The Structure of Graphs Not Topologically Containing the Wagner Graph

The structure of graphs not topologically containing the Wagner graph

John Maharry, Neil Robertson1 Department of Mathematics The Ohio State University Columbus, OH USA [email protected], [email protected]

Abstract A structural characterization of graphs not containing the Wagner graph, also known as V8, is shown. The result was announced in 1979 by the second author, but until now a proof has not been published.

Keywords: Wagner graph, V8 graph, Excluded-minor

1. Introduction Graphs in this paper are finite and have no loops or multiple edges. Given a graph G, one can obtain a contraction K of G by contracting pairwise disjoint connected induced subgraphs to single (distinct) vertices where distinct vertices of K are adjacent if and only if there exists an edge of G with an endvertex in each of the corresponding subgraphs of G. A graph M is a minor of a graph G when M is a contraction of a subgraph of G. We write H ≤m G when H is isomorphic to a minor M of G. Given an edge e ∈ E(G) with endvertices x and y, we define the single-edge contraction to be the graph formed by contracting the edge e to a new vertex, call it z. An equivalent definition of minor inclusion of a graph is H ≤m G if and only if H is isomorphic to a graph obtained by a series of single-edge contractions starting with a subgraph S of G.

1Research funded in part by King Abdulaziz University, 2011 and in part by SERC Visiting Fellowship Research Grant, University of London, 1985

Preprint submitted to Journal Combin. Ser. B December 3, 2015 A single-edge contraction of an edge with at least one endvertex divalent is called a series contraction. If one restricts the single-edge contractions to series contractions, then the resulting graph H is a topological minor of G. In this case, the subgraph S of G is called a subdivision of H in G. We write H ≤t G when H is isomorphic to a topological minor of G. In such a topological inclusion, the edges of H correspond to paths in S whose internal vertices are divalent in S. The corresponding paths in S are called branches. Similarly, the vertices of S that correspond to vertices of H are called nodes. Note that H is isomorphic to the graph of nodes and branches of S we will normally denote this graph by H and S by H0. The subpath of a branch M from vertex a to vertex b on M will be denoted M[a, b], or simply [a, b]. We will use standard notation such as (a, b) if one or both of the endpoints are not included. Deﬁne a separation of G to be an expression G = G1 ∪ G2, where G1, G2 are non-null edge-disjoint subgraphs of G and |E(G1)| ≥ |V (G1 ∩ G2)| ≤ |E(G2)|. A separation where |V (G1 ∩ G2)| = k is called a k-separation.A graph G is k-connected if G has no k0-separation for any k0 < k and |V (G)| > k. A 3-connected graph G is called internally-4-connected if |V (G)| ≥ 4 and there does not exist a 3-separation G = G1∪G2 where |E(G1)| ≥ 4 ≤ |E(G2)|. Note that, unless G is isomorphic to K4, if G contains a vertex that is both in a triangle and of valency three, then G is not internally-4-connected.

2. Main Theorem The Wagner graph is the graph W which is formed by adding to an octagon four edges joining its diagonally opposite pairs of vertices. This graph appears in a theorem of Wagner [20] which states that any graph with no minor isomorphic to K5 can be obtained by 0, 1, 2 and 3-summing beginning with planar graphs and the graph W . The graph W (see Figure 1) is also known as V8 or the M¨obiusladder with four rungs. Note that Wagner’s theorem implies that W is the unique maximal 3-connected nonplanar graph that does not contain K5 as a minor. In a sense, the graph W can be viewed as a graph intermediate between the Kuratowski graph K3,3 and the Petersen graph P (the unique 3-regular graph of girth 5 on 10 vertices). It is straightforward to check that as W is a cubic graph, containing W as a minor is equivalent to containing W under topological inclusion. In this paper, we present an exact characterization of those internally-4- connected graphs that do not contain as a subgraph a subdivision W . This

2 Figure 1: The Wagner Graph W , also known as V8 theorem was originally announced in 1979 by the second author. It has become widely known. However no proof has, until now, appeared in print. As the theorem has recently been used as a starting point for several other theorems, it is time the result is written formally. The proof in this paper is based on a partial hand-written draft by the second author from 1985 and has been streamlined and modiﬁed. The result is central to the proofs of at least four recent results: a characterization of 2-crossing critical graphs [1], a structural characterization of graphs with no Octahedron minor [4], a characterization of the ﬂexibility of graph embeddings on the Projective Plane [12] and a paper showing that all V8-free graphs have a closed 2-cell embedding [15]. Prior to the initial announcement of this result, there were few well- known excluded minor characterizations. The best-known were results for K5, K3,3 and the 3-Prism [20, 21, 3]. More recently, a number of other graphs have been so characterized, including the Cube [11], the Octahedron [4], the complement of P7 [6], the 5- and 6-wheels [7, 8] and several graphs on ten or fewer edges [5]. We now state the Main Theorem of this paper. Here L(K3,3) denotes the line graph of K3,3. Theorem 2.1. If G is an internally-4-connected graph then either G contains a subgraph S which is isomorphic to a subdivision of W or one of the following structural conditions holds: 1. G is planar, 2. G − {x, y} is a cycle, for some adjacent x, y ∈ V (G), 3. G − {w, x, y, z} is edgeless, for some w, x, y, z ∈ V (G), ∼ 4. G = L(K3,3), or

3 5. |V (G)| ≤ 7.

Suppose G is an internally-4-connected graph. If vertices x, y exist such that G − {x, y} is a cycle as in Case 2, then G is called a double wheel. We refer to the rim, hub and spokes of G naturally. We observe that double wheels are planar or not planar according as x, y are nonadjacent or adjacent. It is routine to check that in an internally-4-connected double wheel G, the hub vertices x and y are either both adjacent to all vertices of the rim or to successively alternating vertices of an even length rim. Such double wheels are called complete or alternating, respectively. When nonplanar, complete double wheels exist if |V (G)| ≥ 5 and alternating double wheels exist for even |V (G)| ≥ 8, as internal-4-connectivity fails at lower values. When |V (G)| ≥ 8, a double wheel contains a topological Cube C0 with its rim C0 − {x, y} subdivided and no other edges subdivided. Note that x and y are antipodal nodes of C0. If vertices w, x, y, z exist such that G − {w, x, y, z} is edgeless as in Case 3, then G is called vertex 4-covered and {w, x, y, z} is a vertex 4-cover of G. When |V (G)| ≥ 8, a vertex 4-covered graph contains a Cube C as a subgraph with {w, x, y, z} as a colour class. Conversely, if G contains a subdivided Cube C0, then C = C0 and a vertex 4-cover {w, x, y, z} must be a colour class of C.

Proposition 2.2. W is not contained topologically in G when the conditions of 2.1 hold.

