Planar graphs without 5-cycles and intersecting triangles are (1,1,0) -colorable
aa r X i v : . [ m a t h . C O ] S e p PLANAR GRAPHS WITHOUT 5-CYCLES AND INTERSECTING TRIANGLES ARE (1 , , -COLORABLE RUNRUN LIU † AND XIANGWEN LI † AND GEXIN YU † ‡
Abstract.
A ( c , c , ..., c k )-coloring of G is a mapping ϕ : V ( G )
7→ { , , ..., k } such that for every i, ≤ i ≤ k , G [ V i ] has maximum degree at most c i , where G [ V i ] denotes the subgraph induced by the vertices colored i . Borodin and Raspaud conjecture that every planar graph without 5-cycles and intersecting triangles is(0 , , , , Introduction
Graph coloring is one of the central topics in graph theory. A graph is ( c , c , · · · , c k ) -colorable if thevertex set can be partitioned into k sets V , V , . . . , V k , such that for every i : 1 ≤ i ≤ k the subgraph G [ V i ]has maximum degree at most c i . Thus a (0 , , Conjecture 1.1 (Steinberg, [13]) . All planar graphs without -cycles and -cycles are -colorable. Some relaxations of the Steinberg Conjecture are known to be true. Along the direction suggested byErd˝os to find a constant c such that a planar graph without cycles of length from 4 to c is 3-colorable,Borodin, Glebov, Raspaud, and Salavatipour [4] showed that c ≤
7, and more results similar to those can befound in the survey by Borodin [1]. Another direction of relaxation of the conjecture is to allow some defectsin the color classes. Chang, Havet, Montassier, and Raspaud [6] proved that all planar graphs without 4-cycles or 5-cycles are (2 , , , , , , , , et al. [16, 17].Havel [9] proposed that planar graphs with triangles far apart should be properly 3-colorable, which wasconfirmed in a recent preprint of Dvo¨r´ak, Kr´al and Thomas [7]. Borodin and Raspaud [5] combined the ideasof Havel and Steinberg and proposed the following so called Bordeaux Conjecture in 2003. Conjecture 1.2 (Borodin and Raspaud, [5]) . Every planar graph without intersecting triangles and without -cycles is -colorable. A planar graph without intersecting triangles means the distance between triangles is at least 1. Let d ▽ denote the smallest distance between any pair of triangles in a planar graph. A relaxation of the BordeauxConjecture with d ▽ ≥ d ▽ ≥ d ▽ ≥ Date : April 8, 2019.The second author was supported by the Natural Science Foundation of China (11171129) and by Doctoral Fund of Ministryof Education of China (20130144110001); The third author’s research was supported in part by NSA grant H98230-12-1-0226. sing the relaxed coloring notation, Xu [15] proved that all planar graphs without adjacent triangles and5-cycles are (1 , , G be the family of plane graphs with d ▽ ≥ G are (4 , , , , G are (2 , , G be a graph and H bea subgraph of G . We call ( G, H ) to be superextendable if each (1 , , H can be extended to G so that vertices in G − H have different colors from their neighbors in H ; in this case, we call H to be asuperextendable subgraph. Theorem 1.3.
Every triangle or -cycle of a planar graph in G is superextendable. As a corollary, we have the following relaxation of the Bordeaux Conjecture.
Theorem 1.4.
A planar graph in G is (1 , , -colorable. To see the truth of Theorem 1.4 by way of Theorem 1.3, we may assume that the planar graph containsa triangle C since G is (0 , , G has no triangle. Then color the triangle, and by Theorem 1.3,the coloring of C can be superextended to G . Thus, we get a coloring of G .As many results with similar fashion, we use a discharging argument to prove Theorem 1.3. This argumentconsists of two parts: structures and discharging. After introduce some common notations in Section 2, weshow in Section 3 some useful special structures in a minimal counterexample to the theorem, then inSection 4, we design a discharging process to distribute the charges and use the special structures to reacha contradiction.It should be noted that while the proof of our main theorem shares a lot of common properties with the(2 , ,
0) result in [12], it is much more involved. We have to extend some powerful tools from [15] by Xu,and discuss in detail the structures around 4-vertices and 5-vertices. It would be interesting to know how touse the new tools developed in this paper to improve our result.2.
Preliminaries
In this section, we introduce some notations used in the paper.Graphs mentioned in this paper are all simple. For a positive integer n , let [ n ] = { , , . . . , n } . A k -vertex( k + -vertex, k − -vertex) is a vertex of degree k (at least k , at most k ). The same notation will apply to facesand cycles. We use b ( f ) to denote the vertex sets on f . We use F ( G ) to denote the set of faces in G . An( l , l , . . . , l k )-face is a k -face v v . . . v k with d ( v i ) = l i , respectively. A face f is a pendant -face of vertex v if v is not on f but is adjacent to some 3-vertex on f . A pendant neighbor of a 3-vertex v on a 3-face isthe neighbor of v not on the 3-face.Let C be a cycle of a plane graph G . We use int ( C ) and ext ( C ) to denote the sets of vertices locatedinside and outside C , respectively. The cycle C is called a separating cycle if int ( C ) = ∅ 6 = ext ( C ), and iscalled a nonseparating cycle otherwise. We still use C to denote the set of vertices of C .Let S , S , . . . , S l be pairwise disjoint subsets of V ( G ). We use G [ S , S , . . . , , S l ] to denote the graphobtained from G by identifying all the vertices in S i to a single vertex for each i ∈ { , , . . . , l } .A vertex v is properly colored if all neighbors of v have different colors from v . A vertex v is nicely colored if it shares a color (say i ) with at most max { s i − , } neighbors, where s i is the deficiency allowed for color i ; thus if a vertex v is nicely colored by a color i which allows deficiency s i >
0, then an uncolored neighborof v can be colored by i . . Special configurations
Let (
G, C ) be a minimum counterexample to Theorem 1.3 with minimum σ ( G ) = | V ( G ) | + | E ( G ) | , where C is a triangle or a 7-cycle in G that is precolored. For simplicity, let F k = { f : f is a k -face and b ( f ) ∩ C = ∅} , F ′ k = { f : f is a k -face and | b ( f ) ∩ C | = 1 } , and F ′′ k = { f : f is a k -face and | b ( f ) ∩ C | = 2 } .The following lemmas are shown in [12]. Proposition 3.1 (Prop 3.1 in [12]) . (a) Every vertex not on C has degree at least .(b) A k -vertex in G can have at most one incident -face.(c) No -face and -face in G can have a common edge. Lemma 3.2 (Lemma 3.2 in [12]) . The graph G contains neither separating triangles nor separating -cycles. Lemma 3.3 (Lemma 3.3 in [12]) . If G has a separating -cycles C = v v v v v , then ext ( C ) = { b, c } such that v bc is a -cycle. Furthermore, the -cycle is the unique separating -cycle. Lemma 3.4 (Lemma 3.4 in [12]) . If x, y ∈ C with xy E ( C ) , then xy E ( G ) and N ( x ) ∩ N ( y ) ⊆ C . Lemma 3.5 (Lemma 3.6 in [12]) . Let u, w be a pair of diagonal vertices on a -face. If at most one of u and w is incident to a triangle, G [ { u, w } ] ∈ G . Lemma 3.6 (Lemma 3.7 in [12]) . Let f be a face in F ∪ F ′ . Then(1) if b ( f ) ∩ C = { u } , then each of u and w is incident to a triangle.(2) if f = uvwx ∈ F is a face with d ( u ) = d ( w ) = 3 , then each of v and x is incident to a triangle. By Lemma 3.2, we may assume that C is the boundary of the outer face of G . Lemma 3.7.
In int ( C ) , let v and u be two adjacent -vertices. Then each vertex in ( N ( u ) ∪ N ( v )) \ { u, v } has degree at least .Proof. Suppose to the contrary that v is a neighbor of v that has degree 3. Let G ′ = G − { u, v } . By theminimality of G , ( G ′ , C ) is superextendable. Recolor v properly, and then color u properly. Now v can becolored, or the three neighbors of v are colored differently. In the latter case, 1 or 2 (say 1) is used on u or v . Then we color v with 1, a contradiction. (cid:3) Lemma 3.8.
