Contractible Small Subgraphs in k-connected Graphs

Graphs and Combinatorics (2010) 26:499–511DOI 10.1007/s00373-010-0930-0

ORIGINAL PAPER

Contractible Small Subgraphs in k-connected Graphs

Shinya Fujita · Ken-ichi Kawarabayashi

Published online: 27 April 2010© Springer 2010

Abstract We prove that if G is highly connected, then either G contains a non-sep-arating connected subgraph of order three or else G contains a small obstruction forthe above conclusion. More precisely, we prove that if G is k-connected (with k ≥ 2),then G contains either a connected subgraph of order three whose contraction resultsin a k-connected graph (i.e., keeps the connectivity) or a subdivision of K −

4 whoseorder is at most 6.

Keywords Contractible small subgraphs · Highly connected graph

Mathematics Subject Classification (2000) 05C40 · 05C75

1 Introduction

It is well known that every 3-connected graph of order 5 or more has an edge whosecontraction still results in a 3-connected graph (see [5]). McCuaig and Ota [3] hasextended this result by showing that every 3-connected graph with at least nine verti-ces has a connected graph of order 3 whose contraction results in a 3-connected graph.This result was extended further by Kriesell [2].

Shinya Fujita’s work is supported by the JSPS Research Fellowships for Young Scientists. Ken-ichiKawarabayashi’s research partly supported by Japan Society for the Promotion of Science, Grant-in-Aidfor Scientific Research, by Sumitomo Foundation, by Inamori Foundation and by Kayamori Foundation.

S. Fujita (B)Department of Mathematics, Gunma National College of Technology,580 Toriba, Maebashi, Gunma 371-8530, Japane-mail: [email protected]

K. KawarabayashiNational Institute of Informatics, 2-1-2 Hitotsubashi, Chiyoda-ku, Tokyo 101-8430, Japan

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Our purpose of this paper is to extend these results to a higher connected graph.But, as Thomassen [4] pointed out, there exist infinitely many k-connected k-regulargraphs which do not have an edge whose contraction results in a k-connected graphfor k ≥ 4. Also, as is shown in below, there exist infinitely many k-connected graphswhich do not contain a connected subgraph of order 3 whose contraction results ink-connected for k ≥ 4.

Therefore, we have to set a modest goal in this paper: We would like to consider ak-connected graph which has either some fixed subgraph or a connected subgraph oforder 3 whose contraction results in a k-connected graph. Let us remind that Thom-assen [4] proved that every k-connected graph has either an edge whose contrac-tion results in a k-connected graph or a triangle. This result was further extended byKawarabayashi [1]. In order to achieve our goal, a triangle or a K −

4 (i.e., a graph whichis obtained from K4 by deleting one edge) does not seem enough to exclude. To seethis, we construct the following graph. Let k be an even integer with k ≥ 4 and let Xi

be a graph such that Xi ∼= K k/2 for 1 ≤ i ≤ l where l is a large integer. Considerthe graph G = X1 + X2 + · · · + Xl + X1. It is easy to see that G is k-connected andG does not contain a triangle nor a connected subgraph of order 3 whose contractionresults in a k-connected graph.

The above constructed graph suggests that excluding a triangle as a subgraph isnot enough, but it is easy to see that there are many “theta graphs”. In general, anysubdivision of K −

4 is often called a theta graph. (Note that K −4 itself is also regarded

as a theta graph.) For a theta graph with at most 6 vertices, we often call it a smalltheta graph. As indicated above, we suspect that the obstructions to have a connectedsubgraph of order 3 whose contraction results in a k-connected graph are small thetagraphs. [In fact, there are many theta graphs of order 5 and 6 (and hence they are smalltheta graphs.)] So, our purpose in this paper is to show that if we exclude a small thetagraph, then we can always find a connected subgraph of order 3.

More precisely, we prove the following theorem:

Theorem 1 Let k be an integer with k ≥ 2. If G is k-connected, then G containseither a theta graph of order at most 6 or a connected subgraph of order 3 whosecontraction results in a k-connected graph.

In the conclusion of this theorem, “a theta graph of order at most 6” is best possible.To see this, consider the k-dimensional cube. The graph is k-connected, but it does notcontain a theta graph of order at most 5 nor a connected subgraph of order 3 whosecontraction results in a k-connected.

We only prove Theorem 1 for k ≥ 4, since when k = 2, it is easy, and when k = 3,this follows from the result of McCuaig and Ota [3]. (It is easy to prove that if Gdoes not contain any member of small theta graphs, then it has at least nine vertices.)We shall give a proof of Theorem 1 here, but to do so, we need some definitions andnotations.

All graphs considered in this paper are finite, undirected, and without loops ormultiple edges. For a graph G, V (G), E(G) and δ(G) denote the set of vertices andthe set of edges and the minimum degree of G, respectively. For an edge e ∈ E(G),let G/e be the graph obtained from G by contracting e (and replacing each of theresulting pairs of double edges by a single edge). Let k ≥ 2 be an integer. An edge e

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of a k-connected graph is said to be k-contractible if G/e is still k-connected. A cutsetconsisting of k vertices is called a k-cutset.

2 Notation and Some Lemmas

For a given graph G and x ∈ V (G), we write NG(x) for the neighborhood of V (G)

and let dG(x) := |NG(x)|. For a subgraph H of G, let N (H) := ⋃v∈V (H) NG(v).

For a subset S of V (G), the subgraph induced by S is denoted by 〈S〉. When S ={x1, . . . ,xl}, 〈S〉 is often written as 〈x1, . . . ,xl〉. (Strictly speaking, it should be writ-ten as 〈{x1, . . . ,xl}〉. However, in this paper, we often prefer a simpler way of notationas long as there is no fear of confusion in the context.) Let E(A, B) be the set of edgesbetween A and B where A and B are vertex subsets with A ∩ B = ∅; when A consistsof one vertex, say, A = {v}, we often write E(v, B) in place of E({v}, B). In thispaper, when there is no fear of confusion in the context, a vertex subset X and thesubgraph 〈X〉 are often identified. So, a subgraph H of G is often called a cutset whenV (H) is a cutset in G, and when we write E(A, B), it is possible that A is a vertexsubset and B is a subgraph in G.

