City, University of London Institutional Repository
Citation
:
Howe, J. M. and King, A. (2000). Abstracting Numeric Constraints with Boolean
Functions. Information Processing Letters, 75(1-2), pp. 17-23. doi:
10.1016/S0020-0190(00)00081-8
This is the unspecified version of the paper.
This version of the publication may differ from the final published
version.
Permanent repository link: http://openaccess.city.ac.uk/1696/
Link to published version
:
http://dx.doi.org/10.1016/S0020-0190(00)00081-8
Copyright and reuse:
City Research Online aims to make research
outputs of City, University of London available to a wider audience.
Copyright and Moral Rights remain with the author(s) and/or copyright
holders. URLs from City Research Online may be freely distributed and
linked to.
City Research Online:
http://openaccess.city.ac.uk/
[email protected]
Boolean Funtions
Jaob M. Howe
1
and Andy King
ComputingLaboratory, University of Kent,Canterbury, CT2 7NF
Abstrat
A simple, syntati algorithm for abstrating numeri onstraints for groundness
analysisispresentedandproved orret. Thetehnique usesneitherprojetionnor
temporaryvariables,and plugsa gap intheabstrat interpretation literature.
Key words: Abstrat interpretation; Compilers;Constraint programming.
1 Introdution
Groundness analysis is an important theme of logi programming and
ab-strat interpretation. Groundness analyses identify those program variables
whih at run time will be bound to terms that ontain no variables (ground
terms)[1,3℄.Foronstraintlanguages, likeCLP(R) [7℄, ananalogousproblem
isdeduing whihvariables are denite, that is,ompletelyxed by the store
[2,4,5℄. Groundness and deniteness are strongly related, and groundness is
oftenused for both onepts.
Little has been written about how to abstrat numeri onstraints, that is,
taking a numeri onstraint as input and omputing as output a Boolean
formula that aurately desribes the grounding behaviour of the onstraint.
For example, [4,5℄ just give some example groundness abstrations in a
ta-ble; no algorithmfor alulating an abstration for anarbitrary onstraint is
desribed. In addition, [2℄ also explains the r^ole of temporary variables and
projetion in abstration. The proedure is as follows: rst, a numeri
on-straint, for instane, w = x +y z, is written in three-variable form, for
example, w = x+v;v = yz, where v is a fresh, temporary variable.
Se-ond, table lookup is used to map three-variable forms to Boolean formulae,
1
1
(w^v))^(v (w^x)) and f
2
= (v (y^z)). The grounding behaviour
of the onstraint w =x+v;v =yz is desribed by f
1
^f
2
. Third,
tempo-raryvariablesare removed by projetion, forexample,v anbeeliminatedby
9v:(f 1
^f
2
)=(w (x^y^z))^(x (w^y^z)).
The third step an be omitted but this typially only defers projetion. F
ur-thermore, retainingthe temporaryvariables an degrade the time and spae
eÆieny of many the representations of Boolean funtions [1,3,5,6,8,9℄ that
have been proposed for groundness analysis. Projetion is partiularly
inon-venient for the analysis of [6℄ beause variable elimination is not required
elsewhere in the analysis. This paperaddresses these problems by presenting
asimple,syntatialgorithmforabstrating numerionstraintsthat neither
uses projetion nor introdues temporary variables. The algorithm gives
ab-strations whih are guaranteed to be atleast as preise as those given using
the three variable formmethod.
The paper is organised as follows: Setion 2 presents a semantis for
(non-Herbrand) CLP(R) onstraints. Abstration is alsoformalised. Setion 3
de-tailsthree-variable form. Setion4 explains how abstration an be reast as
the problem of reognising those variables ina numerionstraint whihwill
take a unique value when the others are grounded. Setion 5onludes.
2 Abstrat Interpretation
2.1 Conrete Domain
Let R denotethe real numbers and let
N
denote the set of funtor symbols
of CLP(R) [7℄, f+, , =, , abs , aros , arsin, os , max, min, pow , sing,
where is unary minus. (Binary subtration x y abbreviates x +( y),
thus is not modeled diretly). Let ? 2= R denote a speial symbol reserved
for error handling and let V denote a denumerable set of variables. Let X
?
denote X[f?g. Put = R
?
