus there exists an with such that
Æ Since , it follows that there exist and words Æ 1.
2.
Æ
such that:
Æ and Æ for some Æ Æ
Since , the word leads to states in where appears (Lemma 3.10.4). Particularly,
Æ
Æ
such a state has a -loop (Lemma 3.6.3). From it follows that for all we have
3.
Æ (12)
Since are connected via in , we have Æ , Æ Æ and Æ (Lemma 3.6.2). Assume that Æ , then we would obtain Æ . It would
follow that
Æ Æ Æ Æ Æ Æ Æ This is a contradiction since Æ . It follows that
Æ (13)
Æ
Æ
Æ
From , it follows that . From equation (13) we obtain
Æ (14)
Observe that . Therefore, dfa does not have pattern . Since we can
Æ Æ
Æ Æ
apply Lemma 3.11. From equation (14) we obtain . It follows that . This is a contradiction to equation (12). It follows that .
❑
26
3.4 Inclusion Structure of the classes In this subsection we show that if some initial pattern satisfies a certain (weak) property then the
inclusion holds for all .
Definition 3.16. For initial patterns and we define every there exists an such that for every dfa the following holds:
if and only if for
Æ and all states
1. If appears at , then appears at .
2. If are connected via , then are connected via .
Lemma 3.17. For initial patterns and the following holds.
Proof. Let be initial patterns and such that . Hence for a given there exist such that for every dfa Æ and all states the following holds:
If appears at , then appears at for .
(15)
If (16)
are connected via , then are connected via for .
Define and observe that be a dfa and . We want to show the following:
. Now let Æ (i) If appears at , then appears at .
(ii) If are connected via , then are connected via .
Suppose that appears at , then there are states with such that Æ ,
, Æ for and the states are connected via for .
From (16) it follows that the states are also connected via for . Therefore, appears
at , and we have shown statement (i).
Suppose now that are connected via . By definition, appears at the states and there
for
exist states with such that Æ , , Æ and appears at state for . From statement (i) we obtain that appears at
state and at state . Furthermore, from (15) it follows that also appears at state for .
Hence are connected via , and we have proven statement (ii).
It follows that ❑
. This proves the lemma.
Lemma 3.18. For initial patterns and the following holds.
Proof. Let be initial patterns and such that . For an arbitrary language
with we want to show that . Since , the dfa Æ with
has pattern . Thus there exist , , such that Æ ,
Æ , Æ and the states are connected via . Since , there exists a
such
that the states are connected via . It follows that has also pattern . This
, and hence .
❑
shows 27
Lemma 3.19. For an initial pattern and the following holds.
1. 2.
Proof. From Theorem 3.15 it follows that .
. This also implies
❑
Theorem 3.20. Let be an initial pattern with , then for the following holds.
Proof. From Lemma 3.17 we obtain for . Lemma 3.18 implies for
. From this we conclude for . Together with Lemma 3.19 this proves the
theorem.
❑
3.5 Pattern Iteration remains Starfree
In this subsection we show that the pattern iterator can be considered as a starfree iterator. Let be
an arbitrary initial pattern and recall that denotes the class of
languages. In Theorem 3.26
starfree
if and only if . For the proof of this
we show that for it holds that
theorem we need some auxiliary results on periodic infinite words (cf. Lemma 3.21) and a modification
of Proposition 2.2 (cf. Lemma 3.24) by taking forbidden patterns into account. We also make a remark
on the restriction in Theorem 3.26.
In this subsection we will consider infinite words, where we take over certain notions from finite
words. If with for alphabet letters , then denotes the infinite word
. Let and with for alphabet letters , then
is the –suffix of . For and we use as an abbreviation for
.
, with Lemma 3.21. Let such that there do not exist . If for some , then is a multiple of .
and Proof. Let with as in the lemma. We assume that for some ,
where is not a multiple of . This will lead to a contradiction.
Let and , and note that .
Moreover, let be the –suffix of , and let be the –suffix of . Then we have the following
decomposition of .
Observe that . Otherwise we would obtain which implies
that is a multiple of . This is a contradiction to our assumption.
Note that . It follows that . W.l.o.g. we assume
that . Since and are suffixes of , there exists a such that . This
implies and it follows that . Note that . Thus we have
found , with and . This is a contradiction to the assumption of the
lemma.
❑
28
The following lemma is an extension of Proposition 2.2. By this proposition, we find a permutation
(induced by some word ) in every minimal dfa where is not starfree. Here we show that
this permutation can be chosen in a minimal way, i.e., there do not exist words with and
. Note that this is nontrivial, since we have to prove that the existence of such words indeed induces a permutation of distinct states.
Lemma 3.22. Let Æ be a minimal dfa.
is not starfree there exist , some and distinct states
such that
(i) there do not exist , with and (ii) Æ for (with )
Proof. “ ”: This direction is an immediate consequence of Proposition 2.2.
“ ”: Suppose that is not starfree. Using Proposition 2.2 we choose a shortest ,
some and distinct states such that Æ for (with
). In the following we will show that the violation of condition (i) implies that the choice of was not minimal (which is a contradiction). This shows that satisfies condition (i).
, such that and . This
We choose a shortest word and some with and . Moreover, implies that there do not exist , implies the existence of some such that (simply delete the prefix of ). So we
can apply Lemma 3.21 and obtain that is a multiple of . This means that and
for suitable and with .
Now we consider the sequence of states where Æ for . From and it follows that Æ and Æ for all and
. Suppose that there is some with . It follows that Æ and
Æ . This implies Æ Æ Æ , which is a contradiction to
our assumption. It follows that for all .
Now choose a smallest such that there is some with (such a exists due to the
finiteness of ). We have already seen that . Thus we have found a and a list
of distinct states such that Æ for (with ). Since , this is a contradiction to the choice of the shortest at
the beginning of this proof. Hence our assumption was false, and we conclude that there do not exist
, with and . This proves our lemma.
❑
Lemma 3.23. Let be an initial pattern, , , such that and there do not
exist , with and . Then there exists a such that for all and the following holds:
There exists some with .
The length of is a multiple of , but it is not a multiple of .
Proof. Choose alphabet letters such that . Let Æ be the dfa
where is our fixed alphabet, and , and
Æ is defined as:
1. Æ for and 2. Æ for 29
3. Æ 4. Æ for all remaining arguments First of all let us determine the language accepted by . For this we observe that is the only sink, and
the initial state is the only rejecting state in the dfa . Thus a word is not in if and
only if Æ . Moreover, the only possible path which starts at and which does not lead
to the sink looks as follows:
We go along this path if and only if the input is . Therefore,
aword is not in
if and only if it is of the form . This shows .
Furthermore, is a minimal dfa. Otherwise there would exist different states with
Æ Æ for all . Note that both states have to be different from , since
is an accepting sink and from all other states a rejecting state is reachable. So it must be that and . The length of a shortest non–empty word leading from ( , resp.) to a rejecting
state is equal to the length of the shortest non–empty word which leads from ( , resp.) to ,
since is the only rejecting state. Observe that these lengths equal if we start at , and if we start at . By assumption, these
lengths have to be equal. So we obtain . From it follows that
and which is a contradiction to the choice of and . This shows that is a minimal
dfa.
So we have a minimal dfa , a word and states such that Æ for and Æ . From Proposition 2.2 it follows that is not starfree.
