The Removable Edges and the Contractible Subgraphs of 5-Connected Graphs

Graphs and CombinatoricsDOI 10.1007/s00373-013-1368-y

ORIGINAL PAPER

The Removable Edges and the Contractible Subgraphsof 5-Connected Graphs

Chengfu Qin · Xiaofeng Guo · Kiyoshi Ando

Received: 3 April 2012 / Revised: 19 August 2013© Springer Japan 2013

Abstract An edge of a k-connected graph G is said to be k-removable if G − e isstill k-connected. A subgraph H of a k-connected graph is said to be k-contractible ifits contraction, that is, identification every component of H to a single vertex, resultsagain a k-connected graph. In this paper, we show that there is either a removable edgeor a contractible subgraph in a 5-connected graph which contains an edge with bothendvertices have degree more than five. Thus every edge of minor minimal 5-connectedgraph is incident to at least one vertex of degree 5.

Keywords 5-connected graph · Contractible subgraph · Removable edge ·Minor minimal

Mathematics Subject Classification (2000) 05C40

The project supported by the National Natural Science Foundation of China (No. 11126321, No. 1161006);Natural Sciences Foundation of Guangxi Province (No. 2012GXNSFBA053005); The Scientific ResearchFoundation of Guangxi Education Committee (No.200103YB069).

C. Qin (B)School of Mathematics Science, Guangxi Teachers Education University,Guangxi 530001, Nanning, People’s Republic of Chinae-mail: [email protected]

X. GuoDepartment of Mathematics, Xiamen University, Xiamen 310065, Fujian,People’s Republic of China

K. AndoDepartment of Information and Communication Engineering,The University of Electro-Communications, 1-5-1 Chofugaoku,Chofu, Tokyo 182-8585, Japan

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Graphs and Combinatorics

1 Introduction

A subgraph of a k-connected graph G is said to be k-contractible if its contraction(i.e., identifying each components to a single vertex, removing each of the resultingloops and, finally, replacing each of the resulting duplicate edges by a single edge)results again a k-connected graph. Since Tutte [1] proved his famous wheel theorem,the distribution of k-contractible subgraphs of k-connected graph is an interestingresearch area within graph connectivity theory. The wheel theorem shows that every3-connected graph on more than four vertices contains an edge whose contractionresults again a 3-connected graph. By the wheel theorem, One can give an inductiveproof of Kuratowski’s theorem. The existence of 3-contractible edges led also to someother interesting results in graph theory [2,3].

For k-connected graphs with k ≥ 4, it is difficult to performing an induction proof byusing single edge contraction as there are infinitely many nonisomorphic k-connectedgraphs which do not contain any k-contractible edge. These graphs are said to becontraction critical k-connected.

A set T ⊆ V (G) is called a separating set of a connected graph G, if G − T hasat least two connected components. A separating set with κ(G) vertices is called a κ-separator, here κ(G) stands for the connectivity of G. Clearly, the two endvertices ofedge which is not k-contractible are contained in some k-separators. The contractioncritical 4-connected graphs are characterized, which are two special kinds of graphs.The classification of contraction critically 5-connected graphs seems to be totallyhopeless, although there are several results on local structure of these graphs [4–13].

However, every 4-connected graph on at least seven vertices can be reduced to asmaller 4-connected graph by contracting one or two edges subsequently. So, naturally,the question that whether there exist post integers b and h such that every k-connectedgraph on more than b vertices can be reduced to a more smaller k-connected graph bycontracting less than h edges for every k ≥ 1 is posted [14]. It is holds for k = 1, 2, 3and 4. But for k ≥ 6, such a statement fails since toroidal triangulations of large facewidth is a counterexample [15]. The question is still open for k = 5.

Conjecture 1 [14] There exist b, h such that every 5-connected graph G with at leastb vertices can be contracted to a 5-connected graph H such that 0 < |V (G)|−|V (H)|< h.

The icosahedron shows that b ≥ 13. A k-connected graph which can not be reducedto a smaller k-connected graph by contracting or deleting any number of edges is saidto be a minor minimal k-connected graph. So, in order to deal with Conjecture 1, wemust find out all minor minimal 5-connected graphs. The graph minor Theorem assuresus that there are only finitely many minor minimal 5-connected graphs. Determiningall minor minimal 5-connected graphs should be a hard task, Fijavž [16] posted thefollowing conjecture.

Conjecture 2 [16] Every 5-connected graph contains a minor which isomorphic toone of the graphs K6, K2,2,2,1, C5 ∗ K3, I, ˜I or G0.

Here K6 is the complete graph on six vertices, the graph K2,2,2,1 is obtained froma complete graph on seven vertices by deleting three independent edges, C5 ∗ K3 is

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obtained from a cycle C5 by adding three new vertices and making them adjacentto all vertices of C5. I stand for the icosahedron, ˜I is the graph obtained from I byreplacing the edges of a cycle abcdea induced by the neighborhood of some vertexwith the edges of a cycle abceda and G0 is the graph obtained from the icosahedronby deleting a vertex w, replacing the edge ab of the cycle abcdea induced by theneighborhood of w with the two edges ac and ad, and, finally, identifying b and e.

