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From Automata to Formulas: Convex Integer Polyhedra
Louis Latour
Université de Liège
Institut Montefiore, B28
4000 Liège, Belgium
[email protected]
Abstract
Automata-based representations have recently been investigated as a tool for representing and manipulating sets
of integer vectors. In this paper, we study some structural
properties of automata accepting the encodings (most significant digit first) of the natural solutions of systems of lin
ear Diophantine inequations, i.e., convex polyhedra in
.
Based on those structural properties, we develop an algorithm that takes as input an automaton and generates a
quantifier-free formula that represents exactly the set of integer vectors accepted by the automaton. In addition, our
algorithm generates the minimal Hilbert basis of the linear
system. In experiments made with a prototype implementation, we have been able to synthesize in seconds formulas and Hilbert bases from automata with more than 10,000
states.
1. Introduction
Presburger arithmetic [16], i.e., the first-order additive
theory of integers, is a powerful, and decidable, formalism
that is frequently used. The manipulation of sets defined
in Presburger arithmetic is central to many applications including integer programming problems [18], compiler optimization techniques [15], program analysis tools [20] and
model-checking [6].
The most direct way of algorithmically handling
Presburger-definable sets consists in manipulating Presburger formulas explicitly, using quantifier elimination
techniques. This approach has been successfully implemented in the Omega package [15], which is probably the
most widely used Presburger tool at the present time. Unfortunately, this approach presents a major drawback, its
lack of canonicity. As a result, a formula can be large although the represented set is simple. This is particularly
This paper is an extended version of [13].
relevant for decision procedures involving sets inclusion because those procedures involve a test of emptiness.
Although it has been known for a long time that automata can represent Presburger-definable sets [7, 19], finite automata have only recently been investigated as an efficient data structure for practical applications [23, 4, 2, 11].
This approach presents two main advantages. First, finite
automata theory has been investigated for a long time, and
efficient procedures exist for set operations. Second, automata have a canonical form. Those characteristics make
the automata-based approach particularly suitable for applications involving a lot of set manipulations. However, although efficient algorithms exist for set manipulation, those
may not be sufficient or efficient if one is interested in extracting the information contained in the automata. For example, in the context of software verification, the automatabased approach may be used for computing the set of reachable configurations of a specified system. It may then be
interesting to generate a simple formula representing the
computed set, which could then be used as input for other
tools handling formulas, such as [15]. The formulas could
also be used directly by the developer for understanding
the sometimes hidden relationships between variables, or
as a broad check for unexpected configurations. Finally, it
should be noted that some operations, such as translation,
can be achieved on formula-based representations more efficiently than on automata-based ones. In this context, it may
appear to be more efficient to generate a formula, apply the
operation on the formula and generate (if necessary) the automaton corresponding to the resulting formula.
In this paper, we present an efficient method for generating a non-quantified formula from a finite automaton accepting the encodings of the natural solutions of systems
of linear Diophantine inequations, i.e., convex polyhedra in
. There is already an analysis of the structure of automata
representing numbers in [11], but the analysis is done for
atomic formulas (one equation, one inequation or one congruence), it is more concerned with the number of states,
2.1. Algebra
and it does not address the question of formula generation. To our knowledge, the idea of generating a formula
corresponding to an automaton representing a Presburgerdefinable set has only been investigated in [14]. A major difference between our approach and that of [14] is that in [14],
the sets handled are those corresponding to Boolean formulas whose atoms are linear equations, i.e conjunctions, disjunctions and negations of equation are handled but inequations are not. So, the domain of our methods are not directly
comparable.
A very interesting aspect of our approach is that it is
based directly on the structure of automata. The method developed relies on a combination of automata theory, such as
properties of minimal automata, and, on algebraic properties, such as linear independence between vectors. By exploiting directly the information given by the structure of
automata, our method avoids costly operations such as determinization, and allows us to generate a simple formula
for automata with 10,000 states in seconds. The fine characterization may also be of significant use in the design of
more efficient procedures for the manipulation of automata
representing integer vectors.
Our method generates another canonical representation
of the sets of natural solutions of polyhedra, the minimal integer basis ([18, 1, 17]). A basis is a pair of finite integer
vector sets from which all solutions are generated. This representation has recently been proposed as an efficient representation in the context of model checking [17]. It has also
been of interest in work related to the automatic deduction
framework [1].
Finally, this work together with [11] is a step toward a
dual representation for Presburger-definable sets, formula
and automata-based representations, using the representation leading to the most efficient procedure according to
the operations performed. Indeed, the formulas we generate
are efficiently handled by the method for constructing deterministic automata from quantifier-free formula proposed
in [11].
The paper is organized as follows. In Section 2, we recall notions and definitions used in the sequel. In Section 3,
we consider the particular case of homogeneous system of
linear equations, and we then consider the general case of
polyhedra in Section 4. We present the algorithms relying
on the theorems of previous sections in Section 5 and some
experimental results in Section 5.3. We then conclude in
Section 6.
maLet and be strictly positive integers. An
trix is a rectangular array of numbers with
rows indexed from 0 to
and columns indexed from to
. An integer matrix is a matrix whose coefficients are
integer. A matrix with a single row (column) is called a row
(column) vector and is written . The component of index
of a vector is written
. The vector having all components equal to zero is denoted . For any vector ,
(resp.
) is the sum of the absolute value of the positive (resp. negative) components of . The determinants
of a matrix are the determinants of the greatest square matrices contained in it. A prime matrix is an integer matrix
whose determinants are relatively prime (i.e. whose greatest common divisor is 1).
The set of rational, integer and natural vectors of dimen sion are denoted by
,
and
respectively. Similarly, the set of rational, integer and natural matrices of di
mension
are denoted
,
and
respectively.
A subspace V of
is a subset of
containing and
such that if
, then
,
The linear hull of a set
is the smallest subspace containing and is denoted by lin-hull
. The in
teger linear hull (resp. natural linear hull ) of a set
is the intersection of the linear hull of with
(resp. )
and is denoted by int-lin-hull
(resp. nat-lin-hull
).
Let
. The vectors
are said
to be linearly dependent (over ) if there exist scalars
such that
, with at least
one
. If the vectors are not linearly dependent, they
are linearly independent. The dimension of a set , written dim
, is the cardinality of the largest linearly independent subset of .
The number of linearly independent rows (or columns)
of a matrix is called the rank of , written rank .
An (in)equation is Diophantine if all its coefficients are
rational. In this paper, we systematically multiply Diophantine (in)equations by the lowest common multiple
of the denominators in order to have integer coefficients.
(In)equations can be represented in vector form as
where
,
and is the vector whose components are indeterminate. A linear system of (in)equations is
a set of linear (in)equations
in variables, (
being for equations and for inequations). Linear systems can be represented in matrix form as
, respectively, where is a
integer matrix. When an
(in)equation
is one of the (in)equations of a system
, we write
. A system
obtained from a linear system
by removing some (in)equation(s) is called a subsystem of
.
/
%'&)(*%,+.-0 /
<
% & (;EFE;EG(*%IH:-
K & (;EFK E;E;( K HL-M
OUR T ?@<VA <
!#" $#"
#"
1<>2=&)(324 +.-5 2&6%7&98:24+; %,+.?@<BA <C=
?D<BA < ?@<BA
% & (;EFE;EF(J%IH
N OQH P & K O % OSR <
? A
WE %XZY
O O
R
] W %[X\Y %^X`_ X
>M
WE %aXbY WE %cXdYb- %cXd_
c
%
d
X
_
$e %MXf_ e
%gX_
%0XL_
WB-B Y:-
2. Basic notions
In this section, we recall some notions and definitions
used in this paper. In addition, we present some general theorems used in the sequel. We mainly follow the notation
used in [18].
2
%
linear system % R where
Theorem 4. Let - "! be a prime matrix with #"
Consider
$
the homogeneous
rank ? A R
that
. For all &%0, and W -' , we have the following
- - #"(FE;E;E6(and
suchwhose
? 1 O9R 4A , and .letLet< %7be&theE;E;E*%array
)
G
A
?
%
equivalence :
columns are % & E;E;E*% .
W( mod %*)W+a mod %
If M
, then the determinants of < are all equal to
zero. If 9]Z[ and if the determinants of < are not all
Proof. Suppose W, mod % . Since is prime, acequal to zero, the columns of < are said to form a set of incording to Theorems 2 and 3, there is only one incondependent solutions and if in addition, R Z , they
gruous solution to .%- mod % , and for any solution
form a complete set of independent solutions. If < is prime
_ .- W a(; (;EFmod
E;E6(/% % \. 0 , W _ mod % . Since is a soluand if %7&EFE;E % form a complete set of independent solution, 1
tions, then they are called a fundamental set of solutions.
Suppose now that W2 mod % . Then it is obvious that
We have the following theorem ([21]) :
for any 3 -f' , 3 E W4 mod % , and therefore, U%5
and _ -5
mod % .
Theorem 1. If - #" with rank ? A R
R
are such that the linear system %
has at least one in_
Theorem 5. If 6 & (;EFE;EF(/6 -5 is a complete
set of positive
teger solution, then the set of integer solutions for the linear
independent solutions for some system % R , then there
system is :
exists a fundamental set of positive solutions 687& (;E;EFE;(96:7 for
the same system.
8 OQP & 2 O % O 2 O - Proof. We may suppose that the components of any vector
in 6 & (;E;EFE;(96 -B admit of no common divisor but unity;
where:
for if all components of a vector 6 O are divisible by ; O , the
the set F%7&)(;E;EFE;(*% is a set of fundamental solu- vectors
6 &< ; & (;EFE;solutions
EG(96 < ; forare also% R a complete
set of posiR
tive
independent
.
tions of the homogeneous system %
,
O?> ] $
a sequence
of matrix += , ]
is a particular integer solution of the linear system, suchWethatwilltheconstruct
?
O
>
columns
of 1= are a set ofO@> positive indepen
R and the 1= are> prime, with
%
dent
solutions
for
Let W(J% -M (*Yb-g#(
, a congruence W,E %
>
&
R
such that the columns of 1=? form a fondaA= set67& ofandsolutions
mod expresses that W,E %:LY is divisible by . The
Ynumber
mental
for % R .
is called the modulus. A system of linear con
H>
The idea is to augment the matrix = , which is supgruences is a set of congruences W O E %gY O mod and
posed >to be prime, by a linear combination of the columns
can be represented in matrix form as %^_ mod . A
H
matrix is prime. Let
of (= and 6 H & such that the resulting
system of linear congruences with indeterminates is re>
H
1
=
the
determinants
of
the
matrix
augmented
with 6IH &
dundant if and defective if . Two solutions
B
E
C
D
;
(
F
E
;
E
F
E
/
(
C
H
admit
as
greatest
divisor.
Determine
by the
&
%are& (Jcongruous
% + of a system
of linear congruences %c_ mod system of congruences
if for all , % & % + mod , otherwise,
they are incongruous.
H & F H > F
Col ?G = AHC I6 H &KJALNMOB
Theorem 2. If every determinant of the augmented
ma
*
F
P
D
trix of a redundant system of linear congruences %
H>
_estmod
be divisible by the modulus , while the greatwhich is always soluble from Theorem 2 since P= is prime
divisor of the unaugmented matrix is prime to the modby recursive hypothesis.
ulus, the system is resoluble and admits of only one inconWe can always choose the C F to be negative by substractgruous solution.
B . with the vector 6 such
ing a finite number
> augmented
H & > be (of= H times
Let += Proof. [21].
that
H &
>
H
R
R
6
I
6
H
C
Col R ?G = A/S
&
Theorem 3. If the greatest common divisor of the determiB+Q R P D
nants of the unaugmented matrix of a non-redundant,
non
> is prime. In addition, since the elH
&
defective system of linear congruences %c_ mod is
(
=
By construction,
>
H
not divisible by the modulus , the system is resoluble and
ement of A= are nonnegative by hypothesis,
> isthean elements
H
&
admits of only one incongruous solution.
1
=
of 6 are nonnegative integer.
Finally,
indepen
dent set of solutions of > % R . Indeed, the T first columns
H
Proof. [21].
are the columns of (= forming an independent set of solutions by recursive hypothesis, and 6 is a solution since it
3
$
R $ , there
exist linearly indepen?
