Topological embedding problems.
Topological embedding problems are among the most fundamental problems in computational topology, already emphasized since the early developments of this discipline [8, Section 10]. Their general form is as follows: Given topological spaces and , does there exist an embedding (a continuous, injective map) from to ? Since a finite description of and is needed, typically they are represented as finite simplicial complexes, which are topological spaces obtained by attaching simplices (points, segments, triangles, tetrahedra, etc.) of various dimensions together.
The case where the host space equals (or, almost equivalently, , which can be modeled as a simplicial complex) has been studied the most. The case corresponds to the planarity testing problem, which has attracted considerable interest . The case is much harder, and has only recently been shown to be decidable by Matoušek, Sedgwick, Tancer, and Wagner . The general problem for arbitrary has been extensively studied in the last few years, starting with hardness results by Matoušek, Tancer, and Wagner , and continuing with some algorithmic results in a series of articles; we refer to Matoušek et al. [21, Introduction] for a state of the art.
What about more general choices of ? The case where is a graph is essentially the subgraph homeomorphism problem, asking if contains a subdivision of a graph . This is hard in general, easy when is fixed, and polynomial-time solvable for every fixed , by using graph minor algorithms. The case where is a graph and a 2-dimensional simplicial complex that is homeomorphic to a surface has been much investigated, also in connection to topological graph theory  and algorithms for surface-embedded graphs [11, 7]: The problem is NP-complete, as proved by Thomassen , but Mohar  has proved that it can be solved in linear time if is fixed (in some recent works, the proof has been simplified and the result extended [15, 17]). The case where is a 2-complex and is (a 2-complex homeomorphic to) a surface essentially boils down to the previous case; see Mohar . More recently, Čadek, Krčál, Matoušek, Vokřínek, and Wagner [4, Theorem 1.4] considered the case where the host complex has an arbitrary (but fixed) dimension; they provide a polynomial-time algorithm for the related map extension problem, under some assumptions on the dimensions of and ; in particular, must have trivial fundamental group (because they manipulate in an essential way the homotopy groups of , which have to be Abelian); but the maps they consider need not be embeddings.
Another variation on this problem is to try to embed such that it extends a given partial embedding of (we shall consider such embedding extension problems later). This problem has already been studied in some particular cases; in particular, Angelini, Battista, Frati, Jelínek, Kratochvìl, Patrignani, and Rutter [1, Theorem 4.5] provide a linear-time algorithm to decide the embedding extension problem of a graph in the plane.
In this article, we study the topological embedding problem when is an arbitrary graph , and is an arbitrary two-dimensional simplicial complex (actually, a simplicial complex of dimension at most two—abbreviated as 2-complex below). Formally, we consider the following decision problem:
Input: A graph with vertices and edges, and a 2-complex with simplices.
Question: Does have a topological embedding into ?
(We use the parameters and whenever we need to refer to the input size.) Here are our main results:
The problem Embed is NP-complete.
The problem Embed can be solved in time , where is some computable function of .
As for Theorem 1.1, it is straightforward that the problem is NP-hard (as the case where is a surface is already NP-hard); the interesting part is to provide a certificate checkable in polynomial time when an embedding exists. Note that Theorem 1.2 shows that, for every fixed complex , the problem of deciding whether an input graph embeds into is polynomial-time solvable. Actually, our algorithm is explicit, in the sense that, if there exists an embedding of on , we can provide some representation of such an embedding (in contrast to some results in the theory of graph minors, where the existence of an embedding can be obtained without leading to an explicit construction).
Why do 2-complexes look harder than surfaces?
A key property of the class of graphs embeddable on a fixed surface is that it is minor-closed: Having a graph embeddable on a surface , removing or contracting any edge yields a graph embeddable on . By Robertson and Seymour’s theory, this immediately implies a cubic-time algorithm to test whether a graph embeds on , for every fixed surface . In contrast, the class of graphs embeddable on a fixed 2-complex is, in general, not closed under taking minors, and thus this theory does not apply. For example, let be obtained from two tori by connecting them together with a line segment, and let be obtained from two copies of by joining them together with a new edge ; then embeds into , but the minor obtained from by contracting does not.
Two-dimensional simplicial complexes are topologically much more complicated than surfaces. For example, there exist linear-time algorithms to decide whether two surfaces are homeomorphic (this amounts to comparing the Euler characteristics, the orientability characters, and, in case of surfaces with boundary, the numbers of boundary components), or to decide whether a closed curve is contractible (see Dey and Guha , Lazarus and Rivaud , and Erickson and Whittlesey ). In contrast, the homeomorphism problem for 2-complexes is as hard as graph isomorphism, as shown by Ó Dúnlaing, Watt, and Wilkins . Moreover, the contractibility problem for closed curves on 2-complexes is undecidable; even worse, there exists a fixed 2-complex such that the contractibility problem for closed curves on is undecidable (this is because every finitely presented group can be realized as the fundamental group of a 2-complex, and there is such a group in which the word problem is undecidable, by a result of Boone ; see also Stillwell [29, Section 9.3]).
