3-FACIAL COLOURING OF PLANE GRAPHS

Transcription

1 LABORATOIRE INFORMATIQUE, SIGNAUX ET SYSTÈMES DE SOPHIA ANTIPOLIS UMR FACIAL COLOURING OF PLANE GRAPHS Frédéric Hae, Jean-Sébasien Sereni, Rise Skrekoski Proje MASCOTTE Rappor de recherche ISRN I3S/RR FR Juille 2006 LABORATOIRE I3S: Les Algorihmes / Euclide B 2000 roue des Lucioles B.P Sophia-Anipolis Cedex, France Tél. (33) Télécopie : (33) hp://.i3s.unice.fr/i3s/fr/

2 RÉSUMÉ : Un graphe planaire es l-facialemen k-colorable s il exise une coloraion de ses sommes aec k couleurs elle que les sommes reliés par un chemin facial de longueur au plus l soien colorés différemmen. Nous démonrons que ou graphe planaire es 3-facialemen 11-colorable. Par conséquen, ou graphe planaire 2-connexe don oues les faces on aille au plus 7 es cycliquemen 11-colorable. Ces deux bornes son à une couleur près celles proposées par la conjecure 3l+1 e la conjecure cyclique. MOTS CLÉS : coloraion faciale, coloraion cyclique, graphes planaires ABSTRACT: A plane graph is l-facially k-colourable if is erices can be coloured ih k colours such ha any o disinc erices on a facial segmen of lengh a mos l are coloured differenly. We proe ha eery plane graph is 3-facially 11-colourable. As a consequence, e derie ha eery 2-conneced plane graph ih maximum face-size a mos 7 is cyclically 11-colourable. These o bounds are for one off from hose ha are proposed by he (3l+1)-Conjecure and he Cyclic Conjecure. KEY WORDS : facial colouring, cyclic colouring, planar graphs

3 3-facial colouring of plane graphs Frédéric Hae Jean-Sébasien Sereni Rise Škrekoski Absrac A plane graph is l-facially k-colourable if is erices can be coloured ih k colours such ha any o disinc erices on a facial segmen of lengh a mos l are coloured differenly. We proe ha eery plane graph is 3-facially 11-colourable. As a consequence, e derie ha eery 2-conneced plane graph ih maximum face-size a mos 7 is cyclically 11-colourable. These o bounds are for one off from hose ha are proposed by he (3l+1)- Conjecure and he Cyclic Conjecure. 1 Inroducion The concep of facial colourings, inroduced in [11], exends he ell-knon concep of cyclic colourings. A facial segmen of a plane graph G is a sequence of erices in he order obained hen raersing a par of he boundary of a face. The lengh of a facial segmen is is number of edges. To erices u and of G are l-facially adjacen, if here exiss a facial segmen of lengh a mos l beeen hem. An l-facial colouring of G is a funcion hich assigns a colour o each erex of G such ha any o disinc l-facially adjacen erices are assigned disinc colours. A graph admiing an l-facial colouring ih k colours is called l-facially k-colourable. The folloing conjecure, called (3l + 1)-Conjecure, is proposed in [11]: Conjecure 1 (Král, Madaras and Škrekoski). Eery plane graph is l-facially colourable ih 3l + 1 colours. Obsere ha he bound offered by Conjecure 1 is igh: as shon by Figure 1, for eery l 1, here exiss a plane graph hich is no l-facially 3l-colourable. Conjecure 1 can be considered as a counerpar for l-facial colouring of he folloing famous conjecure by Ore and Plummer [12] concerning he cyclic colouring. A plane graph G is said o be cyclically k-colourable, if i admis a erex colouring ih k colours such ha any pair of erices inciden o a same face are assigned disinc colours. MASCOTTE, I3S (CNRS UNSA) INRIA, 2004 Roue des Lucioles, BP 93, Sophia Anipolis Cedex, France. s: {fhae,sereni}@sophia.inria.fr Deparmen of Mahemaics, Uniersiy of Ljubljana, Jedranska 19, 1111 Ljubljana, Sloenia. Suppored in par by Minisry of Science and Technology of Sloenia, Research Program P bluesky2high@yahoo.com 1

4 l l l Figure 1: The plane graph G l = (V,E): each hread represens a pah of lengh l. The graph G l is no l-facially 3l-colourable: eery o erices are l-facially adjacen, herefore any l-facial colouring mus use V = 3l + 1 colours. Conjecure 2 (Ore and Plummer). Eery plane graph is cyclically denoes he size of a bigges face of G colourable, here Noe ha Conjecure 1 implies Conjecure 2 for odd alues of. The bes knon resul oards Conjecure 2 has been obained by Sanders and Zhao [15], ho proed he bound. Denoe by f c (x) he minimum number of colours needed o cyclically colour eery plane graph of maximum face size x. The alue of f c (x) is knon for x {3,4}: f c (3) = 4 (he problem of finding f c (3) being equialen o he Four Colour Theorem proed in [1]) and f c (4) = 6 (see [3, 5]). I is also knon ha f c (5) {7,8} and f c (6) 10 [6], and ha f c (7) 12 [4]. Conjecure 1 is riially rue for l = 0, and is equialen o he Four Colour Theorem for l = 1. I is open for all oher alues of l. As noed in [11], if Conjecure 1 ere rue for l = 2, i ould hae seeral ineresing corollaries. Besides giing he exac alue of f c (5) (hich ould hen be 7), i ould allo o decrease from 16 o 14 (by applying a mehod from [11]) he upper bound on he number of colours needed o 1-diagonally colour eery plane quadrangulaion (for more deails on his problem, consul [9, 13, 14, 11]). I ould also imply Wegner s conjecure on 2-disance colourings (i.e. colourings of squares of graphs) resriced o plane cubic graphs since colourings of he square of a plane cubic graph are precisely is 2-facial colourings (refer o [10, Problem 2.18] for more deails on Wegner s conjecure). Le f f (l) be he minimum number of colours needed o l-facially colour eery plane graph. Clearly, f c (2l + 1) f f (l). So far, no alue of l is knon for hich his inequaliy is sric. The folloing problem is offered in [11]. Problem 1. Is i rue ha, for eery ineger l 1, f c (2l + 1) = f l (l)? Anoher conjecure ha should be maybe menioned is he so-called 3l-Conjecure proposed in [7], saing ha eery plane riangle-free graph is l-facially 3l-colourable. Similarly as he (3l+1)-Conjecure, if his conjecure ere rue, hen is bound ould be igh and i ould hae seeral ineresing corollaries (see [7] for more deails). I is proed in [11] ha eery plane graph has an l-facial colouring using a mos 18 5 l + 2 colours (and his bound is decreased by 1 for l {2,4}). So, in paricular, eery plane graph has a 3-facial 12-colouring. In his paper, e improe his las resul by proing he folloing heorem