Along this line, we introduce some other abbreviations. For two vertex-disjoint sub-graphs H, H ′ of G, 〈V (H)∪ V (H ′)〉 is often written as H ∪ H ′ briefly; also, we oftenwrite N (H) ∩ H ′ in place of N (H) ∩ V (H ′). In addition, along this line, for a graphG and its subgraph H , G − H means 〈V (G) − V (H)〉. For an edge e = xy ∈ E(G),let V (e) = {x, y}.

Let D′ = {e ∈ E(G) | e is not contained in any triangle in G} and let E ′ = {e ∈E(G) | e is k-contractible and e ∈ D′}. For a path P = x1x2x3 (of order 3), if{x1x2,x2x3} ⊂ E ′, then P is called a good path. In Sect. 3, we proceed to prove ourmain result by assuming that every connected subgraph of order 3 is contained in acutset S with k ≤ |S| ≤ k + 1. So, in what follows, we assume it implicitly. Then, fora good path P , there exists a (k + 1)-cutset which contains P in G. (Note that thereis no k-cutset which contains P in G, because P contains a k-contractible edge.) Iffor every (k + 1)-cutset S which contains P and every connected component H ofG − S, |H | ≥ 3 holds, then P is called a special good path. By the definition of aspecial good path, since G does not contain a small theta graph, one can easily checkthe following fact.

Fact. If a good path P is not a special good path, then there exists a (k + 1)-cutset Swith P ⊂ S and there exists a component H of G − S with |H | = 1.

Also, for a triangle T in G, if T is not contained in any k-cutset, then T is called agood triangle. In this paper, a special good path and a good triangle play an importantrole. Their structures are sometimes discussed along a different line, but sometimesalong the same line. So, for the convenience, when we do not have to consider themseparately, we often call each of them a good element.

We define three sets A1, A2, A3 as follows:

A1 := {S ⊂ G | S is a k-cutset in G such that E(S) ∩ D′ = ∅};A2 := {S ⊂ G | S is a k-cutset such that S contains a triangle};A3 := {S ⊂ G | S is a (k + 1)-cutset in G such that S contains a good element}.

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As a preparation for the proof of our result, we prove the following basic lemmas.

Lemma 1 Let G be a k-connected graph with no small theta graphs. Let Q ∈ A1 ∪A2 ∪ A3, and let H be a connected component of G − Q. Then |H | ≥ k.

Proof By the above fact and by the definition of each Ai (1 ≤ i ≤ 3), it is easy tocheck that |H | ≥ 3 holds because G does not contain a small theta graph. First sup-pose that H contains a triangle T = xyzx. Since G does not contain K −

4 , we haveNG(x)∩ NG(y) = {z}, NG(x)∩ NG(z) = {y}, NG(y)∩ NG(z) = {x}. Hence we have|NG(x)∪ NG(y)∪ NG(z)| ≥ 3(k −2)+3. Since NG(x)∪ NG(y)∪ NG(z) ⊂ H ∪ Q, itfollows that |H | ≥ |NG(x)∪NG(y)∪NG(z)|−|Q| ≥ 3(k−2)+3−(k+1) = 2k−4 ≥k, as claimed. Thus we may assume that H does not contain a triangle. Let xyz be a pathin H . Since G does not contain K −

4 nor K2,3, note that NG(x)∩ NG(y)∩ NG(z) = ∅,|NG(x) ∩ NG(y)| ≤ 1, 1 ≤ |NG(x) ∩ NG(z)| ≤ 2, |NG(y) ∩ NG(z)| ≤ 1. Moreover,if |NG(x)∩ NG(z)| = 2, then NG(x)∩ NG(y) = NG(y)∩ NG(z) = ∅. Hence we have|NG(x) ∪ NG(y) ∪ NG(z)| = |NG(x)| + |NG(y)| + |NG(z)| − |NG(x) ∩ NG(y)| −|NG(x)∩NG(z)|−|NG(y)∩NG(z)| ≥ 3k−3. Since NG(x)∪NG(y)∪NG(z) ⊂ H∪Q,it follows that |H | ≥ |NG(x)∪ NG(y)∪ NG(z)|− |Q| ≥ 2k − 4 ≥ k, as claimed. ��

3 Proof of Theorem 1

Let G be a k-connected graph with no small theta graphs. By contradiction, we mayassume that every connected subgraph of order 3 is contained in a cutset S withk ≤ |S| ≤ k + 1.

Lemma 2 Let S be a (k + 1)-cutset which contains a good path P. If S /∈ A3, thenthere exists a vertex v such that dG(v) = k and 〈{v} ∪ P〉 ∼= C4.

Proof The assertion follows from the definition of A3 because there exists a connectedcomponent H of G − S with |H | = 1. ��Lemma 3 Suppose that there exists a subgraph X such that X ∼= K1,3 and E(X)⊂ E ′.Then, every edge e in E(X) is contained in a special good path.

Proof Take e ∈ E(X). Note that there are two distinct good paths which contain ein X . Hence, for each good path Pi with i = 1, 2 such that e ∈ E(Pi ), there existsa (k + 1)-cutset Si in G. In view of Lemma 2, if Si /∈ A3 for each i = 1, 2, we caneasily find a K2,3 or a theta graph of order 6. This is a contradiction. ��Lemma 4 A1 ∪ A2 ∪ A3 = ∅.

Proof If there is a triangle T in G, then there exists a cutset S with k ≤ |S| ≤ k + 1such that S contains T , which implies A2 ∪ A3 = ∅. Thus we may assume that Gdoes not contain any triangle. Let e be an edge in G. If e is not k-contractible, thenthere exists a k-cutset S which contains e, which implies S ∈ A1. So we may assumethat every edge in G is k-contractible. Then, there exists a subgraph X in G such thatX ∼= K1,3 and E(X) ⊂ E ′. Applying Lemma 3 to X , we can find a cutset S ∈ A3.Hence A1 ∪ A2 ∪ A3 = ∅ holds. ��

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Take Q ∈ A1 ∪ A2 ∪ A3. Let H be a connected component of G − Q. In the fol-lowing argument, we may assume that Q and H are chosen so that |H | is minimum.Also, let W = G − Q − H .

For a cutset Q′ ∈ A1∪ A2∪ A3 with Q′∩ H = ∅, if Q′ satisfies one of the followingproperties (i)–(iii), then we call Q′ a good cutset.

