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.Comment: 9 page

Delcourt, Michelle

Ferber, Asaf

The Electronic Journal of Combinatorics

English

arXiv

Michelle Delcourt

Asaf Ferber

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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

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On a Conjecture of ThomassenMichelle Delcourt∗Department of MathematicsUniversity of IllinoisUrbana, Illinois 61801, U.S.A.delcour2@illinois.eduAsaf FerberDepartment of MathematicsYale UniversityNew Haven, Connecticut, U.S.A., andDepartment of Mathematics,MIT, Cambridge, Massachusetts, U.S.A.asaf.ferber@yale.edu ferbera@mit.eduSubmitted: Oct 18, 2014; Accepted: Jun 11, 2015; Published: Jul 1, 2015Mathematics Subject Classifications: 05C20, 05C40AbstractIn 1989, Thomassen asked whether there is an integer-valued function f(k) such thatevery f(k)-connected graph admits a spanning, bipartite k-connected subgraph. Inthis paper we take a first, humble approach, showing the conjecture is true up to alog n factor.1 IntroductionErdo˝s noticed [4] that any graph G with minimum degree δ(G) at least 2k − 1 containsa spanning, bipartite subgraph H with δ(H) > k. The proof for this fact is obtained bytaking a maximal edge-cut, a partition of V (G) into two sets A and B, such that thenumber of edges with one endpoint in A and one in B, denoted |E(A,B)|, is maximal.Observe that if some vertex v in A does not have degree at least k in G[B], then by movingv to B, one would increase |E(A,B)|, contrary to maximality. The same argument holdsfor vertices in B. In fact this proves that for each vertex v ∈ V (G), by taking such asubgraph H, the degree of v in H, denoted dH(v), is at least dG(v)/2. This will be usedthroughout the paper.Thomassen observed that the same proof shows the following stronger statement.Given a graph G which is at least (2k − 1) edge-connected (that is one must remove atleast 2k−1 edges in order to disconnect the graph), then G contains a bipartite subgraphH for which H is k 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 forvertex-connectivity.∗Research supported by NSF Graduate Research Fellowship DGE 1144245.the electronic journal of combinatorics 22(3) (2015), #P3.2 1Before proceeding to the statement of Thomassen’s conjecture, we remind the readerthat a graph G is said to be k vertex-connected or k-connected if one must remove atleast k vertices from V (G) in order to disconnect the graph (or to remain with one singlevertex). We also let κ(G) denote the minimum integer k for which G is k-connected.Roughly speaking, Thomassen conjectured that any graph with high enough connectivityalso should contain a k-connected spanning, bipartite subgraph. The following appearsas Conjecture 7 in [3].Conjecture 1. For all k, there exists a function f(k) such that for all graphs G, ifκ(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 thefollowing:Theorem 1. For all k and n, and for every graph G on n vertices the following holds.If κ(G) > 1010k3 log n, then there exists a spanning, bipartite subgraph H ⊆ G such thatκ(H) > k.Because of the log n factor, we did not try to optimize the dependency on k in Theorem1. However, it looks like our proof could be modified to give slightly better bounds.2 Preliminary ToolsIn this section, we introduce a number of preliminary results.2.1 Mader’s TheoremThe first tool is the following useful theorem due to Mader [2].Theorem 2. Every graph of average degree at least 4` has an `-connected subgraph.Because we are interested in finding bipartite subgraphs with high connectivity, thefollowing corollary will be helpful.Corollary 1. Every graph G with average degree at least 8` contains a (not necessarilyspanning) bipartite subgraph H which is at least `-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 bipartitegraph G′ with parts A and B has average degree at least 4`. Now, by applying Theorem2 to G′ we obtain the desired subgraph H.the electronic journal of combinatorics 22(3) (2015), #P3.2 22.2 Merging k-connected GraphsWe will also make use of the following easy expansion lemma.Lemma 1. Let H1 and H2 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 H1 andH2 by deleting fewer than