A FRACTIONAL ANALOGUE OF BROOKS THEOREM

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1 A FRACTIONAL ANALOGUE OF BROOKS THEOREM ANDREW D. KING, LINYUAN LU, AND XING PENG Astrt. Let (G) e the mimm egree of grph G. Brooks theorem sttes tht the onl onnete grphs ith hromti nmer χ(g) = (G) + 1 re omplete grphs n o les. We proe frtionl nloge of Brooks theorem in this pper. Nmel, e lssif ll onnete grphs G sh tht the frtionl hromti nmer χ f (G) is t lest (G). These grphs re omplete grphs, o les, C 2 8, C 5 K 2, n grphs hose liqe nmer ω(g) eqls the mimm egree (G). Among the to spori grphs, the grph C 2 8 is the sqre grph of le C 8 hile the other grph C 5 K 2 is the strong prot of C 5 n K 2. In ft, e proe stronger reslt; if onnete grph G ith (G) 4 is not one of the grphs liste oe, then e he χ f (G) (G) Ke ors. Frtionl hromti nmer, Brooks theorem, hromti nmer. AMS sjet lssifitions. 05C72, 05C15 1. Introtion. The hromti nmer of grphs ith one egrees hs een stie for mn ers. Brooks theorem perhps is one of the most fnmentl reslts; it is inle mn tetooks on grph theor. Gien simple onnete grph G, let (G) e the mimm egree, ω(g) e the liqe nmer, n χ(g) e the hromti nmer. Brooks theorem sttes tht χ(g) (G) nless G is omplete grph or n o le. Ree [10] proe tht χ(g) (G) 1 if ω(g) (G) 1 n (G) 0 for some lrge onstnt 0. This eellent reslt s proe proilisti methos, n 0 is t lest hnres. Before this reslt, Boroin n Kostohk [1] me the folloing onjetre. Conjetre [1]: Sppose tht G is onnete grph. If ω(g) (G) 1 n (G) 9, then e he χ(g) (G) 1. If the onjetre is tre, then it is est possile sine there is K 8 -free grph G = C 5 K 3 (tll K 7 -free, see Figre 1) ith (G) = 8 n χ(g) = 8. Here e se the folloing nottion of the strong prot. Gien to grphs G n H, the strong prot G H is the grph ith erte set V (G) V (H), n (, ) is onnete to (, ) if one of the folloing hols = n E(H), E(G) n =, E(G) n E(H). Ree s reslt [10] settle Boroin n Kostohk s onjetre for sffiientl lrge (G), t the ses ith smll (G) re hr to oer sing the proilisti metho. In this pper e onsier frtionl nloge of this prolem. The frtionl hromti nmer χ f (G) n e efine s follos. A -fol oloring of G ssigns set of olors to eh erte sh tht n to jent erties reeie isjoint sets of olors. We s grph G is :-olorle if there is -fol oloring of G in hih eh olor is rn from plette of olors. We refer to sh oloring s n :-oloring. The -fol oloring nmer, enote χ (G), is the smllest integer sh tht G hs n :-oloring. Note Simon Frser Uniersit, Brn, British Colmi, Cn (nre..king@gmil.om). This thor s spporte in prt n EBCO/Eih Postotorl Sholrship n the NSERC Disoer Grnts of Pol Hell n Bojn Mohr. Uniersit of Soth Crolin, Colmi, SC 29208, (l@mth.s.e). This thor s spporte in prt NSF grnt DMS Uniersit of Soth Crolin, Colmi, SC 29208, (peng@milo.s.e). This thor s spporte in prt NSF grnt DMS

2 2 King, L, n Peng Fig The grph C 5 K 3. tht χ 1 (G) = χ(g). It s shon tht χ + (G) χ (G) + χ (G). The frtionl hromti χ nmer χ f (G) is lim (G). B the efinition, e he χ f (G) χ(g). The frtionl hromti nmer n e iee s reltion of the hromti nmer. Mn prolems inoling the hromti nmer n e ske gin sing the frtionl hromti nmer. The frtionl nloge often hs simpler soltion thn the originl prolem. For emple, the fmos ω χ onjetre of Ree [9] sttes tht for n simple grph G, e he ω(g) + (G) + 1 χ(g). 2 The frtionl nloge of ω χ onjetre s proe Mollo n Ree [8]; the tll proe stronger reslt ith eiling remoe, i.e., χ f (G) ω(g) + (G) + 1. (1.1) 2 In this pper, e lssif ll onnete grphs G ith χ f (G) (G). Theorem 1.1. A onnete grph G stisfies χ f (G) (G) if n onl if G is one of the folloing 1. omplete grph, 2. n o le, 3. grph ith ω(g) = (G), 4. C 2 8, 5. C 5 K 2. For the omplete grph K n, e he χ f (K n ) = n n (K n ) = n 1. For the o le C 2k+1, e he χ f (C 2k+1 ) = k n (C 2k+1) = 2. If G is neither omplete grph nor n o le t ontins liqe of sie (G), then e he (G) ω(g) χ f (G) χ(g) (G). (1.2) The lst ineqlit is from Brooks theorem. The seqene of ineqlities oe implies χ f (G) = (G). If G is erte-trnsitie grph, then e he [11] χ f (G) = V (G) α(g),

3 A Frtionl Anloge of Brooks Theorem 3 here α(g) is the inepenene nmer of G. Note tht oth grphs C 2 8 n C 5 K 2 re erte-trnsitie n he the inepenene nmer 2. Ths e he χ f (C 2 8) = 4 = (C 2 8) n χ f (C 5 K 2 ) = 5 = (C 5 K 2 ). C 2 8 C 5 K 2 Fig The grph C 2 8 n C 5 K 2. Atll, Theorem 1.1 is orollr of the folloing stronger reslt. Theorem 1.2. Assme tht onnete grph G is neither C 2 8 nor C 5 K 2. If (G) 4 n ω(g) (G) 1, then e he χ f (G) (G) Remrk: In the se (G) = 3, Hekmn n Thoms [5] onjetre tht χ f (G) 14/5 if G is tringle-free. Htmi n Zh [4] proe χ f (G) for n tringle-free grph G ith (G) 3. The seon n thir thors shoe n improe reslt χ f (G) in the preios pper [7]. Ths e nee onl onsier the ses (G) 4. For n onnete grph G ith sffiientl lrge (G) n ω(g) (G) 1, Ree s reslt [10] χ(g) (G) 1 implies χ f (G) (G) 1. The metho introe in [4] n strengthene in [7], hs strong inflene on this pper. The reers re enorge to re these to ppers [4, 7]. Let f(k) = inf G { (G) χ f (G)}, here the infimm is tken oer ll onnete grphs G ith (G) = k n not one of the grphs liste in Theorem 1.1. Sine χ f (G) ω(g), tking grph ith ω(g) = (G) 1, e he f(k) 1. Theorem 1.2 ss f(k) 2 67 for n k 4. Ree s reslt [10] implies f(k) = 1 for sffiientl lrge k. Hekmn n Thoms [5] onjetre f(3) = 1/5. It is n interesting prolem to etermine the le of f(k) for smll k. Here e onjetre f(4) = f(5) = 1 3. If Boroin n Kostohk s onjetre is tre, then f(k) = 1 for k 9. Theorem 2 is proe intion on k. Bese the proof is qite long, e split the proof into the folloing to lemms. Lemm 1.3. We he f(4) Lemm 1.4. For eh k 6, e he f(k) min { f(k 1), 2} 1. We lso he f(5) min { f(4), 3} 1. It is es to see the omintion of Lemm 1.3 n Lemm 1.4 implies Theorem 1.2. The ie of retion omes from the first thor, ho pointe ot f(k) min { f(k 1), 2} 1 for k 7 se on his reent reslts [6]. The seon n thir thors orginll proe f(k) C k (for some C > 0) sing ifferent metho in the first ersion; the 5 lso proe the retions t k = 5, 6, hih re mh hrer thn the se k 7. We o not kno hether similr retion eists for k = 4. The rest of this pper is orgnie s follos. In setion 2, e ill introe some nottion n proe Lemm 1.4. In setion 3 n setion 4, e ill proe f(4) 2 67.

