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Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions
Discrete Applied Mathematics xxx (xxxx) xxx
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Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions Chao Song, Xue Jin, Chang-Qing Xu
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School of Science, Hebei University of Technology, Tianjin 300401, China
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Article history: Received 12 February 2019 Received in revised form 4 December 2019 Accepted 26 December 2019 Available online xxxx Keywords: Neighbor sum distinguishing total coloring IC-planar graph Discharging method
a b s t r a c t A graph is IC-planar if it admits a drawing in the plane with at most one crossing per edge, such that two pairs of crossing edges share no common end vertex. For a given graph G, a proper total coloring φ : V (G) ∪ E(G) → {1, 2, . . . , k} ∑ is neighbor sum distinguishing if fφ (u) ̸ = fφ (v ) for each uv ∈ E(G), where fφ (v ) = uv∈E(G) φ (uv ) + φ (v ), v ∈ V (G). The smallest integer k in such a coloring of G is the neighbor sum ′′ distinguishing total chromatic number, denoted by χΣ (G). In this paper, by using the ′′ discharging method, we prove that χΣ (G) ≤ max{∆(G) + 3, 10} if G is a triangle free ′′ IC-planar graph and χΣ (G) ≤ max{∆(G) + 3, 13} if G is an IC-planar graph without adjacent triangles, where ∆(G) is the maximum degree of G. © 2020 Elsevier B.V. All rights reserved.
1. Introduction In this paper, all graphs considered are simple, finite and undirected. The terminology and notation used but undefined in this paper can be found in [2]. For a graph G, we denote its vertex set, edge set and maximum degree by V (G), E(G) and ∆(G), respectively. We use dG (v ) to denote the degree of vertex v in G. For a plane graph G, F (G) denotes its face set. A vertex v is a t-vertex (t − -vertex, t + -vertex) if dG (v ) = t (dG (v ) ≤ t , dG (v ) ≥ t). A t-face (t − -face, t + -face) is defined similarly. Let dt (v ) denote the number of t-vertices adjacent to v in graph G. A graph is 1-planar if it can be drawn in the plane so that each edge is crossed by at most one other edge. Zhang et al. [22,23] considered total colorings of 1-planar graph. A graph is IC-planar if it admits a drawing in the plane with at most one crossing per edge, such that two pairs of crossing edges share no common end vertex. Obviously, each IC-planar graph is also a 1-planar graph. Given a graph G and a positive integer k, a proper total k-coloring of G is a mapping φ : V (G) ∪∑ E(G) → {1, 2, . . . , k} such that φ (x) ̸ = φ (y) for each pair of adjacent or incident elements x, y ∈ V (G) ∪ E(G). Let fφ (v ) = uv∈E(G) φ (uv ) + φ (v ), v ∈ V (G). If fφ (u) ̸= fφ (v ) for each uv ∈ E(G), then φ is a neighbor sum distinguishing total k-coloring, or k-tnsd-coloring ′′ for simplicity. The smallest integer k is the neighbor sum distinguishing total chromatic number, denoted by χΣ (G). For k-tnsd-coloring, Pilśniak and Woźniak [10] gave the following conjecture. ′′ Conjection 1.1 ([10]). For any graph G, χΣ (G) ≤ ∆(G) + 3.
Pilśniak and Woźniak also verified that Conjecture 1.1 holds for complete graphs, cycles, bipartite graphs and subcubic graphs. For planar graph G, it was proved that this conjecture holds with ∆(G) ≥ 13 by Li et al. [7], ∆(G) ≥ 11 by Qu et al. [11] and ∆(G) = 10 by Yang et al. [21]. For planar graph G with ∆(G) ≥ 7, it was proved that this conjecture holds ∗ Corresponding author. E-mail address: [email protected] (C.-Q. Xu). https://doi.org/10.1016/j.dam.2019.12.023 0166-218X/© 2020 Elsevier B.V. All rights reserved.
Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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for this kind of graph without 3-cycle by Wang et al. [18], without 4-cycle by Wang et al. [16] and without 5-cycle by Ge et al. [6]. More results about k-tnsd-coloring of planar graphs see [3–5,8,9,14,17,19,20] and so on. For IC-planar graphs, Conjecture 1.1 is confirmed for graphs G with ∆(G) ≥ 14 by Song et al. [12], for triangle free graphs G with ∆(G) ≥ 8 by Song et al. [13] and for adjacent triangles free graphs G with ∆(G) ≥ 12 by Song et al. [13]. In [13], Song et al. proposed the following question. ′′ (G) ≤ ∆(G) + 3 for any triangle free IC-planar graph G with ∆(G) = 7 ? Question ([13]). Is it true that χΣ
We confirm this question and prove the following results in this paper. ′′ Theorem 1.2. Let G be a triangle free IC-planar graph. Then χΣ (G) ≤ max{∆(G) + 3, 10}. ′′ (G) ≤ max{∆(G) + 3, 13}. Theorem 1.3. Let G be an IC-planar graph without adjacent triangles. Then χΣ
Theorem 1.2 solves the above Question in [13]. A planar graph is also an IC-planar graph, so Theorem 1.2 generalizes the result in [18]. Theorem 1.3 shows that Conjecture 1.1 is true for IC-planar graph G without adjacent triangles with ∆(G) ≥ 10, which improves the corresponding result in [13]. 2. Preliminaries In the following, we always assume that every IC-planar graph G is drawn in a plane such that the number of crossings is as small as possible. If G is an IC-planar graph embedded in the plane, we use C (G) to denote the set of all crossings and E0 (G) to denote the non-crossed edges in G. The associated plane graph G× of G is the plane graph such that V (G× ) = V (G) ∪ C (G) and E(G× ) = E0 (G) ∪ {xz , yz |xy ∈ E(G)\E0 (G) and z is the crossing on xy}. Thus in G× , each crossing becomes a vertex of degree four. For convenience, a vertex in G× is false if it is not a vertex of G and true otherwise. A face f in G× is called a false face if it contains a false vertex. One should be clear that every true vertex in G× is adjacent to at most one false vertex. Theorem 2.1 (Combinatorial Nullstellensatz [1]). Let∑ F be an arbitrary field, and let P = P(x1 , x2 , . . . , xn ) be a polynomial in k k n k F[x1 , x2 , . . . , xn ]. Suppose that the degree of P equals i=1 ki , where each ki is a nonnegative integer, and cP (x11 x22 · · · xnn ) ̸ = 0. If S1 , S2 , . . . , Sn ⊆ F with |Si | > ki for 1 ≤ i ≤ n, then there are s1 ∈ S1 , s2 ∈ S2 , . . . , sn ∈ Sn so that P(s1 , s2 , . . . , sn ) ̸ = 0. Lemma 2.2 ([7]). Let S1 , S2 be the sets of integers with |S1 | = m ≥ 2 and |S2 | = n ≥ 2. Let |L| = {x + y| x ∈ S1 , y ∈ S2 , x ̸ = y}. Then |L| ≥ m + n − 3. Moreover, if |S1 | ̸ = |S2 |, then |L| ≥ m + n − 2. 3. Proof of Theorem 1.2 We will prove Theorem 1.2 by contradiction. Assume that G is a counterexample of Theorem 1.2 such that |V (G)|+|E(G)| is as small as possible. By Theorem 1.4 in [13], Conjecture 1.1 holds for any triangle free IC-planar graph with maximum degree at least 8, so we can assume that ∆(G) ≤ 7. By the choice of G, G does not have any 10-tnsd-coloring, while any proper subgraph G′ of G has a 10-tnsd-coloring φ ′ . We will always extend a 10-tnsd-coloring φ ′ of some G′ to a 10-tnsdcoloring of G to get a contradiction. It is worth noting that we will omit the colors of all 3− -vertices in the following proof. Since they have at most 9 forbidden colors and there are 10 colors, they can be colored easily. In the following proof, we use d(v ) to denote dG (v ). It is easy to see that the following claims given by [15] also hold with the graph G in our proof. Claim (1) (2) (3)
3.1 ([15]). In graph G, the following results hold. Each t − -vertex is not adjacent to any (7 − t)− -vertex, where t = 4, 5. If d(v ) = 6, then d2− (v ) ≤ 1. Furthermore, if d2− (v ) = 1, then d3 (v ) = 0. If d2− (v ) = 0, then d3 (v ) ≤ 2. If d(v ) = 7, then d2− (v ) ≤ 2. Furthermore, if d2− (v ) ≥ 1, then d3− (v ) ≤ 2.