U
A
O
6'&)(FE;E;EF(/6 - -cwith
6 > - for all . From
=
"
with rank ? A R
$
F%V - U% R R
R N OP O O O
F
V
%
5
%
& B 6 (!B - E
We have
R %B- #% R E
Indeed, we prove the mutual inclusion :
If % - ,$ then, % R N OQP & 2 O 6 O , 2 O - $ since
dim ? UA R
(otherwise, there would be at least 8Z
linearly independent vectors in ). So, by definition of
R
If ,%V#- % and. #% R . Then, by definition of , % R
, 2 O -Z . For all , 6 O that
N 6 O PO & R 2 O 6 .O Thus,
% R ? N O P & 2 O 6 O A Rimplying
and %
.
T
6 6& HE; EFE/& 6I H
Theorem 6. Let 67& (;EFE;EF> (/6 -0 be linearly
independent.
$ such that
There exists
with rank ? A R .
5
=
"
F%-5 % R N R R O P O O O
F%V-5 $ % N & B 6 with B - ' E
Proof. Let be the :
integer matrix whose columns
are 6 & (FE;EFE;(/6 and let be transpose of . We have
F%-5 % R N OQP & B O 6 O (NR B O F%V-- ' % R 3 (E3 - Let be the row reduced echelon form
of and $
its transpose. Since the rank of is , the rank of $ $
$
is also , and therefore, there are columns of indexes
all the elements of column ? A
?areD
A6zero
(;EFE;EG( except
? element
A such that equals
to
indexes of the
$ . The
A
other columns are J?D
A6(FE;E;E;(9J?D .
Since is generated from by elementary row operations, we have
F%-5 % R 3 ( 3S- F%VR - % R 3 e (!3 e - R RR T , the equations of 2.2. Polyhedra in
Since ? AG(
In this section, the definition space is , so, any scalar
% R 3 e are
or coefficient is considered to be a rational.
e
$
R
A non-empty set M= is a (convex) cone if 2 %'8B6 %9 ? A R 3 FP & e with ] $
whenever %(!6Land 2I(!B" . A cone is polyhe%9 *? A N H D *? A6(T 3 T with ] dral
if
e
R F%V- %V] Therefore, 3 can be eliminated and we have
rational matrix .
%9 *? A R N FH P D& J? AG(T %9 ? T A with ] : $ forThesome
cone generated by the vectors %'& (;EFE;EG(*%
is the set
$ equations into a system,
Organizing these 2 O % O 1 2 O " N
e
e
R
cone F%'& (FE;E;EF(J% R R
R
F%-5 % 3 with 3 -5 F%V- % !
OP &
$
where
matrix with :
A set of vectors in is called a polyhedron if
( R is
aif ?@ - JA!? A6(F EQ(rational
$
R
T
,
J
D
?
:
A
J
?
A
R % %[]:_
(( RR if J-? AG (JT ? ifA6(FEQ(J- ?D ? $ A6(F:EQ( A? $ 5J?A A R ? T , A R .
$
For the following
definitions, we assume that is a polyThe columns *?D
AG (;EFE;EG(J?@ $ A being linearly indehedron F%V- %[]:_ .
pendent, the rank of
is .
The characteristic cone of , denoted by char-cone ? #A ,
Since multiplying the equations by a scalar does
notby is the polyhedral cone
change the solutions, we can multiply each row of
the least common multiple of the denominator
of its ele
6
char-cone ? #A R
$
ments and get the integer ?D A -matrix such that
R 6 - ? 1 6 %V] - # A;?@% 86V- #A F%V- % R ! R %B- % R !
% For any nonempty polyhedron , if % -`
char-cone ? #A %V- char-cone ? #A contains only , then
pointed. Remark that if a polyhedron R %B
is called
R
R
R
%
g
]
_
% " % " , then
is included in %Z-:
it
Theorem
% with dim ? A
$ . - 7.= Let> " such %[that?
#
!
A
=
F
V
%
is pointed.
Indeed,
char-cone
and
T B% -5 such that both % and % belong to char-cone? #A ,
R %B- #% R E
except .
is a linear combination of solutions and it is independent
of the first columns since
is linearly independent of
any linear combination of
.
Proof. Since dim dent vectors
Theorem 6, there exists
such
that
!
4
vector and there exists ; R max E % Theorem 13. Every polyhedron b=
has a unique ba V% If]:_is ,athenon-zero
hyperplane F% E % R ; is a supporting hysis.
perplane of . A subset of is called a face of if eiLet R F%V%[] _ . An inequality WE %[]Y[R
ther
, or is the intersection of with a supporting
is an implicit bounded inequality in if there exist
%
V
:
]
_
hyperplane of . A facet of is a maximal (w.r.t. inclutwo numbers Y
and Y
such that Y
]:WE %[]:Y for all
sion) face distinct from . We have the following theorems
B
%
. We use the following notation :
(see [18]).
D % ]L_ D is the subsystem of % ] _ formed by the
Theorem 8.
is a face of iff is nonempty and
implicit bounded inequalities in ,
e
e
% ]g_ is the subsystem of % ]M_ formed by
R F%V- % R _ all non implicit bounded inequalities.
e e for some subsystem %[]:_ of %[] _ .
Theorem
cone of R %B
14. Let be the characteristic
Theorem 9. Each face of , except itself, is the intersec % ] _ . WE % ] Y - , WE 6% R] _ is an implicit bounded
tion of facets of .
inequality in iff 1E6[
Theorem 10. The dimension of any facet of is one less
Proof. By definition, 1 6, 6\] , and in particular,
than the dimension of .
WE 6 with]Z
W. E In6 addition,
N3 with 2 683 - , 22"Z
. If 6 , then WE ? 2 6083 A can be arbitrarily large
2.3. Polyhedra in
by increasing 2 . Since ? 12 AG?2!60835- #A , WE % ]Y is not
an implicit bounded inequality.
if ? 1 6d- .A;?DWE 6 R A , by Theorem 12,
For any polyhedron , the integer hull of is
A. finite set of vectors W & (FE;EFE;(*W is a Hilbert basis if each ? Y Conversely,
O E 3S]:Y A , and WE %V]:Y
*
(
Y
G
A
?
1
3
#
;
A
@
?
Y
:
]
W
is an implicit bounded inequality.
integer vector _ in cone W & (FE;E;EF(*W is a nonnegative inteTheorem 15. Let be the characteristic cone of a polyheger combination of W&)(FE;E;E;(*W . An integer Hilbert basis is
dron R :
%
an Hilbert basis consisting of integer vectors only.
- D R %]a_ . The natural linear hull of
% .
is %BTheorem 11. Each rational polyhedral cone is gener ated by an integer Hilbert basis. If is pointed there is a
Proof.
From Theorem 14, nat-lin-hull ? .AB= F% D
unique minimal (w.r.t. inclusion) integer Hilbert basis gen%LetR 6 - . with D 6 R . For each row W O of , let
erating .
and by Theorem 14, !3 O KOR O
of
W E 6 . By definition
Proof. [10, 22, 9, 18].
such that 3 O and W O E 3 O J 7? K O (3
A . Let 3 R
F
F
N683 3.- By construction,
, 35, 6830and
0
3
There is a similar theorem for polyhedra within
implying
that
nat-lin-hull
.
Thus,
L
6
?
.
A
6
([18]) :
D R
6
=
nat-lin-hull ? .A . Theorem 12. For any polyhedron
=
D 6 R . So, 13 From Theorem
14, 1 6 ,
there exist a pair of finite sets of natural vectors
D
nat-lin-hull
D 6 R . ? .A , 3 R and nat-lin-hull? .A!= 6[- Q %7& (;EFE;EG(*% ,67& (;EFE;EF(/6 R S , such that
R
R OQP 2 O % O 8 OQP B O 6 O 2 O (9B O - and OQP 2 O7R 0 E Lemma 16. Let := F% - % ]Z . Let & and +
&
&
&
be two faces of .
Note that in the theorem above, the natural vectors 67&)(;EFE;E;(96 R form the integer Hilbert basis generating
& + ) nat-lin-hull? & A nat-lin-hull? + A
the characteristic cone of the polyhedron.
If & + , obviously,
Proof.
A vector % belonging to a polyhedron
is nondecomposable if it cannot be expressed as 3 &$8 3 + where
nat.lin.hull ? &;A = nat.lin.hull +9E
3tic& cone
- ofand .3 If+ 1- ? -I (; EF E; E6A ( where
is the characteris
T & A . SinceT
, % O is non-decomposable
;
A
?
c
6
6
In addition, ? 6cand
+
and if 6 & (FE;E;EF(/6 R is a minimal Hilbert basis of , then the
6 & -, the+ ,only
we have that 6Land 6 ]\ . Since
%\]b6 -
pair ?D% & (FE;EFE;(J% (96 & (FE;E;E;(/6 R A formes a basis of . The vecpossibility is that WE % ] [tors % & (FE;EFE;(J%
are the constants and 6 & (;E;EFE;(96
are
such that
the periods of this basis. We have the following theorem [18, 1, 17] :
? 1 %V- AG?D%V- & WE % R A and ?DW,E 6 A
5
6 -T
& and
nat.lin.hull &
nat.lin.hull +
If & T + , either +\= & , or ? %- AG?D%& and % -T + A .
– If + =
& then
nat.lin.hull + = nat.lin.hull & E
So, nat.lin.hull & T nat.lin.hull + .
and 6 - T + )
– If ? 6 G
A
G
?
6
&
Then, W,E %[] %V] such that
? 1%- AG?D%V- + WE % R A and ?DW,E 6 A
So, 6 - T nat.lin.hull + and
nat.lin.hull & T nat.lin.hull +
So,
nat.lin.hull
2.4. Finite automata
Theorem 17. Let 6 & (FE;E;E;(/6 R be an integer Hilbert basis of
R %[R ]L .R O O
a cone
%B- % N OQP & B R 6 F( %Vwith- B O -5%B*- int-lin-hullO
Proof. By definition
Hilbert Basis, we have
O
F%-5 = % %BR -N OQR P & % ofB RO the6 ON ( integer
with
B
* 2 O - '
OQR P & R 2 O 6 O -5( with
== int-lin-hull
6'& (FE;E;EF(/6 int-lin-hull
Let %B- such that 2&)(FE;EFEG(32 R - with
R
R% O P 2 O 6 O
&
!
O
R
O
O
O is the largest inte2
D2 R where
2
Let 0]gY
R
O
O
gerR smaller than 2 . Let 6 N O P & Y 6 O and 3 R %5 6 R
N O ByP & definition
2 O 6 O . Sinceof the%[- Hilbert
andbase,
3S- 6 , 6O ]f- which
.
im
plies that 6 ]L . So, & (;E;EFEG( R such that
R
R6 OQP O 6 O
&
R
We have therefore % R 6583 R N O P & B O 6 O with B OR
O 8 D2 O - and more generally,
R int-lin-hull 6 & (FE;EFR E;(/6 R int-lin-hull
R F%V- % R OQP 2 O 6 O ( with 2 O -5 R&
R = F%V- % OQP & B O 6 O ( with B O -5*
? ( (9;)( D ( A
is a finite set of states,
is the input alphabet,
; : is the transition function,
D - is the initial state,
= is the set of final states.
e R ; ? ( A for ( e - and Ze - , then we say
If
that there is a transition from to
labeled by . Let
DO (;R EFE;; EF? ( O b& - (
O A ,, then
$& (;EFE;E;( . If 1 - (;EFE;EG( ,
we say that there is a path from D to
labeled by R R $&E;EFE .
For a DFA ? !( (;)( D ( A , the language accepted
from a state , denoted by " ? A is the set of words such
that there is a path labeled by from to for some
7
7
"
?
U
A
"
?
D
A
. The language
accepted
by
,
denoted
,
is
.
e e is e a strongly connected
A set =#
component, (SCC),
e
R
(
if for
all
either
or
there
is a path from
e
T e and e - e , either
to . If fore all $
e there ise no path
from to or there is no path from to , then is maximal.
Let " be an arbitrary language over . For each %5&
- (' ,
the left quotient of " by % is the set
%*)+" R , &- ' %-
\- " We define the equivalence relation /
.'0= 1' 2 1' by
% 3 .54 iff %*)+" R 4/)+"
We have % 3
only if for all .#4 - if" and
- 1' , %-
6- "
exactly when 47
. We denote the equivalence class that
8:9 , i.e.
contains % by %
% ; 8:9 R , 4 2- ' 4 3 .5% If ; is total, i.e., defined for each pair in L<
, then is
a complete DFA. A DFA such that for every state there is
An alphabet is a finite nonempty set of symbols. A word
over an alphabet is a finite sequence of symbols taken
from . The symbol denotes the empty word, i.e., the
word containing no symbol. The length of a word , denoted
, is the number of symbols in . A language over
is a set of words over .