Despite this stark contrast between surfaces and 2-complexes, if we care only on the polynomiality or non-polynomiality, our results show that the complexities of embedding a graph into a surface or a 2-complex are similar: If the host space is not fixed, the problem is NP-complete; otherwise, it is polynomial-time solvable. Compared to the aforementioned hard problems on general 2-complexes, one feature related to our result is that every graph embeds on a 3-book (a complex made of three triangles sharing a common edge); thus, we only need to consider 2-complexes without 3-book, for otherwise the problem is trivial; this significantly restricts the structure of the 2-complexes to be considered. The problem of whether Embed admits an algorithm that is fixed-parameter tractable in terms of the parameter , however, remains open for general complexes, whereas it is the case when restricting to surfaces .
Why is embedding graphs on 2-complexes interesting?
First, let us remark that, if we consider the problem of embedding graphs into simplicial complexes, then the case that we consider, in which the complex has dimension at most two, is the only interesting one, since every graph can be embedded in a single tetrahedron.
We have already noted that the problem we study is more general than the problem of embedding graphs on surfaces. It is indeed quite general, and some other problems studied in the past can be recast as an instance of Embed or as variants of it. For example, the crossing number of a graph is the minimum number of crossings in a (topological) drawing of in the plane. Deciding whether a graph has crossing number at most is NP-hard, but fixed-parameter tractable in , as shown by Kawarabayashi and Reed . This is easily seen to be equivalent to the embeddability of into the complex obtained by removing disjoint disks from a sphere and adding, for each resulting boundary component , two edges with endpoints on whose cyclic order along is interlaced. Of course, the embeddability problem on a 2-complex is more general and contains, for example, the problem of deciding whether there is a drawing of a graph on a surface of genus with at most crossings. In topological graph theory, embeddings of graphs on pseudosurfaces (which are special 2-complexes) have been considered; see Archdeacon [2, Section 5.7] for a survey. Slightly more remotely, a book embedding of a graph (see, e.g., Malik ) is also an embedding of into a particular 2-complex, with additional constraints on the embedding.
Strategy of the proof and organization of the paper.
For clarity of exposition, in most of the paper, we focus on developing an algorithm for the problem Embed (Theorem 1.2). Only at the end (Section 8) we explain why our techniques imply that the problem is in NP. The idea of the algorithm is to progressively reduce the problem to simpler problems. We first deal with the case where the complex contains a 3-book (Section 3). From Section 4 onwards, we reduce Embed to embedding extension problems (EEP), similar to the Embed problem except that an embedding of a subgraph of the input graph is already specified. In Section 4, we reduce Embed to EEPs on a pure 2-complex (in which every segment of the complex is incident to at least one triangle). In Section 5, we further reduce it to EEPs on a surface. In Section 6, we reduce it to EEPs on a surface in which every face of the subgraph is a disk. Finally, in Section 7, we show how to solve EEPs of the latter type using a key component in an algorithm by the third author  to decide embeddability of a graph on a surface.
2.1 Embeddings of graphs into 2-complexes
A 2-complex is an abstract simplicial complex of dimension at most two: a finite set of 0-simplices called nodes, 1-simplices called segments, and 2-simplices called triangles (we use this terminology to distinguish from that of vertices and edges, which we reserve for graphs); each segment is a pair of nodes, and each triangle is a triple of nodes; moreover, each subset of size two in a triangle must be a segment. Each 2-complex corresponds naturally to a topological space, obtained in the obvious way: Start with one point per node in ; connect them by segments as indicated by the segments in ; similarly, for every triangle in , create a triangle whose boundary is made of the three segments contained in that triangle. By abuse of language, we identify with that topological space. To emphasize that we consider the abstract simplicial complex and not only the topological space, we sometimes use the name triangulation or triangulated complex.
In this paper, graphs are finite, undirected, and may have loops and multiple edges. In a similar way as for 2-complexes, each graph has an associated topological space; an embedding of a graph into a 2-complex is an injective continuous map from (the topological space associated to) to (the topological space associated to) .
2.2 Structural aspects of 2-complexes
We say that a 2-complex contains a 3-book if some three distinct triangles share a common segment.
Let be a node of . A cone at is a cyclic sequence of triangles , all incident to , such that, for each , the triangles and share a segment incident with , and any other pair of triangles have only in common. A corner at is an inclusionwise maximal sequence of triangles , all incident to , such that, for each , the triangles and share a segment incident with , and any other pair of triangles have only in common. An isolated segment at is a segment incident to but not incident to any triangle.
If contains no 3-books, the set of segments and triangles incident with a given node of are uniquely partitioned into cones, corners, and isolated segments. We say that is a regular node if all the segments and triangles incident to form a single cone or corner. Otherwise, is a singular node. A 2-complex is pure if it contains no isolated segment, and each node is incident to at least one segment.
2.3 Embedding extension problems and reductions
An embedding extension problem (EEP) is a decision problem defined as follows:
Input: A graph with vertices and edges, a subgraph of with vertices and edges, and an embedding of into a 2-complex with simplices.
Question: Does have an embedding into whose restriction to is ?
To be precise, we will have to explain how we represent the embedding , but this will vary throughout the proof, and we will be more precise about this in subsequent sections. Let us simply remark that, since the complexity of our algorithm is a polynomial of large degree (depending on the complex ) in the size of the input graph, the choice of representation is not very important, because converting between any two reasonable representations is possible in polynomial time.
