On a Conjecture of Thomassen

On a Conjecture of Thomassen

Michelle Delcourt
Department of Mathematics
University of Illinois
Urbana, Illinois 61801, U.S.A.
delcour2@illinois.edu
Research supported by NSF Graduate Research Fellowship DGE 1144245.
Asaf Ferber
Department of Mathematics
Yale University, and
Department of Mathematics,
MIT.
asaf.ferber@yale.edu, and ferbera@mit.edu.
Submitted: Oct 17, 2014; Accepted: XX; Published: XX
Mathematics Subject Classifications: 05C20, 05C40
Submitted: Oct 17, 2014; Accepted: XX; Published: XX
Mathematics Subject Classifications: 05C20, 05C40
Abstract

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

1 Introduction

Erdős noticed [4] that any graph with minimum degree at least contains a spanning, bipartite subgraph with . The proof for this fact is obtained by taking a maximal edge-cut, a partition of into two sets and , such that the number of edges with one endpoint in and one in , denoted , is maximal. Observe that if some vertex in does not have degree at least in , then by moving to , one would increase , contrary to maximality. The same argument holds for vertices in . In fact this proves that for each vertex , by taking such a subgraph , the degree of in , denoted , is at least . This will be used throughout the paper.

Thomassen observed that the same proof shows the following stronger statement. Given a graph which is at least edge-connected (that is one must remove at least edges in order to disconnect the graph), then contains a bipartite subgraph for which is edge-connected. In fact, each edge-cut keeps at least half of its edges. This observation led Thomassen to conjecture that a similar phenomena also holds for vertex-connectivity.

Before proceeding to the statement of Thomassen’s conjecture, we remind the reader that a graph is said to be vertex-connected or -connected if one must remove at least vertices from in order to disconnect the graph (or to remain with one single vertex). We also let denote the minimum integer for which is -connected. Roughly speaking, Thomassen conjectured that any graph with high enough connectivity also should contain a -connected spanning, bipartite subgraph. The following appears as Conjecture 7 in [3].

Conjecture 1

For all , there exists a function such that for all graphs , if , then there exists a spanning, bipartite such that .

In this paper we prove that Conjecture 1 holds up to a factor by showing the following:

Theorem 1

For all and , and for every graph on vertices the following holds. If , then there exists a spanning, bipartite subgraph such that .

Because of the factor, we did not try to optimize the dependency on in Theorem 1. However, it looks like our proof could be modified to give slightly better bounds.

2 Preliminary Tools

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

The first tool is the following useful theorem due to Mader [2].

Theorem 2

Every graph of average degree at least has an -connected subgraph.

Because we are interested in finding bipartite subgraphs with high connectivity, the following corollary will be helpful.

Corollary 1

Every graph with average degree at least contains a (not necessarily spanning) bipartite subgraph which is at least -connected.

Proof:

Let be such a graph and let be a partition of such that is maximal. Observe that , and therefore, the bipartite graph with parts and has average degree at least . Now, by applying Theorem 2 to we obtain the desired subgraph .

2.2 Merging k-connected Graphs

We will also make use of the following easy expansion lemma.

Lemma 1

Let and be two vertex-disjoint graphs, each of which is -connected. Let be a graph obtained by adding independent edges between these two graphs. Then, .

Proof:

Note first that by construction, one cannot remove all the edges between and by deleting fewer than vertices. Moreover, because and are both -connected, each will remain connected after deleting less than vertices. From here, the proof follows easily.

Next we will show how to merge a collection of a few -connected components and single vertices into one -connected component. Before stating the next lemma formally, we will need to introduce some notation. Let be vertex-disjoint -connected graphs, let be a set consisting of vertices which are disjoint to for , and let be a -connected graph on the vertex set . Also let be a -tuple and denote the th element of . Finally, let denote the family consisting of all graphs which satisfy the following:

1. the disjoint union of the elements of is a spanning subgraph of , and

2. for every distinct if , then there exists an edge in between and , and

3. for every , there is a set of independent edges between and distinct vertex sets , where .

Lemma 2

Let be vertex-disjoint graphs, each of which is -connected, and let be a set of vertices for which for every . Let be a -connected graph on the vertex-set , and let . Then, any graph is -connected.

