A 3-color Theorem on Plane Graphs without 5-circuits

Acta Mathematica Sinica, English Series Jun., 2007, Vol. 23, No. 6, pp. 1059–1062 Published online: Sep. 20, 2006 DOI: 10.1007/s10114-005-0851-7 Http://www.ActaMath.com

A 3-color Theorem on Plane Graphs without 5-circuits Bao Gang XU School of Math. & Computer Science, Nanjing Normal University, Nanjing, 210097, P. R. China E-mail: [email protected] Abstract In this paper, we prove that every plane graph without 5-circuits and without triangles of distance less than 3 is 3-colorable. This improves the main result of Borodin and Raspaud [Borodin, O. V., Raspaud, A.: A sufficient condition for planar graphs to be 3-colorable. Journal of Combinatorial Theory, Ser. B, 88, 17–27 (2003)], and provides a new upper bound to their conjecture. Keywords plane graph, circuit, coloring MR(2000) Subject Classification 05C15, 05C78

1 Introduction In 1976, Steinberg (see [1, p. 229] or [2]) conjectured that every plane graph without 4- and 5-circuits is 3-colorable. In 1990, Erd¨ os (also see [1, p. 229]) suggested the following relaxation: Is there an integer k ≥ 5 such that every plane graph without i-circuits for 4 ≤ i ≤ k is 3colorable? Abbott and Zhou [3] showed that k = 11 is acceptable. Sanders and Zhao [4], and Borodin [2] independently, improved that to k = 9. Borodin, et al. [5] again improved this result to k = 7. Recently, we proved [6] that every plane graph without 5- and 7-circuits and without adjacent triangles is 3-colorable. The distance between triangles in a graph is defined as the length of the shortest path between vertices of different triangles. In [7], Borodin and Raspaud proved that if G is a plane graph without 5-circuits and without triangles of distance less than four, then G is 3-colorable, and they conjectured that every plane graph without 5-circuits and without intersecting triangles is 3-colorable. In [8], Xu proved that if a plane graph contains no intersecting triangles and contains no 4-, 5- and 7-circuits, then it is 3-colorable. Let G be the set of plane graphs without 5-circuits and without triangles of distance less than 3. In this paper, we will prove that Theorem 1 Let G be a plane graph in G , f a face of degree 3 or 7. Then, any 3-coloring of f can be extended to G, where a 3-coloring of f is a 3-coloring of the subgraph induced by the vertices on the boundary of f . As a corollary of Theorem 1, every plane graph in G is 3-colorable. To see this, let G be a plane graph in G . By Gr¨ otzsch’s theorem, G is 3-colorable while G contains no triangles. If G contains a triangle T , then we may assign T a 3-coloring φ and then extend φ to both the interior and exterior of T , that also produces a 3-coloring of G. Let G be a plane graph. We use N (v) and d(v) to denote the neighbor set and degree of a vertex v, respectively. Let f be a face of G. b(f ) is the boundary of f in G, and V (f ) is the set of vertices on b(f ). The degree of f , denoted by d(f ), is the number of edges incident with it, where every cut-edge is counted twice. A k-vertex (k-face) is a vertex (face) of degree k, and a k≥ -vertex (k≥ -face) is a vertex (face) of degree at least k. Let C be a circuit of G. We use int(C) and ext(C) to denote the sets of vertices lying inside and outside C, respectively. C is called a separating circuit if both int(C) = ∅ and ext(C) = ∅, and a facial circuit, otherwise. Received November 30, 2004, Accepted July 14, 2005 Supported partially by NSFC 10001035 and 10371055