[
N
and
V
= [V. Let T and T
V
denote
the (ground) and (non-ground)terms generated from and
V
respetively.
A valuation isa total map, :V !R
?
, and the set of valuationsis denoted
. Let D be an interpretation of the symbols of . D(d) =d for all d 2R
? .
Eah symbol f 2
N
of arity n is interpreted by D as a map f : R
n ?
! R
? .
For example, + : R
?
R
?
! R
?
maps numbers d
1
;d
2
2 R to their sum,
otherwise, if either d
1
= ? or d
2
= ?, it maps to ?. The other symbols
in
N
are interpreted in the usual way, exept in the following three ases:
(d=0) = ? for every d 2 R
?
, arsin(d) = ?i :( 1 d 1);aros(d) =
1 2 1 2
and d
1
= d
2
; D interprets d
1
d
2
in the obvious way. Let C denote the
set of onstraints generated by T
V
and , whih is losed under onjuntion
(^), existentialquantiation(9)and renaming().Avaluation, ,naturally
extends to terms and onstraints using the interpretation D. Entailment of
onstraints, j=, is dened by
1
j=
2
i 8 2 : (
1
) ) (
2
). Equivalene,
,isdenedby
1
2
i
1
j=
2
and
2
j=
1
.hC=;j=;^iisa(bounded)meet
semi-lattiewiththebottomandtopelementsfalseandtrue,wherej=and^
are liftedtoequivalene lasses of onstraints. LetP(X)denote the powerset
ofX.Forthepurposesofabstratinterpretation,theonretedomainistaken
tobethe lattie hP(C= );;[;\i.
Inommonwithmanyonstraintsolvers,thesolverofCLP(R)[7℄ispartialin
thesensethatitanonlydetetthesatisability/unsatisabilityofonstraints
that beome linear. To modelthis, let L denote the set of linear onstraints,
that is (where d 2 R), true;false,
P n i=1
d i
x
i
= d,
P n i=1
d i
x
i
d 2 L.
L is losed under onjuntion, existential quantiation and renaming. The
transition system ( (LC)
2
is dened by: h;d^d
0
i ( h
00
;d
0
i i there
exists
0
2 L suh that j= ^
i2I (x
i
= d
i
), d^^
i2I (x
i
=d
i )
0
and
00
^
0
(whereIisapossiblyemptyindexset).Forexample,htrue;(y =xsin (z)^z =
=2)i ( h2z = ;y = xsin(z)i ( h(x = y^2z = );truei. Let (
?
abbreviate zero ormore (transitions. The transitionsystem (is onuent
in that if htrue;i (
? h
0
;d
0
i 6( and htrue;i (
? h
00
;d
00
i 6(, then
0
00
and d
0
d
00
. Let f : L ! L be a map suh that f() =f(
0
) i
0
. This
gives the (deterministi) map ( hf(
0
);d
0
i where htrue;i (
? h
0
;d
0
i 6(.
This isneessary toformulateabstration asa mapping.Let C
=
denote those
2 C of form (t = t
0
) and suh that 6( h
0
;d
0
i with
0
j= false. Let
var(o)denotethe (free) variablesinsyntatiobjeto andletmvar(o) denote
the variable ourrenes in syntati objet o (as a multiset), for example,
mvar(f(u;v;v)) = fu;v;vg. Let M
#
denote the set of singularly ourring
elements of the multiset M.Finally,let M(X)=fMjM
#
2P(X)g.
