Planar Graphs with the Distance of 6^--Cycles at Least 2 from Each Other Are DP-3-Colorable

Abstract

DP-coloring as a generalization of list coloring was introduced by Dvořák and Postle recently. In this paper, we prove that every planar graph in which the distance between 6${}^{-}$-cycles is at least 2 is DP-3-colorable, which extends the result of Yin and Yu [Discret. Math. 2019, 342, 2333–2341].

1. Introduction

All graphs are finite and simple in this paper. Let G be a plane graph and V, E, and F be sets of vertices, edges, and faces of G, respectively. Two faces are adjacent if they have a common edge. For a face $f\in F$, we write $f=[v_1,v_2,\cdots ,v_k]$ when the vertices on f in a cyclic order are $v_1,v_2,\cdots ,v_k$. A k-vertex ($k^{-}$-vertex, $k^{+}$-vertex) is a vertex of degree k (at most k, at least k). A k-face ($k^{-}$-face, $k^{+}$-face) is a face of degree k (at most k, at least k). The notation will be same for cycles. A triangle is a 3-cycle in G. A vertex or an edge of G is triangular when it is on a triangle. We say a chord is triangular in a cycle C if it splits the cycle C into at least one triangle. Let an ($l_1,l_2,\cdots ,l_k$)-face be a k-face $f=[v_1{v}_2\cdots v_k]$ with $d\left(v_i\right)=l_i$. Let ($l_1,l_2$)-edge be an edge $e=v_1{v}_2$ with $d\left(v_i\right)=l_i$. Let $\left|C\right|$ be the length (number of edges) of the cycle C. Let $\left|f\right|$ be the number of edges incident with f. Let $Ext\left(C\right)$ and $Int\left(C\right)$ denote the sets of vertices lying outside and inside of C, respectively. A cycle C is called separating if $Ext\left(C\right)\ne \varnothing$ and $Int\left(C\right)\ne \varnothing$. The distance $d(u,v)$ between two vertices u and v in G is the length (number of edges) of the shortest path between them. The distance $d(C,C^{\prime })$ between two cycles C and $C^{\prime }$ in G is the minimum of the distances between vertices $u\in V\left(C\right)$ and $v\in V\left(C^{\prime }\right)$. A matching of G is a set of independent edges in G.

A proper k-coloring of G is a function $f:V\left(G\right)\to \{1,2,\cdots ,k\}$ such that $f\left(u\right)\ne f\left(v\right)$ for every edge $uv\in E\left(G\right)$. Let $\chi \left(G\right)$, the chromatic number of G, be the smallest k such that G is k-colorable. A list assignment of G is a mapping L that assigns to each vertex $v\in V\left(G\right)$ a list $L\left(v\right)$ of colors. An L-coloring of G is a function $f:V\to {\bigcup }_{v\in V}L\left(v\right)$ such that $f\left(v\right)\in L\left(v\right)$ for every $v\in V$ and $f\left(u\right)\ne f\left(v\right)$ for every edge $uv\in E\left(G\right)$. A graph G is k-choosable if G has a L-coloring for every assignment L with $\left|L\right(v\left)\right|\ge k$. Let ${\chi }_l\left(G\right)$, the choice number of G, be the smallest k such that G is k-choosable.

It is well known that 3-COLORING is NP-complete for planar graphs. This provides motivation for finding some sufficient conditions for 3-coloring of planar graphs. In 1959, Grötzsch [1] proved that planar graphs with no triangles are 3-colorable. In 1969, Havel [2] asked whether there exists or not a constant d such that if G is a planar graph with the distance of triangles at least d, then G is 3-colorable. Borodin and Glebov [3] proved that every planar graph with no 5-cycles and $d=2$ is 3-colorable. Dvor̂ák, Kral, and Thomas [4] showed that for every planar graph $d={10}^{100}$ suffices.

List coloring was introduced as a generalization of proper coloring by Vizing [5] and independently by Erdos, Rubin, and Taylor [6]. Thomassen [7] showed that planar graphs with girth at least 5 are 3-choosable. Dvor̂ák [8] proved that planar graphs with the distance of 4${}^{-}$-cycles from each other at least 26 are 3-choosable.

There are fewer techniques to approach list problems than ordinary coloring. Identifications of vertices are involved in the reduction configurations for ordinary coloring. However, in list coloring, because different vertices have different lists, it is not possible to use identification of vertices. The concept of DP-coloring as a generalization of list coloring, was introduced by Dvořák and Postle [9].

Definition 1.

Let G be a simple graph, and L be a list assignment of $V\left(G\right)$. For each vertex $v\in V\left(G\right)$, let $L_v=\left\{v\right\}\times L\left(v\right)$. For each edge $uv$ in G, let $M_{uv}$ be a partial matching between the sets $L_u$ and $L_v$ and let $\mathcal{M}=\{M_{uv}:uv\in E\left(G\right)\}$, called the matching assignment. The matching assignment is called a k-matching assignment if $L\left(v\right)=\left[k\right]$ for each $v\in V\left(G\right)$.

Definition 2.

A $\mathcal{M}$-coloring of G is a function ϕ that assigns each vertex $v\in V\left(G\right)$ a color $\varphi \left(v\right)\in L\left(v\right)$, such that for every $uv\in E\left(G\right)$, the vertices $(u,\varphi (u\left)\right)$ and $(v,\varphi (v\left)\right)$ are not adjacent in $M_{uv}$. We say that G is $\mathcal{M}$-colorable if such a $\mathcal{M}$-coloring exists.

Definition 3.

The graph G is DP-k-colorable if, for each k-list assignment L and each matching assignment $\mathcal{M}$ over L, it has an $\mathcal{M}$-coloring. The minimum k such that G is DP-k-colorable is the DP-chromatic number of G, denoted by ${\chi }_{DP}\left(G\right)$.

If every $(u,c_1)(v,c_2)\in E\left(M_{u,v}\right)$ satisfies $c_1=c_2$, then $uv\in E\left(G\right)$ is straight in a k-matching assignment $\mathcal{M}$. Dvořák and Postle [9] proved that planar graphs with no cycles of length from 4 to 8 are 3-choosable and noted that ${\chi }_{DP}\left(G\right)\le 3$ if G is a planar graph with no 4${}^{-}$-cycles. Liu and Li [10] proved that planar graphs without adjacent cycles of length at most 8 are 3-choosable. Zhao and Miao [11] proved that every planar graph in which the distance between 5${}^{-}$-cycles is at least 2 is DP-3-colorable. Bernshteyn et al. [12,13,14,15,16] gave some results on DP-coloring. DP-3-colorable planar graphs can be found in [17,18] and DP-4-colorable planar graphs can be found in [19,20,21]. Yin and Yu [22] proved planar graphs with no $\{4,5,6\}$-cycles in which the distance between triangles is at least 2 are DP-3-colorable. We present the following result in this paper.