5 Theorem 1. Eery plane graph is 3-facially 11-colourable. To proe his resul, e shall suppose ha i is false. In Secion 2, e ill exhibi some properies of a minimal graph (regarding he number of erices) hich conradics Theorem 1. Relying on hese properies, e ill use he Discharging Mehod in Secion 3 o obain a conradicion. 2 Properies of (3,11)-minimal graphs Le us sar his secion by inroducing some definiions. A erex of degree d (respeciely a leas d, respeciely a mos d) is said o be a d-erex (respeciely a ( d)-erex, respeciely a ( d)-erex). The noion of a d-face (respeciely a ( d)-face, respeciely a ( d)-face) is defined analogously regarding he size of a face. An l-pah is a pah of lengh l. To faces are adjacen, or neighbouring, if hey share a common edge. A 5-face is bad if i is inciden o a leas four 3-erices. I is said o be ery-bad if i is inciden o fie 3-erices. If u and are 3-facially adjacen, hen u is called a 3-facial neighbour of. The se of all 3-facial neighbours of is denoed by N 3 (). The 3-facial degree of, denoed by deg 3 (), is he cardinaliy of he se N 3 (). A erex is dangerous if i has degree 3 and i is inciden o a face of size hree or four. A 3-erex is safe if i is no dangerous, i.e. i is no inciden o a ( 4)-face. Le G = (V,E) be a plane graph, and U V. Denoe by G 3 [U] he graph ih erex se U such ha xy is an edge in G 3 [U] if and only if x and y are 3-facially adjacen erices in G. If c is a parial colouring of G and u an uncoloured erex of G, e denoe by L c (u) (or jus L(u)) he se {x {1,2,...,11} : for all N 3 (u),c() x}. The graph G 3 [U] is L-colourable if here exiss a proper erex colouring of he erices of G 3 [U] such ha for eery u U holds c(u) L(u). The nex o resuls are used by Král, Madaras and Škrekoski [11]: Lemma 1. Le be a erex hose inciden faces in a plane graph G are f 1, f 2,..., f d. Then ( d ) deg 3 () min( f i,7) 2d, i=1 here f i denoes he size of he face f i. Suppose ha Theorem 1 is false: a (3,11)-minimal graph G is a plane graph hich is no 3-facially 11-colourable, ih V (G) + E(G) as small as possible. Lemma 2. Le G be a (3,11)-minimal graph. Then, (i) G is 2-conneced; (ii) G has no separaing cycle of lengh a mos 7; (iii) G conains no adjacen f 1 -face and f 2 -face ih f 1 + f 2 9; 3

6 (i) G has no erex hose 3-facial degree is less han 11. In paricular, he minimum degree of G is a leas hree; and () G conains no edge u separaing o ( 4)-faces ih deg 3 (u) 11 and deg 3 () 12. In he remaining of his secion, e gie addiional local srucural properies of (3, 11)- minimal graphs. Lemma 3. Le G be a (3,11)-minimal graph. Suppose ha and are o adjacen 3-erices of G, boh inciden o a same 5-face and a same 6-face. Then he size of he hird face inciden o is a leas 7. Proof. By conradicion, suppose ha he size of he las face inciden o is a mos 6. Then, according o Lemma 1, e infer ha deg 3 () 12 and deg 3 () 11, bu his conradics Lemma 2(). A reducible configuraion is a (plane) graph ha canno be an induced subgraph of a (3,11)- minimal graph. The usual mehod o proe ha a configuraion is reducible is he folloing: firs, e suppose ha a (3,11)-minimal graph G conains a prescribed induced subgraph H. Then e conrac some subgraphs H 1,H 2,...,H k of H. Mosly, e hae k 2. This yields a proper minor G of G, hich by he minimaliy of G admis a 3-facial 11-colouring c. The goal is o derie from c a 3-facial 11-colouring c of G, hich ould gie a conradicion. To do so, each non-conraced erex of G keeps is colour c (). Le h i be he erex of G creaed by he conracion of he erices of H i : some erices of H i are assigned he colour c (h i ) (in doing so, e mus ake care ha hese erices are no 3-facially adjacen in G). Las, e sho ha he remaining uncoloured erices can also be coloured. In oher ords, e sho ha he graph G 3 [U] is L-colourable, here for each u U, L(u) is he lis of he colours hich are assigned o no erex in N 3 (u) \ U (defined in Secion 1) and U is he se of uncoloured erices. In mos of he cases, he erices of U ill be greedily coloured. In all figures of he paper, he folloing conenions are used: a riangle represens a 3- erex, a square represens a 4-erex and a circle may be any kind of erex hose degree is a leas he maximum beeen hree and he one i has in he figure. The edges of each subgraph H i are dran in bold, and he circled erices are he erices of U = {,,...}. A dashed edge beeen o erices indicaes a pah of lengh a leas one beeen hose o erices. An (in)equaliy rien in a bounded region denoes a face hose size achiees he (in)equaliy. Las, erices hich are assigned he colour c (h i ) are denoed by,, if a unique subgraph is conraced or by x 1,x 2 for i = 1 and y 1,y 2 for i = 2 if o subgraphs are conraced. Lemma 4. Configuraions in Figures 2, 3 and 4 are reducible. Proof. Le H be an induced subgraph of G. We shall suppose ha H is isomorphic o one of he configuraions saed and derie a ay o consruc a 3-facial 11-colouring of G, a conradicion. 4

7 u 4 u 5 6 u 4 u 5 u 4 6 (L1) (L2) (L3) 4 u 4 (L4) (L5) x 2 y 2 4 u 4 u 5 u 4 x 1 y 1 (L6) (L7) u 5 u 4 u 6 (L8) (L9) Figure 2: Reducible configuraions (L1) (L9). 5

