A structure theorem for graphs with no cycle with a unique chord and its consequences

A structure theorem for graphs with no cycle with a unique chord and its consequences

Abstract

We give a structural description of the class  of graphs that do not contain a cycle with a unique chord as an induced subgraph. Our main theorem states that any connected graph in is either in some simple basic class or has a decomposition. Basic classes are chordless cycles, cliques, bipartite graphs with one side containing only nodes of degree two and induced subgraphs of the famous Heawood or Petersen graph. Decompositions are node cutsets consisting of one or two nodes and edge cutsets called 1-joins. Our decomposition theorem actually gives a complete structure theorem for , i.e. every graph in can be built from basic graphs that can be explicitly constructed, and gluing them together by prescribed composition operations; and all graphs built this way are in .

This has several consequences: an -time algorithm to decide whether a graph is in , an -time algorithm that finds a maximum clique of any graph in and an -time coloring algorithm for graphs in . We prove that every graph in is either 3-colorable or has a coloring with colors where is the size of a largest clique. The problem of finding a maximum stable set for a graph in is known to be NP-hard.

AMS Mathematics Subject Classification: 05C17, 05C75, 05C85, 68R10

Key words: cycle with a unique chord, decomposition, structure, detection, recognition, Heawood graph, Petersen graph, coloring.

1 Motivation

In this paper all graphs are simple. We give a structural characterization of graphs that do not contain a cycle with a unique chord as an induced subgraph. For the sake of conciseness we call this class of graph. Our main result, Theorem 2.1, states that every connected graph in is either in some simple basic class or has a particular decomposition. Basic classes are chordless cycles, cliques, bipartite graphs with one side containing only nodes of degree two and graphs that are isomorphic to an induced subgraph of the famous Heawood or Petersen graph. Our decompositions are node cutsets consisting of one or two nodes or an edge cutset called a 1-join. The definitions and the precise statement are given in Section 2. The proof is given in Section 3. Both Petersen and Heawood graphs were discovered at the end of the XIXth century in the research on the four color conjecture, see [20] and [13]. It is interesting to us to have them both as sporadic basic graphs. Note that our theorem works in two directions: a graph is in  if and only if it can be constructed by gluing basic graphs along our decompositions (this is proved in Section 4). Such structure theorems are stronger than the usual decomposition theorems and there are not so many of them (see [3] for a survey). This is our first motivation.

Our structural characterization allows us to prove properties of classical invariants. We prove in Section 6 that every graph in  satisfies either or (where denotes the chromatic number and denotes the size of a maximum clique). This is a strengthening of the classical Vizing bound . So this class of graphs belongs to the family of -bounded graphs, introduced by Gyárfás [12] as a natural extension of perfect graphs: a family of graphs is -bounded with -binding function if, for every induced subgraph of , . A natural question to ask is: what choices of forbidden induced subgraphs guarantee that a family of graphs is -bounded? Much research has been done in this area, for a survey see [22]. We note that most of that research has been done on classes of graphs obtained by forbidding a finite number of graphs. Since there are graphs with arbitrarily large chromatic number and girth [11], in order for a family of graphs defined by forbidding a finite number of graphs (as induced subgraphs) to be -bounded, at least one of these forbidden graphs needs to be acyclic. Vizing’s Theorem [25] states that for a simple graph , (where denotes the maximum vertex degree of , and denotes the chromatic index of , i.e. the minimum number of colors needed to color the edges of so that no two adjacent edges receive the same color). This implies that the class of line graphs of simple graphs is a -bounded family with -binding function . This special upper bound for the chromatic number is called the Vizing bound. We obtain the Vizing bound for the chromatic number by forbidding a family of graphs none of which is acyclic. Our result is algorithmic: we provide an algorithm that computes an optimal coloring of every graph in . Furthermore, it is easy to see that there exists an algorithm that computes a maximum clique for every graph in ; and it follows from a construction of Poljak [21] that finding a maximum stable set of a graph in  is NP-hard (see Section 7). All this is our second motivation.

