Linkages in Large Graphs of Bounded TreeWidth
Abstract
We show that all sufficiently large connected graphs of bounded treewidth are linked. Thomassen has conjectured that all sufficiently large connected graphs are linked.
lemthm \aliascntresetthelem \newaliascntprothm \aliascntresetthepro \newaliascntcorthm \aliascntresetthecor \newaliascntquethm \aliascntresettheque \newaliascntconjthm \aliascntresettheconj \newaliascntexmthm \aliascntresettheexm
1 Introduction
Given an integer , a graph is linked if for any choice of distinct vertices and of there are disjoint paths in such that the end vertices of are and for . Menger’s theorem implies that every linked graph is connected.
One can conversely ask how much connectivity (as a function of ) is required to conclude that a graph is linked. Larman and Mani [13] and Jung [8] gave the first proofs that a sufficiently highly connected graph is also linked. The bound was steadily improved until Bollobás and Thomason [3] gave the first linear bound on the necessary connectivity, showing that every connected graph is linked. The current best bound shows that connected graphs are also linked [20].
What is the best possible function one could hope for which implies an connected graph must also be linked? Thomassen [22] conjectured that connected graphs are linked. However, this was quickly proven to not be the case by Jørgensen with the following example [23]. Consider the graph obtained from obtained by deleting the edges of a matching of size . This graph is connected but is not linked. Thus, the best possible function one could hope for to imply linked would be . However, all known examples of graphs which are roughly connected but not linked are similarly of bounded size, and it is possible that Thomassen’s conjectured bound is correct if one assumes that the graph has sufficiently many vertices.
In this paper, we show Thomassen’s conjectured bound is almost correct with the additional assumption that the graph is large and has bounded treewidth. This is the main result of this article.
Theorem 1.1.
For all integers and there exists an integer such that a graph is linked if
where is the connectivity of the graph and is the treewidth.
The treewidth of the graph is a parameter commonly arising in the theory of graph minors; we will delay giving the definition until Section 2 where we give a more in depth discussion of how treewidth arises naturally in tackling the problem. The value would be best possible; see Section 8 for examples of arbitrarily large graphs which are connected but not linked.
Our work builds on the theory of graph minors in large, highly connected graphs begun by Böhme, Kawarabayashi, Maharry and Mohar [1]. Recall that a graph contains as a minor if there can be obtained from a subgraph of by repeatedly contracting edges. Böhme et al. showed that there exists an absolute constant such that every sufficiently large connected graph contains as a minor. This statement is not true without the assumption that the graph be sufficiently large, as there are examples of small graphs which are connected but still have no minor [12, 21]. In the case where we restrict our attention to small values of , one is able to get an explicit characterisation of the large connected graphs which do not contain as a minor.
Theorem 1.2 (Kawarabayashi et al. [10]).
There exists a constant such that every 6connected graph on vertices either contains as a minor or there exists a vertex such that is planar.
Jorgensen [7] conjectures that Theorem 1.2 holds for all graphs without the additional restriction to graphs on a large number of vertices. In 2010, Norine and Thomas [19] announced that Theorem 1.2 could be generalised to arbitrary values of to either find a minor in a sufficiently large connected graph or alternatively, find a small set of vertices whose deletion leaves the graph planar. They have indicated that their methodology could be used to show a similar bound of on the connectivity which ensures a large graph is linked.
2 Outline
In this section, we motivate our choice to restrict our attention to graphs of bounded treewidth and give an outline of the proof of Theorem 1.1.
We first introduce the basic definitions of treewidth. A treedecomposition of a graph is a pair where is a tree and is a collection of subsets of indexed by the vertices of . Moreover, satisfies the following properties.

