Generating simple near-bipartite bricks

TL;DR

This paper develops a specific generation procedure for simple near-bipartite bricks, refining existing theories to identify key edges and transformations that preserve their structure.

Contribution

It establishes a new generation theorem for simple near-bipartite bricks, building on previous work and refining the brick generation process for this specific class.

Findings

Every simple near-bipartite brick has an edge whose removal and contraction yields another such brick.

The generation process excludes eight well-defined infinite families of graphs.

The main theorem is derived from a broader result on near-bipartite bricks.

Abstract

A brick is a $3$-connected graph such that the graph obtained from it by deleting any two distinct vertices has a perfect matching. A brick $G$ is near-bipartite if it has a pair of edges $\alpha$ and $\beta$ such that $G-\{\alpha,\beta\}$ is bipartite and matching covered; examples are $K_4$ and the triangular prism $\overline{C_6}$. The significance of near-bipartite bricks arises from the theory of ear decompositions of matching covered graphs. The object of this paper is to establish a generation procedure which is specific to the class of simple near-bipartite bricks. In particular, we prove that every simple near-bipartite brick $G$ has an edge $e$ so that the graph obtained from $G-e$ by contracting each edge that is incident with a vertex of degree two is also a simple near-bipartite brick, unless $G$ belongs to any of eight well-defined infinite families. This is a refinement of the brick generation theorem of Norine and Thomas (Generating Bricks, J. Combin. Theory Ser. B, 2007) which is appropriate for the restricted class of near-bipartite bricks. Earlier, the first author (Generating near-bipartite bricks, J. Graph Theory, 2019) proved a similar generation theorem for (not necessarily simple) near-bipartite bricks; we deduce our main result from this theorem. Our proof is based on the strategy of Carvalho, Lucchesi and Murty (2008) and uses several of their techniques and results. The results presented here also appear in the Ph.D. thesis of the first author.

Full text

The object of this paper is to establish a generation procedure which is specific to the class of simple near-bipartite bricks. In particular, we prove that every simple near-bipartite brick $G$ has an edge $e$ such that the graph obtained from $G-e$ by contracting each edge that is incident with a vertex of degree two is also a simple near-bipartite brick, unless $G$ belongs to any of eight well-defined infinite families. This is a refinement of the brick generation theorem of Norine and Thomas [10] which is appropriate for the restricted class of near-bipartite bricks.

Earlier, the first author proved a similar generation theorem for (not necessarily simple) near-bipartite bricks [7, 6]; we deduce our main result from this theorem. Our proof is based on the strategy of Carvalho, Lucchesi and Murty [4] and uses several of their techniques and results. The results presented here also appear in the Ph.D. thesis of the first author [7].

1. 1 Brick Generation

2. 1.1 Norine-Thomas families

3. 1.2 Near-Bipartite Bricks

4. 1.2.1 $R$-compatible Edges

5. 1.2.2 $R$-thin and Strictly $R$-thin Edges

6. 1.3 Families of $R$-bricks free of Strictly $R$-thin Edges

7. 2 $R$-configurations

8. 2.1 $R$-biwheel configurations

9. 2.2 $R$-ladder configurations

10. 2.3 Corners, Rungs and Spokes

11. 2.4 Two distinct $R$-configurations

12. 2.5 The $R$-biwheel and $R$-ladder Theorems

13. 2.6 Proof Sketch of Theorem 1.8

14. 3 $R$-thin edges

15. 3.1 Multiple Edges in Retracts

16. 3.2 Proof of the $R$-biwheel Theorem

17. 3.3 Proof of the $R$-ladder Theorem

18. 4 Properties of $R$-configurations

19. 4.1 Proof of Proposition 2.3

20. 5 Strictly $R$-thin Edge Theorem

This paper is a sequel to a recent paper of the first author [6]. In the following section, we recall the most important definitions and results from [6], and we explain the contributions of this paper. Please see [6] for any missing definitions. The reader, who is interested in gaining a deeper understanding, should perhaps read the first two sections of [6].

1 Brick Generation

A matching covered graph free of nontrivial tight cuts is called a brace if it is bipartite; otherwise, it is called a brick. Lovász [8] proved the remarkable result that any two tight cut decompositions of a matching covered graph $G$ yield the same list of bricks and braces (except possibly for multiplicities of edges). In particular, any two tight cut decompositions of $G$ yields the same number of bricks; this number is denoted by $b(G)$.

Edmonds, Lovász and Pulleyblank [5] established the following important equivalence.

Theorem 1.1

A graph $G$ is a brick if and only if it is $3$-connected and bicritical.

An edge $e$ of a matching covered graph is removable if $G-e$ is also matching covered; furthermore, it is $b$-invariant if $b(G-e)=b(G)$. A $b$-invariant edge $e$ of a brick $G$ is thin if the retract of $G-e$ is also a brick. (The retract is the graph obtained by contracting each edge that is incident with a vertex of degree two.) Carvalho, Lucchesi and Murty [3] established the following.

Theorem 1.2

[Thin Edge Theorem]*

Every brick distinct from $K_{4}$, $\overline{C_{6}}$ and the Petersen graph has a thin edge.*

As a consequence, every brick may be constructed from one of $K_{4}$, $\overline{C_{6}}$ and the Petersen graph using four elementary “expansion operations” that are described in their paper [3].

In order to establish a recursive procedure for generating simple bricks, one needs the notion of a strictly thin edge. A thin edge $e$ of a simple brick $G$ is strictly thin if the retract of $G-e$ is simple. For each brick shown in Figure 3, its thin edges are indicated by bold lines; it is easily seen that none of these edges is strictly thin.

As an example, consider the Tricorn, shown in Figure 1, which has precisely three thin edges indicated by bold lines; deleting one of them, say $e$, and taking the retract yields the simple odd wheel $W_{5}$. Thus each thin edge of the Tricorn is strictly thin.

Next, we describe infinite families of bricks that do not contain any strictly thin edges. Norine and Thomas [10] proved that these families, together with the Petersen graph, include all the bricks that are free of strictly thin edges; for this reason, we refer to these families as Norine-Thomas families, and we refer to their members as Norine-Thomas bricks.

1.1 Norine-Thomas families

Odd Wheels. The odd wheel $W_{2k+1}$, for $k\geq 1$, is defined to be the join of an odd cycle $C_{2k+1}$ and $K_{1}$. See Figure 3a. The smallest odd wheel is $K_{4}$. If $k\geq 2$, then $W_{2k+1}$ has exactly one vertex of degree $2k+1$, called its hub, and the edges incident at the hub are called its spokes. The remaining $2k+1$ vertices lie on a cycle, called the rim, and they are referred to as rim vertices.

Each member of the remaining four families contains a bipartite matching covered subgraph which is either a ‘ladder’ or a ‘partial biwheel’. These bipartite graphs are also the main building blocks of additional families of bricks which are of interest in Section 1.3. For this reason, we start with a description of these two families of bipartite graphs.

Ladders. Let $x_{0}x_{1}\dots x_{j}$ and $y_{0}y_{1}\dots y_{j}$ be two vertex-disjoint paths, where $j\geq 2$. The graph $K$ obtained by the union of these two paths, and by adding edges $x_{i}y_{i}$ for $0\leq i\leq j$, is called a ladder, and its edges joining $x_{i}$ and $y_{i}$ are referred to as its rungs. See Figure 2. The two rungs $x_{0}y_{0}$ and $x_{j}y_{j}$ are external, and the remaining rungs are internal. We say that $K$ is odd (even) if it has an odd (even) number of rungs.

Partial Biwheels. Let $x_{0}x_{1}\dots x_{2j+1}$ be an odd path, where $j\geq 1$. The graph $K$ obtained by adding two new vertices $u$ and $w$, joining $u$ to vertices in $\{x_{0},x_{2},\dots,x_{2j}\}$, and joining $w$ to vertices in $\{x_{1},x_{3},\dots,x_{2j+1}\}$, is called a partial biwheel; the vertices $x_{0}$ and $x_{2j+1}$ are referred to as its ends, whereas $u$ and $w$ are referred to as its hubs; and an edge incident with a hub is called a spoke. See Figure 2. The two spokes $ux_{0}$ and $wx_{2j+1}$ are external, and the remaining spokes are internal.

When referring to a ladder or to a partial biwheel, say $K[A,B]$, with external rungs/spokes $au$ and $bw$, we adopt the convention that $a,w\in A$ and $b,u\in B$; furthermore, when $K$ is a partial biwheel, $u$ and $w$ shall denote its hubs; as shown in Figure 2. (Sometimes, we may also use subscript notation, such as $A_{i}$, $B_{i}$, $a_{i}u_{i}$ and $b_{i}w_{i}$ where $i$ is an integer, and this convention extends naturally.)

It should be noted that a partial biwheel of order six is also a ladder. However, a partial biwheel of order eight or more has only two vertices of degree two, namely, its ends; whereas every ladder has four such vertices. We remark that, a biwheel, as defined by McCuaig [9], has order at least eight and contains an additional edge joining its ends; and these constitute an important class of braces.

We now proceed to describe the remaining four Norine-Thomas families using ladders and partial biwheels.

Prisms, Möbius Ladders and Truncated Biwheels. Let $H[A,B]$ denote either a ladder or a partial biwheel of order $n$, with external rungs/spokes $au$ and $bw$, and let $G$ be the graph obtained from $H$ by adding two edges, namely, $aw$ and $bu$. If $H$ is an odd ladder then $G$ is a prism and it is denoted by $P_{n}$, see Figure 3b. If $H$ is an even ladder then $G$ is a Möbius ladder and it is denoted by $M_{n}$, see Figure 3f. Finally, if $H$ is a partial biwheel then $G$ is a truncated biwheel and it is denoted by $T_{n}$, see Figure 3c. Note that $\overline{C_{6}}$ is the smallest prism as well as the smallest truncated biwheel. For convenience, we shall consider $K_{4}$ to be the smallest Möbius ladder.

Staircases. Let $K[A_{1},B_{1}]$ denote a ladder of order $n$, with external rungs $a_{1}u_{1}$ and $b_{1}w_{1}$. Then the graph $G$ obtained from $K$, by adding two new vertices $a_{2}$ and $b_{2}$, and by adding five new edges $a_{1}a_{2},u_{1}a_{2},b_{1}b_{2},w_{1}b_{2}$ and $a_{2}b_{2}$, is called a staircase, and it is denoted by $St_{n+2}$. See Figures 3d and 3e.

Using this terminology, the theorem of Norine and Thomas [10] may be stated as follows.

Theorem 1.3

[Strictly Thin Edge Theorem]*

Let $G$ be a simple brick. If $G$ is free of strictly thin edges then $G$ is either the Petersen graph, or it is an odd wheel, a prism, a Möbius ladder, a truncated biwheel or a staircase.*

It should be noted that Norine and Thomas did not state their results in terms of strictly thin edges. Subsequently, Carvalho, Lucchesi and Murty [4] used their Thin Edge Theorem (1.2) to deduce the Strictly Thin Edge Theorem (1.3). The following result of Norine and Thomas [10] is an immediate consequence of Theorem 1.3.

Theorem 1.4

Given any simple brick $G$, there exists a sequence $G_{1},G_{2},\dots,G_{k}$ of simple bricks such that:

1. (i)

$G_{1}$* is a Norine-Thomas brick,* 2. (ii)

$G_{k}:=G$, and 3. (iii)

for $2\leq i\leq k$, there exists a strictly thin edge $e_{i}$ of $G_{i}$ such that $G_{i-1}$ is the retract of $G_{i}-e_{i}$.

The above theorem implies that every simple brick can be generated from one of the Norine-Thomas bricks by means of four expansion operations as described by Carvalho, Lucchesi and Murty (see [3]). These expansion operations are simply the inverse of the operation of deleting a strictly thin edge and then taking the retract.

We remark that Norine and Thomas proved a generalization of Theorem 1.4, which they refer to as the ‘splitter theorem for bricks’, since it is motivated by the splitter theorem for $3$-connected graphs due to Seymour [11]. The notions of thin and strictly thin edges are easily generalized to braces (see [4]). A ‘splitter theorem for braces’ was established by McCuaig [9].

1.2 Near-Bipartite Bricks

A nonbipartite matching covered graph $G$ is near-bipartite if it has a pair $R:=\{\alpha,\beta\}$ of edges such that $H:=G-R$ is matching covered and bipartite. Such a pair $R$ is called a removable doubleton. Furthermore, if $G$ happens to be a brick, we say that $G$ is a near-bipartite* brick*. For instance, $K_{4}$ and $\overline{C_{6}}$ are near-bipartite bricks, and each of them has three distinct removable doubletons. On the other hand, the Petersen graph is not near-bipartite. (A result of Carvalho, Lucchesi and Murty [2] implies that if $G$ is a near-bipartite matching covered graph then $b(G)=1$.)

Observe that the edge $\alpha$ joins two vertices in one color class of $H$, and that $\beta$ joins two vertices in the other color class. Consequently, if $M$ is any perfect matching of $G$ then $\alpha\in M$ if and only if $\beta\in M$. (In particular, neither $\alpha$ nor $\beta$ is a removable edge of $G$.) It is easily verified that every Norine-Thomas brick, except for the odd wheels and for the Petersen graph, is near-bipartite.

The difficulty in using Theorem 1.4 as an induction tool for studying near-bipartite bricks, is that even if $G_{k}:=G$ is a near-bipartite brick, there is no guarantee that all of the intermediate bricks $G_{1},G_{2},\dots G_{k-1}$ are also near-bipartite. For instance, the brick shown in Figure 4a is near-bipartite with a (unique) removable doubleton $R:=\{\alpha,\beta\}$. Although the edge $e$ is strictly thin; the retract of $G-e$, as shown in Figure 4b, is not near-bipartite since it has three edge-disjoint triangles.

In other words, deleting an arbitrary thin edge may not preserve the property of being near-bipartite. In this sense, the Thin Edge Theorem (1.2) and the Strictly Thin Edge Theorem (1.3) are inadequate for obtaining inductive proofs of results that pertain only to the class of near-bipartite bricks.

To fix this problem, the first author started this line of investigation and decided to look for thin edges whose deletion preserves the property of being near-bipartite. Kothari [7, 6] proved a ‘thin edge theorem’ for near-bipartite bricks; in particular, he showed that every near-bipartite brick $G$ distinct from $K_{4}$ and $\overline{C_{6}}$ has a thin edge $e$ such that the retract of $G-e$ is also near-bipartite (see Theorem 1.6). In the present paper, we use this to deduce a ‘strictly thin edge theorem’ for near-bipartite bricks. This is similar to the approach of Carvalho, Lucchesi and Murty [4] — they use their Thin Edge Theorem (1.2) to deduce the Strictly Thin Edge Theorem (1.3) of Norine and Thomas.

As in [7, 6], we find it convenient to fix a removable doubleton $R$ (of the brick under consideration), and then look for a strictly thin edge whose deletion preserves this removable doubleton. To make this precise, we will first define a special type of removable edge which we call ‘$R$-compatible’.

1.2.1 $R$-compatible Edges

We use the abbreviation $R$-graph for a near-bipartite graph $G$ with (fixed) removable doubleton $R$, and we shall refer to $H:=G-R$ as its underlying bipartite graph. In the same spirit, an $R$-brick is a brick with a removable doubleton $R$.

A removable edge $e$ of an $R$-graph $G$ is $R$-compatible if it is removable in $H$ as well. Equivalently, an edge $e$ is $R$-compatible if $G-e$ and $H-e$ are both matching covered. For instance, the graph $St_{8}$ shown in Figure 3d has two removable doubletons $R:=\{\alpha,\beta\}$ and $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$, and its unique removable edge $e$ is $R$-compatible as well as $R^{\prime}$-compatible.

Now, let $G$ denote the $R$-brick shown in Figure 4a, where $R:=\{\alpha,\beta\}$. The thin edge $e$ is incident with an edge of $R$ at a cubic vertex; consequently, $H-e$ has a vertex whose degree is only one, and so it is not matching covered. In particular, $e$ is not $R$-compatible.