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Let u and v be two vertices on C. C[u, v] is the path on C clockwise from u to v, and C(u, v) = C[u, v] \ {u, v}. 2 Structures of a Minimum Counterexample In this section, we always assume that G is a counterexample of Theorem 1 with minimum σ(G) = |V (G)| + |E(G)|, i.e., G ∈ G , G contains a face f of degree 3 or 7 with a 3-coloring that cannot be extended to G, and σ(G) is as small as possible. Without loss of generality, we may assume that f is the unbounded face fo of G. Let C = b(fo ), and suppose φ is a 3-coloring of fo that cannot be extended to G. Then, d(u) ≥ 3 for u ∈ V (G) \ V (C). Since G contains neither 5-circuits nor triangles of distance less than three, and G \ int(C  ) is still in G for any separating circuit C  of G, the following two lemmas immediately follow from the choice of G. Lemma 1 G contains neither separating triangles nor separating 7-circuits. Lemma 2 No 4-face is adjacent to 3-faces. Lemma 3 G is 2-connected. Proof Assume to the contrary that G has a cut vertex x. Let H be a component of G − x ¯ be the subgraph induced by V (H) ∪ {x}. By our choice of with V (H) ∩ V (C) = ∅, and let H G, φ can be extended to a 3-coloring φ of G \ V (H). ¯ has no triangle, then by Gr¨ ¯ admits a 3-coloring φ with φ (x) = If H otzsch’s theorem, H  ¯ has a triangle T , then by Lemma 1, T is a 3-face, we may assign T a 3-coloring and φ (x). If H ¯ with φ (x) = φ (x). In either case φ , together with φ , then extend it to a 3-coloring φ of H yields a 3-coloring of G that extends φ. From Lemma 3, we know that b(f ) is a circuit for every face f . Lemma 4 C is chordless. Proof Assume to the contrary that C has a chord, say uv. Without loss of generality, suppose that |V (C(u, v))| < |V (C(v, u))|. It is certain that C(u, v) is a single vertex, say w. By Lemma 1, uvwu is a facial circuit. Let G be the graph obtained from G \ {w} by inserting into uv a new vertex w colored with φ(w). Then, G ∈ G , σ(G ) < σ(G), the unbounded face of G is a 7-face, and hence we can extend the 3-coloring of C[v, u] ∪ {w } to G . But this provides a 3-coloring of G that extends φ. Lemma 5 G contains no separating 4-circuits. Proof Assume to the contrary that G contains a separating 4-circuit C  . Let H be the graph obtained from the subgraph induced by V (C  ) ∪ int(C  ) by inserting three new vertices into an edge of C  . Certainly, H ∈ G , σ(H) < σ(G) and the boundary of the unbounded face of H is a 7-circuit. By the choice of G, φ can be extended to a 3-coloring φ of G \ int(C  ), and then the restriction of φ on C  can be extended to a 3-coloring φ of H. But φ and φ provide a 3-coloring of G that extends φ. Lemma 6 If x ∈ V (C) has two neighbors, say u, v, on C, then uv ∈ E(C). Proof Assume to the contrary that uv ∈ E(C). Then, |V (C)| = 7. Let P be the shorter path joining u and v on in C. It is clear that 3 ≤ |V (P )| ≤ 4. Since P ∪ {ux, xv} is a circuit of length |V (P )| + 1, |V (P )| = 3, which implies that P ∪ {ux, xv} is a 4-circuit and (E(C)\E(P ))∪{ux, xv} forms a 7-circuit, by Lemma 1 and Lemma 5, E(G) = E(C)∪{ux, xv}, which implies that d(x) = 2, a contradiction. Lemma 7 Let f be a 4-face, and u an arbitrary vertex on b(f ). Then, u is not incident with a 3-face, and there exists a vertex v ∈ N (u) \ V (f ) incident with a 3-face. Proof Let b(f ) = u1 u2 u3 u4 u1 . First we show that ui is not incident with a 3-face for every i. If it is not the case, assume that u1 is incident with a 3-face. Since G contains no triangles of

