Induced subgraphs of graphs with large chromatic number. VIII. Long odd holes
Maria Chudnovsky, Alex Scott, Paul Seymour, Sophie Spirkl

TL;DR
This paper proves a conjecture that graphs with bounded clique number and large chromatic number necessarily contain long odd cycles, advancing understanding of graph structure related to chromatic properties.
Contribution
It confirms Gyarfas's conjecture, establishing that such graphs always contain long odd holes, linking chromatic number to the existence of specific cycle lengths.
Findings
Graphs with bounded clique number and large chromatic number have long odd holes.
The proof confirms a longstanding conjecture in graph theory.
Provides new insights into the structure of complex graphs.
Abstract
We prove a conjecture of Andras Gyarfas, that for all k,t, every graph with clique number at most k and sufficiently large chromatic number has an odd hole of length at least t.
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Induced subgraphs of graphs with large chromatic number.
VIII. Long odd holes
Maria Chudnovsky
Princeton University, Princeton, NJ 08544, USA
Alex Scott
Mathematical Institute, University of Oxford, Oxford OX2 6GG, UK
Paul Seymour
Princeton University, Princeton, NJ 08544, USA
Sophie Spirkl
Princeton University, Princeton, NJ 08544, USA Supported by NSF grant DMS-1550991 and US Army Research Office Grant W911NF-16-1-0404.Supported by ONR grant N00014-14-1-0084 and NSF grant DMS-1265563.Current address: Rutgers University, New Brunswick, NJ 08901, USA
(December 12, 2016; revised )
Abstract
We prove a conjecture of András Gyárfás, that for all , every graph with clique number at most and sufficiently large chromatic number has an odd hole of length at least .
1 Introduction
All graphs in this paper are finite and have no loops or parallel edges. We denote the chromatic number of a graph by , and its clique number (the cardinality of its largest clique) by . A hole in means an induced subgraph which is a cycle of length at least four, and an odd hole is one with odd length. In [4], András Gyárfás proposed three conjectures about the lengths of holes in graphs with large chromatic number and bounded clique number, the following:
1.1
For all there exists such that for every graph , if and then has an odd hole.
1.2
For all there exists such that for every graph , if and , then has a hole of length at least .
1.3
For all there exists such that for every graph , if and then has an odd hole of length at least .
The third evidently contains the other two. The first was proved in [5], and the second in [2], but until now the third has remained open. In this paper we prove the third.
Since this paper was submitted for publication, two of us have proved an even stronger result [7]:
1.4
For all integers (with ), there exists such that for every graph , if and , then has a hole of length modulo .
There have been two other partial results approaching 1.3, in [1, 6]. The second is the stronger, namely:
1.5
For all , there exists such that for every graph , if and then has holes of consecutive lengths (and in particular has an odd hole of length at least ).
One might expect that an analogous result holds for all graphs with , for all fixed , as was conjectured in [6], but this remains open. We can prove that there are two holes both of length at least and with consecutive lengths, by a refinement of the methods of this paper, but we omit the details.
The proof of 1.3 will be by induction on , with fixed, so we may assume that and the result holds for all smaller . In particular there exists such that for every graph , if and has no odd hole of length at least then . We might as well assume that is odd and . Let us say a graph is a -candidate if , and has no odd hole of length at least , and every induced subgraph of with clique number less than is -colourable. We will show that for all there exists such that every -candidate has chromatic number at most .
If is a subgraph of , and not necessarily induced, then adjacency in may be different from in , and we speak of being -adjacent, a -neighbour, and so on, to indicate in which graph we are using adjacency when there may be confusion. If , the subgraph of induced on is denoted by , and we often write for . The distance or -distance between two vertices of is the length of a shortest path between , or if there is no such path. If and is an integer, or denotes the set of all vertices with distance exactly from , and or denotes the set of all with distance at most from . If is a nonnull graph and , we define to be the maximum of taken over all vertices of . (For the null graph we define .)
As in several other papers of this series, the proof of 1.3 examines whether there is an induced subgraph of large chromatic number such that every ball of small radius in it has bounded chromatic number. The proof breaks into three steps, as follows. We first show that for every -candidate , is bounded by a function of . Second, a similar argument shows the same for ; that is, there is a function such that for every -candidate . For the third step, we apply a result of [3], which implies that in these circumstances there is an upper bound on the chromatic number of all -candidates.
