# On the Decidability of Connectedness Constraints in 2D and 3D Euclidean Spaces

**ABSTRACT** We investigate (quantifier-free) spatial constraint languages with equality,

contact and connectedness predicates as well as Boolean operations on regions,

interpreted over low-dimensional 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

interior-connectedness predicate (and without contact) is undecidable over

polygons or regular closed sets in the Euclidean plane, NP-complete over

regular closed sets in three-dimensional Euclidean space, and ExpTime-complete

over polyhedra in three-dimensional Euclidean space.

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**ABSTRACT:**We consider the quantifier-free languages, Bc and Bc0, obtained by augmenting the signature of Boolean algebras with a unary predicate representing, respectively, the property of being connected, and the property of having a connected interior. These languages are interpreted over the regular closed sets of n-dimensional Euclidean space (n greater than 1) and, additionally, over the regular closed polyhedral sets of n-dimensional Euclidean space. The resulting logics are examples of formalisms that have recently been proposed in the Artificial Intelligence literature under the rubric "Qualitative Spatial Reasoning." We prove that the satisfiability problem for Bc is undecidable over the regular closed polyhedra in all dimensions greater than 1, and that the satisfiability problem for both languages is undecidable over both the regular closed sets and the regular closed polyhedra in the Euclidean plane. However, we also prove that the satisfiability problem for Bc0 is NP-complete over the regular closed sets in all dimensions greater than 2, while the corresponding problem for the regular closed polyhedra is ExpTime-complete. Our results show, in particular, that spatial reasoning over Euclidean spaces is much harder than reasoning over arbitrary topological spaces.10/2011; - SourceAvailable from: Matthias Westphal
##### Conference Paper: Transition Constraints: A Study on the Computational Complexity of Qualitative Change

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**ABSTRACT:**Many formalisms discussed in the literature on qualitative spatial reasoning are designed for expressing static spatial constraints only. However, dynamic situations arise in virtually all applications of these formalisms, which makes it necessary to study variants and extensions dealing with change. This paper presents a study on the computational complexity of qualitative change. More precisely, we discuss the reasoning task of finding a solution to a temporal sequence of static reasoning problems where this sequence is subject to additional transition constraints. Our focus is primarily on smoothness and continuity constraints: we show how such transitions can be defined as relations and expressed within qualitative constraint formalisms. Our results demonstrate that for point-based constraint formalisms the interesting fragments are NP-complete in the presence of continuity constraints, even if the satisfiability problem of its static descriptions is tractable.IJCAI; 01/2013

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 Pratt-Hartmann2and 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 (quantifier-free) spatial constraint

languageswith equality,contact and connectedness

predicates, as well as Boolean operations on re-

gions, interpreted over low-dimensional Euclidean

spaces. We show that the complexity of reasoning

varies dramatically depending on the dimension of

thespaceandonthetypeofregionsconsidered. For

example, the logic with the interior-connectedness

predicate (and without contact) is undecidable over

polygons or regular closed sets in R2, EXPTIME-

complete over polyhedra in R3, and NP-complete

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 non-logical 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 best-known 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 RCC8-formula is satisfiable in n-dimensional Euclidean

space, it suffices to demonstrate its satisfiability in any topo-

logical space[Renz, 1998]; for BRCC8-formulas, satisfiabil-

ity in any connected space is enough. This inexpressiveness

Introduction

yields (relatively) low computational complexity: satisfiabil-

ity of BRCC8-, RCC8- and RCC5-formulas over arbitrary

topological spaces is NP-complete; satisfiability of BRCC8-

formulas over connected spaces is PSPACE-complete.

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 [Pratt-Hartmann, 2007].

Unfortunately, little is known about the complexity of topo-

logical constraint satisfaction by non-pathological objects

in low-dimensional Euclidean spaces.

sult[Schaefer et al., r003]in this area showsthat satisfiability

of RCC8-formulas by disc-homeomorphs 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 NP-complete, while for the

other three languages, it is EXPTIME-complete. 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 EXPTIME-hard (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 EXPTIME-complete.

(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 first-order

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, non-emptysets closed in the subspace topology;

X is interior-connected if X◦is connected.

