Coloring graphs with no induced five-vertex path or gem
aa r X i v : . [ m a t h . C O ] J a n Coloring graphs with no induced five-vertex pathor gem ∗ M. Chudnovsky † T. Karthick ‡ P. Maceli § Fr´ed´eric Maffray ¶ January 3, 2019
Abstract
For a graph G , let χ ( G ) and ω ( G ) respectively denote the chromaticnumber and clique number of G . We give an explicit structural descrip-tion of ( P , gem)-free graphs, and show that every such graph G satisfies χ ( G ) ≤ ⌈ ω ( G )4 ⌉ . Moreover, this bound is best possible. Keywords : P -free graphs; Chromatic number; Clique number; χ -boundedness. All our graphs are finite and have no loops or multiple edges. For any integer k ,a k -coloring of a graph G is a mapping φ : V ( G ) → { , . . . , k } such that any twoadjacent vertices u, v in G satisfy φ ( u ) = φ ( v ). A graph is k -colorable if it admitsa k -coloring. The chromatic number χ ( G ) of a graph G is the smallest integer k such that G is k -colorable. A clique in a graph G is a set of pairwise adjacentvertices, and the clique number of G , denoted by ω ( G ), is the size of a maximumclique in G . Clearly χ ( H ) ≥ ω ( H ) for every induced subgraph H of G . A graph G is perfect if every induced subgraph H of G satisfies χ ( H ) = ω ( H ). FollowingGy´arf´as [8], we say that a class of graphs is χ -bounded if there is a function f (called a χ -bounding function ) such that every member G of the class satisfies χ ( G ) ≤ f ( ω ( G )). Thus the class of perfect graphs is χ -bounded with f ( x ) = x .For any integer ℓ we let P ℓ denote the path on ℓ vertices and C ℓ denote thecycle on ℓ vertices. The gem is the graph that consists of a P plus a vertexadjacent to all vertices of the P . A hole ( antihole ) in a graph is an inducedsubgraph that is isomorphic to C ℓ ( C ℓ ) with ℓ ≥
4, and ℓ is the length of thehole (antihole). A hole or an antihole is odd if ℓ is odd. Given a family of graphs ∗ Dedicated to the memory of Professor Fr´ed´eric Maffray. † Princeton University, Princeton, NJ 08544, USA. Supported by NSF grant DMS-1763817.This material is based upon work supported in part by the U. S. Army Research Laboratoryand the U. S. Army Research Office under grant number W911NF-16-1-0404. ‡ Computer Science Unit, Indian Statistical Institute, Chennai Centre, Chennai 600029,India. § Adelphi University, Garden City, NY 11530, USA. ¶ Deceased on August 22, 2018. , a graph G is F -free if no induced subgraph of G is isomorphic to a memberof F ; when F has only one element F we say that G is F -free; when F has twoelements F and F , we simply write G is ( F , F )-free instead of { F , F } -free.Here we are interested on χ -boundedness for the class of ( P , gem)-free graphs.Gy´arf´as [8] showed that the class of P t -free graphs is χ -bounded. Gravieret al. [7] improved Gy´arf´as’s bound slightly by proving that every P t -free graph G satisfies χ ( G ) ≤ ( t − ω ( G ) − . It is well known that every P -free graphis perfect. The preceding result implies that every P -free graph G satisfies χ ( G ) ≤ ω ( G ) − . The problem of determining whether the class of P -free graphsadmits a polynomial χ -bounding function remains open, and the known χ -bounding function f for such class of graphs satisfies c ( ω / log w ) ≤ f ( ω ) ≤ ω ;see [11]. So the recent focus is on obtaining χ -bounding functions for someclasses of P -free graphs, in particular, for ( P , H )-free graphs, for various graphs H . The first author and Sivaram [5] showed that every ( P , C )-free graph G satisfies χ ( G ) ≤ ω ( G ) − , and that every ( P , bull)-free graph G satisfies χ ( G ) ≤ (cid:0) ω ( G )+12 (cid:1) . Fouquet et al. [6] proved that there are infinitely many( P , P )-free graphs G with χ ( G ) ≥ ω ( G ) α , where α = log −
1, and thatevery ( P , P )-free graph G satisfies χ ( G ) ≤ (cid:0) ω ( G )+12 (cid:1) . The second author withChoudum and Shalu [3] studied the class of ( P , gem)-free graphs and showedthat every such graph G satisfies χ ( G ) ≤ ω ( G ). Later Cameron, Huang andMerkel [2] impvroved this result replacing 4 ω with ⌊ ω ( G )2 ⌋ . In this paper weestablish the best possible bound, as follows. Theorem 1
Let G be a ( P , gem)-free graph. Then χ ( G ) ≤ ⌈ ω ( G )4 ⌉ . Moreover,this bound is tight. The degree of a vertex in a graph G is the number of vertices adjacent toit. The maximum degree over all vertices in G is denoted by ∆( G ). Clearlyevery graph G satisfies ω ( G ) ≤ χ ( G ) ≤ ∆( G ) + 1. Reed [12] conjectured thatevery graph G satisfies χ ( G ) ≤ ⌈ ∆( G )+ ω ( G )+12 ⌉ . Reed’s conjecture is still openin general. It is shown in [9] that if a graph G satisfies χ ( G ) ≤ ⌈ ω ( G )4 ⌉ , then χ ( G ) ≤ ⌈ ∆( G )+ ω ( G )+12 ⌉ . So by Theorem 1, we immediately have the followingtheorem. Theorem 2