k vertices. Moreover, because H1 and H2 are both k-connected,each will remain connected after deleting less than k vertices. From here, the proof followseasily.Next we will show how to merge a collection of a few k-connected components andsingle vertices into one k-connected component. Before stating the next lemma formally,we will need to introduce some notation. Let G1, . . . , Gt be t vertex-disjoint k-connectedgraphs, let U = {ut+1, . . . , ut+s} be a set consisting of s vertices which are disjoint toV (Gi) for 1 6 i 6 t, and let R be a k-connected graph on the vertex set {1, . . . , t + s}.Also let X = (G1, . . . Gt, ut+1, . . . , ut+s) be a (t+ s)-tuple and Xi denote the ith elementof X. Finally, let FR := FR(X) denote the family consisting of all graphs G which satisfythe following:(i) the disjoint union of the elements of X is a spanning subgraph of G, and(ii) for every distinct i, j ∈ V (R) if ij ∈ E(R), then there exists an edge in G betweenXi and Xj, and(iii) for every 1 6 i 6 t, there is a set of k independent edges between V (Gi) and kdistinct vertex sets {V (Xj1), . . . , V (Xjk)}, where V (ui) = {ui}.Lemma 2. Let G1, . . . , Gt be t vertex-disjoint graphs, each of which is k-connected, andlet U = {ut+1, . . . , ut+s} be a set of s vertices for which U ∩ V (Gi) = ∅ for every 1 6i 6 t. Let R be a k-connected graph on the vertex-set {1, . . . , t + s}, and let X ={G1, . . . Gt, ut+1, . . . , ut+s}. Then, any graph G ∈ FR(X) is k-connected.Proof. Let G ∈ FR(X), and let S ⊆ V (G) be a subset of size at most k − 1. We wishto show that the graph G′ := G \ S is still connected. Let x, y ∈ V (G′) be two distinctvertices 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 Gi, then because each Gi is k-connected,there is nothing to prove. Moreover, if both x and y are in distinct elements of X whichare 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 theelements containing x and y which goes only through “untouched” elements of X, andhence, 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, andat least one of them is not disjoint with S. Assume x is in some such Xi (y will bethe electronic journal of combinatorics 22(3) (2015), #P3.2 3treated similarly). Using Property (iii) of FR, there is at least one edge between Xi andan untouched Xj. Therefore one can find a path between x and some vertex x′ in anuntouched Xj. This takes us back to the previous case.2.3 Main Technical LemmaA directed graph or digraph is a set of vertices and a collection of directed edges; notethat bidirectional edges are allowed. For a directed graph D and a vertex v ∈ V (D) welet 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, welook at sub-digraphs in an auxiliary digraph.The following is our main technical lemma and is the main reason why we have a log nfactor.Lemma 3. If D is a finite digraph on at most n vertices with minimum out-degreeδ+(D) > (k − 1) dlog ne ,then there exists a sub-digraph D′ ⊆ D such that1. For all v ∈ V (D′) we have d+D′(v) > d+D(v)− (k − 1) dlog ne, and2. κ(U(D′)) > k.Proof. If κ(U(D)) > k, then there clearly is nothing to prove. So we may assume thatκ(U(D)) 6 k− 1. Delete a separating set of size at most k− 1. The smallest component,say C1, has size at most n/2 and for any v ∈ V (C1), every out-neighbor of v is either inV (C1) or in the separating set that we removed, and sod+C1(v) > d+D(v)− (k − 1).We continue by repeatedly applying this step, and note that this process must termi-nate. Otherwise, after at most log n steps we are left with a component which consists ofone single vertex and yet contains at least one edge, a contradiction.3 Highly Connected GraphsWith the preliminaries out of the way, we are now ready to prove Theorem 1.Proof. Let G be a finite graph on n vertices withκ(G) > 1010k3 log n.In order to find the desired subgraph, we first initiate G1 := G and start the followingprocess.the electronic journal of combinatorics 22(3) (2015), #P3.2 4As long as Gi contains a bipartite subgraph which is at least k-connected on at least103k2 log n vertices, let Hi = (Si ∪ Ti, Ei) be such a subgraph of maximum size, and letGi+1 := Gi \ V (Hi). Note that H1 must exist asδ(G1) > 1010k3 log n− 2k > 8000k2 log n,and so by Corollary 1, G1 must contain