Partitioning Planar Graphs Without Specific Cycles into a Forest and a Disjoint Union of Paths

Abstract

In this paper, we show that if a planar graph G satisfies the following conditions: (i) none of its 3-faces is adjacent to a $6^{-}$-face, and (ii) none of its 4-faces is adjacent to a $5^{-}$-face, then $V\left(G\right)$ can be partitioned into two subsets, each induces a forest, while one of the forests has maximum degree of at most $2.$ This result implies that every planar graph (i) with no chordal $7^{-}$-cycles, or (ii) with neither 4- nor 6-cycles, also admits such a partition.

1. Introduction

This paper considers only undirected and simple graphs. A graph G has a $({\mathcal{G}}_1,\dots ,{\mathcal{G}}_k)$-partition into $V_1,\dots ,V_k$ if its vertex set can be partitioned into k sets $V_1,\dots ,V_k$ where $V_i$ is an empty set or the induced subgraph $G\left[V_i\right]$ is in a family of graphs ${\mathcal{G}}_i$ for each $i\in \{1,\dots ,k\}$. This concept generalizes a notion of proper k-vertex coloring (when each family ${\mathcal{G}}_i$ consists of edgeless graphs). Other widely studied families consist of graphs with no edges, graphs with maximum degree at most d, forests with maximum degree at most $d,$ and forests without restricted degree. Such families are denoted by $\mathcal{I},{\Delta }_d,{\mathcal{F}}_d,$ and $\mathcal{F}$, respectively.

Naturally, vertex partition has been investigated in planar graphs with specific girths.

It was shown that every planar graph has an $(\mathcal{I},\mathcal{I},\mathcal{I},\mathcal{I})$-partition [FCT], a $({\Delta }_2,{\Delta }_2,{\Delta }_2)$-partition [1], an $(\mathcal{I},\mathcal{F},\mathcal{F})$-partition [2], and a $({\mathcal{F}}_2,{\mathcal{F}}_2,{\mathcal{F}}_2)$-partition [3].

For each planar graph with girth 4, it was shown that this graph has an $(\mathcal{I},\mathcal{I},\mathcal{I})$-partition [4] and a $({\mathcal{F}}_5,\mathcal{F})$-partition [5].

For each planar graph with girth 5, it was shown that this graph has a $({\Delta }_3,{\Delta }_4)$-partition [6] and a $(\mathcal{I},\mathcal{F})$-partition [7,8]. With an additional condition that this graph has no adjacent 5-cycles, then it has a $({\mathcal{F}}_3,{\mathcal{F}}_3)$-partition [9].

For each planar graph with girth 6, it was shown that this graph has a $({\Delta }_1,{\Delta }_4)$-partition [10], $({\Delta }_2,{\Delta }_2)$-partition [11], a $({\mathcal{F}}_1,{\mathcal{F}}_4)$-partition [12], and a $({\mathcal{F}}_2,{\mathcal{F}}_2)$-partition [13].

For each planar graph with girth 7, it was shown that this graph has an $(\mathcal{I},{\Delta }_4)$-partition [10] and a $(\mathcal{I},{\mathcal{F}}_5)$-partition [14].

For each planar graph with girth 8, it was shown that this graph has an $(\mathcal{I},{\Delta }_2)$-partition [10] and an $(\mathcal{I},{\mathcal{F}}_3)$-partition [14].

For each planar graph with girth 10, it was shown that this graph has an $(\mathcal{I},\mathcal{I},\mathcal{I})$-partition [14] and an $(\mathcal{I},{\mathcal{F}}_2)$-partition [14].

Some results presented in the list above may be redundant since we can partition a forest into two independent sets, whereas ${\mathcal{F}}_d$ is a subfamily of ${\Delta }_d.$ However, for chronological clarity, earlier results for $({\Delta }_{d_1},\dots ,{\Delta }_{d_n})$-partitions are presented.

The search for sufficient conditions for an $(\mathcal{F},\mathcal{F})$-partition of planar graphs has gained attention since Chartrand and Kronk [15] provided an example of a planar graph without such partition.