A third motivation is the detection of induced subgraphs. A subdivisible graph (s-graph for short) is a triple such that is a graph and . The edges in are said to be real edges of while the edges in are said to be subdivisible edges of . A realisation of is a graph obtained from by subdividing edges of into paths of arbitrary length (at least one). The problem is the decision problem whose input is a graph and whose question is ”Does contain a realisation of as an induced subgraph?”. In the discussion below, by “detection problem”, we mean “problem for some fixed s-graph ”. This is restrictive since a lot of detection problems of great interest (such as the detection of odd holes, where a hole is an induced cycle of length at least four) are not of that kind.

Let be the s-graph on nodes with real edges , , and subdivisible edges , . We also define for the s-graph obtained from by subdividing the edge into a path of length and the edge into a path of length . See Fig. 1 where real edges are represented as straight lines and subdivisible edges as dashed lines. The question in Problem can be rephrased as “Does contain a cycle with a unique chord?” or “Is not in ?”. The existence of a polynomial time algorithm was an open question. A consequence of our structural description of  is an -time algorithm for (see Section 5). This is a solution to the recognition problem for the class  and it is interesting for reasons explained below.

   
   
Figure 1: Some s-graphs

Several problem ’s can be solved in polynomial time by non-trivial algorithms (such as detecting pyramids in [2] and thetas in [4]) and others that may look similar at first glance are NP-complete (see [1], [18], and [16] for a survey). A general criterion on an s-graph that decides whether the related decision problem is NP-complete or polynomial would be of interest. Our solution of gives some insight in the quest for such a criterion.

A very powerful tool for solving detection problems is the algorithm three-in-a-tree of Chudnovsky and Seymour (see [4]). This algorithm decides in time whether three given nodes of a given graph are in an induced tree of . In [4] and [16] it is observed that every detection problem for which a polynomial time algorithm is known can be solved easily by a brute force enumeration or by using three-in-a-tree. But as far as we can see, three-in-a-tree cannot be used to solve , so our solution of yields the first example of a detection problem that does not fall under the scope of three-in-a-tree. Is there a good reason for that? We claim that a polynomial time algorithm for exists thanks to what we call degeneracy. Let us explain this. Every statement that we give from here on to the end of the section is under the assumption that PNP.

Degeneracy has to deal with the following question: does putting bounds on the lengths of the paths in realisations of an s-graph affect the complexity of the related detection problem? For upper bounds, the answer can be found in previous research. First, putting upper bounds may turn the complexity from NP-complete to polynomial. This follows from a simple observation: let be any s-graph. A realisation of , where the lengths of the paths arising from the subdivisions of subdivisible edges are bounded by an integer , has a number of nodes bounded by a fixed integer (that depends only on and the size of ). So, such a realisation can be detected in time by a brute force enumeration. But surprisingly, putting upper bounds in another way may also turn the complexity from polynomial to NP-complete: in [2], a polynomial time algorithm for is given, while in [19] it is proved that is NP-complete, where are the s-graphs represented in Figure 2. Note that is usually called the pyramid (or 3PC()) detection problem.

 
Figure 2: Some s-graphs

Can putting lower bounds turn the complexity from polynomial to NP-complete? Our recognition algorithm for shows that the answer is yes since in Section 8 we also prove that the problem is NP-complete. A realisation of is simply a realisation of where every subdivisible edge is subdivided into a path of length at least three. We believe that a satisfactory structural description of the class  of graphs that do not contain a realisation of is hopeless because is NP-complete. So why is there a decomposition theorem for  ? Simply because degenerate small graphs like the diamond (that is the cycle on four nodes with exactly one chord) are forbidden in , not in , and this helps a lot in our proof of Theorem 2.1 (the decomposition theorem for ). This is what we call the degeneracy of the class . It is clear that degeneracy can help in solving detection problems, and our results give a first example of this phenomenon.

