A local strengthening of Reed's ω, Δ, χ conjecture for quasi-line graphs
Maria Chudnovsky, Andrew D. King, Matthieu Plumettaz, Paul Seymour
AA local strengthening of Reed’s ω , ∆, χ conjecture forquasi-line graphs Maria Chudnovsky ∗ , Andrew D. King † , and Matthieu Plumettaz ‡ Department of Industrial Engineering and Operations ResearchColumbia University, New York NYPaul Seymour § Department of MathematicsPrinceton University, Princeton NJOctober 31, 2018
Abstract
Reed’s ω , ∆, χ conjecture proposes that every graph satisfies χ ≤ (cid:100) (∆ + 1 + ω ) (cid:101) ; it is known to hold for all claw-free graphs. In this paper we consider a localstrengthening of this conjecture. We prove the local strengthening for line graphs,then note that previous results immediately tell us that the local strengtheningholds for all quasi-line graphs. Our proofs lead to polytime algorithms for con-structing colourings that achieve our bounds: O ( n ) for line graphs and O ( n m )for quasi-line graphs. For line graphs, this is faster than the best known algorithmfor constructing a colouring that achieves the bound of Reed’s original conjecture. ∗ Supported by NSF grants DMS-1001091 and IIS-1117631. † Corresponding author. Email: [email protected]. Supported by an NSERC PostdoctoralFellowship. ‡ Email: [email protected]. Partially supported by NSF grant DMS-1001091. § Supported by ONR grant N00014-10-1-0680 and NSF grant DMS-0901075. a r X i v : . [ c s . D M ] N ov Introduction
All graphs and multigraphs we consider in this paper are finite. Loops are permitted inmultigraphs but not graphs. Given a graph G with maximum degree ∆( G ) and cliquenumber ω ( G ), the chromatic number χ ( G ) is trivially bounded above by ∆( G ) + 1 andbelow by ω ( G ). Reed’s ω , ∆, χ conjecture proposes, roughly speaking, that χ ( G ) fallsin the lower half of this range: Conjecture 1 (Reed) . For any graph G , χ ( G ) ≤ (cid:6) (∆( G ) + 1 + ω ( G )) (cid:7) . One of the first classes of graphs for which this conjecture was proved is the classof line graphs [7]. Already for line graphs the conjecture is tight, as evidenced by thestrong product of C and K (cid:96) for any positive (cid:96) ; this is the line graph of the multigraphconstructed by replacing each edge of C by (cid:96) parallel edges. The proof of Conjecture 1for line graphs was later extended to quasi-line graphs [5, 6] and later claw-free graphs[5] (we will define these graph classes shortly). In his thesis, King proposed a localstrengthening of Reed’s conjecture. For a vertex v , let ω ( v ) denote the size of thelargest clique containing v . Conjecture 2 (King [5]) . For any graph G , χ ( G ) ≤ max v ∈ V ( G ) (cid:6) ( d ( v ) + 1 + ω ( v )) (cid:7) . There are several pieces of evidence that lend credence to Conjecture 2. First is thefact that the fractional relaxation holds. This was noted by McDiarmid as an extensionof a theorem of Reed [8]; the proof appears explicitly in [5] § Theorem 3 (McDiarmid) . For any graph G , χ f ( G ) ≤ max v ∈ V ( G ) (cid:0) ( d ( v ) + 1 + ω ( v )) (cid:1) . The second piece of evidence for Conjecture 2 is that the result holds for claw-freegraphs with stability number at most three [5]. However, for the remaining classesof claw-free graphs, which are constructed as a generalization of line graphs [3], theconjecture has remained open. In this paper we prove that Conjecture 2 holds for linegraphs. We then show that we can extend this result to quasi-line graphs in the sameway that Conjecture 1 was extended from line graphs to quasi-line graphs in [6]. Ourmain result is: 2 heorem 4.
For any quasi-line graph G , χ ( G ) ≤ max v ∈ V ( G ) (cid:6) ( d ( v ) + 1 + ω ( v )) (cid:7) . Furthermore our proofs yield polytime algorithms for constructing a proper colour-ing achieving the bound of the theorem: O ( n ) time for a line graph on n vertices, and O ( n m ) time for a quasi-line graph on n vertices and m edges.Given a multigraph G , the line graph of G , denoted L ( G ), is the graph with vertexset V ( L ( G )) = E ( G ) in which two vertices of L ( G ) are adjacent precisely if theircorresponding edges in H share an endpoint. We say that a graph G (cid:48) is a line graph iffor some multigraph G , L ( G ) is isomorphic to G (cid:48) . A graph G is quasi-line if every vertex v is bisimplicial , i.e. the neighbourhood of v induces the complement of a bipartitegraph. A graph G is claw-free if it contains no induced K , . Observe that every linegraph is quasi-line and every quasi-line graph is claw-free. In order to prove Conjecture 2 for line graphs, we prove an equivalent statement in thesetting of edge colourings of multigraphs. Given distinct adjacent vertices u and v ina multigraph G , we let µ G ( uv ) denote the number of edges between u and v . We let t G ( uv ) denote the maximum, over all vertices w / ∈ { u, v } , of the number of edges withboth endpoints in { u, v, w } . That is, t G ( uv ) := max w ∈ N ( u ) ∩ N ( v ) ( µ G ( uv ) + µ G ( uw ) + µ G ( vw )) . We omit the subscripts when the multigraph in question is clear.Observe that given an edge e in G with endpoints u and v , the degree of uv in L ( G ) is d ( u ) + d ( v ) − µ ( uv ) −
1. And since any clique in L ( G ) containing e comes fromthe edges incident to u , the edges incident to v , or the edges in a triangle containing u and v , we can see that ω ( v ) in L ( G ) is equal to max { d ( u ) , d ( v ) , t ( uv ) } . Therefore weprove the following theorem, which, aside from the algorithmic claim, is equivalent toproving Conjecture 2 for line graphs: Theorem 5.
