1. Introduction
Pathwidth is a graph parameter of fundamental importance, especially in graph structure theory. The pathwidth of a graph is the minimum integer for which there is a sequence of sets such that for each , for every vertex of , the set is a nonempty interval, and for each edge of , some contains both and .
In the first paper of their graph minors series, Robertson and Seymour [8] proved the following theorem.
1.1.
For every forest , there exists a constant such that every graph with pathwidth at least contains as a minor.
The constant was later improved to (which is best possible) by Bienstock, Robertson, Seymour, and Thomas [1]. A simpler proof of this result was later found by Diestel [6].
Since forests have unbounded pathwidth, 1.1 implies that a minorclosed class of graphs has unbounded pathwidth if and only if it includes all forests. However, these certificates of large pathwidth are not 2connected, so it is natural to ask for which minorclosed classes , does every 2connected graph in have bounded pathwidth?
In 1993, Paul Seymour proposed the following answer (see [5]). A graph is an apexforest if is a forest for some . A graph is outerplanar if it has an embedding in the plane with all the vertices on the outerface. These classes are relevant since they both contain 2connected graphs with arbitrarily large pathwidth. Seymour conjectured the following converse holds.
1.2.
For every apexforest and outerplanar graph there is an integer such that every 2connected graph of pathwidth at least contains or as a minor.
Equivalently, 1.2 says that for a minorclosed class , every 2connected graph in has bounded pathwidth if and only if some apexforest and some outerplanar graph are not in .
The original motivation for conjecturing 1.2 was to seek a version of 1.1 for matroids (see [3]). Observe that apexforests and outerplanar graphs are planar duals (see 2.1). Since a matroid and its dual have the same pathwidth (see [7] for the definition of matroid pathwidth), 1.2 provides some evidence for a matroid version of 1.1.
In this paper we prove 1.2. An independent proof was recently obtained by Dang and Thomas [3]. We actually prove a slightly different, but equivalent version of 1.2. Namely, we prove that there are two unavoidable families of minors for connected graphs of large pathwidth. We now describe our two unavoidable families.
A binary tree is a rooted tree such that every vertex is either a leaf or has exactly two children. For , the complete binary tree of height , denoted , is the binary tree with leaves such that each root to leaf path has edges. It is well known that has pathwidth . Let be the graph obtained from by adding a new vertex adjacent to all the leaves of . See Figure 1. Note that is a connected apexforest, and its pathwidth grows as grows (since it contains ).
Our second set of unavoidable minors is defined recursively as follows. Let be a triangle with a root edge . Let and be copies of with root edges and . Let be a triangle with edges , and . Define by gluing each to along and then declaring as the new root edge. See Figure 2. Note that is a connected outerplanar graph, and its pathwidth grows as grows (since it contains ).
The following is our main theorem.
1.3.
For every integer there is an integer such that every connected graph of pathwidth at least contains or as a minor.
In Section 2, we prove that every apexforest is a minor of a sufficiently large and every outerplanar graph is a minor of a sufficiently large . Thus, Theorem 1.3 implies Seymour’s conjecture.
1.4.
For all integers , there exists an integer such that every connected graph with a minor contains or as a minor.
Our approach is different from that of Dang and Thomas [3], who instead observe that by the Grid Minor Theorem [9], one may assume that has bounded treewidth but large pathwidth. Dang and Thomas then apply the machinery of ‘nonbranching tree decompositions’ that they developed in [4] to prove 1.2.
The rest of the paper is organized as follows. Section 2 proves the universality of our two families. In Sections 3 and 4, we define ‘special’ ear decompositions and prove that special ear decompositions always yield or minors. In Section 5, we prove that a minimal counterexample to 1.4 always contains a special ear decomposition. Section 6 concludes with short derivations of our main results.
2. Universality
This section proves some elementary (and possibly wellknown) results. We include the proofs for completeness.
2.1.
Outerplanar graphs and apexforests are planar duals.
Proof.
Let be an apexforest, where is a forest. Consider an arbitrary planar embedding of . Note that every face of includes (otherwise would contain a cycle). Let be the planar dual of . Let be the face of corresponding to . Since every face of includes , every vertex of is on . So is outerplanar.
Conversely, let be an outerplanar graph. Consider a planar embedding of , in which every vertex is on the outerface . Let be the planar dual of . Let be the vertex of corresponding to . If contained a cycle , then some vertex of on the ‘inside’ of would not be on . Thus is a forest, and is an apexforest. ∎
We now show that Theorem 1.3 implies Seymour’s conjecture, by proving two universality results.
2.2.
Every apexforest on vertices is a minor of .
If is a minor of and , the branch set of is the set of vertices of that are contracted to . 2.2 is a corollary of the following.
2.3.
Every tree with vertices is a minor of , such that each branch set includes a leaf of .
Proof.
We proceed by induction on . The base case is trivial. Let be a tree with vertices. Let be a leaf of . Let be the neighbour of in . By induction, is a minor of , such that each branch set includes a leaf of . In particular, the branch set for includes some leaf of . Note that is obtained from by adding two new leaf vertices adjacent to each leaf of . Let and be the leaf vertices of adjacent to . Extend the branch set for to include and let be the branch set of . For each leaf of , if is in the branch set of some vertex of , then extend this branch set to include one of the new leaves in adjacent to . Now is a minor of , such that each branch set includes a leaf of . ∎
Our second universality result is for outerplanar graphs.
2.4.
Every outerplanar graph on vertices is a minor of .
2.4 is a corollary of the following.
2.5.
Every outerplanar triangulation on vertices is a minor of , such that for every edge on the outerface of , there is a nonroot edge on the outerface of joining the branch sets of and .
Proof.
We proceed by induction on . The base case, , is easily handled as illustrated in Figure 3. Let be an outerplanar triangulation with vertices. Every such graph has a vertex of degree 2, such that if and are the neighbours of , then is an outerplanar triangulation and is an edge on the outerface of . By induction, is a minor of , such that for every edge on the outerface of , there is a nonroot edge on the outerface of joining the branch sets of and . In particular, there is a nonroot edge of joining the branch sets of and . Note that is obtained from by adding, for each nonroot edge on the outerface of , a new vertex adjacent to and . Let the branch set of be the vertex of adjacent to and . Thus contains as a minor. Every edge on the outerface of is one of or , or is on the outerface of . By construction, is a nonroot edge on the outerface of joining the branch sets of and . Similarly, is a nonroot edge on the outerface of joining the branch sets of and . For every edge on the outerface of , where , if is the vertex in adjacent to and , extend the branch set of to include . Now is an edge on the outerface of joining the branch sets for and . Thus for every edge on the outerface of , there is a nonroot edge of joining the branch sets of and . ∎
3. Binary Ear Trees
Henceforth, all graphs in this paper are finite and simple. In particular, after contracting an edge, we suppress parallel edges and loops. If and are graphs, then is the graph with and . If is a subgraph of , then an ear is a path in with its two ends in but with no internal vertex in .
A binary ear tree in a graph is a pair , where is a binary tree, and is a collection of paths of satisfying