We will reduce our original problem to more and more specialized EEPs. We will use the word “reduce” in a somewhat sloppy sense: A decision problem reduces to instances of the decision problem if solving these instances of allows to solve the instance of in time . We will have to be more precise when we consider the NP-completeness of Embed in Section 8.
In Section 6, we will assume some familiarity with surface topology; see, e.g., [25, 29, 5] for suitable introductions under various viewpoints. We recall some basic definitions and properties. A surface is a compact, connected Hausdorff topological space in which every point has a neighborhood homeomorphic to the plane. Every surface is obtained from a sphere by:
either removing open disks and attaching a handle (a torus with an open disk removed) to each resulting boundary component, for an even, nonnegative number called the (Euler) genus of ; in this case, is orientable;
or removing open disks and attaching a Möbius band to each resulting boundary component, for a positive number called the genus of ; in this case, is non-orientable.
A surface with boundary is obtained from a surface by removing a finite set of interiors of disjoint closed disks. The boundary of each disk forms a boundary component of . A possibly disconnected surface is a disjoint union of surfaces. An embedding of into a surface , possibly with boundary, is cellular if each face of the embedding is homeomorphic to an open disk. If is cellularly embedded on a surface with genus and boundary components, with vertices, edges, and faces, then Euler’s formula stipulates that .
An ambient isotopy of a surface with boundary is a continuous family of self-homeomorphisms of such that is the identity.
3 Reduction to complexes containing no 3-book
The following folklore observation allows us to solve the problem trivially if contains a 3-book. We include a proof for completeness.
If contains a 3-book, then every graph embeds into .
Let be a graph. We first draw , possibly with crossings, in general position in the interior of a closed disk . Let be a simple curve in with endpoints on and passing through all crossing points of the drawing of . By perturbing , we can ensure that, in the neighborhood of each crossing point of that drawing, coincides with the image of one of the two edges involved in the crossing. See Figure 1, left.
Let be a closed disk disjoint from . We attach to by identifying with a part of the boundary of . Now, in the neighborhood of each crossing of the drawing of , we push inside the part of the edge coinciding with , keeping its endpoints fixed. See Figure 1, right. This removes the crossings.
So embeds in the topological space obtained from by attaching a part of the boundary of along . But this space embeds in , because contains a 3-book. ∎
4 Reduction to EEPs on a pure 2-complex
Our next task is to reduce the problem Embed to a problem on a pure 2-dimensional complex. More precisely, let EEP-Sing be the problem EEP, restricted to instances where: is a pure 2-complex containing no 3-books; is a set of vertices of ; and is an injective map from to the nodes of such that contains all singular nodes of . In this section, we prove the following result.
Any instance of Embed reduces to instances of EEP-Sing.
First, a definition. Consider a map , where is a set of nodes in containing all singular nodes of . We say that an embedding of respects if, for each , the following holds: If , then is not in the image of ; otherwise, .
In this section, we will need the following intermediate problem:
Input: A graph with vertices and edges, a 2-complex (not necessarily pure) containing no 3-books, with simplices, and a map as above, with domain of size .
Question: Does have an embedding into respecting ?
Any instance of Embed reduces to instances of
By Proposition 3.1, we can without loss of generality assume that contains no 3-books. Let be the graph obtained from by subdividing each edge times, where is the number of singular nodes of . We claim that has an embedding into if and only if has an embedding into such that each singular node of in the image of is the image of a vertex of .
Indeed, assume that has an embedding on . Each time an edge of is mapped, under , to a singular node of , we subdivide this edge and map this new vertex to ; the image of the embedding is unchanged. This ensures that only vertices are mapped to singular nodes. Moreover, there were at most subdivisions, one per singular node. So, by further subdividing the edges until each original edge is subdivided times, we obtain an embedding of to such that only vertices are mapped on singular nodes. The reverse implication is obvious: If has an embedding into , then so has . This proves the claim.
To conclude, for each map from the set of singular vertices of to , we solve the problem whether has an embedding on respecting . The graph embeds on if and only if the outcome is positive for at least one such map . By construction, there are at most such maps, because has size at most . ∎
Embed-Resp reduces to EEP-Sing.
The proof is a bit long, and we refer to Appendix A, but the idea is simple: Because we restrict ourselves to embeddings that respect , and thus specify which vertex of is mapped to each of the singular nodes of , what happens on the isolated segments of is essentially determined. ∎
Before giving the proof, we first note:
5 Reduction to an EEP on a possibly disconnected surface
The previous section led us to an embedding extension problem in a pure 2-complex without 3-book where the images of some vertices are predetermined. Now, we show that solving such an EEP amounts to solving another EEP in which the complex is a surface.