Proof: To show this, suppose for a contradiction that S which is isomorphic to a subdivision of W , is a subgraph of G. For case (1), W , and hence S, is nonplanar. For case (2), if S is a subgraph of G, then S − {x, y} is a subgraph of G − {x, y}, which is a cycle. Hence, the valencies of the (at least) six remaining nodes of S were reduced when x and y were removed. This could only occur if x and y are nodes of S which are adjacent to the other 6 nodes of S. As W has diameter 2, this cannot happen for S. For case (3), note that the edge covering number of a subdivision is at least the edge covering number of the original graph. Since the edge covering number of W is ﬁve, S − {w, x, y, z} cannot be edgeless. For case (4), L(K3,3) does not have disjoint cycles of length at least 4, because the only quadrilateral in L(K3,3), up to isomorphism, leaves two triangles with one common vertex in its (vertex-set) complement. Finally, for case (5), the graphs are too small.

4 Note that the only internally-4-connected graphs on six or fewer vertices are K4,K5,K3,3 and the Octahedron with 0, 1, 2, or 3 diagonals added. Fur- ther, one can verify that there are exactly 28 internally-4-connected graphs on seven vertices. We note that the characterization for graphs with no W -subdivision can be reduced to the case where G is internally-4-connected.

Proposition 2.3. Any graph G with no W -subdivision can be obtained from internally-4-connected graphs with no W -subdivision by standard summing operations.

Proof: Assume G is not internally-4-connected and suppose that G = G1∪G2 is a separation of G supporting this assertion. When |V (G1∩G2)| ≤ 1 it is clear that W ≤t G if and only if W ≤t G1 or W ≤t G2. Assume G is + + 2-connected and |V (G1 ∩ G2)| = 2. Deﬁne G1 and G2 to be G1 and G2, respectively, with an edge added where necessary joining the two common + + vertices. Then G1 , G2 ≤t G and it is easy to see that W ≤t G if and only + + if W ≤t G1 or W ≤t G2 . Now assume G is 3-connected, |V (G1 ∩ G2)| = 3 + + and |E(G1)| ≥ 4 ≤ |E(G2)|. Deﬁne G1 and G2 to be G1 and G2, with new vertices x1 and x2 respectively, adjacent to each of the three vertices of + + G1 ∩ G2 by new edges. Again, the reader can verify that G1 , G2 ≤t G, and + + W ≤t G if and only if W ≤t G1 or W ≤t G2 . Note also, for inductive purposes, that G1,G2 are proper subgraphs of G + + and that G1 and G2 are both properly topologically contained in G. The proof of Theorem 2.1 has two main steps. To begin, in Section 3, we prove the following fact, where L(K3,3)) is the line graph of K3,3. ∼ Theorem 2.4. If G is internally-4-connected, |V (G)| ≥ 8 and G 6= L(K3,3) then W ≤t G or C ≤t G.

Given this theorem, one can assume that G is internally-4-connected and contains a topological Cube. Then in Section 4, we consider the types of ‘bridges’ that attach to the Cube. It will be shown that every bridge of the Cube is either ‘antipodal‘, ‘nodal’ or ‘facial’. In subsections 4.2, 4.3, and 4.4, these types of bridges will be analyzed. In each case, the graph will contain W topologically or will be shown to be planar, isomorphic to a nonplanar double wheel or to have a vertex 4-cover. It seems that the case when |V (G)| = 8 needs to be handled separately. This case is analyzed in Lemma 4.11.

5 Before we begin, we need the notion of ‘bridges’ of a subgraph Q of G. Given a subgraph Q of a graph G, we can define two types of sets of edges in E(G) − E(Q). The first type consists of a singleton edge with both ends in V (Q). The second type of set consists of the edges of a component of G − Q plus any edges joining that component to Q.A bridge of Q in G is a graph induced by the edges in one of these sets or an isolated vertex in G not in Q. A bridge arising from the first type is called an inner bridge, while one arising from the second type is an outer bridge. Bridges of subgraphs are basic entities in graph theory and their elemen- tary properties are well-known (see [19]). Clearly G is the edge-disjoint union of Q and the bridges of Q, the bridges are connected and meet pairwise in vertices of Q only. Given a bridge B of Q in G, the vertices of B ∩ Q are the feet of the bridge B. An often used property is that any two vertices x, y ∈ V (B) can be joined by a path in B which has no internal vertices in Q. Such paths are called Q-avoiding. According to Tutte, a vertex of attachment of a bridge B of Q in G is a vertex in V (B) that is incident with an edge in E(G) − E(B). In several places in the literature, the notions of feet of a bridge and vertices of attachment have become merged. We note that in the case that Q has no isolated vertices, then the notion of feet and vertices of attachment are indeed the same. In general, all vertices of attachment are feet, but not all feet are vertices of attachment. In this article, we will speak in terms of feet of a bridge. Further, we will say that a bridge lands on its feet or is attached to Q at its feet.

3. Proof of Theorem 2.4 In this section, we will prove Theorem 2.4 by way of the following lemma.

Lemma 3.1. If G is an internally-4–connected graph then either G contains two disjoint cycles, each of which contains at least four edges, or |V (G)| ≤ 7, ∼ or G = L(K3,3).

To see that this lemma implies Theorem 2.4, consider an internally-4- connected graph G that contains disjoint cycles Q1 and Q2, each of length at least 4. By internal-4-connectivity, there is a 4-join from Q1 to Q2 in G consisting of disjoint paths P1,P2,P3,P4. The graph Q1 ∪Q2 ∪P1 ∪P2 ∪P3 can be regarded as a subdivided 3-prism with ends Q1,Q2 and sides P1,P2,P3. Now P4 can without loss of generality be assumed to join a vertex strictly

6 between the ends of P1 and P3 on Q1 to either a vertex strictly between the ends of P1 and P3 on Q2 (forming a subdivided C) or a vertex strictly between the ends of P1 and P2 on Q2 (forming a subdivided W ) as required. To begin the proof of Lemma 3.1, we apply Dirac’s theorem [3].

Theorem 3.2 (Dirac). If G is a 3-connected graph then either G contains two disjoint cycles or one of the following structural conditions holds: 1. G − {x} is a cycle for some x ∈ V (G), 2. G − {x, y, z} is edgeless for some x, y, z ∈ V (G), or ∼ 3. G = K5.

When G is internally-4-connected these conditions simplify considerably. In ∼ ∼ ∼ fact (1) becomes G = K4, (2) becomes G = K3,3, and (3) remains G = K5. Thus with only three exceptional graphs disjoint cycles appear. The following lemma allows us to increase the size of one of the cycles.

Lemma 3.3. If G is internally-4-connected and |V (G)| ≥ 7, then G contains disjoint cycles Q1, Q2 where |E(Q2)| ≥ 4.

Proof: By Dirac’s theorem disjoint cycles Q1, Q2 exist. If the lemma fails then Q1, Q2 must be both triangles. But then a vertex x ∈ V (G)−V (Q1∪Q2) exists, and x is joined by three paths to Q1 ∪ Q2 which meet only at x, by Menger’s theorem and the 3-connectivity of G. But then Q2 (say) is joined by two of these paths to x and can be enlarged, disjoint from Q1, to a cycle 0 Q2 of length at least 4, proving the theorem. We will need the following fact in order to prove Lemma 3.5.