Fijavz showed that Conjecture 2 hold for minor minimal 5-connected projectivegraphs. It is true for all graphs on at most 10 vertices and all 5-regular graphs on atmost 12 vertices [16]. So we only need to deal with the 5-connected graph with atleast 10 vertices.

Let G be a graph with κ(G) = 5, T be a 5-separator of G. We say that T is quasi-trivial if T = N (x) for some x ∈ V (G). We call a graph G with κ(G) = 5 is ofsuper 5-connected graph if every 5-separator is quasi-trivial. Further, a graph G withκ(G) = 5 is called essentially 6-connected if for every 5-separator T, G − T hasexactly two components and one has cardinality one.

In [14], Kriesell characterized a special kind of minor minimal 5-connected graphsas following.

Theorem A [14] Let G be a minor minimal 5-connected graph. If G is essentially6-connected, then G has at most 12 vertices.

Let G be a k-connected graph with |V (G)| ≥ k + 2. An edge e of G is said to bek-removable if G − e is still k-connected. If G contains no k-removable edge, thenwe called G is a minimally k-connected graph. Let e = xy be an edge of minimallyk-connected graph G, one see that there is a (k − 1)-separator which separates x andy in G − xy. A k-connected graph with neither k-removable edge nor k-contractibleedge is called a minimally contraction-critically k-connected graph.

Further, Ando et al. [10] showed that, for minimally contraction-critical 5-connectedgraphs, there are two special structures around the edge whose both endvertices havedegree more than five. To state the result we need to introduce the following twoconfigurations.

x y

z

wv

v

u

u2

11

2

A configuration of the first kind

u yx

w

w

w

w

1

2

3

4

A configuration of the second kind

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Configuration of the first kind.

Let G be a 5-connected graph and let e = xy be an edge of G with {x, y}∩V5(G) =∅. An induced subgraph H on eight vertices is called a configuration of the first kindaround e if the following (1), (2), (3) and (4) hold.

1. V (H) = {x, y, z, w, u1, u2, v1, v2}2. Let F = {xy, xz, zw, zu1, zv1, wu1, wu2, wv1, wv2, u1u2, v1v2}. Then F ⊆

E(H) ⊆ F ∪ {xv1, xv2, yu1, yu2}3. {z, w, u1, v1} ⊆ V5(G)

4. There is a 4− separator S of G − xy such that {z, w} ⊆ S and S separates{x, v1, v2} and {y, u1, u2}.

Configuration of the second kind.

Let G be a 5-connected graph and let e = xy be an edge of G with {x, y}∩V5(G) =∅. An induced subgraph H on seven vertices is called a configuration of the secondkind around e if the following (1), (2), (3) and (4) hold.

1. V (H) = {x, y, u, w1, w2, w3, w4}2. Let F = {ywi , uwi |i = 1, 2, 3, 4}. Then F ∪ {xy, yu} ⊆ E(H) ⊆ F ∪ {xy, yu}

∪{xwi |i = 1, 2, 3, 4} ∪ {wiw j |1 ≤ i < j ≤ 4}3. {u, w1, w2, w3, w4} ⊆ V5(G) and d(y) = 6, d(x) ≥ 64. {w1, w2, w3, w4} is a 4− separator of G − xy which separates x and y.

An edge e of G is said to be an edge of the first kind (second kind) if there is aconfiguration of the first kind (second kind, respectively) around e.

Theorem B [10] Let G be a minimally contraction critically 5-connected graph.Suppose that G has an edge e = xy such that V (e)∩ V5(G) = ∅, then around e thereis a configuration of the first kind or the second kind.

Here, by showing that there is a contractible subgraph around the edge of the firstkind and the second kind, we have the following results.

Theorem 1 Let G be a 5-connected graph with |V (G)| ≥ 10. If there is an edge withboth end vertices have degree more than five, then either G has an 5-removable edgeor G has 5-contractible subgraph.

Clearly, a minor minimal k-connected graph must be a minimally contraction-critically k-connected. Thus, as a corollary, we can see that a minor minimal 5-connected graph has the following property.

Theorem 2 Let G be a minor minimal 5-connected graph with |V (G)| ≥ 10. Thenevery edge of G is incident to at least one vertex of degree 5.

2 Terminology

All graphs considered here are supposed to be finite, simple and undirected graphs.For terms not defined here we refer the reader to [17]. Let G = (V (G), E(G)) be a

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graph, where V (G) denotes the vertex set and E(G) the edge set. Let κ(G) denotes thevertex connectivity of G. An edge joining the vertices x and y will be written as xy. Forx ∈ F ⊆ V (G), we define NG(x) = {y|xy ∈ E(G)} and NG(F) = ∪y∈F NG(y)− F .Further, dG(x) = |NG(x)| denotes the degree of x and Vk(G)(V≥k(G)) denotes theset of vertices of degree k(at least k, respectively) in G. For A ⊆ V (G), G[A] denotesthe subgraph induced by A and G − A denotes the graph obtained from G by deletingthe vertices of A together with the edges incident with them. Let G be a k-connectednon-complete graph, T be a k-separator. The union of at least one but not of all thecomponents of G−T is called a T -fragment. A fragment of G is a T -fragment for someseparator T . Let F ⊆ V (G) be a T -fragment. Therefore, F = V (G) − (F ∪ T ) = ∅and F is also a T -fragment with NG(F) = T = NG(F). The set of all k-separators ofG will be denoted by TG . For NG(x), dG(x), NG(F) and TG , we often omit the indexG if it is clear from the context.