Let EEP-Surf be the problem EEP, restricted to instances where the input complex is (homeomorphic to) a possibly disconnected triangulated surface without boundary (which we denote by instead of , for clarity). To represent the embedding in such an EEP instance , it will be convenient to use the fact that, in all our constructions below, the image of every connected component of under will intersect the 1-skeleton of at least once, and finitely many times. (Note that may use some nodes of .) Consider the overlay of the triangulation of and of , the union of the 1-skeleton of and of the image of ; this overlay is the image of a graph on ; each of its edges is either a piece of the image of an edge of or a piece of a segment of ; each of its vertices is the image of a vertex of and/or a node of . By the assumption above on , this overlay is cellularly embedded on , and we can represent it by its combinatorial map [19, 10] (possibly on surfaces with boundary, since at intermediary steps of our construction we will have to consider such surfaces).
In this section, we prove the following proposition.
Any instance of EEP-Sing reduces to an instance of EEP-Surf.
We will first reduce the original EEP to an intermediary EEP on a surface with boundary.
Any instance of EEP-Sing reduces to an instance of EEP in which the considered 2-complex is a possibly disconnected surface with boundary.
The key property that we will use is that, since is pure and contains no 3-books, each singular node is incident to cones and corners only.
Figure 2 illustrates the proof. Let be the instance of EEP-Sing. We first describe the construction of the instance on the possibly disconnected surface with boundary. Let be a singular node; we modify the complex in the neighborhood of as follows. Let be the number of cones at and be the number of corners at . We remove a small open neighborhood of from , in such a way that the boundary of is a disjoint union of circles and arcs. We create a sphere with boundary components. Finally, we attach each circle and arc to a different boundary component of , choosing an arbitrary orientation for each gluing; circles are attached to an entire boundary component of , while arcs cover only a part of a boundary component of . Doing this for every singular node , we obtain a surface (possibly disconnected, possibly with boundary), which we denote by .
We now define , , and from , , and (again, refer to Figure 2). Let be a singular node of and the vertex of mapped on by . In (and thus also ), we add a set of loops with vertex . Let be a point in the interior of ; in , we map to , and we map these loops on in such a way that each loop encloses a different boundary component of (thus, if we cut along these loops, we obtain annuli and one disk).
Finally, we add to (and thus also to ) a set of new edges, each connecting to a new vertex. In , each new vertex is mapped to the boundary component of corresponding to a corner, but not on the corresponding arc.
Let us call and the resulting graphs, and the resulting embedding of . Note that, from the triangulation of with simplices, we can easily obtain a triangulation of with simplices, and that the image of each edge of crosses edges of this triangulation.
There remains to prove that these two EEPs are equivalent; see Appendix B. ∎
We now deduce from the previous EEP the desired EEP on a surface without boundary. Any instance of EEP-Sing on a possibly disconnected surface with boundary reduces to an instance of EEP-Surf.
(Figure 4, in Appendix, illustrates the proof.) Let be an instance of an EEP on a possibly disconnected surface with boundary. We first describe the construction of , the EEP instance on a possibly disconnected surface without boundary.
Let be obtained from by gluing a disk along each boundary component. Let be a boundary component of . If maps at least one vertex to , then we add to (and thus also to ) a new vertex , which we connect, also by a new edge, to each of the vertices mapped to by . We extend by mapping vertex and its incident edges inside . Let us call and the resulting graphs, and the resulting embedding of . For each vertex of on a boundary component, we added to and at most one vertex and one edge. There remains to prove that these two EEPs are equivalent; see Appendix B. ∎
6 Reduction to a cellular EEP on a surface
Let EEP-Cell be the problem EEP, restricted to instances where is a surface and is cellularly embedded and intersects each connected component of .
In this section, we prove the following proposition.
Any instance of EEP-Surf reduces to instances of EEP-Cell.
As will be convenient also for the next section, we do not store an embedding of a graph on a surface by its overlay with the triangulation, as was done in the previous section, but we forget the triangulation. In other words, we have to store the combinatorial map corresponding to , but taking into account the fact that is not necessary cellular: We need to store, for each face of the embedding, whether it is orientable or not, and a pointer to an edge of each of its boundary components (with some orientation information). Such a data structure is known under the name of extended combinatorial map [6, Section 2.2] (only orientable surfaces were considered there, but the data structure readily extends to non-orientable surfaces).
6.1 Reduction to connected surfaces
We first build intermediary EEPs over connected surfaces. Let EEP-Conn be the problem EEP, restricted to instances where is a surface (connected and without boundary) and intersects every connected component of .
Any instance of EEP-Surf reduces to instances of EEP-Conn. More precisely (and this is a fact that will be useful to prove that Embed is in NP, see Theorem 1.1), any instance of EEP-Surf is equivalent to the disjunction (OR) of instances, each of them being the conjunction (AND) of instances of EEP-Conn.
Sketch of proof.
See Appendix C for details. We can embed each connected component of that is planar and disjoint from anywhere. There remains connected components of that are disjoint from . For each of these, we choose a vertex, and we guess in which face of it belongs. We then know which connected component of the surface each connected component of is mapped into. ∎
6.2 The induction
The strategy for the proof of Proposition 6 is as follows. For each EEP from the previous lemma, we will extend to make it cellular, by adding either paths connecting two boundary components of a face of , or paths with endpoints on the same boundary component of a face of in a way that the genus of the face decreases. We first define an invariant that will allow to prove that this process terminates.