Let G be a ( P , gem)-free graph. Then χ ( G ) ≤ ⌈ ∆( G )+ ω ( G )+12 ⌉ .Moreover, this bound is tight. The bounds in Theorem 1 and in Theorem 2 are tight on the followingexample. Let G be a graph whose vertex-set is partitioned into five cliques Q , . . . , Q such that for each i mod 5, every vertex in Q i is adjacent to everyvertex in Q i +1 ∪ Q i − and to no vertex in Q i +2 ∪ Q i − , and | Q i | = q for all i ( q > G is ( P , gem)-free. Moreover, we have ω ( G ) = 2 q , ∆( G ) = 3 q −
1, and χ ( G ) ≥ ⌈ q ⌉ since G has no stable set of size 3.Our proof of Theorem 1 uses the structure theorem for ( P , gem)-free graphs(Theorem 3). Before stating it we recall some definitions.Let G be a graph with vertex-set V ( G ) and edge-set E ( G ). For any twosubsets X and Y of V ( G ), we denote by [ X, Y ], the set of edges that has one2nd in X and other end in Y . We say that X is complete to Y or [ X, Y ]is complete if every vertex in X is adjacent to every vertex in Y ; and X is anticomplete to Y if [ X, Y ] = ∅ . If X is singleton, say { v } , we simply write v iscomplete (anticomplete) to Y instead of writing { v } is complete (anticomplete)to Y . For any x ∈ V ( G ), let N ( x ) denote the set of all neighbors of x in G ; andlet deg G ( x ) := | N ( x ) | . The neighborhood N ( X ) of a subset X ⊆ V ( G ) is theset { u ∈ V ( G ) \ X | u is adjacent to a vertex of X } . If X ⊆ V ( G ), then G [ X ]denote the subgraph induced by X in G . A set X ⊆ V ( G ) is a homogeneousset if every vertex with a neighbor in X is complete to X . Note that in anygem-free graph G , for every v ∈ V ( G ), N ( v ) induces a P -free graph, and hencethe subgraph induced by a homogeneous set in G is P -free.An expansion of a graph H is any graph G such that V ( G ) can be partitionedinto | V ( H ) | non-empty sets Q v , v ∈ V ( H ), such that [ Q u , Q v ] is complete if uv ∈ E ( H ), and [ Q u , Q v ] = ∅ if uv / ∈ E ( H ). An expansion of a graph is a cliqueexpansion if each Q v is a clique, and is a P -free expansion if each Q v inducesa P -free graph. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x G G G G G x x x x x x x x G x x x x x x x x G x x x x x x x x G x x x x x x x x x G x x x x x x x x x G Figure 1: Basic graphsLet G , G , . . . , G be the ten graphs shown in Figure 1. Clearly each of G , . . . , G is ( P , gem)-free. Moreover, it is easy to check that any P -freeexpansion of a ( P , gem)-free graph is ( P , gem)-free.Let H be the class of connected ( P , gem)-free graphs G such that V ( G ) canbe partitioned into seven non-empty sets A , . . . , A such that: • Each A i induces a P -free graph. • [ A , A ∪ A ∪ A ] is complete and [ A , A ∪ A ∪ A ] = ∅ . • [ A , A ∪ A ∪ A ] is complete and [ A , A ∪ A ] = ∅ . • [ A , A ∪ A ] is complete and [ A , A ∪ A ] = ∅ . • [ A , A ∪ A ∪ A ] = ∅ and [ A , A ∪ A ] = ∅ . • The vertex-set of each component of G [ A ] is a homogeneous set.(Adjacency between A and A is not specified, but it is restricted by thefact that G is ( P , gem)-free.) 3ow we can state our structural result. Theorem 3
Let G be a connected ( P , gem)-free graph that contains an induced C . Then either G ∈ H or G is a P -free expansion of either G , G , . . . , G or G . We note that another structure theorem for ( P , gem)-free graphs using a re-cursive construction is given by Brandst¨adt and Kratsch [1]. However, it seemsdifficult to use that theorem to get the bounds derived in this paper. Let G be a connected ( P , gem )-free graph. Since G contains an induced C ,there are five non-empty and pairwise disjoint sets A , ..., A such that for each i modulo 5 the set A i is complete to A i − ∪ A i +1 and anticomplete to A i − ∪ A i +2 .Let A = A ∪ · · · ∪ A . We choose these sets such that A is maximal. From nowon every subscript is understood modulo 5. Let R = { x ∈ V ( G ) \ A | x has noneighbor in A } , and for each i let: Y i = { x ∈ V ( G ) \ A | x is complete to A i , anticomplete to A i − ∪ A i +1 , and x has a neighbor in each of A i − and A i +2 , and x is completeto one of A i − and A i +2 } . Claim 3.1 V ( G ) = A ∪ · · · ∪ A ∪ Y ∪ · · · ∪ Y ∪ R .Proof. Consider any x ∈ V ( G ) \ ( A ∪ R ). For each i let a i be a neighbor of x in A i (if any such vertex exists) and b i be a non-neighbor of x in A i (if any exists).Let L = { i | a i exists } . Then L = ∅ since x / ∈ R . Up to symmetry there arefour cases:(a) L = { i } or { i, i + 1 } for some i . Then x - a i - b i − - b i − - b i − is a P , a contra-diction.(b) L = { i − , i + 1 } or { i − , i, i + 1 } for some i . Then x is complete to A i − ∪ A i +1 , for otherwise we find a P as in case (a). But then x can be addedto A i , contradicting the maximality of A .(c) L = { i, i − , i + 2 } for some i . Then x is complete to A i , for otherwise wefind a P as in case (a), and similarly x must be complete to one of A i − and A i +2 . So x is in Y i .(d) | L | ≥
