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Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph
Discrete Applied Mathematics xxx (xxxx) xxx
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Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph✩ ∗
Ziwen Huang a , , Runrun Liu b , Gaozhen Wang a a b
School of Mathematics and Computer Science & Center of Applied Mathematics, Yichun University, Yichun, Jiangxi, 336000, China School of Mathematics & Statistics, Central China Normal University, Wuhan 430079, China
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Article history: Received 17 July 2018 Received in revised form 31 January 2019 Accepted 6 April 2019 Available online xxxx Keywords: Planar graph Bad cycles Superextension Discharging
a b s t r a c t Let c1 , c2 , . . . , ck be non-negative integers. A graph G is (c1 , c2 , . . . , ck )-colorable if the vertex set of G can be partitioned into k sets V1 , V2 , . . . , Vk , such that G[Vi ], the subgraph induced by Vi , has maximum degree at most ci for i ∈ [k]. In this paper, we prove that every planar graph without triangular 4-cycles is (1, 1, 1)-colorable. Consequently, such planar graphs can be decomposed into a matching and a 3-colorable graph. © 2019 Elsevier B.V. All rights reserved.
1. Introduction All graphs considered in this paper are finite, simple and undirected. Let c1 , c2 , . . . , ck be k non-negative integers. A graph G is (c1 , c2 , . . . , ck )-colorable if the vertex set can be partitioned into k sets V1 , V2 , . . . , Vk , such that G[Vi ], the subgraph induced by Vi , has maximum degree at most ci for i ∈ [k]. Cowen, Cowen and Woodall [9] showed that every planar graph is (2, 2, 2)-colorable. The well-known Four Color Theorem states that every planar graph is (0, 0, 0, 0)colorable [1,2], and the classic result by Grötzsch says that every triangle-free planar graph is (0, 0, 0)-colorable [13]. Garey et al. [12] showed that the problem of deciding whether a planar graph is (0, 0, 0)-colorable is NP-complete. The 3-coloring problem of planar graphs has been extensively studied in the literature, and interested readers may read the recent survey by Borodin [3]. The following are some of the most famous problems in 3-coloring of planar graphs. Conjecture 1.1 (Steinberg’s Conjecture [15,20]). Every planar graph without cycles of length 4 and 5 is (0, 0, 0)-colorable. Conjecture 1.2 (Bordeaux Conjecture [6]). (Weak version) Every planar graph with neither 5-cycles nor intersecting triangles is (0, 0, 0)-colorable. (Strongly version) Every planar graph with neither 5-cycles nor adjacent triangles is (0, 0, 0)-colorable. Conjecture 1.3 (Novosibirsk 3-Color Conjecture(Nsk3CC) [4,5]). Every planar graph without 3-cycles adjacent to cycles of length from 3 to 5 is (0, 0, 0)-colorable. ✩ The first author was partially supported by NSFC, China (No. 11571134, 11861069) and by Department of Education of Jiangxi Province, China (No. GJJ161030). The second author was supported in part by the NSFC, China (No. 11728102). ∗ Corresponding author. E-mail address: [email protected] (Z. Huang). https://doi.org/10.1016/j.dam.2019.04.026 0166-218X/© 2019 Elsevier B.V. All rights reserved.
Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Fig. 1. The bad 6-cycle in G.
Very recently, all except the weak Bordeaux conjecture are disproved by Cohen-Addad et al. [8], who constructed a non-3-colorable planar graph with neither 4- nor 5-cycles. In the spirit of Steinberg’s conjecture, Chen, Wang, Liu and Xu [7] recently proved that planar graphs without cycles of length 4 or 5 are (2, 0, 0)-colorable. Motivated by the Bordeaux Conjectures, Liu, Li, and Yu [17,18] showed that every planar graph with neither 5-cycles nor intersecting triangles is (2, 0, 0)- and (1, 1, 0)-colorable, and Huang, Li and Yu [14] further showed that every planar graph with neither 5-cycles nor adjacent triangles is (1, 1, 0)-colorable. Now we turn to (1, 1, 1)-coloring of planar graphs. Cowen, Goddard and Jesurum [10] proved that the (1, 1, 1)-coloring problems are NP-complete even for planar graphs. For each i ∈ {5, 6, 7}, Lih et al. [16] proved that planar graphs with neither 4- nor i-cycles are (1, 1, 1)-colorable. Dong and Xu [11] extended the result to include i ∈ {8, 9}. In 2013, Wang and Xu [21] proved that planar graphs without 4-cycles are (1, 1, 1)-colorable. In 2015, Wang and Xu [22] proved that planar graphs without 5-cycles are (1, 1, 1)-colorable. This paper is motivated by the Novosibirsk 3-Color Conjecture. Miao, Wang, Zhang, and Zhang [19] proved that planar graphs without 3-cycles adjacent to cycles of length 3 or 5 are (1, 1, 1)-colorable, which improves the result in [23]. In this paper, we further improve the result of Miao et al. [19]. Theorem 1.4.
Every planar graph without triangular 4-cycles is (1, 1, 1)-colorable.
As pointed out in [22,23], a (1, 1, 1)-colorable graph G has a matching M such that G − M is 3-colorable, which offers an approach decomposing G into a matching and a 3-colorable graph. By Theorem 1.4, Corollary 1.5 is immediate. Corollary 1.5.
Every planar graph without triangular 4-cycles can be decomposed into a matching and a 3-colorable graph.
We will extend a coloring on some subgraph of the graph to the whole graph, an approach dates back to the proof of the classic result by Grötzsch [13]. More specifically, we will use the notion of superextension, introduced by Xu [23]. Definition 1.6. Let G be a graph and φ0 be a (1, 1, 1)-coloring of an induced subgraph H of G. We say that φ0 can be superextended to G if φ0 can be extended to a (1, 1, 1)-coloring φ of G so that φ (v ) ̸ = φ (u) whenever uv ∈ E(G) with u ∈ V (H) and v ∈ G \ V (H). Call C bad if C is the boundary of the unbounded face of the subgraph isomorphic to the configuration depicted in Fig. 1. Otherwise, it is good. Let G denote the family of planar graphs without triangular 4-cycles. We shall prove the following stronger result: Theorem 1.7.
Let G ∈ G . Every (1, 1, 1)-coloring of a good cycle of length at most six in G can be superextended to G.
Proof of Theorem 1.4 by Theorem 1.7. By the Grötzsch Theorem, we may assume that G has a 3-cycle, C . By Theorem 1.7, every (1, 1, 1)-coloring of C can be superextended to G, thus we get a (1, 1, 1)-coloring of G. ■ We finish this section with some notations used in this paper. Let [k] denote the set {i ∈ Z : 1 ≤ i ≤ k}. Let N(v ) denote the set of neighbors of a vertex v in G and N [v] = N(v ) ∪ {v}. Let d(v ) be the degree of a vertex v and d(f ) the degree of a face f . Call v a k+ -vertex, a k-vertex and a k− -vertex if d(v ) ≥ k, d(v ) = k and d(v ) ≤ k, respectively. Similar notations are used for cycles and faces. We write f = [v1 v2 . . . vk ] if v1 v2 . . . vk is the boundary walk of f . An (ℓ1 , ℓ2 , . . . , ℓk )-face is a k-face [v1 v2 · · · vk ] with d(vi ) = ℓi for each i ∈ [k]. We use int(C ) and ext(C ) to denote the sets of vertices located inside and outside a cycle C , respectively. A cycle C is called a separating cycle if int(C ) ̸ = ∅ ̸ = ext(C ). 2. Reducible configurations Let (G, C0 ) be a counterexample with minimum σ (G) = |V (G)| + |E(G)|, where C0 is a good 6− -cycle that is precolored with a (1, 1, 1)-coloring φ0 . If C0 is a separating cycle, then C0 can be superextended to both G − ext(C0 ) and G − int(C0 ). Thus, we get a (1, 1, 1)-coloring of G, contrary to the choice of (G, C0 ). Thus, we may assume that C0 is the boundary of the outer face f0 of G in the rest of this paper. Call a face f internal if f ̸ = f0 , and a vertex v internal if it is in int(C0 ). The following are some simple observations about (G, C0 ). Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Fig. 2. A light 4-vertex and a bad 5-vertex.
Proposition 2.1. (a) Every internal vertex has degree at least 3. (b) No 3-face and 4− -face in G can have a common edge. For shortness, let Fk = {f : f is a k-face and b(f ) ∩ V (C0 ) = ∅}, Fk′ = {f : f is a k-face and |b(f ) ∩ V (C0 )| = 1}, and Fk = {f : f is a k-face and |b(f ) ∩ V (C0 )| = 2}. If uv is an edge in G, then v is called a d(v )-neighbor of u. ′′
Lemma 2.2 ([19]). The following are true to G. (1) G has no adjacent internal 3-vertices. (2) G has no (4− , 4− , 4− )-face in F3 . (3) Each internal 4-vertex has at most two internal 3-neighbors and each internal 5-vertex has at most four internal 3-neighbors. Lemma 2.3.