Let G be a multigraph on m edges, and let γ (cid:48) l ( G ) := max uv ∈ E ( G ) (cid:6) max (cid:8) d ( u ) + ( d ( v ) − µ ( vu )) , d ( v ) + ( d ( u ) − µ ( uv )) , ( d ( u ) + d ( v ) − µ G ( uv ) + t ( uv )) (cid:9)(cid:7) . (1) Then χ (cid:48) ( G ) ≤ γ (cid:48) l ( G ) , and we can find a γ (cid:48) l ( G ) -edge-colouring of G in O ( m ) time. G is a minimum counterexample, then characterizing γ (cid:48) l ( G )-edge-colourings of G − e for an edge e . We want an algorithmic result, so we will have to be abit more careful to ensure that we can modify partial γ (cid:48) l ( G )-edge-colourings efficientlyuntil we find one that we can extend to a complete γ (cid:48) l ( G )-edge-colouring of G .We begin by defining, for a vertex v , a fan hinged at v . Let e be an edge incident to v , and let v , . . . , v (cid:96) be a set of distinct neighbours of v with e between v and v . Let c : E \ { e } → { , . . . , k } be a proper edge colouring of G \ { e } for some fixed k . Then F = ( e ; c ; v ; v , . . . , v (cid:96) ) is a fan if for every j such that 2 ≤ j ≤ (cid:96) , there exists some i less than j such that some edge between v and v j is assigned a colour that does notappear on any edge incident to v i (i.e. a colour missing at v i ). We say that F is hingedat v . If there is no u / ∈ { v, v , . . . , v (cid:96) } such that F (cid:48) = ( e ; c ; v ; v , . . . , v (cid:96) , u ) is a fan, wesay that F is a maximal fan . The size of a fan refers to the number of neighbours of thehinge vertex contained in the fan (in this case, (cid:96) ). These fans generalize Vizing’s fans,originally used in the proof of Vizing’s theorem [13]. Given a partial k -edge-colouringof G and a vertex w , we say that a colour is incident to w if the colour appears on anedge incident to w . We use C ( w ) to denote the set of colours incident to w , and we use¯ C ( w ) to denote [ k ] \ C ( w ).Fans allow us to modify partial k -edge-colourings of a graph (specifically thosewith exactly one uncoloured edge). We will show that if k ≥ γ (cid:48) l ( G ), then either everymaximal fan has size 2 or we can easily find a k -edge-colouring of G . We first provethat we can construct a k -edge-colouring of G from a partial k -edge-colouring of G − e whenever we have a fan for which certain sets are not disjoint. Lemma 6.
For some edge e in a multigraph G and positive integer k , let c be a k -edge-colouring of G − e . If there is a fan F = ( e ; c ; v ; v , . . . , v (cid:96) ) such that for some j , ¯ C ( v ) ∩ ¯ C ( v j ) (cid:54) = ∅ , then we can find a k -edge-colouring of G in O ( k + m ) time.Proof. Let j be the minimum index for which ¯ C ( v ) ∩ ¯ C ( v j ) is nonempty. If j = 1 thenthe result is trivial, since we can extend c to a proper k -edge-colouring of G . Otherwise j ≥ j in O ( m ) time. We define e to be e . We then construct afunction f : { , . . . , (cid:96) } → { , . . . (cid:96) − } such that for each i , (1) f ( i ) < i and (2) thereis an edge e i between v and v i such that c ( e i ) is missing at v f ( i ) . We can find thisfunction in O ( k + m ) time by building a list of the earliest v i at which each colour ismissing, and computing f for increasing values of i starting at 2. While doing so wealso find the set of edges { e i } (cid:96)i =2 .We construct a k -edge-colouring c j of G − e j from c by shifting the colour c ( e j )from e j to e f ( j ) , shifting the colour c ( e f ( j ) ) from e f ( j ) to e f ( f ( j )) , and so on, until weshift a colour to e . We now have a k -edge-colouring c j of G − e j such that some colouris missing at both v and v j . We can therefore extend c j to a proper k -edge-colouringof G in O ( k + m ) time. 4 emma 7. For some edge e in a multigraph G and positive integer k , let c be a k -edge-colouring of G − e . If there is a fan F = ( e ; c ; v ; v , . . . , v (cid:96) ) such that for some i and j satisfying ≤ i < j ≤ (cid:96) , ¯ C ( v i ) ∩ ¯ C ( v j ) (cid:54) = ∅ , then we can find v i and v j in O ( k + m ) time, and we can find a k -edge-colouring of G in O ( k + m ) time.Proof. We can easily find i and j in O ( k + m ) time if they exist. Let α be a colourin ¯ C ( v ) and let β be a colour in ¯ C ( v i ) ∩ ¯ C ( v j ). Note that by Lemma 6, we can assume α ∈ C ( v i ) ∩ C ( v j ) and β ∈ C ( v ).Let G α,β be the subgraph of G containing those edges coloured α or β . Everycomponent of G α,β containing v , v i , or v j is a path on ≥ v i or v j is in a component of G α,β not containing v . Exchanging the colours α and β on thiscomponent leaves us with a k -edge-colouring of G − e in which either ¯ C ( v ) ∩ ¯ C ( v i ) (cid:54) = ∅ or ¯ C ( v ) ∩ ¯ C ( v j ) (cid:54) = ∅ . This allows us to apply Lemma 6 to find a k -edge-colouring of G .We can easily do this work in O ( m ) time.The previous two lemmas suggest that we can extend a colouring more easily whenwe have a large fan, so we now consider how we can extend a fan that is not maximal.Given a fan F = ( e ; c ; v ; v , . . . , v (cid:96) ), we use d ( F ) to denote d ( v ) + (cid:80) (cid:96)i =1 d ( v i ). Lemma 8.