for all ,

for all distinct in such that is not a child of , and are internallydisjoint,

for all such that and are the children of ,
and are ears, and has an end that is not an end of nor of .
The main result of this section is the following.
3.1.
For every integer , if has a binary ear tree such that , then contains or as a minor.
Before starting the proof, we first set up notation for a Ramseytype result that we will need.
If and are vertices of a tree , then let denote the unique path in . If is a subdivision of a tree , the vertices of coming from are called original vertices and the other vertices of are called subdivision vertices. Given a colouring of the vertices of with colours , we say that contains a red subdivision of , if it contains a subdivision of such that all the original vertices of are red, and for all with a descendent of , the path is descending. (Here a path is descending if it is contained in a path that starts at the root.) Define to be the minimum integer such that every colouring of with colours contains a red subdivision of or a blue subdivision of . We will use the following easy result.
3.2.
for all integers .
Proof.
We proceed by induction on . As base cases, it is clear that and for all . For the inductive step, assume and let be a coloured copy of . By symmetry, we may assume that the root of is coloured red. Let and be the components of , both of which are copies of . If or contains a blue subdivision of , then so does and we are done. By induction, , so both and contain a red subdivision of . Add the paths from to the roots of these red subdivisions. We obtain a red subdivision of , as desired. ∎
We now prove 3.1.
Proof of 3.1.
Let be a nonleaf vertex of . Let and be the children of . Let and be the ends of . Let and be the ends of . We say that is nested if or . If is not nested, then is split. See Figures 4 and 5. Regarding and as colours, by 3.2, contains a subdivision of or a subdivision of .
3.3.
If contains a subdivision of with no subdivision vertices, then contains as a minor.
Subproof.
We first setup a stronger induction hypothesis. Let be the graph obtained from by deleting its root edge . Define the root path of as , where is the unique common neighbour of and . Let be a path in a graph . We say that a minor in is rooted on if there are two edges and of such that the root path of is obtained from by contracting all edges of except for and . We prove the stronger result that contains a minor rooted on , where is the root of .
We proceed by induction on . The case is clear since is just a path with three vertices. For the inductive step, let and be the children of and let and be the subtrees of rooted at and . By induction, contains a minor rooted on , and contains a minor rooted on .
We prove that and are vertexdisjoint, except possibly at a vertex of (there is at most one such vertex since is split). For each , let be the highest vertex of such that . If and is an end of , then by property 3 of binary ear trees, where is the parent of in . However, this contradicts the maximality of . Therefore, for each , or is an internal vertex of . Define analogously for each and suppose . If , then is an internal vertex of . But then contradicts 2. By symmetry, and . Thus, , as desired.
Let and be the ends of , and and be the ends of . By symmetry, we may assume that the ordering of these points along is either or . Let , and be the vertices of immediately following , and respectively. In the first case let and . In the second case let and . In both cases, extends to a minor rooted on by contracting all edges of . See Figure 6. Since contains a minor, we are done.
∎
We will handle the case when has subdivision vertices at the very end of the proof. For now we switch to the case that contains a subdivision of . Orient each path in inductively as follows. Let be the root of and orient arbitrarily. If has already been oriented and is a child of , then orient so that does not contain a directed cycle. Consider each path in to be oriented from left to right, and thus with left and right ends.
We may consider an original vertex of to be a vertex of (and vice versa). Let be a nonleaf, original vertex of and let and be the children of in . Define to be leftbad if at least one of the left ends of or is equal to the left end of . Define to be leftgood if is not leftbad. Define rightgood and rightbad analogously. Let be the tree obtained from by deleting all leaves. By 3.2, contains a leftgood subdivision of or a leftbad subdivision of . By 3, no vertex of can be both leftbad and rightbad. Therefore, contains a leftgood subdivision of or a rightgood subdivision of . By symmetry, we may assume that has a leftgood subdivision of .
