On trianglefree graphs that do not contain a subdivision of the complete graph on four vertices as an induced subgraph
Abstract
We prove a decomposition theorem for the class of trianglefree graphs that do not contain a subdivision of the complete graph on four vertices as an induced subgraph. We prove that every graph of girth at least 5 in this class is 3colorable.
AMS Classification: 05C75
1 Introduction
Here graphs are simple and finite. We say that contains when is isomorphic to an induced subgraph of . We say that a graph is free if does not contain . For a family of graphs , we say that is free if for every , does not contain . Subdividing an edge of a graph means deleting , and adding a new vertex of degree 2 adjacent to and . A subdivision of a graph is any graph obtained from by repeatedly subdividing edges. Note that is a subdivision of . We say that is an ISK4 of a graph when is an induced subgraph of and is a subdivision of (where denotes the complete graph on four vertices). ISK4 stands for “Induced Subdivision of ”.
In [2], a decomposition theorem for ISK4free graphs is given (see Theorem 2.1) and it is proved that their chromatic number is bounded by a constant . The proof in [2] follows from a theorem by Kühn and Osthus [1], from which it follows that is at least . It is conjectured in [2] that every ISK4free is 4colorable. The goal of this paper is to prove a stronger decomposition theorem for ISK4free graphs, under the additional assumption that they are trianglefree (see Theorems 2.2 and 3.8). We also propose the following conjectures, prove that the first one implies the second one (see Theorem 2.7), and prove both of them for graphs of girth at least 5 (see Theorem 5.5). A complete bipartite graph with partitions of size and , is denoted by .
Conjecture 1.1
Every {triangle, ISK4, }free graph contains a vertex of degree at most 2.
Conjecture 1.2
Every {triangle, ISK4}free graph is 3colorable.
In Section 2, we state several known decomposition theorems, and derive easy consequences of them for our class. In particular we prove that Conjecture 1.1 implies Conjecture 1.2. In Section 3, we prove our main decomposition theorem. In Section 4, we give some properties needed later for the class of chordless graphs (graphs where all cycles are chordless). In Section 5, we prove Conjecture 1.1 for graphs of girth at least 5.
2 Decomposition theorems
In this section, we provide the notation needed to state decomposition theorems for ISK4free graphs, and we state them.
A graph is seriesparallel if no subgraph of is a subdivision of . Clearly, every seriesparallel graph is ISK4free. When is a graph, the line graph of is the graph whose vertexset is and such that two vertices of are adjacent whenever the corresponding edges are adjacent in .
For a graph , when is a subset of , we write instead of . A cutset in a graph is a set of vertices such that is disconnected. A star cutset is a cutset that contains a vertex , called a center of , adjacent to all other vertices of . Note that a star cutset may have more than one center, and that a cutset of size 1 is a star cutset. A double star cutset is a cutset that contains two adjacent vertices and , such that every vertex of is adjacent to or . Note that a star cutset of size at least 2 is a double star cutset.
A path from a vertex to a vertex is refered to as an path. A proper 2cutset of a connected graph is a pair of nonadjacent vertices such that can be partitioned into nonempty sets , and so that: there is no edge between and ; and both and contain an path and neither of nor is a chordless path. We say that is a split of this proper 2cutset. The following is the main decomposition theorem for ISK4free graphs.
Theorem 2.1 (see [2])
If is an ISK4free graph, then is seriesparallel, or is the line graph of a graph of maximum degree at most 3, or has a proper 2cutset, a star cutset, or a double star cutset.
Our first goal is to improve this theorem for trianglefree graphs. Some improvements are easy to obtain (they trivially follow from the absence of triangles). The nontrivial one is done in the next two sections: we show that double star cutsets and proper 2cutsets are in fact not needed. We state the result now, but it follows from Theorem 3.8 that needs more terminology and is slightly stronger.
Theorem 2.2
If is a {triangle, ISK4}free graph, then either is a seriesparallel graph or a complete bipartite graph, or has a clique cutset of size at most two, or has a star cutset.
Proof.
Follows directly from Theorem 3.8. ∎
We state now several lemmas from [2] that we need. A hole in a graph is a chordless cycle of length at least 4. A prism is a graph made of three vertex disjoint paths of length at least 1, , and , with no edges between them except the following: , , , , , . Note that the union of any two of the paths of a prism induces a hole. A wheel is a graph that consists of a hole plus a vertex that has at least three neighbors on .
Lemma 2.3 (see [2])
If is an ISK4free graph, then either is a seriesparallel graph, or contains a prism, or contains a wheel or contains .
A complete tripartite graph is a graph that can be partitioned into three stable sets so that every pair of vertices from two different stable sets is an edge of the graph.
Lemma 2.4 (Lévêque, Maffray and Trotignon [2])
If is an ISK4free graph that contains , then either is a complete bipartite graph, or is a complete tripartite graph, or has a cliquecutset of size at most 3.
We now state the consequences of the lemmas above for trianglefree graphs.
Lemma 2.5
If is a {triangle, ISK4}free graph, then either is a seriesparallel graph, or contains a wheel or contains .
Proof.
Clear from Lemma 2.3 and the fact that every prism contains a triangle. ∎
Lemma 2.6
If is an {ISK4, triangle}free graph that contains , then either is a complete bipartite graph, or has a cliquecutset of size at most 2.
Proof.
Clear from Lemma 2.4 and the fact that complete tripartite graphs and cliquecutsets of size 3 contain triangles. ∎
It is now easy to prove the next theorem.
Proof.
Suppose that Conjecture 1.1 is true. Let be a {triangle, ISK4}free graph. We prove Conjecture 1.2 by induction on . If , the outcome is clearly true.
If contains , then by Lemma 2.6, either is bipartite and therefore 3colorable, or has a cliquecutset . In this last case, we recover a 3coloring of from 3colorings of , …, where are the connected components of .
If contains no , then by Conjecture 1.1 it has a vertex of degree at most 2. By the induction hypothesis, has a 3coloring, and we 3color by giving to a color not used by its two neighbors. ∎
3 Proof of the decomposition theorem
Appendices to a hole
When is a vertex of a graph , denotes the neighborhood of , that is the set of all vertices of adjacent to . We set . When , we set . When is a graph, an induced subgraph of , and a set of vertices disjoint from , the attachment of to is , that we also denote by .
When is a path and , we denote by the subpath of . Let and be two disjoint vertex sets such that no vertex of is adjacent to a vertex of . A path connects and if either and has a neighbor in and a neighbor in , or and one of the two endvertices of is adjacent to at least one vertex in and the other is adjacent to at least one vertex in . is a direct connection between and if in no path connecting and is shorter than . The direct connection is said to be from to if is adjacent to a vertex of and is adjacent to a vertex of .
Let be a hole. A chordless path in is an appendix of if no vertex of has a neighbor in , and one of the following holds:

, and is not an edge, or

, , and .
So is an attachment of to . The two subpaths of are called the sectors of w.r.t. .
Let be another appendix of , with attachment . Appendices and are crossing if one sector of w.r.t. contains , the other contains and .
Lemma 3.1
If is an {ISK4, }free graph, then no two appendices of a hole of can be crossing.
Proof.
Let and be appendices of a hole of , and suppose that they are crossing. Let be the attachment of to , and let be the attachment of to . So and w.l.o.g. appear in this order when traversing . W.l.o.g. is adjacent to , and to .
A vertex of must be adjacent to or coincident with a vertex of , since otherwise induces an ISK4. Note that . Let be the vertex of with lowest index that has a neighbor in , and let (resp. ) be the vertex of with lowest (resp. highest) index adjacent to . Note that is not coincident with a vertex of .
First suppose that . If then induces an ISK4. So . In particular, no vertex of is coincident with a vertex of , and is the only edge between and . If then induces an ISK4. So and by symmetry . Since cannot induce a , w.l.o.g. is not an edge. Let be the subpath of that does not contain and . Then induces an ISK4. Therefore .
If has a unique neighbor in , then induces an ISK4. Let be the sector of w.r.t. that contains . If has exactly two neighbors in , then induces an ISK4. So has at least three neighbors in . In particular, . But then induces an ISK4. ∎
Lemma 3.2
Let be an {triangle, ISK4, }free graph. Let be a hole and , , a chordless path in . Suppose that or 2, or 2, no vertex of has a neighbor in , and . Then is an appendix of .
Proof.
Assume not. If one of or has one neighbor in and the other one has two neighbors in , then induces an ISK4. So , and hence (since is trianglefree) both and are appendices of . By Lemma 3.1, and cannot be crossing. So for a sector of w.r.t. , . But then induces an ISK4. ∎
Wheels
Let be a wheel contained in a graph . A sector is a subpath of whose endvertices are adjacent to and interior vertices are not. Two sectors are consecutive or adjacent if they have an endvertex in common.
Throughout this section we use the following notation for a wheel . We denote by the neighbors of in , appearing in this order when traversing . In this case, we also say that is an wheel. For , denotes the sector of whose endvertices are and (here and throughout this section we assume that indices are taken modulo ).
A path is an appendix of a wheel if the following hold:

