On the Decidability of Connectedness Constraints in 2D and 3D Euclidean Spaces
ABSTRACT We investigate (quantifierfree) spatial constraint languages with equality,
contact and connectedness predicates as well as Boolean operations on regions,
interpreted over lowdimensional Euclidean spaces. We show that the complexity
of reasoning varies dramatically depending on the dimension of the space and on
the type of regions considered. For example, the logic with the
interiorconnectedness predicate (and without contact) is undecidable over
polygons or regular closed sets in the Euclidean plane, NPcomplete over
regular closed sets in threedimensional Euclidean space, and ExpTimecomplete
over polyhedra in threedimensional Euclidean space.
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Conference Proceeding: Recognizing string graphs in NP.
Proceedings on 34th Annual ACM Symposium on Theory of Computing, May 1921, 2002, Montréal, Québec, Canada; 01/2002
Page 1
arXiv:1104.0219v1 [cs.LO] 1 Apr 2011
On the Decidability of Connectedness Constraints in 2D and 3D Euclidean Spaces
Roman Kontchakov1, Yavor Nenov2, Ian PrattHartmann2and Michael Zakharyaschev1
1Department of Computer Science
and Information Systems
Birkbeck College London, U.K.
2School of Computer Science
University of Manchester, U.K.
Abstract
We investigate (quantifierfree) spatial constraint
languageswith equality,contact and connectedness
predicates, as well as Boolean operations on re
gions, interpreted over lowdimensional Euclidean
spaces. We show that the complexity of reasoning
varies dramatically depending on the dimension of
thespaceandonthetypeofregionsconsidered. For
example, the logic with the interiorconnectedness
predicate (and without contact) is undecidable over
polygons or regular closed sets in R2, EXPTIME
complete over polyhedra in R3, and NPcomplete
over regular closed sets in R3.
1
A central task in Qualitative Spatial Reasoning is that of de
terminingwhethersomedescribedspatial configurationis ge
ometrically realizable in 2D or 3D Euclidean space. Typi
cally, such a description is given using a spatial logic—a for
mal languagewhose variables range over (typed)geometrical
entities, and whose nonlogical primitives represent geomet
rical relations and operations involving those entities. Where
thegeometricalprimitivesofthe languagearepurelytopolog
ical in character, we speak of a topological logic; and where
thelogicalsyntaxis confinedtothatofpropositionalcalculus,
we speak of a topological constraint language.
Topological constraint languages have been intensively
studied in Artificial Intelligence over the last two decades.
The bestknown of these, RCC8 and RCC5, employ
variables ranging over regular closed sets in topological
spaces, and a collection of eight (respectively, five) bi
nary predicates standing for some basic topological re
lations between these sets [Egenhofer and Franzosa, 1991;
Randell et al., 1992; Bennett, 1994; Renz and Nebel, 2001].
An important extension of RCC8, known as BRCC8, ad
ditionally features standard Boolean operations on regular
closed sets[Wolter and Zakharyaschev, 2000].
A remarkable characteristic of these languages is their
insensitivity to the underlying interpretation. To show that
an RCC8formula is satisfiable in ndimensional Euclidean
space, it suffices to demonstrate its satisfiability in any topo
logical space[Renz, 1998]; for BRCC8formulas, satisfiabil
ity in any connected space is enough. This inexpressiveness
Introduction
yields (relatively) low computational complexity: satisfiabil
ity of BRCC8, RCC8 and RCC5formulas over arbitrary
topological spaces is NPcomplete; satisfiability of BRCC8
formulas over connected spaces is PSPACEcomplete.
However, satisfiability of spatial constraints by arbi
trary regular closed sets by no means guarantees realiz
ability by practically meaningful geometrical objects, where
connectedness of regions is typically a minimal require
ment [Borgo et al., 1996; Cohn and Renz, 2008].
nected region is one which consists of a ‘single piece.’) It is
easy to write constraints in RCC8 that are satisfiable by con
nected regular closed sets over arbitrary topological spaces
but not over R2; in BRCC8 we can even write formulas satis
fiable by connected regular closed sets over arbitrary spaces
but not over Rnfor any n. Worse still: there exist very
simple collections of spatial constraints (involving connect
edness) that are satisfiable in the Euclidean plane, but only
by ‘pathological’ sets that cannot plausibly represent the re
gions occupied by physical objects [PrattHartmann, 2007].
Unfortunately, little is known about the complexity of topo
logical constraint satisfaction by nonpathological objects
in lowdimensional Euclidean spaces.
sult[Schaefer et al., r003]in this area showsthat satisfiability
of RCC8formulas by dischomeomorphs in R2is still NP
complete, though the decision procedure is vastly more intri
cate than in the general case. In this paper, we investigate the
computational properties of more general and flexible spatial
logicswith connectednessconstraintsinterpretedoverR2and
R3.
We consider two ‘base’ topological constraint languages.
The languageB features = as its onlypredicate, but has func
tion symbols +, −, · denoting the standard operations of fu
sion, complement and taking common parts defined for regu
lar closed sets, as well as the constants 1 and 0 for the entire
space and the empty set. Our second base language, C, ad
ditionally features a binary predicate, C, denoting the ‘con
tact’ relation (two sets are in contact if they share at least one
point). The languageC is a notational variant of BRCC8 (and
thus an extension of RCC8), while B is the analogous exten
sion of RCC5. We add to B and C one oftwo new unarypred
icates: c, representing the property of connectedness, and c◦,
representingthe (stronger)propertyof having a connectedin
terior. We denote the resulting languages by Bc, Bc◦, Cc and
Cc◦. We are interested in interpretations over (i) the regular
(A con
One landmark re
Page 2
closed sets of R2and R3, and (ii) the regular closed poly
hedral sets of R2and R3. (A set is polyhedral if it can be
defined by finitely many bounding hyperplanes.) By restrict
ing interpretations to polyhedra we rule out satisfaction by
pathologicalsets and use the same ‘data structure’ as in GISs.
When interpreted over arbitrary topological spaces, the
complexity of reasoning with these languages is known: sat
isfiability of Bc◦formulas is NPcomplete, while for the
other three languages, it is EXPTIMEcomplete. Likewise,
the 1D Euclidean case is completely solved. For the spaces
Rn(n ≥ 2), however, most problems are still open. All
four languages contain formulas satisfiable by regular closed
sets in R2, but not by regular closed polygons; in R3, the
analogous result is known only for Bc◦and Cc◦. The sat
isfiability problem for Bc, Cc and Cc◦is EXPTIMEhard (in
both polyhedral and unrestricted cases) for Rn(n ≥ 2); how
ever, the only knownupper boundis that satisfiability of Bc◦
formulas by polyhedra in Rn(n ≥ 3) is EXPTIMEcomplete.
(See[Kontchakov et al., 2010b]for a summary.)
This paper settles most of these open problems, reveal
ingconsiderabledifferencesbetweenthe computationalprop
erties of constraint languages with connectedness predi
cates when interpreted over R2and over abstract topologi
cal spaces. Sec. 3 shows that Bc, Bc◦, Cc and Cc◦are all
sensitive to restriction to polyhedra in Rn(n ≥ 2). Sec. 4
establishes an unexpected result: all these languages are un
decidablein 2D, bothin the polyhedralandunrestrictedcases
([Dornheim, 1998] proves undecidability of the firstorder
versionsofthese languages). Sec. 5 resolvestheopenissue of
the complexity of Bc◦over regular closed sets (not just poly
hedra) in R3by establishing an NP upper bound. Thus, Qual
itative Spatial Reasoning in Euclidean spaces proves much
more challenging if connectedness of regions is to be taken
into account. We discuss the obtained results in the context
of spatial reasoning in Sec. 6. Omitted proofs can be foundin
the appendix.
2Constraint Languages with Connectedness
Let T be a topological space. We denote the closure of any
X ⊆ T by X−, its interior by X◦and its boundaryby δX =
X−\X◦. We call X regularclosedif X = X◦−, anddenote
by RC(T) the set of regular closed subsets of T. Where T is
clear from context, we refer to elements of RC(T) as regions.
RC(T) forms a Boolean algebra under the operations X +
Y = X ∪ Y , X · Y = (X ∩ Y )◦−and −X = (T \ X)−.
We write X ≤ Y for X ·(−Y ) = ∅; thus X ≤ Y iff X ⊆ Y .
A subset X ⊆ T is connected if it cannotbe decomposedinto
two disjoint, nonemptysets closed in the subspace topology;
X is interiorconnected if X◦is connected.
Any (n−1)dimensionalhyperplanein Rn, n ≥ 1, bounds
two elements of RC(Rn) called halfspaces. We denote by
RCP(Rn) the Boolean subalgebra of RC(Rn) generated by
the halfspaces, and call the elements of RCP(Rn) (regular
closed) polyhedra. If n = 2, we speak of (regular closed)
polygons. Polyhedramayberegardedas‘wellbehaved’or,in
topologists’ parlance, ‘tame.’ In particular, every polyhedron
has finitely many connected components, a property which is
not true of regular closed sets in general.
The topological constraint languages considered here all
employ a countably infinite collection of variables r1,r2,...
The language C features binary predicates = and C, together
with the individual constants 0, 1 and the function symbols
+, ·, −. The terms τ and formulas ϕ of C are given by:
τ
ϕ
::=
::=
r  τ1+ τ2  τ1· τ2  −τ1  1  0,
τ1= τ2  C(τ1,τ2)  ϕ1∧ ϕ2  ¬ϕ1.
The language B is defined analogously, but without the pred
icate C. If S ⊆ RC(T) for some topological space T, an
interpretation over S is a function ·Imapping variables r to
elements rI∈ S. We extend ·Ito terms τ by setting 0I= ∅,
1I= T, (τ1+ τ2)I= τI
iff τI
C(τ1,τ2) as ‘τ1contacts τ2.’ The relation = is extended to
nonatomic formulas in the obvious way. A formula ϕ is sat
isfiable over S if I = ϕ for some interpretation I over S.
Turning to languages with connectedness predicates, we
define Bc and Cc to be extensions of B and C with the unary
predicate c. We set I = c(τ) iff τIis connected in the topo
logical space under consideration. Similarly, we define Bc◦
and Cc◦to be extensions of B and C with the predicate c◦,
setting I = c◦(τ) iff (τI)◦is connected. Sat(L,S) is the set
of Lformulas satisfiable over S, where L is one of Bc, Cc,
Bc◦or Cc◦(the topological space is implicit in this notation,
but will always be clear fromcontext). We shall be concerned
withSat(L,S), whereS is RC(Rn) orRCP(Rn) forn = 2,3.
