This is an old revision of the document!
* why the problem is interesting
deciding WS1S - done through automata construction (mona)
main problem: quantifier elimination
is there another way of QE? - can we characterize relations in WS1S in some other way?
WS1S too big: look at a smaller subset of the language
our L: characterize unary relations, binary relations,…n-ary
* what was done before
p-recognizability
* what the main idea is
* what you did (precise description)
* the most surprising/interesting things you found
* limitations
* future directions
look at other subsets of WS1S, whole of WS1S
START
Recall the syntax for WS1S:
The language is the language of WS1S with the restriction that there are no free second-order (set) variables.
A w-relation is a relation on first-order variables expressible in
.
The W-Problem
Characterize the set of w-relations expressible in .
We want to characterize the set by proving that a w-relation is recognized by an automaton (with parallel input) that accepts strings of the form ABC where the length of B is completely described by a union of a finite number of sets S(a,b) for some finite number of tuples (a,b) for given constants a and b.
Here, .
Terminology clarification
Define a set to be ultimately periodic iff its characteristic function
is ultimately periodic.
A sequence is ultimately periodic iff there exists
and
such that
.
Let .
Observe that a set is ultimately periodic iff it is a union of a finite number of sets of the form .
proof
- Consider an ultimately periodic set
with some
and
. Consider the set
of numbers
belonging to
, and the set
of numbers
between
and
belonging to
. Then
- We prove the other side of the equivalence by induction.
Base Case: Clearly is ultimately periodic (
,
).
Induction: The union of an ultimately periodic set (,
) and a set of the form
is an ultimately periodic set with
and
TO EXPLAIN WITH MORE DETAIL IF NOT OBVIOUS
Little Problem
Let L be a regular language over an alphabet and
be the set of lengths of words in L. Prove that M is a union of a
finite number of sets S(a,b) for some finite number of tuples (a,b).
Here, .
Proof
We prove the claim by induction on the structure of the regular expression representing the language.
Notation: We say that a set has property F if it is a union of finitely many sets S(a,b) for some tuples (a,b).
We say a language L (or the corresponding regular expression) has property P if its set of lengths M has property F.
Hence, we need to prove that any regular language has property P.
Recall that Regular Expressions over an alphabet are all strings over the alphabet
that can be obtained as follows:
(1) and each member of
is a regular expression.
(2) If and
are regular expressions, then so is their concatenation:
.
(3) If and
are regular expressions, then so is their union:
.
(4) If is a regular expression, then so is its Kleene closure:
.
(5) Nothing is a regular expression unless it follows from (1) through (4).
A Regular Language, is generated as follows:
(1) and
for each
.
(2) If and
are regular expressions, then
.
(3) If and
are regular expressions, then
.
(4) If is a regular expression, then
.
Base Case:
The length of a single character in is 1. Thus the length of any word corresponding to a single character belongs to the set
. Hence any such word has property P.
To prove the inductive step, we need the following lemma:
Lemma 1
The set where
satisfies property F, that is,
is equal to a union of finitely many sets of the form
,
.
Proof of Lemma 1
Consider the following two (mutually disjoint and exhaustive) cases:
For ,
(1) gcd(b,c)=1
(2) gcd(b,c)=d, ,
Case (1): gcd(b,c)=1
Consider the numbers contained in the set
Any such number is congruent to
Letting
range over
, we obtain for each value of
a distinct congruence class
:
The congruence classes are distinct by the following argument: assume there exist and
such that
and
Then
for some
Hence
Since b divides
it must divide
But gcd(b,c)=1 hence b must divide
. But
, hence we must have
Since there are b distinct congruence classes the set
contains all natural numbers beyond
.
This is because of the following theorem from Elementary Number Theory: If gcd(b,c)=1 and n >= (b-1)(c-1). Then bx+cy=n has a
non-negative solution, that is, one in which both x and y are non-negative integers.
There are also finitely many numbers below
that are contained in the set
(For example,
is in the set).
Hence, the set is equal to the union of a finite number of constants (less than
) and the set
. Each constant
can be represented in the set
.
In other words, if gcd(b,c) = 1, then the set
where
satisfies property F.
Case(2): gcd(b,c)=d,
Any number of the form
can be written as
where
and
and
. From (1), we know that the set
satisfies property F. The set
can be obtained from
by multiplying the constant factors (a and b) in each set of
by
and so, satisfies property F.
Hence, if gcd(b,c) = d , then the set where
satisfies property F.
Inductive step:
We prove that the operations union, concatenation and Kleene closure
of two regular expressions satisfying property P yield a regular expression satisfying property P.
Union:
Let and
be two regular expressions satisfying property P. Then, their corresponding sets of lengths
and
satisfy property F. The set of lengths
of
is simply
. Hence
satisfies property P.