So the last question is: can putting lower bounds turn the complexity from NP-complete to polynomial? We do not know the answer. Also, we were not able to solve the following questions: what is the complexity of the problems , , and ? The related classes of graphs are not degenerate enough to allow us to decompose, and they are too degenerate to allow us to find an NP-completeness proof.

A fourth motivation is that our class is related to well studied classes. It is a generalization of strongly balanceable graphs, see [7] for a survey. A bipartite graph is balanceable if there exists a signing of its edges with and so that the weight of every hole is a multiple of 4. A bipartite graph is strongly balanceable if it is balanceable and it does not contain a cycle with a unique chord. There is an excluded induced subgraph characterization of balanceable bipartite graphs due to Truemper [24]. A wheel in a graph consists of a hole and a node that has at least three neighbors in , and the wheel is odd if has an odd number of neighbors in . In a bipartite graph , a 3-odd-path configuration consists of two nonadjacent nodes and that are on opposite sides of the bipartition of , together with three internally node-disjoint -paths, such that there are no other edges in among the nodes of the three paths. A bipartite graph is balanceable if and only if it does not contain an odd wheel nor a 3-odd-path configuration [24]. So a bipartite graph is strongly balanceable if and only if it does not contain a 3-odd-path configuration nor a cycle with a unique chord.

A bipartite graph is restricted balanceable if there exists a signing of its edges with and so that the weight of every cycle is a multiple of 4. Conforti and Rao [8] show that a strongly balanceable graph is either restricted balanceable or has a 1-join, which enables them to recognize the class of strongly balanceable graphs (they decompose along 1-joins, and then directly recognize restricted balanceable graphs). A bipartite graph is 2-bipartite if all the nodes in one side of the bipartition have degree at most 2. Yannakakis [26] shows that a restricted balanceable graph is either 2-bipartite or has a 1-cutset or a 2-join consisting of two edges (this is an edge cutset that consists of two edges that have no common endnode), and hence obtains a linear time recognition algorithm for restricted balanceable graphs.

We note that the basic graphs from our decomposition theorem that do not have any of our cutsets, and are balanceable, are in fact 2-bipartite.

Class is contained in another well studied class of graphs, the cap-free graphs (where a cap is a graph that consists of a hole and a node that has exactly two neighbors on this hole, and these two neighbors are adjacent) [6]. In [6] cap-free graphs are decomposed with 1-amalgams (a generalization of a 1-join) into triangulated graphs and biconnected triangle-free graphs together with at most one additional node that is adjacent to all other nodes of the graph. This decomposition theorem is then used to recognize strongly even-signable and strongly odd-signable graphs in polynomial time, where a graph is strongly even-signable if its edges can be signed with 0 and 1 so that every cycle of length with at most one chord has even weight and every triangle has odd weight, and a graph is strongly odd-signable if its edges can be signed with 0 and 1 so that cycles of length 4 with one chord are of even weight and all other cycles with at most one chord are of odd weight.

2 The main theorem

We say that a graph contains a graph if is isomorphic to an induced subgraph of . A graph is -free if it does not contain . For , denotes the subgraph of induced by . A cycle in a graph is a sequence of nodes , that are distinct except for the first and the last node, such that for , is an edge and is an edge (these are the edges of ). An edge of with both endnodes in is called a chord of if it is not an edge of . One can similarly define a path and a chord of a path. In this paper we will only use what is in literature known as chordless paths, so for the convenience, in this paper (like in [5]) we define a path as follows: a path in a graph is a sequence of distinct nodes such that for , is an edge and these are the only edges of that have both endnodes in . Such a path is also called a -path. A hole is a chordless cycle of length at least four. A triangle is a cycle of length 3. A square is a hole of length 4. A cycle in a graph is Hamiltonian is every node of the graph is in the cycle. Let us define our basic classes:

The Petersen graph is the graph on nodes so that and both induce a with nodes in their natural order, and such that the only edges between the ’s and the ’s are , , , , . See Fig. 3.

Figure 3: Four ways to draw the Petersen graph

The Heawood graph is the graph on so that is a Hamiltonian cycle with nodes in their natural order, and such that the only other edges are , , , , , , . See Fig. 4.