To illustrate, consider the Bc◦formulas ϕkgiven by
1+ τI
2, etc. We write I = τ1= τ2
1∩ τI
1= τI
2, and I = C(τ1,τ2) iff τI
2?= ∅. We read
?
1≤i≤k
?c◦(ri)∧(ri?= 0)?∧
?
i<j
?c◦(ri+rj)∧(ri·rj= 0)?. (1)
One can show that ϕ3is satisfiable over RCP(Rn), n ≥ 2,
but not over RCP(R), as no three intervals with nonempty,
disjoint interiors can be in pairwise contact. Also, ϕ5is sat
isfiable over RCP(Rn), for n ≥ 3, but not over RCP(R2),
as the graph K5is nonplanar. Thus, Bc◦is sensitive to the
dimension of the space. Or again, consider the Bc◦formula
?
1≤i≤3
c◦(ri) ∧ c◦(r1+r2+r3) ∧
?
2≤i≤3
¬c◦(r1+ri). (2)
One can show that (2) is satisfiable over RC(Rn), for any
n ≥ 2 (see, e.g., Fig. 1), but not over RCP(Rn). Thus
Bc◦is sensitive to tameness in Euclidean spaces.It is
r1
r2
r3
Figure 1: Three regions in RC(R2) satisfying (2).
known [Kontchakov et al., 2010b] that, for the Euclidean
plane, the same is true of Bc and Cc: there is a Bcformula
satisfiable over RC(R2), but not over RCP(R2). (The exam
ple required to show this is far more complicated than the
Bc◦formula (2).) In the next section, we prove that any of
Page 3
Bc, Cc and Cc◦contains formulas satisfiable over RC(Rn),
for every n ≥ 2, but only by regions with infinitely many
components. Thus, all four of our languages are sensitive to
tameness in all dimensions greater than one.
3
Fix n ≥ 2 and let d0,d1,d2,d3be regions partitioning Rn:
Regions with Infinitely Many Components
??
0≤i≤3di= 1?
∧
?
0≤i<j≤3(di· dj= 0).
(3)
We construct formulas forcing the dito have infinitely many
connected components. To this end we require nonempty
regions aicontained in di, and a nonempty region t:
?
0≤i≤3
?(ai?= 0) ∧ (ai≤ di)?
∧
(t ?= 0).
(4)
The configuration of regions we have in mind is depicted in
Fig. 2, where components of the diare arranged like the lay
ers of an onion. The ‘innermost’ component of d0 is sur
rounded by a component of d1, which in turn is surrounded
by a componentof d2, and so on. The regiont passes through
every layer, but avoids the ai. To enforce a configuration of
thissort, weneedthefollowingthreeformulas,for0 ≤ i ≤ 3:
c(ai+ d⌊i+1⌋+ t),
¬C(ai,d⌊i+1⌋· (−a⌊i+1⌋)) ∧ ¬C(ai,t),
¬C(di,d⌊i+2⌋),
(5)
(6)
(7)
where ⌊k⌋ = k mod4. Formulas (5) and (6) ensure that each
component of aiis in contact with a⌊i+1⌋, while (7) ensures
that no component of dican touch any component of d⌊i+2⌋.
a1
d1
a0
d0
a3
d3
a2
d2
a1
d1
a0
d0
t
...
Figure 2: Regions satisfying ϕ∞.
Denote by ϕ∞ the conjunction of the above constraints.
Fig.2shows howϕ∞canbesatisfied overRC(R2). By cylin
drification, it is also satisfiable over any RC(Rn), for n > 2.
The arguments of this section are based on the following
property of regular closed subsets of Euclidean spaces:
Lemma 1 If X ∈ RC(Rn) is connected, then every compo
nent of −X has a connected boundary.
The proof of this lemma, which follows from a result
in [Newman, 1964], can be found in Appendix A. The result
fails for other familiar spaces such as the torus.
Theorem 2 There is a Ccformula satisfiable over RC(Rn),
n ≥ 2, but not by regions with finitely many components.
Proof. Let ϕ∞be as above. To simplify the presentation, we
ignore the difference between variables and the regions they
standfor, writing,forexample,aiinsteadofaI
asequenceofdisjointcomponentsXiofd⌊i⌋andopensets Vi
connecting Xito Xi+1(Fig. 3). By the first conjunct of (4),
i. We construct
let X0be a componentof d0containingpoints in a0. Suppose
Xihas been constructed. By (5) and(6), Xiis in contactwith
a⌊i+1⌋. Using (7) and the fact that Rnis locally connected,
one can find a component Xi+1of d⌊i+1⌋which has points
in ai+1, and a connected open set Visuch that Vi∩ Xiand
Vi∩ Xi+1are nonempty, but Vi∩ d⌊i+2⌋is empty.
...
X3
X2
X1
X0
V2
V1
V0
Figure 3: The sequence {Xi,Vi}i≥0generatedby ϕ∞. (Si+1
and Ri+1are the ‘holes’ of Xi+1containing Xiand Xi+2.)
To see that the Xiare distinct, let Si+1and Ri+1be the
componentsof −Xi+1containingXiand Xi+2, respectively.
It suffices to show Si+1⊆ S◦
Vimust intersect δSi+1. Evidently,δSi+1⊆ Xi+1⊆ d⌊i+1⌋.
Also, δSi+1 ⊆ −Xi+1; hence, by (3) and (7), δSi+1 ⊆
di∪d⌊i+2⌋. By Lemma 1, δSi+1is connected, and therefore,
by (7), is entirely contained either in d⌊i⌋or in d⌊i+2⌋. Since
Vi∩δSi+1?= ∅andVi∩d⌊i+2⌋= ∅, wehaveδSi+1?⊆ d⌊i+2⌋,
so δSi+1 ⊆ di. Similarly, δRi+1 ⊆ di+2. By (7), then,
δSi+1∩δRi+1= ∅, andsinceSi+1andRi+1are components
of the same set, they are disjoint. Hence, Si+1⊆ (−Ri+1)◦,
and since Xi+2 ⊆ Ri+1, also Si+1 ⊆ (−Xi+2)◦.
Si+1lies in the interior of a component of −Xi+2, and since
δSi+1⊆ Xi+1⊆ Si+2, that component must be Si+2.
Now we show how the Ccformula ϕ∞can be transformed
to Cc◦ and Bcformulas with similar properties. Note first
that all occurrences of c in ϕ∞have positive polarity. Let
ϕ◦
In Fig. 2, the connected regions mentioned in (5) are in
fact interiorconnected;henceϕ◦
Since interiorconnectedness implies connectedness, ϕ◦
tails ϕ∞, and we obtain:
Corollary 3 ThereisaCc◦formulasatisfiableoverRC(Rn),
n ≥ 2, but not by regions with finitely many components.
To construct a Bcformula,we observe that all occurrences
of C in ϕ∞are negative. We eliminate these using the pred
icate c. Consider, for example, the formula ¬C(ai,t) in (6).
By inspection of Fig. 2, one can find regions r1, r2satisfying
i+2. Note that the connected set
So,
❑
∞be the result of replacing them with the predicate c◦.
∞is satisfiable overRC(Rn).
∞en
c(r1) ∧ c(r2) ∧ (ai≤ r1) ∧ (t ≤ r2) ∧ ¬c(r1+ r2).
On the other hand, (8) entails ¬C(ai,t). By treating all other
noncontact relations similarly, we obtain a Bcformula ψ∞
that is satisfiable over RC(Rn), and that entails ϕ∞. Thus:
Corollary 4 There is a Bcformula satisfiable over RC(Rn),
n ≥ 2, but not by regions with finitely many components.
Obtaining a Bc◦analogue is complicated by the fact that
we must enforce noncontactconstraints using c◦(rather than
c). In the Euclidean plane, this can be done using planarity
constraints; see Appendix A.
Theorem 5 There is a Bc◦formula satisfiable over RC(R2),
but not by regions with finitely many components.
(8)
Page 4
Theorem 2 and Corollary 4 entail that, if L is Bc or Cc,
then Sat(L,RC(Rn)) ?= Sat(L,RCP(Rn)) for n ≥ 2. The
orem 5 fails for RC(Rn) with n ≥ 3 (Sec. 5). However, we
knowfrom(2)thatSat(Bc◦,RC(Rn)) ?= Sat(Bc◦,RCP(Rn))
for all n ≥ 2. Theorem 2 fails in the 1D case; moreover,
Sat(L,RC(R)) = Sat(L,RCP(R)) only in the case L = Bc
or Bc◦[Kontchakov et al., 2010b].
4
Let L be any of Bc, Cc, Bc◦or Cc◦. In this section, we show,
via a reduction of the Post correspondence problem (PCP),
that Sat(L,RC(R2)) is r.e.hard,andSat(L,RCP(R2)) is r.e.
complete.An instance of the PCP is a quadruple w =
(S,T,w1,w2) where S and T are finite alphabets, and each
wiis a word morphism from T∗to S∗. We may assume that
S = {0,1} and wi(t) is nonempty for any t ∈ T. The in
stance w is positive if there exists a nonempty τ ∈ T∗such
that w1(τ) = w2(τ). The set of positive PCPinstances is
known to be r.e.complete. The reduction can only be given
in outline here: full details are given in Appendix B.
To deal with arbitrary regular closed subsets of RC(R2),
we use the technique of ‘wrapping’ a region inside two big
ger ones. Let us say that a 3region is a triple a = (a, ˙ a,¨ a) of
elements of RC(R2) such that 0 ?= ¨ a ≪ ˙ a ≪ a, where r ≪ s
abbreviates ¬C(r,−s). It helps to think of a = (a, ˙ a,¨ a)
as consisting of a kernel, ¨ a, encased in two protective lay
ers of shell. As a simple example, consider the sequence
of 3regions a1,a2,a3depicted in Fig. 4, where the inner
most regions form a sequence of externally touching poly
gons. When describing arrangements of 3regions, we use
Undecidability in the Plane
a1
a2
a3
˙ a1
˙ a2
˙ a3
¨ a1
¨ a2
¨ a3
Figure 4: A chain of 3regions satisfying stack(a1,a2,a3).
the variable r for the triple of variables (r, ˙ r, ¨ r), taking the
conjuncts ¨ r ?= 0, ¨ r ≪ ˙ r and ˙ r ≪ r to be implicit. As with
ordinary variables, we often ignore the difference between 3
region variables and the 3regions they stand for.