,

for all , there exists such that both ends of are contained in , and

for all , the subset induces a connected subtree of .
The sets in are sometimes called the bags of the decomposition. The width of the decomposition is , and the treewidth of is the minimum width of a treedecomposition.
Robertson and Seymour showed that if a connected graph contains as a minor, then it is linked [16]. Thus, when one considers connected graphs which are not linked, one can further restrict attention to graphs which exclude a fixed clique minor. This allows one to apply the excluded minor structure theorem of Robertson and Seymour [17]. The structure theorem can be further strengthened if one assumes the graph has large treewidth [5]. This motivates one to analyse separately the case when the treewidth is large or bounded. The proofs of the main results in [1] and [10] similarly split the analysis into cases based on either large or bounded treewidth.
We continue with an outline of how the proof of Theorem 1.1 proceeds. Assume Theorem 1.1 is false, and let be a connected graph which is not linked. Fix a set such that there do not exist disjoint paths where the ends of are and for all . Fix a treedecomposition of of minimal width .
We first exclude the possibility that has a high degree vertex. Assume is a vertex of of large degree. By Property 3 in the definition of a treedecomposition, if we delete the set of vertices from , the resulting graph must have at least distinct connected components. By the connectivity of , each component contains internally disjoint paths from a vertex to distinct vertices in . If the degree of is sufficiently large, we conclude that the graph contains a subdivision of for some large value . We now prove that that if a graph contains such a large complete bipartite subdivision and is connected, then it must be linked (Section 7).
We conclude that the tree does not have a high degree vertex, and consequently contains a long path. It follows that the graph has a long path decomposition, that is, a treedecomposition where the tree is a path. As the bags of the decomposition are linearly ordered by their position on the path, we simply give the path decomposition as a linearly ordered set of bags for some large value . At this point in the argument, the pathdecomposition may not have bounded width, but it will have the property that is bounded, and this will suffice for the argument to proceed. Section 3 examines this path decomposition in detail and presents a series of refinements allowing us to assume the path decomposition satisfies a set of desirable properties. For example, we are able to assume that is the same for all , . Moreover, there exist a set of disjoint paths starting in and ending in . We call these paths the foundational linkage and they play an important role in the proof. A further property of the path decomposition which we prove in Section 3 is that for each , , if there is a bridge connecting two foundational paths in in , then for all , , there exists a bridge connecting the same foundational paths in . This allows us to define an auxiliary graph with vertex set and two vertices of adjacent in if there exists a bridge connecting them in some .
Return to the linkage problem at hand; we have terminals and which we would like to link appropriately, and is our path decomposition with the foundational linkage running through it. Let the set be labeled . As our path decomposition developed in the previous paragraph is very long, we can assume there exists some long subsection such that no vertex of is contained in for some large value . By Menger’s theorem, there exist paths linking to the set . We attempt to link the terminals by continuing these paths into the subgraph induced by the vertex set . More specifically, we extend the paths along the foundational paths and attempt to link up the terminals with the bridges joining the various foundational paths in each of the . By construction, the connections between foundational paths are the same in for all , ; thus we translate the problem into a token game played on the auxiliary graph . There each terminal has a corresponding token, and the desired linkage in will exist if it is possible to slide the tokens around in such a way to match up the tokens of the corresponding pairs of terminals. The token game is rigorously defined in Section 4, and we present a characterisation of what properties on will allow us to find the desired linkage in .
The final step in the proof of Theorem 1.1 is to derive a contradiction when doesn’t have sufficient complexity to allow us to win the token game. In order to do so, we use the high degree in and a theorem of Robertson and Seymour on crossing paths. We give a series of technical results in preparation in Section 5 and Section 6 and present the proof of Theorem 1.1 in Section 7.
3 Stable Decompositions
In this section we present a result which, roughly speaking, ensures that a highly connected, sufficiently large graph of bounded treewidth either contains a subdivision of a large complete bipartite graph or has a long path decomposition whose bags all have similar structure.
Such a theorem was first established by Böhme, Maharry, and Mohar in [2] and extended by Kawarabayashi, Norine, Thomas, and Wollan in [9], both using techniques from [14]. We shall prove a further extension based on the result by Kawarabayashi et al. from [9] so our terminology and methods will be close to theirs.