Concatenation:
Let and
be two regular expressions satisfying property P. Then, their corresponding sets of lengths
and
satisfy property F. The set of lengths of
is the finite set
This set contains elements of the form
By Lemma 1,
satisfies property F. Set Q therefore, also satisfies property F. Thus,
satisfies property P.
Kleene Closure:
Let be a regular expression satisfying property P. Let the corresponding set of lengths be
that satisfies property F.
where
is a set of the form S(a,b) and
is finite. The regular expression
can contain any number of repetitions (including zero) of any of the sub-regular expressions. Hence, the possible lengths of words in
is given by
where
. Each term of this summation is of the form
where
.
Consider one such term: . For different values of
, we get terms of the form:
Any number of the form
can also be obtained from
by choosing the constants appropriately:
Thus, the term
can be written as
(Example: Consider the regular expression: . The set of lengths of words corresponding to its language is
. The corresponding set for
is
.)
So, where
. From lemma 1, we know that the sum of any two terms in this summation is a set that satisfies property F.
Thus,
is the sum of finitely many sets: each set satisfying property F. We can repeatedly combine the terms of all these sets and by the argument in the case for concatenation, we know that the resulting set also satisfies property F. Hence,
satisfies property F.
Since, any regular expression can be constructed only by the application of the above three steps: Union, Concatenation and Kleene Closure, the language corresponding to any regular expression satisfies property P.
The unary case for the W-problem
We consider the special case of a unary relation on natural numbers corresponding to a formula with only one free variable
.
This formula defines a set
Claim: Set S is ultimately periodic.
Proof:
Given formula , there exists and automaton A with input alphabet
which accepts the string
iff
We define the automaton on
which accepts the string
iff
Note that
.
Note that the language corresponding to is regular iff the language corresponding to
is regular. Indeed, if
is regular then
is the concatenation of the regular expressions
and
and hence is regular. Conversely, if
is regular, then the language corresponding to
is the intersection of the languages corresponding to the regular expressions
and
, and hence is regular.
Furthermore, accepts string
iff
accepts strings
.
Consider the language of
By the Little Problem, the set of lengths
of words in
is a finite union of linear sets
For all
, the set
is the set of lengths of a subset of the words accepted by
, all of the form
. Since
,
is a subset of the integers
such that
is accepted by A, i.e. a subset of S. Furthermore, the finite disjoint union
contains exactly all lengths of words accepted by
, hence
contains exactly all integers
such that
is accepted by A, i.e.
.
S is thus a finite union of linear sets Hence by definition, S is ultimately periodic.
The binary case for the W-problem Consider a binary relation in
and the corresponding formula
with two free variables
and
.
defines a set
. Elements of the set
are recognized by an automaton
with parallel inputs and input alphabet
. The language L of A is a subset of the regular language corresponding to
. Thus the input string to the automaton is of the form
or
or
.
Claim A pair belongs to some set
for some
iff the sets
of possible lengths of the exponents
are ultimately periodic.
Proof
- Only if: Let
belong to
. There exists an automaton A with parallel inputs which accepts the corresponding input string.
W.l.o.g, assume Then the input string can be of one of the two forms:
(a)
(b)
case (a): We want to show that the sets ,
and
are ultimately periodic. Due to the little problem, it is sufficient to show that the language generated by the unary alphabet
:
is regular.
We show that
is regular by constructing an accepting automaton,
. Let
be the automaton that accepts the original input string (a). Let
be the accepting run in
. In
, there exist unique states
and
such that
. Define
. Similarly there exist unique
such that
. Define
.
contains all the states of
that are in
between
and
. And,
is the restriction of
to the transitions in
between
and
.
. Thus the automaton is defined unambiguously.
We can similarly construct automata for the languages and
. In the former case, the start state corresponds to
and the final state is the state (in
) from which there is a transition on consuming
. In the latter case, the final state corresponds to
and the start state is that state in
that is reached after consuming
.
The construction of the automata for case © is along the same lines.
If: If is an ultimately periodic set. then the expression
where
belongs to a unary alphabet and
is a regular expression. This is because for each set
one can construct a regular expression (ppp…a times) concatenated with the kleene closure of (ppp…b times). Since
is the finite union of sets of the form
, one can construct a regular expression which is the union of all the regular expressions formed as described above. Thus,
is a regular expression for
.
We now extend our alphabet to include
,
and
. The above argument for
still holds. (a), (b) and © are all regular expressions formed by concatenation. Thus, the input strings are accepted by finite automata. They are satisfying interpretations of some formula in
because (a) can be written as
, and (b) is simply
. © is symmetric to (a).
The above reasoning can be extended to the n-ary case. The input string has n non-zero elements and n+1 expressions of the form . A tuple
belongs to an n-ary relation iff the possible values of each exponent
forms an ultimately periodic set.