Theorem 1.

Let G be a planar graph in which the distance between 6${}^{-}$-cycles is at least 2. Let $C_0$ be a ${10}^{-}$-cycle in G. Then, for every DP-3-coloring ${\varphi }_0$ of $C_0$, there exists a DP-3-coloring of G whose restriction to $C_0$ is ${\varphi }_0$.

Corollary 1.

Every planar graph in which the distance between 6${}^{-}$-cycles is at least 2 is DP-3-colorable.

Proof. 

Let G be a planar graph. By Dvořák and Postle [9], if G is 4${}^{-}$-cycle free then G is DP-3-colorable. So, we may assume that G contains a $4^{-}$-cycle and the $4^{-}$-cycle can be precolored. Then, G has a DP-3-coloring extended from the coloring of the $4^{-}$-cycle by Theorem 1 when the distance between 6${}^{-}$-cycles is at least 2 in G. □

2. Proof of Theorem 1

To prove Theorem 1, we use the reductio ad absurdum. Let G be a counterexample with the least number of vertices to Theorem 1. If G is 10${}^{-}$-cycle free then G is DP-3-colorable by Dvořák and Postle [9]. So, we may assume that G contains a ${10}^{-}$-cycle $C_0$.

The following Lemma 1 to Lemma 8 are about some crucial properties of the minimal counterexample G.

Lemma 1.

If $v\in V(G-C_0)$, then $d\left(v\right)\ge 3$.

Proof. 

Let $v\in V(G-C_0)$ and $d\left(v\right)\le 2$. Because G is a minimal counterexample, we can first extend ${\varphi }_0$ of $C_0$ to $V\left(G\right)-\left\{v\right\}$. Then we can select a color $\varphi \left(v\right)$ for v such that $(v,\varphi \left(v\right))(u,\varphi \left(u\right))\notin E\left(M_{uv}\right)$ for each neighbor u of v. Therefore, G has been colored, a contradiction. □

Lemma 2.

$C_0$ is the boundary of the outer face.

Proof. 

First, we show that $C_0$ is not separating. For otherwise, if $C_0$ is separating, G can be colored by extending the coloring of $C_0$ to both $Int\left(C_0\right)$ and $Ext\left(C_0\right)$, a contradiction. Therefore, either $Int\left(C_0\right)$ or $Ext\left(C_0\right)$ is empty. Then we may assume that $Ext\left(C_0\right)$ is empty without loss of generality. So $C_0$ is the boundary of the outer face. □

Lemma 3.

There exist no separating ${10}^{-}$-cycles.

Proof. 

By Lemma 2, $C_0$ is not a separating ${10}^{-}$-cycle. Let $C\ne C_0$ be a separating ${10}^{-}$-cycle in G. Because G is a minimal counterexample, we can first extend ${\varphi }_0$ of $C_0$ to $G-Int\left(C\right)$. Then the coloring of the cycle C can be extended to $Int\left(C\right)$. Therefore, G has been colored, a contradiction. □

Lemma 4.

Let C be a cycle in G. If $\left|C\right|\le 7$, then C has no chord. If C, $8\le \left|C\right|\le 10$, has a chord e, then either e is triangular, or e splits C into a 7-cycle and a 4-cycle when $\left|C\right|=9$, or e splits C into a 8-cycle and a 4-cycle when $\left|C\right|=10$, or e splits C into a 7-cycle and a 5-cycle when $\left|C\right|=10$.

Proof. 

As the distance between $6^{-}$ cycles is at least 2 in G, C cannot have a chord if $\left|C\right|\le 7$. If $\left|C\right|=8$, then C can only have a triangular chord. If $\left|C\right|=9$, then either e is triangular or e splits C into a 7-cycle and a 4-cycle. If $\left|C\right|=10$, then either e is triangular or e splits C into an 8-cycle and a 4-cycle, or e splits C into a 7-cycle and a 5-cycle. □

Lemma 5.

$C_0$ has no chord.

Proof. 

If $C_0$ has a chord e, then e must be one of the cases described in Lemma 4. Because G has no separating ${10}^{-}$-cycles by Lemma 3, G has no other vertices except the vertices on $C_0$. Then the coloring of $C_0$ is a coloring of G, a contradiction. □

Lemma 6

([18]). Let $k\ge 3$ and H be a subgraph of G. If the vertices of H can be ordered as $v_1,v_2,\cdots ,v_l$ such that the following hold

(1) 

$v_1{v}_l\in E\left(G\right)$, and$v_1$has no neighbor in$G-H$,

(2) 

$d\left(v_l\right)\le k$ and $v_l$ has at least one neighbor outside of H,

(3) 

for each$2\le i\le l-1$,$v_i$has at most$k-1$neighbors in$G\left[v_1,\cdots ,v_{i-1}\right]\bigcup (G-H)$,

then a DP-k-$coloring$of$G-H$can be extended to a DP-k-$coloring$of G.

A vertex is internal if it is not incident with $C_0$ and a face is internal if it contains no vertex on $C_0$.

Lemma 7.

Let f be an internal 7-face in G. If all vertices on f are vertices with degree 3, then f cannot be adjacent to an internal $6^{-}$-face $f_1$ such that all vertices on $f_1$ are 3-vertices.

Proof. 

Let $f=\left[v_1{v}_2{w}_1{w}_2{w}_3{w}_4{w}_5\right]$ and $f_1=[v_1{v}_2\cdots v_i](i\in \{3,4,5,6\})$ such that $v_1{v}_2$ is the common edge of f and $f_1$, and all vertices on f and $f_1$ are vertices with degree 3. Let $H=\{v_1,w_5,w_4,w_3,w_2,w_1,v_2,v_3,\cdots ,v_i\}$$(i\in \{3,4,5,6\left\}\right)$. Order the vertices in H as $v_1,w_5,w_4,w_3,w_2,$$w_1,v_2,v_3,\cdots ,v_i$$(i\in \{3,4,5,6\left\}\right)$. Since f and $f_1$ are internal faces, no vertex in H is on $C_0$. Because G is a minimal counterexample, we can first extend ${\varphi }_0$ of $C_0$ to $G-H$. Then by Lemma 6, the coloring of $G-H$ can be extended to a coloring of G, a contradiction. □

Lemma 8.