The brick shown in Figure 5 has two distinct removable doubletons $R:=\{\alpha,\beta\}$ and $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$. Its edges $e$ and $f$ are both $R^{\prime}$-compatible, but neither of them is $R$-compatible.

Observe that, if $e$ is an $R$-compatible edge of an $R$-graph $G$, then $R$ is a removable doubleton of $G-e$, whence $G-e$ is also near-bipartite and thus $b(G-e)=1$. Consequently, every $R$-compatible edge is $b$-invariant.

Furthermore, as shown in [7, 6], if $e$ is an $R$-compatible edge of an $R$-brick $G$ then the unique brick $J$ of $G-e$ is also an $R$-brick; in particular, $J$ is near-bipartite. The following is a special case of a theorem of Carvalho, Lucchesi and Murty [1].

Theorem 1.5

[$R$-compatible Edge Theorem]*

Every $R$-brick distinct from $K_{4}$ and $\overline{C_{6}}$ has an $R$-compatible edge.*

In [1], they proved a stronger result. In particular, they showed the existence of an $R$-compatible edge in $R$-graphs with minimum degree at least three. (They did not use the term ‘$R$-compatible’.) Using the notion of $R$-compatibility, we now define a type of thin edge whose deletion preserves the property of being near-bipartite.

1.2.2 $R$-thin and Strictly $R$-thin Edges

A thin edge $e$ of an $R$-brick $G$ is $R$-thin if it is $R$-compatible. Equivalently, an edge $e$ is $R$-thin if it is $R$-compatible as well as thin, and in this case, the retract of $G-e$ is also an $R$-brick.

As noted earlier, the graph $St_{8}$, shown in Figure 3d, has two removable doubletons $R$ and $R^{\prime}$. Its unique removable edge $e$ is $R$-thin as well as $R^{\prime}$-thin. Using the $R$-compatible Edge Theorem (1.5) of Carvalho, Lucchesi and Murty, the following ‘thin edge theorem’ was proved by Kothari [7, 6].

Theorem 1.6

[$R$-thin Edge Theorem]*

Every $R$-brick distinct from $K_{4}$ and $\overline{C_{6}}$ has an $R$-thin edge.*

Each $R$-compatible edge of an $R$-brick may be associated with two integer parameters — rank and index — as defined in [6]. In fact, Kothari proved the following stronger result that immediately implies the $R$-thin Edge Theorem (1.6) since the rank and index are bounded quantities.

Theorem 1.7

Let $G$ be an $R$-brick which is distinct from $K_{4}$ and $\overline{C_{6}}$, and let $e$ denote an $R$-compatible edge of $G$. Then one of the following alternatives hold:

• •

either $e$ is $R$-thin,

• •

or there exists another $R$-compatible edge $f$ such that:

1. (i)

$f$* has an end each of whose neighbours in $G-e$ lies in a barrier of $G-e$, and* 2. (ii)

${\sf rank}(f)+{\sf index}(f)>{\sf rank}(e)+{\sf index}(e)$.

An $R$-thin edge $e$ of a simple $R$-brick $G$ is strictly $R$-thin if it is strictly thin. In other words, a strictly $R$-thin edge $e$ is one which is $R$-compatible as well as strictly thin; and in this case, the retract of $G-e$ is also a simple $R$-brick.

For instance, let $G$ denote the $R$-brick shown in Figure 6(a), where $R:=\{\alpha,\beta\}$. The retract of $G-e$ is the truncated biwheel $T_{8}$ shown in Figure 6(b); consequently, $e$ is strictly $R$-thin.

Recall that the Norine-Thomas bricks are precisely those simple bricks which are free of strictly thin edges. In particular, every $R$-brick, which is a member of the Norine-Thomas families, is free of strictly $R$-thin edges. A natural question arises as to whether there are any simple $R$-bricks, different from the Norine-Thomas bricks, which are also free of strictly $R$-thin edges. It turns out that there indeed are such bricks; we have already encountered two examples in Figures 4a and 5, as explained below.

Let $G$ denote the $R$-brick, shown in Figure 4a, where $R:=\{\alpha,\beta\}$ is its unique removable doubleton. It can be checked that $G$ has precisely four strictly thin edges, depicted by bold lines; these are similar under the automorphisms of the graph. As noted earlier, if $e$ is any of these edges, then $e$ is not $R$-compatible; furthermore, the retract of $G-e$ is isomorphic to the graph shown in Figure 4b, which is not even near-bipartite as it has three edge-disjoint triangles. Thus, the generation of $G$ using the Norine-Thomas procedure cannot be achieved within the class of near-bipartite bricks.

Now, let $G$ denote the brick shown in Figure 5; it has two removable doubletons $R:=\{\alpha,\beta\}$ and $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$. It may be verified that $G$ has precisely two strictly thin edges, namely $e$ and $f$, each of which is $R^{\prime}$-compatible but neither is $R$-compatible. In particular, $G$ is free of strictly $R$-thin edges; in this sense it is similar to the graph in Figure 4a. On the other hand, $G$ has strictly $R^{\prime}$-thin edges; if $e$ is any such edge then the retract of $G-e$ is a simple near-bipartite brick with removable doubleton $R^{\prime}$. In this sense, $G$ is different from the graph in Figure 4.

1.3 Families of $R$-bricks free of Strictly $R$-thin Edges

We will introduce seven infinite families of simple $R$-bricks which are free of strictly $R$-thin edges, and are different from the Norine-Thomas families. The members of these will be described using their specific bipartite subgraphs, each of which is either a ladder or a partial biwheel; see Figure 2. The occurrence of these subgraphs may be justified as follows. Let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges. If $e$ is any $R$-thin edge of $G$, at least one end of $e$ is cubic and the retract of $G-e$ has multiple edges. These strictures can be used to deduce that $G$ contains either a ladder or a partial biwheel, or both, as subgraphs.

In our descriptions of these families, we use $\alpha$ and $\beta$ to denote the edges of the (fixed) removable doubleton $R$. Apart from $R$, a member may have at most one removable doubleton which will be denoted as $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$. We adopt the notational conventions stated in Section 1.1. (Recall that a partial biwheel of order six is also a ladder; for this reason, some of our families overlap.)

Pseudo-Biwheels. Let $K[A_{1},B_{1}]$ denote a partial biwheel, of order at least eight, and with external spokes $a_{1}u_{1}$ and $b_{1}w_{1}$. Then the graph $G$ obtained from $K$, by adding two new vertices $a_{2}$ and $b_{2}$, and by adding five new edges $\alpha:=a_{1}a_{2},\alpha^{\prime}:=u_{1}a_{2},\beta:=b_{1}b_{2},\beta^{\prime}:=w_{1}b_{2}$ and $a_{2}b_{2}$, is called a pseudo-biwheel. Figure 5 shows the smallest pseudo-biwheel.

It is worth comparing the above with our desription of staircases in Section 1.1. Although a pseudo-biwheel $G$ is free of strictly $R$-thin edges, the two external spokes of $K$, namely $a_{1}u_{1}$ and $b_{1}w_{1}$, are both strictly $R^{\prime}$-thin.

In order to describe the members of the remaining six families, we need two (sub)graphs. For $i\in\{1,2\}$, let $K_{i}[A_{i},B_{i}]$ denote either a ladder or a partial biwheel with external rungs/spokes $a_{i}u_{i}$ and $b_{i}w_{i}$, such that $K_{1}$ and $K_{2}$ are disjoint.

Double Biwheels, Double Ladders and Laddered Biwheels of Type I. Let the graph $G$ be obtained from $K_{1}\cup K_{2}$, by adding edges $\alpha:=a_{1}a_{2}$ and $\beta:=b_{1}b_{2}$, by identifying vertices $u_{1}$ and $u_{2}$, and by identifying vertices $w_{1}$ and $w_{2}$. There are three possibilities depending on the graphs $K_{1}$ and $K_{2}$. In the case in which $K_{1}$ and $K_{2}$ are both partial biwheels, $G$ is a double biwheel of type I. Likewise, in the case in which $K_{1}$ and $K_{2}$ are both ladders, $G$ is a double ladder of type I. Finally, when one of $K_{1}$ and $K_{2}$ is a partial biwheel and the other one is a ladder, $G$ is a laddered biwheel of type I.

A member of any of these families has a unique removable doubleton $R$, and is free of strictly $R$-thin edges. The graph in Figure 4a is the smallest member of each of these families, although its drawing is suggestive of a double biwheel. Figure 7a shows a double ladder. A laddered biwheel is obtained from the graph in Figure 7b by identifying $u_{1}$ with $u_{2}$, and likewise, $w_{1}$ with $w_{2}$.

Double Biwheels, Double Ladders and Laddered Biwheels of type II. Let the graph $G$ be obtained from $K_{1}\cup K_{2}$, by adding four edges, namely, $\alpha:=a_{1}a_{2}$, $\beta:=b_{1}b_{2}$, $\alpha^{\prime}:=u_{1}w_{2}$ and $\beta^{\prime}:=w_{1}u_{2}$. As before, we have three possibilities. In the case in which $K_{1}$ and $K_{2}$ are both partial biwheels of order at least eight, $G$ is a double biwheel of type II. Likewise, in the case in which $K_{1}$ and $K_{2}$ are both ladders, $G$ is a double ladder of type II. Finally, when one of $K_{1}$ and $K_{2}$ is a partial biwheel of order at least eight, and the other one is a ladder, $G$ is a laddered biwheel of type II.

A member of any of these families has two removable doubletons $R$ and $R^{\prime}$, and it is free of strictly $R$-thin edges. However, a double biwheel or a laddered biwheel as shown in Figure 8 has strictly $R^{\prime}$-thin edges; these are the external spokes of a partial biwheel of order at least eight as depicted by the bold lines in the figure.

On the other hand, a double ladder, as shown in Figure 9, is free of strictly $R^{\prime}$-thin edges as well. This may be explained as follows. Every double ladder is cubic, and it has precisely four strictly thin edges; these are the external rungs of the two ladders, depicted by bold lines in the figure. One end of any such edge, say $e$, is incident with an edge of $R$ and the other end is incident with an edge of $R^{\prime}$; since each end of $e$ is cubic, it is neither $R$-compatible nor $R^{\prime}$-compatible.

Using a strengthening (see Theorem 1.7) of the $R$-thin Edge Theorem (1.6), we will prove that the seven families described above and four of the Norine-Thomas families are the only simple $R$-bricks which are free of strictly $R$-thin edges.

Theorem 1.8

[Strictly $R$-thin Edge Theorem]*

Let $G$ be a simple $R$-brick. If $G$ is free of strictly $R$-thin edges then $G$ belongs to one of the following infinite families:*

1. (i)

Truncated biwheels 2. (ii)

Prisms 3. (iii)

Möbius ladders 4. (iv)

Staircases 5. (v)

Pseudo-biwheels 6. (vi)

Double biwheels of type I 7. (vii)

Double ladders of type I 8. (viii)

Laddered biwheels of type I 9. (ix)

Double biwheels of type II 10. (x)

Double ladders of type II 11. (xi)

Laddered biwheels of type II

We present a proof of the above theorem in Section 5. As mentioned earlier, our proof is inspired by the proof of the Strictly Thin Edge Theorem (1.3) given by Carvalho et al. [4], and uses several of their results and techniques.

We shall denote by $\mathcal{N}$ the union of all of the eleven families which appear in the statement of Theorem 1.8. The following is an immediate consequence.

Theorem 1.9

Given any simple $R$-brick $G$, there exists a sequence $G_{1},G_{2},\dots,G_{k}$ of simple $R$-bricks such that:

1. (i)

$G_{1}\in\mathcal{N}$, 2. (ii)

$G_{k}:=G$, and 3. (iii)

for $2\leq i\leq k$, there exists an $R$-thin edge $e_{i}$ of $G_{i}$ such that $G_{i-1}$ is the retract of $G_{i}-e_{i}$.

In other words, every simple $R$-brick can be generated from some member of $\mathcal{N}$ by repeated application of the expansion operations such that at each step we have a simple $R$-brick.

Finally, recall that members of three of the aforementioned families do have strictly $R^{\prime}$-thin edges, where $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$ in our description of these families; these are pseudo-biwheels, double biwheels of type II and laddered biwheels of type II. In view of this, we say that a strictly thin edge $e$ of a simple near-bipartite brick $G$ is compatible if it is $R$-compatible for some removable doubleton $R$. We thus have the following theorem (with eight infinite families) alluded to in the abstract.

Theorem 1.10

Let $G$ be a simple near-bipartite brick. If $G$ is free of compatible strictly thin edges then $G$ belongs to one of the following infinite families:

1. (i)

Truncated biwheels 2. (ii)

Prisms 3. (iii)

Möbius ladders 4. (iv)

Staircases 5. (v)

Double biwheels of type I 6. (vi)

Double ladders of type I 7. (vii)

Laddered biwheels of type I 8. (viii)

Double ladders of type II

Four of the families in the above theorem are Norine-Thomas families; these are free of strictly thin edges. As we did in Figure 4, it may be verified that if $G$ is a member of any of the remaining four families and $e$ is any strictly thin edge of $G$ then the retract $J$ of $G-e$ is not near-bipartite. (For example, consider the graph $G$ and edge $e$ shown in Figure 9, and let $J$ be the retract of $G-e$. It can be checked that $J$ has four odd cycles, $C_{0},C_{1},C_{2}$ and $C_{3}$, such that $C_{1},C_{2}$ and $C_{3}$ are edge-disjoint with $C_{0}$, and furthermore, there is no single edge which belongs to all three of them.)

For the rest of this paper, our goal is to present a complete proof of Theorem 1.8. This result and its proof also appear in the Ph.D. thesis of the first author [7, Chapter 6].

Organization of this paper:

In Section 2, we define two special types of subgraphs, namely, an ‘$R$-biwheel configuration’ and an ‘$R$-ladder configuration’, we state a few theorems related to these configurations without proofs, and we conclude with a proof sketch of Theorem 1.8. In Sections 3 and 4, we prove the technical results and theorems that are stated in Section 2. Finally, in Section 5, we provide a complete proof of Theorem 1.8.

2 $R$-configurations

Recall the definitions of ladders and partial biwheels from Section 1.1. In our descriptions of the eleven families that appear in the statement of Theorem 1.8, we constructed their members using either one or two disjoint bipartite matching covered graphs, each of which is either a ladder or a partial biwheel, and thereafter, adding a few vertices and/or edges and possibly identifying two pairs of vertices. As we will see, these constructions are indicative of how these graphs appear in our proof of Theorem 1.8. In this section, we will define two special types of subgraphs, namely, an ‘$R$-biwheel configuration’ and an ‘$R$-ladder configuration’; we will conclude with a proof sketch of Theorem 1.8.

For the rest of this paper, we adopt the following notational and figure conventions.

Notation 2.1

For a simple $R$-brick $G$, we shall denote by $H[A,B]$ the underlying bipartite graph $G-R$. We let $\alpha$ and $\beta$ denote the constituent edges of $R$, and we adopt the convention that $\alpha:=a_{1}a_{2}$ has both ends in $A$, whereas $\beta:=b_{1}b_{2}$ has both ends in $B$. We denote by $V(R)$ the set $\{a_{1},a_{2},b_{1},b_{2}\}$. Furthermore, in all of the figures, the hollow vertices are in $A$, and the solid vertices are in $B$.

We will also adopt the following notational conventions for a subgraph which is either a ladder or a partial biwheel.

Notation 2.2

When referring to a subgraph $K$ of $H$, such that $K$ is either a ladder or a partial biwheel with external rungs/spokes $au$ and $bw$, we adopt the convention that $a,w\in A$ and $b,u\in B$; furthermore, when $K$ is a partial biwheel, $u$ and $w$ shall denote its hubs; as shown in Figures 10 and 12. (We may also use subscript notation, such as $a_{i}u_{i}$ and $b_{i}w_{i}$ where $i$ is an integer, and this convention extends naturally.)