We begin this section with a general Lemma about nested separations. Let be a graph. A separation of is an ordered pair of sets such that . If is a separation of , then is called its separator and its order. Two separations and of are called nested if either and or and . In the former case we write and in the latter . This defines a partial order on all separations of . A set of separations is called nested if the separations of are pairwise nested, that is, is a linear order on . To avoid confusion about the order of the separations in we do not use the usual terms like smaller, larger, maximal, and minimal when talking about this linear order but instead use left, right, rightmost, and leftmost, respectively (we still use successor and predecessor though). To distinguish from we say ‘left’ for the former and ‘strictly left’ for the latter (same for and right).
If and are both separations of , then so are and and a simple calculation shows that the orders of and sum up to the same number as the orders of and . Clearly each of and is nested with both, and .
For two sets we say that a separation of is an – separation if and . If and are – separations in , then so are and . Furthermore, if and are – separations of of minimum order, say , then so are and as none of the latter two can have order less than but their orders sum up to .
Lemma \thelem.
Let be a graph and . If for every there is an – separation of of minimal order with in its separator, then there is a nested set of – separations of minimal order such that their separators cover .
Proof.
Let be a maximal nested set of – separations of minimal order in (as is finite the existence of a leftmost and a rightmost element in any subset of is trivial). Suppose for a contradiction that some is not contained in any separator of the separations of .
Set and . Clearly and . Moreover, if and are both nonempty, then the rightmost element of is the predecessor of the leftmost element of in . Loosely speaking, and contain the separations of “on the left” and “on the right” of , respectively, and and are the separations of and whose separators are “closest” to .
By assumption there is an – separation of minimal order in with . Set
(but if and if ). As , , and are all – separations of minimal order in so must be and . Moreover, we have and thus .
By construction we have and . To verify that we need to show and . All required inclusions follow from and . So by transitivity is right of all elements of and left of all elements of , in particular, it is nested with all elements of , contradicting the maximality of the latter. ∎
We assume that every path comes with a fixed linear order of its vertices. If a path arises as an – path, then we assume it is ordered from to and if a path arises as a subpath of some path , then we assume that is ordered in the same direction as unless explicitly stated otherwise.
Given a vertex on a path we write for the initial subpath of with last vertex and for the final subpath of with first vertex . If and are both vertices of , then by or we mean the subpath of that ends in and and is ordered from to or from to , respectively. By we denote the path with inverse order.
Let be a set of disjoint paths in some graph . We do not distinguish between and the graph formed by uniting these paths; both will be denoted by . By a path of we always mean an element of , not an arbitrary path in .
Let be a graph. For a subgraph an bridge in is a connected subgraph such that is edgedisjoint from and either is a single edge with both ends in or there is a component of such that consists of all edges that have at least one end in . We call a bridge trivial in the former case and nontrivial in the latter. The vertices in and are called the attachments and the inner vertices of , respectively. Clearly an bridge has an inner vertex if and only if it is nontrivial. We say that an bridge attaches to a subgraph if has an attachment in . Note that bridges are pairwise edgedisjoint and each common vertex of two bridges must be an attachment of both.
A branch vertex of is a vertex of degree in and a segment of is a maximal path in such that its ends are branch vertices of but none of its inner vertices are. An bridge in is called unstable if some segment of contains all attachments of , and stable otherwise. If an unstable bridge has at least two attachments on a segment of , then we call a host of and say that is hosted by . For a subgraph we say that two segments of are bridge adjacent or just bridge adjacent in if contains an bridge that attaches to both.
If a graph is the union of its segments and no two of its segments have the same end vertices, then it is called unambiguous and ambiguous otherwise. It is easy to see that a graph is unambiguous if and only if all its cycles contain a least three branch vertices. In our application will always be a union of disjoint paths so its segments are precisely these paths and is trivially unambiguous.
Let be unambiguous. We say that is a rerouting of if there is a bijection from the segments of to the segments of such that every segment of has the same end vertices as (and thus is unique by the unambiguity). If contains no edge of a stable bridge, then we call a proper rerouting of . Clearly any rerouting of the unambiguous graph has the same branch vertices as and hence is again unambiguous.
The following Lemma states two observations about proper reroutings. The proofs are both easy and hence we omit them.
Lemma \thelem.
Let be a proper rerouting of an unambiguous graph and let be as in the definition. Both of the following statements hold.