Proposition 3.4. Let G be an internally-4-connected graph with at least 8 vertices. Suppose G is the disjoint union of a triangle T and an induced cycle P plus some edges joining T and P . Then G contains a quadrilateral Q such that G − V (Q) contains a cycle and is connected.

Proof: There are two cases: Either P contains two adjacent cubic vertices or it has two non-cubic vertices with a common adjacent vertex on P . To see this, pick a vertex on P , cubic if possible, and consider its neighbors. In the ﬁrst case, suppose the cycle P has adjacent cubic vertices u, v. Then u and v have adjacent vertices v, v0, v00 and u, u0, u00 respectively with

7 u, u0, v, v0 on P and u00, v00 on T . As u, v are cubic and G is internally- 4-connected, it follows that u00 6= v00. Also, we have that u, v, v00, u00 is a quadrilateral Q and V (T ) = {x, u00, v00}. Then x is not adjacent to either u or v and hence, since x has valency at least four in G, it has valency at least two in M = G − V (Q). The graph M − x is a path in P from u0 to v0, so M is connected and also contains a cycle, as required. In the second case, we may assume that the non-cubic vertices u, v of P exist. Both u and v are adjacent to a vertex w on P and a vertex x on T (by the pigeonhole principle). Note that u and v are also adjacent to distinct u0, v0(6= w) on P , respectively. Then u, x, v, w is the quadrilateral Q and M contains the edge T − x with endvertices y, z. Note that u0 and v0 are both adjacent to vertices of T − x, otherwise they would violate internal-4- connection by forming a triangle-triad pair. This implies that M is connected and contains a cycle. Now we can further analyze the disjoint cycles from Lemma 3.3.

Lemma 3.5. Suppose G is internally-4-connected and |V (G)| ≥ 8. Then G contains disjoint subgraphs Q1, Q2 with |E(Q1)| ≥ 4 ≤ |E(Q2)| such that Q2 is a cycle, and Q1 is either a cycle or a triangle plus one pendant edge.

0 Proof: Applying Lemma 3.3, we obtain disjoint cycles Q1,Q2 where |E(Q2)| ≥ 4. We can assume that Q2 is chosen to be as small as possi- 0 0 ble. If |E(Q1)| ≥ 4 the lemma is satisﬁed, thus Q1 may be assumed to be a 0 0 triangle. Also if a vertex of Q1 is adjacent to a vertex not in Q1 ∪ Q2 then 0 a pendant edge can be added to Q1 to form Q1 as required for Lemma 3.5. 0 Thus Q1 is joined only to vertices of Q2 in G. We consider three cases. 0 Case 1: There is an outer bridge B of Q2 in G that does not contain Q1. 0 Then there is a vertex x ∈ V (B) − V (Q1 ∪ Q2) and x is joined to Q2 by three paths that meet only at x. By the internal-4-connectivity of G there is a vertex y of Q2, not an endvertex of one of the three paths, that is adjacent 0 to a vertex z of Q1 by an edge a ∈ E(G). Using the two paths from x with endvertices nearest to y and the segment of Q2 joining these endvertices but 0 not going through y we obtain a cycle Q2 of length ≥ 4 and can add the edge 0 0 a to Q1 to form the triangle plus pendant edge Q1 disjoint from Q2. Case 2: There is an inner bridge B of Q2 in G and no outer bridge except 0 0 the one containing Q1. Now V (G) = V (Q1 ∪ Q2) and B is a single edge. 0 00 Then Q2 ∪ B has two cycles Q2, Q2 distinct from Q2. As Q2 was chosen as

8 0 00 small as possible, neither Q2 nor Q2, can be of length four or more. This contradicts the fact that |V (G)| ≥ 8. Case 3: There is only one bridge B of Q2 in G. Then B contains Q1 and is 0 an outer bridge. Further, V (G) = V (Q1 ∪ Q2) and Q2 is an induced cycle in 0 G. As |V (G)| ≥ 8 and Q1 is a triangle, |V (Q2)| ≥ 5. However, Proposition 3.4 implies |V (Q2)| = 4, a contradiction which completes the proof of Case 3. It is now convenient to break the proof of Lemma 3.1 into three cases as described in the following lemma, using a conﬁguration derived from Lemma 3.6. After the proof of Lemnma 3.6, it will be shown that conﬁgurations H1 ∼ and H2 lead to disjoint cycles of length at least 4 unless G = L(K3,3) and that H3 always leads to the required disjoint cycles. In Figure 2, the bold edges in the drawings of H1,H2, and H3, are not subdivided edges, while the other edges may or may not be subdivided.

Lemma 3.6. If G is internally-4-connected and |V (G)| ≥ 8, then either G contains disjoint cycles of length ≥ 4, or one of the three subgraphs H1,H2,H3 in Figure 2 must be contained in G as a subdivision.

x' x'=a' x' a' a a x x x

y z y z y z a y' z' y' z' y' z' x'' x'' x'' H H1 H2 3

Figure 2: Three cases in the proof of Proposition 3.1

Proof: We may assume that disjoint subgraphs Q1, Q2 of G exist, with Q1 a triangle xyz plus pendant edge xa, and Q2 a cycle of length ≥ 4. By Menger’s theorem and the 4-connectivity of G we have four disjoint paths P1,P2,P3,P4 joining vertices of Q1 to vertices of Q2. Without loss of gener- 0 0 00 0 ality, the paths must be joined to the cycle Q2 at x , z , x , y in the manner

9 x'

P2 a x P1

y z y' P3 P4 z' x'' J'

Figure 3: A 4-join between Q1 and Q2 with no disjoint cycles of length at least 4. shown in Figure 3 to create the subgraph J 0 of G. Otherwise, the required disjoint cycles would exist. 0 Consider the subgraph J ⊂ J which is the union of the cycle Q2, the triangle xyz and the three paths P2,P3 and P4. Let B be the bridge of J in 00 G that contains the vertex a (and the path P1 from a to x ). By internal-4- connection, B must have at least two other feet on J, including x. If one of the feet is on the path P2 − {x}, then either subdivision H1 or subdivision H2 is obtained. If any feet of B are contained in the interior of the path in 0 0 0 0 0 Q2 from x to y avoiding z , or the interior of the path in Q2 from x to z avoiding y0 then we again have the required disjoint cycles. Finally, if there is a foot of B on the interior of the path P3yzP4, the disjoint cycles arise again with Q2 as one of the cycles. Hence, we can assume that any foot of B on J either lands on the path 0 0 0 P of Q2 from y to z avoiding x or it lands on x. The bridge B must have at least three feet on P , call them y00, x00, z00. Pick the foot y00 (resp. z00) as close as possible to y0 (resp. z0). Assume that y0 6= y00. Then there exist disjoint cycles; a cycle from y00 through the bridge B, to z00, x00 and back to y00, and another cycle x, x0, y0, y. So we can assume both y0 = y00 and z0 = z00. Note that the paths (x00, y0) and (x00, z0) must both be single edges. We claim that the interior of the bridge B is the single vertex a. To see this 0 consider a shortest J-avoiding path Py0 in B from y to the path P1. Similarly consider a path Pz0 . Let Py0 ∩ P1 = {vy} and let Pz0 ∩ P1 = {vz}. If there are no disjoint cycles (of length at least 4) in G, then vy = vz. Further, each of 0 0 00 the paths (y , vy), (z , vz), and (vy, x ) is a single edge.

