 # Coloring graphs with no induced five-vertex path or gem

For a graph G, let χ(G) and ω(G) respectively denote the chromatic number and clique number of G. We give an explicit structural description of (P_5,gem)-free graphs, and show that every such graph G satisfies χ(G)<5ω(G)/4. Moreover, this bound is best possible.

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## 1 Introduction

All our graphs are finite and have no loops or multiple edges. For any integer , a -coloring of a graph is a mapping such that any two adjacent vertices in satisfy . A graph is -colorable if it admits a -coloring. The chromatic number of a graph is the smallest integer such that is -colorable. A clique in a graph is a set of pairwise adjacent vertices, and the clique number of , denoted by , is the size of a maximum clique in . Clearly for every induced subgraph of . A graph is perfect if every induced subgraph of satisfies . Following Gyárfás , we say that a class of graphs is -bounded if there is a function (called a -bounding function) such that every member of the class satisfies . Thus the class of perfect graphs is -bounded with .

For any integer we let denote the path on vertices and denote the cycle on vertices. The gem is the graph that consists of a plus a vertex adjacent to all vertices of the . A hole (antihole) in a graph is an induced subgraph that is isomorphic to () with , and is the length of the hole (antihole). A hole or an antihole is odd if

is odd. Given a family of graphs

, a graph is -free if no induced subgraph of is isomorphic to a member of ; when has only one element we say that is -free; when has two elements and , we simply write is ()-free instead of -free. Here we are interested on -boundedness for the class of (, gem)-free graphs.

Gyárfás  showed that the class of -free graphs is -bounded. Gravier et al.  improved Gyárfás’s bound slightly by proving that every -free graph satisfies . It is well known that every -free graph is perfect. The preceding result implies that every -free graph satisfies . The problem of determining whether the class of -free graphs admits a polynomial -bounding function remains open, and the known -bounding function for such class of graphs satisfies ; see . So the recent focus is on obtaining -bounding functions for some classes of -free graphs, in particular, for ()-free graphs, for various graphs . The first author and Sivaram  showed that every ()-free graph satisfies , and that every (, bull)-free graph satisfies . Fouquet et al.  proved that there are infinitely many ()-free graphs with , where , and that every ()-free graph satisfies . The second author with Choudum and Shalu  studied the class of (, gem)-free graphs and showed that every such graph satisfies . Later Cameron, Huang and Merkel  impvroved this result replacing with . In this paper we establish the best possible bound, as follows.

###### Theorem 1

Let be a (, gem)-free graph. Then . Moreover, this bound is tight.

The degree of a vertex in a graph is the number of vertices adjacent to it. The maximum degree over all vertices in is denoted by . Clearly every graph satisfies . Reed  conjectured that every graph satisfies . Reed’s conjecture is still open in general. It is shown in  that if a graph satisfies , then . So by Theorem 1, we immediately have the following theorem.

###### Theorem 2

Let be a (, gem)-free graph. Then . Moreover, this bound is tight.

The bounds in Theorem 1 and in Theorem 2 are tight on the following example. Let be a graph whose vertex-set is partitioned into five cliques such that for each , every vertex in is adjacent to every vertex in and to no vertex in , and for all (). It is easy to check that is (, gem)-free. Moreover, we have , , and since has no stable set of size .

Our proof of Theorem 1 uses the structure theorem for (, gem)-free graphs (Theorem 3). Before stating it we recall some definitions.

Let be a graph with vertex-set and edge-set . For any two subsets and of , we denote by , the set of edges that has one end in and other end in . We say that is complete to or is complete if every vertex in is adjacent to every vertex in ; and is anticomplete to if . If is singleton, say , we simply write is complete (anticomplete) to instead of writing is complete (anticomplete) to . For any , let denote the set of all neighbors of in ; and let . The neighborhood of a subset is the set . If , then denote the subgraph induced by in . A set is a homogeneous set if every vertex with a neighbor in is complete to . Note that in any gem-free graph , for every , induces a -free graph, and hence the subgraph induced by a homogeneous set in is -free.