Claim 3.2. In graph G, if d(v ) = 4, then d4 (v ) ≤ 1. Proof. Suppose to the contrary that there exists a 4-vertex v which is adjacent to two 4-vertices v1 , v2 . Since G is triangle free, v1 is not adjacent to v2 . Let N(v ) = {v1 , v2 , v3 , v4 }, N(v1 ) = {v, u2 , u3 , u4 } and N(v2 ) = {v, w2 , w3 , w4 }. Consider G′ = G − vv1 − vv2 , then G′ admits a 10-tnsd-coloring φ ′ . We delete the colors of v, v1 , v2 . Let S1 , S2 , S3 , S4 , S5 be the sets of available colors of vv1 , vv2 , v, v1 , v2 , respectively. It is easy to know that |S1 | ≥ 10 − 5 = 5, |S2 | ≥ 10 − 5 = 5, |S3 | ≥ 10 − 4 = 6, |S4 | ≥ 10 − 6 = 4, |S5 | ≥ 10 − 6 = 4.∏We associate vv1 , vv2 , v, v1 , v2 with the variables ∏ x1 , x2 , x3 , x4 , x5 , 4 respectively. Then we consider P1 (x1 , x2 , x3 , x4 , x5 ) = 1≤i
∏ + x4 − φ ′ (v1 ) + fφ ′ (v1 ) − fφ ′ (uq )) 4r =2 (x2 + x5 − φ ′ (v2 ) + fφ ′ (v2 ) − fφ ′ (wr ))(x2 + x3 + ∑ ∑ φ ′ (vv3 ) + φ ′ (vv4 ) − x4 − 4s=2 φ ′ (v1 us ))(x1 + x3 + φ ′ (vv3 ) + φ ′ (vv4 ) − x5 − 4t =2 φ ′ (v2 wt )). We have cP1 (x41 x42 x43 x34 x25 ) = 4. By Theorem 2.1, there exists xi ∈ Si , for i = 1, 2, . . . , 5 such that P1 (x1 , x2 , x3 , x4 , x5 ) ̸ = 0. We can color vv1 , vv2 , v, v1 , v2 with x1 , x2 , x3 , x4 , x5 respectively to obtain a 10-tnsd-coloring of G, a contradiction. x3 + φ ′ (vv3 ) + φ ′ (vv4 ) − fφ ′ (vp ))
∏4
q=2 (x1
Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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Note that in the forthcoming proofs with the arguments similar to Claim 3.2, we will omit some obvious but tedious statements. Moreover, such as in Claim 3.2, the coefficient of the monomial x41 x42 x43 x34 x25 in the expansion of ∏ P1 (x1 , x2 , x3 , x4 , x5 ) is equal to that of the same monomial in the polynomial Q = 1≤i
Let T be the graph obtained from G by removing all 2− -vertices and T × be the associated plane graph of T . Since G is triangle free, each vertex in T × is incident with at most one false 3-face. In the following, we also use dt (v ) to denote the number of t-vertices adjacent to v in graph G, d(v ) to denote the degree of v in graph G. By Claims 3.1 and 3.3, we have the following facts. Fact 3.1.
δ (T × ) ≥ 3. dT × (v ) = d(v ), for 3 ≤ d(v ) ≤ 5. If d(v ) ≥ 6, then dT × (v ) ≥ 5.
Fact 3.2. (1) If dT × (v ) = 5, then d3 (v ) ≤ 1. Moreover, if d3 (v ) = 1, then dT × (v ) = d(v ) and d4 (v ) = 0. (2) Each 3-face in T × is a false (3, 4, 5+ )-face or a false (4, 4+ , 4+ )-face. Noting that G is triangle free, in T × , a 3-face intersects a 3-face if they share a false vertex. A 3-face is bad if it intersects a 3-face, otherwise, it is good. The false vertex incident with a 3-face f is denoted by cf . Let Nf (cf ) be the set of neighbors of cf which are not incident with f . In T × , a 3-vertex is good if each of its neighbors with degree 5 (if exists) in T × is not incident with any 3-face, otherwise, it is bad. A bad 3-vertex v is of type 1 if v is not incident with any (3, 4, 5)-face (see (a) and (b) in Fig. 1). A bad 3-vertex v is of type 2 if v is incident with a bad (3, 4, 5)-face f ∗ = v cf ∗ u, where f ∗ intersects a 3-facef ⋄ (see (c) in Fig. 1). A bad 3-vertex v is of type 3 if v is incident with a good (3, 4, 5)-face f ∗ = v cf ∗ u (see (d) in Fig. 1). In Fig. 1, we draw a vertex v in black if it has no other neighbors than the ones already depicted, and a vertex v in white if it might have more neighbors than the ones shown in∑ the figure. ∑ × Since T × is a plane graph, by Euler’s formula, we have v∈V (T × ) (dT × (v ) − 4) + f ∈F (T × ) (dT (f ) − 4) = −8. In the following proof, we will use the discharging method to obtain a contradiction. We define the charge function w : V (T × ) ∪ F (T × ) −→ R by w (v ) = dT × (v ) − 4 if v ∈ V (T × ) and w (f ) = dT × (f ) − 4 if f ∈ F (T × ). Next we will give some discharging rules to get a new charge function w ′ after the discharging. After the discharging, the total charge remains unchanged. In the following, we will show that w ′ (x) ≥ 0 for each x ∈ V (T × ) ∪ F (T × ), which leads to a contradiction.
Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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Fig. 1. Illustration of type 1, type 2 and type 3 bad 3-vertices.
The discharging rules are defined as follows. The degree of vertex or face is its degree in T × . R1: Every good 3-vertex v receives 31 from each of its neighbors in G. R2: Every type 1 bad 3-vertex receives 1 from f ∗ (f ∗ must be a (4, 5, 5+ )-face by Facts 3.1–3.2(1)); every type 2 bad 3-vertex receives 1 from f ⋄ (f ⋄ must be a (4, 5+ , 5+ )-face by Claim 3.1 and Facts 3.1–3.2(1)); every type 3 bad 3-vertex receives 12 from each vertex in Nf ∗ (cf ∗ ) (see Fig. 1). R3: Every 5+ -vertex v gives 1 to its incident 3-face (if exists). R4: Every bad (4, 4, 4)-face receives 1 from its intersecting 3-face f ⋄ (f ⋄ must be a (4, 5+ , 5+ )-face by Claims 3.1–3.2 and Facts 3.1–3.2(1)), every good (4, 4, 4)-face f receives 21 from each vertex in Nf (cf ). In the following, a t-face means a t-face in T × . We use n3 (v ) to denote the number of 3-faces incident with v in T × . We have the following fact. Fact 3.3. (1) If dT × (v ) = 5, then v gives at most 1 away. (2) If dT × (v ) = l (l = 6, 7), then v gives at most 1 + 31 d3 (v ) away. Proof. (1) If dT × (v ) = 5, clearly, n3 (v ) ≤ 1. If n3 (v ) = 1, by R1-R4, v gives at most 1 away. If n3 (v ) = 0, by Claim 3.4 and Facts 3.1–3.2(1), there is no path P = v1 v2 v3 v4 in T × such that dT × (v1 ) = dT × (v4 ) = 3 and dT × (v2 ) = dT × (v3 ) = 5. And by Fact 3.2(1), R1, R2 and R4, v gives at most 21 away. (2) If dT × (v ) = l (l = 6, 7), clearly, n3 (v ) ≤ 1. If n3 (v ) = 1, by R1-R4, v gives at most 1 + 31 d3 (v ) away. If n3 (v ) = 0, by R1, R2 and R4, v gives at most 21 + 13 d3 (v ) away. Now we will verify the new charge of each x ∈ V (T × ) ∪ F (T × ). We first consider the new charge of each f ∈ F (T × ). If d(f ) = 3. By Fact 3.2(2), f is a false (3, 4, 5+ )-face or a false (4, 4+ , 4+ )-face. If f is a (3, 4, 5+ )-face or a (4, 4, 4+ )-face. By R2 and R4, f does not give any charge away. By R3 and R4, w ′ (f ) = 3 − 4 + 1 = 0, or w ′ (f ) = 3 − 4 + 2 × 12 = 0. If f is a (4, 5, 5)-face, by Facts 3.1–3.2 and Claim 3.4, there is no path P = v1 v2 v3 v4 in T × such that dT × (v1 ) = dT × (v4 ) = 3 and dT × (v2 ) = dT × (v3 ) = 5. There is at most one 5-vertex incident with f adjacent to a 3-vertex. By Fact 3.2(1), if f intersects a (4, 4, 4)-face, then the two 5-vertices incident with f are not adjacent to any 3-vertex. By R2 and R4, f gives at most 1 away. So by R2-R4, w ′ (f ) ≥ 3 − 4 + 2 × 1 − 1 = 0. If f is a (4, 5, 6+ )-face, by Fact 3.2(1), the 5-vertex incident with f is adjacent to at most one 3-vertex, and if it is adjacent to a 3-vertex, then it is not adjacent to any 4-vertex. That is, if f intersects a (4, 4, 4)-face, then the 5-vertex incident with f is not adjacent to any 3-vertex. By R2 and R4, f gives at most 1 away. So by R2-R4, w ′ (f ) ≥ 3 − 4 + 2 × 1 − 1 = 0. If f is a (4, 6+ , 6+ )-face, noting that a 3-face intersects at most one 3-face, by R2 and R4, f gives at most 1 away. By R2-R4, w ′ (f ) ≥ 3 − 4 + 2 × 1 − 1 = 0. If d(f ) = l (l ≥ 4), no rule applies to f . w ′ (f ) = w (f ) = l − 4 ≥ 0. Next we will consider the new charge of each v ∈ V (T × ). If v is a false vertex of T × , then dT × (v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Next we will consider each true vertex v in T × according to its degree in G. Note that dT × (v ) = d(v ) − d2− (v ). Suppose that d(v ) = 3. By Fact 3.1, dT × (v ) = d(v ) = 3. If v is a good 3-vertex, by R1, w ′ (v ) = 3 − 4 + 3 × 13 = 0; if v is a bad 3-vertex of type 1 or type 2, by R2, w ′ (v ) = 3−4+1 = 0; if v is a bad 3-vertex of type 3, by R2, w ′ (v ) = 3−4+2× 21 = 0. Suppose that d(v ) = 4. By Fact 3.1, dT × (v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Suppose that d(v ) = 5. By Fact 3.1, dT × (v ) = d(v ) = 5. By Fact 3.3(1), w ′ (v ) ≥ 5 − 4 − 1 = 0. Suppose that d(v ) = 6. By Claim 3.1(2), d2− (v ) ≤ 1. If d2− (v ) = 1, then dT × (v ) = 5, by Fact 3.3(1), w ′ (v ) ≥ 5 − 4 − 1 = 0. If d2− (v ) = 0, then dT × (v ) = 6 and d3 (v ) ≤ 2. By Fact 3.3(2), w ′ (v ) ≥ 6 − 4 − 1 − 2 × 31 = 13 . Suppose that d(v ) = 7. By Claim 3.1(3), d2− (v ) ≤ 2. If d2− (v ) = 2, then dT × (v ) = 5, by Fact 3.3(1), w ′ (v ) ≥ 5 − 4 − 1 = 0. If d2− (v ) = 1, then dT × (v ) = 6 and d3 (v ) ≤ 1. By Fact 3.3(2), w ′ (v ) ≥ 6 − 4 − 1 − 1 × 31 = 32 . If d2− (v ) = 0, then dT × (v ) = 7. By Fact 3.1 and Claim 3.5, d3 (v ) ≤ 5. By Fact 3.3(2), w ′ (v ) ≥ 7 − 4 − 1 − 5 × 31 = 13 . Now we get that for each x ∈ V (T × ) ∪ F (T × ), w ′ (x) ≥ 0. Thus we obtain a contradiction. This completes the proof of Theorem 1.2. Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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4. Proof of Theorem 1.3 We will prove Theorem 1.3 by contradiction. Assume that G is a counterexample of Theorem 1.3 such that |V (G)|+|E(G)| is as small as possible. By Theorem 1.5 in [13], Conjecture 1.1 holds for any IC-planar graph G without adjacent triangles with ∆(G) ≥ 12. So we can assume that ∆(G) ≤ 11. Set k = max{∆(G) + 3, 13}. By the choice of G, any proper subgraph G′ of G has a k-tnsd-coloring φ ′ . We will always extend a k-tnsd-coloring φ ′ of some G′ to a k-tnsd-coloring of G to get a contradiction. It is worth noting that we will omit the colors of all 4− -vertices in the following proof. Since they have at most 12 forbidden colors and k ≥ 13, they can be colored easily. In the following, we use d(v ) to denote dG (v ). It is easy to see that the following claims given by [21] also hold with the graph G in our proof. Claim 4.1 ([21]). Every 4− -vertex is not adjacent to any 6− -vertex in G. Claim 4.2 ([21]). Every 5− -vertex is not adjacent to any 5− -vertex in G. Claim 4.3 ([21]). If d(v ) = 7, then d4− (v ) ≤ 1. Claim 4.4. For each 8-vertex v of G, we have d2− (v ) ≤ 1. Moreover, (1) if d2− (v ) = 1, then d3 (v ) = 0. (2) If d2− (v ) = 0, then d3 (v ) ≤ 2. Proof. We will prove d2− (v ) ≤ 1 and (1) at the same time. Suppose to the contrary that there exists an 8-vertex v which is adjacent to a 2− -vertex v1 and a 3− -vertex v2 . Set N(v ) = {v1 , v2 , . . . , v8 }. Consider G′ = G − vv1 − vv2 , then G′ has a k-tnsd-coloring φ ′ . We delete the colors of v1 , v2 . Let S1 , S2 be the sets of available colors of vv1 , vv2 , respectively. It is easy to know that |S1 | ≥ 13 − 8 = 5, |S2 | ≥ 13 − 9 = 4. By Lemma 2.2, |L| ≥ 5 + 4 − 2 = 7(> 6), so we have at least 7 available colors sets for {vv1 , vv2 }. Therefore, we can color vv1 , vv2 properly such that the sum at v is distinct from the sums at v3 , v4 , . . . , v8 , respectively. Noting that v1 , v2 are 3− -vertices, thus we can obtain a k-tnsd-coloring of G, a contradiction. Now we prove (2). Suppose to the contrary that there exists an 8-vertex v with d2− (v ) = 0 and d3 (v ) ≥ 3. Assume that d(vi ) = 3, for i = 1, 2, 3. Consider G′ = G − vv1 − vv2 − vv3 , then G′ has a k-tnsd-coloring φ ′ . We delete the colors of v1 , v2 , v3 . Let S1 , S2 , S3 be the sets of available colors of vv1 , vv2 , vv3 , respectively. It is easy to know that |Si | ≥ 13 − 8 = 5,∑for i = 1∏ , 2, 3. We associate vv1 , vv2 , vv3 with the variables x1 , x2 , x3 , respectively. Then we consider 3 P5 (x1 , x2 , x3 ) = ( k=1 xk )5 1≤i
Now we prove (2). Suppose to the contrary that there exists a 9-vertex v with d2− (v ) = 0 and d3 (v ) ≥ 4. Assume that d(vi ) = 3, for i = 1, 2, 3, 4. Consider G′ = G − vv1 − vv2 − vv3 − vv4 , then G′ admits a k-tnsd-coloring φ ′ . We delete the colors of v, v1 , v2 , v3 , v4 . Let S1 , S2 , S3 , S4 , S5 be the sets of available colors of vv1 , vv2 , vv3 , vv4 , v respectively. It is easy to know that |Si | ≥ 13 − 7 = 6, for i = 1, 2, 3, 4, |S5 | ≥ 13 − 10 = 3. We∑associate∏ vv1 , vv2 , vv3 , vv4 , v with 5 the variables x1 , x2 , x3 , x4 , x5 , respectively. Then we consider P7 (x1 , x2 , x3 , x4 , x5 ) = ( k=1 xk )5 1≤i
For each 10-vertex v of G, if d2− (v ) ≥ 1 and d3− (v ) ≥ 2, then d4− (v ) ≤ 4.
Proof. Suppose to the contrary that there exists a 10-vertex v which is adjacent to a 2− -vertex v1 , a 3− -vertex v2 and three 4− -vertices v3 , v4 , v5 . Consider G′ = G − {vv1 , vv2 , vv3 , vv4 , vv5 }, then G′ admits a k-tnsd-coloring φ ′ . We delete the colors of v1 , v2 , . . . , v5 . Let Si be the sets of available colors of vvi , for i = 1, 2, . . . , 5, respectively. It is easy to know that |S1 | ≥ 13 − 7 = 6, |S2 | ≥ 13 − 8 = 5, |Si | ≥ 13 − 9 = 4, for i = 3, 4 , 5. We associate vvi with the ∑ ∏ 5 variables xi , for i = 1, 2, . . . , 5, respectively. Then we consider P8 (x1 , x2 , x3 , x4 , x5 ) = ( k=1 xk )5 1≤i
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have cP8 (x51 x42 x33 x24 x5 ) = 1. By Theorem 2.1 and noting that v1 , v2 , . . . , v5 are 4− -vertices, we can obtain a k-tnsd-coloring of G, a contradiction. Claim 4.7.
For each 11-vertex v of G, if d2− (v ) ≥ 1, then d3− (v ) ≤ 4.
Proof. Suppose to the contrary that there exists an 11-vertex v which is adjacent to a 2− -vertex v1 and four 3− -vertices v2 , v3 , v4 , v5 . Consider G′ = G − {vv1 , vv2 , vv3 , vv4 , vv5 }, then G′ admits a k-tnsd-coloring φ ′ . We delete the colors of v, v1 , v2 , . . . , v5 . Let S1 , S2 , S3 , S4 , S5 , S6 be the sets of available colors of vv1 , vv2 , vv3 , vv4 , vv5 , v respectively. It is easy to know that |S1 | ≥ 14 − 7 = 7, |Si | ≥ 14 − 8 = 6, for i = 2, 3, 4, 5, |S6 | ≥ 14 − 12 = 2. We associate vv 3 , vv4 , vv5 , v with the variables xi , for i = 1, 2, . . . , 6, respectively. Then we consider P9 (x1 , x2 , x3 , x4 , x5 , x6 ) = ∑16, vv2 , vv∏ ( k=1 xk )6 1≤i
If dT × (v ) = l, then n3 (v ) ≤ ⌈ 2l ⌉.