10 If a = vy, we have the required H3 structure. So we assume that a 6= vy. In this case, consider the subgraph J¯ ⊂ G obtained from J by adding the 0 00 0 ¯ ¯ three edges vyy , vyx and vyz . Let B be the bridge of J that contains the vertex a. To avoid disjoint cycles, the feet of B¯ must be confined to the path 0 00 0 P and the vertex vy. Yet, any foot on y , x or z would result in such cycles. Hence B¯ cannot have any internal vertices. This implies that a is in fact 0 0 00 equal to vy. In this case it is clear that each path, (a, y ), (a, z ), (a, x ) must be a single edge. Hence we have the requirements for H3. Now we complete the proof of Lemma 3.1 by considering the following three cases. Case 1: Configuration H1 exists in G: Consider the triangle T on {x, y, z} in H1. By the internal-4-connectivity it follows there is another path from y to V (G) \{x, z, y0}. If this path lands on either the vertex a or along the 0 00 0 path of Q2 from x to x through z , the disjoint cycles are obtained. Thus 00 0 there must be a path from y to a vertex y on the path of Q2 from y to x0, where y00 could be identified with x0. Moreover, it can be assumed that y, y0, y00 is a triangle as otherwise, there would be the required disjoint cycles. Repeating this argument at a0, z0 and a, similar triangles a0x0w; zz0z00 and ax00w0 can be found. Without loss of generality y0, y00, x0, w, z0, z00, x00, w0 occur in circular order around Q2 (not necessarily all distinct). The above argument also leads to vertices x, y, z, a, a0 all having valency 4 in G and the 0 bridge of Q2 containing x, y, z, a, a is completely determined in G. At this point, if there is another bridge of Q2, we can easily determine the required disjoint cycles. Assuming that there is no other bridge of Q2, then it can be seen that y00 = x0, w = z0, z00 = x00 and w0 = y0 from which it follows that ∼ G = L(K3,3). See Figure 4.

x' a' w a y'' x

y z

y' z'' z' w' x''

00 0 0 00 00 0 0 Figure 4: L(K3,3) is obtained when y = x , w = z , z = x and w = y

11 Case 2: Configuration H2 exists in G: Repeating the connectivity argument of Case 1 along with the fact that xx0a and xyz are triangles, there must be three pairwise disjoint paths from a to Q2, from y to Q2 and from z to Q2 that are each internally disjoint from {a, x, y, z}. Let us call the ends of these 00 00 0 00 0 00 joins on Q2, a , x ; y , y ; z , z , respectively. One can find disjoint cycles of length at least four using these paths, unless the vertices x0, z00, z0, a00, x00, y0, y00 0 00 0 00 appear in that cyclic order on Q2. Further it must be that yy y and zz z form triangles. See Figure 5. Focusing on the cycle through x0, z00, z, y, y00 and the triangles xax0 and 00 00 00 0 00 00 0 00 ax a and the paths x, y; x , y , y ; and a , z , z we obtain a repeat of H2. 0 00 0 00 The preceding argument tells us that the segments y , x and z , a on Q2 not through x0 must be degenerate, i.e. x00 = y0 and a00 = z0. Also the argument that established the triangles yy0y00, zz0z00 and aa00x00 gives x, a, y, z all 4-valent in G so that the bridge of Q2 containing the triangle xyz is determined. Now the five specified triangles axx0; xyz; yy0y00; zz0z00; and aa00x00 form a subgraph K which is attached to edges of G not in K only at x0, y00, z00. Moreover, as |V (G)| ≥ 8 at least two of these vertices are distinct (say x0 6= y00). Then y00 is incident with two edges not in K, whence x0, y00, z00 are all distinct (G is ∼ 3-connected) and form a triangle, determining G = L(K3,3).

x'=a'

y'' a z'' x

y z y' z' a'' x''

0 00 00 0 00 00 Figure 5: L(K3,3) is obtained when y = x , a = z and the edge (z , y ) exists

Case 3: Conﬁguration H3 exists in G (but not H1 or H2). Consider the 00 subgraph J of G consisting of the the cycle Q2, the four edges incident with a and the path P2. Now let U be the union of the triangle T = xyz and 0 0 the paths P3 and P4. Clearly, U − {x, y , z } is contained in a bridge B3 00 0 0 of J . By connectivity, there must be a foot of B3 other than x, y and z . As a has valency exactly four and there is no foot on P2, there are (up to

12 0 00 symmetry) only three options for the remaining feet of the bridge B3; x , x 00 0 0 or a vertex z in the interior of the path from x to z on Q2. If there is a path 0 0 0 00 in B3 from U − {x, y , z } to either x or z , then there are disjoint cycles, 0 00 0 one of which is a, z , x , y . On the other hand, consider a path in B3 from U − {x, y0, z0} to x00. If the path leaves U − {x, y0, z0} from w on the path from z to z0, then there are disjoint cycles through w, z0, a, x00 and through 0 0 x, y, y , x . This completes the proof of Lemma 3.1.

4. Bridges on a Cube To proceed with the proof of Theorem 2.1 using Theorem 2.4, we may assume without loss of generality that G is internally-4-connected and C ≤t G. Using the condition W 6≤t G it is then incumbent to show conditions (1), (2) or (3) of the main theorem apply. In the rest of this paper, we will use C0 to denote a subdivided Cube contained in G as a subgraph in witness of the inclusion C ≤t G. This is generic notation as we will often re-embed C into G.

4.1. Types of Bridges A bridge of C0 that has all of its feet on a single branch of C0 is said to be conﬁned to a branch of C0 and is called a local bridge. The ﬁrst step in the analysis of the bridges of the Cube is to normalize the (subdivided) Cube C0 as embedded in G so that there are no local bridges of C0, a standard argument used often in such situations. Such embeddings are called proper embeddings of a graph H in G. Proofs of this fact can be found for example in [19] and [10].

Lemma 4.1. If G is a 3-connected graph and H0 ⊆ G is a subdivision of a trivalent 3-connected graph H, then it is possible to ﬁnd a subgraph H00 ⊆ G, that is also a subdivision of H such that H00 has the same nodes as H0 and no bridge of H00 in G has all its feet in one branch of H00.