Proof:

Let , and let be a subset of size at most . We wish to show that the graph is still connected. Let be two distinct vertices in ; we show that there exists a path in connecting to . Towards this end, we first note that if both and are in the same , then because each is -connected, there is nothing to prove. Moreover, if both and are in distinct elements of which are also disjoint from , then we are also finished, as follows. Because is -connected, if we delete all of the vertices in corresponding to elements of which intersect , the resulting graph is still connected. Therefore, one can easily find a path between the elements containing and which goes only through “untouched” elements of , and hence, there exists a path connecting and .

The remaining case to deal with is when and are in different elements of , and at least one of them is not disjoint with . Assume is in some such ( will be treated similarly). Using Property of , there is at least one edge between and an untouched . Therefore one can find a path between and some vertex in an untouched . This takes us back to the previous case.

2.3 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 and a vertex we let denote the out-degree of . We let denote the underlying graph of , that is the graph obtained by ignoring the directions in and merging multiple edges. In order to find the desired spanning, bipartite -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 factor.

Lemma 3

If is a finite digraph on at most vertices with minimum out-degree

 δ+(D)>(k−1)⌈logn⌉,

then there exists a sub-digraph such that

1. For all we have , and

2. .

Proof:

If , then there clearly is nothing to prove. So we may assume that . Delete a separating set of size at most . The smallest component, say , has size at most and for any , every out-neighbor of is either in or in the separating set that we removed, and so

 d+C1(v)⩾d+D(v)−(k−1).

We continue by repeatedly applying this step, and note that this process must terminate. Otherwise, after at most steps we are left with a component which consists of one single vertex and yet contains at least one edge, a contradiction.

3 Highly Connected Graphs

With the preliminaries out of the way, we are now ready to prove Theorem 1.

Proof:

Let be a finite graph on vertices with

 κ(G)⩾1010k3logn.

In order to find the desired subgraph, we first initiate and start the following process.

As long as contains a bipartite subgraph which is at least -connected on at least vertices, let be such a subgraph of maximum size, and let . Note that must exist as

 δ(G1)⩾1010k3logn−2k⩾8000k2logn,

and so by Corollary 1, must contain a -connected bipartite subgraph of size at least .

Let be the sequence obtained in this manner, and note that all the ’s are vertex disjoint with and . Observe that if is spanning, then there is nothing to prove. Therefore, suppose for a contradiction that is not spanning. Let be the subset of remaining after this process; note that it might be the case that . Because each is a bipartite, -connected subgraph of of maximum size and is connected, we show that the following are true:

1. For every , there are less than independent edges between and , and

2. for every and , the number of edges in between and , denoted by , is less than , and

3. for every , there exists a set consisting of exactly independent edges, each of which has exactly one endpoint in .

Indeed, for showing , note that if there are at least independent edges between to , by pigeonhole principle, at least of them are between the same part of (say ) and the same part of (say ). Therefore, the graph obtained by joining to with this set of at least edges is a -connected (by Lemma 1), bipartite graph and is larger than , contrary to the maximality of .

For showing , note that if there are at least between and then there are at least edges incident with touch the same part of , and let be a set of such edges. Second, we mention that joining a vertex of degree at least to a -connected graph trivially yields a -connected graph. Next, since all the edges in are touching the same part, the graph obtained by adding to and to , will also be bipartite. This contradicts the maximality of .

For , note first that since is not spanning, using we conclude that in the construction of the bipartite subgraphs in the process above,

 δ(G2)⩾1010k3logn−2k⩾8000k2logn.

Therefore, using Corollary 1, it follows that contains a bipartite subgraph of size at least which is also -connected.

Therefore, the process does not terminate at this point, and exists (that is, ). It also follows that for each we have . Next, note that is connected, and that each is of size at least . For each , consider the bipartite graph with parts and and with the edge-set consisting of all the edges of which touch both of these parts. Using König’s Theorem (see [5], p. 112), it follows that if there is no such of size , then there exists a set of strictly fewer than vertices that touch all the edges in this bipartite graph (a vertex cover). By deleting these vertices, one can separate what is left from and its complement, contrary to the fact that is connected.