A deterministic finite automaton (DFA) is a quintuple
where
a path from the initial state to and a path from to an accepting state is called a reduced DFA. It is always possible
to reduce an automaton without affecting its accepted language, except when the accepted language is empty, but this
particular case is easily dealt with.
It is well known that for any DFA , there exists exactly
one complete DFA
, with
>=@? A R ? (! (;)( D ( A
6
" ?UA R
"$?>=@? A A , such that 1 & (
+ - ("$? )&6A R " ? +FA') )
& R
Number Decision Diagram (NDD) representing . Our vector encoding scheme that we use here is slightly different
from the one proposed in [5, 2], in which the negative integers are encoded by means of an -complement scheme,
the first symbol of any encoding being therefore treated as
a sign digit.
It is known for a long time ([7, 19, 5]) that the sets definable in Presburger arithmetic ([16]), i.e., the first-order theory can be represented by finite-state automata
using an encoding scheme similar (using -complements for
negative numbers) to the one that has just been discussed.
Since a polyhedron can be defined as a conjunction of linear inequations, the language of the encodings of the natural
solutions
can be represented by an NDD. A description of efficient algorithms for constructing automata from
linear (in)equations is given in [11].
F+
this paper, we consider reduced minimal automata. Except
. If the latter property holds, the automaton is minimal. In
%
for the automata whose language is empty, for any minimal
automaton, there exists a unique reduced minimal which is
obtained by reducing the complete minimal automaton.
? ( (9;)( ( A
"
R 8 9 -1' ,
; ?J 8 9 ( A R ( 8 9 for all 8 9 -2
R 8 9 ,
R 8 9 -" .
R
Theorem 18. The minimal (complete) automaton
accepting a language is unique up to an
isomorphism and we have :
and
-&
DU( 8U(;]
,
%[-
"
%
4
% )" R 4)"
From the above theorem, it is clear that if
for some language , then, and label paths to the same
state in the minimal automaton.
2.5. Automata-based
ral vector sets
representation
of
3. Structure of automata representing systems of homogeneous equations
In this section, we consider NDDs accepting the encodings of the natural solutions of linear systems of homogeneous equations
, i.e, natural linear hulls, and we
show that those NDDs have a special structure.
% R Lemma 20. Let R ? (! (9; ( D ( A be the reduced minimal NDD accepting the encodings in base
% ofR the natuR
F
\
%
% . Let
ral solutions of the cone
%5- ? A' .
1 4 Q % 4 - " ? UA8) 4 - "$? A/S ) %5- " ? UA
natu-
Here, we explain how finite-state machines can represent
sets of integer vectors. The main idea consists of establishing a mapping between vectors and words. Our encoding
scheme of vectors is based on the positionalal expression
, according to which an encodof numbers in a base
ing of a natural is a word
, where each
digit belongs to the finite alphabet
and . Note
that if
is an encoding
of , then
, with
, is also an encoding of .
, one
In order to encode a vector
simply reads synchronously one digit from the encodings of
all its components, provided that these encodings share the
same length. The result can then be viewed as a word over
the alphabet
. The symbol
will be written for
short. The fact that the word encodes the natu
ral vector is denoted dec
. Note that for any
, there exists one and only one satis
, and similarly to the case of a single
fying dec
natural, if
,
is an encoding of a vector
then
, with
, is also an encoding of . According to our vector encoding scheme, we have the following
theorem :
% 7 & + & D
R (; (;E;EFE6(/% 0
RO & O O
N O P% D % 5%V- ? A'
R ? 4 & (4 + (;E;EFEG(4 A
Proof. We will prove the sufficient and necessary conditions separately.
1 4 % 4S-"$? A iff 4 - " ? UA S . If one chooses
4 R Q V4 - "$? UA since dec ?; A R , implying
% -" ? UA , i.e. %5-"$? UA .
Suppose
% - $" ? UA . Let
4
?
A ' and
now that
d
R
%and
(J% (J% %
A
, dec ?4 A R % %
with dec ?
dec ?% 4 A R % . Since ?G% % - % " ? 8:U%A , byA R defini % .
tion, % R and % R
So, %-4 -$
" ? UA iff 4 -$" ? UA .
R ?D
(;EFE;EG(3
A6( ?D
(*
(FE;EFEG(; A6(;EFE;EG(FE;EFEF? %S
(;EFE;EG(9% A
? % (FE;-aE;E;? (* A A'
%
% A R %
?
%Z- ? A' R
% ?Z% % - A ? % A'
% % %\-
Suppose
, then,
that
R ? (! 9( ; ( D ( A
R F% - % % R N
D
Proof. Since M- "$? UA , the state D is accepting. From
Lemma 20, 1 % - "$? A , % 4 -"$? UA ) 4 -" ? UA , so
%)6" R 4 4a- "$? UA . Therefore, the languages accepted from any accepting states are the same, which means
Lemma 21. Let
be the reduced minimal NDD accepting the encodings in base
of the natu ral solutions of the cone . The
state is the unique accepting state of , and there is only
one maximal SCC in .
%7(4 R - ? A' , if
% dec ?% 4 A , then
%% R R % % ?%A8 % % R ?4 A
Let \=
be a set of natural vectors. If the language
"$? A containing all the encodings of all the vectors in is
regular, then any finite-state automaton accepting "$?
A is a
Theorem 19. For all words
dec
dec
and
, .
%
7
& +
_
D %+&
%&
D +
% & A R % R R ?% + A R
?
% 1 4S- ? A ' S ? % % 8 % ?4 AJA _ R ?G% % % 8
Q ?4 A*A S Q
? 1 4 A;)?%,& &R4[- + "$? A ) % + 4B_ "$? UAJA
_
Suppose
that _ is associated to a state of and that
-? R A ' and %a- satisfy dec ?
A R % and
% _ . Since is reduced, there is a path labeled
by 4B-Z? A ' from to the unique accepting state D
and with dec ? 4 A R %
% such that for all %
labeling a path from D to , if dec ? % A R % , from
the
section, % R _ and
% % 8 % R
previous
R
R
% _8 % , i.e. % % % _ 8 . This
by
% means,
definition of % , that R %
and there
is path labeled by 4 from D to D . Since there is at
most one state labeled by a given _ , the path labeled
by reaches . So,
for all encodings of the positive
%
of % R _ there is a path from D to
solutions labeled by .
D
D Lemma 22. Let R ? (! (9;)( D ( A be the reduced minimal NDD accepting the encodings in base
natuR F%:- % % Rof Nthe . There
ral solutions of the cone
is exactly one loop (with no cycle) labeled by a sequence of
in and this is a loop formed by a transition labeled by
from
D to itself.
that there is one single accepting state since the NDD is reduced minimal, and this state is . Since is the only
initial state and the only accepting state and since is reduced, there is only one maximal SCC in .
Suppose there are two states and both labeled by
the same label . Since is reduced, there is a paths
from to labeled by
and one from to la and dec
beled by . If dec
, then, by hypothesis,
. So,
dec
dec
implying that
. Since is minimal, . So, for any ,
there is at most one state of the maximal SCC labeled
by .
Q 1 4c-? A 'S Q 4c- " ? UA2) 4 - "$D ? UA S
H
RT D
%
D
4
%
*
(
% R
R
?
%
A
%
?
4
A
%
? % % 8L% A R ? % H % U8L% A R % H 4
D
% R ?G% ? % H A % A R R D
D
Lemma 23. Let R ? !( (9;)( D ( A be the reduced min
Lemma 24. Let R ? (! (9; ( D ( A be the reduced minimal NDD accepting the encodings in base % of the natuR % - % R N . Let imal NDD accepting the encodings
in base
ral solutions of the cone
% ofR the natu$ be the dimension of . We can
R
F
%
% ! .SCC
ral
solutions
of
the
cone
For
associate one vector _ to
each
symbol
%
in
and
each
state
of
the
maximal
each state of the maximal SCC such that :
e
of , there is at most one state e of the maximal SCC such
Any path from
from to labeled by % .
D to is labeled by an encoding of a that there is a transition
solution of % R _ .
For any _ , there is at most one state of the maximal Proof. Let %V- with dec ?%A R % and let _ be the
vector associated to . If exist two states & and + reachlabeled by _ .
ing by a transition labeled by % , the vector _ & and _+ as SCC
R
For all encoding of the positive solutions of %
_
sociated to )& and + respectively satisfy :
there is a path from D to labeled by .
(1)
% _9&8 % R _ R % _'+!8 % Proof. From Lemma 21, D is the only accepting state in .
Let & and + be two paths from D to labeled by %,& This implies that _9& R _+ and from Theorem 26, & R + .
and %+ respectively, encoding
two positive integer vec tors % and % . Since is reduced, there is a
Lemma 25. Let R ? (! (9; ( D ( A be the reduced minpath from to the unique accepting state D . Suppose
imal NDD accepting the encodings in base
% R of the natu-$
that this path is labeled by 4 - ? A' and that % R
ral
solutions
of
the
cone
F
%
% ! . Let
satisfies dec ? 4 A R % . Then, there are two paths labe the dimension of . Each state of the non-trivial strongly
beled by % & 4 and % + 4 from D to D such that % ,
connected component of has % incoming transitions.
% - with dec
% & 4 A R % satisfy :
?
$ , there exists $ natural vectors
Proof. Since dim ? UA R
R
R
R % R ?G% % 8 % A
7% & (FE;E;exists
EF(J% linearly
independent
in . From Theorem 6,
> $ such
%
G
?
%
8 % A
%
=
"
there
M
with
rank ? A R % A R ? % % A and % R % . that F % - U% R N R F% - % R
So, ? % O P & B O % O withR B O - ' . So, % - % R N R
Therefore, all paths from D to are labeled
by enN
R
codings of solutions of the same system %
_.
F%V- U% N
accept all encodings of the solutions, we
Proof. Since
must have
,
which means that there is a loop labeled by from to itself.
Suppose that there is another loop labeled by
going through the state and suppose is reachable from the initial state via a path labeled by and that
there is a path from to labeled
by . If and dec
with dec
. Then, we have
. If one considers the
path labeled by
also leading to , we have
.
This implies that
. The only possibil
since
ity is that but this would mean that is the only accepting state according to Lemma 21.
8
Proof. Direct consequence of Lemmas 21,22 23 and 25.
%7#& % (;EFE;R EG(*% R 6
;
(
F
E
;
E
G
E
9
(
6
#
%
&
The particular case of a system of homogeneous equa6 & (;- EFE;EF(/6 "
tions is important because, as we will show, any reduced
minimal NDD representing a general polyhedron has maximal strongly connected components with the same strucD
ture as a reduced minimal NDD representing an homoge_
neous system of equations.
R
D
#%
We will now define notions used to characterize some
states sharing characteristics similar to the ones of the ini
R
F%-5R U% _ R
state of an NDD representing a natural linear hull.
F%V-5 % dec ? A,8 N O P & B O 6 O , B O -5* . tial
Definition 27. Let R ? (! (9; ( D ( A be the reduced
Let %7( 4S-[? A ' , ( -2 , and %I(J%- such that
minimal NDD accepting the encodings in base % of
the natural solutions of the polyhedron := %B%[] _ .
dec ? % A R
dec ?
A,8U%
(2)
A state is a zero-state if it has a loop of size 1 labeled by
dec ? 4 A R
dec ?
A,8U%
(3)
. A state is a spatial-state if the language of the words la Obviously, dec ? % A
dec ? % A mod %#)
#
%
beling paths from to itself is exactly the set of all the enof the encoding scheme,
.% mod % . Bydec definition
codings of the vectors of a natural linear hull.
A mod % and similarly, dec ?4 R A2
?
dec ?% A2
From the characterization of automata representing natdec ? A mod % , and dec ? A* dec ? A mod %+) .
ural
linear hulls, we can deduce the following theorem :
)
#
% U% mod % . Since is prime, acSo, R
Theorem 28. Let R ? (! (;)( D ( A be the reduced
cording to Theorem 4, we have #% .U% mod % if and
minimal NDD accepting the encodings in base
% PL% mod % . We have therefore proven that :
only if %[the] nat % of
_ .
ural solutions of the polyhedron := %B R ) % 1L% mod %)E
In each maximal strongly connected component of there
is at most one spatial-state.
%
This means that the last letter of any encodings of 6 8(U
Based on the previous theorem, we can now partition the
%
are identical if and only if K Y mod % .
and 6 8#
maximal strongly connected components of an NDD into
Since 1%M- ? (FE;E; E;(*
AG(F? (;EFE;EG(F AG(;E;EFEG(F?G% (FE;EFEG(/% those that have a spatial state and those that don’t.