2.1 $R$-biwheel configurations

Let $K$ be a subgraph of $H$ such that $K$ is a partial biwheel with external spokes $au$ and $bw$; see Figure 10. We say that $K$ is an $R$-biwheel* configuration* of $G$ if it satisfies the following conditions:

1. (i)

in $G$, the hubs $u$ and $w$ are both noncubic, and every other vertex of $K$ is cubic, 2. (ii)

the ends of $K$, namely $a$ and $b$, both lie in $V(R)$, and, 3. (iii)

in $G$, every internal spoke of $K$ is an $R$-thin edge whose index is one.

A pseudo-biwheel, as shown in Figure 11, has two removable doubletons $R:=\{\alpha,\beta\}$ and $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$. The subgraph $K$, depicted by solid lines, is an $R$-biwheel configuration. (To see this, note that every internal spoke of $K$ is an $R$-thin edge of index one.) However, $K$ is not an $R^{\prime}$-biwheel configuration because its ends $a$ and $b$ are not incident with edges of $R^{\prime}$.

2.2 $R$-ladder configurations

Let $K$ be a subgraph of $H$ such that $K$ is a ladder with external rungs $au$ and $bw$; see Figure 12. We say that $K$ is an $R$-ladder* configuration* of $G$ if it satisfies the following conditions:

1. (i)

in $G$, every vertex of $K$, except possibly for $u$ and $w$, is cubic, 2. (ii)

the vertices $a$ and $b$ both lie in $V(R)$, and, 3. (iii)

in $G$, every internal rung of $K$ is an $R$-thin edge whose index is two.

A prism of order $n$ has $\frac{n}{2}$ removable doubletons. If $R:=\{\alpha,\beta\}$ is a fixed removable doubleton of a prism $G$ of order ten or more, then the graph $H=G-R$ is itself an $R$-ladder configuration, as shown in Figure 13. (An analogous statement holds for Möbius ladders of order eight or more.)

2.3 Corners, Rungs and Spokes

We shall often need the flexibility of referring to a subgraph $K$ which is either an $R$-ladder configuration or an $R$-biwheel configuration, and in this case, we simply write that $K$ is an $R$-configuration. Additionally, we may also state that $K$ has external rungs/spokes $au$ and $bw$ (possibly with subscript notation); in this case, we implicitly adopt the conventions stated in Notation 2.2, and we refer to $a,u,b$ and $w$ as the corners of $K$. Furthermore, as shown in Figures 10 and 12, we will assume that $a,b\in V(R)$. We refer to $u$ and $w$ as the free corners of $K$; these may lie in $V(R)$ as in Figure 13, or they may not lie in $V(R)$ as in Figure 11. Observe that any vertex of $K$, which is not a corner, does not lie in $V(R)$.

For any two distinct rungs/spokes of an $R$-configuration $K$, say $e$ and $f$, we say that $e$ and $f$ are consecutive, or equivalently, that $e$ is consecutive with $f$, whenever an end of $e$ which is not a free corner is adjacent with an end of $f$ which is also not a free corner. Clearly, each internal rung (spoke) is consecutive with two rungs (spokes); whereas each external rung (spoke) is consecutive with only one rung (spoke) and the latter is internal. Now, let $e$ denote an internal rung (spoke) of $K$, and let $f$ and $g$ denote the two rungs (spokes) with which $e$ is consecutive. By definition, $e$ is an $R$-thin edge of $G$. Observe that $f$ and $g$ are multiple edges in the retract of $G-e$; consequently, $e$ is not strictly thin.

2.4 Two distinct $R$-configurations

A laddered biwheel of type II, as shown in Figure 14, has two removable doubletons $R:=\{\alpha,\beta\}$ and $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$. Observe that the graph obtained by removing the edge set $R\cup R^{\prime}$ has two connected components, of which one is an $R$-ladder configuration with external rungs $a_{1}u_{1}$ and $b_{1}w_{1}$, and the other is an $R$-biwheel configuration with external spokes $a_{2}u_{2}$ and $b_{2}w_{2}$. In this case, the two $R$-configurations are vertex-disjoint.

On the other hand, a double ladder of type I, as shown in Figure 15, has only one removable doubleton $R:=\{\alpha,\beta\}$ and it has two $R$-ladder configurations which share their free corners $u_{1}$ and $w_{1}$, but are otherwise vertex-disjoint. One of these is depicted by dashed lines, and it has external rungs $a_{1}u_{1}$ and $b_{1}w_{1}$, whereas the other one has external rungs $a_{2}u_{1}$ and $b_{2}w_{1}$.

The reader is advised to check that members of all of the eleven families that appear in Theorem 1.8, except for $K_{4}$ and $\overline{C_{6}}$, have either one or two $R$-configurations for an appropriately chosen removable doubleton $R$. (The choice of $R$ matters only in the case of three families, namely, pseudo-biwheels, double biwheels of Type II and laddered biwheels of Type II. Figure 11 shows a pseudo-biwheel and its two removable doubletons.)

In order to sketch a proof of Theorem 1.8, we will require a few results which are stated next; their proofs will appear in later sections. In particular, the following proposition states that two distinct $R$-configurations are either vertex-disjoint, or they have the same free corners but are otherwise vertex-disjoint; its proof appears in Section 4.1.

Proposition 2.3

[$R$-configurations are Almost Disjoint]*

Let $G$ be a simple $R$-brick, and let $K_{1}$ denote an $R$-configuration with free corners $u_{1}$ and $w_{1}$. If $K_{2}$ is any $R$-configuration distinct from $K_{1}$, then precisely one of the following statements holds:*

1. (i)

$K_{1}$* and $K_{2}$ are vertex-disjoint, or,* 2. (ii)

$u_{1}$* and $w_{1}$ are the free corners of $K_{2}$, and $K_{2}$ is otherwise vertex-disjoint with $K_{1}$.*

By the above proposition, the only vertices that can be possibly shared between two distinct $R$-configurations are their respective free corners. The remaining two corners of each $R$-configuration lie in $V(R)$. Since $|V(R)|=4$, we immediately have the following consequence.

Corollary 2.4

A simple $R$-brick has at most two distinct $R$-configurations. $\Box$

For instance, if $G$ is a Norine-Thomas brick or if it is a pseudo-biwheel then it has only one $R$-configuration. On the other hand, if $G$ is a double biwheel or a double ladder or a laddered biwheel, then it has two $R$-configurations, say $K_{1}$ and $K_{2}$. Furthermore, if $G$ is of type II then $K_{1}$ and $K_{2}$ are vertex-disjoint as in Proposition 2.3(i); whereas, if $G$ is of type I then $K_{1}$ and $K_{2}$ have the same free corners but they do not have any other vertices in common as in Proposition 2.3(ii).

2.5 The $R$-biwheel and $R$-ladder Theorems

It is easily verified that if $G$ is any $R$-brick in $\mathcal{N}$, then every $R$-thin edge of $G$ lies in an $R$-configuration. Here, we state two theorems which show that this is not a coincidence.

Now, let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges. Given any $R$-thin edge $e$ of $G$, we may invoke one of these theorems (depending on the index of $e$) to find an $R$-configuration $K$ containing the edge $e$. In particular, if the index of $e$ is one, we apply Theorem 2.5 and in this case $K$ is an $R$-biwheel configuration; whereas, if the index of $e$ is two, we apply Theorem 2.6 and in this case $K$ is an $R$-ladder configuration.

Theorem 2.5

[$R$-biwheel Theorem]*

Let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges, and let $e$ denote an $R$-thin edge whose index is one. Then $G$ contains an $R$-biwheel configuration, say $K$, such that $e$ is an internal spoke of $K$.*

The proof of the above theorem appears in Section 3.2, and it is along the same lines as the proof of [4, Theorem 4.6].

Given the statement of Theorem 2.5, one would expect that, likewise, if $e$ is an $R$-thin edge whose index is two then $G$ contains an $R$-ladder configuration, say $K$, such that $e$ is an internal rung of $K$. Unfortunately, this is not true, in general. Consider the double ladder of type I, shown in Figure 15; $e$ is an $R$-thin edge of index two, and although it is part of an $R$-ladder configuration, it is not a rung of that ladder. We instead prove the following slightly weaker statement concerning $R$-thin edges of index two.

Theorem 2.6

[$R$-ladder Theorem]*

Let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges, and let $e$ denote an $R$-thin edge whose index is two. Then $G$ contains an $R$-ladder configuration, say $K$, such that $e\in E(K)$.*

The proof of the above theorem appears in Section 3.3 and it is significantly longer than that of the $R$-biwheel Theorem (2.5). These two theorems (2.5 and 2.6) are central to our proof of the Strictly $R$-thin Edge Theorem (1.8).

2.6 Proof Sketch of Theorem 1.8

As in the statement of the theorem, let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges. Our goal is to show that $G$ is a member of one of the eleven infinite families which appear in the statement of the theorem, that is, to show that $G\in\mathcal{N}$. We adopt Notation 2.1.

We may assume that $G$ is different from $K_{4}$ and $\overline{C_{6}}$, and thus, by the $R$-thin Edge Theorem (1.6), $G$ has an $R$-thin edge, say $e_{1}$. Depending on the index of $e_{1}$, we invoke either the $R$-biwheel Theorem (2.5) or the $R$-ladder Theorem (2.6) to deduce that $G$ has an $R$-configuration, say $K_{1}$, such that $e_{1}\in E(K_{1})$. We shall let $a_{1}u_{1}$ and $b_{1}w_{1}$ denote the external rungs/spokes of $K_{1}$, and adjust notation so that $u_{1}$ and $w_{1}$ are its free corners.

We will show that either $u_{1}$ and $w_{1}$ both lie in $V(R)$, or otherwise neither of them lies in $V(R)$. In the former case, we will conclude that $G$ is either a prism or a Möbius ladder or a truncated biwheel, and we are done.

Now suppose that $u_{1},w_{1}\notin V(R)$. In this case, we will show that either $G$ is a staircase or a pseudo-biwheel, and we are done; or otherwise, $G$ has an $R$-compatible edge which is not in $E(K_{1})$. In the latter case, we will apply Theorem 1.7 to deduce that $G$ has an $R$-thin edge, say $e_{2}$, which is not in $E(K_{1})$. Depending on the index of $e_{2}$, we may once again use either the $R$-biwheel Theorem (2.5) or the $R$-ladder Theorem (2.6) to conclude that $G$ has an $R$-configuration, say $K_{2}$, such that $e_{2}\in E(K_{2})$.

By Proposition 2.3, either $K_{1}$ and $K_{2}$ are vertex-disjoint, or otherwise $K_{2}$ has the same free corners as $K_{1}$ but is otherwise vertex-disjoint with $K_{1}$. In the latter case, we will conclude that $G$ is either a double biwheel or a double ladder or a laddered biwheel, each of type I, and we are done.

Now suppose that $K_{1}$ and $K_{2}$ are vertex-disjoint. We will argue that either $G$ is a double biwheel or a double ladder or a laddered biwheel, each of type II, and we are done; or otherwise, $G$ has an $R$-compatible edge which is not in $E(K_{1}\cup K_{2})$. In the latter case, we will once again apply Theorem 1.7 to conclude that $G$ has an $R$-thin edge, say $e_{3}$, which is not in $E(K_{1}\cup K_{2})$. As usual, depending on the index of $e_{3}$, we invoke either the $R$-biwheel Theorem (2.5) or the $R$-ladder Theorem (2.6) to deduce that $G$ has an $R$-configuration, say $K_{3}$, such that $e_{3}\in E(K_{3})$.

We have thus located three distinct $R$-configurations in the brick $G$, namely, $K_{1},K_{2}$ and $K_{3}$. However, this contradicts Corollary 2.4, and completes the proof sketch of the Strictly $R$-thin Edge Theorem (1.8).

3 $R$-thin edges

Here, we will prove the $R$-biwheel Theorem (2.5) and the $R$-ladder Theorem (2.6). Our proofs are inspired by the work of Carvalho et al. [4]. In the next section, we will review conditions under which an $R$-thin edge is not strictly thin, and we will state a few key lemmas (3.8, 3.9 and 3.10) from [4] which are used in our proofs. Before that, we state a few preliminary facts; their proofs may be found in [7, Chapter 4].

The removable edges of a bipartite graph satisfy the following ‘exchange property’.

Proposition 3.1

Let $H$ denote a bipartite matching covered graph, and let $e$ denote a removable edge of $H$. If $f$ is a removable edge of $H-e$, then:

1. (i)

$f$* is removable in $H$, and* 2. (ii)

$e$* is removable in $H-f$. *$\Box$

A matching covered subgraph $K$ of a matching covered graph $H$ is conformal if the graph $H-V(K)$ has a perfect matching; equivalently; $K$ is conformal if each perfect matching of $K$ extends to a perfect matching of $H$. The following is a generalization of Proposition 3.1 that is easily proved using the theory of ear decompositions (see [7]).

Proposition 3.2

Let $K$ be a conformal matching covered subgraph of a bipartite matching covered graph $H$. Let $e$ denote a removable edge of $K$. Then $e$ is removable in $H$ as well. $\Box$

The following is a useful characterization of non-removable edges in bipartite graphs.

Proposition 3.3

Let $H[A,B]$ denote a bipartite matching covered graph on four or more vertices. An edge $e$ of $H$ is non-removable if and only if there exist partitions $(A_{0},A_{1})$ of $A$ and $(B_{0},B_{1})$ of $B$ such that $|A_{0}|=|B_{0}|$ and $e$ is the only edge joining a vertex in $B_{0}$ to a vertex in $A_{1}$. $\Box$

This fact yields the following corollary.

Corollary 3.4

Suppose that $Q$ is a $4$-cycle of a bipartite matching covered graph $H$, and let $e$ and $f$ denote two nonadjacent edges of $Q$. If $f$ is admissible in $H-e$ then $e$ is removable in $H$. $\Box$

3.1 Multiple Edges in Retracts

Throughout this section, $G$ is a simple $R$-brick, and we adopt Notation 2.1. Furthermore, we shall let $e$ denote an $R$-thin edge which is not strictly thin, and $J$ the retract of $G-e$. Since $e$ is not strictly thin, $J$ is not simple, and we shall let $f$ and $g$ denote two multiple (parallel) edges of $J$. It should be noted that since $J$ is also an $R$-brick, neither edge of $R$ is a multiple edge of $J$. In particular, $f$ and $g$ do not lie in $R$.

We denote the ends of $e$ by letters $y$ and $z$ with subscripts $1$; that is, $e:=y_{1}z_{1}$. Adjust notation so that $y_{1}\in A$ and $z_{1}\in B$. If either end of $e$ is cubic, then we denote its two neighbours in $G-e$ by subscripts [math] and $2$. For example, if $y_{1}$ is cubic then $N(y_{1})=\{z_{1},y_{0},y_{2}\}$.

As $G$ is simple, it follows that $J$ has a contraction vertex which is incident with both $f$ and $g$. We infer that one end of $e$, say $y_{1}$, is cubic, and that $f$ is incident with $y_{0}$, and $g$ is incident with $y_{2}$. See Figure 16. As noted earlier, $f\notin R$; consequently, $e$ and $f$ are nonadjacent. Likewise, $e$ and $g$ are nonadjacent.

We will consider two separate cases depending on whether the edges $f$ and $g$ are adjacent (in $G$) or not. In the case in which they are adjacent, we shall denote their common end by $w$, as shown in Figure 17a. Now suppose that $f$ and $g$ are nonadjacent. Since they are multiple (parallel) edges of $J$, we infer that both ends of $e$ are cubic, and that $f$ and $g$ join the two contraction vertices of $J$. This proves the following proposition; see Figure 17b.

Proposition 3.5

Suppose that $f$ and $g$ are nonadjacent in $G$. Then the following hold:

1. (i)

each end of $e$ is cubic, 2. (ii)

consequently, the index of $e$ is two, and 3. (iii)

*one of $f$ and $g$ is incident with $z_{0}$ whereas the other one is incident with $z_{2}$. *$\Box$

In view of statement (iii), whenever $f$ and $g$ are nonadjacent, we shall assume without loss of generality that $f:=y_{0}z_{0}$ and $g:=y_{2}z_{2}$, as shown in Figure 17b.

