Sar coloring of pare graph Yeha B Daniel W. Cranon Mickaël Monaier André Rapad Weifan Wang Abrac A proper coloring of he erice of a graph i called a ar coloring if he nion of eery o color clae indce a ar fore. The ar chromaic nmber χ (G) i he malle nmber of color reqired o obain a ar coloring of G. In hi paper, e dy he relaionhip beeen he ar chromaic nmber χ (G) and he maimm aerage degree Mad(G) of a graph G. We proe ha: 1. If G i a graph ih Mad(G) < 26, hen χ(g) 4. 2. If G i a graph ih Mad(G) < 18 and girh a lea 6, hen χ(g) 5. 3. If G i a graph ih Mad(G) < 8 and girh a lea 6, hen χ(g) 6. 3 Thee rel are obained by proing ha ch graph admi a pariclar decompoiion ino a fore and ome independen e. 1 Inrodcion Le G be a graph. A proper ere coloring of G i an aignmen c of ineger (or label) o he erice of G ch ha c() c() if he erice and are adjacen in G. A k-coloring i a proper ere coloring ing k color. The minimm k ch ha G ha a k-coloring i called he chromaic nmber of G, denoed by χ(g). A proper ere coloring of a graph i acyclic if here i no bicolored cycle in G. In oher ord, he graph indced by he nion of eery eery o color clae i a fore. The acyclic chromaic nmber, denoed by χ a (G), of a graph G i he malle ineger k ch ha G ha an acyclic k-coloring. The acyclic coloring of graph a inrodced by Grünbam in [3]. Finally, a proper ere coloring of a graph i called a ar coloring if he nion of eery o color clae indce a ar fore. The ar chromaic nmber, denoed by χ (G), i he malle ineger k ch ha G ha a ar k-coloring. Deparemen of Mahemaic, Zhejiang Normal Unieriy, Jinha 321004, P. R. China, yhb@zjn.cn DIMACS, Rger Unieriy, Picaaay, Ne Jerey 08854, dcrano@dimac.rger.ed LaBRI UMR CNRS 5800, Unierié Bordea I, 33405 Talence Cede, France, monai@labri.fr LaBRI UMR CNRS 5800, Unierié Bordea I, 33405 Talence Cede, France, rapad@labri.fr Deparemen of Mahemaic, Zhejiang Normal Unieriy, Jinha 321004, P. R. China, f@zjn.cn 1
Sar coloring hae recenly been ineigaed by Ferin, Rapad, and Reed [2], Nešeřil and Oana de Mendez [4], and Alberon e al. [1]. Alberon e al. proed an inereing relaionhip beeen χ, χ, and χ a of G and ome rcre of G (called F -redcion), hich gie he folloing corollary. Corollar [1] Le G be a planar graph. 1. If g(g) 3, hen χ (G) 20. 2. If g(g) 5, hen χ (G) 16. 3. If g(g), hen χ (G) 9. Here g(g) i he girh of G, i.e. he lengh of a hore cycle in G. Moreoer, Alberon e al. gae a planar graph G ih χ (G) = 10 and hey obered ha for any girh here ei a planar graph G ih χ (G) > 3. In hi paper, e dy he relaionhip beeen ar coloring and maimm aerage degree. The maimm aerage degree of G, denoed by Mad(G), i defined a: { } 2 E(H) Mad(G) = ma, H G. V (H) Or main rel i he folloing heorem. Theorem 2 Le G be a graph. 1. If Mad(G) < 26, hen χ (G) 4. 2. If Mad(G) < 18 and g(g) 6, hen χ (G) 5. 3. If Mad(G) < 8 3 and g(g) 6, hen χ (G) 6. I i ell-knon ha if G i a planar graph, hen Mad(G) < 2 g(g) g(g) 2. Hence, e obain he folloing corollary. Corollary 3 Le G be a planar graph. 1. If g(g) 13, hen χ (G) 4. 2. If g(g) 9, hen χ (G) 5. 3. If g(g) 8, hen χ (G) 6. We hold noe ha par 2 and 3 of Corollary 3 ere obained independenly by Timmon [5] and ha he proof of Theorem 2 in he preen paper e echniqe imilar o [5]. Or proof are baed on an eenion of an oberaion by Alberon e al. [1] (Lemma 5.1). They noiced ha if here ei a pariion of he erice of a graph G ch ha: V (G) = F I and he folloing o condiion hold: 2