Additionally, exploring conditions for an $({\mathcal{F}}_{d_1},{\mathcal{F}}_{d_2})$-partition is also of particular interest since it may require stronger conditions than that of an $(\mathcal{F},\mathcal{F})$-partition. For example, since every planar graph with girth 4 is 3-degenerate, it has an $(\mathcal{F},\mathcal{F})$-partition. However, Montassier and Ochem [16] constructed, for each $d_1$ and $d_2$, a planar graph with girth 4 having no $({\Delta }_{d_1},{\Delta }_{d_2})$-partitions. As a result, an existence of $(\mathcal{F},\mathcal{F})$-partition cannot guarantee to be strengthened into any $({\mathcal{F}}_{d_1},{\mathcal{F}}_{d_2})$-partition.

To further study planar graphs of girth $4,$ an $({\mathcal{F}}_d,\mathcal{F})$-partition (where the condition of bounded degree of one family is relaxed to be unbounded) was considered instead. Dross et al. [5] showed that every planar graph of girth 4 has an $({\mathcal{F}}_d,\mathcal{F})$-partition when $d=5.$ The case for $d\le 4$ is still open. Note that the case $d=0,$, if positive, will imply Grötzsch’s result [4] (an $(\mathcal{I},\mathcal{I},\mathcal{I})$-partition). Now, Feghali and Šámal [17] answer the following question for $d=3.$

Apart from girth, certain conditions related to cycles have been explored. Sittitrai and Nakprasit [18] proved that every planar graph with neither 4- nor 5-cycles admits both a $({\Delta }_3,{\Delta }_5)$-partition and a $({\Delta }_4,{\Delta }_4)$-partition. Later, Cho et al. [19] strengthened the result by establishing an $({\mathcal{F}}_3,{\mathcal{F}}_4)$-partition.

Several results concerning planar graphs neither 4- nor 6-cycles are as follows. Wang and Chen [20] showed that such a graph admits an $(\mathcal{I},\mathcal{I},\mathcal{I})$-partition if it also has no 8-cycles, while Kang et al. [21] showed a similar result if it has no 9-cycles. Liu and Yu [22] improved both results by establishing an $(\mathcal{I},\mathcal{F})$-partition for such graphs.

Without additional requirements, Nakprasit et al. [23] showed that every planar graph with neither 4- nor 6-cycles has a $({\Delta }_3,{\Delta }_4)$-partition. Recently, Hu and Huang [24] improved the result into an (${\mathcal{F}}_3,{\mathcal{F}}_4$)-partition. Furthermore, Wang and Xu [25] proved that every planar graph with neither 4- nor 6-cycles has a $({\Delta }_2,\mathcal{I},\mathcal{I})$-partition. Furthermore, Huang et al. and Sittitrai et al. [26,27] independently improved the result in the following theorem.

Theorem 1

([26,27]). Every planar graph with neither 4- nor 6-cycles has an (${\mathcal{F}}_2,\mathcal{F}$)-partition.

Another condition for a planar graph to have an (${\mathcal{F}}_2,\mathcal{F}$)-partition was provided by Liu and Wang as follows.

Theorem 2

([28]). Every planar graph with neither 3-cycles nor chordal 6-cycles has an (${\mathcal{F}}_2,\mathcal{F}$)-partition.

The following theorem, which is the main result of this work, can be shown to imply Theorems 1 and 2.

Theorem 3.

If a planar graph G satisfies the following conditions: (i) none of its 3-faces is adjacent to a $6^{-}$-face, and (ii) none of its 4-faces is adjacent to a $5^{-}$-face, then it has an (${\mathcal{F}}_2,\mathcal{F}$)-partition.

We organize this paper as follows. In Section 2, we investigate configurations of the minimal counterexample to the main theorem. In Section 3, we employ a discharging method, with the knowledge of configurations, to prove the main theorem. In Section 4, we discuss the implications of the main result, including Theorems 1 and 2. Additionally, we propose a conjecture related to partitioning planar graphs with neither 4- nor 6-cycles.

2. Structures of Minimal Counterexamples

We use the following notation in this paper. A k-vertex, $k^{+}$-vertex, and $k^{-}$-vertex is a vertex of degree k, at least k, and at most k, respectively. We also have a similar notation for faces. A k-neighbor of v is a vertex with degree k adjacent to v.

