On graphs with no induced five-vertex path or paraglider

03/27/2019 ∙ by Shenwei Huang, et al. ∙ 0

Given two graphs H_1 and H_2, a graph is (H_1, H_2)-free if it contains no induced subgraph isomorphic to H_1 or H_2. For a positive integer t, P_t is the chordless path on t vertices. A paraglider is the graph that consists of a chorless cycle C_4 plus a vertex adjacent to three vertices of the C_4. In this paper, we study the structure of (P_5, paraglider)-free graphs, and show that every such graph G satisfies χ(G)<3/2ω(G) , where χ(G) and ω(G) are the chromatic number and clique number of G, respectively. Our bound is attained by the complement of the Clebsch graph on 16 vertices. More strongly, we completely characterize all the (P_5, paraglider)-free graphs G that satisfies χ(G)> 3/2ω(G). We also construct an infinite family of (P_5, paraglider)-free graphs such that every graph G in the family has χ(G)=3/2ω(G) -1. This shows that our upper bound is optimal up to an additive constant and that there is no (3/2-ϵ)-approximation algorithm to the chromatic number of (P_5, paraglider)-free graphs for any ϵ>0.

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1 Introduction

Graphs in this paper are simple and finite. Given a positive integer , we denote the path on vertices by , and we denote the complete graph on vertices by . For an integer , is the cycle on vertices. A paraglider is the graph that consists of a plus a vertex adjacent to three vertices of the . Given two graphs and , we denote by the disjoint union of and , and by the join of and . The union of copies of the same graph will be denoted by ; for example denotes the graph that consists in two disjoint copies of . The complement of a graph is denoted by . 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.

For any integer , a -coloring of a graph is a mapping such that whenever and are adjacent in . 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 . Obviously for every induced subgraph of . A graph is perfect if every induced subgraph of satisfies . Chudnovsky et al. [8] showed that a graph is perfect if and only if it does not contain an odd hole or an odd antihole as an induced subgraph, and is known as the Strong Perfect Graph Theorem (SPGT). A class of graphs is said to be -bounded [16] if there is a function (called a -binding function) such that every satisfies . For instance, the class of perfect graphs is -bounded with identity function as the -binding function. In fact, several classes of graphs are known to be -bounded; see [12, 14, 17, 18].

Gyárfás [16] studied the -boundedness for the class of -free graphs, and showed that every -free graph has . It is well known that for , -free graphs are perfect. The problem of determining whether the class of -free graphs () admits a polynomial -binding function remains open, and seems to be difficult even when . Moreover, the existence of polynomial -binding function for the class of -free graphs () would imply the Erdös-Hajnal conjecture for -free graphs; see [5]. The best known -binding function for the class of -free graphs satisfies ; see [13]. Here we are interested in -binding functions for the class of (, )-free graphs, for various graphs . Recently, Brause et al. [2] showed that the class of (, )-free graphs does not admit a linear -binding function. It follows that the class of (, )-free graphs, where is any -free graph with independence number , does not admit a linear -binding function. Thus it is interesting to the study of -boundedness for the class of (, )-free graphs where . Choudum et al. [3] showed that every (, )-free graph satisfies , and that every (, )-free graph satisfies . It is shown in [11, 18] that every (, diamond)-free graph satisfies , and in [2] that every (, paw)-free graph satisfies . Chudnovsky and Sivaram [7] showed that every (, )-free graph satisfies . Fouquet et al. [9] proved that there are infinitely many ()-free graphs with , where , and that every ()-free graph satisfies . Very recently, Chudnovsky et al. [6] showed that every (, )-free graph satisfies . We refer to a recent comprehensive survey of Schiermeyer and Randerath [18] for more results.

In this paper, we study the structure of the class of (, paraglider)-free graphs, and show that every such graph satisfies . Our bound is attained by the complement of the well-known -regular Clebsch graph on 16 vertices. More strongly, we completely characterize all the (, paraglider)-free graphs that satisfies . We also construct an infinite family of (, paraglider)-free graphs such that every graph in the family has . This shows that our upper bound is optimal up to an additive constant, and that there is no -approximation algorithm to the chromatic number of (, paraglider)-free graphs for any . Moreover, our results generalizes the results known on the existence of linear -binding functions for (, )-free graphs, (, paw)-free graphs, (, diamond)-free graphs, and for (, paraglider)-free graphs [4].

2 Notations and Preliminaries

We use standard notation and terminology. In a graph , the neighborhood of a vertex is the set ; we drop the subscript when there is no ambiguity. The non-neighborhood of a vertex is the set , and is denoted by . A vertex is universal if it is adjacent to all other vertices. Two non-adjacent vertices and in a graph are comparable if or . For any and , we let . Let be a subset of . We denote by the subgraph induced by in . For simplicity, we write instead of . Further if is singleton, say , we write instead of . 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 . We say that a subgraph of is dominating if every vertex in is a adjacent to a vertex in . A clique-cutset of a graph is a clique in such that has more connected components than . An atom is a connected graph without a clique-cutset.