Let f = uvw be a face in F . Then each of the following holds.(1) If d ( u ) = d ( v ) = 3 , then d ( w ) ≥ .(2) If f is a (3 , , + ) -face, then each pendant neighbor of u or v is either on C or has degree at least .(3) If f is a (3 , , -face, then the pendant neighbor of u is either on C or has degree at least andat least one of the neighbors (not on f ) of each -vertex is either on C or has degree at least .Consequently, a -vertex cannot be incident to a (3 , , -face and a (3 , , , + ) -face from F .Proof. (1) Suppose otherwise that f = uvw is a (3 , , − )-face. Let G ′ = G − { u, v } . It follows that σ ( G ′ ) < σ ( G ). By the minimality of G , ( G ′ , C ) is superextendable. Recolor w properly and then color u properly. Then v can be colored, or N ( v ) contains three different colors. In the latter case, 1 or 2 (say 1) isused on u or w , then we can color v with 1, a contradiction.(2) Let f = uvw be a (3 , , + )-face. Let u ′ be the pendant neighbor of u . Assume that u ′ is not on C .Suppose otherwise that d ( u ′ ) = 3. By Lemma 3.7, each vertex in ( N ( u ) ∪ N ( v )) \{ u, v } has degree at least4. So d ( u ′ ) ≥
4, a contradiction.(3) Let f = uvw be a (3 , , u ′ be the pendant neighbor of u . Assume that u ′ is not on C .Suppose otherwise that d ( u ′ ) = 3. By the minimality of G , ( G − { u, u ′ } , C ) is superextendable. Color u ′ properly. Then u can be colored, or N ( u ) contains three different colors. In the latter case, if u ′ is colored ith 1 or 2, then we color u with the color of u ′ . Thus, we may assume that u ′ is colored with 3, and assumethat v, w are colored with 1 , v not on f arecolored with 1 and 2, then we recolor v with 3 and color u with 1. So, we may assume that they are coloredwith 1 and 3. Similarly, we may assume that the neighbors of w not on f are colored with 2 and 3. Now weswitch the color of v and w , and color u with 1, a contradiction.Now let v , v be the two neighbors of v not on f . Suppose otherwise that d ( v ) = d ( v ) = 3 and v , v V ( C ). By the minimality of G , ( G − { u, v, w, v , v } , C ) is superextendable. We properly color v , v , w and u in order. Then v can be properly colored, or N ( v ) has three different colors. In the lattercase, only one vertex in { u, w, v , v } is colored with 1 or 2 (say 1), so we color v with 1, a contradiction. (cid:3) Lemma 3.9.
Let v be a k -vertex with N ( v ) = { v i : i ∈ [ k ] } . Then each of the following holds.(1) For k = 4 , if v is incident to a (3 , , -face f = v vv from F and a -face f = vv uv from F with d ( u ) = 3 , then both v and v are incident to triangles. Consequently, f cannot be a (3 , , , + ) -face.(2) For k = 5 , let v be incident to a (3 , − , -face f = v v v from F and two -faces f = vv uv and f = vv wv from F . If d ( u ) = d ( w ) = 3 , then at least two vertices in { v , v , v } are incident totriangles.Proof. (1) Suppose otherwise that at most one vertex in { v , v } is incident to a triangle. Let G ′ = G [ { v , v } ]and v ′ be the new vertex. By Lemma 3.5, G ′ ∈ G . Then ( G ′ − { v, v , v , u } , C ) is superextendable. Wecolor v and v with the color of v ′ , then properly color v , v , u in order. Then v can be properly colored,or N ( v ) has three different colors. In the latter case, 1 or 2 (say 1) is used on v or v , so we color v with 1,a contradiction.(2) Suppose otherwise that at most one vertex in { v , v , v } is incident to a triangle. Let G ′ = G [ { v , v , v } ], and let v ′ be the new vertex. By lemma 3.5, G ′ ∈ G . Then ( G ′ − { v, u, w } , C ) is su-perextendable. Color v , v , v with the color on v ′ , and then properly color u and w since d ( u ) = d ( w ) = 3.We uncolor v , v and then recolor v , v properly in the order. Then v can be properly colored, or N ( v )has three different colors. In the latter case, 1 or 2 (say 1) is used on v or v , so we can color v with 1, acontradiction. (cid:3) We first prove the following useful lemma.
Lemma 3.10.
Let v be a -vertex in int ( C ) with N ( v ) = { v i : i ∈ [4] } . If v is incident to two -faces thatshare an edge, then there is no t -path from v i to v i +2 with t ∈ { , , , } , where the subscripts of v are takenmodulo .Proof. As v is incident to two 4-faces that share an edge, in any embedding, v i and v i +2 cannot be in thesame face, for otherwise, they will be in a separating 4-cycle, contrary to Lemma 3.3. Suppose otherwisethat P is a t -path from v i to v i +2 with t ∈ { , , , } . Consider cycle C = v i P v i +2 vv i . If t = 1 or 5, then C is a 3- or 7-cycle separating v i +1 and v i +3 , a contradiction to Lemma 3.2; if t = 2, then C is a 4-cycleseparating v i +1 and v i +3 , a contradiction to Lemma 3.3; if t = 3, then C is a 5-cycle, a contradiction to G ∈ G . (cid:3) Let v be a 4-vertex with its neighbor v , v , v , v in the clockwise order in the embedding. Then v iscalled ( v i , v i +2 ) -behaved if at most one of v i and v i +2 is incident to a triangle. Lemma 3.11.
Let v be a -vertex in int ( C ) with N ( v ) = { v i : i ∈ [4] } . Then each of the following holds.(1) If v is incident to two -faces f i = vv i u i v i +1 and f i +1 = vv i +1 u i +1 v i +2 with f i , f i +1 ∈ F , andat most one of { v i , v i +1 , v i +2 } is incident to a triangle, then d ( u i ) ≥ or d ( u i +1 ) ≥ , where thesubscripts of u and v are taken modulo 4.
2) If v is incident to two -faces f i = vv i u i v i +1 and f i +2 = vv i +2 u i +2 v i +3 with f i , f i +2 ∈ F , and atmost one vertex from each of { v i , v i +1 } and { v i +2 , v i +3 } is incident to a triangle, then d ( u i ) ≥ or d ( u i +2 ) ≥ , where the subscripts of u and v are taken modulo 4.(3) The vertex v is incident to at most one (3 , , , + ) -face from F .(4) Let v be incident to two -faces that share an edge. If v is ( v , v ) -behaved and ( v , v ) -behaved, thennone of the -faces can be (3 , , , + ) -face.Proof. (1) By symmetry we assume that i = 1. Suppose otherwise that d ( u ) = d ( u ) = 3. Let G ′ = G [ { v , v , v } ]. Since at most one vertex in { v , v , v } is incident to a triangle, by Lemma 3.5, G ′ ∈ G .Thus, ( G ′ , C ) is superextendable. Color v , v and v with the color of the resulting vertex of identificationand then we can recolor u , u and v properly, a contradiction.(2) By symmetry we assume that i = 1. Suppose otherwise that d ( u ) = d ( u ) = 3. Let G ′ = G [ { v , v } , { v , v } ]. Let v ′ and v ′′ be the new vertices by identifying v with v , and v with v , respectively.Since at most one vertex from each of { v , v } and { v , v } is incident to a triangle, by Lemma 3.5, G ′ ∈ G .Thus ( G ′ , C ) is superextendable. Color v , v with the color of v ′ and color v , v with the color of v ′′ , thenwe can recolor v , u and u properly, a contradiction.(3) Suppose otherwise that v is incident to at least two (3 , , , + )-faces f , f ∈ F . If f and f share anedge, let f = vv u v and f = vv u v , then d ( u ) = d ( u ) = 3. We first show that d ( v ) ≥
4. Assume that d ( v ) = 3. Since u and v are two adjacent 3-vertices in int ( C ), so by Lemma 3.7, ( N ( u ) ∪ N ( v )) \{ u , v } has degree at least 4, which implies that d ( u ) ≥
4, a contradiction. Thus f is a (3 , , , + )-face with d ( u ) = d ( v ) = 3 and f is a (3 , , , + )-face with d ( u ) = d ( v ) = 3. By Propositin 3.1(c), none of v and v is incident to a triangle. So by (1), d ( u ) ≥ d ( u ) ≥
4, a contradiction to d ( u ) = d ( u ) = 3. If f and f do not share an edge, then it contradicts to (2).(4) Assume that v is incident to a (3 , , , + )-face f . Then by symmetry d ( v ) = d ( u ) = 3 or d ( u ) = d ( v ) = 3. First we assume that d ( v ) = d ( u ) = 3. Let G ′ = G − v and H = G ′ [ { v , v } ]. By Lemma 3.10,there is no t -path from v to v with t ∈ { , , , } , so H contains no 5-cycle and no new triangles, in additionto the fact that G is ( v , v )-behaved, H has no intersecting triangles, therefore H ∈ G . Thus ( H, C ) issuperextendable. Color v and v with the color of the new vertex, then v can be colored properly, or N ( v )has three different colors. Consider the latter case. Recolor u , v properly in the order. If v is colored with1 or 2, then we color v with the color of v ; if v is colored 3, then color v with 3 and recolor v with thecolor of u . In either case, we reach a contradiction. Similar to the above argument, v cannot be incident toa (3 , , , + )-face with d ( u ) = d ( v ) = 3. (cid:3) For k = 4 ,
5, we call a k -vertex in int ( C ) to be poor if it is incident to k F . If a k -vertex isnot poor, then we call it rich . Lemma 3.12.