2.2 Abstrat Domain
Groundness and groundness dependenies are often represented by Boolean
funtions[1,3℄.LetXdenoteanitesetofvariables,andletBool
X
denotethe
setofBooleanformulaeoverX.Eahf 2Bool
X
representsanjXj-aryBoolean
funtion, so funtion and formulaare used interhangeably. The formula ^Y
is sometimes written Y. A Boolean funtion f is positive i X j= f. Let
Pos
X
denote the set of positive Boolean funtions over X, augmented with
thelogialonstantfalse. hPos
X
;j=;_;^iisaompletelattie{theabstrat
domain.9fy
1
;:::;y
n
g:f abbreviates 9y
1
:::9y
The onretisationmap, X
:Pos
X
!P(C=), details how formulae represent
onstraints.Conretisation is dened by
X
(f)=f[℄
j8
0
2C:(^
0
)6false)assign
X
(^
0
)j=fg
Here, assign
X
:C!Bool
X
is given by assign
X
() =Y^(^f:y j y 2X nYg)
and Y = fy 2 X j 9d 2 R: ( h
0
;d
0
i^
0
j= (y = d)g. The abstration
map,
X
:P(C=)!Pos
X
,isdened by
X
(C)=^ff 2Pos
X
jC
X
(f)g.
X
() abbreviates
X (f[℄
g).
Proposition 1
X
;
X
form a Galois insertion.
Proof. Bythe denition of
X
, thereisa Galoisonnetion.Toshowthat it
isaninsertion,itissuÆienttodemonstratethat
X
isinjetive.Supposethat
X
(f 1
) =
X (f
2
) and f
1
6= f
2
. Then there exists g = X
1
^(^f:xjx 2 X
2 g),
where X =X
1
[X
2
and X
1
\X
2
=;, suh that (without lossof generality)
g j= f
1
, but g 6j= f
2
. Let = ^fx = 1 j x 2 X
1
g. Piking
0
= true, it
is seen that assign
X
(^
0
) = g. Thus [℄
2
X (f
1
), but [℄
=
2
X (f
2
). A
ontradition. Therefore
X
is injetive. 2
3 Computing Abstrations With Projetion
C is not nite and thus
X (
X
) annot be interpreted as an algorithm for
omputing
X
(f) (
X
()) forarbitrary f 2Pos
X
(2C). This motivatesthe
translation of aonstraint intothree-variableform.
Denition 2 C
2
is the least binary relation suh that:
(1) ( ^ t = t
0
) ( ^x = t ^ x = t
0
) if t 62 V [R, t
0
62 V [R and
x62var(^t=t
0 );
(2) (^t = t
0
) (^t = t
00
^y = t
i
) if t 2 V [R, t
0
= f(t
1
;:::;t
i
;:::;t
n ),
t 00
=f(t
1
;:::;y;:::;t
n
), where t
i
62V [R, and y62var(^t=t
0 );
(3) (^t=t
0
) (^t
0
=t) if t62V [R and t
0
2V [R.
Proposition 3 nitely terminates.
Example 4 (x=((sin(y)=2)z)+7) (x=u+7^u=(sin(y)=2)z)
(x=u+7^u =vz^v =sin(y)=2) (x=u+7^u =vz^v =w=2^w=
Three-variable groundnessabstration forCLP(R) tbl X () tbl X ()
t=dt
0
V $V
0
t=t
0 t 00 V (V 0 ^V 00 )
t=t
0 +t 00 f 2 (V;V 0 ;V 00
) t= t
0
V $V
0
t=t
0
=d V $V
0
t=d=t
0
V $V
0
t=t
0 =t 00 f 1 (V;V 0 ;V 00
) t=pow(t
0
;t
00
) V (V
0
^V
00 )
t=os (t
0
) V V
0
t=sin(t
0
) V V
0
t=aros (t
0
) V $V
0
t=arsin(t
0
) V $V
0
t=min(t
0
;t
00
) V (V
0
^V
00
) t=max(t
0
;t
00
) V (V
0
^V
00 )
t=abs(t
0
) V V
0
tt
0 true
where d 2 R n f0g; t;t
0
;t
00
2 R [V; V = var(t);V
0
= var(t
0
);V
00
= var(t
00 );
f 1
(x;y;z)=(x (y^z))^(y (x^z))andf
2
(x;y;z) =f
1
(x;y;z)^(z (x^y))
Aonstraint2C issaidtobeinthree-variableformi6 .Let
?
abbrevi-atezeroormoreredutions.Thisleadstothefollowingabstrationtehnique:
Denition 5 The abstration map
tvf X
:C!Pos
X
is dened as follows.