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 .

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, is a -free expansion if each induces a -free graph, and is a perfect expansion if each induces a perfect graph. By a classical result of Lovász [15], any perfect expansion of a perfect graph is perfect. In particular, any -free expansion of a perfect graph is perfect.

Figure 1: Some special graphs

Let be five graphs as shown in Figure 1.

Let be the class of graphs such that can be partitioned into five sets such that:

  • , (where ), are cliques, is a perfect matching, say and .

  • and are perfect.

  • , are complete, and .

  • is complete.

  • There exists an injective function such that for each vertex , is anti-complete to , and is complete to .

  • No other edges in .

Clearly, the graphs and belong to . See Section 4 for more examples.

We will use the following theorem of Brandstädt and Hoàng [1].

Theorem 1 ([1])

Let be a (, paraglider)-free atom that has no universal or pair of comparable vertices. Then either is or every induced in is dominating.

3 Structure of (, paraglider)-free graphs

In this section, we prove the following structure theorem for the class of (, paraglider)-free graphs.

Theorem 2

Let be a (, paraglider)-free atom with no universal or pair of comparable vertices. Then one of the following hold:

  • is an induced subgraph of the complement of the Clebsch graph.

  • is a -free expansion of .

  • has a stable set such that either is good or is perfect.

  • .

Proof. If is , then is a stable set such that is perfect. If is perfect, then any color class in a -coloring of is a good stable set. So we may assume that is not , and is not perfect. Now since a -free graph contains no hole of length at least , and a paraglider-free graph contains no antihole of length at least , it follows by the Strong Perfect Graph Theorem [8] that contains a hole of length . That is, contains a as an induced subgraph. Now the theorem follows from Theorem 1, and from Theorems 4, 5, 6 and Theorem 7 given below.

In the next theorem, we make some general observations about the situation when a (, paraglider)-free graph contains a hole (which must have length ).

Theorem 3

Let be any (, paraglider)-free graph that contains a with vertex-set and . Suppose that is an atom and has no pair of comparable vertices. Let:

Moreover, let , , , and . Then the following properties hold for all , :

  1. is a dominating induced subgraph of and .

  2. (a) . If , then we denote by .
    (b) , for every ; so is an independent set.
    (c) is complete, and , for every .

  3. (a) is an independent set.
    (b) is complete.
    (c) .
    (d) If , then .

  4. (a) is -free. Hence is a complete multi-partite graph.
    (b) is complete.
    (c) If , then is a matching.
    (d) If , then and are cliques.
    (e) If and if is complete, then . More generally, if is complete, then and .

  5. (a) . If , then we denote by .
    (b) .
    (c) is complete.

  6. (a) is complete.
    (b) .

  7. (a) is complete.
    (b) .

  8. (a) is a clique.
    (b) is complete.

  9. Suppose that . Then:
    (a) The sets , , are empty.
    (b) , and are empty.

  10. Let and . Then the following hold:
    (a) If and are adjacent, then is either complete or anti-complete to .
    (b) If and are not adjacent, then is adjacent one of , .
    (c) If and are adjacent, then is adjacent to one of , .
    (d) is a stable set.

  11. Suppose that is -free. Then , and hence .

Proof. Let be the given graph with vertex-set and edge-set .

Proof of (R1). Since has no clique cut-set, by Theorem 1, is dominating, and so every vertex in has a neighbor in . Now (R1) follows since is -free. Indeed if a vertex has exactly one neighbor (say, ) or has exactly two neighbors that are consecutive (say, and ) in , then ---- is a .

Proof of (R2). : Otherwise, for any two vertices and in , either or induces a paraglider. So (a) holds.
: Suppose to the contrary that there are adjacent vertices and . Now if , then induces a paraglider, and if , then induces a paraglider, a contradiction. This proves item .
: Pick a vertex and a vertex . Up to symmetry, we may assume that . If , then , for otherwise induces a paraglider. If , then , for otherwise induces a paraglider. If , then , for otherwise ---- is a . Since this holds for any and , it proves item (c).

Proof of (R3). : Otherwise, for any two adjacent vertices and in , induces a paraglider. So (a) holds.
: Suppose not, and let and be not adjacent. Then ---- is a . So (b) holds.
: We may assume that . We first claim that is a matching. Suppose not. Then, up to symmetry, we may assume that there exist vertices and such that . By (a), . But then induces a paraglider. So is a matching. Now, if , then there exist matching edges , say and with and . By (a), we have and . But then ---- is a , a contradiction. This proves item (c).