4 4 King, L, n Peng 2. Proof of Lemm 1.4. In this pper, e se the folloing nottion. Let G e simple grph ith erte set V (G) n ege set E(G). The neighorhoo of erte in G, enote Γ G (), is the set {: E(G)}. The egree G () of is the le of Γ G (). The inepenent set (or stle set) is set S sh tht no ege ith oth ens in S. The inepenene nmer α(g) is the lrgest sie of S mong ll the inepenent sets S in G. When T V (G), e se α G (T) to enote the inepenene nmer of the ine sgrph of G on T. Let (G) e the mimm egree of G. For n to erte-sets S n T, e efine E G (S, T) s { E(G) : S n T }. Wheneer G is ler ner ontet, e ill rop the ssript G for simpliit. If S is sset of erties in G, then ontrting S mens repling erties in S single ft erte, enote S, hose inient eges re ll eges tht ere inient to t lest one erte in S, eept eges ith oth ens in S. The ne grph otine ontrting S is enote G/S. This opertion is lso knon s ientifing erties of S in the litertre. For ompleteness, e llo S to e single erte or een the empt set. If S onl onsists of single erte, then G/S = G; if S =, then G/S is the nion of G n n isolte erte. When S onsists of 2 or 3 erties, for oneniene, e rite G/ for G/{, } n G/ for G/{,, }; the ft erte ill e enote n, respetiel. Gien to isjoint ssets S 1 n S 2, e n ontrt S 1 n S 2 seqentill. The orer of ontrtions oes not mtter; let G/S 1 /S 2 e the reslte grph. We se G S to enote the sgrph of G ine V (G) S. In orer to proe Lemm 1.4, e nee se the folloing theorems e to King [6]. Theorem 2.1 (King [6]). If grph G stisfies ω(g) > 2 3 ( (G) + 1), then G ontins stle set S meeting eer mimm liqe. Theorem 2.2 (King [6]). For positie integer k, let G e grph ith erties prtitione into liqes V 1,..., V r. If for eer i n eer V i, hs t most min{k, V i k} neighors otsie V i, then G ontins stle set of sie r. Lemm 2.3. Sppose tht G is onnete grph ith (G) 6 n ω(g) 5. Then there eists n inepenent set meeting ll ine opies of K 5 n C 5 K 2. Proof: We first sho tht there eists n inepenent set meeting ll opies of K 5. If G ontins no K 5, then this is triil. Otherise, e n ppl Theorem 2.1 to get the esire inepenent set sine ω(g) > 2 3 ( (G) + 1) is stisfie. No e proe the Lemm ontrition. Sppose the Lemm is flse. Let G e minimm onteremple (ith the smllest nmer of erties). For n inepenent set I, let C(I) e the nmer of ine opies of C 5 K 2 in G I. Among ll inepenent sets hih meet ll opies of K 5, there eists one sh inepenent set I sh tht C(I) is minimie. Sine C(I) > 0, there is n ine op of C 5 K 2 in G I; e se H to enote it. In C 5 K 2, there is niqe perfet mthing sh tht ientifing the to ens of eh ege in this mthing reslts C 5. An ege in this niqe mthing is lle nonil ege. We efine ne grph G s follos: First e ontrt ll nonil eges in H to get C 5, here its erties re lle ft erties. Seon e fie eges trning the C 5 into K 5. Osere tht eh erte in this C 5 n he t most to neighors in G H n (G ) 6. We ill onsier the folloing for ses. Cse 1: There is K 6 in the ne grph G. Sine the originl grph G is K 6 -free, the K 6 is forme the folloing to possile s. Sse 1: This K 6 ontins 5 ft erties. B the smmetr of H, there is n ine C 5 in H sh tht the erties in C 5 ontin ommon neighor erte in G \ V (H), see Figre 2.1. Sine H is K 5 -free, e n fin, in this C 5 sh tht, is non-ege. Let I := (I \ {}) {, }; I is lso n inepenent set. Osere tht is not in n K 5 in G I. Ths the set I is lso n inepenent set n meets eer K 5 in G. Sine C 5 K 2 is 5-reglr grph, n op of C 5 K 2 ontining mst ontin t lest one of n.

5 Ths, C(I ) < C(I). Contrition! A Frtionl Anloge of Brooks Theorem 5 Fig Sse 1. Fig Sse 1. Sse 1: This K 6 ontins 4 ft erties. Let, e the other to erties. B the smmetr of H, there is niqe to onnet n to H s shon Figre 2.2. Sine is n ege, one of n is not in I. We ssme I. Let {, } Γ G () V (H) s shon in Figre 2.2 n I = I \ {} {, }. Osere tht I is n inepenent set n is not in K 5 in G I. Ths I is n inepenent set meeting eh K 5 in G. Sine eh C 5 K 2 ontining mst ontin one of n. Ths C(I ) < C(I). Contrition! Cse 2: There is K 5 interseting H ith 4 erties. Let e the erte of this K 5 t not in H, see Figre 2.3. We he to sses. Sse 2: The erte hs nother neighor in H t not in this K 5. Sine H is K 5 -free, e n selet erte in this K 5 sh tht is not n ege of G. Let I := I \ {} {, }. Note tht is not in K 5 in G I, n I is n inepenent set. Ths I is n inepenent set meeting eh K 5 in G. Sine n C 5 K 2 ontining mst ontin one of n, e he C(I ) < C(I). Contrition! Sse 2: All neighors of in H re in this K 5. Let e n erte in this K 5 other thn, n I := I \ {} {}. In this se, there is onl one K 5 ontining. Ths, I is lso n inepenent set meeting eer op of K 5 in G. Osere tht Γ G () \ {} is isonnete. If H = C 5 K 2, then Γ G () H is onnete. Ths is not in C 5 K 2 in G I n C(I ) < C(I). Contrition! Fig Cse 2. Fig Cse 3. Cse 3: There is n ine sgrph H isomorphi to C 5 K 2 sh tht H n H re interseting, see Figre 2.4. Sine V (H) V (H ) n H H, e n fin nonil ege of H n nonil ege of H sh tht V (H ) n V (H). If is non-ege, then let I := I \ { } {}. It is es to hek I is still n inepenent set. We lso osere tht n possile K 5 ontining mst lso ontin. Ths, I meets eer op of K 5 in G. We he in no C 5 K 2 in G I sine is not n ege. We therefore get C(I ) < C(I). Contrition! If is n ege, then loll there re to K 5 interseting t,, n ; s the other for erties re 1, 2, 1, 2, here to liqes