Let us now focus on the case in which $f$ and $g$ are adjacent, as shown in Figure 17a. We remark that, in this case, the index of $e$ is not determined; that is, its index could be either one or two depending on the degree of its end $z_{1}$. Instead, we are able to say something about the degree of $w$.

Proposition 3.6

Suppose that $f$ and $g$ are adjacent in $G$, and let $w$ be their common end. Then $w$ has degree four or more.

Proof: First suppose that $w$ is not a neighbour of $z_{1}$. In this case, $w$ is not affected by the bicontractions in $G-e$. Consequently, $w$ is a vertex of the brick $J$, whence it has at least three distinct neighbours. Since $f$ and $g$ are multiple edges, $w$ has degree four or more.

Now suppose that $w$ is a neighbour of $z_{1}$. Observe that the neighbours of $y_{1}$ are precisely $y_{0},y_{2}$ and $z_{1}$; each of which is adjacent with $w$. See Figure 17a. Note that, if $w$ is cubic, then its neighbourhood is the same as that of $y_{1}$; and in this case, $\{y_{0},y_{2},z_{1}\}$ is a barrier of the brick $G$; this is absurd. Thus $w$ has degree four or more. $\Box$

Note that $f$ and $g$, being multiple edges of $J$, are both $R$-thin in $J$. We shall now examine conditions under which one of them, say $f$, fails to be $R$-thin in $G$. This may be the case for three different reasons; firstly, $f$ is non-removable in the bipartite graph $H=G-R$; secondly, $f$ is non-removable in $G$; and thirdly, $f$ is removable in $G$ but it is not thin.

We begin with the situation in which $f$ is non-removable in $H$. Note that, if an end of $f$ is cubic (in $G$) and if it also lies in $V(R)$, then it has degree two in $H$, rendering $f$ non-removable. We will now argue that the converse also holds.

Lemma 3.7

The edge $f$ is non-removable in $H$ if and only if it has a cubic end which lies in $V(R)$.

Proof: Suppose that $f$ has no cubic end which lies in $V(R)$. Consequently, each end of $f$ has degree two or more in $H-f$. Furthermore, since $e$ and $f$ are nonadjacent, each end of $f$ has degree two or more in $H-e-f$ as well. We will argue $H-e-f$ is matching covered, that is, $f$ is removable in $H-e$. The exchange property (Proposition 3.1) then implies that $f$ is also removable in $H$.

Note that $f$ is a multiple edge of $J-R$, whence $J-R-f$ is matching covered. Note that any graph obtained from a matching covered graph by means of bi-splitting a vertex is also matching covered; see [7, Section 1.5.2]. We will argue that $H-e-f$ may be obtained from $J-R-f$ by means of bi-splitting one or two vertices.

Note that $J$ is obtained from $G-e$ by means of bicontracting one or two vertices (of degree two); likewise, $J-R$ may be obtained from $H-e$ by means of bicontractions. Conversely, $H-e$ may be obtained from $J-R$ by means of bi-splitting one or two vertices; these are the contraction vertices of $J$. As noted earlier, since each end of $f$ has degree two or more in $H-e-f$, we may similarly obtain $H-e-f$ from $J-R-f$ by means of bi-splitting the same vertices. As discussed above, $H-e-f$ is matching covered; consequently, $f$ is removable in $H$. $\Box$

We now turn to the situation in which $f$ is non-removable in $G$. For convenience, we will state two lemmas (3.8 and 3.9), depending on the index of $e$. These appear in the work of Carvalho et al. [4, Lemma 4.2] as a single lemma. (In their work, they deal with the more general context in which $e$ is a thin edge of a brick $G$, which need not be near-bipartite.)

The first lemma (3.8) considers the scenario in which the index of $e$ is one. By Proposition 3.5(ii), $f$ and $g$ are adjacent; and by Proposition 3.6, their common end $w$ is non-cubic.

Lemma 3.8

[4]*

Suppose that the index of $e$ is one. If $f$ is non-removable in $G$ then $f$ has a cubic end which is adjacent with both ends of $e$. (In particular, the cubic end of $f$ lies in $V(R)$.) $\Box$*

As $w$ is non-cubic, $y_{0}$ is the cubic end of $f$, and it is adjacent with $z_{1}$, as shown in Figure 18a. Clearly, the edge joining $y_{0}$ and $z_{1}$ is none other than $\beta$.

The situation in Lemma 3.8 arises in truncated biwheels, as shown in Figure 18. Note that, every perfect matching which contains $e$ also contains $f$, rendering $f$ non-removable.

The second lemma (3.9) deals with the scenario in which the index of $e$ is two, that is, each end of $e$ is cubic.

Lemma 3.9

[4]* Suppose that the index of $e$ is two. If $f$ is non-removable in $G$ then the following hold:*

1. (i)

each end of $f$ is cubic, 2. (ii)

consequently, $f$ and $g$ are nonadjacent, and 3. (iii)

the ends of $f$ have a common neighbour.

(In particular, one of the ends of $f$ is cubic and it also lies in $V(R)$.)* $\Box$*

By statement (i), each end of $f$ is cubic; thus $f$ and $g$ are nonadjacent (see Proposition 3.6). By Proposition 3.5, and as per our notation, $f=y_{0}z_{0}$ and $g=y_{2}z_{2}$, as shown in Figure 19a. By statement (iii), $y_{0}$ and $z_{0}$ have a common neighbour, say $x$. Clearly, one of $xy_{0}$ and $xz_{0}$ is an edge of $R$, depending on whether $x$ lies in $A$ or in $B$; however, these cases are symmetric. Adjust notation so that $x\in B$; thus $xy_{0}$ is the edge $\beta$. Using the fact that $G$ is free of nontrivial barriers, it is easily verified that $x$ is not an end of $g$.

The situation in Lemma 3.9 is observed in staircases, as shown in Figure 19b. The edge $f$ is non-removable since every perfect matching which contains $e$ also contains $f$.

Finally, we turn to the case in which $f$ is removable in $G$ but it is not thin. This is handled by Lemma 3.10 which appears in the work of Carvalho et al. [4, Lemma 4.3].

Lemma 3.10

[4]* If $f$ is removable in $G$ but it is not thin then the following hold:*

1. (i)

the index of $e$ is two, 2. (ii)

$f$* and $g$ are adjacent and their common end $w$ is not adjacent with any end of $e$,* 3. (iii)

$g$* is a thin edge, and* 4. (iv)

$N(y_{0})\subseteq N(z_{1})\cup\{w\}$*; recall that $y_{0}$ is the other end of $f$, and $z_{1}$ is the end of $e$ not adjacent with $y_{0}$. *$\Box$

The lemma concludes that the index of $e$ is two; that is, its end $z_{1}$ is cubic, and as per our notation, the neighbours of $z_{1}$ are precisely $y_{1},z_{0}$ and $z_{2}$. Furthermore, it concludes that $f$ and $g$ are adjacent and that their common end $w$ is distinct from each of $z_{0}$ and $z_{2}$, as shown in Figure 20a. Another consequence which may be inferred from their proof is that all of the neighbours of $y_{0}$ lie in the set $N(z_{1})\cup\{w\}=\{w,y_{1},z_{0},z_{2}\}$. (This is not stated explicitly in the statement of [4, Lemma 4.3].) Since $y_{0}$ has degree at least three, we may adjust notation so that $y_{0}$ is adjacent with $z_{0}$, and it may or may not be adjacent with $z_{2}$.

The situation in Lemma 3.10 is best illustrated by a double ladder of type I in which at least one of the two $R$-ladder configurations is of order eight, as shown in Figure 20b. The edge $e$ is $R$-thin; deleting it and taking the retract yields the staircase $St_{10}$ with multiple edges, two of which are $f$ and $g$. It may be verified that both $f$ and $g$ are removable, but of them only $g$ is thin.

3.2 Proof of the $R$-biwheel Theorem

In this section, we prove the $R$-biwheel Theorem (2.5); our proof is along the same lines as that of [4, Theorem 4.6]. Before that, we need one more lemma pertaining to the structure of $R$-thin edges of index one (in an $R$-brick which is free of strictly $R$-thin edges).

Lemma 3.11

Let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges, $e$ an $R$-thin edge whose index is one, and $y_{1}$ the cubic end of $e$. Let $y_{0}$ and $y_{2}$ denote the neighbours of $y_{1}$ in $G-e$. Then $y_{0}$ and $y_{2}$ are both cubic, and they have a common neighbour $w$ which is non-cubic. Let $f:=wy_{0}$ and $g:=wy_{2}$. Furthermore, the following statements hold:

1. (i)

if $f$ is not $R$-compatible then $y_{0}\in V(R)$, and 2. (ii)

if $f$ is $R$-compatible then it is $R$-thin and its index is one.

(Similar statements also apply to $g$.)

Proof: Let $J$ denote the retract of $G-e$, that is, $J$ is obtained from $G-e$ by bicontracting the vertex $y_{1}$. By hypothesis, $e$ is not strictly thin, whence $J$ has multiple edges. This implies that $G$ has a vertex $w$, distinct from $y_{1}$, that is adjacent to both $y_{0}$ and $y_{2}$, as shown in Figure 17a. As in the statement of the lemma, let $f:=wy_{0}$ and $g:=wy_{2}$. By Proposition 3.6, $w$ has degree four or more.

First consider the case in which $f$ is not $R$-compatible. That is, either $f$ is not removable in $H$ or it is not removable in $G$, and it follows from Lemma 3.7 or from Lemma 3.8, respectively, that the end $y_{0}$ of $f$ is cubic and it lies in $V(R)$.

Now consider the case in which $f$ is $R$-compatible. Since the index of $e$ is one, Lemma 3.10 implies that $f$ is thin, whence it is $R$-thin. By hypothesis, $f$ is not strictly $R$-thin. Consequently, the end $y_{0}$ of $f$ is cubic, and the index of $f$ is one. Applying a similar argument to the edge $g$, we may conclude that $y_{2}$ is also cubic. $\Box$

Proof of the $R$-biwheel Theorem (2.5): As in the statement of the theorem, let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges, and let $e$ denote an $R$-thin edge whose index is one. Our goal is to show that $G$ has an $R$-biwheel configuration of which $e$ is an internal spoke.

As in the statement of Lemma 3.11, we let $y_{1}$ denote the cubic end of $e$, and $y_{0}$ and $y_{2}$ the neighbours of $y_{1}$ in $G-e$. By the lemma, $y_{0}$ and $y_{2}$ are both cubic, and they have a common neighbour $w$ which is non-cubic. We denote by $u$ the non-cubic end of $e$, as shown in Figure 21. Observe that $y_{0}y_{1}y_{2}$ is a path in $H-\{u,w\}$.

We let $P:=v_{1}v_{2}\dots v_{j}$, where $j\geq 3$, be a path of maximum length in the graph $H-\{u,w\}$ that has the following properties (see Figure 22):

1. (i)

$y_{1}$ is an internal vertex of $P$, 2. (ii)

every vertex of $P$ is cubic in $G$; furthermore, if it lies in $A$ then it is adjacent with $u$, and if it lies in $B$ then it is adjacent with $w$, and 3. (iii)

for every internal vertex $v_{i}$ of $P$, the edge that joins $v_{i}$ to one of $u$ and $w$ is $R$-thin of index one.

(Note that the path $y_{0}y_{1}y_{2}$ shown in Figure 21 satisfies all of the above properties; thus such a path $P$ exists.)

We adjust notation so that $v_{1}$ lies in $B$ as shown in Figure 22. It should be noted that the other end of $P$, namely $v_{j}$, may lie in $A$ or in $B$, depending on whether $P$ is an odd path or even. We shall let $K$ denote the subgraph of $H$, which has vertex set $V(P)\cup\{u,w\}$ and edge set $E(P)\cup\{v_{i}w:1\leq i\leq j{\rm\leavevmode\nobreak\ and\leavevmode\nobreak\ }i{\rm\leavevmode\nobreak\ odd}\}\cup\{v_{i}u:1\leq i\leq j{\rm\leavevmode\nobreak\ and\leavevmode\nobreak\ }i{\rm\leavevmode\nobreak\ even}\}$.

Our goal is to show that $K$ is an $R$-biwheel configuration. To this end, we need to establish two additional properties of the path $P$: first, that it is an odd path; and second, that both its ends $v_{1}$ and $v_{j}$ lie in $V(R)$.

We begin by arguing that the two ends of $P$ are nonadjacent (in $G$). Suppose not, that is, say $v_{1}v_{j}$ is an edge of $G$. Since each vertex of $P$ is cubic, it follows that $V(G)=V(K)$; since otherwise $\{u,w\}$ is a $2$-vertex-cut of $G$, and we have a contradiction. Since $G$ has an even number of vertices, $P$ is of odd length. Furthermore, either $G$ is the same as $K$, or otherwise, $G$ has an additional edge joining $u$ and $w$. In both cases, the graph $G$ is bipartite; this is absurd. Thus $v_{1}$ and $v_{j}$ are nonadjacent.

Now, let $f$ denote the edge $v_{1}w$. We will argue that $f$ is not $R$-compatible, and then use this fact to deduce that $v_{1}\in V(R)$. Suppose instead that $f$ is $R$-compatible. Applying Lemma 3.11(ii), with $v_{2}u$ playing the role of $e$, we conclude that $f$ is $R$-thin and its index is one. Let $v_{0}$ denote the neighbour of $v_{1}$ which is distinct from $v_{2}$ and $w$; note that $v_{0}\in A$. By the preceding paragraph, $v_{0}$ is distinct from $v_{j}$, and since each vertex of $P$ is cubic, $v_{0}$ is not in $V(P)$. Applying Lemma 3.11 again, this time with $f$ playing the role of $e$, we deduce that $v_{0}$ is cubic. Furthermore, $v_{0}$ and $v_{2}$ have a common neighbour whose degree is four or more; thus $v_{0}$ is adjacent with $u$. Observe that the path $v_{0}v_{1}P$ contradicts the maximality of $P$. We conclude that $f=v_{1}w$ is not $R$-compatible. By Lemma 3.11(i), the cubic end $v_{1}$ of $f$ lies in $V(R)$.

A similar argument shows that $v_{j}$ lies in $V(R)$. Since $v_{1}$ and $v_{j}$ are nonadjacent, one of them lies in $A$ and the other one lies in $B$. (As per our notation, $v_{1}\in B$ and $v_{j}\in A$.) In particular, $P$ is an odd path, and thus $K$ is an $R$-biwheel configuration. Observe that by property (i) of the path $P$, the end $y_{1}$ of $e$ is an internal vertex of $P$, whence $e$ is an internal spoke of $K$, as desired. This completes the proof of Theorem 2.5. $\Box$

3.3 Proof of the $R$-ladder Theorem

Here, we prove the $R$-ladder Theorem (2.6); its proof is significantly longer than that of the $R$-biwheel Theorem. In its proof, we will need two lemmas (3.12 and 3.13), each of which pertains to the structure of $R$-thin edges of index two (in an $R$-brick which is free of strictly $R$-thin edges); these lemmas correspond to two cases that appear in the proof of Theorem 2.6.

Lemma 3.12

Let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges and $e:=y_{1}z_{1}$ an $R$-thin edge whose index is two. Let $y_{0}$ and $y_{2}$ denote the neighbours of $y_{1}$ which are distinct from $z_{1}$, and let $z_{0}$ and $z_{2}$ denote the neighbours of $z_{1}$ which are distinct from $y_{1}$. Suppose that $y_{1}$ is the only common neighbour of $y_{0}$ and $y_{2}$, and that $z_{1}$ is the only common neighbour of $z_{0}$ and $z_{2}$. Then there are precisely two (nonadjacent) edges, say $f$ and $g$, between $\{y_{0},y_{2}\}$ and $\{z_{0},z_{2}\}$. Adjust notation so that $f:=y_{0}z_{0}$ and $g:=y_{2}z_{2}$. Furthermore, the following statements hold:

1. (i)

if $f$ is not $R$-compatible then an end of $f$ is cubic and it lies in $V(R)$, and 2. (ii)

if $f$ is $R$-compatible then it is $R$-thin and its index is two.