From and we obtain that has pattern . By Definition 3.5, there exist
, , such that Æ , Æ , Æ and the states
are connected via . Note that and are different from , since rejecting states are reachable
from and (e.g. the state Æ Æ is rejecting). So we have and for suitable .
We consider now an arbitrary state . By the construction of , the following holds for all
words :
Æ if and only if is a prefix of .
(17)
Let and . By Lemma 3.10, the states are also connected via and .
Æ
Moreover, by Lemma 3.6 we have (for this it is necessary that ), Æ ,
Æ and Æ Æ . From (17) it follows that is a prefix of and ÆÆ . Furthermore, we have , since Æ . This yields the
equation , and we obtain . From the definition of it follows that
where . This shows , and the first statement of
the lemma follows.
From our assumption and Lemma 3.21 it follows that the length of is a multiple of . Suppose
that the length of is a multiple of . Then from the construction of it follows that Æ . This is a contradiction to the choice of and such that Æ and Æ . This
shows the second statement of the lemma.
❑
Lemma 3.24. Let Æ be a minimal dfa, an initial pattern and such that
. is not starfree if and only if there exist a , some and distinct states
such that are connected via for (with ).
30
Proof. The if–part follows from Lemma 2.2, since Æ for and (Lemma 3.6.2).
Let and assume that is not starfree (if then set ). By Lemma 3.22,
there exist , some and distinct states such that (i) there do not exist
, with and and (ii) Æ for (with
). By Lemma 3.23, there
for it holds that (i) there
exists a such that
with and (ii) the length of is a multiple of , but it is not a multiple
exists a with of . Hence and for suitable , (note that is not possible, since by Lemma 3.6.4).
Now we consider the sequence of states where Æ for . Suppose
that there is some with . Note that
Æ Æ Æ Æ for . It follows that
Æ Æ Æ Æ Æ By the choice of the states this implies that is a multiple of . From it
follows that is a multiple of . This is a contradiction to the choice of . We conclude that
for all .
From the sequence we choose an earliest such that there is some with . We
have already seen that . Thus we have found a list of distinct states such that Æ for (with ). From Lemma 3.10.6 it follows that
the states are connected via for . This proves the only–if–part of our
lemma.
❑
Lemma 3.25. If
for some initial pattern and , then
.
Proof. Let be an initial pattern and such that . Moreover, let Æ be
. By Lemma 3.24,
a minimal dfa such that is not starfree, we will show that there exist a , some and distinct states such that are connected
via for (with ).
Now define and observe that . First of all we will show that the states
, are connected via . From Lemma 3.6.2 it follows that Æ for . Hence
Æ and the states , are connected via . This shows that appears at , and
analogously we obtain that appears at . Moreover, Æ and appears at state (Lemma 3.6.1). This shows that , are connected via . Analogously one shows for that the states , are connected via .
such that Æ Since is minimal, there exist with and a word Æ . Since the states form a cycle, there exist also an with such that Æ and Æ . Furthermore, there exists a with Æ (by
the minimality of ). Since we have already seen that , are connected via , it follows that has
, and it follows that .
❑
pattern . This shows 31
Theorem 3.26. Let be an initial pattern and .
Proof. Let be an initial pattern and . It suffices to show that implies for all
. By Lemma 3.25, this implication holds for all .
If then also , since is closed under complementation. From Lemma 3.19.2
it follows that . If we use this argument repeatedly, we obtain for
all .
❑
Remark 3.27. It is necessary that we assume in Theorem 3.26. In fact, this theorem does not
hold for . For this end we consider a two letter alphabet and the initial pattern . With the help of the known forbidden pattern for level of the Straubing–
Thérien hierarchy (cf. Figure 4) we observe that . In contrast, we will see that
. To see this we have a look at the following dfa Æ .
It is easy to see that is a minimal dfa, and that it is not permutationfree. By Proposition 2.2, this
. It remains to show that , i.e., we have to see that does not have
implies pattern .
Assume that there are states , words and a such that Æ , Æ ,
Æ and the states are connected via . Choose suitable and such
that . So we have and Æ . Note that
, since rejecting states are reachable from and . It follows that and Æ are
alternating sequences of letters , since all other words lead to the sink .
Note that for all , and assume that there exists a with .
It follows that either is equal to the last letter of or it is equal to the first letter of . This is
a contradiction to the fact that is an alternating sequences of letters . It follows that Æ where , since Æ . W.l.o.g. we assume that . Hence Æ Æ , and
it follows that Æ and Æ . Therefore, at least one of the states and does not
have a Æ -loop. This is a contradiction to our assumption, and it follows that does not have pattern
. This shows .
Together we see that and since which is easy to verify.
3.6 Decidability of the Pattern Classes
In this subsection we treat the decidability aspects of the pattern classes. It will turn out that is decidable in nondeterministic logarithmic space whenever a certain decision problem for the initial pattern 32
is decidable in these space bounds. Note that the decidability of the pattern classes has to depend on the
initial pattern, since an undecidable set (which can be easily constructed) leads to undecidable pattern
classes.
We start with the definition of two problems addressing the question for the existence of paths (cf.
Definition 3.28) and patterns (cf. Definition 3.29) which appear simultaneously in a dfa. In the Lemmas 3.30 and 3.31 we investigate the decidability of these problems, and at the end this leads to the
desired results.
Definition 3.28. Let . We define R EACH to be the set of pairs such that:
1.
2.
3.
Æ is a dfa
with There exists a word such that Æ for all .
Definition 3.29. Let be an initial pattern, and . We define PATTERN to be the set of
all triples such that the following holds:
1.
2.
3.
4.
Æ is a dfa
with with There exists a such that for all and (ii) ! are connected via .
and all ! it holds that (i) appears at We want to make precise how we think of a dfa as an input to a Turing machine. Therefore, we
give an explicit encoding as follows. Using the three–letter alphabet " we want to encode a dfa
Æ . For this we fix arbitrary
orderings on the sets , and , such that we obtain
, (one of them is the starting state ) and .
Moreover, we identify the elements , of the sets , with their index numbers , (resp.). Now we
can encode in the following way.
Æ Æ Æ Æ Æ Æ " " Æ
" "
The sets , , in the Definitions 3.28 and 3.29 are encoded analogous to (we encode a pair
! by ). Taking the respective codes together (separated by " signs), we obtain
codes for and . It is easy to see that on input of a word from " , we can check
in deterministic logarithmic space whether this word is a valid representation of a pair (of a triple
, resp.). Therefore, in forthcoming investigation of algorithms we may assume that all inputs
are valid representations.
Lemma 3.30. For we have R EACH #$.
Proof. We use a slight modification of the algorithm solving the graph accessibility problem. If then we are done. Otherwise we assign the elements of to program variables and (some may take the same value if ). Now we guess a word letter by letter, and we
simultaneous follow the paths which start at and which are labeled with . Moreover, in each
step we guess whether we have already reached the end of , and if so, we check whether for all
.
❑
33
We will consider oracle machines working in nondeterministic logarithmic space which have the
following access to the oracle. The machine has an (read–only) input tape, a (write–only) query tape and
a (read–write) working tape which is bounded logarithmically in the input size. Furthermore, from the
moment where we write the first letter on the query tape, we are not allowed to make nondeterministic
branches until we ask the oracle. After doing this we obtain the corresponding answer and the query tape
is empty. Using this model, introduced in [RST82], we can prove the following lemma. We assume that
the machine represents a single state of a dfa on its working tape in binary by the index number of the
state. Hence the space needed to do this is bounded logarithmically in the input size.