P1. G[F] i a fore P2. G[I] i an independen e ch ha for all, y I, d G (, y) > 2, hen G ha a ar 4-coloring; e can conrc hi coloring by ar 3-coloring G[F] ih color 1, 2, and 3, (ee Oberaion 5) and coloring G[I] ih color 4. We eend hi idea by ing many independen e aifying P2 in a cerain ene. In Secion 2 e proe Theroem 2.1, in Secion 3 e proe Theorem 2.2, and in Secion 4 e proe Theorem 2.3. In Secion 5, e offer conclding remark. Some noaion: We e V (G) and E(G) o denoe repeciely he e of erice and edge of a graph G; e e δ(g) o denoe he minimm degree of G. For a ere, e denoe by d() i degree. A ere of degree k (rep. a lea k, a mo k) i called a k-ere (rep. k-ere, k-ere). A k-pah i a pah of lengh k (nmber of edge). A k-hread i a (k + 1)-pah hoe k inernal erice are of degree 2. A k i1,,i k -ere ih i 1 i k i a k-ere ha i he iniial ere of k hread of lengh i 1,, i k. Le S V (G); e denoe by G[S] he bgraph of G indced by he erice of S. Le and y be o erice of G; e denoe by d G (, y) he diance in G beeen and y, i.e. he lengh (nmber of edge) of a hore pah beeen and y in G. In or diagram, a ere i colored black if all of i inciden edge are dran; a ere i colored hie if i may hae more inciden edge han hoe dran. We rie S = S 1 S 2 S 3 o denoe ha he S i pariion he e S. 2 Graph ih Mad(G) < 26 In hi ecion, e proe ha eery graph G ih Mad(G) < 26 i ar 4-colorable. Lemma 4 refer o an independen e I ch ha for all, y I, d G (, y) > 2; e call ch a e 2-independen. Lemma 4 If Mad(G) < 26, hen here ei a ere pariion V (G) = I F ch ha F indce a fore and I i a 2-independen e. Proof. Le G be a conereample ih he fee erice. We may ame ha G i conneced and ha no ere of degree 1, ince ch a ere cold be added o F. We begin by proing for claim abo rcre ha m no appear in G. In each cae, le H be he forbidden rcre, le G = G H, and le V (G ) = I F be he deired pariion of V (G ), hich i garaneed by he minimaliy of G. We conclde ih a dicharging argmen, hich ho ha G m conain a lea one of hee forbidden rcre. Thi conradicion implie ha no conereample ei, o he lemma i re. Claim 1: G ha no 2 1,1 -ere. Sppoe i a 2 1,1 -ere in G. Conider he pariion for he graph obained by remoing and i neighbor from G. We eend hi pariion o G. If any of he 3
econd neighbor of are in I, hen add and i neighbor o F. Oherie, add o I and add he neighbor of o F. Claim 2: G ha no 3 1,1,1 -ere adjacen o a 2 0,1 -ere. Sppoe i a 3 1,1,1 -ere adjacen o a 2 0,1 -ere y. Conider he pariion for he graph obained by remoing, y, and heir neighbor from G. If boh of he econd neighbor of ha are no adjacen o y are in I, hen add all he remoed erice o F. If boh of hee econd neighbor are in F, hen add o I and add all oher erice o F. Th, e may ame ha one econd neighbor of i in F, and he oher i in I. If he econd neighbor of y no adjacen o i in I, hen add all erice o F. Oherie, add y o I and all oher erice o F. Claim 3: G ha no 4 1,1,1,1 -ere adjacen o for 2 0,1 -erice. Sppoe i ch a 4 1,1,1,1 -ere in G. Conider he pariion for he graph obained by remoing, he neighbor of, and he econd neighbor of. To eend he pariion o G, add o I and add he neighbor and econd neighbor of o F. We no ho ha G canno hae cerain ype of cycle. We an J o be he maimm bgraph of G ch ha all i erice eiher are 2 0,1 -erice or are 3 0,1,1 - erice ha are adjacen o a lea one 2 0,1 -ere; frher, if d G () = 3, hen e an d J () 2. To form J, le A be he e of 2 0,1 -erice and of 3 0,1,1 -erice ha are adjacen o o 2 0,1 -erice. Add o A all 3 0,1,1 -erice ha are adjacen o one 2 0,1 -ere and anoher 3 0,1,1 -ere, hich i ielf adjacen o a lea one 2 0,1 -ere. Le J be he bgraph indced by A. Claim 4: Eery componen