For k ≥ 3, define the formula stack(a1,...,ak) by
?
1≤i≤k
c(˙ ai+ ¨ ai+1+ ··· + ¨ ak)
∧
?
j−i>1
¬C(ai,aj).
Thus,
stack(a1,a2,a3).
our proof. If stack(a1,...,ak) holds, then any point p0in
the inner shell ˙ a1 of a1 can be connected to any point pk
in the kernel ¨ akof akvia a Jordan arc γ1···γkwhose ith
segment, γi, never leaves the outer shell aiof ai. Moreover,
each γiintersects the inner shell ˙ ai+1of ai+1, for 1 ≤ i < k.
This technique allows us to write Ccformulas whose sat
isfying regions are guaranteed to contain various networks of
arcs, exhibiting almost any desired pattern of intersections.
thetriple of3regionsinFig.4satisfies
This formula plays a crucial role in
Now recall the construction of Sec. 3, where constraints on
the variablesd0,...,d3were usedto enforce‘cyclic’patterns
of components. Using stack(a1,...,ak), we can write a for
mula with the property that the regions in any satisfying as
signment are forced to contain the pattern of arcs having the
form shown in Fig. 5. These arcs define a ‘window,’ contain
ζ1
η1
ζ2
η2
ζ3
η3
ζn
ηn
Figure 5: Encoding the PCP: Stage 1.
ing a sequence {ζi} of ‘horizontal’ arcs (1 ≤ i ≤ n), each
connected by a corresponding‘vertical arc,’ ηi, to some point
on the ‘top edge.’ We can ensure that each ζiis included in a
region a⌊i⌋, and each ηi(1 ≤ i ≤ n) in a region b⌊i⌋, where
⌊i⌋ now indicates i mod 3. By repeating the construction, a
second pair of arcsequences,{ζ′
be established, but with each η′
edge.’ Again, we can ensure each ζ′
a′
can ensure that the final horizontal arcs ζnand ζ′
others) are joined by an arc ζ∗lying in a region z∗. The cru
i} and {η′
iconnecting ζ′
iis included in a region
⌊i⌋(1 ≤ i ≤ n′). Further, we
i} (1 ≤ i ≤ n′) can
ito the ‘bottom
⌊i⌋and each η′
iin a region b′
n′ (but no
ζ′
1
η′
1
ζ′
2
η′
2
ζ′
3
η′
3
ζ′
n
η′
n
ζ∗
Figure 6: Encoding the PCP: Stage 2.
cial step is to match up these arcsequences. To do so, we
write ¬C(a′
for all i, j (0 ≤ i,j < 3, i ?= j). A simple argument based
on planarity considerations then ensures that the upper and
lower sequences of arcs must cross (essentially) as shown in
Fig. 6. In particular, we are guaranteed that n = n′(without
specifying the value n), and that, for all 1 ≤ i ≤ n, ζiis
connected by ηi(and also by η′
Having established the configuration of Fig. 6, we write
(bi≤ l0+ l1) ∧ ¬C(bi· l0,bi· l1), for 0 ≤ i < 3, ensuring
that each ηiis included in exactly one of l0, l1. These inclu
sionsnaturallydefineawordσ overthealphabet{0,1}. Next,
we write Ccconstraints which organize the sequences of arcs
{ζi} and{ζ′
blocksofarcscanthenbeputin1–1correspondenceusinges
sentially the same constructionused to put the individualarcs
in 1–1 correspondence. Each pair of corresponding blocks
can now be made to lie in exactly one region from a collec
tiont1,...,tℓ. We thinkofthetjas representingtheletters of
the alphabet T, so that the labelling of the blocks with these
elements defines a word τ ∈ T∗. It is then straightforward
to write noncontact constraints involving the arcs ζiensur
ing that σ = w1(τ) and noncontact constraints involving the
i,bj) ∧ ¬C(ai,b′
j) ∧ ¬C(bi+ b′
i,bj+ b′
j+ z∗),
i) to ζ′
i.
i} (independently)intoconsecutiveblocks. These
Page 5
arcs ζ′
of all the foregoing Ccformulas. Thus, if ϕwis satisfiable
overRC(R2), then w is a positive instance of the PCP. On the
other hand, if w is a positive instance of the PCP, then one
can construct a tuple satisfying ϕwover RCP(R2) by ‘thick
ening’ the above collections of arcs into polygons in the ob
vious way. So, w is positive iff ϕwis satisfiable over RC(R2)
iff ϕwis satisfiable over RCP(R2). This shows r.e.hardness
of Sat(Cc,RC(R2)) and Sat(Cc,RCP(R2)). Membership of
the latter problem in r.e. is immediate because all polygons
may be assumed to have vertices with rational coordinates,
and so may be effectively enumerated. Using the techniques
of Corollaries 3–4 and Theorem 5, we obtain:
iensuring that σ = w2(τ). Let ϕwbe the conjunction
Theorem 6 For L ∈ {Bc◦,Bc,Cc◦,Cc}, Sat(L,RC(R2)) is
r.e.hard, and Sat(L,RCP(R2)) is r.e.complete.
The complexity of Sat(L,RC(R3)) remains open for the
languages L ∈ {Bc,Cc◦,Cc}. However, as we shall see in
the next section, for Bc◦it drops dramatically.
5
In this section, we consider the complexity of satisfying Bc◦
constraints by polyhedra and regular closed sets in three
dimensional Euclidean space. Our analysis rests on an im
portant connection between geometrical and graphtheoretic
interpretations. We begin by briefly discussing the results
of[Kontchakov et al., 2010a]for the polyhedral case.
Recall that every partial order (W,R), where R is a transi
tive and reflexive relation on W, can be regarded as a topo
logicalspace by takingX ⊆ W to be openjust in case x ∈ X
and xRy imply y ∈ X. Such topologies are called Aleksan
drov spaces. If (W,R) contains no proper paths of length
greater than 2, we call (W,R) a quasisaw (Fig. 8). If, in ad
dition, no x ∈ W has more than two properRsuccessors, we
call (W,R)a 2quasisaw. Thepropertiesof2quasisawswe
need are as follows[Kontchakov et al., 2010a]:
– satisfiability of Bcformulas in arbitrary topological
spaces coincides with satisfiability in 2quasisaws, and
is EXPTIMEcomplete;
– X ⊆ W is connected in a 2quasisaw (W,R) iff it is
interiorconnected in (W,R).
The following construction lets us apply these results to the
problem Sat(Bc◦,RCP(R3)). Say that a connected partition
in RCP(R3) is a tuple X1,...,Xkof nonempty polyhedra
having connected and pairwise disjoint interiors, which sum
to the entire space R3. The neighbourhood graph (V,E) of
this partition has vertices V = {X1,...,Xk} and edges E =
{{Xi,Xj}  i ?= j and (Xi+ Xj)◦is connected} (Fig. 7).
One can show that every connected graph is the neighbour
hood graph of some connected partition in RCP(R3). Fur
thermore, every neighbourhood graph (V,E) gives rise to
a 2quasisaw, namely, (W0∪ W1,R), where W0 = V ,
W1 = {zx,y  {x,y} ∈ E}, and R is the reflexive closure
of {(zx,y,x),(zx,y,y)  {x,y} ∈ E}. From this, we see
that (i) a Bc◦formula ϕ is satisfiable over RCP(R3) iff (ii)
ϕ is satisfiable over a connected 2quasisaw iff (iii) the Bc
formula ϕ•, obtained from ϕ by replacing every occurrence
Bc◦in 3D
X1
X2
X3
X4
X5
X6
X1
X2
X3
X4
X5
X6
Figure 7: A connectedpartition and its neighbourhoodgraph.
ofc◦withc, issatisfiableoveraconnected2quasisaw. Thus,
Sat(Bc◦,RCP(R3)) is EXPTIMEcomplete.
The picture changes if we allow variables to range over
RC(R3) ratherthanRCP(R3). NotefirstthattheBc◦formula
(2) is not satisfiable over 2quasisaws, but has a quasisaw
model as in Fig. 8. Some extra geometrical work will show
x1
x2
x3
z
R
R
R
W1= depth 1
W0= depth 0
Figure 8: A quasisaw model I of (2): rI
i= {xi,z}.
now that (iv) a Bc◦formula is satisfiable over RC(R3) iff (v)
it is satisfiable over a connected quasisaw. And as shown
in [Kontchakov et al., 2010a], satisfiability of Bc◦formulas
in connected spaces coincides with satisfiability over con
nected quasisaws, and is NPcomplete.
Theorem 7 The problem Sat(Bc◦,RC(R3)) is NPcomplete.
Proof. From the precedingdiscussion, it suffices to show that
(v) implies (iv) for any Bc◦formula ϕ. So suppose A = ϕ,
with A based on a finite connected quasisaw (W0∪ W1,R),
where Wicontains all points of depth i ∈ {0,1} (Fig. 8).
Without loss of generality we will assume that there is a spe
cial point z0of depth 1 such that z0Rx for all x of depth 0.
We show how A can be embedded into RC(R3).
Takepairwisedisjointclosed ballsB1
pairwise disjoint open balls Dz, for all z of depth 1 except z0
(we assume the Dzare disjoint from the B1
closure of the complement of all B1
We expand the B1
(A) the Bxform a connected partition in RC(R3), that is,
they are regular closed and sum up to R3, and their inte
riors are nonempty, connected and pairwise disjoint;
(B) every point in Dzis either in the interior of some Bx
with zRx, or on the boundaryof all of the Bxwith zRx.
The required Bxare constructed as follows. Let q1,q2,...
be an enumerationof all the points in the interiors of Dzwith
rationalcoordinates. Forx ∈ W0, we set Bxto betheclosure
of the infinite union?∞
sets Bk
that the Bk
q1,q2,... that is not in any Bk
of some Dz. Take an open ball Cqiin the interior of Dz
centred in qiand disjoint from the Bk
x, forxofdepth0,and
x). Let Dz0be the
xand Dz.
xto sets Bxin such a way that
k=1(Bk
x)◦, where the regular closed
xare defined inductivelyas follows (Fig. 9). Assuming
xare defined, let qi be the first point in the list
xyet. So, qiis in the interior
x. For each x ∈ W0with
Page 6
zRx, expand Bk
connecting it to B1
are disjoint from the rest of the Bk
Bk+1
x
. Consider a function f that maps regular closed sets
xby a closed ball in Cqiand a closed ‘rod’
xin such a way that the ball and the rod
x; the result is denoted by
Bx1
Bx2
Bx3
Dz1
Cq
q
Figure 9: Filling Dz1with Bxi, for z1Rxi, i = 1,2,3.