Let v be a poor -vertex with N ( v ) = { v i : i ∈ [4] } and four incident -faces f i = vv i u i v i +1 for i ∈ [4] , where the subscripts of v and u are taken modulo . Furthermore, v is ( v , v ) -behaved. If either d ( v ) = 3 or d ( v ) = 4 and v is ( u , u ) -behaved, then d ( v ) ≥ , or d ( v ) = 4 and v is not ( u , u ) -behaved.Proof. Suppose to the contrary that d ( v ) = 3 or d ( v ) = 4 and v is ( u , u )-behaved.Consider that d ( v ) = d ( v ) = 3. Let G ′ = G − v and H = G ′ [ { v , v } ]. By Lemma 3.10, there is no t -path from v to v with t ∈ { , , , } . It follows that H contains no 5-cycle and no new triangles. Inaddition to the fact that v is ( v , v )-behaved, H has no intersecting triangles. Therefore, H ∈ G . Thus( H, C ) is superextendable. Color v and v with the color of the new vertex, and recolor v , v properly.Then 1 or 2 (say 1) is used on v or v . Now color v with 1, a contradiction.By symmetry, consider that d ( v ) = 3 and d ( v ) = 4. Let G ′ = G −{ v, v } and let H = G ′ [ { v , v } , { u , u } ].Let v ′ and v ′ be the new vertices by identifying v with v and u with u , respectively. As above, there is o 5-cycle or new 3-cycle containing v ′ or v ′ . Furthermore, if there is a 3-cycle, 5-cycle containing v ′ and v ′ , then there is a 2-path or a 4-path from { v , v } to { u , u } , thus there is 5-cycle or separating 7-cyclein G , a contradiction. Therefore, H ∈ G . Note that now ( H, C ) is superextendable. Color v , v with thecolor of v ′ and color u , u with the color of v ′ , then properly color v , v . Now v can be colored, or N ( v )contains three different colors. In the latter case, 1 or 2 (say 1) is used on v or v , then color v with 1, acontradiction.Consider d ( v ) = d ( v ) = 4. Let G ′ = G − { v, v , v } , and let H = G ′ [ { u , u } , { v , v } , { u , u } ]. Let v ′ , v ′ , v ′ be the new vertices by identifying u with u , v with v and u with u , respectively. As shownabove, there is no 3-cycle or 5-cycle containing one of v ′ , v ′ , v ′ , or the pairs in { v ′ , v ′ } , { v ′ , v ′ } , { v ′ , v ′ } . Ifthere is a 3-cycle or 5-cycle containing v ′ , v ′ and v ′ then there is 1- or 3-path from v ′ to v ′ or a 2-pathfrom v ′ to v ′ , but in either case, there is a 5-cycle or a separating 7-cycle, a contradiction. Thus, ( H, C )is superextendable. Color the vertices with the color of their resulting vertex, respectively, then color v , v properly. Now v can be colored, or N ( v ) contains three different colors. In the latter case, 1 or 2 (say 1) isused on v or v , then we color v with 1, a contradiction. (cid:3) Lemma 3.13.
Let v be a poor -vertex with N ( v ) = { v i : i ∈ [5] } and five incident -faces f i = vv i u i v i +1 for i ∈ [5] , where the subscripts of u and v are taken modulo 5. Suppose that at most one vertex in N ( v ) isincident with a triangle. Then each of the following holds.(1) If d ( u i ) = d ( v i ) = 3 for some i ∈ [5] , then d ( u j ) ≥ for j ∈ [5] − { i } .(2) At most two vertices in { u i : i ∈ [5] } have degree .(3) Let d ( u i ) = 3 . If v j has degree 3 or is a 4-vertex with ( u j − , u j ) -behaved, then d ( v k ) ≥ , or d ( v k ) = 4 and v k is not ( u k − , u k ) -behaved, where { j, k } = { i − , i + 2 } .Proof. (1)Without loss of generality, We may assume that i = 1. By Lemma 3.7, d ( u ) ≥
4. Supposeotherwise that d ( u j ) = 3 for some j = 1 ,
5. Let H = G ′ [ { v j , v j +1 , v j +3 } ], where G ′ = G − v . By Lemma 3.5and 3.12, H ∈ G . So ( H, C ) is superextendable. In G ′ , color v j , v j +1 , v j +3 with the color of the resultingvertex, and uncolor u j , u , v and recolor them properly in the order, we get a desired coloring of G ′ . Now v can be properly colored, or N ( v ) contains three different colors. In the latter case, if v is colored with 1 or2, then color v with the color of v ; if v is colored with 3, then color v with 3 and recolor v with the colorof u , a contradiction.(2) Suppose otherwise that at least three vertices in { u i : i ∈ [5] } have degree 3. By symmetry, u i , u i +1 , u i +2 have degree 3 or u i , u i +1 , u i +3 have degree 3 for some i ∈ [5]. We may assume that i = 1. Let d ( u ) = d ( u ) = d ( u ) = 3. Consider H = G [ { v , v , v , v } ]. By Lemma 3.5, H ∈ G . So ( H, C ) is superex-tendable. In G , color v , v , v , v with the color of the resulting vertex and recolor u , u , u properly and fi-nally color v properly, a contradiction. Let d ( u ) = d ( u ) = d ( u ) = 3. Consider H = G [ { v , v , v } , { v , v } ].Let v ′ and v ′′ be the resulting vertices by identifying v , v , v and v , v , respectively. By Lemma 3.5, H ∈ G .So ( H, C ) is superextendable. In G , color v , v , v with the color of v ′ and color v , v with the color of v ′′ and recolor u , u , u properly, and now v can be properly colored, a contradiction.(3)Without loss of generality, We assume that i = 2 and d ( u ) = 3. Let H = G [ { v , v } ] − u and theresulting vertex be v ′ . By symmetry, let j = i − k = i + 2 = 4. Suppose to the contrary that d ( v ) = 3 or d ( v ) = 4 and v is ( u , u )-behaved. By the proof of Lemma 3.12, we can get a desired coloringof H and the color of v is different from the color of v ′ . Then we color v and v with the color of v ′ andcolor u properly, a contradiction. (cid:3) Discharging Procedure
In this section, we will finish the proof of the main theorem by a discharging argument. Let the initialcharge of vertex v ∈ G be µ ( v ) = 2 d ( v ) −
6, and the initial charge of face f = C be µ ( f ) = d ( f ) − ( C ) = d ( C ) + 6. Then X v ∈ V ( G ) µ ( v ) + X f ∈ F ( G ) µ ( f ) = 0 . We will use the following special 4-faces from F in the discharging. • A (3 , , , special if none of the 4-vertices is incident to triangles. • A (3 , , , weak if exactly one of the 4-vertices is incident to a triangle. • A (3 , , , , , , , , , special if the 5-vertices on the face are poor. • A (4 , , , special if the 5-vertex and the 4-vertices adjacent to the 5-vertex are poor. • A (4 , , , special if the 4-vertices and 5-vertices are poor. • A 4-face is rich if it contains two rich 5-vertices or 6 + -vertices.The discharging rules are as follows.(R1) Let v C . Then v gives charges in the following ways:(R1.1) d ( v ) = 4(R1.1.1) If v is rich, then v gives to each incident (3 , , F and 1 to other 3-facesfrom F , to each pendant 3-face from F , 1 to each incident (3 , , , + )-face from F .Furthermore, if v is incident to a triangle, then v gives to its incident 4-face (other than(3 , , , + )-face from F ); if v is not incident to a triangle, then v distributes the remainingcharges only to other incident 4-faces form F evenly.(R1.1.2) If v is poor, then v gives max { , − w ( f ) | Q | } to f , where v is on 4-face f and Q is the set ofpoor 4-vertices on f , and w ( f ) is the weight that f receives from vertices not in Q .(R1.2) d ( v ) = 5(R1.2.1) If v is rich, then v gives 2 to each incident (3 , − , F , and to other incident3-faces from F , to each pendant 3-face from F . Furthermore, if v is incident to a triangle,then v gives 1 to its incident 4-face; if v is not incident to a triangle, then v distributes theremaining charges only to other incident 4-faces form F evenly.(R1.2.2) If v is poor, then v gives 1 to each incident (3 , , , + )-face or (3 , , , , , , to each incident special (3 , , , , , , , , , , , , , , , , , , to each other incident4-face.(R1.3) Each 6 + -vertex gives 2 to each incident 3-face, to each pendant 3-face, and distributes theremaining charges to incident 4-faces evenly.(R2) Each v ∈ C gives to each pendant face from F , 1 to each incident face from F ′′ , to each incidentface from F ′′ or F ′ , and 3 to each incident face from F ′ .(R3) C gives 2 to each 2-vertex on C , to each 3-vertex on C , and 1 to each 4-vertex on C . In addition,if C is a 7-face with six 2-vertices, then it gains 1 from the incident face.The following useful facts are from the rules. Lemma 4.1.