% A letters , #&- % -such thatand% O labels
.% R a path
_ , there
are exactly
Definition 29. A spatial strongly connected component
from D to for
(written spatial SCC) of representative system F%Vsome % , i.e. there are exactly % incoming transitions in ,
minimal NDD R ? (S(9;)( D ( A
#is % a R maximal
! of a reduced
and each transition has a different label.
strongly connected component of hav
Theorem 26. Let R ? !( (9; ( D ( A be the reduced
ing exactly one spatial-state , which is therefore refered
minimal NDD accepting the encodings in base
associated linear hull
R F%[- % % R ofthe . Letnat-$ isto as % the- spatial-state
R of , .whose
ural solutions of the cone
#
%
The dimension of a spatial
be the dimension of .
SCC is the dimension of the natural linear hull of its repsystem. If in addition the automaton R
has one and only one maximal strongly connected resentative
? )& ( !( F+ - (9; ( , 1 ( - A , where
are the states of R and
component.
1
, ; ? )& ( A R
+ ; ? )& ( A + ,
D is the unique accepting state of .
is isomorphic to the reduced minimal NDD accepting the
There is one and only one loop composed of transi- encodings of F%- #% R , is entirely spatial.
tions labeled solely by . This loop is from D to itself
4. Structure of automata representing polyheand is labeled by , i.e., is of size 1.
dra
Each state of can be associated with an integer
vector _ such that the encodings of the natural soluIn the sequel, R ? S!( (9;)( D ( A will be the reduced
tions of % R _ are the labels of the paths from D to
minimal NDD accepting the encodings in base % of the nat the state .
ural solutions of a polyhedron := %L Each state of has %0 incoming transitions, and all will denote the characteristic cone of . "$? # A and%L] " ? _. A ,
incoming transitions of a given state have a different
will denote the language of the encodings of the elements
form a complete set of independent solutions
of . From Theorems 5, there exist a fondamental set of natural solutions
of . Let
be the matrix whose columns are the vectors
. By definition, is prime.
Let be a state of the strongly connected component and
let be a path from to . From Theorem 26, we can
associate to such that for all encodings of all natural
solutions of the systems , there is a path from to
, and from Theorem 1, we have
label.
in 9
and respectively.
Lemma 33. Any loop in
labeled by a sequence of ’s
must be of length 1. Therefore, the state associated to the
loop is a zero-state according to our definition.
& +
%H %4
% f-$" ? #A % H I -#T R "$? #A
T
% (J% ? %A %
?
A R %
% H H % H % #8:H %
% ,H T % ,! % H 8%
% H `"$ ? #H A % , -" T ? #A ?G% % 8 % A!] _
?G% ,! WE %[% ]:Y[8 - % A%[] ]:_ _
WH E ? I% H % 8 % A ] Y
WE ? %
% 8 % A Y
H
Substracting (5) from (4), we get % ? ,A%).A WE % :
.
Since %) Z
, WE % Z
. Inserting this result in Equation 4, we have:
? 1T - A Q % H 8 WE % :] Y S
The last formula is not valid, proving that it is not possible
to have a loop labeled by of length larger than 1.
Theorem 34. Let = be the set of states from
which there is a path to a zero-state labeled by for some
5- . For all %0-V? A' , % labels a path from D to a state
.- iff %5- "$? .A .
Let %0H - " ? .A . From Theorem
Proof.
32, 1 4 - "$? #A ,
1toT an- accepting
, % 4 - state.
"$? #A and % H 4 labels a path from D
Let be the state reached via the
path labeled by % from D . So, one can follow from a
path of arbitrary length labeled by a sequence in .
Since the number of states of is finite, there must be
loop labeled by a sequence of reachable
from via a
path labeled by for some [. From Lemma 33,
the loop is a self-loop from a zero-state of . So, for
"$? .A , for some 5- , % labels a path from
any %5
the initial to a zero-state.
Let
be a state
e such that there is a path from to a
zero-state of labeled by for some . So, for
all % labeling a path frome D to , % labels a path
T . Then,
%
]a
from D to the zero-state . Suppose
there is an inequality W,E %c] Y in %g] _ such that
WE % g
. Since
path labeled
e is reduced, there ise abeing
by from to an accepting state.
a zero H - "$? #A for all T- and according
state, % c2
to
theorem 19,we have
Y " W,?G% H % 8 % AK" % H 8 WE % E
W E % ]ZY W % ,W - represents
the sum
.
WE % ] J ?4 A IR?DY9(;% UW ;A 1 4- ? A' and % WE ? % % 8 % A] WE % ]J I?DY9(;UW ;AE
Proof. The result is a consequence of the encoding scheme.
Indeed, we have 1W - , 1 4 -[? A' ,
%
WE % ] ?G% A % W R ? % : A;W % is sufficiently negative, WE % cannot offSo, if WE ?DWE % A .
set %
Lemma 31. Let WE % ]LY and let W, represents the sum
of the absolute value of the negative elements of W . Let % . If WE % " JR ,?DY9( W;A , then 1 4 -? A' and % with dec ? 4 A
% ,
W,E ?G% % 8 % AK"WE % "J ?@Y9( W ;A
Lemma 30. Let
and let
of the positive elements of . Let , then
If with dec
,
. being
Proof. Suppose there exists a loop of size
reduced and minimal, there are two states and of the
loop reachable from the initial state of by a path labeled
by and
respectively such that there is a word with
for all . Let
and
and dec
such that dec
.
By Theorem 19,
the vectors corresponding to the words
are
and
those corresponding to the
. Since
words
are
and ,
and
. This means that there is an inequality
such that
(4)
(5)
Proof. Similar proof as in Lemma 30.
The close relationship between the solutions of a polyhedron and those of its characteristic cone, in particular, the
property that one can always add a solution of the characteristic cone to a solution of the polyhedron and get another
solution of the polyhedron, also appear when considering
the encodings in a NDD context, as the Theorem 32 emphasizes. Exploiting the nature of the relationship between the
encodings of the solutions of a polyhedron and those of
its characteristic cone leads us to a method for generating a deterministic NDD representing in linear time from
the reduced minimal NDD representing .
. and
So, WE % ] Y H % ] _ such that WE % ?
!
T
FA ? %- W? E % A' A %J H I
?@Y(-T W"$!? ;#A .A S From
Lemma 31, ? 1 and
? 4 - " ? #A*A Q %-4 -T "$? #A/S Q
%5-[? A ' .
%5-" ? .A8) Q 1 4 - " ? #A S Q %-4 -"$? #A S E
Proof. Let M
% -$" ? .A and 4 -" ? #A . Let % R (J% such that dec ?4 A R % and dec
?%] A _ . % So .
We
have
therefore
and
%
]
%
?G% % 8 % A]_ and %-4 -$" ? #A .
Let % -T $" ? .A and dec ?% A R % . We have % ]T
Theorem 32. Let
The following lemmas, mentioned in [8, 11], are important for understanding the structure of NDDs representing
inequations. they mean that for some prefixes, an inequation may be verified (or falsified) for all suffixes.
10
T % ]L
%5-"$? .A
Lemma 36. Any zero-state of
spatial SCC.
By choosing sufficiently large, one can violate the
last inequation. So,
and
.
D
? %A R
?4 + A R % ,1 2 ( 2 -0
2 1 + C
\-V? - A ' S % -" 4 ? & #A8+ )% 47
Q
Q
"$? .A
D
R ?S!( (9; ( D ( A
%
4 & R4 +
?4 & A %
R 2 &C 8
C
R
-?4 "$A ? #A S %
1 WE %V]YV- %[]_
If WE %+ , then % ?DW,E % A J 7?DY9(;U,W ;A ,
If WE % :
, then % ?DWE % A J '?@Y9(FUW ;A ,
If WE % :
, then % ?DWE % A J '?@Y9(FUW ;A ,
e e A R % . By definition,
?
%
Let % R % 4 & 4 + and dec
%tion,
R e % %'8+% % 8+% % . By construc% labels a path from D to . Therefore, proving that
Q 1 \-[? A ' S Q % \- "$? #A8) %-4 \- " ? #A9S
is equivalent to proving that
1Q \- ? A ' S Q % e \- "$? #A8) % e 47
\- "$? #A/S
%b]f_ . Since % labels a path from
Let WE %b] Y^34, WE % ]M
. So, either
WD E % to a zero-state,
or WE % R from
. Theorem
If W,E % . Then % ? WE % A(J I?@Y(;UW FA and
by Lemma 30,
WE % J 7?DY9(;UW ;A
WE ? % % 8 % AKJ I?DY9(;U,W ;A
WE ? % % 8 % % 8 % AKJ I?DY9(;U,W ;A
If W,E % R , since % 4 & and % 4 + label paths from the
initial state of to , by Theorem 34, WE % ] and
WE % ] . R
– If WE % ,
W
E
R . Then, WE % R and
%
Wand
,E ?2 &;% 8 24+G% A R . So, W,E ?G% % 8 % )A R WE % RR WE ? % % 8% % 8 % A
WE ? % % 48 % A
– If WE % or WE % Lemma
. Then,30, WE % J 7?DY9(;UW ;A . Applying
W E ? % % 8 % A J '?@Y9(FUW FA
W,E ?G% % 8'% % 8 % AKJ I?DY9(;UW ;A
% ) "$? #A R % % ) "$? #A
?
1
A Q % \- " ? #A8)R % %-
\-"$? #A S R
?
A %
?%A
% \-[? A'
Suppose
%-
\-$" ? #A
%
Proof. Let be a zero-state. There is a path from to la
beled by if and only if
, which is
equivalent to
. Let
with dec
, and let dec
.
8 %A ] _ . According to Theorem 34
%G? % ]% . Therefore
?G% %98P% %8 % A!] ?G% %98 % A!]:_
By
Theorem 19, %-% is an encoding of % % 8
% % 8 % and we have % %-
\-"$? #A .
Suppose
%-
-T "$? #A
Since there is a simple loop labeled by from to ,
we have for all T :
?G% H % 8 % A ]T _
(6)
If one choose T R % 8Z , Equation 6 becomes
? % & % 8 %A ]T _
(7)
According
34,
% & " to % Theorem
8% ,% we have] : . Since
?G% & % A!] ? % % 8 % % A
?G% & % 8.% A] ?G% % 8*% % 8 % A
Therefore,
?G% % 8 % % 8 % A ]T _
By
of % Theorem 19, %-% is an encoding
% % 8 % and we have % %-
-T "$? #A . % 8
%%
I( U-
Lemma 35. For any zero-state in , if is reachable from
by a path labeled by , then there is a path from to labeled by .
%
is the spatial-state of a
Proof. Let be a zero-state of . Let be a word labeling
. Let and be
a path from to and dec
to with dec
words labeling
two
paths
from
and dec
.
We will show that
such
that ,
for
all
with
dec
,
we
have
. Once this is
proven, the result is immediate since we would have shown
that the words labeling paths from to are exactly the encodings of vectors forming a natural linear hull.
Let such that
,
Based on the previous theorem, we can generate in linear time a deterministic NDD
accepting performing a backward search, starting from all zero-states, and following only transitions labeled by . The states reached are
exactly those in
, and
.
We can refine Lemma 33 by proving that any zero-state
is a spatial-state, and, characterizing the spatial SCCs even
further, we show that the spatial SCCs are entirely spatial
and their representative system is the linear hull of a face of
.
11
So, we have
We now prove that the spatial SCC form a particular lattice.
WE ?G% % 8 % A ] Y)WE ? % % 8'% % 8 %A!] Y Theorem 38. Let & and + be spatial SCC whose spatialthe same reasoning for all inequalities,
states are & and + respectively and whose representative
? %Applying
R and %:- + % R
%
are F%&
% 8 %A!] _ ) ? e% % 8*% %e 8 %A!]:_ systems
. Let& " and? & A and
"$? + A be the language accepted from the
We have then proved that % 47
f+
" ? #A) % f$- "$? #A . state
respectively. The following assertions are
+
Since is arbitrary,
equivalent :
e
e
) % 4 - "$? #A S
1. a path from & to F+
Q 1 \-[? A ' S Q % - " ? #A8
2. F%V& % R .= F%V- + % R 3. "$? )&GA#
= "$? + A
Theorem 37. Any zero-state of is a spatial-state, and
it is the spatial-state of a spatial SCC whose representative
Proof. We will prove that 1 2 3 1.
R satisfies
U
%
system F%V Suppose that there is a path from & to F+ labeled by .
F%V- U% R #= nat-lin-hull(C),
For any %'(4 -[? A' such that % labels a path from D
for each path labeled by % from D to , if dec ? %A R
to )& and 4 labels a path from )& to itself, 1 I( ,
% 4 R labels a path D toR + . If dec ? %A R % ,
% , then U% R .
dec ?4 A
% and dec ?