It can be checked that both Petersen and Heawood graph are in . Note that since the Petersen graph and the Heawood graph are both vertex-transitive, and are not themselves a cycle with a unique chord, to check that they are in , it suffices to delete one node, and then check that there is no cycle with a unique chord. Also the Petersen graph has girth 5 so a cycle with a unique chord in it must contain at least 8 nodes. The Heawood graph has girth 6 so a cycle with a unique chord in it must contain at least 10 nodes. For the Petersen graph, deleting a node yields an Hamiltonian graph, and it is easy to check that it does not contain a cycle with a unique chord. For the Heawood graph, it is useful to notice that deleting one node yields the Petersen graph with edges subdivided.

Figure 4: Four ways to draw the Heawood graph

Let us define our last basic class. A graph is strongly 2-bipartite if it is square-free and bipartite with bipartition where is the set of all degree 2 nodes of and is the set of all nodes of with degree at least 3. A strongly 2-bipartite graph is clearly in because any chord of a cycle is an edge linking two nodes of degree at least three, so every cycle in a strongly 2-bipartite graph is chordless.

We now define cutsets used in our decomposition theorem:

  • A 1-cutset of a connected graph is a node such that can be partitioned into non-empty sets , and , so that there is no edge between and . We say that is a split of this 1-cutset.

  • A proper 2-cutset of a connected graph is a pair of non-adjacent nodes , both of degree at least three, such that can be partitioned into non-empty sets , and so that: , ; there are no edges between and ; and both and contain an -path. We say that is a split of this proper 2-cutset.

  • A 1-join of a graph is a partition of into sets and such that there exist sets satisfying:

    • , ;

    • and ;

    • there are all possible edges between and ;

    • there are no other edges between and .

    We say that is a split of this 1-join. The sets are special sets with respect to this 1-join.

    1-Joins were first introduced by Cunningham [9]. In our paper we will use a special type of a 1-join called a proper 1-join: a 1-join such that and are stable sets of of size at least two. Note that a square admits a proper 1-join.

Our main result is the following decomposition theorem:

Theorem 2.1

Let be a connected graph that does not contain a cycle with a unique chord. Then either is strongly 2-bipartite, or is a hole of length at least 7, or is a clique, or is an induced subgraph of the Petersen or the Heawood graph, or has a 1-cutset, a proper 2-cutset, or a proper 1-join.

The following intermediate results are proved in the next section. Theorem 2.1 follows from Theorems 2.3 and 2.4 (more precisely, it follows from 2.4 for square-free graphs, and from 2.3 for graphs that contain a square).

Theorem 2.2

Let be a connected graph that does not contain a cycle with a unique chord. If contains a triangle then either is a clique, or one node of the maximal clique that contains this triangle is a 1-cutset of .

Theorem 2.3

Let be a connected graph that does not contain a cycle with a unique chord. Suppose that contains either a square, the Petersen graph or the Heawood graph. Then either is the Petersen graph or is the Heawood graph or has a 1-cutset or a proper 1-join.

Theorem 2.4

Let be a connected square-free graph that does not contain a cycle with a unique chord. Then either is strongly 2-bipartite, or is a hole of length at least 7, or is a clique or is an induced subgraph of the Petersen or the Heawood graph, or has a 1-cutset or a proper 2-cutset.

3 Proof of Theorems 2.2, 2.3 and 2.4

We first need two lemmas:

Lemma 3.1

Let be a graph in , a hole of and a node of . Then has at most two neighbors in , and these two neighbors are not adjacent.

Proof.

If has at least three neighbors in , then contains a subpath with exactly three neighbors of and induces a cycle of with a unique chord, a contradiction. If has two neighbors in , they must be non-adjacent for otherwise is a cycle with a unique chord. ∎

In a connected graph two nodes and form a 2-cutset if is disconnected.

Lemma 3.2

Let be a connected graph that has no 1-cutset. If is a 2-cutset of and is an edge, then .

