Plane augmentation of plane graphs to meet parity constraints
PPlane augmentation of plane graphs to meet parity constraints
J.C. Catana ∗ A. Garc´ıa † J. Tejel ‡ J. Urrutia § Abstract
A plane topological graph G = ( V, E ) is a graph drawn in the plane whose vertices are pointsin the plane and whose edges are simple curves that do not intersect, except at their endpoints.Given a plane topological graph G = ( V, E ) and a set C G of parity constraints, in which every vertexhas assigned a parity constraint on its degree, either even or odd, we say that G is topologicallyaugmentable to meet C G if there exits a plane topological graph H on the same set of vertices, suchthat G and H are edge-disjoint and their union is a plane topological graph that meets all parityconstraints.In this paper, we prove that the problem of deciding if a plane topological graph is topologicallyaugmentable to meet parity constraints is N P -complete, even if the set of vertices that must changetheir parities is V or the set of vertices with odd degree. In particular, deciding if a plane topo-logical graph can be augmented to a Eulerian plane topological graph is N P -complete. Analogouscomplexity results are obtained, when the augmentation must be done by a plane topological perfectmatching between the vertices not meeting their parities.We extend these hardness results to planar graphs, when the augmented graph must be planar,and to plane geometric graphs (plane topological graphs whose edges are straight-line segments). Inaddition, when it is required that the augmentation is made by a plane geometric perfect matchingbetween the vertices not meeting their parities, we also prove that this augmentation problem is
N P -complete for plane geometric trees and paths.For the particular family of maximal outerplane graphs, we characterize maximal outerplanegraphs that are topological augmentable to satisfy a set of parity constraints. We also provide apolynomial time algorithm that decides if a maximal outerplane graph is topologically augmentableto meet parity constraints, and if so, produces a set of edges with minimum cardinality.
Keywords:
Plane topological graphs, plane geometric graphs, planar graphs, augmentation prob-lems,
N P -complete problems, outerplanar graphs. An embedding of a graph G = ( V, E ) on the plane is a drawing of G on the plane in which its verticesare represented by points and its edges by simple curves joining pairs of adjacent vertices of G . We shallrefer to these embeddings as topologial graphs . A geometric graph is a topological graph whose edges arerepresented by straight-line segments. A topological graph is plane if the curves representing its edgesdo not intersect, except at their endpoints. Two plane topological graphs G = ( V, E ) and H = ( V, E (cid:48) )on the same set of vertices are called compatible if their union is a plane topological graph.Let G be a plane topological graph with n vertices v , . . . , v n . Suppose that each vertex v i of G hasbeen assigned a parity constraint c i , that is, c i is either even or odd . The set C G = { c , c , ..., c n } willbe called the set of parity constraints of G . The plane topological augmentation problem of G to meet C G = { c , c , ..., c n − } is that of finding a plane topological graph H such that G and H are compatible ∗ [email protected] . Universidad Nacional Aut´onoma de M´exico, Mexico. † [email protected] . IUMA, Universidad de Zaragoza, Spain. ‡ [email protected] . IUMA, Universidad de Zaragoza, Spain. § [email protected] . Universidad Nacional Aut´onoma de M´exico, Mexico.This project has received funding from the European Union’s Horizon 2020 researchand innovation programme under the Marie Sk(cid:32)lodowska-Curie grant agreement No734922. a r X i v : . [ c s . C G ] J u l (a) vP (b) P P (c) Figure 1: Plane geometric augmentation problems to meet a set of parity constraints. Vertices in R aredepicted as red points and the rest as blue points. We use thick lines for the given plane geometric graphand dashed lines for the added edges.and edge-disjoint, and the degree of every vertex in G ∪ H meets its parity constraint in C G . If thereexists such a graph H , then we say that G is topologically augmentable to meet C G .Analogously, given a plane geometric graph G and a set C G of parity constraints, the plane geometricaugmentation problem of G to meet C G is that of finding a plane geometric graph H such that G and H are compatible and edge-disjoint, and the degree of every vertex in G ∪ H meets its parity constraintin C G . If there exists such a graph H , then we say that G is geometrically augmentable to meet C G .Hereafter we denote by R ( C G ) (or simply R if there is no ambiguity) the set of vertices in G notsatisfying their parity constraints in C G , that is the set of vertices in G that must change their parities.For simplicity in further reasoning, a vertex v i ∈ R ( C G ) is called a red vertex , otherwise it is called a bluevertex . We say that an edge of G is a red edge (respectively blue edge ), if both of its endpoints are redvertices (respectively blue). An edge of G is called a red-blue edge if its endpoints have different colors.Figure 1 shows some examples of plane geometric augmentation problems to meet a set of parityconstraints. In Figure 1a, a plane geometric tree T is augmented to a Eulerian plane geometric graph(all vertices have even degree), by adding a plane geometric perfect matching between the vertices withodd degree in T . Figure 1b shows a plane geometric path P such that there is only one way of changingthe parities of all vertices except v . In Figure 1c, the parities of the ten red vertices cannot all be changedsimultaneously. These vertices define two empty convex pentagons P and P , and the parity of onlytwo vertices in each pentagon can be changed using diagonals. This construction can be generalized byadding more empty convex pentagons, and the resulting n -vertex graph is a plane geometric graph suchthat the parity constraints of at most n vertices can be satisfied in any plane augmentation.Note that if a plane topological graph G can be augmented to a plane topological graph G (cid:48) = G ∪ H to satisfy a set C G of parity constraints, the degree in H of all red vertices in G is odd, while the degreeof all blue vertices is even. It follows that the number of red vertices of G must be even since any graphhas an even number of odd vertices. Moreover, if H has as few edges as possible, then H is a forest.Otherwise, if a connected component of H contains a cycle C , all of the edges of C could be removedfrom H without changing the parities of the vertices in G ∪ H . Therefore, the number of edges in H is at least | R ( C G ) | /
2. If H has | R ( C G ) | / H is a perfect matching between the vertices in R ( C G ). Related previous work
Graph augmentation is a family of problems in which one would like to add new edges, ideally as fewas possible, to a given graph, in such a way, that some desired property is achieved. Connectivityaugmentation is one of the most studied augmentation problems, due to a wide range of applications indesigning reliable networks and database systems.Connectivity augmentation in abstract graphs was first studied by Eswaran and Tarjan in 1976 (andindependently by Plesn´ık at the same time [27]). They obtained two polynomial time algorithms tooptimally augment a graph to a 2-connected or a 2-edge-connected graph [17]. In the same paper,Eswaran and Tarjan also studied the problem of augmenting weighted oriented graphs to achieve strong2onnectivity, proving that this problem is
N P -complete. Since then, numerous papers about this topichave been published, studying efficient algorithms to increase the vertex- and edge-connectivity (see forexample [31]) or giving approximation algorithms for hard versions of the problem (see for example [24]).The reader can consult [18] for a broad overview on these connectivity augmentation problems.For planar graphs, Kant and Bodlaender [23] proved that finding a minimal set of edges that makes aplanar connected graph planar and 2-connected is