4. Then { x, a i , a i +1 , a i +2 , a i +3 } induces a gem for some i , a contra-diction. ♦ Claim 3.2
For each i , G [ A i ] and G [ Y i ] are P -free.Proof. Since G is gem-free, the claim follows by the definitions of A i and Y i . ♦ Claim 3.3
For each i we have [ Y i − , Y i +1 ] = ∅ .Proof. Pick any y ∈ Y i − and z ∈ Y i +1 . We know that y has neighbors a i +1 ∈ A i +1 and a i +2 ∈ A i +2 , and z has a neighbor a i − ∈ A i − . Then yz / ∈ E ( G ), forotherwise { a i − , z, a i +1 , a i +2 , y } induces a gem, a contradiction. ♦
4e say that a vertex in Y i is pure if it is complete to A i − ∪ A i +2 , and theset Y i is pure if every vertex in Y i is pure. Claim 3.4
Suppose that there exists a pure vertex in Y i for some i . Then Y i ispure.Proof. We may assume that i = 1 and let p ∈ Y be pure. Suppose to thecontrary that there exists a vertex y ∈ Y and is not pure, say y has a non-neighbor b ∈ A . Moreover, by the definition of Y , y has a neighbor a ∈ A .Then b a / ∈ E ( G ), for otherwise b - a - y - a - a is a P for any a ∈ A and a ∈ A . Also, for any a ∈ A and a ∈ A , since { a , y, a , b , p } does notinduce a gem, we have py / ∈ E ( G ). But, then for any a ∈ A , { b , p, a , y, a } induces a gem, a contradiction. ♦ Claim 3.5
For each i , we have: either [ Y i , A i +2 ] is complete or [ Y i , A i − ] iscomplete.Proof. We may assume that i = 1. Suppose to the contrary that there existvertices y and y in Y such that y has a non-neighbor b ∈ A and y hasa non-neighbor b ∈ A . By the definition of Y , y is complete to A , andhas a neighbor a ∈ A . Likewise, y is complete to A , and has a neighbor a ∈ A . Then a b / ∈ E ( G ), for otherwise b - a - y - a - a is a P for any a ∈ A and a ∈ A . Also, for any a ∈ A , since { a , y , a , b , y } does not inducea gem, we have y y / ∈ E ( G ). But, then { b , y , a , y , a } induces a gem, acontradiction. ♦ Claim 3.6
Suppose that [ Y i , A i − ] is complete for some i . Let A ′ i +2 = N ( Y i ) ∩ A i +2 and A ′′ i +2 = A i +2 \ A ′ i +2 . Then (i) [ A ′ i +2 , A ′′ i +2 ] = ∅ , and (ii) [ Y i , A ′ i +2 ] iscomplete.Proof. ( i ): Suppose to the contrary that there are adjacent vertices p ∈ A ′ i +2 and q ∈ A ′′ i +2 . Pick a neighbor of p in Y i , say y . Clearly yq / ∈ E ( G ). Then forany a i ∈ A i and a i − ∈ A i − , q - p - y - a i - a i − is a P , a contradiction. This provesitem ( i ).( ii ): Suppose to the contrary that there are non-adjacent vertices y ∈ Y i and p ∈ A ′ i +2 . Pick a neighbor of p in Y i , say y ′ . By the definition of Y i , y has aneighbor in A ′ i +2 , say q . Pick any a i − ∈ A i − , a i − ∈ A i − and a i ∈ A i . Now, pq / ∈ E ( G ), for otherwise p - q - y - a i - a i − is a P . Also, yy ′ / ∈ E ( G ), for otherwise { p, a i − , y, a i , y ′ } induces a gem. Then since { y, q, y ′ , p, a i − } does not inducea gem, qy ′ / ∈ E ( G ). But then p - y ′ - a i - y - q is a P , a contradiction. This provesitem ( ii ). ♦ Claim 3.7
Suppose that Y i − and Y i +2 are both non-empty for some i . Let A − i = N ( Y i − ) ∩ A i and A + i = N ( Y i +2 ) ∩ A i . Then:(a) [ Y i − , Y i +2 ] is complete, A − i ∩ A + i = ∅ , and [ A − i , A + i ] = ∅ ,(b) [ A i \ ( A − i ∪ A + i ) , A − i ∪ A + i ] = ∅ ,(c) [ Y i − , A i +1 ∪ A − i ] and [ Y i +2 , A i − ∪ A + i ] are complete, d) Y i − ∪ Y i +1 = ∅ ,(e) Any vertex in Y i is pure,(f ) One of the sets A i \ ( A − i ∪ A + i ) and Y i is empty.Proof. Pick any y ∈ Y i − and z ∈ Y i +2 . So y has neighbors a i − ∈ A i − , a i +1 ∈ A i +1 and a i ∈ A i , and z has neighbors a i +2 ∈ A i +2 , a i − ∈ A i − and b i ∈ A i .( a ): Now yz ∈ E ( G ), for otherwise y - a i +1 - a i +2 - z - a i − is a P . Since this holdsfor arbitrary y, z , we obtain that [ Y i − , Y i +2 ] is complete. Then za i / ∈ E ( G ),for otherwise { a i , a i +1 , y, z, a i − } induces a gem, and similarly yb i / ∈ E ( G ). Inparticular a i = b i ; moreover a i b i / ∈ E ( G ), for otherwise a i - b i - z - a i +2 - a i − is a P . Since this holds for any y, z, a i , b i , it proves item (a).( b ): Suppose that there are adjacent vertices u ∈ A i \ ( A − i ∪ A + i ) and v ∈ A − i ∪ A + i ,say v ∈ A − i . Then u - v - y - a i − - a i +2 is a P , a contradiction. This proves item (b).( c ): Since y and z are not complete to A i (by ( a )), by Claim 3.5, [ Y i − , A i +1 ]and [ Y i +2 , A i − ] are complete. Also, by Claim 3.6(ii), [ Y i − , A − i ] and [ Y i +2 , A + i ]are complete. This proves item (c).( d ): If Y i − = ∅ then, by a similar argument as in the proof of ( c ) (withsubscripts shifted by 1), [ Y i − , A i ] should be complete, which it is not. So Y i − = ∅ , and similarly Y i +1 = ∅ . This proves item (d).( e ): Consider any x ∈ Y i and suppose that it is not pure; up to symmetry x hasa non-neighbor b ∈ A i +2 and is complete to A i − . By Claim 3.3 we know that xz / ∈ E ( G ). Then a i - x - a i − - b - z is a P . This proves item (e).( f ): Suppose that there are vertices b ∈ A i \ ( A − i ∪ A + i ) and u ∈ Y i . By thedefinition of Y i , we know that bu ∈ E ( G ), and by Claim 3.3, uy, uz / ∈ E ( G ).Then by item (c) and item (e), for any a i − ∈ A i − , b - u - a i − - y - z is a P , acontradiction. This proves item (f). ♦ Claim 3.8