There are no separating good 6− -cycles in G.
Proof. Suppose to the contrary that G contains a separating good 6− -cycle C . Thus, C is inside of C0 . By the minimality of (G, C0 ), the (1, 1, 1)-coloring of C0 can be superextended to G − int(C ). After that, C is colored. Again, the coloring of C can be superextended to G − ext(C ), this implies that C0 can be superextended to G, a contradiction. □ Lemma 2.4.
The following are true to C0 .
(1) C0 is chordless. (2) Let P = xyz be a 2-path in G with x, z ∈ C0 and y ∈ int(C0 ). Then xz ∈ E(G). (3) Let P = xyy′ z be a 3-path in G with x, z ∈ C0 and y, y′ ∈ int(C0 ). Then P together with one of the two paths of C0 between x and z forms a 4- or 5-cycle. Proof. (1) Let e be a chord of C0 . Since |C0 | ≤ 6, C0 + e contains two 5− -cycles. By Lemma 2.3, V (G) = V (C0 ), thus we get a (1, 1, 1)-coloring of G, a contradiction. Let P1 , P2 be the two paths of C0 between x and z such that |P1 | ≤ |P2 |. Let Di = P ∪ Pi for i ∈ [2] with |D1 | ≤ |D2 |. (2) Suppose to the contrary that xz ∈ / E(G). Then |D1 | = |D2 | = 5 or |D1 | = 4, |D2 | = 6. In the former case, since d(y) ≥ 3 by Proposition 2.1(a) and G ∈ G , D1 or D2 is a separating 5-cycle, contrary to Lemma 2.3. In the latter case, D1 is actually a 4-face by Lemma 2.3. Since C0 is not bad, y has a neighbor y0 inside of D2 . If D2 is a bad 6-cycle, then by Lemma 2.3 y, y0 are two adjacent internal 3-vertices in G, contrary to Lemma 2.2(1). So D2 is a good separating 6-cycle, contrary to Lemma 2.3. (3) Suppose otherwise that |D1 | ≥ 6. Since |C0 | ≤ 6, we have |D1 | = |D2 | = 6. By Proposition 2.1(a) and Lemma 2.2(1), we may assume that d(y) ≥ 3 and d(y′ ) ≥ 4. By Proposition 2.1(b), D1 or D2 is a separating good 6-cycle, contrary to Lemma 2.3. □ Let an internal 4− -vertex v be incident to a 3-face f = [uvw]. Call v light, if either d(v ) = 3 or d(v ) = 4 and v has two internal 3-neighbors not on f . Call f = [uvw] special if f is a (4, 4− , 5)-face with v light. We call a 5-vertex bad if it is incident to two 3-faces f1 , f2 in F3 such that one of which is special and has an internal 3-neighbor not on f1 and f2 , see Fig. 2. Lemma 2.5. Let f = [uvw] be a face in F3 with a light 4− -vertex v . If the (1, 1, 1)-coloring φ of C0 can be superextended to G − v such that either one of u and w is properly colored or both u, w have the same color, then φ can be superextended to G. Proof. Properly recolor each vertex in N(v ) − V (f ) if d(v ) = 4. We may assume that N(v ) has three different colors. If d(v ) = 3 or d(v ) = 4 and N(v ) − V (f ) have the same color, then one of u and w , say u, is properly colored. We can get a (1, 1, 1)-coloring of G by coloring v with the color of u. Otherwise, we can also get a (1, 1, 1)-coloring of G by coloring v with a color in the color set of N(v ) − V (f ) that appears once in N(v ). □ Lemma 2.6.
Let v be a vertex incident to a special 3-face f . Then,
Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Fig. 3. The forbidden structures of Lemma 2.7 when u is bad.
(1) if d(v ) = 5, then v has at most one internal 3-neighbor not on f and cannot be on a (5, 4− , 4− )-face in F3 distinct from f; (2) if d(v ) = 6, then v cannot be on two (6, 4− , 4− )-faces in F3 distinct from f . Proof. (1) Let N(v ) = {vi : 1 ≤ i ≤ 5} and f = [vv1 v2 ] be a special 3-face with v1 light. Suppose otherwise that either d(v3 ) = d(v4 ) = 3 or f ′ = [vv3 v4 ] is a (5, 4− , 4− )-face in F3 . Let S = {v, v1 , v2 , v3 , v4 }. By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G − S. We color each vertex in {v2 , v3 , v4 } with a color different from the colors of its neighbors in G − S. Observe that there is a color appearing at most once on the set {v2 , v3 , v4 } other than the color of v5 , and we assign such a color to v . Then either v, v2 have the same color or v2 is properly colored. By Lemma 2.5, φ can be superextended to G, a contradiction. (2) Let N(v ) = {vi : 1 ≤ i ≤ 6} and f = [vv1 v2 ] be a special 3-face with v1 light. Suppose otherwise that [vv3 v4 ] and [vv5 v6 ] are both (6, 4− , 4− )-faces in F3 . By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G − N [v]. We color each vertex in N(v ) − v1 with a color different from the colors of its neighbors in G − N [v]. Observe that there is a color appearing at most once on the set N(v ) − v1 , and we assign such a color to v . Then either v, v2 have the same color or v2 is properly colored. By Lemma 2.5, φ can be superextended to G, a contradiction. □ Lemma 2.7.
Every internal non-special 3-face f = [uvw] has at most one light 4− -vertex or bad 5-vertex.
Proof. First suppose that both u and v are light. By Proposition 2.1(a), Lemma 2.2(1) and Lemma 2.5, d(u) = d(v ) = 4. Let N(v ) = {v1 , v2 , u, w} and N(u) = {u1 , u2 , v, w}. Let G′ = G − {u, u1 , u2 , v, v1 , v2 }. By the minimality of (G, C0 ), φ can be superextended to G′ . Properly color v1 , v2 , u1 , u2 . If N(u) or N(v ), say N(u) has at most two colors, then we can properly color u, and by Lemma 2.5, φ can be superextended to G, a contradiction. So we may assume that φ (w ) = 1 and {u1 , u2 }, {v1 , v2 } have the same color set {2, 3}. Then we color u with 2 and v with 3, a contradiction. Now we suppose that u is a bad 5-vertex and v is a light 4− -vertex or bad 5-vertex. By Lemma 2.6(1), d(v ) = 4 if v is light, see Fig. 3. let [uu1 u2 ] be the special 3-face with u2 light and u3 be the 3-neighbor of u. Let v1 , v2 be the two 3-neighbors of v not on f if d(v ) = 4, let [vv1 v2 ] be the special 3-face with v2 light and v3 be the 3-neighbor of v if d(v ) = 5. Let S = {u, u1 , u2 , u3 }. By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G − S. Properly color u1 and u3 . We will show that in any case, u can be colored without changing the color of u1 , which implies that φ can be superextended to G by Lemma 2.5, a contradiction. We may assume that u1 and w have different colors, for otherwise, we can either properly color u or color u with the color of u3 , By symmetry assume that u1 is colored with 1 and w is colored with 2. Then u3 is colored with 1 and v together with a neighbor are colored with 3, for otherwise, we can color u with 1 or 3. If v is light, then properly recolor v1 , v2 . Then v can be either properly recolored or recolored with 1. Thus, we can color u with 3. If v is bad, then we uncolor v, v2 and properly recolor the two 3-neighbors of v2 not on [vv1 v2 ] if d(v2 ) = 4. Then recolor v1 , v3 properly and recolor v2 different from the colors of its neighbors not on [vv1 v2 ]. We may assume that v1 , v2 , v3 have color set {1, 1, 3}, for otherwise, we can color v with 1 or 3 and color u with 3. If v2 is colored with 3, then we can recolor v2 with 1 or 2. If v2 is colored with 1, then we can recolor v2 with 2 or 3. In any case, we can color v with 1 or 3 and color u with 3. □ The following are two local structures obtained by a graph operation, that is, identifying two vertices in G. Let x(y) denote the resulting vertex by identifying vertices x and y in G. Remark 2.8. In order to use the minimality of (G, C0 ), we make G a smaller graph by a graph operation. In doing so, we should avoid the following (1) (2) (3) (4) (5)
identifying two vertices of C0 ; creating an edge between two vertices of C0 ; creating loops or multi-edges; creating 4-cycles adjacent to triangles; making C0 bad.
Lemma 2.9.
Every internal 4-face in G has at most one 3-vertex.
Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Proof. Suppose otherwise that f = [uvw x] is an internal 4-face with two 3-vertices. By Lemma 2.2(1), we may assume that d(u) = d(w ) = 3. Let G′ be the graph obtained from G by identifying v and x. We claim that this operation satisfies Remark 2.8(1)–(5). (1) Suppose to the contrary that v, x ∈ C0 . By Lemma 2.3, v x ∈ / E(G). By Lemma 2.4(2), u, w ∈ C0 , contrary to the hypothesis that f is internal. (2) Suppose otherwise that v ∈ C0 and x has a neighbor x′ on C0 . If u ∈ C0 , then by Lemma 2.4(2), that ux′ ∈ E(G), which implies a 4-face f adjacent to a 3-face [uxx′ ] in G, a contradiction. So, u ∈ int(D). By symmetry, w ∈ int(D). Then one of the two path on C0 between v and x′ together with P1 = v uxx′ or P2 = vw xx′ yields a separating 4- or 5-cycle by Lemma 2.4(3), a contradiction. (3) Suppose otherwise, then there is a path of length 1 or 2 connecting v and x in G. Then this path together with vw x or v ux yields a separating 4− -cycle, contrary to Lemma 2.3. (4) We may assume that vw x or v ux is not on a bad 6-cycle, for otherwise, the new 4-cycle created by the identification cannot be adjacent to a 3-cycle since each bad 6-cycle cannot be adjacent to 3-faces. By Lemma 2.3, there are no paths of length 3 or 4 between v and x. So no new 3- or 4-cycles can be created. Thus G′ ∈ G . (5) Suppose otherwise that the operation makes C0 bad, the C0 is also bad in G, contrary to our assumption. Now by the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G′ . Then we color v and x with the color of v (x) and recolor u and w properly, a contradiction. □ Lemma 2.10. Let v be a 4-vertex incident to four face fi (i ∈ [4]) in a cyclic order and f1 = [vv1 u1 v2 v] is a 4-face. If f3 is a 4-face or a 5-face, then v has no internal 3-neighbor. Proof. Suppose to the contrary that v is adjacent to an internal 3-vertex vi . Let G′ be the graph obtained from G by identifying vi−1 and vi+1 of G − v (index module 4). We claim that this operation satisfies Remark 2.8 (1)-(5). (1) Suppose to the contrary that both vi−1 , vi+1 ∈ C0 . Then vi−1 vvi+1 is a path of length 2 such that v ∈ int(C0 ). By Lemma 2.4(2), vi−1 vi+1 ∈ E(G). So vi−1 vi+1 vvi−1 is a separating 3-cycle, contrary to Lemma 2.3. (2) Suppose otherwise that vi−1 ∈ C0 and vi+1 has a neighbor vi′+1 on C0 . Then vi−1 vvi+1 vi+1′ is a path of length 3 with v, vi+1 ∈ int(C0 ). Then one of the two paths on C0 between vi−1 and vi′+1 together with v, vi+1 yields a separating 4- or 5-cycle by Lemma 2.4(3), a contradiction. (3) Suppose otherwise, then there is a path of length 1 or 2 connecting vi−1 and vi+1 in G. Then this path together with v yields a separating 4− -cycle, contrary to Lemma 2.3. (4) We may assume that vi−1 vvi+1 is not a part of a bad 6-cycle, for otherwise, the new 4-cycle created by the identification cannot be adjacent to a 3-cycle since each bad 6-cycle cannot be adjacent to 3-faces. By Lemma 2.3, there are no paths of length 3 or 4 between v and x. So no new 3-or 4-cycles can be created. Thus G′ ∈ G . (5) Suppose otherwise that the operation makes C0 bad, then int(C0 ) contains exactly one 3-vertex and no other vertices in G′ since G′ also has no separating cycles. But vi is a 2-vertex in G′ , a contradiction. By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G′ . Then we color vi−1 and vi+1 with the color of vi−1 (vi+1 ), and properly recolor vi . Now v can be either properly colored or has the same color with vi , a contradiction. □ 3. Discharging procedure In this section, we will finish the proof of Theorem 1.7 by a discharging argument. Let the initial charge of internal vertex or face x in G be µ(x) = d(x) − 4 and µ(f0 ) = d(f0 ) + 4. Then
∑
µ(x) = 0.
x∈V (G)∪F (G)
By defining appropriate discharging rules and redistributing charges accordingly, we will show that each x ∈ F (G) ∪ V (G) other than f0 has final charge µ∗ (x) ≥ 0 and µ∗ (f0 ) > 0, which yields a contradiction. The discharging rules are defined as follows: (R1) Let v be an internal 3-vertex with N(v ) = {v1 , v2 , v3 } and fi be the face containing vi vvi+1 (index module 3). 4 Then v gets 51 from f1 and 25 from v3 if d(f1 ) = 3, d(f2 ) = d(f3 ) = 5; v gets 51 from f1 and 15 from v3 if 1 2 d(f1 ) = 3, d(f2 ) ≥ 5, d(f3 ) ≥ 6; v gets 3 from each incident vertex if d(f1 ) = d(f2 ) = d(f3 ) = 4; v gets 5 from v2 and 15 from each of v1 and v3 if d(f1 ) = d(f2 ) = 4, d(f3 ) ≥ 5; v gets 51 from each incident vertex if d(f1 ) ≥ 4, d(f2 ), d(f3 ) ≥ 5. (R2) Every 3-face gives 51 to each incident light 4-vertex. (R3) The 3-faces mentioned in this rule are all from F3 . Every (3, 5+ , 5+ )-face gets 35 from each incident 5+ -vertex; every special 3-face gets 1 from the incident 5+ -vertex and 15 from the non-light 4-vertex; every (4, 4+ , 5+ )-face gets 51 from each incident non-light 4-vertex or bad 5-vertex and 35 from each incident non-bad 5-vertex; every Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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(5+ , 5+ , 5+ )-face gets 15 from each incident bad 5-vertex and (1 − c) evenly from incident non-bad vertices, where c is the charge given by the incident bad 5-vertex. d(f )−4 (R4) Each 4+ -face f ̸ = f0 gives d(f ) to each incident internal vertex and gives the surplus charge to f0 . (R5) Each vertex on f0 gives its initial charge to f0 and f0 gives a 3-face in F3′ ∪ F3′′ . Lemma 3.1.
6 5
to each 3-face in F3′ ∪ F3′′ ,
2 5
to each 3-neighbor not on
Each internal face f has non-negative final charge.
Proof. By (R4), µ∗ (f ) = 0 if d(f ) ≥ 4. So we may assume that f is a 3-face. By Lemma 2.4(1), |V (f ) ∩ C0 | ≤ 2. If |V (f ) ∩ C0 | ≥ 1, then f is incident to at most one internal light 4− -vertex by Lemma 2.7. So f gets 56 from f0 and gives 1 to the incident light 4− -vertex by (R1),(R2) and (R5). So µ∗ (f ) ≥ −1 + 56 − 15 = 0. Assume that |V (f ) ∩ C0 | = 0. 5 Then f cannot be a (4− , 4− , 4− )-face by Lemma 2.2(2). Note that f has at most one light 4− -vertex or bad 5-vertex by Lemma 2.7. If f is a (3, 5+ , 5+ )-face, then µ∗ (f ) ≥ −1 + 35 · 2 − 15 = 0 by (R1) and (R3). If f is a (4− , 4− , 5+ )-face, then by (R3) µ∗ (f ) ≥ −1 + 15 + 1 − 15 = 0 when f is special, and µ∗ (f ) ≥ −1 + 15 · 2 + 35 = 0 otherwise. If f is a (4, 5+ , 5+ )-face, then by (R2) and (R3) µ∗ (f ) ≥ −1 + 15 · 2 + 35 = 0 when f has a bad 5-vertex, and µ∗ (f ) ≥ −1 + 35 · 2 − 51 = 0 otherwise. If f is a (5+ , 5+ , 5+ )-face, then by (R3) µ∗ (f ) ≥ 0. □ Lemma 3.2.
Each vertex v in G has non-negative final charge.