For some edge e in a multigraph G and integer k ≥ ∆( G ) , let c be a k -edge-colouring of G − e and let F be a fan. Then we can extend F to a maximal fan F (cid:48) = ( e ; c ; v ; v , v , . . . , v (cid:96) ) in O ( k + d ( F (cid:48) )) time.Proof. We proceed by setting F (cid:48) = F and extending F (cid:48) until it is maximal. To this endwe maintain two colour sets. The first, C , consists of those colours appearing incidentto v but not between v and another vertex of F (cid:48) . The second, ¯ C F (cid:48) , consists of thosecolours that are in C and are missing at some fan vertex. Clearly F (cid:48) is maximal if andonly if ¯ C F (cid:48) = ∅ . We can perform this initialization in O ( k + d ( F )) time by countingthe number of times each colour in C appears incident to a vertex of the fan.Now suppose we have F (cid:48) = ( e ; c ; v ; v , v , . . . , v (cid:96) ), along with sets C and ¯ C F (cid:48) , whichwe may assume is not empty. Take an edge incident to v with a colour in ¯ C F ; call itsother endpoint v (cid:96) +1 . We now update C by removing all colours appearing between v and v (cid:96) +1 . We update ¯ C F (cid:48) by removing all colours appearing between v and v (cid:96) +1 , andadding all colours in C ∩ ¯ C ( v (cid:96) +1 ). Set F (cid:48) = ( e ; c ; v ; v , v , . . . , v (cid:96) +1 ). We can performthis update in d ( v (cid:96) +1 ) time; the lemma follows.We can now prove that if k ≥ γ (cid:48) l ( G ) and we have a maximal fan of size 1 or at least3, we can find a k -edge-colouring of G in O ( k + m ) time. Lemma 9.
For some edge e in a multigraph G and positive integer k ≥ γ (cid:48) l ( G ) , let c be a k -edge-colouring of G − e and let F = ( e ; c ; v ; v ) be a fan. If F is a maximal fanwe can find a k -edge-colouring of G in O ( k + m ) time. roof. If ¯ C ( v ) ∩ ¯ C ( v ) is nonempty, then we can easily extend the colouring of G − e toa k -edge-colouring of G . So assume ¯ C ( v ) ∩ ¯ C ( v ) is empty. Since k ≥ γ (cid:48) l ( G ) ≥
1, ¯ C ( v )is nonempty. Therefore there is a colour in ¯ C ( v ) appearing on an edge incident to v whose other endpoint, call it v , is not v . Thus ( e ; c ; v ; v , v ) is a fan, contradictingthe maximality of F . Lemma 10.
For some edge e in a multigraph G and positive integer k ≥ γ (cid:48) l ( G ) , let c be a k -edge-colouring of G − e and let F = ( e ; c ; v ; v , v , . . . , v (cid:96) ) be a maximal fan with (cid:96) ≥ . Then we can find a k -edge-colouring of G in O ( k + m ) time.Proof. Let v denote v for ease of notation. If the sets ¯ C ( v ) , ¯ C ( v ) , . . . , ¯ C ( v (cid:96) ) are notall pairwise disjoint, then using Lemma 6 or Lemma 7 we can find a k -edge-colouringof G in O ( m ) time. We can easily determine whether or not these sets are pairwisedisjoint in O ( k + m ) time. Now assume they are all pairwise disjoint; we will exhibita contradiction, which is enough to prove the lemma.The number of missing colours at v i , i.e. | ¯ C ( v i ) | , is k − d ( v i ) if 2 ≤ i ≤ (cid:96) , and k − d ( v i ) + 1 if i ∈ { , } . Since F is maximal, any edge with one endpoint v andthe other endpoint outside { v , . . . , v (cid:96) } must have a colour not appearing in ∪ (cid:96)i =0 ¯ C ( v i ).Therefore (cid:32) (cid:96) (cid:88) i =0 k − d ( v i ) (cid:33) + 2 + (cid:32) d ( v ) − (cid:96) (cid:88) i =1 µ ( v v i ) (cid:33) ≤ k. (2)Thus (cid:96)k + 2 − (cid:96) (cid:88) i =1 µ ( v v i ) ≤ (cid:96) (cid:88) i =1 d ( v i ) . (3)But since k ≥ γ (cid:48) l ( G ), (1) tells us that for all i ∈ [ (cid:96) ], d ( v i ) + ( d ( v ) − µ ( v v i )) ≤ k (4)Thus substituting for k tells us (cid:96) (cid:88) i =1 d ( v ) + 2 d ( v i ) − µ ( v v i )2 + 2 − (cid:96) (cid:88) i =1 µ ( v v i ) ≤ (cid:96) (cid:88) i =1 d ( v i ) . So 2 + (cid:96)d ( v ) − (cid:96) (cid:88) i =1 µ ( v v i ) ≤
02 + (cid:96)d ( v ) ≤ (cid:96) (cid:88) i =1 µ ( v v i ) (cid:96) d ( v ) < d ( v ) . (cid:96) ≥ Lemma 11.