The vertex-set of each component of G [ R ] is a homogeneous set,and hence each component of G [ R ] is P -free.Proof. Suppose that a vertex-set of a component T of G [ R ] is not homogeneous.Then there are adjacent vertices u, t ∈ V ( T ) and a vertex y ∈ V ( G ) \ V ( T ) with yu ∈ E ( G ) and yt / ∈ E ( G ). By Claim 3.1 we have y ∈ Y i for some i . Then t - u - y - a i - a i +1 is a P , for any a i ∈ A i and a i +1 ∈ A i +1 , a contradiction. ♦ Claim 3.9
Suppose that there is any edge ry with r ∈ R and y ∈ Y i . Then y ispure and Y i − ∪ Y i +1 = ∅ . Moreover if any of the sets Y i − , Y i +2 is non-emptythen exactly one of them is non-empty, and R is complete to that non-empty setand to Y i .Proof. Consider any edge ry with r ∈ R and y ∈ Y i . So y has a neighbor a j ∈ A j for each j ∈ { i, i − , i + 2 } . If y is not pure, then up to symmetry y has a non-neighbor b ∈ A i − , and then r - y - a i - a i − - b is a P for any a i − ∈ A i − , a contradiction. So y is pure, and by Claim 3.7 Y i − ∪ Y i +1 = ∅ . Nowsuppose up to symmetry that there is a vertex z ∈ Y i +2 . By Claim 3.3 we6ave yz / ∈ E ( G ). Then rz ∈ E ( G ), for otherwise r - y - a i +2 - z - a i − is a P , forany a i − ∈ A i − ∩ N ( z ). Now by the same argument as above, z is pure, andby Claim 3.7 Y i +1 ∪ Y i +3 = ∅ . Since this holds for any z , the vertex r is completeto Y i +2 , and then by symmetry r is complete to Y i ; and by Claim 3.8 and thefact that G is connected, R is complete to Y i ∪ Y i +2 . ♦ It follows from the preceding claims that at most three of the sets Y , ..., Y are non-empty, and if R = ∅ then at most two of Y , ..., Y are non-empty. Hencewe have the following cases:(A) R = ∅ and Y ∪ Y ∪ Y = ∅ . Any of Y , Y may be non-empty.We may assume that both Y and Y are not empty, [ Y , A ] is complete and[ Y , A ] is complete. (Otherwise, using Claims 3.2, 3.5 and 3.6, it followsthat G is a P -free expansion of either G , G , . . . , G or G .) Suppose thereexists y ∈ Y that has a non-neighbor a ∈ A , and there exists y ∈ Y that has a non-neighbor a ∈ A , then for any a ∈ A , a - y - a - a - y isa P in G , a contradiction. So either Y is pure or Y is pure. Then byClaims 3.2, 3.5 and 3.6, we see that G is a P -free expansion of G or G .(B) R = ∅ and Y , Y are both non-empty.Then Claims 3.2 and 3.7 implies that G is a P -free expansion of either G , G or G .(C) R = ∅ and exactly one of Y , ..., Y is non-empty, say Y is non-empty.In this case, we show that G ∈ H as follows: Since R = ∅ , there existsa vertex r ∈ R and a vertex y ∈ Y such that ry ∈ E ( G ). Then byClaim 3.9, y is a pure vertex of Y . So, by Claim 3.4, Y is pure, and henceby Claims 3.2 and 3.8, we see that G ∈ H .(D) R = ∅ and exactly two of Y , ..., Y are non-empty.In this case, by Claims 3.8 and 3.9 and up to symmetry we may assume that Y and Y are non-empty, all vertices in Y ∪ Y are pure, and [ R, Y ∪ Y ] iscomplete. Moreover, since G is gem-free, G [ R ] is P -free. So by Claim 3.2, G is a P -free expansion of G .This completes the proof of Theorem 3. (cid:3) 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 G , we say that a stable set is good if it meets every clique of size ω ( G ). Moreover, we say that a clique K in G isa t -clique of G if | K | = t .We use the following theorem often. Theorem 4 ([10])
Let G be a graph such that every proper induced subgraph G ′ of G satisfies χ ( G ′ ) ≤ ⌈ ω ( G ′ ) ⌉ . Suppose that one of the following occurs:(i) G has a vertex of degree at most ⌈ ω ( G ) ⌉ − .(ii) G has a good stable set. iii) G has a stable set S such that G \ S is perfect.(iv) For some integer t ≥ the graph G has t stable sets S , . . . , S t such that ω ( G \ ( S ∪ · · · ∪ S t )) ≤ ω ( G ) − ( t − .Then χ ( G ) ≤ ⌈ ω ( G ) ⌉ . Given a graph G and a proper homogeneous set X in G , let G/X be the graphobtained by replacing X with a clique Q of size ω ( X ) (i.e., G/X is obtainedfrom G \ X and Q by adding all edges between Q and the vertices of V ( G ) \ X that are adjacent to X in G ). Lemma 1 ([9])
In a graph G let X be a proper homogeneous set such that G [ X ] is P -free. Then ω ( G ) = ω ( G/X ) and χ ( G ) = χ ( G/X ) . Moreover, G hasa good stable set if and only if G/X has a good stable set.