6 6 King, L, n Peng re { 1, 2,,, } n { 1, 2,,, }, see Figre 2.4. Let I = I { 1, 1 } \ { }. Note tht I is n inepenent set n is not in K 5 in G I. Ths I is n inepenent set meeting eh K 5 in G. Osere tht n opies of C 5 K 2 ontining mst ontin one of 1 n 1 ; e he C(I ) < C(I). Contrition!. Cse 4: This is the remining se, G is K 6 -free. We he ω(g ) 5 n V (G ) < V (G). B the minimlit of G, there is n inepenent set I of G meeting eer op of K 5 n C 5 K 2. In I, there is niqe erte of the K 5 otine from ontrting nonil eges of H. Let e the nonil ege orresponing to. Let I = I \ {} {}, e get n inepenent set I of G. Note tht n H \ {} is not in n K 5 of G I Cse 2 s ell s not in n C 5 K 2 of G I Cse 3. Ths I hits eh K 5 in G n C(I ) = 0. Contrition! The folloing lemm etens Theorem 2.1 hen ω(g) = 4; similr reslt s proe inepenentl in [2]. Lemm 2.4. Let G e onnete grph ith (G) 5 n ω(g) 4. If G C 2l+1 K 2 for some l 2, then there is n inepenent set I hitting ll opies of K 4 in G. Proof: We ill proe it ontrition. If the lemm is flse, then let G e minimm onteremple. If G is K 4 -free, then there is nothing to proe. Otherise, e onsier the liqe grph C(G), hose ege set is the set of ll eges ppering in some op of K 4. Bese of (G) = 5, here re ll possile onnete omponent of C(G). 1. C t K 2 for t 4. If this tpe ors, then eer erte in C t K 2 hs egree 5; ths, this is the entire grph G. If t is een, then e n fin n inepenent set I meeting eer K 4. If t is o, then it is impossile to fin sh n inepenent set. Hoeer, this grph is ele from the ssmption of the Lemm. 2. P t K 2 for t 3. In this se, ll internl erties he egree 5 hile the for en erties he egree 4. Consier ne grph G hih is otine eleting ll internl erties n ing for eges to mke the for en erties s K 4. It is es to hek (G ) 5 n ω(g ) 4. Sine G < G, there is n inepenent set I of G meeting eer op of K 4 in G. Note tht there is etl one en erte in I. Osere tht n one en erte n e etene into miml inepenent set meeting eer op of K 4 in P t K 2. Ths, e n eten I to n inepenent set I of G sh tht I meets eer op of K 4 in G. Hene, this tpe of omponent oes not or in C(G). 3. There re for other tpes liste in Figre 2.5. Fig All tpes of omponents in the liqe grph C(G). For eh omponent C i in C(G), let V i e the set of ommon erties in ll K 4 s of C i ; for the leftmost figre in Figre 2.5, V i is the set of ll 4 erties; for the mile to figres, V i is the set of ottom three erties; for the rightmost figre, V i onsists of the left-ottom erte n the mile-ottom erte. Note tht ll V i s re pirise isjoint. Let G e the ine sgrph of G on i V i. Note tht G oes not ontins n erte in C i \ V i. B heking eh tpe, e fin ot tht for eh i n eh V i, hs t most min{2, V i 2} neighors otsie V i in G (not in G!). Appling Theorem 2.2 to G, e onle tht there eists n inepenent set I of G meeting eer V i ; ths I meets eer K 4 in G. Contrition!

7 A Frtionl Anloge of Brooks Theorem 7 Lemm 2.5. Let G e onnete grph ith (G) 5 n ω(g) 4. If G C 2l+1 K 2 for some l 2, then there eists n inepenent set meeting ll ine opies of K 4 n C8 2. Proof: We ill se proof ontrition. Sppose the Lemm is flse. Let G e minimm onteremple (ith the smllest nmer of erties). For n inepenent set I, let C(I) e the nmer of ine opies of C8 2 in G I. Among ll inepenent sets hih meet ll opies of K 4, there eists n inepenent set I sh tht C(I) is minimie. Sine C(I) > 0, let H e op of C8 2 in G I. The erties of H re liste i for i Z 8 ntilokise sh tht i j is n ege of H if n onl if i j 2. The erte i+4 is the ntipoe of i for n i Z 8. Cse 1: There eists erte V (H) sh tht hs fie neighors in H. B the Pigeonhole Priniple, Γ() ontins pir of ntipoes. Withot loss of generlit, s 0, 4 Γ(). If the other three neighors of o not form tringle, then e let I := I \ {} { 0, 4 }; note tht is not in n K 4 of G I. Ths I is n inepenent set meeting eer op of K 4. Sine eer op of C8 2 ontining mst ontin one of 0 n 4, e he C(I ) < C(I). Contrition! Hene, the other three neighors of mst form tringle. Withot loss of generlit, e n ssme tht the three neighors re 1, 2, n 3. No e let I := I \ {} { 0, 3 }; note tht K 4 G I. Ths I is lso n inepenent set meeting eer op of K 4 of G. Sine eer op of C8 2 ontining mst ontin one of 0 n 3, e he C(I ) < C(I). Contrition! Cse 2: There eists erte V (H) sh tht hs etl for neighors in H. Sine H is K 4 -free, e n fin i, j Γ() V (H) sh tht i j is non-ege. Let I := I \ {} { i, j }; I is lso n inepenent set. Note tht Γ() \ { i, j } n not e tringle, is not in n K 4 G I. Ths I meets eer op of K 4. Sine eer op of C8 2 ontining mst ontin one of i n j, e he C(I ) < C(I). Contrition! Cse 3: There eists erte V (H) sh tht hs etl three neighors in H. If the 3 neighors o not form tringle, then hoose i, j Γ() V (H) sh tht i j is non-ege. Note tht Γ() \ { i, j } n not e tringle; is not in n K 4 G I. Let I := I \ {} { i, j }; I is lso n inepenent set meeting eer op of K 4. Sine eer op of C8 2 ontining mst ontin one of i n j, e he C(I ) < C(I). Contrition! Else, the three neighors form tringle; let i e one of them n I := I \ {} { i }; is not in n K 4 G I. Ths I is n inepenent set meeting eer op of K 4. Note tht Γ() \ { i } hs onl to erties in H. The ine grph on Γ() \ { i } is isonnete. Hoeer, for n erte in H = C8 2, the sgrph ine Γ G() H is P 4. There is no C8 2 in G I ontining. Ths, C(I ) < C(I). Contrition! Cse 4: Eer erte otsie H n he t most 2 neighors in H. We ientif eh pir of ntipoes of H to get ne grph G from G. After ientifing, H is trne into K 4 ; here the erties of this K 4 re referre s ft erties. Sse 4: G C 2l+1 K 2. Osere (G ) 5. We lim G is K 5 -free. Sppose not. Sine eer erte in H hs t most one neighor otsie H, then eh ft erte n he t most to neighors otsie H. Rell tht the originl grph G is K 5 -free. If G hs some K 5, then this K 5 ontins either 3 or 4 ft erties. Let e one of the other erties in this K 5. We get hs t lest three neighors in H. Hoeer, this is oere Cse 1, Cse 2, or Cse 3. Ths, G is K 5 -free. Sine G < G, the minimlit of G, G hs n inepenent set I meeting eer op of K 4 n C8 2 in G. There is etl one ft erte in I. No repling this ft erte its orresponing pir of ntipol erties, e get n inepenent set I ; e ssme the pir of ntipol erties re 2 n 6. It is es to hek tht I is n inepenent set of G. Net e lim n V (H) \ { 2, 6 } is neither in K 4 V (G) I nor in C8 2 V (G) I. Sppose there is some sh tht K 4 G I. Rell eh V (H) hs t most one neighor otsie H n H

