Abstract 1 Introduction 2 Preliminaries 3 NP-Hardness of Outerthickness 4 NP-hardness of 𝓕-Thickness 5 Concluding Remarks References

Determining the Outerthickness of Graphs Is NP-Hard

Pin-Hsian Lee ORCID National Taiwan University, Taipei, Taiwan    Te-Cheng Liu ORCID Academia Sinica, Taipei, Taiwan    Meng-Tsung Tsai ORCID Academia Sinica, Taipei, Taiwan
Abstract

We give a short, self-contained, and easily verifiable proof that determining the outerthickness of a general graph is NP-hard. This resolves a long-standing open problem on the computational complexity of outerthickness.

Moreover, our hardness result applies to a more general covering problem P,k, defined as follows. Let be a proper graph class. Let k1 be an integer parameter. Given an undirected simple graph G=(V,E), the task is to cover the edge set E(G) by at most k subsets E1,,Ek such that each subgraph (V(G),Ei) for i[k] belongs to . Note that if is monotone (in particular, when is the class of all outerplanar graphs), any such cover can be converted into an edge partition by deleting overlaps; hence, in this case, covering and partitioning are equivalent.

Our result shows that for every proper graph class that satisfies all of the following conditions: (a) is closed under topological minors, (b) is closed under 1-sums, and (c) contains a cycle of length 3, the problem P,k is NP-hard for every integer k3. In particular:

  • For equal to the class of all outerplanar graphs, our result settles the long-standing open problem on the complexity of determining outerthickness.

  • For equal to the class of all planar graphs, our result complements Mansfield’s NP-hardness result (1983) for the thickness, which applies only to the case k=2.

It is also worth noting that each of the three conditions above is necessary. If is the class of all eulerian graphs, then condition (a) fails. If is the class of all pseudoforests, then condition (b) fails. If is the class of all forests, then condition (c) fails. For each of these three classes , the problem P,k is solvable in polynomial time for every integer k3, showing that none of the three conditions can be dropped unless P=NP.

Keywords and phrases:
outerthickness, outerplanar graphs, edge partition
Category:
Track A: Algorithms, Complexity and Games
Funding:
Meng-Tsung Tsai: This research was supported in part by the National Science and Technology Council under contract NSTC 114-2221-E-001-023.
Copyright and License:
[Uncaptioned image] © Pin-Hsian Lee, Te-Cheng Liu, and Meng-Tsung Tsai; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation Problems, reductions and completeness
Related Version:
Full Version: https://arxiv.org/abs/2602.07607
Acknowledgements:
We want to thank the anonymous reviewers for their helpful comments.
Editors:
Sayan Bhattacharya, Danupon Nanongkai, Michael Benedikt, and Gabriele Puppis

1 Introduction

Guy (1974) [16] defined the outerthickness of a graph G as the minimum number of parts in an edge partition of G such that each part induces an outerplanar subgraph; that is, a planar graph admitting an embedding in the plane in which every node lies on the outer face. Outerthickness is an analogue of thickness, except that each part is required to be outerplanar rather than merely planar. Although Mansfield (1983) proved that determining the thickness of a graph is NP-hard [24], the complexity of determining the outerthickness has remained unsettled [2, 3, 18, 23, 27, 28, 29, 30].

The algorithmic study of outerthickness is motivated in part by a conjecture of Chartrand et al. (1971) [9], who proposed, among other things, that every planar graph admits an edge partition into two outerplanar graphs; equivalently, that every planar graph has outerthickness at most 2. Elmallah and Colbourn (1988) [13] cited this conjecture and addressed it for 4-node-connected planar graphs by presenting an algorithm that produces such a partition into two outerplanar subgraphs. Although they did not explicitly analyze the running time, using the state of the art of the required subroutines at the time [20, 15], their algorithm can be implemented in O(n3) time, where n denotes the number of nodes in the input graph (and we use this convention throughout the paper).

Subsequently, Heath (1991) claimed a proof of Chartrand et al.’s conjecture and gave an O(n2logn)-time algorithm to partition the edges of any planar graph into at most two outerplanar graphs [17]. Later, Gonçalves (2005) established the same statement with an optimal O(n)-time construction [14]. It is worth noting that Gonçalves remarked that some earlier claimed proofs were later found to be incorrect, and subsequent work attribute the resolution of Chartrand et al.’s conjecture to Gonçalves [11, 32, 10, 4].