(i) Q′ ∈ A1 and (H∪Q)∩Q′ contains an edge e such that e ∈ D′ and H∩V (e) = ∅.(ii) Q′ ∈ A2 and (H ∪ Q) ∩ Q′ contains a triangle T such that H ∩ V (T ) = ∅.

(iii) Q′ ∈ A3 and (H ∪ Q) ∩ Q′ contains either a special good path P such thatE(H) ∩ E(P) = ∅ (i.e., |H ∩ V (P)| ≥ 2) or a good triangle T such thatE(H) ∩ E(T ) = ∅ (i.e., |H ∩ V (T )| ≥ 2).

In the rest of the proof, if there exists a good cutset Q′ ∈ A1 ∪ A2 ∪ A3, we use thefollowing notation:

Let H ′ and W ′ be unions of some connected components of G − Q′ such thatH ′ = ∅, W ′ = ∅, H ′ ∩ W ′ = ∅ and G − Q′ = 〈V (H ′) ∪ V (W ′)〉. In this stage,note that the roles of H ′ and W ′ are symmetric, but we will break them in the laterargument (just before Lemma 11).

Let H1, H2 and H3 denote H ∩ H ′, H ∩ Q′ and H ∩ W ′, respectively. Also, letW1, W2 and W3 denote W ∩ H ′, W ∩ Q′ and W ∩ W ′, respectively. Let Q1, Q2 andQ3 denote Q ∩ H ′, Q ∩ Q′ and Q ∩ W ′, respectively.

In view of Lemma 1, note that |H | ≥ k, |H ′| ≥ k, |W | ≥ k, |W ′| ≥ k. Also, notethat H2 = ∅.

Lemma 5 Suppose that there exists a good cutset Q′ ∈ A1 ∪ A2 ∪ A3. Then, thefollowing statements hold:

(i) If H1 = ∅ and H2 ∪ Q1 ∪ Q2 is a cutset with |H2 ∪ Q1 ∪ Q2| ≤ k + 1, then|H2 ∪ Q1 ∪ Q2| = k + 1 and Q′ ∈ A1 ∪ A2.

(ii) If H3 = ∅ and H2 ∪ Q2 ∪ Q3 is a cutset with |H2 ∪ Q2 ∪ Q3| ≤ k + 1, then|H2 ∪ Q2 ∪ Q3| = k + 1 and Q′ ∈ A1 ∪ A2.

Proof Suppose that H2 ∪ Q1 ∪ Q2 is a cutset with |H2 ∪ Q1 ∪ Q2| ≤ k + 1. Firstsuppose that |H2 ∪ Q1 ∪ Q2| = k + 1. Then we may assume that Q′ ∈ A3. However,this contradicts the minimality of |H | (because then we have H2 ∪ Q1 ∪ Q2 ∈ A3).Hence we may assume that |H2 ∪ Q1 ∪ Q2| = k. Note that then Q′ /∈ A3. If Q′ ∈ A2,then H2 ∪ Q1 ∪ Q2 ∈ A2 holds because (H ∪ Q) ∩ Q′ contains a triangle. However,this contradicts the minimality of |H |. Thus we may assume that Q′ ∈ A1. This meansthat H2 ∪ Q1 ∪ Q2 is a good cutset in A1. However, this contradicts the minimalityof |H |. Thus (i) holds. We can similarly prove (ii). ��Lemma 6 Suppose that there exists a good cutset Q′ ∈ A1 ∪ A2 ∪ A3. Then thefollowing two statements hold:

(i) If H1 = ∅, then W3 = ∅.(ii) If H3 = ∅, then W1 = ∅.

Proof To prove (i), suppose that H1 = ∅ and W3 = ∅. Then by the minimality of|H |, we see that |H1 ∪ H2| ≤ |Q3|. Since H1 = ∅, we have |H2| < |Q3|. Then,|H2 ∪ Q1 ∪ Q2| < |Q| = k + 1, which contradicts Lemma 5(i). We can similarlyprove (ii). ��

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Lemma 7 Suppose that there exists a good cutset Q′ ∈ A1 ∪ A2 ∪ A3. Then Q, Q′ ∈A3 holds. Further, if there exists a special good path P in Q, either P ∩ H ′ = ∅ orP ∩ W ′ = ∅ holds.

Proof First we claim that Q ∈ A3. By contradiction, assume for a while that Q ∈A1 ∪ A2. We claim that H1 = H3 = ∅. Suppose that H1 = ∅. Then by the minimalityof |H |, we see that |H2 ∪ Q1 ∪ Q2| ≥ k + 1 or |H2 ∪ Q1 ∪ Q2| ≥ k + 2 accord-ing as Q′ ∈ A1 ∪ A2 or Q′ ∈ A3. This implies |Q2 ∪ Q3 ∪ W2| ≤ k − 1. HenceW3 = ∅, which contradicts Lemma 6(i). Thus we have H1 = ∅. We can similarlyobtain H3 = ∅ from Lemma 6(ii). Thus H1 = H3 = ∅, as claimed. Now we haveH1 = H3 = ∅ and |H2| = |H | ≥ k. Note that |Q2 ∪ W2| ≤ 1. Suppose that W1 = ∅.Then we have |Q1| ≥ k − 1. This together with |Q2 ∪ W2| ≤ 1 implies W3 = ∅.Then |W ′| = |Q3| < k. This is a contradiction. Thus W1 = ∅. Arguing similarly, wehave W3 = ∅. Then |W | = |W2| < k, which contradicts the minimality of |H |. HenceQ ∈ A3 holds.