For an edge e of G, a fragment A of G is said to be a fragment with respect to e ifV (e) ⊆ N (A). For a set of edge F ⊆ E(G), we say A is a fragment with respect toF if A is a fragment with respect to some e ∈ F .

The following terminologies was first introduced by Ando and Iwase [9]. Let Abe a fragment of a 5-connected graph and let S = N (A). Let x be a vertex of S andy ∈ N (x) ∩ A. A vertex z is said to be an admissible vertex of (x, y; A) if both of thefollowing two conditions hold.

(1) z ∈ N (x) ∩ N (y) ∩ S ∩ V5(G); (2) |N (z) ∩ A| ≥ 2

Further, if |N (z) ∩ A| = 1, then z is said to be strongly admissible.A vertex z is said to be an (strongly) admissible vertex of (x; A), if z is an (strongly)

admissible vertex of (x, y; A) for some y ∈ N (x) ∩ A. Let Ad(x, y; A) (resp.Ad(x; A)) stand for the set of admissible vertices of (x, y; A) (resp. (x; A)).

The following properties of fragments are well known (for the proof see [18]), wewill use them without any further reference.

Let T, T ′ ∈ TG , and F, F ′ be the T, T ′-fragment of G, respectively. If F ∩ F ′ = ∅,then

|F ∩ T ′| ≥ |F ′ ∩ T |, |F ′ ∩ T | ≥ |F ∩ T ′| (1)

If F ∩ F ′ = ∅ = F ∩ F ′, then both F ∩ F ′ and F ∩ F ′ are fragments of G, andN (F ∩ F ′) = (F ′ ∩ T ) ∪ (T ′ ∩ T ) ∪ (F ∩ T ′). If F ∩ F ′ = ∅ and F ∩ F ′ is not afragment of G, then F ∩ F ′ = ∅ and

|F ∩ T ′| > |F ′ ∩ T |, |F ′ ∩ T | > |F ∩ T ′| (2)

3 Some Lemmas

In [9], Ando and Iwase showed the following two lemmas which are heavily used inthe proof of the main theorem.

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Lemma 1 [9] Let G be a contraction-critical 5-connected graph. Let A be a fragmentwith x ∈ N (A) such that |A| ≥ 3 and |A| ≥ 2. If |N (x) ∩ A| = 1, Suppose|N (x) ∩ A| = 1, say N (x) ∩ A = {y}, then the following (1) and (2) holds.

1. Ad(x, y; A) = ∅.2. If d(y) ≥ 6, then there is a strongly admissible vertex of (x, y; A).

Lemma 2 [9] Let G be a contraction-critical 5-connected graph. Let A be a fragmentsuch that |A| ≥ 2, |A| = 2 and A ∩ V≥6(G) = ∅. Then the following (1), (2) and (3)

holds.

1. |{x ∈ N (A)|Ad(x; A) = ∅}| ≥ 4.2. For any x ∈ N (A), if Ad(x; A) = ∅, then N (x) ∩ N (A) = ∅.3. |N (A) ∩ V5(G)| ≥ 4.

Lemma 3 Let G be a contraction critical 5-connected graph, A = {x, y} is a fragmentof G. Let B be a fragment with respect to xz, where z ∈ N (A). If N (A)−{z} ⊆ N (y),then A ⊆ N (B).

Proof Assume that A ⊆ N (B), we may let y ∈ A ∩ B. Then we can see thatB ∩ N (A) = ∅, since N (A) − {z} ⊆ N (y). Further, we observe that B ∩ A = ∅since |A| = 2. On the other hand, the fact that A ∩ N (B) = ∅ and B ∩ N (A) = ∅show that B∩ A = ∅. It follows that B = ∅, a contradiction. This contradiction assuresus that A ⊆ N (B). � Lemma 4 [10] Let A be a fragment of a k-connected graph G, S ⊆ N (A). If |N (S)∩A| < |S|, then A = N (S) ∩ A.

Lemma 5 [10] Let G be a minimally contraction critically 5-connected graph. Sup-pose that G has an edge e = xy such that V (e)∩V5(G) = ∅ and x or y is contained ina fragment of cardinality 2, then there is a configuration of the second kind around e.

4 Proof of Theorem 1

Let G be a 5-connected graph with |V (G)| ≥ 10. If G has a removable edge or acontractible edge, the we are done. So we may assume that G has neither a removableedge nor a contractible edge. Thus, G is minimally contraction-critically 5-connectedgraph. Let e = xy be an edge of G with both endvertices have degree more thanfive. Then, by Theorem B, there is a configuration of the first kind or the second kindaround e. So, in the following, we will complete the proof of Theorem 1 by showingthat there is a contractible subgraph around the configuration of the first kind and thesecond kind.