As in the previous case, we assume that all vertices of are original vertices, and then reduce to this case at the end of the proof. Let be a nonleaf vertex of and and be the children of in . Let be the first vertex of that is a left end of either or of . Let be the last edge of incident to a left end of either or . If is a leaf vertex of , then let be any internal vertex of and be the last edge of incident to . Let and . Since every vertex of is nested, contains two components and such that contains all left ends of and contains all right ends of . It is easy to see that contains a subdivision of whose set of original vertices is ; see Figure 7. By construction, each leaf of is incident to an edge in . Also, is clearly connected. Therefore, after contracting all edges of , contains a minor.
To complete the proof, it suffices to consider the case that or contain subdivision vertices. Let be either or . Suppose that and are original vertices of with a descendent of , and is a path in such that is a subdivision vertex for all . It is easy to see that there exists a path in such that has the same ends as and ; see Figure 8. Thus, we may replace by and suppress all vertices in . ∎
4. Binary Pear Trees
In order to prove our main theorem, we need something slightly more general than binary ear trees, which we now define. A binary pear tree in a graph is a pair , where is a binary tree, and is a collection of pairs of paths of satisfying

and for all ,

for all , if contains an internal vertex of , then is a descendent of or of the sibling of ,

for all such that and are the children of ,
and are ears, has an end that is not an end of nor of , no internal vertex of is in , and no internal vertex of is in .
Note that if is a binary ear tree, then is a binary pear tree. We now prove the following converse.
4.1.
If has a binary pear tree , then has a minor such that has a binary ear tree .
Proof.
Let be a binary pear tree in , where . We prove the stronger result that there exist and such that is a minor of , is a binary ear tree in , and for all leaves of . Arguing by contradiction, suppose that this is not true. Among all counterexamples, choose such that is minimum. This clearly implies that . For , let .
Let and be leaves of with a common parent . Let be the set of vertices that are an internal vertex of or of . By 2, for all . Therefore, if , then is a smaller counterexample. Thus, and are edgedisjoint. If , then is a smaller counterexample, unless is a ear or has one end on and the other end on .
Suppose is a ear. The unique cycle in must contain an edge of . Otherwise, we obtain a smaller counterexample by deleting from and rerouting through . Since is not incident to an internal vertex of , . Let be the unique path in containing and having the same ends as . Since and have the same ends, we can use to extend to a path with the same ends as . Observe that contains a ear such that has the same ends as and . There is also a unique ear in such that . Let , where if , and if . By 3, at least one end of is not an end of . Therefore, at least one end of is not an end of , and so is a pear tree in . However, clearly must avoid some edge of . Therefore, is a smaller counterexample.
It follows that every edge of has one end on and the other end on . By symmetry, every edge of has one end on and the other end on . In particular, and .
Suppose and . Recall that is the set of vertices that are an internal vertex of or of . Let and . By minimality, has a minor such that has a binary ear tree such that for all leaves of . Since is a leaf of , we have . Let and be such that for all , , and . It is clear that satisfies conditions 1 and 3 for binary ear trees. Since , it is easy to see that also satisfies condition 2 for binary ear trees.
If and , then has one end on and the other end on . Since , this contradicts 3.
By symmetry, the last case is and . By 3, there must exist and such that and have a common end not on . Let be the end of on and be the end of on . If , then , which contradicts the fact that and are edgedisjoint. Thus, . Since is not a ear and is not a ear, by symmetry we may assume that is not an end of . We then obtain a smaller counterexample by deleting from and replacing by . ∎
5. Finding Binary Pear Trees
The main result of this section is the following.
5.1.
For all integers and , if is a minorminimal connected graph containing a subdivision of and is a binary tree of height at most , then either contains as a minor, or contains a binary pear tree .
We proceed via a sequence of lemmas.
5.2.
If is a minorminimal connected graph containing a subdivision of , then every subdivision of in is a spanning tree.
Proof.
Let be a subdivision of in . We use the wellknown fact that for all , at least one of or is connected. Therefore, if some edge of has an end not in , then or is a connected graph containing a subdivision of , which contradicts the minorminimality of . ∎
For a vertex in a rooted tree , let be the subtree of rooted at .
5.3.
Let and let be a with root . Suppose that a nonempty subset of vertices of are marked. Then