Lemma 3.31. Let be an initial pattern, then PATTERN #$PATTERN for each and
each .
Proof. In Table 1 we describe a nondeterministic algorithm having access to a R EACH oracle and to
a PATTERN oracle. The notations in this table are adopted from the Figures 1 and 2. We will
show that this algorithm works in logarithmic space and decides PATTERN . By Lemma 3.30 we have
R EACH #$. Since the access to an #$ oracle does not rise the power of an #$ machine, i.e.,
#$ #$ [RST82, Sze87, Imm88], we can go without the R EACH oracle and obtain the desired
algorithm.
First of all we want to observe that the algorithm accesses the oracle in the way as described above.
For this we only have to consider step 4. Since on the one hand we have already computed the sets , and (they are stored on the working–tape) and on the other hand is stored on the input–tape, we can
actually write down the queries and without making any nondeterministic branches.
Let us analyse the space on the working–tape which is needed on input . Note that our
program uses only a constant number of variables (this number can be bounded by a function of " ,
and is a constant). Moreover, all variables except contain index numbers of states of ,
which can be stored in logarithmic space. Each of the variables contains a set consisting of at
most % (pairs of) index numbers of states. Note also that we can produce the encoding of the queries as
needed for the oracle input deterministically with a logarithmic space bound on the working tape. This
shows that our algorithm works in logarithmic space.
In the remaining part of this proof we will show that our algorithm decides PATTERN . First of
all we want to see that the computation has an accepting path if PATTERN . For this let
be a witnessing pattern (cf. Definition 3.29.4). We denote the involved
states of the appearance of at as in Figure 1, and we denote the involved states of the connection of
! via as in Figure 2. Now consider that path of the computation where we carry out exactly passes of the loop and where we guess the states # $ % & Æ at the beginning of the
-th pass of the loop (starting with pass ). It can be easily verified that this is an accepting path.
Now suppose that the computation on input has an accepting path, and fix one of
these paths. Choose such that on this path the loop is passed times. Note that in each
pass of the loop we receive positive answers to the queries R EACH and PATTERN (otherwise the fixed path would be rejecting). It follows that for each pass there
exists a word witnessing R EACH , and there exists a pattern witnessing PATTERN . Now define . Using the states
# $ % & Æ which were guessed at the beginning of the -th pass of the loop, we can
and (ii) all ! with ! are connected via .
❑
verify that (i) appears at all Corollary 3.32. Let be an initial pattern such that PATTERN #$ for all
PATTERN #$ for each and each .
. Then
Proof. We prove this by induction on . The induction base is by assumption. The induction step follows
❑
from Lemma 3.31 and the fact that #$ #$ [RST82, Sze87, Imm88].
34
½
½
½
½
¼
¼
¼
¼
½
Figure 1: Example for an appearance of a pattern at some state .
½
½
½
½
½
¼
¼
½
½
¼
¼
¼
Æ¼
½
Æ
¼
¼
½
¼
¼
¼
Æ½
½
½
½
½
Æ Figure 2: Example for a connection of two states ! via a pattern.
35
½
Step,
Label
1.
2.
Command
Remark
Let , and let ! such that
and ! ! .
Note that are bounded by the constant .
We have to decide whether there is a such that (i) appears at all (Figure 1) and
(ii) all ! are connected via (Figure 2).
For and let:
$ Æ 3.
loop:
4.
% !
!
Guess states # $ for , states
% & Æ for and let:
# $ % & Æ $ # % Æ & The guessed states correspond to Figures 1 and
2. In the -th pass of this loop (starting with
pass ) the variables # $ % & Æ correspond to # $ % & Æ (respectively).
Moreover, at the beginning of the -th pass we have a corresponand
dence between $ % Æ $ % Æ (resp.). Using and we will ask the oracle whether there is
a pattern which connects (and appears at)
the guessed states. With we will test the
existence of a word (cf. Figures 1 and 2).
Ask the following queries and reject when If at least one negative answer is given then
the states guessed at the previous step do not
a negative answer is given.
correspond to a pattern from .
R EACH
5.
Variables marked with ‘start’ contains the
starting point from where we have to guess and
check the next fragment of the pattern.
PATTERN For and let:
$ $
Æ Æ
Here we set the next starting points.
% %
6.
Jump nondeterministically to loop or to Guess whether we have already checked the
exit.
right number of fragments of the pattern, i.e.,
whether the number of passes equals .
7.
exit:
Accept if and only if the following con- It remains to check whether the guessed loops
ditions hold for all and have reached their starting points, and whether
:
the path which was guessed via leads from
! to .
$ % !
Æ Table 1: An algorithm which decides PATTERN on input of a dfa and sets
and with .
36
Æ Corollary 3.33. Let be an initial pattern such that PATTERN #$ for each . Then for a
fixed it is decidable in nondeterministic logspace whether a given dfa has pattern .
Proof. On input Æ we guess states and check whether and . Now we test REACH and REACH which is possible in nondeterministic logspace by Lemma 3.30. It remains to check whether or not
PATTERN which is also possible in #$ by Corollary 3.32.
❑
37
4 Consequences for Concatenation Hierarchies
From now on we consider two special initial patterns. We will see that the classes of languages being defined by these patterns are closely related to the dot–depth hierarchy and the Straubing–Thérien
hierarchy.
Definition 4.1. We define the following initial patterns.
It is easy to see that and are indeed initial patterns.
4.1 Some Auxiliary Results
We start with some easy to see results that make handling the lower levels of the pattern classes easier.
Proposition 4.2. Let . Let only if Æ Æ .
Æ be a dfa and . Then appears at if and
with and and for all . First assume appears at . Then Æ Æ by Lemma 3.6.3.
Conversely, let Æ Æ and define the witnessing states as Æ , ' Æ and Æ ' for , and ' Æ . One easily verifies with help of
these states that appears at . Just observe that two states ' are connected via some pattern
if and only if Æ '.
❑
Proof. Let
Lemma 4.3. Let such that and have the same first letter and .
Then there is some such that for all dfa Æ and the following holds.
Æ
If and are connected via , then and are connected via and for .
Proof. Let for some and . Because we
can write for all as . We can assume that . Now define with for , and for and . By Definition 3.1 we have and with Definition 3.2 we see that .
Æ
Observe with Definition 3.4 that and that . It remains to show for some dfa Æ and that and are connected via under the assumption that and are connected via . From the latter we have by
Æ
Definition 3.1 that Æ and Æ Æ . Hence, Æ Æ and
Æ Æ Æ , and by Proposition 4.2 we see that appears at and at . With the
for and definition of witnessing states Æ , Æ we get from Definition 3.4 and Æ that and are connected via . Note that appears at
❑
every state .
4.2 Inclusion Relations and Lower Levels
Lemma 4.4. The following holds.
1. 2. 3. 38
4. Proof. To see the first statement, let and define . Obviously,
. Let Æ be a dfa and . Clearly, Æ , so appears at .
Now suppose that are connected via . In particular Æ , so are connected via
. This shows .