By Claims 4.1–4.7, we have the following fact. Fact 4.2. (1) δ (T × ) ≥ 3. dT × (v ) = d(v ), for 3 ≤ d(v ) ≤ 6. If d(v ) ≥ 7, then dT × (v ) ≥ 6. (2) Each 3-face in T × is a (3, 4, 7+ )-face, a (3, 7+ , 7+ )-face, a (4, 4, 7+ )-face, a (4, 5+ , 6+ )-face or a (5+ , 6+ , 6+ )-face. × Since T × is a plane graph, by Euler’s formula, we have v∈V (T × ) (dT × (v ) − 4) + f ∈F (T × ) (dT (f ) − 4) = −8. In the following proof, we will use the discharging method to obtain a contradiction. We define the charge function w : V (T × ) ∪ F (T × ) −→ R by w (v ) = dT × (v ) − 4 if v ∈ V (T × ) and w (f ) = dT × (f ) − 4 if f ∈ F (T × ). Next we will give some discharging rules to get a new charge function w ′ after the discharging. After the discharging, the total charge remains unchanged. In the following, we will show that w ′ (x) ≥ 0 for each x ∈ V (T × ) ∪ F (T × ), which leads to a contradiction. The discharging rules are defined as follows. The degree of vertex or face is its degree in T × . R1: Every 3-vertex v receives 31 from each of its neighbors in G. R2: Every 5-vertex v gives 83 to each incident (4, 5, 6+ )-face. R3: Every 6+ -vertex v gives 1 to each incident (3, 4, 7+ )-face and (4, 4, 7+ )-face, gives 85 to each incident (4, 5, 6+ )-face, gives 21 to every other incident 3-faces. By the discharging rules, we obtain the following fact.
∑
Fact 4.3.
∑
If dT × (v ) = l (l ≥ 7), then v gives at most 1 + 12 (⌈ 2l ⌉ − 1) to its incident 3-faces.
Proof. Since T is an IC-planar graph, every 7+ -vertex v in T × is incident with at most two false 3-faces, where at most one of them belongs to a false (3, 4, 7+ )-face, a false (4, 4, 7+ )-face or a false (4, 5, 6+ )-face by Claim 4.2 and Fact 4.2(1). By Fact 4.1 and R3, v gives at most 1 + 12 (⌈ 2l ⌉ − 1) to its incident 3-faces. Now we will verify the new charge of each x ∈ V (T × ) ∪ F (T × ). We first consider the new charge of each f ∈ F (T × ). If d(f ) = 3, by Fact 4.2(2), f is a (3, 4, 7+ )-face, a (3, 7+ , 7+ )-face, a (4, 4, 7+ )-face, a (4, 5+ , 6+ )-face or a (5+ , 6+ , 6+ )face. If f is a (3, 4, 7+ )-face or a (4, 4, 7+ )-face, by R3, w ′ (f ) = 3 − 4 + 1 = 0. If f is a (4, 5, 6+ )-face, by R2 and R3, w ′ (f ) = 3 − 4 + 38 + 85 = 0. If f is a (3, 7+ , 7+ )-face or a (4+ , 6+ , 6+ )-face, by R3, w ′ (f ) ≥ 3 − 4 + 2 × 12 = 0. If d(f ) = l (l ≥ 4), no rule applies to f . w ′ (f ) = w (f ) = l − 4 ≥ 0. Next we will consider the new charge of each v ∈ V (T × ). If v is a false vertex of T × , then dT × (v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Next we will consider each true vertex v in T × according to its degree in G. Note that dT × (v ) = d(v ) − d2− (v ). Suppose that d(v ) = 3. By Fact 4.2(1), dT × (v ) = d(v ) = 3. By R1, w ′ (v ) = 3 − 4 + 3 × 13 = 0. Suppose that d(v ) = 4. By Fact 4.2(1), dT × (v ) = d(v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Suppose that d(v ) = 5. By Claim 4.2 and Fact 4.2(1), dT × (v ) = d(v ) = 5 and v is incident with at most two (4, 5, 6+ )-faces. By R2, w ′ (v ) ≥ 5 − 4 − 2 × 83 = 41 . Suppose that d(v ) = 6. By Claim 4.1, dT × (v ) = d(v ), d3 (v ) = d4 (v ) = 0. By Claim 4.2 and Fact 4.2(1), a 5-vertex is not adjacent to any 5-vertex in T × , so v is incident with at most one (4, 5, 6)-face noting that v is adjacent to at most one false 4-vertex. By Fact 4.1, n3 (v ) ≤ 3. By R3, w ′ (v ) ≥ 6 − 4 − 85 − 2 × 12 = 38 . Suppose that d(v ) = 7. By Claim 4.3, d2− (v ) ≤ 1. If d2− (v ) = 1, then dT × (v ) = 6, d3 (v ) = d4 (v ) = 0. We can verify that w′ (v ) ≥ 0 similarly as the above proof for d(v ) = 6. If d2− (v ) = 0, then dT × (v ) = 7. By Claim 4.3 and Fact 4.1, d3 (v ) ≤ 1 and n3 (v ) ≤ 4. By R1 and Fact 4.3, w ′ (v ) ≥ 7 − 4 − 31 − 1 − 3 × 12 = 16 . Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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Suppose that d(v ) = 8. By Claim 4.4, d2− (v ) ≤ 1. If d2− (v ) = 1, then dT × (v ) = 7 and d3 (v ) = 0. By Fact 4.3, w′ (v ) ≥ 7 − 4 − 1 − 3 × 21 = 12 . If d2− (v ) = 0, then dT × (v ) = 8 and d3 (v ) ≤ 2. By Fact 4.1, n3 (v ) ≤ 4. By R1 and Fact 4.3, w ′ (v ) ≥ 8 − 4 − 2 × 13 − 1 − 3 × 12 = 56 . Suppose that d(v ) = 9. By Claim 4.5, d2− (v ) ≤ 2. If d2− (v ) ≥ 1, then d3 (v ) ≤ 1. If d2− (v ) = i, then dT × (v ) = 9 − i, where i = 1, 2. We can verify that w ′ (v ) ≥ 0 similarly as the above proof for d(v ) = 9 − i, for i = 1, 2. If d2− (v ) = 0, then dT × (v ) = 9 and d3 (v ) ≤ 3. By Fact 4.1, n3 (v ) ≤ 5. By R1 and Fact 4.3, w ′ (v ) ≥ 9 − 4 − 3 × 31 − 1 − 4 × 12 = 1. Suppose that d(v ) = 10. By Claim 4.6, d2− (v ) ≤ 4. If d2− (v ) = i, then d3 (v ) ≤ 4 − i, dT × (v ) = 10 − i, where i = 1, 2, 3, 4. We can verify that w ′ (v ) ≥ 0 similarly as the above proof for d(v ) = 10 − i, for i = 1, 2, 3, 4. If d2− (v ) = 0, then dT × (v ) = 10. We consider the following two cases: Case 1. d3 (v ) = 10. By Claim 4.1 and Fact 4.2(1), a 3-vertex is not adjacent to any 3-vertex in T × , so n3 (v ) ≤ 1. By R1 and R3, w ′ (v ) ≥ 10 − 4 − 10 × 31 − 1 = 53 . Case 2. d3 (v ) ≤ 9. By Fact 4.1, n3 (v ) ≤ 5. By R1 and Fact 4.3, w ′ (v ) ≥ 10 − 4 − 9 × 13 − 1 − 4 × 21 = 0. Suppose that d(v ) = 11. By Claim 4.7, d2− (v ) ≤ 4, and if d2− (v ) = i, then d3 (v ) ≤ 4 − i and dT × (v ) = 11 − i, for i = 1, 2, 3, 4. We can verify that w ′ (v ) ≥ 0 similarly as the above proof for d(v ) = 11 − i, for i = 1, 2, 3, 4. If d2− (v ) = 0, then dT × (v ) = 11. We consider the following two cases: Case 1. d3 (v ) = 11. By Claim 4.1 and Fact 4.2(1), a 3-vertex is not adjacent to any 3-vertex in T × , so n3 (v ) ≤ 1. By R1 and R3, w ′ (v ) ≥ 11 − 4 − 11 × 13 − 1 = 73 . Case 2. d3 (v ) ≤ 10. By Fact 4.1, n3 (v ) ≤ 6. By R1 and Fact 4.3, w ′ (v ) ≥ 11 − 4 − 10 × 13 − 1 − 5 × 12 = 16 . Now we get that for each x ∈ V (T × ) ∪ F (T × ), w ′ (x) ≥ 0. Thus we obtain a contradiction. This completes the proof of Theorem 1.3. CRediT authorship contribution statement Chao Song: Writing - original draft, Writing - review & editing. Xue Jin: Writing - original draft. Chang-Qing Xu: Methodology, Supervision, Writing - review & editing. Acknowledgments The authors would like to express their thanks to the referees for their valuable comments and suggestions that greatly improve the format and correctness of the manuscript. This work was supported by National Natural Science Foundation of China (11671232). References [1] N. Alon, Combinatorial nullstellensatz, Combin. Probab. Comput. 8 (1999) 7–29, http://dx.doi.org/10.1017/S0963548398003411. [2] J. Bondy, U. Murty, Graph Theory with Applications, North-Holland, NewYork, 1976. [3] X. Cheng, D. Huang, G. Wang, J. Wu, Neighbor sum distinguishing total colorings of planar graphs with maximum degree ∆, Discrete Appl. Math. 190 (2015) 34–41, http://dx.doi.org/10.1016/j.dam.2015.03.013. [4] L. 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Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions Chao Song, Xue Jin, Chang-Qing Xu
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School of Science, Hebei University of Technology, Tianjin 300401, China
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Article history: Received 12 February 2019 Received in revised form 4 December 2019 Accepted 26 December 2019 Available online xxxx Keywords: Neighbor sum distinguishing total coloring IC-planar graph Discharging method
a b s t r a c t A graph is IC-planar if it admits a drawing in the plane with at most one crossing per edge, such that two pairs of crossing edges share no common end vertex. For a given graph G, a proper total coloring φ : V (G) ∪ E(G) → {1, 2, . . . , k} ∑ is neighbor sum distinguishing if fφ (u) ̸ = fφ (v ) for each uv ∈ E(G), where fφ (v ) = uv∈E(G) φ (uv ) + φ (v ), v ∈ V (G). The smallest integer k in such a coloring of G is the neighbor sum ′′ distinguishing total chromatic number, denoted by χΣ (G). In this paper, by using the ′′ discharging method, we prove that χΣ (G) ≤ max{∆(G) + 3, 10} if G is a triangle free ′′ IC-planar graph and χΣ (G) ≤ max{∆(G) + 3, 13} if G is an IC-planar graph without adjacent triangles, where ∆(G) is the maximum degree of G. © 2020 Elsevier B.V. All rights reserved.