of J i a ree or a cycle. Sppoe J conain a cycle ih ere e C hich conain a ere r ih d J (r) = 3. Chooe ch a cycle o ha C i minimal. By minimaliy C indce a chordle cycle in G. Le V (G C) = F I be a ere pariion of G C, here e may ame ha a 1-ere in G C or a 2-ere adjacen o a 1-ere i in F. We eend hi pariion o all of G in o phae. In phae 1, e ill form I and F by cceiely adding he erice of C o I and F o ha I i a 2-independen e in G, and he only poible cycle indced by F i C ielf. In phae 2, e modify I and F, if neceary, o ha I i a 2-independen e and F i a acyclic. Before beginning phae 1, e alo need o definiion. For conenience, e aign a cyclic orienaion o C. If erice and are in C F, hen i a predeceor of if he direced pah along C from o i enirely in F. We ay ha F i ellbehaed, if for all erice, ih (F C), (F C), and E(G), e hae eiher ha i a leaf in F or ha ha a mo o predeceor, each of hich i a 2-ere in G. We no ho hy i i efl o hae F ell-behaed. Aiming for a conradicion, ame B i a cycle in F in hich B C. Le and be he endpoin of a maimal componen of B C. Wiho lo of generaliy, ame i a predeceor of. Le be he neighbor of no on C. Since lie on B, i no a leaf in F. Since F i ell-behaed, he o predeceor of hae degree 2 in G. Thi conradic he fac ha d G () = 3. Obere ha a he ar of phae 1 e 4
hae F = F and I = I o ha F i riially ell-behaed. We ill enre ha F remain ell-behaed hrogho phae 1, o ha a he end of he phae he only poible cycle indced by F i C. (1) 1 1 (2) y (3) Figre 1: The hree key configraion of Claim 4. To eece phae 1, obere ha he immediae neighborhood (in G) of each 3 0,1,1 - ere on C m appear a in Figre 1.1 or Figre 1.2, here he arro indicae C; recall ha he black erice in he figre hae all neighbor hon, b hie erice may hae addiional neighbor (in G). One by one, e eamine he neighborhood of each 3-ere on C proceeding in he order a e enconer hem on C, and aign each ere eiher o I or F. If C conain a ere ih a 3-neighborhood a in Figre 1.2, hen e begin a ch a. Oherie e chooe a 3-ere a in Figre 1.1 for hich, if poible, I. Sppoe fir ha he 3-neighborhood of look like in Figre 1.1. If or i in I, hen add,, o F. If and are no in I, hen add o I and, o F. Thi keep F ell-behaed: If i aigned o I, hen ha no predeceor. If i no aigned o I, hen i m be he cae ha or i in I. If I, hen ha o predeceor, boh of hich are 2-erice in G, herea if I, hen F i riially 5
ell-behaed. Sppoe hen ha he neighborhood of look like in Figre 1.2. If F, hen add o I and,, 1 o F. If I, and 1 F, hen add 1 o I, and,, o F. If, 1 I, hen add,,, 1 o F. To ee ha F i ell-behaed, obere ha hen F, hen i aigned o I, o / F and 1 ha no predeceor. If I, hen he neighbor of no in C i a leaf of F. Frhermore, if alo 1 I, hen he neighbor of 1 no in C i a leaf in F, herea if 1 F, hen 1 / F. Thi complee phae 1. If F i acyclic, hen e hae eended he pariion o G o e may ame ha F conain a cycle. Since F i ell-behaed, hi cycle m be C. We correc hi problem in phae 2. All of C lie in F preciely hen for each 3-ere on C a in Figre 1.1 e hae I, and for each 3-ere a in Figre 1.2 e hae, 1 I. To form I and F, recall ha C conain a ere r ih d J (r) = 3; ha i r i adjacen o a 3 0,1,1 -ere and o o 2 0,1 -erice. If C rn hrogh = r a in Figre 1.1, hen he immediae neighborhood of m look like Figre 1.3. In hi cae e m hae I. If alo y I, hen le I = I +, herea if y F, hen le I = I + +. In he remaining cae C rn hrogh r a in Figre 1.2. If r =, hen i a 2-ere ih I, hich conradic or ampion ha degree 2 erice adjacen o leae in G C are in F, and h F. If r = 1, hen e