is an appendix of ,

each of the sectors of w.r.t. properly contains a sector of , and

has at most one neighbor in .
Lemma 3.3
Let be an ISK4free graph. Let be an appendix of a wheel of , and let be a sector of w.r.t. . Then contains at least three neighbors of . In particular, contains at least two sectors of .
Proof.
Let be the attachment of to . Since is an appendix of , contains at least two neighbors of . Suppose contains exactly two neighbors of . If has a neighbor in , then induces an ISK4. So does not have a neighbor in . Since is an appendix of , properly contains a sector of and so w.l.o.g. is not adjacent to . Let be the other sector of w.r.t. , and let be the neighbor of in that is closest to . Note that since cannot induce an ISK4, , and hence and is not an edge. Let be the subpath of . Then induces an ISK4. ∎
A wheel of is proper if vertices are one of the following types:

type 0: ;

type 1: ;

type 2: and for some sector of , .
Lemma 3.4
Let be a {triangle, ISK4}free graph. If is a wheel of with fewest number of vertices, then is a proper wheel.
Proof.
Let . It follows from the following two claims that is of type 0, 1 or 2 w.r.t. , and hence that is proper.
(1) For every sector of , .
Otherwise, induces a wheel with fewer vertices than , a contradiction. This proves (1).
(2) For some sector of , .
Assume otherwise, and choose so that has a neighbor in and in , is not contained in a sector of and is minimized. W.l.o.g. and (since the case when is symmetric to the case when ). Let (resp. ) be the neighbor of in that is closest to (resp. ). Let (resp. ) be the neighbor of in that is closest to (resp. ). Let be the subpath of that contains . Note that by the choice of , vertex has no neighbor in the interior of and . Since is trianglefree, is not an edge, and if then is not an edge.
First suppose that has at least two neighbors in . Then by (3.4), has exactly two neighbors in . If then (by the choice of ), and hence induces an ISK4. So . If is an edge, then is not an edge (since is trianglefree) and hence induces an ISK4. So is not an edge. But then together with induces an ISK4.
Therefore has exactly one neighbor in , and by symmetry it has exactly one neighbor in . If then induces an ISK4. So . If contains at least three neighbors of , then (since and ) induces a wheel with center that has fewer vertices than , a contradiction. Therefore contains exactly two neighbors of . But then and hence together with induces an ISK4. This proves (2).
∎
Lemma 3.5
Let be a {triangle, ISK4, }free graph. Let be a proper wheel of with fewest number of spokes. If has an appendix, then is a 4wheel.
Proof.
Assume has an appendix. Note that by Lemma 3.3, if is an appendix of then each of the sectors of w.r.t. contains at least two sectors of . If has an appendix such that one of the sectors of w.r.t. this appendix is for some , then let be such an appendix and . Otherwise, let be an appendix of such that for a sector of w.r.t. , there is no appendix of such that properly contains a sector of w.r.t. . Assume further that such a is chosen so that is minimized. Let be the other sector of w.r.t. . W.l.o.g. we may assume that contains , , and does not contain . Let be the attachment of to such that and . Let and w.l.o.g. assume that is adjacent to , and to . Let be the subpath of , and the subpath of . Let be the hole induced by . Note that since , is a wheel.
(3) is a proper wheel.
Let and assume that is not of type 0, 1 or 2 w.r.t. . Note that . Since is a proper wheel, must have a neighbor in . Let . Let (resp. ) be the neighbor of in that is closest to (resp. ). Note that sectors of are also sectors of .
First suppose that . Since is proper, cannot have a neighbor in both and . If has a neighbor in , then by Lemma 3.2 and since has a neighbor in , contains an appendix of such that a sector of w.r.t. is properly contained in , contradicting our choice of . So does not have a neighbor in , and by symmetry it does not have a neighbor in . Since is not of type 0, 1 or 2 w.r.t. , , and since is trianglefree, is not an edge. Let be the chordless path from to in that contains . Since the length of cannot be less than the length of , by the choice of , there is a vertex such that and are edges. Since is trianglefree, is not adjacent to , and hence has exactly two neighbors in . Since is not of type 2 w.r.t. , must be adjacent to . But then induces an ISK4.
Therefore must have a neighbor in . Since is proper, for some , . Suppose that is of type 2 w.r.t. . W.l.o.g. is not adjacent to . Let be the vertex of with lowest index adjacent to . If then, since is trianglefree, not both and can be adjacent to , and hence induces an ISK4. So , and hence induces an ISK4. Therefore is of type 1 w.r.t. . Recall that by our assumption that has a neighbor in , is not adjacent to nor . But then contains an ISK4. This proves (3).
So is a proper wheel. Since it cannot have fewer sectors than and by Lemma 3.3, it follows that , and . But then by the choice of , , and hence is a 4wheel. ∎
A short connection between sectors and of is a chordless path , , in such that the following hold:

, ,

, , and

the only vertex of that may have a neighbor in is .
Lemma 3.6
Let be a {triangle, ISK4, }free graph. Let be a proper wheel of with fewest number of spokes. Then has no short connection.
Proof.
Suppose has a short connection . Assume that and are chosen so that is minimized. W.l.o.g. is adjacent to and to . Let be the subpath of , and let be the subpath of . Let be the subpath of that contains . Let be the hole induced by .
(4)
If then induces an ISK4. If then induces an ISK4. Therefore, . This proves (4).
(5) has no appendix.
(6) Vertex has a neighbor in .
Assume not. If has no neighbor in , then induces an ISK4. So w.l.o.g. is adjacent to . Then is not an edge, since is trianglefree. If is not adjacent to , then induces an ISK4. So is adjacent to , and hence is not an edge. But then induces an ISK4. This proves (6).
(7) is a proper wheel.
By (3.6) is a wheel. Assume it is not proper and let be such that it is not of type 0, 1 or 2 w.r.t. . Note that , and hence is of type 0, 1 or 2 w.r.t. . It follows that must have a neighbor in . Let (resp. ) be the vertex of with lowest (resp. highest) index adjacent to .
First suppose that . Then . If does not have a neighbor in , then has at least three neighbors in and hence is a short connection of that contradicts our choice of . So has a neighbor in . W.l.o.g. . Since is proper, . If has a unique neighbor in , then (since is trianglefree and has at least two neighbors in ) and hence is a short connection of that contradicts our choice of . So is of type 2 w.r.t. . If then and hence induces an ISK4. So . If is an edge then (since is trianglefree and has at least three neighbors in ) and hence is a short connection of that contradicts our choice of . So is not an edge. If has no neighbor in and , then let . Otherwise let be the chordless path from to in . Then induces an ISK4.
Therefore, for some , has a neighbor in . Since is proper, . If let , if let , and otherwise let be a chordless path in from to a vertex of that is adjacent to (note that such a vertex exists by (3.6)). By Lemma 3.2 is an appendix of . In particular, is of type 1 w.r.t. . Let be the neighbor of in . Since by (3.6) cannot be an appendix of , w.l.o.g. and . Note that and . Let be the vertex of with lowest index adjacent to (such a vertex exists by (3.6)). If then induces an ISK4. So . If If then together with the subpath of induces an ISK4. So . But then induces an ISK4. This proves (7).
By (3.6) is a proper wheel that has fewer sectors than , a contradiction. ∎
Lemma 3.7
Let be a {triangle, ISK4, }free graph. Let be a proper wheel of with fewest number of spokes. Then for every , is a star cutset separating from .
Proof.
Assume not. Then w.l.o.g. there is a direct connection from to in . Note that the only vertices of that may have a neighbor in the interior of are and . Since is proper, and and are of type 1 or 2 w.r.t. . Let be such that .
First suppose that no vertex of has a neighbor in . By Lemma 3.2, is an appendix of . In particular, and are both of type 1 w.r.t. . If then is an appendix of . It follows from Lemma 3.5 that and . But then, since has a neighbor in , and contradict Lemma 3.3. So , and hence is a short connection, contradicting Lemma 3.6. Therefore, a vertex of has a neighbor in .
Let be the vertex of with highest index adjacent to a vertex of . W.l.o.g. is an edge. We now show that if has two neighbors in , then has a neighbor in . Assume not. Then has two neighbors in and does not have a neighbor in . Since is of type 2 w.r.t. , is not an edge. But then and (where is the vertex of with lowest index adjacent to ) contradict Lemma 3.2.
Suppose . If has two neighbors in , then induces an ISK4. So, has a unique neighbor in . If does not have a neighbor in , then has a unique neighbor in and hence is a short connection of , contradicting Lemma 3.6. So has a neighbor in . Let be such a neighbor with highest index. Then is a short connection of , contradicting Lemma 3.6. So . If then either is a short connection of contradicting Lemma 3.6, or contains an ISK4.