Proof. By (R5) µ∗ (v ) ≥ 0 if v ∈ C0 . So we assume that v ∈ int(C0 ). By Proposition 2.1, d(v ) ≥ 3. Let N(v ) = {vi : 1 ≤ i ≤ 4} and fi be the face containing vi , v, vi+1 (index module 4). Let tk be the number of incident k-faces of v for k = 3, 4. 4 (1) d(v ) = 3. If v is on a 3-face, then by (R1) µ∗ (v ) ≥ −1 + 15 + min{ 15 · 2 + 25 , 15 + 13 + 15 } = 0. Otherwise, note that 1 1 2 + ∗ v gets at least 5 from each incident 5 -face by (R4). So µ (v ) ≥ −1 + min{ 3 · 3, 5 + 51 · 3, 15 · 5} = 0 by (R1). (2) d(v ) = 4. Then t3 ≤ 2 since G contains no triangular 4-cycles. If t3 = 2, then v is incident to two 5+ -faces and v is not light by Lemma 2.2(1). So µ∗ (v ) ≥ 2 · 51 − 2 · 51 = 0 by (R3)(R4). If t3 = 1, then by symmetry let d(f1 ) = 3. If d(f3 ) = 4, then at most one of v3 , v4 is an internal 3-vertex by Lemma 2.9, which implies that v is not light. So µ∗ (v ) ≥ 2 · 51 − 51 − 15 = 0 by (R1)(R3) and (R4). If d(f3 ) ≥ 5, then v gets 51 from f1 when both v3 , v4 are 3-vertices by (R2). So µ∗ (v ) ≥ 3 · 51 − max{2 · 25 − 51 , 25 + 51 } = 0 by (R1)-(R4). Assume that t3 = 0. Note that v has at most two internal 3-neighbors by Lemma 2.2(3). If t4 = 0, then µ∗ (v ) ≥ 4 · 51 − 2 · 25 = 0 by (R1) and (R4). If d(fi ) = 4, d(fi+2 ) = 4, 5 for i = 1, 2, then by Lemma 2.10, v has no internal 3-neighbor. So µ∗ (v ) = µ(v ) = 0. So we may assume that d(f1 ) = 4, d(f3 ) ≥ 6. If one of f2 , f4 , say f2 is also a 4-face, then by symmetry, d(f4 ) ≥ 6. If 4 = 0 by (R1) and (R4). Otherwise, v gives d(v2 ) = 3, then by Lemma 2.9, d(v1 ), d(v3 ) ≥ 4. so µ∗ (v ) ≥ 2 · 31 − 25 − 15 4 1 7 7 ∗ at most 15 + 5 = 15 to at most two internal 3-neighbors. So µ (v ) ≥ 2 · 13 − 15 > 0 by (R1) and (R4). If both f2 , f4 1 4 1 + ∗ are 5 -faces, then µ (v ) ≥ 2 · 5 + 3 − 2 · 15 > 0 by (R1) and (R4). (3) d(v ) = 5. Then t3 ≤ 2. Let t3 = 2. By symmetry let d(f1 ) = d(f3 ) = 3. Then f2 , f4 , f5 are all 5+ -faces. By Lemma 2.6, v is on at most one special 3-face. If none of f1 , f3 is special, then µ∗ (v ) ≥ 1 + 3 · 15 − 2 · 35 − 25 = 0 by (R1)(R3)(R4). If one of f1 , f3 are special, thenµ∗ (v ) ≥ 1 + 3 · 51 − 1 − 25 − 51 = 0 if v is bad, and µ∗ (v ) ≥ 1 + 3 · 15 − 1 − 53 = 0 otherwise. Let t3 = 1, say d(f1 ) = 3. Then f2 , f5 are both 5+ -faces. If all of v3 , v4 , v5 are internal 3-vertices, then by Lemma 2.9 f3 , f4 are both 5+ -faces and f1 cannot be special by Lemma 2.6(1). So µ∗ (v ) ≥ 1 + 4 · 15 − 53 − 3 · 52 = 0 by (R1)(R3)(R4). So we may assume that at most two of v3 , v4 , v5 are internal 3-vertices. Note that v has at most one internal 3-neighbor not on f1 if f1 is special by Lemma 2.6(1). So µ∗ (v ) ≥ 1 + 25 − max{1 + 25 , 35 + 52 · 2} = 0 by (R1)(R3)(R4). Let t3 = 0. Then v has at most four internal 3-neighbors by Lemma 2.2(3). If t4 ≤ 2, then v is on at least three 5+ -faces. So µ∗ (v ) ≥ 1 + 3 · 15 − 4 · 25 = 0 by (R1)(R4). So we may assume that t4 ∈ {3, 4, 5}. By Lemma 2.9, v is adjacent to at most two internal 3-vertices if t4 = 5, at most three internal 3-vertices if t4 = 3, 4, So µ∗ (v ) ≥ 1 − max{2 · 52 , 3 · 25 − 15 } = 0 by (R1)(R4). (4) d(v ) = 6. Then t3 ≤ 3. If t3 = 3, then by Lemma 2.6(2), at least one incident 3-face of v is not special, so µ∗ (v ) ≥ 2 + 3 · 51 − 2 · 1 − 35 = 0. If t3 = 2, then either v is on four 5+ -faces or v has at most one internal 3-neighbor not on any 3-face incident to v by Lemma 2.9. So µ∗ (v ) ≥ 2 + 3 · 51 − 2 · 1 − max{2 · 25 − 15 , 52 } = 0 by (R1)(R3)(R4). If t3 = 1, then either v is on at least three 5+ -faces or v has at most three internal 3-neighbors not on any 3-face incident to v by Lemma 2.9. So µ∗ (v ) ≥ 2 + 2 · 15 − 1 − max{4 · 25 − 15 , 52 · 3} = 0 by (R1)(R3)(R4). If t3 = 0, then either v is on at least two 5+ -faces or has at most five internal 3-neighbors. So µ∗ (v ) ≥ 2 − max{6 · 25 − 15 · 2, 52 · 5} = 0 by (R1)(R3)(R4).
Finally, let d(v ) ≥ 7. Since G has no triangular 4-cycle, the number of incident 5+ -faces of v is at least f3 . By (R1), (R3), and (R4), we have µ∗ (v ) ≥ d(v ) − 4 + 51 t3 − t3 − 25 (d(v ) − 2t3 ) ≥ 53 d(v ) − 4 > 0. □ To complete our proof, we will check the final charge of f0 and show that µ∗ (f0 ) > 0. Let f3 be the number of 3-faces in F3′ ∪ F3′′ and g3 be the number of 3-vertices that get charge from f0 . Let b be the charge that f0 gets from its adjacent Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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5+ -faces. Let E(C0 , V (G) − C0 ) be the set of edges between C0 and V (G) − C0 and let e be its size. Then by (R4)(R5),
µ∗ (f0 ) = d(f0 ) + 4 +
∑
(d(v ) − 4) −
v∈f0
= d(f0 ) + 4 +
6 5
f3 −
∑
2 5
(d(v ) − 2) − 2d(f0 ) −
v∈f0
6
2
5 4
5 3
5
5
g3 + b 6 5
f3 −
2 5
g3 + b
= 4 − d(f0 ) + e − f3 − g3 + b ′
= 4 − d(f0 ) + e + f3 + g3 + b where e′ is the number of edges in E(C0 , V (G) − C0 ) that are not on any 3-faces and not incident to any 3-vertices that get charge from f0 . Now if d(f0 ) = 3, then µ∗ (D) ≥ 4 − 3 = 1 > 0; while if d(f0 ) = 4, then µ∗ (f0 ) ≤ 0 only if e′ = f3 = g3 = 0, which implies that V (G) = V (f0 ), a contradiction. If d(f0 ) = 5, then µ∗ (f0 ) ≤ 0 only if e′ + 54 f3 + 53 g3 + b ≤ 1. So e′ + f3 + g3 ≤ 1. Actually, e′ + f3 + g3 = 1 since G ̸ = C0 . In this case, f0 is adjacent to a 5+ -face that shares at least five vertices with f0 . By (R4), b ≥ 15 · 5 = 1. So µ∗ (f0 ) ≥ 4 − 5 + 35 + 1 > 0. So assume that d(f0 ) = 6. Then µ∗ (f0 ) ≤ 0 only if e′ + 54 f3 + 35 g3 + b ≤ 2. Then e′ ≤ 2. If e′ = 2, then f3 = g3 = 0. So f0 shares at least four vertices with a 5+ -face. Thus, µ∗ (f0 ) ≥ −2 + 2 + 15 · 4 > 0. If e′ = 1, then f3 + g3 ≤ 1. So f0 shares 6 vertices with 5+ -faces by Lemma 2.4(2). So µ∗ (f0 ) ≥ −2 + 1 + 15 · 6 > 0. If e′ = 0, then 1 ≤ f3 + g3 ≤ 3. If f3 + g3 = 1, then f0 shares at least 6 vertices with 8+ -faces. So µ∗ (f0 ) ≥ −2 + 53 + 12 · 6 > 0. Note that the face adjacent to a 3-face must be a 5+ -face since G ∈ G and every 2-vertex on f0 must be on a 5+ -face by Lemma 2.4(2). If f3 + g3 ≥ 2, then f0 shares at least five vertices with 5+ -faces. So µ∗ (f0 ) ≥ −2 + 35 · 2 + 51 · 5 > 0. Acknowledgments The authors are grateful to the anonymous referees for their helpful suggestions and comments which lead to a great improvement of this paper, and thank to Prof Gexin Yu for his suggestions on the revised version. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23]
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Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph✩ ∗
Ziwen Huang a , , Runrun Liu b , Gaozhen Wang a a b
School of Mathematics and Computer Science & Center of Applied Mathematics, Yichun University, Yichun, Jiangxi, 336000, China School of Mathematics & Statistics, Central China Normal University, Wuhan 430079, China
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Article history: Received 17 July 2018 Received in revised form 31 January 2019 Accepted 6 April 2019 Available online xxxx Keywords: Planar graph Bad cycles Superextension Discharging
a b s t r a c t Let c1 , c2 , . . . , ck be non-negative integers. A graph G is (c1 , c2 , . . . , ck )-colorable if the vertex set of G can be partitioned into k sets V1 , V2 , . . . , Vk , such that G[Vi ], the subgraph induced by Vi , has maximum degree at most ci for i ∈ [k]. In this paper, we prove that every planar graph without triangular 4-cycles is (1, 1, 1)-colorable. Consequently, such planar graphs can be decomposed into a matching and a 3-colorable graph. © 2019 Elsevier B.V. All rights reserved.