Proof.

Suppose is a 2-cutset of , and is an edge. Let be the connected components of . Since is connected and has no 1-cutset, for every , both and have a neighbor in . Let be the graph obtained from by removing the edge . So for every , there is an -path in whose interior nodes are contained in . Then is a cycle with a unique chord, and hence . ∎

If is any induced subgraph of and is a subset of nodes of , the attachment of over is the set of all nodes of that have at least one neighbor in . When clear from context we do not distinguish between a graph and its node set, so we also refer to the attachment of over .

Proof of Theorem 2.2

Suppose contains a triangle, and let be a maximal clique of that contains this triangle. In fact, is unique or otherwise contains a diamond. If and if no node of is a 1-cutset of then let be a connected induced subgraph of , whose attachment over contains at least two nodes, and that is minimal with respect to this property. So, is a path with one end adjacent to , the other end adjacent to and induces a chordless cycle. If has length zero, then its unique node (say ) must have a non-neighbor since is maximal. Hence, induces a diamond, a contradiction. If has length at least one then let be any node of . Then the hole induced by and node contradict Lemma 3.1. This proves Theorem 2.2.

Proof of Theorem 2.3

Claim 1

We may assume that is triangle-free.

Proof.

Clear by Theorem 2.2 (note that cannot be a clique). ∎

Claim 2

We may assume that is square-free.

Proof.

Assume contains a square. Then contains disjoint sets of nodes and such that and are both stable graphs, and every node of is adjacent to every node of . Let us suppose that is chosen to be maximal with respect to this property. If then is a proper 1-join of , so we may assume that there are nodes in .

(1) Every component of has neighbors only in or only in .

Else, let us take a connected induced subgraph of , whose attachment over contains nodes of both and , and that is minimal with respect to this property. So is a path, no interior node of which has a neighbor in and there exists , such that . By Claim 1, , has no neighbor in and has no neighbor in . By maximality of , has a non-neighbor and has a non-neighbor . Now, is a cycle with a unique chord (namely ), a contradiction. This proves (1).

From (2), it follows that has a proper 1-join with special sets . ∎

Now, we just have to prove the following two claims:

Claim 3

If contains the Petersen graph then the theorem holds.

Proof.

Let be a set of ten nodes of so that has adjacencies like in the definition of the Petersen graph. We may assume that there are some other nodes in for otherwise the theorem holds.

(2) A node of has at most one neighbor in .

Otherwise contains a triangle or a square, contrary to Claims 1, 2. This proves (2).

Here below, we use symmetries in the Petersen graph to shorten the list of cases. First, the Petersen graph is edge-transitive, so up to an automorphism, all edges are equivalent. But also, it is “distance-two-transitive”, meaning that every induced is equivalent to every other induced . To see this, it suffices to check that every induced is included in an induced and that removing any always yields the same graph.

(3) The attachment of any component of over contains at most one node.

Else, let be a connected induced subgraph of whose attachment over contains at least two nodes, and that is minimal with respect to this property. By minimality and up to symmetry, is a path with one end adjacent to (and to no other node of by (3)), one end adjacent to (and to no other node of ). Moreover, no interior node of has a neighbor in . If then is a cycle with a unique chord, a contradiction. If then is a cycle with a unique chord, a contradiction again. This proves (3).

From (3) it follows that has a 1-cutset. ∎

Claim 4

If contains the Heawood graph then the theorem holds.

Proof.

Let be a set of fourteen nodes of so that has adjacencies like in the definition of the Heawood graph. We may assume that there are some other nodes in for otherwise the theorem holds.

(4) A node of has at most two neighbors in .

Suppose that some node in has at least two neighbors in . Since the Heawood graph is vertex-transitive we may assume . By Claims 1 and 2, cannot be adjacent to a node at distance 1 or 2 from , namely to any of . So, the only other possible neighbors are . But these four nodes are pairwise at distance two in , so by Claim 2, can be adjacent to at most one of them. This proves (4).

(5) The attachment of any component of over contains at most one node.