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distance less than three, each vertex in (N (u2 ) ∪ N (u4 )) \ {u1 } is not incident with a 3-face. Let H be the graph obtained by identifying u2 and u4 . Then, H ∈ G and σ(H) < σ(G). If we can prove that this identification does not damage the pre-coloring φ, then we can extend φ to H and then to G, a contradiction. We claim that, at most one of u2 and u4 is on C. If it is not the case, assume that {u2 , u4 } ⊂ V (C). Since d(fo ) = 4 and G is chordless by Lemma 4, at least one of u1 and u3 , say u1 , is not on C. By Lemma 6, u2 u4 ∈ E(C) so that G contains two adjacent triangles u1 u2 u4 u1 and u3 u2 u4 u3 , which contradicts the choice of G. If {u2 , u4 } ∩ V (C) = ∅, or just one of u2 and u4 is on C and another has no neighbor on C, we are done. So, we assume that u2 ∈ V (C), u4 ∈ V (C) and u4 has a neighbor w on C. Since u4 ∈ V (C), if {u1 , u3 } ∩ V (C) = ∅, say u1 ∈ V (C), then by Lemma 6, wu1 ∈ E(C) that produces a 3-face wu1 u4 w adjacent to f , which contradicts Lemma 2. So, {u1 , u3 } ∩ V (C) = ∅. Since δ(G) ≥ 3 and G contains no 5-circuits, C ∪ b(f ) ∪ {wu4 } contains either a separating 4-circuit or a separating 7-circuit, which contradicts Lemma 1 or Lemma 5. Therefore, the identification of u2 and u4 does not damage the pre-coloring φ. In the following, we need to show only that there is a 3-face that s s has distance 1 from ui for every i. If it is not so, by symmetry we s@s s s may assume that every 3-face has distance at least 2 from u2 . The s @s same arguments as above ensure that we can extend φ to the graph obtained by identifying u2 and u4 , and then to G. s s s s @s Lemma 7 tells us that a 4-face and the 3-faces around it must be s @s as shown in Figure 1. Let f be a k-face of G, k ≥ 6, and u1 uvv1 a segment on b(f ). If Figure 1 k = 7, then G \ {u, v} + u1 v1 contains no 5-circuits. Lemma 8 For k = 7, if a k-face f is adjacent to a 3-face at an edge uv with {u, v}∩V (C) = ∅, then either u or v is a 4≥ -vertex. Proof Assume to the contrary that d(u) = d(v) = 3. Let u1 uvv1 be the segment on b(f ), and w the vertex adjacent to both u and v. It is certain that k = 6 or k ≥ 8. Since G contains no 5-circuits, if {u1 , v1 } ⊂ V (C), then C ∪ (u1 uvv1 ) ∪ (uwv) contains separating 7-circuits that contradicts Lemma 1. Therefore, at most one of u1 and v1 is on C. Let G = G \ {u, v} + u1 v1 . Then, G ∈ G and σ(G ) < σ(G). φ can be extended to a 3-coloring φ of G . Since φ (u1 ) = φ (v1 ), {1, 2, 3}\{φ (u1 ), φ (w)} = {1, 2, 3}\{φ (v1 ), φ (w)}, and hence φ can be extended to G. Lemma 9 If a 6-face f is adjacent to a 3-face at an edge uv with u ∈ V (C), then v is a 4≥ -vertex. Proof By Lemma 4, v ∈ V (C). Assume to the contrary that d(v) = 3. Let v1 = u be another neighbor of v on b(f ). By Lemma 6, v1 is not on C and v1 has no neighbor on C. Let G be the graph obtained from G \ {v} by identifying u and v1 . Then, G ∈ G and σ(G ) < σ(G). φ is not damaged by the identification, and can be extended to G and then to G sequentially. If we assign every vertex of an even circuit a list of two colors, then we can color this circuit no matter what the lists are. Therefore: Lemma 10 If f is an even face not adjacent to fo , then b(f ) has a 4≥ -vertex. 3 Proof of Theorem 1 This is done by contradiction. If the conclusion is not valid, then we choose a counterexample G and its unbounded face fo is as defined at the beginning of Section 2. Let r(v) = d(v) 3 −1  d(f ) for v ∈ V (G), and r(f ) = 6 − 1 for f ∈ F (G). By Euler’s formula, x∈V (G)∪F (G) r(x) =   d(v) d(f ) v∈V (G) ( 3 − 1) + f ∈F (G) ( 6 − 1) = −2. A face f adjacent to fo is called a light-face if the vertices in V (f ) ∩ V (fo ) are all 3-vertices. We will modify r to a new weight r ∗ according to the following discharging rules:

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R-1 Let f be a 3- or 4-face, v a 4≥ -vertex on b(f ). v transfers to f : (a) 13 if f is a 3-face, or a 4-face incident with a unique 4≥ -vertex. (b) 16 if f is a 4-face incident with at least two 4≥ -vertices. R-2 Every bounded 7≥ -face transfers 16 to every adjacent 3-face. R-3 fo transfers 13 toevery incident 2-vertex, or light-3-face, or light-4-face. If we can show that x∈V (G)  F (G) r ∗ (x) > −2, then this contradiction to   ∗   x∈V (G) F (G) r (x) = x∈V (G) F (G) r(x) = −2 will complete our proof. Claim 1 r ∗ (f ) ≥ 0 for every f = fo . Proof Let f be a 3-face. If f is incident with at least two 4≥ -vertices, then by (R-1) (a), r ∗ (f ) ≥ r(f ) + 2 · 13 > 0. If the number of 4≥ -vertices on b(f ) is at most one, then by Lemmas 8 and 9, and by (R-2) and (R-3), either f is incident with a 4≥ -vertex and adjacent to at least one 7-face from which f receives at least 16 so that r ∗ (f ) ≥ r(f ) + 13 + 16 = 0, or f is adjacent to three 7≥ -faces so that r ∗ (f ) ≥ r(f ) + 3 · 16 = 0. Let f be a 4-face. If f is incident with a 4≥ -vertex, then r ∗ (f ) ≥ r(f ) + 2 · 16 = r(f ) + 13 = 0 by (R-1). If f is not incident with a 4≥ -vertex, then f is adjacent to fo by Lemma 10, so r ∗ (f ) ≥ r(f ) + 13 = 0 by (R-3). If f is a 6-face, then r ∗ (f ) = r(f ) ≥ 0. If f is a 7≥ -face, then f is adjacent to at most d(f ) ) 1

4  3-faces, and r ∗ (f ) ≥ r(f ) − d(f 4  · 6 ≥ 0. Claim 2 r ∗ (v) ≥ 0 for every vertex v. Proof If v is a 2-vertex then v ∈ V (C) and r ∗ (v) = r(v) + 13 = 0. If v is a 3-vertex then r ∗ (v) = r(v) = 0. Let v be a 4≥ -vertex. If v is incident with neither a 3-face nor a 4-face, then r ∗ (v) = r(v) > 0. If v is incident with a 3-face, then, by Lemma 7, v is not incident with 4-faces, and hence r ∗ (v) = r(v) − 13 ≥ 0. In the following, we assume that v is incident with some 4-faces, and hence is not incident with a 3-face. By Lemma 7, at least one edge incident with v is not incident with any 4-faces. Let the faces incident with v be f0 , f1 , . . . , fd(v)−1 clockwise around v. Without loss of generality, assume that both f0 and fd(v)−1 are 6≥ -faces, and f1 is a 4-face. By (R1 ), f2i−1 and f2i totally receive . Therefore, v transfers out at most 13 · d(v)−2 , and at most 13 from v, 1 ≤ i ≤ d(v)−2 2 2 d(v)−2 d(v)−4 1 ∗ r (v) ≥ r(v) − 3 · 2  ≥  6 ≥ 0. By Claim 1 and Claim 2, x∈V (G)  F (G) r ∗ (x) ≥ r ∗ (fo ). Let s be the number of 2-vertices on C, t the number of light-3-faces and light-4-faces. Since each light-face shares at least two common 3-vertices with fo , s + 2t ≤ d(fo ), by Lemmas 3 and 5, s ≤ d(fo ) − 2 if t = 0, and s ≤ d(fo ) − 3 if t ≥ 1. Therefore, s + t ≤ d(fo ) − 2, and r ∗ (fo ) ≥ r(fo ) − (s + t) · 13 ≥ d(fo ) o )−5 − 1 − d(fo3)−2 = −2d(f > −2. This completes the proof of Theorem 1. 5 15 References [1] Steinberg, R.: The state of the three color problem, Quo Vadis. Graph Theory? J. Gimbel, J. W. Kennedy & L. V. Quintas (eds). Ann Discrete Math., 55, 211–248 (1993) [2] Borodin, O. V.: Structural properties of plane graphs without adjacent triangles and an application to 3-colorings. Journal of Graph Theory, 21(2), 183–186 (1996) [3] Abbott, H. L., Zhou, B.: On small faces in 4-critical graphs. Ars Combin., 32, 203–207 (1991) [4] Sanders, D. P., Zhao, Y.: A note on the three color problem. Graphs and Combinatorics, 11, 91–94 (1995) [5] Borodin, O. V., et al.: Planar graphs without cycles of length from 4 to 7 are 3-colorable. Journal of Combinatorial Theory, Ser. B, 93, 303–311 (2005) [6] Xu, B.: On 3-colorable plane graphs without 5- and 7-circuits. Journal of Combinatorial Theory, Ser. B, in press [7] Borodin, O. V., Raspaud, A.: A sufficient condition for planar graphs to be 3-colorable. Journal of Combinatorial Theory, Ser. B, 88, 17–27 (2003) [8] Xu, B.: On 3-colorings of plane graphs. Acta Math. Appl. Sinica, 20, 597–604 (2004)