(Similar statements also apply to $g$.)

Proof: Let $J$ denote the retract of $G-e$, that is, $J$ is obtained from $G-e$ by bicontracting vertices $y_{1}$ and $z_{1}$. By hypothesis, $e$ is not strictly thin, whence $J$ has multiple edges. Also, as stated in the assumptions, $y_{1}$ is the only common neighbour of $y_{0}$ and $y_{2}$, and likewise, $z_{1}$ is the only common neighbour of $z_{0}$ and $z_{2}$. It follows that there are precisely two nonadjacent edges between $\{y_{0},y_{2}\}$ and $\{z_{0},z_{2}\}$, as shown in Figure 17b. As in the statement, adjust notation so that $f:=y_{0}z_{0}$ and $g:=y_{2}z_{2}$.

First consider the case in which $f$ is not $R$-compatible. That is, either $f$ is not removable in $H$ or it is not removable in $G$, and it follows from Lemma 3.7 or from Lemma 3.9, respectively, that an end of $f$ is cubic and it lies in $V(R)$.

Now consider the case in which $f$ is $R$-compatible. Since $f$ and $g$ are nonadjacent, Lemma 3.10 implies that $f$ is thin, whence it is $R$-thin. It remains to argue that the index of $f$ is two. Suppose to the contrary that an end of $f$, say $z_{0}$, is non-cubic. By hypothesis, $f$ is not strictly $R$-thin, whence its other end $y_{0}$ is cubic. Using the fact that $y_{1}$ is the only common neighbour of $y_{0}$ and $y_{2}$, it is easily verified that the retract of $G-f$ has no multiple edges, that is, $f$ is strictly $R$-thin; this contradicts the hypothesis. Thus, each end of $f$ is cubic, whence the index of $f$ is two. $\Box$

Lemma 3.13

Let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges and $e:=y_{1}z_{1}$ an $R$-thin edge whose index is two. Let $y_{0}$ and $y_{2}$ denote the neighbours of $y_{1}$ which are distinct from $z_{1}$, and let $z_{0}$ and $z_{2}$ denote the neighbours of $z_{1}$ which are distinct from $y_{1}$. Suppose that $y_{0}$ and $y_{2}$ have a common neighbour $w$ which is distinct from $y_{1}$. Let $f:=y_{0}w$ and $g:=y_{2}w$. Then $w$ is non-cubic and is distinct from each of $z_{0}$ and $z_{2}$. Furthermore, $f$ and $g$ are both removable, $y_{0}$ and $y_{2}$ are both cubic, and the following statements hold:

1. (i)

one of $f$ and $g$ is $R$-compatible; adjust notation so that $f$ is $R$-compatible; 2. (ii)

$f$* is not thin, and its cubic end $y_{0}$ is adjacent with (exactly) one of $z_{0}$ and $z_{2}$; and,* 3. (iii)

$g$* is thin but it is not $R$-compatible, and its cubic end $y_{2}$ lies in $V(R)$.*

Proof: Note that $f$ and $g$ are multiple edges in the retract $J$ of $G-e$. Since $f$ and $g$ are adjacent, by Proposition 3.6, their common end $w$ is non-cubic. Consequently, by Lemma 3.9, $f$ and $g$ are both removable. Note that $y_{0}$ and $y_{2}$ are nonadjacent, since otherwise $e$ is non-removable. In particular, at least one of $y_{0}$ and $y_{2}$ does not lie in $V(R)$. By Lemma 3.7, at least one of $f$ and $g$ is $R$-compatible.

We now argue that $w$ is distinct from each of $z_{0}$ and $z_{2}$. Suppose not, and assume without loss of generality that $w=z_{0}$. By Lemma 3.10(ii), $f$ and $g$ are both thin; in particular, at least one of them is $R$-thin. Adjust notation so that $f$ is $R$-thin. By hypothesis, $f$ is not strictly $R$-thin, whence the retract of $G-f$ has multiple edges; consequently, the end $y_{0}$ of $f$ is cubic. Let $v$ denote the neighbour of $y_{0}$ which is distinct from $y_{1}$ and $z_{0}$. Furthermore, as $f$ is not strictly $R$-thin, we infer that $v$ and $y_{1}$ have a common neighbour which is distinct from $y_{0}$; by Proposition 3.6, such a common neighbour is non-cubic. Since $z_{1}$ is cubic, we infer that $y_{2}$ is non-cubic. By Lemma 3.7, $g$ is $R$-compatible. As noted earlier, $g$ is thin; whence $g$ is $R$-thin. Since each end of $g$ is non-cubic, $g$ is strictly $R$-thin, contrary to the hypothesis. Thus $w$ is distinct from each of $z_{0}$ and $z_{2}$; see Figure 23.

Let us review what we have proved so far. We have shown that $y_{0}$ and $y_{2}$ are not both adjacent with $z_{0}$. An analogous argument shows that $y_{0}$ and $y_{2}$ are not both adjacent with $z_{2}$. By symmetry, $z_{0}$ and $z_{2}$ are not both adjacent with $y_{0}$; likewise, $z_{0}$ and $z_{2}$ are not both adjacent with $y_{2}$. In summary, there are at most two edges between $\{y_{0},y_{2}\}$ and $\{z_{0},z_{2}\}$; and if there are precisely two such edges then they are nonadjacent.

Now we argue that $y_{0}$ and $y_{2}$ are both cubic. Suppose instead that $y_{0}$ is non-cubic; then, by Lemma 3.7, $f$ is $R$-compatible. Note that since each end of $f$ is non-cubic, if $f$ is thin then it is strictly $R$-thin, contrary to the hypothesis. So it must be the case that $f$ is not thin. By Lemma 3.10(iv), $N(y_{0})\subseteq N(z_{1})\cup\{w\}=\{z_{0},z_{2},y_{1},w\}$. As $y_{0}$ is non-cubic, it must be adjacent with each of $z_{0}$ and $z_{2}$; however, this contradicts what we have already established in the preceding paragraph. We conclude that $y_{0}$ and $y_{2}$ are both cubic.

As noted earlier, at least one of $f$ and $g$ is $R$-compatible. As in statement (i) of the lemma, adjust notation so that $f$ is $R$-compatible. We will now argue that $f$ is not thin.

Suppose instead that $f$ is thin. Let $v$ denote the neighbour of $y_{0}$ which is distinct from $y_{1}$ and $w$. By hypothesis, $f$ is not strictly $R$-thin, whence $v$ and $y_{1}$ have a common neighbour which is distinct from $y_{0}$; by Proposition 3.6, such a common neighbour is non-cubic. However, this is not possible as each neighbour of $y_{1}$ is cubic. Thus, $f$ is not thin. An analogous argument shows that if $g$ is $R$-compatible then $g$ is not thin.

Since $f$ is removable but it is not thin, by Lemma 3.10(iv), $N(y_{0})\subseteq N(z_{1})\cup\{w\}=\{z_{0},z_{2},y_{1},w\}$. It follows from our previous observation that $y_{0}$ is adjacent with exactly one of $z_{0}$ and $z_{2}$; adjust notation so that $y_{0}$ is adjacent with $z_{0}$. This proves statement (ii).

Also, by Lemma 3.10, one of $f$ and $g$ is thin; as per our notation, $g$ is thin. Consequently, $g$ is not $R$-compatible. By Lemma 3.7, the cubic end $y_{2}$ of $g$ lies in $V(R)$. This proves statement (iii), and we are done. $\Box$

Proof of the $R$-ladder Theorem (2.6): As in the statement of the theorem, let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges, and let $e$ denote an $R$-thin edge whose index is two. We shall let $y_{1}$ and $z_{1}$ denote the ends of $e$, where $y_{1}\in A$ and $z_{1}\in B$. Furthermore, we let $y_{0}$ and $y_{2}$ denote the neighbours of $y_{1}$ which are distinct from $z_{1}$, and likewise, we let $z_{0}$ and $z_{2}$ denote the neighbours of $z_{1}$ which are distinct from $y_{1}$.

Our goal is to show that $G$ has an $R$-ladder configuration which contains the edge $e$. As mentioned earlier, we will consider two separate cases which correspond to the situations in Lemmas 3.12 and 3.13, respectively.

Case 1: $y_{1}$ is the only common neighbour of $y_{0}$ and $y_{2}$, and likewise, $z_{1}$ is the only common neighbour of $z_{0}$ and $z_{2}$.

By Lemma 3.12, there are precisely two nonadjacent edges between $\{y_{0},y_{2}\}$ and $\{z_{0},z_{2}\}$. Adjust notation so that $y_{0}z_{0}$ and $y_{2}z_{2}$ are edges of $G$, as shown in Figure 24. Observe that the graph in the figure is a ladder of which $e$ is an internal rung; furthermore, it is a subgraph of $H$.

We let $K$ be a subgraph of $H$ of maximum order that has the following properties:

1. (i)

$K$ is a ladder and $e$ is an internal rung of $K$, and 2. (ii)

every internal rung of $K$ is an $R$-thin edge whose index is two.

Note that the subgraph $K$ is either an odd ladder or an even ladder; see Figure 25. We shall denote by $au$ and $bw$ the external rungs of $K$ such that $a,w\in A$ and $b,u\in B$, as shown in the figure. It follows from property (ii) of $K$ that each of its vertices, except possibly $a,u,b$ and $w$, is cubic in $G$.

Remark 3.14

Note that, if $|V(K)|=6$ then $K$ is the same as the subgraph of $H$ shown in Figure 24; in particular, $\{u,b\}=\{y_{0},y_{2}\}$, and likewise, $\{w,a\}=\{z_{0},z_{2}\}$; consequently, by our hypothesis, $y_{1}$ is the only common neighbour of $u$ and $b$, and likewise, $z_{1}$ is the only common neighbour of $w$ and $a$.

Our goal is to show that $K$ is an $R$-ladder configuration. To this end, we need to establish that $a$ and $b$ (or likewise, $u$ and $w$) are both cubic in $G$ and they lie in $V(R)$.

Now, let $f$ denote the edge $au$. We will argue that $f$ is not $R$-compatible, and then use this fact to deduce that one of the ends of $f$ is cubic and it lies in $V(R)$. As shown in Figure 25, let $s_{2}$ denote the neighbour of $u$ in $K$ which is distinct from $a$, and likewise, let $t_{2}$ denote the neighbour of $a$ in $K$ which is distinct from $u$.

Suppose instead that $f$ is $R$-compatible. By Lemma 3.12(ii), with $s_{2}t_{2}$ playing the role of $e$, we conclude that $f$ is $R$-thin and its index is two. We shall let $s_{0}$ denote the neighbour of $u$ which is distinct from $s_{2}$ and $a$, and likewise, let $t_{0}$ denote the neighbour of $a$ which is distinct from $t_{2}$ and $u$. Note that $s_{0}\in A$ and $t_{0}\in B$. It is easily seen that if $s_{0}$ is the same as $w$ then $V(K)\cap A$ is a (nontrivial) barrier of $G$; this is absurd as $G$ is a brick. Thus $s_{0}\neq w$, and likewise, $t_{0}\neq b$. It follows that $s_{0},t_{0}\notin V(K)$.

We will use the fact that $f$ is not strictly $R$-thin to deduce that $s_{0}$ and $t_{0}$ are adjacent; this will help us contradict the maximality of $K$. First suppose that $s_{0}$ and $s_{2}$ have a common neighbour $x$ which is distinct from $u$. By Proposition 3.6, $x$ is non-cubic. Observe that, if $|V(K)|\geq 8$ then every neighbour of $s_{2}$ is cubic; and if $|V(K)|=6$ then $b$ is the only neighbour of $s_{2}$ which is possibly non-cubic. We conclude that $|V(K)|=6$ and that $x=b$. Now, $s_{0}$ is a common neighbour of $u$ and $b$; this contradicts the hypothesis (see Remark 3.14). We conclude that $u$ is the only common neighbour of $s_{0}$ and $s_{2}$. An analogous argument shows that $a$ is the only common neighbour of $t_{0}$ and $t_{2}$. It follows that $s_{0}$ and $t_{0}$ are adjacent, as $f$ is not strictly thin. Now, let $K^{\prime}$ denote the subgraph of $H$ obtained from $K$ by adding the vertices $s_{0}$ and $t_{0}$, and the edges $us_{0},s_{0}t_{0}$ and $t_{0}a$; then $K^{\prime}$ contradicts the maximality of $K$.

We thus conclude that $f=au$ is not $R$-compatible. Consequently, by Lemma 3.12(i), with $s_{2}t_{2}$ playing the role of $e$, at least one of $a$ and $u$ is cubic and it also lies in $V(R)$. Adjust notation so that $a$ is cubic and it lies in $V(R)$.

An analogous argument shows that at least one of $b$ and $w$ is cubic and it lies in $V(R)$; we claim that $b$ must satisfy both of these properties. Suppose not; then $w$ is cubic and it lies in $V(R)$; this means that the edge $\alpha$ of $R$ joins the vertices $a$ and $w$. Observe that $\{b,u\}$ is a $2$-vertex cut of $G$; this is absurd as $G$ is a brick.

We have shown that $a$ and $b$ both are cubic and they lie in $V(R)$. Thus $K$ is an $R$-ladder configuration. Observe that, by property (i) of $K$, the edge $e$ is an internal rung of $K$. In particular, $e$ is an edge of $K$, as desired.

Case 2: $y_{0}$ and $y_{2}$ have a common neighbour which is distinct from $y_{1}$, or likewise, $z_{0}$ and $z_{2}$ have a common neighbour which is distinct from $z_{1}$.

As shown in Figure 26, assume without loss of generality that $y_{0}$ and $y_{2}$ have a common neighbour, say $w$, which is distinct from $y_{1}$. We let $f:=y_{0}w$ and $g:=y_{2}w$. We invoke Lemma 3.13 to infer the following: $w$ is non-cubic and it is distinct from each of $z_{0}$ and $z_{2}$; whereas $y_{0}$ and $y_{2}$ are both cubic; $f$ and $g$ are both removable edges. Furthermore, adjusting notation as in the lemma, $f$ is $R$-compatible but it is not thin and its cubic end $y_{0}$ is adjacent with one of $z_{0}$ and $z_{2}$. Assume without loss of generality that $y_{0}$ is adjacent with $z_{0}$. The edge $g$ is thin but it is not $R$-compatible and its cubic end $y_{2}$ lies in $V(R)$. As per our notation, $y_{2}$ is an end of $\beta$; we shall let $x$ denote the other end of $\beta$.

We will consider two subcases. In the first one, we assume that $z_{0}$ is cubic and it lies in $V(R)$; and in the second case, we assume that either $z_{0}$ is non-cubic or it is not in $V(R)$.

Case 2.1: $z_{0}$ is cubic and it lies in $V(R)$.

In this case, we shall denote by $K$ the subgraph whose vertex set is $\{z_{0},z_{1},y_{0},y_{1},w,y_{2}\}$ and edge set is $\{e,y_{1}y_{2},g,f,y_{0}z_{0},z_{0}z_{1},y_{0}y_{1}\}$. Observe that $K$ is a ladder of order six and it is a subgraph of $H$; furthermore, two of its corners, namely $y_{2}$ and $z_{0}$, are cubic and they both lie in $V(R)$. To complete the proof in this case, we will show that $K$ is an $R$-ladder configuration; for this, we only need to prove that the internal rung $y_{0}y_{1}$ is $R$-thin and its index is two.

We begin by showing that $y_{0}y_{1}$ is $R$-compatible, that is, $y_{0}y_{1}$ is removable in $H$ as well as in $G$. Here, we will not require the hypothesis that $z_{0}$ is cubic and it lies in $V(R)$.

Claim 3.15

The edge $y_{0}y_{1}$ is $R$-compatible.