1. Introduction In this paper, all graphs considered are simple, finite and undirected. The terminology and notation used but undefined in this paper can be found in [2]. For a graph G, we denote its vertex set, edge set and maximum degree by V (G), E(G) and ∆(G), respectively. We use dG (v ) to denote the degree of vertex v in G. For a plane graph G, F (G) denotes its face set. A vertex v is a t-vertex (t − -vertex, t + -vertex) if dG (v ) = t (dG (v ) ≤ t , dG (v ) ≥ t). A t-face (t − -face, t + -face) is defined similarly. Let dt (v ) denote the number of t-vertices adjacent to v in graph G. A graph is 1-planar if it can be drawn in the plane so that each edge is crossed by at most one other edge. Zhang et al. [22,23] considered total colorings of 1-planar graph. A graph is IC-planar if it admits a drawing in the plane with at most one crossing per edge, such that two pairs of crossing edges share no common end vertex. Obviously, each IC-planar graph is also a 1-planar graph. Given a graph G and a positive integer k, a proper total k-coloring of G is a mapping φ : V (G) ∪∑ E(G) → {1, 2, . . . , k} such that φ (x) ̸ = φ (y) for each pair of adjacent or incident elements x, y ∈ V (G) ∪ E(G). Let fφ (v ) = uv∈E(G) φ (uv ) + φ (v ), v ∈ V (G). If fφ (u) ̸= fφ (v ) for each uv ∈ E(G), then φ is a neighbor sum distinguishing total k-coloring, or k-tnsd-coloring ′′ for simplicity. The smallest integer k is the neighbor sum distinguishing total chromatic number, denoted by χΣ (G). For k-tnsd-coloring, Pilśniak and Woźniak [10] gave the following conjecture. ′′ Conjection 1.1 ([10]). For any graph G, χΣ (G) ≤ ∆(G) + 3.
Pilśniak and Woźniak also verified that Conjecture 1.1 holds for complete graphs, cycles, bipartite graphs and subcubic graphs. For planar graph G, it was proved that this conjecture holds with ∆(G) ≥ 13 by Li et al. [7], ∆(G) ≥ 11 by Qu et al. [11] and ∆(G) = 10 by Yang et al. [21]. For planar graph G with ∆(G) ≥ 7, it was proved that this conjecture holds ∗ Corresponding author. E-mail address: [email protected] (C.-Q. Xu). https://doi.org/10.1016/j.dam.2019.12.023 0166-218X/© 2020 Elsevier B.V. All rights reserved.
Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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for this kind of graph without 3-cycle by Wang et al. [18], without 4-cycle by Wang et al. [16] and without 5-cycle by Ge et al. [6]. More results about k-tnsd-coloring of planar graphs see [3–5,8,9,14,17,19,20] and so on. For IC-planar graphs, Conjecture 1.1 is confirmed for graphs G with ∆(G) ≥ 14 by Song et al. [12], for triangle free graphs G with ∆(G) ≥ 8 by Song et al. [13] and for adjacent triangles free graphs G with ∆(G) ≥ 12 by Song et al. [13]. In [13], Song et al. proposed the following question. ′′ (G) ≤ ∆(G) + 3 for any triangle free IC-planar graph G with ∆(G) = 7 ? Question ([13]). Is it true that χΣ
We confirm this question and prove the following results in this paper. ′′ Theorem 1.2. Let G be a triangle free IC-planar graph. Then χΣ (G) ≤ max{∆(G) + 3, 10}. ′′ (G) ≤ max{∆(G) + 3, 13}. Theorem 1.3. Let G be an IC-planar graph without adjacent triangles. Then χΣ
Theorem 1.2 solves the above Question in [13]. A planar graph is also an IC-planar graph, so Theorem 1.2 generalizes the result in [18]. Theorem 1.3 shows that Conjecture 1.1 is true for IC-planar graph G without adjacent triangles with ∆(G) ≥ 10, which improves the corresponding result in [13]. 2. Preliminaries In the following, we always assume that every IC-planar graph G is drawn in a plane such that the number of crossings is as small as possible. If G is an IC-planar graph embedded in the plane, we use C (G) to denote the set of all crossings and E0 (G) to denote the non-crossed edges in G. The associated plane graph G× of G is the plane graph such that V (G× ) = V (G) ∪ C (G) and E(G× ) = E0 (G) ∪ {xz , yz |xy ∈ E(G)\E0 (G) and z is the crossing on xy}. Thus in G× , each crossing becomes a vertex of degree four. For convenience, a vertex in G× is false if it is not a vertex of G and true otherwise. A face f in G× is called a false face if it contains a false vertex. One should be clear that every true vertex in G× is adjacent to at most one false vertex. Theorem 2.1 (Combinatorial Nullstellensatz [1]). Let∑ F be an arbitrary field, and let P = P(x1 , x2 , . . . , xn ) be a polynomial in k k n k F[x1 , x2 , . . . , xn ]. Suppose that the degree of P equals i=1 ki , where each ki is a nonnegative integer, and cP (x11 x22 · · · xnn ) ̸ = 0. If S1 , S2 , . . . , Sn ⊆ F with |Si | > ki for 1 ≤ i ≤ n, then there are s1 ∈ S1 , s2 ∈ S2 , . . . , sn ∈ Sn so that P(s1 , s2 , . . . , sn ) ̸ = 0. Lemma 2.2 ([7]). Let S1 , S2 be the sets of integers with |S1 | = m ≥ 2 and |S2 | = n ≥ 2. Let |L| = {x + y| x ∈ S1 , y ∈ S2 , x ̸ = y}. Then |L| ≥ m + n − 3. Moreover, if |S1 | ̸ = |S2 |, then |L| ≥ m + n − 2. 3. Proof of Theorem 1.2 We will prove Theorem 1.2 by contradiction. Assume that G is a counterexample of Theorem 1.2 such that |V (G)|+|E(G)| is as small as possible. By Theorem 1.4 in [13], Conjecture 1.1 holds for any triangle free IC-planar graph with maximum degree at least 8, so we can assume that ∆(G) ≤ 7. By the choice of G, G does not have any 10-tnsd-coloring, while any proper subgraph G′ of G has a 10-tnsd-coloring φ ′ . We will always extend a 10-tnsd-coloring φ ′ of some G′ to a 10-tnsdcoloring of G to get a contradiction. It is worth noting that we will omit the colors of all 3− -vertices in the following proof. Since they have at most 9 forbidden colors and there are 10 colors, they can be colored easily. In the following proof, we use d(v ) to denote dG (v ). It is easy to see that the following claims given by [15] also hold with the graph G in our proof. Claim (1) (2) (3)
3.1 ([15]). In graph G, the following results hold. Each t − -vertex is not adjacent to any (7 − t)− -vertex, where t = 4, 5. If d(v ) = 6, then d2− (v ) ≤ 1. Furthermore, if d2− (v ) = 1, then d3 (v ) = 0. If d2− (v ) = 0, then d3 (v ) ≤ 2. If d(v ) = 7, then d2− (v ) ≤ 2. Furthermore, if d2− (v ) ≥ 1, then d3− (v ) ≤ 2.