Let $b\left(f\right)$ denote the boundary walk of a face f. A vertex v and a face f are incident if v is on $b\left(f\right).$ The degree of a face $f,$ denoted by $d\left(f\right),$ is the number of the boundary walk of a face $f.$ If a vertex v is not incident to a face f but is adjacent to a 3-vertex u on $b\left(f\right),$ then f is a pendent face of a vertex v and v is a pendent neighbor of u (with respect to f).

We use $n_i\left(v\right)$ to denote the number of incident i-faces of a vertex v, and use $m_i\left(v\right)$ to denote the number of pendent i-faces of a vertex v. We use $i^{-}$ in the previous notation when considering $i^{-}$-faces.

A 3-vertex u incident to a k-face f for $k\in \{3,4,5\}$ is a terrible 3-vertex (see Figure 1) of f if u has a pendent $4^{-}$-neighbor with respect to $f.$ A 3-face f is a poor 3-face (see Figure 2) if f is incident to two terrible 3-vertices.

Suppose $G^{\prime }$ is a (not necessarily proper) subgraph of G, and $V\left(G^{\prime }\right)$ is partitioned into $V_1$ and $V_2$. If a vertex u in $V_i$ ($i\in \{1,2\}$) is a neighbor of v (where v may be not in $V\left(G^{\prime }\right)$, then we call u a $V_i$-neighbor of v. Furthermore, v is called $V_1$-saturated if $v\in V_1$ and v has at least two $V_1$-neighbors.

Conditions (i) and (ii) in Theorem 3 are not considered in the next three lemmas. The only assumption in those lemmas is that a planar graph G is a minimal graph (subject to the number of vertices and edges) without an (${\mathcal{F}}_2,\mathcal{F}$)-partition. In contrast, Lemma 4 is only concerned with Conditions (i) and (ii).

Lemma 1.

If G is a minimal graph without an (${\mathcal{F}}_2,\mathcal{F}$)-partition, then each vertex in G is a $3^{+}$-vertex.

Proof. 

We show the contradiction only for the case of a 2-vertex u in G, as the remaining case is more straightforward. By minimality, $G-u$ has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2$.

Let v and w are two adjacent vertices of $u.$ If $v\in V_1$ or $w\in V_1$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2\cup \left\{u\right\}$, a contradiction. If $v,w\in V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{u\right\}$ and $V_2,$ a contradiction. □

Lemma 2.

If G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2,$ then we can assume that (i) each 3-vertex in $V_1$ is not $V_1$-saturated, and (ii) each 2-vertex is in $V_1$ but has no $V_1$-neighbors, or is in $V_2$ with at most one $V_2$-neighbor.

Proof. 

Assume G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2.$ (i) We can reduce the 3-vertices in $V_1$ that are $V_1$-saturated as follows. If v is a $3^{-}$-vertex that is $V_1$-saturated, then v has at most one $V_2$-neighbor. Consequently, G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1-\left\{v\right\}$ and $V_2\cup \left\{v\right\}$ with a decreasing number of $3^{-}$-vertices that is $V_1$-saturated. The process can be continued until each 3-vertex is not $V_1$-saturated.

(ii) We say that a 2-vertex v has the undesired property if v is in $V_1$ and has a $V_1$-neighbor, or is in $V_2$ and has two $V_2$-neighbors. From the process above, each 2-vertex in $V_1$ has at most one $V_1$-neighbor.

Consider a 2-vertex v in $V_1$ with one $V_1$-neighbor and one $V_2$-neighbor, say u. We obtain a new (${\mathcal{F}}_2,\mathcal{F}$)-partition of G as follows. If u is a 2-vertex with one $V_2$-neighbor, then we have a new (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{u\right\}-\left\{v\right\}$ and $V_2\cup \left\{v\right\}-\left\{u\right\}$, otherwise we have a new (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup -\left\{v\right\}$ and $V_2\cup \left\{v\right\}$. In either case, G has a new (${\mathcal{F}}_2,\mathcal{F}$)-partition with the reduced number of undesired 2-vertices.

Consider a 2-vertex v in $V_2$ with two $V_2$-neighbors. Then, G has a new (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{v\right\}$ and $V_2-\left\{v\right\}$ with the reduced number of undesired 2-vertices.