Else, let be a connected induced subgraph of whose attachment over contains at least two nodes, and is minimal with respect to this property. By minimality and up to symmetry, is a path, possibly of length zero, with one end adjacent to , one end adjacent to where and no interior node of has neighbors in . Note that by assumption, if is of length at least one then no end of can have more than one neighbor in , because such an end would contradict the minimality of . So, are the only nodes of that have neighbors in (when is of length zero this holds by (4).

If then is a cycle with a unique chord. If then is a cycle with a unique chord. If then is a cycle with a unique chord. If then is a cycle with a unique chord. If then is a cycle with a unique chord. If then is a cycle with a unique chord. If then is a cycle with a unique chord. If then is a cycle with a unique chord. In every case, there is a contradiction. This proves (5).

From (4) it follows that has a 1-cutset. ∎

This proves Theorem 2.3.

Proof of Theorem 2.4

We consider a graph containing no cycle with a unique chord and no square. So:

Claim 1

is square-free.

Our proof now goes through thirteen claims, most of them of the same kind: if some basic graph is an induced subgraph of , then either and so itself is basic, or some nodes of must be attached to in a way that entails a proper 2-cutset. At the end of this process there are so many induced subgraphs forbidden in that we can prove that is strongly 2-bipartite.

Claim 2

We may assume that is triangle-free.

Proof.

Clear by Theorem 2.2. ∎

Claim 3

We may assume that does not contain the Petersen graph.

Proof.

Clear by Theorem 2.3. Note that cannot admit a proper 1-join since it is square-free. ∎

Claim 4

We may assume that does not contain the Heawood graph.

Proof.

Clear by Theorem 2.3. ∎

Claim 5

We may assume that does not contain the following configuration: three node-disjoint paths , and , of length at least two and with no edges between them. There are four more nodes . The only edges except those from the paths are .

Proof.

Let . Nodes are called here the branch nodes of . It is convenient to notice that can be obtained by subdividing the edges of any induced matching of size three of the Petersen graph. Note also that either is the Heawood graph with one node deleted (when all have length two), or has a proper 2-cutset (when one of the paths is of length at least three, the proper 2-cutset is formed by the ends of that path). Hence we may assume that there are nodes in .

(6) A node of has neighbors in at most one of the following sets: .

Let be a node of . Note that are pairwise at distance two in , so by Claims 1 and 2, can be adjacent to at most one of them.

Suppose first that . Then can be adjacent to none of by Claims 1 and 2. If is adjacent to some other branch-node of distinct from , then we may assume that is adjacent to one of (say up to symmetry), but then is not adjacent to so is a cycle with a unique chord, a contradiction. Hence we may assume that has a neighbor in the interior of or (say up to symmetry) for otherwise the claim holds. By Lemma 3.1 and since is a hole, we note that has exactly two neighbors in , namely and . If then is a cycle with a unique chord, a contradiction. So and is a cycle with a unique chord (namely ), a contradiction again. Hence we may assume that , and symmetrically .

Suppose now that . Then cannot be adjacent to any other branch-node of by the discussion above. So we may assume that has a neighbor in the interior of (say up to symmetry) for otherwise our claim holds. Now we define to be the neighbor of along closest to and observe that is a cycle with a unique chord, a contradiction. Therefore we may assume that is not adjacent to any branch-node of .

We may suppose now that has neighbors in the interior of at least two of the paths among (say w.l.o.g.) for otherwise our claim holds. Let (resp. ) be a neighbor of in (resp. ). Since induces a hole, by Lemma 3.1, . If then is a cycle with unique chord. So and symmetrically, . If has no neighbor in then is a cycle with a unique chord, so must have at least one neighbor in . By the same discussion that we have done above on we can prove that . Now we observe that is the Heawood graph, contradicting Claim 4. This proves (6).

(7) The attachment of any component of is included in one of the sets .