Unlike the substantial progress on planar graphs, there has been little progress on determining outerthickness for general graphs. Mäkinen and Poranen (2004) conjectured in [27, 29] that deciding the outerthickness of a general graph is NP-hard. Since then, the open status of this problem has been repeatedly highlighted in subsequent work [28, 3, 23, 30], and was reiterated as recently as 2023 by Hlinený and Masarík [18] and 2024 by Balko et al. [2]. We further note that Balko et al. prove NP-hardness for computing the uncrossed number under the assumption that computing outerthickness is NP-hard [2].

In this paper, we settle the computational complexity of determining the outerthickness of general graphs, a question that has remained open since outerthickness was introduced in 1974. Formally, we consider the following decision problem.

Problem 1 (Outerthickness).
  • Input: An n-node undirected simple graph G=(V,E).

  • Parameter: An integer k1.

  • Question: Does G admit an edge partition E=E1E2Ek such that (V,Ei) is outerplanar for every i[k]?

We use k-OuterThickness(G) to denote Problem 1 with input graph G and parameter k for short. Our main result is the following.

Theorem 2.

For every integer k3, k-OuterThickness(G) for general graphs G is NP-complete.

Moreover, our hardness result applies to a more general covering problem P,k, defined as follows. Let be a proper graph class, where a graph class is called proper if it does not contain all graphs. Let k1 be an integer parameter. Given an undirected simple graph G=(V,E), the task is to cover the edge set E(G) by at most k subsets E1,,Ek such that each subgraph (V(G),Ei) belongs to . Note that if is monotone, meaning that whenever G and H is a subgraph of G, we have H, then any such cover can be converted into an edge partition by deleting overlaps. Hence, for monotone classes , covering and partitioning are equivalent. In particular, the class of all outerplanar graphs is monotone.

We generalize the NP-hardness of k-OuterThickness(G) to P,k for every proper that satisfies the three conditions stated below:

Theorem 3.

For every proper graph class that satisfies all of the following conditions:

  1. (a)

    is closed under topological minors111A graph class is closed under topological minors if for every G and every graph H that is a topological minor of G, we have H. Here H is a topological minor of G if some subdivision of H appears as a subgraph of G (equivalently, H can be obtained from a subgraph of G by repeatedly smoothing degree-2 nodes). To smooth a degree-2 node means to remove the node and join its two neighbors by an edge.,

  2. (b)

    is closed under 1-sums222A graph class is closed under 1-sums if for every G1,G2 and every choice of nodes v1V(G1) and v2V(G2), the graph obtained from the disjoint union G1G2 by identifying v1 and v2 into a single node also belongs to ., and

  3. (c)

    contains a cycle of length 3,

the problem P,k is NP-hard for every integer k3.

In our NP-hardness reduction, we need an indicator array A,q that records, for each graph with at most q nodes, whether it belongs to , where q is a function of k. Since there are 2O(q2) such graphs, A,q is a bit string of length 2O(q2). Thus, the existence of the appropriate array is sufficient to establish, nonconstructively, the existence of a valid NP-hardness reduction, even if there is no uniform procedure for computing this array from and q. However, if one requires the reduction to be constructible within some bounded amount of resources for algorithmic purposes, then one must impose an additional condition on to ensure that A,q can be computed within those resource bounds.

Moreover, if membership in can be decided by an algorithm M in polynomial time, then P,kNP. A certificate is an edge covering of E(G) by k subsets Ei for i[k], and we can verify in polynomial time that (V(G),Ei) for every i[k] by running M on each part. Hence the certificate certifies that the -thickness of G is at most k. As a result, Theorem 3 also yields NP-completeness if membership in can be decided in polynomial time.

As a corollary of Theorem 3, by setting to be the class of all planar graphs, we obtain Corollary 4, which complements Mansfield’s NP-hardness result (1983) for the thickness [24], established only for the case k=2.

Corollary 4.

For every integer k3, k-Thickness(G) for general graphs is NP-complete.