a k-connected bipartite subgraph of size at least103k2 log n.Let H1, . . . , Ht be the sequence obtained in this manner, and note that all the Hi’sare vertex disjoint with κ(Hi) > k and |V (Hi)| > 103k2 log n. Observe that if H1 isspanning, then there is nothing to prove. Therefore, suppose for a contradiction that H1is not spanning. Let V0 := V (Gt+1) = {v1, . . . , vs} be the subset of V (G) remaining afterthis process; note that it might be the case that V0 = ∅. Because each Hi is a bipartite,k-connected subgraph of Gi of maximum size and G is 1010k3 log n connected, we showthat the following are true:(a) For every 1 6 i < j 6 t, there are less than 4k independent edges between Hi andHj, and(b) for every j > i and v ∈ V (Gj), the number of edges in G between v and Hi, denotedby dG(v, V (Hi)), is less than 2k, and(c) for every 1 6 i 6 t, there exists a set Mi consisting of exactly 103k2 log n independentedges, each of which has exactly one endpoint in Hi.Indeed, for showing (a), note that if there are at least 4k independent edges betweenHi to Hj, by pigeonhole principle, at least k of them are between the same part of Hi(say Si) and the same part of Hj (say Sj). Therefore, the graph obtained by joining Hito Hj with this set of at least k edges is a k-connected (by Lemma 1), bipartite graphand is larger than Hi, contrary to the maximality of Hi.For showing (b), note that if there are at least 2k between v and Hi then there areat least k edges incident with v touch the same part of Hi, and let F be a set of k suchedges. Second, we mention that joining a vertex of degree at least k to a k-connectedgraph trivially yields a k-connected graph. Next, since all the edges in F are touchingthe same part, the graph obtained by adding v to V (Hi) and F to E(Hi), will also bebipartite. This contradicts the maximality of Hi.For (c), note first that since H1 is not spanning, using (b) we conclude that in theconstruction of the bipartite subgraphs H1, . . . , Ht in the process above,δ(G2) > 1010k3 log n− 2k > 8000k2 log n.Therefore, using Corollary 1, it follows that G2 contains a bipartite subgraph of size atleast 103k2 log n which is also k-connected.Therefore, the process does not terminate at this point, and H2 exists (that is, t > 2).It also follows that for each 1 6 i 6 t we have |V (G) \ V (Hi)| > 103k2 log n. Next, notethat G is 1010k3 log n connected, and that each Hi is of size at least 103k2 log n. For eachthe electronic journal of combinatorics 22(3) (2015), #P3.2 5i, consider the bipartite graph with parts V (Hi) and V (G) \ V (Hi) and with the edge-setconsisting of all the edges of G which touch both of these parts. Using Ko¨nig’s Theorem(see [5], p. 112), it follows that if there is no such Mi of size 103k2 log n, then there existsa set of strictly fewer than 103k2 log n vertices that touch all the edges in this bipartitegraph (a vertex cover). By deleting these vertices, one can separate what is left from Hiand its complement, contrary to the fact that G is 1010k3 log n connected.In order to complete the proof, we wish to reach a contradiction by showing that onecan either merge few members of {H1, . . . , Ht} with vertices of V0 into a k-connectedcomponent or find a k-connected component of size at least 103k2 log n which is containedin V0. In order to do so, we define an auxiliary digraph, using a special subgraph G′ ⊆ G,and use Lemmas 3 and 2 to achieve the desired contradiction. We first describe how tofind G′.First, we partition V0 into two sets, say A and B, whereA ={v ∈ V0 : dG(v,t⋃i=1V (Hi))> 104k3 log n},and observe that, using (b), since A ⊆ V0, any vertex a ∈ A must send edges to more than104k3 log n/(2k) = 5000k2 log ndistinct elements in X := {H1, . . . , Ht, v1, . . . , vs}. For each 1 6 i 6 t, let Mi be a set asdescribed in (c). Observe that, using (b), each such Mi touches more than103k2 log n/(4k) = 250k log ndistinct elements of X\ {Hi}. Let M ′i ⊆ Mi be a subset of size exactly 250k log n suchthat each pair of edges in M ′i touches two distinct elements of X\ {Hi}, which of courseare distinct from Gi. Recall that Hi = (Si ∪ Ti, Ei) for every 1 6 i 6 t.For Y := {S1, . . . , St, T1, . . . , Tt, v1, . . . , vs}, letΦ : Y → {L,R}be a mapping, generated