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

Graphs in this paper are finite and simple in the sense that they have no loops or multiple edges. Such graphs are determined by their vertex- adjacency relation. 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 [18] 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 is also known as V8 or the M¨obiusladder with four rungs. 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

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 October 4, 2013 Figure 1: The Wagner Graph W , also known as V8 to be the graph formed by contracting the edge e to a new vertex, call it z. Formally, the result of the single-edge contraction is the graph G/e with vertex set V (G/e) := (V (G) \{x, y}) ∪ {z} and edge set E(G/e) := {uv ∈ E(G)|{u, v} ∩ {x, y} = ∅} ∪ {zw|xw ∈ E(G) \{e} or yw ∈ E(G) \{e}}. 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. A single-edge contraction of an edge with at least one end-vertex 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, which we call branches. Similarly, the vertices of S that correspond to vertices of H are called nodes. Note that if the edge e is contained in a triangle T of S then the two edges of E(T ) \{e} become one edge of S/e. Otherwise, when no such (single-edge) contraction is made, the resulting graph S/e is topologically equivalent to S, i.e. the underlying topological representations of S and S/e are homeomorphic. Note that Wagner’s theorem implies that W is the unique maximal 3- connected graph which does not contain K5 as a minor. There are no maximal 3-connected planar graphs under the relation of minor inclusion. 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). In fact the chain K3,3 cubic graph, containing W as a minor is equivalent to containing W as

2 a topological subgraph. In this paper, we present an exact characterization of those graphs which do not contain a subgraph topologically equivalent to W . This theorem was originally announced in 1979 by the second author. It has become widely known. However no proof has, until now, ever 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 has been streamlined and modified and is based on a partial hand-written draft by the second author from 1985. The result is central to the proofs of at least four recent results: a characterization of the flexibility of graph embeddings on the Projective Plane [10], a characterization of 2-crossing critical graphs [1], a structural characterization of graphs with no Octahedron minor by Ding [4] and a paper showing that all V8-free graphs have a closed 2-cell embedding [13]. 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 [18, 19, 3]. More recently, a number of other graphs have been so characterized, including the Cube [9], the Octahedron [4], the 5- and 6-wheels [6, 7] and several graphs on ten or fewer edges [5].

1. Introduction Consider the problem of finding a subdivision of W contained in a graph G, i.e. concretely showing that W ≤t G. First, by a standard argument we can reduce to the case where G is ‘internally-4-connected’. 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)|. Suppose |V (G1 ∩ G2)| ≤ 3. When |V (G1 ∩ G2)| ≤ 1 it is clear that W ≤t G if and only if W ≤t G1 or W ≤t G2. If such separations do not occur and |V (G)| ≥ 3 then G is said to be 2-connected. Assume | ∩ | + + this and suppose 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 . Again, if such separations do not occur and |V (G)| ≥ 4 then G is 3-connected. Assume this and suppose |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

3 ≤ + ≤ + 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. When a graph G does not admit separations of the type described above and |V (G)| ≥ 4, it is said to be internally-4-connected. It is readily seen that W itself is internally-4-connected and that K4 is the smallest such graph. We can also see that all the above remarks go through with similar proofs when W is replaced by the graph C of the 3-dimensional cube. Moreover, when an internally-4-connected graph G contains disjoint circuits Q1 and Q2, each of length at least 4, then Menger’s theorem can be applied to produce a 4-join from Q1 to Q2, i.e. four vertex-disjoint simple paths from Q1 to Q2. These can be seen (Cor. 2.6) to produce either W or C embedded in G. It is thus of interest to characterize the internally-4-connected graphs which do not contain disjoint circuits of length ≥ 4. Note that if a vertex is both in a triangle and of valency three, this violates the conditions of internal-4-connection except in the case of K4.

Proposition 1.1. If G is an internally-4–connected graph then either G has disjoint cicuits Q , Q with |E(Q )| ≥ 4 ≤ |E(Q )|, or |V (G)| ≤ 7, or ∼ 1 2 1 2 G = L(K3,3).

Proof of this proposition is postponed until the main result of the paper is stated. However, by our earlier remarks, it shows that if G is not too small and is not isomorphic to one particular exceptional graph, L(K3,3), the line graph of K3,3, then it contains either W or C as an embedded subgraph. It may be noted that there are a fairly large number of ‘small’ internally-4-connected graphs G. The only internally-4-connected graphs on six or fewer vertices are K4,K5,K3,3 and the Octahedron with k = 0, 1, 2, 3 diagonals added. On seven vertices, one can verify that there are exactly 28 internally-4-connected graphs. When V = 8 we have either W or C as a spanning subgraph, by the above argument, and so there are just the many supergraphs of these which do not contain an incident triad-triangle pair in the class of graphs.

Theorem 1.2 (Main). 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 hold: 1. G is planar,

4 2. G − {x, y} is a circuit, 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 5. |V (G)| ≤ 7.

Note that clearly W is not contained topologically in G when these conditions hold. 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 circuit. 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 there were two vertices of S which cover all of the nodes of S. By inspection, this cannot happen for W . Hence, it cannot happen for any subdivision of W . 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 circuits 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. It will be shown after the proof of the main theorem that graphs that satisfy condition (2) are exactly the (non-planar) double-hubbed wheels. Moreover, as G is internally-4-connected, either the spokes ‘alternate’ as one traverses the rim or are complete to both hubs. The graphs that satisfy condition (3) are called vertex-4-coverable.