An expansion of a graph is any graph such that can be partitioned into non-empty sets , , such that is complete if , and if . An expansion of a graph is a clique expansion if each is a clique, and is a -free expansion if each induces a -free graph.

Let be the ten graphs shown in Figure 1. Clearly each of is (, gem)-free. Moreover, it is easy to check that any -free expansion of a (, gem)-free graph is (, gem)-free.

Let be the class of connected (, gem)-free graphs such that can be partitioned into seven non-empty sets such that:

• Each induces a -free graph.

• is complete and .

• is complete and .

• is complete and .

• and .

• The vertex-set of each component of is a homogeneous set.

• (Adjacency between and is not specified, but it is restricted by the fact that is (, gem)-free.)

Now we can state our structural result.

###### Theorem 3

Let be a connected (, gem)-free graph that contains an induced . Then either or is a -free expansion of either , , …, or .

We note that another structure theorem for (, gem)-free graphs using a recursive construction is given by Brandstädt and Kratsch . However, it seems difficult to use that theorem to get the bounds derived in this paper.

## 2 Proof of Theorem 3

Let be a connected -free graph. Since contains an induced , there are five non-empty and pairwise disjoint sets such that for each modulo  the set is complete to and anticomplete to . Let . We choose these sets such that is maximal. From now on every subscript is understood modulo . Let has no neighbor in , and for each let:

 Yi = {x∈V(G)∖A∣x is complete to Ai, % anticomplete to Ai−1∪Ai+1, and x has a neighbor in each of Ai−2 and Ai+2, and x is complete to one of Ai−2 and Ai+2}.
###### Claim 3.1

.

Proof. Consider any . For each let be a neighbor of in (if any such vertex exists) and be a non-neighbor of in (if any exists). Let exists. Then since . Up to symmetry there are four cases:
(a) or for some . Then ---- is a , a contradiction.
(b) or for some . Then is complete to , for otherwise we find a as in case (a). But then can be added to , contradicting the maximality of .
(c) for some . Then is complete to , for otherwise we find a as in case (a), and similarly must be complete to one of and . So is in .
(d) . Then induces a gem for some , a contradiction.

###### Claim 3.2

For each , and are -free.

Proof. Since is gem-free, the claim follows by the definitions of and .

###### Claim 3.3

For each we have .

Proof. Pick any and . We know that has neighbors and , and has a neighbor . Then , for otherwise induces a gem, a contradiction.

We say that a vertex in is pure if it is complete to , and the set is pure if every vertex in is pure.

###### Claim 3.4

Suppose that there exists a pure vertex in for some . Then is pure.

Proof. We may assume that and let be pure. Suppose to the contrary that there exists a vertex and is not pure, say has a non-neighbor . Moreover, by the definition of , has a neighbor . Then , for otherwise ---- is a for any and . Also, for any and , since does not induce a gem, we have . But, then for any , induces a gem, a contradiction.

###### Claim 3.5

For each , we have: either is complete or is complete.

Proof. We may assume that . Suppose to the contrary that there exist vertices and in such that has a non-neighbor and has a non-neighbor . By the definition of , is complete to , and has a neighbor . Likewise, is complete to , and has a neighbor . Then , for otherwise ---- is a for any and . Also, for any , since does not induce a gem, we have . But, then induces a gem, a contradiction.

###### Claim 3.6

Suppose that is complete for some . Let and . Then (i) , and (ii) is complete.

Proof. : Suppose to the contrary that there are adjacent vertices and . Pick a neighbor of in , say . Clearly . Then for any and , ---- is a , a contradiction. This proves item .