Claim 3.2. In graph G, if d(v ) = 4, then d4 (v ) ≤ 1. Proof. Suppose to the contrary that there exists a 4-vertex v which is adjacent to two 4-vertices v1 , v2 . Since G is triangle free, v1 is not adjacent to v2 . Let N(v ) = {v1 , v2 , v3 , v4 }, N(v1 ) = {v, u2 , u3 , u4 } and N(v2 ) = {v, w2 , w3 , w4 }. Consider G′ = G − vv1 − vv2 , then G′ admits a 10-tnsd-coloring φ ′ . We delete the colors of v, v1 , v2 . Let S1 , S2 , S3 , S4 , S5 be the sets of available colors of vv1 , vv2 , v, v1 , v2 , respectively. It is easy to know that |S1 | ≥ 10 − 5 = 5, |S2 | ≥ 10 − 5 = 5, |S3 | ≥ 10 − 4 = 6, |S4 | ≥ 10 − 6 = 4, |S5 | ≥ 10 − 6 = 4.∏We associate vv1 , vv2 , v, v1 , v2 with the variables ∏ x1 , x2 , x3 , x4 , x5 , 4 respectively. Then we consider P1 (x1 , x2 , x3 , x4 , x5 ) = 1≤i
∏ + x4 − φ ′ (v1 ) + fφ ′ (v1 ) − fφ ′ (uq )) 4r =2 (x2 + x5 − φ ′ (v2 ) + fφ ′ (v2 ) − fφ ′ (wr ))(x2 + x3 + ∑ ∑ φ ′ (vv3 ) + φ ′ (vv4 ) − x4 − 4s=2 φ ′ (v1 us ))(x1 + x3 + φ ′ (vv3 ) + φ ′ (vv4 ) − x5 − 4t =2 φ ′ (v2 wt )). We have cP1 (x41 x42 x43 x34 x25 ) = 4. By Theorem 2.1, there exists xi ∈ Si , for i = 1, 2, . . . , 5 such that P1 (x1 , x2 , x3 , x4 , x5 ) ̸ = 0. We can color vv1 , vv2 , v, v1 , v2 with x1 , x2 , x3 , x4 , x5 respectively to obtain a 10-tnsd-coloring of G, a contradiction. x3 + φ ′ (vv3 ) + φ ′ (vv4 ) − fφ ′ (vp ))
∏4
q=2 (x1
Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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Note that in the forthcoming proofs with the arguments similar to Claim 3.2, we will omit some obvious but tedious statements. Moreover, such as in Claim 3.2, the coefficient of the monomial x41 x42 x43 x34 x25 in the expansion of ∏ P1 (x1 , x2 , x3 , x4 , x5 ) is equal to that of the same monomial in the polynomial Q = 1≤i
Let T be the graph obtained from G by removing all 2− -vertices and T × be the associated plane graph of T . Since G is triangle free, each vertex in T × is incident with at most one false 3-face. In the following, we also use dt (v ) to denote the number of t-vertices adjacent to v in graph G, d(v ) to denote the degree of v in graph G. By Claims 3.1 and 3.3, we have the following facts. Fact 3.1.
δ (T × ) ≥ 3. dT × (v ) = d(v ), for 3 ≤ d(v ) ≤ 5. If d(v ) ≥ 6, then dT × (v ) ≥ 5.
Fact 3.2. (1) If dT × (v ) = 5, then d3 (v ) ≤ 1. Moreover, if d3 (v ) = 1, then dT × (v ) = d(v ) and d4 (v ) = 0. (2) Each 3-face in T × is a false (3, 4, 5+ )-face or a false (4, 4+ , 4+ )-face. Noting that G is triangle free, in T × , a 3-face intersects a 3-face if they share a false vertex. A 3-face is bad if it intersects a 3-face, otherwise, it is good. The false vertex incident with a 3-face f is denoted by cf . Let Nf (cf ) be the set of neighbors of cf which are not incident with f . In T × , a 3-vertex is good if each of its neighbors with degree 5 (if exists) in T × is not incident with any 3-face, otherwise, it is bad. A bad 3-vertex v is of type 1 if v is not incident with any (3, 4, 5)-face (see (a) and (b) in Fig. 1). A bad 3-vertex v is of type 2 if v is incident with a bad (3, 4, 5)-face f ∗ = v cf ∗ u, where f ∗ intersects a 3-facef ⋄ (see (c) in Fig. 1). A bad 3-vertex v is of type 3 if v is incident with a good (3, 4, 5)-face f ∗ = v cf ∗ u (see (d) in Fig. 1). In Fig. 1, we draw a vertex v in black if it has no other neighbors than the ones already depicted, and a vertex v in white if it might have more neighbors than the ones shown in∑ the figure. ∑ × Since T × is a plane graph, by Euler’s formula, we have v∈V (T × ) (dT × (v ) − 4) + f ∈F (T × ) (dT (f ) − 4) = −8. In the following proof, we will use the discharging method to obtain a contradiction. We define the charge function w : V (T × ) ∪ F (T × ) −→ R by w (v ) = dT × (v ) − 4 if v ∈ V (T × ) and w (f ) = dT × (f ) − 4 if f ∈ F (T × ). Next we will give some discharging rules to get a new charge function w ′ after the discharging. After the discharging, the total charge remains unchanged. In the following, we will show that w ′ (x) ≥ 0 for each x ∈ V (T × ) ∪ F (T × ), which leads to a contradiction.
Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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Fig. 1. Illustration of type 1, type 2 and type 3 bad 3-vertices.
The discharging rules are defined as follows. The degree of vertex or face is its degree in T × . R1: Every good 3-vertex v receives 31 from each of its neighbors in G. R2: Every type 1 bad 3-vertex receives 1 from f ∗ (f ∗ must be a (4, 5, 5+ )-face by Facts 3.1–3.2(1)); every type 2 bad 3-vertex receives 1 from f ⋄ (f ⋄ must be a (4, 5+ , 5+ )-face by Claim 3.1 and Facts 3.1–3.2(1)); every type 3 bad 3-vertex receives 12 from each vertex in Nf ∗ (cf ∗ ) (see Fig. 1). R3: Every 5+ -vertex v gives 1 to its incident 3-face (if exists). R4: Every bad (4, 4, 4)-face receives 1 from its intersecting 3-face f ⋄ (f ⋄ must be a (4, 5+ , 5+ )-face by Claims 3.1–3.2 and Facts 3.1–3.2(1)), every good (4, 4, 4)-face f receives 21 from each vertex in Nf (cf ). In the following, a t-face means a t-face in T × . We use n3 (v ) to denote the number of 3-faces incident with v in T × . We have the following fact. Fact 3.3. (1) If dT × (v ) = 5, then v gives at most 1 away. (2) If dT × (v ) = l (l = 6, 7), then v gives at most 1 + 31 d3 (v ) away. Proof. (1) If dT × (v ) = 5, clearly, n3 (v ) ≤ 1. If n3 (v ) = 1, by R1-R4, v gives at most 1 away. If n3 (v ) = 0, by Claim 3.4 and Facts 3.1–3.2(1), there is no path P = v1 v2 v3 v4 in T × such that dT × (v1 ) = dT × (v4 ) = 3 and dT × (v2 ) = dT × (v3 ) = 5. And by Fact 3.2(1), R1, R2 and R4, v gives at most 21 away. (2) If dT × (v ) = l (l = 6, 7), clearly, n3 (v ) ≤ 1. If n3 (v ) = 1, by R1-R4, v gives at most 1 + 31 d3 (v ) away. If n3 (v ) = 0, by R1, R2 and R4, v gives at most 21 + 13 d3 (v ) away. Now we will verify the new charge of each x ∈ V (T × ) ∪ F (T × ). We first consider the new charge of each f ∈ F (T × ). If d(f ) = 3. By Fact 3.2(2), f is a false (3, 4, 5+ )-face or a false (4, 4+ , 4+ )-face. If f is a (3, 4, 5+ )-face or a (4, 4, 4+ )-face. By R2 and R4, f does not give any charge away. By R3 and R4, w ′ (f ) = 3 − 4 + 1 = 0, or w ′ (f ) = 3 − 4 + 2 × 12 = 0. If f is a (4, 5, 5)-face, by Facts 3.1–3.2 and Claim 3.4, there is no path P = v1 v2 v3 v4 in T × such that dT × (v1 ) = dT × (v4 ) = 3 and dT × (v2 ) = dT × (v3 ) = 5. There is at most one 5-vertex incident with f adjacent to a 3-vertex. By Fact 3.2(1), if f intersects a (4, 4, 4)-face, then the two 5-vertices incident with f are not adjacent to any 3-vertex. By R2 and R4, f gives at most 1 away. So by R2-R4, w ′ (f ) ≥ 3 − 4 + 2 × 1 − 1 = 0. If f is a (4, 5, 6+ )-face, by Fact 3.2(1), the 5-vertex incident with f is adjacent to at most one 3-vertex, and if it is adjacent to a 3-vertex, then it is not adjacent to any 4-vertex. That is, if f intersects a (4, 4, 4)-face, then the 5-vertex incident with f is not adjacent to any 3-vertex. By R2 and R4, f gives at most 1 away. So by R2-R4, w ′ (f ) ≥ 3 − 4 + 2 × 1 − 1 = 0. If f is a (4, 6+ , 6+ )-face, noting that a 3-face intersects at most one 3-face, by R2 and R4, f gives at most 1 away. By R2-R4, w ′ (f ) ≥ 3 − 4 + 2 × 1 − 1 = 0. If d(f ) = l (l ≥ 4), no rule applies to f . w ′ (f ) = w (f ) = l − 4 ≥ 0. Next we will consider the new charge of each v ∈ V (T × ). If v is a false vertex of T × , then dT × (v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Next we will consider each true vertex v in T × according to its degree in G. Note that dT × (v ) = d(v ) − d2− (v ). Suppose that d(v ) = 3. By Fact 3.1, dT × (v ) = d(v ) = 3. If v is a good 3-vertex, by R1, w ′ (v ) = 3 − 4 + 3 × 13 = 0; if v is a bad 3-vertex of type 1 or type 2, by R2, w ′ (v ) = 3−4+1 = 0; if v is a bad 3-vertex of type 3, by R2, w ′ (v ) = 3−4+2× 21 = 0. Suppose that d(v ) = 4. By Fact 3.1, dT × (v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Suppose that d(v ) = 5. By Fact 3.1, dT × (v ) = d(v ) = 5. By Fact 3.3(1), w ′ (v ) ≥ 5 − 4 − 1 = 0. Suppose that d(v ) = 6. By Claim 3.1(2), d2− (v ) ≤ 1. If d2− (v ) = 1, then dT × (v ) = 5, by Fact 3.3(1), w ′ (v ) ≥ 5 − 4 − 1 = 0. If d2− (v ) = 0, then dT × (v ) = 6 and d3 (v ) ≤ 2. By Fact 3.3(2), w ′ (v ) ≥ 6 − 4 − 1 − 2 × 31 = 13 . Suppose that d(v ) = 7. By Claim 3.1(3), d2− (v ) ≤ 2. If d2− (v ) = 2, then dT × (v ) = 5, by Fact 3.3(1), w ′ (v ) ≥ 5 − 4 − 1 = 0. If d2− (v ) = 1, then dT × (v ) = 6 and d3 (v ) ≤ 1. By Fact 3.3(2), w ′ (v ) ≥ 6 − 4 − 1 − 1 × 31 = 32 . If d2− (v ) = 0, then dT × (v ) = 7. By Fact 3.1 and Claim 3.5, d3 (v ) ≤ 5. By Fact 3.3(2), w ′ (v ) ≥ 7 − 4 − 1 − 5 × 31 = 13 . Now we get that for each x ∈ V (T × ) ∪ F (T × ), w ′ (x) ≥ 0. Thus we obtain a contradiction. This completes the proof of Theorem 1.2. Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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4. Proof of Theorem 1.3 We will prove Theorem 1.3 by contradiction. Assume that G is a counterexample of Theorem 1.3 such that |V (G)|+|E(G)| is as small as possible. By Theorem 1.5 in [13], Conjecture 1.1 holds for any IC-planar graph G without adjacent triangles with ∆(G) ≥ 12. So we can assume that ∆(G) ≤ 11. Set k = max{∆(G) + 3, 13}. By the choice of G, any proper subgraph G′ of G has a k-tnsd-coloring φ ′ . We will always extend a k-tnsd-coloring φ ′ of some G′ to a k-tnsd-coloring of G to get a contradiction. It is worth noting that we will omit the colors of all 4− -vertices in the following proof. Since they have at most 12 forbidden colors and k ≥ 13, they can be colored easily. In the following, we use d(v ) to denote dG (v ). It is easy to see that the following claims given by [21] also hold with the graph G in our proof. Claim 4.1 ([21]). Every 4− -vertex is not adjacent to any 6− -vertex in G. Claim 4.2 ([21]). Every 5− -vertex is not adjacent to any 5− -vertex in G. Claim 4.3 ([21]). If d(v ) = 7, then d4− (v ) ≤ 1. Claim 4.4. For each 8-vertex v of G, we have d2− (v ) ≤ 1. Moreover, (1) if d2− (v ) = 1, then d3 (v ) = 0. (2) If d2− (v ) = 0, then d3 (v ) ≤ 2. Proof. We will prove d2− (v ) ≤ 1 and (1) at the same time. Suppose to the contrary that there exists an 8-vertex v which is adjacent to a 2− -vertex v1 and a 3− -vertex v2 . Set N(v ) = {v1 , v2 , . . . , v8 }. Consider G′ = G − vv1 − vv2 , then G′ has a k-tnsd-coloring φ ′ . We delete the colors of v1 , v2 . Let S1 , S2 be the sets of available colors of vv1 , vv2 , respectively. It is easy to know that |S1 | ≥ 13 − 8 = 5, |S2 | ≥ 13 − 9 = 4. By Lemma 2.2, |L| ≥ 5 + 4 − 2 = 7(> 6), so we have at least 7 available colors sets for {vv1 , vv2 }. Therefore, we can color vv1 , vv2 properly such that the sum at v is distinct from the sums at v3 , v4 , . . . , v8 , respectively. Noting that v1 , v2 are 3− -vertices, thus we can obtain a k-tnsd-coloring of G, a contradiction. Now we prove (2). Suppose to the contrary that there exists an 8-vertex v with d2− (v ) = 0 and d3 (v ) ≥ 3. Assume that d(vi ) = 3, for i = 1, 2, 3. Consider G′ = G − vv1 − vv2 − vv3 , then G′ has a k-tnsd-coloring φ ′ . We delete the colors of v1 , v2 , v3 . Let S1 , S2 , S3 be the sets of available colors of vv1 , vv2 , vv3 , respectively. It is easy to know that |Si | ≥ 13 − 8 = 5,∑for i = 1∏ , 2, 3. We associate vv1 , vv2 , vv3 with the variables x1 , x2 , x3 , respectively. Then we consider 3 P5 (x1 , x2 , x3 ) = ( k=1 xk )5 1≤i
Now we prove (2). Suppose to the contrary that there exists a 9-vertex v with d2− (v ) = 0 and d3 (v ) ≥ 4. Assume that d(vi ) = 3, for i = 1, 2, 3, 4. Consider G′ = G − vv1 − vv2 − vv3 − vv4 , then G′ admits a k-tnsd-coloring φ ′ . We delete the colors of v, v1 , v2 , v3 , v4 . Let S1 , S2 , S3 , S4 , S5 be the sets of available colors of vv1 , vv2 , vv3 , vv4 , v respectively. It is easy to know that |Si | ≥ 13 − 7 = 6, for i = 1, 2, 3, 4, |S5 | ≥ 13 − 10 = 3. We∑associate∏ vv1 , vv2 , vv3 , vv4 , v with 5 the variables x1 , x2 , x3 , x4 , x5 , respectively. Then we consider P7 (x1 , x2 , x3 , x4 , x5 ) = ( k=1 xk )5 1≤i
For each 10-vertex v of G, if d2− (v ) ≥ 1 and d3− (v ) ≥ 2, then d4− (v ) ≤ 4.
Proof. Suppose to the contrary that there exists a 10-vertex v which is adjacent to a 2− -vertex v1 , a 3− -vertex v2 and three 4− -vertices v3 , v4 , v5 . Consider G′ = G − {vv1 , vv2 , vv3 , vv4 , vv5 }, then G′ admits a k-tnsd-coloring φ ′ . We delete the colors of v1 , v2 , . . . , v5 . Let Si be the sets of available colors of vvi , for i = 1, 2, . . . , 5, respectively. It is easy to know that |S1 | ≥ 13 − 7 = 6, |S2 | ≥ 13 − 8 = 5, |Si | ≥ 13 − 9 = 4, for i = 3, 4 , 5. We associate vvi with the ∑ ∏ 5 variables xi , for i = 1, 2, . . . , 5, respectively. Then we consider P8 (x1 , x2 , x3 , x4 , x5 ) = ( k=1 xk )5 1≤i
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have cP8 (x51 x42 x33 x24 x5 ) = 1. By Theorem 2.1 and noting that v1 , v2 , . . . , v5 are 4− -vertices, we can obtain a k-tnsd-coloring of G, a contradiction. Claim 4.7.
For each 11-vertex v of G, if d2− (v ) ≥ 1, then d3− (v ) ≤ 4.
Proof. Suppose to the contrary that there exists an 11-vertex v which is adjacent to a 2− -vertex v1 and four 3− -vertices v2 , v3 , v4 , v5 . Consider G′ = G − {vv1 , vv2 , vv3 , vv4 , vv5 }, then G′ admits a k-tnsd-coloring φ ′ . We delete the colors of v, v1 , v2 , . . . , v5 . Let S1 , S2 , S3 , S4 , S5 , S6 be the sets of available colors of vv1 , vv2 , vv3 , vv4 , vv5 , v respectively. It is easy to know that |S1 | ≥ 14 − 7 = 7, |Si | ≥ 14 − 8 = 6, for i = 2, 3, 4, 5, |S6 | ≥ 14 − 12 = 2. We associate vv 3 , vv4 , vv5 , v with the variables xi , for i = 1, 2, . . . , 6, respectively. Then we consider P9 (x1 , x2 , x3 , x4 , x5 , x6 ) = ∑16, vv2 , vv∏ ( k=1 xk )6 1≤i
If dT × (v ) = l, then n3 (v ) ≤ ⌈ 2l ⌉.