It may be worth noting that Mansfield’s NP-hardness proof for the thickness proceeds via a reduction from Planar-3SAT [22] due to Lichtenstein (1982), which is technical. In contrast, our hardness proof uses other source problems: for k=3, we reduce from the chromatic index problem for 3-regular graphs due to Holyer (1981) [19], and for every integer k3, we reduce from the chromatic index problem for k-regular graphs due to Leven and Galil (1983) [21].

Table 1: The complexity status of problem P,k for some graph classes . Note that eulerian graphs here including disconnected ones, and tw(G) denotes the treewidth of G. The results highlighted in blue are proved in Theorem 3, and NPC stands for NP-complete. The NP-hardness of Peulerian,2 follows by observing its equivalence to 3-edge-colorability for 3-regular graphs, as implicitly noted in the proof of Proposition 4 of [26].
Cond. (a) Cond. (b) Cond. (c) k=1 k=2 each k3
forests Y Y N P P [12] P [12]
pseudo-forests Y N Y P P [12] P [12]
eulerian graphs N Y Y P NPC [26] P [1]
cacti Y Y Y P NPC
outerplanar graphs Y Y Y P [25] NPC
planar graphs Y Y Y P [20] NPC [24] NPC
graphs of tw(G)t Y Y Y P [5] NPC

As illustrated in Table 1, each of the three conditions mentioned in Theorem 3 is necessary. If is the class of all eulerian333We follow Catlin’s convention [8] and use lowercase intentionally and consistently. graphs (including disconnected ones), then condition (a) fails. If is the class of all pseudoforests, that is, the class of all undirected graphs in which each connected component contains at most one cycle, then condition (b) fails. If is the class of all forests, then condition (c) fails. For each of these three classes , the problem P,k is solvable in polynomial time for every integer k3 (the case Peulerian,k is due to Alon and Tarsi [1], and the cases Pforests,k and Ppseudoforests,k follow from Edmonds [12]), showing that none of the three conditions can be dropped unless P=NP. In addition to the three example problems above, we found that the class of all graphs whose edge sets can be partitioned into at most a constant number of forests, that is, graphs with arboricity bounded by a constant, also serves as a useful counterexample when attempting to replace the three conditions with weaker ones; see Section 5 for details.

1.1 Paper Organization

In Section 2, we introduce our notation. In Section 3, we present a proof showing that outerthickness is NP-hard, thereby settling the long-standing open question on its computational complexity. Then, in Section 4, we prove our main theorem (Theorem 3), which in particular establishes Theorem 2 for every integer k3. We conclude with remarks in Section 5.

2 Preliminaries

All graphs and subgraphs considered in this paper are finite, undirected, and simple. For a graph G, we write V(G) and E(G) for its node set and edge set, respectively. For convenience, we sometimes use an edge subset EE(G) to denote the subgraph of G induced by E, namely (V(G),E).

Let 𝒫 be a graph property. A graph G is said to be edge-maximal with respect to 𝒫 if G satisfies 𝒫, but for every pair of non-adjacent nodes u,vV(G), the graph obtained by adding the edge {u,v} to G does not satisfy property 𝒫.

For every integer m1, we use the notation [m] to denote the set of integers {1,,m}. The reduction source of our problem is defined below.

Theorem 5 (Edge-Coloring [19, 21]).

Let k3 be an integer. A proper k-edge-coloring of an undirected graph G=(V,E) is a mapping c:E{1,2,,k} such that c(e)c(f) for every pair of edges e,fE that share an endnode. For every k3, given a k-regular undirected simple graph G, deciding whether G admits a proper k-edge-coloring is NP-complete.

3 NP-Hardness of Outerthickness

Given an input instance G of Edge-Coloring on 3-regular graphs, defined in Theorem 5, we compute a label function φ defined below and an auxiliary graph H in polynomial time. Then, we obtain a supergraph G from joining G and H using the label function φ such that G admits a 3-edge-coloring if and only if G has an edge-partition into 3 outerplanar subgraphs (Lemma 7). This proves the NP-hardness stated in Theorem 2 for k=3. Moreover, 3-OuterThickness(G)NP: a certificate is an edge partition of E(G) into 3 subsets Ei for i[3], and we can verify in polynomial time that (V(G),Ei) is outerplanar for every i[3] by running the linear-time outerplanarity test of Mitchell [25] on each part. Hence the certificate certifies that the outerthickness of G is at most 3.