Let f be an internal 7-face in G. Let $f_1$ be an internal $6^{-}$-face which is adjacent to f. If except one vertex of f, all other vertices on f and $f_1$ are 3-vertices, then each of following holds:

(a) 

If f contains a (3, 4)-edge and is adjacent to another internal $6^{-}$-face $f_2$ with the common (3, 4)-edge, then $f_2$ has another vertex with degree at least 4.

(b) 

f cannot be adjacent to another internal $6^{-}$-face $f_2$ such that all vertices on $f_2$ are 3-vertices.

Proof. 

Let $f=[v_1{v}_2\cdots v_7]$ and $f_1=[v_1{v}_2{w}_1\cdots w_i]$$(i\in \{1,2,3,4\left\}\right)$ with the common (3, 3)-edge $v_1{v}_2$. Since the $6^{-}$-cycles in G are at a distance of at least 2 from each other, by symmetry we may assume that the edge $v_4{v}_5$ is on $f_2$ and $f_2=[v_4{v}_5{u}_1\cdots u_j]$$(j\in \{1,2,3,4\left\}\right)$.

(a)

Suppose otherwise that all vertices in $\{u_1\cdots u_j\}$$(j\in \{1,2,3,4\left\}\right)$ are vertices with degree 3. If $d\left(v_4\right)=4$, then Let H be the set of vertices listed as: $v_2,v_3,v_4,u_j,\cdots ,u_1,v_5,$$v_6,v_7,v_1,w_i,\cdots ,w_1$$(i\in \{1,2,3,4\left\}\right)$ and $(j\in \{1,2,3,4\left\}\right)$. If $d\left(v_5\right)=4$, then Let H be the set of vertices listed as: $v_1,v_7,v_6,v_5,u_1,\cdots ,u_j,v_4,v_3,v_2,$$w_1,\cdots ,w_i$$(i\in \{1,2,3,4\left\}\right)$ and $(j\in \{1,2,3,4\left\}\right)$. Since f, $f_1$ and $f_2$ are internal faces, no vertex in H is on $C_0$. Because G is a minimal counterexample, we can first extend ${\varphi }_0$ of $C_0$ to $G-H$. Then by Lemma 6, the coloring of $G-H$ can be extended to a coloring of G, a contradiction.

(b)

Suppose otherwise that $f_2$ is an internal $6^{-}$-face and that all vertices on $f_2$ are 3-vertices. Since f has six vertices with degree 3, by symmetry we assume that $d\left(v_6\right)=3$. Let u be the neighbor of $u_1$ not on $f_2$. We can rename the lists of vertices in $\{u_1,v_5,v_4,v_6,v_7\}$ such that each edge in $\{u{u}_1,u_1{v}_5,v_4{v}_5,v_5{v}_6,v_6{v}_7\}$ is straight. Consider the graph $G^{\prime }$ obtained from $G-\{v_6,v_5,v_4,u_1,\cdots ,u_j\}$$(j\in \{1,2,3,4\left\}\right)$ by identifying $v_7$ and u. We claim that no new loops, multiple edges or cycles with length 3, 4, 5 or 6 are created. Otherwise, there is a $\{1,2,3,4,5,6\}$-path from $v_7$ to u in G, which together with $v_6,v_5,u_1$ forms a cycle C, $5\le d\left(C\right)\le 10$. Since $f_2$ is a $6^{-}$-face, C cannot be a $6^{-}$-cycle.

• If $v_4$ is in $Int\left(C\right)$ see Figure 1a, then C is a separating $\{7,8,9,10\}$-cycle, it is a contradiction to Lemma 3.
• If $v_4$ is not in $Int\left(C\right)$ see Figure 1b. Since f is a 7-cycle and $d\left(v_6\right)=3$, by Lemma 3 and 4, $v_6$ must be incident with an edge e in $Int\left(C\right)$. The other end vertex of e is either on C or not. If it is on C, then e is a chord of C. By Lemma 4, $8\le d\left(C\right)\le 10$ and e is on a $5^{-}$-cycle $C^{\prime }$. Then the distance between $C^{\prime }$ and $f_2$ is at most 1, a contradiction. If it is not on C, then it is in $Int\left(C\right)$. So C must be a separating $\{7,8,9,10\}$-cycle, it is a contradiction to Lemma 3. Because none of $v_7$ and u is on a $6^{-}$-cycle, the $6^{-}$-cycles in $G^{\prime }$ are at a distance of at least 2 from each other. Now, we claim that no new chord in $C_0$ is formed in $G^{\prime }$. For otherwise, u is on $C_0$ and $v_7$ is adjacent to a vertex $v_7^{\prime }$ on $C_0$, then there is a path between $v_7^{\prime }$ and u on $C_0$ with length at most five, which forms a $\{6,7,8,9,10\}$-cycle with $u_1,v_5,v_6,v_7$. Similar to the proof above, it does not occur.
  Since $C_0$ is still the boundary of the outer face of the embedding of $G^{\prime }$, the coloring of $C_0$ can be extended to $G^{\prime }$ by minimality of G. Now keep the colors of all vertices in $G^{\prime }$ and color $v_7$ and u with the color of the identified vertex. Now color $v_6$, and then color $u_1$ with the color of $v_6$. We can do this because the edges in $\{u{u}_1,u_1{v}_5,v_4{v}_5,v_5{v}_6,v_6{v}_7\}$ are straight and the color of $v_6$ is different from the color of $v_7$ and u. If $|f_2|=3$, then we color $v_4,v_5$ in the order. If $|f_2|=4,5$ or 6, then we color $u_2,\cdots ,u_j,v_4,v_5$$(j\in \{2,3,4\left\}\right)$ in the order. Then we obtain a coloring of G, a contradiction.