: Suppose to the contrary that there are non-adjacent vertices and . Pick a neighbor of in , say . By the definition of , has a neighbor in , say . Pick any , and . Now, , for otherwise ---- is a . Also, , for otherwise induces a gem. Then since does not induce a gem, . But then ---- is a , a contradiction. This proves item .

###### Claim 3.7

Suppose that and are both non-empty for some . Let and . Then:

1. is complete, , and ,

2. ,

3. and are complete,

4. ,

5. Any vertex in is pure,

6. One of the sets and is empty.

Proof. Pick any and . So has neighbors , and , and has neighbors , and .

: Now , for otherwise ---- is a . Since this holds for arbitrary , we obtain that is complete. Then , for otherwise induces a gem, and similarly . In particular ; moreover , for otherwise ---- is a . Since this holds for any , it proves item (a).

: Suppose that there are adjacent vertices and , say . Then ---- is a , a contradiction. This proves item (b).

: Since and are not complete to (by ), by Claim 3.5, and are complete. Also, by Claim 3.6(ii), and are complete. This proves item (c).

: If then, by a similar argument as in the proof of (with subscripts shifted by ), should be complete, which it is not. So , and similarly . This proves item (d).

: Consider any and suppose that it is not pure; up to symmetry has a non-neighbor and is complete to . By Claim 3.3 we know that . Then ---- is a . This proves item (e).

: Suppose that there are vertices and . By the definition of , we know that , and by Claim 3.3, . Then by item (c) and item (e), for any , ---- is a , a contradiction. This proves item (f).

###### Claim 3.8

The vertex-set of each component of is a homogeneous set, and hence each component of is -free.

Proof. Suppose that a vertex-set of a component of is not homogeneous. Then there are adjacent vertices and a vertex with and . By Claim 3.1 we have for some . Then ---- is a , for any and , a contradiction.

###### Claim 3.9

Suppose that there is any edge with and . Then is pure and . Moreover if any of the sets is non-empty then exactly one of them is non-empty, and is complete to that non-empty set and to .

Proof. Consider any edge with and . So has a neighbor for each . If is not pure, then up to symmetry has a non-neighbor , and then ---- is a for any , a contradiction. So is pure, and by Claim 3.7 . Now suppose up to symmetry that there is a vertex . By Claim 3.3 we have . Then , for otherwise ---- is a , for any . Now by the same argument as above, is pure, and by Claim 3.7 . Since this holds for any , the vertex is complete to , and then by symmetry is complete to ; and by Claim 3.8 and the fact that is connected, is complete to .

It follows from the preceding claims that at most three of the sets are non-empty, and if then at most two of are non-empty. Hence we have the following cases:

1. and . Any of may be non-empty.
We may assume that both and are not empty, is complete and is complete. (Otherwise, using Claims  3.2, 3.5 and 3.6, it follows that is a -free expansion of either or .) Suppose there exists that has a non-neighbor , and there exists that has a non-neighbor , then for any , ---- is a in , a contradiction. So either is pure or is pure. Then by Claims 3.23.5 and 3.6, we see that is a -free expansion of or .

2. and are both non-empty.
Then Claims 3.2 and 3.7 implies that is a -free expansion of either , or .

3. and exactly one of is non-empty, say is non-empty.
In this case, we show that as follows: Since , there exists a vertex and a vertex such that . Then by Claim 3.9, is a pure vertex of . So, by Claim 3.4, is pure, and hence by Claims 3.2 and 3.8, we see that .

4. and exactly two of are non-empty.
In this case, by Claims 3.8 and 3.9 and up to symmetry we may assume that and are non-empty, all vertices in are pure, and is complete. Moreover, since is gem-free, is -free. So by Claim 3.2, is a -free expansion of .

This completes the proof of Theorem 3.

## 3 Bounding the chromatic number

A stable set is a set of pairwise non-adjacent vertices. We say that two sets meet if their intersection is not empty. In a graph , we say that a stable set is good if it meets every clique of size . Moreover, we say that a clique in is a -clique of if .