8 8 King, L, n Peng is K 4 -free; there is some V (H) sh tht hs t lest three neighors in H. This is lre onsiere Cse 1, Cse 2, or Cse 3. We re left to sho tht C 2 8 G I for eh V (H) \ { 2, 6 }. If not, there eists op H of C 2 8 in G I ontining. Note H is 4-reglr, n erte in H n he t most one neighor in I ; in prtilr, 0, 4. Withot loss of generlit, e ssme = 3. Then there is erte V (H) sh tht 3 is n ege, see Figre 2.6. Osere tht the neighorhoo of eh erte of n ine C 2 8 is is P 4. Sine 1 4 n 1 5 re to non-eges, e he 1 eing n ege. Osere Γ G ( 1 ) = { 7, 0, 2, 3, }. Sine 2 H, e he 0 H ; 0 hs to neighors ( 2 n 6 ) otsie H, ontrition! Therefore, I meets eer op of K 4 n C 2 8 in G. Contrition! Fig Sse 4. Sse 4: G = C 2l+1 K 2. The grph G n e reoere from G. It onsists of n ine sgrph H = C8 2 n n ine sgrph P 2l 1 K 2. For eh erte in H, there is etl one ege onneting it to one of the for en erties of P 2l 1 K 2 ; for eh en erte of P 2l 1 K 2, there re etl to eges onneting to the erties in H. First, e tke n mimm inepenent set I of P 2l 1 K 2. Osere tht I hs etl to en points of P 2l 1 K 2 ; so I hs etl for neighors in H. In the remining for erties of H, there eists non-ege i j sine H is K 4 -free. Let I := I { i, j }. Clerl I is n inepenent set of G meeting eer op of K 4 n C8. 2 Contrition! We re re to proe Lemm 1.4. Proof of Lemm 1.4: We nee proe for k 5 n n onnete grph G ith (G) = k n ω(g) k 1 stisfies { χ f (G) k min f(k 1), 1 }. (2.1) 2 If ω(g) k 2, then ineqlit (1.1), e he χ f (G) (G) + ω(g) k 1 2. Ths, ineqlit (2.1) is stisfie. From no on, e ssme ω(g) = (G) 1. For (G) = k 6 n ω(g) = k 1, the onition ω(g) > 2 3 ( (G) +1) is stisfie. B Theorem 2.1, G ontins n inepenent set meeting eer mimm liqe. Eten this inepenent set to miml inepenent set n enote it I. Note tht (G I) k 1 n ω(g I) k 2. Cse 1: k 7. From the efinition of f(k 1), e he χ f (G I) (G I) f(k 1). Ths, χ f (G) χ f (G I) + 1 k 1 f(k 1) + 1 = k f(k 1). Ths, e he f(k) min{f(k 1), 1/2}.

9 A Frtionl Anloge of Brooks Theorem 9 Cse 2: k = 6. B Lemm 2.3, e n fin n inepenent set meets eer op of K 5 n C 5 K 2 ; e eten this inepenent set s miml inepenent set I. Note tht G I ontins no ine sgrph isomorphi C 5 K 2. We he χ f (G I) 5 f(5); it implies χ f (G) 6 f(5). Ths, f(6) min{f(5), 1/2} n e re one. Cse 3: k = 5. If G = C 2l+1 K 2 for some l 3; then G is erte-trnsitie n α(g) = l. It implies tht χ f (G) = V (G) α(g) = l If G C 2l+1 K 2, then Lemm 2.5, e n fin n inepenent set meeting eer op of K 4 n C8 2 ; e eten it s miml inepenent set I. Note tht G I ontins no ine sgrph isomorphi C8. 2 We he χ f (G I) 4 f(4); it implies χ f (G) 5 f(4). Ths, f(3) min{f(4), 1/3} n e re finishe. 3. The se (G) = 4. To proe f(4) 2 67, e ill se n pproh hih is similr to those in [4, 7]. We ill onstrt olorle ilir grphs, n from these olorings e ill onstrt 134-fol oloring of G sing 532 olors. It sffies to proe tht the minimm onteremple oes not eist. Let G e grph ith the smllest nmer of erties n stisfing 1. (G) = 4 n ω(g) 3; 2. χ f (G) > ; 3. G C8 2. B the minimlit of G, eh erte in G hs egree either 4 or 3. To proe Lemm 1.3, e ill sho χ f (G) , hih gies s the esire ontrition. For gien erte in V (G), it is es to olor its neighorhoo Γ G () sing 2 olors. If G () = 3, then e pik non-ege S from Γ G () n olor the to erties in S sing olor 1. If G () = 4 n α(γ G ()) 3, then e pik n inepenent set S in Γ G () of sie 3 n ssign the olor 1 to eh erte in S. If G () = 4 n α(γ G ()) = 2, then e pik to isjoint non-eges S 1 n S 2 from Γ G () ; e ssign olor 1 to eh erte in S 1 n olor 2 to eh erte in S 2. The folloing Lemm shos tht G hs ke propert, hih eentll implies tht this lol oloring sheme orks simltneosl for in lrge sset of V (G). Lemm 3.1. For eh V (G) ith G () = 4 n α(γ G ()) = 2, there eist to erte-isjoint non-eges S 1 (), S 2 () Γ G () stisfing the folloing propert. If e ontrt S 1 () n S 2 (), then the reslting grph G/S 1 ()/S 2 () ontins neither K5 nor G 0. Here K5 is the grph otine from K 5 remoing one ege n G 0 is the grph shon in Figre 3.1. The proof of this lemm is qite long n e ill present its proof in setion 4. For eh erte in G, e ssoite smll set of erties S() selete from Γ G () s follos. If G () = 3, then let S() e the enpoints of non-ege in Γ G () n lel the erties in S() s 1; if G () = 4 n α(γ G ()) 3, then let S() e n inepenent set of sie 3 in Γ G () n lel ll erties in S() s 1; if G () = 4 n α(γ G ()) = 2, then let S() = S 1 () S 2 (), here S 1 () n S 2 () re grntee Lemm 3.1; e lel the erties in S 1 () s 1 n the erties in S 2 () s 2. For n V (G), e he S() = 2, 3, or 4. The folloing efinitions epen on the hoie of S( ), hih is ssme to e fie throgh this setion. For G n j {1, 2, 3}, e efine N j G () = { there is pth 0... j 2 in G of length j sh tht 0 S() n j 2 S()}. We no efine N j G () for j {4, 5, 7}; eh Nj G () is sset of the jth neighorhoo of. For j = 4, N 4 G () if G() = 4, α(γ G ()) = 2, n re onnete s shon in