□

Let $f_0$ be the outer face of the embedding of G. We are now ready to present a discharging procedure. We set the initial charge of every vertex $v\in V\left(G\right)$ to be $\mu \left(v\right)=2d\left(v\right)-6$, of every face $f\ne f_0$ in our fixed plane drawing of G to be $\mu \left(f\right)=\left|f\right|-6$, and set $\mu \left(f_0\right)=\left|f_0\right|+6$. Then ${\sum }_{x\in V\bigcup F}\mu \left(x\right)=0$ by Euler’s Formula.

let $F_k=\{f:$$V\left(f\right)\bigcap V\left(C_0\right)\ne \varnothing$ and f be a k-face}. A 7-face f is special when f is in $F_7$ and adjacent to two internal $6^{-}$-faces. We say a vertex v is rich to a $7^{+}$-face f when v is on f and not on a $5^{-}$-face which is adjacent to f.

The discharging rules:

(R1): If v is an internal 4-vertex and on a $5^{-}$-face f, then v gives $\frac{3}{2}$ to f.

(R2): If v is an internal $5^{+}$-vertex, then v gives $\frac{3}{2}$ to its incident $5^{-}$-face if any and $\frac{1}{2}$ to its incident 7-face if any.

(R3): Each $7^{+}$-face f ($f\ne f_0$) gives $\frac{1}{2}$ to its adjacent internal $(3,3,4^{+})$-face if any, $\frac{1}{8}$ to its adjacent internal $(3,3,3,4^{+})$-face if any, 1 to its adjacent internal $(3,3,3)$-face if any, $\frac{1}{2}$ to its adjacent internal $(3,3,3,3)$-face if any and $\frac{1}{5}$ to its adjacent internal $(3,3,3,3,3)$-face if any.

(R4): Each internal 7-face receives $\frac{1}{2}$ from its incident rich 4-vertex.

(R5): After (R3) and (R4), each $7^{+}$-face gives all its remaining charge to $f_0$.

(R6): The outer face $f_0$ receives $\mu \left(v\right)$ from each $v\in C_0$, gives 1 to each special 7-face if any, 3 to each face in $F_3$ if any, 2 to each face in $F_4$ if any and 1 to each face in $F_5$ if any.

Let ${\mu }^{*}\left(x\right)$ denote the final charge of $x\in V\bigcup F$. To lead to a contradiction, we will prove that ${\mu }^{*}\left(x\right)\ge 0$ for all $x\in V\bigcup F\backslash \left\{f_0\right\}$ and ${\mu }^{*}\left(f_0\right)>0$.

Lemma 9.

For all $v\in V$, ${\mu }^{*}\left(v\right)\ge 0$.

Proof. 

Since $f_0$ receives $\mu \left(v\right)$ from each $v\in C_0$ by (R6) whether $\mu \left(v\right)$ is positive or not, ${\mu }^{*}\left(v\right)=0$ when v is on $C_0$. Let v be an internal vertex in G, then by Lemma 1 $d\left(v\right)\ge 3$. If $\mathit{d}\left(\mathit{v}\right)=\mathbf{3}$, then ${\mu }^{*}\left(v\right)=2d\left(v\right)-6=0$.

Because the $6^{-}$-cycles in G are at a distance of at least 2 from each other, each vertex can be incident with at most one $6^{-}$-face. Let $\mathit{d}\left(\mathit{v}\right)=\mathbf{4}$. If v is on a $5^{-}$-face f, then v gives $\frac{3}{2}$ to f and $\frac{1}{2}$ to its incident 7-face when v is rich to the 7-face by (R1) and (R4). If v is not on a $5^{-}$-face, then by (R4) v gives at most $\frac{1}{2}$ to each incident face. Thus, ${\mu }^{*}\left(v\right)\ge 2d\left(v\right)-6-max\{\frac{3}{2}+\frac{1}{2},\frac{1}{2}\times 4\}=0$.

Let $\mathit{d}\left(\mathit{v}\right)\ge \mathbf{5}$. By (R2), v gives $\frac{3}{2}$ to its incident $5^{-}$-face if any and at most $\frac{1}{2}$ to each other incident face. Thus, ${\mu }^{*}\left(v\right)\ge 2d\left(v\right)-6-\frac{3}{2}-\frac{1}{2}\times (d\left(v\right)-1)>0$. □

Lemma 10.

For all $f\in F-\left\{f_0\right\}$, ${\mu }^{*}\left(f\right)\ge 0$.

Proof. 

Let $\left|\mathit{f}\right|=\mathbf{3}$. If $V\left(f\right)\bigcap V\left(C_0\right)\ne \varnothing$, then by (R6) f receives 3 from $f_0$, so ${\mu }^{*}\left(f\right)=\left|f\right|-6+3=0$. Now let $V\left(f\right)\bigcap V\left(C_0\right)=\varnothing$. If $V\left(f\right)$ contains at least two $4^{+}$-vertices, then by (R1) and (R2) f receives $\frac{3}{2}$ from each of the $4^{+}$-vertices. So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{3}{2}\times 2=0$. If $V\left(f\right)$ contains exactly one $4^{+}$-vertex, then f receives $\frac{3}{2}$ from the $4^{+}$-vertex and receives $\frac{1}{2}$ from each of its adjacent $7^{+}$-faces by (R1), (R2) and (R3). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{3}{2}+\frac{1}{2}\times 3=0$. If f is an internal $(3,3,3)$-face, then f receives 1 from each of the adjacent $7^{+}$-faces by (R3). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+1\times 3=0$.

Let $\left|\mathit{f}\right|=\mathbf{4}$. If $V\left(f\right)\bigcap V\left(C_0\right)\ne \varnothing$, then by (R6) f receives 2 from $f_0$, so ${\mu }^{*}\left(f\right)=\left|f\right|-6+2=0$. Now let $V\left(f\right)\bigcap V\left(C_0\right)=\varnothing$. If $V\left(f\right)$ contains at least two $4^{+}$-vertices, then f receives $\frac{3}{2}$ from each of the $4^{+}$-vertices by (R1) and (R2). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{3}{2}\times 2>0$. If $V\left(f\right)$ contains exactly one $4^{+}$-vertex, then f receives $\frac{3}{2}$ from the $4^{+}$-vertex and receives $\frac{1}{8}$ from each adjacent $7^{+}$-face by (R1), (R2) and (R3). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{3}{2}+\frac{1}{8}\times 4=0$. If f is an internal $(3,3,3,3)$-face, then f receives $\frac{1}{2}$ from each adjacent $7^{+}$-face by (R3). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{1}{2}\times 4=0$.