N P -complete. Later, Rutter and Wolff [28] extendedthis result by showing the
N P -completeness of the same problem replacing 2-connected by 2-edge-connected. For plane geometric graphs, these problems remain hard even if restricted to trees [28]. Thereader is referred to [1, 6, 7, 20, 25, 28, 29] for different results about connectivity augmentation problemsin plane geometric graphs, and the survey [21] for more details and related topics. Recent research oncompatible plane graphs can be found in [2–4, 19, 22].Augmentation to meet parity constraints has also been studied in the literature, as a weaker versionof the more general classic problem of augmenting a graph to meet a given degree sequence. Parityconstraints characterize some types of graphs such as Eulerian graphs or maximal planar graphs that are3-colorable (in both cases, connected graphs where all vertices have even degree). For abstract graphs,Dabrowski et al. [13] presented a polynomial algorithm to solve the augmentation problem to meet a setof parity constraints. This result extends the polynomial algorithm for the particular case of augmentinga graph to Eulerian [10]. Dabrowski et al. [13] also provided a polynomial algorithm to obtain a graphmeeting the parity constraints, when addition and removal of edges are allowed. More results on editinggraphs to make them Eulerian are available [11–13, 15]. The relevance of Eulerian graphs relies on awide range of applications to some problems in different areas such as route inspection problems, DNAfragmentation problems, scheduling problems, or the design of CMOS VLSI circuits. The reader canalso see [14, 32] for other related results.The geometric setting of the problem was also been studied [5,9]. Given a point set S in the plane anda set C S of parity constraints on the points of S , Aichholzer et al. [5] proved that it is always possibleto build up a plane tree and a 2-connected outerplanar graph satisfying all parity constraints, and apointed pseudotriangulation that satisfy all but at most three parity constraints. For triangulations,they showed examples in which a linear number of parity constraints cannot be satisfied, and theyprovided triangulations satisfying about n of the parity constraints. Later, Alvarez [9] showed that itis always possible to build up an even triangulation (a triangulation having all its vertices with evendegree) on top of a point set S ∪ S (cid:48) , where S is a n -point set in general position and S (cid:48) a set of Steinerpoints of at most n + c points, with c being a positive integer constant. Our results
In this paper, we study the plane augmentation problem to meet parity constraints. In Section 2, wefocus on the topological plane augmentation problem. Given a plane topological graph G = ( V, E ) anda set C G of parity constraints, we show that deciding if G is topologically augmentable to meet C G is N P -complete, even if R ( C G ) coincides with V or is the set of vertices with odd degree in G . In particular,deciding if a plane topological graph can be augmented to a Eulerian plane topological graph is N P -complete. The same complexity results are obtained, when the augmentation must be done by a planetopological perfect matching between the red vertices. Hence, deciding if G admits a plane topologicalperfect matching, compatible with G and edge-disjoint, is also N P -complete. As a side result, we alsoextend these
N P -completeness results to planar graphs.Section 3 is devoted to the geometric augmentation problem to meet parity constraints. As a con-sequence of the results shown in Section 2, we can prove the
N P -completeness of all the problemspreviously described, when the input graph and the augmenting graph are plane and geometric. Inaddition, when it is required that the geometric augmentation is made by a plane geometric perfectmatching between the red vertices, we also prove that deciding if a plane geometric tree (or path) isgeometrically augmentable to meet parity constraints is
N P -complete. In particular, the two followingproblems are also
N P -complete: Deciding if a plane geometric tree (or path) admits a plane geometricperfect matching, compatible and edge-disjoint, and deciding if a plane geometric tree is geometricallyaugmentable to Eulerian by a plane geometric perfect matching between the vertices with odd degree.Due to the hardness of the plane topological augmentation problem in general, in Section 4 we addressthis problem for a particular family of plane topological graphs, the family of maximal outerplane graphs.For this family, we characterize maximal outerplane graphs that are topological augmentable to satisfy3opological augmentationDecision MatchingAny
R R = V Eulerian Any
R R = V EulerianPlane topological graph
NP NP NP NP NP NP
Maximal outerplane graph
P P P P P P
Geometric augmentationDecision MatchingAny
R R = V Eulerian Any
R R = V EulerianPlane geometric graph
NP NP NP NP NP NP
Plane geometric tree ? ? ? NP NP NP
Plane geometric path ? ? ? NP NP P
Table 1: Summary of our results for the plane augmentation problem to meet a set of parity constraints.We distinguish if the augmentation is topological or geometric and if the input graph G ( V, E ) is aug-mentable by an arbitrary graph (Decision) or by a plane perfect matching between the red vertices(Matching). We also distinguish the cases that the set R of red vertices is arbitrary, coincides with V oris the set of odd vertices in G (Eulerian).all parity constraints, and we provide an O ( n ) time algorithm to find a minimum plane topologicalaugmentation, if it exists. Given a maximal outerplane graph, we also show that there is always a planetopological matching such that the union of the maximal outerplane graph and the matching satisfies allbut at most four parity constraints. Table 1 summarizes our results for the plane augmentation problemto meet parity constraints. In this section we address the plane topological augmentation problem to meet a set of parity constraints,that is, given a plane topological graph G = ( V, E ) and a set C G of parity constraints, we look for aplane topological graph H such that G and H are compatible and edge-disjoint, and all parity constrainsare satisfied in G ∪ H . We show the N P -completeness of several variants of this problem and we extendthese results to planar graphs.We first recall the Planar 3-SAT problem. This problem will play a key role in the forthcomingtheorems. Let Φ be a Boolean formula in conjunctive normal form and let F Φ be the incidence graph ofΦ, that is, the bipartite graph whose vertices are the variables and the clauses of Φ and an edge connectsa variable and a clause if and only if the variable (negated or unnegated) occurs in the clause. A formulais a 3-SAT formula if every clause contains at most three literals, where a literal is either a variable(called positive literal) or the negation of a variable (called negative literal). The Planar 3-SAT problemasks whether a given 3-SAT formula Φ is satisfiable, assuming that F Φ is planar. This problem has beenshown to be N P -complete [26].Given a formula Φ, suppose that F Φ is planar and F is a plane embedding of F Φ . In this case, observethat if l , l , . . . , l k is the set of literals of a variable x in Φ, then the clockwise cyclic order around x of the clauses adjacent to x in F also implies an order for the literals of x , that is, two literals l i and l i +1 are consecutive if the clauses in F where they occur are consecutive in the cyclic order around x .In the rest of this section, L x = { l , l , . . . , l k } will denote the ordered list of literals of a variable x ,according to a plane embedding F . Figure 2 shows a plane embedding of F Φ for the 3-SAT formulaΦ = (¯ x ∨ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ), where ¯ x means the negation of x . The (cyclic) orderedlist of literals of x is L x = { x , ¯ x , x } .We next show the N P -completeness of the plane topological augmentation problem.
Theorem 2.1.
Let G = ( V, E ) be a plane topological graph, and let C G be a set of parity constraints.Then, the problem of deciding if G is topologically augmentable to meet C G is N P -complete.Proof.