2. Disjoint Circuits There are three cases to be considered in the proof of the main theorem: first, the case where the disjoint circuits of length at least 4 do not exist; second, the case where C ≤t G and G is nonplanar; and third, the case where G is planar. Thus, in addition to Proposition 1.1, it is necessary to have a strong planarity lemma (Lemma 3.7) to work with in this context. In order to start the proof of Proposition 1.1, we apply Dirac’s theorem [3], stated below

Theorem 2.1 (Dirac). If G is a 3-connected graph then either G contains disjoint circuits Q1, Q2 or one of the following structural conditions holds:

5 1. G − {x} is a circuit 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) remains G = K4, (2) becomes G = K3,3, and (3) becomes G = K5. Thus with only three exceptional graphs disjoint circuits appear. It is possible to do a little better easily.

Lemma 2.2. If G is internally-4-connected and |V (G)| ≥ 7, then G contains disjoint circuits Q1, Q2 where |E(Q2)| ≥ 4.

Proof: By Dirac’s theorem disjoint circuits 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 circuit ′  Q2 of length at least 4, proving the theorem. Another intermediate stage can be established, with somewhat more ef- fort. We begin with a discussion 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 set of 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 [17]). 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 V (B) ∩ V (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 a 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 on 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

6 that in the case that Q has no isolated vertices, then the notion of feet and vertices of attachment are indeed the same. In this paper, the subgraphs that lead to the bridges discussed never have isolated vertices. We will say that a bridge lands on its feet or is attached to Q at its feet. Now we can further analyze the disjoint circuits from Lemma 2.2.

Lemma 2.3. 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 circuit, and Q1 is either a circuit or a triangle plus one pendant edge.

′ Proof: Applying Lemma 2.2, we obtain disjoint circuits Q1,Q2 where |E(Q2)| ≥ 4. We can assume that Q2 is chosen to be as small as possible. If | ′ | ≥ ′ E(Q1) 4 the lemma is satisfied, thus Q1 may be assumed to be a triangle. ′ ′ ∪ Also if a vertex of Q1 is adjacent to a vertex not in Q1 Q2 then a pendant ′ ′ edge can be added to Q1 to form Q1 as required for Lemma 2.3. Thus Q1 is joined only to vertices of Q2 in G. We consider three cases. ′ Case 1: There is an outer bridge B of Q2 in G which does not contain Q1. ∈ − ′ ∪ Then there is a vertex x V (B) V (Q1 Q2) and x is joined to Q2 by three paths which meet only at x. By the internal-4-connectivity of G there is a vertex y of Q2, not an end vertex of one of the three paths, which is adjacent ′ ∈ to a vertex z of Q1 by an edge a E(G). Using the two paths from x with end-vertices nearest to y and the segment of Q2 joining these endvertices but ′ ≥ not going through y we obtain a circuit Q2 of length 4 and can add the ′ ′ edge 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 ′ ′ ∪ the one containing Q1. Now V (G) = V (Q1 Q2) and B is a single edge. ∪ ′ ′′ Then Q2 B has two circuits Q2, Q2 distinct from Q2. As Q2 was chosen ′ ′′ as 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 ′ ∪ an outer bridge. Further, V (G) = V (Q1 Q2) and Q2 is an induced circuit | | ≥ ′ in G. As V (G) 8 and Q1 is a triangle, the following proposition will complete the proof of Case 3.

Proposition 2.4. 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 circuit P plus some edges joining T and P . Then G contains a quadrilateral Q such that G − V (Q) = M contains a circuit C and is connected.

7 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 circuit P has adjacent cubic vertices u, v. Then u and v have adjacent vertices v, v′, v′′ and u, u′u′′ respectively with u, u′, v, v′ on P and u′′, v′′ on T . As u, v are cubic and G is internally- 4-connected, it follows that u′′ ≠ v′′. Also, we have that u, v, v′′, u′′ is a quadrilateral Q and V (T ) = {x, u′′, v′′}. 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 u′ to v′, so M is connected and also contains a circuit, 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 u′, v′(≠ 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 u′ and v′ 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 circuit.  This completes the proof of Lemma 2.3.  It is now convenient to break the proof of Proposition 1.1 into three parts, using a configuration derived from Lemma 2.3.

Lemma 2.5. If G is internally-4-connected and |V (G)| ≥ 8, then either G contains disjoint circuits of length ≥ 4, or one of the following three subgraphs H1,H2,H3 must be contained in G as a subdivision (see Fig.2). Note that, in the drawings of H1,H2, and H3, the bold edges are not subdivided edges, while the other edges may or may not be subdivided.

Proof: We may assume that disjoint subgraphs Q1, Q2 of G exist, with Q1 a triangle xyz plus pendant edge xa, and Q2 a circuit 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- ′ ′ ′′ ′ ality, the paths must be joined to the circuit Q2 at x , z , x , y in the manner shown in Fig. 3 to create the subgraph J ′ of G. Otherwise, the required disjoint circuits would exist. ′ Consider the subgraph J ⊂ J which is the union of the circuit Q2, the triangle xyz and the three paths P2,P3 and P4. Let B be the bridge of J in

8 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 1.1

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 circuits of length at least 4.