Lemma 6 Let G be a contraction-critical 5-connected graph with |G| ≥ 10, A ={u, y} be a fragment of G, N (A) = {x, w1, w2, w3, w4}. Suppose {x, y}∩ V5(G) = ∅and xu ∈ E(G). Then

1. G[{w1, w2, w3, w4}] ∼= 2K2.

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2. Suppose G[{wi , wi+1}] ∼= K2 for i = 1, 3. Then |N (x) ∩ {wi , wi+1}| ≤ 1 fori = 1, 3.

Proof Let W = {w1, w2, w3, w4} and S = N (A). Lemma 2 assures us that W ⊆V5(G).

(1) Assume G[W ] ∼= 2K2. Let ˜S = {t ∈ S|Ad(t; A) = ∅}.At first we show that x ∈ ˜S. For otherwise, assume Ad(x; A) = ∅. Then Lemma 2

assures us that ˜S = W and N (x)∩ S = ∅. Let B be a fragment with respect to xy andlet T = N (B). Then, since S − {x} ⊆ N (u), we observe that u ∈ T , which impliesthat A ∩ B = A ∩ B = ∅. Hence neither S ∩ B nor S ∩ B is empty. We show that|S∩ B| ≥ 2. Suppose |S∩ B| = 1, say S∩ B = {wi }. Since |S∩ B| < |T ∩ A|, we haveA∩ B = ∅, which implies B = {wi } and wi ∈ N (x)∩ S. This contradicts the previousassertion that N (x) ∩ S = ∅. Thus we see that |S ∩ B| ≥ 2. Further, by symmetry,we obtain |S ∩ B| ≥ 2. It follows that |S ∩ B| = |S ∩ B| = 2, say S ∩ B = {w1, w2}and S ∩ B = {w3, w4}. Then, since ˜S = W , we see that wi ∈ Ad(wi+1; A) andwi+1 ∈ Ad(wi ; A) for i = 1, 3. Hence we have G[W ] ∼= 2K2, which contradicts ourassumption. Now it is shown that x ∈ ˜S.

Next we show that wi ∈ ˜S for i = 1, 2, 3, 4. Without loss of generality, assumew1 ∈ ˜S, then Lemma 2 assures us that ˜S = {x, w2, w3, w4} and N (w1) ∩ S = ∅.Let B1 be a fragment with respect to w1u and let T1 = N (B1). Now, similar to thediscussion of last paragraph, we can observe that |S ∩ B1| = |S ∩ B1| = 2, sayS ∩ B1 = {x, w2} and S ∩ B1 = {w3, w4}. It follows that x ∈ Ad(w2; A) sinceN (w1) ∩ S = ∅. Hence we have d(x) = 5, a contradiction. Thus w1 ∈ ˜S and, bysymmetry, wi ∈ ˜S for i = 2, 3, 4. It follows that S = ˜S.

Since S = ˜S, without loss of generality we may assume that w1 ∈ Ad(x; A) andw2 ∈ Ad(w1; A). Now, since d(w1) = 5, we can see that N (w1) ∩ S = {x, w2} and|N (w1) ∩ A| = 1.

We show that N (w2)∩ {w3, w4} = ∅. Assume w3 ∈ N (w2). Then we observe thatN (w2) ∩ S = {w1, w3} and |N (w2) ∩ A| = 1. Now we see that w3 ∈ Ad(w4; A),since N (w2) ∩ S = {w1, w3}, N (w1) ∩ S = {x, w2} and Ad(w4; A) = ∅. Thisimplies that N (w3) ∩ S = {w2, w4} and |N (w3) ∩ A| = 1. Since |A| = 2, |A| ≥ 3and |N (w2) ∩ A| = 1, by applying Lemma 1 we can see that Ad(w2; A) = ∅. Thiscontradicts the fact that N (w2)∩S = {w1, w3}, |N (w1)∩A| = 1 and |N (w3)∩A| = 1.So w3 ∈ N (w2). Similarly, w4 ∈ N (w2). Thus N (w2) ∩ {w3, w4} = ∅.

Now we obtain N ({w1, w2}) ∩ {w3, w4} = ∅. Since S = ˜S, we observe thatw3 ∈ Ad(w4; A) and w4 ∈ Ad(w3; A). Hence G[W ] ∼= 2K2, a contradiction. Thiscontradiction completes the proof of (1).

(2) We show that |N (x) ∩ {wi , wi+1}| ≤ 1 for i = 1, 3. Assume that xw1 ∈ E(G)

and xw2 ∈ E(G). Now we can observe that N (w1) ∩ S = {w2, x}, N (w2) ∩ S ={w1, x}, |N (w1) ∩ A| = 1 and |N (w2) ∩ A| = 1. Now applying Lemma 1, we cansee Ad(w1; A) = ∅. This implies that d(x) = 5, a contradiction. Thus |N (x) ∩{w1, w2}| ≤ 1. By symmetry, we have |N (x) ∩ {w3, w4}| ≤ 1. Hence (2) is provedand the proof of Lemma 6 is completed. �

Let G be a contraction critical 5-connected graph, e be an edge of the second kind inG. The following Lemma shows that there are two edges around e whose contractionresults again a 5-connected graph.