Continue this process until we have no undesired 2-vertices. Note that each $3^{-}$-vertex in $V_1$ still is not $V_1$-saturated by this process. Thus, the proof is complete. □

Lemma 3.

A planar graph G which is a minimal graph without an (${\mathcal{F}}_2,\mathcal{F}$)-partition has the following properties.

(i)

Each 3-vertex is adjacent to a $5^{+}$-vertex.

(ii)

Each 5-vertex is not incident to a poor 3-face.

(iii)

Each 6-vertex is incident to at most one poor 3-face.

Proof. 

(i) Suppose to the contrary that a 3-vertex v is adjacent to three $4^{-}$-vertices. By minimality, $G-v$ has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2$.

If v has no $V_1$-neighbors, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{v\right\}$ and $V_2.$ If v has at least two $V_1$-neighbors, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2\cup \left\{v\right\}.$

It remains to consider that v has exactly one $V_1$-neighbor u. By Lemma 2, u is not $V_1$-saturated. Consequently, G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{v\right\}$ and $V_2.$ This contradiction completes the proof.

(ii) Suppose to the contrary that a poor 3-face f is incident to a 5-vertex v. Let $b\left(f\right)=v{v}_1{v}_2.$ From the definition, $v_1$ and $v_2$ are terrible 3-vertices with pendent $4^{-}$-neighbors, say $v_1^{\prime }$ and $v_2^{\prime }$, respectively.

Consider $G-\{v_1,v_2\}$ with an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2.$ Note that each of v, $v_1^{\prime }$, and $v_2^{\prime }$ is a $3^{-}$-vertex in $G-\{v_1,v_2\}.$ By Lemma 2, each of them is not $V_1$-saturated.

-

Suppose $v\in V_1$.

If $v_1^{\prime }$ or $v_2^{\prime }$ is in $V_1$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2\cup \{v_1,v_2\},$ a contradiction.

If $v_1^{\prime }$ and $v_2^{\prime }$ are in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{v_1\right\}$ and $V_2\cup \left\{v_2\right\}$ since v is not $V_1$-saturated, a contradiction.

-

Suppose $v\in V_2$.

If $v_1^{\prime }$ and $v_2^{\prime }$ are in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \{v_1,v_2\}$ and $V_2,$ a contradiction.

If $v_1^{\prime }$ is in $V_1$ and $v_2^{\prime }$ is in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{v_2\right\}$ and $V_2\cup \left\{v_1\right\},$ a contradiction.

If $v_1^{\prime }$ and $v_2^{\prime }$ are in $V_1$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1\cup \left\{v_1\right\}$ and $V_2\cup \left\{v_2\right\}$ since $v_1^{\prime }$ is not $V_1$-saturated, a contradiction.

(iii) Suppose to the contrary that v is incident to two poor 3-faces with the boundary walks $v{v}_1{v}_2$ and $v{v}_3{v}_4.$ From the definition, $v_i$ is a terrible 3-vertices with pendent $4^{-}$-neighbors, say $v_i^{\prime }$, for $i\in \{1,2,3,4\}$.

Consider $G^{\prime }=G-\{v_1,v_2,v_3,v_4\}$ with an (${\mathcal{F}}_2,\mathcal{F}$)-partition into $V_1$ and $V_2.$ Note that $v_i^{\prime }$ or v, where $i\in \{1,2,3,4\},$ is a $3^{-}$-vertex in $G^{\prime }$. By Lemma 2, these vertices are not $V_1$-saturated. Additionally, if $v\in V_1,$ then it has no $V_1$-neighbors.

We show that we can extend an (${\mathcal{F}}_2,\mathcal{F}$)-partition to G as follows.

If $v,v_1^{\prime },$ and $v_3^{\prime }$ are in $V_1,$ then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition of G into $V_1$ and $V_2\cup \{v_1,v_2,v_3,v_4\}$.

If v and $v_1^{\prime }$ are in $V_1$ while $v_3^{\prime }$ and $v_4^{\prime }$ are in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition of G into $V_1\cup \{v_3,v_4\}$ and $V_2\cup \{v_1,v_2\}$.

If v is in $V_1$ while $v_1^{\prime },v_2^{\prime },v_3^{\prime },$ and $v_4^{\prime }$ are in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition of G into $V_1\cup \{v_2,v_4\}$ and $V_2\cup \{v_1,v_3\}$.