2 Gradings
Let be a graph. We say a grading of is a sequence of subsets of , pairwise disjoint and with union . If is such that for we say the grading is -colourable. We say that is earlier than (with respect to some grading ) if and where . We need some lemmas about gradings. The first is:
2.1
Let and let be a -colourable grading of a graph with . Then there exist subsets of with the following properties:
- •
* are both connected;*
- •
every vertex in is earlier than every vertex in ;
- •
some vertex in has a neighbour in ; and
- •
.
Proof. Let us say is right-active if there exists such that
- •
is connected;
- •
is earlier than every vertex in ;
- •
has a neighbour in ; and
- •
.
Let be the set of all vertices of that are not right-active, and . Since , either or . Suppose that , and let be a component of with maximum chromatic number. Choose with minimum such that . Since , it follows that . Let be a component of with maximum chromatic number; consequently . Choose with a neighbour in (this exists since is connected). But then is right-active with , a contradiction.
This proves that , and so . Let be a component of with maximum chromatic number. Choose with maximum such that , and choose . Since is right-active, there exists as in the definition of “right-active” above; and then setting satisfies the theorem. This proves 2.1.
Next, we need:
2.2
Let and let be a -colourable grading of a graph . Let be a subgraph of (not necessarily induced) with . Then there is an edge of , and a subset of , such that
- •
* is connected;*
- •
* are both earlier than every vertex in ;*
- •
exactly one of has a -neighbour in ; and
- •
.
Proof. We claim:
(1) There exist with the following properties:
- •
* and are connected;*
- •
every vertex of is earlier than every vertex of ;
- •
some vertex of has no -neighbour in , and some vertex of has a -neighbour in ; and
- •
.
By 2.1 applied to , there exist such that
- •
are both connected;
- •
every vertex in is earlier than every vertex in ;
- •
some vertex in has an -neighbour in ; and
- •
.
If some vertex of has no -neighbour in , then (1) holds, so we may assume that every vertex of has a -neighbour in . Choose and let be the set of vertices in that are -adjacent to . Let be a component of with maximum chromatic number. Since , it follows that . If some vertex of has a -neighbour in then (1) holds with , so we assume not. Choose with a -neighbour in . Then is connected, and some vertex in has a -neighbour in it, namely . We may therefore assume that every vertex of has a -neighbour in , and hence is -adjacent to , since no vertex of has a -neighbour in . But this is impossible since . This proves (1).
Let be as in (1). Since some vertex of has a -neighbour in and some vertex of has no -neighbour in , and is connected, it follows that there is an edge of with such that exactly one of has a -neighbour in ; and so the theorem holds. This proves 2.2.
We also need the following:
2.3
Let be a graph, let , let , and let , such that is connected, has a neighbour in , and . Then there is an induced path of where , and a subset of , with the following properties:
- •
;
- •
* is connected;*
- •
* has a neighbour in , and have no neighbours in ; and*
- •
.
Proof. We proceed by induction on ; the result holds if , so we assume that and the result holds for . Consequently there is an induced path of where , and a subset of , such that
- •
;
- •
is connected;
- •
has a neighbour in , and have no neighbours in ; and
- •
.
Let be the set of neighbours of , and let be the vertex set of a component of , chosen with maximum (there is such a component since ). Let be a neighbour of with a neighbour in . Then and satisfy the theorem. This proves 2.3.
If is a graph and , we say that covers if and every vertex in has a neighbour in . Let be a graph, and let , where covers . Let . For we say that is earlier than (with respect to the enumeration ). For , let be minimum such that are adjacent; we call the earliest parent of . An edge of is said to be square (with respect to the enumeration ) if the earliest parent of is nonadjacent to , and the earliest parent of is nonadjacent to . Let , and let be a grading of . We say the enumeration of and the grading are compatible if for all with earlier than , the earliest parent of is earlier than the earliest parent of .
2.4
Let be a graph, and let , where covers . Let every induced subgraph of with have chromatic number at most . Let the enumeration of and the grading of be compatible. Let be the subgraph of with vertex set and edge set the set of all square edges. Let be -colourable; then .