Let be an embedding of a graph on a surface . The cellularity defect of is the non-negative integer
Some obvious remarks: can contain isolated vertices; by convention, each of them counts as a boundary component of the face of it lies in. With this convention, every face of has at least one boundary component, except in the very trivial case where is empty. This implies that is a cellular embedding if and only if .
The following lemma reduces an EEP to EEPs with a smaller cellularity defect.
Any instance of EEP-Conn reduces to instances
of EEP-Conn where .
The reduction does not depend on the size of ; furthermore, the new graph is obtained from the old one by adding exactly one edge and no vertex.
Proof of Proposition 6.
We first apply Lemma 6.1, obtaining instances of EEP-Conn. To each of these EEPs, we apply recursively Lemma 6.2 until we obtain cellular EEPs. The cellularity defect of the initial instance is , being at most the genus of plus , because each boundary component of a face of is incident to at least one edge of (and each edge accounts for at most two boundary components in this way) or to one isolated vertex of . Thus, the number of instances of EEP-Cell at the bottom of the recursion tree is , in which the size of the graph is and the surface has at most simplices. ∎
6.3 Proof of Lemma 6.2
There remains to prove Lemma 6.2. The proof uses some standard notions in surface topology, homotopy, and homology; we refer to textbooks and surveys [25, 29, 5]. We only consider homology with coefficients.
Let be a surface with a single boundary component and let be a path with endpoints on the boundary of . If we contract this boundary component to a single point, the path becomes a loop, which can be null-homologous or non-null-homologous. We employ the same adjectives null-homologous and non-null-homologous for the path . Recall that, if is simple, it separates if and only if it is null-homologous. The reversal of a path is denoted by . The concatenation of two paths and is denoted by .
Let be a surface with boundary, let be a point in the interior of , and let , , and be points on the boundary of . For each , let be a path connecting to . Let , , and . Let be a possible property of the paths , among the following ones:
“the endpoints of lie on the same boundary component of ”;
“ is null-homologous” (if has a single boundary component).
Then the following holds: If both and have property , then so does .
This is a variant on the 3-path condition from Mohar and Thomassen [25, Section 4.3]. The first item is immediate. The second one follows from the fact that homology is an algebraic condition: The concatenation of two null-homologous paths is null-homologous, and removing subpaths of the form from a path does not affect homology. ∎
Proof of Lemma 6.2.
Since , there must be a face of with either (1) several boundary components, or (2) a single boundary component but positive genus. We will consider each of these cases separately, but first introduce some common terminology.
Let be an arbitrary spanning forest of rooted at . This means that is a subgraph of that is a forest with vertex set such that each connected component of contains exactly one vertex of , its root. The algorithm starts by computing an arbitrary such forest in linear time.
For each vertex of , let be the unique root in the same connected component of as , and let be the unique path connecting to . If and are two vertices of , let be the graph obtained from by adding one edge, denoted , connecting and . (This may be a parallel edge if and were already adjacent in , but in such a situation when we talk about edge we always mean the new edge.) Let be the unique path between and in that is the concatenation of , edge , and .
Case 1: has several boundary components.
Assume that has a solution . We claim that, for some vertices and of , the embedding extends to an embedding of in which the image of the path lies in and connects two distinct boundary components of .
Indeed, let be a curve drawn in connecting two different boundary components of . We can assume that it intersects the boundary of exactly at its endpoints, at vertices of . We can deform so that it intersects only at the images of vertices, and never in the relative interior of an edge. We can, moreover, assume that is simple (except perhaps that its endpoints coincide on ). Let be the vertices of encountered by , in this order. We denote by the part of between vertices and . We claim that, for some , we have that connects two different boundary components of : Otherwise, by induction on , applying the first case of Lemma 6.3 to the three paths , , and , we would have that, for each , has its endpoints on the same boundary component of , which is a contradiction for (for which the curve is ). So let be such that connects two different boundary components of ; letting and , and embedding edge as , gives the desired embedding of . This proves the claim.
The strategy now is to guess the vertices and and the way the path is drawn in , and to solve a set of EEPs where is chosen as an appropriate extension of . Let us first assume that is orientable. One subtlety is that, given and , there can be several essentially different ways of embedding inside , if there is more than one occurrence of and on the boundary of . So we reduce our EEP to the following set of EEPs: For each choice of vertices and of , and each occurrence of and on the boundary of , we consider the EEP where extends and maps to an arbitrary path in connecting the chosen occurrences of and on the boundary of .
It is clear that, if one of these new EEPs has a solution, the original EEP has a solution. Conversely, let us assume that the original EEP has a solution; we now prove that one of these new EEPs has a solution. By our claim above, for some choice of and , some EEP has a solution, for some mapping inside and connecting different boundary components of . In that mapping, connects two occurrences of and inside . We prove that, for these choices of occurrences of and , the corresponding EEP described in the previous paragraph, , has a solution as well. These two EEPs are the same except that the path may be drawn differently in and , although they connect the same occurrences of and on the boundary of . Under , the face is transformed into a face that has the same genus and orientability character as , but one boundary component less. The same holds, of course, for . Moreover, the ordering of the vertices on the boundary components of the new face is the same in and . Thus, there is a homeomorphism of that keeps the boundary of fixed pointwise and such that . This homeomorphism, extended to the identity outside , maps any solution of to a solution of , as desired.