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Lemma 7 Let G be a contraction-critical 5-connected graph with |G| ≥ 10, A ={u, y} be a fragment of G, N (A) = {x, w1, w2, w3, w4}. Further, let d(u) = d(wi ) =5(i = 1, 2, 3, 4), {x, y}∩V5(G) = ∅, xu ∈ E(G) and w1w2 ∈ E(G), w3w4 ∈ E(G).

Then there are two edges around e whose contraction results again a 5-connectedgraph.

Proof By Lemma 6, we know that |N (x)∩{w1, w2}| ≤ 1 and |N (x)∩{w3, w4}| ≤ 1.Without loss of generality, assume xw2 ∈ E(G) and xw4 ∈ E(G). We contract twoedges uw1, yw2 and we denote G1 the resulting graph. We denote the vertices of G1that get from the edges uw1 and yw2 by [uw1] and [yw2], respectively. We prove thatG1 is 5-connected. Assume κ(G1) ≤ 4, where κ(G1) stands for the connectivity ofG1. � Claim 7.1 κ(G1) = 4.

Proof Assume κ(G1) ≤ 3. Then G1 has a separating set T1, which contains both[uw1] and [yw2]. Let T be the separating set of G corresponding to T1. Hence T isa 5-separator such that {w1, w2, y, u} ⊆ T . Let B be a T − fragment of G. Since{y, u} ⊆ T , we observe that A ∩ B = A ∩ B = ∅, which implies that neitherS ∩ B nor S ∩ B is empty. Since {w1, w2} ⊆ T , without loss of generality, wemay assume that {w3, w4} ∩ (S ∩ B) = ∅, which implies that S ∩ B = {x} sincew3w4 ∈ E(G). Then, since |S ∩ B| < |T ∩ A|, we see that A ∩ B = ∅, which impliesthat B = {x} and d(x) = 5, a contradiction. This contradiction complete the proof ofClaim 7.1. �

By Claim 7.1, we know κ(G1) = 4. Let T1 be a separating set of G1 with cardinalityfour and let T be the separating set of G which corresponds to T1. We observe that{[uw1], [yw2]} ∩ T1 = ∅.

Claim 7.2 {[uw1], [yw2]} ⊆ T1.

Proof Assume {[uw1], [yw2]} ⊆ T1. Therefore, T is a 5-separator of G such thateither T ∩ {u, w1, y, w2} = {u, w1} or T ∩ {u, w1, y, w2} = {y, w2}. Let B be aT -fragment of G.

We show that T ∩ {u, w1, y, w2} = {y, w2}. For otherwise, assume T ∩{u, w1, y, w2} = {u, w1}, which implies that T ∩ {y, w2} = ∅. Without loss ofgenerality we may assume that y ∈ B, which implies A ∩ B = ∅ since A = {y, u}.Furthermore, since S ⊆ N (y), we observe that S ∩ B = ∅, which implies that B = ∅,a contradiction. This contradiction assures us that T ∩ {u, w1, y, w2} = {y, w2}.

Without lose of generality, assume that u ∈ B, which implies A ∩ B = ∅ and{w1, u} ⊆ B since T ∩ {u, w1, y, w2} = {y, w2}. Moreover, since S − {x} ⊆ N (u),we observe that S ∩ B = {x}, which implies that N (y)∩ B = {x}. Since d(x) ≥ 6 andw2x ∈ E(G), we observe that |N (x)∩(B∩ A)| ≥ 2, which implies that |B| ≥ 3. Since|B| ≥ 3, |B| ≥ 3 and N (y) ∩ B = {x}, Lemma 1 assures us that Ad(y, x; B) = ∅.Since xw2 ∈ E(G) and S ∩ T ⊆ {w2, w3, w4}, we see that Ad(y, x; B) ⊆ {w3, w4}.Since xw4 ∈ E(G), we have w3 ∈ Ad(y, x; B). Further, Lemma 1 assures us thatw3 is the strongly admissible vertex of (y, x; B), since d(x) ≥ 6. This implies thatN (w3) ∩ B = {u}. This assures us that w4 ∈ S ∩ T and S ∩ T = {w2, w3, w4}. Now

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we observe that |(S ∩ B)∪ (S ∩ T )∪ (A ∩ T )| = 5 and |A ∩ T | = 1, say A ∩ T = {v}.Now the fact that N (w3) ∩ B = {u} implies that N (w3) ∩ (A ∩ B) = ∅. Then, since|(S ∩ B)∪ (S ∩ T )∪ (A ∩ T )| = 5 and N (w3)∩ (A ∩ B) = ∅, the 5-connectedness ofG assures us that A∩ B = ∅. Further, we observe that B ∩ S = {w1} and B = {u, w1}.Recall that, by Lemma 6, we know that N (w1) ∩ {w3, w4} = ∅. Since A ∩ B = ∅and B = {u, w1}, N (w1) ⊆ {w2, v, u, y}, which contradicts the 5-connectedness ofG. This contradiction proves Claim 7.2. �

By Claim 7.2, we know that {[uw1], [yw2]} ⊆ T1, which implies that T is aseparating set of G with cardinality six and {u, w1, y, w2} ⊆ T . Let B be a nonemptyunion of components of G −T such that B = G −T − B = ∅. Since A = {u, y} ⊆ T ,we observe that A ∩ B = A ∩ B = ∅, which implies that neither S ∩ B nor S ∩ B isempty. Since {w1, w2} ⊆ V5(G), w1w2 ∈ E(G) and {y, u} ⊆ N (wi ) for i = 1, 2, weobserve that |N (wi ) ∩ (B ∪ B)| ≤ 2 for i = 1, 2.