Let $\left|\mathit{f}\right|=\mathbf{5}$. If $V\left(f\right)\bigcap V\left(C_0\right)\ne \varnothing$, then by (R6) f receives 1 from $f_0$, so ${\mu }^{*}\left(f\right)=\left|f\right|-6+1=0$. Now let $V\left(f\right)\bigcap V\left(C_0\right)=\varnothing$. If $V\left(f\right)$ contains a $4^{+}$-vertex, then f receives $\frac{3}{2}$ from the $4^{+}$-vertex by (R1) and (R2). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{3}{2}>0$. If f is an internal (3, 3, 3, 3, 3)-face, then f receives $\frac{1}{5}$ from each adjacent $7^{+}$-face by (R3). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6+\frac{1}{5}\times 5=0$.

Let $\left|\mathit{f}\right|=\mathbf{6}$, by our rules f sends out nothing, so ${\mu }^{*}\left(f\right)=\left|f\right|-6=0$.

Let $\left|\mathit{f}\right|\ge \mathbf{7}$. By (R3) f needs to give $\frac{1}{2}$ to its adjacent internal $(3,3,4^{+})$-faces if any, $\frac{1}{8}$ to its adjacent internal $(3,3,3,4^{+})$-faces if any, 1 to its adjacent internal $(3,3,3)$-faces if any, $\frac{1}{2}$ to its adjacent internal $(3,3,3,3)$-faces if any and $\frac{1}{5}$ to its adjacent internal $(3,3,3,3,3)$-faces if any. Since the distance between $6^{-}$-cycles is at least 2, f is adjacent to at most $\lfloor \frac{\left|f\right|}{3}\rfloor$ internal $6^{-}$-faces. If $\left|\mathit{f}\right|\ge \mathbf{8}$, then ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1\times \lfloor \frac{\left|f\right|}{3}\rfloor \ge 0$.

Let $\left|\mathit{f}\right|=\mathbf{7}$. Since the distance between $6^{-}$-cycles is at least 2, f is adjacent to at most 2 internal $6^{-}$-faces. Let $V\left(f\right)\bigcap V\left(C_0\right)\ne \varnothing$. If f is adjacent to at most one internal $6^{-}$-face, then by R(3) f gives at most 1 to the adjacent $6^{-}$-face if any. So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1=0$. If f is special, then f gives at most 1 to the adjacent $6^{-}$-faces and receives 1 from $f_0$, so ${\mu }^{*}\left(f\right)=\left|f\right|-6-2+1=0$. Now let $V\left(f\right)\bigcap V\left(C_0\right)=\varnothing$. If f is adjacent to at most one internal $6^{-}$-face, then by R(3) f gives at most 1 to the adjacent $6^{-}$-face if any. So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1=0$. Let f be adjacent to two internal $6^{-}$-faces. If none of the $6^{-}$-faces is a $(3,3,3)$-face, then by R(3) f gives at most $\frac{1}{2}\times 2$ to the $6^{-}$-faces, so ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1=0$. If one of the $6^{-}$-faces has at least two $4^{+}$-vertices, then the $6^{-}$-face receives nothing from f by (R3), so ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1=0$. So, we assume that f is adjacent to a $(3,3,3)$-face $f_1$ and another $6^{-}$-face $f_2$ that $f_2$ has at most one $4^{+}$-vertex. By Lemma 7, f has at least one $4^{+}$-vertex.

• If $f_2$ shares a $(3,4^{+})$-edge with f, then f contains another $4^{+}$-vertex $v^{\prime }$. For otherwise, $f_2$ has at least two $4^{+}$-vertices by Lemma 8(1), a contradiction. Since $\left|f\right|=7$ and the distance between $6^{-}$-cycles is at least 2, f can be adjacent to at most two $6^{-}$-faces. So, if $d\left(v^{\prime }\right)=4$, then $v^{\prime }$ must be rich and gives $\frac{1}{2}$ to f by (R4). If $d\left(v^{\prime }\right)\ge 5$, then $v^{\prime }$ gives $\frac{1}{2}$ to f by (R2). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1-\frac{1}{2}+\frac{1}{2}=0$.
• If $f_2$ shares a $(3,3)$-edge with f. If all vertices on $f_2$ are 3-vertices, then f contains at least two $4^{+}$-vertices. For otherwise, $f_2$ cannot be a $6^{-}$-face that all vertices on $f_2$ are 3-vertices by Lemma 8(2). Since $\left|f\right|=7$ and the distance between $6^{-}$-cycles is at least 2, f can be adjacent to at most two $6^{-}$-faces. So, if one of the $4^{+}$-vertices is a 4-vertex, then it must be rich and gives $\frac{1}{2}$ to f by (R4). If one of the $4^{+}$-vertices is a $5^{+}$-vertex, then it gives $\frac{1}{2}$ to f by (R2). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1\times 2+\frac{1}{2}\times 2=0$. If $f_2$ contains a $4^{+}$-vertex, then f has a $4^{+}$-vertex $v^{″}$ because $f_1$ is a (3, 3, 3)-face by Lemma 7. Since $\left|f\right|=7$ and the distance between $6^{-}$-cycles is at least 2, f can be adjacent to at most two $6^{-}$-faces. So, if $d\left(v^{″}\right)=4$, then $v^{″}$ must be rich and gives $\frac{1}{2}$ to f by (R4). If $d\left(v^{″}\right)\ge 5$, then $v^{″}$ gives $\frac{1}{2}$ to f by (R2). So ${\mu }^{*}\left(f\right)\ge \left|f\right|-6-1\times 2+\frac{1}{2}\times 2=0$.

□

Lemma 11.

${\mu }^{*}\left(f_0\right)>0$.

Proof. 