If $v_1^{\prime }$ and $v_3^{\prime }$ are in $V_1$ while v is in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition of G into $V_1\cup \{v_2,v_4\}$ and $V_2\cup \{v_1,v_3\}$.

If $v_1^{\prime }$ is in $V_1$ while $v,v_3^{\prime },$ and $v_4^{\prime }$ are in $V_2$, then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition of G into $V_1\cup \{v_2,v_3,v_4\}$ and $V_2\cup \left\{v_1\right\}$.

If $v,v_1^{\prime },v_2^{\prime },v_3^{\prime }$ and $v_4^{\prime },$ are in $V_2,$ then G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition of G into $V_1\cup \{v_1,v_2,v_3,v_4\}$ and $V_2$.

Each of the remaining cases is symmetric to one of the previous cases. Thus, G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition. This contradiction completes the proof. □

Lemma 4.

If G satisfies Conditions (i) and (ii) in Theorem 3, then each k-vertex v has bounds on $m_i\left(v\right)$ and $n_i\left(v\right)$ as follows.

(i)

$n_{4^{-}}\left(v\right)\le \lfloor {\displaystyle \frac{k}{2}}\rfloor$.

(ii)

$n_5\left(v\right)\le \left\{\begin{array}{cc}0,\hfill & if{m}_{4^{-}}\left(v\right)+2n_{4^{-}}\left(v\right)=k,\hfill \\ k,\hfill & if{m}_{4^{-}}\left(v\right)+2n_{4^{-}}\left(v\right)=0,\hfill \\ k-m_{4^{-}}\left(v\right)-2n_{4^{-}}\left(v\right)-1,\hfill & \mathit{otherwise}.\hfill \end{array}\right.$

(iii)

$m_{4^{-}}\left(v\right)\le k-2n_{4^{-}}\left(v\right)$.

(iv)

$m_5\left(v\right)\le k-m_{4^{-}}\left(v\right)-2n_{4^{-}}\left(v\right)$.

Proof. 

Let $v_1,\dots ,v_k$ be the neighbors of a k-vertex $v.$

Define the following sets: A: = the set of $v_i$ where $v_i$ and v share the same incident $4^{-}$-face.

B: = the set of $v_i$ where v is a pendent neighbor of $v_i$ with regard to a $4^{-}$-face.

C: = the set of $v_i$ where $v_i$ and v share the same incident 5-face.

D: = the set of $v_i$ where v is a pendent neighbor of $v_i$ with regard to a 5-face.

The following properties arise directly from the definitions of the four sets and Condition (ii) of G which states that none of its 4-faces is adjacent to a $5^{-}$-face.

Property (a): The sets $A,B,C,$ and D are mutually disjoint.

Property (b): If $v_i\in A$, then $v_i$ is incident to exactly one 4-face of v. On the other hand, a 4-face of v is incident to two neighbors of $v.$ Thus, $\left|A\right|=2n_{4-}\left(v\right).$

Property (c): Since a 5-face incident to v is also incident to two neighbors of v, it follows that $n_5\left(v\right)\le \left|C\right|-1$ when $1\le n_5\left(v\right)\le k-1$.

From Properties (a) and (b), we deduce that

$$k\ge 2\left|A\right|+\left|B\right|+\left|C\right|+\left|D\right|\ge 2n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)+n_5\left(v\right)+m_5\left(v\right).$$

The inequalities imply (i), (iii), and (iv), as well as the first two cases of (ii), immediately.

It follows from Property (c) that the second inequality becomes strict when $1\le n_5\left(v\right)\le k-1$. This implies the last case of (ii) to complete the proof. □

3. Proof of Theorem 3

Suppose G is a minimal counterexample to the theorem. The discharging process is defined as follows. The initial charge of a vertex v is given by $\theta \left(v\right)=2d\left(v\right)-6,$ and the initial charge of a face f is given by $\theta \left(f\right)=d\left(f\right)-6$. By Euler’s formula $\left|V\right(G\left)\right|-\left|E\right(G\left)\right|+\left|F\right(G\left)\right|=2$ and the Handshaking Lemma, we obtain that

$${\displaystyle \sum _{v\in V\left(G\right)}\theta \left(v\right)+\sum _{f\in F\left(G\right)}\theta \left(f\right)=-12.}$$

Now, we establish a new charge ${\theta }^{*}\left(x\right)$ for each $x\in V\left(G\right)\cup F\left(G\right)$ by redistributing charge from one element to another. The total sum of the new charge ${\theta }^{*}\left(x\right)$ remains $-12.$ If, after the redistribution, the final charge satisfies ${\theta }^{*}\left(x\right)\ge 0$ for all $x\in V\left(G\right)\cup F\left(G\right)$, we obtain a contradiction, thereby completing the proof.