8 L1. Suppose ha H is isomorphic o he configuraion (L1) of Figure 2. Denoe by H 1 he subgraph induced by he bold edges. Conrac he erices of H 1, hereby creaing a ne erex h 1. By minimaliy of G, le c be a 3-facial 11-colouring of he obained graph. Assign o each erex x no in H 1 he colour c (x), and o each of,, he colour c (h 1 ). Obsere ha no o erices among,, are 3-facially adjacen in G, oherise here ould be a ( 7)- separaing cycle in G, hereby conradicing Lemma 2(ii). According o Lemma 1, deg 3 ( ) 15, deg 3 (u i ) 14 if i {2,3} and deg 3 (u i ) 11 if i {4,5}. Noe ha any o erices of U = {,,...,u 5 } are 3-facially adjacen, ha is G 3 [U] K 5. Hence, he number of coloured 3-facial neighbours of is a mos 11, i.e. N 3 ( )\{,,u 4,u 5 } 11. Moreoer, a leas o of hem are assigned he same colour, namely and. Therefore, L( ) 1. For i {2,3}, he erex u i has a mos 10 coloured 3-facial neighbours. Furhermore, a leas o 3-facial neighbours of are idenically coloured, namely and. Thus, L( ) 2. No, obsere ha a leas hree 3-facial neighbours of are coloured he same, namely, and. Hence, L( ) 3. For i {4,5}, he erex u i has a mos 7 coloured 3-facial neighbours. Thus, L(u 4 ) 4, and because a leas o 3-facial neighbours of u 5 are idenically coloured ( and ), L(u 5 ) 5. So, he graph G 3 [U] is greedily L-colourable, according o he ordering,,,u 4,u 5. This allos us o exend c o a 3-facial 11-colouring of G. L2. Suppose ha H is isomorphic o he configuraion (L2) of Figure 2. Le c be a 3-facial 11-colouring of he minor of G obained by conracing he bold edges ino a single erex h 1. Le c(x) = c (x) for eery erex x h 1. Define c() = c() = c() = c (h 1 ). The obained colouring is sill 3-facial since no o erices among,, are 3-facially adjacen in G by Lemma 2(ii). Noe ha G 3 [U] K 5. In paricular, each erex u i has four uncoloured 3-facial neighbours. By Lemma 1, deg 3 ( ) 15,deg 3 (u i ) 14 if i {2,3} and deg 3 (u i ) 11 if i {4,5}. Moreoer, each of and has a leas o 3-facial neighbours coloured he same; for, hese erices are, and for hey are,. So, here exiss a leas one colour hich is assigned o no erex of N 3 ( ) and a leas o colours assigned o no erex of N 3 ( ). Also, has a leas hree 3-facial neighbours coloured he same, namely, and, hence a leas hree colours are assigned o no erex of N 3 ( ). Therefore, L( ) 1, L( ) 2 and L( ) 3. Furhermore, L(u 4 ) 4 and L(u 5 ) 5 because and are boh 3-facial neighbours of u 5. So G 3 [U] is L-colourable, and hence G is 3-facially 11-colourable. L3. Suppose ha H is isomorphic o he configuraion (L3) of Figure 2. Conrac he bold edges ino a ne erex h 1, and le c be a 3-facial 11-colouring of he obained graph. This colouring can be exended o a 3-facial 11-colouring c of G as follos: firs, le c() = c() = c() = c (h 1 ). Noe ha no o of hese erices can be 3-facially adjacen in G ihou conradicing Lemma 2(ii). By Lemma 1, deg 3 ( ) 14, deg 3 ( ) 13 and for i {3,4}, deg 3 (u i ) 12. Obsere ha G 3 [U] K 4. Moreoer, each of,, has a se of o 3-facial neighbours coloured by c (h 1 ). These ses are {,}, {,} and {,} for, and, respeciely. Thus, L( ) 1, L( ) 2 and L( ) 3. Also L(u 4 ) 4 because u 4 has a leas hree idenically coloured 3-facial neighbours, namely, and. Hence, G 3 [U] is L-colourable, so G is 3-facially 11-colourable. 6

9 L4. Le c be a 3-facial 11-colouring of he graph obained by conracing he bold edges ino a ne erex h 1. Define c(x) = c (x) if x / {,,, } and c() = c() = c (h 1 ). Obsere ha and canno be 3-facially adjacen in G since G has no small separaing cycle according o Lemma 2(ii). By Lemma 1, deg 3 ( ) 12 and deg 3 ( ) 11. Furhermore, boh and hae o 3-facial neighbours idenically coloured, namely and. Moreoer, and are 3-facially adjacen, hence L( ) 1 and L( ) 2. Therefore, c can be exended o a 3-facial 11-colouring of G. L5. Firs, obsere ha since G is a plane graph, if N 3 () hen / N 3 ( ). So, by symmery, e may assume ha and are no 3-facially adjacen in G. No, conrac he bold edges ino a ne erex h 1. Again, denoe by c a 3-facial 11-colouring of he obained graph, and define c o be equal o c on all erices of V (G) \ {,,,,,,u 4 }. Le c() = c() = c() = c (h 1 ). Noe ha he parial colouring c is sill 3-facial due o he aboe assumpion. The graph G 3 [U] is isomorphic o K 4, and according o Lemma 1, deg 3 (u i ) 12 for all i {1,2,3,4}. Moreoer, for i {2,3}, he erex u i has a leas o 3-facial neigbhours ha are coloured he same, namely and. Las, he erex u 4 has a leas hree such 3-facial neighbours, namely,,. Therefore, L( ) 2, L(u i ) 3 for i {2,3} and L(u 4 ) 4. So, G 3 [U] is L-colourable, and hence G is 3-facially 11-colourable. L6. The same remark as in he preious configuraion allos us o assume ha / N 3 (). Again, he graph obained by conracing he bold edges ino a ne erex h 1 admis a 3-facial 11-colouring c. As before, define a 3-facial 11-colouring c of he graph induced by V (G) \ U. Then, for eery i {1,2,3,4},deg 3 (u i ) 12 and G 3 [U] K 4. Thus, L( ) 2 and L( ) 2. Remark ha has a leas o idenically coloured 3-facial neighbours, namely and, so L( ) 3. Las, he erex u 4 has a leas hree such neighbours, hence L(u 4 ) 4. Therefore, he graph G 3 [U] is L-colourable, and so he graph G admis a 3-facial 11-colouring. L7. Le H 1 be he pah x 1 u 5 x 2, H 2 he pah y 1 u 4 y 2 and c a 3-facial colouring of he graph obained from G by conracing each pah H i ino a erex h i. Noice ha c (h 1 ) c (h 2 ). For eery / V (H 1 ) V (H 2 ), le c() = c (). Obsere ha x 1 and x 2 canno be 3-facially adjacen in G, oherise G ould hae a separaing ( 7)-cycle, conradicing Lemma 2(iii). Noe ha he same holds for y 1 and y 2 ; herefore defining c(x 1 ) = c(x 2 ) = c (h 1 ) and c(y 1 ) = c(y 2 ) = c (h 2 ) yields a parial 3-facial 11-colouring of G, since c (h 1 ) c (h 2 ). I remains o colour he erices of U = {,,...,u 5 }. Noe ha G 3 [U] K 5. According o Lemma 2(ii), deg 3 ( ) 15 and deg 3 (u i ) 12 if i 2. The number of coloured 3-facial neighbours of, i.e. is number of 3- facial neighbours in V (G) \ {,,u 4,u 5 }, is a mos 11 because each u i ih i 2 is a 3-facial neighbour of. Furhermore, has o 3-facial neighbours coloured ih he same colour, namely x 1 and x 2. Hence, L( ) 1. The erex has four uncoloured 3-facial neighbours, so L( ) 3. For i {3,4}, he erex u i has a leas o 3-facial neighbours coloured he same, namely x 1,x 2 for, and y 1,y 2 for u 4, so L(u i ) 4. Finally, obsere ha u 5 has o pairs of idenically coloured 3-facial neighbours; he firs pair being x 1,x 2 and he second y 1,y 2. Thus, L(u 5 ) 5, hence he graph G 3 [U] is L-colourable, hich yields a conradicion. 7

10 4 u 4 u 4 u 5 (L10) (L11) 4 u 5 u 1 u 4 u 5 u 6 4 u 4 (L12) (L13) u 4 u 5 u 4 u 5 u 4 u 5 4 (L14) (L15) (L16) Figure 3: Reducible configuraions (L10) (L16). 8