Suppose otherwise that ${\mu }^{*}\left(f_0\right)\le 0$. Let $E(G-C_0,C_0)$ denote the set of edges between $G-C_0$ and $C_0$. Let $E^{\prime }$ be the set of edges which are in $E(G-C_0,C_0)$ but not on $6^{-}$-faces. Let $e^{\prime }$ be the number of edges in $E^{\prime }$. Let x be the charges that $f_0$ receives by (R5), so $x\ge 0$. Let ${\ell }_i$ be the number of faces in $F_i$$(i\in \{3,4,5,6\left\}\right)$. Since $C_0$ has no chord by Lemma 5, each face in $F_3$, $F_4$, $F_5$ and $F_6$ has at least two edges in $E(G-C_0,C_0)$. Let ${\ell }_7$ be the number of special 7-faces. By (R5) and (R6),

$$\begin{array}{cc}\hfill {\mu }^{*}\left(f_0\right)& =|f_0|+6+\sum _{v\in C_0}(2d\left(v\right)-6)-3{\ell }_3-2{\ell }_4-{\ell }_5-{\ell }_7+x\hfill \\ \hfill & =|f_0|+6+\sum _{v\in C_0}2(d\left(v\right)-2)-2d\left(C_0\right)-3{\ell }_3-2{\ell }_4-{\ell }_5-{\ell }_7+x\hfill \\ \hfill & =6-|f_0|+2|E(G-C_0,C_0)|-3{\ell }_3-2{\ell }_4-{\ell }_5-{\ell }_7+x\hfill \\ \hfill & \ge 6-|f_0|+4{\ell }_3+4{\ell }_4+4{\ell }_5+4{\ell }_6+2e^{\prime }-3{\ell }_3-2{\ell }_4-{\ell }_5-{\ell }_7+x\hfill \\ \hfill & =6-|f_0|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-{\ell }_7+x\hfill \end{array}\left(1\right)$$

Equality holds when each $6^{-}$-face in $F_3$, $F_4$, $F_5$ and $F_6$ contains two edges in $E(G-C_0,C_0)$.

We consider the cases.

Case 1. If G has a special 7-face f, then ${\ell }_7>0$ and $E(G-C_0,C_0)\bigcap E\left(f\right)\ne \varnothing$. Because f is adjacent to two internal $6^{-}$-faces and the distance between $6^{-}$-cycles is at least 2, so each edge in $E(G-C_0,C_0)\bigcap E\left(f\right)$ is in $E^{\prime }$ and f shares exactly one vertex or one edge with $C_0$. So ${\mathit{e}}^{\prime }\ge {\mathbf{\ell }}_{\mathbf{7}}$.

• Let${\mathit{e}}^{\prime }={\mathbf{\ell }}_{\mathbf{7}}>\mathbf{0}$. If f is a special 7-face, then each edge in $E(G-C_0,C_0)\bigcap E\left(f\right)$ is in $E^{\prime }$ and $|E(G-C_0,C_0)\bigcap E\left(f\right)|=2$. So ${\ell }_3={\ell }_4={\ell }_5={\ell }_6=0$. For otherwise if $f^{\prime }$ is a $6^{-}$-face in $F_i$$(i\in \{3,4,5,6\left\}\right)$, then there must be two $8^{+}$-faces adjacent to $f^{\prime }$ and containing vertices of $C_0$, then $e^{\prime }>{\ell }_7$, a contradiction. Since each special 7-face shares exactly one vertex or one edge with $C_0$, ${\ell }_7\ge \left|f_0\right|$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+2e^{\prime }-{\ell }_7+x=6-\left|f_0\right|+{\ell }_7+x>0$, a contradiction.
• Let${\mathit{e}}^{\prime }>{\mathbf{\ell }}_{\mathbf{7}}>\mathbf{0}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-{\ell }_7+x\ge 6-|f_0|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2({\ell }_7+1)-{\ell }_7+x=8-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+{\ell }_7+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{8}$, then ${\mu }^{*}\left(f_0\right)\ge 8-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+{\ell }_7+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\mu }^{*}\left(f_0\right)\ge 8-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+{\ell }_7+x=-1+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+{\ell }_7+x$. Since ${\mu }^{*}\left(f_0\right)\le 0$, ${\ell }_{\mathbf{7}}\le \mathbf{1}$. Recall ${\ell }_7>0$, so ${\mathbf{\ell }}_{\mathbf{7}}=\mathbf{1}$. Because $e^{\prime }>{\ell }_7=1$ and $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-{\ell }_7+x=6-9+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-1+x$, so $e^{\prime }=2$ and ${\ell }_3={\ell }_4={\ell }_5={\ell }_6=x=0$. It follows that $C_0$ is adjacent to a $9^{+}$-face f which has at least 7 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-9}{3}\rceil >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\mu }^{*}\left(f_0\right)\ge 8-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+{\ell }_7+x=-2+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+{\ell }_7+x$. Since ${\mu }^{*}\left(f_0\right)\le 0$, ${\mathbf{\ell }}_{\mathbf{7}}\le \mathbf{2}$. If ${\mathbf{\ell }}_{\mathbf{7}}=\mathbf{1}$, because $e^{\prime }>{\ell }_7=1$ and $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-{\ell }_7+x=6-10+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-1+x$, then $e^{\prime }=2$ and ${\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x\le 1$. If ${\ell }_3=1$, then ${\ell }_4={\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least 3 consecutive 2-vertices and is adjacent to the 3-face in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor >0$ by (R3), a contradiction. If ${\ell }_3=0$, then ${\ell }_4={\ell }_5={\ell }_6=0$, $x\le 1$ and $C_0$ is adjacent to a ${10}^{+}$-face f which has at least 8 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-10}{3}\rceil >2$ by (R3), a contradiction. If ${\mathbf{\ell }}_{\mathbf{7}}=\mathbf{2}$, because $e^{\prime }>{\ell }_7$ and $e^{\prime }>{\ell }_7=2$ and $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-{\ell }_7+x=6-10+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }-2+x$, then $e^{\prime }=3$ and ${\ell }_3={\ell }_4={\ell }_5={\ell }_6=x=0$. Thus, the two 7-faces in $F_7$ must share an edge in $E(G-C_0,C_0)$. Then $C_0$ is adjacent to a $8^{+}$-face f which has at least 7 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-8}{3}\rceil >0$ by (R3), a contradiction.

Case 2. If G has no special 7-faces, then ${\ell }_7=0$. Recall that ${\mathit{e}}^{\prime }\ge \mathbf{0}$

• Let${\mathit{e}}^{\prime }=\mathbf{0}$. By (1), $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$, ${\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x\le \left|f_0\right|-6$. Since $|f_0|\le 10$ and the distance between $6^{-}$-cycles is at least 2, ${\mathbf{\ell }}_{\mathbf{3}}\le \mathbf{3}$.