A % , we have from TheProof. If is a zero-state, then from Lemma 36, is the
orem 37 :
spatial-state of the spatial SCC . Let %'( 4 be the labels
of a path from D to and from to respectively. Let
?
%
% 8 % J % 8 % A R +
%
and dec ? 4 A R % .
dec ? %A R %
R
Since and are arbitrary, this implies that + From Theorem 34, since % and %-4 label two paths
from
R
R
. By definition of a spatial SCC,
+ %encodings
+ % of the natural
D% Ato]a zero-state,
%'( % 4 - " ? .A , % ] and ? % % 8
all
solution of +F
%
R la.
bels a path from + to itself. Since 4 was chosen arbiBy Theorem 12, %'& (;EFE;EG(*% ,67& (FE;E;EF(/6 R such that
trarily,
we have :
R
R O P 2 O % O 8 O P B O 6 O 2 O (/B O - and O P 2 O R R
F
V
%
.
&
%
.
=
B
%
+;% R &
&
&
Suppose that F% - .&6 % R =F% - where 6 & (;E;EF EG(/6 R is the Hilbert basis of the characteristic
labels of path
+G% R D to . Let& and%,& ( %+ +respectively.
-\? A ' beWethewill
K %V] .
cone F%Vfrom
show that
K
So, & (;E;EFE R such that
R
$
"
?
A
=
$
"
?
A
?
%
A
%
? % + A R
.
Let
dec
,
dec
&
&
+
R K
R
% and dec ? A % . From Theorem 34, % ]
% R OQP & O 6 O
% andA0] %_ . Since
] . If% % & ]- "$, ? #?GA %, then,
? % % 8
% 8
and & (;EFE;E R such that
? #A . By hypothe%
% & % 8 % R A ] _ and+;% % + R%,!& gand- " therefore,
R
sis,
.
by definition of the spatial SCC, there is a loop labeled by %7&
% % 8 % R OQP & O 6 O
from + to itself. So, if %+ %,!& f#
there is
- "$accepting
? #A , thenstate
a
path
labeled
by
from
to
an
and
+
So,
.
Since
is
any
word
accepted
from
%
+
b
"
?
#
A
&,
R
we
have
.
$
"
?
)
G
&
A
=
#
$
"
?
F
+
A
% R O P & O 6 O % %
Suppose "$? &GA = "$? F+ A . We will show that there is a
R
path labeled by %+ from )& to + . If " ? )&GA = "$? +;A ,
R O P ? O % K O A6 O
then %,& ) "$? #A = % +() "$? #A and %,&% + ) "$? #AS=
&
%+ %+ )+" ? #A . From Theorem 37, there is a loopR from
+ to itself labeled by %+ . So, %+ %(+ ) " ? #A %(+ )
By Theorem 17, % - int-lin-hull ? .A , and since % ,
"$? #A , and %,&!%+3)6"$? #A5= %+3) "$? #A . From The% Finally,
- nat-lin-hull
?
.
A
.
- "$? .A and from Theorem 32, 1 Corem 34, % & from Lemma 35, there is a loop labeled by %
from to itself. So, by definition of , we have #% R
?
A S Q %+
- " ? #A %&!% + -"$? #A9Q S . So,
.
%+ )"$? #A += %,!& %+ )"$? #A . We have therefore proven
12
%+/)" ? #A R %&% +>)" ? #A
D
N F*P & that% . By
T R J % F - & , and, TUF %and R % ,R implying
1 (;E;EFE6( W O E % R and therefore
1 - (;EFE;E6( )A Q W O E ?G% H % 8 % A R S (9)
if W H F E % R , then
FW E 1 % S -2 F8, and (;EFifE;WE6(JF 2E %, byconstruction,
F
, then W E ? % % 8 % A .
Thus, we have,
1 - ,8Z (;EFE;EG(* )A Q W F E ? % H % 8 % AK S (10)
If dec ? 4 A R % , then, since #% R , 4 labels a loop
rooted at , and by definition of a spatial-state, there is a
H
loop labeled by rooted at . So, % R 4 labels a path
R
% , then % R
from
initial state to and if dec ? %A
the
H
%and from% (10),
8L% 1 , so,- from
(9), 1 - (;E;EFEG( W O E % R ,
,8L (;EFE;EG(* !W F E % .
Theorem 41. Let be a spatial SCC, its spatial-state
and F%bsystem.
Let
OW %L] _! , -R # (;% E;EFR EG( , beitsthe representative
inequalities of %L]\_
O E % R and let
such that U%
FW %[] _! , S- 8L (F%E;EFEG-(J2 , be the W remaining
inequalities of %L]\_ . Let " ? A be the language accepted from
that
. Since
is mini
mal, the state reached from via a path labeled by
leads to the same state as the path labeled by ,
i.e. and there is a path from to labeled by .
% &% +
+
Let
construction,
,
%+
%+
& +
% ]L_ :- ? A'
W
E
%
L
]
Y
% ? 4 A R %
4
,
W
E
%
b
1
%
M
?
A D % ? %A R % %
WE % : % ] _ and let 4d- ? A'
Proof. Let W E % ] Y and %such that dec ? 4 A R % and WE %
.
%
R . Then,
From Theorem 34, WE % a] . Suppose WE WE ?G% % U8% A
g
. But % 4 labels a path from the initial
state to , and by Theorem 34, WE ?G% %80% A]
. There
% P:
is a violation, and the only possibility is that W,E Lemma
SCC, its spatial-state and
YF%LO , - - 40. (F E;#EFLetE6% ( R , bebe atheitsspatial
representative system.
Let W O %Z]
O E % R andof let%:W ]ZO %M_ ] suchY O , that
# 8% ZR (FE;EFBE;(J2%M- , be the remaining
W inequalities
inequalities of %[]:_ .
There exists a path labeled by % from D to and a vector
dec ? %A R % such that 1 -. (;EFE;EG( %W O E % - R with
and 1 -V I8L (J W F E % .
Proof. Let be the label of a path from the initial state to
and let % - with dec ? A R % . By Theorem 37,
#% R , and there is a loop rooted at labeled by , so,
? 1 -2 (;EFE;EG( AG? W O E % R AGE
(8)
34, ? 1
- - 8and
Let T & (FE;E;EF(T
%(; EFInE;EG (*(;addition,
EFE;)EGAG(*? %W F E % - from
] such
Theorem
A . that
1 - '8 (FE;EFEG(J2 such that W F E % M
, % R H
and T F is such that, % ? W F E % A(#J I?@Y F (;UW F FA .
?
A
'
From Lemma 30, 1 4>and % %
with
dec ?4 A R
, W F E ? % H % 8% A J I?DY9(;U,W F ;A .
1 -2 48: (;EFE;EG(* such that W F E % R , T F R and
% F is determined as follows.
Since - 8B (FR E;EFE3(* ,
by definition, % and
F % R T . So, if decF ? R Fsuch
R that
FF , #F % labels
Wrooted
A
%
a loop
, W % (g
otherwise
at . Since W E % Lemma 39 woulde be violated. Similarly, Lemma 39
- 78 (;EFE;EG(* , W F E % A
\
imposes that , 1
WtiveF E % system,
f
. Finally, by definition of the representa
OW E % R . #% R , implying that 1 - (FE;E;E;( ,
By definition
of the
O,
W
R
%B- #% %B#- = O W O E % R N
Let 4 - ? A' such & that 1 (FE;E;E;( ? W O E dec ?4 A R A . By Lemma 40,
there is a path from the initial state of to
labeled by % such that 1 - (FE;EFEG( WW FO E E decdec ? ?%%AKA R and 1 d- 8 (;E;EFEG(J2
So, T =@? A such that 1T," T =@? A we
H W O E H dec ?%A R F
(
;
E
F
E
G
E
(
%
have 1
and 1 C- 8f (FE;EFEG(J2% W F E dec ? %A#
J 7By?DY F definition
(;U;W F ;A of 4 and by Lemma 30, we
- (FE;E;E6( W O E ? % H dec ? %AS8
have 1
1. This is proven in Lemma 40.
2. In order to prove the equality, we prove the mutual inclusion.
Proof.
1. there exists a path
labeled by a vector with dec
,
and
,
% fromR D to and
%
1W O E % - 1 : (F
E;E;E;( W O E % R 1 ?% A - I8%a such
(FE;EFEG(J2that,
R R % - O & W O E % R
#
%
2. F%:
N ,
O & W O E dec ?%A ]
3. "$? A R , % -? A' _! .
. We have
Lemma 39. Let be a spatial SCC whose spatial-state is
. For all inequalities
, if and with dec
such that labels a loop
rooted at and
, then
, , if labels a path from to then
with dec
.
13
O & W O E6 R 6 - % - -L? 8^A' (;EFE;EF(J2 ? 6 1 - K- F R(FE;E;E;W ( F E 6 A;?DW O ? E A ?%A R A
? 1 - ? %I8A5- (FE;EFEG(J2K A;?DR W F E 8 J?% AF A ? K F A
3 ? 1 - - 58 (FE;3 E;E;(JR 2 AG6g? W 8 F E 3 K % & A
? 1 - K % (;EF- E;E6( )AG? W O E 3 R A 6 3 6 - 6 ? A
nat-lin-hull ? A %B- O & W O E % R ! (13)
A and?%AC1 8 - dec8 ? 4 (FA*E;A EFEG(J2
FW E ?G% ? H 4 F AJA R dec
J 7This?@Y (;implies
U;W F ;that
A
% H )+"$? #A R % H 4 ) " ? #A
Since the NDD is minimal, the paths e labeled
H H
by % and % 4 reach the same state, . Since
there is a simple loop labeled by from to ,
% H R labels
path from the initial state to , so,
e anda there
is a path labeled by 4 from to . So, by definition of the representative system, U% R and
%B- #% R F%V- O W O E % R N
&
3. By Lemma 40, there is a path from the initial state
of to labeled by % and !TZsuch that 1 ( W? %O AKE dec
? %A R F andF 1 - 78 (FE;EFEG(J2
% H W (;WeFE;EFE E6decnow
J
7?@Y (;U;W ;A .
prove the mutual inclusion.
Let
we have
? % dec- ?" %A? 8 A . Bydec definition,
?
J
A
A
]
1W O E - dec (F? E;
E;AE;( ] _!W OE .dec ? %A R , 1 - _ .(FE;Since
E;E;( Let - ? A ' with 1 - (;EFE;EG( have 1 - (;E;EFEG( WW OO EE ? % decH ? AVdec]_!?% A. We
dec ?
AJA ]
8
_
and by Lemma
30, 1
- 8 (;E;EFEG(J2
WJ F E7?G% ?@Y H F (; U ;W F dec
? %A 8 H dec ? AJA ;A . Thus, % - "$? #A .
Since there is a loop labeled by rooted at ,
% \- " ? #A , i.e., - "$? A .
dec
Let
such that .
nat-lin-hull
. For each
We will prove that
, let
. By Theorem
41,
such that
dec
and dec
. This means
that dec
.
Let
. By conand let
such that
struction,
and
, implying that
.
Since
, we have that
int-lin-hull , and
given that
,
nat-lin-hull
. We have therefore proven that
From (11), (12) and (13), we have
? A R F%V- #% R (14)
e , !T - e 8 (FE;FE EG(J2 suche that
Finally,e 1
, dec ? %A - T
and
?% 1-%aT "$- ? e A AG.? W H E % R A . So, 1
nat-lin-hull
Theorem 42. For any spatial SCC , the representative
system of is the positive linear hull of a face
of and there exists a path labeled by from
the
initial
state
to the spatial-state of such that
and
,
.
%
%-2" ? A
e
1
e
% -T " ? A
Proof. Let be a spatial SCC, be its spatial-state and
WF%O %M- ] Y O , - U% R(;EFE;E6( be, beitstherepresentative
system.
Let
inequalities of %g] _
such that ? #% R :% and let
W O E % R inequaliWtiesF %:of] Y %[F , ]:-_ . From
78 Theorem
(;E;EFEG(J2 ,41,beA the remaining
F%V- U% R R F%V- O & W O E % R (11)
N
R F% - % ] O & W O E % R ! .
Let
Clearly, is a face of char-cone ? #A , and
nat-lin-hull ? A= F%VO & W O E % R N (12)
" ? AeR $
e
e
e
? A ' a% ]g_ ?
A ]C _ R ? (! % (; ] ( _ ( A
1 ? aP -( (9; P ; ( ? )P & ( ( P A A R + ) ; ? )& ( A R 1 )&+ ( +f- P R
D D
e% R %
P
P -^P ? A'
D
P
P
'
1
%
?