The plane topological augmentation problem is trivially in
N P since, given a set of edges E (cid:48) ,verifying if G (cid:48) = ( V, E ∪ E (cid:48) ) is a plane topological graph meeting C G takes polynomial time.4 x x x ( x ∨ ¯ x ∨ ¯ x )(¯ x ∨ x ∨ ¯ x ) ( x ∨ ¯ x ∨ ¯ x ) Figure 2: A plane embedding of F Φ for the formula Φ = (¯ x ∨ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ).The proof is based on a reduction from Planar 3-SAT. Given a 3-SAT formula Φ and a planeembedding F of F Φ , we build in polynomial time a plane topological graph G Φ and a set C G Φ ofparity constraints such that G Φ is topologically augmentable to meet C G if and only if Φ is sat-isfiable. The graph G Φ is built from F , replacing clause vertices by clause gadgets, variable ver-tices by variable gadgets and edges by pairs of wire gadgets. Figure 5 illustrates G Φ for the formulaΦ = (¯ x ∨ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ); the plane embedding of F Φ shown in Figure 2.Without loss of generality, we can assume that every clause in Φ consists of three literals. The basic gadget in our reduction (see Figure 3a) consists of a plane topological graph that has only twopossible plane topological augmentations to change the parities of all red interior vertices. These twoaugmentations are plane matchings that match all red vertices of the gadget except two. The negative augmentation (depicted in the figures with red dashed edges) does not match the exterior vertices v r and v l (see Figure 3a) and the positive augmentation (depicted with blue dashed edges) does not matchthe exterior vertices v t and v b (Figure 3b).A literal gadget is obtained by the union of two basic gadgets joined as shown in Figure 3b. Asbefore, one can easily verify that there are only two possible plane topological augmentations (planematchings) to change the parities of all red interior vertices. Note that if we choose the (negative)positive augmentation for one of the two basic gadgets, then we must also choose the (negative) positiveaugmentation for the other one. Figure 3b illustrates the positive augmentation of a literal gadget, inwhich positive augmentations for the basic gadgets are used in the augmentation, and where the exteriorvertices v t and v (cid:48) t remain without changing their parities. In the negative augmentation of a literal gadget,negative augmentations for the two basic gadgets appear and v l and v (cid:48) r remain without changing theirparities. The two exterior red vertices v l , v (cid:48) r will be used to concatenate literal gadgets corresponding tothe same variable, one after another. The two exterior red vertices v t , v (cid:48) t , called the output of the literalgadget, will be used to connect the literal gadget to a clause gadget (not defined yet).Observe that in the literal gadget shown in Figure 3b, the two bottom red vertices, each in a differentbasic gadget, are enclosed in a quadrilateral face, in such a way that if both vertices are free, then theyare forced to join each other in order to meet their parity constraints. This kind of structure that consistsof eight vertices (only two of them being red) and encloses its two red vertices is called a wire gadget .In Figure 3b, the wire gadget consists of the three bottom vertices of each basic gadget, plus two extrablue vertices in the middle of the gadget to define the quadrangular face. Wire gadgets will be also usedto connect literal gadgets to clause gadgets.Let L x = { l , l , . . . , l k } be the ordered list of literals of a variable x , according to F . We define the variable gadget associated with x as a sequence of k literal gadgets joined as follows: If two consecutiveliterals l i and l i +1 are both positive or negative, then their corresponding literal gadgets are joined asillustrated in Figure 4a. Otherwise, they are joined as illustrated in Figure 4b. In this way, the sameaugmentation for a literal gadget is transmitted to the following one in the first case, and the oppositeaugmentation is transmitted to the following literal gadget in the second case. We also join the leftmostexterior vertex of the first literal gadget to the rightmost exterior vertex of the last literal gadget with awire gadget, when l and l k are both positive or negative. If l is positive and l k is negative or viceversa,then we join these two red vertices with a double wire gadget , that is, a gadget obtained by gluing twowire gadgets (see the variable gadget corresponding to x in Figure 5).By construction, it is straightforward to see again that there are only two possible plane topologicalaugmentations to change the parities of all red interior vertices in a variable gadget. In addition, all5 a) (b) (c) Figure 3: (a) The basic gadget. It admits only two possible plane augmentations of its red interiorvertices. The negative augmentation is shown with red dashed edges. (b) A literal gadget with a positiveaugmentation. (c) A clause gadget. (a) (b)
Figure 4: (a) Union of two literal gadgets when their corresponding consecutive literals in L x are bothpositive or negative. (b) Union of two literal gadgets when one of their corresponding literals is positiveand the other negative.literal gadgets corresponding to positive literals in L x must have the same literal gadget augmentation,while all literal gadgets corresponding to negative literals must have the opposite one. See Figure 5. Theaugmentation of a variable gadget with positive augmentations for the literal gadgets corresponding topositive literals in L x is called positive , and negative otherwise.A clause gadget is a graph as shown in Figure 3c, whose main part is a decagon having all its vertices,say from v to v , colored in red. We identify the vertices v and v as the first input, v and v as thesecond input, and v and v as the third input. An input will be connected to the output of a literalgadget as we will explain later.From F , we build G Φ as follows. A clause in F is replaced by a clause gadget. A variable x in F is replaced by a variable gadget, where the order of the literal gadgets corresponds to the order of theliterals in L x . An edge in F is replaced by two wire gadgets connecting the red vertices of an input tothe red vertices of an output, Finally, the exterior of every gadget is triangulated. See Figure 5. C G Φ is chosen such that R ( C G Φ ) is the set of red vertices of G Φ . Note that G Φ is topological and plane andany edge that is added to G Φ must be inside a gadget.From this construction, one can easily prove that G Φ is topologically augmentable to meet C G if andonly if Φ is satisfiable. Suppose first that Φ is satisfiable. We can augment G Φ as follows. If a variable x takes the “true” value, then we add to G Φ a positive augmentation for the corresponding variablegadget. Otherwise, we add a negative augmentation. In a positive augmentation of a literal gadget, thetwo vertices of its output do not meet their parity constraints and must be connected to the two verticesof an input through the wire gadgets. Since Φ is satisfiable, at least one literal in each clause is true,implying that the two red vertices of the output of this true literal are connected to the two vertices, say v , v , of one of the inputs of the clause. Thus, we can add a star centered at v to meet the parities ofthe red vertices in the clause gadget not yet meeting their parities. Hence, G Φ is augmentable.On the contrary, suppose that G Φ is topologically augmentable to meet C G . A variable gadgetcorresponding to a variable x can be augmented only by a positive or negative augmentation. This6igure 5: The graph G Φ for the formula Φ = (¯ x ∨ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ) ∧ ( x ∨ ¯ x ∨ ¯ x ) and the planeembedding F of F Φ shown in Figure 2. The exterior of the gadgets is triangulated. The plane topologicalaugmentation shown in the figure correspond to the following assignment ( x , x , x , x ) = ( T, T, F, T ).implies the “true” value for x in the first case and the “false” value in the second case. On the otherhand, the parities of the ten red vertices of a clause gadget cannot be changed using only edges inside thedecagon. The decagon and the addition of any set of non-crossing diagonals inside it define a biconnectedouterplanar graph. It is well known that a biconnected outerplanar graph always has at least two verticesof degree 2. Thus, if there is a plane augmentation for a clause gadget, then at least two vertices mustbe connected to output vertices through the wire gadgets to meet their parities. Given a gadget clausecorresponding to a clause c , if a vertex of an input, say v , is connected in G Φ to a vertex of an output,necessarily v is also connected to the same output. Hence, the literal gadget containing this output willhave a positive augmentation in G Φ , implying that c has at least a true literal. Therefore, Φ is satisfiableand the theorem follows.In the previous theorem, R ( C G ) ⊂ V . Let us strengthen the previous result to the special case R ( C G ) = V . Theorem 2.2.
Let G = ( V, E ) be a plane topological graph and let C G be a set of parity constraints.Suppose that R ( C G ) = V . Then, the problem of deciding if G is topologically augmentable to meet C G is N P -complete.Proof.