For the second statement let with and and
for all . Define Æ . By Lemma 3.6.4 we have
. Again, let Æ be a dfa and . First assume that
appears at . By Lemma 3.6.3 we have Æ Æ , and hence also appears at . Now suppose
are connected via . Then appears at and at , and so does . Furthermore, Æ by Lemma 3.6.2, so are connected via . This shows .
Taking together the first and second statement, we get statement 3. The last statement has the same
proof as the second one.
❑
Theorem 4.5. For the following holds.
1.
2.
3.
4.
Proof. We have and by Lemma 4.4. By Lemma 3.17 this implies and
for all . Now the first two statements follow from Lemma 3.18. From Lemma 4.4 we
❑
also know and . Just apply Theorem 3.20 to get the remaining two statements.
The reader may compare these inclusion relations with Propositions 2.7 and 2.8. However, the connections between pattern classes and classes of concatenation hierarchies are even closer (which explains
the naming of the initial patterns).
Theorem 4.6. For the following holds.
1. 2. 3. 4. 5. 6. Proof. To show statements 1, 2 and 4 we recall forbidden pattern characterizations from [PW97]. For
statement 5 we recall such a result from [GS00].
Statement 1: The following is shown in [PW97].
(a) Let be a minimal dfa with . Then if and only if does not
have a subgraph in its transition graph as depicted in Figure 4 with and ( ) being
an accepting (rejecting) state.
39
This is [PW97, Theorem 8.5] after rewriting their notations, together with Theorem 2.20.
Now suppose and let Æ be the minimal dfa with . Assume
, such that
that has pattern , hence by Definition 3.5 there are , Æ , Æ , Æ and Æ . But this is a subgraph exactly as stated
in (a) above, contradicting . Hence does not have pattern , so .
Conversely, let and let Æ be the minimal dfa with . Assume
that has a subgraph in its transition graph as stated in (a). Then and we see with
Definition 3.5 that has pattern since are connected via , contradicting . So or by (a). The latter is not possible since by definition of .
(b) Let be a minimal dfa with . Then if and only if does not have a
subgraph in its transition graph as depicted in Figure 7 with , , and ( )
being an accepting (rejecting) state.
This is [PW97, Theorem 8.15] after rewriting their notations (recall also by Theorem 2.20 that we talk
about the same class of languages ).
Suppose and let Æ be the minimal dfa with . Assume that
has pattern , hence by Definition 3.5 there are , , such that Æ Æ , Æ , Æ and Æ Æ . But
this is a subgraph exactly as stated in (b) above, with , , and ,
contradicting . Hence does not have pattern , so .
that has a subgraph in its transition graph as stated in (b). Then and we see with
Definition 3.5 that has pattern since are connected via , contradicting . So .
(c) Let be a minimal dfa with . Then if and only
if does not have a subgraph in its transition graph as depicted in Figure 7 with ,
and () being an accepting (rejecting) state.
This is [PW97, Theorem 8.9] after rewriting their notations, together with Theorem 2.20.
Suppose and let Æ be the minimal dfa with . Assume
that has pattern and let with and and for all . By Definition 3.5 there are and such that Æ , Æ , Æ and are connected via . Observe that
Æ and by Lemma 3.6.4. Since appears at we
get from Lemma 3.6.3 that Æ Æ . The same holds for . Moreover, it holds that Æ by Lemma 3.6.2. So with , Æ , Æ Æ and we find in the transition
graph of a subgraph as described in (c). This contradicts . Hence does not have pattern
, so .
with that has a subgraph in its transition graph as stated in (c). So let such that Æ Æ , Æ Æ , Æ and Æ (cf.
Figure 7). We want to obtain a contradiction to . Note with Definition 3.5 that it suffices to
show that and are connected via some . In order to achieve this, we want to provide the
prerequisites of Lemma 4.3. Observe that because otherwise which is not possible.
Since also . Let and . Then and . Moreover, and start with the same letter and are connected
via . So we obtain from Lemma 4.3 that there is some such that are connected via . It
follows that , contradicting our assumption. Hence or for some by (c). The latter is not possible since by definition of .
Statement 5: The following is shown in [GS00].
40
(d) Let be a dfa with . Then if and only if does not have a subgraph in
, , and ( )
its transition graph as depicted in Figure 10 with , being an accepting (rejecting) state.
This is [GS00, Theorem 6] (recall also by Theorem 2.20 that we talk about the same class of languages
). We first equivalently restate the appearance of some as certain reachability conditions in the
transition graph of a dfa. We show afterwards that this is equivalent to the appearance of the subgraph
mentioned in (d). Let Æ be a dfa and let . Then are connected via if and only if
(!) there are , and ! for such that
1. Æ 2.
3.
4.
5.
6.
and Æ ,
and ,
Æ and Æ for ,
Æ and Æ for ,
! Æ , ! and Æ! ! for ,
Æ for where is either ! or .
The conditions (!) are displayed in Figure 8. Only a few observations are needed since this is a straightforward translation of Definition 3.4. To see the only-if–part let and with and for . All statements can be verified
with help of the witnessing states from Definition 3.4. For the if–part let with for . Then and . Just consider Definition 3.4 again and
observe from statements 3 and 6 that for the states and are connected via , and
from statement 6 that appears at ! .
Now we want to see that the occurrences of the subgraphs in Figures 8 and 10 imply one another.
This is an easy transformation of patterns, which we carry out next. First suppose and let
Æ be a dfa with . Then we find in the transition graph of a subgraph as
and for , , and stated in (d) with , ( ) being an accepting (rejecting) state. It suffices to ensure the reachability conditions (!) to see that
are connected via some , and hence . Therefore, define
,
and for ,
2. for , and
3. for .
4. Then and with the last definition we also ensure that . Moreover, let
1. , , and ! ,
and Æ for ,
2. Æ and Æ for ,
3. Æ for .
4. ! Æ 1.
41
Since Æ Æ , Æ Æ and Æ Æ ! we can verify all conditions (!). Note that Æ Æ Æ and also
Æ Æ Æ for .
Conversely, suppose and let Æ be a dfa with . Then we find in
the transition graph of a subgraph as given in Figure 8, while we keep the notations from (!). We want
to show that we find a subgraph as given in Figure 10. Therefore, define
,
and ,
2. , and for .
3. . Since Æ Æ , Æ Æ and Æ Then Æ , Æ Æ we find a subgraph as given in Figure 10 with these definitions. So by (d) we obtain that . This completes the proof of statement 5.
1.
Statement 3, 6: This follows immediately from statement 1 and 4, together with Lemma 2.24,
❑
Theorem 3.15 and the monotony of the operators and .
4.3 Pattern Classes are Starfree
Our pattern hierarchies exhaust the starfree languages.
Theorem 4.7. It holds that
.
. It holdsthat Proof. From the first two statements of Theorem 4.5 we get
by Theorem 4.6 and Proposition 2.9. So we may apply Theorem
3.26 to obtain
. Conversely, we get from Proposition 2.9 and Theorem 4.6 that . ❑
4.4 The Hierarchy of Pattern Classes is Strict
We want to show the strictness of the two hierarchies and in a certain way, namely we
take witnessing languages from [Tho84] that were used there to separate the classes of the dot–depth
hierarchy. As remarked in [Tho84], these languages can also be used to show that the Straubing–Thérien
hierarchy is strict. A first proof of strictness was given in [BK78] using similar languages. We could also
do our separation here with these languages, but to facilitate the exposition we stick to [Tho84] since
there the dot–depth hierarchy was defined exactly as we did here (namely, not taking into account).