10 10 King, L, n Peng s 1 s 2 p q r Fig The grph G 0. Figre 3.2; otherise NG 4 () =. In Figre 3.2, is onnete to one of the to erties in S 2 (). Similrl, in Figre 3.3 n 3.3, erte is onnete to grop of erties mens it is onnete to n erte in this grop. For j = 5, NG 5 () if G() = 4, α(γ G ()) = 2 for {, } n n re onnete s shon in Figre 3.3; otherise NG 5 () =. For j = 7, NG 7 () if G() = 4, α(γ G ()) = 2 for {, } n n re onnete s shon in Figre 3.4; otherise NG 7 () =. S () S 1 () 2 S 1 () S () S 1 () S () 2 2 S 1 () S 2 () S 1 () S 2 () e f Fig th neighorhoo. Fig th neighorhoo. Fig th neighorhoo. Note tht for j {1, 2, 3, 5, 7}, N j G () if n onl if Nj G (); t this oes not hol for j = 4. We he the folloing lemm. Lemm 3.2. For V (G) sh tht G () = 4 n α(γ G ()) = 2, e he NG 1 () NG 2 () N3 G () N4 G () N5 G () N7 G () 96. Proof: It is ler tht N1 G () N2 G () NG 3 () = 36. We net estimte N4 G (). In Figre 3.2, osere tht is onnete to one erte of S 2 () n Γ G (). For fie, there re t most for hoies for, t most three hoies for, n t most three hoies for. Therefore, e he NG 4 () = 36. Let s estimte NG 5 (). In Figre 3.3, for fie, e he for hoies for n to hoies for. Fi. Assme Γ G () \ {} = {,, }. Let T 1 = {, }, T 2 = {, }, n T 3 = {, }. We he the folloing lim. Clim There re t most three NG 5 () sh tht for eh e he Γ G() Γ G () = T i for some 1 i 3 s shon in Figre 3.3. Proof of the lim: For eh 1 i 3, there re t most three NG 5 () sh tht Γ G () Γ G () = T i s shon in Figre 3.3 sine eh erte in T i hs t most three

11 A Frtionl Anloge of Brooks Theorem 11 neighors other thn. If the lim is flse, then there is 1 i j 3 sh tht Γ G () Γ G ( i ) = T i n Γ G () Γ G ( i ) = T i for some i, i N5 G (), n Γ G() Γ G ( j ) = T j for some j NG 5 (), here i, i, j re istint. Withot loss of generlit, e ssme Γ G () Γ G ( 1 ) = Γ G () Γ G ( 1 ) = T 1 for 1, 1 N5 G (), n Γ G() Γ G ( 2 ) = T 2 for some 2 NG 5 (), see Figre 3.5. Osere tht Γ G() = { 1, 1, 2, }. Sine Γ G () Γ G ( 1 ) = T 1 s shon in Figre 3.3, n one of s neighors form S i ( 1 ) for some i {1, 2}; e ssme it is S 1 ( 1 ). Note {, 1, 1 } Γ G(). Ths S 1 ( 1 ) = {, 2 } n 2 Γ G ( 1 ). Similrl, e n sho S 1 ( 1) = {, 2 } n 2 Γ G ( 1). No, osere tht Γ G ( 2 ) = { 1, 1,, }. Sine Γ G () Γ G ( 2 ) = T 2 s shon in Figre 3.3, n one of neighors of 2 form S i ( 2 ) for some i {1, 2}; e ssme i = 1. Bese { 1, 1} Γ G (), then S 1 ( 2 ) = {, }. Hoeer, n re not is in the sme inepenent set in the efinition of NG 5 (), see Figre 3.3. This is ontrition n this se n not hppen. The lim follos S 1 () S 2 () Fig The pitre for the lim. Therefore, NG 5 () = 24. In Figre 3.4, for fie, e he to hoies for the ege e, one hoie for, to hoies for, n three hoies for the ege f. Fi. B onsiering the egrees of the enpoints of f, there is t most one f n t most one NG 7 () sh tht Γ G(f) Γ G () = 4 s shon in Figre 3.4. Therefore, e he NG 7 () = 4. Lst, e estimte NG 5() N7 G (). If there is some N7 G (), then e osere tht there re t most fie s (see Figre 3.3). We get the nmer of NG 5 () is t most 5 3 = 15. In this se, e he If NG 7 () =, then lso e he Therefore N 5 G() N 7 G() < 24. N 5 G() N 7 G() 24. N 1 G() N 2 G() N 3 G() N 4 G() N 5 G() N 7 G() = 96. Bse on the grph G, e efine n ilir grph G on erte set V (G). The ege set is efine s follos: E(G ) preisel if either NG 1 () N2 G () N3 G () N4 G () NG 5 () N7 G (), or N4 G (). We he the folloing lemm. Lemm 3.3. The grph G is 133-olorle. Proof: Let σ e n inresing orer of V (G ) stisfing the folloing onitions. 1: For n sh tht G () = 3 n G () = 4, e he σ() < σ(). 2: For n sh tht G () = G () = 4, α(γ G ()) 3, n α(γ G ()) = 2, e he σ() < σ(). We ill olor V (G ) oring to the orer σ. For eh, e he the folloing estimte on the nmer of olors forien to se for.