For k ∈ { , , . . . , } , let G k be the class of graphs that are P -free expan-sions of G k , and let G ∗ k be the class of graphs that are clique expansions of G k . Let H ∗ be the class of graphs G ∈ H such that, with the notation as inSection 1, the five sets A , A , . . . , A , and the vertex-set of each component of G [ A ] are cliques.The following lemma can be proved using Lemma 1, and the proof is verysimilar to that of Lemma 3.3 of [9], so we omit the details. Lemma 2
For every graph G in G i ( i ∈ { , . . . , } ) (resp. G in H ) there is agraph G ∗ in G ∗ i ( i ∈ { , . . . , } ) (resp. G ∗ in H ∗ ) such that ω ( G ) = ω ( G ∗ ) and χ ( G ) = χ ( G ∗ ) . Moreover, G has a good stable set if and only if G ∗ has a goodstable set. By Lemma 2 and Theorem 3, to prove Theorem 1, it suffices to consider theclique expansions of G , G . . . , G and the members of H ∗ . Theorem 5
Let G be a clique expansion of either G , . . . , G or G , and as-sume that every induced subgraph G ′ of G satisfies χ ( G ′ ) ≤ ⌈ ω ( G ′ ) ⌉ . Then χ ( G ) ≤ ⌈ ω ( G ) ⌉ .Proof. Throughout the proof of this theorem, we use the following notation: Let q = ω ( G ). Suppose that G is a clique expansion of H ∈ { G , . . . , G } . So thereis a partition of V ( G ) into | V ( H ) | non-empty cliques Q , . . . , Q | V ( H ) | , where Q i corresponds to the vertex x i of H . We write, e.g., Q instead of Q ∪ Q , Q instead of Q ∪ Q ∪ Q , etc. For each i ∈ { , . . . , | V ( H ) |} we call x i one vertexof Q i . Moreover if | Q i | ≥ x ′ i one vertex of Q i \ { x i } . Recall that if G has a good stable set, then we can conclude the theorem using Theorem 4(ii).(I) Suppose that G is a clique expansion of G . (We refer to [9, 10] for alternateproofs.) We may assume that | Q i | ≥
2, for each i ∈ { , . . . , } , otherwise if | Q | = 1 (say), then G \ { x } is perfect, as it is a clique expansion of P , andwe can conclude with Theorem 4(iii). Let X be a subset of V ( G ) obtained bytaking two vertices from Q i for each i ∈ { , . . . , } . Then since χ ( G [ X ]) = 5and ω ( G \ X ) = q −
4, by hypothesis, we have χ ( G ) ≤ ⌈ ω ( G \ X ) ⌉ + 5 ≤ ⌈ q ⌉ .8II) Suppose that G is a clique expansion of G . Then { x , x , x } is a goodstable set of G , and we can conclude with Theorem 4(ii).(III) Suppose that G is a clique expansion of G . Suppose that | Q | ≤ | Q | .By hypothesis we can color G \ Q with ⌈ q ⌉ colors. Since Q is complete to Q ∪ Q , which is equal to N ( Q ), we can extend this coloring to Q , usingfor Q the colors used for Q . Therefore let us assume that | Q | > | Q | . Itfollows that | Q | > | Q | , so Q is not a q -clique. Likewise we may assumethat | Q | > | Q | , and consequently Q is not a q -clique. Then: Q is a q -clique, for otherwise { x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x } is a good stable set of G .The above properties imply that there is an integer a with 1 ≤ a ≤ q − | Q | = | Q | = | Q | = a and | Q | = | Q | = q − a . Since | Q | > | Q | ,we have a ≥
2. Since q = | Q | = 2 a , a = q . So q is even, q ≥ | Q | = | Q | = | Q | = | Q | = | Q | = q ≥ { x , x , x } , { x ′ , x } , { x , x , x ′ } , { x , x ′ } and { x ′ , x ′ } . It is easy to see that their union U meets every q -clique four times.It follows that ω ( G \ U ) = q −
4, and we can conclude using Theorem 4(iv).(IV) Suppose that G is a clique expansion of either G or G . Suppose that | Q | ≤ | Q | . By hypothesis we can color G \ Q with ⌈ q ⌉ colors. Since Q iscomplete to Q ∪ Q , which is equal to N ( Q ), we can extend this coloring to Q ,using for Q the colors used for Q . Therefore let us assume that | Q | > | Q | .It follows that | Q | > | Q | , so Q is not a q -clique. Likewise we may assumethat | Q | > | Q | (for otherwise any ⌈ q ⌉ -coloring of G \ Q can be extended to Q ), and consequently Q is not a q -clique. Now if G is a clique expansion of G , then { x , x , x } is a good stable set of G , and if G is a clique expansionof G , then { x , x , x , x } is a good stable set of G . In either case, we canconclude the theorem with Theorem 4(ii).(V) Suppose that G is a clique expansion of G . Suppose that | Q | ≤ | Q | .By hypothesis we can color G \ Q with ⌈ q ⌉ colors. Since Q is complete to Q ∪ Q ∪ Q , which is equal to N ( Q ), we can extend this coloring to Q , usingfor Q the colors used for Q . Therefore let us assume that | Q | > | Q | . Itfollows that | Q | > | Q | and | Q | > | Q | , and consequently Q and Q are not q -cliques. Likewise we may assume that | Q | > | Q | (for otherwise any ⌈ q ⌉ -coloring of G \ Q can be extended to Q ), and consequently Q is not a q -clique. Then: Q is a q -clique, for otherwise { x , x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x } is a good stable set of G . Q is a q -clique, for otherwise { x , x , x } is a good stable set of G .Now we claim that Q is not a q -clique. Suppose not. Then the above propertiesimply that there is an integer a with 1 ≤ a ≤ q − | Q | = | Q | = | Q | = a and | Q | = | Q | = | Q | = q − a . Since | Q | = | Q | + 2( q − a ) ≤ q , wehave | Q | ≤ a − q . Also, since | Q | = | Q | + 2 a ≤ q , we have | Q | ≤ q − a .9owever, 2 ≤ | Q | ≤ ( q − a ) + (2 a − q ) = 0 which is a contradiction. So Q isnot a q -clique. Then { x , x , x } is a good stable set of G , and we can concludethe theorem with Theorem 4(ii). (cid:3) Theorem 6