For some edge e in a multigraph G and positive integer k ≥ γ (cid:48) l ( G ) , let c be a k -edge-colouring of G − e . Then we can find a k -edge-colouring of G in O ( k + m ) time. As we will show, this lemma easily implies Theorem 5. We approach this lemma byconstructing a sequence of overlapping fans of size two until we can apply a previouslemma. If we cannot do this, then our sequence results in a cycle in G and a set of partial k -edge-colourings of G with a very specific structure that leads us to a contradiction. Proof.
We postpone algorithmic considerations until the end of the proof.Let v and v be the endpoints of e , and let F = ( e ; c ; v ; v , u , . . . , u (cid:96) ) be amaximal fan. If |{ u , . . . , u (cid:96) }| (cid:54) = 1 then we can apply Lemma 9 or Lemma 10. Moregenerally, if at any time we find a fan of size three or more we can finish by applyingLemma 10. So assume { u , . . . , u (cid:96) } is a single vertex; call it v .Let ¯ C denote the set of colours missing at v in the partial colouring c , and takesome colour α ∈ ¯ C . Note that if α does not appear on an edge between v and v then we can find a fan ( e ; c ; v ; v , v , u ) of size 3 and apply Lemma 10 to completethe colouring. So we can assume that α does appear on an edge between v and v .Let e denote the edge between v and v given colour α in c . We constructa new colouring c of G − e from c by uncolouring e and assigning e colour α .Let ¯ C denote the set of colours missing at v in the colouring c . Now let F =( e ; c ; v ; v , v ) be a maximal fan. As with F , we can assume that F exists and isindeed maximal. The vertex v may or may not be the same as v .Let α ∈ ¯ C be a colour in ¯ C . Just as α appears between v and v in c , wecan see that α appears between v and v . Now let e be the edge between v and v having colour α in c . We construct a colouring c of G − e from c by uncolouring e and assigning e colour α .We continue to construct a sequence of fans F i = ( e i , c i ; v i +1 ; v i , v i +2 ) for i =0 , , , . . . in this way, maintaining the property that α i +2 = α i . This is possiblebecause when we construct c i +1 from c i , we make α i available at v i +2 , so the set ¯ C i +2 (the set of colours missing at v i +2 in the colouring c i +2 ) always contains α i . We continueconstructing our sequence of fans until we reach some j for which v j ∈ { v i } j − i =0 , whichwill inevitably happen if we never find a fan of size 3 or greater. We claim that v j = v and j is odd. To see this, consider the original edge-colouring of G − e and note thatfor 1 ≤ i ≤ j − α appears on an edge between v i and v i +1 precisely if i is odd, and α appears on an edge between v i and v i +1 precisely if i is even. Thus since the edgesof colour α form a matching, and so do the edges of colour α , we indeed have v j = v j odd. Furthermore F = F j . Let C denote the cycle v , v , . . . , v j − . In eachcolouring, α and α both appear ( j − / C , in a near-perfect matching.Let H be the sub-multigraph of G consisting of those edges between v i and v i +1 for0 ≤ j ≤ j − j ). Let A be the set of colours missing on at leastone vertex of C , and let H A be the sub-multigraph of H consisting of e and thoseedges receiving a colour in A in c (and therefore in any c i ).Suppose j = 3. If some colour is missing on two vertices of C in c , c , or c , wecan easily find a k -edge-colouring of G since any two vertices of C are the endpointsof e , e , or e . We know that every colour in ¯ C appears between v and v , andevery colour in ¯ C appears between v and v = v . Therefore | E ( H A ) | = | A | + 1.Our construction tells us that every colour in ¯ C appears between v and v , and everycolour in ¯ C appears between v and v = v . Therefore2 γ (cid:48) l ( G ) ≥ d G ( v ) + d G ( v ) + t G ( v v ) − µ G ( v v )= d H A ( v ) + d H A ( v ) + 2( k − | A | ) + t G ( v v ) − µ G ( v v ) ≥ d H A ( v ) + d H A ( v ) + 2( k − | A | ) + t H A ( v v ) − µ H A ( v v ) ≥ | E ( H A ) | + 2( k − | A | ) > | A | + 2( k − | A | ) = 2 k This is a contradiction since k ≥ γ (cid:48) l ( G ). We can therefore assume that j ≥ β be a colour in A \ { α , α } . If β is missing at two consecutive vertices v i and v i +1 then we can easily extend c i to a k -edge-colouring of G . Bearing in mind thateach F i is a maximal fan, we claim that if β is not missing at two consecutive verticesthen either we can easily k -edge-colour G , or the number of edges coloured β in H A isat least twice the number of vertices at which β is missing in any c i .To prove this claim, first assume without loss of generality that β ∈ ¯ C . Since β isnot missing at v , β appears on an edge between v and v for the same reason that α does. Likewise, since β is not missing at v j − , β appears on an edge between v j − and v j − . Finally, suppose β appears between v and v , and is missing at v in c . Thenlet e β be the edge between v and v with colour β in c . We construct a colouring c (cid:48) from c by giving e colour β and giving e β colour α (i.e. we swap the colours of e β and e ). Thus c (cid:48) is a k -edge-colouring of G − e in which β is missing at both v and v . We can therefore extend G − e to a k -edge-colouring of G . Thus if β is missing at v or v j − we can easily k -edge-colour G . We therefore have at least two edges of H A coloured β for every vertex of C at which β is missing, and we do not double-countedges. This proves the claim, and the analogous claim for any colour in A also holds.Now we have j − (cid:88) i =0 µ H A ( v i v i +1 ) = | E ( H A ) | > j − (cid:88) i =0 ( k − d G ( v i )) . (5)8herefore taking indices modulo j , we have j − (cid:88) i =0 (cid:0) d G ( v i ) + µ H A ( v i +1 v i +2 ) (cid:1) > jk. (6)Therefore there exists some index i for which d G ( v i ) + µ H A ( v i +1 v i +2 ) > k. (7)Therefore k ≥ d G ( v i ) + µ G ( v i +1 v i +2 ) > k. (8)This is a contradiction, so we can indeed find a k -edge-colouring of G . It remains toprove that we can do so in