It also follows from the previous paragraph that the cellularity defect decreases by one. To conclude this case, we note that the number of new EEPs is : indeed, there are possibilities for the choice of (or ), and possibilities for the choice of the occurrence of (or ) on the boundary of .
If is non-orientable, the same argument works, except that there are two possibilities for the cyclic ordering of the vertices along the new boundary component of the new face: If we walk along one of the boundary components of (in an arbitrary direction), use , and walk along the other boundary component of , we do not know in which direction this second boundary component is visited. So we actually need to consider two EEPs for each choice of , , and occurrences of and , instead of one. The rest is unchanged.
Case 2: has a single boundary component and positive genus. The proof is very similar to the previous case, the main difference being that, instead of paths in connecting different boundary components of , we now consider paths in that are non-null-homologous. See Appendix C.∎
7 Solving a cellular EEP on a surface
There is a function such that every instance of EEP-Cell can be solved in time .
This is essentially the main result of Mohar . The algorithm in  makes reductions to even more specific EEPs, and one feature that is needed for bounding the number of new instances is that the embedded subgraph satisfies some connectivity assumptions. Our Appendix D provides more details. ∎
We can finally prove our main theorems. First, let us prove that we have an algorithm with complexity :
Proof of Theorem 1.2.
Finally, we prove that the Embed problem is NP-complete:
Proof of Theorem 1.1.
We give a detailed proof in Appendix E. The problem Embed is NP-hard because deciding whether an input graph embeds into an input surface is NP-hard . It is in NP because (assuming contains no 3-books), a certificate that an instance is positive is given by a certificate that some instance of EEP-Surf is positive (see the proof of Proposition 5). Such an instance has a certificate, given by the combinatorial map of a supergraph of cellularly embedded on (see Section 6), that can be checked in polynomial time. ∎
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Appendix A Omitted proof from Section 4
Proof of Lemma 4.
Formally, we describe a set of transformations on , , and . The invariant is that they preserve the existence or non-existence of an embedding of into respecting .
Step 1. We start by dissolving all the degree-two vertices of that are not in the image of . (We recall that dissolving a degree-two vertex , incident to edges and , means removing and replacing and with a single edge .) It is clear that the original graph has an embedding on respecting if and only if the new graph (still called ) has an embedding on respecting .
Step 2. If a singular node is not used, then we can remove it without affecting the embeddability of . However, removing a vertex from a 2-complex does not yield a 2-complex. We thus define a 2-complex that has the same properties. Let be a singular node of a 2-complex . Let be the set of triangles and segments of incident with , uniquely partitioned into , where each is either a cone, a corner, or an isolated segment. The withdrawal of from is the complex obtained by doing the following operation for each : We first create a new node , and then replace, in each triangle and edge of , the node by .
For every singular node of such that and incident to at least two segments, we withdraw from . For each created node , we let . The fact that embeds, or not, on respecting is preserved: Indeed, if embeds on the original complex respecting , then this corresponds to an embedding on the complex obtained by withdrawing , and avoiding the s, thus respecting ; conversely, if embeds on the complex obtained by withdrawing , respecting , then it avoids the s, and, after identifying together the s to a single point , this embedding avoids , and thus respects .
Step 3. At this point, every singular node with is incident to exactly one segment, and to no triangle. For each isolated segment of such that and are both different from , and contains an edge connecting and , we remove from and remove from . We need to prove that this operation does not affect the (non-)existence of an embedding of respecting . First, assume that, initially, there was an embedding of on respecting ; then either segment is used by , in which case clearly there is still an embedding after this operation, or edge does not use segment at all, in which case we can first modify by embedding edge on segment and by moving the edges on on the space where was before, so we are now in the previous case. Conversely, if after this operation has an embedding on respecting , trivially it is also the case before.
Step 4. For each isolated segment of such that and are both different from , but contains no edge connecting and , we do the following. In , we remove and add a new segment where and are new nodes; we also extend by letting . Finally, if contains at least one edge of the form , where has degree one and is not in the image of , we remove a single such edge; similarly, if it contains at least one edge of the form , where has degree one and is not in the image of , we remove a single such edge. This operation does not affect our invariant, for similar reasons; for example, if initially there was an embedding of respecting , then segment can only contain edges of that are themselves connected components of , which we can re-embed on the new segment , or edges of the form or , where and have degree one and are not in the image of .
Step 5. For each isolated segment where is different from but , we remove from and add a new segment where and are new nodes; we also extend by letting . Finally, if contains at least one edge of the form , where has degree one and not in the image of , we remove a single such edge. The invariant is preserved, for reasons similar to the previous case.
Step 6. Now, every segment of the complex (still called ) is incident to one or two triangles, except perhaps some segments that are themselves connected components of and whose endpoints are not in the domain of . If there is at least one such segment, we remove all of them from , and remove all edges from that are themselves connected components of and such that and are not in the image of ; as above, this does not affect whether embeds into respecting .
Step 7. Finally, for each node of incident to no segment, such that is different from , we do the following: If has degree zero, we remove and ; otherwise, we immediately return that does not embed into respecting . For each node of that is incident to no segment and such that , we remove , and remove a single degree-zero vertex of not in the image of , if one exists.