Since neither S ∩ B nor S ∩ B is empty and {w1, w2} ⊆ T , we observe that either{w3, w4}∩(S ∩ B) = ∅ or {w3, w4}∩(S ∩ B) = ∅. Without loss of generality we mayassume that {w3, w4}∩(S∩ B) = ∅, which implies that {w3, w4}∩ B = ∅ and S∩ B ={x}, since w3w4 ∈ E(G). Since x ∈ V≥6(G), A ∩ B = ∅ and N (x) ∩ {w2, u} = ∅,we see that |N (x) ∩ (A ∩ B)| ≥ 2, which implies that |A ∩ B| ≥ 2. Since neitherA ∩ B nor A ∪ B is empty, we observe that A ∩ B is a fragment of G. Since A ∩ B isa fragment of G, we see that N (wi ) ∩ (A ∩ B) = ∅ for i = 1, 2.

Claim 7.3 Either N (w1) ∩ B = ∅ or N (w2) ∩ B = ∅.

Proof Suppose N (w1) ∩ B = ∅. We show N (w2) ∩ B = ∅. Since S ∩ B ⊆{w3, w4}, N (w1) ∩ {w3, w4} = ∅ and |N (w1) ∩ (B ∪ B)| ≤ 2, we observe that|N (w1) ∩ (A ∩ B)| = |N (w1) ∩ (A ∩ B)| = 1. Since N (w1) ∩ (A ∩ B) = ∅,we observe that N (w1) ∩ (S ∩ B) = ∅, which implies w1x ∈ E(G). Then, sinceN (x) ∩ {w1, w2, u} = ∅, and x ∈ V≥6(G), we see that |N (x) ∩ (A ∩ B)| ≥ 3, whichimplies that |A∩ B| ≥ 3. Since |A∩ B| ≥ 3, |A∪ B| ≥ 2 and |N (w1)∩ (A ∩ B)| = 1,applying Lemma 1 with the roles A and x replaced by A ∩ B and w1, respectively,we see that Ad(w1; A ∩ B) = ∅. Since N (w1) ∩ N (A ∩ B) = {w2}, w2 is an admis-sible vertex of Ad(w1; A ∩ B), which implies |N (w2) ∩ (A ∩ B)| ≥ 2 and henceN (w2) ∩ B = ∅. Now Claim 7.3 is proved. �

By Claim 7.3, we know that either N (w1)∩ B = ∅ or N (w2)∩ B = ∅. Let {1, 2} ={i, j} and suppose N (wi ) ∩ B = ∅. Let B ′ = B ∪ {wi } and T ′ = N (B ′) = T − {wi }.Then we observe that |T ′| = 5 and neither B ′ nor B ′ = B is empty. Hence T ′ is a5-separator of G and B ′ is a T ′-fragment of G. We observe that N (w j )∩ (A ∩ B) = ∅since x ∈ S ∩ B, G[{w1, w2, w3, w4}] ∼= 2K2 and N (w j ) ∩ B = ∅.

Since N (w j ) ∩ (A ∩ B ′) = N (w j ) ∩ (A ∩ B) is not empty, we see that |N (w j ) ∩(A ∩ B ′)| = 1, say N (w j ) ∩ (A ∩ B ′) = {v}. Since A ∩ B ′ = ∅, we know that|S ∩ B ′| ≥ |A ∩ T ′| = |{y, u}| = 2, which implies |B ′| = |A ∩ B ′| + |S ∩ B ′| ≥ 3.Since |B ′| ≥ 3, |B ′| ≥ 2 and |N (w j )∩ B ′| = 1, applying Lemma 1 with the roles A, xand y replaced by B ′, w j and v, respectively, we see that Ad(w j , v; B ′) = ∅. HoweverN (w j )∩ T ′ = {y, u} and N (v)∩{y, u} = ∅, since v ∈ A. Hence Ad(w j , v; B ′) = ∅,which contradicts the previous assertion. This is the final contradiction and the proofof Lemma 7 is completed. �

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Next lemma shows that there are two edges around the edge of the second kindwhich contraction results again a 5-connected graph.

Lemma 8 Let G be contraction critical 5-connected graph and e = xy be the edgeof the first kind but not the second kind.

Then there are two edges around e whose contraction result again a 5-connectedgraph.