Any (n−1)-dimensionalhyperplanein Rn, n ≥ 1, bounds

two elements of RC(Rn) called half-spaces. We denote by

RCP(Rn) the Boolean subalgebra of RC(Rn) generated by

the half-spaces, and call the elements of RCP(Rn) (regular

closed) polyhedra. If n = 2, we speak of (regular closed)

polygons. Polyhedramayberegardedas‘well-behaved’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

non-atomic 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 L-formulas 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 non-empty,

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 non-planar. 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 Bc-formula

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 non-empty

regions aicontained in di, and a non-empty 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 Cc-formula 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 non-empty, 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 Cc-formula ϕ∞can be transformed

to Cc◦- and Bc-formulas 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 interior-connected;henceϕ◦

Since interior-connectedness 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 Bc-formula,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

non-contact relations similarly, we obtain a Bc-formula ψ∞

that is satisfiable over RC(Rn), and that entails ϕ∞. Thus:

Corollary 4 There is a Bc-formula 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 non-contactconstraints 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 non-empty for any t ∈ T. The in-

stance w is positive if there exists a non-empty τ ∈ T∗such

that w1(τ) = w2(τ). The set of positive PCP-instances 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 3-region 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 3-regions a1,a2,a3depicted in Fig. 4, where the inner-

most regions form a sequence of externally touching poly-

gons. When describing arrangements of 3-regions, we use

Undecidability in the Plane

a1

a2

a3

˙ a1

˙ a2

˙ a3

¨ a1

¨ a2

¨ a3

Figure 4: A chain of 3-regions 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 3-regions 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 Cc-formulas whose sat-

isfying regions are guaranteed to contain various networks of

arcs, exhibiting almost any desired pattern of intersections.

thetriple of3-regionsinFig.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 arc-sequences,{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 arc-sequences. 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 Cc-constraints 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 non-contact constraints involving the arcs ζiensur-

ing that σ = w1(τ) and non-contact 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 Cc-formulas. 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 graph-theoretic

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 quasi-saw (Fig. 8). If, in ad-

dition, no x ∈ W has more than two properR-successors, we

call (W,R)a 2-quasi-saw. Thepropertiesof2-quasi-sawswe

need are as follows[Kontchakov et al., 2010a]:

– satisfiability of Bc-formulas in arbitrary topological

spaces coincides with satisfiability in 2-quasi-saws, and

is EXPTIME-complete;

– X ⊆ W is connected in a 2-quasi-saw (W,R) iff it is

interior-connected 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 non-empty 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 2-quasi-saw, 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 2-quasi-saw 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, issatisfiableoveraconnected2-quasi-saw. Thus,

Sat(Bc◦,RCP(R3)) is EXPTIME-complete.

The picture changes if we allow variables to range over

RC(R3) ratherthanRCP(R3). NotefirstthattheBc◦-formula

(2) is not satisfiable over 2-quasi-saws, but has a quasi-saw

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 quasi-saw 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 quasi-saw. And as shown

in [Kontchakov et al., 2010a], satisfiability of Bc◦-formulas

in connected spaces coincides with satisfiability over con-

nected quasi-saws, and is NP-complete.

Theorem 7 The problem Sat(Bc◦,RC(R3)) is NP-complete.

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 quasi-saw (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 non-empty, 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 non-emptiness or non-internal-connectedness

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 ‘face-contact’ 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 non-contact 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

NP-complete, which contrasts with EXPTIME-completeness

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 disc-homeomorphs

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 non-trivial 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.

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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, non-empty,

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 interior-connected;henceϕ◦

Since interior-connectedness 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 Bc-formula 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 Bc-formula 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 non-degenerate 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 non-degenerate.

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 non-contact 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 quasi-bounded 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 quasi-bounded

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 quasi-boundedpolygons,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 sub-regions. We denote the sub-region whose boundary

is disjoint from α0by U0, and the other sub-regionwe 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 cross-cut 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 non-overlapping con-

0except at its endpoints, and hence it is a cross-cut in

0since

the boundary of U0 is disjoint from α0. So, α0,1α0,2α0,3

divides U′

whose boundarycontains β1by W0, and the other sub-region

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 cross-cut 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

cross-cut in V0or W0, and since the boundary of V0is dis-

joint from β1it has to be a cross-cut 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 sub-regions. We denote the sub-region

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 twosub-regionsV1and

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 non-contact 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 3-regions

Then, for every point

satisfying

1∪ {p1}, i.e., the arc β1[p0,p1] is either

1, or is an end-cut 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 3-region (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 ‘de-activate’

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 non-empty, then X ∩ t1

and X ∩ (t2+ ··· + tℓ) partition X into non-empty, 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 PCP-instance w = ({0,1},T,w1,w2)

be given, where T is a finite alphabet, and wi: T∗→ {0,1}∗

a word-morphism (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 non-empty. 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 Cc-literals, 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 3-regions depicted

in Fig. 20, assigned to the 3-region 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 3-regions 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 3-regions r and r′are drawn as

not being in contact. (Remember, r is the outer-most shell of

the 3-region 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 3-regions (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 3-regions satisfying (20)–(22). The

arrangement of components of the ai,j and bi,j repeats an

indeterminate number of times. The 3-regions a, b and one

component of a0,3are shown in dotted lines. The 3-regions

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 Cc-formulas 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 3-region

variables, let z be an ordinary region-variable, 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 3-regions 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

left-hand 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 re-drawn,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 re-drawn).

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 ‘re-packaging’

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 · Jun 1, 2014
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