Proof: Note that $y_{0}y_{1}$ is removable in the subgraph $K$. We will argue that $K$ is a conformal subgraph of $H$, and then use Proposition 3.2 to deduce that $y_{0}y_{1}$ is removable in $H$.

Let $M$ be any perfect matching of $H$ which contains the edge $z_{0}z_{1}$. Since $M$ does not contain $\alpha$ or $\beta$, it is easily verified that $M\cap E(K)$ is a perfect matching of $K$, whence $K$ is a conformal subgraph of $H$; consequently, $y_{0}y_{1}$ is removable in $H$.

To conclude that $y_{0}y_{1}$ is removable in $G$, we will show that $G-y_{0}y_{1}$ has a perfect matching $M$ which contains both $\alpha$ and $\beta$. Let $N$ be a perfect matching of $G-\{z_{1},x\}$; such a perfect matching exists as $G$ is a brick; note that $\alpha\in N$ and $\beta\notin N$. Clearly, either $y_{1}y_{2}\in N$ or $g\in N$. If $y_{1}y_{2}\in N$, we let $M:=(N-y_{1}y_{2})+e+\beta$. On the other hand, if $g\in N$ then $y_{0}y_{1}\in N$, and we let $M:=(N-g-y_{0}y_{1})+e+f+\beta$. In either case, $M$ is the desired perfect matching, and this completes the proof. $\Box$

We now proceed to show that $y_{0}y_{1}$ is an $R$-thin edge. To this end, we will use the characterization of $R$-thin edges in terms of barriers given by [6, Proposition 2.9].

Claim 3.16

The edge $y_{0}y_{1}$ is $R$-thin, and its index is two.

Proof: Observe that, since $y_{0}$ and $y_{1}$ are both cubic, $G-y_{0}y_{1}$ has two maximal nontrivial barriers; one of them, say $S_{A}$, is a subset of $A$ and it contains $z_{0}$ and $w$; the other one, say $S_{B}$, is a subset of $B$ and it contains $z_{1}$ and $y_{2}$. In particular, the index of $y_{0}y_{1}$ is two.

We will argue that $S_{A}=\{z_{0},w\}$; our argument does not use the fact that $w$ is non-cubic, and it may be mimicked to show that $S_{B}=\{z_{1},y_{2}\}$; thereafter, we apply [6, Proposition 2.9] to infer that $y_{0}y_{1}$ is $R$-thin.

Note that $w$ is in the barrier $S_{A}$. Now, let $v$ be any vertex in $A-\{z_{0},w\}$. We will show that $(G-y_{0}y_{1})-\{w,v\}$ has a perfect matching $M$; this would imply that $v$ is not in the barrier $S_{A}$. Let $N$ be a perfect matching of $G-\{w,v\}$; note that $\beta\in N$ and $\alpha\notin N$. If $y_{0}y_{1}\notin N$ then let $M:=N$, and we are done. Now suppose that $y_{0}y_{1}\in N$. By our hypothesis, $z_{0}$ is cubic and it lies in $V(R)$; this means that the three edges incident at $z_{0}$ are $z_{0}y_{0},z_{0}z_{1}$ and $\alpha$. Since, $y_{0}y_{1}\in N$ and $\alpha\notin N$ and $v\neq z_{0}$, we conclude that $z_{0}z_{1}\in N$. Now, $M:=(N-y_{0}y_{1}-z_{0}z_{1})+y_{0}z_{0}+e$ is the desired perfect matching. We conclude that $S_{A}=\{z_{0},w\}$. As discussed in the preceding paragraph, this completes the proof. $\Box$

We have shown that the only internal rung of $K$, namely $y_{0}y_{1}$, is an $R$-thin edge whose index is two. As discussed earlier, $K$ is indeed an $R$-ladder configuration, and since it contains $e$, this completes the proof in this case (2.1).

Case 2.2: Either $z_{0}$ is non-cubic or it does not lie in $V(R)$, possibly both.

As per our notation, $z_{0}\in A$; it follows from the hypothesis of this case that $z_{0}$ has at least one neighbour which lies in $B-\{z_{1},y_{0}\}$; we shall let $u$ denote such a neighbour of $z_{0}$, as shown in Figure 27. Observe that $u$ is distinct from $y_{2}$; however, it is possible that $u=x$.

In this case, we will prove that $z_{0}z_{1}$ is an $R$-thin edge whose index is two; in particular, $z_{0}$ is cubic and $z_{0}\notin V(R)$. (If not, we will find a strictly $R$-thin edge contrary to the hypothesis.) Thereafter, we argue that $u$ is adjacent with $z_{2}$; this establishes a certain symmetry between $y_{0},y_{1},y_{2},w$ and $z_{0},z_{1},z_{2},u$, respectively; see Figure 29. We shall exploit this to deduce that $y_{0}y_{1}$ is an $R$-thin edge (whose index is two), and that $z_{2}$ is cubic and it lies in $V(R)$. In the end, we will find an $R$-ladder configuration of order eight whose internal rungs are $y_{0}y_{1}$ and $z_{0}z_{1}$.

Our first step is to show that $z_{0}z_{1}$ is $R$-compatible, that is, $z_{0}z_{1}$ is removable in $H$ as well as in $G$.

Claim 3.17

The edge $z_{0}z_{1}$ is $R$-compatible.

Proof: Note that $y_{0}y_{1}z_{1}z_{0}y_{0}$ is a $4$-cycle containing the edges $y_{0}y_{1}$ and $z_{0}z_{1}$. We will show that $y_{0}y_{1}$ is admissible in $H-z_{0}z_{1}$, and then invoke Corollary 3.4 to deduce that $z_{0}z_{1}$ is removable in $H$.

We need to show that $H-z_{0}z_{1}$ has a perfect matching $M$ which contains $y_{0}y_{1}$. Let $N$ be any perfect matching of $H-\{u,y_{1}\}$; such a perfect matching exists by [6, Proposition 2.1]. Observe that $g\in N$; consequently, $y_{0}z_{0}\in N$. Now, $M:=(N-y_{0}z_{0})+uz_{0}+y_{0}y_{1}$ is the desired perfect matching. As discussed above, $z_{0}z_{1}$ is removable in $H$.

To conclude that $z_{0}z_{1}$ is removable in $G$, we will show that $G-z_{0}z_{1}$ has a perfect matching $M$ which contains both $\alpha$ and $\beta$. Let $N$ be any perfect matching of $G$ which contains $\alpha$ and $\beta$. If $z_{0}z_{1}\notin N$ then let $M:=N$, and we are done. Now suppose that $z_{0}z_{1}\in N$. Observe that $y_{0}y_{1}\in N$; furthermore, $M:=(N-y_{0}y_{1}-z_{0}z_{1})+e+y_{0}z_{0}$ is the desired perfect matching. This completes the proof. $\Box$

We proceed to prove that $z_{0}z_{1}$ is an $R$-thin edge whose index is two. As we did in Claim 3.16, we will use the characterization of $R$-thin edges given by [6, Proposition 2.9]. However, here we need more general arguments since we do not know the degree of $z_{0}$.

Claim 3.18

The edge $z_{0}z_{1}$ is $R$-thin, and its index is two.

Proof: Observe that, since $z_{1}$ is cubic, $G-z_{0}z_{1}$ has a maximal nontrivial barrier, say $S_{A}$, which is a subset of $A$ and contains $y_{1}$ and $z_{2}$. We will first prove that $S_{A}=\{y_{1},z_{2}\}$.

Let $v$ be any vertex in $A-\{y_{1},z_{2}\}$. We will show that $(G-z_{0}z_{1})-\{z_{2},v\}$ has a perfect matching $M$; this would imply that $v$ is not in the barrier $S_{A}$. Let $N$ be a perfect matching of $G-\{z_{2},v\}$; note that $\beta\in N$ and $\alpha\notin N$. If $z_{0}z_{1}\notin N$ then let $M:=N$, and we are done. Now suppose that $z_{0}z_{1}\in N$, and observe that $y_{0}y_{1}\in N$; consequently, $M:=(N-z_{0}z_{1}-y_{0}y_{1})+e+y_{0}z_{0}$ is the desired perfect matching. Thus, $S_{A}=\{y_{1},z_{2}\}$.

Since $z_{0}z_{1}$ is $R$-compatible, by [6, Lemma 2.8], either $S_{A}$ is the only maximal nontrivial barrier of $G-z_{0}z_{1}$, or $G-z_{0}z_{1}$ has another maximal nontrivial barrier, say $S_{B}$, which is a subset of $B$. We now argue that, in the former case, $z_{0}z_{1}$ is strictly $R$-thin, contrary to the hypothesis.

Suppose that $S_{A}$ is the only maximal nontrivial barrier of $G-z_{0}z_{1}$; in this case, the index of $z_{0}z_{1}$ is one. By [6, Proposition 2.9], $z_{0}z_{1}$ is $R$-thin. Also, $z_{0}$ is non-cubic, since otherwise its two neighbours distinct from $z_{1}$ would lie in a barrier. Observe that, since $z_{1}$ is the only common neighbour of $y_{1}$ and $z_{2}$, the retract of $G-z_{0}z_{1}$ is simple, and thus $z_{0}z_{1}$ is strictly $R$-thin; this is a contradiction.

It follows that $G-z_{0}z_{1}$ has a maximal nontrivial barrier, say $S_{B}$, which is a subset of $B$; in particular, the index of $z_{0}z_{1}$ is two. By [6, Lemma 2.8], $z_{0}$ is isolated in $(G-z_{0}z_{1})-S_{B}$; that is, in $G-z_{0}z_{1}$, every neighbour of $z_{0}$ lies in the barrier $S_{B}$. In particular, $u,y_{0}\in S_{B}$. We will prove that $S_{B}=\{u,y_{0}\}$.

Let $v$ be any vertex in $B-\{u,y_{0}\}$. We will show that $(G-z_{0}z_{1})-\{u,v\}$ has a perfect matching $M$; this would imply that $v$ is not in the barrier $S_{B}$. Let $N$ be a perfect matching of $G-\{u,v\}$; note that $\alpha\in N$ and $\beta\notin N$. If $z_{0}z_{1}\notin N$ then let $M:=N$, and we are done. Now suppose that $z_{0}z_{1}\in N$. If $y_{0}y_{1}\in N$ then $M:=(N-z_{0}z_{1}-y_{0}y_{1})+e+y_{0}z_{0}$ is the desired perfect matching. Now suppose that $y_{0}y_{1}\notin N$; then $f,y_{1}y_{2}\in N$, and $M:=(N-z_{0}z_{1}-f-y_{1}y_{2})+y_{0}z_{0}+g+e$ is the desired perfect matching. Thus, as discussed above, $v\notin S_{B}$; consequently, $S_{B}=\{u,y_{0}\}$. In particular, $z_{0}$ is cubic. Furthermore, by [6, Proposition 2.9], $z_{0}z_{1}$ is $R$-thin. $\Box$

We have shown that $z_{0}z_{1}$ is an $R$-thin edge and its index is two; in particular, both its ends are cubic. The three neighbours of $z_{0}$ are $y_{0},z_{1}$ and $u$; see Figure 27.

By hypothesis, $z_{0}z_{1}$ is not strictly $R$-thin; whence the retract of $G-z_{0}z_{1}$ has multiple edges. Observe that $z_{1}$ is the only common neighbour of $y_{1}$ and $z_{2}$. Consequently, at least one of the following must hold: either $u$ and $y_{0}$ have a common neigbour which is distinct from $z_{0}$, or $u$ and $z_{2}$ are adjacent. We shall rule out the former case by arriving at a contradiction.

Suppose that $u$ and $y_{0}$ have a common neighbour which is distinct from $z_{0}$; this is true if and only if $u$ is adjacent with $w$. We now invoke Lemma 3.13, with $z_{0}z_{1}$ playing the role of $e$, with $u$ playing the role of $y_{2}$, and with $uw$ playing the role of $g$; see Figure 28a, and compare with Figure 27. The lemma implies that $u$ is a cubic vertex, and since $f$ is $R$-compatible, $uw$ is thin but it is not $R$-compatible; furthermore, $u\in V(R)$. In particular, $u$ is an end of $\beta$ which implies that $u=x$; see Figures 27 and 28b. Note that all of the labelled vertices in Figure 28b are pairwise distinct; furthermore, each of them except $w$ and possibly $z_{2}$, is cubic. Since $z_{2}$ has at least one neighbour in $B$ which is distinct from $z_{1}$, the graph has more vertices; consequently, $\{w,z_{2}\}$ is a $2$-vertex cut of $G$; this is a contradiction.

We have shown that $z_{0}$ is the only common neighbour of $u$ and $y_{0}$; as discussed earlier, this implies that $u$ and $z_{2}$ are adjacent; see Figure 29. Note that $u$ is now a common neighbour of $z_{0}$ and $z_{2}$, and it is distinct from $z_{1}$; this establishes a symmetry between $y_{0},y_{1},y_{2},w$, and $z_{0},z_{1},z_{2},u$, respectively. We invoke Lemma 3.13 to conclude that $u$ is non-cubic, whereas $z_{2}$ is cubic and it lies in $V(R)$. Using arguments analogous to those in the proofs of Claims 3.17 and 3.18, we conclude that $y_{0}y_{1}$ is an $R$-thin edge (whose index is two).

Now, let $K$ denote the subgraph which consists of all of the labelled vertices shown in Figure 29, and all of the edges between those vertices which are shown in the figure. Note that $K$ is an $R$-ladder configuration, and since it contains $e$, this completes the proof of the $R$-ladder Theorem (2.6). $\Box$

4 Properties of $R$-configurations

In this section, we prove a few results pertaining to $R$-configurations. These are used in our proof of the Strictly $R$-thin Edge Theorem (1.8), which appears in the next section. We will find the following consequence of [6, Lemma 2.3] useful; its proof may be found in [7].

Corollary 4.1

Let $G$ be an $R$-brick, and let $H:=G-R$. Then for any vertex $b$, at most two edges of $\partial_{H}(b)$ are non-removable in $H$.

For the rest of this section, $G$ is a simple $R$-brick, and we adopt Notation 2.1; furthermore, $K_{1}$ is an $R$-configuration with external rungs/spokes $a_{1}u_{1}$ and $b_{1}w_{1}$. As usual, $u_{1}$ and $w_{1}$ are the free corners of $K_{1}$; see Figure 30.

Note that $K_{1}$ is either a ladder or a partial biwheel. In either case, it is easily verified that the graph obtained from $K_{1}$ by adding two edges, one joining $a_{1}$ and $b_{1}$, and another joining $u_{1}$ and $w_{1}$, is a brace. This fact, in conjunction with the characterization of braces provided by [7, Proposition 4.12], yields the following easy observation.

Proposition 4.2

The following statements hold:

1. (i)

for every pair of distinct vertices $v_{1},v_{2}\in A\cap V(K_{1})$, the graph $K_{1}-\{b_{1},u_{1},v_{1},v_{2}\}$ has a perfect matching; and likewise, 2. (ii)

*for every pair of distinct vertices $v_{1},v_{2}\in B\cap V(K_{1})$, the graph $K_{1}-\{a_{1},w_{1},v_{1},v_{2}\}$ has a perfect matching. *$\Box$

In the following lemma, we prove some conformality properties of $R$-configurations; these are useful in subsequent lemmas to show that a certain edge is $R$-compatible.

Lemma 4.3

The following statements hold:

1. (i)

$u_{1}$* lies in $V(R)$ if and only if $w_{1}$ lies in $V(R)$,* 2. (ii)

$K_{1}$* is a conformal matching covered subgraph, and* 3. (iii)

the subgraph induced by $E(K_{1})\cup R$ is conformal.

Proof: First, we prove (i). Suppose instead that $u_{1}\in V(R)$ and $w_{1}\notin V(R)$; that is, $u_{1}=b_{2}$, whereas $w_{1}$ and $a_{2}$ are distinct; see Figure 31. For $X:=V(K_{1})-w_{1}$, note that every edge in $\partial(X)$, except for $\alpha$, is either incident with $u_{1}$ or with $w_{1}$. Recall that if $M$ is any perfect matching, then $\alpha\in M$ if and only if $\beta\in M$. Using these facts, it is easy to see that $\partial(X)$ is a tight cut; this is a contradiction.