Proof. Let be a stable set of . We claim that . Suppose not. Since is -colourable, there is a partition of such that is stable in for and . Consequently there exists such that . From the choice of , it follows that . Let be a clique with . Choose maximum such that , and let . Let be the earliest parent of . Since , there exists nonadjacent to . Since , and are adjacent, it follows that , and so is earlier than , from the choice of . Since the enumeration and the grading are compatible, the earliest parent of is earlier than the earliest parent of , and in particular is nonadjacent to ; and so is a square edge, contradicting that . This proves that . Since can be partitioned into sets that are stable in , it follows that . This proves 2.4.
Combining these lemmas yields the main result of this section:
2.5
Let be a graph, and let , where covers . Let every induced subgraph of with have chromatic number at most . Let be a -colourable grading of , and let the enumeration of be compatible with this grading. Let be integers. Let
[TABLE]
Then there is an induced path of , such that
- •
;
- •
* is a square edge;*
- •
* are both earlier than all of ; and*
- •
let be the earliest parent of respectively; then for each , the -distance between and is at least .
Proof. Let . Thus
[TABLE]
Let be the subgraph of with edge set the square edges; then by 2.4,
[TABLE]
By 2.2, applied to and , there is a square edge , and a subset of , such that
- •
is connected;
- •
are both earlier than every vertex in ;
- •
exactly one of is adjacent in to a member of ; and
- •
.
Let where has a neighbour in . Since , 2.3 implies that there is an induced path of where , and a subset of , with the following properties:
- •
;
- •
is connected;
- •
has a neighbour in , and have no neighbours in ; and
- •
.
Let be the earliest parent of respectively, and let . Thus . Since , there exists such that the -distance between and each member of is at least . Let be an induced path of between and ; then satisfies the theorem. This proves 2.5.
3 Little bounding balls
In this section we carry out the first two steps of the proof, showing that the chromatic number of every -candidate can be bounded in terms of , and then can be bounded in terms of . More precisely we first prove the following.
3.1
Let be a -candidate; then .
Proof. We may assume that . Since some component of has the same chromatic number, we may assume that is connected. Choose some vertex, and for let be the set of vertices with -distance from the vertex. There exists such that . Since , it follows that . Let be the vertex set of a component of with maximum chromatic number; and choose with a neighbour in . For let be the set of vertices in with -distance from . Choose such that . Again, since , it follows that .
The set of vertices in with -distance at most three from has chromatic number at most ; so there is a set with such that every vertex in has -distance at least four from . Let be the set of vertices in with a neighbour in , and let be the set of vertices in with a neighbour in . Thus every vertex in has -distance at least three from .
Let . Thus is connected, and there are no edges between and . Let be the set of vertices in with no neighbour in . Let be the set of vertices in with a neighbour in such that the -distance between is odd, and let be the set where this distance is even. Every vertex in has a neighbour in at least one of ; let be the set of vertices in with a neighbour in for .
(1) .
Take an enumeration of , and for let be the set of vertices such that is adjacent to and nonadjacent to . Then is a grading of , and is -colourable (indeed, -colourable, but we need the bound), and the enumeration is compatible with it. Suppose that . By 2.5 with and , there is an induced path of , such that
- •
;
- •
is a square edge;
- •
are both earlier than all of ; and
- •
let be the earliest parent of respectively; then for each , the -distance between and is at least .
Since , there are induced paths of between and between respectively, both of length . Let be adjacent to . There is an induced path between with interior in , since . Now is nonadjacent to since the -distance between and is at least four. It follows that is an induced path between , and is an induced path between , of the same length. Also is nonadjacent to since has -distance four from all these vertices. Choose minimum such that is adjacent to . Now are both nonadjacent to since are both earlier than ; and since is nonadjacent to and is nonadjacent to (because is a square edge) it follows that and are both induced paths, joining and respectively. So the union of with is a hole, and the union of with is a hole; and these holes differ in length by one. Since they both have length more than , this is impossible. This proves (1).
(2) For ,
Enumerate the vertices of in increasing order of -distance from , breaking ties arbitrarily; that is, take an enumeration of where for the -distance between in is at most that between . For each , choose minimum such that some vertex in is adjacent both to and to . (Such a value of exists from the definition of .) We call the earliest grandparent of . For , let be the set of vertices in with earliest grandparent . Thus every vertex in has -distance two from , and so ; and it follows that is a -colourable grading of (indeed, it is -colourable, since ). Take an enumeration of such that vertices with earlier neighbours in come first; that is, such that for , there is a neighbour of such that for every neighbour of in . It follows that and are compatible.