  • Let φ:E(G) be a labeling function, where ={1,2,,||} is a constant-size (not necessarily minimum-size) label set, such that for every path of length at most 3 in G, the edges on the path receive pairwise distinct labels under φ. The existence of φ, as well as a construction running in time polynomial in the input size of G, is guaranteed by Proposition 11.

  • We pick a sufficiently large constant C such that every C-node graph444By a C-node graph, we mean a graph with C nodes. with outerthickness 3 contains an (not necessarily maximum) independent set of α|| nodes. Then we construct a C-node graph H that is edge-maximal with outerthickness 3; that is, adding any edge not in E(H) increases the outerthickness of H. Given the promise of C, let w1,w2,,wα be the α nodes in an independent set in H. The existence of H, as well as a construction running in time constant in the input size of G, is guaranteed by Proposition 12.

We initialize G by taking a copy of G and a copy of H with disjoint node sets, and then identifying G with their union. Then, for each edge e{u,v}E(G), we add the edges {u,wφ(e)} and {v,wφ(e)} to G, as illustrated in Figure 1. Note that G remains a simple graph: since any two edges incident to the same node in G receive distinct labels under φ, no parallel edges are created.

Figure 1: An illustration of the construction of G. For each edge e{u,v}E(G), we add the edges {u,wφ(e)} and {v,wφ(e)} to G. In this example, we have ={1,2,,6}, and we depict the added edges corresponding to e with φ(e){1,2}.

We need the following observation for our main lemma.

Observation 6.

Let z1 be an integer. Let Q be a graph that is edge-maximal with outerthickness z, and let E1,E2,,Ez be an edge-partition of E(Q) into z outerplanar subgraphs. Let P be a path with endpoints x and x such that V(P)V(Q)=. For any nodes y,yV(Q), define

Sy,yE1P{{x,y},{x,y}}.
  1. (a)

    If y=y, then Sy,y is outerplanar.

  2. (b)

    If yy and {y,y}E(Q), then Sy,y is not outerplanar.

Proof.

(a) Fix an outerplanar embedding of E1 in which all nodes, and in particular y, lie on the outer face. Draw the path P in the outer face without crossings, and then connect both end-nodes of P to y by drawing the two edges within the outer face so that they intersect the embedding of E1 only at y. This yields an outerplanar embedding of Sy,y, as illustrated in Figure 2.

(b) If yy and {y,y}E(Q), suppose for a contradiction that Sy,y is outerplanar. Every node of P has degree 2 in Sy,y. Smoothing a degree-2 node in an outerplanar graph preserves outerplanarity: given an outerplanar embedding in which all nodes lie on the outer face, smoothing such a node yields an embedding in which all remaining nodes still lie on the outer face. Hence we may smooth all nodes of P in Sy,y while preserving the outerplanarity of Sy,y, thereby replacing the union of path P and {{x,y},{x,y}} in Sy,y by the single edge {y,y}. It follows that Q{{y,y}} has outerthickness z. Since {y,y}E(Q), this adds an edge to Q, contradicting the assumption that Q is edge-maximal with outerthickness z. Therefore, Sy,y is not outerplanar.

Figure 2: An illustration of Sy,y with Q=K4{e} and z=1.

We are ready to prove the main lemma.

Lemma 7.

G admits a 3-edge-coloring if and only if G has outerthickness at most 3.

Proof.

If G admits a 3-edge-coloring, then E(G) can be partitioned into three matchings M1,M2, and M3. Let i[3] and . Then the edges e{u,v}Mi with φ(e)= give rise to a collection Ti, of triangles, each on nodes {u,v,w}, where the only node shared among these triangles is the common node w. Since Mi is a matching, no triangle in Ti, shares a node with any triangle in Ti, for . Let H1,H2, and H3 be an edge partition of H into three outerplanar subgraphs. By Observation 6(a), for each i[3], the subgraph

HiTi,

is outerplanar. Thus, G has outerthickness at most 3.