Sup-pose (h
0
;d
0
i. If
0
j=false then
tvf X
()=false,else
tvf X ()= tvf X 0 ( 0 ^d 0 ) where tvf X 0 ()= 8 > > < > > : tvf X 0 ( 0 )^ tvf X 0 ( 00
) if =
0
^
00
9(var( 0
)nvar()):
tvf X 0 ( 0 ) if ? 0
6 and 6=
00 ^ 000 tbl X () otherwise.
The redution ( h
0
;d
0
i an be performed using CLP(R) mahinery [7℄.
Table 1 denes
tbl X
() for a (non-ompound) three-variable onstraint . To
see that
tvf X
is well-dened, let
?
1
6 and
?
2
6 . A renaming
: Y
1
! Y
2
exists with Y
1
= var(
1
) n var(), Y
2
= var(
2
) n var() and
( 1
)=
2
. Moreover, 9Y
2 : tvf X ( 2
)= 9(Y
1 ): tvf X (( 1
)) =9(Y
1 ):( tvf X ( 1 ))= 9Y 1 : tvf X ( 1
)sineisbijetiveand var((
tvf X
( 1
)))n(Y
1
)=var(
tvf X
( 1
))nY
1 .
Intuitively,
tvf X
is well-dened sine any extra variables introdued by are
eliminated. Observe
X
()j=
tvf X
() for all 2C.
Proposition 6 Table 1 is safe, that is,
X
() j=
tbl X
(), where is in
three-variable form,6=
0
^
00
and var()X.
Proof. Safety isdemonstratedonlyforthe asex=yz;otherasesmay be
treated similarly. Let fx;y;zg X. Assume, for the sake of a ontradition,
that thereexists
0
2C suhthat assign
X
(x=yz^
0
)6j=x (y^z).Thus
assign X
(x=yz^
0
)=(:x)^y^z andhtrue;x=yz^
0 i( ? h 00 ;d 00 iwhere 00
j=(y =d
1
)^(z =d
2
).Hene h
00 ;d 00 i( ? h 000 ;d 000 i where 000
j=(x=d
1
d
2 )
whihontraditsassign
X
(x=yz^
0
Observe that X
6=
X
sine
X
(x y(1=x) = 0) = true and
X (x
y(1=x)) = (y x). Moreover, it should be noted that the table does not
aurately desribe the grounding behaviour of some unusual (and spei)
onstraints inthree-variable form. For example,
X
(x=min(y;y))=x$y,
whereas
tbl X
(x = min(y;y)) = x y. In pratie it is expeted that suh
onstraints will not our, however, the table ould be extended to inlude
these extra ases.
4 Computing Abstrations Without Projetion
Abstration may be reast as the problem of reognising those variables in
a onstraint whih will take a unique value when the other variables in the
expression are grounded. This is ahieved preisely by the map det
. The
approah is formulated in terms of approximationsto det
, and is at least as
aurate as the three-variable formmethod.
Denition 7 The map det
: C
=
! P(V) is given by x 2 det
() i 8 :
var()nfxg!R:9!d 2R
?
:()(x=d).
An abstration map ould be dened in terms of det
. However, omputing
thismayrequirenon-trivialsymbolimanipulationof.Forexample,det
(x=
y(1=z))=fx;yg requires the reognitionthat y(1=z)y=z. To build an
abstration map in terms of a simple pattern reogniser (together with ()
det
is approximated by a lass of maps Det.
Denition 8 Det istheleastsetof mapsdet:C
=
!P(V)thatsatisfy(where
f 2
N ):
safety: det () det
();
preision1: if =(x=t) and 6 , then mvar()
#
\det
()det ();
preision2: if y2det (x=t) and y2= var(t
0
), then y2det (t=t
0 );
preision3: let = (t = f(t
1
;:::;t
i
;:::;t
n
)) and
0
= (t = f(t
1
;:::;y;:::;t
n )),
with y 2= var(), if y 2 det (
0
), then det(y = t
i
) nvar(
0
) det () and
det( 0
)nvar(y =t
i
)det() .