contains a subdivision of , all of whose leaves are marked, or

there exist a vertex and a child of such that has at least one marked vertex but has none, and is at distance at most from .
Proof.
A vertex in is good if there is a marked vertex in , and is bad otherwise. Let be the subtree of induced by vertices at distance at most from in . If each leaf of is good, then for each such leaf we can find a marked vertex in , and is a subdivision with all leaves marked, as required by 1. Now assume that some leaf of is bad. Let be the bad vertex closest to on the path. Since some vertex in is marked, is good. Thus . Moreover, the parent of is good, by our choice of . Also, is at distance at most from . Therefore, and satisfy 2. ∎
Our main technical tools are 5.4 and 5.5 below, which are lemmas about connected graphs containing a subdivision of as a spanning tree. In order to state them, we need to introduce some definitions and notation that we will use in this setting. For each vertex , let be the number of original nonleaf vertices on the path , where is any leaf of . We stress the fact that subdivision vertices are not counted when computing . We also use the shorthand notation when and are clear from the context. For , we say that sees if for some and . If is a path with ends and , and is a path with ends and , then let be the walk that follows from to and then follows from to .
A path of is special if , and are the ends of , and is a child of such that and . A vertex is safe for an special path if satisfies the following properties:

the parent of is in ;

;

;

does not see , and

if is an original vertex and is its child distinct from , then either or does not see .
5.4.
Let . Let be a minorminimal connected graph containing a subdivision of . Let be a subdivision of in , with , and be a child of . Then, either contains a minor, or there is a special path and two distinct safe vertices for such that:

,

,

sees ,

does not see , and

sees if is an original vertex and is its child distinct from .
Proof.
By 5.2, is a spanning tree of . Color red each vertex of that sees a vertex in . Observe that there is at least one red vertex. Indeed, must see , for otherwise would be a cut vertex separating from in .
Let be the complete binary tree obtained from by iteratively contracting each edge of the form with a subdivision vertex and the child of into vertex . Declare to be colored red after the edge contraction if at least one of was colored red beforehand.
If contains a subdivision of with all leaves colored red, then so does . Therefore, contains as a minor, because induces a connected subgraph of which is vertexdisjoint from and which sees all the leaves of . Thus, by 5.3, we may assume there is a vertex of and a child of with such that has at least one red vertex but has none. Going back to , we deduce that there is a vertex of and a child of with such that has at least one red vertex but has none. To see this, choose as the deepest red vertex in the preimage of . Note that or could be subdivision vertices.
If is an original vertex, let denote the child of distinct from . Since is not a cut vertex of , one of the two subtrees and sees . If does not see , then has no red vertex and sees . Therefore, by exchanging and if necessary, we guarantee that the following two properties hold when exists.
sees and has no red vertex.  (1) 
We iterate this process in . Color blue each vertex of that sees a vertex in . There is at least one blue vertex, since otherwise would be a cut vertex of separating from . Defining similarly as above, if contains a subdivision of with all leaves colored blue, then has a minor. Applying 5.3 and going back to , we may assume there is a vertex of and a child of with such that has at least one blue vertex but has none.
We now define the special path , and identify two distinct safe vertices for . To do so, we will need to consider different cases. In all cases, the end will be a vertex of seen by a (carefully chosen) blue vertex in , thus , and the path will be such that . Note that the end of satisfies , as desired.
Before proceeding with the case analysis, we point out the following property of . If is an edge of such that contains a subdivision of , then is not connected by the minorminimality of , and it follows that is a cutset of . Note that this applies if is an edge of such that at least one of is a subdivision vertex, or if
Comments
There are no comments yet.