Otherwise, G does not admit a 3-edge-coloring. Suppose E(G) can be partitioned into three outerplanar subgraphs Ei for i[3], the following claims hold.

Claim 8.

Some path of length 2 in G is contained entirely in an outerplanar subgraph Ei for some i[3].

Proof.

Now E(G) is partitioned into three outerplanar subgraphs Ei for i[3]. Restricting this partition to E(G)E(G), the induced parts E(G)Ei for i[3] cannot all be matchings; otherwise, they would yield a 3-edge-coloring of G. Therefore, for some i[3], the subgraph Ei contains a path of length 2 in G.

Claim 9.

For each node vV(G), let wv,j for j[3] be neighbors of v in H. Then, the edges {v,wv,j} for j[3] have to be assigned to pairwise distinct outerplanar subgraphs Ei for i[3].

Proof.

The nodes wv,j for j[3] are pairwise distinct, since the three edges of G incident to v receive distinct labels in by the construction of φ. Recall that w1,w2,,wα (and hence wv,j for j[3]) form an independent set.

Suppose that, for some i[3], the subgraph (V(G),Ei) contains at least two edges from {{v,wv,j}:j[3]}, say without loss of generality {v,wv,1} and {v,wv,2}. Let

E(E(H)Ei){{v,wv,j}:j[2]}.

By Observation 6(b), the graph (V(H){v},E) is not outerplanar: apply the observation with (Q,E1,P,y,y)6=(H,Ei,(v),wv,1,wv,2)here. On the other hand, E is a subgraph of the outerplanar graph Ei, a contradiction.

Consequently, the three edges {v,wv,j} for j[3] must be assigned to three distinct outerplanar subgraphs Ei for i[3].

Claim 10.

For each edge e{x,y}E(G), the three edges {x,wφ(e)}, {y,wφ(e)}, {x,y} have to be assigned to the same outerplanar subgraph Ei for some i[3].

Proof.

Since E1,E2,E3 form a partition of E(G), we may assume w.l.o.g. that {x,y}E1. Let

Wx{wx,1,wx,2,wφ(e)}andWy{wy,1,wy,2,wφ(e)}

denote the sets of nodes in H that are adjacent to x and y in G, respectively. By the construction of φ, each of Wx and Wy consists of three distinct nodes, and WxWy={wφ(e)}. By Claim 9, there exist edges {x,w} with wWx and {y,w} with wWy that both lie in E1.

If ww, then by Observation 6(b) the subgraph induced by the union of the path P(w,x,y,w) and E1E(H) is not outerplanar: apply the observation with (Q,E1,P,y,y)6=(H,E1,(x,y),w,w)here.

Therefore w=w=wφ(e), or equivalently, the three edges {x,wφ(e)}, {y,wφ(e)}, and {x,y} are all assigned to E1.

By Claim 8, we assume w.l.o.g. that E1 contains the path of length 2, denoted by (a,b,c) with a,b,cV(G) and e1{a,b},e2{b,c}E(G). By Claim 10, E1 contains the four edges {a,wφ(e1)}, {b,wφ(e1)}, {b,wφ(e2)}, {c,wφ(e2)} as well. However, by Claim 9, {b,wφ(e1)} and {b,wφ(e2)} have to be assigned to different Ei’s, a contradiction.

As a result, E(G) cannot be partitioned into three outerplanar subgraphs Ei for i[3] or, equivalently, G has outerthickness greater than 3. This completes the proof.

3.1 Proofs of Deferred Claims

Proposition 11.

Let k3 be an integer. Let G be a k-regular graph. There exists a labeling function φ:E(G) with ||2k(k1)+1, computable in polynomial time, such that the edges on any path in G of length at most 3 receive pairwise distinct labels.

Proof.

In a k-regular graph G, let e={u,v} be an edge. Any other edge fe that can appear together with e on a path of length at most 3 must either be incident to u or v (at most 2(k1) choices), or be incident to one of the 2(k1) neighbors of u and v (at most 2(k1)2 additional choices). Hence, for each edge e, the number of edges fe that can be contained with e in some path of length at most 3 is at most 2k(k1). Therefore, we can greedily assign to each edge e a label from {1,2,,2k(k1)+1} so that no such edge f receives the same label as e; since at most 2k(k1) labels are forbidden at each step, at least one label remains available. The running time of the above greedy labeling is linear in the input size.