The proof is again by reduction from Planar 3-SAT. Given a 3-SAT formula Φ and a planeembedding F of F Φ , we build in polynomial time a plane topological graph G (cid:48) Φ such that R ( G (cid:48) Φ ) = V and G (cid:48) Φ is topologically augmentable to meet C G (cid:48) Φ if and only if Φ is satisfiable. Equivalently, we provethat G (cid:48) Φ is topological augmentable if and only if G Φ is topologically augmentable, where G Φ is thegraph built in Theorem 2.1.From G Φ , we show how to change all blue vertices in G Φ to red vertices in G (cid:48) Φ , keeping the rest ofthe vertices as red vertices. Each blue vertex in G Φ is adjacent to at least one triangular face, so we cando a mapping of each blue vertex in G Φ to one of its adjacent triangular faces in G Φ .Let ∆ = ( v , v , v ) be a triangular face in G Φ . We have the following two cases: Only one bluevertex, say v , is assigned to ∆, or more than one blue vertex is assigned to ∆. In the first case, wecan change the color of v and preserve the parities of v , v , by adding three red vertices v (cid:48) , v (cid:48) , v (cid:48) andconnecting them as shown in Figure 6a. Observe that in any plane augmentation, v (cid:48) and v (cid:48) are forcedto join each other to meet their parities. In the same way, v is forced to join v (cid:48) , assuming that v isred. In the second case, more than one blue vertex is assigned to ∆, suppose that v and v are blue.We can change the color of v by adding the previous construction in ∆, and the color of v by addingthe construction in the triangle ( v , v (cid:48) , v ). If v is also blue, then we add again the construction in thetriangle ( v (cid:48) , v (cid:48) , v ) to change the color of v . 7 v v v (cid:48) v (cid:48) v (cid:48) (a) u v i +1 v i (b) Figure 6: (a) Changing the color of v by adding three red vertices in a triangular face ∆. (b) Thetriangle ( v i , v i +1 , u ) with the addition of 5 blue vertices and 14 edges, to replace the duplicated edge e (cid:48) i = ( v i , v i +1 ).Obviously, the graph G (cid:48) Φ obtained after applying the previous rules to all triangular faces of G Φ ,is plane and satisfying R ( G (cid:48) Φ ) = V . In addition, it is straightforward to see that G (cid:48) Φ is topologicallyaugmentable if and only if G Φ is topologically augmentable, since all red vertices in G (cid:48) Φ that do notbelong to G Φ must be matched between them to change their parities in any augmentation. Thus, thetheorem follows.A similar reasoning applies to prove that the problem of deciding if a plane topological graph can beaugmented to a Eulerian plane topological graph is N P -complete. Before proving this result, we recallthe concept of T -join. Let G = ( V, E ) be a graph and let T ⊆ V . A subset E (cid:48) ⊆ E of edges is a T -join if the set of vertices with odd degree in the subgraph induced by E (cid:48) is precisely T . If G is connectedand | T | is even, then G always has a T -join. In this case, a T -join of minimum size consists of | T | / T , and can be computed in O ( n ) [16]. Theorem 2.3.
Let G be a plane topological graph. Then, the problem of deciding if G can be augmentedto a Eulerian plane topological simple graph is N P -complete.Proof.
Given a 3-SAT formula Φ and a plane embedding F of F Φ , let G Φ = ( V, E ) be the graph built inTheorem 2.1. We show that we can transform G Φ into a plane topological graph G (cid:48) Φ , having all its redvertices with odd degree and all its blue vertices with even degree, in such a way that G (cid:48) Φ is topologicallyaugmentable if and only if G Φ is topologically augmentable. We recall that to obtain a Eulerian graph,we have to change the parities of the odd degree vertices.Let V (cid:48) be the set of vertices with odd degree in G Φ . Let T be the set of vertices v i ∈ V , such that v i is red and has even degree or v i is blue and has odd degree. The number of vertices in T is even because | R ( G Φ ) | and | V (cid:48) | are even. G Φ is connected, so we can find a T -join of minimum size G T in O ( n ) time,which consists of | T | / T .We claim that any triangular face ∆ = ( v i , v j , v k ) of G Φ has at most one common edge with G T .Suppose to the contrary that ∆ has at least two common edges, say ( v i , v j ) , ( v j , v k ). If they belong tothe same path P , ( v i , v j ) , ( v j , v k ) , P of G T , then replacing the two common edges by the third edgewe obtain a shorter path P , ( v i , v k ) , P , contradicting the minimality of G T . If they belong to differentpaths, P , ( v i , v j ) , P and P (cid:48) , ( v j , v k ) , P (cid:48) , then P (cid:48) , P and P , ( v i , v k ) , P (cid:48) also connect the same endpointsand are shorter, again a contradiction. Hence, the claim follows.Let G (cid:48) be the multigraph obtained by adding G T to G Φ , so G (cid:48) contains duplicated edges. Note thateach edge of G Φ is adjacent to at least a triangular face and that the vertices with odd degree in G (cid:48) areprecisely the red vertices. By the previous claim, each duplicated edge in G (cid:48) can be mapped to a differenttriangular face and embedded into it. Suppose that e (cid:48) i = ( v i , v i +1 ) is a duplicated edge embedded insidethe triangular face ( v i , v i +1 , u ). Then, we can replace e (cid:48) i by 5 blue vertices and 14 edges and connect themas shown in Figure 6b. Observe that the 5 added vertices have even degree, and v i , v i +1 , u keep theirparities. By doing this replacement for every duplicated edge of G (cid:48) , we obtain a new plane topologicalsimple graph G (cid:48) Φ such that its red vertices are precisely the ones with odd degree. By construction,it is straightforward to verify that G (cid:48) Φ is topologically augmentable if and only if G Φ is topologicallyaugmentable, so the theorem follows. 8 a) (b) Figure 7: (a) A clause gadget receiving three negative values. (b) A clause gadget receiving at least onepositive value.Next, we present analogous results to the previous ones, when the added graph is required to bea plane topological perfect matching M between the red vertices. Such a matching, if it exists, is theoptimal way of augmenting a graph, since any graph with 2 k red vertices cannot be augmented with lessthan k edges. Theorem 2.4.
Let G = ( V, E ) be a plane topological graph and let C G be a set of parity constraints.Then, the problem of deciding if G is topologically augmentable to meet C G , by a plane topological perfectmatching M between the vertices in R ( C G ) , is N P -complete. Proof.
The proof is based again on a reduction from Planar 3-SAT. Given a 3-SAT formula Φ and aplane embedding F of F Φ , we build in polynomial time a plane topological graph G Φ and a set C G Φ of parity constraints, such that G Φ is topologically augmentable to meet C G Φ , by a plane topologicalperfect matching between the red vertices of G Φ , if and only if Φ is satisfiable.We construct G Φ in a similar way as described in Theorem 2.1, replacing some of its gadgets asfollows, so that G Φ is topological and plane. Variable gadgets are the same as described in the proofof Theorem 2.1. A clause gadget is now a subgraph as shown in Figure 7a, whose main part consistsof a 14-gon having eight red vertices. We identify the vertices v , v as the first input, v , v as thesecond output and v , v as the third input. The vertices in a input can change their parities by only twooptions, either by adding an edge to a red vertex in the wire gadget or being adjacent to a red vertexinside the face of the clause. The remaining two red vertices in the clause, v and v , can change theirparities only by the addition of edges lying on the face of the clause. Wire gadgets connecting literalsgadgets and clause gadgets are now replaced by double wire gadgets. Thus, if a positive augmentationis used for a literal gadget, then the only way of changing the parities of the vertices in the output ofthis literal gadget is connecting them to the middle red vertices in the double wire gadgets. This impliesthat the vertices in the corresponding input must change their parities by adding edges in the face ofthe clause to which they belong. On the contrary, if a negative augmentation is used for a literal gadget,then the red vertices in the middle of the double wire gadgets must change their parities by connectingthem to the vertices of the input. Finally, the exterior of the gadgets is triangulated.Observe that if a clause gadget receives three negative values (negative augmentations in its corre-sponding literal gadgets), then there is no way of changing the parities of the vertices v and v with lessthan two edges, as the clause gadget shown in Figure 7a depicts. Otherwise, if at least one input receivesa positive value (a positive augmentation in its corresponding literal gadget), then the remaining redvertices in the clause can meet their parities by doing a plane matching between them. An example isshown in Figure 7b.The new graph G Φ obtained after this replacement is clearly plane. In addition, by the previousdiscussion, one can easily verify that G Φ is topologically augmentable by a plane topological perfectmatching between the red vertices if and only if Φ is satisfiable, hence the theorem follows. Theorem 2.5.
Let G = ( V, E ) be a plane topological graph. Then, the problem of deciding if there This result was presented at the 29th Canadian Conference on Computational Geometry, Ottawa (2017) [8]. xists a plane topological perfect matching M such that G and M are compatible and edge-disjoint is N P -complete.Proof.