X ⊆ W to RC(R3) so that f(X) is the union of all Bx, for
x of depth 0 in X. By (A), f preserves +, ·, −, 0 and 1.
Define an interpretation I over RC(R3) by rI= f(rA). To
show that I = ϕ, it remains to prove that X◦is connected iff
(f(X))◦is connected (details are in Appendix C).
❑
The remarkably diverse computational behaviour of Bc◦
over RC(R3), RCP(R3) and RCP(R2) can be explained as
follows. To satisfy a Bc◦formula ϕ in RC(R3), it suffices
to find polynomially many points in the regions mentioned in
ϕ (witnessing nonemptiness or noninternalconnectedness
constraints), and then to ‘inflate’ those points to (possibly in
ternally connected) regular closed sets using the technique of
Fig. 9. By contrast, over RCP(R3), one can write a Bc◦
formulaanalogousto (8) stating that two internallyconnected
polyhedra do not share a 2D face. Such ‘facecontact’ con
straints can be used to generate constellations of exponen
tially many polyhedra simulating runs of alternating Tur
ing machines on polynomial tapes, leading to EXPTIME
hardness. Finally, over RCP(R2), planarity considerations
endow Bc◦with the extra expressive power required to en
force full noncontact constructs (not possible in higher di
mensions), and thus to encode the PCP as sketched in Sec. 4.
6
This paper investigated topological constraint languages fea
turing connectedness predicates and Boolean operations on
regions.Unlike their less expressive cousins, RCC8 and
RCC5, such languages are highly sensitive to the spaces
over which they are interpreted, and exhibit more challeng
ing computational behaviour. Specifically, we demonstrated
that the languages Cc, Cc◦and Bc contain formulas satisfi
able over RC(Rn), n ≥ 2, but only by regions with infinitely
many components. Using a related construction, we proved
that the satisfiability problem for any of Bc, Cc, Bc◦and Cc◦,
interpreted either over RC(R2) or over its polygonal subal
gebra, RCP(R2), is undecidable. Finally, we showed that
the satisfiability problemfor Bc◦, interpretedoverRC(R3), is
NPcomplete, which contrasts with EXPTIMEcompleteness
for RCP(R3). The complexity of satisfiability for Bc, Cc and
Cc◦over RC(Rn) or RCP(Rn) for n ≥ 3 remains open. The
obtained results rely on certain distinctive topological prop
erties of Euclidean spaces. Thus, for example, the argument
Conclusion
of Sec. 3 is based on the property of Lemma 1, while Sec. 4
similarly relies on planarity considerations. In both cases,
however, the moral is the same: the topological spaces of
most interest for Qualitative Spatial Reasoning exhibit spe
cial characteristics which anytopologicalconstraintlanguage
able to express connectedness must take into account.
The results of Sec. 4 pose a challenge for Qualitative Spa
tial Reasoning in the Euclidean plane. On the one hand, the
relatively low complexity of RCC8 over dischomeomorphs
suggests the possibility of usefully extending the expressive
power of RCC8 without compromising computational prop
erties. On the other hand, our results impose severe limits
on any such extension. We observe, however, that the con
structions used in the proofs depend on a strong interaction
betweenthe connectednesspredicatesandthe Booleanopera
tions onregularclosed sets. We believethat byrestrictingthis
interaction one can obtain nontrivial constraint languages
with more acceptable complexity. For example, the exten
sion of RCC8 with connectedness constraints is still in NP
for both RC(R2) and RCP(R2)[Kontchakov et al., 2010b].
Acknowledgments. This work was partially supported by
the U.K. EPSRC grants EP/E034942/1 and EP/E035248/1.
References
[Bennett, 1994] B. Bennett. Spatial reasoning with proposi
tional logic. In Proc. of KR, pages 51–62. 1994.
[Borgo et al., 1996] S. Borgo, N. Guarino,and C. Masolo. A
pointless theory of space based on strong connection and
congruence. In Proc. of KR, pages 220–229. 1996.
[Cohn and Renz, 2008] A. Cohn and J. Renz.
spatial representationand reasoning. In Handbookof Kno
wledge Representation, pages 551–596. Elsevier, 2008.
[Dornheim, 1998] C. Dornheim. Undecidabilityof plane po
lygonal mereotopology. In Proc. of KR. 1998.
[Egenhofer and Franzosa, 1991] M. Egenhofer and R. Fran
zosa. Pointset topological spatial relations. International
J. of GeographicalInformationSystems, 5:161–174,1991.
[Kontchakov et al., 2010a] R.
Hartmann, F. Wolter, and M. Zakharyaschev.
logics with connectedness predicates. Logical Methods in
Computer Science, 6(3), 2010.
[Kontchakov et al., 2010b] R. Kontchakov, I. PrattHart
mann, and M. Zakharyaschev. Interpreting topological
logics over Euclidean spaces. In Proc. of KR. 2010.
[Newman, 1964] M.H.A. Newman. Elements of the Topol
ogy of Plane Sets of Points. Cambridge, 1964.
[PrattHartmann, 2007] I. PrattHartmann. Firstorder mere
otopology. In Handbook of Spatial Logics, pages 13–97.
Springer, 2007.
[Randell et al., 1992] D. Randell, Z. Cui, and A. Cohn. A
spatial logic based on regions and connection. In Proc. of
KR, pages 165–176. 1992.
[Renz and Nebel, 2001] J. Renz and B. Nebel.
methods for qualitative spatial reasoning. J. Artificial In
telligence Research, 15:289318,2001.
Qualitative
Kontchakov,I. Pratt
Spatial
Efficient
Page 7
[Renz, 1998] J. Renz. A canonical model of the region con
nection calculus. In Proc. of KR,‘pages 330–341. 1998.
[Schaefer et al., r003] M. Schaefer,
D.ˇStefankoviˇ c. Recognizing string graphs in NP. J. of
Computer and System Sciences, 67:365–380,2003.
[Wolter and Zakharyaschev, 2000] F. Wolter and M. Za
kharyaschev. Spatial reasoning in RCC8 with Boolean re
gion terms. In Proc. of ECAI, pages 244–248. 2000.
E. Sedgwick,and
Page 8
A
First we give detailed proofs of Lemma 1 and Theorem 2.
Theorem 8 ([Newman, 1964]) If X is a connected subset of
Rn, then every connected component of Rn\ X has a con
nected boundary.
Regions with infinitely many components
Lemma 1. If X ∈ RC(Rn) is connected, then every compo
nent of −X has a connected boundary.
Proof. Let Y be a connected component of −X. Suppose
that the boundary β of Y is not connected, and let β1and β2
be two sets separating β: β1and β2are disjoint, nonempty,
closed subsets of β whose union is β. We will show that Y is
not connected. We have Y = (?
I, where the Ziare distinct connectedcomponentsof Rn\X.
By Theorem8,‘the boundariesαiof Ziare connectedsubsets
of β, for each i ∈ I. Hence, either αi ⊆ β1or αi ⊆ β2,
for otherwise αi∩ β1and αi∩ β2would separate αi. Let
Ij= {i ∈ I  αi⊆ βj} and Yj= (?
Clearly, Y1and Y2are closed, and Y = Y1∪ Y2. Hence, it
suffices to show that Y1and Y2are disjoint. We know that,
for j = 1,2,
i∈IZi)−, for some index set
i∈IjZi)−, for j = 1,2.
Yj= (
?
i∈Ij
αi)
−∪
?
i∈Ij
Zi.
Clearly,?
that (?
β1and β2, respectively. Finally, (?
are disjoint, for j,k = 1,2, as subsets of the boundary and
the interior of Y , respectively. So, Y is not connected, which
is a contradiction.
i∈I1Ziand?
i∈I1αi)−and (?
i∈I2Ziare disjoint. We also know
i∈I2αi)−are disjoint, as subsets of
i∈Ijαi)−and?
i∈IkZi
❑
Theorem 2. If I is an interpretation over RC(Rn) such that
I = ϕ∞, then every dI
Proof. To simplify presentation, we ignore the difference be
tween variables and the regions they stand for, writing, for
example, ai instead of aI
We construct a sequence of disjoint components Xiof d⌊i⌋
and open sets Vi connecting Xi to Xi+1 (Fig. 3). By the
first conjunct of (4), let X0be a component of d0containing
points in a0. Suppose Xihas been constructed, for i ≥ 0.
By (5) and (6), there exists a point q ∈ Xi∩ a⌊i+1⌋. Since
q / ∈ b⌊i+1⌋∪ d⌊i+2⌋∪ d⌊i+3⌋, and because Rnis locally
connected, there exists a connected neighbourhood Vi of q
such that Vi∩ (b⌊i+1⌋∪ d⌊i+2⌋∪ d⌊i+3⌋) = ∅, and so,
by (3), Vi ⊆ d⌊i⌋+ a⌊i+1⌋. Further, since q ∈ a⌊i+1⌋,
Vi∩ a⌊i+1⌋◦?= ∅. Take X′
that intersects Viand Xi+1the component of d⌊i+1⌋contain
ing X′
To see that the Xiare distinct, let Si+1and Ri+1be the
componentsof −Xi+1containingXiand Xi+2, respectively.
It suffices to show Si+1⊆ S◦
Vimust intersect δSi+1. Evidently,δSi+1⊆ Xi+1⊆ d⌊i+1⌋.
Also, δSi+1 ⊆ −Xi+1; hence, by (3) and (7), δSi+1 ⊆
di∪d⌊i+2⌋. By Lemma 1, δSi+1is connected, and therefore,
by (7), is entirely contained either in d⌊i⌋or in d⌊i+2⌋. Since
ihas infinitely many components.
i. We also set bi = di· (−ai).
i+1to be a component of a⌊i+1⌋
i+1.
i+2. Note that the connected set
Vi∩δSi+1?= ∅andVi∩d⌊i+2⌋= ∅, wehaveδSi+1?⊆ d⌊i+2⌋,
so δSi+1 ⊆ di. Similarly, δRi+1 ⊆ di+2. By (7), then,
δSi+1∩δRi+1= ∅, andsinceSi+1andRi+1are components
of the same set, they are disjoint. Hence, Si+1⊆ (−Ri+1)◦,
and since Xi+2 ⊆ Ri+1, also Si+1 ⊆ (−Xi+2)◦.