Now, we prove (ii). Since $K_{1}$ is either a ladder or a partial biwheel, it is matching covered. To show that $K_{1}$ is conformal, we will display a perfect matching $M$ of $G-V(K_{1})$. Let $N$ be a perfect matching of $H$ which contains $a_{1}u_{1}$; observe that $M:=N-E(K_{1})$ is the desired perfect matching.

Note that, if $u_{1},w_{1}\in V(R)$, then (iii) follows immediately from (ii). Now suppose that $u_{1},w_{1}\notin V(R)$, and let $N$ be a perfect matching of $G-\{a_{2},w_{1}\}$; note that $\beta\in N$. A simple counting argument shows that $M:=N-E(K_{1})-R$ is a perfect matching of $G-V(K_{1})-V(R)$; and this proves (iii). $\Box$

In the following two lemmas, apart from other things, we show that under certain circumstances there exists an $R$-compatible edge which is not in $K_{1}$.

Lemma 4.4

Suppose that $u_{1},w_{1}\notin V(R)$. Then at most one edge of $\partial(u_{1})-E(K_{1})$ is not $R$-compatible. (An analogous statement holds for $w_{1}$.)

Proof: Note that, by Corollary 4.1, at most two edges of $\partial(u_{1})$ are non-removable in $H$; one of these is $a_{1}u_{1}$. Consequently, at most one edge of $\partial(u_{1})-E(K_{1})$ is non-removable in $H$. To complete the proof we will show that if $e$ is any removable edge of $H$ such that $e\in\partial(u_{1})-E(K_{1})$, then $e$ is removable in $G$ as well; for this, it suffices to show a perfect matching $M$ which contains $\alpha$ and $\beta$ but does not contain $e$.

Let $M_{1}$ be a perfect matching of $G-V(K_{1})-V(R)$; such a perfect matching exists by Lemma 4.3(iii). Let $M_{2}$ be a perfect matching of $K_{1}-\{a_{1},b_{1}\}$; since $K_{1}$ is bipartite matching covered, such a perfect matching exists by [6, Proposition 2.1]. Now, $M:=M_{1}\cup M_{2}\cup R$ is the desired perfect matching alluded to above, and this completes the proof. $\Box$

Lemma 4.5

Suppose that $u_{1},w_{1}\notin V(R)$. If $|\partial(u_{1})-E(K_{1})|\leq 1$ and $|\partial(w_{1})-E(K_{1})|\leq 1$ then the following statements hold:

1. (i)

$u_{1}$* and $w_{1}$ are nonadjacent,* 2. (ii)

$\partial(u_{1})-E(K_{1})$* has exactly one member, say $\alpha^{\prime}$, and likewise, $\partial(w_{1})-E(K_{1})$ has exactly one member, say $\beta^{\prime}$,* 3. (iii)

$\alpha$* and $\alpha^{\prime}$ are adjacent if and only if $\beta$ and $\beta^{\prime}$ are adjacent,* 4. (iv)

if $\alpha$ and $\alpha^{\prime}$ are nonadjacent then at most one edge of $\partial(v)-\alpha^{\prime}$ is not $R$-compatible, where $v$ denotes the end of $\alpha^{\prime}$ which is distinct from $u_{1}$; an analogous statement holds for $\beta$ and $\beta^{\prime}$.

Proof: We first verify (i) and (ii). Observe that, if $u_{1}$ and $w_{1}$ are adjacent, or, if the sets $\partial(u_{1})-E(K_{1})$ and $\partial(w_{1})-E(K_{1})$ are both empty, then $\{a_{1},b_{1}\}$ is a $2$-vertex cut of $G$; this is absurd. This proves (i). Note that, if only one of $\partial(u_{1})-E(K_{1})$ and $\partial(w_{1})-E(K_{1})$ is nonempty then $H$ has a cut-edge; this is a contradiction. This proves (ii). As in the statement, let $\alpha^{\prime}$ denote the only member of $\partial(u_{1})-E(K_{1})$; and likewise, let $\beta^{\prime}$ denote the only member of $\partial(w_{1})-E(K_{1})$. See Figure 32.

We now show that (iii) holds. Suppose instead that $\beta$ and $\beta^{\prime}$ are adjacent, whereas $\alpha$ and $\alpha^{\prime}$ are nonadjacent. In particular, $\beta^{\prime}$ has ends $w_{1}$ and $b_{2}$. We let $T:=B-V(K_{1})-b_{2}$, and note that $T$ is nonempty. Furthermore, all of the neigbours of $T$ lie in the set $S:=A-V(K_{1})$; consequently, $S$ is a nontrivial barrier of $G$; this is absurd.

We now proceed to prove (iv). Suppose that $\alpha$ and $\alpha^{\prime}$ are nonadjacent; and as in the statement of the lemma, let $v$ denote the end of $\alpha^{\prime}$ which is distinct from $u_{1}$. By (iii), $\beta$ and $\beta^{\prime}$ are also nonadjacent. We will first argue that at most one edge of $\partial(v)-\alpha^{\prime}$ is non-removable in $H$.

Observe that $\{\alpha^{\prime},\beta^{\prime}\}$ is a $2$-cut of $H$; thus, neither $\alpha^{\prime}$ nor $\beta^{\prime}$ is removable in $H$. By Corollary 4.1, at most two edges of $\partial(v)$ are non-removable in $H$; one of these is $\alpha^{\prime}$. Consequently, at most one edge of $\partial(v)-\alpha^{\prime}$ is non-removable in $H$. To complete the proof we will show that if $e$ is any removable edge of $H$ such that $e\in\partial(v)-\alpha^{\prime}$, then $e$ is removable in $G$ as well; for this, it suffices to show a perfect matching $M$ which contains $\alpha$ and $\beta$ but does not contain $e$.

Let $M_{1}$ be any perfect matching of $G-\{a_{2},v\}$; note that $\beta\in M_{1}$. A simple counting argument shows that $\beta^{\prime}$ lies in $M_{1}$ as well. Now, let $M_{2}$ be a perfect matching of $K_{1}-\{a_{1},u_{1},b_{1},w_{1}\}$; such a perfect matching exists due to Proposition 4.2. Observe that $M:=(M_{1}-E(K_{1}))\cup M_{2}\cup\{\alpha,\alpha^{\prime}\}$ is the desired perfect matching alluded to above. As discussed, this completes the proof. $\Box$

In the previous two lemmas, we have shown that under certain circumstances there exists an $R$-compatible edge which is not in $K_{1}$. However, in the proof of the Strictly $R$-thin Edge Theorem (1.8), we will be interested in finding an $R$-thin edge which is not in $K_{1}$. To do so, we will choose an $R$-compatible edge appropriately, and use Theorem 1.7, in conjunction with the following lemma, to argue that the chosen edge is indeed $R$-thin.

Lemma 4.6

Suppose that $u_{1},w_{1}\notin V(R)$. Let $e$ denote an $R$-compatible edge which does not lie in $E(K_{1})$, let $S$ denote a nontrivial barrier of $G-e$, and $I$ the set of isolated vertices of $(G-e)-S$. Then the following statements hold:

1. (i)

$S\cap V(K_{1})$* contains at most one vertex, and* 2. (ii)

$I\cap V(K_{1})$* is empty.*

Proof: Since $e$ is $R$-compatible, $S$ is a subset of one of the two color classes of $H$; assume without loss of generality that $S\subset A$. To establish (i), we will show that if $v_{1}$ and $v_{2}$ are any two distinct vertices in $V(K_{1})\cap A$, then $(G-e)-\{v_{1},v_{2}\}$ has a perfect matching $M$.

Let $M_{1}$ be a perfect matching of $(H-e)-\{v_{1},b_{2}\}$ where $b_{2}$ is the end of $\beta$ which is not in $V(K_{1})$; such a perfect matching exists by [6, Proposition 2.1] as $H-e$ is matching covered. A simple counting argument shows that $M\cap\partial(V(K_{1}))$ contains only one edge, and this edge is incident with the free corner $u_{1}$. Let $M_{2}$ be a perfect matching of $K_{1}-\{b_{1},u_{1},v_{1},v_{2}\}$; such a perfect matching exists due to Proposition 4.2. Observe that $M:=(M_{1}-E(K_{1}))+M_{2}+\beta$ is the desired perfect matching of $(G-e)-\{v_{1},v_{2}\}$, and this proves (i).

We now deduce (ii) from (i). Suppose to the contrary that $I\cap V(K_{1})$ is nonempty, and let $x$ denote any of its members. Observe that $x$ is adjacent with at least two vertices in $V(K_{1})$, and each of these must lie in $S$; this contradicts (i), and completes the proof. $\Box$

4.1 Proof of Proposition 2.3

As in the statement of the proposition, let $G$ be a simple $R$-brick, and let $K_{1}$ be an $R$-configuration with external rungs/spokes $a_{1}u_{1}$ and $b_{1}w_{1}$, where $u_{1}$ and $w_{1}$ denote the free corners of $K_{1}$; see Figure 30. Suppose that $G$ has an $R$-configuration $K_{2}$ which is distinct from $K_{1}$; that is, $K_{1}$ and $K_{2}$ are not identical subgraphs of $G$. We assume that $K_{1}$ and $K_{2}$ are not vertex-disjoint. Our goal is to deduce that $u_{1}$ and $w_{1}$ are the free corners of $K_{2}$, and that $K_{2}$ is otherwise vertex-disjoint with $K_{1}$.

We first argue that $u_{1},w_{1}\notin V(R)$. Note that every vertex of $K_{1}$, except possibly $u_{1}$ and $w_{1}$, is cubic in $G$. Consequently, if $u_{1},w_{1}\in V(R)$ then $V(G)=V(K_{1})$, since otherwise $\{u_{1},w_{1}\}$ is a $2$-vertex cut of $G$; furthermore, either $G$ is precisely the graph induced by $E(K_{1})\cup R$, or otherwise, $G$ has one additional edge joining $u_{1}$ and $w_{1}$; in either case, it is easily seen that $K_{1}$ is the only subgraph with all the properties of an $R$-configuration; this contradicts the hypothesis. By Lemma 4.3(i), $u_{1},w_{1}\notin V(R)$.

Claim 4.7

Let $z_{1}$ be any vertex of $K_{1}$ which is distinct from $u_{1}$ and $w_{1}$. If $z_{1}\in V(K_{2})$ then every edge of $K_{1}$ which is incident with $z_{1}$ lies in $E(K_{2})$.

Proof: Assume that $z_{1}\in V(K_{2})$. First consider the case in which $z_{1}\in\{a_{1},b_{1}\}$. Note that the degree of $z_{1}$ in $H$ is two; consequently, both edges of $H$ incident with $z_{1}$ lie in $E(K_{2})$.

Now consider the case in which $z_{1}\notin\{a_{1},b_{1}\}$. Note that $z_{1}$ is cubic. Observe that, for an $R$-configuration $K$, any vertex of $K$, which is not one of its corners, is cubic in $K$ as well as in $G$. Thus, it suffices to show that $z_{1}$ is not a corner of $K_{2}$.

Suppose instead that $z_{1}$ is a corner of $K_{2}$. As $z_{1}\notin V(R)$, it is a free corner. Since $z_{1}$ is cubic, $K_{2}$ is an $R$-ladder configuration. Also, $z_{1}$ must be adjacent with a corner of $K_{2}$ which lies in $V(R)$; such a corner is either $a_{1}$ or $b_{1}$. Adjust notation so that $z_{1}$ is adjacent with $a_{1}$; thus, both edges of $H$ incident with $a_{1}$ lie in $E(K_{2})$. Note that $a_{1}z_{1}$ is an external rung of $K_{2}$. Also, since $u_{1}$ is not a corner of $K_{2}$, it is cubic in $K_{2}$ and in $G$. We infer that $K_{1}$ is also an $R$-ladder configuration; see Figure 33.

Let $y_{1}$ denote the neighbour of $u_{1}$ in $K_{1}$ which is distinct from $a_{1}$, and let $v$ denote the third neighbour of $u_{1}$. Note that $y_{1},v\in V(K_{2})$. Since $|\partial(u_{1})-E(K_{1})|=1$, Lemma 4.5(i) implies that $v$ is distinct from $w_{1}$. Since $K_{2}$ is a ladder, $a_{1}z_{1}$ lies in a $4$-cycle of $K_{2}$; this implies that $y_{1}z_{1}\in E(K_{2})$. Note that $u_{1}y_{1}$ is an internal rung of $K_{2}$.

Let $y_{2}$ denote the neighbour of $y_{1}$ which is distinct from $u_{1}$ and $z_{1}$. Note that $y_{2}\in V(K_{2})$. Since $a_{1}z_{1}$ and $u_{1}y_{1}$ are rungs of $K_{2}$, it must be the case that $v$ and $y_{2}$ are adjacent and the edge joining them is a rung of $K_{2}$; however, it is easily seen that $v$ and $y_{2}$ are nonadjacent. We thus have a contradiction. This completes the proof of Claim 4.7. $\Box$

We will now use Claim 4.7 to deduce that, since $K_{1}$ and $K_{2}$ are distinct $R$-configurations, the only vertices of $K_{1}$ which may lie in $K_{2}$ are its free corners (that is, $u_{1}$ and $w_{1}$).

Suppose instead that $(V(K_{1})-\{u_{1},w_{1}\})\cap V(K_{2})$ is nonempty. Since $K_{1}-\{u_{1},w_{1}\}$ is connected, Claim 4.7 implies that $V(K_{1})\subseteq V(K_{2})$ and $E(K_{1})\subseteq E(K_{2})$. Furthermore, since $|V(K_{1})\cap V(K_{2})|\geq 6$, the set $(V(K_{2})-\{u_{2},w_{2}\})\cap V(K_{1})$ is also nonempty, where $u_{2}$ and $w_{2}$ denote the free corners of $K_{2}$. By symmetry, $V(K_{2})\subseteq V(K_{1})$ and $E(K_{2})\subseteq E(K_{1})$. We conclude that $K_{1}$ and $K_{2}$ are identical subgraphs of $G$; contrary to our hypothesis.

Thus, each member of $V(K_{1})\cap V(K_{2})$ is a free corner of $K_{1}$, and by symmetry, it is a free corner of $K_{2}$ as well. By our hypothesis, $V(K_{1})\cap V(K_{2})$ is nonempty; thus, at least one of $u_{1}$ and $w_{1}$ is a free corner of $K_{2}$. Adjust notation so that $u_{1}$ is a free corner of $K_{2}$. To complete the proof, we will show that $w_{1}$ is also a free corner of $K_{2}$.

Suppose not, that is, say $V(K_{1})\cap V(K_{2})=\{u_{1}\}$, and let $w_{2}$ denote the free corner of $K_{2}$ distinct from $u_{1}$. Observe that the ends $a_{2}$ of $\alpha$ and $b_{2}$ of $\beta$ both lie in $V(K_{2})$; see Figure 34. Furthermore, $|B-V(K_{1}\cup K_{2})|=|A-V(K_{1}\cup K_{2})|+1$. We shall let $T:=B-V(K_{1}\cup K_{2})$. Since every vertex of $K_{1}\cup K_{2}$, except possibly $u_{1},w_{1}$ and $w_{2}$, is cubic, all neighbours of $T$ lie in the set $S:=(A-V(K_{1}\cup K_{2}))\cup\{w_{1},w_{2}\}$. Consequently, $S$ is a nontrivial barrier of $G$; this is absurd.

Thus, $u_{1}$ and $w_{1}$ are the free corners of $K_{2}$, and $K_{2}$ is otherwise vertex-disjoint with $K_{1}$. This completes the proof of Proposition 2.3. $\Box$

5 Strictly $R$-thin Edge Theorem

As in the statement of the theorem (1.8), let $G$ be a simple $R$-brick which is free of strictly $R$-thin edges. Our goal is to show that $G$ is a member of one of the eleven infinite families which appear in the statement of the theorem, that is, to show that $G\in\mathcal{N}$. We adopt Notation 2.1.