O ( k + m ) time.Given the colouring c i , we can construct the fan F i = ( e i , c i ; v i +1 ; v i , v i +2 ) anddetermine whether or not it is maximal in O ( k + d ( F i )) time. If it is not maximal,we can complete the k -edge-colouring of G in O ( m ) time; this will happen at mostonce throughout the entire process. Therefore we will either complete the colouringor construct our cycle of fans F , . . . , F j − in O ( (cid:80) j − i =0 ( k + d ( F i ))) time. This is notthe desired bound, so suppose there is an index i for which k > d ( F i ). In this case wecertainly have two intersecting sets of available colours in F i , so we can apply Lemma6 or 7 when we arrive at F i , and find the k -edge-colouring of G in O ( k + m ) time.If no such i exists, then jk = O ( (cid:80) j − i =0 ( d ( F i ))) = O ( m ), and we indeed complete theconstruction of all fans in O ( k + m ) time.Since each F i is a maximal fan, in c there must be some colour β / ∈ { α , α } missing at two consecutive vertices v i and v i +1 , otherwise we reach a contradiction.We can find this β and i by going around the cycle of fans and comparing ¯ C i and ¯ C i +1 ,and since this is trivial if | ¯ C i | + | ¯ C i +1 | > d ( v i ) + d ( v i +1 ) we can find β and i in O ( k + m )time, after which it is easy to construct the k -edge-colouring of G from c i .We now complete the proof of Theorem 5. Proof of Theorem 5.
Order the edges of
G e , . . . , e m arbitrarily and let k = γ (cid:48) l ( G ),which we can easily compute in O ( nm ) time. For i = 0 , . . . , m , let G i denote thesubgraph of G on edges { e j | j ≤ i } . Since G is empty it is vacuously k -edge-coloured.Given a k -edge-colouring of G i , we can find a k -edge-colouring of G i +1 in O ( k + m )time by applying Lemma 11. Since k = γ (cid:48) l ( G ) = O ( m ), the theorem follows.This gives us the following result for line graphs, since for any multigraph G wehave | V ( L ( G )) | = | E ( G ) | : 9 heorem 12. Given a line graph G on n vertices, we can find a proper colouring of G using γ l ( G ) colours in O ( n ) time. This is faster than the algorithm of King, Reed, and Vetta [7] for γ ( G )-colouringline graphs, which is given an improved complexity bound of O ( n / ) in [5], § We now leave the setting of edge colourings of multigraphs and consider vertex colour-ings of simple graphs. As mentioned in the introduction, we can extend Conjecture2 from line graphs to quasi-line graphs using the same approach that King and Reedused to extend Conjecture 1 from line graphs to quasi-line graphs in [6]. We do notrequire the full power of Chudnovsky and Seymour’s structure theorem for quasi-linegraphs [2]. Instead, we use a simpler decomposition theorem from [3].
We wish to describe the structure of quasi-line graphs. If a quasi-line graph does notcontain a certain type of homogeneous pair of cliques, then it is either a circular intervalgraph or built as a generalization of a line graph – where in a line graph we wouldreplace each edge with a vertex, we now replace each edge with a linear interval graph.We now describe this structure more formally. A linear interval graph is a graph G = ( V, E ) with a linear interval representation ,which is a point on the real line for each vertex and a set of intervals, such thatvertices u and v are adjacent in G precisely if there is an interval containing bothcorresponding points on the real line. If X and Y are specified cliques in G consistingof the | X | leftmost and | Y | rightmost vertices (with respect to the real line) of G respectively, we say that X and Y are end-cliques of G . These cliques may be empty.Accordingly, a circular interval graph is a graph with a circular interval representa-tion , i.e. | V | points on the unit circle and a set of intervals (arcs) on the unit circle suchthat two vertices of G are adjacent precisely if some arc contains both correspondingpoints. Circular interval graphs are the first of two fundamental types of quasi-linegraph. Deng, Hell, and Huang proved that we can identify and find a representationof a circular or linear interval graph in O ( m ) time [4]. We now describe the second fundamental type of quasi-line graph.10 ue e e e e e e S e S e S e S e S e S e S e X e X e X e X e X e Y e Y e Y e Y e C u C v Figure 1: We compose a set of strips { ( S e , X e , Y e ) | e ∈ E ( H ) } by joining them togetheron their end-cliques. A hub clique C u will arise for each vertex u ∈ V ( H ).A linear interval strip ( S, X, Y ) is a linear interval graph S with specified end-cliques X and Y . We compose a set of strips as follows. We begin with an underlyingdirected multigraph H , possibly with loops, and for every every edge e of H we take alinear interval strip ( S e , X e , Y e ). For v ∈ V ( H ) we define the hub clique C v as C v = (cid:16)(cid:91) { X e | e is an edge out of v } (cid:17) ∪ (cid:16)(cid:91) { Y e | e is an edge into v } (cid:17) . We construct G from the disjoint union of { S e | e ∈ E ( H ) } by making each C v aclique; G is then a composition of linear interval strips (see Figure 1). Let G h denotethe subgraph of G induced on the union of all hub cliques. That is, G h = G [ ∪ v ∈ V ( H ) C v ] = G [ ∪ e ∈ E ( H ) ( X e ∪ Y e )] . Compositions of linear interval strips generalize line graphs: note that if each S e satisfies | S e | = | X e | = | Y e | = 1 then G = G h = L ( H ). A pair of disjoint nonempty cliques (
A, B ) in a graph is a homogeneous pair of cliques if | A | + | B | ≥
3, every vertex outside A ∪ B is adjacent to either all or none of A , and11very vertex outside A ∪ B is adjacent to either all or none of B . Furthermore ( A, B ) is nonlinear if G contains an induced C in A ∪ B (this condition is equivalent to insistingthat the subgraph of G induced by A ∪ B is a linear interval graph). Chudnovsky and Seymour’s structure theorem for quasi-line graphs [3] tells us thatany quasi-line graph not containing a clique cutset is made from the building blockswe just described.