ge a imilar conradicion for 1. In any cae, I i 2-independen and F = V (G) I i no acyclic, finihing he proof of Claim 4. No e proe ha eery graph G ih minimm degree 2 and Mad(G) < 26 conain one of he configraion forbidden by Claim 1 4. Le G be a conereample, i.e. a graph ih δ(g) 2, ih Mad(G) < 26, and conaining none of he configraion forbidden by Claim 1 4. We e a dicharging argmen ih iniial charge ω() = d() a each ere and ih he folloing for dicharging rle (R1) (R4), here A and J refer o he ere e and bgraph from Claim 4. We rie ω () o denoe he change a each ere afer e apply he dicharging rle. If i a 3-ere, hen a 2-ere i nearby if i adjacen o, or if and hae a common neighbor of degree 2 (in G). Noe ha he dicharging rle do no change he m of he charge. To complee he proof, e ho ha ω () 26 for all V (G); hi lead o he folloing obio conradicion: 26 26 V (G) V (G) V (G) ω () = V (G) Hence no conereample can ei. The dicharging rle are defined a follo. (R1) Each 3-ere gie 2 o each nearby 2-ere. V (G) ω() = 2 E(G) Mad(G) < 26 V (G) V (G). (R2) Each 3 0,1,1 -ere receie 1 from i neighbor of degree a lea 3, nle hi neighbor i in A in hich cae no charge i ranferred. 6
(R3) If d J () = 3, hen ge 1 from he bank. (R4) If / A and i adjacen o k 2 0,1 -erice, hen gie k o he bank. We no erify ha ω () 26 for each V (G). Noe ha each leaf in J i a 2 0,1 -ere. Hence, by (R4), he bank receie 1 for each leaf of J. By (R3), he bank gie aa for each 3-ere of J. Claim 4 implie ha J ha aerage degree a mo 2, o J conain a lea a many leae a 3-erice; hence, he bank ha nonnegaie charge. Cae d() = 2 By (R1), receie 2 from each nearby 3-ere, o ω () = 2+2 2 = 26. Cae d() = 3 If i a 3-ere, hen e conider eeral poibiliie: If i a 3 1,1,1 -ere, hen by Claim 2, i no adjacen o a 2 0,1 -ere. Hence, doe no end any charge o he bank. In hi cae, ω () = 3 3 2 = 2. If i a 3 0,1,1 -ere adjacen o o 2 0,1 -erice, hen A. Le be he neighbor of ha i no a 2-ere. If E(J), hen d J () = 3. By (R3), ill ge 1 2 from he bank. By (R1), end a oal of 4 o nearby 2-erice. Th ω () = 3 + 1 4 2 = 26. If / E(J), hen / A, o ha by (R2), receie 1 from. Therefore he ame calclaion applie o. If i a 3 0,1,1 -ere adjacen o a mo one 2 0,1 -ere, hen gie a mo 1 o he bank. Therefore ω () 3 3 2 1 = 26. If i a 3-ere adjacen o a mo one 2-ere, hen ω () 3 2 2 2 1 1 = 26. Cae d() = 4 If i a 4-ere, hen by Claim 3, i adjacen o a mo hree 2 0,1 -erice, o ω () 4 2 3 1 = 2. Cae d() 5 If i a 5-ere, hen ω () d() 4 d() 1 d() = 6 30 d(). Thi implie ha Mad(G) 26, hich proide he needed conradicion. Oberaion 5 If F i a fore, hen χ (G) 3. In each componen T of F, arbirarily chooe a roo r; color each ere of T ih color (1 + d T (r, )) (mod 3). Proof of Theorem 2.1 Le G be a graph ih Mad(G) < 26. By Lemma 4, here ei a ere pariion V (G) = F I. No e color: F ih color 1, 2, and 3 (a in Oberaion 5), and I ih color 4. Thi prodce a ar 4-coloring of G.
3 Graph ih Mad(G) < 18 and g(g) 6 In hi ecion, e proe ha eery graph G ih Mad(G) < 18 and g(g) 6 i ar 5-colorable. The proof in hi ecion and he ne ecion cloely reemble hoe in Secion 2. The main difference i ha in boh hi ecion and he ne, e pli he argmen ino o lemma, raher han one. In hi ecion, e fir proe ha each graph of inere m conain one of 14 configraion; e e hi lemma o proe a rcral decompoiion rel, hich in rn lead o he ar coloring rel. The oline of Secion 4 i imilar. Lemma 6 A graph G ih Mad(G) < 18 and g(g) 6 conain one of he folloing 14 configraion (ee Figre 2 and 3): 1. A 1-ere. 2. A 2 1,1 -ere. 3. A 3 1,1,1 -ere. 4. A 3 