′′ G which 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. 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 Q2 from x to y ′ ′ ′ ′ avoiding z , or the interior of the path in Q2 from x to z avoiding y then we again have the required disjoint circuits. Finally, if there is a foot of B on the interior of the path P3yzP4, the disjoint circuits arise again with Q2 as one of the circuits. Hence, we can assume that all feet of B on J land on the path P of Q2 from y′ to z′ avoiding x′ or they land on the vertex x. The bridge B must have at least three feet on P , call them y′′, x′′, z′′. Pick the foot y′′ (resp. z′′) as close as possible to y′ (resp. z′). Assume that y′ ≠ y′′. Then there exist disjoint circuits; a circuit from y′′ through the

9 bridge B, to z′′, x′′ and back to y′′, and another circuit x, x′, y′, y. So we can assume both y′ = y′′ and z′ = z′′. Note that the paths (x′′, y′) and (x′′, z′) must both be single edges. We claim that the interior of the bridge B is the single vertex a. To see this ′ consider a shortest J-avoiding path Py′ in B from y to the path P1. Similarly consider a path Pz′ . Let Py′ ∩ P1 = {vy} and let Pz′ ∩ P1 = {vz}. If there are no disjoint circuits (of length at least 4) in G, then vy = vz. Further, each of ′ ′ ′′ the paths (y , vy), (z , vz), and (vy, x ) is a single edge. If a = vy, we have the required H3 structure. So we assume that a ≠ vy. In this case, consider the subgraph J¯ ⊂ G obtained from J by adding the ′ ′′ ′ ¯ ¯ three edges vyy , vyx and vyz . Let B be the bridge of J that contains the vertex a. To avoid disjoint circuits, the feet of B¯ must be confined to the ′ ′′ ′ path P and the vertex vy. Yet, any foot on y , x or z would result in such circuits. Hence B¯ cannot have any internal vertices. This implies that a is in ′ ′ ′′ fact 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.  To complete the proof of Proposition 1.1, it will be shown that configu- ∼ rations H1 and H2 lead to disjoint circuits of length ≥ 4 unless G = L(K3,3) and that H3 always leads to the required disjoint circuits. 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, y′}. If this path lands on either the vertex a or along the path ′ ′′ ′ of Q2 from x to x through z , the disjoint circuits are obtained. Thus there ′′ ′ ′ must be a path from y to a vertex y on the path of Q2 from y to x , where y′′ could be identified with x′. Moreover, it can be assumed that y, y′, y′′ is a triangle as otherwise, there would be the required disjoint circuits. Repeating this argument at a′, z′ and a, similar triangles a′x′w; zz′z′′ and ax′′w′ can be found. Without loss of generality y′, y′′, x′, w, z′, z′′, x′′, w′ occur in circular order around Q2 (not necessarily all distinct). The above argument also leads to vertices x, y, z, a, a′ all having valency 4 in G and the ′ 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 circuits. Assuming that there is no other bridge of Q2, then it can be seen that y′′ = x′, w = z′, z′′ = x′′ and w′ = y′ from which it follows that ∼ G = L(K3,3). See Fig. 4. Case 2: Configuration H2 exists in G: Repeating the connectivity argu- ment of Case 1 along with the fact that xx′a and xyz are triangles, there must be three 2-joins from a to Q2, from y to Q2 and from z to Q2 that are

10 x' a' w a y'' x

y z

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

′′ ′ ′ ′′ ′′ ′ ′ Figure 4: L(K3,3) is obtained when y = x , w = z , z = x and w = y each internally disjoint from {a, x, y, z}. Let us call the ends of these 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 x′, z′′, z′, a′′, x′′, y′, y′′ 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 circuit through x′, z′′, z, y, y′′ and the triangles xax′ and ′′ ′′ ′′ ′ ′′ ′′ ′ ′′ ax a and the paths x, y; x , y , y ; and a , z , z we obtain a repeat of H2. ′ ′′ ′ ′′ The preceding argument tells us that the segments y , x and z , a on Q2 not through x′ must be degenerate, i.e. x′′ = y′ and a′′ = z′. Also the argument that established the triangles yy′y′′, zz′z′′ and aa′′x′′ 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 axx′; xyz; yy′y′′; zz′z′′; and aa′′x′′ form a subgraph K which is attached to edges of G not in K only at x′, y′′, z′′. Moreover, as |V (G)| ≥ 8 at least two of these vertices are distinct (say x′ ≠ y′′). Then y′′ is incident with two edges not in K, whence x′, y′′, z′′ are all distinct (G is ∼ 3-connected) and form a triangle, determining G = L(K3,3). 