1. Introduction All graphs considered in this paper are finite, simple and undirected. Let c1 , c2 , . . . , ck be k non-negative integers. A graph G is (c1 , c2 , . . . , ck )-colorable if the vertex set can be partitioned into k sets V1 , V2 , . . . , Vk , such that G[Vi ], the subgraph induced by Vi , has maximum degree at most ci for i ∈ [k]. Cowen, Cowen and Woodall [9] showed that every planar graph is (2, 2, 2)-colorable. The well-known Four Color Theorem states that every planar graph is (0, 0, 0, 0)colorable [1,2], and the classic result by Grötzsch says that every triangle-free planar graph is (0, 0, 0)-colorable [13]. Garey et al. [12] showed that the problem of deciding whether a planar graph is (0, 0, 0)-colorable is NP-complete. The 3-coloring problem of planar graphs has been extensively studied in the literature, and interested readers may read the recent survey by Borodin [3]. The following are some of the most famous problems in 3-coloring of planar graphs. Conjecture 1.1 (Steinberg’s Conjecture [15,20]). Every planar graph without cycles of length 4 and 5 is (0, 0, 0)-colorable. Conjecture 1.2 (Bordeaux Conjecture [6]). (Weak version) Every planar graph with neither 5-cycles nor intersecting triangles is (0, 0, 0)-colorable. (Strongly version) Every planar graph with neither 5-cycles nor adjacent triangles is (0, 0, 0)-colorable. Conjecture 1.3 (Novosibirsk 3-Color Conjecture(Nsk3CC) [4,5]). Every planar graph without 3-cycles adjacent to cycles of length from 3 to 5 is (0, 0, 0)-colorable. ✩ The first author was partially supported by NSFC, China (No. 11571134, 11861069) and by Department of Education of Jiangxi Province, China (No. GJJ161030). The second author was supported in part by the NSFC, China (No. 11728102). ∗ Corresponding author. E-mail address: [email protected] (Z. Huang). https://doi.org/10.1016/j.dam.2019.04.026 0166-218X/© 2019 Elsevier B.V. All rights reserved.
Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Fig. 1. The bad 6-cycle in G.
Very recently, all except the weak Bordeaux conjecture are disproved by Cohen-Addad et al. [8], who constructed a non-3-colorable planar graph with neither 4- nor 5-cycles. In the spirit of Steinberg’s conjecture, Chen, Wang, Liu and Xu [7] recently proved that planar graphs without cycles of length 4 or 5 are (2, 0, 0)-colorable. Motivated by the Bordeaux Conjectures, Liu, Li, and Yu [17,18] showed that every planar graph with neither 5-cycles nor intersecting triangles is (2, 0, 0)- and (1, 1, 0)-colorable, and Huang, Li and Yu [14] further showed that every planar graph with neither 5-cycles nor adjacent triangles is (1, 1, 0)-colorable. Now we turn to (1, 1, 1)-coloring of planar graphs. Cowen, Goddard and Jesurum [10] proved that the (1, 1, 1)-coloring problems are NP-complete even for planar graphs. For each i ∈ {5, 6, 7}, Lih et al. [16] proved that planar graphs with neither 4- nor i-cycles are (1, 1, 1)-colorable. Dong and Xu [11] extended the result to include i ∈ {8, 9}. In 2013, Wang and Xu [21] proved that planar graphs without 4-cycles are (1, 1, 1)-colorable. In 2015, Wang and Xu [22] proved that planar graphs without 5-cycles are (1, 1, 1)-colorable. This paper is motivated by the Novosibirsk 3-Color Conjecture. Miao, Wang, Zhang, and Zhang [19] proved that planar graphs without 3-cycles adjacent to cycles of length 3 or 5 are (1, 1, 1)-colorable, which improves the result in [23]. In this paper, we further improve the result of Miao et al. [19]. Theorem 1.4.
Every planar graph without triangular 4-cycles is (1, 1, 1)-colorable.
As pointed out in [22,23], a (1, 1, 1)-colorable graph G has a matching M such that G − M is 3-colorable, which offers an approach decomposing G into a matching and a 3-colorable graph. By Theorem 1.4, Corollary 1.5 is immediate. Corollary 1.5.
Every planar graph without triangular 4-cycles can be decomposed into a matching and a 3-colorable graph.
We will extend a coloring on some subgraph of the graph to the whole graph, an approach dates back to the proof of the classic result by Grötzsch [13]. More specifically, we will use the notion of superextension, introduced by Xu [23]. Definition 1.6. Let G be a graph and φ0 be a (1, 1, 1)-coloring of an induced subgraph H of G. We say that φ0 can be superextended to G if φ0 can be extended to a (1, 1, 1)-coloring φ of G so that φ (v ) ̸ = φ (u) whenever uv ∈ E(G) with u ∈ V (H) and v ∈ G \ V (H). Call C bad if C is the boundary of the unbounded face of the subgraph isomorphic to the configuration depicted in Fig. 1. Otherwise, it is good. Let G denote the family of planar graphs without triangular 4-cycles. We shall prove the following stronger result: Theorem 1.7.
Let G ∈ G . Every (1, 1, 1)-coloring of a good cycle of length at most six in G can be superextended to G.
Proof of Theorem 1.4 by Theorem 1.7. By the Grötzsch Theorem, we may assume that G has a 3-cycle, C . By Theorem 1.7, every (1, 1, 1)-coloring of C can be superextended to G, thus we get a (1, 1, 1)-coloring of G. ■ We finish this section with some notations used in this paper. Let [k] denote the set {i ∈ Z : 1 ≤ i ≤ k}. Let N(v ) denote the set of neighbors of a vertex v in G and N [v] = N(v ) ∪ {v}. Let d(v ) be the degree of a vertex v and d(f ) the degree of a face f . Call v a k+ -vertex, a k-vertex and a k− -vertex if d(v ) ≥ k, d(v ) = k and d(v ) ≤ k, respectively. Similar notations are used for cycles and faces. We write f = [v1 v2 . . . vk ] if v1 v2 . . . vk is the boundary walk of f . An (ℓ1 , ℓ2 , . . . , ℓk )-face is a k-face [v1 v2 · · · vk ] with d(vi ) = ℓi for each i ∈ [k]. We use int(C ) and ext(C ) to denote the sets of vertices located inside and outside a cycle C , respectively. A cycle C is called a separating cycle if int(C ) ̸ = ∅ ̸ = ext(C ). 2. Reducible configurations Let (G, C0 ) be a counterexample with minimum σ (G) = |V (G)| + |E(G)|, where C0 is a good 6− -cycle that is precolored with a (1, 1, 1)-coloring φ0 . If C0 is a separating cycle, then C0 can be superextended to both G − ext(C0 ) and G − int(C0 ). Thus, we get a (1, 1, 1)-coloring of G, contrary to the choice of (G, C0 ). Thus, we may assume that C0 is the boundary of the outer face f0 of G in the rest of this paper. Call a face f internal if f ̸ = f0 , and a vertex v internal if it is in int(C0 ). The following are some simple observations about (G, C0 ). Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Fig. 2. A light 4-vertex and a bad 5-vertex.
Proposition 2.1. (a) Every internal vertex has degree at least 3. (b) No 3-face and 4− -face in G can have a common edge. For shortness, let Fk = {f : f is a k-face and b(f ) ∩ V (C0 ) = ∅}, Fk′ = {f : f is a k-face and |b(f ) ∩ V (C0 )| = 1}, and Fk = {f : f is a k-face and |b(f ) ∩ V (C0 )| = 2}. If uv is an edge in G, then v is called a d(v )-neighbor of u. ′′
Lemma 2.2 ([19]). The following are true to G. (1) G has no adjacent internal 3-vertices. (2) G has no (4− , 4− , 4− )-face in F3 . (3) Each internal 4-vertex has at most two internal 3-neighbors and each internal 5-vertex has at most four internal 3-neighbors. Lemma 2.3.