A
P
P; ? DP R ( %A R P
; ? ; (P %?A P ( %A R P 1 % c- - ? A' `;- P ? -( %A R P
D4`- ? A ' ; P ? DP ( 4 A R P
1
P
P
; ? (4 A R
e E ; P ? %? A D R (4 A e ER ? 4 AP
P ? D ( 4 A R T P
% e ) "$? A R R T 4 )0 e " ? A
;
$e EE ? ?% %P A A R ER T e E ?4 A ? 47
A -f% ? A '
D P 47
; ?% ( 4 A 4 R T
Lemma 44. Let R nat-lin.hull ? .A and "$? UA be the encodings of the positive integer solutions of . % - " ? .A
such that :
? 1 4S- " ? UA;? % 4 - "$? .AJA
Proof. Let be a spatial SCC of , let be its spatial
state. From Theorem
41, we know that
dec
where
is a sub
where
system of
. Let ,
is the set of states of
in and ,
. Let
be the reduced minimal NDD accept
ing the encodings of the solutions satisfying
. We
will prove that and
are isomorphic.
By definition of the spatial SCC, for any
,
rooted at in
if and
there is a loop labeled by
only if there is a loop labeled by rooted in . Thus,
by definition of and
,
, with
with
if and only if . Finally,
, if with with
and , then,
,
if and only if
. Indeed, if
, then, accord
ing to Theorem 26,
dec
dec
. Therefore,
. Conversely, if
,
dec
dec
and
such
that
dec
dec
, i.e.,
labels a loop
rooted at in
but
does not. So, and can not label paths to the same state in and .
Corollary 43. Any spatial SCC is entirely spatial.
14
& (;EFE;EF(/6 R (96 & (FE;EFE;(/6
6
6 & (;EFE;EG(96 R 6 & (;EFE;EG(96 B F& (FE;E;E/B FR -R 6 F & R N OR P && B FO 6 + O - (;E;EFEG( R
R
1% B - 2 & (;EFE;? E;B( 2 & (F- E;E;EF( B ( B & (;EFE;E;( B A
% R F* P & 2 F 6 F
R
R F* P 2 F ? OQP B FO 6 O A
R& & F R OQP ? F*P 2 F B O AH6 O
& &
Let 4 =@? A be the length of the minimal encoding of % in
base % . By definition of the encoding scheme and since all
vectors of
Hilbert basis are positive, we have
2 O So,
] % N F* theP . 2 integer
&R F B FO N] %OR P O B .
% B & 6 . Bybeconstruction,
% Z- .
Let Let % (4 -[? A' , % (J% such that dec ?% A R
% , dec ? 4 A R % , dec ? % 4 A R % .
% 8 % . We have:
From Theorem 19, % R % R R
% O P & Q % ? B A78Z? F* P & 2 F B FO A/S 6 O
for some 2&)(FE;EFEG(32 and for each -2 (;E;EFEG( ,
% ? B A78Z? F* P & 2 F B FO A " % B$8a? F* P & 2 F B FO A
" % B ? F* P & 2 F B FO A " Therefore, % and % @4 - " ? .A . Since 4 is arbitrary,
satisfies
the
theorem.
%
Proof. From Theorem 11, there exist
such that
and are the
Hilbert basis for and since they are both polyhedral cone.
,
From Theorem 17, for each with
.
max
Let such that :
, - Let
4 &)(4 +BR - "$? UA , and let % R (J% (*% - suchR that
dec ? %9A
% , dec ?4 &FA % and dec ?4 +FA % .
By definition of % (4 & (4 + ,
? 1 T - A Q 4 & 4 +H - "$? UA % 4 &4 +H -"$? A S (16)
show that
?We% will
% 8 % A " ?G% % 8 % % 8 % A
H-"? A
Let WE %V]:Y[[
%
]
_
. Since =
and % 4 & 4 + according to (16), we have
H & O H H
WE ? % % 8 % % 8 O P D % % A ]:
(17)
that WE ? % % 8c
%
A W,E ?G% % 8
Suppose
% % 8[% A by. Then,
by
multiplying both members of the
F
lastF inequation
and then adding to both members
%
N O P D & % O F W E % , weF find
that 1 F - & ,O WE ? % F % .& > 8 % % F 8 & > N O P D % W,FOQP E % OA Therefore,
WE ? % 8= the sequence
% 8A% = of terms
% 8 N D % F W E % A
F% % 8 N FOQP D & % O WE % A with R W,
E ?G(;% (; EFE; E6(;E is% a se-8
quence of strictly increasing integers. However, from (17)
it is bounded by , which is not possible. So,
WE ? % % 8U% A ":WE ? % % 8 % % 8U% A (18)
Since WE %:]Y was chosen arbitrarily in %]Z_ , we have
proven that
?G% % 8 % AK" ? % % 8 % % 8 % A (19)
- " ? #A . By definition and if
Suppose now that % 4 & +
one takes (19) into account, we have
_ " ? % % 8 % % 8 % A
" Q % ?G% % 8 % A78 % S
" Q % ?G% % 8% % 8 % AI 8 % S
" Q % % 8% % 8% % 8 % S
So, % 4 & 4 + - " ? #A . Since 4 & ( 4 + where chosen arbitrarily in " ? UA , we have
Q 1 4 & ( 4 + - "$? UA/S Q % 4 & - "$? #A % 4 & 4 + - " ? #A/S
The next theorem shows that there is a surjection be%%0-"$"$? ? A A "$? A
tween the faces of and the spatial SCC.
Theorem 46. For each face of , there exists one and
one spatial SCC whose spatial-state is reachable
Q 1 4 & (4 + -"$? A S Q % 4 &!
-"$? #A %-4 &4 + \-"$? #A S only
from the initial state of by a path labeled by % with
Proof. By Theorem 32, 1 % - "$? A9S 1 4B- "$? #A/S % 4% -R "$? A , and whose representative system is F%a- :
Q
Q
Q
"$? #A9S . Since is itself a cone (whose characteristic cone #% such that
is itself by definition), by Lemma 44, % - "$? A such
= %B- #% R that
(15)
where R nat-lin-hull ? A .
Q 1 4 - "$? UA/S Q % 4 - "$? A/S
Lemma 45. Let be a face of and Let be the natural
and be the language of the
linear hull of . Let
encodings in base of and respectively.
There exists
such that
15
% e -T $" ? #A
$" ? A
R.R
1Q %- "$? A S Q 1 % 4-&"" ?? #AA S Q %-4 - " ? #? A SA
%-
- "$? #A
and , violating our assumption. So, there is
at most one spatial SCC reachable via a path labeled by a
word of
and such that .
Proof. Let be a face of and let nat-lin-hull
.
By Theorem 32,
.
This shows that from any word in
, there is a suf
fix such that
. From Lemma 45, there exists
such that
% - "$? A
% 4 &!4 + \-"$? #A S
Q 1 4 &)(4 + -$" ? UA S Q % 4 &
-"$? #A "$? A
e
e
F%V-
#% R E
By Theorem 46, there is one and only one spatial SCC
associated to each face of char-cone ? #A .
The representative system of the spatial SCC associnat-lin-hull
#% R "$? A
" ? UA
= F%V-
e
e
%e
R C
E
!
D
C
]
+
C
D
&
C D$2C & ] D 1C D "
*" ,
a
$
4" with R
Lemma 48. Let
D R
and of %B ]L% ]and andeachisinequality
of
F%V %- ] is an% inequality
irredundant
D % R %[]L . in %[- There is a bijection between the rows of
the facets
is associated and
of such that the row W of
to the facet
of , with
R F%V- D % R %V]L WE % R N
D R
% WE % R N nat-lin-hull ? A R F%V
Proof.
of are given by R F%- D % R All the facets
R
%[]L WE % N , for some row W of .
Suppose now that there are two rows W & and W + of
such that
R F%V- DD % R and %V] and W4& E % R N
R F%V- % R and %V] and W + E % R N
Then,
D
R Q %V-5 D % R %[] SW + E % R S W4&)E % (22)
R Q %V-5 % R %[] SW & E % R S W + E % (23)
Let
be the matrix
without the rows corresponding
to W & and W + .
D R
W
E
M
%
]
Since
is not redundant
in %g-Z
&
%6 ]
D
W + E 6%V]]L
, then,
6
such that 6 R R K . W
E
6
&
W & is a(i.e.,
row of
14,
D 3 R and by Theorem
NW43& E By3 - R construction,
such that 3
3
\
]
) and
. This implies, given (22), that W + E 3 R T ee - "$? A
%
F%V- % R = %B- e % R e
Since e R T , - ? A ' such that either %
T
T "$? #A and % e T -- "$"$? ? ##A A .
and % - "$? #A , or % - "$? #A and
Without
loss of generality, suppose % % e - "$? #A . This means that the path labeled by from
leads to a non-accepting state. Since for all encodings
4 of the integer solutions in nat-lin-hulle ? A there is a loop
frome to labeled by 4 and since % - nat-lin-hull ? A ,
% % -T " ? #A . We have
? % dec ? % AI8% dec ? % e A78 dec ? AJA ]T _
dec ? %9A ] , and
But,
,
% % since
? % decdec ? %?% 9 AA7]L8 % .] So, dec? % ? % dece AI8 ? % dec
e AI8 ?
decA*A ? A*A
Thus,
? % dec ? % e AI8 dec ? A*A ]:T _
e
Suppose now that there is another spatial SCC whose
zero-state is reachable from the intitial state of
via
a path labeled by with and whose represen tative system
satisfies:
(21)
ated to a face of the characteristic cone of a polyhedron
is not necessarily equal to the positive linear hull of . Indeed, the spatial SCC whose representative system is the
positive linear hull of can be somehow merged
with the
spatial SCC associated to a face with
. For an ex
ample, consider the face
in the poly
.
hedron
%
#% R ? A!= F%V
4 & e 4 &!4 + e
4 & (4 + -"$? UA e R e
"$? .A =e "$? A e
"$? A " ? A
04 - "$? UA
4
A 4 O -&"$? UA S
4 &!4 +EFE;E4 H EFE;E
?
1
Q
4& 4+ 4H
M-$" ? A
F%[-
%U% - R " N
? A
%
4 & (4 + - "$? A 4 & 4 + 4
4\-"$? UA
%
Definition 47. The spatial SCC associated to a face ,
denoted , is the spatial SCC whose spatial-state is
reachable from the initial state of
by a path labeled
by with
and whose representative system is
such that
(20)
Let be the state reached via a path from the initial state
of
labeled by . Since
, there exists a partial order on the states reachable
from by paths labeled by words such that
based on the languages accepted from those states. Indeed,
if and are reachable from via paths labeled by
and
respectively, with
, by (20),
. Since the NDD is minimal, either or . So, there must be a state such that
for all
, there is a loop rooted at labeled by .
Otherwise, there would be an infinite path from labeled
by
such that
and the states
reached via , , . . . , . . . , would be all different, which
is impossible since the number of states is finite. Finally,
. By Theothere is a loop from to since
rem 37, is therefore the zero-state of a spatial SCC whose
representative system is
. By construction, is reachable via a path labeled by an encoding of
, and since there is a loop rooted at for each words
of , we have
U% R = F%V-
16
? AJA R $ ? A
e
e
e
eA R
?
? A
e
R
$ $
Theorem 51. Let R dim$ ? .A and let & (;EFE;E;( be the
spatial SCC of dimension c . Let & (FE;E;EF(J be
D such
R
%
that the representative system of O is % O
O
R
E % ! and g= F%V- E %V]
! .
R F%V- O O E %B] D % R &
D R
Proof.
Let -B$." such that R % %V] is an% ineach inequality of
equality %Vof] %[where
D R
and is irredundant in %V-5
]
%
%V]L .
From Lemma
facet of satisfies R F%V- D % R 48,%[any
R D R W is a rowR of ,
Z
]
W
E
%
N where
R
F%V% WE % SCC
!asso.
and nat-lin-hull ? A
From Lemma 50, the dimension of the spatial
$ $
ciated to the facets of are$ of dimension or : and all
spatial SCCs of dimension 0 are associated to one facet.
can be partitioned according to the folSo, the rows of
lowing map.