The vertices and faces mentioned in this lemma are disjoint from C .(1) If a -vertex is incident to a triangle, then it gives to each incident (3 , , , + ) - or (3 , , , + ) -face,and at least to each other -face.(2) Each rich -vertex gives at least to each incident -face, and if it is not incident to (3 , , , + ) -face,then it gives at least to each incident -face.(3) Let f = uvwx be a (3 , , , -face with w not incident with a triangle. Then each rich -vertex on b ( f ) gives at least to f .(4) A rich -vertex gives at least to each incident -face. Moreover, if such a 5-vertex is incident to atriangle that is not a (3 , − , -face, then it gives at least two each incident -face. A + -vertex gives t least to each incident -face. Moreover, if such a + -vertex is incident to a triangle, then it givesat least to each incident -face.(5) Let v be a poor -vertex on a -face f . Then v gives at most to each incident (3 , , , + ) -face, at most to each incident (3 , , , -face, at most to a (4 , + , + , + ) -face that is adjacent to a triangle andat most to each other incident -face.Proof. (1) By (R1.1.1), we just need to show that when v is incident to a (3 , , , + ) or a (3 , , , + )-face, v cannot be incident to a (3 , , v be a rich 4-vertex, note that v is incident to at most three 4-faces. Suppose that v is incidentto exactly one 4-face f . So if v is incident to a triangle, then by (R1.1.1), it gives at least 2 − = to f ;if v is not incident to a triangle but adjacent to pendant triangles, then it gives at least 2 − · = 1 to f ;otherwise, v gives at least 2 to f .Let v be incident to exactly two 4-faces. Since G has no 5-cycle, v is not incident to a triangle. If v isnot adjacent to a pendant triangle, then it gives at least 1 to each incident 4-face. Let v be adjacent to apendant triangle. If v is not incident to (3 , , , + )-face, then by (R1.1.1), v gives − = to each 4-face;if v is incident to a (3 , , , + ), then by Lemma 3.11(3), it is incident to exactly one (3 , , , + )-face. By(R1.1.1), v gives 2 − − to the other 4-face.If v is incident to exactly three 4-faces, by Lemma 3.11(3), it is incident to at most one (3 , , , + )-face.If v is incident to a (3 , , , + )-face, then by (R1.1.1), it gives at least − = to each incident 4-face,otherwise, v gives at least to each incident 4-face.(3) By symmetry suppose that v or w is rich 4-vertices. By Lemma 3.11(1) and (4) v or w cannot beincident to a (3 , , , + )-face that share an edge with f since w is not incident to a triangle. By (R1.1.1) v or w gives at least to f .(4) Let v be a rich 5-vertex that is incident to t ≤ s pendant 3-faces. Then v is incidentto at most (5 − t − s −
1) 4-faces. By (R1.2.1), v gives at least − t − s − t − s − ≥ v is incident to a triangle that is not a (3 , − , v gives at least − − s − − s − ≥ to each incident 4-face.Similarly, if v is a t -vertex with t ≥ t ≤ s pendant 3-faces, then v is incident to at most ( t − t − s ) 4-faces. By (R1.3), v gives at least t − − t − st − t − s = t − t − s +( t − t − t − s ≥ t = 1, then v is incident to at most ( t − s −
3) 4-faces. In this case, v gives at least (2 t − − − st − s − ≥ t − t − ≥ to each incident 4-face.(5) First assume that f is a (3 , , , + )-face with d ( x ) ≥
4. If x is also a poor 4-vertex, then by (R1.1.2)both x and v give 1 to f . If x is not a poor 4-vertex, then by (R1.1.1),(R1.2.2) and (4), x gives at least 1 to f . In either case, by (R1.1.2) v gives at most 1 to f .Second, assume that f is a (3 , , , v is poor, the 4-vertex not adjacent to 3-vertex on f isnot incident to a triangle. By (3) and (R1.1.2), v gives at most to f .Next, assume that f = vuwx is a (4 , + , + , + )-face that is adjacent to a triangle. Let w be incident toa triangle. If d ( w ) = 4, then both u and x are rich. By (2), (4) and (R1.2.2), u and by (1) x each gives atleast to f and w gives at least to f . So by (R1.1.2) v gives at most to f . If d ( w ) = 5, then u or x is not poor. We assume, without loss of generality, that x is not poor. By (2)(4) and (R1.2.2), x gives atleast to f . In this case, u and v may be both poor. It follows by (4) and (R1.1.2) that v gives at most − − = to f . If d ( w ) ≥
6, then by (4) w gives at least to f . In this case, each of u, x and v may bepoor. By (R1.1.2), v gives at most − = < to f . Now by symmetry let u be incident to a triangle.Then d ( u ) ≥
5. It follows that either d ( u ) ≥ w is not poor. In the former case, similarly, we can showthat v gives at most < . In the latter case, by (2)(4) and (R1.2.2) w gives at least to f and u gives atleast 1 to f . Note that x may be poor. Thus by (R1.1.2), v gives at most − − = to f . inally, assume that f is a 4-face which is neither (3 , , , + ) nor (3 , , , f is at most two. Since f is not (3 , , , + )-face, the number of 3-vertices on f isat most one. First consider that f contains no 3-vertex. If f is a rich 4-face, then by (4) each of the tworich 5-vertices or 6 + -vertices gives at least 1 to f . In this case, by (R1.1.2) v gives 0 to f . If f is not a rich4-face, then by (2) (4) and (R1.2.2), each of 4 + -vertices on f not in Q gives at least to f , where Q is theset of poor 4-vertices on f . By (R1.1.2), v gives at most − (4 −| Q | ) | Q | = to f .Next consider that f contains one 3-vertex. Since f is not (3, 4, 4, 4), it contains at least one 5 + -vertex.On the other hand, since f contains one 3-vertex and one 4-vertex v , f contains at most two 5 + -vertices.Assume first that f contains exactly two 5 + -vertices. If both 5 + -vertices are rich 5-vertices or 6 + -vertices,by (4), each of them gives 1 to f . By (R1.1.2), v gives 0 to f . If exactly one of 5 + -vertex is poor 5-vertex.Then by (4) and (R1.2.2), the poor 5-vertex gives at least to f and the other 5 + -vertex gives at least 1to f . Thus, by (R1.1.2) v gives at most to f . Thus, we may assume that both of the 5 + -vertices must bepoor 5-vertices. It follows that f is a special (3 , , ,
5) or (3 , , , v givesat most 2 − · = to f .Thus, assume that f contains one 5 + -vertex. It follows that f is a (3 , , , + ) or (3 , , + , + -vertex is not poor 5-vertex, then by (4), it gives at least 1 to f . If the other 4-vertex is rich, then by (2),it gives to f . Thus, by (R1.1.2), v gives at most to f . If the other 4-vertex is poor, then by (R1.1.2)again, v gives at most to f . Thus, we may assume that the 5 + -vertex is a poor 5-vertex. In this case, f isa special (3 , , , , , , , , , v givesat most max { − , − · } = to f . (cid:3) Now we shall show that each x ∈ V ( G ) ∪ F ( G ) other than C has final charge µ ∗ ( x ) ≥ µ ∗ ( C ) > int ( C ). Note that int ( C ) contains no 2 − -vertices by Proposition 3.1. As3-vertices in int ( C ) is not involved in the discharging process, they have final charge 2 · − + -vertices have nonnegative final charges. Thus, we are left with 4-vertices and 5-vertices in int ( C ).In Lemmas 4.2 -4.5, when we discuss the case that v is a poor k -vertex for k = 4 ,
5, we assume that N ( v ) = { v i : i ∈ [ k ] } and f i = vv i u i v i +1 for i ∈ [ k ] be the k incident 4-faces of v (the subscripts of u and v are taken modulo k ). We further assume that v , v , . . . , v k are in the clockwise order in the embedding. Lemma 4.2.