There are no separating good 6− -cycles in G.
Proof. Suppose to the contrary that G contains a separating good 6− -cycle C . Thus, C is inside of C0 . By the minimality of (G, C0 ), the (1, 1, 1)-coloring of C0 can be superextended to G − int(C ). After that, C is colored. Again, the coloring of C can be superextended to G − ext(C ), this implies that C0 can be superextended to G, a contradiction. □ Lemma 2.4.
The following are true to C0 .
(1) C0 is chordless. (2) Let P = xyz be a 2-path in G with x, z ∈ C0 and y ∈ int(C0 ). Then xz ∈ E(G). (3) Let P = xyy′ z be a 3-path in G with x, z ∈ C0 and y, y′ ∈ int(C0 ). Then P together with one of the two paths of C0 between x and z forms a 4- or 5-cycle. Proof. (1) Let e be a chord of C0 . Since |C0 | ≤ 6, C0 + e contains two 5− -cycles. By Lemma 2.3, V (G) = V (C0 ), thus we get a (1, 1, 1)-coloring of G, a contradiction. Let P1 , P2 be the two paths of C0 between x and z such that |P1 | ≤ |P2 |. Let Di = P ∪ Pi for i ∈ [2] with |D1 | ≤ |D2 |. (2) Suppose to the contrary that xz ∈ / E(G). Then |D1 | = |D2 | = 5 or |D1 | = 4, |D2 | = 6. In the former case, since d(y) ≥ 3 by Proposition 2.1(a) and G ∈ G , D1 or D2 is a separating 5-cycle, contrary to Lemma 2.3. In the latter case, D1 is actually a 4-face by Lemma 2.3. Since C0 is not bad, y has a neighbor y0 inside of D2 . If D2 is a bad 6-cycle, then by Lemma 2.3 y, y0 are two adjacent internal 3-vertices in G, contrary to Lemma 2.2(1). So D2 is a good separating 6-cycle, contrary to Lemma 2.3. (3) Suppose otherwise that |D1 | ≥ 6. Since |C0 | ≤ 6, we have |D1 | = |D2 | = 6. By Proposition 2.1(a) and Lemma 2.2(1), we may assume that d(y) ≥ 3 and d(y′ ) ≥ 4. By Proposition 2.1(b), D1 or D2 is a separating good 6-cycle, contrary to Lemma 2.3. □ Let an internal 4− -vertex v be incident to a 3-face f = [uvw]. Call v light, if either d(v ) = 3 or d(v ) = 4 and v has two internal 3-neighbors not on f . Call f = [uvw] special if f is a (4, 4− , 5)-face with v light. We call a 5-vertex bad if it is incident to two 3-faces f1 , f2 in F3 such that one of which is special and has an internal 3-neighbor not on f1 and f2 , see Fig. 2. Lemma 2.5. Let f = [uvw] be a face in F3 with a light 4− -vertex v . If the (1, 1, 1)-coloring φ of C0 can be superextended to G − v such that either one of u and w is properly colored or both u, w have the same color, then φ can be superextended to G. Proof. Properly recolor each vertex in N(v ) − V (f ) if d(v ) = 4. We may assume that N(v ) has three different colors. If d(v ) = 3 or d(v ) = 4 and N(v ) − V (f ) have the same color, then one of u and w , say u, is properly colored. We can get a (1, 1, 1)-coloring of G by coloring v with the color of u. Otherwise, we can also get a (1, 1, 1)-coloring of G by coloring v with a color in the color set of N(v ) − V (f ) that appears once in N(v ). □ Lemma 2.6.
Let v be a vertex incident to a special 3-face f . Then,
Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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Fig. 3. The forbidden structures of Lemma 2.7 when u is bad.
(1) if d(v ) = 5, then v has at most one internal 3-neighbor not on f and cannot be on a (5, 4− , 4− )-face in F3 distinct from f; (2) if d(v ) = 6, then v cannot be on two (6, 4− , 4− )-faces in F3 distinct from f . Proof. (1) Let N(v ) = {vi : 1 ≤ i ≤ 5} and f = [vv1 v2 ] be a special 3-face with v1 light. Suppose otherwise that either d(v3 ) = d(v4 ) = 3 or f ′ = [vv3 v4 ] is a (5, 4− , 4− )-face in F3 . Let S = {v, v1 , v2 , v3 , v4 }. By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G − S. We color each vertex in {v2 , v3 , v4 } with a color different from the colors of its neighbors in G − S. Observe that there is a color appearing at most once on the set {v2 , v3 , v4 } other than the color of v5 , and we assign such a color to v . Then either v, v2 have the same color or v2 is properly colored. By Lemma 2.5, φ can be superextended to G, a contradiction. (2) Let N(v ) = {vi : 1 ≤ i ≤ 6} and f = [vv1 v2 ] be a special 3-face with v1 light. Suppose otherwise that [vv3 v4 ] and [vv5 v6 ] are both (6, 4− , 4− )-faces in F3 . By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G − N [v]. We color each vertex in N(v ) − v1 with a color different from the colors of its neighbors in G − N [v]. Observe that there is a color appearing at most once on the set N(v ) − v1 , and we assign such a color to v . Then either v, v2 have the same color or v2 is properly colored. By Lemma 2.5, φ can be superextended to G, a contradiction. □ Lemma 2.7.
Every internal non-special 3-face f = [uvw] has at most one light 4− -vertex or bad 5-vertex.
Proof. First suppose that both u and v are light. By Proposition 2.1(a), Lemma 2.2(1) and Lemma 2.5, d(u) = d(v ) = 4. Let N(v ) = {v1 , v2 , u, w} and N(u) = {u1 , u2 , v, w}. Let G′ = G − {u, u1 , u2 , v, v1 , v2 }. By the minimality of (G, C0 ), φ can be superextended to G′ . Properly color v1 , v2 , u1 , u2 . If N(u) or N(v ), say N(u) has at most two colors, then we can properly color u, and by Lemma 2.5, φ can be superextended to G, a contradiction. So we may assume that φ (w ) = 1 and {u1 , u2 }, {v1 , v2 } have the same color set {2, 3}. Then we color u with 2 and v with 3, a contradiction. Now we suppose that u is a bad 5-vertex and v is a light 4− -vertex or bad 5-vertex. By Lemma 2.6(1), d(v ) = 4 if v is light, see Fig. 3. let [uu1 u2 ] be the special 3-face with u2 light and u3 be the 3-neighbor of u. Let v1 , v2 be the two 3-neighbors of v not on f if d(v ) = 4, let [vv1 v2 ] be the special 3-face with v2 light and v3 be the 3-neighbor of v if d(v ) = 5. Let S = {u, u1 , u2 , u3 }. By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G − S. Properly color u1 and u3 . We will show that in any case, u can be colored without changing the color of u1 , which implies that φ can be superextended to G by Lemma 2.5, a contradiction. We may assume that u1 and w have different colors, for otherwise, we can either properly color u or color u with the color of u3 , By symmetry assume that u1 is colored with 1 and w is colored with 2. Then u3 is colored with 1 and v together with a neighbor are colored with 3, for otherwise, we can color u with 1 or 3. If v is light, then properly recolor v1 , v2 . Then v can be either properly recolored or recolored with 1. Thus, we can color u with 3. If v is bad, then we uncolor v, v2 and properly recolor the two 3-neighbors of v2 not on [vv1 v2 ] if d(v2 ) = 4. Then recolor v1 , v3 properly and recolor v2 different from the colors of its neighbors not on [vv1 v2 ]. We may assume that v1 , v2 , v3 have color set {1, 1, 3}, for otherwise, we can color v with 1 or 3 and color u with 3. If v2 is colored with 3, then we can recolor v2 with 1 or 2. If v2 is colored with 1, then we can recolor v2 with 2 or 3. In any case, we can color v with 1 or 3 and color u with 3. □ The following are two local structures obtained by a graph operation, that is, identifying two vertices in G. Let x(y) denote the resulting vertex by identifying vertices x and y in G. Remark 2.8. In order to use the minimality of (G, C0 ), we make G a smaller graph by a graph operation. In doing so, we should avoid the following (1) (2) (3) (4) (5)
identifying two vertices of C0 ; creating an edge between two vertices of C0 ; creating loops or multi-edges; creating 4-cycles adjacent to triangles; making C0 bad.
Lemma 2.9.
Every internal 4-face in G has at most one 3-vertex.