We assume in this subsection that . In fact, we will separate the instances of our hierarchies defined for this alphabet, see also Remark 4.14 below. Let us first define a family of patterns
" .
Definition 4.8. Let Æ be a dfa and let . We say that has pattern " if and
only there are states such that for it holds that Æ and
Æ .
Lemma 4.9. Let . Then there exist such that for every dfa for every occurrence of " in it holds that
1. the states and are connected via , and
2. the states and are connected via .
42
Æ and
Proof. We show the lemma by induction on . For the induction base let and define & and & . Obviously, & & . We can apply Lemma 4.3 to see that in fact there are
such that if two states are connected via & or & in a dfa , then they are connected via or ,
respectively. Note that the definition of and does not depend on . So it suffices to do the following.
Let some states witness that some dfa Æ has pattern " . Then it
holds that Æ Æ and Æ Æ , so & and & both appear at and .
Since Æ and Æ we see that the states and are connected via &, and that the
states and are connected via & . This shows the induction base.
Now assume that there exist & & for some having the properties stated in the lemma
. Obviously,
and we want to show the lemma for . Define & and & &
. Suppose we are given a dfa Æ and states that
witness an occurrence of pattern " in . Since the states and also the states show that has pattern " we can apply the induction hypothesis and obtain that
and are connected via &,
(b) and are connected via & ,
(c) and are connected via & and
(d) and are connected via & .
Since & appears at and Lemma 3.8 provides us with the following.
and
(e) and are connected via &
.
(f) and are connected via &
Let us verify that and are connected via . Since Æ and because and are connected
via & by (b) we get that appears at . Similarly, since Æ and because and are
connected via & by (d) we get that appears at . Moreover, Æ and because & appears at by (b) we obtain that and are connected via .
Now we want so see that and are connected via . We get that appears at because
Æ , the states and are connected via & by (c), Æ , and since and are
by (f). Similarly, we get that appears at because Æ , the states and
connected via &
are connected via & by (a), Æ , and since and are connected via & by (e). Finally,
appears at by (e), we
because Æ and & appears at by (a), and Æ and &
❑
see that and are connected via .
(a)
Next we recall the definition of a particular family of languages of from [Tho84]. Denote for
by ( ) the number of occurrences of (, resp.) in . Now define for the
language to be the set of words such that and for every prefix of it
holds that . It was shown in [Tho84] that (these languages were denoted
as there).
Definition 4.10. Let . Define Æ with and
Æ and Æ for ,
Æ and Æ , and
Æ Æ .
It is easy to see that . We turn to our hierarchy classes.
43
Lemma 4.11. Let . Then Proof. As already mentioned, we get from [Tho84] that . So and by Theorem 4.6 we obtain .
Lemma 4.12. Let . Then by Proposition 2.7
❑
Proof. Let . Then we have to show that has pattern . Let us first show that there is some
such that and are connected via . If we define we get and since
Æ , Æ and Æ we obtain that and are connected via . Now
we apply Lemma 4.3 to see that in fact there is some & such that the states and are connected
via &. Finally, define and to witness that has pattern .
Now suppose . Then we have to show that has pattern . Note that has pattern " .
So from Lemma 4.9 we obtain that there exists & such that the states and are connected via
&. Now define & &. Then and we show as before that and are connected
via . Note that & appears at and so we have by Lemma 3.8 that the states and are connected via
&.
We get that appears at because Æ , the states and are connected via &, Æ , and since and are connected via &. Clearly, by Proposition 3.12 we have that appears at
because Æ Æ . Finally, because Æ and & appears at , and
Æ and & appears at also by Proposition 3.12, we see that and are connected via
. Finally, define and to witness that has pattern .
❑
So we obtain the following theorem.
Theorem 4.13. Let . Then it holds that
1.
2.
and
.
Proof. Taking together Lemma 4.11 and Lemma 4.12 we see that . By Theorem 4.5 we
have . It follows that is a witness for , which is statement 1.
On the other hand we have by Theorem 4.5 that . So is a witness for .
Note from Theorem 4.6 that and and that is known, e.g., [SW98].
So we get also statement 2.
❑
Remark 4.14. Suppose we deal with some alphabet such that , e.g., ( ( for some . If we define such that Æ ( for and for all , we still find
the desired patterns and can show Lemma
4.12. This means on the language side that we intersect the
( . A look at [Tho84] makes clear that
expressions for with this does not increase the dot–depth, so Lemma 4.11 also holds.
4.5 Decidability of the Pattern Classes and Next we see that our hierarchies of pattern classes structure the class of starfree languages in a decidable
way. We can determine the membership to a hierarchy class even in an efficient way.
Theorem 4.15. Fix some . On input of a dfa
space whether or not is in ( , resp.).
it is decidable in nondeterministic logarithmic
Proof. It holds that PATTERN
PATTERN #$ for each . To see this observe that due to the
definition of the initial pattern the problems PATTERN
and PATTERN are just reachability problems
very similar to R EACH , which can be solved in #$ (cf. Lemma 3.30). Now the theorem follows from
Corollary 3.33.
❑
44
Since membership to is decidable, this yields an algorithm to determine the minimal such that
( , resp.).
is in
4.6 Lower Bounds and a Conjecture for Concatenation Hierarchies
We subsume the inclusion structure of concatenation and pattern hierarchies in Figure 3. Inclusions
hold from bottom to top, and doubled lines stand for equality. Observe the structural similarities w.r.t.
inclusion in each hierarchy.
starfree
Figure 3: Concatenation Hierarchies and Forbidden Pattern Classes
In fact, the inclusions and establish a lower bound algorithm for the
dot–depth of a given language. This follows from the fact that the pattern hierarchies are decidable (cf.
Theorem 4.15). In the previous subsections, some necessary conditions as strictness and the upper bound
were shown to hold for the pattern classes. So at this point we have no evidence against the following
conjecture.
Conjecture 4.16. For all it holds that
1. and
45
2. .
Note that this conjectures an effective characterization of all levels of the dot–depth hierarchy
and the Straubing–Thérien hierarchy. As stated in Theorem 4.6 the conjecture is true for . To
get a better impression of this conjecture, the reader may compare Definitions 3.1 and 3.4 with Figures 4,
5, 6 and Figures 7, 8, 9.
If we look at the witnessing language from Lemma 4.11 again we see that it is in but not
in . So which shows that the pattern class captures but not . We may
conclude that our pattern classes are not ‘too big’.
46
5
is decidable for two–letter alphabets
We show in this section that in case of a two–letter alphabet. To do so, we prove a more
general result: Whenever for some and arbitrary alphabets is known, then it holds
that in case of a two–letter alphabet. Due to the result from [GS99], which is
valid for arbitrary alphabets and restated in Theorem 4.6, it then follows in particular that in
the two–letter case. Let in this section.
We use the observation made in Proposition 2.3 to encode the alternating blocks of ’s and ’s in a
word from over some larger alphabet. For this end, we look for the remainder of this section at some
fixed but arbitrary minimal permutationfree dfa Æ and set . Nothing new
happens in if it reads more that consecutive letters ’s or ’s from the input.
5.1 Changing the Alphabet
and . We define an operation which maps words
to words where . Note that we can write every as for some and factors of maximal length such that .