Let${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{3}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x=9-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{8}$, then ${\mu }^{*}\left(f_0\right)\ge 9-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\ell }_4={\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least one 2-vertex and is adjacent to two 3-faces in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_4={\ell }_5={\ell }_6=0$ and $x\le 1$. It follows that $C_0$ is adjacent to three $7^{+}$-faces and each $7^{+}$-face f contains at least one 2-vertex and is adjacent to two 3-faces in $F_3$. So $x\ge 3\times \left[\right|f|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor ]\ge 3$ by (R3), a contradiction.

Let${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{2}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+2+2{\ell }_4+3{\ell }_5+4{\ell }_6+x=8-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{7}$, then ${\mu }^{*}\left(f_0\right)\ge 8-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{8}$, then ${\ell }_4={\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least 2 consecutive 2-vertices and is adjacent to two 3-faces in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-6}{3}\rfloor >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\ell }_4={\ell }_5={\ell }_6=0$ and $x\le 1$. It follows that $C_0$ is adjacent to two $7^{+}$-faces that each $7^{+}$-face f contains at least one 2-vertex and is adjacent to two 3-faces in $F_3$. So $x\ge 2\times \left[\right|f|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor ]\ge 2$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_5={\ell }_6=0$ and ${\ell }_4\le 1$. If ${\ell }_4=1$, then $x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least one 2-vertex and is adjacent to two 3-faces in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor >0$ by (R3), a contradiction. If ${\ell }_4=0$, then $x\le 2$. It follows that $C_0$ is adjacent to a $7^{+}$-face $f_1$ which has at least one 2-vertex and is adjacent to two 3-faces in $F_3$, and a $8^{+}$-face $f_2$ which has at least three 2-vertices and is adjacent to two 3-faces in $F_3$. So $x\ge |f_1|-6-\lfloor \frac{|f_1|-5}{3}\rfloor +\left|f_2\right|-6-\lfloor \frac{|f_1|-7}{3}\rfloor \ge 3$ by (R3), a contradiction.

Let${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{1}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+1+2{\ell }_4+3{\ell }_5+4{\ell }_6+x=7-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{6}$, then ${\mu }^{*}\left(f_0\right)\ge 7-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{7}$, then ${\ell }_4={\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $8^{+}$-face f which has at least 5 consecutive 2-vertices and is adjacent to the 3-face in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-8}{3}\rfloor >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{8}$, then ${\ell }_4={\ell }_5={\ell }_6=0$ and $x\le 1$. It follows that $C_0$ is adjacent to a $9^{+}$-face which has at least 6 consecutive 2-vertices and is adjacent to the 3-face in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-9}{3}\rfloor \ge 3$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\ell }_4\le 1$. If ${\ell }_4=1$, then then ${\ell }_5={\ell }_6=x=0$. It follows that $C_0$ is adjacent to a $7^{+}$-face f which has at least 2 consecutive 2-vertices and is adjacent to the 3-face in $F_3$ and the 4-face in $F_4$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-6}{3}\rfloor >0$ by (R3), a contradiction. If ${\ell }_4=0$, then ${\ell }_5={\ell }_6=0$ and $x\le 2$. It follows that $C_0$ is adjacent to a ${10}^{+}$-face which has at least 7 consecutive 2-vertices and is adjacent to the 3-face in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-10}{3}\rfloor \ge 4$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_6=0$ and ${\ell }_5\le 1$. If ${\ell }_5=1$, then ${\ell }_4=x=0$. It follows that $C_0$ is adjacent to a $7^{+}$-face f which has at least 2 consecutive 2-vertices and is adjacent to the 3-face in $F_3$ and the 5-face in $F_5$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-6}{3}\rfloor >0$ by (R3), a contradiction. If ${\ell }_5=0$, then ${\ell }_4\le 1$. If ${\ell }_4=1$, then $x\le 1$. It follows that $C_0$ is adjacent to two $7^{+}$-faces and each $7^{+}$-face f contains at least one 2-vertex and is adjacent to the 3-face in $F_3$ and the 4-face in $F_4$. So $x\ge 2\times \left[\right|f|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor ]\ge 2$ by (R3), a contradiction. If ${\ell }_4=0$, then $x\le 3$ and $C_0$ is adjacent to a ${11}^{+}$-face f which has at least 8 consecutive 2-vertices and is adjacent to the 3-face in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-11}{3}\rfloor >5$ by (R3), a contradiction.

Let${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{0}$, then ${\ell }_4+{\ell }_5+{\ell }_6>0$. For otherwise that $G=C_0$. Since $d\left(C_0\right)\le 10$ and $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$, ${\mathbf{\ell }}_{\mathbf{4}}\le \mathbf{2}$.

Let ${\mathbf{\ell }}_{\mathbf{4}}=\mathbf{2}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+4+3{\ell }_5+4{\ell }_6+x=10-\left|f_0\right|+3{\ell }_5+4{\ell }_6+x$. If $|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{9}$, then ${\mu }^{*}\left(f_0\right)\ge 10-\left|f_0\right|+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least 2 consecutive 2-vertices and is adjacent to the 4-faces in $F_4$. Thus, $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-6}{3}\rfloor >0$ by (R3), a contradiction.

Let${\mathbf{\ell }}_{\mathbf{4}}=\mathbf{1}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+2+3{\ell }_5+4{\ell }_6+x=8-\left|f_0\right|+3{\ell }_5+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{7}$, then ${\mu }^{*}\left(f_0\right)\ge 8-\left|f_0\right|+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{8}$, then ${\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $8^{+}$-face f which has at least 5 consecutive 2-vertices and is adjacent to the 4-face in $F_4$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-8}{3}\rfloor >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\ell }_5={\ell }_6=0$, $x\le 1$ and $C_0$ is adjacent to a $9^{+}$-face f which has at least 6 consecutive 2-vertices and is adjacent to the 4-face in $F_4$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-9}{3}\rfloor >1$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_5={\ell }_6=0$, $x\le 2$ and $C_0$ is adjacent to a ${10}^{+}$-face f which has at least 7 consecutive 2-vertices and is adjacent to the 4-face in $F_4$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-10}{3}\rfloor >2$ by (R3), a contradiction.

Let${\mathbf{\ell }}_{\mathbf{4}}=\mathbf{0}$, then ${\ell }_5+{\ell }_6>0$. For otherwise that $G=C_0$. Since $d\left(C_0\right)\le 10$ and $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+3{\ell }_5+4{\ell }_6+x$, ${\mathbf{\ell }}_{\mathbf{5}}\le \mathbf{1}$.