Si+1lies in the interior of a component of −Xi+2, and since
δSi+1⊆ Xi+1⊆ Si+2, that component must be Si+2.
So,
❑
Now we extend the result to the language Cc◦. All occur
rences of c in ϕ∞have positive polarity. Let ϕ◦
of replacing them with the predicate c◦. In the configura
tion of Fig. 2, all connected regions mentioned in ϕ∞are in
fact interiorconnected;henceϕ◦
Since interiorconnectedness implies connectedness, ϕ◦
tails ϕ∞in a common extension of Cc◦and Cc. Hence:
Corollary 3.
There is a Cc◦formula satisfiable over
RC(Rn), n ≥ 2, but not by regions with finitely many compo
nents.
∞be the result
∞is satisfiable overRC(Rn).
∞en
a2
b2
a1
b1
a0
b0
a3
b3
a2
b2
a1
b1
a0
b0
t
.. .
s
Figure 10: Satisfying ϕc
¬C(a0,b1,s,t) and ϕc
¬C(a0,b2,s,t).
To extendTheorem2 to the languageBc, notice that all oc
currences of C in ϕ∞are negative. We shall eliminate these
using only the predicate c. We use the fact that, if the sum
of two connected regions is not connected, then they must be
disjoint. Consider the formula
ϕc
¬C(r,s,r′,s′) := c(r + r′) ∧ c(s + s′)
∧¬c((r + r′) + (s + s′)).
Note that ϕc
¬C(ai,t) with ϕc
clearly satisfiable by the regions on Fig. 2. Further, we re
place ¬C(ai,b⌊i+1⌋) with ϕc
on Fig. 10, there exists a region s satisfying this formula. In
stead of dealing with ¬C(di,di+2), we consider the equiva
lent:
¬C(r,s,r′,s′) implies ¬C(r,s).
¬C(ai,t,a0+ a1+ a2+ a3,t), which is
We replace
¬C(ai,b⌊i+1⌋,s,t). As shown
¬C(ai,b⌊i+2⌋) ∧ ¬C(bi,a⌊i+2⌋)∧
¬C(ai,a⌊i+2⌋) ∧ ¬C(bi,b⌊i+2⌋).
We replace ¬C(ai,b⌊i+2⌋) by ϕc
is satisfiable by the regions depicted on Fig. 10.
ignore ¬C(bi,a⌊i+2⌋), because it is logically equivalent
to ¬C(ai,b⌊i+2⌋), for different values of i.
¬C(ai,a⌊i+2⌋) by ϕc
isfiable by the regions depicted on Fig. 11. The fourth con
junct is then treated symmetrically. Transforming ϕ∞in the
way just described, we obtain a Bcformula ϕc
plies ϕ∞(in the language Cc) and which is satisfiable by the
arrangement of RC(Rn). Hence, we obtain the following:
¬C(ai,b⌊i+2⌋,s,t), which
We
We replace
¬C(ai,a⌊i+2⌋,a′
i,a′
⌊i+2⌋), which is sat
∞, which im
Page 9
...
a0
a′
0
a3
a2
a′
2
a1
a0
b
a′
0
a3
b3
a2
b2
a′
2
a1
b1
a0
b0
a′
0
Figure 11: Satisfying ϕc
¬C(a0,a2,a′
0,a′
2).
Corollary 4.
RC(Rn), n ≥ 2, but not by regions with finitely many compo
nents.
There is a Bcformula satisfiable over
The only remaining task in this section is to prove Theo
rem 5. The construction is similar to the one developed in
Sec. 4, and as such uses similar techniques. We employ the
following notation. If α is a Jordan arc, and p, q are points on
α such that q occurs after p, we denote by α[p,q] the segment
of α from p to q. Consider the formula stack◦(a1,...,an)
given by:
?
1≤i<n
(c◦(ai+ ··· + an) ∧ ai· ai+1= 0) ∧
?
j−i>1
¬C(ai,aj)
This formula allows us to construct sequences of arcs in the
following sense:
Lemma 9 Suppose that the condition stack◦(a1,...,an)
obtains, n > 1. Then every point p1∈ a◦
to every point pn∈ a◦
that for all i (1 ≤ i < n), each segment αi⊆ (ai+ ai+1)◦is
a nondegenerate Jordan arc starting at some point pi∈ a◦
1can be connected
nby a Jordanarc α = α1···αn−1such
i.
Proof. By c◦(a1+ ··· + an), let α′
be a Jordan arc connecting p1to pn(Fig. 12). By the non
contact constraints, α′
one such point. For 2 ≤ i < n we suppose α1,...,αi−2,
α′
c◦(ai+ ··· + an), let α′′
arc connecting p′
can intersect α1···αi−2α′
Let pi−1be the first point of α′
the initial segment of α′
final segment of α′′
αn−1, and to this end, we simply set αn−1:= α′
that pi, 2 ≤ i < n, are as required, note that pi∈ αi∩ αi−1.
By the disjoint constraintspimust be in ai. If piwas in δ(ai),
it would also have to be in δ(ai−1) and δ(ai+1), which is
forbidden by the disjoint constraints. Hence pi ∈ a◦
i ≤ n. Given ai· ai+1= 0, 1 ≤ i < n, this also guarantees
that the arcs αiare nondegenerate.
1⊆ (a1+ ··· + an)◦
1has to contain points in a◦
2. Let p′
2be
i−1and p′
ito have been defined, and proceedas follows. By
i⊆ (ai+ ··· + an)◦be a Jordan
ito pn. By the noncontact constraints, α′′
i−1only in its final segment α′
i−1lying on α′
i−1ending at pi−1; and let α′
istarting at pi−1. It remains only to define
i
i−1.
i; let αi−1be
ibe the
n−1. To see
i, 1 ≤
❑
p1
p2
α1
α′
1
p′
2
α′′
2
α′
1
α2
...
αn−2
pn−1
pn−1′
α′
n−2
α′′
n−1
pn
αn−1
Figure 12: The constraint stack◦(a1,...,an) ensures the ex
istence of a Jordan arc α = α1···αn−1which connects a
point p1∈ a◦
1to a point pn∈ a◦
n.
Consider nowthe formulaframe◦(a0,...,an−1) givenby:
?
0≤i<n
?c◦(ai) ∧ c◦(ai+ a⌊i+1⌋) ∧ ai?= 0?∧
?
j−i>1
ai· aj= 0,
where ⌊k⌋ denotes k mod n. This formula allows us to con
struct Jordan curves in the plane, in the following sense:
Lemma 10 Let n ≥ 3, and suppose frame◦(a0,...,an−1).
Then there exist Jordan arcs α0, ..., αn−1 such that
α0...αn−1is a Jordan curve lying in the interior of a0+
··· + an−1, and αi⊆ (ai+ a⌊i+1⌋)◦, for all i, 0 ≤ i < n.
Proof. For all i (0 ≤ i < n), pick p′
Jordan arc α′
i (2 ≤ i ≤ n), let p⌊i⌋be the first point of αi−1lying on
α⌊i⌋, and let p′′
(2 ≤ i < n), let αi= α′
let α′′
from p0to p′′
let α0= α′′
verifythatthearcs α0, ..., αn−1havetherequiredproperties.
i∈ a◦
i, and pick a
i⊆ (ai+ a⌊i+1⌋)◦from pito p⌊i+1⌋. For all
1be the first point of α′
i[pi,pi+1], let α′′
0denote the section of α′
1. Now let p1be the first point of α′′
0[p0,p1], and let α1= α′′
0lying on α′
1= α′
1. For all i
1[p′′
1,p2], and
0(in the appropriate direction)
0lying on α′′
1,
1[p1,p2]. It is routine to
❑
We will now show how to separate certain types of regions
in the language Bc◦. We make use of Lemma 10 and the
following fact.
Lemma 11 [Newman, 1964, p. 137] Let F, G be disjoint,
closed subsets of R2such that R2\ F and R2\ G are con
nected. Then R2\ (F ∪ G) is connected.
M1
P0
µ0
P1
µ1
P2
µ2
τ2
τ1
τ0
n0
n2
n1
Figure 13: The Jordan curve Γ = τ0τ1τ2separating m1from
m2.
We say that a region r is quasibounded if either r or −r is
bounded. We can now prove the following.
Page 10
Lemma 12 There exists a Bc◦formula η∗(r,s, ¯ v) with the
following properties: (i) η∗(r,s, ¯ v) entails ¬C(r,s) over
RC(R2); (ii) if the regions r and s can be separated by a Jor
dan curve, then there exist polygons ¯ v such that η∗(τ1,τ2, ¯ v);
(iii) if r, s are disjoint polygons such that r is quasibounded
andR2\(r+s) is connected,then there exist polygons ¯ v such
that η∗(τ1,τ2, ¯ v).
Proof. Let ¯ v be the tuple of variables (t0,...,t5,m1,m2),
and let η∗(r,s, ¯ v) be the formula
frame◦(t0,...,t5) ∧ r ≤ m1∧ s ≤ m2∧
(t0+ ... + t5) · (m1+ m2) = 0 ∧
?
i=1,3,5
j=1,2
c◦(ti+ mj).
Property (i) follows by a simple planarity argument. By
frame◦(t0,...,t5) and Lemma 10, let αi, for 0 ≤ i ≤ 5,
be such that Γ = α0···α5 is a Jordan curve included in
(t0+ ··· + t5)◦. Further, let τi = α2iα2i+1, 0 ≤ i ≤ 2
(Fig.13). Note that all points in a2i+1, 0 ≤ i ≤ 2, that
are on Γ are on τi. By c◦(t2i+1+ m1), 0 ≤ i ≤ 2, let
µi⊆ (m1+t2i+1)◦be a Jordanarc with endpointsM1∈ m◦
and Ti ∈ τi∩ t◦
sect only at their common endpoint M1, so that they divide
the residual domain of Γ which contains M1into three sub
domainsni,for0 ≤ i ≤ 2. TheexistenceofapointM2∈ m2
in any ni, 0 ≤ i ≤ 2, will contradict c◦(t2i+1+m2). So, m2
must be contained entirely in the residual domain of Γ not
containing M1. Similarly, all points in m1must lie in the
residual domain of Γ containing M1. It follows that m1and
m2are disjoint, and by r ≤ m1and s ≤ m2, that r and s are
disjoint as well. For Property (ii), let Γ be a Jordan curve sep
arating r and s. Now thicken Γ to form an annular element of
RCP(R2), still disjoint from r and s, and divide this annulus
into the three regions t0,...,t5as shown (up to similar situa
tion) in Fig. 14. Choose m1and m2to be the connectedcom
ponents of −(t0+ ··· + t5) containing r and s, respectively.