By Claims 4.1–4.7, we have the following fact. Fact 4.2. (1) δ (T × ) ≥ 3. dT × (v ) = d(v ), for 3 ≤ d(v ) ≤ 6. If d(v ) ≥ 7, then dT × (v ) ≥ 6. (2) Each 3-face in T × is a (3, 4, 7+ )-face, a (3, 7+ , 7+ )-face, a (4, 4, 7+ )-face, a (4, 5+ , 6+ )-face or a (5+ , 6+ , 6+ )-face. × Since T × is a plane graph, by Euler’s formula, we have v∈V (T × ) (dT × (v ) − 4) + f ∈F (T × ) (dT (f ) − 4) = −8. In the following proof, we will use the discharging method to obtain a contradiction. We define the charge function w : V (T × ) ∪ F (T × ) −→ R by w (v ) = dT × (v ) − 4 if v ∈ V (T × ) and w (f ) = dT × (f ) − 4 if f ∈ F (T × ). Next we will give some discharging rules to get a new charge function w ′ after the discharging. After the discharging, the total charge remains unchanged. In the following, we will show that w ′ (x) ≥ 0 for each x ∈ V (T × ) ∪ F (T × ), which leads to a contradiction. The discharging rules are defined as follows. The degree of vertex or face is its degree in T × . R1: Every 3-vertex v receives 31 from each of its neighbors in G. R2: Every 5-vertex v gives 83 to each incident (4, 5, 6+ )-face. R3: Every 6+ -vertex v gives 1 to each incident (3, 4, 7+ )-face and (4, 4, 7+ )-face, gives 85 to each incident (4, 5, 6+ )-face, gives 21 to every other incident 3-faces. By the discharging rules, we obtain the following fact.
∑
Fact 4.3.
∑
If dT × (v ) = l (l ≥ 7), then v gives at most 1 + 12 (⌈ 2l ⌉ − 1) to its incident 3-faces.
Proof. Since T is an IC-planar graph, every 7+ -vertex v in T × is incident with at most two false 3-faces, where at most one of them belongs to a false (3, 4, 7+ )-face, a false (4, 4, 7+ )-face or a false (4, 5, 6+ )-face by Claim 4.2 and Fact 4.2(1). By Fact 4.1 and R3, v gives at most 1 + 12 (⌈ 2l ⌉ − 1) to its incident 3-faces. Now we will verify the new charge of each x ∈ V (T × ) ∪ F (T × ). We first consider the new charge of each f ∈ F (T × ). If d(f ) = 3, by Fact 4.2(2), f is a (3, 4, 7+ )-face, a (3, 7+ , 7+ )-face, a (4, 4, 7+ )-face, a (4, 5+ , 6+ )-face or a (5+ , 6+ , 6+ )face. If f is a (3, 4, 7+ )-face or a (4, 4, 7+ )-face, by R3, w ′ (f ) = 3 − 4 + 1 = 0. If f is a (4, 5, 6+ )-face, by R2 and R3, w ′ (f ) = 3 − 4 + 38 + 85 = 0. If f is a (3, 7+ , 7+ )-face or a (4+ , 6+ , 6+ )-face, by R3, w ′ (f ) ≥ 3 − 4 + 2 × 12 = 0. If d(f ) = l (l ≥ 4), no rule applies to f . w ′ (f ) = w (f ) = l − 4 ≥ 0. Next we will consider the new charge of each v ∈ V (T × ). If v is a false vertex of T × , then dT × (v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Next we will consider each true vertex v in T × according to its degree in G. Note that dT × (v ) = d(v ) − d2− (v ). Suppose that d(v ) = 3. By Fact 4.2(1), dT × (v ) = d(v ) = 3. By R1, w ′ (v ) = 3 − 4 + 3 × 13 = 0. Suppose that d(v ) = 4. By Fact 4.2(1), dT × (v ) = d(v ) = 4. No rule applies to v , w ′ (v ) = 4 − 4 = 0. Suppose that d(v ) = 5. By Claim 4.2 and Fact 4.2(1), dT × (v ) = d(v ) = 5 and v is incident with at most two (4, 5, 6+ )-faces. By R2, w ′ (v ) ≥ 5 − 4 − 2 × 83 = 41 . Suppose that d(v ) = 6. By Claim 4.1, dT × (v ) = d(v ), d3 (v ) = d4 (v ) = 0. By Claim 4.2 and Fact 4.2(1), a 5-vertex is not adjacent to any 5-vertex in T × , so v is incident with at most one (4, 5, 6)-face noting that v is adjacent to at most one false 4-vertex. By Fact 4.1, n3 (v ) ≤ 3. By R3, w ′ (v ) ≥ 6 − 4 − 85 − 2 × 12 = 38 . Suppose that d(v ) = 7. By Claim 4.3, d2− (v ) ≤ 1. If d2− (v ) = 1, then dT × (v ) = 6, d3 (v ) = d4 (v ) = 0. We can verify that w′ (v ) ≥ 0 similarly as the above proof for d(v ) = 6. If d2− (v ) = 0, then dT × (v ) = 7. By Claim 4.3 and Fact 4.1, d3 (v ) ≤ 1 and n3 (v ) ≤ 4. By R1 and Fact 4.3, w ′ (v ) ≥ 7 − 4 − 31 − 1 − 3 × 12 = 16 . Please cite this article as: C. Song, X. Jin and C.-Q. Xu, Neighbor sum distinguishing total coloring of IC-planar graphs with short cycle restrictions, Discrete Applied Mathematics (2020), https://doi.org/10.1016/j.dam.2019.12.023.
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Suppose that d(v ) = 8. By Claim 4.4, d2− (v ) ≤ 1. If d2− (v ) = 1, then dT × (v ) = 7 and d3 (v ) = 0. By Fact 4.3, w′ (v ) ≥ 7 − 4 − 1 − 3 × 21 = 12 . If d2− (v ) = 0, then dT × (v ) = 8 and d3 (v ) ≤ 2. By Fact 4.1, n3 (v ) ≤ 4. By R1 and Fact 4.3, w ′ (v ) ≥ 8 − 4 − 2 × 13 − 1 − 3 × 12 = 56 . Suppose that d(v ) = 9. By Claim 4.5, d2− (v ) ≤ 2. If d2− (v ) ≥ 1, then d3 (v ) ≤ 1. If d2− (v ) = i, then dT × (v ) = 9 − i, where i = 1, 2. We can verify that w ′ (v ) ≥ 0 similarly as the above proof for d(v ) = 9 − i, for i = 1, 2. If d2− (v ) = 0, then dT × (v ) = 9 and d3 (v ) ≤ 3. By Fact 4.1, n3 (v ) ≤ 5. By R1 and Fact 4.3, w ′ (v ) ≥ 9 − 4 − 3 × 31 − 1 − 4 × 12 = 1. Suppose that d(v ) = 10. By Claim 4.6, d2− (v ) ≤ 4. If d2− (v ) = i, then d3 (v ) ≤ 4 − i, dT × (v ) = 10 − i, where i = 1, 2, 3, 4. We can verify that w ′ (v ) ≥ 0 similarly as the above proof for d(v ) = 10 − i, for i = 1, 2, 3, 4. If d2− (v ) = 0, then dT × (v ) = 10. We consider the following two cases: Case 1. d3 (v ) = 10. By Claim 4.1 and Fact 4.2(1), a 3-vertex is not adjacent to any 3-vertex in T × , so n3 (v ) ≤ 1. By R1 and R3, w ′ (v ) ≥ 10 − 4 − 10 × 31 − 1 = 53 . Case 2. d3 (v ) ≤ 9. By Fact 4.1, n3 (v ) ≤ 5. By R1 and Fact 4.3, w ′ (v ) ≥ 10 − 4 − 9 × 13 − 1 − 4 × 21 = 0. Suppose that d(v ) = 11. By Claim 4.7, d2− (v ) ≤ 4, and if d2− (v ) = i, then d3 (v ) ≤ 4 − i and dT × (v ) = 11 − i, for i = 1, 2, 3, 4. We can verify that w ′ (v ) ≥ 0 similarly as the above proof for d(v ) = 11 − i, for i = 1, 2, 3, 4. If d2− (v ) = 0, then dT × (v ) = 11. We consider the following two cases: Case 1. d3 (v ) = 11. By Claim 4.1 and Fact 4.2(1), a 3-vertex is not adjacent to any 3-vertex in T × , so n3 (v ) ≤ 1. By R1 and R3, w ′ (v ) ≥ 11 − 4 − 11 × 13 − 1 = 73 . Case 2. d3 (v ) ≤ 10. By Fact 4.1, n3 (v ) ≤ 6. By R1 and Fact 4.3, w ′ (v ) ≥ 11 − 4 − 10 × 13 − 1 − 5 × 12 = 16 . Now we get that for each x ∈ V (T × ) ∪ F (T × ), w ′ (x) ≥ 0. 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