On a Conjecture of Thomassen

In 1989, Thomassen asked whether there is an integer-valued function f(k) such that every f(k)-connected graph admits a spanning, bipartite $k$-connected subgraph. In this paper we take a first, humble approach, showing the conjecture is true up to a log n factor.

Conjecture 1 For all k, there exists f (k) such that for all graphs G, if κ(G) ≥ f (k), then there exists a spanning, bipartite H ⊆ G such that κ(H) ≥ k.
In this paper we prove that Conjecture 1 holds up to a log n factor by showing the following: Theorem 1 For all k and n, and for every graph G on n vertices the following holds. If κ(G) ≥ 10 10 k 3 log n, then there exists a spanning, bipartite subgraph H ⊆ G such that κ(H) ≥ k.
Because of the log n factor, we did not try to optimize the dependency on k in Theorem 1. However, it looks like our proof could be modified to give slightly better bounds.

Preliminary Tools
In this section, we introduce a number of preliminary results.

Mader's Theorem
The first tool is the following useful theorem due to Mader [2].
Theorem 2 Every graph of average degree at least 4ℓ has an ℓ-connected subgraph.
Because we are interested in finding bipartite subgraphs with high connectivity, the following corollary will be helpful.
Corollary 1 Every graph G with average degree at least 8ℓ contains a (not necessarily spanning) bipartite subgraph H which is at least ℓ-connected.
Proof: Let G be such a graph and let V (G) = A ∪ B be a partition of V (G) such that |E(A, B)| is maximal. Observe that |E(A, B)| ≥ |E(G)|/2, and therefore, the bipartite graph G ′ with parts A and B has average degree at least 4ℓ. Now, by applying Theorem 2 to G ′ we obtain the desired subgraph H.

Merging k-connected Graphs
We will also make use of the following easy expansion lemma.
Lemma 1 Let H 1 and H 2 be two vertex-disjoint graphs, each of which is k-connected. Let H be a graph obtained by adding k independent edges between these two graphs. Then, κ(H) ≥ k.
Proof: Note first that by construction, one cannot remove all the edges between H 1 and H 2 by deleting fewer than k vertices. Moreover, because H 1 and H 2 are both k-connected, each will remain connected after deleting less than k vertices. From here, the proof follows easily.
Next we will show how to merge a collection of a few k-connected components and single vertices into one k-connected component. Before stating the next lemma formally, we will need to introduce some notation. Let G 1 , . . . , G t be t vertex-disjoint k-connected graphs, let U = {u t+1 , . . . , u t+s } be a set consisting of s vertices which are disjoint to V (G i ) for 1 ≤ i ≤ t, and let R be a k-connected graph on the vertex set {1, . . . , t + s}. Also let X = {G 1 , . . . G t , u t+1 , . . . , u t+s } be an ordered set and X i denote the ith element of X. Finally, let F R := F R (X) denote the family consisting of all graphs G which satisfy the following: , then there exists an edge in G between X i and X j , and Lemma 2 Let G 1 , . . . , G t be t vertex-disjoint graphs, each of which is k-connected, and let U = {u t+1 , . . . , u t+s } be a set of s vertices for which U ∩ V (G i ) = ∅ for every 1 ≤ i ≤ t. Let R be a k-connected graph on the vertex-set {1, . . . , t + s}, and let X = {G 1 , . . . G t , u t+1 , . . . , u t+s }. Then, any graph G ∈ F R (X) is k-connected.
Proof: Let G ∈ F R (X), and let S ⊆ V (G) be a subset of size at most k − 1. We wish to show that the graph G ′ := G \ S is still connected. Let x, y ∈ V (G ′ ) be two distinct vertices in G ′ ; we show that there exists a path in G ′ connecting x to y. Towards this end, we first note that if both x and y are in the same G i , then because each G i is k-connected, there is nothing to prove. Moreover, if both x and y are in distinct elements of X which are also disjoint from S, then we are also finished, as follows. Because R is k-connected, if we delete all of the vertices in R corresponding to elements of X which intersect S, the resulting graph is still connected. Therefore, one can easily find a path between the elements containing x and y which goes only through "untouched" elements of X, and hence, there exists a path connecting x and y.
The remaining case to deal with is when x and y are in different elements of X, and at least one of them is not disjoint with S. Assume x is in some such X i (y will be treated similarly). Using Property (iii) of F R , there is at least one edge between X i and an untouched X j . Therefore one can find a path between x and some vertex x ′ in an untouched X j . This takes us back to the previous case.