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Proof. Suppose otherwise that f = [uvw x] is an internal 4-face with two 3-vertices. By Lemma 2.2(1), we may assume that d(u) = d(w ) = 3. Let G′ be the graph obtained from G by identifying v and x. We claim that this operation satisfies Remark 2.8(1)–(5). (1) Suppose to the contrary that v, x ∈ C0 . By Lemma 2.3, v x ∈ / E(G). By Lemma 2.4(2), u, w ∈ C0 , contrary to the hypothesis that f is internal. (2) Suppose otherwise that v ∈ C0 and x has a neighbor x′ on C0 . If u ∈ C0 , then by Lemma 2.4(2), that ux′ ∈ E(G), which implies a 4-face f adjacent to a 3-face [uxx′ ] in G, a contradiction. So, u ∈ int(D). By symmetry, w ∈ int(D). Then one of the two path on C0 between v and x′ together with P1 = v uxx′ or P2 = vw xx′ yields a separating 4- or 5-cycle by Lemma 2.4(3), a contradiction. (3) Suppose otherwise, then there is a path of length 1 or 2 connecting v and x in G. Then this path together with vw x or v ux yields a separating 4− -cycle, contrary to Lemma 2.3. (4) We may assume that vw x or v ux is not on a bad 6-cycle, for otherwise, the new 4-cycle created by the identification cannot be adjacent to a 3-cycle since each bad 6-cycle cannot be adjacent to 3-faces. By Lemma 2.3, there are no paths of length 3 or 4 between v and x. So no new 3- or 4-cycles can be created. Thus G′ ∈ G . (5) Suppose otherwise that the operation makes C0 bad, the C0 is also bad in G, contrary to our assumption. Now by the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G′ . Then we color v and x with the color of v (x) and recolor u and w properly, a contradiction. □ Lemma 2.10. Let v be a 4-vertex incident to four face fi (i ∈ [4]) in a cyclic order and f1 = [vv1 u1 v2 v] is a 4-face. If f3 is a 4-face or a 5-face, then v has no internal 3-neighbor. Proof. Suppose to the contrary that v is adjacent to an internal 3-vertex vi . Let G′ be the graph obtained from G by identifying vi−1 and vi+1 of G − v (index module 4). We claim that this operation satisfies Remark 2.8 (1)-(5). (1) Suppose to the contrary that both vi−1 , vi+1 ∈ C0 . Then vi−1 vvi+1 is a path of length 2 such that v ∈ int(C0 ). By Lemma 2.4(2), vi−1 vi+1 ∈ E(G). So vi−1 vi+1 vvi−1 is a separating 3-cycle, contrary to Lemma 2.3. (2) Suppose otherwise that vi−1 ∈ C0 and vi+1 has a neighbor vi′+1 on C0 . Then vi−1 vvi+1 vi+1′ is a path of length 3 with v, vi+1 ∈ int(C0 ). Then one of the two paths on C0 between vi−1 and vi′+1 together with v, vi+1 yields a separating 4- or 5-cycle by Lemma 2.4(3), a contradiction. (3) Suppose otherwise, then there is a path of length 1 or 2 connecting vi−1 and vi+1 in G. Then this path together with v yields a separating 4− -cycle, contrary to Lemma 2.3. (4) We may assume that vi−1 vvi+1 is not a part of a bad 6-cycle, for otherwise, the new 4-cycle created by the identification cannot be adjacent to a 3-cycle since each bad 6-cycle cannot be adjacent to 3-faces. By Lemma 2.3, there are no paths of length 3 or 4 between v and x. So no new 3-or 4-cycles can be created. Thus G′ ∈ G . (5) Suppose otherwise that the operation makes C0 bad, then int(C0 ) contains exactly one 3-vertex and no other vertices in G′ since G′ also has no separating cycles. But vi is a 2-vertex in G′ , a contradiction. By the minimality of (G, C0 ), the (1, 1, 1)-coloring φ of C0 can be superextended to G′ . Then we color vi−1 and vi+1 with the color of vi−1 (vi+1 ), and properly recolor vi . Now v can be either properly colored or has the same color with vi , a contradiction. □ 3. Discharging procedure In this section, we will finish the proof of Theorem 1.7 by a discharging argument. Let the initial charge of internal vertex or face x in G be µ(x) = d(x) − 4 and µ(f0 ) = d(f0 ) + 4. Then
∑
µ(x) = 0.
x∈V (G)∪F (G)
By defining appropriate discharging rules and redistributing charges accordingly, we will show that each x ∈ F (G) ∪ V (G) other than f0 has final charge µ∗ (x) ≥ 0 and µ∗ (f0 ) > 0, which yields a contradiction. The discharging rules are defined as follows: (R1) Let v be an internal 3-vertex with N(v ) = {v1 , v2 , v3 } and fi be the face containing vi vvi+1 (index module 3). 4 Then v gets 51 from f1 and 25 from v3 if d(f1 ) = 3, d(f2 ) = d(f3 ) = 5; v gets 51 from f1 and 15 from v3 if 1 2 d(f1 ) = 3, d(f2 ) ≥ 5, d(f3 ) ≥ 6; v gets 3 from each incident vertex if d(f1 ) = d(f2 ) = d(f3 ) = 4; v gets 5 from v2 and 15 from each of v1 and v3 if d(f1 ) = d(f2 ) = 4, d(f3 ) ≥ 5; v gets 51 from each incident vertex if d(f1 ) ≥ 4, d(f2 ), d(f3 ) ≥ 5. (R2) Every 3-face gives 51 to each incident light 4-vertex. (R3) The 3-faces mentioned in this rule are all from F3 . Every (3, 5+ , 5+ )-face gets 35 from each incident 5+ -vertex; every special 3-face gets 1 from the incident 5+ -vertex and 15 from the non-light 4-vertex; every (4, 4+ , 5+ )-face gets 51 from each incident non-light 4-vertex or bad 5-vertex and 35 from each incident non-bad 5-vertex; every Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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(5+ , 5+ , 5+ )-face gets 15 from each incident bad 5-vertex and (1 − c) evenly from incident non-bad vertices, where c is the charge given by the incident bad 5-vertex. d(f )−4 (R4) Each 4+ -face f ̸ = f0 gives d(f ) to each incident internal vertex and gives the surplus charge to f0 . (R5) Each vertex on f0 gives its initial charge to f0 and f0 gives a 3-face in F3′ ∪ F3′′ . Lemma 3.1.
6 5
to each 3-face in F3′ ∪ F3′′ ,
2 5
to each 3-neighbor not on
Each internal face f has non-negative final charge.
Proof. By (R4), µ∗ (f ) = 0 if d(f ) ≥ 4. So we may assume that f is a 3-face. By Lemma 2.4(1), |V (f ) ∩ C0 | ≤ 2. If |V (f ) ∩ C0 | ≥ 1, then f is incident to at most one internal light 4− -vertex by Lemma 2.7. So f gets 56 from f0 and gives 1 to the incident light 4− -vertex by (R1),(R2) and (R5). So µ∗ (f ) ≥ −1 + 56 − 15 = 0. Assume that |V (f ) ∩ C0 | = 0. 5 Then f cannot be a (4− , 4− , 4− )-face by Lemma 2.2(2). Note that f has at most one light 4− -vertex or bad 5-vertex by Lemma 2.7. If f is a (3, 5+ , 5+ )-face, then µ∗ (f ) ≥ −1 + 35 · 2 − 15 = 0 by (R1) and (R3). If f is a (4− , 4− , 5+ )-face, then by (R3) µ∗ (f ) ≥ −1 + 15 + 1 − 15 = 0 when f is special, and µ∗ (f ) ≥ −1 + 15 · 2 + 35 = 0 otherwise. If f is a (4, 5+ , 5+ )-face, then by (R2) and (R3) µ∗ (f ) ≥ −1 + 15 · 2 + 35 = 0 when f has a bad 5-vertex, and µ∗ (f ) ≥ −1 + 35 · 2 − 51 = 0 otherwise. If f is a (5+ , 5+ , 5+ )-face, then by (R3) µ∗ (f ) ≥ 0. □ Lemma 3.2.
Each vertex v in G has non-negative final charge.