0,0,2 -ere. 5. A 3 0,1,1 -ere adjacen o a 3 0,1,1 -ere. 6. A 3 0,0,1 -ere adjacen o o 3 0,1,1 -ere.. A 4 1,1,1,2 -ere. 8. A 4 0,2,2,2 -ere adjacen o a 3-ere. 9. A 4 0,1,1,1 -ere adjacen o a 3 0,1,1 -ere. 10. A 4 0,1,1,1 -ere adjacen o a 4 0,2,2,2 -ere.. A 4 0,0,2,2 -ere adjacen o o 4 0,2,2,2 -ere. 12. A 4 0,0,2,2 -ere adjacen o a 4 0,2,2,2 -ere and a 3 0,1,1 -ere. 13. A 5 1,2,2,2,2 -ere. 14. A 5 0,2,2,2,2 -ere adjacen o a 4 0,2,2,2 -ere. Proof. Le G be a conereample, i.e. a graph ih Mad(G) < 18, ih g(g) 6, and conaining none of he configraion of Lemma 6. We e a dicharging argmen ih iniial charge ω() = d() a each ere ere, and ih he folloing for dicharging rle (R1) (R4), hich decribe ho o rediribe he charge. We rie ω () o denoe he charge a each ere afer e apply he dicharging rle. Noe ha he dicharging rle do no change he m of he charge. To complee he proof, e ho ha ω () 18 for all V (G); hi lead o he folloing obio conradicion: 18 18 V (G) V (G) ω () V (G) = ω() = 2 E(G) Mad(G) < 18 V (G) V (G) V (G) V (G). Hence no conereample can ei. 8
y z (4) (1) y z p q r y p q r (5) r y z p () z y r q p o n (8) y z o n (6) (3) y (2) Figre 2: (1 of 2) The fir 8 of he 14 naoidable configraion in Lemma 6. 9
o p q r z 2 z 3 y 3 3 y 4 4 y 2 2 5 y 5 (9) y z 1 6 z 1 y 6 (10) 1 1 1 1 r1 1 1 r 1 1 2 3 2 3 y 2 y 3 1 2 3 2 3 y 2 y 3 2 z 4 5 4 5 y 4 y 5 2 z p 1 q 1 r 2 6 6 y 6 r 2 p 2 2 2 q 2 () (12) 4 4 3 y 3 3 y 4 4 5 y 5 3 r 3 r 4 4 3 1 1 2 2 y 2 2 y 2 2 1 2 r 2 2 1 r 1 3 3 (13) 1 1 (14) y 3 Figre 3: (2 of 2) The la 6 of he 14 naoidable configraion in Lemma 6. 10
The dicharging rle are defined a follo. Each 3-ere gie: (R1) 2 o each 2-ere ha i no adjacen o a 2-ere. (R2) 4 o each 2-ere ha i adjacen o a 2-ere ( i a 2 0,1-ere). (R3) 1 o each 3 0,1,1-ere. (R4) 2 o each 4 0,2,2,2-ere. To complee he proof, i ffice o erify ha ω () 18 for all V (G). Cae d() = 2 If i a 2 0,1 -ere, hen receie 4 from i neighbor of degree a lea 3. Hence, ω () = 2+ 4 = 18. Oherie, i adjacen o o 3-erice, hich each gie 2 o. Hence ω () = 2 + 2 2 = 18. Cae d() = 3 By Lemma 6.4, i no inciden o a 2-hread. If i a 3 0,1,1 -ere, hen gie 2 o each adjacen 2-ere and receie 1 from i hird neighbor (ince hi hird neighbor i neiher a 2-ere, nor a 3 0,1,1 -ere, nor a 4 0,2,2,2 -ere). Hence, ω () = 3 2 2 + 1 = 18. If i a 3 0,0,1 -ere, hen i adjacen o a mo one 3 0,1,1 -ere and o no 4 0,2,2,2 -ere. Hence, ω () 3 2 1 = 18. Finally, ame ha i no adjacen o a 2-ere. Since i no adjacen o a 4 0,2,2,2 -ere, ω () 3 3 1 = 18. Cae d() = 4 By Lemma 6.8, i inciden o a mo hree 2-hread. If i inciden o eacly hree 2-hread, hen he forh neighbor of i neiher a 2-ere, nor a 3 0,1,1 -ere, nor a 4 0,2,2,2 -ere; hence, hi forh neighbor gie 2 o. Th ω () = 4 3 4 + 2 = 18. Ame ha i inciden o eacly o 2-hread. If he hird (or forh) neighbor of i a 2-ere or a 4 0,2,2,2 -ere, hen he final neighbor of doe no receie charge from (by Lemma 6. and 6.9 6.12). In hi cae, ω () = 4 2 4 2 = 18. I i alo poible ha one or boh of he remaining neighbor of are 3 0,1,1 -erice. In hi cae, ω () 4 2 4 2 1 = 18. If i inciden o a mo one 2-hread, hen ω () 4 4 3 2 = 18. Cae d() = 5 By Lemma 6.13, i inciden o a mo for 2-hread. If i inciden o eacly for 2-hread, hen i fifh neighbor i neiher a 2- ere, nor a 4 0,2,2,2 -ere. Hence, ω () = 5 4 4 1 = 18. If i inciden o a mo hree 2-hread, hen ω () 5 3 4 2 2 = 19. Cae d() 6 Triially, ω () d() d() 4 = 3 d() 18. Thi implie ha Mad(G) 18, hich proide he needed conradicion. Lemma If G i a graph ih Mad(G) < 18 and g(g) 6, hen here ei a ere pariion V (G) = F I 1 I 2 ch ha:
P1. F indce a fore, P2. I 1 i an independen e ch ha for all, y I 1, d G[F I1](, y) > 2, and P3. I 2 i an independen e ch ha for all, y I 2, d G (, y) > 2. Thi decompoiion a fir inrodced by Timmon in [5]. Proof. Le G be a conereample ih he fee erice. By Lemma 6, G conain one of he 14 configraion hon in Figre 2 and 3. In each configraion, le N denoe he e of black erice. In each cae, e delee a bgraph H and pariion V (G H) ino e F, I 1, and I 2 ha aify P 1, P 2, and P 3. We no ho ho o add he deleed erice o F, I 1, and I 2 o ha he reling e ill aify P 1, P 2, and P 3. We conider each of he 14 configraion in Figre 2 and 3. (1) Le G = G. By he minimaliy of G, he ere e of G i he dijoin nion of e F, I 1, and I 2 ha aify P1, P2, and P3. No add o F. (2) Le G = G\N. If, y F, hen add and o F and add o I 1. Oherie, add,, o F. (3) Le G = G\N. If, y, F, hen add,, z o F and add o I 1. If eacly o of he erice, y, are in F, hen.l.o.g., y F and I 1 (rep. I 2 ); add,, z o F and add o I 2 (rep. I 1). If a mo one of he erice, y, i in F, hen add,,, z o F. (4) Le G = G \ {, }. Conider he cae, F. If y, z I 1 I 2 hen add and o F. Oherie: if y, z I 2 F, hen add o F and o I 1 ; imilarly, if y, z I 1 F, hen add o F and add o I 2. Finally, if / F or / F, hen add and o F. (5) Le G = G \ N. If, p, y, F, hen add, q,, o F, add o I 1, and add r o I 2. If eacly hree of, p, y, are in F, hen W.l.o.g.,, p, y F and I 1; add o I 1 and, p, r,, o F. The remaining cae are eaier and are lef o he reader. (6)-(9) Thee eay cae are lef o he reader. (10) Le G = G \ N. We conider o cae: (i) If a lea o of y 4, y 5, y 6 are in independen e, hen add o an independen e and add he oher erice o he fore; (ii) If a mo one of y 4, y 5, y 6 i in an independen e, hen add o he oher independen e and add all he oher erice o he fore. () Le G = G \ N. Add and o I 1, add z o I 2, and add all he oher erice o he fore. (12) Le G = G \ N. We j conider he cae here q 1 and q 2 are in F ; he remaining cae are eaier. Add and o I 1, add z o I 2, and add he oher erice o he fore. 12
(13) Le G = G \ N. Add o an independen e ha doen conain y 5 and add all he oher erice o he fore. (14) Le G = G \ N. Add o I 1, add o I 2, and add all oher oher erice o he fore. Proof of Theorem 2.2 Le G be a graph ih Mad(G) < 18 and g(g) 6. By Lemma, here ei a ere pariion V (G) = F I 1 I 2. No e color: G[F] ih he color 1, 2, and 3 (a in Oberaion 5), G[I 1 ] ih he color 4, and G[I 2 ] ih he color 5. Thi prodce a ar 5-coloring of G. Oberaion 8 The condiion g(g) 6 can be dropped. Since here are no 3-hread, e j e hi ampion o enre ha hie and black erice are diinc in he configraion in Figre 2 and 3. 4 Graph ih Mad(G) < 8 3 and g(g) 6 In hi ecion, e proe ha eery graph G ih Mad(G) < 8 3 6-colorable. and g(g) 6 i ar Lemma 9 A graph G ih Mad(G) < 8 3 and g(g) 6 conain one of he folloing 8 configraion: 1. A 1-ere. 2. A 2 0,1 -ere adjacen o a 4-ere. 3. A 3 0,1,1 -ere adjacen o a 3-ere. 4. A k 1,1,2,,2 -ere. 5. A k 0,2,,2 -ere adjacen o a 3-ere. 6. A k 0,1,2,,2 -ere adjacen o k 0,1,1,2,,2 -ere.. A k 0,0,2,,2 -ere adjacen o o 3 0,1,1 -erice. 8. A k 0,0,2,,2 -ere adjacen o a k 0,1,1,2,,2-ere and o a k 0,1,2,2,,2 -ere. 13
y 2 z r y 3 (1) (2) y 2 (3) z n 2 y n 2 n 2 1 1 z n 2 y n 2 n 2 1 z 2 y 2 2 z 1 1 (4) 2 2 z 2 y 2 2 z 1 1 (5) 2 2 z n 2 z 2 z 1 y n 2 y 2 n 2 2 1 3 3 1 1 2 2 r m 3 m 3 m 3 r 2 2 2 r 1 1 1 z n 2 z 2 z 1 y n 2 y 2 n 2 2 r 2 1 2 1 1 1 (6) 3 () 2 3 z n 2 y n 2 n 2 z 2 2 y 2 z 1 1 2 2 a l 2 b l 2 c l 2 a 2 b 2 c 2 a 1 b 1 c 1 r m 3 m 3 m 3 r 2 2 2 1 r 1 1 1 1 (8) Figre 4: The 8 naoidable configraion in Lemma 9. 14