A set of three branches of C0 centered at n is called the triad of branches centered at n. When it does not cause confusion, we sometimes refer to this as simply a triad. Suppose for the triad of branches centered at n, a bridge B of C0 in G has at least one attachment distinct from the center on each branch of a triad. We say that such a bridge B is a nodal bridge or a nodal bridge over n and that it covers the triad of branches at n. If a bridge that

13 covers a triad has no attachments outside the triad, we say that such a bridge is confined to the triad. In fact, if a bridge is not confined to a triad it could cover several triads. The term nodal refers to the fact that we will use such a bridge to switch nodes to obtain a new subdivided Cube. We define x to be near the node n if x = n or x is interior to a branch of C0 that is incident to n. We call B an antipodal bridge of C0 if for some antipodal pair of nodes of C0, n and n0, B has a foot near n and a foot near n0. Given a subdivided Cube C0 in a graph G, a long-diagonal of C0 is an inner bridge of C0 in a graph G with feet that are antipodal nodes of the Cube. Long-diagonals are antipodal bridges. We will prove the converse in Lemma 4.4 under the assumption that W is not a topological subgraph of G. Given a bridge B of C0 in G, where C0 is properly embedded in G, if C0 ∪B is a planar graph, then we will call B a planar bridge of C0. Otherwise B is a nonplanar bridge of C0. A bridge B that has all of its attachments on the branches of a quadrilateral Q of C will be called a facial bridge belonging to Q or simply a facial bridge. Note that B is not a local bridge and, as C0 has a unique planar embedding, a facial bridge belongs to a unique face of C0. If a bridge does not belong to any face, we say that it is a nonfacial bridge. We define a Y -graph, an X-graph or an H-graph to be a graph topologically equivalent to the corresponding capital letter. We can then define a Y -bridge, an X-bridge or an H-bridge as a bridge of C0 in G isomorphic to the corresponding graph with the feet of the bridge being the pendant vertices of the graph. It is important to note that in a 3-connected graph the edges in these bridges are in fact not subdivided. Note that, by Tutte [19], if B is a bridge of a subgraph Q of G with at least three feet x, y, z on Q, then there exists a Y -graph, Y in G with pendant vertices x, y, z and no other vertices on Q.

Theorem 4.2. If C0 is embedded in a graph G and B is a bridge of C0, then B is either antipodal, nodal or facial.

Proof: Let x be a foot of a bridge B of C0 and n a node of C0. A node n is clean with respect to B if no attachment of B is near n. Let {a, b, c, d} be a colour class of C. We adopt the notation that for a node n of C0, n0 is the antipodal node. We suppose that B is not antipodal and show that B is either nodal or facial.

14 Since B is not antipodal, if B is attached near a node n, then the antipodal node n0 must be clean with respect to B. Therefore, for each pair of antipodal nodes of C0 at least one of them is clean with respect to B. This leads to three cases up to symmetry: either (1) a, b, c0, d0 are clean, (2) a, b, c, d are clean or (3) a, b, c, d0 are clean and a0, b0, c0, d are not clean. In case (1), the bridge B is facial on a0, b, c0, d. In case (3) the bridge B is nodal over node d and conﬁned to the branches of d. In case (2), B is facial if it is attached at at most two of the nodes and nodal if it is attached at at least three of the nodes. 4.2. Antipodal Bridges In this and the following two subsections, we examine the three possible types of bridges on the Cube; antipodal, nodal and facial. To begin, we consider the possible positions of two feet of a bridge on C0.

Proposition 4.3. If C0 is properly embedded in a 3-connected graph G and B is a bridge of C0, then B has two feet, x, y, that are not conﬁned to just one branch of C. Moreover, x and y can be joined by a C0-avoiding path in B to give a subgraph C+ properly embedded in G of one of the eight types depicted in Figure 6.

x x x y x

y y y

(1) (2) (3) (4)

x x x x y

y (5) y (6) (8) y (7)

Figure 6: Eight diﬀerent ways of adjoining an arc to C0 inside G.

Proof: Choose the feet x and y not on the same branch as follows. If the bridge B has three feet which are nodes of C0 then two of these nodes x, y are

15 nonadjacent in C0 as the Cube has no triangles. Any bridge B with a foot x internal to a branch has another foot y external to the branch. Any bridge B with only two feet x, y which are both nodes of C0, has x, y nonadjacent in C0 as C0 is properly embedded. In the three cases, there exists a C0-avoiding path P in B with endvertices x and y. The proof is completed by a trivial running through of the various cases, up to the symmetries of the Cube, for a pair of vertices x, y which may 0 be on the nodes or branches of C , not both on the same branch. We remark from the diagrams (1),(2),...,(8) of Figure 6 that C0 ∪ xy is planar only in diagrams (1), (2), (3), and (4). Furthermore, we see that W is contained in C0 ∪ xy in each of the diagrams (6),(7) and (8). The arc L in diagram (5) is a long-diagonal of the Cube C0.

Lemma 4.4. Suppose G is internally-4-connected and C0 is a properly embedded subdivided Cube in G. Let n and n0 be antipodal nodes of C0. Let B be an antipodal bridge of C0 in G on n and n0. Then either B is a long-diagonal 0 with endvertices n and n or W ≤t G.

Proof: Let m and m0 be attachments of B near n and n0 respectively. The subdivided Cube union the path within B from m to m0 must contain one of the graphs of diagrams (5), (6), (7) or (8) from Proposition 4.3. Of these only the graph in (5) does not contain W , hence n and n0 are attachments of B. If B is not a long-diagonal, then there must be a third attachment of B on C0. Without loss of generality, this third attachment is either on the interior of a branch incident with n or on a branch not incident with either 0 n or n . In either case, the reader can verify that W ≤t G. From now on, we can assume that whenever W 6≤t G, any antipodal bridge of C0 is a long-diagonal.

4.3. Nodal Bridges A series of lemmas now will establish that if there exists a nodal bridge 0 B of C , then either W ≤t G or G−{w, x, y, z} is edgeless, where {w, x, y, z} is a colour class of C.

Lemma 4.5. Suppose G is internally-4-connected and C0 is a subdivided Cube in G. Let B be a nodal bridge conﬁned to the triad of branches centered at a node n of C0. Then there exists a subdivided Cube C00 in G, a node n0

16 of C00 and a bridge B0 of C00 which is an unconﬁned nodal bridge of C00 at n0. Moreover, C00 diﬀers from C0 only on the triad at n above the extreme attachments of B0 at n0.