The onditions in denition 8 relate to the abstration map in the following
way. The safety ondition ensures
X
() j=
det X
() if det 2 Det. Preision 1
guarantees
det X
() j=
tvf X
() for a non-ompound three-variable onstraint
and the ompositionality properties of preision 2 and preision 3 (with
preision1) ensures that
det X
()j=
tvf X
() for arbitrary .
Proposition 9 det
Proof. Onlypreision3isnon-trivial.Suppose thatx2det (y =t i
)nvar(
0 ).
It is demonstrated that x 2 det
(). Sine y 2 det
() and x 62 var(
0
), and
given : var()nfxg !R, it an be seen that there is a unique d 2 R suh
that (
0
)(y =d). Put
0
=[fy7!dg. Hene:
(t=f(t
1
;:::;t
i
;:::;t
n ))
0
(t=f(t
1
;:::;t
i
;:::;t
n ))
0
(t =f(t
1
;:::;y;:::;t
n
)^y =t
i )
0
(y =t
i )
x=d
0
sine x2det
(y =t
i )
Hene x2det
(). Similarly,the seond ondition holds. 2
An abstrationmap, parameterised by det , an now bedened.
Denition 10 Theabstrationmap
det X
:C!Pos
X
isdenedasfollows.
Sup-pose(h
0
;d
0
i.If
0
j=false then
det X
()=false,else
det X ()= det X 0 ( 0 ^d 0 ) where det X 0 () = 8 > > < > > :
true if =(tt
0 ) det X 0 ( 0 )^ det X 0 ( 00
) if =(
0 ^ 00 ) ^ v2det()
(v (var()nfvg)) otherwise
Theorem 11 If 2C and det2Det, then
X
()j=
det X
()j=
tvf X
().
Proof. The rst entailmentisestablished by demonstratingtheany Boolean
formulathat isentailed by
det X
isalsoentailed by
X
. Theseond entailment
isestablished by demonstratingthat eah redution resultsin aonstraint
whose
det X
abstration is not stronger than that of the previous onstraint.
The base ase demonstrates that
det X
()j=
tvf X
() for inthe lookup table.
Consider 2 C suh that ( h
0
;d
0
i. If
0
j=false, the result is immediate.
Letdet2Det.
To show
X
()j=
det X (), onsider 0 ^d 0 = 1 ^:::^ n .If X ( i
)j=
det X 0 ( i ), then X
() j=
det X
(). Suppose, for a ontradition, that
X (
i
) 6j=
det X 0 ( i ), for some i
. Then there exists f
x
= x Y, where x 2 det (
i
) and Y =
var( i
)nfxg, suh that 9
00 2 C:assign X ( i ^ 00
) 6j= f
x
and
i
^
00
6 false.
Hene assign X ( i ^ 00
) j= (:x) ^Y. Thus for every y 2 Y there is e 2 R
suh that i ^ 00 ( ? h 000 ;d 000 i and 000
j= (y = e); indeed, it may be assumed
that 8y 2 Y:9e 2 R:
0 00
j= (y = e). By safety, sine x 2 det (
i
), for some
e 0 2 R, i ^ 000
j= (x = e
0 ). Therefore, i ^ 00 ( ? h 000 ;d 000 i ( ? h 0000 ;d 0000 i and 0000
j=(x=e
0
). A ontradition. Thus
X (
i
)j=f
x
and the result follows.
To show
det X
() j=
tvf X (). Let 0 ^d 0 = 1 ^:::^ n
. It is enough to show
that
det X
( i
)j=
tvf X
( i
ase, onsider =(t=f(t 1
;:::;t
n
)),wheref 2
N
and 6 (the ase is
obvious). Suppose
tvf X
( 00
) j=(x Y), where x 2= Y. By inspetion of Table
1, x 2 (mvar(
00 ))
#
. By Proposition 6,
X (
00
) j=
tvf X ( 00 ). Hene X ( 00
) j=
(x Y). Let :var(
00
)nfxg!R and put
000
=^fy=(y)jy2Yg. Hene
either 00 ^ 000 ( ? h 0000 ;d 0000 i and 0000
j= false, in whih ase x 2 det
( 00 ), sine ( 00
) (x =?), or
00 ^ 000 ( h 00000 ;d 00000 i and 00000
j=(x =e), in whih
ase (x = e) (
00
^
000
) (
00
) and x 2 det
(
00
). Sine x 2 det
(
00
), by
preision 1, x 2 det(
00
). By inspetion of Table 1, Y = var(
00
)nfxg. Hene
det X
( 00
)j=
tvf X
( 00
).