Theorem 13.
Any quasi-line graph containing no clique cutset and no nonlinear ho-mogeneous pair of cliques is either a circular interval graph or a composition of linearinterval strips.
To prove Theorem 4, we first explain how to deal with circular interval graphs andnonlinear homogeneous pairs of cliques, then move on to considering how to decomposea composition of linear interval strips.
We can easily prove Conjecture 2 for circular interval graphs by combining previouslyknown results. Niessen and Kind proved that every circular interval graph G satisfies χ ( G ) = (cid:100) χ f ( G ) (cid:101) [9], so Theorem 3 immediately implies that Conjecture 2 holds forcircular interval graphs. Furthermore Shih and Hsu [11] proved that we can optimallycolour circular interval graphs in O ( n / ) time, which gives us the following result: Lemma 14.
Given a circular interval graph G on n vertices, we can γ l ( G ) -colour G in O ( n / ) time. There are many lemmas of varying generality that tell us we can easily deal withnonlinear homogeneous pairs of cliques; we use the version used by King and Reed [6]in their proof of Conjecture 1 for quasi-line graphs:
Lemma 15.
Let G be a quasi-line graph on n vertices containing a nonlinear homo-geneous pair of cliques ( A, B ) . In O ( n / ) time we can find a proper subgraph G (cid:48) of G such that G (cid:48) is quasi-line, χ ( G (cid:48) ) = χ ( G ) , and given a k -colouring of G (cid:48) we can find a k -colouring of G in O ( n / ) time. It follows immediately that no minimum counterexample to Theorem 4 contains anonlinear homogeneous pair of cliques. 12 .4 Decomposing: Clique cutsets
Decomposing graphs on clique cutsets for the purpose of finding vertex colourings isstraightforward and well understood.For any monotone bound on the chromatic number for a hereditary class of graphs,no minimum counterexample can contain a clique cutset, since we can simply “pastetogether” two partial colourings on a clique cutset. Tarjan [12] gave an O ( nm )-timealgorithm for constructing a clique cutset decomposition tree of any graph, and notedthat given k -colourings of the leaves of this decomposition tree, we can construct a k -colouring of the original graph in O ( n ) time. Therefore if we can γ l ( G )-colour anyquasi-line graph containing no clique cutset in O ( f ( n, m )) time for some function f ,we can γ l ( G )-colour any quasi-line graph in O ( f ( n, m ) + nm ) time.If the multigraph H contains a loop or a vertex of degree 1, then as long as G isnot a clique, it will contain a clique cutset. A canonical interval 2-join is a composition by which a linear interval graph is attachedto another graph. Canonical interval 2-joins arise from compositions of strips, and canbe viewed as a local decomposition rather than one that requires knowledge of a graph’sglobal structure as a composition of strips.Given four cliques X , Y , X , and Y , we say that (( V , X , Y ) , ( V , X , Y )) is an interval 2-join if it satisfies the following: • V ( G ) can be partitioned into nonempty V and V with X ∪ Y ⊆ V and X ∪ Y ⊆ V such that for v ∈ V and v ∈ V , v v is an edge precisely if { v , v } is in X ∪ X or Y ∪ Y . • G | V is a linear interval graph with end-cliques X and Y .If we also have X and Y disjoint, then we say (( V , X , Y ) , ( V , X , Y )) is a canonicalinterval 2-join . The following decomposition theorem is a straightforward consequenceof the structure theorem for quasi-line graphs: Theorem 16.
Let G be a quasi-line graph containing no nonlinear homogeneous pairof cliques. Then one of the following holds. • G is a line graph • G is a circular interval graph • G contains a clique cutset • G admits a canonical interval 2-join. γ l ( G ). Given a canonical interval 2-join (( V , X , Y ) , ( V , X , Y )) in G with an appropriate partitioning V and V , let G denote G | V , let G denote G | V and let H denote G | ( V ∪ X ∪ Y ). For v ∈ H wedefine ω (cid:48) ( v ) as the size of the largest clique in H containing v and not intersectingboth X \ Y and Y \ X , and we define γ jl ( H ) as max v ∈ H (cid:100) d G ( v ) + 1 + ω (cid:48) ( v ) (cid:101) (herethe superscript j denotes join ). Observe that γ jl ( H ) ≤ γ l ( G ). If v ∈ X ∪ Y , then ω (cid:48) ( v ) is | X | + | X | , | Y | + | Y | , or | X ∩ Y | + ω ( G | ( X ∪ Y )).The following lemma is due to King and Reed and first appeared in [5]; we includethe proof for the sake of completeness. Lemma 17.