Suppose that . By 2.5 with and , there is an induced path of , such that
- •
;
- •
is a square edge;
- •
are both earlier than all of ; and
- •
let be the earliest parent of respectively; then for each , the -distance between and is at least .
Let be the earliest parent of , and let be the earliest grandparents of respectively. It follows that are edges. Since both occur in the enumeration before , and is the earliest grandparent of , there are induced paths of between and between respectively, with lengths of the same parity as (because ), such that has no neighbours in . There is an induced path between with interior in , using only two vertices of (neighbours of respectively). Now since , they both have -distance at least four from ; and so both have -distance at least three from . Moreover both have -distance at least four from and hence at least three from . Consequently they both have -distance at least two from the vertices of in (it is to arrange this that we need to control the chromatic number of balls of radius three); and so both have no neighbours in . It follows that is an induced path between , and is an induced path between , and they have lengths of the same parity. Now is nonadjacent to since the -distance between and is at least four. Choose minimum such that is adjacent to . Also, are nonadjacent to since are earlier than ; and since is square, the union of with and the union of with are holes of opposite parity, both of length more than , which is impossible. This proves (2).
From (1) and (2), we deduce that . But , so . Since , and , we deduce that
[TABLE]
This proves 3.1.
Next we bound in terms of , as follows.
3.2
Let be a -candidate; then .
Proof. Let be a vertex such that , and let be the set of vertices with -distance from , for . Fix a -colouring of , and for each choose a path of length three from to , and let be the colours of its second and third vertex respectively. Choose colours , and let be the set of such that and . Let be the set of vertices in with colour and with a neighbour in with colour . Consequently covers , and any two vertices in are joined by an induced path of even length with interior in .
Let be some enumeration of , and for let be the set of such that is adjacent to but not to . Thus is a -colourable grading of compatible with . Suppose that . Then by 2.5 with and there is an induced path of , such that
- •
;
- •
is a square edge;
- •
are both earlier than all of ; and
- •
let be the earliest parent of respectively; then for each , the -distance between and is at least .
It follows that are nonadjacent to ; let be adjacent to , and choose minimum such that is adjacent to . Since has -distance at least three from each of , it follows that is nonadjacent to all these vertices. Choose minimum such that is adjacent to ; then and are induced paths both of length more than and with lengths of opposite parity. Let be one of them with odd length. There is an induced path of even length joining the ends of with interior in ; and its union with is an odd hole of length more than , which is impossible.
This proves that . Since this holds for every choice of , and there are only such choices, it follows that But , so . From 3.1, it follows that . This proves 3.2.
4 Multicovers
In this section we combine 3.2 with a result of [3] to deduce 1.3, and for this we need some definitions. If are disjoint subsets of the vertex set of a graph , we say
- •
is complete to if every vertex in is adjacent to every vertex in ; and
- •
is anticomplete to if every vertex in nonadjacent to every vertex in .
(If we say is complete to instead of , and so on.)
Let , let be some set of neighbours of , and let be disjoint from , such that is anticomplete to and covers . In this situation we call a cover of in . For , a multicover of in is a family such that
- •
is stable;
- •
for each is a cover of ;
- •
for all distinct , is anticomplete to (and in particular all the sets are pairwise disjoint).
Its length is ; and the multicover is stable if each of the sets is stable.
4.1
For all , suppose that , and for every induced subgraph of with . If admits a multicover with length of a set with , then admits a stable multicover contained in with length , of some subset with .
Proof. Let be a multicover in of length , of a set with . For each , since has clique number less than , this subgraph is -colourable; choose some such colouring, with colours (for each ). For each , let such that for each , some neighbour of in has colour . There are only possibilities for , so there is a function and a subset with , such that for all . For each , let be the set of vertices in with colour ; then is a stable multicover of . This proves 4.1.
Let be a multicover of in . It is said to be -crested if there are vertices and vertices of , all distinct, with the following properties:
- •
and the vertices do not belong to ;
- •
for and each , is adjacent to , and there are no other edges between the sets and ;
- •
for and each , is adjacent to , and there are no other edges between and
- •
are pairwise nonadjacent;
- •
for all distinct and all , is nonadjacent to .
Note that may be adjacent to . We say the multicover is stably -crested if for and all distinct , are nonadjacent. Theorem 2.1 of [2] (setting of the theorem to be ) implies:
4.2
For all there exist with the following property. Let be a graph with , such that for every induced subgraph of with . Let be a stable multicover in of some set , such that and . Then there exist with , and with , and a stable multicover of contained in that is 1-crested.