Proof Let S be a 4-separator of G − xy which separating {x, v1, v2} and {y, u1, u2}.Let A be the component of (G − xy) − S which contains x, v1, v2 and let A =(G − xy) − S − A. Clearly, {y, u1, u2} ⊆ A. Let Ax = A − {x}, Sx = S ∪ {x}and Ax = A. Let Ay = A, Sy = T ∪ {y} and Ay = A − {y}. Then Sx and Sy are5-separators of G, Ax and Ay are a Sx -fragment and a Sy-fragment, respectively.

We contract two edges v1z and v2w and we denote G1 the resulting graph. Wedenote the vertices of G1 that get from contracting edges v1z and v2w by [v1z] and[v2w], respectively. We prove that G1 is 5-connected. Assume κ(G1) ≤ 4.

Claim 8.1 κ(G1) = 4.

Proof Assume κ(G1) ≤ 3. Let T1 be a separating set of G1 with cardinality threeand let T be the separating set of G corresponding to T1. Then {v1, v2, z, w} ⊆ T and|T | = 5. Let B be a T -fragment of G. Since u1u2 ∈ E(G), we observe that either{u1, u2} ∩ B = ∅ or {u1, u2} ∩ B = ∅. Without loss of generality we may assumethat {u1, u2} ∩ B = ∅. Then, since N (w) = {z, v1, v2, u1, u2}, {z, v1, v2} ⊆ T and{u1, u2} ∩ B = ∅, we see that N (w) ∩ B = ∅, which contradicts the 5-connectednessof G. This contradiction proves Claim 8.1. �

By Claim 8.1, we know κ(G1) = 4. Let T1 be a 4-separator of G1 and let T be theseparating set of G which corresponds to T1.

Claim 8.2 {[v1z], [v2w]} ⊆ T1.

Proof Assume {[v1z], [v2w]} ⊆ T1. Then T is a 5-separator of G such that eitherT ∩ {v1, v2, z, w} = {v1, z} or T ∩ {v1, v2, z, w} = {v2, w}. Let B be a T -fragmentof G.

We show T ∩{v1, v2, z, w} = {v2, w}. Suppose T ∩{v1, v2, z, w} = {v1, z}. Sincev2w ∈ E(G), without loss of generality we may assume {v2, w} ⊆ B. The fact{v2, w} ⊆ B implies {u1, u2} ∩ B = ∅, since {u1, u2} ⊆ N (w). Since N (z) ∩ B = ∅and {w, v1, u1}∩ B = ∅, we observe that {x, y}∩ B = ∅. Now the fact {x, y}∩ B = ∅implies that {x, y} ∩ B = ∅, since xy ∈ E(G). Clearly, v2 ∈ Ax ∩ B, which impliesthat Ax ∩ B = ∅. Hence we observe that N (z) ∩ (Ax ∩ B) = ∅, which implies that|(Ax ∩ T ) ∪ (Sx ∩ T ) ∪ (Sx ∩ B)| ≥ 6 and Ax ∩ B = ∅. Since {x, y} ∩ B = ∅ andAx ∩ B = ∅, we observe that {x, y} ∩ (Sx ∩ B) = ∅, which implies Sx ∩ B = ∅.Further, the fact that x ∈ Tx and y ∈ Ax shows that x ∈ Sx ∩ B.

We show |Sx ∩ B| = 1. Suppose |Sx ∩ B| ≥ 2. Then, since |(Ax ∩ T ) ∪ (Sx ∩T ) ∪ (Sx ∩ B)| ≥ 6, we see that |Ax ∩ T | ≥ |Sx ∩ B| + 1 ≥ 3, which implies|Ax ∩ T | ≤ 1. Then, since |Ax ∩ T | < |Sx ∩ B|, we observe that Ax ∩ B = ∅, whichimplies |Ax | = |Ax ∩ T | = 1. This contradicts the fact that {y, u1, u2} ⊆ Ax and it isshown that |Sx ∩ B| = 1. It follows that Sx ∩ B = {x} since x ∈ Sx ∩ B.

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We show that |Sx∩B| ≤ |Ax∩T |. Suppose |Sx∩B| > |Ax∩T |, then Ax∩B = ∅ andAx = Ax ∩T , which implies that |Ax | ≤ 1. This contradicts the fact that {y, u1, u2} ⊆Ax . Thus |Sx ∩ B| ≤ |Ax ∩ T |. Since N (z)∩ (Ax ∩ B) = ∅ and |Sx ∩ B| ≤ |Ax ∩ T |,we see that Ax ∩ B = ∅, which implies B = Sx ∩ B = {x}. This contradicts the factthat x ∈ V6(G). This contradiction shows that T ∩ {v1, v2, z, w} = {v2, w}.