Let $\tau (v\to x)$ represent the charge transferred from a vertex v to both an incident face and a pendent face x. Similarly, let $\tau (f\to y)$ represent the charge transferred from a face f to an incident vertex y.

The rules for discharging these charges are articulated as follows.

(R1)

Let v be a 3-vertex.

$\tau (v\to x)={\displaystyle \frac{1}{3}}$ if x is its incident 3-face.

(R2)

Let v be a 4-vertex.

$\tau (v\to x)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{incident}{4}^{-}\text{-}\mathrm{face},\hfill \\ {\displaystyle \frac{1}{2}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{incident}5\text{-}\mathrm{face}.\hfill \end{array}\right.$

(R3)

Let v be a $5^{+}$-vertex.

$\tau (v\to x)=\left\{\begin{array}{cc}{\displaystyle \frac{5}{3}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{incident}\mathrm{non}\text{-}\mathrm{poor}3\text{-}\mathrm{face},\hfill \\ {\displaystyle \frac{7}{3}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{incident}\mathrm{poor}3\text{-}\mathrm{face},\hfill \\ {\displaystyle \frac{5}{3}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{incident}4\text{-}\mathrm{face},\hfill \\ {\displaystyle \frac{1}{2}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{incident}5\text{-}\mathrm{face},\hfill \\ {\displaystyle \frac{2}{3}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{pendent}{4}^{-}\text{-}\mathrm{face},\hfill \\ {\displaystyle \frac{1}{4}},\hfill & \mathrm{if}x\mathrm{is}\mathrm{its}\mathrm{pendent}5\text{-}\mathrm{face}.\hfill \end{array}\right.$

(R4)

Let f be a $7^{+}$-face.

$\tau (v\to y)={\displaystyle \frac{1}{6}}$ if y is its incident 3-vertex that is also incident to a 3-face.

To complete the proof, we need to show that the resulting ${\theta }^{*}\left(x\right)\ge 0$ for each $x\in V\left(G\right)\cup F\left(G\right)$. It is clear that ${\theta }^{*}\left(z\right)\ge 0$ when z is a 3-vertex not incident to any 3-faces or when z is a 6-face.

CASE 1: Let v be a 3-vertex incident to a 3-face.

Recall Condition (i) of G that none of its 3-faces is adjacent to a $6^{-}$-face. It follows that v is incident to two $7^{+}$-faces and one 3-face. Consequently, ${\theta }^{*}\left(v\right)=\theta \left(v\right)-{\displaystyle \frac{1}{3}}+2\times {\displaystyle \frac{1}{6}}=0$ by (R1) and (R4).

CASE 2: Let v be a 4-vertex.

By (R2), v loses charge $n_{4^{-}}\left(v\right)\times 1+n_5\left(v\right)\times {\displaystyle \frac{1}{2}}$. Using Lemmas 4 (i) and (ii), we consider the following subcases:

-

If $n_{4^{-}}\left(v\right)=0$, then $n_5\left(v\right)\le 4$. Thus, ${\theta }^{*}\left(v\right)\ge \theta \left(v\right)-4\times {\displaystyle \frac{1}{2}}=0.$

-

If $n_{4^{-}}\left(v\right)=1$, then $n_5\left(v\right)\le 1$. Thus, ${\theta }^{*}\left(v\right)\ge \theta \left(v\right)-1\times 1-1\times {\displaystyle \frac{1}{2}}>0.$

-

If $n_{4^{-}}\left(v\right)=2$, then $n_5\left(v\right)=0$. Thus, ${\theta }^{*}\left(v\right)=\theta \left(v\right)-2\times 1=0.$

CASE 3: Let v be a $5^{+}$-vertex and $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)=0$.