Proof. Or proof i imilar o he proof of Lemma 4 and 6. Le G be a conereample. We e a dicharging argmen ih iniial charge ω() = d() a each ere ere, and ih he folloing hree dicharging rle (R1) (R3), hich decribe ho o rediribe he charge. We rie ω () o denoe he charge a each ere afer e apply he dicharging rle. Noe ha he dicharging rle do no change he m of he charge. To complee he proof, e ho ha ω () 8 3 for each V (G); hi lead o he folloing obio conradicion: 8 8 3 3 V (G) V (G) Hence no conereample can ei. V (G) ω () V (G) = ω() = 2 E(G) Mad(G) < 8 V (G) V (G) V (G) 3. The dicharging rle are defined a follo. (R1) Each 5-ere gie 2 3 o each adjacen 2-ere hich i adjacen o a 2- ere ( i a 2 0,1 -ere). (R2) Each 3-ere gie 1 3 o each adjacen 2-ere hich i no adjacen o a 2- ere. (R3) Each 4-ere gie 1 3 o each adjacen 3 0,1,1-ere and 1 3 o each adjacen 5 0,2,2,2,2 -ere. We call 3 0,1,1 -erice and 5 0,2,2,2,2 -erice ligh erice. If a ere i he iniial or final ere of a hread, e call i an end of he hread. By Lemma 9.1., δ(g) 2. Cae d() = 2 If i no adjacen o any 2-erice, hen i receie 1 3 from each neighbor by (R2); o, ω () = 2 + 2 1 3 = 8 3. If i a 2 0,1-ere, hen i i adjacen o a 5-ere. By (R3), gie 2 3 o, o ω () = 2 + 2 3 = 8 3. Cae d() = 3 By Lemma 9.2 and he girh condiion, i no he end of a 2-hread. Moreoer, by Lemma 9.3, i he end of a mo o 1-hread. If i he end of eacly o 1-hread, hen i i adjacen o a 4-ere, by Lemma 9.3. So, ω () = 3 2 1 3 + 1 3 = 8 3 by (R2) and (R3). If i he end of a mo one 1-hread, hen ω () 3 1 3 = 8 3. Cae d() = 4 By (R2) and (R3), ω () 4 4 1 3 = 8 3. Cae d() = 5 By Lemma 9.4, i he end of a mo for 2-hread. If i he end of eacly for 2-hread ( i ligh), hen i i neiher adjacen o a 3-ere, by Lemma 9.4 and 9.5, nor o a 5 0,2,2,2,2 -ere, by Lemma 9.6. Hence, receie 1 3 by (R3). So, ω () = 5 4 2 3 + 1 3 = 8 3. If i he end of eacly hree 2-hread, hen i adjacen o a mo one ligh ere, by Lemma 9. and 9.8. So, ω () 5 3 2 3 1 3 8 3. Finally, if i he end of a mo o 2-hread, hen ω () 5 2 2 3 3 1 3 8 3. 15
Cae d() 6 By Lemma 9.4, i he end of a mo d() 1 2-hread. If i he end of eacly d() 1 2-hread, hen i i neiher adjacen o a 3- ere, by Lemma 9.4 and 9.5, nor o a 5 0,2,2,2,2 -ere, by Lemma 9.6; hence, ω () = d() (d() 1) 2 3 8 3. If i he end of eacly d() 2 2-hread, hen i adjacen o a mo one ligh ere, by Lemma 9. and 9.8. So, ω () d() (d() 2) 2 3 1 3 3. Finally, if i he end of a mo d() 3 2-hread, hen ω () d() (d() 3) 2 3 3 1 3 3. Thi implie ha Mad(G) 8 3, hich proide he needed conradicion. Lemma 10 Le G be a graph ih Mad(G) < 8 3 ere pariion V (G) = F I 1 I 2 I 3 ch ha: and g(g) 6. Then here ei a P1. F indce a fore. P2. I i i an independen e ch ha for all, y I i, d G (, y) > 2 for i = 1, 2, 3. Proof. Le G be a conereample ih he fee erice. By Lemma 9, G conain one of he 8 configraion in Figre 4. In each configraion, le N denoe he e of black erice. In each cae, e delee a bgraph H and pariion V (G H) ino e F, I 1, I 2, and I 3 ha aify P 1 and P 2. We no ho ho o add he deleed erice o F, I 1, I 2, and I 3 o ha he reling e ill aify P 1 and P 2. We conider each of he 8 configraion in Figre 4: (1) Le G = G. By he minimaliy of G, here ei a ere pariion V (G ) = F I 1 I 2 I 3 aifying P1 and P2. Add o F. (2) Le G = G \ {, } and V (G ) = F I 1 I 2 I 3. If a lea one of he erice and belong o an independen e, hen add and o F. So ame ha, F. If all hree y i are in diinc independen e, hen add and o F. Oherie, add o he fore and add o