12 12 King, L, n Peng 1: For sh tht G () = 3, the nmer of olors forien to se for is t most NG 1 () N2 G () N3 G () = 39. 2: For sh tht G () = 4 n α(γ G ()) 3, the nmer of olors forien to se for is t most NG 1 () N2 G () N3 G () = 39. 3: For sh tht G () = 4 n α(γ G ()) = 2, the nmer of olors forien to se for is t most NG 1() N2 G () N3 G () N4 G () N5 G () N7 G () + N4 G () = 132 Lemm 3.2. Therefore, the gree lgorithm shos G is 133-olorle. Let X e olor lss of G. We efine ne grph G(X) the folloing proess. 1. For eh X, if S() = 2 or S() = 3, then e ontrt S() s single erte, elete the erties in Γ G () \S(), n keep lel 1 on the ne erte; if S() = 4, i.e., S() = S 1 () S 2 (), then e ontrt S 1 () n S 2 () s single erties n keep their lels. After tht, e elete X. Let H e the reslting grph. 2. Note tht Γ H () Γ H () = n there is no ege from Γ H () to Γ H () for n, X s X is olor lss. 3. We ientif ll erties ith lel i s single erte i for i {1, 2}. Let G(X) e the reslte grph. We he the folloing lemm on the hromti nmer of G(X). Lemm 3.4. The grph G(X) is 4-olorle for eh olor lss. We postpone the proof of this lemm ntil the en of this setion n proe Lemm 1.3 first. Proof of Lemm 1.3: B Lemm 3.3, there is proper 133-oloring of G. We ssme V (G ) = V (G) = 133 i=1 X i, here X i is the i-th olor lss. For eh i {1,..., 133}, Lemm 3.4 shos G(X i ) is 4-olorle; let i : V (G(X i )) T i e proper 4-oloring of the grph G(X i ). Here T 1, T 2,..., T 133 re pirise isjoint; eh of them onsists of 4 olors. For i {1,...,133}, the 4-oloring i n e iee s 4-oloring of G\X i sine eh erte ith lel j reeies the olor i ( i ) for j = 1, 2 n eh remoe erte hs t most three neighors in G \ X i. No e rese the nottion i to enote this 4-oloring of G \ X i. For eh X i, e he ΓG() i () 2. We n ssign to nse olors, enote the set Y (), to. We efine f i : V (G) P(T i ) (the poer set of T i ) stisfing { {i ()} if V (G) \ X f i () = i, Y () if X i. Osere tht eh erte in X i reeies to olors from f i n eer other erte reeies one olor. Let σ : V (G) P( k i=1 T i) e mpping sh tht σ() = m i=1 f i(). It is es to erif σ is 134-fol oloring of G sh tht eh olor is rn from plette of 532 olors; nmel e he χ f (G) = The proof of Lemm 1.3 is finishe. Before e proe Lemm 3.4, e nee the folloing efinitions. A lok of grph is miml 2-onnete ine sgrph. A Glli tree is onnete grph in hih ll loks re either omplete grphs or o les. A Glli forest is grph ll of hose omponents re Glli trees. A k-glli tree (forest) is Glli tree (forest) sh tht the egree of ll erties re t most k 1. A k-ritil grph is grph G hose hromti nmer is k n eleting n erte n erese the hromti nmer. Glli shoe the folloing Lemm. Lemm 3.5. [3] If G is k-ritil grph, then the sgrph of G ine on the erties of egree k 1 is k-glli forest.

13 A Frtionl Anloge of Brooks Theorem 13 Proof of Lemm 3.4: We se proof ontrition. Sppose tht G(X) is not 4-olorle. The onl possile erties in G(X) ith egree greter thn 4 re the erties 1 n 2, hih re otine ontrting the erties ith lel 1 n 2 in the intermeite grph H. The simple gree lgorithm shos tht G(X) is ls 5-olorle. Let G (X) e 5-ritil sgrph of G(X). Appling Lemm 3.5 to G (X), the sgrph of G (X) ine on the erties of egree 4 is 5-Glli forest F. The erte set of F m ontin 1 or 2. Delete 1 n 1 from F if F ontins one of them. Let F e the reslting Glli forest. (An ine sgrph of Glli forest is still Glli forest.) The Glli forest F is not empt. Let T e onnete omponent of F n B e lef lok of T. The lok B is either liqe or n o le from the efinition of Glli tree. Let e erte in B. As hs t most to neighors ( 1 n 2 ) otsie F in G(X), e he F () 2. If is not in other loks of F, then e he B () 2. It follos tht B 3. Sine B is sgrph of G n G is K 4 -free, the lok B is n o le. Let 1 2 e n ege in B sh tht 1 n 2 re not in other loks. The egree reqirement implies i j re eges in G(X) for ll i, j {1, 2}. For i = 1, 2, there re erties i, i X stisfing S( i ) Γ G ( i ) n S( i ) Γ G ( i ) ; moreoer either S( i ) = 4 or S( i ) = 4 sine one of its neighorhoo hs lel 2. Withot loss of generlit, e ssme S( i ) = 4 for i {1, 2}. If i i, then i N 4 G ( i), i.e., i Γ G ( i ); this ontrits X eing olor lss. Ths e he i = i n S( i ) = 4 for i {1, 2}. For {i, j} = {1, 2}, if i j, then i N 5 G ( i), i.e., i Γ G ( i ); this is ontrition of X eing olor lss. Ths e he 1 = 2 = 1 = 2. Let enote this ommon erte oe. Then G () = 4 n α(γ G ()) = 2. Let 0 e the onl erte in B shre other loks. Sine B 0 is onnete, the rgment oe shos there is ommon for ll eges in B 0. If Γ G(X) ( 0 ) { 1, 2 }, the there is some erte 0 X sh tht S( 0 ) Γ G ( 0 ). B the similr rgment, e lso he 0 =. Therefore, epens onl on B. In the sense tht for n X n n B, if S() Γ G (), then =. The lok B is n o le s e mentione oe. Sppose B = 2r + 1. Let 0, 1,..., 2r e the erties of B in li orer n 0 e the onl erte hih m e shre other lok. Let X e the erte etermine B. Rell G () = 4 n α(γ G ()) = 2. Eh erte in Γ() n he t most 2 eges to B. We get 4r E(B, Γ()) 8. (3.1) We he r 2. The lok B is either C 5 or K 3. We lim oth 0 1 n 0 2 re non-eges of G(X). If B = C 5, then ineqlit (3.1) implies tht 0 hs no neighor in Γ() n the lim hols. If B = K 3, then the lim lso hols; otherise B {S 1 (), S 2 ()} forms K 5 in G/S 1 ()/S 2 (), hih is ontrition to Lemm 3.1. Let 1 n 2 e the to neighors of 0 in other loks of F. If 1 n 2 re in the sme lok, then this lok is n o le; otherise, 0 1 n 0 2 re in to ifferent loks. The nion of non-lef loks of T is Glli-tree, enote T. The rgment oe shos eer lef lok of T mst e n o le. Let C e sh lef lok of T. No C is n o le, n C is onnete to C 1 lef loks of T. Let B n B e to lef loks of T sh tht B C is jent to B C. Withot loss of generlit, e m ssme B is the one e onsiere efore. B the sme rgment, B is n o le of sie