Proposition 12.

For each integer t1, there exists a constant C such that every C-node graph H with outerthickness 3 contains an independent set of size at least αt. Moreover, such a graph H and an independent set of size α in H can be computed in constant time.

Proof.

Consider an arbitrary constant C such that

(C2)>3(2C3). (1)

Every C-node outerplanar graph has at most 2C3 edges, and hence any graph with outerthickness 3 has at most 3(2C3) edges, implying that the complete graph KC has outerthickness greater than 3.

Let H be any C-node graph with outerthickness 3. Such a subgraph exists: the empty graph has outerthickness 0, while KC has outerthickness greater than 3, and adding a single edge can increase the outerthickness by at most one.

By the Caro–Wei bound [7, 31], every C-node m-edge graph has an independent set of size at least C2C+2m. Since H has m3(2C3) edges, we obtain that H has an independent set of size at least

C2C+6(2C3)>t for some sufficiently large C. (2)

If we set C>13t, then Equations 1 and 2 both hold.

Finally, since C is a constant, we can compute such an H and an independent set of size at least t in constant time by brute force, using an outerplanarity testing algorithm [25].

It may be worth noting that a simpler proof of Proposition 12 can be obtained from the fact that every outerplanar graph is 3-colorable. However, we retain the current proof because it generalizes to the proof of Lemma 13.

4 NP-hardness of 𝓕-Thickness

We generalize the NP-hardness proof for outerthickness in Section 3 to every proper graph class that satisfies all of the following conditions:

  1. (a)

    is closed under topological minors,

  2. (b)

    is closed under 1-sums, and

  3. (c)

    contains a cycle of length 3.

By Condition (a), is monotone. Thus, any covering of the edge set of an input graph for the problem P,k can be converted into an edge partition by deleting overlaps; hence, in this setting, covering and partitioning are equivalent. Thus, in this section, we treat P,k as a partition problem: Let be a proper graph class. We define the -thickness of an undirected simple graph G, denoted by θ(G), to be the minimum integer k such that E(G) admits a partition into at most k subsets E1,E2,,Ek with (V(G),Ei) for all i[k]. Thus, the problem P,k asks, given an undirected simple graph G, whether θ(G)k.

Our NP-hardness reduction proceeds as follows. Let k3 be an integer. Given an input instance G of Edge-Coloring on k-regular graphs, defined in Theorem 5, we compute a label function φ defined below and an auxiliary graph H in polynomial time. Then, we obtain a supergraph G from joining G and H using the label function φ such that G admits a k-edge-coloring if and only if G has θ(G)k (Lemma 16). This proves the NP-hardness claimed in Theorem 3 for every integer k3.

  • Let φ:E(G) be a labeling function, where ={1,2,,||} is a constant-size (not necessarily minimum-size) label set, such that for every path in G of length at most 3, the edges on the path receive pairwise distinct labels under φ. The existence of such a function φ, together with a construction running in time polynomial in the input size of G, is guaranteed by Proposition 11. The resulting label set satisfies ||=2k(k1)+1=O(1), as desired.

  • We pick a sufficiently large constant C such that every C-node graph with -thickness k contains an (not necessarily maximum) independent set of α|| nodes. Then we construct a C-node graph H that is edge-maximal with -thickness k. Given the promise of C, let w1,w2,,wα be the α nodes in an independent set in H. The existence of H, as well as a construction running in time constant in the input size of G, is shown in Lemma 13.

Lemma 13.

For each integer 1, there exists a constant C such that every C-node graph H that is edge-maximal with -thickness k contains an independent set of at least α nodes. Such an H and an independent set in H of α nodes can be computed in constant time.

Proof.

Since is proper and closed under topological minors, there exists an integer t such that the complete graph Kt. By a theorem of Bollobás and Thomason [6], there is a constant Dt such that every graph F with more than Dt|V(F)| edges contains a subdivision of Kt, and hence does not belong to . Therefore, every F satisfies |E(F)|Dt|V(F)|.

Choose an arbitrary constant C such that

(C2)>k(DtC). (3)
Claim 14.