Ramsey’s theorem applied to the vertices together with 4.2 yields (under the same hypotheses) that such a stable multicover exists which is stably 1-crested; and combining this with 4.1 yields:
4.3
For all there exist with the following property. Let be a graph with , such that for every induced subgraph of with . Let be a multicover in of some set , such that and . Then there exist with , and with , and a stable multicover of contained in that is stably 1-crested.
Repeated application of this yields:
4.4
For all there exist with the following property. Let be a graph with , such that for every induced subgraph of with . Let be a multicover in of some set , such that and . Then there exist with , and with , and a stable multicover of contained in that is stably -crested.
We deduce:
4.5
For all , there exist such that if satisfies:
- •
;
- •
* for every induced subgraph of with ; and*
- •
* admits a multicover of length at least of a set with ;*
then has an odd hole of length at least .
Proof. Let satisfy 4.4, with both replaced by and with . Let be as in the theorem; so admits a stably -crested stable multicover of a set , where and . Let and the vertices be as in the definition of stable -crested. Choose distinct . Let be a multiple of four. There is an induced path between of length such that . Since , there is a -clique . Choose with as many neighbours in as possible. By exchanging if necessary, we may assume that . Now is not complete to since ; so there exists nonadjacent to . Choose adjacent to . From the choice of , there exists adjacent to and not to . If are adjacent let be the path , and otherwise let be the path . Thus is induced, and has length three or five, and the union of and is an odd hole of length more than . This proves 4.5.
Let denote the set of nonnegative integers, let be a nondecreasing function, and let be an integer. We say a graph is -controlled if for every induced subgraph of . Consequently 3.2 implies that:
4.6
Every -candidate is -controlled where is the function
[TABLE]
We need the following, a consequence of theorem 9.7 of [3]. That involves “trees of lamps”, but we do not need to define those here; all we need is that a cycle of length is a tree of lamps. (Note that what we call a “multicover” here is called a “strongly-independent 2-multicover” in that paper, and indexed in a slightly different way.)
4.7
Let , and let be non-decreasing. Then there exists with the following property. Let be a graph such that
- •
;
- •
* is -controlled;*
- •
* does not admit a multicover of length of a set with chromatic number more than ; and*
- •
* has no hole of length .*
Then .
Finally, we can prove 1.3, which we restate:
4.8
For all there exists such that for every graph , if and then has an odd hole of length at least .
Proof. As we saw in section 1, it suffices to prove an upper bound on the chromatic number of all -candidates, for all . Let satisfy 4.5, and let be as in 4.6. Let satisfy 4.7, with replaced by .
Now let be a -candidate. If admits a multicover of length of a set with chromatic number more than , then by 4.5, has an odd hole of length at least , which is impossible. Thus does not contain such a multicover. Since is -controlled by 4.6, it follows from 4.7 that . This proves 4.8.
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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- 2[2] Maria Chudnovsky, Alex Scott and Paul Seymour, “Induced subgraphs of graphs with large chromatic number. III. Long holes”, Combinatorica , 37 (2017), 1057–72.
- 3[3] Maria Chudnovsky, Alex Scott and Paul Seymour, “Induced subgraphs of graphs with large chromatic number. V. Chandeliers and strings”, submitted for publication, ar Xiv:1609.00314 .
- 4[4] András Gyárfás, “Problems from the world surrounding perfect graphs”, Proceedings of the International Conference on Combinatorial Analysis and its Applications , (Pokrzywna, 1985), Zastos. Mat. 19 (1987), 413–441.
- 5[5] Alex Scott and Paul Seymour, “Induced subgraphs of graphs with large chromatic number. I. Odd holes”, J. Combinatorial Theory, Ser. B , 121 (2016), 68–84.
- 6[6] Alex Scott and Paul Seymour, “Induced subgraphs of graphs with large chromatic number. IV. Consecutive holes”, J. Combinatorial Theory, Ser B. 132 (2018), 180–235, ar Xiv:1509.06563 .
- 7[7] Alex Scott and Paul Seymour, “Induced subgraphs of graphs with large chromatic number. X. Holes with specific residue”, Combinatorica , to appear, ar Xiv:1705.04609 .