14 14 King, L, n Peng 2r + 1 ith r {1, 2}. Let 0, 1,..., 2r e the erties of B n 0 e the onl erte in B C. For i in {1, 2,..., 2r } n j in {1, 2}, i j re eges in G (X). Similrl, there eists erte X ith G ( ) = 4 n α(γ G ( )) = 2 sh tht E( i, S 1 ( )) 1 n E( i, S 2 ( )) 1. We mst he = ; otherise NG 7 (), i.e., Γ G (), n this ontrits the ft tht X is olor lss in D. No e he E(Γ(), B) 4r n E(Γ(), B ) 4r. B onting the egrees of erties in Γ() in G, e he 4r + 4r We get r = r = 1. Both B n B re K 3 s. In this se, G/S 1 ()/S 2 () ontins the grph G 0, see figre 3.1. This ontrits Lemm 3.1. We n fin the esire ontrition, so the lemm follos. 4. Proof of Lemm 3.1. In this setion, e ill proe Lemm 3.1. We first reie Lemm from [7]. Lemm 4.1. Let G e grph. Sppose tht G 1 n G 2 re to sgrphs sh tht G 1 G 2 = G n V (G 1 ) V (G 2 ) = {, }. 1. If is n ege of G, then e he 2. If is not n ege of G, then e he χ f (G) = m{χ f (G 1 ), χ f (G 2 )}. χ f (G) m{χ f (G 1 ), χ f (G 2 + ), χ f (G 2 /)}, here G 2 + is the grph otine from G 2 ing ege n G 2 / is the grph otine from G 2 ontrting {, }. Proof of Lemm 3.1: Rell tht G is onnete K 4 -free grph ith minimm nmer of erties sh tht G C8 2 n χ f(g) > Note tht G is 2-onnete. We ill proe it ontrition. Sppose Lemm 3.1 fils for some erte in G. Osere Γ G () is one of the grphs in Figre 4.1. Here e ssme Γ G () = {,,, }. Throgh the proof of the lemm, let S 1 n S 2 e to erte-isjoint inepenent sets in Γ G (), H e tringle in V (G)\({} Γ G ()), then s (S 1, S 2, H) is triple if {S 1, S 2, H} ontins K5 in G/S 1/S 2. Γ G () = P 4 Γ G () = 2 e Γ G () = C 4 Fig Three possile ses of Γ G (). If Γ G () = P 4, then {, } n {, } is the onl pir of isjoint non-eges. There is tringle H ith V (H) = {,, } sh tht ({, }, {, }, H) is triple. Note tht E({,,, }, {,, }) = 5 or 6. B n ehstie serh, the ine sgrph of G on {,,,,,,, } is one of the folloing si grphs (see Figre 4.2). If Γ G () = 2 e, then ({, }, {, }) n ({, }, {, }) re to pirs of isjoint noneges. B onsiering the egrees of erties in Γ G (), there is onl one tringle H ith

15 A Frtionl Anloge of Brooks Theorem 15 H 1 H 2 H 3 H 4 H 5 H 6 Fig If Γ G () = P 4, then there re si possile ine sgrphs. V (H) = {,, } sh tht ({, }, {, }, H) n ({, }, {, }, H) re to triples. B n ehstie serh, the ine sgrph of G on {,,,,,,, } is one of the folloing three grphs (see Figre 4.3). H 7 H 8 H 9 Fig If Γ G () = 2 e, then there re three possile ine sgrphs. It sffies to sho tht G nnot ontin H i for 1 i 9. Sine ll erties in H 1 (n H 2 ) he egree 4, H 1 (n H 2 ) is the entire grph G. Osere tht H 1 is isomorphi to C 2 8 n H 2 is 11:3-olorle (see Figre 4.4). Contrition! In H 7, the erte is the onl erte ith egree less thn 4. If H 7 is not the entire grph G, then is t erte of G. This ontrits the ft tht G is 2-onnete. Ths G = H 7. The grph H 7 is 11:3-olorle s shon Figre 4.4. Contrition! No e onsier the se H 3. Note H 3 + is the grph H 2. We he χ f (H 3 ) χ f (H 2 ) 11/3. The grph H 3 mst e proper ine sgrph of G, n the pir {, } is erte t of G. Let G e the ine sgrph of G eleting ll erties in H 3 t,. We ppl Lemm 4.1 to G + ith G 1 = H 3 + = H 2 n G 2 = G +. We he χ f (G + ) m{χ f (H 2 ), χ f (G + )}. Note χ f (H 2 ) 11/3 n 11/3 < χ f (G) χ f (G + ). We he χ f (G) χ f (G + ). Both

16 16 King, L, n Peng st 569 st st 489 st4 123 H 2 H 7 Fig The grph H 2 n H 7 re 11:3-olorle. n he t most 2 neighors in G +. Ths G + is K 4 -free; G + C8 2 n hs feer erties thn G. This ontrits to the minimlit of G. Note H 5 + = H 2. The se H 5 is similr to the se H 3. Note tht H 4, H 6, n H 8 re isomorphi to eh other. It sffies to sho G oes not ontin H 4. Sppose tht H 4 is proper ine sgrph of G. Let G 1 e the ine sgrph of G eleting ll erties in H 4. Note C8 2 is not proper sgrph of n grph in G 4. We he G 1 C8 2. Note tht n he egree 3 hile other erties in H 4 he egree 4. Sine G is 2-onnete, hs niqe neighor, enote, in V (G 1 ). Similrl, hs niqe neighor, enote, in V (G 1 ). Osere tht the pir {, } forms erte t of G. Let G 2 e the ine grph of G on V (H 4 ) {, }. Appling Lemm 4.1 to G ith G 1 n G 2, e he χ f (G) m{χ f (G 1 ), χ f (G 2 + ), χ f (G 2 /)}. Figre 4.5 shos χ f (G 2 + ) n χ f (G 2 /) re t most 11/ s t s t t 12s 45t 12s G 2 + G 2 / Fig Cse H 4 : oth grph G 2 + n G 2 / re 11:3-olorle. Sine χ f (G) > 11/3, e he χ f (G) χ f (G 1 ). No G 1 is K 4 -free n hs mimm egree t most 4; G 1 hs feer erties thn G. This ontrits the minimlit of G. If G = H 4, then χ f (H 4 ) 11/3, sine H 4 is sgrph of G 2 + in Figre 4.5. No e onsier the lst se H 9. First, e ontrt,, into ft erte enote. We rite G/ for the grph fter this ontrtion. Osere tht {, } is erte-t of G/. Let G 4 n G 4 e to onnete sgrphs of G/ sh tht G 4 G 4 = G/, G 4 G 4 = {, }, n {, } G 4. Note tht G 4 is 11:3 olorle, see Figre 4.6. No Lemm 4.1, e he χ f (G/) m{χ f (G 4 ), χ f (G 4 )}. As {,, } is n inepenent set, eh :-oloring of G/ gies n :-oloring of G, tht is χ f (G/) χ f (G) > 11/3. The grph G 4 is 11:3-olorle; see Figre 4.6. Ths e he χ f (G 4 ) χ f(g/) χ f (G). It is es to hek tht G 4 hs mimm egree 4, K 4-free,