For every F and eE(F), θ(F{e})θ(F)+1.

Proof.

Since C3 (Condition (c)) and is closed under topological minors (Condition (a)), we have P2. Thus, we have θ(F{e})θ(F)+θ({e})θ(F)+1.

Since θ()=0, the family

𝒮{SKC:θ(S)k}

is nonempty. Let S be a member of 𝒮 with maximum number of edges. Thus S is edge-maximal with -thickness at most k. Here is why. For every edge eE(KC)E(S) we must have θ(S{e})>k, otherwise S{e}𝒮 contradicts the choice of S. Moreover, by Claim 14, we have θ(S)=k.

By the Caro–Wei bound [7, 31], every C-node m-edge graph has an independent set of size at least C2C+2m. Since H has mk(DtC) edges, we obtain that H has an independent set of size at least

C2C+2k(DtC)> for some sufficiently large C. (4)

If we set C>(1+2kDt), then Equations 3 and 4 both hold.

Finally, since C is a constant, we can compute such an H and an independent set of size at least in constant time by brute force, using the indicator array A,q for qC mentioned in Section 1.

Observation 15.

Let z1 be an integer. Let Q be a graph that is edge-maximal with -thickness z, and let E1,E2,,Ez be an edge-partition of E(Q) into z subgraphs in . Let P be a path with endpoints x and x such that V(P)V(Q)=. For any nodes y,yV(Q), define

Sy,yE1P{{x,y},{x,y}}.

If yy and {y,y}E(Q), then Sy,y does not belong to .

Proof.

If yy and {y,y}E(Q), suppose for contradiction that Sy,y belongs to . Every node of P has degree 2 in Sy,y. Since is closed under topological minors (Condition (a)), smoothing a degree-2 node preserves the membership in . Hence we may smooth all nodes of P in Sy,y while preserving that Sy,y, thereby replacing the union of path P and {{x,y},{x,y}} in Sy,y by the single edge {y,y}. It follows that Q{{y,y}} has -thickness z. Since {y,y}E(Q), this adds an edge to Q, contradicting the assumption that Q is edge-maximal with -thickness z. Therefore, Sy,y does not belong to .

We initialize G by taking a copy of G and a copy of H with disjoint node sets, and then identifying G with their union. Then, for each edge e{u,v}E(G), we add the edges {u,wφ(e)} and {v,wφ(e)} to G. Note that G remains a simple graph: since any two edges incident to the same node in G receive distinct labels under φ, no parallel edges are created.

We are ready to prove the key lemma.

Lemma 16.

G admits a k-edge-coloring if and only if G has θ(G)k.

Proof.

If G admits a k-edge-coloring, then E(G) can be partitioned into k matchings Mi for i[k]. Let i[k] and . The edges e{u,v}Mi with φ(e)= give rise to a collection Ti, of triangles, each on nodes {u,v,w}, where the only node shared among these triangles is the common node w. Since Mi is a matching, no triangle in Ti, shares a node with any triangle in Ti, for . Since H is edge-maximal with -thickness k, H has an edge partition into k subgraphs in , denoted by Hi for i[k]. Since is closed under 1-sums (Condition (b)) and contains C3 (Condition (c)), for each i[k], the subgraph

HiTi,

is contained in . To see why, for every , we attach each triangle in Ti, to Hi by taking a 1-sum that identifies the node w. Repeating this 1-sum operation for all triangles in Ti, for all , yields exactly HiTi,, which therefore lies in . This argument holds for every i[k], so G has -thickness at most k.

Otherwise, G does not admit a k-edge-coloring. Suppose E(G) can be partitioned into k subgraphs in , denoted by Ei for i[k], the following claims hold.

Claim 17.

Some path of length 2 in G is contained entirely within a subgraph Ei (in ) for some i[k].

Proof.

Now E(G) is partitioned into k subgraphs Ei in for i[k]. Restricting this partition to E(G)E(G), the induced parts E(G)Ei for i[k] cannot all be matchings; otherwise, they would yield a k-edge-coloring of G. Therefore, for some i[k], the subgraph Ei contains a path of length 2 in G.

Claim 18.