For the indutive ase, suppose
i
00 ?
000
6 . By hypothesis,
det X
( 00
)j=
9(var( 000
)nvar(
00 )): tvf X ( 000
), so it is enough to show
det X
( i
) j= 9(var(
00 )n var( i )): det X ( 00 ).
(1) Toshow
det X
(t =t
0
)j=9x:
det X
(x=t ^ x =t
0
).Sine9x:
det X
(x=t ^ x =t
0 )
= 9x:(
det X
(x=t)^
det X
(x=t
0 ))
= 9x:((^
v2det(x=t)
v (var(x =t)nfvg))
^(^
u2det(x=t 0
)
u (var(x=t
0
)nfug)))=f
Suppose f j= (y Y). To show ^
v2det(t=t 0
)
(v var(t = t
0
)nfvg) j=
(y Y), suppose, without loss of generality, y 2 det (x = t
0
). Sine
x2 det (x=t
0
), y 2= var(x =t
0
). Bypreision 2,y 2det (t =t
0
) and thus
the result follows.
(2) Toshow
det X
(t=t
0
)=9y:
det X
(t=t
00
^y=t
i
),wheret
0
=f(t
1
;:::;t
i
;:::;t
n ), t 00 = f(t 1
;:::;y;:::;t
n
) and y 2= var(t
0
). Sine 9y:
det X
(t = t
00
^ y = t
i )
= 9y:(
det X
(t=t
00
)^
det X
(y=t
i )) = 9y:((^ v2det(t=t 00 )
v (var(t=t
00
)nfvg)))
^(^
u2det(y=t i
)
u (var(y =t
i
)nfug)))
Usingpreision3,the resultan beestablishedanalogouslytothe
previ-ous ase.
(3) To show
det X
(t = t
0 ) = det X (t 0
= t), where t
0
2 R [V and t 2= R [V.
Immediate. 2
Next, a spei map in det
1
2Det is desribed. The map syntatially
iden-ties those variables that our one in a numeri onstraint expression and
whih take a unique value when the variablesare grounded.
Denition 12 The map det
1
: C
=
! P(V) is dened by: det
1
(t = t
0
) =
(det 1
(t)[det
1 (t
0 ))
#
, where det
1
:T
V
ExamplegroundnessabstrationsforX=fw;x;y;zg i 0 i tvf X ( i ) det 1 X ( i ) det X ( i ) X ( i )
w=x(y+z)
1 f 1 f 1 f 1 f 1
w=x+w x=0 f
2 f 2 f 2 f 2
w=x=0 false f
3 f 3 f 3 f 3
w=x+x=w
4 f 4 f 4 f 5 f 5
(wx^abs(x)w)
5 f 4 f 4 f 4 f 6 det 1 (t)=
ftg if t 2V
det 1
(t 0
) if t = t; t =dt
0
; t=t
0
=d ort =d=t
0
det 1
(t 0
)[det
1 (t
00
) if t =t
0 +t 00 det 1 (t 0
)[det
1 (t
0
)[det
1 (t
00
)[det
1 (t
00
) if t =t
0
t
00
; t=pow(t
0
;t
00 )
t=min(t
0
;t
00
) ort =max(t
0 ;t 00 ) det 1 (t 0
)[det
1 (t
00
)[det
1 (t
00
) if t =t
0 =t 00 det 1 (t 0
)[det
1 (t
0
) if t =abs(t
0
); t=os (t
0
)or t=sin(t
0 ) det 1 (t 0
) if t =aros(t
0
)or t=arsin(t
0 )
Proposition 13 det
1
2Det.