Now we shall prove that for any special good path P in Q, either H ′ ∩ P = ∅ orW ′ ∩ P = ∅. Let P = v1v2v3 be a special good path in Q. By the definition of a goodpath, it follows that {v1v2, v2v3} ⊂ E ′. Suppose that H ′ ∩ P = ∅ and W ′ ∩ P = ∅.Note that then |Q2 ∩ P| = 1. We may assume that Q1 ∪ Q2 contains e = v1v2 andQ2 ∪ Q3 contains e′ = v2v3. Now we show H1 = H3 = ∅. Suppose that H1 = ∅.Then, in view of Lemma 5(i), note that |Q1 ∪ Q2 ∪ H2| ≥ k + 1 and the equalityholds only if Q′ ∈ A1 ∪ A2. This implies that |Q2 ∪ Q3 ∪ W2| ≤ k. Since Q2 ∪ Q3contains e′, this forces W3 = ∅. However, this contradicts Lemma 6(i). Thus H1 = ∅,and we can similarly obtain H3 = ∅. Then we have |H2| = |H | ≥ k. Since now|Q2 ∩ P| = 1, this implies that either |Q1| ≤ k/2 or |Q3| ≤ k/2 holds. By symmetry,we may assume that |Q1| ≤ k/2. This forces W1 = ∅ because |Q1 ∪ Q2 ∪ W2| < k.Then by the minimality of |H |, we have |Q1| ≥ k. This is a contradiction. Thus wehave H ′ ∩ P = ∅ or W ′ ∩ P = ∅.

We shall prove Q′ ∈ A3. Suppose that Q′ ∈ A1 ∪ A2.By the symmetry of the roles of H ′ and W ′, we may assume that if there exists a

special good path P in Q, then W ′ ∩ P = ∅, and if there exists a good triangle T , thenW ′ ∩ T = ∅. Then note that Q1 ∪ Q2 contains P or T .

First suppose that H1 = ∅. Since Q1∪Q2 contains P or T , if |H2∪Q1∪Q2| = k+1,then H2 ∪ Q1 ∪ Q2 can be a cutset in A3, which contradicts the minimality of |H |.Thus we have |H2 ∪ Q1 ∪ Q2| ≥ k + 2. Hence |Q2 ∪ Q3 ∪ W2| < k. This impliesW3 = ∅, which contradicts Lemma 6(i). Thus we have H1 = ∅. Next suppose thatH3 = ∅. By Lemma 6(ii), we may assume that W1 = ∅. Since P or T is not containedin any k-cutset, |Q1 ∪ Q2 ∪ W2| ≥ k + 1. This forces |H2 ∪ Q2 ∪ Q3| = k, whichcontradicts Lemma 5(ii).

Thus we have H1 = H3 = ∅. Then |H2| ≥ k, and note that Q2 ∪ W2 = ∅. Sincenow it follows that Q1 contains P or T , |Q3| ≤ k − 3. This together with W2 = ∅implies W3 = ∅. However, then we have |W ′| = |Q3| ≤ k −3, a contradiction. HenceQ′ ∈ A3 holds. ��Lemma 8 There exists a good cutset Q′ ∈ A1 ∪ A2 ∪ A3.

Proof Take an edge e ∈ E(H). If e /∈ D′, then there exists a triangle T which con-tains e. This implies there exists a good cutset Q′ ∈ A2 ∪ A3. Thus we may assume

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that e ∈ D′. If e /∈ E ′, then there exists a good cutset Q′ ∈ A1 which contains e.Thus we may assume that e ∈ E ′. Since e is taken arbitrarily, in view of the previousargument, we may assume that E(H) ⊂ E ′. If H contains a subgraph X such thatX ∼= K1,3, then applying Lemma 3 to X , we can find a special good path P such thatE(H) ∩ E(P) = ∅. This implies that there is a good cutset in A3. Hence we mayassume that H is isomorphic to a path or a cycle. We claim that H contains a pathwith 5 vertices. To show this, we have only to consider the case where k = 4 and H isisomorphic to a path or a cycle with 4 vertices because |H | ≥ k. Then by the degreecondition, we have |E(H, Q)| ≥ 8. Since |Q| ≤ k + 1 = 5 and E(H) ⊂ E ′(⊂ D′),we can easily find a small theta graph in 〈H ∪ Q〉, a contradiction. Thus we mayassume that H contains a path P∗ = p1 p2 p3 p4 p5, as claimed. Let P ′

i be the subpathof P∗ such that P ′

i = pi pi+1 pi+2 for i = 1, 2, 3. We claim that P∗ contains a specialgood path, which implies that there is a good cutset in A3. Suppose that each Pi withi = 1, 2 is a good path but it is not a special good path. In view of Lemma 2, for each P ′

iwith i = 1, 2, there exists a vertex vi with vi = pi+1 such that vi pi , vi pi+2 ∈ E(G).Since E(H) ⊂ E ′, we see that v1 = v2. To avoid the existence of a small theta graph,this forces v1 = p4, v2 = p1. Thus we have p1 p4 ∈ E(G). If P ′

3 is not a special goodpath, then by Lemma 2, there is a vertex v3 such that v3 p3, v3 p5 ∈ E(G). However,then 〈V (P∗) ∪ {v3}〉 contains a small theta graph because p1 p4 ∈ E(G). This is acontradiction. Hence P ′

3 is a special good path. Thus P∗ contains a special good path,as claimed. This completes the proof of the lemma. ��

By Lemmas 7 and 8, note that we have Q ∈ A3 and every good cutset is in A3. Let Cbe a good element in Q, and fix it in the rest of the proof. For an edge e ∈ E(H) whichis contained in a good element, there exists a good cutset Q′ ∈ A3 which contains e.Then we say that e creates a good cutset. Also, a good element which contains e isdenoted by Ce. Put

E∗ := {e ∈ E(H) | e creates a good cutset}.

In view of Lemmas 7 and 8, it is easy to check that

E(H) − E∗ ⊂ E ′ and (E(H) ∪ E(H, Q)) ∩ D′ ⊂ E ′. (1)

Here we observe the neighbor of an edge e ∈ E(H) − E∗.

Lemma 9 Let e be an edge in E(H)− E∗. Then, for each x ∈ V (e) with dG(x)−1 =m, one of the followings holds:

(i) m is even and 〈NG(x) − V (e)〉 ∼= mK2 holds, and each triangle in 〈{x} ∪(NG(x) − V (e))〉 is a good triangle.

(ii) m is odd and 〈NG(x) − V (e)〉 ∼= K1 ∪ (m − 1)K2 holds, and each triangle in〈{x} ∪ (NG(x) − V (e))〉 is a good triangle, and moreover, there exists exactlyone good path which contains e in 〈{x} ∪ NG(x)〉.