Conclusion. Now, has no node that is itself a connected component; each of its segments is incident to one or two triangles. Also, maps each singular node of to a vertex of . It may map some other nodes of to , but such nodes are non-singular and, if an embedding uses them, a slight perturbation will avoid them, so we can remove the nodes such that from the domain of without affecting the result. Now, we have an EEP as specified in the statement of Proposition 4.
Finally, it is easy to check that, in each of the seven steps above, the numbers of vertices and edges of do not increase, and the number of simplices of increase by at most a multiplicative constant. Moreover, the domain of also does not increase (it increases when we withdraw singular nodes, but the images of the new nodes are , and such nodes are later removed from the domain of ). ∎
Appendix B Omitted proofs from Section 5
End of proof of Lemma 5.
There remains to prove that the two EEPs constructed are equivalent.
Let us first prove that any solution of the EEP instance yields a solution of the EEP instance . Let be a singular node of , and let be the vertex of mapped to . We need to modify locally in the neighorhood of that is removed when transforming to . Without loss of generality, up to an ambient isotopy of that does not move , we can assume that the image of intersects exactly in straight line segments having as one endpoint. To build a solution of , we first remove the part of inside , and we reconnect to each point of the image of lying on the boundary of , by paths on ; this is certainly possible because each point of not in the image of can be connected to by a path that does not meet (except at ). Thus, has a solution.
Conversely, let be a solution of the EEP ; we build a solution of . As above, let be a point of that was obtained from a singular node of . We now show that we can assume that does not enter , except for , , , and the edges in incident to :
First, the part of lying in the connected component of minus the image of that is a disk corresponds to a planar subgraph of , connected to the rest of by only; we can re-embed this planar subgraph of in . Next, consider a cone at . The part of enclosed by the corresponding loop of is an annulus; the situation is as on Figure 3, top, and, by an ambient isotopy of , we can push the part of that lies in the annulus outside it, except for those edges touching . Finally, consider a corner at , and the annulus that is the part of enclosed by the corresponding loop in . The local picture for this part of is as shown on Figure 3, bottom, and similarly by an ambient isotopy of we can push the part of on outside it, except for those edges touching .
Now, a solution of can be obtained by the following procedure, for each of the singular nodes : (1) Remove the sphere , together with the image of inside ; (2) add the neighborhood of ; (3) reconnect to the points on the image of that lie on the boundary of . ∎
End of proof of Lemma 5.
Clearly, any solution of yields a solution of
. Conversely, let be a solution of ; we build a solution of . Let be a boundary of . There are two cases:
If no vertex of is mapped to , then maps outside the closure of , so, by an ambient isotopy of that does not move , we can push outside .
Otherwise, the disk is split into sectors by the edges incident to ; the image of by does not enter the interior of each sector, and intersects the boundary of a sector exactly along the image of and of two edges incident to . Thus, by an ambient isotopy that keeps the image of fixed, we can push the image of out of each sector.
After doing this for every boundary component , we obtain that the restriction of to is a solution of . ∎
Appendix C Omitted proofs from Section 6
Proof of Lemma 6.1.
We start by removing the connected components of that are planar and disjoint from . (Testing planarity takes linear time .) This does not change the solution of the EEP, because such connected components can be embedded on an arbitrarily small planar portion of (provided is non-empty, but otherwise the original EEP-Surf instance can be solved trivially). So without loss of generality, every connected component of disjoint from is non-planar. Let be an arbitrary set of vertices, one per non-planar connected component of disjoint from . Without loss of generality, the number of vertices in is at most the genus of , which is at most , because otherwise the initial EEP has no solution.
Let ; every connected component of intersects . For each vertex of , we guess the face of it has to be embedded in, and extend accordingly, by adding the images of in ; let be the resulting embedding of . By the previous paragraph, the number of these guesses is at most . It is clear that the initial EEP has a solution if and only if one of these EEPs has a solution.
These EEPs are almost of the form announced in the lemma, except that can be disconnected. However, in any solution of this EEP, we know the connected component of each connected component of has to embed in, because each connected component of intersects . We can thus reformulate the EEP as the conjunction of several EEPs, one per connected component of . (Of course, we can discard the connected components of disjoint from .) ∎
End of proof of Lemma 6.2.
There remains to prove Case 2: has a single boundary component and positive genus.
The proof is very similar to Case 1, the main difference being that, instead of paths in connecting different boundary components of , we now consider paths in that are non-null-homologous (if is orientable) or one-sided (if is non-orientable). For brevity we call such paths essential. Observe that Lemma 6.3 is similarly valid for the property “ is one-sided” (if has a single boundary component).
Assume that has a solution . We claim that, for some vertices and of , the embedding extends to an embedding of in which the image of the path lies in and is essential. The proof is similar in spirit to the corresponding claim in Case 1: We let be an essential curve in intersecting the boundary of exactly at its endpoints; we can assume similarly that it is simple and intersects only vertices of , in the order . For some , must be essential, by induction and by Lemma 6.3; this gives an embedding of .