In this case, the set of red vertices of G = ( V, E ) is precisely V , and we look for a plane topologicalperfect matching M that is compatible and disjoint. From the graph G Φ built in the previous theorem,observe that we only need to show how to change the color of every blue vertex in G Φ to build a newgraph G (cid:48) Φ , such that G Φ is topologically augmentable by a plane topological perfect matching betweenthe its red vertices, if and only if, G (cid:48) Φ is topologically augmentable by a plane topological perfect matchingbetween its vertices. As in the proof of Theorem 2.2, this can be done using the gadget shown in Figure 6ato change the color of every blue vertex in G Φ . Theorem 2.6.
Let G = ( V, E ) be a plane topological graph. Then, the problem of deciding if G canbe augmented to a Eulerian plane topological simple graph, by the addition of a plane topological perfectmatching M between the vertices with odd degree in G , is N P -complete.Proof.
In this case, the set of vertices with odd degree is the set of red vertices, and we look for a planetopological perfect matching M between the vertices with odd degree. To prove the N P -completenessof this problem, we proceed as in the proof of Theorem 2.3. From the graph G Φ built in the proof ofTheorem 2.4, we add a minimum T -join to G Φ , where T is the set of red vertices with even degree andthe set of blue vertices with odd degree, and we replace every duplicated edge by the gadget shown inFigure 6b. In the resulting graph G (cid:48) Φ , every vertex with odd degree is red and every vertex with evendegree is blue. It is straightforward to see that G Φ admits a plane topological perfect matching betweenits red vertices, if and only, if G (cid:48) Φ admits a plane topological perfect matching between its vertices withodd degree, so the theorem holds.To finish this section, we point out that the previous results can be extended to planar graphs, usingthe Whitney’s theorem on 3-connected planar graphs. Theorem 2.7.
Let G = ( V, E ) be a planar graph and let C G be a set of parity constraints. Then, theproblem of deciding if there exists a planar graph H on the same vertex set V , such that G and H areedge-disjoint, G (cid:48) = G ∪ H is planar and G (cid:48) meets C G , is N P -complete. The problem is also
N P -completewhen H must be a perfect matching between the vertices in R ( C G ) . In addition, the problem remains N P -complete in both variants, even when R ( C G ) = V or R ( C G ) is the set of vertices with odd degree in G .Proof. It can be easily checked that the graph G Φ built in the proof of Theorem 2.1 is 3-connected.By Whitney’s theorem, a 3-connected planar graph has a unique embedding (up to the choice of theunbounded face). Hence, we can consider G Φ as a planar graph that is augmentable if and only if itsunique plane embedding is topologically augmentable. Therefore, the first part of the theorem holds. Inaddition, the different graphs G Φ or G (cid:48) Φ built in the proofs of Theorems 2.2, 2.3, 2.4, 2.5 and 2.6 are also3-connected. Thus, the rest of the theorem also holds. In this section we address the plane geometric augmentation problem to meet a set of parity constraints,that is, given a plane geometric graph G = ( V, E ) and a set C G of parity constraints, we look for a planegeometric graph H such that G and H are compatible and edge-disjoint, and all parity constraints aresatisfied in G ∪ H . We extend the N P -completeness results obtained for plane topological graphs toplane geometric graphs. Moreover, some of these variants remain
N P -complete even on more restrictivegeometric graph families, like trees or paths.The following theorem is a consequence of Tutte’s theorem on convex faces of 3-connected planargraphs [30].
Theorem 3.1.
Let G = ( V, E ) be a plane geometric graph and let C G be a set of parity constraints.Then, the problem of deciding if G is geometrically augmentable to meet C G is N P -complete. Theproblem is also
N P -complete when the augmenting graph must be a plane geometric perfect matchingbetween the vertices in R ( C G ) . In addition, the problem remains N P -complete in both variants, evenwhen R ( C G ) = V or R ( C G ) is the set of vertices with odd degree in G . a) (b) Figure 8: (a) Spiral gadget. (b) Arrow gadget.
Proof.
By Tutte’s theorem on convex faces of triconnected planar graphs, all graphs G Φ and G (cid:48) Φ builtin Theorems 2.1, 2.2, 2.3, 2.4, 2.5 and 2.6, have an embedding where all edges are straight-line segmentsand all faces are convex. Then, they are geometrically augmentable to meet parity constraints if andonly if Φ is satisfiable.In particular, from this theorem, deciding if a plane geometric graph can be geometrically augmentedto Eulerian is N P -complete, even when the added graph is required to be a plane geometric perfectmatching between the vertices with odd degree.Some previous variants remain
N P -complete for plane geometric trees and paths, as we show in thefollowing theorems.
Theorem 3.2.
Let T = ( V, E ) be a plane geometric tree and let C T be a set of parity constraints. Then,the problem of deciding if T is geometrically augmentable by a plane geometric matching M between thevertices in R ( G T ) is N P -complete.Proof.
We explain how to transform the graph G Φ built in Theorem 2.4 into a tree T Φ . As shownin [28], to break a cycle one can replace any edge ( a, b ) of a cycle by the spiral gadget shown in Figure 8a.Repeating this operation, we can break all cycles in G Φ , to obtain a plane geometric tree T Φ . Notice thatthe two red vertices in each spiral gadget are forced to join each other in any minimum augmentation.Therefore, G Φ is geometrically augmentable with a plane geometric perfect matching between its redvertices, if and only if, T Φ is geometrically augmentable with a plane geometric perfect matching betweenits red vertices, so the theorem follows. Theorem 3.3.
Let T = ( V, E ) be a plane geometric tree. Then, the problem of deciding if there existsa plane geometric perfect matching M , such that T and M are compatible and edge-disjoint, is N P -complete.Proof.
Note that in this case, the set of red vertices is V . To prove that this problem is N P -complete,we only need to show how to change the color of each blue vertex in the tree T Φ built in the previoustheorem, such that all new red vertices in the resulting plane geometric tree match between them in anyminimum augmentation.Consider an edge ( a, b ) ∈ T Φ , with a and b being blue vertices. We can change the colors of a and b by using what we call an arrow gadget . We add to ( a, b ) two nested paths, each one consisting ofthree red vertices, placed at distance (cid:15) on each side of the edge as shown in Figure 8b. The vertices ofthe path on the upper side of the edge, say u, u (cid:48) , u (cid:48)(cid:48) , are placed in such a way that they cannot see thevertices of the path in the lower side, say v, v (cid:48) , v (cid:48)(cid:48) . Moreover, the upper path is placed in such a waythat u (cid:48) is visible only from a, b , and u (cid:48)(cid:48) is visible only from u, a . Symmetrically for the case of the lowerpath. Observe that u (cid:48)(cid:48) is forced to join u and v (cid:48)(cid:48) is forced to join v , because otherwise the parities of theremaining vertices in the gadget are not met. Thus, the vertices in an arrow gadget must be matchedbetween them in any perfect matching between the red vertices.Consider now an edge ( a, b ) ∈ T Φ with a being red and b being blue. We can subdivide ( a, b ) intotwo edges ( a, a (cid:48) ) and ( a (cid:48) , b ) and mark a (cid:48) as blue. Then, we can replace the edge ( a (cid:48) , b ) by an arrowgadget as described previously. Iterating this replacement for all blue-blue edges and all blue-red edgesin T Φ , we obtain a plane geometric tree T (cid:48) with all its vertices marked as red. Clearly, T (cid:48) has a planegeometric perfect matching, compatible and edge-disjoint with T (cid:48) , if and only if, T Φ can be geometricallyaugmented by a plane geometric perfect matching between its red vertices.11 j v i (a) v jk v i v j v j v j (b) v j v jk v j v j u u (cid:48) w w (cid:48) u u k − (c) Figure 9: (a) A geometric tree T Φ . (b) A polygon Q Φ obtained by traversing the unbounded face of T Φ .(c) Reconfiguration of v j .In the following theorem, we establish the hardness of transforming a plane geometric tree intoEulerian, by adding a plane geometric matching between the vertices with odd degree. Theorem 3.4.