Main Technical Lemma
A directed graph or digraph is a set of vertices and a collection of directed edges; note that bidirectional edges are allowed. For a directed graph D and a vertex v ∈ V (D) we let d + D (v) denote the out-degree of v. We let U(D) denote the underlying graph of D, that is the graph obtained by ignoring the directions in D and merging multiple edges. In order to find the desired spanning, bipartite k-connected subgraph in Theorem 1, we look at sub-digraphs in an auxiliary digraph.
The following is our main technical lemma and is the main reason why we have a log n factor.
Proof: If κ(U(D)) ≥ k, then there clearly is nothing to prove. So we may assume that κ(U(D)) ≤ k − 1. Delete a separating set of size at most k − 1. The smallest component, say C 1 , has size at most n/2 and for any v ∈ V (C 1 ), every out-neighbor of v is either in V (C 1 ) or in the separating set that we removed, and so We continue by repeatedly applying this step, and note that this process must terminate. Otherwise, after at most log n steps we are left with a component which consists of one single vertex and yet contains at least one edge, a contradiction.

Highly Connected Graphs
With the preliminaries out of the way, we are now ready to prove Theorem 1.
Proof: Let G be a finite graph on n vertices with κ(G) ≥ 10 10 k 3 log n.
In order to find the desired subgraph, we first initiate G 1 := G and start the following process.
As long as G i contains a bipartite subgraph which is at least k-connected on at least 10 3 k 2 log n vertices, let H i = (S i ∪ T i , E i ) be such a subgraph of maximum size, and let G i+1 := G i \ V (H i ).
Let H 1 , . . . , H t be the sequence obtained in this manner, and note that all the H i 's are vertex disjoint with κ(H i ) ≥ k and |V (H i )| ≥ 10 3 k 2 log n. Observe that if H 1 is spanning, then there is nothing to prove. Therefore, suppose for a contradiction that H 1 is not spanning. Let V 0 = {v 1 , . . . , v s } be the subset of V (G) remaining after this process; note that it might be the case that V 0 = ∅. Because each H i is a bipartite, k-connected subgraph of G i of maximum size and G is 10 10 k 3 log n connected, we show that the following are true: Indeed, for showing (a), note that if there are at least 4k independent edges between H i to H j , by pigeonhole principle, at least k of them are between the same part of H i (say S i ) and the same part of H j (say S j ). Therefore, the graph obtained by joining H i to H j with this set of at least k edges is a k-connected (by Lemma 1), bipartite graph and is larger than H i , contrary to the maximality of H i .
For showing (b), note first that at least k edges incident with v touch the same part of H i , and let F be a set of k such edges. Second, we mention that joining a vertex of degree at least k to a k-connected graph trivially yields a k-connected graph. Next, since all the edges in F are touching the same part, the graph obtained by adding v to V (H i ) and F to E(H i ), will also be bipartite. This contradicts the maximality of H i .
For (c), note first that since H 1 is not spanning, using (b) we conclude that in the construction of the bipartite subgraphs H 1 , . . . , H t in the process above, δ(G 2 ) ≥ 10 10 k 3 log n − 2k ≥ 8000k 2 log n.
Therefore, using Corollary 1, it follows that G 2 contains a bipartite subgraph of size at least 10 3 k 2 log n which is also k-connected.