Let G be a graph on n vertices and suppose G admits a canonical interval2-join (( V , X , Y ) , ( V , X , Y )) . Then given a proper l -colouring of G for any l ≥ γ jl ( H ) , we can find a proper l -colouring of G in O ( nm ) time. Since γ jl ( H ) ≤ γ l ( G ), this lemma implies that no minimum counterexample toTheorem 4 contains a canonical interval 2-join.It is easy to see that a minimum counterexample cannot contain a simplicial vertex(i.e. a vertex whose neighbourhood is a clique). Therefore in a canonical interval 2-join(( V , X , Y ) , ( V , X , Y )) in a minimum counterexample, all four cliques X , Y , X ,and Y must be nonempty. Proof.
We proceed by induction on l , observing that the case l = 1 is trivial. We beginby modifying the colouring so that the number k of colours used in both X and Y in the l -colouring of G is maximal. That is, if a vertex v ∈ X gets a colour that isnot seen in Y , then every colour appearing in Y appears in N ( v ). This can be donein O ( n ) time. If l exceeds γ jl ( H ) we can just remove a colour class in G and applyinduction on what remains. Thus we can assume that l = γ jl ( H ) and so if we applyinduction we must remove a stable set whose removal lowers both l and γ jl ( H ).We use case analysis; when considering a case we may assume no previous caseapplies. In some cases we extend the colouring of G to an l -colouring of G in onestep. In other cases we remove a colour class in G together with vertices in G suchthat everything we remove is a stable set, and when we remove it we reduce γ jl ( v ) forevery v ∈ H ; after doing this we apply induction on l . Notice that if X ∩ Y (cid:54) = ∅ andthere are edges between X and Y we may have a large clique in H which containssome but not all of X and some but not all of Y ; this is not necessarily obvious butwe deal with it in every applicable case.Case 1. Y ⊆ X . 14 is a circular interval graph and X is a clique cutset. We can γ l ( H )-colour H in O ( n / ) time using Lemma 14. By permuting the colour classes we canensure that this colouring agrees with the colouring of G . In this case γ l ( H ) ≤ γ jl ( H ) ≤ l so we are done. By symmetry, this covers the case in which X ⊆ Y .Case 2. k = 0 and l > | X | + | Y | .Here X and Y are disjoint. Take a stable set S greedily from left to right in G .By this we mean that we start with S = { v } , the leftmost vertex of X , andwe move along the vertices of G in linear order, adding a vertex to S wheneverdoing so will leave S a stable set. So S hits X . If it hits Y , remove S alongwith a colour class in G not intersecting X ∪ Y ; these vertices together make astable set. If v ∈ G it is easy to see that γ jl ( v ) will drop: every remaining vertexin G either loses two neighbours or is in Y , in which case S intersects everymaximal clique containing v . If v ∈ X ∪ Y then since X and Y are disjoint, ω (cid:48) ( v ) is either | X | + | X | or | Y | + | Y | ; in either case ω (cid:48) ( v ), and therefore γ jl ( v ),drops when S and the colour class are removed. Therefore γ jl ( H ) drops, and wecan proceed by induction.If S does not hit Y we remove S along with a colour class from G that hits Y (and therefore not X ). Since S ∩ Y = ∅ the vertices together make a stableset. Using the same argument as before we can see that removing these verticesdrops both l and γ jl ( H ), so we can proceed by induction.Case 3. k = 0 and l = | X | + | Y | .Again, X and Y are disjoint. By maximality of k , every vertex in X ∪ Y hasat least l − G . Since l = | X | + | Y | we know that ω (cid:48) ( X ) ≤| X | + | Y | − | X | and ω (cid:48) ( Y ) ≤ | X | + | Y | − | Y | . Thus | Y | ≥ | X | and similarly | X | ≥ | Y | . Assume without loss of generality that | Y | ≤ | X | .We first attempt to l -colour H − Y , which we denote by H , such that everycolour in Y appears in X – this is clearly sufficient to prove the lemma sincewe can permute the colour classes and paste this colouring onto the colouring of G to get a proper l -colouring of G . If ω ( H ) ≤ l − | Y | then this is easy: we can ω ( H )-colour the vertices of H , then use | Y | new colours to recolour Y and | Y | vertices of X . This is possible since Y and X have no edges between them.Define b as l − ω ( H ); we can assume that b < | Y | . We want an ω ( H )-colouringof H such that at most b colours appear in Y but not X . There is someclique C = { v i , . . . , v i + ω ( H ) − } in H ; this clique does not intersect X because | X ∪ X | ≤ l − | Y | ≤ l − | Y | < l − b . Denote by v j the leftmost neighbourof v i . Since γ jl ( v i ) ≤ l , it is clear that v i has at most 2 b neighbours outside C ,15nd since b < | Y | ≤ | X | we can be assured that v i / ∈ X . Since ω ( H ) > | Y | , v i / ∈ Y .We now colour H from left to right, modulo ω ( H ). If at most b colours appearin Y but not X then we are done, otherwise we will “roll back” the colouring,starting at v i . That is, for every p ≥ i , we modify the colouring of H by giving v p the colour after the one that it currently has, modulo ω ( H ). Since v i has at most2 b neighbours behind it, we can roll back the colouring at least ω ( H ) − b − ω ( H ) − b proper colourings of H .Since v i / ∈ Y the colours on Y will appear in order modulo ω ( H ). Thus thereare ω ( H ) possible sets of colours appearing on Y , and in 2 b + 1 of them thereare at most b colours appearing in Y but not X . It follows that as we roll backthe colouring of H we will find an acceptable colouring.Henceforth we will assume that | X | ≥ | Y | .Case 4. 