Each -vertex v ∈ int ( C ) has nonnegative final charge.Proof. First suppose that v is rich. Note that when v is incident with a 3-face, it is incident with at mostone 4-face and at most one 3-face, since G has no 5-cycle and intersecting 3-cycle. By Lemma 3.11(3), v is incident to at most one (3 , , , + )-face from F . So by (R1.1.1), v gives out more than 2 only if v isincident to a (3 , , F and a (3 , , , + )-face from F , which is impossible by Lemma 3.9 (1), ora (3 , , F and two pendant 3-faces from F , which is also impossible by Lemma 3.8. So v givesout at most 2, and its final charge is at least 2 · − − v is poor. We distinguish the following two cases. Case 1. N ( v ) has at least two vertices incident to triangles.Assume that N ( v ) has at least three vertices incident to triangles, without loss of generality, that each of v , v , v is incident with a triangle. Since G contains no 5-cycle, d ( v i ) ≥ i ∈ [3]. By Lemma 4.1(4), v i for i ∈ [3] gives at least 1 to each incident 4-face. By (R1.1.2), v gives 0 to f and f , and at most 1 to f and f , respectively. Thus, µ ∗ ( v ) = 2 − · N ( v ) has exactly two verticesincident with triangles.First let the two vertices be v and v . By Lemma 4.1(4), f gets at least 2 from v and v . By (R1.1.2), v gives 0 to f . Since only each of v and v is incident with a 3-face, v is ( v , v )-behaved and ( v , v )-behaved. y Lemma 3.11(4) none of f i with i ∈ [4] is a (3 , , , + )-face. Thus v gives at most to each of f , f and f by Lemma 4.1(5). Thus, µ ∗ ( v ) ≥ − · = 0.Then, by symmetry let the two vertices be v and v . Since G has no 5-cycle, d ( v ) ≥ d ( v ) ≥
5. Itfollows that none of f i for i ∈ [4] is a (3 , , , , , , + )-face, then by Lemma 4.1(5), v gives at most to each f i . Thus, µ ∗ ( v ) ≥ − · = 0. So we may assume that f is a (3 , , , + )-face,i.e., d ( u ) = d ( v ) = 3. By Lemma 3.7, d ( u ) ≥
4. By Lemma 3.11(1) and (2), d ( u ) , d ( u ) ≥
4. This impliesthat only one of f i , where i ∈ [4], is a (3 , , , + )-face.Let d ( v ) ≥
4. By Lemma 4.1 (5), v gives at most to each of f and f , at most 1 to f and to f .Thus, µ ∗ ( v ) ≥ − − − · = 0.Let d ( v ) = 3. By Lemma 4.1(5), v gives at most to f and f , respectively. If d ( v ) = 5, then u is richsince G is 5-cycle free. By Lemma 4.1(2), u gives at least to f . Note that the 5-vertex v is incident toa 3-face and two 4-faces, d ( v ) = d ( v ) = 3 and at most one vertex in { u , v, u } is incident with a triangle.By Lemma 3.9 (2), the triangle incident with v cannot be a (3 , − , v gives atleast to each of f and f . Thus, v gives at most 2 − − = to f and 2 − = to f . If d ( v ) ≥ v gives at least to each of f and f . Thus, v gives at most 2 − = to f andat most − = to f . Therefore, µ ∗ ( v ) ≥ − · − max { + , + } = 0. Case 2. N ( v ) has at most one vertex incident with a triangle.In this case, v is ( v , v )-behaved and ( v , v )-behaved. It follows by Lemma 3.11(4) that no 4-facesincident to v is a (3 , , , + )-face. On the other hand, if v is not incident to a (3 , , , v gives at most to each incident 4-face. Thus µ ∗ ( v ) ≥ − · = 0. Therefore, we mayassume that v is incident to a (3 , , , f such that d ( u ) = 3 or d ( v ) = 3. Claim.
We may assume that none of f , f , f is a (3 , , , Proof of Claim.
We may assume that d ( v ) = 3. For otherwise, let d ( u ) = 3. Then by Lemma 3.11(1)and (2), d ( u i ) ≥ i = 2 , ,
4. Since d ( v ) = 4 and v is ( u , u )-behaved, by Lemma 3.12, d ( v ) ≥ d ( v ) = 4 and v is ( u , u )-behaved implies that d ( v ) ≥
4. Thus, each f i is a (4 , + , + , + )-facefor i = 2 , , d ( v ) ≥ d ( v ) ≥
4. Moreover, since d ( v ) = 3 and v is a poor4-vertex and ( v , v )-behaved, by Lemma 3.12 either d ( v ) = 4 and v is not ( u , u )-behaved or d ( v ) ≥ f and f is a (3 , , , f is a (3 , , , µ ∗ ( v ) ≥ v is ( v , v )-behaved, and d ( v ) = d ( v ) = d ( v ) = 4, by Lemma 3.12, v is not ( u , u )-behavedor v is not ( u , u )-behaved. By symmetry, we assume that v is not ( u , u )-behaved. This means thateach of { u , u } is incident to a triangle. So f is a (4 , + , + , + )-face that is adjacent to a triangle. So byLemma 4.1(5) v gives at most to f . As d ( u ) = 4 and u is incident to a triangle, by Lemma 4.1(1) u gives at least to f and v has at most three incident 4-faces. By Claim 4.1(3) v gives at least to f .So by (R1.1.2), v gives at most 2 − − = to f . Note that v gives at most to f and to f byLemma 4.1(5). Thus µ ∗ ( v ) ≥ − − − − = 0. This proves our claim.Now we are ready to complete our proof. By Lemma 4.1, v gives at most to f and to each of f and f . In order to show that µ ∗ ( v ) ≥
0, we just need to show that v gives at most to f .We may assume that d ( v ) ≥
5. Note that d ( v ) = 3, or if d ( u ) = 3, then u is not incident with atriangle by Proposition 3.1(c) and hence v is ( u , u )-behaved. It follows by Lemma 3.12 that d ( v ) = 4and v is not ( u , u )-behaved or d ( v ) ≥
5. But in the former case, that means both u and u are incidentto triangles. By Lemma 4.1(5), v gives at most to f . Therefore, we may assume the latter is true, thatis, d ( v ) ≥ ssume first that v is a poor 4-vertex. Then the four 4-faces incident to v are f , f , f and f , where f = v v ′ u ′ u and f = v v ′ u ′ u . As d ( u ) = 3 or d ( u ) = 4 and u is ( u ′ , v )-behaved, and v is ( v, v ′ )-behaved, by Lemma 3.12 d ( u ) = 4 and u is not ( u ′ , v )-behaved or d ( u ) ≥
5. In the former case, since v is incident with a triangle and d ( v ) ≥
5, by Lemma 4.1(4), f gains at least 1 from v ; If u is a poor4-vertex, By (R1.1.2), v gives at most − = to f ; If u is rich, u gives at least to f , thus by (R1.1.2), v gives at most − − = to f . In the latter case, if at least one of u and v is a rich 5-vertex or 6 + -vertex,then by (R1.2.2) and Lemma4.1(4) v gives at most − − = to f ; thus, we may assume that both u and v are poor 5-vertices, but it follows that f is a special (4 , , , v gives − · = to f .Now we assume that v is a rich 4-vertex. Then v is incident to at most three 4-faces.We first show that v cannot be incident to a (3 , , , + )-face. Suppose otherwise that v is incident tosuch 4-face. Note that f and f are not (3 , , , + )-face. Thus assume that v is incident to a (3 , , , + )-face f that share an edge with f or f . Let N ( v ) = { u , v, u , v ′ } . If d ( u ) = 3, then v is ( u , u )-behavedand ( v, v ′ )-behaved, thus by Lemma 3.11(4) f cannot be a (3 , , , + )-face, a contradiction. If d ( v ) = 3,then d ( u ) = 4, thus by Lemma 3.11(1) and (2), f cannot be a (3 , , , + )-face, a contradiction.Thus by Lemma 4.1(2), v gives at least to f . Now we consider the degree of u . Recall that d ( v ) ≥ v is a poor 4-vertex. If u is a 3-vertex, then by Lemma 4.1(4) or (R1.2.2), v gives at least 1 to f , thusby (R1.1.2), v gives at most 2 − − = to f . If u is a rich 4-vertex or d ( u ) ≥
5, then by Lemma 4.1(2)(4)and (R1.2.2), v gives at most 2 − · − = to f . Finally let u be a poor 4-vertex. If v is a poor 5-vertex,then f is a special (4 , , , v gives to f ; If v is not a poor 5-vertex, then Lemma4.1(4), v gives at least 1 to f . Thus, by (R1.1.2) v gives at most max { − − , − − } = ≤ to f . (cid:3) In order to prove that 5-vertices have nonnegative charges (Lemma 4.5), we first handle two special casesin Lemmas 4.3 and 4.4.
Lemma 4.3.