For each node vV(G), let wv,j for j[k] be neighbors of v in H. Then, the edges {v,wv,j} for j[k] have to be assigned to pairwise distinct subgraphs Ei (in ) for i[k].

Proof.

The nodes wv,j for j[k] are pairwise distinct, since the k edges in G incident to v receive distinct labels in by the construction of φ. Recall that w1,w2,,wα (and hence wv,j for j[k]) form an independent set.

Suppose that, for some i[k], the subgraph Ei contains at least two edges from {{v,wv,j}:j[k]}, say without loss of generality {v,wv,1} and {v,wv,2}. Let

E(E(H)Ei){{v,wv,j}:j[2]}.

By Observation 15, the graph (V(H){v},E) does not belong to : indeed, apply the observation with

(Q,E1,P,y,y)15=(H,Ei,(v),wv,1,wv,2)here.

On the other hand, E is a subgraph of Ei, which lies in . This contradicts Condition (a), since is closed under topological minors (and hence under taking subgraphs).

Consequently, the k edges {v,wv,j} for j[k] must be assigned to k distinct subgraphs Ei for i[k].

Claim 19.

For each edge e{x,y}E(G), the three edges {x,wφ(e)}, {y,wφ(e)}, {x,y} have to be assigned to the same subgraph Ei (in ) for some i[k].

Proof.

Since Ei for i[k] form a partition of E(G), we may assume w.l.o.g. that {x,y}E1. Let

Wx{wx,1,,wx,k1,wφ(e)} and Wy{wy,1,,wy,k1,wφ(e)}

denote the sets of nodes in H that are adjacent to x and y in G, respectively. By the construction of φ, each of Wx and Wy consists of k distinct nodes, and WxWy={wφ(e)}. By Claim 18, there exist edges {x,w} with wWx and {y,w} with wWy that both lie in E1.

If ww, then by Observation 15 the subgraph induced by the union of the path P(w,x,y,w) and E1E(H) does not belong to : apply the observation with

(Q,E1,P,y,y)15=(H,E1,(x,y),w,w)here.

Therefore w=w=wφ(e), or equivalently, the three edges {x,wφ(e)}, {y,wφ(e)}, and {x,y} are all assigned to E1.

By Claim 17, we assume w.l.o.g. that E1 contains the path of length 2, denoted by (a,b,c) with a,b,cV(G) and e1{a,b},e2{b,c}E(G). By Claim 19, E1 contains the four edges {a,wφ(e1)}, {b,wφ(e1)}, {b,wφ(e2)}, {c,wφ(e2)} as well. However, by Claim 18, {b,wφ(e1)} and {b,wφ(e2)} have to be assigned to different Ei’s, a contradiction.

As a result, E(G) cannot be partitioned into k subgraphs Ei in for i[k] or, equivalently, G has -thickness greater than k. This completes the proof.

As a result, Theorem 3 follows.

5 Concluding Remarks

Finally, we remark on the case k=2. It is known that every planar graph has outerthickness at most 2 due to Gonçalves (2005) [14], and hence 2-OuterThickness(G) for planar graphs contains only Yes-instances. For general graphs, however, the complexity of 2-OuterThickness(G) was, to the best of our knowledge, open at the time this paper was submitted for review. We later found that 2-OuterThickness(G) is NP-hard as well; the proof is included in the arXiv version of this paper. More broadly, in the context of our hardness framework in Theorem 3, any proper graph class satisfying Conditions (a)–(c) for which the corresponding problem P,k with k=2 is NP-hard.

It is also worth noting that Conditions (a)–(c) in Theorem 3 are necessary even for k=2, in the following sense: relaxing any one of them allows graph classes for which P,2 is solvable in polynomial time.

For instance, if is the class of forests, then Condition (c) fails, and P,k for k=2 is in P by the same algorithm as in the case k3. Similarly, if is the class of pseudoforests, then Condition (b) fails, and P,k for k=2 also is in P by the same algorithm as for k3. Finally, if is the class of graphs of bounded arboricity, then Condition (a) is not satisfied555For example, the 1-subdivision of K5 has arboricity at most 2, whereas its topological minor K5 has arboricity 3., and the corresponding partitioning problem P,2 is in P by Edmonds’ matroid-partitioning algorithm [12].

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