Every hosted bridge has a unique host. For each segment of the segment of is contained in the union of and all bridges hosted by .

For every stable bridge there is a stable bridge with . Moreover, if attaches to a segment of , then attaches to .
Note that Section 3 (ii) implies that no unstable bridge contains an edge of a stable bridge. Together with (i) this means that being a proper rerouting of an unambiguous graph is a transitive relation.
The next Lemma is attributed to Tutte; we refer to [9, Lemma 2.2] for a proof^{7}^{7}7 To check that Lemma 2.2 in [9] implies our Section 3 note that if is obtained from by “a sequence of proper reroutings” as defined in [9], then by transitivity is a proper rerouting of according to our definition. And although not explicitly included in the statement, the given proof shows that no trivial bridge can be unstable. .
Lemma \thelem.
Let be a graph and unambiguous. There exists a proper rerouting of in such that if is an bridge hosted by some segment of , then is nontrivial and there are vertices such that the component of that contains is disjoint from .
This implies that the segments of are induced paths in as trivial bridges cannot be unstable and no two segments of have the same end vertices.
Let be a graph. A set of disjoint paths in is called a linkage. If with , then a set of disjoint – paths in is called an – linkage or a linkage from to . Let be an ordered tuple of subsets of . Then is the length of , the sets with are its bags, and the sets with are its adhesion sets. We refer to the bags with as inner bags. When we say that a bag of contains some graph , we mean . Given an inner bag of , the sets and are called the left and right adhesion set of , respectively. Whenever we introduce a tuple as above without explicitly naming its elements, we shall denote them by where is the length of . For indices we use the shortcut .
The tuple with the following five properties is called a slim decomposition of .

and every edge of is contained in some bag of .

If , then .

All adhesion sets of have the same size.

No bag of contains another.

contains a – linkage.
The unique size of the adhesion sets of a slim decomposition is called its adhesion. A linkage as in (L5) together with an enumeration of its paths is called a foundational linkage for and its members are called foundational paths. Each path contains a unique vertex of every adhesion set of and we call this vertex the vertex of that adhesion set. For an inner bag of the vertex in the left and right adhesion set of are called the left and right vertex of , respectively. Note that is allowed to contain trivial paths so may be nonempty.
The enumeration of a foundational linkage for is a formal tool to compare arbitrary linkages between adhesion sets of to by their ‘induced permutation’ as detailed below. When considering another foundational linkage for we shall thus always assume that it induces the same enumeration as on , in other words, and start on the same vertex.
Suppose that is a slim decomposition of some graph with foundational linkage . Then any bridge in is contained in a bag of , and this bag is unique unless is trivial and contained in one or more adhesion sets.
We say that a linkage in a graph is attached if each path of is induced in and if some nontrivial bridge attaches to a nontrivial path of , then either attaches to another nontrivial path of or there are at least trivial paths of such that contains a bridge (which may be different from ) attaching to and .
We call a pair of a slim decomposition of and a foundational linkage for a regular decomposition of attachedness of if there is an integer such that the axioms (L6), (L7), and (L8) hold.

is attached in for all inner bags of .

A path is trivial if is trivial for some inner bag of .

For every , if some inner bag of contains a bridge attaching to and , then every inner bag of contains such a bridge.
The integer is not unique: A regular decomposition of attachedness has attachedness for all integers . Note that satisfies (L7) if and only if every vertex of either lies in at most two bags of or in all bags. This means that either all foundational linkages for satisfy (L7) or none.
The next Theorem follows^{8}^{8}8 The statement of Lemma 3.1 in [9] only asserts the existence of a minor isomorphic to rather than a subdivision of like we do. But its proof refers to an argument in the proof of [14, Theorem 3.1] which actually gives a subdivision. from the Lemmas 3.1, 3.2, and 3.5 in [9].
Theorem 3.1 (Kawarabayashi et al. [9]).
For all integers there exists an integer with the following property. If is a connected graph of treewidth less than with at least vertices, then either contains a subdivision of , or has a regular decomposition of length at least , adhesion at most , and attachedness .
Note that [9] features a stronger version of Theorem 3.1, namely Theorem 3.8, which includes an additional axiom (L9). We omit that axiom since our arguments do not rely on it.
Let be a slim decomposition of adhesion and length for a graph . Suppose that is a linkage from the left adhesion set of to the right adhesion set of for two indices and with . The enumeration of induces an enumeration of where is the path of starting in the left vertex of . The map such that ends in the right vertex of for is a permutation because is a linkage. We call it the induced permutation of . Clearly the induced permutation of is the composition of the induced permutations of , , …, . For any permutation of and any graph on we write to denote the graph . For a subset we set .
Keep in mind that the enumerations induces on linkages as above always depend on the adhesion set where the considered linkage starts. For example let be as above and for some index with set . Then need not be the same as . More precisely, we have where denotes the induced permutation of .
For some subgraph of the bridge graph of in , denoted , is the graph with vertex set in which is an edge if and only if and are bridge adjacent in . Any bridge in that attaches to and is said to realise the edge . We shall sometimes think of induced permutations as maps between bridge graphs.
For a slim decomposition of length of with foundational linkage we define the auxiliary graph . Clearly for each inner bag of and if is regular, then by (L8) we have equality.
Set and . Given a subgraph and some foundational linkage for , we write for the graph obtained by deleting from the union of and those bridges in inner bags of that realise an edge of or attach to but to no path of . For a subset we write instead of . Note that . Hence and are the same graph and we denote it by .
A regular decomposition of a graph is called stable if it satisfies the following two axioms where .