0 < k < | X | .Take a stable set S in G − X greedily from left to right. If S hits Y , weremove S from G , along with a colour class from G intersecting X but not Y .Otherwise, we remove S along with a colour class from G intersecting both X and Y . In either case it is a simple matter to confirm that γ jl ( v ) drops for every v ∈ H as we did in Case 2. We proceed by induction.Case 5. k = | Y | = | X | = 1.In this case | X | = k = 1. If G is not connected then X and Y are bothclique cutsets and we can proceed as in Case 1. If G is connected and containsan l -clique, then there is some v ∈ V of degree at least l in the l -clique. Thus γ jl ( H ) > l , contradicting our assumption that l ≥ γ jl ( H ). So ω ( G ) < l . Wecan ω ( G )-colour G in linear time using only colours not appearing in X ∪ Y ,thus extending the l -colouring of G to a proper l -colouring of G .Case 6. k = | Y | = | X | > k is not minimal. That is, suppose there is a vertex v ∈ X ∪ Y whose closed neighbourhood does not contain all l colours in the colouring of G .Then we can change the colour of v and apply Case 4. So assume k is minimal.Therefore every vertex in X has degree at least l + | X | −
1. Since X ∪ X isa clique, γ jl ( H ) ≥ l ≥ ( l + | X | + | X | + | X | ), so 2 | X | ≤ l − k . Similarly,2 | Y | ≤ l − k , so | X | + | Y | ≤ l − k . Since there are l − k colours not appearingin X ∪ Y , we can ω ( G )-colour G , then permute the colour classes so that nocolour appears in both X ∪ Y and X ∪ Y . Thus we can extend the l -colouringof G to an l -colouring of G . 16hese cases cover every possibility, so we need only prove that the colouring canbe found in O ( nm ) time. If k has been maximized and we apply induction, k willstay maximized: every vertex in X ∪ Y will have every remaining colour in its closedneighbourhood except possibly if we recolour a vertex in Case 6. In this case theoverlap in what remains is k −
1, which is the most possible since we remove a vertexfrom X or Y , each of which has size k . Hence we only need to maximize k once.We can determine which case applies in O ( m ) time, and it is not hard to confirm thatwhenever we extend the colouring in one step our work can be done in O ( nm ) time.When we apply induction, i.e. in Cases 2, 4, and possibly 6, all our work can be donein O ( m ) time. Since l < n it follows that the entire l -colouring can be completed in O ( nm ) time. We can now prove an algorithmic version of Theorem 4.
Theorem 18.
Let G be a quasi-line graph on n vertices and m edges. Then we canfind a proper colouring of G using γ ( G ) colours in O ( n m ) time.Proof. We proceed by induction on n . As already explained, we need only considergraphs containing no clique cutsets since n m ≥ nm . We begin by applying Lemma 15at most m times in order to find a quasi-line subgraph G (cid:48) of G such that χ ( G ) = χ ( G (cid:48) ),and given a k -colouring of G (cid:48) , we can find a k -colouring of G in O ( n m ) time. Wemust now colour G (cid:48) .If G (cid:48) is a circular interval graph we can determine this and γ l ( G )-colour it in O ( n / )time. If G (cid:48) is a line graph we can determine this in O ( m ) time using an algorithm ofRoussopoulos [10], then γ l ( G )-colour it in O ( n ) time. Otherwise, G (cid:48) must admit acanonical interval 2-join. In this case Lemma 6.18 in [5], due to King and Reed, tellsus that we can find such a decomposition in O ( n m ) time.This canonical interval 2-join (( V , X , Y ) , ( V , X , Y )) leaves us to colour the in-duced subgraph G of G (cid:48) , which has at most n − γ l ( G )-colouring of G we can γ l ( G )-colour G (cid:48) in O ( nm ) time, then reconstruct the γ l ( G )-colouring of G in O ( n m ) time. The induction step takes O ( n m ) time and re-duces the number of vertices, so the total running time of the algorithm is O ( n m ). Remark:
With some care and using more sophisticated results on decomposing quasi-line graphs (see [1]), we believe it should be possible to reduce the running time of theentire algorithm to O ( m ). 17 eferences [1] M. Chudnovsky and A. D. King. Optimal antithickenings of claw-free trigraphs.Submitted. Arxiv preprint 1110.5111, 2011.[2] M. Chudnovsky and P. Seymour. Claw-free graphs VII. Quasi-line graphs. Sub-mitted.[3] M. Chudnovsky and P. Seymour. The structure of claw-free graphs. In B. S.Webb, editor, Surveys in Combinatorics , volume 327 of
London MathematicalSociety Lecture Note Series . Cambridge University Press, 2005.[4] X. Deng, P. Hell, and J. Huang. Linear-time representation algorithms for propercircular-arc graphs and proper interval graphs.
SIAM Journal on Computing ,25:390–403, 1996.[5] A. D. King.
Claw-free graphs and two conjectures on ω , ∆ , and χ . PhD thesis,McGill University, October 2009.[6] A. D. King and B. A. Reed. Bounding χ in terms of ω and ∆ for quasi-line graphs. Journal of Graph Theory , 59(3):215–228, 2008.[7] A. D. King, B. A. Reed, and A. Vetta. An upper bound for the chromatic numberof line graphs.
Eur. J. Comb. , 28(8):2182–2187, 2007.[8] M. Molloy and B. Reed.
Graph Colouring and the Probabilistic Method . Springer,2000.[9] T. Niessen and J. Kind. The round-up property of the fractional chromatic numberfor proper circular arc graphs.
J. Graph Theory , 33:256–267, 2000.[10] N. D. Roussopoulos. A max { m, n } algorithm for determining the graph H fromits line graph G . Information Processing Letters , 2:108–112, 1973.[11] W.-K. Shih and W.-L. Hsu. An O ( n / ) algorithm to color proper circular arcs. Discrete Applied Mathematics , 25:321–323, 1989.[12] R. E. Tarjan. Decomposition by clique separators.