11 L8. We conrac he bold edges ino a ne erex h 1, ake a 3-facial 11-colouring of he graph obained, and define a 3-facial 11-colouring c of V (G)\U as usual. By Lemma 1, deg 3 (u i ) 15 if i {1,2}, deg 3 (u i ) 12 if i {3,4,5} and deg 3 (u 6 ) 11. Moreoer, G 3 [U] K 6. As, and are coloured he same, and {,} N 3 (u i ) for i {2,5}, {,} N 3 (u 4 ) and {,} N 3 (u 5 ), e obain L(u i ) i for eery i {1,2,3,4,5,6}. Thus, he graph G 3 [U] is L-colourable, and hence G admis a 3-facial 11-colouring. L9. We conrac he bold edges ino a ne erex, ake a 3-facial 11-colouring of he graph obained, and define a 3-facial 11-colouring of V (G)\U as usual. Then, G 3 [U] K 2. Moreoer, deg 3 ( ) 12 and deg 3 ( ) 11. Furhermore, {,} N 3 (u i ) for i {1,2}. Thus, e infer L(u i ) i for i {1,2}. Therefore, G 3 [U] is L-colourable. L10. We conrac he bold edges ino a ne erex, ake a 3-facial 11-colouring of he graph obained, and define a 3-facial 11-colouring of V (G)\U as usual. Then, G 3 [U] K 4. Moreoer, deg 3 ( ) 13, deg 3 ( ) 12 and deg 3 (u i ) 11 for i {3,4}. Furhermore, {,} N 3 (u i ) for i {1,4}. Thus, e infer L(u i ) 2 for i {1,2}, and L(u i ) i for i {3,4}. Therefore, G 3 [U] is L-colourable. L11. We conrac he bold edges ino a ne erex h 1, ake a 3-facial 11-colouring of he graph obained, and define a 3-facial 11-colouring c of V (G)\U as usual. By Lemma 1, deg 3 ( ) 15 and deg 3 (u i ) 11 if i {2,3,4,5}. Moreoer, G 3 [U] K 5. As and are coloured he same, and {,} N 3 (u i ) for i {1,4,5}, e obain L( ) 1, L(u i ) 4 if i {2,3} and L(u i ) 5 if i {4,5}. Thus, he graph G 3 [U] is L-colourable, and hence G admis a 3-facial 11-colouring. L12. Le c be a 3-facial 11-colouring of he graph G obained by conracing he bold edges ino a ne erex h 1. Define c(x) = c (x) for eery erex x V (G) V (G ), and le c() = c() = c (h 1 ). By Lemma 1, deg 3 (u i ) 15 for i {1,2} and deg 3 (u i ) 11 for i {3,4,5}. Moreoer, G 3 [U] K 6. Hence, L( ) 1 and L(u i ) i for i {3,4,5}. As and are coloured he same, and {,} N 3 (u i ) for i {2,6}, e infer ha L( ) 2 and L(u 6 ) 6. Thus, he graph G is 3-facially 11-colourable. L13. Le us define he parial 3-facial 11-colouring c as alays, regarding he bold edges and he erices and. From Lemma 1 e ge deg 3 ( ) 15, deg 3 (u i ) 12 for i {2,3,4} and deg 3 (u 5 ) 11. Moreoer, since G 3 [U] K 5 and {,} N 3 (u i ) for i {1,4,5}, e obain L( ) 1, L(u i ) 3 for i {2,3}, L(u 4 ) 4 and L(u 5 ) 5. Therefore, G 3 [U] is L- colourable. L14. Define he parial 3-facial 11-colouring c as usual, regarding he bold edges and he erices and. By Lemma 1, deg 3 ( ) 15 and deg 3 (u i ) 11 for i {2,3,4,5}. Moreoer, since G 3 [U] K 5 and {,} N 3 (u i ) for i {1,5}, e obain L( ) 1, L(u i ) 4 for i {2,3,4} and L(u 5 ) 5. Therefore, G 3 [U] is L-colourable. 9

12 L15. Le us define he parial 3-facial 11-colouring c as alays, regarding he bold edges and he erices and. Again, G 3 [U] K 5. From Lemma 1 e ge deg 3 ( ) 15 and deg 3 (u i ) 11 if i {2,3,4,5}. Moreoer, since {,} N 3 (u i ) for i {1,5}, e obain L( ) 1, L(u i ) 4 for i {2,3,4} and L(u 5 ) 5. Therefore, G 3 [U] is L-colourable. L16. Define he parial 3-facial 11-colouring c as alays, regarding he bold edges and he erices, and. Then, G 3 [U] K 5 and deg 3 (u i ) 15 for i {1,2}, deg 3 (u i ) 12 for i {3,4} and deg 3 (u 5 ) 11. Moreoer, noice ha {,} N 3 (u i ) for i {1,4}, {,,} N 3 ( ) and {,} N 3 (u 5 ). Thus, e obain L( ) 1, L( ) 2, L( ) 3, L(u 4 ) 4 and L(u 5 ) 5. Therefore, G 3 [U] is L-colourable. L17. Define he parial 3-facial 11-colouring c as alays, regarding he bold edges and he erices, and. Then, G 3 [U] K 5 and deg 3 (u i ) 15 for i {1,2}, deg 3 ( ) 12 and deg 3 (u i ) 11 for i {4,5}. Moreoer, noice ha {,} N 3 (u i ) for i {1,5}, {,,} N 3 ( ) and {,} N 3 ( ). Thus, e obain L( ) 1, L( ) 2, L(u i ) 4 for i {3,4} and L(u 5 ) 5. Therefore, G 3 [U] is L-colourable. L18. Le us define he parial 3-facial 11-colouring c as alays, regarding he bold edges and he erices and. Then, G 3 [U] K 3, deg 3 ( ) 13 and deg 3 (u i ) 11 for i {2,3}. Moreoer, {,} N 3 (u i ) for i {1,2,3}. Thus, e obain L( ) 1 and L(u i ) 3 for i {2,3}. Therefore, G 3 [U] is L-colourable. L19. Again, G 3 [U] K 5 and deg 3 (u i ) 15 for i {1,2} hile deg 3 (u i ) 11 for i {3,4,5}. Furhermore, {,} N 3 (u i ) for i {1,3,4}, {,} N 3 (u 5 ) and {,,} N 3 ( ). Thus, e deduce L( ) 1, L( ) 2 and L(u i ) 5 for i {3,4,5}. Therefore, G 3 [U] is L- colourable. L20. Here, G 3 [U] K 6. Also, deg 3 (u i ) 15 for i {1,2,3}, deg 3 (u 4 ) 13 and deg 3 (u i ) 11 for i {5,6}. Furhermore, {,} N 3 (u i ) for i {1,6}, {,,} N 3 ( ) and {,} N 3 (u i ) for i {2,4}. Thus, e infer L(u i ) 2 for i {1,2}, L( ) 3, L(u 4 ) 4, L(u 5 ) 5 and L(u 6 ) 6. Therefore, G 3 [U] is L-colourable. L21. Again G 3 [U] K 6. Also, deg 3 (u i ) 15 for i {1,2,3}, deg 3 (u i ) 12 for i {4,5} and deg 3 (u 6 ) 11. Furhermore, {,} N 3 (u i ) for i {1,5}, {,,} N 3 ( ) and {,} N 3 (u i ) for i {2,6}. Thus, e infer L(u i ) 2 for i {1,2} and L(u i ) i for i {3,4,5,6}. Therefore, G 3 [U] is L-colourable. L22. In his case, G 3 [U] K 6. Also, deg 3 (u i ) 13 for i {1,2,3,4} and deg 3 (u i ) 12 for i {5,6}. Furhermore, {,} N 3 (u i ) for i {4,5}, {,,} N 3 (u 6 ) and {,} N 3 (u i ) for i {2,3}. Thus, e infer L( ) 3, L(u i ) 4 for i {2,3,4}, L(u 5 ) 5 and L(u 6 ) 6. Therefore, G 3 [U] is L-colourable. 10