17 A Frtionl Anloge of Brooks Theorem 17 n it is not C8 2. Hene G 4 mst ontin K 4. Otherise, it ontrits the minimlit of G. Seon, e ontrt {,, } into ft erte n enote the grph G/. Let G 5 n G 5 e to onnete sgrphs of G/ sh tht G 5 G 5 = G/, G 5 G 5 = {, }, n {, } G 5. Note tht G 5 is 11:3-olorle; see Figre 4.6. B similr rgment, G 5 mst ontin K st st 456 9st st 145 5st st t 236 p 149 G 4 G 5 G 6 Fig Cse H 9 : the grphs G 4, G 5, n G 6 re 11:3-olorle. The remining se is tht oth G 4 n G 5 he K 4 hen e ontrt n. Sine the originl grph G is K 4 -free, the K 4 in G 4 (n in G 5 ) mst ontin the ft erte (or ), respetiel. Note tht eh of the for erties,,, hs t most one ege leing H 9. There mst e tringle p in G n these for otr eges re onnete to some element of {,, p}. The grph G/ mst ontin the sgrph G 6 s rn in Figre 4.6. Note tht {, } is erte-t in G/. Let G 6 n G 6 e to onnete sgrphs of G/, hih stisf G 6 G 6 = G, G 6 G 6 = {, }, n G 6. B Lemm 4.1, e he χ f (G/) m{χ f (G 6 ), χ f (G 6)}. Note tht G 6 is 11:3-olorle; see Figre 4.6. We lso he χ f (G/) χ f (G) > We otin χ f (G 6) χ f (G/) χ f (G). Osere tht G 6 is sgrph of G. We rrie t ontrition of the minimlit of G. If Γ G () = C 4, then the onl possile hoie for the to inepenent sets re {, } n {, }. If there is some tringle H sh tht ({, }, {, }, H) is triple, then e he E(Γ G (), H) 5. Hoeer, E(Γ G (), H) 4. This is ontrition. Ths the lemm follos in this se. We n selet to erte isjoint non-eges S 1 n S 2 sh tht the grph G/S 1 /S 2 ontins no K 5. For these prtilr S 1 n S 2, if G/S 1 /S 2 ontins no G 0, then Lemm 3.1 hols. Withot loss of generlit, e ssme tht G/S 1 /S 2 oes ontin G 0. Let s i = S i for i = 1, 2. Osere tht oth s 1 n s 2 he for neighors,, p, q other thn in G 0. It follos tht On the one hn, e he E(G S1 S 2 ) = 1 2 E(S 1 S 2, {,, p, q}) 8. ( S 1 S 2 () E(S 1 S 2, {,, p, q}) 4 1 (16 8 4) 2 = 2. )

18 18 King, L, n Peng On the other hn, α(γ()) = 2 implies G S1 S 2 ontins t lest to eges. Ths, e he Γ G () = 2 e. Lel the erties in Γ G (),,, s in Figre 4.1. We ssme n re eges hile,,, re non-eges. Osere tht eh erte in {,, p, q} hs etl to neighors in {,,, }. If one erte, s, hs to neighors forming non-ege, s, then e n hoose S 1 = {, } n S 2 = {, }. It is es to hek tht G/S 1/S 2 ontins neither G 0 nor K 5. We re one in this se. In the remining se, e n ssme tht for eh erte in {,, p, q}, the neighors of in {,,, } ls form n ege. Up to releling erties, there is onl one rrngement for eges eteen {,, p, q} n {,,, }; see the grph H 10 efine in Figre 4.7. The st p q s 1st 238 r 47t 569 Fig H 10 n n 11:3-oloring of H 10. grph H 10 is 11:3-olorle s shon in Figre 4.7. Sine χ f (G) > 11/3, H 10 is proper sgrph of G. Note in H 10, eer erties eept n r hs egree 4; oth n r he egree 3. Ths, {, r} is erte t of G. Let G 1 = H 10 n G 2 e the sgrph of G eleting erties in {,,,,, p, q,, }. Appling Lemm 4.1 ith G 1 n G 2 efine oe, e he χ f (G) m{χ f (G 1 ), χ f (G 2 )}. Sine χ f (G) > 11/3 n χ f (G 1 ) 11/3 (see Figre 4.7), e mst he χ f (G 2 ) χ f (G). Note tht G 2 hs feer nmer of erties thn G. This ontrits the minimlit of G. Therefore, the lemm follos. REFERENCES [1] O. Boroin n A. Kostohk, On n pper on on grph s hromti nmer, epening on the grph s egree n ensit, J. Com. Theor Ser. B, 23 (1977), [2] D. Christofies, K. Ers, n A.D. King, A note on hitting mimm n miml liqes ith stle set, Smitte. Ari preprint , (2011). [3] T Glli, Kritishe grphen I, Mgr T. Ak. Mt. Ktt ó Int. Köl, 8 (1963), [4] H. Htmi n X. Zh, The frtionl hromti nmer of grphs of mimm egree t most three, SIAM. J. Disrete Mth., 24 (2009), [5] C. C. Hekmn n R. Thoms, A ne proof of the inepenene rtio of tringle-free i grphs, Disrete Mth., 233 (2001), [6] Anre D. King, Hitting ll mimm liqes ith stle set sing lopsie inepenent trnsersls, J. Grph Theor, 67-4 (2011), [7] L. L n X. Peng, The frtionl hromti nmer of tringle-free grphs ith 3, smitte. [8] M. Mollo n B. Ree, Grph oloring n the proilisti metho, olme 23 of Algorithms n Comintoris, Springer-Verlg, Berlin, [9] B. Ree, ω,, n χ, J. Grph Theor, 27-4 (1998), [10] B. Ree, A strengthening of Brooks thoerem, J. Com. Theor Ser. B, 76 (1999),

19 A Frtionl Anloge of Brooks Theorem 19 [11] E. R. Sheinermn n D. H. Ullmn, Frtionl grph theor, A rtionl pproh to the theor of grphs, Wile-Intersi. Ser. Disrete Mth. Optim, John Wile & Sons, In, Ne York, 1997.