Proof The assertion follows from (1), Lemma 3, and from the assumption that G doesnot contain a small theta graph. ��

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Lemma 10 One of the followings holds:

(i) There exists a connected subgraph M1 in H such that E(M1)⊂ E∗ and |M1|≥k.(ii) There exist two vertex-disjoint connected subgraphs M1, M2 in H such that

E(M1 ∪ M2) ⊂ E∗ and |M1| + |M2| ≥ k + 1.(iii) k = 4 and there exist three independent edges ei ∈ E∗ with i = 1, 2, 3.

Proof If E(H) = E∗, then by letting H = M1, we get a desired subgraph in (i) because|H | ≥ k. So we may assume that E(H)− E∗ = ∅. Take e = xy ∈ E(H)− E∗(⊂ E ′).For z ∈ {x, y}, put

Az = {v ∈ NG(z)|zv is contained in a good triangle}.

Since e ∈ D′, note that Ax ∩ Ay = ∅. By the definition, note that for any v ∈ Az ,zv is contained in a good triangle. Since G does not contain K −

4 , note that for any pairv, v′ ∈ Az and the corresponding good triangles T, T ′ caused by v, v′, respectively,if T = T ′ then T ∩ T ′ = {z} holds. By Lemma 9, we see that

for z ∈ {x, y}, |Az | ≥ k − 2.

Also, for any u ∈ Ax, w ∈ Ay , note that 〈u,x, y, w〉 ∼= P4 and NG(u) ∩ NG(w) = ∅because ux is contained in a triangle and G does not contain a small theta graph. Thuswe have y /∈ NG(u),x /∈ NG(w). For z ∈ {x, y}, here we observe the neighbour ofAz . Since G does not contain a small theta graph, we see that for any pair u, w ∈ Az ,NG(u) ∩ NG(w) = {z} and for each v ∈ Az , |E(v, Az − v)| = 1. Hence, in viewof Lemma 3, we see that for each v ∈ Az ∩ H , there are at least k − 4 edges inE(v, NG(v) − (Az ∪ {z})) such that each of them is contained in a good element. Inview of this, for each u ∈ Az ∩ H , put

Buz = {v ∈ NG(u) − (Az ∪ {z})|uv is contained in a good element} and bu

z = |Buz |.

Then it follows that buz ≥ k−4. Note that for any distinct u, w ∈ Az ∩H , Bu

z ∩Bwz = ∅

because uz is contained in a triangle and G does not contain a small theta graph. Also,for each z ∈ {x, y}, put hz = |Az ∩ H |. In the following argument, we divide the proofinto two cases:

Case 1 k ≥ 5Suppose that hx ≤ 1 and hy ≤ 1. Then, since |Az ∩ Q| ≥ k − 3 for each z ∈ {x, y}

and Ax ∩ Ay = ∅, it follows that |Ax ∩ Q| + |Ay ∩ Q| ≥ 2k − 6 ≥ k − 1 and theequality holds only if k = 5. Since |Q − C | = k − 2, we have (Ax ∪ Ay) ∩ C = ∅.Then, there is a triangle T = u1u2u3u1 such that u1 ∈ {x, y}, u2 ∈ C, u3 /∈ {x, y}.Note that if C is a good path, C is isomorphic to a path of order 3 in G and every edgein C is not contained in any triangle in G because E(C) ⊂ E ′(⊂ D′). Since G doesnot contain K −

4 , this together with the observation on C implies u3 /∈ C . This implies|E({x, y}, C)| ≤ 1 since otherwise 〈C ∪ T ∪ {x, y}〉 contains a small theta graph.

Hence it follows that |E({x, y}, Q)| = |E({x, y}, C)| + |E({x, y}, Q − C)| ≤1+(k−2) = k−1. By the above estimation on |Ax∩ Q|+|Ay ∩ Q|, the equality holds

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and we have k = 5. Also, note that now we have dG(x) = dG(y) = 5, hx = hy = 1and there exist two vertices u, v ∈ V (G) such that ux, vy ∈ E ′. However, this con-tradicts Lemma 9(i).

Hence, in the following argument, we may assume that hx ≥ 2. We consider twovertex-disjoint subgraphs M1 = 〈{x} ∪ (H ∩ (Ax ∪ (∪v∈Ax∩H Bv

x)))〉 and M2 =〈{y} ∪ (H ∩ Ay)〉. We claim that M1 and M2 are the desired subgraph in (ii) (whenM2 = {y}, we claim that M1 is the desired subgraph in (i)). To see this, it suffices toshow that |M1|+|M2| ≥ k+1 because each Mi contains a spanning tree by edges in E∗.By the above definitions, note that |M1| = hx+1+|(∪v∈Ax∩H Bv

x)∩H |, |M2| = 1+hy .Since Q ∩ Ax, Q ∩ Ay, (

⋃v∈Ax∩H Bv

x) ∩ Q are disjoint subsets of Q, the cardinalityof their union is at most k + 1. Also, since k ≥ 5, by Lemma 9, we have |Ax| ≥ 4 and|Ay| ≥ 4. Consequently, |M1| + |M2| ≥ 2 + 2 · 4 + hx(k − 4) − (k + 1) ≥ k + 1holds, as claimed.

Case 2 k = 4We may assume that there exist at most two independent edges of E∗ in H . (Oth-

erwise, (iii) holds.)First suppose that max{hx, hy} ≥ 2, say, hy ≥ 2 holds, and take u, w ∈ Ay ∩ H .

Put M = 〈y, u, w〉. We may assume that E∗ ⊂ E(M) since otherwise we can easilyget the desired subgraph satisfying (i) or (ii) (because |M | = 3). So we have hx = 0.Since we get |Ax| = |Ax ∩ Q| ≥ k − 2 = 2 (as is shown in Case 1) and each edge ofE(x, Ax) is contained in a (good) triangle, arguing similarly as in the proof where weintroduced T = u1u2u3u1 and considered the structure in 〈{x, y} ∪ T ∪ C〉 in Case 1,we see that |Ax ∩ C | ≤ 1 and hence |Ax ∩ (Q − C)| ≥ 1 and the equality holds onlyif E(y, C) = ∅ because G does not contain a small theta graph and when C is a goodpath, every edge in C is not contained in a triangle.