If is a linkage from the left to the right adhesion set of some inner bag of , then its induced permutation is an automorphism of .

If is a linkage from the left to the right adhesion set of some inner bag of , then every edge of with one end in is also an edge of .
Given these definitions we can further expound our strategy to prove the main theorem: We will reduce the given linkage problem to a linkage problem with start and end vertices in for some stable regular decomposition of length . The stability ensures that we maximised the number of edges of , i.e. no rerouting of will give rise to new bridge adjacencies. We will focus on a subset and show that the minimum degree of forces a high edge density in , leading to a high number of edges in . Using combinatoric arguments, which we elaborate in Section 4, we show that we can find linkages using segments of and bridges in to realise any matching of start and end vertices in , showing that is in fact linked.
We strengthen Theorem 3.1 by the assertion that the regular decomposition can be chosen to be stable. We like to point out that, even with the left out axiom (L9) included in the definition of a regular decomposition, Theorem 3.2 would hold. By almost the same proof as in [9] one could also obtain a stronger version of (L8) stating that for every subset of if some inner bag of contains a bridge attaching every path of but to no path of , then every inner bag does.
Theorem 3.2.
For all integers there exists an integer with the following property. If is a connected graph of treewidth less than with at least vertices, then either contains a subdivision of , or has a stable regular decomposition of length at least , adhesion at most , and attachedness .
Before we start with the formal proof let us introduce its central concepts: disturbances and contractions. Let be a regular decomposition of a graph . A linkage is called a twisting disturbance if it violates (L10) and it is called a bridging disturbance if it violates (L11). By a disturbance we mean either of these two and a disturbance may be twisting and bridging at the same time. If the referred regular decomposition is clear from the context, then we shall not include it in the notation and just speak of a disturbance. Note that a disturbance is always a linkage from the left to the right adhesion set of an inner bag of .
Given a disturbance in some inner bag of which is neither the first nor the last inner bag of , it is not hard to see that replacing with yields a foundational linkage for such that properly contains and we shall make this precise in the proof. As the auxiliary graph can have at most edges, we can repeat this step until no disturbances (with respect to the current decomposition) are left and we should end up with a stable regular decomposition, given that we can somehow preserve the regularity.
This is done by “contracting” the decomposition in a certain way. The technique is the same as in [2] or [9]. Given a regular decomposition of length of some graph and a subsequence of , the contraction of along is the pair defined as follows. We let with ,
, and (with the induced enumeration).
Lemma \thelem.
Let be the contraction of a regular decomposition of some graph of adhesion and attachedness along the sequence . Then the following two statements hold.

is a regular decomposition of length of of adhesion and attachedness , and .

The decomposition is stable if and only if none of the inner bags of contains a disturbance.
Proof.
The first statement is Lemma 3.3 of [9]. The second statement follows from the fact that an inner bag of contains a disturbance if and only if one of the bags of with contains a disturbance (unless has no inner bag, that is, ). The “if” direction is obvious and for the “only if” direction recall that the induced permutation of is the composition of the induced permutations of the with and every bridge in is also a bridge and hence must be contained in some bag with . ∎
Let be a linkage in a graph and denote the trivial paths of by . Let be the union of with a proper rerouting of obtained from applying Section 3 to in . We call a bridge stabilisation of in . The next Lemma tailors Section 3 and Section 3 to our application.
Lemma \thelem.
Let be a linkage in a graph . Denote by the trivial paths of and let be a bridge stabilisation of in . Let and be paths of and let and be the unique paths of with the same end vertices as and , respectively. Then the following statements hold.

is contained in the union of with all bridges in that attach to but to no other path of .