Case 3: Configuration H3 exists in G (but not H1 or H2). Consider of sub- ′′ graph J of G consisting of the the circuit Q2, the four edges incident with a and the path P2. Now let U be the union of the triangle T = xyz and the ′ ′ ′′ paths P3 and P4. Clearly, U − {x, y , z } is contained in a bridge B3 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 symmetry) ′ ′′ only three options for the remaining feet of the bridge B3; x , x or a vertex ′′ ′ ′ z in the interior of the path from x to z on Q2. If there is a path in B3 from U − {x, y′, z′} to either x′ or z′′, then there are disjoint circuits, one of which

11 x'=a'

y'' a z'' x

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

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

′ ′′ ′ ′ ′ is a, z , x , y . On the other hand, consider a path in B3 from U − {x, y , z } to x′′. If the path leaves U − {x, y′, z′} from w on the path from z to z′, then there are disjoint circuits through w, z′, a, x′′ and through x, y, y′, x′. This completes the proof of Proposition 1.1.  ∼ Corollary 2.6. If G is internally-4-connected, |V (G)| ≥ 8 and G ̸= L(K3,3) then W ≤t G or C ≤t G.

Proof: Under these hypotheses, disjoint circuits Q1,Q2 of length ≥ 4 exist, and by the 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 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. 

3. Bridges on a Cube To proceed with the proof of Theorem 1.2 using Corollary 2.6, we may assume without loss of generality that G is internally-4-connected, |V (G)| ≥ ∼ 8, G ̸= L(K3,3) and C ≤t G but W ̸≤t G. Using the condition W ̸≤t G it is incumbent to show conditions (1), (2) or (3) of the main theorem apply. In the rest of this paper, we will use C′ to denote the subdivided cube contained in G as a subgraph in witness of the inclusion C ≤t G.

12 Here we need to define several types of bridges of C′. In what follows, a Y -graph, an X-graph or an H-graph is 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 C′ 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 [17], 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. Note that a Y -bridge is a triad of G. A set of three branches of C′ centered at n is called the triad of branches centered at n. Suppose for the triad of branches centered at n, a bridge B of C′ 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 at n. Note that a bridge that covers a triad could either be confined to the triad if it has no attachments outside the triad, or not confined. 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. A bridge is attached to a triad centered at a node n of C′ if B has an attachment on n or on the interior of at least one of the branches incident with n. We call B an antipodal bridge of C′ if for some antipodal pair of nodes of C′, n and n′, B is attached to both the triad centered at n and the triad centered at n′. Given a subdivided cube C′ in a graph G, a long diagonal of C′, is an inner bridge of C′ 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 3.3. The first step in the analysis of the bridges of the cube is to normalize the (subdivided) cube C′ as embedded in G so that no bridge of C′ is confined to a single branch of C′, a standard argument used often in such situations. Such embeddings are called proper embeddings of a graph H in G. A proof and application of this are shown for example in page 121 of [17].

Lemma 3.1. If G is a 3-connected graph and H′ ⊂ G is a subdivision of a trivalent 3-connected graph H, then it is possible to find a subgraph H′′ ⊂ G, which is also a subdivision of H such that no bridge of H′′ in G has all its feet in one branch of H′′.

13 From now on, we assume that C′ is chosen to be properly embedded in G. Corollary 3.2. If C′ is properly embedded in a 3-connected graph G and B is a bridge of C′, then B has two feet, x, y, which are not confined to just one branch of C. Moreover, x and y can be joined by a C′-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 x

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

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

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

Proof: Choose the feet x and y not on the same branch as follows. If the bridge B has three feet with are nodes of C′ then two of these nodes x, y are non adjacent in C′ 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 both nodes of C′, has x, y nonadjacent as C′ is properly embedded. In the three cases, there exists a C′-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 be on the nodes or branches of C′, not both on the same branch.  Remarks: We see from the diagrams of Figure 6 that C′ ∪ xy is planar in cases (1), (3), (5), and (6). Furthermore, we see that W ≤t (4) and (4) ≤m (7), (4) ≤m (8), whence W ≤t (7), W ≤t (8). The arc L in case (2) is a long diagonal of the cube C′.

14 Given a bridge B of C′ in G, where C′ is properly embedded in G, if C′ ∪B is a planar graph, then we will call B a planar bridge of C′. Otherwise B is a nonplanar bridge of C′. A bridge B which has all of its attachments on the branches of a quadrilateral Q of C′ will be called a facial bridge belonging to Q. Note that, as B is not a local bridge and C′ has a unique planar embedding, a facial bridge belongs to a unique face of C′. If a bridge does not belong to any face, we say that it is a nonfacial bridge. A series of lemmas now will establish that nonfacial bridges of C must (after a slight modification) conform to the bridges allowed in parts (2) and (3) of Theorem 1.2.

Lemma 3.3. Suppose G is internally-4-connected and C′ is a properly em- bedded subdivided cube in G. Let n and n′ be antipodal nodes of C′. Let B be an antipodal bridge of C′ in G on n and n′. Then either B is a long diagonal ′ with endvertices n and n or W ≤t G.

Proof: Let m and m′ be attachments of B in the triads centered at n and n′ respectively. The subdivided cube union the path within B from m to m′ must contain one of (2), (4), (7) or (8) from Cor. 3.2. Of these only (2) does not contain W , hence n and n′ are attachments of B. If B has a third attachment on C′, then it can be checked that G would contain (4) and hence W ≤t G.  From now on, we can assume that since W ̸≤t G, any antipodal bridge is a long diagonal.