Let T = ( V, E ) be a plane geometric tree. Then, the problem of deciding if T can beaugmented to a Eulerian plane geometric graph, by the addition of a plane geometric perfect matching M between the vertices with odd degree in T , such that T and M are compatible and edge-disjoint, is N P -complete.Proof.
Given a 3-SAT formula Φ and a plane embedding F of F Φ , we can construct a plane topologicalgraph as described in Theorem 2.4, that can be transformed into a plane geometric graph G Φ with allfaces being convex. Then, we apply the same technique used in the proof of Theorem 2.3, and we add a T -join of minimum size to G Φ , where T consists of all red vertices with even degree and all blue verticeswith odd degree. After that, we replace every duplicated edge as indicated in Figure 6b, to obtain anew plane geometric graph G (cid:48) Φ such that all vertices with odd degree are red and all vertices with evendegree are blue.We now transform G (cid:48) Φ into a tree T Φ by adding spiral gadgets to break cycles, as described in theproof of Theorem 3.2. Note that all vertices with odd degree in T Φ are red and all vertices with evendegree are blue. Clearly, by construction, G Φ can be augmented by a plane geometric perfect matchingbetween its red vertices, if and only if, T Φ can be augmented by a plane geometric perfect matchingbetween the vertices with odd degree. Therefore, the theorems follows.For plane geometric paths, we can prove the following results. Theorem 3.5.
Let P = ( V, E ) be a plane geometric path and let C P be a set of parity constraints.Then, the problem of deciding if P is geometrically augmentable to meet C G , by a plane geometricperfect matching M between the vertices of R ( C P ) , is N P -complete.Proof.
Given a 3-SAT formula Φ and a plane embedding F of F Φ , we build a plane geometric tree T Φ as described in Theorem 3.2. We recall that Φ is satisfiable if and only if T Φ can be augmented to meetparity constraints by a plane geometric perfect matching between the vertices in R ( T Φ ). From T Φ , weconstruct a plane geometric path P Φ as follows.It is well known that a simple polygon Q Φ can be built from T Φ by traversing the boundary ofthe unbounded face of T Φ , placing a copy infinitesimally close of a vertex every time it is visited, andconnecting the copies according the traversal order, as shown in Figures 9a and 9b. We start such atraversal at an arbitrary leaf u i , and any copy of a vertex keeps the color of the vertex. For a non-leafvertex, we can assume that the copies of the vertex are placed on a circumference of radius (cid:15) around thevertex.For each red vertex v j ∈ T Φ with degree k ≥
2, we apply the following transformation to itscorresponding vertices in Q Φ . An example of this transformation is shown in Figure 9c for a vertex12 a) Figure 10: A geometric path consisting of 22 disk-shaped vertices and 22 cross-shaped vertices.with degree 4 in T Φ . Let C = { v j , v j , ..., v jk } be the set of vertices in Q Φ corresponding to a redvertex v j ∈ T Φ , and assume that the clockwise order of these points along the boundary of Q Φ is . . . , w , v j , w (cid:48) , . . . , w , v j , w (cid:48) , . . . , w k , v jk , w (cid:48) k , . . . For l = 1 , . . . , k −
1, we add a narrow wedge v jl , u l , u (cid:48) l inside Q Φ such that: (i) u l is placed very closeto the edge ( w l +1 , v jl +1 ), and (ii) u (cid:48) l is placed very close to v jl so that v jl cannot be connected to w (cid:48) l , but v jl can be connected to any other red vertex that was previously visible from it through the exterior of Q Φ . See Figure 9c. At vertex v jk we do the same operation, but also adding before the wedge a zig-zagpath with most of the points in convex position and k − u , u , . . . , u k − , being red, in sucha way that: (i) the only red vertices that are visible from a red vertex in the zig-zag belong to C , (ii) u , u , . . . , u k − are placed inside the circle of radius (cid:15) centered at v j so that every vertex v jl ∈ C , with1 ≤ l < k , can see all of them, and (iii) v jk is visible from at least one of u , u , . . . , u k − .Observe that the addition of these narrow wedges implies that a vertex v jl ∈ C cannot be connectedto any other red vertex v j (cid:48) l (cid:48) / ∈ C through the interior of Q Φ , and that the red vertices in the zig-zag pathmust be matched to k − v j in any augmentation. Thus, one of the copies of v j is alwaysfree to join any other red vertex in the rest of the graph through the exterior of Q Φ . Q Φ can be obviously transformed into a plane geometric path P Φ , by duplicating a blue vertex tobreak the boundary of Q Φ . Then, it is straightforward to see that P Φ admits a plane geometric perfectmatching between its red vertices, if and only if, T Φ can be augmented by a plane geometric perfectmatching between its red vertices. Hence, the theorem holds. Theorem 3.6.
Let P = ( V, E ) be a plane geometric path. Then, the problem of deciding if there exists aplane geometric perfect matching M such that P and M are compatible and edge-disjoint is N P -complete.Proof.
Note that this is the case in which the set of red vertices coincides with V . Also note that weonly need to show how to change the color of each blue vertex in the path P Φ built in the previous proof,without interfering with any possible augmentation of the red vertices of P Φ .Consider an edge ( a, b ) ∈ P Φ , with a and b being blue vertices. We replace the edge ( a, b ) by thegadget shown in Figure 10, which is a path connecting a and b , so that the colors of a and b are changedto red. In this gadget, we identify two sets of red vertices, the 22 cross-shaped vertices and the 22disk-shaped vertices. Observe that the disk-shaped vertices are placed in such a way that they are eitheradjacent or they are not visible to each other. Therefore, it is not possible to join two disk-shapedvertices with an edge. Observe as well that cross-shaped vertices are the only vertices of the graph thatare visible from disk-shaped vertices. This implies that the cross-shaped vertices are forced to join thedisk-shaped vertices in any matching, because otherwise there would be red disk-shaped vertices notmeeting their parity constraints.Consider now an edge ( a, b ) ∈ P Φ , with a being red and b being blue. We can subdivide ( a, b ) intotwo edges ( a, a (cid:48) ) and ( a (cid:48) , b ) and mark a (cid:48) as blue. Then, we apply the previous gadget to ( a (cid:48) , b ). By13epeating this operation for all blue edges and all red-blue edges of P Φ , we obtain a path P (cid:48) with all itsvertices marked as red. The proof follows since, if we are able to augment P (cid:48) with a plane geometricperfect matching, then we are able to find a plane geometric perfect matching between the red verticesof P Φ , and viceversa.To finish this section, we remark that deciding if a plane geometric path can be augmented to aEulerian plane geometric graph by adding a matching between its odd degree vertices can be clearlysolved in linear time. The only vertices of a path with odd degree are its endpoints, and checking if thesegment connecting these two endpoints crosses the path can be done in linear time. To the best of our knowledge, the family of plane topological trees is so far the only family of planetopological graphs, for which the plane topological augmentation problem to meet parity constraints ispolynomial [8]. In this section we show that this problem is also polynomial for the family of maximalouterplane graphs.An outerplanar graph G is a graph that has a plane embedding where all of its vertices belong to theunbounded face of the embedding. We shall refer to these embeddings as outerplane graphs . A graph G is a maximal outerplanar graph if it is not possible to add any edge to it, such that the resulting graphis still outerplanar. A maximal outerplane graph (for short a MOP) is an outerplane graph of a maximalouterplanar graph.For the sake of clarity, in the rest of this section we will assume that all the vertices of a MOP G are placed on an unit circle C , and that they are labeled from v to v n in clockwise order. We alsoassume that all of the edges of G are arcs of C connecting consecutive vertices, or straight-line segmentscontained in the interior of C called diagonals . Therefore, G consists of n arcs and n − v i , v j ] the set of vertices { v i , v i +1 , . . . , v j } (mod n ), and we call [ v i , v j ] an interval of G .Given a MOP G and a set C G of parity constraints, recall that a vertex v ∈ G is red if its paritymust change to meet its parity constraint (that is, v ∈ R ( C G )), and blue otherwise. Further, recall thatan edge of G is red (respectively blue), if both of its endpoints are red vertices (respectively blue). Anedge of G is red-blue if its endpoints have different colors. Consider two red-blue diagonals of G . Wesay that they are parallel if there is a straight line that separates their red endpoints from their blueendpoints. For instance, the red-blue diagonals ( v m , v i ) and ( v l , v k ) in Figure 11c are parallel, and thered-blue diagonals ( v l , v i ) and ( v j , v k ) in Figure 11b are not. Note that the cyclic order of the colors ofthe four endpoints is red, red, blue, blue if the two red-blue diagonals are parallel, and red, blue, red,blue if they are not.In the following theorem we characterize the MOPs that are topologically augmentable. Theorem 4.1.