We say that a graph has a wheel decomposition if for some wheel , for every , is a cutset separating from . We say that such a wheel decomposition is w.r.t. wheel . Note that if a graph has a wheel decomposition, then it has a star cutset.
Theorem 3.8
If is a {triangle, ISK4}free graph, then either is a seriesparallel graph or a complete bipartite graph, or has a clique cutset of size at most two, or has a wheel decomposition.
Proof.
Assume is not seriesparallel nor a complete bipartite graph. By Lemma 2.5 contains a wheel or . By Lemma 2.6 if contains a then it has a clique cutset of size at most two. So we may assume that does not contain a . So contains a wheel. By Lemma 3.4 contains a proper wheel, and hence by Lemma 3.7 has a wheel decomposition. ∎
Theorem 3.9
If is a {triangle, ISK4, }free graph, then either is seriesparallel or has a wheel decomposition.
Proof.
The following corollary is needed in the next section.
Corollary 3.10
If is an ISK4free graph of girth at least 5, then either is seriesparallel or has a star cutset.
Proof.
Follows directly from Theorem 3.9 because contains a cycle of length 4. ∎
4 Chordless graphs
A graph is chordless if no cycle in has a chord. Chordless graphs were introduced in [2] as roots of wheelfree line graphs, and it is a surprise to us that we need them here for a completely different reason in a very similar class. A graph is sparse if every edge is incident to at least one vertex of degree at most 2. A sparse graph is chordless because any chord of a cycle is an edge between two vertices of degree at least three. Recall that proper 2cutsets are defined in Section 2.
Theorem 4.1 (see [3])
If is a 2connected chordless graph, then either is sparse or admits a proper 2cutset.
The following theorem is mentioned in [3] without a proof, and we need it in the next section. So, we prove it for the sake of completeness.
Theorem 4.2
In every cycle of a 2connected chordless graph that is not a cycle, there exist four vertices that appear in this order and such that have degree 2 and b, d have degree at least 3.
Proof.
We prove the result by induction on . If is sparse (in particular, if ), then it is enough to check that every cycle of contains at least two vertices of degree at least 3, because these vertices cannot be adjacent in a sparse graph. But this true, because a cycle with all vertices of degree 2 must be the whole graph (since is connected), and a cycle with a unique vertex of degree 3 cannot exists in a 2connected graph (the vertex of degree at least 3 would be a cutvertex).
So, by Theorem 4.1 we may assume that has a proper 2cutset with split . We now build two blocks of decompositions of as follows. The block is obtained from by adding a marker vertex adjacent to and , and the block is obtained from by adding a marker vertex adjacent to and . By the definition of proper 2cutsets, . Also, and are chordless and 2connected. So we may apply the induction hypothesis to the blocks of decomposition.
Let be a cycle of . If , then is a cycle of , so by the induction hypothesis we get four vertices in . We now check that a vertex has degree 2 in if and only if it has degree 2 in . This is obvious, except if . But in this case, because of and because lies in a cycle of that does not contain , has degree at least 3 in both and . This proves our claim. It follows that we obtain by the induction hypothesis the condition that we need for the degrees of , , and . The proof is similar when .
We may now assume that has vertices in and . It follows that edge wise partitions into a path whose interior is in and a path whose interior is in . We apply the induction hypothesis to and . So, we get four vertices , , and in and they have degree 2, , 2, respectively (in ). These four vertices are in and have the degrees we need (in ), except possibly when . In this case, we may assume w.l.o.g. that or , and we find in place of a vertex of degree 2 in , where . This vertex exists by the induction hypothesis applied to and . ∎
5 Degree 2 vertices
We need the following application of Menger’s theorem.