′ Lemma 3.4. Suppose G is internally-4-connected, W ̸≤t G, and C is a properly embedded subdivided cube in G. Let B be a nonfacial bridge of C′ in G, then either B is a long diagonal or B is a nodal bridge.

Proof: As a first case, assume that B is attached at an internal vertex, n, of some branch (w, z′) of C′ with C′ labelled as in Fig. 7. Since (4),(7) and (8) all contain the Wagner graph, all the remaining feet of B are confined ′ ′ to the two faces of the cube containing n, say Q1 = (w, z , y, x ) and Q2 = (w, z′, x, y′). Note that, as B is a nonfacial bridge of C′, B must have a foot ′ ′ m1 on Q1 \(w, z ) and a foot m2 on Q2 \(w, z ). It is straightforward to check that either B is a nodal bridge centered at w or z′ or that B is antipodal and hence, a long diagonal. As the second case, assume that B is only attached at nodes of C′. Sup- pose first that B is attached at adjacent nodes w, z′ of C′ with C′ labelled

15 x' y

w z'

y' x z w'

Figure 7: Labeling of Nodes of a Cube as in Fig. 7. Clearly, B must have at least one other foot, m a node of C′. ′ If m is antipodal to w or z , then we obtain W ≤t G by Lemma 3.3 since B is not a long diagonal. Hence neither w′ nor z is a foot of B. If B is not a nodal bridge centered at z′, then at least one of y or x is a not foot of B. By symmetry, we can assume that x is not an attachment of B. Similarly, as B is not a nodal bridge centered at w, we can assume that at least one of y′ or x′ is a not foot of B. As B is nonfacial, B must be have an attachment at y and y. This implies that B is antipodal, but not a long diagonal. Hence G must contain a W -subdivision. Finally, we can assume that B does not have feet which are adjacent nodes of C′. Without loss of generality, assume that w is a foot of B with C′ labeled again as in Fig. 7. The only possible feet of B are z, x, y or w′. If w′ is a foot of B, then there can be no other feet and B is a long diagonal. As B is nonfacial, then it must have feet on at least three of w, x, y and z. Hence it is a nodal bridge. 

Lemma 3.5. Suppose G is internally-4-connected and C′ is a properly em- bedded subdivided cube in G. Let B be a nodal bridge centered at n of C′. ′′ Then either W ≤t G or G contains a subdivided cube C with an X-bridge on a colour class of C′′.

′ ′ Proof: Let n1, n2 and n3 be the nodes of C adjacent to n and let n be ′ 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 ′ possible. There will be a Y -graph Y in B meeting C exactly in x1, x2, x3 and centered at x′. We will consider two cases; either B is confined to the triad of branches centered at n or it is not. First we consider the case where it is not

16 ′ confined. For notation, assume that the 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. Note that B is not an antipodal bridge, ′ ′ ′ hence w cannot be on n1 or the interior of any branch of C incident with n1. ′ ′ Similar statements can be made about n2 and n3. Therefore, the attachment ′ w must be n . Again as B is not antipodal, the foot x1 cannot be in the interior of the branch (n, n1). So we can conclude that x1 = n1. Similarly, x2 = n2 and x3 = n3. As shown in Fig. 8, if the bridge B with exactly four ′ attachments n1, n2, n3, n contains an H-bridge, then W ≤t G. Therefore, B is an X-bridge and the conclusion of the lemma holds.

′ Figure 8: An H-bridge on a colour class of C leading to W ≤t G.

Now we suppose that the bridge B is confined to the triad of branches centered at n. Then there will be a Y -graph Y ′ in C′ centered at n and endvertices x1, x2, x3. Switching the two Y -graphs yields a second subdivided cube C′′ which with a little care will satisfy the bridge condition of Lemma 3.1. Choosing Y in B can be done without any bridge of Y in B confined to a single branch of the Y -graph, using the standard argument from the proof of Lemma 3.1. Note that there are exactly three types of bridges of C′′: proper subgraphs of B, the bridge B′ of C′′ which contains Y , and bridges which were also bridges of C′. The only other changes in a bridge occur ′ ′′ ′ ′′ for the bridge B of C containing Y 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 C′, hence are of the required type. If the new bridge B′ is still confined to a triad of branches of C′′ (necessarily including the Y -graph Y ) it will have a new attachment on a subpath (x1, n1), (x2, n2), or (x3, n3). The switching can be repeated using ′ ′ ′ 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 C′). Otherwise there is an attachment w, of the new bridge B’ which is not on the triad of (the new) C′ at n. This will be a nodal bridge that covers

17 the triad of branches but is not confined to it. By the proof of the first case of this lemma, B is an X-bridge on a colour class of C′. This completes the proof of Lemma 3.5.  Now we consider the possible bridges of C′ given the existence of at least one X-bridge.

Lemma 3.6. Suppose G is internally-4-connected, and that C′ is a properly embedded subdivided cube in G. Suppose there exists an X-bridge X of C′ with its attachments on a colour class w, x, y, z of nodes in the cube C′. Then either W ≤t G or G \{w, x, y, z} is edgeless.