Suppose that v is a poor 5-vertex and N ( v ) has no vertex incident to a triangle. If f i = u i v i +1 vv i is a (3 , , , -face, then µ ∗ ( v ) ≥ .Proof. By symmetry, let i = 1. First we show that none of f and f is a special (4 , , , , , , f is a special (4 , , , , , , , , , , , , v is poor and d ( v ) = d ( u ) = 4. By Lemma 3.12, v must be incident to atriangle, a contradiction. It follows that if f (or f ) is a (4 + , + , + , v gives at most to it.By Lemma 3.13 (2), at most two vertices in u i with i ∈ [5] are 3-vertices. Since d ( u ) = 3, at most oneof u and u is a 3-vertex. By symmetry we consider the following two cases.Assume first that d ( u ) ≥ d ( u ) ≥
4. If min { d ( v ) , d ( v ) } ≥
4, then v gives at most to f and f and at most 1 to each other incident 4-face, thus µ ∗ ( v ) ≥ − · − · d ( v ) = 3. By Lemma 3.6(2) d ( v ) ≥
4. Since d ( u ) = 3, by Lemma 3.13(1), d ( u ) ≥
4. We claimthat d ( u ) ≥
4, for otherwise, since d ( v ) = 3, by Lemma 3.13(3) d ( v ) = 4 and v is not ( u , u )-behaved,or d ( v ) ≥
5, which is contrary to our assumption that d ( v ) = 4 and v is ( u , u )-behaved(note that u cannot be incident to a triangle). Since d ( v ) = 3, applying Lemma 3.13 (3) to u , we get d ( v ) = 4 and v isnot ( u , u )-behaved or d ( v ) ≥
5. In the former case, f is a (4 , , + , u incident to a triangleand f is a (4 , , + , + )-face with u incident to a triangle, then by (R1.2.2), v gives at most to each of f and f , thus, µ ∗ ( v ) ≥ − · − · = 0. Consider the latter case now. As the argument above, v gives atmost to f . Note that f is a (4 + , + , + , f is not special (4 , , , v gives at most to f , and it follows that µ ∗ ( v ) ≥
0; thus, we may assume that f is a special (4 , , , v and u are both poor 4-vertices. Since v and v are not incident to triangles, applying emma 3.12 to v , we have d ( u ) ≥
5, so f is a (3 , , , + )-face. By (R1.2.2), v gives at most to each of f and f , so µ ∗ ( v ) ≥ − · − · − = 0.Assume now by symmetry that d ( u ) = 3 and d ( u ) ≥
4. By Lemma 3.13(1), d ( v ) ≥
4. By Lemma 3.13(2) d ( u i ) ≥ i = 2 , ,
4. Since d ( v ) = 4 and v is ( u , u )-behaved, by Lemma 3.13 (3) (with i = 5), weget d ( v ) ≥ d ( v ) = 4 and v is not ( u , u )-behaved. If d ( v ) ≥
4, then v gives at most to each of f and f and to f by (R1.2.2), thus µ ∗ ( v ) ≥
0. So let d ( v ) = 3. By Lemma 3.13 (3) (with i = 1), we get d ( v ) ≥ d ( u ) = 3, v is not ( u , u )-behaved). Since d ( v ) ≥ d ( v ) = 4 but both u and u areincident to triangles, f is a (4 + , + , , + )-face but not a special (4 , , , f is a (3 , + , + , , , , v gives at most to f and to each f and f . So µ ∗ ( v ) ≥ − · − · − = 0. (cid:3) Lemma 4.4.
Suppose that v is a poor 5-vertex and N ( v ) has no vertex incident to a triangle. If f i = v i u i v i +1 v is a special (3 , , , -face, then µ ∗ ( v ) ≥ .Proof. By symmetry, let i = 1. By Lemma 4.3 we may assume that v is not incident to a (3 , , , d ( v ) = 3, by Lemma 3.6(2) d ( v ) ≥
4. Since f is a special (3 , , , u is not incident to atriangle, thus at most one vertex in { u , v, u } is incident to a triangle, so by applying Lemma 3.11(1) to v , d ( v ) ≥ f is neither a special (4 , , , , , , v is poor and d ( v ) = d ( u ) = d ( u ) = 4, and none of v , v, v is incident to a triangle, a contradictionto Lemma 3.12. It follows that if f is a (4 , , + , + )-face, then by (R1.2.2), v gives at most to f .We may also assume that d ( u ) ≥
4. Suppose otherwise that d ( u ) = 3. By Lemma 3.13 (1) d ( u i ) ≥ i = 2 , ,
4. Since f is a special (3 , , , u is not incident with a triangle. Applying Lemma 3.13(3) to u , d ( v ) ≥
4, so by (R1.2.2), v gives at most to each f and f . Note that v gives at most 1 / f , as it is a (4 , , + , + )-face. But now µ ∗ ( v ) ≥ − · − · − = 0.Now we consider the following four cases depending on the degree of u and u .Let d ( u ) = d ( u ) = 3. By Lemma 3.7 d ( v ) ≥
4. By Lemma 3.13(3) (with i = 4) d ( v ) ≥ i = 3) d ( v ) ≥
5. It follows that for i ∈ { , , , } , f i is a (3 + , + , , + )-face, so by (R1.2.2), v gives at most to f i . Thus, µ ∗ ( v ) ≥ − − · = 0.Let d ( u ) = 3 and d ( u ) ≥
4. Since d ( v ) = 4 and v is ( u , u )-behaved, by Lemma 3.13(3) (with i = 3),either d ( v ) = 4 and v is not ( u , u )-behaved or d ( v ) ≥
5, then f and f are (3 , + , + , , , , v gives at most to each of f and f . If d ( u ) ≥
4, then v gives atmost to f which is a (4 , , + , + )-face, so µ ∗ ( v ) ≥ − · − · − = 0. Thus, we assume that d ( u ) = 3.As f cannot be a (3 , , , d ( v ) ≥
5, so f and f are (3 , + , , + )-faces, then by (R1.2.2), v givesat most to each of f and f . We conclude that µ ∗ ( v ) ≥ − − · = 0.Let d ( u ) ≥ d ( u ) = 3. By Lemma 3.13(3) (with i = 4), d ( v ) = 4 and v is not ( u , u )-behaved or d ( v ) ≥
5. In the former case, f is a (4 , , , + )-face with u incident to a triangel and f is a (3 + , + , , u incident to a triangle; In the latter case, each of f and f is a (3 + , + , , + )-face; so by (R1.2.2) v gives at most to each of f and f . If d ( u ) = 3, then by Lemma 3.13 (3) (with i = 2), d ( v ) ≥
5, thus f is a (3 , , , + )-face, f is a (4 + , + , , + )-face and f is a (3 , + , , + )-face, so by (R1.2.2), v gives atmost to f and at most to each f and f , therefore, µ ∗ ( v ) ≥ − · − · − = 0. Now we assumethat d ( u ) ≥
4. If d ( v ) = 4, then at most one of { u , v, u } is incident with a triangle, so by applyingLemma 3.11 (1) to v , we have d ( v ) ≥
4; as v is not incident to (3 , , , d ( v ) ≥
5. Now, f is a (4 , , + , + )-face, f is a (4 + , + , , + )-face, and f is a (3 , , , + )-face, so by(R1.2.2), v gives at most to f and to each of f and f . It follows that µ ∗ ( v ) ≥ − · − · − = 0.If d ( v ) ≥
5, then by (R1.2.2), v gives at most to each of f and f , and gives at most to f which is a(4 , , + , + )-face. We conclude that µ ∗ ( v ) ≥ − · − · − = 0. e are left to consider the case that d ( u ) ≥ d ( u ) ≥ d ( u ) ≥
4. Note that v gives to f which is a (4 , , + , + )-face. If d ( v ) ≥
4, theneach of f and f is a (4 + , + , + , v gives to each of f and f , it follows that µ ∗ ( v ) ≥ − · − · − = 0. So let d ( v ) = 3. Then both f and f are (3 , + , + , d ( v ) = 4and v is ( u , u )-behaved, then at most one of { u , u , v } is incident with a triangle, and d ( v ) = d ( v ) = 3,a contradiction to Lemma 3.11(1). This means either d ( v ) = 4 and both u and u are incident to trianglesor d ( v ) ≥
5. Thus, none of f and f is a special (3 , , , v gives at most to each of f and f . Thus, we also have µ ∗ ( v ) ≥ − · − · − = 0.Thus, we may assume that d ( u ) = 3. By Lemma 3.13 (3) (with i = 2), either d ( v ) = 4 and v is not( u , u )-behaved or d ( v ) ≥
5. In the former case, both f and f are (4 , + , + , , , , , , , u and u are incident with triangles; by (R1.2.2), v give atmost to f and f , thus, µ ∗ ( v ) ≥ − − · = 0. So consider the latter case that d ( v ) ≥
5. We claimthat f is not a special (4 , , , d ( v ) = d ( u ) = 4 and v is poor, but none of v and v is incident to triangles, and d ( u ) = 3, a contradiction to Lemma 3.12. It follows by (R1.2.2) that v gives at most to f . If f is not a special (4 , , , v gives at most to f , whichimplies that µ ∗ ( v ) ≥ − − · = 0. Thus, we assume that f is a special (4 , , , d ( u ) = d ( v ) = 4 and v is poor. By Lemma 3.12, d ( u ) ≥
5. It follows that f is a (3 , , , + )-face. By(R.1.2.2), v gives at most to each of f and f . Therefore, µ ∗ ( v ) ≥ − · − · − = 0. (cid:3) Lemma 4.5.