Let G = ( V, E ) be a MOP and let C G be a set of parity constraints. Then, G istopologically augmentable to meet C G if one of the following conditions is satisfied:(i) G contains a blue diagonal.(ii) G contains two non-parallel red-blue digonals.(iii) There exist two parallel red-blue diagonals ( v i , v m ) , ( v k , v l ) ∈ E such that v i , v k are blue, v l , v m arered (possibly with v l = v m ), and there exists a vertex v j of degree two such that the order of thesevertices along C is v i , v j , v k , v l , v m in clockwise direction.Otherwise, G is non-augmentable.Proof. We first prove that if G satisfies at least one of (i), (ii) or (iii), then it is topologically augmentable.Case (i): Suppose that G contains a blue diagonal ( v i , v j ). Then, each of the intervals A = [ v j +1 , v i − ]and B = [ v i +1 , v j − ] contains at least one vertex. If one of these intervals, say A , contains only bluevertices, then we join a blue vertex in A to all of the evenly many red vertices in B , and we are done.Suppose then that each of A and B contains at least one red vertex. In the clockwise traversal of C ,let v ∈ A and v (cid:48) ∈ B be the red vertices that precede v i and v j , respectively. Note that these verticesare not necessarily v i − and v j − . One can easily verify that joining v to all red vertices in B (except14 a) (b) (c) Figure 11: Topologically augmentable MOPs. (a) A blue diagonal exists. (b) Two non-parallel red-bluediagonals exist. (c) Two parallel red-blue diagonals and a degree-2 vertex exist.possibly v (cid:48) ) and joining v (cid:48) to all red vertices in A (except possibly v ), every red vertex will meet itsparity constraint in the resulting graph. See Figure 11a.Case (ii): Suppose that G contains two non-parallel edges ( v i , v l ) and ( v j , v k ) such that v i and v k areblue vertices, and v j and v l are red vertices. Assume without loss of generality that the relative orderof these vertices along C is v i , v j , v k , v l , see Figure 11b. In this case, we can proceed as in Case (i), as ifthe blue edge ( v i , v k ) was an edge of G .Case (iii): We can connect v i − to each red vertex in [ v i +1 , v j − ] and v k +1 to each red vertex in[ v j +1 , v k − ], see Figure 11c. Note that after adding these edges, v i − and v k +1 could meet their paritiesor not. Then, we connect all red vertices in [ v k +1 , v i − ] to v j , including v i − and v k +1 if they do notmeet their parities after the previous addition of edges. Observe that it does not matter what color v j has, the parities of all vertices in G are met (see Figure 11c).We now prove that if G is topologically augmentable, then it must satisfy at least one of the threeprevious conditions.Suppose that G satisfies neither (i), nor (ii), nor (iii). Then, G must satisfy the following conditions.1. By (i), G can have only red or red-blue diagonals.2. By (ii), G cannot have non-parallel red-blue diagonals. This implies that there is an interval U in C containing all of the blue endpoints of all red-blue diagonals of G , and none of the red endpointsof these diagonals. Assume that U is the shortest such interval and that, relabeling the vertices ifnecessary, U = [ v , v i ]. See Figure 12a. Observe that if G does not have red-blue diagonals, then U is empty, and if G contains only one red-blue diagonal, then U consists of one blue vertex. Alsoobserve that D = [ v i +1 , v n ] must always contain red vertices, regardless if U is empty or not.We claim that if D contains a blue vertex v r , then it must have degree two in G . To prove thisclaim, we distinguish whether U is empty or not. Assume that U is non-empty and suppose thatthere is a diagonal ( v r , v s ) incident to v r in G . Thus, v s must be a red vertex, because otherwisea blue diagonal would exist and G would satisfy (i). In addition, v s cannot lie between v i +1 and v r clockwise, for otherwise ( v r , v s ) is a red-blue diagonal and thus v r would belong to U . Finally,if v s lies between v r and v clockwise, then non-parallel blue-red diagonals would exist, and thus G would satisfy (ii). Hence, v r must have degree two in G . Assume now that U is empty, so thereare no red-blue diagonals. Since G does not contain either blue diagonals, then no diagonal can beincident to v r and the claim follows.As a consequence of this claim, there are no consecutive blue vertices in D , that is, if v r is a bluevertex in D and v r − and v r +1 belong to D , then v r − and v r +1 are red vertices, and they areadjacent in G .3. Since G does not satisfy (iii), all vertices in U have degree at least three. Therefore, there is nodiagonal connecting two vertices in U , because otherwise a vertex of degree 2 would exist in U .See Figure 12a. 15 a) (b) (c) Figure 12: (a) A non-augmentable MOP with blue-red diagonals. (b) Isolating a red vertex when twovertices in D are connected. (c) Connecting red vertices from D to a vertex in U .Suppose that G is augmentable to meet C G and let H = ( V, E (cid:48) ) be a plane topological graph thataugments G . We show that H cannot exist if G does not satisfy (i), (ii), or (iii).We first prove that there is no edge in H connecting two vertices in D . Assume to the contrarythat such edges exist. Among all of them, we take the edge e = ( v l , v m ) ∈ E (cid:48) such that the interval[ v l , v m ] belongs to D and is the shortest one. Note that v l and v m can neither be consecutive vertices,nor be at distance two if there is a vertex of degree two between them, since joining them with an edgewould cause duplicated edges in G ∪ H . See Figure 12b. Thus, [ v l , v m ] contains at least one red vertexdifferent from v l and v m , since any blue vertex in D must have degree two in G and there are no twoconsecutive blue vertices in D . By the minimality of [ v l , v m ], one can easily verify that the parities ofthe red vertices in [ v l , v m ] cannot all be met. Therefore, there is no edge in H connecting two vertices in D . In particular, if U is empty, this result implies that H cannot exist, since there is no way of meetingall parity constraints by adding edges incident to vertices in D .Hence, we assume that U is non-empty in the rest of the proof, so G does contain parallel red-bluediagonals. Let D R = { v p , v p (cid:48) , ..., v q } be the set of red vertices in D that are adjacent in G to at least onevertex in U , where p < p (cid:48) < . . . < q . Note that D R is non-empty because there are red-blue diagonals.Consider a vertex v s ∈ D R . By the third condition, there is no diagonal in G connecting two verticesin U , so the set of vertices in U adjacent to v s forms an interval [ v l s , v r s ] along C , 1 ≤ l s ≤ r s ≤ i , seeFigure 12c. Since v s is red and there are no edges in H connecting two vertices in D , then H mustcontain at least one edge connecting v s to a vertex w in U , in order to meet the parity constraint of v s .In addition, w cannot be in [ v l s , v r s ] as any vertex in this interval is already adjacent to v s .Consider now v p and the interval [ v l p , v r p ]. Note that v r p = v i . In order to meet its parity constraint, v p must be adjacent to a vertex v t p with t p < l p , i.e. v t p lies to the left of v l p . In a recursive way, we canprove that any v s in D R must be adjacent to a vertex v t s in U to the left of the interval [ v l s , v r s ]. Thisalso applies to v q , but the neighbours of v q in U form an interval that starts at v , so v t q must be to theleft of v and thus it cannot be in U , which is a contradiction. It follows that H cannot exist and if G is topologically augmentable, then it must satisfy at least one of (i), (ii), or (iii).It is clear that given a MOP G , we can check in linear time if each one of (i),(ii), and (iii) is satisfied.Thus, we can decide if G is topologically augmentable to meet a set of parity constraints in O ( n ) time. Inaddition, it is easy to see that the set of edges added to G in the proof of Theorem 4.1 is not necessarilyone with minimum size. We next provide a polynomial time algorithm to compute a minimum set ofedges to augment a MOP G (if it exists) to meet a set of parity constraints. Theorem 4.2.