Call this the -factorization of and observe that this factorization is unique due to the maximality
Let
condition.
Definition 5.1. Let and let ( ( ( with
( for and .
for some be the -factorization of . Then
We say that ( has type if ( and it has type if ( . Clearly, if and only
if ( has type and if and only if ( has type . For later use, denote by ) * (*) the type
of the first (last, resp.) letter of * .
Without further reference, we will make use of the following arguments. If for
some is the -factorization of then , which follows directly from
the above definition. Just observe that ( ( ( if and only if ( for . On the
other hand, if we know for some that ( ( ( for some then the -factorization
of has exactly factors.
In the remainder of this subsection we give a number of easy to see Propositions that will be helpful
in later proofs and that state basic properties. First observe that we encode the length of in ( .
Proposition 5.2. Let and let be the -factorization of for some .
Moreover, let ( ( ( for ( . If ( then and if ( then
for and .
Proof. We show that if ( , the other case is completely analogously. Because (
has type it must be that for some , and it holds that by definition. On one
❑
hand, we get , on the other hand we have .
Next we see that this coding does what we intended, namely that can not distinguish words having the
same representation over .
Proposition 5.3. Let .
If then Æ Æ for all .
47
Proof. By assumption, ( ( ( for some and ( . Let and let be the -factorizations of and , respectively. Fix some and some
with . We argue that Æ Æ . Assume w.l.o.g. that ( is of type and hence
. If ( with then and hence Æ Æ . If ( then and with . By Proposition 2.3 we have that Æ Æ Æ .
It follows inductively that Æ Æ for all and . Hence,
Æ Æ .
❑
Corollary 5.4. Let .
Then it holds that
for all with .
Proof. We only need to observe for
Æ Æ .
that from
we have by Proposition 5.3 that
❑
Infinitely many words from can appear in the range of the mapping # . The maximality
condition in -factorizations leads to alternations of letter types in .
Definition 5.5. Let * words of . For * define * for
some * .
be the set of well–formed
By definition, none of the sets is empty.
Proposition 5.6. Let * (
( for .
( (
for some . Then * if and only if ( $
Proof. As mentioned before, if * for some then the letters ( alternate between and due to the maximality condition in the -factorization of . Conversely, we may take with if ( and if ( . Then is an -factorization of with ( . So ( ( ( and hence
* .
❑
Clearly, factors of well–formed words are again well–formed.
Proposition 5.7. Let * *
** for some * * and * . Then * .
Proof. Let * ( ( ( for some and suppose * ( ( ( for some with
. Since * we have by Proposition 5.6 that ( $ ( for .
❑
This holds in particular for .
On the other hand, we can concatenate words from
formed.
Proposition 5.8. Let if the concatenation of their images is well–
such that . Then .
Proof. Suppose and for are the -factorizations
of and , respectively. Then ( ( ( and ( ( ( for some ( with
. Moreover, it holds that ( for . Since we have by Proposition 5.6 that in particular ( and ( have different type. It follows that $ . So the -factorization of is . Hence,
( ( ( ( .
❑
48
Finally, let us denote for further reference that we can take factors of that respect the
and obtain the respective factor of .
Proposition 5.9. Let ( ( ( for ( ( ( ( .
-factorization
and let be the -factorization of for some . If
then for all with it holds that Proof. We have ( for and so ( ( ( . Since
is an -factorization we have .
❑
Remark 5.10. The mapping # induces in a natural way an equivalence relation on that
depends on , i.e., define for that +, if and only if . Denote by the
equivalence class including and by the set of all equivalence classes. Then and are isomorphic and +, has an infinite index on . Moreover, it holds that for all .
In fact, Proposition 5.3 shows that can not distinguish equivalent words, so +, is a refinement of the
syntactical congruence.
5.2 Automata Construction
We can take words from as inputs to a dfa which respects the acceptance behaviour of .
Moreover, we want the automaton to reject, whenever it finds two consecutive letters of the same type in
the input (see Proposition 5.6). Therefore, we make the following canonical construction. We define
Æ
with
a. %,
b. ,
c. Æ% ( % for all ( ,
d. Æ Æ for all e. Æ Æ for all f. Æ % for all ,
and ,
g. Æ Æ for all h. Æ Æ for all i. Æ % for all ,
and ,
and and
and .
. Note that we simulate in the first components and store the type of the last input
We set letter in the second components of the states of . Let us make the relation between and precise.
Proposition 5.11. Let and Æ for some . Let ( ) be the first
(last, respectively) letter of . For with and it holds that Æ .
49
Proof. Let with be the -factorization of . Moreover, let ( ( (
and define to be the type of ( . Hence, consists of alternating ’s and ’s, and in particular,
and . For and define if ( and if ( . By Proposition 5.2 we have and so Æ Æ for all by
Proposition 5.3. Assume that with and . We show inductively that
Æ ( ( Æ for .
Induction base. Let and assume w.l.o.g. that . Hence, ( and for
some , and by assumption. First suppose . Since we require that we only make an assertion in case . We conclude from Proposition 2.3 and d. in the definition of
that
Æ ( Æ Æ Æ Æ Now assume that . Then we see with Proposition 2.3 and h. in the definition of that
Æ ( Æ Æ Æ Æ The case of is completely analogously and involves e. and g. from the definition of .
Induction step. Suppose we have Æ ( ( Æ for some with and we want to show this for . Let us assume w.l.o.g. that . Hence, ( , and
, ( , for some .
We conclude with the hypothesis, with Proposition 2.3 and with g. in the definition of that for
we have
Æ (
( ( ÆÆ ( ( ( ÆÆ ( ÆÆ Æ Æ Æ Æ Æ Æ Æ The case of is completely analogously and involves h. from the definition of . This completes
the induction and it follows that
Æ Æ (
( Æ Æ ❑
Proposition 5.12. Suppose Æ * for * and with and . Then * and Æ for all .
50
,
Proof. Let * ( ( ( for some and define again to be the type of ( . Note that if then Æ ( % due to f. and i. in the definition of , and Æ * % due to c. So .
Now assume that * . Then we get from Proposition 5.6 that there is some minimal with
such that ( ( or ( ( . So and ( ( ( . Hence, there
is some with first letter and last letter such that ( ( and there is some with Æ . Because we have all prerequisites of Proposition 5.11 and obtain from it that
Æ ( ( ( . From f. and i. in the definition of we see that Æ ( %
since , and from c. we get Æ * %, a contradiction. This shows that * .
Now let . Then * and has first letter and last letter . There is some such
that Æ . Since we know that we can apply Proposition 5.11 again and obtain
Æ * Æ So Æ .
❑
Since % we get from Proposition 5.12 that following Proposition.
Proposition 5.13. It holds that .
More precisely, we have the
.
. Then Æ * - for some and - by b. in the
Proof. Suppose * definition of . We can apply Proposition 5.12 to see * . So there exists some with
* . Again by Proposition 5.12 we obtain from Æ .
Conversely, suppose . Then Æ for some and by Proposition 5.11 we get
Æ for some . So .
❑
5.3 Transformation of Patterns
Let . We show that if has pattern then even has pattern . The following lemma
shows this for all patterns connecting some states not equal %. This ensures that the pattern contains
only well–formed words. In the following, we consider with respect to , and with respect to
.
Lemma 5.14. Let and such that are connected via in for some
with and . Then there exists some such that
for all with and the following holds.