Each -vertex v ∈ int ( C ) has nonnegative final charge.Proof. If v is rich, then by (R1.2.1), v gives at most 2 + max { · , · } = 4 to incident triangles and pendant3-faces and incident 4-faces, thus its final charge must be nonnegative. Thus, we may assume that v is poor.We may further assume that some vertex in N ( v ) is incident to a triangle. Suppose otherwise. ByLemmas 4.3 and 4.4, we may assume that v is not incident to a (3 , , , , , , v is not incident to a (3 , , , + )-face, then by (R1.2.2), µ ∗ ( v ) ≥ − · >
0, so by symmetry, we assumethat f = u v vv is a (3 , , , + )-face. By Lemma 3.6(2), d ( v ) ≥
4. For i ∈ { , , , } , d ( u i ) ≥ f i cannot be a (3 , , , + )-face, so by (R1.2.2), v gives at most 3 / f i . It followsthat µ ∗ ( v ) ≥ − − × Case 1. N ( v ) has at least two vertices incident to triangles.If v i and v i +1 for some i ∈ [5] are incident to triangles, then f i is rich and by (R1.2.2) v gives 0 to f i andat most 1 to each other 4-face. Thus, µ ∗ ( v ) ≥ − v and v are incident with triangles. If d ( v ) ≥
4, then by (R1.2.2), v gives at most to each of f and f , andgives at most 1 to each of f , f and f , so µ ∗ ( v ) ≥ − · − · = 0. Thus, we may assume that d ( v ) = 3.If min { d ( u ) , d ( u ) } ≥
4, then each of f and f is a (3 , + , , + )-face but not a special (3 , , , v gives at most to each of f and f , therefore, µ ∗ ( v ) ≥ − · − · = 0. Thus, by symmetry,assume that d ( u ) = 3. By Lemma 3.7 d ( u ) ≥
4. Note that v gives at most to f . If one of f and f , say f , is not (3 , , , + )-face, then f is a (3 , , , + )-face, so by (R1.2.2), v gives to f , therefore, µ ∗ ( v ) ≥
0. Thus, we may assume that f and f are (3 , , , + )-faces. It follows that d ( v ) = d ( v ) = 3. ByLemma 3.6(2), each of u and v is incident with a triangle, a contradiction. Case 2. N ( v ) has exactly one vertex incident to a triangle.We assume, without loss of generality, that v is incident with a triangle. If neither f nor f is a(3 , , , + )-face, then by (R1.2.2), v gives at most to each of them. This implies that µ ∗ ( v ) ≥ − · − · =0. Thus, by symmetry we may assume that f is a (3 , , , + )-face. It follows that d ( u ) = d ( v ) = 3. ByLemma 3.13(1), d ( u i ) ≥ i ∈ [4]. By Lemma 3.6(2), d ( v ) ≥
4. By (R1.2.2) v gives at most to f . e may assume that d ( v ) = 4, for otherwise, both f and f are (3 + , + , + , v gives at most to each of them, so µ ∗ ( v ) ≥ − − · − · v ,we get either both u and u are incident to triangles or d ( v ) ≥
4. In the former case, none of f and f is aspecial (3 , , , v gives at most to each of them, so µ ∗ ( v ) ≥ − − · − · d ( v ) ≥
4. If f is neither a special (4 , , , , , , v gives at most to f , so µ ∗ ( v ) ≥ − · − · = 0. Thus, we may assume that f is aspecial (4 , , , , , , v gives at most to f . By the definition ofspecial (4 , , , , , , v is poor and d ( v ) = d ( u ) = 4. Note that no vertex in { v , v, v } is incident to a triangle. By Lemma 3.12, d ( u ) ≥
5. So f is a (3 , , , + )-face and by (R1.2.2) v gives atmost to f . Thus, µ ∗ ( v ) ≥ − (2 · · + ) = 0. (cid:3) Now we consider the case v ∈ C . Lemma 4.6.
Each v ∈ C has nonnegative final charge.Proof. We consider the following cases according to the degree of v . For l = 3 ,
4, by Lemma 3.4 each l -face f in G satisfies that | b ( f ) ∩ C | ≤ | b ( f ) ∩ C | = 2, f and C share a common edge.(1) d ( v ) = 2. By (R3), µ ∗ ( v ) = 2 × − d ( v ) = 3. Then v could be incident with at most one triangle from F ′′ or has at most one pendant 3-facefrom F ′ . By (R2) and (R3), µ ∗ ( v ) ≥ × − − + = 0.(3) d ( v ) = 4. Assume first that v is incident with a 3-face f . If f ∈ F ′ , then by (R2) and (R3), µ ∗ ( v ) =2 − f ∈ F ′′ , then it could be incident to at most one 4-face from F ′′ or adjacent to at mostone pendent 3-face from F . By (R2) and (R3), µ ∗ ( v ) ≥ − − >
0. Thus, we may assume v is not incident to a 3-face. By Lemma 3.6 (1), v is not incident face from F ′ . Thus, we assume that v is incident with k ≤ F ′′ . Then v is adjacent to at most 2 − k pendent 3-faces from F .By (R2) and (R3), µ ∗ ( v ) ≥ − k − (2 − k ) ≥ d ( v ) = k ≥
5. If v is not incident with any 3-face, then by Lemma 3.6, v is not incident face from F ′ , so by(R2), µ ∗ ( v ) ≥ k − − · ( k − ≥ >
0. Thus, we first assume that v is incident with a face from F ′ . Let s be the number of 4-faces in F ′ incident with v . If s = 0, then by (R2), µ ∗ ( u ) ≥ k − − ( k − − ≥ s ≥
1, then s ≤ k −
5. By (R2), µ ∗ ( v ) ≥ k − − − s − ( k − s −
4) = k − s − ≥ . Next, weassume that v is incident with a face from F ′′ . If s = 0, then by (R2), µ ∗ ( v ) ≥ k − − ( k − − ≥ ;if s ≥
1, then s ≤ k −
4. By (R2), µ ∗ ( v ) ≥ k − − s − − ( k − s −
3) = k − s − ≥ s − ≥ (cid:3) Then we consider faces. As G contains no 5-faces, and 6 + -faces other than C are not involved in thedischarging procedure, we only need to show that C , and 3-faces and 4-faces other than C have nonnegativecharges. Lemma 4.7.
Each -face f = C has nonnegative final charge.Proof. Note that f has initial charge 3 − −
3. By Lemma 3.4 | b ( f ) ∩ C | ≤
2. If | b ( f ) ∩ C | = 1, then by(R2), µ ∗ ( f ) ≥ − | b ( f ) ∩ C | = 2, then by (R2), µ ∗ ( f ) ≥ − × b ( f ) ∩ C = ∅ . Let f = uvw with corresponding degrees ( d , d , d ). Let x ′ be the pendant neighbor of x on a 3-face f. By Lemma 3.8 (1), we only need to check the following cases:(1) f is a (3 , , + )-face. By Lemma 3.8 (2), u ′ and v ′ are either on C or have degree at least 4. By (R1.1.1)and (R1.2.1), f receives from each of u ′ and v ′ . By (R1.2.1) and (R1.3), f receives 2 from w . Thus, µ ∗ ( f ) = − × f is a (3 , , u ′ is either on C or has degree at least 4. By (R1.1.1)and (R1.2.1), f receives from u ′ . By (R1.1.1), f receives from each of v and w . Thus, µ ∗ ( f ) = − × = 0.(3) f is a (3 , , f receives 1 from v and 2 from w . Thus, µ ∗ ( f ) = − f is a (3 , , f receives from each of v and w . Thus, µ ∗ ( f ) ≥ − × f is a (3 , + , + )-face. By (R1.1.1), (R1.2.1) and (R1.3), f receives at least 1 from v and 2 from w .Thus, µ ∗ ( f ) ≥ − f is a (4 + , + , + )-face. By (R1.1.1),(R1.2.1) and (R1.3), f receives at least 1 from each of u, v and w .Thus, µ ∗ ( f ) ≥ − × (cid:3) Lemma 4.8.