Else let be a connected induced subgraph of whose attachment overlaps two of the sets, and is minimal with respect this property. By (5), is a path of length at least one, with ends and (resp. ) has neighbors in exactly one set (resp. ) of . Moreover, no interior node of has a neighbor in . If and then let be the neighbor of closest to along . We observe that is a cycle with a unique chord, a contradiction. Every case where there is an edge between and is symmetric, so we may assume that there is no edge between and . If and then let (resp. ) be the neighbor of (resp. of ) closest to along (resp. to along ). If and then is a cycle with a unique chord, a contradiction. So up to symmetry we may assume . Hence is a cycle with a unique chord, a contradiction. So up to symmetry we may assume that and . But then is a cycle with a unique chord, a contradiction. This proves (7).

By (5), either some component of attaches to a node of and there is a 1-cutset, or some component attaches to one of (say up to symmetry), and is a proper 2-cutset. ∎

Claim 6

We may assume that does not contain the following configuration: three node-disjoint paths , and , of length at least two, and such that the only edges between them are , , , , and .

Proof.

Note that is a hole on six nodes. Also either is the Petersen graph with one node deleted (when have length two), or has a proper 2-cutset (when one of the paths is of length at least three, the 2-cutset is formed by the ends of that path). Hence we may assume that there are nodes in .

(8) A node of has neighbors in at most one of the sets .

Let be a node of . Note that has neighbors in at most one of the following sets: , , , for otherwise contains a triangle or a square, contradicting Claims 1 and 2.

If has at least two neighbors among then we may assume by the paragraph above . Since and both induce holes, every node in is in a hole going through . So, by Lemma 3.1 has no other neighbors in . Hence, from here on, we assume that has at most one neighbor among .

If then we may assume that has neighbors in one of , say up to symmetry. Let be a neighbor of . Then by Lemma 3.1, since induces a hole, and are the only neighbors of in . So, is a cycle with a unique chord, a contradiction. Hence we may assume that has no neighbors among .

If has neighbors in the interior of at most one of our claim holds, so let us suppose that has neighbors in the interior of and the interior of . Since induces a hole, by Lemma 3.1, has a unique neighbor and a unique neighbor . If has no neighbor in then is a cycle with a unique chord, a contradiction. So has a neighbor that is unique by Lemma 3.1. If then is a cycle with a unique chord, a contradiction. So, up to a symmetry we may assume . If then is a cycle with a unique chord, a contradiction, so . By the same argument, we can prove that . If then we observe that is the Petersen graph, contradicting Claim 3. If then we observe that the three paths , , and nodes have the same configuration as those in Claim 5, a contradiction. So, we may assume that and . But then, is a cycle with a unique chord, a contradiction. This proves (8).

(9) The attachment of any component of is included in one of the sets .

Else let be a connected induced subgraph of , whose attachment overlaps two of the sets, and is minimal with this property. By (6), is a path of length at least one, with ends and no interior node of has a neighbor in . We may assume that has neighbors only in and only in . Let (resp. ) be the neighbor of (resp. of ) closest to along (resp. to along ). If and then is a cycle with a unique chord, a contradiction, so we may assume . Let be the neighbor of closest to along . If then is a cycle with a unique chord a contradiction. So, and is a cycle with a unique chord, a contradiction. This proves (9).

By (6) one of , , is a proper 2-cutset of . ∎

Claim 7

We may assume that does not contain the following configuration: four node-disjoint paths , , and , of length at least two, and such that the only edges between them are , , , , , , and .

Proof.

Either is obtained from the Heawood graph by deleting two adjacent nodes (when have length two), or has a proper 2-cutset (when one of the paths is of length at least three, the 2-cutset is formed by the ends of that path). Hence we may assume that there are nodes in .

(10) A node of has at most two neighbors in .

For suppose that a node of has at least three neighbors in . Since every pair of path from can be embedded in a hole (for instance, or are holes, and the other cases are symmetric), by Lemma 3.1, the neighbors of lie on three or four paths and every path contains at most one neighbor of .