an independen e ha doe no conain any y i. (3) Le G = G \ {,, z}. If hree or more of he erice, r,, y 2 belong o independen e, hen add,, z o he fore. So ppoe ha a mo o of hee erice belong o independen e. Add o an independen e ha doe no conain any of, r,, y 2 ; add all he oher erice o he fore. I may happen ha and are in he ame independen e; in hi cae, e can eiher add,, z o he fore or e can moe o he fore. (4) Le G = G \N. Add o an independen e ha conain neiher 1 nor 2 and add all he oher erice o he fore. (5) Le G = G \ {black erice ecep }. Add o an independen e ha conain neiher 1 nor 2 and add all he oher erice o he fore (moe o he fore if neceary). 16
(6) Le G = G\N. Add o an independen e ha conain neiher 2 nor 3, add o an independen e ha conain neiher nor 1, and add he all he oher erice o he fore. () Le G = G\N. Add r o an independen e ha conain neiher 1 nor 2, and add o an independen e ha conain neiher 1, nor 2, nor r; if here i no choice for, hen add o he fore. Add o an independen e ha conain neiher r nor, and add all he oher erice o he fore. (8) Le G = G \ N. Add o an independen e ha conain neiher 1 nor 2, add o an independen e ha conain neiher nor 3, and add o an independen e ha conain neiher nor. Finally, add all he oher erice o he fore. Proof of Theorem 2.3 Le G be a graph ih Mad(G) < 8 3 and g(g) 6. By Lemma 10, here ei a ere pariion V (G) = F I 1 I 2 I 3. No e color: F ih he color 1, 2, and 3 (a in Oberaion 5), I 1 ih he color 4, I 2 ih he color 5, and I 3 ih he color 6. Thi prodce a ar 6-coloring of G. Oberaion The condiion g(g) 6 can be dropped. Since here are no 3- hread, e j e hi ampion o enre ha hie and black erice are diinc in he configraion in Figre 4. 5 Conclion In hi paper, e proed ha if G i a graph ih Mad(G) < 26 18 (rep. Mad(G) < and g(g) 6, Mad(G) < 8 3 and g(g) 6), hen χ (G) 4 (rep. 5, 6). Are hee rel opimal? Le define a fncion f a follo. Definiion 12 Le f : N R be defined by Obere ha: f(n) = inf{mad(h) χ (H) > n} f(1) = 1. Acally, if G conain an edge, hen χ (G) > 1; oherie, G i a independen e, o χ (G) = 1 and Mad(G) = 0. Noe alo ha: 1
f(2) = 3 2. If G conain a pah of lengh 3, hen χ (G) > 2 and Mad(G) 3 2. Oherie, G i a ar fore, o χ (G) 2 and Mad(G) < 2. Finally, noe ha: f(3) = 2. If Mad(G) < 2, hen G i a fore and χ (G) = 3. In conra, if G i a 5-cycle, hen Mad(G) = 2 and χ (G) = 4. Problem 13 Wha i he ale of f(n) for n = 4, 5, 6? y z Figre 5: A graph G ih Mad(G) = 18 and χ (G) > 4. From Theorem 2.1, e kno ha: 26 f(4). The graph G, in Figre 5, ha Mad(G) = 18 and χ (G) > 4, o G gie an pper bond on f(4). Hence: 26 = 182 198 f(4) = 18 Finally, obere ha here ei a ar 4-colorable graph G ih Mad(G) = 5 2 ha doe no admi a pariion V = F I ch ha for all, y I, d G (, y) > 2. See Figre 6. 6 Acknoledgemen Thank o Craig Timmon and André Kündgen for helpfl dicion. 18
Figre 6: A ar 4-colorable graph G ih Mad(G) = 5 2 ha ha no pariion F I ch ha for all, y I, d G (, y) > 2. Reference [1] M.O. Alberon, G.G. Chappell, H.A. Kieread, A. Kündgen, and R. Ramamrhi. Coloring ih no 2-colored P 4. The Elecronic Jornal of Combinaoric, (1):R26, 2004. [2] G. Ferin, A. Rapad, and B. Reed. On ar coloring of graph. LNCS, 2204:140 153, 2001. [3] B. Grünbam. Acyclic coloring of planar graph. Irael J. Mah., 14:390 408, 193. [4] J. Nešeřil and P. Oana de Mendez. Coloring and homomorphim of minor cloed clae. Dicree and Compaional Geomery: The Goodman-Pollack Fechrif, 651 664, 2003. [5] C. Timmon. Sar coloring planar graph. Maer hei, California Sae Unieriy San Marco, May 200. 19