Proof. By (R5) µ∗ (v ) ≥ 0 if v ∈ C0 . So we assume that v ∈ int(C0 ). By Proposition 2.1, d(v ) ≥ 3. Let N(v ) = {vi : 1 ≤ i ≤ 4} and fi be the face containing vi , v, vi+1 (index module 4). Let tk be the number of incident k-faces of v for k = 3, 4. 4 (1) d(v ) = 3. If v is on a 3-face, then by (R1) µ∗ (v ) ≥ −1 + 15 + min{ 15 · 2 + 25 , 15 + 13 + 15 } = 0. Otherwise, note that 1 1 2 + ∗ v gets at least 5 from each incident 5 -face by (R4). So µ (v ) ≥ −1 + min{ 3 · 3, 5 + 51 · 3, 15 · 5} = 0 by (R1). (2) d(v ) = 4. Then t3 ≤ 2 since G contains no triangular 4-cycles. If t3 = 2, then v is incident to two 5+ -faces and v is not light by Lemma 2.2(1). So µ∗ (v ) ≥ 2 · 51 − 2 · 51 = 0 by (R3)(R4). If t3 = 1, then by symmetry let d(f1 ) = 3. If d(f3 ) = 4, then at most one of v3 , v4 is an internal 3-vertex by Lemma 2.9, which implies that v is not light. So µ∗ (v ) ≥ 2 · 51 − 51 − 15 = 0 by (R1)(R3) and (R4). If d(f3 ) ≥ 5, then v gets 51 from f1 when both v3 , v4 are 3-vertices by (R2). So µ∗ (v ) ≥ 3 · 51 − max{2 · 25 − 51 , 25 + 51 } = 0 by (R1)-(R4). Assume that t3 = 0. Note that v has at most two internal 3-neighbors by Lemma 2.2(3). If t4 = 0, then µ∗ (v ) ≥ 4 · 51 − 2 · 25 = 0 by (R1) and (R4). If d(fi ) = 4, d(fi+2 ) = 4, 5 for i = 1, 2, then by Lemma 2.10, v has no internal 3-neighbor. So µ∗ (v ) = µ(v ) = 0. So we may assume that d(f1 ) = 4, d(f3 ) ≥ 6. If one of f2 , f4 , say f2 is also a 4-face, then by symmetry, d(f4 ) ≥ 6. If 4 = 0 by (R1) and (R4). Otherwise, v gives d(v2 ) = 3, then by Lemma 2.9, d(v1 ), d(v3 ) ≥ 4. so µ∗ (v ) ≥ 2 · 31 − 25 − 15 4 1 7 7 ∗ at most 15 + 5 = 15 to at most two internal 3-neighbors. So µ (v ) ≥ 2 · 13 − 15 > 0 by (R1) and (R4). If both f2 , f4 1 4 1 + ∗ are 5 -faces, then µ (v ) ≥ 2 · 5 + 3 − 2 · 15 > 0 by (R1) and (R4). (3) d(v ) = 5. Then t3 ≤ 2. Let t3 = 2. By symmetry let d(f1 ) = d(f3 ) = 3. Then f2 , f4 , f5 are all 5+ -faces. By Lemma 2.6, v is on at most one special 3-face. If none of f1 , f3 is special, then µ∗ (v ) ≥ 1 + 3 · 15 − 2 · 35 − 25 = 0 by (R1)(R3)(R4). If one of f1 , f3 are special, thenµ∗ (v ) ≥ 1 + 3 · 51 − 1 − 25 − 51 = 0 if v is bad, and µ∗ (v ) ≥ 1 + 3 · 15 − 1 − 53 = 0 otherwise. Let t3 = 1, say d(f1 ) = 3. Then f2 , f5 are both 5+ -faces. If all of v3 , v4 , v5 are internal 3-vertices, then by Lemma 2.9 f3 , f4 are both 5+ -faces and f1 cannot be special by Lemma 2.6(1). So µ∗ (v ) ≥ 1 + 4 · 15 − 53 − 3 · 52 = 0 by (R1)(R3)(R4). So we may assume that at most two of v3 , v4 , v5 are internal 3-vertices. Note that v has at most one internal 3-neighbor not on f1 if f1 is special by Lemma 2.6(1). So µ∗ (v ) ≥ 1 + 25 − max{1 + 25 , 35 + 52 · 2} = 0 by (R1)(R3)(R4). Let t3 = 0. Then v has at most four internal 3-neighbors by Lemma 2.2(3). If t4 ≤ 2, then v is on at least three 5+ -faces. So µ∗ (v ) ≥ 1 + 3 · 15 − 4 · 25 = 0 by (R1)(R4). So we may assume that t4 ∈ {3, 4, 5}. By Lemma 2.9, v is adjacent to at most two internal 3-vertices if t4 = 5, at most three internal 3-vertices if t4 = 3, 4, So µ∗ (v ) ≥ 1 − max{2 · 52 , 3 · 25 − 15 } = 0 by (R1)(R4). (4) d(v ) = 6. Then t3 ≤ 3. If t3 = 3, then by Lemma 2.6(2), at least one incident 3-face of v is not special, so µ∗ (v ) ≥ 2 + 3 · 51 − 2 · 1 − 35 = 0. If t3 = 2, then either v is on four 5+ -faces or v has at most one internal 3-neighbor not on any 3-face incident to v by Lemma 2.9. So µ∗ (v ) ≥ 2 + 3 · 51 − 2 · 1 − max{2 · 25 − 15 , 52 } = 0 by (R1)(R3)(R4). If t3 = 1, then either v is on at least three 5+ -faces or v has at most three internal 3-neighbors not on any 3-face incident to v by Lemma 2.9. So µ∗ (v ) ≥ 2 + 2 · 15 − 1 − max{4 · 25 − 15 , 52 · 3} = 0 by (R1)(R3)(R4). If t3 = 0, then either v is on at least two 5+ -faces or has at most five internal 3-neighbors. So µ∗ (v ) ≥ 2 − max{6 · 25 − 15 · 2, 52 · 5} = 0 by (R1)(R3)(R4).
Finally, let d(v ) ≥ 7. Since G has no triangular 4-cycle, the number of incident 5+ -faces of v is at least f3 . By (R1), (R3), and (R4), we have µ∗ (v ) ≥ d(v ) − 4 + 51 t3 − t3 − 25 (d(v ) − 2t3 ) ≥ 53 d(v ) − 4 > 0. □ To complete our proof, we will check the final charge of f0 and show that µ∗ (f0 ) > 0. Let f3 be the number of 3-faces in F3′ ∪ F3′′ and g3 be the number of 3-vertices that get charge from f0 . Let b be the charge that f0 gets from its adjacent Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.
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5+ -faces. Let E(C0 , V (G) − C0 ) be the set of edges between C0 and V (G) − C0 and let e be its size. Then by (R4)(R5),
µ∗ (f0 ) = d(f0 ) + 4 +
∑
(d(v ) − 4) −
v∈f0
= d(f0 ) + 4 +
6 5
f3 −
∑
2 5
(d(v ) − 2) − 2d(f0 ) −
v∈f0
6
2
5 4
5 3
5
5
g3 + b 6 5
f3 −
2 5
g3 + b
= 4 − d(f0 ) + e − f3 − g3 + b ′
= 4 − d(f0 ) + e + f3 + g3 + b where e′ is the number of edges in E(C0 , V (G) − C0 ) that are not on any 3-faces and not incident to any 3-vertices that get charge from f0 . Now if d(f0 ) = 3, then µ∗ (D) ≥ 4 − 3 = 1 > 0; while if d(f0 ) = 4, then µ∗ (f0 ) ≤ 0 only if e′ = f3 = g3 = 0, which implies that V (G) = V (f0 ), a contradiction. If d(f0 ) = 5, then µ∗ (f0 ) ≤ 0 only if e′ + 54 f3 + 53 g3 + b ≤ 1. So e′ + f3 + g3 ≤ 1. Actually, e′ + f3 + g3 = 1 since G ̸ = C0 . In this case, f0 is adjacent to a 5+ -face that shares at least five vertices with f0 . By (R4), b ≥ 15 · 5 = 1. So µ∗ (f0 ) ≥ 4 − 5 + 35 + 1 > 0. So assume that d(f0 ) = 6. Then µ∗ (f0 ) ≤ 0 only if e′ + 54 f3 + 35 g3 + b ≤ 2. Then e′ ≤ 2. If e′ = 2, then f3 = g3 = 0. So f0 shares at least four vertices with a 5+ -face. Thus, µ∗ (f0 ) ≥ −2 + 2 + 15 · 4 > 0. If e′ = 1, then f3 + g3 ≤ 1. So f0 shares 6 vertices with 5+ -faces by Lemma 2.4(2). So µ∗ (f0 ) ≥ −2 + 1 + 15 · 6 > 0. If e′ = 0, then 1 ≤ f3 + g3 ≤ 3. If f3 + g3 = 1, then f0 shares at least 6 vertices with 8+ -faces. So µ∗ (f0 ) ≥ −2 + 53 + 12 · 6 > 0. Note that the face adjacent to a 3-face must be a 5+ -face since G ∈ G and every 2-vertex on f0 must be on a 5+ -face by Lemma 2.4(2). If f3 + g3 ≥ 2, then f0 shares at least five vertices with 5+ -faces. So µ∗ (f0 ) ≥ −2 + 35 · 2 + 51 · 5 > 0. Acknowledgments The authors are grateful to the anonymous referees for their helpful suggestions and comments which lead to a great improvement of this paper, and thank to Prof Gexin Yu for his suggestions on the revised version. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23]
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Please cite this article as: Z. Huang, R. Liu and G. Wang, Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph, Discrete Applied Mathematics (2019), https://doi.org/10.1016/j.dam.2019.04.026.