Therefore, the process does not terminate at this point, and H 2 exists (that is, t ≥ 2). It also follows that for each 1 ≤ i ≤ t we have |V (G) \ V (H i )| ≥ 10 3 k 2 log n. Next, note that G is 10 10 k 3 log n connected, and that each H i is of size at least 10 3 k 2 log n. For each i, consider the bipartite graph with parts V (H i ) and V (G) \ V (H i ) and with the edge-set consisting of all the edges of G which touch both of these parts. Using König's Theorem (see [5], p. 112), it follows that if there is no such M i of size 10 3 k 2 log n, then there exists a set of strictly fewer than 10 3 k 2 log n vertices that touch all the edges in this bipartite graph (a vertex cover). By deleting these vertices, one can separate what is left from H i and its complement, contrary to the fact that G is 10 10 k 3 log n connected.
In order to complete the proof, we wish to reach a contradiction by showing that one can either merge few members of {H 1 , . . . , H t } with vertices of V 0 into a k-connected component or find a k-connected component of size at least 10 3 k 2 log n which is contained in V 0 . In order to do so, we define an auxiliary digraph, based off of a special subgraph G ′ ⊆ G, and use Lemmas 3 and 2 to achieve the desired contradiction. We first describe how to find G ′ .
First, we partition V 0 into two sets, say A and B, where and observe that, using (b), any vertex a ∈ A must send edges to at least Finally, define G ′ as the spanning bipartite graph of G obtained by deleting all of the edges within A and for distinct i and j, the edges between H i and H j which are not contained in M ′ i ∪ M ′ j . Recall by construction, using Φ we generated labels at random; therefore, by using Chernoff bounds (for instance see [1]), one can easily check that with high probability the following hold: Note that here we relied on the luxury of losing the log n factor for using Chernoff bounds, but it seems like we could easily handle this "cleaning process" completely by hand. Now we are ready to define our auxiliary digraph D. To this end, we first orient edges (perhaps not all of them) of G ′ in the following way: For every 1 ≤ i ≤ t, we orient all of the edges in E(G ′ )∩M ′ i out of H i . We orient all of the edges between A and ∪ t i=1 V (H i ) out of A. We orient edges between B and ∪ t i=1 V (H i ) arbitrarily, and we orient edges within A ∪ B in both directions. Now, we define D to be the digraph with vertex set V (D) = X, and − → xy ∈ E(D) if and only if there exists an edge between x and y in G ′ which is oriented from x to y.
In order to complete the proof, we first note that D is a digraph on at most n vertices with out-degree δ + (D) > (k − 1)⌈log n⌉. This follows immediately from Properties (i)-(iii) as well as the way we oriented the edges. Therefore, one can apply Lemma 3 to find a sub-digraph Note that by construction, every pair of edges which are oriented out of some H i must be independent and go to different components. Using Property 1. above combined with the fact that δ + (D ′ ) ≥ d + D (v) −(k −1) ⌈log n⌉ ≥ k, we may conclude that the subgraph G ′′ ⊆ G ′ induced by the union of all the components in V (D ′ ) satisfies G ′′ ∈ F U (D ′ ) (V (D ′ )). Applying Lemma 2 with X = V (D ′ ) and R = U(D ′ ), it follows that κ(G ′′ ) ≥ k.
In order to obtain the desired contradiction, we consider the following two cases: Case 1: V (G ′′ ) contains V (H i ) for some i. We note that this case is actually impossible because it would contradict the maximality of the minimal index i for which V (H i ) ⊆ V (G ′′ ).
Case 2: V (G ′′ ) ⊆ A ∪ B. We note that in this case, there must be at least one vertex b ∈ B ∩ V (G ′′ ). Indeed, G ′′ is k-connected, and there are no edges within A. Now, it follows from Properties 1. and (ii) above that It thus follow that |V (G ′′ )| ≥ 10 4 k 3 log n. Combining this observation with the facts that G ′′ is k-connected and V (G ′′ ) ⊆ A ∪ B, we obtain a contradiction. This case can not arise because G ′′ should have been included as one of the bipartite subgraphs {H 1 , . . . , H t }.
This completes the proof.