Let${\mathbf{\ell }}_{\mathbf{5}}=\mathbf{1}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+3+4{\ell }_6+x=9-\left|f_0\right|+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{8}$, then ${\mu }^{*}\left(f_0\right)\ge 9-\left|f_0\right|+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\ell }_6=x=0$ and $C_0$ is adjacent to a $8^{+}$-face f which has at least 5 consecutive 2-vertices and is adjacent to the 5-face in $F_5$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-8}{3}\rfloor >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_6=0$, $x\le 1$ and $C_0$ is adjacent to a $9^{+}$-face f which has at least 6 consecutive 2-vertices and is adjacent to the 5-face in $F_5$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-9}{3}\rfloor >1$ by (R3), a contradiction.

Let${\mathbf{\ell }}_{\mathbf{5}}=\mathbf{0}$, then ${\ell }_6>0$. For otherwise that $G=C_0$. Since $d\left(C_0\right)\le 10$ and $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+4{\ell }_6+x$, ${\mathbf{\ell }}_{\mathbf{6}}=\mathbf{1}$. If $|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{9}$, then ${\mu }^{*}\left(f_0\right)\ge 10-\left|f_0\right|+x>0$, a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then $x=0$ and $C_0$ is adjacent to a $8^{+}$-face f which has at least 5 consecutive 2-vertices and is adjacent to the 6-face in $F_6$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-8}{3}\rfloor >0$ by (R3), a contradiction.

• Let${\mathit{e}}^{\prime }>\mathbf{0}$. By (1), $0\ge {\mu }^{*}\left(f_0\right)\ge 6-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }+x$, ${\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2e^{\prime }+x\le \left|f_0\right|-6$. Since $|f_0|\le 10$, ${\mathit{e}}^{\prime }\le \mathbf{2}$.

Let${\mathit{e}}^{\prime }=\mathbf{2}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+4+x=10-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{9}$, then ${\mu }^{*}\left(f_0\right)\ge 10-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_3={\ell }_4={\ell }_5={\ell }_6=x=0$. It follows that $C_0$ is adjacent to a $7^{+}$-face f which has at least 4 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-7}{3}\rceil >0$ by (R3), a contradiction.

Let${\mathit{e}}^{\prime }=\mathbf{1}$. By (1), ${\mu }^{*}\left(f_0\right)\ge 6-|f_0|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+2+x=8-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x$. If$|{\mathit{f}}_{\mathbf{0}}|\le \mathbf{7}$, then ${\mu }^{*}\left(f_0\right)\ge 8-\left|f_0\right|+{\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x>0$, a contradiction. If$|{\mathit{f}}_{\mathbf{0}}|=\mathbf{8}$, then ${\ell }_3={\ell }_4={\ell }_5={\ell }_6=x=0$. It follows that $C_0$ is adjacent to a $9^{+}$-face f which has at least 7 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-9}{3}\rceil >0$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{9}$, then ${\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x\le 1$. So ${\ell }_3\le 1$. If ${\ell }_3=1$, then ${\ell }_4={\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least 3 consecutive 2-vertices and is adjacent to the 3-face in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor >0$ by (R3), a contradiction. If ${\ell }_3=0$, then ${\ell }_4={\ell }_5={\ell }_6=0$ and $x\le 1$. It follows that $C_0$ is adjacent to a ${10}^{+}$-face f which has at least 8 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-10}{3}\rceil >1$ by (R3), a contradiction. If $|{\mathit{f}}_{\mathbf{0}}|=\mathbf{10}$, then ${\ell }_3+2{\ell }_4+3{\ell }_5+4{\ell }_6+x\le 2$. So ${\mathbf{\ell }}_{\mathbf{3}}\le \mathbf{2}$. If ${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{2}$, then ${\ell }_4={\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least one 2-vertex and is adjacent to the 3-faces in $F_3$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor >0$ by (R3), a contradiction. If ${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{1}$, then ${\ell }_4={\ell }_5={\ell }_6=0$ and $x\le 1$. It follows that $C_0$ is adjacent to two $7^{+}$-faces. If one of the $7^{+}$-faces f is a $8^{+}$-face and contains at least 5 consecutive 2-vertices, then $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-7}{3}\rfloor \ge 2$ by (R3), a contradiction. For otherwise that all of $7^{+}$-faces contain at least 3 consecutive 2-vertices and are adjacent to the 3-face in $F_3$, then $x\ge 2\times \left[\right|f|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor ]\ge 2$ by (R3), a contradiction. If ${\mathbf{\ell }}_{\mathbf{3}}=\mathbf{0}$, then $2{\ell }_4+3{\ell }_5+4{\ell }_6+x\le 2$. So ${\mathbf{\ell }}_{\mathbf{4}}\le \mathbf{1}$. If ${\mathbf{\ell }}_{\mathbf{4}}=\mathbf{1}$, then ${\ell }_5={\ell }_6=x=0$ and $C_0$ is adjacent to a $7^{+}$-face f which has at least 3 consecutive 2-vertices and is adjacent to the 4-face in $F_4$. So $x\ge \left|f\right|-6-\lfloor \frac{\left|f\right|-5}{3}\rfloor >0$ by (R3), a contradiction. If ${\mathbf{\ell }}_{\mathbf{4}}=\mathbf{0}$, then ${\ell }_5={\ell }_6=0$, $x\le 2$ and $C_0$ is adjacent to a ${11}^{+}$-face f which has at least 9 consecutive 2-vertices. So $x\ge \left|f\right|-6-\lceil \frac{\left|f\right|-11}{3}\rceil >2$ by (R3), a contradiction. □

Proof 

(Proof of Theorem 1). From Lemmas 9–11, ${\sum }_{x\in V\bigcup F}{\mu }^{*}\left(x\right)>0$, a contradiction to Euler’s Formula. Thus, the counterexample G cannot exist. So, Theorem 1 is true. □

3. Conclusions

The coloring theory of graphs is useful in many fields, such as discrete mathematics, allocation of wireless communication channels, combinatorial optimization, computer theory.

It is well known that 3-COLORING is NP-complete for planar graphs. This provides motivation for finding some sufficient conditions for 3-coloring of planar graphs. DP-coloring is a stronger version of list coloring. Proving a planar graph to be DP-3-colorable is harder than proving a planar graph to be 3-colorable.

It is unknown if there exists a planar graph in which the distance between 6${}^{-}$-cycles at least 1 is not DP-3-colorable.