1. If appears at in , then appears at in .
2. If are connected via in , then are connected via in .
Proof. The proof is by induction on .
Induction base. Let . Then . * for some . * . Since are connected via it follows that Æ . for all and Æ * . We obtain
from Proposition 5.12 that . * for . So we can consider and , and there are and with . and *. Note that and define . Then
(w.r.t. ), and and start with the same letter. Moreover, it holds that . To see
this suppose . Then all letters of . have the same type and . , a contradiction. It
follows that . Now Lemma 4.3 provides some such that Æ and if
any two states in are connected via then they are also connected via .
To see statement 1 suppose appears at in . Then Æ . and we get from
Proposition 5.12 that Æ . So Æ Æ and Proposition 4.2 shows that appears at in
.
51
For statement 2 assume that are connected via in . By definition, Æ . , Æ . and Æ * . We get from Proposition 5.12 that
Æ , Æ and Æ Æ . So are connected via in and we have already seen that they are also connected via .
Induction step. Suppose the lemma holds for some and we want to show that is also holds
for . So let * * with * and for . By
assumption, there are which are connected via , and in particular, appears at .
So there are states , ' for such that
Æ * , ' ,
Æ
' * for , and
the states ' are connected via for .
Æ
Æ
Note that from Lemma 3.6 we obtain that for * * we have Æ and
Æ ' for . Because of c. in the definition of and since % it follows
that the states ' for are not %. So we can rewrite these states as and
' ' for , and for suitable ' and . Due to the construction of the latter are not . Now the induction hypothesis provides for each some such that for all
with and the following holds.
(H1) If appears at in , then appears at in .
(H2) If are connected via in , then are connected via in .
Æ
Moreover, we get from Proposition 5.12 that . So with Proposition 5.7 we see that * for . In particular, there are such that * for . Define now
and observe that .
We turn to statement 1. Suppose appears at in . Then there are states , ' with
such that
,
Æ * , '
Æ
' * for , and
With the same argument as above, we can rewrite these states as and ' ' for
, and for suitable ' and . From Proposition 5.12 and by the induction
hypothesis (H2) we obtain that
Æ , ' ,
Æ ' for , and
This shows that appears at in .
Next we want to prove statement 2. Therefore, assume that are connected via in
. Then appears at and also at , and there are states ! with such that
! Æ * ,
! ,
52
Æ
! * ! for , and
appears at state ! for .
As before, none of the states ! for can be %, so we can rewrite them as ! ! for
, and for suitable ! and . From Proposition 5.12 and by the induction hypothesis
(H1) we obtain that
! Æ ,
! ,
Æ ! ! for , and
appears at state ! for .
Together with the appearance of at and at , which follows from statement 1, this shows that are connected via in .
❑
Lemma 5.15. Let . Suppose has pattern witnessed by and / 0 such that
Æ / ,
Æ
0 ,
Æ
0 , and
the states are connected via .
Then / 0 and / 0 .
Proof. Let * * for some with * and . Recall that
Æ * * and that Æ Æ by Lemma 3.6. So / 0 / Æ 0 and hence, both
words are in by Proposition 5.13. Because we can look at the occurrences of the in for
witnessing that appears at . Again by Lemma 3.6 we see that if appears at some state
Æ
in then this state has a -loop. So we have
/
/
* * * Æ
Æ
Æ
Æ
Æ
Æ
* * * 0
0
and (18)
(19)
Note that none of these factors is the empty word (since ), and that the types of every two
consecutive letters alternate. To argue that / 0 / * * * 0 we need to observe that
also in this word the types of every two consecutive letters alternate. It suffices to show for that * which can be seen from Eq. (18). We obtain fromÆEq. (19) that ) Æ . Now
for some with . Then * ) , a contradiction to Eq. (19).
assume * ❑
Theorem 5.16. Let . If has pattern then has pattern .
Proof. Suppose that has pattern . So there exist , / 0 , such that
Æ / ,
Æ
0 ,
Æ 0 , and
53
the states are connected via .
0 and Æ 0 . We have / , otherwise there is a non–empty loop Set Æ
at , which is not possible by construction. Moreover, we may suppose that 0 since
Æ
0 and Æ 0 . From
otherwise we take 0 and obtain with Lemma 3.6 that Æ
Lemma 5.15 we get that / 0 and / 0 . In particular, Proposition 5.7 says that any
prefix of these words is in . Let such that / and 0 .
We want to argue that none of the (not necessarily strict) prefixes of / 0 and / 0 leads to % in
. To see this let * be an arbitrary such prefix and let such that *. Clearly,
if we give as an input to we end up in some state . Now Proposition 5.11 tells us that
Æ Æ * % for some . So we can rewrite the occurring states as
Æ
and ,
and for suitable with , , and . From Proposition 5.12
and by Lemma 5.14 we obtain that
Æ ,
Æ ,
Æ , and
the states are connected via some This shows that has pattern .
.
Corollary 5.17. Let . If does not have pattern then ❑
.
Proof. By Theorem 5.16 we know that if does not have pattern then does not have pattern .
, since .
❑
So 5.4 Transformation of Expressions
Recall that we still deal with the two–letter alphabet and our fixed permutationfree dfa . We want
to show in this subsection that for and for every language 1 with 1 and
1 there exists some 1 with 1 such that 1 $ 1 for all
. We do this by an appropriate transformation of expressions. First, we observe the following
two propositions.
Proposition 5.18. Let * . Then and $ * for all .
Proof.
Let * ( ( ( for some . Recall that by Definition 5.5 it holds that * . So we immediately have $ * for all and it remains to show
. For ( define
( and ( and ( ( ( and set * ( ( ( . Note with Corollary 2.21.1 that * since we can
substitute subsequent letters - - with - - by - - . We show that *.
54
Let be given and assume . Then * and hence has the -factorization
with factors of maximal length. By Proposition 5.9 we have
( for . Now fix some with and suppose w.l.o.g. that ( for some
. Then for some where . If then and
( . If then and ( . We put all factors together
and see that *.
Now suppose *. Hence, we can write as with ( ( ( , respectively) for . The definition of each ( is such that ( ( ( ,
respectively) implies ( for . On the other hand, the types of the ( alternate between
and since * . So and we can apply Proposition 5.8 to obtain
( ( ( *, i.e., .
❑
Proposition 5.19. It holds that .
Proof. Consider the following definition.
. Clearly, and we want to show . Let * ( ( ( for some be
given and recall from Proposition 5.6 that * if and only if ( $ ( for .
So if every two consecutive letters in * have different type, then * is in none of the sets subtracted from
in the definition of . Conversely, if two consecutive letters ( ( in * have the same type, then
(( and * .
❑
With the following Lemmas 5.20 and 5.21 we prepare the proof of Lemma 5.22. In particular,
Lemma 5.21 will serve as a part of the induction base there.
Lemma 5.20. Let * * * for some and let that * , * , * for . For the language
1 * there exists some language
.
* 1 with
* 1 such that
such
*
1$
1 for all
Proof. We define the transformation of 1 as
1 with
for . Clearly, 1 and we obtain from Proposition 5.18 that for all
. If we choose a representation for each as given by Proposition 2.23.1 we see after
distributing the occurring unions over the concatenations that also 1 has a representation which
satisfies Proposition 2.23.1. So 1 .
Now let and be the -factorization of . Moreover, let ( ( ( with ( .