For Property (iii), it is routine using Lemma 11 to show that
there exists a piecewise linear Jordan curve Γ in R2\ (r + s)
separating r and s.
1
2i+1. We may assume that these arcs inter
❑
r
r
r
ss
t0
t1
t2
t3
t4
t5
Γ
m1
m2
Figure 14: Separating disjoint polygons by an annulus.
Lemma 13 There exists a Bc◦formula η(r,s, ¯ v) with the
following properties:
(i) η(r,s, ¯ v) entails ¬C(r,s) over
RC(R2); (ii) if r, s are disjoint quasiboundedpolygons,then
there exist polygons ¯ v such that η(τ1,τ2, ¯ v).
Proof. Let η(r,s, ¯ v) be the formula
r = r1+ r2∧ s = s1+ s2∧
?
1≤i≤2
1≤j≤2
η∗(ri,sj, ¯ ui,j),
where η∗is the formula given in Lemma 12. Property (i) is
then immediate. For Property (ii), it is routine to show that
there exist polygons r1, r2such that r = r1+ r2and R2\ ri
is connected for i = 1,2; let s1, s2be chosen analogously.
Then for all i (1 ≤ i ≤ 2) and j (1 ≤ j ≤ 2) we have
ri∩sj= ∅ and, by Lemma 11, R2\(ri+sj) connected. By
Lemma 12, let ¯ ui,jbe such that η∗(ri,sj, ¯ ui,j).
❑
We are now ready to prove:
Theorem 5.
RC(R2), butonlyby regionswith infinitelymanycomponents.
Proof. We first write a Cc◦formula, ϕ∗
properties, and then show that all occurrences of C can be
eliminated. Note that ϕ∗
constructed for the proof of Corollary 3.
Let s, s′, a, a′, b, b′, ai,jand bi,j(0 ≤ i < 2, 1 ≤ j ≤ 3)
be variables. The constraints
There is a Bc◦formula satisfiable over
∞with the required
∞is not the same as the formula ϕ◦
∞
frame◦(s,s′,b,b′,a,a′)
stack◦(s,bi,1,bi,2,bi,3,b)
stack◦(b⌊i−1⌋,2,ai,1,ai,2,ai,3,a)
stack◦(a⌊i−1⌋,2,bi,1,bi,2,bi,3,b)
are evidently satisfied by the arrangement of Fig. 15.
(9)
(10)
(11)
(12)
b0,1
b0,2
b0,3
a0,1
a0,2
a0,3
b1,1
b1,2
b1,3
a1,1
a1,2
a1,3
b0,1
b0,2
b0,3
a0,1
a0,2
a0,3
s
s′
a′
b
b′
a
Figure 15: A tuple of regions satisfying (9)–(12): the pattern
of components of the ai,jand bi,jrepeats forever.
Let ϕ∗
juncts
∞be the conjunction of (9)–(12) as well as all con
r · r′= 0,
(13)
where r and r′are any two distinct regions depicted on
Fig. 15. Note that the regions ai,j and bi,j have infinitely
many connected components. We will now show that this is
true for every satisfying tuple of ϕ∗
By (9), we can use Lemma 10 to construct a Jordan curve
Γ = σσ′ββ′αα′whose segments are Jordan arcs lying in
the respective sets (s + s′)◦, (s′+ b)◦, (b + b′)◦, (b′+ a)◦,
(a + a′)◦, (a′+ s)◦. Further, let σ0 = σσ′, β0 = ββ′and
α0 = αα′(Fig. 16a). Note that all points in s, a and b that
are on Γ are on σ0, α0and β0, respectively. Let o′
∞.
0∈ σ0∩
Page 11
s◦, and let q∗∈ β0∩ b◦. By (10) and Lemma 9 we can
connecto′
lie intherespectivesets (s + b0,1)◦, (b0,1+ b0,2+ b0,3)◦and
(b + b0,3)◦(Fig. 16b). Let o0be the last point on β′
on σ0and let β0,1be the final segment of β′
Similarly, let q0be the first point on β′
β0,3be the initial segment of β′
arc β0,1β0,2β0,3divides one of the regions boundedby Γ into
two subregions. We denote the subregion whose boundary
is disjoint from α0by U0, and the other subregionwe denote
by U′
0to q∗bya Jordanarcβ′
0,1β0,2β′
0,3whosesegments
0,1that is
0,1starting at o0.
0,3that is on β0and let
0,3ending at q0. Hence, the
0. Let β1:= β0,3β0[q0,r] ⊆ (b + b0,3+ b1,3)◦.
q
p
r
σ0
β0
α0
(a) The arcs α0, β0and σ0.
o′
0
o0
q∗
q0
β0,1
β0,2
β0,3
U0
U′
0
(b) The regions U0and U′
0.
U0
e′
0
e0
p∗
p0
α0,1
α0,2
α0,3
V0
W0
(c) The regions V0and W0.
q0
p0
o0
e0
U0
V0
α0,2
W0
β1
α1
(d) The regions redrawn.
o′
1
o1
q∗
q1
β1,2
V0
U1
U′
1
α1
(e) The regions U1and U′
1.
U1
e′
1
e1
p∗
p1
α1,1
α1,2
α1,3
V1
W1
(f) The regions V1and W1.
Figure 16: Establishing infinite sequences of arcs.
We will now construct a crosscut α0,1α0,2α0,3in U′
0∈ β0,2∩ b0,2◦and p∗∈ α0∩ a◦. By (11) and Lemma 9
we can connect e′
whose segments lie in the respective sets (b0,2+ a0,1)◦,
(a0,1+ a0,2+ a0,3)◦and (a + a0,3)◦(Fig. 16c). Let e0be
the last point on α′
nal segment of α′
first point on α′
segment of α′
straints, α0,1α0,2α0,3does not intersect the boundaries of U0
and U′
one of these regions. Moreover,that regionhas to be U′
0. Let
e′
0to p∗by a Jordan arc α′
0,1α0,2α′
0,3
0,1that is on β0,2and let α0,1be the fi
0,1starting at e0. Similarly, let p0be the
0,3that is on α0 and let α0,3 be the initial
0,3ending at p0. By the nonoverlapping con
0except at its endpoints, and hence it is a crosscut in
0since
the boundary of U0 is disjoint from α0. So, α0,1α0,2α0,3
divides U′
whose boundarycontains β1by W0, and the other subregion
we denote by V0. Let α1 := α0,3α0[p0,r] (Fig 16d). Note
that α1⊆ (a + a0,3+ a1,3)◦.
We can now forget about the region U0, and start con
structing a crosscut β1,1β1,2β1,3 in W0.
β′
α0,2 ∩ a◦
ments are contained in the respective sets (a0,2+ b1,1)◦,
(b1,1+ b1,2+ b1,3)◦and (b + b1,3)◦. As before, we choose
β1,1
⊆ β′
⊆ β′
β1,1β1,2β1,3with its endpoints removed is disjoint from the
boundaries of V0and W0. Hence β1,1β1,2β1,3has to be a
crosscut in V0or W0, and since the boundary of V0is dis
joint from β1it has to be a crosscut in W0(Fig. 16e). So,
β1,1β1,2β1,3 separates W0 into two regions U1 and U′
that the boundary of U1 is disjoint from α1.
β1,3β1[q1,r] ⊆ (b + b0,3+ b1,3)◦. Now, we can ignore the
region V0, and reasoning as before we can construct a cross
cutα1,1α1,2α1,3in U′
W1.
0into two subregions. We denote the subregion
As before, let
1,1β1,2β′
1,3be a Jordan arc connecting a point o′
0,2to a point q∗∈ β1 ∩ b◦
1
∈
isuch that its seg
1,1and β1,3
1,3so that the Jordan arc
1so
Let β2 :=
1dividingit into twosubregionsV1and
b0,2
a0,2
b0,2
a0,2
s
b′
Figure 17: Separating a0,2from b0,2by a Jordan curve.
Evidently, this process continues forever. Now, note that
by construction and (13), W2icontains in its interior β2i+1,2
together with the connected component c of b1,2which con
tains β2i+1,2. On the other hand, W2i+2is disjoint from c,
and since Wi⊆ Wj, i > j, b1,2has to have infinitely many
connected components.
So far we know that the Cc◦formula ϕ∗
many components. Now we replace every conjunct in ϕ∗
the form ¬C(r,s) by η∗(r,s, ¯ v), where ¯ v are fresh variables
each time. The resulting formulaentails ϕ∗
to show that it is still satisfiable. By Lemma 12 (ii), it suffices
to separate by Jordancurves everytwo regionson Fig. 15 that
are required to be disjoint. It is shown on Fig. 17 that there
exists a curve which separates the regions b0,2and a0,2. All
other noncontact constraints are treated analogously.
∞forces infinitely
∞of
∞, so we onlyhave
❑
BUndecidability of Bc and Cc in the
Euclidean plane
In this section, we prove the undecidability of the problems
Sat(L,RC(R2)) and Sat(L,RCP(R2)), for L any of Bc, Cc,
Page 12
Bc◦or Cc◦. We begin with some technical preliminaries,
again employingthe notationfrom the proofof Theorem5: if
α is a Jordan arc, and p, q are points on α such that q occurs
after p, we denote by α[p,q] the segment of α from p to q.
For brevity of exposition, we allow the case p = q, treating
α[p,q] as a (degenerate) Jordan arc.
Our first technical preliminary is to formalize our earlier
observations concerning the formula stack(a1,...,an), de
fined by:
?
1≤i≤n
c(˙ ai+ ¨ ai+1+ ··· + ¨ an) ∧
?
j−i>1
¬C(ai,aj).
Lemma 14 Let
stack(a1,...,an), for n ≥ 3.
p0 ∈ ˙ a1 and every point pn ∈ ¨ an, there exist points
p1,...,pn−1and Jordan arcs α1,...,αnsuch that:
(i) α = α1···αnis a Jordan arc from p0to pn;
(ii) for all i (0 ≤ i < n), pi∈ ˙ ai+1∩ αi; and
(iii) for all i (1 ≤ i ≤ n), αi⊆ ai.