3 7 2

8 1

6 4 5 9 10

′ Figure 9: Two X-bridges on distinct colour classes of C leading to W ≤t G.

Proof: Suppose there exists a second X-bridge X′ on C′ in G. If the ′ color class for X is not the same as that for X, then W ≤t G. In fact, G contains a V10-subdivision as seen in Figure 9. So we can assume that all of the X-bridges of C′ share a color class, {w, x, y, z}. Suppose that there is a facial bridge B of C′ belonging to the face (x′, w, z′, y). If there is a C′-avoiding path in B with one end in the interior of the path (x′, w, z′) and one end in the interior of (z′, y, x′), then this path can be contracted to an edge with ends w and y. This graph can be seen to contain W (See the first graph in Figure 10). So without loss of generality, there exists a facial bridge with attachments only on the two branches (x′, w) and (w, z′). (See the second graph in Figure 10). This bridge violates internal-4-connectivity, unless there is another bridge of C′ with attachments on the triad of branches centered at w. If there exists a facial bridge with such attachments that also has attachments off of that triad of branches, then a path within that bridge can be contracted to a edge between two black vertices in the boundary of a face of C′. Similar to above, this would lead to a W -graph contained

18 in G. Hence the union of all facial bridges with attachments on the triad of branches centered at a must be entirely on one side of the separation x′, y′, z′. This would violate internal-4-connection unless there was a non-facial bridge crossing that separation. The only possible such bridge is a long-diagonal from w to w′. So assume that the long diagonal from w to w′ exists in G as well as at least one facial bridge with attachments only on the two branches (x′, w) and (w, z′). In this case, we can find a rerouted cube C′′ with node w′′. The bridge of C′′ containing the edge from w to w′ is an antipodal bridge, but not a long diagonal. (See the third graph in Figure 10). This contradiction implies a W subdivision in G. If no facial bridges exist, the the only possible bridges of G are the X-bridge and any long-diagonals. Even if all four long diagonals exist, G \ w, x, y, z remains edgeless. 

x' y x' y x' y

w z' w z' w w'' z'

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

Figure 10: An X-bridge on C′ with various bridges attached at vertex ’a’

At this point, we can assume that the only nonfacial bridges of C′ are long diagonals. Now we need to consider the facial bridges. To this end a planarity lemma (Lem. 3.7) can be stated which is very useful in this type of argument, essentially characterizing when a circuit 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 C. Jung [8]. It was rediscovered several times including by Robertson and Chakravarti [12], relating to Chakravarti’s Ph.D. thesis in 1976 [2], by Seymour [14], by Thomassen [16] and by Siloach[15] each in 1980. Mohar also proved an algorithmic version of the result in 1994 [11]. First, let us explain some technical terms. A diagonal of a circuit 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

19 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 ′ ′ Q by paths P1,P2,P3 (possibly degenerate) with endvertices (x, x ); (y, y ); (z, z′), respectively, and having x′, y′, z′ on Q.

Lemma 3.7 ([8, 12, 2, 14, 16, 15]). Let G be a subdivision of a 3-connected graph and Q be a circuit 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.

Evidently 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.

Corollary 3.8. Let G be an internally-4-connected graph and Q be a circuit of G with |E(Q)| ≥ 4. Let H be the union of some (perhaps all) of the facial bridges of C′ belonging to Q in G. Then either Q ∪ H is planar with Q bounding a face in a plane embedding, or H contains a pair of crossed diagonals of Q.

Proof: By Lemma 3.7 either the above corollary holds or there is a tri- pod T in H which is 3-linked by paths P1,P2,P3 to Q. By the internal-4- connection of G, using the augmenting path version of Menger’s theorem and |V (T )| ≥ 4 ≤ |V (Q)|, with E(T ∩ Q) = ∅, we have a 4-join from T to Q using the endvertices of P1,P2,P3. Without loss of generality we may assume ′ ′ ′ ′ P1,P2,P3,P4 are this 4-join and that the respective endvertices w , x , y , z 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 if T is in one bridge, and otherwise T is the union of two ′ ′ ′ Y -graphs 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 we consider the the facial bridges of C′. Let C˜ be C′ union the set of all facial bridges of C′. If C˜ is planar, then as G is not planar there must be a nonfacial bridge of C′. By earlier arguments, all such nonfacial bridges must be long diagonals of C′. This case will be considered in Lemmas 3.10, 3.11 and 3.12. Here we consider the case where C˜ is nonplanar.

20 Suppose that for each face Q of the embedding of C′, Q union the set H of bridges of C′ 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 G, filling in each face of C′ in the process. Hence, as G is nonplanar, there must exist a face Q of C′ where this embedding is not possible. By Cor. 3.8, a pair L1,L2 of crossed diagonals for some face Q of C′ must exist.

Lemma 3.9. Suppose G is an internally-4-connected graph and C′ is a prop- erly 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 properly em- bedded subdivided cube C′′ in G with a nodal bridge.

(1) (2) (3) (4) x2 y x2 y1 1

x2 x2 y x x1 y2 y 1 1 y1 2 x x1 1 y2 y2

y y1 y2 2

y2 y1 y1 x2 y1 x2 y x2 2 x2 x1 x1 x1 x1

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

Figure 11: Different cases for crossed diagonals on a face of a cube.