Example 14 Consider =(x+v =xy+z=w)and suppose (htrue;i.
Observethat det
1
()=(det
1
(x+v)[det
1
(xy+z=w))
#
=(det
1
(x)[det
1 (v)[ det 1 (x)[det 1 (x)[det 1
(y)[det
1 (y)[det 1 (z)[det 1 (w)[det 1 (w)) #
=fv;w;w;x;
x;x;y;y;zg
#
=fv;zg.Hene
det1
X
()=(v w^x^y^z)^(z v^w^x^y).
Example 15 Table 2 details the abstrations for various onstraints where
f 1
= w (x^y ^z), f
2
= x, f
3
= false, f
4
= true f
5
= x w, f
6 =
w$ x. The abstration algorithmsare dened interms of (. It is assumed
that i ( h 0 ;d 0 i and 0 i = 0 ^d 0
. In pratie, ( is evaluated by posting the
onstraint to the store and then retrieving it. The net eet is to evaluate
groundterms and togroup together like terms.For example,2x=os()
z+x+max(2;3)(hx= z+3;truei.This(isused inthis example.Note
that
X (
i
) j=
det
X (
i
) j=
det 1 X
( i
) j=
tvf X
( i
). The abstrations for
1
;
2
;
3
all agree, illustrating that all methods give good auray. The abstrations
of
4
show that det
1
an stillbe strengthened.
X (
5
) shows that, in general,
systemsofinequations needtobeonsideredtoomputethebest abstration.
Note that if a weaker ( were used, a stronger det ould be dened to give
abstrationsof a similarstrength. The three variableform methodwould not
This paper has desribed a simple algorithm for abstrating numeri
on-straints. This method does not introdue temporary variables, utilises
avail-ableCLP(R)mahinery,andisatleastaspreiseasthethree-variablemethod.
Whilstotherworkshavegiven preisedenitionsofabstration,theyhavenot
addressedhowtoeÆientlyomputethemap.Thispaperplugsthishole.The
algorithman be easilyimplemented and has been used with the analyser in
[8℄. Future work will look at the more general ase of mixing Herbrand and
linear onstraints.
Referenes
[1℄T. Armstrong, K. Marriott, P. Shahte, and H. Sndergaard. Two Classes of
BooleanFuntionsforDependenyAnalysis.SieneofComputerProgramming,
31(1):3{45, 1998.
[2℄N. Baker and H. Sndergaard. Deniteness Analysis for CLP(R). Australian
Computer SieneCommuniations, 15(1):321{332, 1993.
[3℄M. Codishand B. Demoen. Analyzing Logi Programs Using \Prop"-ositional
Logi Programsand aMagi Wand. Journal of Logi Programming, 25(3):249{
274,1995.
[4℄M. Gara de la Banda. Independene, Global Analysis, and Parallelism in
DynamiallySheduled Constraint LogiProgramming. PhDthesis,Universidad
Politeniade Madrid,1994.
[5℄M. Gara de la Banda, M. Hermenegildo, M. Bruynooghe, V. Dumortier,
G. Janssens, and W. Simoens. GlobalAnalysis of Constraint Logi Programs.
ACM Transations on Programming Languages and Systems, 18(5):564{614,
1996.
[6℄A. Heaton, M. Abo-Zaed, M. Codish, and A. King. A Simple Polynomial
GroundnessAnalysis forLogi Programs. Journal of Logi Programming, 2000.
Forthoming.
[7℄N. Heintze, J. Jaar, S. Mihaylov, P. Stukey, and R. Yap. The CLP(R)
Programmer's Manual Version1.1, 1991.
[8℄J. M. Howe and A. King. Implementing Groundness Analysis with Denite
BooleanFuntions.InG.Smolka,editor,EuropeanSymposiumonProgramming,
volume 1782 of Leture Notes in Computer Siene, pages 200{214.
Springer-Verlag, 2000.
[9℄A.King,J. Smaus,and P.Hill. QuotientingShare forDependenyAnalysis. In
S.D. Swierstra, editor, European Symposium on Programming, volume 1576 of