Let (FE;E;E6( be the indices of the rows of such that
the representative system of the spatial SCC (of dimen$
W E % R N
sion ) associated to the facet %c D R
O
O
R
R
is
% W E % ! and let
the indices of the rows of such that
the78 representative
(;EFE;EGF(*% - be system
of the spatial SCC (of dimen$ W E % R ! is
sion ) associated to the facet F%R F%\- D % R D . RFor the latters,
according to
O
%
E
S
W
E
g
%
] . So, we
Lemma 49, %MQ
have the following result :
R F%V- D % R O W O E %[]
N (24)
&
D R
O
O E % R R
Let O such that
a
%
%
D
O
R
R
R
! and M= %B
NO E From
%V] %B
!Theorem
- . 15,% dim ?UUW A ER % $ . So,
P -= > " such that according to Theorem 7,
R %B- P % R E
O must beO linearly
The
of
W
P , vector
P % W ofO % theR rows
$ sinceindependent
otherwise, dim ? A R
and
R F%V- DP % R R F%V- % R D
. Since K ( B-
, ? 6 8 K 3 A3
W
+
E
1
3
L
D
K
R
W & E ? 6.8 K 3 A R , and? E6BW +8 E ? 6#3 A 8 K 3 A , .?But68 thisK 3 violates
A ]b ,
(23).
there is a bijection between the facets and the rows
So, such
that the facet is associated
to the row W of
of
D % R %V]L WE % R N .
iff R F%V D -Z 3" , - " with R
Lemma
Let
49.
D
R and %B ]L and each inequality of
F% %- ] is an% inequality
is irredundant
D % R of %[]% . ] and
in %B-5
D R
R
Let
of , with
%
%V] be a facet
R ! where W is a row of . % ,
W
E
%
If the representative system
D R of the spatial SCC
associD% R
ated to is F%[, then ?D%V%
A WE %[]
D R
T WR E %a]g
. Then,
Proof. Suppose ?D%Z
A
%
' and % 5- with dec ? 4 A % such that
4BD % - ?R A and
WE % :
.
Let be the spatial-state of and let -b? A ' be
a word labeling a path from to an accepting state of .
Since is associated to , %f-? A ' and % such that % labels a path
the inital state of to D % from
%
R 0W,E % R . Since is
and dec ? %A R a spatial-state,
D there is a loop rooted at labeled by and
since % R , there is another loop rooted at labeled
A
%-4 H R
#- "$? #A/S , but this is not posby 4 . So, ? 1T Q
sible. Indeed, since W,E % and WE % , !T =@? A such
H H % A
J?@Y9( WFA
that ? 1 T T =@? A A;? % WE ?G% % 8g
and by Lemma 31, WE %
JT =@? A ?DY9( ,W;A implyingQ that?G% %- 4 H % -8^T % "$ ? A#8^A for% ST,
"
$
Lemma
Let be the dimension of the cone % D % R 50. . Then,
the dimensions of$ the spatial
SCC asso$
ciated to the facets of are either or , and all spa$
tial SCC of dimension are associated to one and only
one facet of .
and since
,
and we have
?
of
By Theorem 42, there exists a face
D % R nat-lin-hull ? .A R F%V$ . According
We have dim ? .A R dim ? nat-lin-hull ? .A*A R
$ , and
to Theorem 10, for each facet of , dim ? A R
$
therefore, dim ? nat-lin-hull ? AJA R
and the dimension
$
of the spatial SCC associated to is at least . By
Theorem 37, the representative system of all spatial SCC is
included in nat-lin-hull
dimension of
$ ? .A , so,$ the maximal
a spatial SCC is , and the dimension
of
a
spatial
SCC $
associated to a facet is either or . $
Let be a spatial SCC of dimension . The
$ di-.
mension of any proper face of a facet is at most Proof. From Theorem 15,
dim nat-lin-hull
and the representative
system of is equal to nat-lin-hull
. This
is there
fore a facet
of . If such that
is another
facet of the characteristic cone and is also associ
ated to , then nat-lin-hull
nat-lin-hull
. Ac
. There is therefore one and
cording to Lemma 16,
only one facet associated to a spatial SCC if the dimension of is
.
such that
17
R F%V- DP % R W O % R N
5.1. Generation of a formula for the characteristic
R F%V- % R W O % R !
cone
R O
Based on Theorems 51 and 52, it is possible to generSimilarly,
O mustW O must
be linearly independent of the rows of
ate a formula for the characteristic cone from the reduced
P . However,
linearly dependent of the set
minimal NDD representing a polyhedron. The algorithm
beP and
O
formed by the rows$ of
, e otherwise, the dimenP
works as follows. It first extracts the spatial SCCs in linear
R
sion of dim ? O A since, if e R %V-0 e %
time.
Indeed, one simply has to check the states having a
$
So, O % forR some
W 2O % &)(FR E;E;E;
N(3 2 , O = (320- and, dim ? A R . loop labeled
by and then compute in linear time the maximal strongly connected components to which those states
belong. The dimension of the spatial SCC is provided by
OW 'R F* P 2 O W FP 8 2 O
the number of incoming transitions from states in the same
SCC (for any state of the SCC) and is therefore computable
& P
time. From the set of spatial $SCCs, one of them,
P
. Since W O and the rows of in D constant
where W F are the rows of
,
has
a
maximal
dimension, equal to D , which is the di P are linearly independent, 2 R T . So, 1 % , P % R mension of the characteristic
$
cone. It suffices to extract D
Q
1 % - , linearly independent vectors whose encodings label a loop
2 O E %:W O ]E % R. Since
S b= 2 FO % E %d ]O E %:
. ]aIn
Nparticular,
, 2 "a
and 1% , we
from the spatial-state of the spatial SCC to itself. From those
have
vectors, according
6 and 7, we can generate a
P with to$ Theorems
P % R W O E % R S C O E %B] D
matrix
rows, such that
Q P % R B O E % R S W O E % ] . There P R
Similarly,
P R
F
V
%
% R %B- D % R E
Q
O
R
fore, %B% F%V- E %V ] !P % R W O E %[]:
! .
P , F%V- D % R O E %[]
N R
The computation of the linearly independent vectors is curBy definition of
D R
renty
testing all possible vectors in increasing order
$
%B% W O E %[]
N of normdoneuntilby one
finds D linearly independent ones.
It can
$
containbe proved
that,
given
a
spatial
SCC
of
dimension
e
$
Applying the same reasoning to all - (FE;E;E;( , given
ing states, we can always find a set of linearly inde (24), we find that
pendent vectors whose components are smaller than %
,
D
i.e.
the
lengths
of
their
minimal
encodings
are
smaller
than
R F%VW O E %[] % R e . The idea is to compute the minimal Hilbert basis of the
O
&
cone corresponding to the representative system of the spaR F%V- O O E %[] D % R tial
SCC.$ The number of linearly independent vectors in the
&
basis is , and the minimal encoding of each element of the
basis labels paths with no cycle from the spatial state to itself. (Otherwise, if % 47
is the encoding of a vector in the
$ $
Theorem 52. Let R dim ? .A . For all spatial SCCs , the
basis and % 47
labels a loop rooted at the spatial state and
dimension of is at most . Moreover,
there is one and only
4 is an inner loop, then % is also an element of the cone,
$
one spatial SCC of
dimension
and
its
representative syswith
dec ?% A ] dec ?%-4 A , and dec ?% 4 A cannot be
D R
tem is F%V% .
in the minimal Hilbert basis of the cone.) Note that the exponential upper bound does not seem to be met in practice
Proof. From Theorem
46, there is one spatial SCC
e
$
as shown in the experimental results given in Section 5.3.
of$ e dimension
associated to the cone
itself
$
Performing$ the same operations for the spatial
SCCs O
with
"
. Since
=
, by $ definition, we have
$
O
R
R
S generates arrays with D 8 rows.
dim nat-lin-hull ? .A S
dim ? .A
, and for all proper
$ . Thus, from The- ofAtdimension
Q
least one of them, O , is linearly independent from the
faces of$ e , dim$ nat-lin-hull ? A/S R Q and there is no other spatial SCC of rowsO of D . We then identify a path O from the spatial-state
orem 42,
$
of towards the spatial-state of D . If the label % O of O is
dimension "
.
such that dec ? % O A R % and O E % :
, one
needs to re
O
verse
the
sign
of
the
components
of
.
Let
5. Algorithms
( be R the O ar- .
ray whose rows are the computed O , i.e.,
The characteristic cone is then
We address in this section the actual generation of a formula and of the basis corresponding to the polyhedron
F%V- P % R %V]
!E
represented by the reduced minimal NDD .
18
P (*_ F (J% F is
F
_
PO R _ FP _ O ] _ F % O "
_
%F
P (*_ F (J% F F
_
redundant,
P
_ H (*_ (*% H - ? ,( ,( A such that
_ FPP R R HP _ PH _ F ] O _ H % F " %IH . Therefore,
_ O _ H _ O ]_ H % "% H .
Definition 55. From Lemma 54, we deduce that for any set
of labels , there exists a unique set of labels denoted irr ? A
containing exactly the irredundant labels of . A set of labels containing no redundant labels is minimal. The operation of generating irr ? !A from is a minimization.
P % R % ]d be a
Theorem 56. Let ?D%IAP
formula corresponding to the characteristic cone of . For
, let R be the set of labels of such
each state
that for P all % labeling a path from D to , P there exists a
label
in R such that
dec ?%A R
P_ O _ O (Jdec_ O ( ? %decA R ?_ %O A , and
there is no other label in R .
=@? A R irr ? R A .
Let R
R =@? A is unique and finite.
Proof. From Equation 25 and according to the definition of
R , if #- , then the range of values taken by the first component of the label is finite. If this is the case for the accept-
P % R %B] ?
@
I
%
A
Q %7& (;EFE;EG(*% 67& (;EF% E;EF(/6 R S ?@%IA
?@%V% O A %B% O " S (25)
Q
5.2. Generation of basis and formula
If one is given a formula
such that the vectors verifying
are those in , and if
,
is the basis of , then a (nonquantified) formula for is given by
In the following, we present a method which does not require neither a product of automata nor a determinization.
For this, we introduce a computable labeling scheme of the
states of . After the label of each state has been computed,
the constants and the periods of the basis are directly extracted from the labels of particular states as it will be shown
in Theorem 57.
The idea of the labeling scheme is a generalization of associating a label to each state of an NDD representing
natural linear hull as mentioned in Theorem 26, i.e., the labels of a state will correspond to the right-hand side of
equations and inequations verified by the vectors whose encodings label paths from to .
PO_ PO (*O_ O (*O % O and
_ (J_ (J% is re-
From the definition, we have the following Lemma.
_ FP (J_ F (J% F R=@? A
_ FP (J_ F (J% F must be minimal with respect to . So, for any state , there
are a finite number of non-redundant labels among the labels that share the same value for the first component. Since
the range of the latter is finite, the set of non-redundant labels, i.e.
, is finite and unique.
Lemma 54. If is a set of labels and
are redundant in , then
dundant in .
_ = %A ]L_ =
?
%
D
_ P (*_ & (J%7& _ P (J_ + (J%,+ ) ?@_ & ] _ + %7&" %,+ A6E
Given the bounds, there exists a unique finite set of min
imum labels with respect to the ordering (i.e., a label &
is minimal with respect to
if there is no label D+ with
redundant
D& + ). Comparing
the ordering with the definition of
labels, we can see that any non-redundant label
D P
%L]M be the
R
Definition 53. Let ?D%IA
%
formula
cone. Let P (J
be the number of rows of
P and of a(respectively),
and let be the dimension
P
P -5 of% .,
A label of is a triplet _ (J_ (J_ such that _
_ _ P (J-_ (J_ and
a label
P 6 ]6 _ -5 satisfies
if _ 6 -R _ P . A vector
6
_
P"\ . If
P P
a label _ O (*_ O (*% O is redundant in is aP set of labels,
if _ F (J_ F (*% F -L? I(3I(3 )A such that _ O R _ F
O ] _ F % O "% F . If a label is not redundant, it is called
_irredundant.
ing states, it must be the case for any state since there is always a path from any state to an accepting state (since is
reduced). For similar reason, there is an upper bound on the
second component,
i.e, there is an integer vector
with
dec
for any word labeling a path rooted
at . Finally, the third component must be a natural vector
and is therefore a lower bound. For any given state, the labels whose first component has one particular value can be
partially ordered in the following way :
_
%[- !? 6 - #A;? N3S- A;?@% R 6 83 A
Unfortunately, the worst case complexity
for the creation of
the corresponding NDD is ? A and the practical
cost is too large, as we experimented.
So, the last step for generating a formula for is to compute the constants of the basis of .
Our first idea relies on the construction of an NDD accepting exactly the encodings of the constants. Intuitively,
the required automata operations are those corresponding
to the following Presburger formula with two predicates
testing the membership to and (corresponding to the
NDDs and
):
Proof. If this does not hold, then, the label
the only label in such that
.