Proof of (R4). :  Suppose to the contrary that contains an induced with vertex-set and edge-set . Then induces a paraglider, which is a contradiction. So (a) holds.
: Suppose not, and let and be not adjacent. Then ---- is a . So (b) holds.
: Suppose not. We may assume, up to symmetry, that and such that . Then or induces a paraglider, a contradiction. This proves item (c).
: Let . Suppose to the contrary that there are non-adjacent vertices and in . By (b), . But then induces a paraglider which is a contradiction. So is a clique. Likewise, is a clique. This proves item (d).
:  This follows by item (c).

Proof of (R5). : Otherwise, for any two vertices and in , either or induces a paraglider.
: Suppose not, and let and be adjacent. Then induces a paraglider.
: Suppose not, and let and be not adjacent. Then induces a paraglider.

Proof of (R6). : Suppose not, and let and be non-adjacent. By symmetry, we may assume that . Now if , then induces a paraglider, and if , then ---- is a , a contradiction. This proves item (a).
: Suppose not, and let and be adjacent. By symmetry, we may assume that . Now if , then ---- is a , and if , then induces a paraglider, a contradiction. This proves item (b).

Proof of (R7). : Suppose not. Up to symmetry, we may assume that there are non-adjacent vertices and . Now if , then induces a paraglider, and if , then ---- is a , a contradiction. This proves item (a).
: Suppose not, and let and be adjacent. Then induces a paraglider.

Proof of (R8). Suppose not, and let and be non-adjacent. If , then induces a paraglider. So let us assume that . Then there exist , modulo  such that and . But then induces a paraglider, a contradiction.

Proof of (R9). Let .
: Suppose to the contrary that there exists a vertex . First suppose that . Then since ---- or ---- is not a , we have . But then or induces a a paraglider. So . Likewise, . Next suppose that . Then since does not induce a paraglider, we have . But then ---- is a . So . Likewise, . Next suppose that . Then since does not induce a paraglider, . But then ---- is a . So . Finally, suppose that . Up to symmetry, we may assume that . Then since does not induce a paraglider, we have . But then induces a paraglider, a contradiction. This proves item (a).
: Suppose that there is an edge in one of the listed sets. If and , then by (R2:c), we have ; and then ---- is a . If and , then by (R2:c), we have ; and then ---- is a . If and , then induces a paraglider. These contradictions show that (b) holds.

Proof of (R10). : Suppose not. Up to symmetry, we may assume that and . Then either or induces a paraglider, a contradiction. So (a) holds.
: Otherwise, ---- is a .
: Otherwise, induces a paraglider.
: This follows by item (c).

Proof of (R11). Suppose to contrary that there are adjacent vertices and . We may assume, up to symmetry, that . Now induces an , a contradiction. So, .

Now we show that . Suppose to the contrary that and let , say for some . We claim that and are comparable. Since is -free, . Now by the preceding point, by the definition of , and by (R3),(R6), (R8:b), and since is dominating, we see that , and . So, and , and hence we conclude that and are comparable, a contradiction. So (R11) holds.

This completes the proof of Theorem 3.

Theorem 4

Let be a (, paraglider)-free atom with no universal vertex. Suppose that contains . Then has a stable set such that is a bipartite graph or a bull. In particular, is perfect.

Proof. Let be the given graph with vertex-set and edge-set . First suppose that contains an . Consider the graph as shown in Figure 1 and let . We use the same notation as in Theorem 3 and use the properties in Theorem 3. Then by (R2:a), and . Moreover, by (R9:a), , and . Then since , any vertex in is a universal vertex of (by (R8)), and hence . Also, by (R9:b), and . Now, let us define , , and . Then by (R3:a) and (R2:c), the set is a stable set. Also, by the preceding points and (R3:a), we see that , and and are stable sets. Hence is bipartite.

Suppose that contains no . Consider the graph as shown in Figure 1 and let . We use the same notation as in Theorem 3 and use the properties in Theorem 3. Since has no , by (R2:a), . Then by (R9), the sets , , and are empty. Then since , any vertex in is a universal vertex of (by (R8)), and hence . Also, if there are adjacent vertices and , then induces an . So . Likewise, .

Suppose that . Then let us define , , and . Then by (R3:a) and (R2:c), the set is a stable set. Also, by the preceding points and (R3:a), we see that , and and are stable sets. Hence is bipartite.

So let us assume that , and by (R5:a), . Then by (R9:b), . Now we claim that , for . Suppose not. Up to symmetry, we may assume that there exists a vertex . If , then by (R2:c) and (R6:a), we have and . But then ---- is a . If , then since does not induce a paraglider, . But then ---- is a . So, we conclude that , for . Now, by (R3:a) and (R6:b), the set is a stable set such that