Let G be a MOP with n vertices, and let C G be a set of parity constraints. It takes O ( n ) time to find a plane topological graph H of minimum size such that G is topologically augmentable by H to meet C G .Proof. We compute an optimum edge set, if it exits, by dynamic programming. Given an interval [ v i , v j ],we define C [ i, j ] as the size of a smallest set of edges having both endpoints in [ v i , v j ] such that this set isa plane topological graph that does not cross G and meets the parity constraints of all vertices in [ v i , v j ]16or ∞ if such a set of edges does not exist). If v denotes the color exchange of v for any vertex v , weanalogously define C [ i, j ], C [ i, j ] and C [ i, j ] as the sizes of the optimum solutions as described previously,for the intervals [ v i , v j ], [ v i , v j ] and [ v i , v j ], respectively.We also define D [ i, j ], D [ i, j ], D [ i, j ] and D [ i, j ] as the size of an optimum solution for the intervals[ v i , v j ], [ v i , v j ], [ v i , v j ] and [ v i , v j ], respectively, assuming that the edge ( v i , v j ) belongs to the solution.We next show how to compute C [ i, j ] and D [ i, j ]. The table entries C [ i, j ], C [ i, j ], C [ i, j ], D [ i, j ], D [ i, j ]and D [ i, j ] are computed in a similar manner.To compute C [ i, j ], we look at the last vertex v k ∈ [ v i , v j ] in clockwise order (if it exists) connected to v i in an optimum solution for the interval [ v i , v j ]. If v k does not exist because v i is not connected to anyother vertex of the interval, then C [ i, j ] = C [ i + 1 , j ] if v i is blue, or C [ i, j ] = ∞ if v i is red. If v k exists,then the optimum solution can be computed by solving two subproblems. On the one hand, finding theminimum solution for the interval [ v i , v k ] assuming that the edge ( v i , v k ) belongs to the solution, and onthe other hand, finding the minimum solution for the interval [ v k , v j ]. Observe that, if v k is blue, thenwe can decide either to keep v k as blue in both subproblems or change its color in both subproblems,and if v k is red, then we can choose v k as red in one of the subproblems and blue in the other one. Wedo this analysis only if the edge ( v i , v k ) does not belong to G . Summarizing, we have: C [ i, j ] = min (cid:40) C [ i + 1 , j ] if v i is blue ∞ if v i is redmin i +2 ≤ k ≤ j ∞ if ( v i , v k ) ∈ E min { D [ i, k ] + C [ k, j ] , D [ i, k ] + C [ k, j ] } if v k is blue and ( v i , v k ) (cid:54)∈ E min { D [ i, k ] + C [ k, j ] , D [ i, k ] + C [ k, j ] } if v k is red and ( v i , v k ) (cid:54)∈ E (1) The analysis to compute D [ i, j ] is similar. Assuming that the edge ( v i , v j ) belongs to an optimumsolution, either v i is not connected to any other vertex of the interval [ v i , v j ] except for v j , or there existsa vertex v k (different from v j ) that is the last vertex of the interval connected to v i . An optimum solutioncan be computed by finding an optimum solution for the interval [ v i , v k ] assuming that ( v i , v k ) belongsto the solution, finding an optimum solution for the interval [ v k , v j ], and adding the edge ( v i , v j ). D [ i, j ] = min (cid:40) C [ i + 1 , j ] + 1 if v i is red ∞ if v i is bluemin i +2 ≤ k Let G = ( V, E ) be a MOP, and let C G be a set of parity constraints. Then, there existsa plane topological matching M such that G and M are compatible and edge-disjoint, and G ∪ M meetsall but at most four parity constraints.Proof. Suppose that R ( C G ) = V . The geometric complete graph K n , with n vertices placed in convexposition, can be decomposed into n/ P , ..., P n/ as follows: P = { v , v , v n , v , v n − , ... } , P = { v , v , v , v , v n , ... } , ..., P n/ = { v n/ , v n/ , v n/ − , ... } . That is to say, P i +1 is obtained byrotating the vertices of P i one position to the right, as shown in Figure 13a.We number the edges of each zig-zag path P i from 1 to n − 1, starting at the edge ( v i , v i +1 ). Observethat each path contains two maximal matchings, one of size n , given by the odd numbered edges (solidsegments in Figure 13b), and the second of size n − 1, given by the even numbered edges (dashed segmentsin Figure 13b). In total, K n has n pairwise disjoint maximal matchings.Since G has n − M i , such that none of its edgesis a diagonal of G . If M i is formed by odd (solid) edges, then there exist two edges on the boundary ofthe unbounded face of G in common with M i , so M i \ E spans n − G . If M i is formed byeven (dashed) edges, then M i ∩ E = ∅ and M i spans n − G since | M i | = n − 1. Notice thatthe edges of M i can be drawn in the unbounded face of G without crossings, hence the theorem followswhen R ( C G ) = V .It is straightforward to see that when R ( C G ) ⊂ V , we can apply the same analysis on the subgraphinduced by the red vertices to prove the theorem. In this paper we have addressed the plane topological augmentation problem to meet parity constraintsand we have proved that deciding if a plane topological graph is topologically augmentable is computa-tionally hard, under several assumptions on the input graph, the vertices that must change their parities,18nd the size of the augmentation. We have obtained analogous results for the plane geometric augmen-tation problem to meet parity constraints. We have also proved the hardness of some of these variantsfor plane geometric trees and paths. Table 1 summarizes these results.For the family of MOPs, we have characterized the MOPs that are topologically augmentable to meetparity constraints, and we have given an O ( n ) time algorithm to augment them with the minimumnumber of edges, if that is the case. We have also shown that there is always a topological augmentationby a plane topological matching, satisfying all but at most four parity constraints.Finally, we conclude this paper with the following conjecture about the complexity of deciding if aplane geometric tree is geometrically augmentable. Conjecture 5.1. Let T = ( V, E ) be a plane geometric tree and let C T be a set of parity constraints.Then, the problem of deciding if T is geometrically augmentable is N P -complete. Acknowledgments Alfredo Garc´ıa, Javier Tejel and Jorge Urrutia were supported by H2020-MSCA-RISE project 734922- CONNECT. Research of Alfredo Garc´ıa and Javier Tejel was also supported by project MTM2015-63791-R (MINECO/FEDER) and by project Gobierno de Arag´on E41-17R (FEDER). Research of JorgeUrrutia was also supported by UNAM project PAPIITIN102117. Research of Juan Carlos Catana wassupported by Consejo Nacional de Ciencia y Tecnolog´ıa, Mexico. References [1] M. Abellanas, A. Garc´ıa, F. Hurtado, J. 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