Let G be a clique expansion of G , and assume that every inducedsubgraph G ′ of G satisfies χ ( G ′ ) ≤ ⌈ ω ( G ′ ) ⌉ . Then χ ( G ) ≤ ⌈ ω ( G ) ⌉ .Proof. Since G is a clique expansion of G there is a partition of V ( G ) intoeight non-empty cliques Q , ..., Q , where Q i corresponds to the vertex x i of G . We write, e.g., Q instead of Q ∪ Q ∪ Q , etc. Let q = ω ( G ). For each i ∈ { , ..., } we call x i one vertex of Q i . Moreover if | Q i | ≥ x ′ i onevertex of Q i \ { x i } , and if | Q i | ≥ x ′′ i one vertex of Q i \ { x i , x ′ i } .Suppose that | Q | ≤ | Q | . By hypothesis we can color G \ Q with ⌈ q ⌉ colors. Since Q is complete to Q ∪ Q , which is equal to N ( Q ), we can extendthis coloring to Q , using for Q the colors used for Q . Therefore let us assumethat | Q | > | Q | ; and similarly, that | Q | > | Q | . It follows that | Q | < | Q | ,so Q is not a q -clique of G . Likewise Q is not a q -clique of G . Therefore all q -cliques of G are in the set Q = { Q , Q , Q , Q , Q , Q , Q } .If Q is not a q -clique, then { x , x , x } is a good stable set of G , and wecan conclude using Theorem 4(ii). Therefore we may assume that Q , andsimilarly Q , is a q -clique of G .If Q is not a q -clique, then { x , x , x } is a good stable set of G , and we canconclude using Theorem 4(ii). Therefore we may assume that Q is a q -cliqueof G .If Q is not a q -clique, then { x , x , x } is a good stable set of G , and wecan conclude using Theorem 4(ii). Therefore we may assume that Q , andsimilarly each of Q , Q and Q , is a q -clique of G .Hence Q is precisely the set of all q -cliques of G . It follows that thereare integers a, b, c with a = | Q ], b = | Q | , c = | Q | , a + b + c = q , andthen | Q | = q − a , | Q | = a , | Q | = q − b , | Q | = b , hence | Q | = c . Since q = | Q | = 2 c , it must be that q is even and c = q , so | Q | = | Q | = q .Since each of Q , Q , Q is non-empty we have q ≥
3, and since q is even, q ≥
4. Hence | Q | , | Q | ≥ x ′ and x ′ exist). Since Q and Q are non-empty, and | Q | = q , we have a < q , so | Q | = q − a > q , so | Q | ≥ x ′ and x ′′ exist). Likewise | Q | ≥ x ′ and x ′′ exist). We observe that the clique Q satisfies | Q | = a + b = q ≤ q − q ≥
4. Likewise | Q | ≤ q − { x , x , x } , { x , x , x } , { x ′ , x , x ′ } , { x ′ ,x , x ′ } and { x ′′ , x ′′ } . It is easy to see that their union U meets every q -clique(every member of Q ) four times, and that it meets each of Q and Q twice.It follows (since | Q | , | Q | ≤ q −
2) that ω ( G \ U ) = q −
4, and we can concludeusing Theorem 4(iv). (cid:3)
Theorem 7
Let G be a clique expansion of either G , G or G , and assumethat every induced subgraph G ′ of G satisfies χ ( G ′ ) ≤ ⌈ ω ( G ′ ) ⌉ . Then χ ( G ) ≤⌈ ω ( G ) ⌉ .Proof. Let q = ω ( G ). 10I) First suppose that G is a clique expansion of G . So there is a partitionof V ( G ) into eight non-empty cliques Q , ..., Q , where Q i corresponds to thevertex x i of G . We write, e.g., Q instead of Q ∪ Q ∪ Q , etc. For each i ∈ { , ..., } we call x i one vertex of Q i .Suppose that | Q | ≤ | Q | . By hypothesis we can color G \ Q with ⌈ q ⌉ colors. Since Q is complete to Q ∪ Q ∪ Q , which is equal to N ( Q ), we canextend this coloring to Q , using for Q the colors used for Q . Therefore letus assume that | Q | > | Q | ; and similarly, that | Q | > | Q | . It follows that | Q | > | Q | , so Q is not a q -clique of G . Likewise | Q | > | Q | , so Q isnot a q -clique of G , and similarly Q is not a q -clique.If Q is not a q -clique, then { x , x , x } is a good stable set of G , and wecan conclude with Theorem 4(ii). Hence we may assume that Q , and similarly Q , is a q -clique. Also Q is a q -clique, for otherwise { x , x , x } is a goodstable set, and similarly Q is a q -clique.Now we claim that Q is not a q -clique of G . Suppose not. Then theabove properties imply that there is an integer a with 1 ≤ a ≤ q − | Q | = | Q | = | Q | = a and | Q | = | Q | = | Q | = q − a . However wehave q ≥ | Q | > a and q ≥ | Q | > q − a ), hence 2 q > a + 2( q − a ), acontradiction. So Q is not a q -clique of G . But, then { x , x } is a good stableset of G , and we can conclude the theorem with Theorem 4(ii).(II) Now suppose that G is a clique expansion of G . Let Q be the cliquethat corresponds to x in the clique expansion. As in the case of G we mayassume that Q , Q and Q are not q -cliques. Likewise we may assume that | Q | > | Q | (for otherwise any ⌈ q ⌉ -coloring of G \ Q can be extended to Q ),and consequently Q is not a q -clique; and similarly Q is not a q -clique.Then Q is a q -clique, for otherwise { x , x , x } is a good stable set, andsimilarly Q is a q -clique. Also Q is a q -clique, for otherwise { x , x , x , x } is a good stable set; and similarly Q is a q -clique. And Q is a q -clique, forotherwise { x , x } is a q -clique.The properties given in the preceding paragraph imply that q is even andthat | Q | = | Q | = | Q | = | Q | = | Q | = q . We now distinguish two cases.First suppose that q = 4 k for some k ≥