Proof: Suppose that the bridge B is nodal and confined to the triad of branches centered at n. There will be a Y -graph Y in B meeting C0 exactly 0 0 0 in x1, x2, x3 and centered at a vertex x of B − V (C ). Now, let Y be the 0 0 Y -graph in C centered at n with endvertices x1, x2, x3. Switching the two Y -graphs yields a second subdivided Cube C00. Note that there are exactly three types of bridges of C00: proper subgraphs of B, the bridge B0 of C00 which contains Y 0, and bridges of C00 which were also bridges of C0. The only other changes in a bridge occur for the bridge B0 of C00 containing Y 0 00 which meets C in x1, x2, x3 and so is not confined to one branch. The unchanged bridges are still associated with the same branches they were in G with respect to C0, hence are of the required type. If the new bridge B0 is still confined to a triad of branches of C00 (necessarily including the Y -graph Y ) it will, by internal-4-connectivity, have a new attachment on a subpath (x1, n1), (x2, n2), or (x3, n3). The switching can be repeated using 0 0 0 vertices x1, x2, x3 at least one below the old attachments, hence shortening the complementary part of the Cube (which will be a subgraph of the original Cube C0). Otherwise there is an attachment w, of the new bridge B’ that is not on the triad of (the new) C0 at n. This will be a nodal bridge that covers the triad of branches but is not confined to it.

Lemma 4.6. Suppose G is internally-4-connected and C0 is a subdivided Cube in G. Let B be an unconﬁned nodal bridge of C0 centered at a node n. 0 Then either W ≤t G or B is an X-bridge on a colour class of C .

0 0 Proof: Let n1, n2 and n3 be the nodes of C adjacent to n and let n be 0 the antipodal node of C . Let x1, x2, x3 be the vertices of attachment of B on the branches (n, n1]; (n, n2]; (n, n3]; respectively, as close to n1, n2, n3 as 0 possible. There will be a Y -graph Y in B meeting C exactly in x1, x2, x3 and centered at a vertex x0 of B − V (C0). For notation, assume that the 0 0 0 0 0 nodes of C antipodal to n, n1, n2, n3 are n , n1, n2, n3 respectively. As B is not conﬁned to the triad of branches it has a foot w outside of that triad. 0 Note that B is not an antipodal bridge, hence w cannot be near n1. Similar 0 0 statements can be made about n2 and n3. Therefore, the attachment w must 0 be n . Again as B is not antipodal, the foot x1 cannot be in the interior of

17 the branch (n, n1). So we can conclude that x1 = n1. Similarly, x2 = n2 and 0 x3 = n3. If the bridge B which has exactly four attachments n1, n2, n3, n contains an H-bridge, then one can check that W ≤t G. Therefore, B is an X-bridge and the conclusion of the lemma holds. Now we consider the possible bridges of C0 given the existence of at least one X-bridge.

Lemma 4.7. Suppose G is internally-4-connected, and that C0 is a subdivided Cube in G. Suppose there exists an X-bridge X of C0 with its attachments on a colour class w, x, y, z of nodes in the Cube C0. Then either W ≤t G or G − {w, x, y, z} is edgeless.

Proof: First, assume that G has an X-bridge, X with feet {w, x, y, z} and suppose that C0 has a second bridge which has at least two feet in the other colour class {w0, x0, y0, z0}. Then without loss of generality, the graph in Figure 7, which can be seen to contain a W topologically, is a minor of G. So we assume that no such bridge of C0 exists.

x y' w' z

y x' z' w

Figure 7: An X-bridge on C0 with a link graph on w0y0

Let b be the center node of the X-bridge X. Consider the Cube C00 obtained from C0 by exchanging the triad of branches centered at w of C0 0 0 0 00 with b and its edges to {x , y , z }. Now the bridge Bw of C which contains w is a nodal bridge with respect to that Cube. By Lemmas 4.5 and 4.6, Bw 0 is an X-bridge or Y -bridge. Hence the branches of C incident with Bw are single edges. Repeating the swapping of nodes at x, y and z implies that all of the branches of C0 are in fact edges. Finally, suppose a bridge B of C0 is a facial bridge. As B cannot have two feet in the colour class {w0, x0, y0, z0} and each face of C0 is a quadrilateral, B

18 must be a single edge bridge between two vertices of {w, x, y, z}. Any such bridge and any antipodal bridge of C0 is allowed as G − {w, x, y, z} would remain edgeless even with these bridges included. 4.4. Facial Bridges At this point, we can assume that the only nonfacial bridges of C0 are long-diagonals. Now we need to consider the facial bridges. To this end a planarity lemma (Lemma 4.8) can be stated which is very useful in this type of argument, essentially characterizing when a cycle in a graph can be embedded in the boundary of a disk, with the graph embedding on the disk. This lemma seems to have originated in 1968 due to H. Jung [9]. It was rediscovered several times including by Robertson and Chakravarti [14, 2], by Seymour [16], by Siloach[17] and by Thomassen [18]. Mohar also proved an algorithmic version of the result in 1994 [13]. First, let us explain some technical terms. A diagonal of a cycle Q is an arc of G meeting Q exactly at its endvertices, and two diagonals are crossed when Q ∪ L1 ∪ L2 is a subdivision of K4.A tripod of G is a subdivision of K2,3, and a tripod T is linked onto Q when T meets Q at most in vertices x, y, z of the 3-element colour class of nodes in K2,3, and T is 3-linked to 0 0 0 Q by paths P1,P2,P3 (possibly degenerate) [x, x ]; [y, y ]; [z, z ], respectively, and having x0, y0, z0 on Q.

Lemma 4.8 ([2, 9, 14, 16, 17, 18]). Let G be a subdivision of a 3-connected graph and Q be a cycle of G. Then either G embeds in the plane with Q bounding a face of the embedding or Q has a pair of crossed diagonals L1,L2 or there is a tripod of G linked onto Q in G.

It is clear that in the latter two cases, the required plane embedding does not occur, and so the conditions are necessary and suﬃcient for nonexistence of such an embedding.

Proposition 4.9. Let G be an internally-4-connected graph, C0 be a properly embedded subdivided Cube in G and Q be a facial cycle of C0. Let H be the union of some (perhaps all) of the facial bridges of C0 belonging to Q in G. Then either Q ∪ H is planar with Q bounding a face in a plane embedding, H contains a pair of crossed diagonals of Q or there exists a subdivided Cube C00 with a nodal bridge.