Each -face f = C has nonnegative final charge.Proof. Let f = uvwx with corresponding degrees ( d , d , d , d ). Note that f has initial charge 4 − − | b ( f ) ∩ C | ≤
2. If | b ( f ) ∩ C | = 1, say u ∈ b ( f ) ∩ C , then by (R2), u gives to f ; ByLemma 3.6 each of u and w is incident to a triangle, so d ( w ) ≥ w gives at least min { , , − } = to f ; So µ ∗ ( f ) ≥ − + >
0. If | b ( f ) ∩ C | = 2, then by (R2), µ ∗ ( f ) ≥ − × b ( f ) ∩ C = ∅ . If some vertex on b ( f ) is poor 4-vertex, thenby (R1.1.2), the poor 4-vertex will give enough charges to f to make its final charge to be 0. So we assumethat each 4-vertex on b ( f ) is rich. By Lemma 3.6 (2), we only need to consider the following 4-faces.(1) f is a (3 , , + , + )-face. By (R1.1.1), (R1.2) and (R1.3), each 4 + -vertex gives at least 1 to f . Thus, µ ∗ ( f ) ≥ − × f is a (3 , + , , + )-face. Then by Lemma 3.6(2), and both v and x are incident to a triangle. ByLemma 4.1 (1), f receives at least 1 from each of v and x , so µ ∗ ( f ) ≥ − × f is a (3 , , , w is not incident to a triangle, then by Lemma 4.1(3), each of the rich 4-verticesgives at least to f . Thus, µ ∗ ( f ) ≥ − · = 0. Let w be incident to a triangle f . Note that w is richand none of v and x is poor. If f is a (3 , , v and x is incidentwith a triangle. In this case, by Lemma 4.1(1), each of v, w and x gives at least to f . This impliesthat µ ∗ ( f ) ≥ − · >
0. Thus, assume that f is not a (3 , , w gives at least 1 to f and each of v and x gives at least to f . Thus, µ ∗ ( f ) ≥ − · = 0.(4) f is a (3 , , , + )-face. First we assume that x is a 6 + -vertex or x is a rich 5-vertex, then by Lemma 4.1(2) and (4), f receives at least 1 from x and from each of v and w , thus µ ∗ ( f ) ≥ − × x is a poor 5-vertex. If none of the two 4-vertices is incident to a triangle, then f isa special (3 , , , f receives at least 1 from x and from eachof v and w . If both of the two 4-vertices are incident to triangles, then by Lemma 4.1 (2) and (R1.2.2), f gets at least from each of v and w and from x . If exactly one of the two 4-vertices (say v ) isincident to a triangle, then f is a weak (3 , , , f receives atleast from each of v and x and from w . In both cases, µ ∗ ( f ) ≥ − max { · , · } = 0.(5) f is a (3 , , + , f receives at least 1 from w and fromeach of v and x . Thus, µ ∗ ( f ) ≥ − × f is a (3 , , + , + )-face or (3 , + , , + )-face. By (R1.2.2) and Lemma 4.1 (2) and (4), f receives at least from each of the two 5 + -vertices, and from the 4-vertex. Thus, µ ∗ ( f ) = − · = 0.(7) f is a (3 , + , + , + )-face. If at least one vertex is a rich 5-vertex or a 6 + -vertex, then by Lemma 4.1(4)and (R1.2.2), f gets at least 1 from the vertex and at least from each of the other 5-vertices. Itfollows that µ ∗ ( f ) ≥ − · = 0. Thus, we may assume that all are poor 5-vertices. In this ase, by (R1.2.2), f is special (3 , , , f receives from each of the 5-vertices. Thus, µ ∗ ( f ) ≥ − · > f is a (4 + , + , + , + )-face. If f is rich, then f contains at least two rich 5 + vertices or 6 + -vertices. ByLemma 4.1(4), µ ∗ ( f ) ≥ − · f is not rich. By Lemma 4.1(2) eachrich 4-vertex gives at least to f . By Lemma 4.1(4) and (R1.2.2) each 5 + -vertex gives at least to f . Note that each 4 + -vertex on f is not poor 4-vertex. Thus, f receives at least from each vertex on b ( f ). So µ ∗ ( f ) ≥ − · = 0. (cid:3) Now we consider the outer-face C . Let t i be the number of i -vertices on C , then d ( C ) ≥ t + t + t .Note that d ( C ) ∈ { , } . By (R3), µ ∗ ( C ) = d ( C ) + 6 − t − t − t ≥ d ( C ) + 6 −
32 ( t + t + t ) − t ≥ d ( C ) + 6 − d ( C ) − t − d ( C )2 − t . If d ( C ) = 3 or t ≤
5, then µ ∗ ( C ) ≥
0. Thus, we may assume that d ( C ) = 7 and ( t , t , t ) ∈{ (6 , , , (7 , , } . If t = 7, then G = C and it is trivially superextendable. If t = 6 and t = 1, then by(R3), C gains 1 from the adjacent face which has degree more than 7. Thus, µ ∗ ( C ) ≥ > d ( C ) = 7 and t = 5 and t = 2(the two 3-vertices must be adjacent andhas a common neighbor not on C ) in which there must be a face other than C having degree more than 7.Thus the face has positive final charge. Therefore, P x ∈ V ( G ) ∪ F ( G ) µ ∗ ( x ) >
0, a contradiction.
References [1] O. V. Borodin, Colorings of plane graphs: A survey.
Discrete Math. , (2013) 517–539.[2] O. V. Borodin and A. N. Glebov, A sufficient condition for planar graphs to be 3-colorable, Diskret Anal Issled Oper. (2004) 3–11 (in Russian)[3] O. V. Borodin and A. N. Glebov, Planar graphs with neither 5-cycles nor close 3-cycles are 3-colorable, J. Graph Theory , (2011), 1–31.[4] O. V. Borodin, A. N. Glebov, A. R. Raspaud, and M. R. Salavatipour. Planar graphs without cycles of length from 4 to 7are 3-colorable. J. of Combin. Theory , Ser. B, (2005), 303–311.[5] O. V. Borodin and A. Raspaud, A sufficient condition for planar graphs to be 3-colorable, J. Combin. Theory , Ser B, (2003), 17–27.[6] G. Chang, F. Havet, M. Montassier, and A. Raspaud, Steinberg’s Conjecture and near colorings, manuscript.[7] Z. Dv¨or´ak, D. Kr´al and R. Thomas, Coloring planar graphs with triangles far apart, Mathematics ArXiV, arXiv:0911.0885,2009.[8] H. Gr¨otzsch, Ein dreifarbensatz f u r dreikreisfreienetze auf der kugel. Math.-Nat.Reihe , (1959), 109–120.[9] I. Havel, On a conjecture of Grunbaum, J. Combin. Theory , Series B, (1969) 184–186.[10] O. Hill, D. Smith, Y. Wang, L. Xu, and G. Yu, Planar graphs without 4-cycles and 5-cycles are (3 , , DiscreteMath. , (2013) 2312–2317.[11] O. Hill, and G. Yu, A relaxation of Steinberg’s Conjecture, SIAM J. of Discrete Math. , (2013) 584–596.[12] R. Liu, X. Li, and G. Yu, A relaxation of the Bordeaux Conjecture, http://arxiv.org/abs/1407.5138. Submitted.[13] R. Steinberg, The state of the three color problem. Quo Vadis, Graph Theory?, Ann. Discrete Math. (1993), 211–248.[14] B. Xu, A 3-color theorem on plane graph without 5-circuits, Acta Math Sinica , (2007) 1059–1062.[15] B. Xu, On (3 , ∗ -coloring of planar graphs, SIAM J. Disceret Math. , (2008), 205–220.[16] L. Xu, Z. Miao, and Y. Wang, Every planar graph with cycles of length neither 4 nor 5 is (1 , , J Comb.Optim. , DOI 10.1007/s10878-012-9586-4.[17] L. Xu and Y. Wang, Improper colorability of planar graphs with cycles of length neither 4 nor 6 (in Chinese),
Sci SinMath , (2013), 15-24.[18] C. Yang, and C. Yerger, The Bordeaux 3-color Conjecture and Near-Coloring, preprint. Department of Mathematics, Huazhong Normal University, Wuhan, 430079, China
E-mail address : [email protected] ‡ Department of Mathematics, The College of William and Mary, Williamsburg, VA, 23185, USA.
E-mail address : [email protected]@wm.edu