Suppose that uw /∈ E(G). Then, by the definition of Ax, there exists two distinctvertices u′, w′ ∈ Q such that 〈y, u, u′〉 ∼= K3, 〈y, w,w′〉 ∼= K3. Since G does notcontain a small theta graph, it follows that |C ∩ {u′, w′}| ≤ 1. This forces |(Q − C) ∩{u′, w′}| ≥ 1. Since |Q − C | = 2, this contradicts the above observation concern-ing |Ax ∩ (Q − C)|. Thus we have uw ∈ E(G). Since |E(x, Q)| ≥ 2 and G doesnot contain a small theta graph, we see that |E({y, u, w}, Q)| ≤ 3 and hence thereexist two distinct vertices xi ∈ H with i = 1, 2 such that E(xi , {y, u, w}) = ∅ andE(xi , {y, u, w}) ⊂ E ′ − E∗. By the symmetry of the roles of x and xi (for i = 1, 2),we see that |E(xi , Q)| ≥ 2 for i = 1, 2. Consequently, we can easily find a smalltheta graph in G. This is a contradiction.

Finally suppose that hx ≤ 1, hy ≤ 1. In the following argument, as we have definedAx or Ay for e = xy, for each e′ = pq ∈ E(H) − E∗, we also define Ap, h p orAq , hq similarly, and we may assume that h p ≤ 1, hq ≤ 1 holds for e′ = pq.

Suppose that E(x, V (G −x))∩ E ′ = {xy} or E(y, V (G − y))∩ E ′ = {xy} holds,say, E(x, V (G − x)) ∩ E ′ = {xy}. Then by Lemma 9, we have dG(x) ≥ 5. Thistogether with the assumptions hx ≤ 1, E(x, V (G − x)) ∩ E ′ = {xy} implies that|E(x, Q)| ≥ 3 and each edge in E(x, Q) is contained in a triangle. This implies that|E(x, C)| = 1 and E(y, C) = ∅ (because |Q − C | = 2 and if C is a good path, everyedge in C is not contained in a triangle and there is no small theta graph in G). Thus wehave |E(x, Q −C)| = 2. Since xy is not contained in a triangle, we get E(y, Q) = ∅.However, since hy ≤ 1 and e /∈ E∗, this forces dG(y) ≤ 3, a contradiction.

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Hence, we may assume that there exists two distinct good paths Px, Py with Px ∩Py = {x, y}. Put Px = x′xy and Py = xyy′. In view of Lemma 2, since G does notcontain a small theta graph, it follows that x′y′ ∈ E(G) (i.e., 〈x′,x, y, y′〉 ∼= C4) anddG(x′) = dG(y′) = 4. In the following argument, we may assume that e has beenchosen from E(H) − E∗ so that

|{x′, y′} ∩ H | is as large as possible. (2)

Under the choice of e, we divide the proof into the following three subcases:Subcase 2.1 x′, y′ ∈ H .In this case, we have x′y′ ∈ E ′ because (E(H) ∪ E(H, Q)) ∩ D′ ⊂ E ′ by (1).

Since xy /∈ E∗, for z ∈ {x, y}, if a good path contained in 〈{x′, y′} ∪ ({x, y} − z)〉is a special good path, dG(z) ≥ 5 must hold. By Lemma 9, this forces dG(z) ≥ 6.Since hz ≤ 1, it follows that |E(z, Q)| ≥ 3 and each edge of E(z, Q) is containedin a triangle. This implies that |E(z, C)| = 1 and |E(z, Q − C)| = 2. Then, usingthe fact that each edge of E(z, Q) is contained in a triangle and the assumption thatG does not contain a small theta graph, we can easily get E(C, {x, y} − z) = ∅.Hence we have E(Q, {x, y} − z) = ∅ (because |E(z, Q − C)| = 2 and xy is notcontained in a triangle). However, then it follows that for the vertex z′ in {x, y} − z,hz′ ≥ 2, a contradiction. Hence, for each good path contained in 〈x′,x, y, y′〉, it is nota special good path. Then, note that each edge in 〈x′,x, y, y′〉 is contained in E ′ − E∗.Since hz ≤ 1 holds for each z ∈ {x′,x, y, y′} and

∑z∈{x′,x,y,y′} hz ≤ 2 (otherwise

we have (iii)), we can easily find a small theta graph in 〈{x′,x, y, y′} ∪ Q〉 because|E({x′,x, y, y′}, Q)| ≥ 6. This is a contradiction.

Subcase 2.2 |{x′, y′} ∩ H | = 1.By symmetry, we may assume that x′ ∈ H and y′ ∈ Q. Suppose that hy = 0.

Then it follows that |E(y, Q)| ≥ 3. Hence, arguing similarly as in the proof of Sub-case 2.1 where we get E(Q, {x, y} − z) = ∅, we have E(x, Q) = ∅. However, thisforces hx ≥ 2, a contradiction. Hence we have hy = 1. Since |E(y, Q)| ≥ 2, it iseasy to check that |E(x, Q)| ≤ 1 (because any edge in E(x, Q) is contained in atriangle and |NG(x) ∪ NG(y) ∩ Q| ≤ 3). This forces hx = 1. If x′x ∈ E∗, thenM1 = 〈{x′,x} ∪ Ax〉 and M2 = 〈{y} ∪ Ay〉 are the desired subgraphs in (ii). Thuswe may assume that x′x /∈ E∗. Then, by the symmetry of the roles of x′x and xy,we can define Ax′ and hx′ , and we may assume that hx′ ≤ 1. If hx′ = 0, then wehave |E(x′, Q)| ≥ 3. Since |E({x′,x, y}, Q)| ≥ 6, we can easily find a small thetagraph in 〈{x′,x, y} ∪ Q〉, a contradiction. So we may assume that hx′ = 1. Since wehave assumed that there are at most two independent edges in E(H) ∩ E∗, this forcesNG(x′) ∩ NG(y) ∩ H = ∅. However, then 〈(NG(x′) ∩ NG(y)) ∪ {x′,x, y}〉 containsa theta graph of order 5, a contradiction.