Suppose is adjacent to one of the ’s, say . Then by Lemma 3.1 has at most one neighbor in since is a hole. So up to symmetry we assume that has a neighbor in , no neighbor in , and so must have a neighbor in . By Lemma 3.1 applied to the hole and node , must be in the interior of . If then is a cycle with a unique chord. So and contains a square, a contradiction to Claim 1. Hence we may assume that has no neighbors among the ’s.

Up to symmetry we assume that has neighbors , , . These neighbors are unique and are in the interior of their respective paths. So is a cycle with a unique chord (namely ), a contradiction. This proves (10).

(11) The attachment of any component of is included in one of the sets .

Else let be a connected induced subgraph of , whose attachment overlaps two of the sets, and is minimal with respect this property. By the choice of , the following hold. is a path, possibly of length zero, with ends , and we may assume up to symmetry that has neighbors in and that has neighbors in or in . No interior node of has neighbors in . If then has neighbors only in and only in or in . If then by (7) has neighbors only in or only in .

If has neighbors in then let be the neighbor of closest to along and be the neighbor of closest to along . If and then is a cycle with a unique chord, so up to symmetry we may assume . But then, is a cycle with a unique chord, a contradiction.

So has neighbors in . We claim that has a unique neighbor in , that is in the interior of , and that has length two. Else, up to the symmetry between and we may assume that the neighbor of closest to along is not and is not adjacent to . Let be the neighbor of closest to along . If then is a cycle with a unique chord. So . If is not adjacent to , then is a cycle with unique chord, a contradiction. So is adjacent to . In particular, . By (7), and are the only neighbors of in . But then is a cycle with unique chord, a contradiction. So, our claim is proved. Similarly, it can be proved that has a unique neighbor in , that this neighbor is in the interior of , and that has length two. We observe that the three paths , , and nodes have the same configuration as those in Claim 5, a contradiction. This proves (11).

By (7), one of , , , is a proper 2-cutset. ∎

Claim 8

We may assume that does not contain the following configuration: five paths , , , , , node disjoint except for their ends, of length at least two, and such that is a hole and the only edges between this hole and are , , and .

Proof.

We put . Either is the Heawood graph with three nodes inducing a deleted (when the five paths have length two), or has a proper 2-cutset (when one of the paths is of length at least three, the 2-cutset is formed by the ends of that path). Hence we may assume that there are nodes in .

(12) A node of has at most two neighbors in .

Let be a node of and suppose that has more than two neighbors in . By Lemma 3.1 and since is a hole, has at most two neighbors among these paths. So, must have one neighbor in , and this neighbor is unique since the union of with any of the other paths yields a hole. For the same reason, has a unique neighbor in exactly two paths among . So there are two cases up to symmetry: either has neighbors in two paths among that have a common end, or has neighbors in two paths among that have no common ends.

In the first case, we may assume that has neighbors , and . Note that and , for otherwise or would contain two neighbors of . Suppose . If then is a cycle with a unique chord, a contradiction. If then by Claim 1, and hence is a cycle with a unique chord, a contradiction. So . But then, since does not contain a square, and hence is a cycle with a unique chord, a contradiction.

In the second case, we may assume that has neighbors in , in and in . If then the previous case applies. Hence we may assume and symmetrically . So, is a cycle with a unique chord (namely ). This proves (12).

(13) The attachment of any component of is included in one of the sets .

Else let be a connected induced subgraph of , whose attachment is not contained in one of the sets, and is minimal with respect to this property. By the choice of the following hold. is a path, possibly of length zero, with ends , where has neighbors in one of the sets that we denote by , and has neighbors in another one, say . No interior node of has neighbors in . If then has neighbors only in and only in . If then by (8) has neighbors only in .

If then up to symmetry we may assume . Let be the neighbor of closest to along . If then let be the neighbor of closest to along . Then is a cycle with a unique chord. If then let be the neighbor of closest to along . Then is a cycle with a unique chord. So, , and symmetrically .

If , are paths with a common end then we may assume and . Let be the neighbor of closest to along and the neighbor of closest to along . We note that for otherwise the attachment of is a single path or