1. Hence ⌈ q ⌉ = 5 k . Let A, B, C, D, E be five disjoint sets of colors, each of size k . We color the ver-tices in Q with the colors from A ∪ B , the vertices in Q with C ∪ D , thevertices in Q with E ∪ A , the vertices in Q with B ∪ C , and the vertices in Q with D ∪ E . Thus we obtain a 5 k -coloring of G [ Q ∪ Q ∪ Q ∪ Q ∪ Q ].We can extend it to the rest of the graph as follows. Since Q is a clique, and | Q | = q = 2 k , we have | Q | + | Q | ≤ k , hence either | Q | ≤ k or | Q | ≤ k .Likewise, we have either | Q | ≤ k or | Q | ≤ k . This yields (up to symmetry)three possibilities:(i) | Q | ≤ k and | Q | ≤ k . Then we can color Q with colors form E , Q withcolors from D , Q with colors from C ∪ D , and Q with colors from A ∪ E .(ii) | Q | ≤ k and | Q | ≤ k . Then we can color Q with colors form E , Q withcolors from D ∪ E , Q with colors from C ∪ D , and Q with colors from A . (Thecase where | Q | ≤ k and | Q | ≤ k is symmetric.)(iii) | Q | ≤ k and | Q | ≤ k . Then we can color Q and Q with colors from D ∪ E , Q with colors from C , and Q with colors from A .Now suppose that q = 4 k + 2 for some k ≥
1. Hence ⌈ q ⌉ = 5 k + 3. Let11 , B, C, D, E and { z } be six disjoint sets of colors, with | A | = | B | = | C | = k and | D | = | E | = k + 1. So these are 5 k + 3 colors. We color the vertices in Q with the colors from C ∪ D , the vertices in Q with A ∪ E , the vertices in Q with B ∪ D , the vertices in Q with C ∪ E , and the vertices in Q with A ∪ B ∪ { z } . Thus we obtain a 5 k + 3-coloring of G [ Q ∪ Q ∪ Q ∪ Q ∪ Q ].We can extend it to the rest of the graph as follows. Since Q is a clique,and | Q | = q = 2 k + 1, we have | Q | + | Q | ≤ k + 1, hence either | Q | ≤ k or | Q | ≤ k (and in any case max {| Q | , | Q |} ≤ k ). Likewise, we have either | Q | ≤ k or | Q | ≤ k (and max {| Q | , | Q |} ≤ k ). This yields (up to symmetry)three possibilities:(i) | Q | ≤ k and | Q | ≤ k . Then we can color Q with colors form B , Q withcolors from A , Q with colors from A ∪ E , and Q with colors from B ∪ D .(ii) | Q | ≤ k and | Q | ≤ k . Then we can color Q with colors form B , Q withcolors from A ∪ B , Q with colors from A ∪ E , and Q with colors from D . (Thecase where | Q | ≤ k and | Q | ≤ k is symmetric.)(iii) | Q | ≤ k and | Q | ≤ k . Then we can color Q and Q with colors from A ∪ B , Q with colors from E , and Q with colors from D .(III) Finally suppose that G is a clique expansion of G . We view G asthe graph with nine vertices u , ..., u and edges u i u i +1 and u i u i +3 for each i modulo 9. For each i let Q i be the clique of G that corresponds to u i , and let x i be one vertex of Q i . As usual we write e.g. Q instead of Q ∪ Q , etc. Wemake two observations. Observation 1 : If for some i the three cliques Q i,i +1 , Q i +1 ,i +2 and Q i +2 ,i +3 are not q -cliques, then { x i +4 , x i +6 , x i +8 } is a good stable set of G , and we canconclude using Theorem 4(ii). ⋄ Observation 2 : If for some i we have | Q i − | ≤ q and | Q i +1 | ≤ q , then | Q i | ≥ q .Indeed suppose (for i = 1) that | Q | ≤ q , | Q | ≤ q and | Q | < q . Then Q and Q are not q -cliques, so, by Observation 1, we may assume that Q and Q are q -cliques. Hence | Q | ≥ q , and consequently | Q | ≤ q and | Q | ≤ q ;and similarly | Q | ≥ q , and consequently | Q | ≤ q and | Q | ≤ q . But then Q , Q and Q are not q -cliques, so we can conclude as in Observation 1. ⋄ Now, since Q is a clique, we have | Q i | ≤ q for some i ∈ { , , } ; andsimilarly | Q j | ≤ q for some j ∈ { , , } , and | Q k | ≤ q for some k ∈ { , , } .Up to symmetry this implies one the following three cases:(a) | Q | , | Q | , | Q | ≤ q . Then we can conclude using Observation 2.(b) | Q | , | Q | , | Q | ≤ q . Then Q is not a q -clique, so, by Observation 1, wemay assume that one of Q and Q , say Q is a q -clique. Hence | Q | ≥ q ,and consequently | Q | ≤ q . But then we are in case (a) again.(c) | Q | , | Q | , | Q | ≤ q . By Observation 2 we have | Q | ≥ q and | Q | ≥ q , andconsequently | Q | ≤ q and | Q | ≤ q . Then Q , Q and Q are like in case (b).This completes the proof of the theorem. (cid:3) H ∗ Recall that H ∗ is the class of graphs G ∈ H such that, with the notation as inSection 1, the five sets A , A , . . . , A , and the vertex-set of each component of G [ A ] are cliques. 