Proof. Since ˙ a1+ ¨ a2+ ··· + ¨ anis a connected subset of
(a1+ ˙ a2+ ··· + ˙ an)◦, let β1be a Jordan arc connecting p0
to pnin (a1+ ˙ a2+ ··· + ˙ an)◦. Since a1is disjoint from all
the aiexcept a2, let p1be the first point of β1lying in ˙ a2,
so β1[p0,p1] ⊆ a◦
included in a◦
p0 = p1.) Similarly, let β′
to pnin (a2+ ˙ a3+ ··· + ˙ an)◦, and let q1be the last point
of β′
α1 = β1[p0,p1], and β2 = β′
are v1and pn. Otherwise, we have q1 ∈ a◦
construct an arc γ1 ⊆ a◦
β′
only at its endpoints, p1and v1(upper diagram in Fig. 18).
Let α1= β1[p0,p1]γ1, and let β2= β′
Since β2 contains a point p2 ∈ ˙ a3, we may iterate this
procedure, obtaining α2,α3,...αn−1,βn. We remark that
αiand αi+1have a single point of contact by construction,
while αiand αj (i < j − 1) are disjoint by the constraint
¬C(ai,aj). Finally, we let αn = βn (lower diagram in
Fig. 18).
a1,...,an
be 3regions
Then, for every point
satisfying
1∪ {p1}, i.e., the arc β1[p0,p1] is either
1, or is an endcut of a◦
2be a Jordan arc connecting p1
1. (We do not rule out
2lying on β1[p0,p1]. If q1 = p1, then set v1 = p1,
2. so that the endpoints of β2
1. We can now
1∪ {p1} from p1to a point v1on
2[q1,pn], such that γ1 intersects β1[p0,p1] and β′
2[q1,pn]
2[v1,pn].
❑
In fact, we can add a ‘switch’ w to the formula
stack(a1,...,an), in the following sense. If w is a region
variable, consider the formula stackw(a1,...,an)
¬C(w · ˙ a1,(−w) · ˙ a1) ∧ stack((−w) · a1,a2,...,an),
where w ·a denotes the 3region (w ·a,w· ˙ a,w·¨ a). The first
conjunct of stackw(a1,...,an) ensures that any component
of ˙ a1is either included in w or included in −w. The sec
ond conjunct then has the same effect as stack(a1,...,an)
for those components of ˙ a1 included in −w.
p ∈ ˙ a1· (−w), we can find an arc α1···αnstarting at p,
with the properties of Lemma 14. However, if p ∈ ˙ a · w, no
such arc need exist. Thus, w functions so as to ‘deactivate’
the formula stackw(a1,...,an) for any component of ˙ a1in
cluded in it.
That is, if
p0
q1
v1
p1
β1
β1
β1
β′
2
β′
2
γ1
β2
vn−2
qn−1
vn−1
pn−1
pn
βn−1
βn−1
βn−1
β′
n
β′
n
γn−1
βn
Figure 18: Proof of Lemma 14.
AsafurtherapplicationofLemma14, considertheformula
frame(a0,...,an) given by:
stack(a0,...,an−1) ∧ ¬C(an,a1+ ... + an−2)∧
c(˙ an) ∧ ˙ a0· ˙ an?= 0 ∧ ¨ an−1· ˙ an?= 0.
This formula allows us to construct Jordan curves in the
plane, in the following sense:
Lemma 15 Letn ≥ 3,andsupposeframe(a0,...,an). Then
there exist Jordan arcs γ0, ..., γnsuch that γ0...γnis a
Jordan curve, and γi⊆ ai, for all i, 0 ≤ i ≤ n.
Proof. By stack(a0,...,an−1), let α0,...,αn−1be Jordan
arcs in the respective regions a0,...,an−1such that, α =
α0···αn−1is a Jordan arc connecting a point p′∈ ˙ a0· ˙ an
to a point q′∈ ¨ an−1· ˙ an (see Fig. 19). Because ˙ anis a
connected subset of the interior of an, let αn ⊆ a◦
arc connecting p′and q′. Note that αndoes not intersect αi,
for 1 ≤ i < n − 1. Let p be the last point on α0that is on
αn(possibly p′), and q be the first point on αn−1that is on
αn(possibly q′). Let γ0be the final segment of α0starting
at p. Let γi := αi, for 1 ≤ i ≤ n − 2. Let γn−1be the
initial segment of αn−1ending at q. Finally, take γnto be
the segment of αnbetween p and q. Evidently, the arcs γi,
0 ≤ i ≤ n, are as required.
(14)
nbe an
❑
p′
α0
α1
...
q′
αn−1
αn−2
p′
p
γ0
γ1= α1
...
q′
q
γn−1
γn−2= αn−2
γn
Figure 19: Establishing a Jordan curve.
Our final technical preliminary is a simple device for la
belling arcs in diagrams.
Lemma 16 Suppose r, t1, ..., tℓare regions such that
?
1≤i<j≤ℓ
(r ≤ t1+ ··· + tℓ) ∧
¬C(r · ti,r · tj),
(15)
Page 13
and let X be a connected subset of r. Then X is included in
exactly one of the ti, 1 ≤ i ≤ ℓ.
Proof. If X ∩ t1and X ∩ t2are nonempty, then X ∩ t1
and X ∩ (t2+ ··· + tℓ) partition X into nonempty, non
intersecting sets, closed in X.
❑
When (15) holds, we may think of the regions t1,...,tℓas
‘labels’ for any connected X ⊆ r—and, in particular, for any
Jordan arc α ⊆ r. Hence, any sequence α1,...,αnof such
arcs encodes a word over the alphabet {t1,...,tℓ}.
The remainder of this section is given over to a proof of
Theorem 6. For L ∈ {Bc◦,Bc,Cc◦,Cc}, Sat(L,RC(R2)) is
r.e.hard, and Sat(L,RCP(R2)) is r.e.complete.
We havealreadyestablishedthe upperbounds;we consider
here only the lower bounds, beginning with an outline of our
proof strategy. Let a PCPinstance w = ({0,1},T,w1,w2)
be given, where T is a finite alphabet, and wi: T∗→ {0,1}∗
a wordmorphism (i = 1,2). We call the elements of T tiles,
and, for each tile t, we call w1(t) the lower word of t, and
w2(t) the upper word of t. Thus, w asks whether there is
a sequence of tiles (repeats allowed) such that the concate
nation of their upper words is the same as the concatenation
of their lower words. We shall henceforth restrict all (upper
and lower) words on tiles to be nonempty. This restriction
simplifies the encoding below, and does not affect the unde
cidability of the PCP.
We define a formula ϕwconsisting of a large conjunction
of Ccliterals, which, for ease of understanding,we introduce
in groups. Wheneverconjunctsare introduced,it can be read
ily checkedthat—providedw is positive—theyare satisfiable
by elements of RCP(R2). (Figs. 20 and 22 depict part of a
satisfying assignment; this drawing is additionally useful as
an aid to intuition throughout the course of the proof.) The
main object of the proof is to show that, conversely, if ϕwis
satisfied by any tuple in RC(R2), then w must be positive.
Thus, the following are equivalent:
1. w is positive;
2. ϕwis satisfiable over RCP(R2);
3. ϕwis satisfiable over RC(R2).
This establishes the r.e.hardness of Sat(L,RC(R2)) and
Sat(L,RCP(R2)) for L = Cc; we then extend the result to
the languages Bc, Cc◦and Bc◦.
The proof proceeds in five stages.
Stage 1. In the first stage, we define an assemblage of arcs
that will serve as a scaffolding for the ensuing construction.
Consider the arrangement of polygonal 3regions depicted
in Fig. 20, assigned to the 3region variables s0,...,s9,
s′
this arrangement can be made to satisfy the following formu
las:
8,...,s′
1, d0,...,d6as indicated. It is easy to verify that
frame(s0,s1,...,s8,s9,s′
(s0≤˙t0) ∧ (s9≤¨t6),
stack(d0,...,d6).
8,...,s′
1),
(16)
(17)
(18)
s′
1
s′
7
s′
8
s9
s0
s′
s′
5
s′
s′
3
2
4
s1
s2
s3
s4
s5
s7
s8
s′
6
s6
d1
d2
d3
d4
d5
d6
d0
Figure 20: A tuple of 3regions satisfying (16)–(18). The 3
regions d0and d6are shown in dotted lines.
γ4
γ′
6
γ6
˜ o2
o2
γ′
7γ′
8
γ9
γ8γ7
γ5,...,γ1
γ′
1,...,γ′
5
˜ q1
γ0
o1
˜ o1
χ1
χ2
χ3
p∗
q∗
Figure 21: The arcs γ0,...,γ9and χ1,...χ3.
And trivially, the arrangementcan be made to satisfy any for
mula
¬C(r,r′)
for which the corresponding 3regions r and r′are drawn as
not being in contact. (Remember, r is the outermost shell of
the 3region r, and similarly for r′.) Thus, for example, (19)
includes ¬C(s0,d1), but not ¬C(s0,d0) of ¬C(d0,d1).
Now suppose
s0,...,s9,
any collection of 3regions (not necessarily polygo
nal) satisfying (16)–(19).
let γ0,...,γ9,γ′
1
be Jordan arcs included in
the respective regions s0,...,s9,s′
Γ = γ0···γ9· γ′
γihave opposite directions). We select points ˜ o1on γ0and
˜ o2on γ9(see Fig. 21). By (17), ˜ o1 ∈˙t0and ˜ o2 ∈¨t6. By
Lemma 14 and (18), let ˜ χ1, χ2, ˜ χ3 be Jordan arcs in the
respective regions
(19)
s′
8,...,s′
1,
d0,...,d6
is
By Lemma 15 and (16),
8,...,γ′
8,...,s′
1, such that
iand
8···γ′
1is a Jordan curve (note that γ′
(d0+ d1),(d2+ d3+ d4),(d5+ d6)
such that ˜ χ1χ2˜ χ3is a Jordan arc from ˜ o1to ˜ o2. Let o1be the
last point of ˜ χ1lying on Γ, and let χ1be the final segment
of ˜ χ1, starting at o1. Let o2be the first point of ˜ χ3lying
on Γ, and let χ3be the initial segment of ˜ χ3, ending at o2.
By (19), we see that the arc χ = χ1χ2χ3intersects Γ only in
its endpoints, and is thus a chord of Γ, as shown in Fig. 21.