19 Proof: By Lemma 4.8 either the above corollary holds or there is a tripod T in H which is 3-linked by paths P1,P2,P3 to Q. If such a tripod exists, either there is a 4-join from T to Q using the endvertices of P1,P2,P3 or there is a 3-separation of Q ∪ H separating the tripod from Q. In the case where there is a 4-join, without loss of generality we may assume P1,P2,P3,P4 are this 4-join and that the respective endvertices w0, x0, y0, z0 on Q are in the circular order of Q, where the vertices x, y and z may be permuted in T . It is now possible to route disjoint crossed paths L1,L2, with L1 containing P1,P3 and L2 containing P2,P4 onto Q. We use the internal-4-connectivity of G to ﬁnd P1,P2,P3,P4 in G linking T to Q. But these are all in the same bridge of Q if T is in one bridge, and otherwise T is the union of two Y -graphs 0 0 0 Y1,Y2 with endvertices x1, x2, x3 on Q in distinct bridges B1,B2, respectively, and P1,P2,P3,P4 are in B1 ∪ B2. In either case the crossed diagonals are in Q ∪ H, as required. Now consider the case where there is a 3-separation {a, b, c} in Q ∪ H separating T from Q. Since G is internally-4-connected, it must be that two of a, b, c are on the same branch of C0 or adjacent branches of C0. Assume that a and b are on Q. Either c is also on Q or there is a path from c to Q disjoint from T with endvertex c0. Let f, g be the nodes of the tripod T with paths to a, b, c. There are, up to symmetry, four cases. In the ﬁrst two cases, a and b are on the same branch of Q and c0 is either on an adjacent branch or not. In the last two cases, a and b are on adjacent branches and c0 is either also on one of those branches or not. See the four diagrams in Figure 8. In each of the four cases, there exists a subdivided Cube C00 with 00 00 f as a node of C and the bridge of C containing g will be nodal.

c' 2 2 1 2 2 1 1 1 c' c' c' a a c c c c g g f f f g f g b b 4 b 3 4 b 3 4 3 4 3 a a

Figure 8: Diﬀerent cases of a tripod attached on a face of a Cube.

Now we consider the facial bridges of C0. Let C˜ be the union of C0 and the set of all facial bridges of C0. If C˜ is planar, then as G is not planar

20 there must be a nonfacial bridge of C0. By the assumptions of this section, all such nonfacial bridges must be long-diagonals of C0. This case will be considered in Lemmas 4.11, 4.12 and 4.13. Here we consider the case where C˜ is nonplanar. Suppose that for each face Q of the embedding of C0, Q union the set H of bridges of C0 that belong to Q form a planar graph Q ∪ H with Q on the outer face. Then evidently these embeddings can be combined to a planar embedding of all of C˜. Assuming that C˜ is nonplanar and there is no subdivided Cube with a nodal bridge in G, Proposition 4.9 implies a pair 0 L1,L2 of crossed diagonals for some face Q of C must exist.

Lemma 4.10. Suppose G is an internally-4-connected graph and C0 is a properly embedded subdivided Cube in G . Let L1,L2 be a pair of crossed diagonals of a face Q of the Cube. Then either W ≤t G or there exists a subdivided Cube C00 in G with a nodal bridge.

1' 2' 3' 1' 2' 1' 2 2 2 1 1 1 2'

( I ) 4' 4' ( II ) ( III ) 4' 3' 3' 4 3 4 3 4 3

Figure 9: Diﬀerent cases for crossed diagonals on a face of a Cube.

Proof: Let 1, 2, 3, 4 be the nodes of the face Q of C0 and 10, 30 and 20, 40 be the ends of the crossed diagonals L1 and L2. We will show that, up to symmetry, the ends of the crossed diagonals land in the intervals shown in either Case (I),(II) or (III) depicted in Figure 9. As C0 is properly embedded in G with no local bridges, we can assume that it is not the case that all four ends of the diagonals lie between consecutive nodes of Q. If three of the four ends lie on a single branch, then we obtain a conﬁguration as in Case (I). Similarly, if no branch of C0 contains more than one end, then we obtain Case (III). Hence we can assume that 1 ≤ 10 < 20 ≤ 2 < 30 < 40 < 1 (where ‘less than’ follows the clockwise cyclic order along Q). Further, we have either both 30 < 4 and 3 < 40 or the crossed diagonals are in a conﬁguration as in

21 Case (II). In the former case, it follows that the intervals (1, 10), (2, 20), (3, 30) and (4, 40) are pairwise disjoint. This again leads to Case (III). 0 Let B1,B2 be the bridges of C containing L1,L2 respectively. If L1,L2 are crossed diagonals on Q then Case (III) easily implies a W minor contained in G. Suppose that L1,L2 are crossed as in Case (I). There are two possible 0 cases: either B = B1 = B2 or B1 6= B2. If B1 6= B2, then, as C is properly embedded, B1 must have a foot not contained in the branch (1, 2). This leads either to Case (II) or Case (III) unless the only other foot of B is at 40. If 0 0 that occurs, there is a vertex 5 in B1 with disjoint paths from 5 to 1 , 3 and 0 00 0 4 in B1. Another subdivided Cube C can be obtained from C by switching 2 to node 5. One can check that the bridge of C00 that contains 20 is nodal in C00 on node 5. In the case where B = B1 = B2, there exists a path from L1 to L2 contained in the bridge B. Suppose the ends of this path are x and y as in Figure 10. Then there exists another subdivided cube C00 with x as one of the nodes. This new Cube is obtained by rerouting the branch of C0 from 2 to 4 along the path (2, 30, x, y, 40, 3). Note that C00 can be modiﬁed, as before, to obtain a Cube with the same nodes as C0 which is properly embedded in G. The bridge of C00 which contains the vertex 20 will be a nodal bridge as it covers the branches of C00 centered at the node x.

1' 2' 3' 2 1

x y 4' ( I )

4 3

Figure 10: Crossed Diagonals contained in a single bridge

Finally, we consider Case (II). In this case, whether L1 and L2 are in the same bridge of C0 or not, we can obtain a Cube C00 by rerouting the branch of C0 from 1 to 2 along the path (1, 40, 20, 2). Note that 40 is a node of C00. Again, C00 can be modiﬁed so that it is properly embedded in G while maintaining the same nodes. The bridge of C00 containing 10 will be a nodal bridge covering the triad of branches centered at 40. Hence the conclusion of the lemma holds.

22 4.5. Planar Bridges and long-diagonals Now we can assume that we are in the case where C˜ is planar, all of the nonfacial bridges of C0 in G must be long-diagonals and at least one such long-diagonal must exist. First, we assume that C0 contained in G is isomorphic to a Cube (not subdivided). In Lemmas 4.12 and 4.13, we consider the case where the Cube is subdivided.

Lemma 4.11. Suppose G is internally-4-connected and has the Cube C as a subgraph with a long-diagonal L and no nodal bridge. Then all bridges are link-graphs, and G is either a double wheel, vertex 4-covered, or has W as a subgraph.

Proof: Suppose G is internally-4-connected, that C is a subgraph of G isomorphic to a Cube and that C has at least one long-diagonal bridge L with ends x and x0 as in Figure 11. If there are no other bridges of C0 then the conclusion holds. 0 If all of the bridges of C are long-diagonals then G is a subgraph of K4,4 and either colour class of the nodes of C0 will satisfy the conditions of the theorem. So we can assume that there is a planar bridge of C0. Note that as C is not a subdivided Cube, any planar bridge has two, three or four feet in the boundary of some face of C. If there are two feet, it is a link-graph. If there were three feet, the triad and triangle would contradict internal-4- connectivity. In the last case, if the bridge had attachments on all four nodes of the face, we can find a different Cube in C˜ and the long-diagonal L is not antipodal with respect to the new Cube. Hence, the graph would contain W . So we can assume that all the planar bridges are link-graphs. Let one colour class, say w, x, y, z, be called the ‘black’ vertices and let the other class w0, x0, y0, z0 be called the ‘white’ vertices. Either there exists a link-graph bridge Bw with both ends in the white colour class and a second bridge Bb with both ends in the black colour class, or one of the colour classes satisfies the condition of the theorem. A bridge Bw is either incident with L or its endpoints are in the neighborhood of an end of L. First consider the case where a link-graph exists in the neighborhood of L, say y0z0 (see the first graph of Figure 11). In this case, internal-4-connectivity of G is violated if w is a cubic vertex. Hence there is another edge incident with w. It is straightforward to check that if w is incident with x, y or z, then G contains W . Therefore, we can assume that the long-diagonal ww0 exists. Note that at this point, the white vertices still form a covering set.