Subcase 2.3 {x′, y′} ∩ H = ∅.In this case, since 2 ≤ |E({x, y}, Q)| ≤ 4 and hx ≤ 1, hy ≤ 1, it is easy to

check that hx = hy = 1 holds. Put Az ∩ H = {az} for z ∈ {x, y}. Note thatax = ay because e ∈ D′. Since it is easy to check that |E({ax,x, y}, Q)| ≤ 5,this implies that there exists a vertex v ∈ H − {x, y, ax, ay} with vax ∈ E(G). Ifvax ∈ E∗, then the subgraphs M1 = 〈v, ax,x〉 and M2 = 〈y, ay〉 satisfy (ii). So wemay assume that vax /∈ E∗. Then, by the choice of e (see the condition (2)) and by

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the symmetry of the roles of e and vax, we see that |E(ax, Q)| ≥ 2. However, thenwe have |E({ax,x, y}, Q)| ≥ 6, and hence we can easily find a small theta graph inG. This is a contradiction. ��

By Lemma 7, C is contained in either Q1 ∪ Q2 or Q2 ∪ Q3. Without loss ofgenerality, we may assume that C is contained in Q1 ∪ Q2. We prove the followinglemmas.

Lemma 11 H3 = ∅ and H1 = ∅.

Proof First assume H3 = ∅. By the minimality of H , we have |H2∪Q2∪Q3| ≥ k+2.Then |Q1 ∪ Q2 ∪ W2| ≤ k. Since now Q1 ∪ Q2 ∪ W2 contains a good element, thisimplies W1 = ∅. However, this contradicts Lemma 6(ii). Thus H3 = ∅ holds. Nextassume that H1 = ∅. Since H3 = ∅, |H2| ≥ k. Then either |Q1 ∪ Q2 ∪ W2| < kor |Q2 ∪ Q3 ∪ W2| < k holds, which means either W1 = ∅ or W3 = ∅ holds. IfW1 = ∅, then by the minimality of |H |, |Q1| ≥ k, which implies W3 = ∅. Similarly,if W3 = ∅, then we can obtain W1 = ∅. Consequently, W1 = W3 = ∅ holds. Then|W | = |W2| < k. This contradicts the minimality of |H |. ��Lemma 12 |Q1 ∪ Q2 ∪ H2| = k + 2 and |W2 ∪ Q2 ∪ Q3| = k. (Consequently|H2| = |Q3| + 1.)

Proof By Lemma 11, now we have H1 = ∅ and H3 = ∅. Then by Lemma 6(i),W3 = ∅. Hence by the minimality of |H |, |Q1 ∪ Q2 ∪ H2| ≥ k + 2. By the connectiv-ity of G, |W2 ∪ Q2 ∪ Q3| ≥ k. But 2k +2 = ∑3

i=1 |Qi |+|W2|+|Q2|+|H2| = |Q1 ∪Q2 ∪ H2| + |W2 ∪ Q2 ∪ Q3|, hence the equalities hold. Thus the lemma holds. (Also,since |Q| = |Q1 ∪ Q2| + |Q3| = k + 1, the assertion implies |H2| = |Q3| + 1.) ��

Since |H2| ≥ 2, we have |Q3| ≥ 1. Next, we prove the following lemmas.

Lemma 13 |N (U ) ∩ H | ≥ |U | + 1 for all nonempty subsets U of Q − C.

Proof Suppose there exists a nonempty subset U of Q − C with |N (U ) ∩ H | ≤ |U |.In this case, |U | ≤ |Q − C | ≤ k − 2 < k ≤ |H |. So, H − N (U ) = ∅. Then,(Q − U ) ∪ (N (U ) ∩ H) is a (k + 1)-cutset containing C and separating H − N (U )

from W ∪U . Then, since |H−N (U )|< |H |, this contradicts the minimality of |H |. ��By Lemma 13, since N (Q3) ∩ H ⊂ H2 and |H2| = |Q3| + 1, we have N (Q3) ∩

H = H2. By Lemmas 7–13, we can obtain the following fact:For each edge ei ∈ E(H) with 1 ≤ i ≤ |E(H)|, if ei creates a good cutset, then let

Qi be the good cutset in A3 such that Qi contains Cei . Let Hi , W i be some connectedcomponents in G − Qi such that Hi = ∅, W i = ∅ and G − Qi = 〈V (Hi )∪ V (W i )〉.Let Hi

1, Hi2 and Hi

3 denote H ∩ Hi , H ∩ Qi and H ∩ W i , respectively. Also, letW i

1, W i2 and W i

3 denote W ∩ Hi , W ∩ Qi and W ∩ W i , respectively. Let Qi1, Qi

2 andQi

3 denote Q ∩ Hi , Q ∩ Qi and Q ∩ W i , respectively. We may assume C ∈ Qi1 ∪ Qi

2.Then Hi

3 = ∅, |Hi2| = |Qi

3| + 1 and N (Qi3) ∩ H = Hi

2. Note that we do not defineQi nor other notations (Hi , W i , . . . ) for the edge ei which can not create any goodcutset.

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In view of Lemmas 7, 8 and 10, we may assume that there exists a subgraph Mwith E(M) = {e1, e2, . . . , em} and m ≤ |E(H)| such that E(M) ⊂ E∗ and M (andk) satisfies one of the following properties:

(P1) M consists of one connected component with |M | ≥ k.(P2) M consists of two connected components M ′, M ′′ with |M ′| ≥ 2, |M ′′| ≥ 2

and |M ′| + |M ′′| ≥ k + 1.(P3) k = 4 and M consists of three independent edges (i.e., |M | = 6).

If (P2) holds, we may assume that E(M ′) = {e1, . . . , et } and E(M ′′) ={et+1, . . . , em} (note that E(M ′) = ∅ and E(M ′′) = ∅). If (P1) holds, wemay assume that E(M) = {e1, e2, . . . , em} is chosen so that (N (Q j

3) ∩ H) ∩(⋃ j−1

i=1 N (Qi3) ∩ H) = ∅ holds for j = 2, . . . , m, and if (P2) holds, then we may

assume that E(M ′) = {e1, . . . , et } and E(M ′′) = {et+1, . . . , em} are chosen so that(N (Q j

3)∩ H)∩ (⋃ j−1

i=1 N (Qi3)∩ H) = ∅ holds for j = 2, . . . , t and (N (Q j

3)∩ H)∩(⋃ j−1

i=t+1 N (Qi3) ∩ H) = ∅ holds for j = t + 2, . . . , m. Moreover, if there exists an

integer j with t + 1 ≤ j ≤ m such that (N (Q j3) ∩ H) ∩ (

⋃ti=1 N (Qi

3) ∩ H) = ∅, wemay assume that j is chosen so that j = t + 1.