according to the following random process:Let X1, . . . , Xt, Y1, . . . , Ys ∼ Bernoulli(1/2) be mutually independent random vari-ables. For each 1 6 i 6 t, if Xi = 1, then let Φ(Si) = L and Φ(Ti) = R. Otherwise,let Φ(Si) = R and Φ(Ti) = L. For every 1 6 j 6 s, if Yj = 1, then let Φ(vj) = L, andotherwise Φ(vj) = R. Now, delete all of the edges between two distinct elements of Ywhich receive the same label according to Φ.Finally, define G′ as the spanning bipartite graph of G obtained by deleting all ofthe edges within A and for distinct i and j, the edges between Hi and Hj which are notcontained in M ′i ∪M ′j.Recall by construction, using Φ we generated labels at random; therefore, by usingChernoff bounds (for instance see [1]), one can easily check that with high probability thefollowing hold:the electronic journal of combinatorics 22(3) (2015), #P3.2 6(i) For every 1 6 i 6 t, each set M ′i ∩ E(G′) touches at least (say) 120k log n otherelements of X, and(ii) for each b ∈ B, the degree of b into A∪B is at least (say) dG′(b, A∪B) > 105k3 log n,and(iii) for each vertex a ∈ A, there exist edges between a and ∪ti=1V (Hi) that touch atleast (say) 2000k2 log n distinct members of {H1, . . . , Ht}.Note that here we relied on the luxury of losing the log n factor for using Chernoffbounds, but it seems like we could easily handle this “cleaning process” completely byhand.Now we are ready to define our auxiliary digraph D. To this end, we first orientedges (again, bidirectional edges are allowed, and un-oriented edges are considered asbidirectional) of G′ in the following way:For every 1 6 i 6 t, we orient all of the edges in E(G′)∩M ′i out of Hi. We orient all ofthe edges between A and ∪ti=1V (Hi) out of A. We orient edges between B and ∪ti=1V (Hi)arbitrarily, and we orient the remaining 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 andonly 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 with high probability D is a digraphon at most n vertices with out-degree δ+(D) > (k − 1)dlog ne. This follows immediatelyfrom Properties (i)-(iii) as well as the way we oriented the edges. Therefore, one canapply Lemma 3 to find a sub-digraph D′ ⊆ D such that1. For all v ∈ V (D′) we have d+D′(v) > d+D(v)− (k − 1) dlog ne, and2. κ(U(D′)) > k.In fact, with high probability, δ+(D) > 120k log n > k + (k− 1) dlog ne . Note that byconstruction, every pair of edges which are oriented out of some Hi must be independentand go to different components. Using Property 1. above combined with the fact thatδ+(D′) > δ+(D)−(k−1) dlog ne > k, we may conclude that the subgraph G′′ ⊆ G′ inducedby the union of all the components in V (D′) satisfies G′′ ∈ FU(D′)(V (D′)). ApplyingLemma 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 (Hi) for some i. We note that this case is actually impos-sible because it would contradict the maximality of Hi for the minimal index i such thatV (Hi) ⊆ V (G′′).Case 2: V (G′′) ⊆ A∪B. We note that in this case, there must be at least one vertexb ∈ B ∩ V (G′′). Indeed, G′′ is k-connected, and there are no edges within A. Now, itfollows from Properties 1. and (ii) above thatd+D′(b) > d+D(b)− (k − 1)dlog ne > 104k3 log n.the electronic journal of combinatorics 22(3) (2015), #P3.2 7Thus, it follows that |V (G′′)| > 104k3 log n. Combining this observation with the factsthat G′′ is k-connected and V (G′′) ⊆ A∪B, we obtain a contradiction. This case can notarise becauseG′′ should have been included as one of the bipartite subgraphs {H1, . . . , Ht}.This completes the proof.AcknowledgmentsThe authors would like to thank the anonymous referees for valuable comments. Thesecond author would also like to thank Andrzej Grzesik, Hong Liu and Cory Palmer forfruitful 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 n-fach zusammenha¨ngender Teilgraphen in Graphen genu¨gendgrosser Kantendichte, Abh. Math. Sem. Univ. Hamburg 37 (1972), pp. 86–97.[3] C. Thomassen. Configurations in graphs of large minimum degree, connectivity, orchromatic number in Combinatorial Mathematics: Proceedings of the Third Inter-national 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.the electronic journal of combinatorics 22(3) (2015), #P3.2 8

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