12 heorem 8 Let G ∈ H ∗ and assume that every induced subgraph G ′ of G satisfies χ ( G ′ ) ≤ ⌈ ω ( G ′ ) ⌉ . Then χ ( G ) ≤ ⌈ ω ( G ) ⌉ .Proof. Let q = ω ( G ). Let T , T , . . . , T k be the components of G [ A ]. Foreach i ∈ { , . . . , } and for each j ∈ { , . . . , k } : let x i be one vertex of A i , andlet t j be one vertex of V ( T j ). Moreover if | A i | ≥ x ′ i one vertex of A i \ { x i } , if | V ( T i ) | ≥ t i one vertex of V ( T i ) \ { t i } , if | V ( T i ) | ≥ t i one vertex of V ( T i ) \ { t i , t i } , and if | V ( T i ) | ≥ t i one vertex of V ( T i ) \ { t i , t i , t i } .Suppose that | A | ≤ ω ( G [ A ]). Then by hypothesis, G \ A can be coloredwith ⌈ q ⌉ colors, and since A is complete to A ∪ A which is equal to N ( A ),we can extend this coloring to A by using the colors of A on A . So we mayassume that | A | > ω ( G [ A ]). Likewise, | A | > ω ( G [ A ]). So it follows that noclique of A ∪ A is a q -clique of G .Now consider the stable set S := { x , x , t , . . . , t k } . We may assume that S is not a good stable set of G (otherwise, we can conclude with Theorem 4(ii)). Sothere is a maximum clique Q of G contained in A ∪ A ∪ A . Further, it followsthat for every maximum clique Q of G with Q ∩ S = ∅ , we have A ∪ A ⊂ Q .If A ∪ A is not a q -clique, then { x , x , t , . . . , t k } is a good stable set of G , and we can conclude using Theorem 4(ii). So we may assume that A ∪ A is a q -clique of G . Likewise, A ∪ A is a q -clique of G .If A ∪ A is not a q -clique, then { x , x , t , . . . , t k } is a good stable set of G , and we can conclude using Theorem 4(ii). So we may assume that A ∪ A is a q -clique of G . Likewise, A ∪ A is a q -clique of G .The above properties imply that there is an integer a with 1 ≤ a ≤ q − | A | = | A | = | A | = a and | A | = | A | = q − a . Moreover, every q -clique of G either contains A i ∪ A i +1 , for some i ∈ { , . . . , } , i modulo 5, orcontains T j , for some j ∈ { , . . . , k } .Now if | V ( T j ) | ≤ a , for some j , then by hypothesis, G \ V ( T j ) can becolored with ⌈ q ⌉ colors. Since | A ∪ A | = 2 a , V ( T j ) is anticomplete to A ∪ A , N ( V ( T j )) ⊆ A , and since A is complete to A ∪ A , we can extend this coloringto V ( T j ) by using the colors of A ∪ A on V ( T j ). So, we may assume that, foreach j ∈ { , . . . , k } , | V ( T j ) | > a .If a = 1, then deg G ( x ) = 2 ≤ ⌈ q ⌉ −
1, and we can conclude with Theo-rem 4(i). So we may assume that a ≥ j ∈ { , . . . , k } , we have | V ( T j ) | ≥
4. Also, since q − a >ω ( G [ A ]), we have q − a ≥ { x , x , t , t . . . , t k } , { x ′ , x , t , t . . . , t k } , { x , x ′ , t , t , . . . , t k } , { x ′ , x , t , t . . . , t k } , and { x ′ , x ′ } . It is easy to see thattheir union U meets every q -clique of G four times. It follows that ω ( G \ U ) = q −
4, and we can conclude using Theorem 4(iv). (cid:3)
Proof of Theorem 1 . If G is perfect, then χ ( G ) = ω ( G ) and the theoremholds. So we may assume that G is not perfect, and that G is connected. Sincea P -free graph contains no hole of length at least 7, and a gem-free graphcontains no antihole of length at least 7, it follows from the Strong PerfectGraph Theorem [4] that G contains a hole of length 5. That is, G contains a C as an induced subgraph. Now, the result follows directly from Theorem 3,Lemma 2, and Theorems 5, 6, 7 and 8. (cid:3) cknowledgements . The second author would like to thank Mathew C. Fran-cis for helpful discussions. References [1] Brandst¨adt, A., Kratsch, D.: On the structure of ( P , gem)-free graphs. DiscreteApplied Mathematics 145 (2005) 155–166.[2] Cameron, K., Huang, S., Merkel, O.: An improved χ -bound for ( P , gem )-freegraphs. Private communication.[3] Choudum, S. A., Karthick, T., Shalu, M. A.: Perfect coloring and linearly χ -bounded P -free graphs. Journal of Graph Theory 54 (2007) 293–306.[4] Chudnovsky, M., Seymour, P., Robertson, N., Thomas, R.: The strong perfectgraph theorem. Annals of Mathematics 164 (2006) 51–229.[5] Chudnovsky, M., Sivaraman, V.: Perfect divisibility and 2-divisibility. Journal ofGraph Theory, available online.[6] Fouquet, J. L., Giakoumakis, V., Maire, F., Thuillier, H.: On graphs without P and P . Discrete Mathematics 146 (1995) 33–44.[7] Gravier, S., Ho`ang, C.T., Maffray, F.: Coloring the hypergraph of maximal cliquesof a graph with no long path. Discrete Mathematics 272 (2003) 285–290.[8] Gy´arf´as, A. Problems from the world surrounding perfect graphs. ZastosowaniaMatematyki Applicationes Mathematicae 19 (1987) 413–441.[9] Karthick, T., Maffray, F.: Coloring (gem, co-gem)-free graphs. Journal of GraphTheory 89 (2018) 288–303.[10] Karthick, T., Maffray, F.: Square-free graphs with no six-vertex induced path.Extended abstract in: Proceedings of 10th International Colloquium on GraphTheory and Combinatorics (ICGT-2018), Lyon, France (2018). Available onarXiv:1805.05007 [cs.DM].[11] Kierstead, H. A., Penrice, S.G., Trotter, W.T.: On-line and first-fit coloring ofgraphs that do not induce P . SIAM Journal of Discrete Mathematics 8 (1995)485–498.[12] Reed. B.: ω, ∆ and χ . Journal of Graph Theory 27 (1998) 177–212.. Journal of Graph Theory 27 (1998) 177–212.