In order to complete the proof, we wish to reach a contradiction by showing that one can either merge few members of with vertices of into a -connected component or find a -connected component of size at least which is contained in . In order to do so, we define an auxiliary digraph, using a special subgraph , and use Lemmas 3 and 2 to achieve the desired contradiction. We first describe how to find .

First, we partition into two sets, say and , where

 A={v∈V0:dG(v,t⋃i=1V(Hi))⩾104k3logn},

and observe that, using , since , any vertex must send edges to more than

 104k3logn/(2k)=5000k2logn

distinct elements in . For each , let be a set as described in . Observe that, using , each such touches more than

 103k2logn/(4k)=250klogn

distinct elements of . Let be a subset of size exactly such that each pair of edges in touches two distinct elements of , which of course are distinct from . Recall that for every .

For , let

 Φ:Y→{L,R}

be a mapping, generated according to the following random process:

Let be mutually independent random variables. For each , if , then let and . Otherwise, let and . For every , if , then let , and otherwise . Now, delete all of the edges between two distinct elements of which receive the same label according to .

Finally, define as the spanning bipartite graph of obtained by deleting all of the edges within and for distinct and , the edges between and which are not contained in .

Recall by construction, using 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:

1. For every , each set touches at least (say) other elements of , and

2. for each , the degree of into is at least (say) , and

3. for each vertex , there exist edges between and that touch at least (say) distinct members of .

Note that here we relied on the luxury of losing the 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 . To this end, we first orient edges (again, bidirectional edges are allowed, and un-oriented edges are considered as bidirectional) of in the following way:

For every , we orient all of the edges in out of . We orient all of the edges between and out of . We orient edges between and arbitrarily, and we orient the remaining edges within in both directions.

Now, we define to be the digraph with vertex set , and if and only if there exists an edge between and in which is oriented from to .

In order to complete the proof, we first note that with high probability is a digraph on at most vertices with out-degree . This follows immediately from Properties - as well as the way we oriented the edges. Therefore, one can apply Lemma 3 to find a sub-digraph such that

1. For all we have , and

2. .

In fact, with high probability, Note that by construction, every pair of edges which are oriented out of some must be independent and go to different components. Using Property above combined with the fact that , we may conclude that the subgraph induced by the union of all the components in satisfies . Applying Lemma 2 with and , it follows that .

In order to obtain the desired contradiction, we consider the following two cases:

Case 1: contains for some . We note that this case is actually impossible because it would contradict the maximality of for the minimal index such that .

Case 2: . We note that in this case, there must be at least one vertex . Indeed, is -connected, and there are no edges within . Now, it follows from Properties  and above that

 d+D′(b)⩾d+D(b)−(k−1)⌈logn⌉⩾104k3logn.

Thus, it follows that . Combining this observation with the facts that is -connected and , we obtain a contradiction. This case can not arise because should have been included as one of the bipartite subgraphs .

This completes the proof.

Acknowledgments. The authors would like to thank the anonymous referees for valuable comments. The second author would also like to thank Andrzej Grzesik, Hong Liu and Cory Palmer for fruitful discussions in a previous attempt to attack this problem.

References

• [1] N. Alon and J. H. Spencer, The probabilistic method, second edition, Wiley-Interscience Series in Discrete Mathematics and Optimization, Wiley-Interscience, New York, 2000.
• [2] W. Mader, Existenz -fach zusammenhängender Teilgraphen in Graphen genügend grosser Kantendichte, Abh. Math. Sem. Univ. Hamburg 37 (1972), pp. 86–97.
• [3] C. Thomassen. Configurations in graphs of large minimum degree, connectivity, or chromatic number in Combinatorial Mathematics: Proceedings of the Third International Conference (New York, 1985). Ann. New York Acad. Sci., Vol. 555, 1989, pp. 402–412.
• [4] C. Thomassen. Paths, circuits and subdivisions. in Selected topics in graph theory, 3, Academic Press, San Diego, CA, 1988, pp. 97–131.
• [5] D. B. West. Introduction to graph theory. Prentice Hall, Inc., Upper Saddle River, NJ, 1966.
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