13 u 5 4 u 4 u 4 u 5 (L17) (L18) (L19) u 6 u 5 5 (L20) u 4 4 u 5 (L21) u 6 u 4 u 5 u 6 (L22) u 4 u (L23) (L24) Figure 4: Reducible configuraions (L17) (L24). 11

14 L23. In his case, G 3 [U] K 3. Also, deg 3 (u i ) 12 for i {1,2,3}. Moreoer, {,,} N 3 (u i ) for i {1,2,3}. Thus, e infer L(u i ) 3 for i {1,2,3}. Therefore, G 3 [U] is L- colourable. L24. Define he parial colouring c as alays, regarding he bold edges and he erex. Remark ha G 3 [U] is isomorphic o he complee graph on four erices minus one edge K 4, since / N 3 ( ) (because he face has size a leas 8). By Lemma 1, deg 3 (u i ) 11 for eery i {1,2,3,4}. Thus, L(u i ) 2 for i {1,2} and L(u i ) 3 for i {3,4}. Hence, he graph G 3 [U] is L-colourable. This asserion can be direcly checked, or seen as a consequence of a heorem independenly proed by Borodin [2] and Erdős, Rubin and Taylor [8] (see also [16]), saing ha a conneced graph is degree-choosable unless i is a Gallai ree, ha is each of is blocks is eiher complee or an odd cycle. Corollary 1. Eery (3,11)-minimal graph G has he folloing properies: (i) Le f 1, f 2 be o 5-faces of G ih a common edge xy. Then, x and y are no boh 3-erices. (ii) Le f be a 7-face hose eery inciden erex is a 3-erex. If f is adjacen o a 3-face, hen eery oher face adjacen o f is a ( 7)-face. (iii) If o adjacen dangerous erices do no lie on a same ( 4)-face, hen none of hem is inciden o a 3-face. (i) To dangerous erices inciden o a same 6-face are no adjacen. () There canno be four consecuie dangerous erices inciden o a same ( 6)-face. (i) A ery-bad face is adjacen o a leas hree ( 7)-faces. (ii) A bad face is adjacen o a leas o ( 7)-faces. Proof. (i) By Lemma 2(), deg 3 (x) + deg 3 (y) 23. By Lemma 1, he 3-facial degree of a 3-erex inciden o o 5-faces is a mos 11. Hence a leas one of x and y is a ( 4)-erex. (ii) Firs noe ha, according o Lemma 2(iii), he faces adjacen o boh f and he 3-face has size a leas 7. Hence, f is adjacen o a mos four ( 6)-faces. No, he asserion direcly follos from he reducibiliy of he configuraions (L1) and (L2) of Figure 2. (iii) This follos from he reducibiliy of he configuraion (L4) of Figure 2. (i) Suppose he conrary, and le x and y be o such erices. By Lemma 2(iii), a 6-face is no adjacen o a 3-face, hence boh x and y are inciden o a 4-face. Then, deg 3 (x) 11 and deg 3 (y) 11, hich conradics Lemma 2(). () Suppose ha he asserion is false. Then, according o he hird iem of his corollary, he graph G mus conain he configuraion (L5) or (L6) of Figure 2, hich are boh reducible. 12

15 (i) Le f be a ery-bad face. By he firs iem of his corollary and Lemma 3, o adjacen ( 6)-faces canno be boh adjacen o f. Hence, f is adjacen o a mos o such faces. (ii) Le f be a bad face, and denoe by α i, i {1,2,3,4,5} is inciden erices in clockise order. Wihou loss of generaliy, assume ha, for eery i {1,2,3,4}, α i is a dangerous erex. For i {1,2,3,4}, denoe by f i he face adjacen o f and inciden o boh α i and α i+1. According o he firs iem of his corollary and Lemma 3, a mos o faces among f 1, f 2, f 3, f 4 can be ( 6)-faces. This concludes he proof. 3 Proof of Theorem 1 Suppose ha Theorem 1 is false, and le G be a (3,11)-minimal graph. We shall ge a conradicion by using he Discharging Mehod. Here is an oerie of he proof: each erex and face is assigned an iniial charge. The oal sum of he charges is knon o be negaie by Euler s Formula. Then, some redisribuion rules are defined, and each erex and face gies or receies some charge according o hese rules. The oal sum of he charges is no changed during his sep, bu a he end e shall sho, by case analysis, ha he charge of each erex and each face is non-negaie, a conradicion. Iniial charge. Firs, e assign a charge o each erex and face. For eery V (G), e define he iniial charge ch() = d() 4, here d() is he degree of he erex in G. Similarly, for eery f F(G), here F(G) is he se of faces of G, e define he iniial charge ch( f ) = r( f ) 4, ih r( f ) he lengh of he face f. By Euler s formula he oal sum is ch() + ch( f ) = 8. V (G) f F(G) Rules. We use he folloing discharging rules o redisribue he iniial charge. Rule R1. A ( 5)-face sends 1/3 o each of is inciden safe erices and 1/2 o each of is inciden dangerous erices. Rule R2. A ( 7)-face sends 1/3 o each adjacen 3-face. Rule R3. A ( 7)-face sends 1/6 o each adjacen bad face. 13