We may assume that $G$ is different from $K_{4}$ and $\overline{C_{6}}$, and thus, by the $R$-thin Edge Theorem (1.6), $G$ has an $R$-thin edge, say $e_{1}$. Depending on the index of $e_{1}$, we invoke either the $R$-biwheel Theorem (2.5) or the $R$-ladder Theorem (2.6) to deduce that $G$ has an $R$-configuration, say $K_{1}$, such that $e_{1}\in E(K_{1})$. We shall let $a_{1}u_{1}$ and $b_{1}w_{1}$ denote the external rungs/spokes of $K_{1}$, and adjust notation so that $u_{1}$ and $w_{1}$ are its free corners. See Notation 2.2 and Figure 30. Note that $a_{1}$ is an end of $\alpha$ and $b_{1}$ is an end of $\beta$.

By Lemma 4.3, either both free corners $u_{1}$ and $w_{1}$ lie in $V(R)$, or otherwise, neither of them lies in $V(R)$; let us first deal with the former case.

Claim 5.1

If $u_{1},w_{1}\in V(R)$ then $G$ is either a prism, or a Möbius ladder or a truncated biwheel.

Proof: Suppose that $u_{1},w_{1}\in V(R)$; that is, $\alpha=a_{1}w_{1}$ and $\beta=b_{1}u_{1}$. Since every vertex of $K_{1}$ is cubic in $G$, except possibly $u_{1}$ and $w_{1}$, we conclude that $V(G)=V(K_{1})$ as otherwise $\{u_{1},w_{1}\}$ is a $2$-vertex cut of $G$. Furthermore, either $G$ is precisely the graph induced by $E(K_{1})\cup R$, or otherwise, $G$ has one additional edge joining $u_{1}$ and $w_{1}$. In the latter case, $u_{1}w_{1}$ is a strictly $R$-thin edge, contrary to the hypothesis. In the former case, observe that: if $K_{1}$ is an $R$-biwheel configuration, as shown in Figure 30a, then $G$ is a truncated biwheel; if $K_{1}$ is an $R$-ladder configuration of odd parity, as shown in Figure 30b, then $G$ is a prism; and if $K_{1}$ is an $R$-ladder configuration of even parity, as shown in Figure 30c, then $G$ is a Möbius ladder. $\Box$

We may thus assume that neither $u_{1}$ nor $w_{1}$ lies in $V(R)$. Consequently, the end $a_{2}$ of $\alpha$ and the end $b_{2}$ of $\beta$ are both in $V(G)-V(K_{1})$.

Claim 5.2

Either $G$ is a staircase or a pseudo-biwheel, or otherwise, $G$ has an $R$-compatible edge which is not in $E(K_{1})$.

Proof: We begin by noting that, if $|\partial(u_{1})-E(K_{1})|\geq 2$, then by Lemma 4.4, some edge of $\partial(u_{1})-E(K_{1})$ is $R$-compatible, and we are done; an analogous argument applies when $|\partial(w_{1})-E(K_{1})|\geq 2$.

Now suppose that $|\partial(u_{1})-E(K_{1})|\leq 1$ and that $|\partial(w_{1})-E(K_{1})|\leq 1$. By Lemma 4.5 (i) and (ii), $u_{1}$ and $w_{1}$ are nonadjacent; furthermore, $\partial(u_{1})-E(K_{1})$ has a single element, say $\alpha^{\prime}$; likewise, $\partial(w_{1})-E(K_{1})$ has a single element, say $\beta^{\prime}$; see Figure 32. We let $R^{\prime}:=\{\alpha^{\prime},\beta^{\prime}\}$. By (iii) of the same lemma, $\alpha$ and $\alpha^{\prime}$ are adjacent if and only if $\beta$ and $\beta^{\prime}$ are adjacent.

First consider the case in which $\alpha$ and $\alpha^{\prime}$ are nonadjacent, and as in the statement of Lemma 4.5(iv), let $v$ denote the end of $\alpha^{\prime}$ which is distinct from $u_{1}$; note that $v\notin V(K_{1})$. By the lemma, $\partial(v)-\alpha^{\prime}$ contains an $R$-compatible edge, and we are done.

Now suppose that $\alpha$ and $\alpha^{\prime}$ are adjacent; whence $\beta$ and $\beta^{\prime}$ are also adjacent. Note that $\alpha^{\prime}=u_{1}a_{2}$ and $\beta^{\prime}=w_{1}b_{2}$. Every vertex of $K_{1}$, except possibly $u_{1}$ and $w_{1}$, is cubic in $G$; furthermore, $\partial(u_{1})-E(K_{1})=\{\alpha^{\prime}\}$, and likewise, $\partial(w_{1})-E(K_{1})=\{\beta^{\prime}\}$. We infer that $V(G)=V(K_{1})\cup\{a_{2},b_{2}\}$ as otherwise $\{a_{2},b_{2}\}$ is a $2$-vertex cut of $G$. Furthermore, since each of $a_{2}$ and $b_{2}$ has degree at least three, there is an edge joining them; and $G$ is precisely the graph induced by $E(K_{1})\cup R\cup R^{\prime}\cup\{a_{2}b_{2}\}$. Observe that if $K_{1}$ is an $R$-biwheel configuration of order at least eight then $G$ is a pseudo-biwheel, and otherwise, $G$ is a staircase. $\Box$

We may thus assume that $G$ has an $R$-compatible edge which is not in $E(K_{1})$. We will now use Theorem 1.7 and Lemma 4.6 to deduce that $G$ has an $R$-thin edge which is not in $E(K_{1})$.

Claim 5.3

$G$* has an $R$-thin edge, say $e_{2}$, which is not in $E(K_{1})$.*

Proof: Among all $R$-compatible edges which are not in $E(K_{1})$, we choose one, say $e_{2}$, such that ${\sf rank}(e_{2})+{\sf index}(e_{2})$ is maximum; we intend to show that $e_{2}$ is $R$-thin. Suppose not; then, by Theorem 1.7, with $e_{2}$ playing the role of $e$, there exists another $R$-compatible edge $f$ such that (i) $f$ has an end each of whose neighbours in $G-e_{2}$ lies in a (nontrivial) barrier $S$ of $G-e_{2}$, and (ii) ${\sf rank}(f)+{\sf index}(f)>{\sf rank}(e_{2})+{\sf index}(e_{2})$.

Let $I$ denote the set of isolated vertices of $(G-e_{2})-S$. Condition (i) above implies that $f$ has one end in $I$ and another end in $S$. By Lemma 4.6, with $e_{2}$ playing the role of $e$, the set $I\cap V(K_{1})$ is empty. Since $f$ has one end in $I$, we infer that $f$ is not in $E(K_{1})$; this, combined with condition (ii) above, contradicts our choice of $e_{2}$. We thus conclude that $e_{2}$ is $R$-thin. $\Box$

Now, depending on the index of $e_{2}$, we invoke either the $R$-biwheel Theorem (2.5) or the $R$-ladder Theorem (2.6) to deduce that $G$ has an $R$-configuration, say $K_{2}$, such that $e_{2}\in E(K_{2})$. As $e_{2}$ is not in $E(K_{1})$ but it is in $E(K_{2})$, the $R$-configurations $K_{1}$ and $K_{2}$ are clearly distinct. By Proposition 2.3: either $K_{1}$ and $K_{2}$ are vertex-disjoint; or otherwise, $u_{1}$ and $w_{1}$ are the free corners of $K_{2}$, and $K_{2}$ is otherwise vertex-disjoint with $K_{1}$. In either case, the end $a_{2}$ of $\alpha$ and the end $b_{2}$ of $\beta$ are the two corners of $K_{2}$ which are distinct from its free corners. Let us first deal with the case in which $K_{1}$ and $K_{2}$ are not vertex-disjoint; Figure 35 shows an example in which $K_{1}$ and $K_{2}$ are both $R$-biwheel configurations.

The proof of the following claim closely resembles that of Claim 5.1.

Claim 5.4

If $K_{1}$ and $K_{2}$ are not vertex-disjoint then $G$ is either a double biwheel or a double ladder or a laddered biwheel, each of type I.

Proof: As noted above, $u_{1}$ and $w_{1}$ are the free corners of $K_{2}$, and $K_{2}$ is otherwise vertex-disjoint with $K_{1}$. Consequently, the external rungs/spokes of $K_{2}$ are $a_{2}u_{1}$ and $b_{2}w_{1}$; see Figure 35. Since every vertex of $K_{1}\cup K_{2}$ is cubic in $G$, except $u_{1}$ and $w_{1}$, we infer that $V(G)=V(K_{1})\cup V(K_{2})$, as otherwise $\{u_{1},w_{1}\}$ is a $2$-vertex cut of $G$. Furthermore, either $G$ is precisely the graph induced by $E(K_{1}\cup K_{2})\cup R$, or otherwise, $G$ has one additional edge joining $u_{1}$ and $w_{1}$. In the latter case, $u_{1}w_{1}$ is a strictly $R$-thin edge, contrary to the hypothesis. In the former case, observe that if $K_{1}$ and $K_{2}$ are both $R$-biwheel configurations then $G$ is a double biwheel of type I; likewise, if $K_{1}$ and $K_{2}$ are both $R$-ladder configurations then $G$ is a double ladder of type I; finally, if one of $K_{1}$ and $K_{2}$ is an $R$-ladder configuration and the other one is an $R$-biwheel configuration then $G$ is a laddered biwheel of type I. $\Box$

We may thus assume that $K_{1}$ and $K_{2}$ are vertex-disjoint; and we shall let $a_{2}u_{2}$ and $b_{2}w_{2}$ denote the external rungs/spokes of $K_{2}$; in particular, $u_{2}$ and $w_{2}$ denote the free corners of $K_{2}$. Figure 36 shows an example in which $K_{1}$ is an $R$-ladder configuration and $K_{2}$ is an $R$-biwheel configuration.

We now find the remaining three families, or show the existence of an $R$-compatible edge which is not in $E(K_{1}\cup K_{2})$; the proof is similar to that of Claim 5.2.

Claim 5.5

Either $G$ is a double biwheel or a double ladder or a laddered biwheel, each of type II, or otherwise, $G$ has an $R$-compatible edge which is not in $E(K_{1}\cup K_{2})$.

Proof: We begin by noting that, if $|\partial(u_{1})-E(K_{1})|\geq 2$, then by Lemma 4.4, some edge of $\partial(u_{1})-E(K_{1})$ is $R$-compatible, and since $u_{1}\notin V(K_{2})$, such an edge is not in $E(K_{2})$, and we are done; an analogous argument applies when $|\partial(w_{1})-E(K_{1})|\geq 2$, or when $|\partial(u_{2})-E(K_{2})|\geq 2$ or when $|\partial(w_{2})-E(K_{2})|\geq 2$.

Now suppose that, for $i\in\{1,2\}$, $|\partial(u_{i})-E(K_{i})|\leq 1$ and $|\partial(w_{i})-E(K_{i})|\leq 1$; by Lemma 4.5 (i) and (ii), $u_{i}$ and $w_{i}$ are nonadjacent; furthermore, each of these inequalities holds with equality. Let $\alpha^{\prime}$ denote the only member of $\partial(u_{1})-E(K_{1})$, and let $\beta^{\prime}$ denote the only member of $\partial(w_{1})-E(K_{1})$.

First consider the case in which either $w_{2}$ is not an end of $\alpha^{\prime}$ or $u_{2}$ is not an end of $\beta^{\prime}$. Assume without loss of generality that $w_{2}$ is not an end of $\alpha^{\prime}$; thus the end of $\alpha^{\prime}$ distinct from $u_{1}$, say $v$, is not in $V(K_{1}\cup K_{2})$. By Lemma 4.5(iv), $\partial(v)-\alpha^{\prime}$ contains an $R$-compatible edge; such an edge is not in $E(K_{1}\cup K_{2})$, and we are done.

Now suppose that $w_{2}$ is an end of $\alpha^{\prime}$ and $u_{2}$ is an end of $\beta^{\prime}$. Note that every vertex of $K_{1}\cup K_{2}$, except possibly $u_{1},w_{1},u_{2}$ and $w_{2}$, is cubic in $G$; furthermore, $\partial(u_{1})-E(K_{1})=\partial(w_{2})-E(K_{2})=\{\alpha^{\prime}\}$, and likewise, $\partial(w_{1})-E(K_{1})=\partial(u_{2})-E(K_{2})=\{\beta^{\prime}\}$. We conclude that $V(G)=V(K_{1}\cup K_{2})$ and $E(G)=E(K_{1}\cup K_{2})\cup R\cup R^{\prime}$. Observe that: if $K_{1}$ and $K_{2}$ are both $R$-biwheel configurations then $G$ is a double biwheel of type II; likewise, if $K_{1}$ and $K_{2}$ are both $R$-ladder configurations then $G$ is a double ladder of type II; finally, if one of $K_{1}$ and $K_{2}$ is an $R$-ladder configuration and the other one is an $R$-biwheel configuration then $G$ is a laddered biwheel of type II. $\Box$

We may thus assume that $G$ has an $R$-compatible edge which is not in $E(K_{1}\cup K_{2})$. We will now use Theorem 1.7 and Lemma 4.6 to deduce that $G$ has an $R$-thin edge which is not in $E(K_{1}\cup K_{2})$. The proof is almost identical to that of Claim 5.3, except that now we have to deal with two $R$-configurations instead of just one.

Claim 5.6

$G$* has an $R$-thin edge, say $e_{3}$, which is not in $E(K_{1}\cup K_{2})$.*

Proof: Among all $R$-compatible edges which are not in $E(K_{1}\cup K_{2})$, we choose one, say $e_{3}$, such that ${\sf rank}(e_{3})+{\sf index}(e_{3})$ is maximum; we intend to show that $e_{3}$ is $R$-thin. Suppose not; then, by Theorem 1.7, with $e_{3}$ playing the role of $e$, there exists another $R$-compatible edge $f$ such that (i) $f$ has an end each of whose neighbours in $G-e_{3}$ lies in a (nontrivial) barrier $S$ of $G-e_{3}$, and (ii) ${\sf rank}(f)+{\sf index}(f)>{\sf rank}(e_{3})+{\sf index}(e_{3})$.

Let $I$ denote the set of isolated vertices of $(G-e_{3})-S$. Condition (i) above implies that $f$ has one end in $I$ and another end in $S$. By Lemma 4.6, with $e_{3}$ playing the role of $e$, the set $I\cap V(K_{1})$ is empty; likewise, the set $I\cap V(K_{2})$ is empty. Since $f$ has one end in $I$, we infer that $f$ is not in $E(K_{1}\cup K_{2})$; this, combined with condition (ii) above, contradicts our choice of $e_{3}$. We thus conclude that $e_{3}$ is $R$-thin. $\Box$

Now, depending on the index of $e_{3}$, we invoke either the $R$-biwheel Theorem (2.5) or the $R$-ladder Theorem (2.6) to deduce that $G$ has an $R$-configuration, say $K_{3}$, such that $e_{3}\in E(K_{3})$. As $e_{3}$ is not in $E(K_{1}\cup K_{2})$ but it is in $E(K_{3})$, the $R$-configuration $K_{3}$ is distinct from each of $K_{1}$ and $K_{2}$. We have thus located three distinct $R$-configurations in the brick $G$; namely, $K_{1},K_{2}$ and $K_{3}$. However, this contradicts Corollary 2.4, and completes the proof of the Strictly $R$-thin Edge Theorem (1.8). $\Box$

Acknowledgments: This work commenced in April 2015 when the first author was a Ph.D. candidate, and the second author was visiting the University of Waterloo for a fortnight. We are thankful to Joseph Cheriyan who helped facilitate this research visit, and participated in several of our discussions. We are indebted to both Cláudio L. Lucchesi and U. S. R. Murty for their constant guidance and support, and for sharing their invaluable insights.