Now we have T ∩{v1, v2, z, w} = {v2, w}. Since v1z ∈ E(G), we may assume that{v1, z} ⊆ B, which implies {x, y}∩B = ∅, since {x, y} ⊆ N (z). Since {x, y}∩B = ∅,we observe that Ax ∩ B = Ay ∩ B. Since v1 ∈ Ax ∩ B ⊆ Ay ∩ B, we see that neitherAx ∩B nor Ay∩B is empty. Since N (w)∩B = ∅, we observe that {u1, u2}∩(Ax ∩B) =∅, which implies Ax ∩ B = Ay ∩ B is not empty. Since none of Ax ∩ B, Ay ∩ B andAx ∩ B, Ay ∩ B is empty, we see that both Ax ∩ B and Ay ∩ B are fragments of G.We observe that y ∈ (Sy ∩ T ) ∪ (Sy ∩ B), since y ∈ Sy and {x, y} ∩ B = ∅. We showx ∈ Ay ∩ B. Suppose x ∈ Ay ∩ B. Then we observe that N ({z, w})∩(Ay ∩ B) = {v1}.Since Ay ∩ B is a fragment of G and N ({z, w})∩ (Ay ∩ B) = {v1}, applying Lemma 4with the roles of A and S replaced by Ay ∩ B and {z, w}, respectively, we see thatAy∩B = {v1} and y ∈ N (v1), which contradicts the fact thatv1 ∈ A and y ∈ A. Now itis shown that x ∈ Ay∩B, which implies that x ∈ Sx∩B. Since Ax∩B is a fragment of Gand N ({z, w})∩(Ax ∩B) = {v1}, again applying Lemma 4, we see that Ax ∩B = {v1}.It follows that Ay ∩ B = {v1, x} since Ay = Ax ∪ {x}. Since Ay ∩ B is fragment of Gand xw ∈ E(G), we have |N (x)| = |{v1}|+|(Sy ∩B)∪(Sy ∩T )∪(Ay ∩T )−{w}| ≤ 5,which contradicts the fact that x ∈ V6(G). This contradiction complete the proof ofClaim 8.2. �

By Claim 8.2 we know that {[v1z], [v2w]} ⊆ T1. Then T is a separating set ofG with cardinality six such that {v1, v2, z, w} ⊆ T . Let B be a nonempty union ofcomponents of G − T such that G − B − T = ∅ and let B = G − B − T . Since Tis a separating set of G with cardinality six, we observe that neither N ({z, w}) ∩ Bnor N ({z, w}) ∩ B is empty. Since xy ∈ E(G), we see that either {x, y} ∩ B = ∅or {x, y} ∩ B = ∅ is empty. Without loss of generality we may assume that {x, y} ∩B = ∅. Then, since N ({z, w}) = {x, y, v1, v2, u1, u2}, {x, y, v1, v2} ∩ B = ∅ andN ({z, w}) ∩ B = ∅, we see that {u1, u2} ∩ B = ∅, which implies that B ∩ Ay = ∅.Since N (w) = {z, v1, v2, u1, u2}, {z, v1, v2} ⊆ T and {u1, u2} ∩ B = ∅, we observethat N (w) ∩ B = ∅ . Hence T ′ = T − {w} is a 5-separator of G. Then B ′ = B andB ′ = B ∪ {w} are T ′-fragments of G. Note that Ay ∩ B ′ = Ay ∩ B and Ay ∩ B ′ = ∅.

We show x ∈ B ′. Assume x ∈ B ′. Since {v1, v2} ⊆ T ′, w ∈ B ′ and x ∈ B ′, weobserve that N (z) ∩ B ′ = {y}, which implies |B ′| ≥ 3 since N (y) ∩ {v1, v2} = ∅.Since |B ′| ≥ 3, |B ′| ≥ 3, N (z) ∩ B ′ = {y} and y ∈ V6(G), applying Lemma 1with the roles of A and x replaced by B ′ and z, respectively, we see that there isa strongly admissible vertex of (z, y; B ′). Since N (z) ∩ T ′ ∩ V5(G) ⊆ {v1, v2, u1}and {v1, v2} ∩ N (y) = ∅, u1 is a strongly admissible vertex of (z, y; B ′), whichimplies u1 ∈ T ′. Since u1 ∈ T ′ and {u1, u2} ∩ B = ∅, we observe that u2 ∈ B ′ andN (u1) ∩ B ′ = {w, u2}, which contradicts the previous assertion that u1 is a stronglyadmissible vertex of (z, y; B ′). This contradiction proves x ∈ B ′.

Now we know that neither Ay ∩ B ′ nor Ay ∩ B ′ is empty, which implies that Ay ∩ B ′is a fragment of G. Since v1 ∈ T ′ and N (y) ∩ Ay = {x}, we see that N (z) ∩ N (y) ∩

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(Ay∩B ′) = {x}. Since Ay∩B ′ is a fragment of G and (N (z)∪N (y))∩(Ay∩B ′) = {x},applying Lemma 4 with the roles S and A replaced by {z, y} and Ay ∩ B ′, respectively,we see that Ay ∩ B ′ = {x}, which contradicts the fact that d(x) ≥ 6. This is the finalcontradiction and the proof of Lemma 8 is completed. � Remark Base on Theorem A and Theorem 1, we can show that Conjecture 2 holdsfor minor minimal super 5-connected graphs. But it is too length to discuss this topichere for which need so much technical details. So strike to balance the length of thepaper and the amount of technical details, we lay it aside in an other papers.

Acknowledgments We would like to thank the anonymous referees for their kind help and valuablesuggestions.

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The Removable Edges and the Contractible Subgraphs of 5-Connected Graphs