55
Assume first that 1. We can write as
* - . 2 * - .2 * - .2 *
with - , 2 and . for . Let such that - (
, . (
(
and 2 (
for , and * (
(
for (set and ). We apply Proposition 5.9 and get * for . By Proposition 5.18 this implies that for . Also
by Proposition 5.9 we see that (
(
- . 2 with and for . Now fix some with . If then . If
we may assume w.l.o.g. that and hence . But since the latter are
factors of maximal length and because there must be some with and .
So . We put all factors together and see that 1.
Now assume that 1 and write as
with for and for . We get from Proposition 5.18 that *
for .
Now fix some with and let ) ) ) for some and ) .
First suppose . Then ) has type and ) has different type . So and ) ) ) . If we may assume w.l.o.g. that . Then ) and ) have both type . By
definition of 1 we get . It follows that there is some factor from in and
hence, there must be some with such that ) has type . In particular, we see that and so ) ) ) .
Define 3 for . We have just shown that 3 for and obtain
*
* 3 * 3 * 1
Now we argue that * is well–formed for which it is sufficient to observe that * ) 3 , that
3 ) *, and that for it holds that 3 ) * and * ) 3 . But all this is clear since we have assumed in the lemma that * , * , * for .
Finally, we apply repeatedly Proposition 5.7 and Proposition 5.8 to * and obtain
So * 1.
❑
Let ,
4
and 4 instead of . We write for short instead of
4.
Lemma 5.21. Let 1 with 1 and 1 . Then there exists some language
1 with 1 such that 1 $ 1 for all .
Proof. By assumption we have 1 , hence we can write 1 as a finite union of languages 1 of the
form
1 *
* * *
for and * for (see Lemma 2.17). Let 5 denote the finite union of all
letters in . We can assume w.l.o.g. that for we have * . To see this assume first that
* . Then it must be that and we rewrite the first as 5 5 , and distribute the
56
occurring unions in 5 over the concatenation. If * we can do a similar thing. Finally, if * for we write as 5 and continue as before.
Moreover, we can write each as 5 and again, distribute the occurring unions
in 5 over the concatenation. So we get as a first step that 1 is a finite union of languages 1 of the form
1 * * *
for some and * for . Now we distribute the remaining unions and find that 1 is a finite union of languages 1 of the form
1 * * * *
(20)
for some , letters and * for .
Observe that if for some 1 the condition
* * * (21)
is true, then 1 , which can be seen as follows. Suppose . witnesses that (21) holds and let us
further assume w.l.o.g. that . * for some fixed with . (the cases . * and . * are completely analogous). Then also any other . * is in since . and
. have the same sequence of types of letters from . Now if there is some % 1 then we
have by definition of 1 that there is some . * such that . is a factor of % ,
contradicting Proposition 5.7.
Suppose that for the sets 1 1 1 of the form as stated in (20) occur in the finite union
describing 1. We turn to the description of 1 now. Let 6 be the set of indices such that
1 does not satisfy (21) for all 6 and set 6 6 . As pointed out before, we have
1 for all 6 . Recall that 1 is a prerequisite of this lemma. So we have
1
since 1 1 ! 1 ! 1 because ! 1 ! 1 Note here that if we only intersect each 1 with instead of making the distinction with condition (21), then we can not show that the resulting language is in . Since for all 6 the sets 1 do
not fulfil (21) it follows that we can apply Lemma 5.20 to each of them. Denote for all 6 by 1 the languages of with 1 provided by Lemma 5.20 and define
1 1 1
!
The class is closed under intersection by definition, so
One easily verifies that
1
and 1 for all 6
1 for all 6
1 for all 6
1
1 . Let be given.
since since 1 $ 1 by 5.20
❑
57
Lemma 5.22. Let 1 with 1 and let .
1. If 1 2.
then there exists some language
1 for all .
1 $ If 1 then there exists some language
1 $ 1 for all .
1 1 with
with
1 1 such that
such that
Proof. We prove both statements simultaneously by induction on .
Induction base. Let . Then statement 1 holds by Lemma 5.21 and we have to show statement 2.
Recall from Lemma 2.24.2 that . So 1 can be written as a finite union of languages
1 for which in turn there are languages for some such that
1 with for . From 1 we get 1 for each member 1 of the union.
With Proposition 5.7 we see that for each 1 it holds that for all . Let us define the
transformation of to be the set from provided by statement 1. Now set
1 and define 1 to be the union of all 1 going over all occurring 1 .
Clearly, 1 since . Let us verify that 1 . Since for we have 1 and also 1 . So 1 by
Lemma 2.24.1.
It remains to show that 1 $ 1 for all . We can restrict ourselves to a
single member of the union and show 1 $ 1 for all . Let and
be the -factorization of . Moreover, let ( ( ( with
( .
Assume first that 1 . We can write as * * * with * for . Let
such that * (
(
for (set ). We
can apply Proposition 5.9 and get * for . From statement 1 we obtain
that implies for . If we put these pieces
together we get 1 .
Now assume conversely that 1 . Let with for . We
get from statement 1 that for . So 1 . We apply
Proposition 5.7 and Proposition 5.8 to see that 1 .
Induction step. Assume the lemma holds for some and we want to show that it also holds
for . First suppose that 1 and define 1 1 . Since and because
by Proposition 5.19, we obtain from the closure of under intersection
(see Lemma 2.25) that 1 . Moreover, we have
1 So we see that
1 1
1
1
1 1 (22)
since 1 because of Eq. Since 1 is a subset of , we can apply the induction hypothesis of statement 2. Denote by 1 the
language of with 1 provided by the hypothesis and define
1 58
1 Note that 1 1
.
Let be given. One easily verifies that
and 1
1
1
1
since since 1 $ 1 by hypothesis
This completes the induction step for statement 1 and we turn to statement 2.
Let 1 with 1 be given and recall from Lemma 2.24.2 that . Now we can proceed exactly as in the proof of the induction base for statement 2. Only
observe that we can use what we have just shown for in the induction step for statement 1. So
there exists some language 1 with 1 such that 1 $ 1 for all
.
❑
5.5 Connecting Hierarchies
Recall that we started with .
now apply Lemma 5.22 to Lemma 5.23. Let . If via the construction of . We
and obtained then .
Proof. By Proposition 5.13 we have , we get from Lemma 5.22 a language
assume let be given. Now we have
. Since we
, so in particular . To show
with by Proposition 5.13
by Lemma 5.22
❑
As the main result of this section we have the following theorem.
Theorem 5.24. Let . Suppose it holds that
for a two–letter alphabet.
for arbitrary alphabets. Then
Proof. The inclusion is from Theorem 4.6 and holds unconditionally. For the other
inclusion let . By Theorem 4.7 and Theorem 2.1 there is some minimal permutationfree dfa
with and which does not have pattern . So we can do the construction of and we
. By assumption, . We apply Lemma 5.23 to see
apply Corollary 5.17 to obtain .
❑
From Theorem 4.6 we know that in particular
Corollary 5.25. Let . Then
for arbitrary alphabets.
.
The following is an immediate consequence of Theorem 4.15.
Corollary 5.26. Let and .
The question is decidable.
59
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61
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62
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65

Concatenation Hierarchies and Forbidden Patterns Christian Glaßer Heinz Schmitz Theoretische Informatik

Transcription

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