However,
since
is
?@%IA( P % R %g]^
R= H >
R=@? A
T
R R=@? A R R = D > Given a formula
corresponding to the characteristic cone of , the algorithm
generating is a fixpoint computation. Let denotes the set of labels of the state generated at step .
We start by setting for each state ,
19
R=@? A R R = D >
P (*_ (J% _
P
P
% _ 8 H & > ?%A6(9% _ 8 R = H & > =@? A
R= R =@? A
T
R
in and
P 6 O ( P % 6F O ( (96O % F is(J% redundant
F
P 6O R P % F R 6O ] % F, such6O that"
%thatF "f P 6 VR % F: R T . But
then, 6 R 6O % F is such
R T . Since
6
]
6
M
"
6
R
T
O
O
6Z- , 6 and 6Z] 6 , 6 is not a period, violating our assumption.
Conversely, if a vector % F is not a period of the baF F - F . Then,
sis, then, % F - T . Indeed, suppose
%
P
F% must be extracted
F
from a label %
(
%
J
(
%
P
F R F % F ] % F "
irr ? A , with
%
F
and % - . Since % is not a 6period,
by defO
R T such
inition, there is at least one period that
P P % F ? R % P - %O )% )F AG] ?D% F R %P O % O % F8b"% % O A .. Thus,
Since
dundant
6 O ( 6 O (/6 O - , % F ( % F (*% F is reand does not belong to irr ? A , violating our
assumption.
By construction and by definition of R ,
for
the label
P % each
O ( constant
O P (*O% O % O - O of O R theforT basis,
state
%
some accepting
% and( % P (*% %F ( - % F irr(*% ? F R - A , irrit ? is re-R A
. If R
dundant in
P % O R P % F % O ] % F % O "
such that
%6 F R % O% O % R T F - % F andand6 % R T F -. Thus, % )O is . decomposBut then,
able and is not a constant, violating our assumption.
Note finally
ap that the only constant which could
R
T
O
pear in
is
,
so,
for
all
constant
%
,
%O - R .
F
Conversely, if a vector
R % is not a constant of
`
F
T
- % F-T , except
the basis, then, %
if is a conR
stant, in which
case
. Suppose
R
that % F and % F is not constant of the bathe label from which % F
sis of . By construction,
P
F
F (*% F - irr ? R A
is extracted must be %
(
%
and it must be the label of an accepting state. So,
Since % F is not a constant, it is decompos%ableF - and. there
are a constant %O R T % F and an elO 8 F % .
that % F R ement % P of P such
%
F
F
R
O
O
Thus,
%
% % ] of % O % label% path"
. Since % O , the encodings
%toO accepting
P % O ( % O (*% O - R .
states
and P
is redundant in R , exF
F
F
%
(
%
J
(
%
Therefore, cept if %O R . In
this case, % F , >=
,
P
R
= P and % F ( % F (*% F - . Since
F ( % F (J% F is not redundant
R
%
the label in ,
F
it is also
in and % . Thus,
% F -T R not redundant
.
R ,( ,( e
>R= H %
?%A6(9% % 8 ?%A R= H & >
D
except for where . Then,
at each step of the computation, we propagate the labels,
from to ,
i.e., if there is a transition labeled by
then, for each label
,
we
add the
label
dec
dec
dec
to
the set . After all states have been processed at step
, we merge with , identify the redundant labels and remove them from and from . If all labels generated at one given step are redundant, then the algorithm terminates.
This algorithm always terminates. Indeed, since the third
component of each label, , is strictly increasing at each iteration step, a label generated at step cannot make a label
generated at an earlier step redundant, and the number of labels in is strictly increasing for all . Given that the final number of labels in is finite according to Theorem 56, the fixpoint is reached in a finite number of steps.
The way we have defined the labels is closely
related to
the notion of minimal base.
If
dec
is redun dant, then the vector dec
dec
cannot be a
constant or a period of the basis. In addition, any encoding of a constant (resp. period) labels a path from to a
state in (resp.
, as defined in Theorem 34). The following theorem emphasizes this relationship.
%
R=@? A
?_ %P A(*8 _ ( ?4 A?%A %
R=@? A
Theorem 57. Let where
with
RR
( =
R R
F%V
R 6[-
_ P (J_ GA ?* _ P (J_
_ P (J_ AG?* _ P J( _
does not appear in
of
R
(*% - irr ? R A (
(96 - irr ? A E
The periods of the basis are given by R . ,Similarly,
the constants
of
the
basis
of
are
given
by
if
- . R , then the constants are given by ? R except
A
Proof. From Theorem 56, for any state of , R =@? A is finite, unique and corresponds to irr ? R A where R is such that
- such that
for all % labeling a path from D to and
%
P
dec ?%P A R % , there
exists a label _ O (J_ O (*% in R such
P
R
O
_
R _ O , and there is no other lathat
%
%
R
bel in .
We will first prove that are the periods of the basis of and then prove that the constants of the basis of
R if - R , then the constants are
are given by
R , except
given by ?
A By Theorem 34, the set of words labelling paths from
D to d- are the encodings
of the elements
R,
of . So, by construction
and
by
definition
P % F ( % F (*% F - R forofsome
1state
% F - char.cone
,
O
- . Suppose that
a period 6 of the basis
?
?
and let is defined as in Theorem 34. Let
D
R R R R@= ? A
T
R=@? A
the label
there is a label
The cost of the labeling algorithm depends on the number of elements in the basis. In the worst case scenario, this
. Then, by construction,
20
The following table shows the results of our experiments.
The columns give successively the name of the formula, the
number of states of the corresponding NDD, the number of
constants and periods in the basis, the time (in seconds) and
memory (in megabytes) required for the generation of a formula corresponding to the characteristic cone, and the time
and the memory required for the computation of the bases,
i.e., the labeling algorithm.
5.3. Experimental results
The algorithms presented in this paper have been implemented within the LASH library [12]. Note that the algorithms have been slightly modified in order to use the serial encoding as presented in [3], which significantly decreases the running time. By using the serial encoding, we
simplify the transition relation at the expense of additional
states. As a rule of thumb, the number of states is multiplied by the number of components of the represented vectors, and the number of transition can be exponentially decreased.
The following formulas have been converted into NDDs
over which we have run our implementation. Note that the
, , , , , have been taken from [1],
formulas
( is an example which is not
efficiently in [1] be handled
cause their pruning criterion
does not apply).
& +
3 >?
#CD
F?
#PD
#QD
V
3
>
#C
F
P
Q
V
! + ,"# $.$%-/-0 1$%$%'2&&(34))& *
! "-0 '+: $%5/+ 687
"-97/7$%5& ; <=7 *
/ &/; '5/ $8-9: $.-/-0 3 )7
"-0@5/$ 68
!
"-9$.-9A =-/-0 + (B + *
E/! -9: : A =7
, 4)7 *
, ! + ' 3 C .-97
J5/' $ A$%> C
GHH
$2 )&
! K&#$% #$2& C 3'& C 6.
HI
! J $% > 6L 7
-97 M N
! J5/J#$2 $ $%A@$ 5/'3(&B
(
3
GHI ! J 6. ! "1 @34B=-97& O
! "1<8#&
J! 5/ $%@5/ > C @ C )7 M N
)7
GHI
! K ,'@31$$ ;68
O
! R-9 =-/-0 $% 3 7 6.7
J! 5/ $%@5/ > C @ C + B-97 M N
GHI
! K ,@31$ B- O
! "-9J, '2 -/$%-0;#S$26L3; 6T-U
M NN
#$2 5/> $%&<! 8-97
! @
N
3
$
.
9
7
J
.
6
W
5
GHH ! "-/-0 : @ $= > )7
! R1 8#&
HI
O
! J1$ ;&3S6.
+ #! $% '2&> > .-97 M NNN
@! 3 '; .-97
NN
GHH
HH
! J' @681$% > 7
!
HI
O
! K&"#1$%5/$%; >)B7 -9
X <Y / ! J Y)/ ! SY)/Z
number is exponential in the number of states. This is the
case for example if one takes the set of all vectors whose
components are smaller than a given power of the (encoding) base.
[ \[
425
198
11609
889
3439
165
8465
2368
99094
132619
43777
7496
21
Basis
Const.
Per.
69
214
29
39
1
26
1
567
8
21
1
15
25
37
95
26
1
246
1040
246
25541
922
88
19
32768
0
Labeling
t (s)
Mb
0.1
0
0.0
0
0.1
3
0.3
1
0.2
1
0.0
0
0.3
5
0.1
1
7.0
86
10.5
112
537.8
116
0.1
2
19.5
5
char-cone
t (s)
Mb
0.1
0
0.1
0
2.1
5
0.1
0
0.2
1
0.0
0
0.1
8
0.2
8
10.5
32
12.5
34
3.7
10
0.8
2
0.1
0
From these results, we can see that the algorithm perform
very well on most examples, especially the generation of the
formula corresponding to the characteristic cone, which is
done in seconds in all cases. As mentioned earlier, although
we can not (yet) prove a better complexity than exponential with respect to the number of states, the actual time and
space cost seem almost linear in the number of states. Regarding the generation of the basis, our algorithm performs
well in most cases, the exceptions being
and
. The
latter corresponds to the worst case scenario for which the
number of elements in the basis is exponential in the number of states of the NDD. For the former, it appears that almost all computation time is spent in the manipulation of
the labels (union, difference and inclusion test) and this can
be very costly if there are both many elements in the basis and many states.
&3&
M NN
N
O
&
6. Conclusion
As we would expect, NDDs constructed from conjunctions of linear inequalities have a very particular structure.
In this paper we have thoroughly investigated this structure. The fine characterization given in Section 3 and 4 is
at the heart of the method presented in Section 5 for generating a formula corresponding to the language accepted
by the NDD. In addition, the method also generates another
canonical representation, the basis.
The experimental results are very encouraging, the generation of formulas and bases corresponding to NDDs with
21
more than 10,000 states can be achieved in seconds. A better algorithm for the generation of the constants of the basis
would allow to handle even larger NDDs, since the generation of the formula corresponding to the characteristic cone
is done in seconds in all cases.
Our ultimate goal is to be able to generate a formula for
NDDs representing any Presburger definable set. We have
already generalized the result of this paper for polyhedra in
and we are currently working on a generalization which
would incorporate congruences.
[13] L. Latour. From automata to formulas: Convex integer polyhedra. In Proceedings of 19th IEEE Symposium on Logic in
Computer Science LICS04. to appear.
[14] J. Leroux. Algorithmique de la vérification des systèmes
à compteurs: approximation et accélération. PhD Thesis,
Ecole Normale Supérieure de Cachan, Cachan, France, 2003.
[15] The Omega Project: Frameworks and algorithms for the analysis and transformation of scientific programs. Available at
http://www.cs.umd.edu/projects/omega/.
[16] M. Presburger. Über die Volständigkeit eines gewissen Systems der Arithmetik ganzer Zahlen, in welchem die Addition als einzige Operation hervortritt. In Comptes Rendus du
Premier Congrès des Mathématiciens des Pays Slaves, pages
92–101, Warsaw, Poland, 1929.
[17] T. Rybina and A. Voronkov. Using canonical representations
of solutions to speed up infinite-state model checking. In
Proceedings of the 14th International Conference on Computer Aided Verification, number 2404 in Lecture Notes In
Computer Science, pages 386–400. Springer-Verlag, 2002.
[18] A. Schrijver. Theory of Linear and Integer Programming.
John Wiley and Sons, New-York, 1986.
[19] A. L. Semenov. Presburgerness of predicates regular in two
number systems. Siberian Mathematical Journal, 18:289–
299, 1977.
[20] T. R. Shiple, J. H. Kukula, and R. K. Ranjan. A comparison
of Presburger engines for EFSM reachability. In Proceedings of the 10th Intl. Conf. on Computer-Aided Verification,
volume 1427 of Lecture Notes in Computer Science, pages
280–292, Vancouver, June/July 1998. Springer-Verlag.
[21] H. J. S. Smith. On systems of linear indeterminate equations
and congruences. Philosophical Transactions of the Royal
Society of London, 151:293–326, 1861.
[22] J. van der Corput. Konstruktion der Minimalbasis für
spezielle Diophantische Systeme von linear-homogenen
Gleichungen und Ungleichungen. In Proceedings Koninlijke Akademie van Wetenschappen te Amsterdam, volume 34,
pages 515–523, 1931.
[23] P. Wolper and B. Boigelot. An automata-theoretic approach
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Springer-Verlag.
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22