Proof: In Figure 11 a list is compiled of all of the different ways in which L1,L2 can be crossed diagonals on Q taking symmetries of the cube into account. In the figure, the cross-hatched paths indicate that the vertex could be anywhere along that part of the boundary of the face including at the branch vertex. Similarly, the arrows indicate that the crossed path intersects the face strictly in the interior of the branch. In the figure, L1,L2 are given endvertices x1, y1 and x2, y2, respectively.

21 In (1), an internal vertex of each branch of C′ in Q is used in the endver- tices of L1 and L2. In (2) and (3), L1 and L2 are confined to three branches ′ of C an no fewer. In (2), both L1 and L2 have a vertex in the interior of the ′ middle of the three branches of C . While in (3), only L1 meets the interior of the middle branch. In (4) and (5), the four endvertices of L1 and L2 are ′ confined to nonadjacent branches of C . In (4) neither L1 nor L2 lies on a ′ single branch of C . While in (5), L2 does. In (6) and (7), the four ends of L1 ′ and L2 are confined to adjacent branches of C . In (6) neither L1 nor L2 lies ′ on a single branch of C . While in (7), L1 does. The final case (8), occurs ′ when all four vertices of L1 and L2 are confined to a single branch of C . We can remark that in cases (1),(2),(3) and (4) evidently W ≤t G. Also, because of the normalization condition the bridge containing L1 in (8) will have an attachment not on the branch containing x1, y1; and hence L1 (or ′ possibly L2 if it is in the same bridge) can be replaced by an arc L1 such that ′ L1,L2 are crossed diagonals, and such that the endvertices are not all on one branch. This leaves the three cases where lengthier arguments are necessary. In Case (6), whether L1 and L2 are in the same bridge or not, we can ′ reroute the branch of C from n to m, by going along the path from n to x2 ′′ to y2 to m to obtain a new subdivided cube C . By earlier arguments about rerouting cubes, C′′ can be assumed to have no local bridges. Moreover, ′′ the bridge of C containing y1 will be a nodal bridge covering the triad of branches centered at x2. Hence the conclusion of the lemma holds. In cases (5) and (7), either L1 and L2 are in a single bridge or not. When ′ L1,L2 are in the same bridge B of C , there is a third arc L joining L1,L2 at internal vertices x, y as in Fig. 12. In both case(5) and case (7), we can ′ reroute the branch of C from n to m by following the path n, y1, x, y, y2, m to obtain the cube C′′ with the node n of C′ replaced by the node x in C′′. ′′ Note that the bridge of C that contains the vertex x2 will be a nodal bridge as it covers the triad of branches centered at x. Thus the conclusion of the lemma holds in both of these cases. It remains to consider the cases where (5) and (7) have L1,L2 in distinct ′ bridges B1,B2 of C . Since B1 is not a local bridge, it has an attachment w not on the branch containing the endvertices of L1. If w ≠ y2 in (5) or w is ′ not on the branch of C containing y2 in (7) then we can easily reduce to case (2) and we have W ≤t G. Thus w = y2 in case (5) and w is on the branch of C containing y2 in (7) can be assumed, and also it can be assumed that this situation cannot be avoided. If w ≠ y2 in case (7) we can reduce to case (6) which also gives W ≤t G. So we can assume in both cases that the only

22 y n m n 2 m

y1 x y1 x y x y x 2 y 2 2 x1 x1

(5) (7)

Figure 12: Rerouting the cube to find a nodal bridge

attachment of B1 not on the original branch is at y2. Again rerouting the node of C′ at n (there are two choices of n in case (5)) to pass through x (the internal vertex of L linked to y2) the earlier nodal bridge situation reoccurs, with y, internal to a branch of C′′ at x. This completes the proof of Lemma 3.9.  Now we can assume that we are in the case where C˜ is planar, all of the nonfacial bridges of C′ in G must be long diagonals and at least one such long diagonal must exist.

Lemma 3.10. Suppose G is internally-4-connected, that C′ is a properly embedded subdivided cube in G and C˜ is planar. Let G have a long diagonal L (at least one), and suppose that all the bridges of C′ are attached only at nodes, i.e. C′ is actually equal to a cube C. Then all bridges are link-graphs, and either G is a double wheel, G is 4-vertex coverable, or W ≤t G.

Remark: It remains to dispose of the case where planar bridges arise at- tached at internal vertices of the branches of C′. These cases will be resolved by Lemmas 3.11 and 3.12. 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 x′ as in Fig 14. If there are no other bridges of C′ then the conclusion holds. ′ 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 C′ will satisfy the conditions of the theorem. So we can assume that there is a planar bridge of C′. 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.