We use the following theorem often.

###### Theorem 4 ()

Let be a graph such that every proper induced subgraph of satisfies . Suppose that one of the following occurs:

1. has a vertex of degree at most .

2. has a good stable set.

3. has a stable set such that is perfect.

4. For some integer the graph has stable sets such that .

Then .

Given a graph and a proper homogeneous set in , let be the graph obtained by replacing with a clique of size (i.e., is obtained from and by adding all edges between and the vertices of that are adjacent to in ).

###### Lemma 1 ()

In a graph let be a proper homogeneous set such that is -free. Then and . Moreover, has a good stable set if and only if has a good stable set.

For , let be the class of graphs that are -free expansions of , and let be the class of graphs that are clique expansions of . Let be the class of graphs such that, with the notation as in Section 1, the five sets , and the vertex-set of each component of are cliques.

The following lemma can be proved using Lemma 1, and the proof is very similar to that of Lemma 3.3 of , so we omit the details.

###### Lemma 2

For every graph in () (resp. in ) there is a graph in () (resp. in ) such that and . Moreover, has a good stable set if and only if has a good stable set.

By Lemma 2 and Theorem 3, to prove Theorem 1, it suffices to consider the clique expansions of and the members of .

### 3.1 Coloring clique expansions

###### Theorem 5

Let be a clique expansion of either or , and assume that every induced subgraph of satisfies . Then .

Proof. Throughout the proof of this theorem, we use the following notation: Let . Suppose that is a clique expansion of . So there is a partition of into non-empty cliques , where corresponds to the vertex of . We write, e.g., instead of , instead of , etc. For each we call one vertex of . Moreover if we call one vertex of . Recall that if has a good stable set, then we can conclude the theorem using Theorem 4(ii).

(I) Suppose that is a clique expansion of . (We refer to [9, 10] for alternate proofs.) We may assume that , for each , otherwise if (say), then is perfect, as it is a clique expansion of , and we can conclude with Theorem 4(iii). Let be a subset of obtained by taking two vertices from for each . Then since and , by hypothesis, we have .

(II) Suppose that is a clique expansion of . Then is a good stable set of , and we can conclude with Theorem 4(ii).

(III) Suppose that is a clique expansion of . Suppose that . By hypothesis we can color with colors. Since is complete to , which is equal to , we can extend this coloring to , using for the colors used for . Therefore let us assume that . It follows that , so is not a -clique. Likewise we may assume that , and consequently is not a -clique. Then:

is a -clique, for otherwise is a good stable set of .

is a -clique, for otherwise is a good stable set of .

is a -clique, for otherwise is a good stable set of .

is a -clique, for otherwise is a good stable set of .

is a -clique, for otherwise is a good stable set of .

The above properties imply that there is an integer with such that and . Since , we have . Since , . So is even, and .

Now consider the five stable sets , , , and . It is easy to see that their union meets every -clique four times. It follows that , and we can conclude using Theorem 4(iv).

(IV) Suppose that is a clique expansion of either or . Suppose that . By hypothesis we can color with colors. Since is complete to , which is equal to , we can extend this coloring to , using for the colors used for . Therefore let us assume that . It follows that , so is not a -clique. Likewise we may assume that (for otherwise any -coloring of can be extended to ), and consequently is not a -clique. Now if is a clique expansion of , then is a good stable set of , and if is a clique expansion of , then is a good stable set of . In either case, we can conclude the theorem with Theorem 4(ii).

(V) Suppose that is a clique expansion of . Suppose that . By hypothesis we can color with colors. Since is complete to , which is equal to , we can extend this coloring to , using for the colors used for . Therefore let us assume that . It follows that and , and consequently and are not -cliques. Likewise we may assume that (for otherwise any -coloring of can be extended to ), and consequently is not a -clique. Then:

is a -clique, for otherwise is a good stable set of .

is a -clique, for otherwise