If and are bridge adjacent in and one of them is nontrivial, then and are bridge adjacent in .

Let be the set of end vertices of the paths of . If is an integer such that for every vertex of there is an – fan of size , then is attached.
Proof.

The statement follows directly from Section 3 (ii) if and are both nontrivial so we may assume that and is nontrivial. By assumption there is a – path in . Clearly contains the end vertices of . On the other hand, by (i) it is clear that . We claim that . Since is internally disjoint from all its inner vertices are inner vertices of some bridge . If is stable or unstable but not hosted by any path of (that is, it has at most one attachment), then Section 3 implies that no path of contains an inner vertex of and that our claim follows. If is hosted by a path of , then this path must clearly be and thus by Section 3 (i) as claimed. Hence contains a – path that is internally disjoint from as desired.

Clearly all paths of are induced in , either because they are trivial or by Section 3. Let be a nontrivial hosted bridge and let be the nontrivial path of to which it attaches. Then by Section 3 there are vertices and on and a separation of such that , , and apart from the inner vertices of all vertices of are in , in particular, . But has an inner vertex which must be in . So by assumption there is an – fan of size in and thus also an – fan of size . It is easy to see that this can gives rise to the desired bridge adjacencies in .
∎
Proof of Theorem 3.2.
We will trade off some length of a regular decomposition to gain edges in its auxiliary graph. To quantify this we define the function by where and call a regular decomposition of a graph valid if it has adhesion at most , attachedness , and length at least where is the number of edges in the complement of that are incident with at least one nontrivial path of .
Set and let be the integer returned by Theorem 3.1 when invoked with parameters , , , and . We claim that the assertion of Theorem 3.2 is true for this choice of . Let be a connected graph of treewidth less than with at least vertices and suppose that does not contain a subdivision of . Then by the choices of and the graph has a valid decomposition (the foundational linkage has at most paths so there can be at most nonedges in the auxiliary graph). Among all valid decompositions of pick such that the number of edges of is maximal and denote the length of by .
We may assume that for any integer with one of the consecutive inner bags of contains a disturbance. If not, then by Section 3, the contraction of along the sequence is a stable regular decomposition of of length , adhesion at most , and attachedness as desired.
Claim 3.2.1.
Let with . Then the graph contains a linkage from the left adhesion set of to the right adhesion set of such that is a proper supergraph of , the induced permutation of is the identity, and is attached in .
Proof.
There are indices , such that for we have . For each one of the at least consecutive inner bags contains a disturbance by our assumption. Let be the bag of that contains and let be the bridge stabilisation of in .
If is a twisting disturbance, then so is as they have the same induced permutation. If is a bridging disturbance, then so is by Section 3 (ii). The set of end vertices of is the union of both adhesion sets of and clearly for every vertex there is an – fan of size in as is connected. So by Section 3 (iii) the linkage is attached in .
For every denote the induced permutation of by . Since the symmetric group has order at most we can pick^{9}^{9}9 Let be a group of order and . Then of the products for , two must be equal by the pigeon hole principle, say with . This means , where is the neutral element of . indices and with such that .
Let be the linkage from the left adhesion set of to the right adhesion set of in obtained from by replacing with for all . Of all the restrictions of to the bags only with need not induce the identity permutation. However, the composition of their induced permutations is the identity by construction and therefore the induced permutation of is the identity.
To see that is a supergraph of note that so and coincide on and hence by (L8) we have
It remains to show that contains an edge that is not in . Set , , and . If is a bridging disturbance, then contains an edge that is not in . Since and coincide on all bags prior to (down to ) we must have .
If is a twisting disturbance, then , in particular, comes before (there is at least one bag between and , namely ). This means and hence we have
On the other hand, the induced permutation of the restriction of to all bags prior to is and thus . But is not an automorphism of and therefore contains an edge that is not in as desired. This concludes the proof of Claim 3.2.1 ∎
To exploit Claim 3.2.1 we now contract subsegments of consecutive inner bags of into single bags. We assumed earlier that is not stable so the number of nonedges of is at least (if is complete there can be no disturbances). Set . As is valid, its length is at least . Let be the contraction of along the sequence defined by for