16 Rule R4. A 6-face sends 1/12 o each adjacen ery-bad face. Rule R5. A ( 5)-erex gies 2/3 o an inciden face f if and only if here exis o 3-faces boh inciden o and boh adjacen o f. (Noe ha he size of such a face f is a leas 7.) We shall proe no ha he final charge ch (x) of eery x V (G) F(G) is non-negaie. Therefore, e obain a conradicion. 8 = ch() + V (G) ch( f ) = f F(G) ch () + V (G) ch ( f ) 0, f F(G) Final charge of erices. Firs, as noiced in Lemma 2(i), G has minimum degree a leas hree. Le be an arbirary erex of G. We ill proe ha is final charge ch () is non-negaie. In order o do so, e consider a fe cases regarding is degree. So, suppose firs ha is a 3- erex. If is a safe erex, hen by Rule R1 is final charge is ch () = = 0. Similarly, if is dangerous, hen ch () = = 0. If is a 4-erex hen i neiher receies nor sends any charge. Thus, ch () = ch() = 0. Finally, suppose ha is of degree d 5. Noice ha may send charge only by Rule R5. This may occur a mos d/2 imes if d is een, and a mos d/2 1 imes if d is odd (since o 3-faces are no adjacen). Thus, ch () d 4 d 2 2 3, hich is non-negaie if d 6. For d = 5, ch () > 0. Final charge of faces. Le f be an arbirary face of G. Denoe by fce and bad he number of 3-faces and he number of bad faces adjacen o f, respeciely. Denoe by sfe and dgs he number of safe erices and he number of dangerous erices inciden o f, respeciely. We ill proe ha he final charge ch ( f ) of f is non-negaie. In order o do so, e consider a fe cases regarding he size of f. f is a 3-face. I is adjacen only o ( 7)-faces by Lemma 2(iii). Thus, by Rule R2, f receies 1/3 from each of is hree adjacen faces, so e obain ch ( f ) = 0. f is a 4-face. I neiher receies nor sends any charge. Thus, ch ( f ) = ch( f ) = 0. f is a 5-face. Then, f is adjacen only o ( 5)-faces due o Lemma 2(iii). So a 5-face may send charge only o is inciden 3-erices, hich are all safe. Consider he folloing cases regarding he number sfe of such erices. sfe 3: Then, ch () = 0. sfe = 4: In his case, f is a bad face. According o Corollary 1(ii), a leas o of he faces ha are adjacen o f hae size a leas 7. Thus, according o Rule R3, f receies 1/6 from a leas o of is adjacen faces. Hence, e conclude ha ch 16 () = 0. 14

17 sfe = 5: Then f is a ery-bad face, and so, according o Corollary 1(i), a leas hree faces adjacen o f hae size a leas 7. Moreoer, all faces adjacen o f hae size a leas 6 by Lemma 2(iii) and Corollary 1(i). By Rules R3 and R4, i follos ha he neighbouring faces of f send a leas 4 1/6 o f, hich implies ha ch 13 () = 0. f is a 6-face. By Lemma 2(iii), fce = 0. Denoe by bd number of ery-bad faces adjacen o 12 1 f. The final charge of f is 2 dgs sfe 13 bd 12 due o Rules R1 and R4. According o Corollary 1(i), o dangerous erices on f canno be adjacen so here are a mos hree dangerous erices on f. Obsere also ha bd sfe/2 by Corollary 1(i) and because a ery-bad face adjacen o f is inciden o o safe erices of f. Le us consider he final charge of f regarding is number of dangerous erices. dgs = 3: Since a safe erex is no inciden o a ( 4)-face, here is a mos one safe erex inciden o f, i.e. sfe 1. Thus, bd = 0, and hence, ch ( f ) > 0. dgs = 2: Then, sfe 3. Le us disinguish o cases according o he alue of sfe. sfe = 3: Noice ha bd = 0, oherise i ould conradic he reducibiliy of (L3). Hence, ch ( f ) = 0. sfe 2: In his case, here is a mos one ery-bad face adjacen o f, so ch ( f ) > 0. dgs = 1: Then, sfe 4 and bd 1 because (L3) is reducible. So, ch ( f ) > 0. dgs = 0: If sfe 5 hen, because (L3) is reducible, bd = 0, herefore ch ( f ) = 0. And, if sfe 4, hen bd 2, so ch ( f ) > f is a 7-face. The final charge of f is a leas 3 dgs 12 (fce + sfe) bad 16. According o Corollary 1(), four dangerous erices canno be consecuie on f, hence here canno be more han fie dangerous erices on f. Denoe by α 1,α 2,...,α 7 he erices of f in clockise order. Le D be he se of dangerous erices of f, so dgs = D. We shall look a he final charge of f, regarding is number dgs of dangerous erices. dgs = 5: Up o symmery, D = {α 1,α 2,α 3,α 5,α 6 }. Suppose firs ha α 5 and α 6 are no inciden o a same ( 4)-face. Then, here can be neiher a safe erex inciden o f nor a bad face adjacen o f, because a safe erex is no inciden o a ( 4)-face, and also a bad face is no adjacen o a ( 4)-face. Moreoer, by Corollary 1(iii), here is no 3- face adjacen o f. Therefore, ch ( f ) > 0. No, if α 5 and α 6 are inciden o a same ( 4)-face, hen he erex α 4 mus be a ( 4)-erex by he reducibiliy of (L7), and because i is no a dangerous erex. Hence, here is no safe erex and no bad face adjacen o f, so is charge is ch ( f ) > 0. 15

18 dgs = 4: We consider seeral subcases, according o he relaie posiion of he dangerous erices on f. Recall ha, by Corollary 1(), here are a mos hree consecuie dangerous erices. Wihou loss of generaliy, e only need o consider he folloing hree possibiliies: D = {α 1,α 2,α 3,α 5 }: The charge of f is ch 13 ( f ) = 1 (fce+sfe) bad 16. Moreoer, sfe 2, bad 1 and fce + sfe 3 by Corollary 1(iii) and because a safe erex is no inciden o a ( 4)-face. So, ch ( f ) is negaie if and only if sfe = 2,bad = 1 and fce = 1. Bu in his case, he obained configuraion is (L8), hich is reducible. D = {α 1,α 2,α 4,α 5 }: As a bad face is neiher adjacen o a ( 4)-face nor inciden o a dangerous erex, e ge bad 1. Obsere also ha, as α 3 is no dangerous, i has degree a leas four by he reducibiliy of (L7) and (L11). Thus, sfe 2. Suppose firs ha bad = 1, hen sfe is one or o. According o he reducibiliy of (L10), e infer sfe + fce 2. Hence, ch ( f ) > 0. Suppose no ha bad = 0. We hae fce 3 and sfe 2. If fce = 3 hen sfe = 0, and if fce = 2, hen sfe 1 according o he reducibiliy of (L12). So, fce + sfe 3. Therefore, ch ( f ) (fce + sfe) D = {α 1,α 2,α 4,α 6 }: In his case, here is no bad face adjacen o f. Furhermore, by Corollary 1(iii), fce 3 and sfe 2, as he dangerous erices α 4 and α 6 preen a leas one non-dangerous erex from being safe. Obsere ha fce +sfe 5 since oherise i ould conradic he reducibiliy of (L13). According o he reducibiliy of (L13), if fce + sfe = 4 hen fce = 3 and no o 3-faces hae a common erex. Hence, he obained configuraion is isomorphic o (L14) or (L15), hich are boh reducible. So, fce + sfe 3 and hus ch ( f ) 3 2 (fce + sfe) dgs = 3: Again, e consider seeral subcases according o he relaie posiion of he dangerous erices on f. D = {α 1,α 2,α 3 }: Then fce+sfe 3 by Corollary 1(iii), and bad 2. Thus, ch ( f ) > 0. D = {α 1,α 2,α 4 }: Then, fce 4. We shall no examine he siuaion according o each possible alue of fce. fce = 4: Necessarily, sfe 1 and bad = 0. No, if sfe = 0, hen ch ( f ) > 0. And, if sfe = 1, hen he safe erex mus be α 3. Moreoer, α 5 mus be a ( 5)-erex because (L9) is reducible. Hence, f is inciden o α 5 beeen o 3-faces, so by Rule R5 he erex α 5 gies 2 3 o f. Thus, ch ( f ) > 0. fce = 3: Suppose firs ha one of he dangerous erices is inciden o a 4-face. Necessarily, sfe 1 and bad 1. Thus, ch ( f ) = 0. Suppose no ha no dangerous erex is inciden o a 4-face. In paricular, sfe 2. If sfe = 2 hen he obained configuraion conradics he reducibiliy 16