By the choice of et+1, notice that if (N (Qt+13 ) ∩ H) ∩ (

⋃ti=1 N (Qi

3) ∩ H) = ∅,then (

⋃ti=1 Qi

3) ∩ (⋃m

i=t+1 Qi3) = ∅. Also, if (P3) holds, since |Q − C | = 2, we

may assume that E(M) = {e1, e2, e3} and V (e1) ∪ V (e2) ⊂ N (Q13) ∩ H , and put

M ′ = 〈V (e1) ∪ V (e2)〉, M ′′ = 〈V (e3)〉.We prove the following lemma.

Lemma 14 Let j be an integer with 2 ≤ j ≤ m. If (N (Qh3) ∩ H) ∩ (

⋃h−1i=1 N (Qi

3) ∩H) = ∅ for each h with 2 ≤ h ≤ j , then |⋃ j

i=1 N (Qi3) ∩ H | ≤ |⋃ j

i=1 Qi3| + 1.

Proof We prove by induction on j . Suppose j = 2. Since N (Q13) ∩ H = H1

2 and

N (Q23) ∩ H = H2

2 , we may assume that H12 ∩ H2

2 = ∅. If Q13 ∩ Q2

3 = ∅, then

|(N (Q13) ∪ N (Q2

3)) ∩ H | = |H12 ∪ H2

2 | = |H12 | + |H2

2 | − |H12 ∩ H2

2 | ≤ |Q13| +

1 + |Q23| + 1 − 1 = |Q1

3 ∪ Q23| + 1. If Q1

3 ∩ Q23 = ∅, then by Lemma 13, we have

|H12 ∩H2

2 | ≥ |N (Q13∩Q2

3)∩H | ≥ |Q13∩Q2

3|+1, and hence |(N (Q13)∪N (Q2

3))∩H | ≤|H1

2 |+|H22 |−(|Q1

3∩Q23|+1) = |Q1

3|+1+|Q23|+1−(|Q1

3∩Q23|+1) = |Q1

3 ∪Q23|+1.

Thus the result follows.Assume j ≥ 3. If Q j

3 ⊂ ⋃ j−1i=1 Qi

3, the result follows by the induction hypothesis.

Assume Q j3 ⊂ ⋃ j−1

i=1 Qi3, and let R = Q j

3 ∩ ⋃ j−1i=1 Qi

3 and S = (N (Q j3) ∩ H) ∩

(⋃ j−1

i=1 N (Qi3) ∩ H). Then |S| ≥ |R| + 1 by the assumption of the lemma or by

Lemma 13 according as R = ∅ or R = ∅. Hence we have |⋃ ji=1 N (Qi

3) ∩ H | ≤| ⋃ j−1

i=1 Qi3| + 1 + |Q j

3| + 1 − |R| − 1 = |⋃ ji=1 Qi

3| + 1. ��First suppose that M satisfies (P1). Then, there exists an edge e j+1 joining from

⋃ ji=1 N (Qi

3) ∩ M to M − ⋃ ji=1 N (Qi

3) when M − ⋃ ji=1 N (Qi

3) = ∅. Also, e j+1is contained in some good element Ce j+1 and Ce j+1 is contained in some good cutsetQ j+1 ∈ A3. Hence we have M ⊂ ⋃m

i=1 N (Qi3) ∩ H and by Lemma 14, we have

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|M | ≤ |⋃mi=1 Qi

3|+ 1 ≤ |Q −C |+ 1 ≤ k − 1. Since |M | ≥ k, this is a contradiction.Thus we may assume that M satisfies (P2) or (P3), i.e., M consists of two vertex-disjoint subgraphs M ′ and M ′′ (these are connected subgraphs unless k = 4 and Mconsists of three independent edges). If (N (Qt+1

3 ) ∩ H) ∩ (⋃t

i=1 N (Qi3) ∩ H) = ∅,

by Lemma 14, we have |⋃mi=1 N (Qi

3) ∩ H | ≤ | ⋃mi=1 Qi

3| + 1, and hence |M | =|M ′| + |M ′′| ≤ | ⋃m

i=1 N (Qi3)∩ H | ≤ |⋃m

i=1 Qi3| + 1 ≤ |Q − C | + 1 ≤ k − 1. Since

|M | ≥ k by Lemma 10, this is a contradiction.Suppose that (N (Qt+1

3 ) ∩ H) ∩ (⋃t

i=1 N (Qi3) ∩ H) = ∅. Then, recall that

(⋃t

i=1 Qi3) ∩ (

⋃mi=t+1 Qi

3) = ∅. Also, in this case, by the symmetry of the rolesof M ′ and M ′′, we can conclude from Lemma 14 that | ⋃m

i=t+1 N (Qi3) ∩ H | ≤

| ⋃mi=t+1 Qi

3| + 1. Since M ′ ⊂ ⋃ti=1 N (Qi

3) ∩ H and M ′′ ⊂ ⋃mi=t+1 N (Qi

3) ∩ H ,consequently, it follows that |M | = |M ′| + |M ′′| ≤ | ⋃t

i=1 Qi3|+ 1 +| ⋃m

i=t+1 Qi3|+

1 ≤ |Q − C | + 2 = k. This contradicts (P2) or (P3). This completes the proof ofTheorem 1. ��Acknowledgment We would like to thank an anonymous referee for careful reading and helpful com-ments.

References

1. Kawarabayashi, K.: Contractible edges and triangles in k-connected graphs. J. Combin. Theory Ser.B 85, 207–221 (2002)

2. Kriesell, M.: Contractible subgraphs in 3-connected graphs. J. Combin. Theory Ser. B 80, 32–48 (2000)3. McCuaig, W., Ota, K.: Contractible triples in 3-connected graphs. J. Combin. Theory Ser. B 60, 308–

314 (1994)4. Thomassen, C.: Non-separating cycles in k-connected graphs. J. Graph Theory 5, 351–354 (1981)5. Tutte, W.: How to draw a graph? Proc. London Math. Soc. 13, 743–767 (1963)

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