23 If there were three feet, the triad and triangle would contradict internal-4- connectivity. While if the bridge had attachments on all four nodes of the face, we 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.

x' y w z'

y' x

z w'

Figure 13: The ’black’ and ’white’ color classes on a Cube

Let one color class, say w, x, y, z, be called the ‘black’ vertices and let the other class w′, x′, y′, z′ be called the ‘white’ vertices. See Fig. 13. 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 in Bw is either incident with L or its end points are in the neighborhood of an end of L. First consider the case where a link-graph exists in the neighborhood of L, say y′z′ (see the first graph of Fig 14). In the 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 ww′ exists. Note that at this point, the white vertices still form a covering set. 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. In the second case, we can assume that no link-graph lies in the neighbor- hood of the endpoints of a long diagonal. Here without loss of generality, we ′ ′ can assume that a link-graph in Bw is x y . Again, note that if w were cubic, internal-4-connectivity would be violated. Therefore either wx or ww′ exists (wz and wy are in the neighborhood of L). Considering ww′ as a second long diagonal L′, we can conclude as in the previous paragraph since x′y′ is in the neighborhood of L′. The only remaining case is where wx exists in G. In

24 this case, there must be another edge incident with z′. The other endpoint of this edge can only be x′ as edges z′w′, z′x′ and z′y′ would all lead to the first case. Now consider the link-graph leaving y. This analysis leads to link graphs from both x′ and x to each vertex on the six-cycle (w, z′, y, w′, z, y′). Hence G is a double wheel. 

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

y' x y' x

z w' z w'

Figure 14: Possible antipodal and facial bridges on a Cube

This leaves the final case where C˜ is planar, the only nonfacial bridges are long diagonals and the cube C′ 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 C′ incident with a long diagonal L will be called a spoke branch on L. The six branches of C′ 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 ′ (planar) bridge of C , say B1, has a foot internal to a spoke branch. The second case is where no bridge of C′ has a foot internal to any spoke branch. See Figure 15.

Lemma 3.11. Suppose G is internally-4-connected, and that C′ is a properly embedded subdivided cube in G. Let C′ have at least one long diagonal L with ′ ˜ endvertices x, x . Further assume that C is planar and that some bridge B1 of C′ is attached at an internal vertex a of some spoke branch of C′ on L. Then W ≤t G.

′ Proof: Suppose B1 is a bridge of C with an attachment, a, on the spoke branch between x and y′ that is embedded in the face {x, y′, w, z′} as labelled in the graph in Figure 15. Note that if B1 also has an attachment along the

25 z'

w y

B1 x w' y' B2 z

x'

Figure 15: Possible planar bridges on a Cube with a long diagonal path {w, x} through z′, then G contains a W -subdivision. Therefore all of ′ 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 y′ and belonging to either face {x, y′, z, w′} or {y′, z, x′, w}. Therefore, unless there exists a long diagonal from y′ to y, the vertices w, x and z form a vertex cut-set of G and if the part containing y′ contains any other vertex, this contradicts internal-4-connection. If the long diagonal (y, y′) does exist, then one can ′ ′′ reroute the branch of the cube C through B1 to obtain C . In this cube, the bridge containing (y, y′) is antipodal but not a long diagonal. This implies the existence of a W -subdivision in G. Therefore no such bridge B1 exists. 

Lemma 3.12. Suppose G is internally-4-connected, C′ is a properly embed- ded subdivided cube in G and that C˜ is planar. Let C′ have at least one long diagonal L with endvertices x, x′. Further, suppose that every bridge B of C′ is attached only at x or y and along the rim branches of C′ around L. Then ′ either G − {x, x } is isomorphic to a circuit or W ≤t G.

Proof: We assume that there is a bridge B2 which has an attachment in ′ ′ 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, x′ or (z, w′]. 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 (y′, z) and (z, w′), then we find a second cube C′′ where x′ is not a node, L is a nonplanar bridge but no longer a

26 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 ′ x and along the branch [y , z]. (See bridge B3 in Fig. 16). 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 16. 

z' z'

w y w y

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

B4

x' x'

Figure 16: A Cube with two ’Rim Bridges’ contains a W -subdivision

Therefore the only possible bridges on C′ are long diagonals and inner bridges of C′ with attachments on either x or y and on the interiors of the rim branches. This implies that either G − {x, x′} is a circuit or there is at least one more long diagonal on C′. Let H be the (possibly subdivided) Hexagon in C′ − {x, x′} and let n be a node of C′ on H. We say that n is fixed if there is no cube C′′ of G where n is not a node of C′′. Note that if there is a long diagonal L′ attached at n (and at its antipodal node n′ on the hexagon) then n must be fixed. If it were possible to obtain a second subdivided cube C′′ contained in G so that n is no longer a node of C′′ then the bridge of C′′ in G that contains the edge (n, n′) is an antipodal bridge, but not a long diagonal of C′′. It would then follow from Lemma 3.3 that W ≤t G. Suppose there is a vertex a in the interior of a branch of H incident with 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 x′. However, this would violate internal-4- connectivity as a would be both cubic and contained in a triangle. Therefore,

27 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 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 C′ other than L from x to x′. Therefore, G − {x, x′} is a circuit as required in the main theorem. In this case, in order to maintain internal-4- connectivity, it is clear that either the spokes incident with x and y alternate as one traverses the rim vertices of the cube or they pair off with all the rim vertices. These two cases give rise to either the (nonplanar) Alternating Wheels ADWn for n ≥ 3 or the (nonplanar) Complete Double Wheels CDWn for n ≥ 4. This completes the analysis of the bridges of C′ and hence the proof of Theorem 1.2. 

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