19 of (L19). Hence, sfe 1 and bad 1. Therefore, ch ( f ) = 0. fce = 2: We shall proe ha sfe 2. This is clear if α 1 and α 2 are no inciden o a same 3-face. So, e may assume ha he edge α 1 α 2 lies on a 3-face. Bu hen e obain he inequaliy due o he reducibiliy of (L19) and (L20). Using Corollary 1(i) and sfe 2, e infer ha bad 1. Hence, ch ( f ) = 0. fce = 1: Then sfe 3 and bad 2. If sfe = 3 and bad = 2, he obained configuraion conradics he reducibiliy (L20) or (L21). So, ch ( f ) = 0. fce = 0: Again, sfe 3 and bad 2, so ch 13 ( f ) > 0. D = {α 1,α 2,α 5 }: As in he preious case, fce 4 and e look a all he possible cases according o he alue of fce. Since a bad face is no inciden o a dangerous erex, noice ha only edges α 3 α 4 and α 6 α 7 can be inciden o a bad face. In paricular, bad 2. fce = 4: In his case, sfe = 0 and bad = 0. Therefore, ch ( f ) = > 0. fce = 3: If one of he dangerous erices is inciden o a 4-face hen sfe = 0, hence bad = 0. Thus, ch ( f ) So no, e infer ha sfe canno be 2, oherise i ould conradic he reducibiliy of (L16). Therefore, sfe is a mos one, and so bad 1 by Corollary 1(i). Thus, ch ( f ) = 0. fce = 2: According o he reducibiliy of (L16) and (L17), sfe 2. As ch ( f ) = (fce + sfe) 13 bad 16, e deduce ch ( f ) < 0 if and only if sfe = 2 and bad = 2. In his case, he obained configuraion is (L18), hich is reducible. fce = 1: Because (L16) and (L17) are reducible, sfe 2. So, ch ( f ) > 0. fce = 0: Then sfe 3, and so ch ( f ) > 0. D = {α 1,α 3,α 5 }: In his case, sfe 2 since a safe erex is no inciden o a ( 4)-face, and bad 1, since a bad face canno be inciden o a dangerous erex. Moreoer, fce 4. Le us examine he possible cases regarding he alue of fce. fce = 4: Obsere ha sfe 1 and bad = 0. Noe also one of α 2,α 3,α 6,α 7 is adjacen o a dangerous erex, and inciden o f beeen o riangles. Hence, by he reducibiliy of (L9), i has degree a leas fie, and by Rule R5, i sends 2 3 o f. Thus, ch ( f ) > 0. fce = 3: If sfe 1 hen ch ( f ) = 0. And, if sfe = 2 hen, up o symmery, he o safe erices are eiher α 6 and α 7, or α 2 and α 6. In he former case, one of α 2,α 4 is inciden o f a he inersecion of o 3-faces. Furhermore, i mus be a ( 5)-erex due o he reducibiliy of (L9). In he laer case, he same holds for α 4 due o he reducibiliy of (L9). Hence, in boh 17

20 cases he face f receies 2/3 from one of is inciden erices by Rule R5. Recall ha bad 1, and herefore, ch ( f ) > 0. fce 2: As sfe 2 and bad 1, e infer ha ch ( f ) = 0. dgs = 2: Again, e consider seeral subcases, regarding he posiion of he dangerous erices on f. D = {α 1,α 2 }: Obsere ha bad 3, and according o Corollary 1(iii), fce + sfe 6. We consider hree cases, according o he alue of fce + sfe. fce + sfe = 6: All he erices inciden o f hae degree hree, and f is adjacen o a 3-face. Thus, by Corollary 1(ii), f is no adjacen o any ( 6)-face. In paricular, no bad face is adjacen o f, i.e. bad = 0. Hence, ch ( f ) = 0. fce + sfe = 5: If bad 2, hen ch 13 ( f ) = 0. Oherise, bad = 3. Noe ha he edge α 1 α 2 mus be inciden o a ( 4)-face. If his face is of size four, hen e obain configuraion (L22). Suppose no ha his face is of size hree. Since here is no hree consecuie bad faces around f, e can assume ha each of he edges α 3 α 4 and α 6 α 7 lies on a bad face. By he reducibiliy of (L18), e conclude ha α 3 and α 7 hae degree a leas four. Bu hen, fce + sfe < 5. fce + sfe 4: In his case, ch ( f ) > 0. D = {α 1,α 3 } or D = {α 1,α 4 }: Again fce+sfe 6, and e consider o cases regarding he alue of fce+sfe. Since a bad face is no inciden o a dangerous erex, e infer ha bad 3. fce + sfe = 6: Suppose firs ha D = {α 1,α 3 }. Le P 1 = α 1 α 2 α 3 and P 2 = α 3 α 4 α 5 α 6 α 7 α 1. In order o assure fce + sfe = 6, obsere ha all edges of P 1 are inciden o 3-faces and all inner erices of P 2 are safe, or ice-ersa. Thus, α 2 or α 4 is a ( 5)-erex by he reducibiliy of (L9). Hence, i gies 3 2 o f by Rule R5. Therefore, ch ( f ) > 0. Suppose no ha D = {α 1,α 4 }. Similarly as aboe, one can sho ha α 2 or α 5 is a ( 5)-erex ha donaes 2 3 o f. Hence, ch ( f ) > 0. fce+sfe 5: Noice ha bad 2. Therefore, ch ( f ) = 0. dgs = 1: Then fce + sfe 6 and, by Corollary 1(i), e infer ha bad 3. So, ch ( f ) = 0. dgs = 0: By Corollary 1(i), fce + sfe 7 and bad 4. So, ch ( f ) = 0. f is an 8-face. Because (L4) and (L23) are reducible, here canno be hree consecuie dangerous erices on f. Hence, dgs 5. Denoe by α i, i {1,2,...,8}, he erices inciden o f in clockise order, and le D be he se of dangerous erices inciden o f. 18