If is orientable, we reduce the original EEP to the following set of EEPs: For each choice of vertices and of , and each occurrence of and on the boundary of , we consider the EEP where extends and maps to an arbitrary path in connecting the chosen occurrences of and on the boundary of in such a way that is non-null-homologous. As before, the only subtlety is to prove that, if we have two EEPs and such that are not embedded exactly in the same way in and , but are non-null-homologous in and connect the same occurrences of and on the boundary of , then these EEPs are equivalent. This follows from the fact that the image of in cuts into a face that is an orientable surface with two boundary components and with (Euler) genus that of minus two (because is non-null-homologous in both embeddings, and thus non-separating in both cases). Moreover, the ordering of the vertices along the boundary components of the new face is the same in both and . Thus, as before, there is a homeomorphism of that keeps the boundary of fixed pointwise and such that , which similarly concludes. It also follows that the cellularity defect decreases by two. The number of these new EEPs is, similarly, .
If is non-orientable and the genus of is either one or even, then by Euler’s formula, a similar argument can be used: Regardless of the way we draw as a path connecting the chosen occurrences of and on the boundary of in such a way that is one-sided, cutting along results in a surface whose topology is uniquely determined (it is a disk if the genus of is one, and a non-orientable surface with genus that of minus one, if the genus of is even). Moreover, the ordering of the vertices along the boundary of the new face is uniquely determined. The cellularity defect decreases by one, and the same argument as above concludes.
Finally, if is non-orientable and its genus is odd and at least three
is non-orientable and its genus is odd and at least three, then cutting along such a one-sided path results in a surface in which the ordering of the vertices along the single boundary component is uniquely determined, but this surface, with genus that of minus one, can be orientable or not. Thus, for each choice of vertices and of , and each occurrence of and on the boundary of , we actually need to consider two EEPs, one in which is mapped to a path that cuts into an orientable surface, and one in which is mapped to a path that cuts into a non-orientable surface. The rest of the argument is unchanged. ∎
Appendix D Omitted proof from Section 7
When solving EEPs on surfaces, a useful property of the embedded subgraph is the following one. We say that a subgraph of has property (E) if has no local bridges. This means that every path of , all of whose vertices have degree at most two in , is an induced path in , and every connected component of is adjacent to a vertex in . In fact, what we need is a weaker version of property (E) where we first prescribe a subset of vertices of , where contains all vertices whose degree in is different from two, and possibly a constant number of vertices whose degree in is equal to two. Then we say that has property (E) with respect to if the above property holds for every path that is disjoint from .
Proof of Proposition 7.
This is essentially the main result from . The algorithm from  first reduces the problem to an instance such that satisfies property (E) with respect to a subset that contains all vertices of of degree different from two (all these are also vertices in of degree different from two) plus a constant number of vertices of degree two in , where this constant number is bounded from above in terms of the genus of , which is itself bounded from above by . For this purpose,  relies on another paper . After achieving this property, the paper  reduces the EEP to a constant number (where the constant depends on ) of “simple” extension problems [24, Section 4], which are then solved in [24, Theorem 5.4]. ∎
Appendix E Omitted proofs from Section 8
End of proof of Theorem 1.2.
Consider an instance of Embed. Proposition 3.1 allows to discard the 2-complexes containing a 3-book. Proposition 4 reduces the problem to instances of EEP-Sing. Proposition 5 reduces each such instance into an instance of EEP-Surf. Proposition 6 reduces that instance into instances of EEP-Cell. Proposition 7 shows that each such instance can be solved in time . ∎
Proof of Theorem 1.1.
Let us first prove that the problem is NP-hard. The following problem Graph-Genus is NP-hard: Given a graph and an integer , decide whether embeds on the orientable surface of genus . This almost immediately implies that Embed is NP-hard; the only subtlety is that in Graph-Genus, is specified in binary, thus more compactly than a triangulated surface of genus (and thus triangles). To be very precise, given an instance of Graph-Genus, we transform it in polynomial time into an equivalent instance of Embed as follows: If has at most edges, then we transform it into a constant-size positive instance of Embed (every graph with edges embeds on the orientable surface of genus ); otherwise, we consider the instance where is a 2-complex that is an orientable surface of genus ; since has at least edges, the transformation takes polynomial time in the size of .
We now prove that the problem Embed belongs to NP. The case where contains a 3-book is trivial; let us assume that it is not the case. The proof of Proposition 5 shows that an Embed instance is positive if and only if at least one instance of EEP-Surf, among of them, is positive. The certificate indicates which of these instances is positive (this requires a polynomial number of bits), together with a certificate that this instance is indeed positive (see below). To check this certificate, the algorithm builds the corresponding instance of EEP-Surf (as done in Section 5—this takes polynomial time) and checks the certificate.
Here is a way to provide a certificate for an instance of EEP-Surf. In Section 6, we have proved that, if we have an instance of EEP-Surf, then there exists a cellular embedding (in the form of a combinatorial map) of a graph containing , and such that extends . Moreover, is obtained from by adding a number of edges that is , where is the size of the original complex. (Recall that in the instance of EEP-Surf, the size of is .) The cellular embedding of , given as a combinatorial map, is the certificate that is positive: Given and this certificate, we can in polynomial time check that contains , that the restriction of to is indeed , and that the combinatorial map of is indeed an embedding on . ∎