23 So there must be at least one edge with both endpoints among the black vertices. It is again simple to verify that any of these six edges forces a W to be contained in G.

x' y x' y w z' w z'

y' x y' x

z w' z w'

Figure 11: Possible antipodal and facial bridges on a Cube

In the second case, we can assume that no link graph lies in the neighborhood of the endpoints of a long-diagonal. If there are at least three long-diagonals, there can then be no link graphs and G is vertex 4-covered. If there are two long-diagonals L = xx0 and M = yy0, the only possible link graphs are xy0 and x0y. Suppose xy0 is in G, then as w cannot be both in a triangle and cubic in an internally-4-connected graph, w must have another edge. See the second graph in Figure 11. However each possible edge would be a long diagonal ww0 or in the neighborhood of the endpoints of L or M. Finally, if there is only one long diagonal, then G \{x, x0} is a hexagon and G is a double wheel. This leaves the ﬁnal case where C˜ is planar, the only nonfacial bridges are long-diagonals, and the Cube C0 has at least one subdivided branch. It will turn out that, if G has no W -subdivision, then there will be only one long-diagonal and this case will lead to the double-wheel structure with the long-diagonal from one hub of the wheel to the other. With that in mind, a branch of C0 incident with a long-diagonal L will be called a spoke branch on L. The six branches of C0 that are not incident with L will be called the rim branches around L. The union of the rim branches around L will simply be called the rim around L. We will analyze this situation in two cases. The ﬁrst case is where some 0 (planar) bridge of C , say B1, has a foot internal to a spoke branch. The second case is where no bridge of C0 has a foot internal to any spoke branch. See Figure 12.

24 z'

w y

B1 x w' y' B2 z

Figure 12: Possible planar bridges on a Cube with a long-diagonal

Lemma 4.12. Suppose G is internally-4-connected, and that C0 is a properly embedded subdivided Cube in G. Let C0 have at least one long-diagonal L with 0 ˜ endvertices x, x . Further assume that C is planar and that some bridge B1 of C0 is attached at an internal vertex of some spoke branch of C0 on L. Then W ≤t G.

0 Proof: Suppose B1 is a bridge of C with an attachment, a, on the spoke branch between x and y0 that is embedded in the face x, y0, w, z0 as labelled in the graph in Figure 12. Note that if B1 also has an attachment along the path {w, x} through z0, then G contains a W -subdivision. Therefore all of 0 the attachments of B1 are conﬁned to the path {w, x} through y (possibly including the ends w and x). A similar statement can be made about any bridge with an attachment internal to a branch incident with y0 and belonging to either face x, y0, z, w0 or y0, z, x0, w. Therefore, unless there exists a long- diagonal from y0 to y, the vertices w, x and z form a vertex cut-set of G and if the part containing y0 contains any other vertex, this contradicts internal- 4-connection. If the long-diagonal (y, y0) does exist, then one can reroute 0 00 the branch of the Cube C through B1 to obtain C . In this Cube, the bridge containing (y, y0) is antipodal but not a long-diagonal. This implies the existence of a W -subdivision in G. Therefore no such bridge B1 exists.

Lemma 4.13. Suppose G is internally-4-connected, C0 is a properly embedded subdivided Cube in G and C˜ is planar. Let C0 have at least one long- diagonal L with endvertices x, x0. Further, suppose that every bridge B of C0

25 is attached only at x or y and along the rim branches of C0 around L. Then 0 either G − {x, x } is isomorphic to a cycle or W ≤t G.

Proof: We assume that there is a bridge B2 that has an attachment in 0 0 the interior of branch (z, y ). As C is properly embedded, B2 must also have attachments not on this branch. Hence there must be attachments of B2 on either x, x0 or (z, w0]. This bridge is either an inner bridge or an outer bridge. First, we assume it is an outer bridge. If an outer bridge has attachments on internal vertices on both branches (y0, z) and (z, w0), then we ﬁnd a second Cube C00 where x0 is not a node, L is a nonplanar bridge but no longer a long-diagonal. Hence G contains a W -subdivision. Therefore if the bridge B2 is an outer bridge, without loss of generality, it can be attached only on 0 x and along the branch [y , z]. (See bridge B3 in Figure 13). By internal-4- connectivity, there must then be another bridge B4 attached ‘outside’ of the rim in between the extreme attachments of B3 as well as at y. This can be seen to imply a W subdivision in Figure 13.

z' z'

w y w y

x x y' y' B3 w' w' z z

x' x'

Figure 13: A Cube with two ‘Rim Bridges’ contains a W -subdivision

Therefore the only possible bridges on C0 are long-diagonals and inner bridges of C0 with attachments on either x or x0 and on the interiors of the rim branches. This implies that either G − {x, x0} is a cycle or there is at least one more long-diagonal on C0. Let H be the (possibly subdivided) Hexagon in C0 − {x, x0} and let n be a node of C0 on H. We say that n is ﬁxed if there is no Cube C00 of G where n is not a node of C00. Note that if there is a long-diagonal L0 attached at n (and at its antipodal node n0 on the hexagon) then n must be ﬁxed. If it were possible to obtain a second

26 subdivided Cube C00 contained in G so that n is no longer a node of C00 then the bridge of C00 in G that contains the edge (n, n0) is an antipodal bridge, but not a long-diagonal of C00. It would then follow from Lemma 4.4 that W ≤t G. Suppose there is a vertex a in the interior of a branch of H incident with with a fixed node n. Choose a to be a neighbor of n in G. Assume n is also adjacent to the hub x, then a cannot also be adjacent to hub x (or the node n could be replaced with the node a to form a new subdivided Cube). The vertex a must then be incident with the other hub x0. However, this would violate internal-4-connectivity as a would be both cubic and contained in a triangle. Therefore, any branch incident with a fixed node must be a single edge. Moreover, the neighbors of a fixed node n on H must also be fixed by this argument. So if any node of H is fixed, then H must be isomorphic to 0 C6. Then C is not subdivided, contrary to assumption in this lemma. This implies that no node of H is fixed, hence there can be no long- diagonal of C0 other than L from x to x0. Therefore, G − {x, x0} is a cycle as required in the main theorem. This completes the analysis of the bridges of C0 and hence the proof of Theorem 2.1.

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