21 dgs = 5: Up o symmery, D = {α 1,α 2,α 4,α 5,α 7 }. Since a bad face is no inciden o a dangerous erex, necessarily bad = 0. For i {1,4}, denoe by f i he face adjacen o f and inciden o boh α i and α i+1. Since (L24) is reducible, a mos one of f 1 and f 4 is a 3-face. Furhermore, a mos o of α 3,α 6,α 8 can be safe erices, since a leas one of α 6,α 8 is a ( 4)-erex. Therefore, fce 2, sfe 2 and so, ch ( f ) > 0. dgs = 4: Up o symmery, he se of dangerous erices is {α 1,α 2,α 4,α 5 }, {α 1,α 2,α 5,α 6 }, {α 1,α 2,α 4,α 6 }, {α 1,α 2,α 4,α 7 } or {α 1,α 3,α 5,α 7 }. In any case, bad 2 and fce + sfe 5. Hence, ch ( f ) = 0. dgs = 3: Then, fce + sfe 6 and bad 3. So, ch ( f ) = 0. dgs = 2: Then, fce+sfe 7, and by Corollary 1(i), bad 4. Thus, ch ( f ) = 0. dgs = 1: Again, fce + sfe 7 and bad 4, so ch ( f ) > 0. dgs = 0: By Corollary 1(i), bad 5. So, ch ( f ) > 0. f is a ( 9)-face. Le f be a k-face ih k 9, and denoe by,,...,u dgs he dangerous erices on f in clockise order. Denoe by f i he ( 4)-face inciden o u i. The facial segmen P = u i j u i+1 of f beeen u i and u i+1 (in clockise order) is of one of he fie folloing ypes: (a) if j 1, 1 is no inciden o f i and j is no inciden o f i+1 ; (b) if j 1, 1 is inciden o f i and j is inciden o f i+1 ; (c) if j 1 and no of ype (a) or (b); (d) if j = 0 and boh f i and f i+1 are he same 3-face; and (e) if j = 0 and no of ype (d). We denoe by α he number of pahs of ype (a), β he number of pahs of ype (b), γ he number of pahs of ype (c), δ he number of pahs of ype (d) and ε he number of pahs of ype (e). Noe ha a pah of ype (d) or (e) is of lengh one. Obsere ha he folloing holds: Claim 1. α + β + γ + δ + ε = dgs. We no bound he number of safe erices and 3-faces. Claim 2. fce + sfe k α γ ε. 19

22 For each l-pah P of ype (a),(c) or (e) he number of safe erices on P plus he number of 3-faces hich share an edge ih P is a mos l 1. Indeed, for any pah of one of hese ypes, here are a mos l faces differen from f and inciden o an edge of he pah, bu a leas one of hem is no a ( 4)-face. There are l 1 erices on he pah, so a mos l 1 safe erices. Furhermore, eery ( 4)-face preens a leas one erex from being safe. Obsere also ha an l-pah of ype (b) or (d) conribues for a mos l, hich hus yields Claim 2. We disinguish o kinds of pahs of ype (e): a pah of ype (e) is of ype (e 0 ) if is edge is no inciden o a 4-face. Oherise, i is of ype (e 1 ). Le ε i be he number of pahs of ype (e i ), i {0,1}. Claim 3. bad k 2dgs + δ + ε 1. Firs, remark ha each dangerous erex preens is o inciden edges on f from belonging o a bad face, since no bad face is inciden o a dangerous erex. By he reducibiliy of (L23), here canno be hree consecuie dangerous erices on f, so i only remains o consider o consecuie dangerous erices, i.e. pahs of ype (d) or (e). A pah of ype (d) or (e 1 ) preens exacly hree edges of f from being inciden o a bad face. Eery 1-pah of ype (e 0 ) preens a leas four edges of f from being inciden o a bad face. To see his, consider a pah u 4 u 5 u 6, here is a 1-pah of ype (e 0 ). Clearly, none of,, u 4 is inciden o a bad face. We claim ha a leas one of u 4 u 5,u 5 u 6 is no inciden o a bad face. Oherise, if u 4 u 5 is inciden o a bad face, hen by Lemma 2(iii), u 4 mus be a ( 4)-erex. Hence, by Corollary 1(i), u 5 u 6 is no inciden o a bad face. As no hree dangerous erices are consecuie of f, his proes Claim 3. Claim 4. α β + ε 0 = δ + ε 1. Associae each dangerous erex u i ih is inciden ( 4)-face f i. Each pah of ype (a) conains no face f i, so does each pah of ype (e 0 ); each pah of ype (c) conains exacly one face f i, and each pah of ype (b),(d) or (e 1 ) conains exacly o faces f i (here a face is couned ih is mulipliciy, i.e. once for each dangerous erex of f inciden o i). So, dgs = γ + 2(β + δ + ε 1 ), and hence α + β + γ + δ + ε = γ + 2(β + δ + ε 1 ), hich gies Claim 4. So, by Claims 1 4, e ge ch ( f ) = k 4 dgs 1 1 (fce + sfe) 2 3 bad 1 6 k 4 dgs 2 k α γ ε k 2dgs + δ + ε = k 2 4 dgs 6 + α + γ + ε 0 + ε 1 δ 3 6 = k (α β + ε 0) + γ 6 = k δ + ε 1 + γ 6 k 2 4 δ δ 3 δ 3

23 According o Corollary 1(iii) and he reducibiliy of (L24), here are a leas o erices beeen any o pahs of ype (d). So, δ 4 k. Therefore, one can conclude ha References ch ( f ) k 2 k 24 4 = k > 0. [1] K. Appel and W. Haken. Eery planar map is four colorable, olume 98 of Conemporary Mahemaics. American Mahemaical Sociey, Proidence, RI, Wih he collaboraion of J. Koch. [2] O. V. Borodin. Crierion of chromaiciy of a degree prescripion (in Russian). In Absracs of IV All-Union Conf. on Theoreical Cyberneics (Noosibirsk), pages , [3] O. V. Borodin. Soluion of he Ringel problem on erex-face coloring of planar graphs and coloring of 1-planar graphs. Meody Diskre. Analiz., 41:12 26, 108, [4] O. V. Borodin. Cyclic coloring of plane graphs. Discree Mah., 100(1-3): , Special olume o mark he cenennial of Julius Peersen s Die Theorie der regulären Graphs, Par I. [5] O. V. Borodin. A ne proof of he 6 color heorem. J. Graph Theory, 19(4): , [6] O. V. Borodin, D. P. Sanders, and Y. Zhao. On cyclic colorings and heir generalizaions. Discree Mah., 203(1-3):23 40, [7] Z. Dořák, R. Škrekoski, and M. Tancer. Lis-colouring squares of sparse subcubic graphs. Technical Repor IMFM-(2005)-PS-985, Uniersiy of Ljubljana, Sloenia, [8] P. Erdős, A. L. Rubin, and H. Taylor. Choosabiliy in graphs. In Proceedings of he Wes Coas Conference on Combinaorics, Graph Theory and Compuing (Humbold Sae Uni., Arcaa, Calif., 1979), Congress. Numer., XXVI, pages , Winnipeg, Man., Uilias Mah. [9] M. Horňák and S. Jendro l. On some properies of 4-regular plane graphs. J. Graph Theory, 20(2): , [10] T. R. Jensen and B. Tof. Graph coloring problems. Wiley-Inerscience Series in Discree Mahemaics and Opimizaion. John Wiley & Sons, Inc., Ne-York, A iley- Inerscience Publicaion. [11] D. Krá l, T. Madaras, and R. Škrekoski. Cyclic, diagonal and facial colorings. European J. Combin., 26(3-4): ,

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