A word is required concerning the generality of this dia
gram. The reader is to imagine the figure drawn on a spheri
cal canvas, of which the sheet of paper or computer screen in
front of him is simply a small part. This sphere represents the
Page 14
a1,4
a1,5
a1,6
b1,4
b1,5
b1,6
b1,2
b1,1
b1,3
a1,1 a1,2
a1,3
b2,6
b2,5
b2,4
a2,3
a2,4
a2,5
a2,6
b2,1 b2,2
b2,3
a2,1
a2,2
b0,6
b0,5
a0,3
a0,4
a0,5
a0,6
b0,1
b0,2
b0,3
a0,1 a0,2
b0,4
s′
6
s3
s6
a
b
d3
Figure 22: A tuple of 3regions satisfying (20)–(22). The
arrangement of components of the ai,j and bi,j repeats an
indeterminate number of times. The 3regions a, b and one
component of a0,3are shown in dotted lines. The 3regions
s3, s6, s′
6and d3are as in Fig 22, but not drawn to scale.
plane with a ‘point’ at infinity, under the usual stereographic
projection. We do not say where this point at infinity is, other
than that it never lies on a drawn arc. In this way, a diagram
in which the spherical canvas is divided into n cells repre
sents n different configurations in the plane—one for each of
the cells in which the point at infinity may be located. For
example, Fig .21 represents three topologically distinct con
figurations in R2, and, as such, depicts the arcs γ0,...,γ9,
γ′
All diagrams in this proof are to be interpreted in this way.
We stress that our ‘spherical diagrams’ are simply a conve
nient device for using one drawing to represent several pos
sible configurations in the Euclidean plane: in particular, we
are interested only in the satisfiability of of Ccformulas over
RCP(R2) and RC(R2), not over the regular closed algebra of
any otherspace! For ease of reference,we refer to the the two
rectangles in Fig .21 as the ‘upper window’ and ‘lower win
dow’, it being understood that these are simply handy labels:
in particular, either of these ‘windows’ (but not both) may be
unbounded.
1,...,γ′
8, χ1, χ2, χ3and points o1, o2in full generality.
Stage 2. In this stage, we we construct two sequences of
arcs, {ζi}, {ηi} of indeterminate length n ≥ 1, such that the
members of the former sequence all lie in the lower window.
Here and in the sequel, we write ⌊k⌋ to denote k modulo 3.
Let a, b, ai,jand bi,j(0 ≤ i < 3, 1 ≤ j ≤ 6) be 3region
variables, let z be an ordinary regionvariable, and consider
the formulas
(s6≤ ¨ a) ∧ (s′
stackz(a⌊i−1⌋,3,bi,1,...,bi,6,b),
stack(bi,3,ai,1,...,ai,6,a).
6≤¨b) ∧ (s3≤ ˙ a0,3),
(20)
(21)
(22)
The arrangement of polygonal 3regions depicted in Fig. 22
(with z assigned appropriately) is one such satisfying assign
ment. We stipulate that (19) applies now to all regions de
picted in either Fig 20 or Fig 22. Again, these additional
constraints are evidently satisfiable.
It will be convenient in this stage to rename the arcs γ6
and γ′
edge of the lower window, and µ0the top edge of the upper
6as λ0and µ0, respectively. Thus, λ0forms the bottom
window. Likewise, we rename γ3as α0, forming part of the
lefthand side of the lower window. Let ˜ q1,1be any point of
α0, p∗any point of λ0, and q∗any point of µ0(see Fig. 21).
By (20), then, ˜ q1,1∈ ˙ a0,3, p∗∈ ¨ a, and q∗∈¨b. Adding the
constraint
¬C(s3,z),
further ensures that ˜ q1,1∈ −z. By Lemma 14 and (21), we
may draw an arc˜β1from ˜ q1,1to q∗, with successive segments
˜β1,1, β1,2, ..., β1,5,˜β1,6lyingin the respectiveregionsa0,3+
b1,1, b1,2, ..., b1,5, b1,6+b; further,wecanguaranteethatβ1,2
contains a point ˜ p1,1∈˙b1,3. Denote the last point of β1,5by
q1,2. Also, let q1,1be the last point of˜β1lying on α0, and q1,3
the first point of˜β1lying on µ0Finally, let β1be the segment
of˜β1between q1,1and q1,2; and we let µ1be the segment of
˜β1from q1,2to q1,3followed by the final segment of µ0from
q1,3. (Fig. 23a). By repeatedly using the constraints in (19),
it is easy to see that that β1together with the initial segment
of µ1up to q1,3form a chord of Γ. Adding the constraints
c(b0,5+ d3),
and taking into account the constraints in (19) ensures that β1
and χ lie in the same residual domain of Γ, as shown. The
wiggly lines indicate that we do not care about the exact po
sitions of ˜ q1,1or q∗; otherwise, Fig. 23a) is again completely
general. Note that µ1lies entirely in b1,6+ b, and hence cer
tainly in the region
b∗= b + b0,6+ b1,6+ b2,6.
Recall that ˜ p1,1 ∈˙b1,3, and p∗∈ ¨ a.
and(22), we maydrawanarc ˜ α1from ˜ p1,1to p∗, with succes
sive segments ˜ α1,1, α1,2, ..., α1,5, ˜ α1,6lying in the respec
tive regions b1,3+ a1,1, a1,2, ..., a1,5a1,6+ a; further, we
can guarantee that the segment lying in a1,3contains a point
˜ q2,1∈ ˙ a1,3. Denote the last point of α1,5by p1,2. Also, let
p1,1be the last point of ˜ α1lying on β1, and p1,3the first point
of ˜ α1lying on λ0. From (19), these points must be arranged
as shown in Fig. 23b. Let α1be the segment of ˜ α1between
p1,1and p1,2. Noting that (19) entails
By Lemma 14
¬C(a1,k,s0+ s9+ d0+ ··· + d5)
we can be sure that α1lies entirely in the ‘lower’ window,
whence β1crosses the central chord, χ, at least once. Let o1
be the first such point (measured along χ from left to right).
Finally, let λ1be the segment of ˜ α1between p1,2and p1,3,
followed by the final segment of λ0from p1,3. Note that λ1
lies entirely in a1,6+ a, and hence certainly in the region
a∗= a + a0,6+ a1,6+ a2,6.
We remark that, in Fig. 23b, the arcs β1and µ1have been
slightly redrawn,for clarity. The regionmarkedS1may now
be forgotten, and is suppressed in Figs. 23c and 23d.
By construction, the point ˜ q2,1lies in some component of
˙ a1,3, and, from the presence of the ‘switching’ variable z
in (22), that component is either included in z or included
in −z. Suppose the latter. Then we can repeat the above
construction to obtain an arc˜β2from ˜ q2,1to q∗, with succes
sive segments˜β2,1, β2,2, ..., β2,5,˜β2,6lying in the respective
1 ≤ k ≤ 6,
Page 15
q1,3
µ0
q∗
q1,2
S1
χ
β1
µ1
˜ q1,1
q1,1
(a) The arc β1.
µ1
q1,2
β1
q1,1
˜ p1,1
λ0
α1
p1,1
λ1
S1
p1,3
p∗
p1,2
˜ q2,1
R1
q1,3
(b) The arc α1.
q1,2
p1,1
p1,2
q2,2
q2,3
q2,1
˜ q2,1
µ1
R1
S2
p1,3
β2
λ1
µ2
(c) The arc β2.
µ2
q2,2
p2,3
λ2
S2
q2,3
R2
χ
p1,3
β2
α2
˜ q3,1
p2,3
(d) The arc α2.
Figure 23: Construction of the arcs {αi} and {βi}
α1
β1
β2
β3
α2
α3
αn
βn
Figure 24: The sequences of arcs {αi} and {βi}.
regions a1,3+ b2,1, b2,2, ..., b2,5, b2,6+ b; further, we can
guarantee that β2,2contains a point ˜ p2,1∈˙b2,3. Denote the
last point of β2,5by q2,2. Also, let q2,1be the last point of˜β2
lying on α1, and q2,3the first point of˜β2lying on µ1. Again,
we let β2be the segment of˜β2between q2,1and q2,2; and we
let µ2be the segment of˜β2from q2,1to q2,3, followed by the
final segment of µ1from q2,3. Note that µ2lies in the set b∗.
It is easy to see that β2must be drawn as shown in Fig. 23c:
in particular,β2cannotenter the interiorof the regionmarked
R1. For, by construction, β2can have only one point of con
tact with α1, and the constraints (19) ensure that β2cannot
intersect any other part of δR1; since q∗∈ a is guaranteed to
lie outside R1, we evidently have β2⊆ −R1. This observa
tion having been made, R1may now be forgotten.
Symmetrically, we construct the arc ˜ α2 ⊆ b1,3+ a2,1+
··· + a2,6+ a, and points p2,1, p2,2, p2,3, together with the
arcs arcs α2and λ2, as shown in Fig. 23d (where the region
R1has been suppressed and the region S2slightly redrawn).
Again, we know from (19) that α2lies entirely in the ‘lower’
window, whence β2must cross the central chord, χ, at least
once. Let o2be the first such point (measured along χ from
left to right).
This process continues, generating arcs βi ⊆ a⌊i−1⌋,3+
b⌊i⌋,1+ ··· + b⌊i⌋,5and αi⊆ b⌊i⌋,3+ a⌊i⌋,1+ ··· + a⌊i⌋,5,
as long as αi contains a point ˜ qi,1 ∈ −z. That we even
tually reach a value i = n for which no such point exists
follows from (19). For the conjuncts ¬C(bi,j,dk) (j ?= 5)
together entail oi∈ b⌊i⌋,5, for every i such that βiis defined;
and these points cycle on χ through the regions b0,5, b1,5and
b2,5. If there were infinitely many βi, the oiwould have an
accumulation point, lying in all three regions, contradicting,
say, ¬C(b0,5,b1,5). The resulting sequence of arcs and points
is shown, schematically, in Fig. 24.
We finish this stage in the construction by ‘repackaging’
the arcs {αi} and {βi}, as illustrated in Fig. 25. Specifically,
for all i (1 ≤ i ≤ n), let ζibe the initial segment of βiup
to the point pi,1followed by the initial segment of αiup to
the point qi+1,1; and let ηibe the final segment of βifrom the
point pi,1:
ζi= βi[qi,1,pi,1]αi[pi,2,qi+1,1]
ηi= βi[pi,1,qi,2].
The final segment ofαifromthe point qi+1may be forgotten.
Defining, for 0 ≤ i < 3,
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 Available from Roman Kontchakov · Oct 25, 2012
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