By Lemmas 4 (ii) and (iv), we have $n_5\left(v\right)\le d\left(v\right)$ and $m_5\left(v\right)\le d\left(v\right).$ Thus,

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& \ge \theta \left(v\right)-d\left(v\right)\times {\displaystyle \frac{1}{2}}-d\left(v\right)\times {\displaystyle \frac{1}{4}}\hfill \\ \hfill & =2d\left(v\right)-6-d\left(v\right)\times {\displaystyle \frac{3}{4}}\hfill \\ \hfill & ={\displaystyle \frac{5}{4}}\times d\left(v\right)-6\hfill \\ \hfill & >0.\hfill \end{array}$$

by (R3) and the fact that $d\left(v\right)\ge 5$.

CASE 4: Let v be a 5-vertex with $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)>0$.

By Lemma 3 (ii), v is not incident to any poor 3-face. Consequently, v gives a charge at most ${\displaystyle \frac{5}{3}}$ to each incident $4^{-}$-face. By (R3), we have the final charge of v as

$${\theta }^{*}\left(v\right)=2d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{5}{3}}+m_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}+n_5\left(v\right)\times {\displaystyle \frac{1}{2}}+m_5\left(v\right)\times {\displaystyle \frac{1}{4}}.$$

CASE 4.1: Suppose $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)=d\left(v\right)=5$.

By Lemma 4, we obtain $n_{4^{-}}\left(v\right)\le 2$ and $m_{4^{-}}\left(v\right)=d\left(v\right)-2\times n_{4^{-}}\left(v\right)$. Also, $n_5\left(v\right)=m_5\left(v\right)=0$. Substituting these into the equation for ${\theta }^{*}\left(v\right)$, we obtain

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& =2d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{5}{3}}-(d\left(v\right)-2\times n_{4^{-}}\left(v\right))\times {\displaystyle \frac{2}{3}}\hfill \\ \hfill & ={\displaystyle \frac{4}{3}}\times d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{3}}\hfill \\ \hfill & \ge {\displaystyle \frac{4}{3}}\times 5-6-2\times {\displaystyle \frac{1}{3}}\hfill \\ \hfill & =0.\hfill \end{array}$$

CASE 4.2: Suppose $0<2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)<d\left(v\right)=5$.

By Lemma 4, we obtain $n_{4^{-}}\left(v\right)\le 2$, $n_5\left(v\right)\le d\left(v\right)-2n_{4^{-}}\left(v\right)-m_{4^{-}}\left(v\right)-1$, and $m_5\left(v\right)\le d\left(v\right)-2\times n_{4^{-}}\left(v\right)-m_{4^{-}}\left(v\right)$. Substituting these into the equation for ${\theta }^{*}\left(v\right)$, we obtain

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& \ge 2d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{5}{3}}-m_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}-(d\left(v\right)-2n_{4^{-}}\left(v\right)-m_{4^{-}}\left(v\right)-1)\times {\displaystyle \frac{1}{2}}\hfill \\ \hfill & -(d\left(v\right)-2n_{4^{-}}\left(v\right)-m_{4^{-}}\left(v\right))\times {\displaystyle \frac{1}{4}}\hfill \\ \hfill & ={\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{6}}+m_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{12}}\hfill \\ \hfill & \ge {\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{6}}\hfill \\ \hfill & \ge {\displaystyle \frac{5}{4}}\times 5-{\displaystyle \frac{11}{2}}-2\times {\displaystyle \frac{1}{6}}\hfill \\ \hfill & >0.\hfill \end{array}$$

CASE 5: Let v be a 6-vertex with $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)>0$.

By Lemma 3 (iii), v is incident to at most one poor 3-face. We only need to consider the case where v incident to exactly one poor 3-face, denoted by f. In this case, v gives a charge of ${\displaystyle \frac{7}{3}}$ to the 3-face f, and ${\displaystyle \frac{5}{3}}$ to the other $4^{-}$-faces.

By (R3), we calculate ${\theta }^{*}\left(v\right)$ as follows:

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& =2d\left(v\right)-6-{\displaystyle \frac{7}{3}}-(n_{4^{-}}\left(v\right)-1)\times {\displaystyle \frac{5}{3}}-m_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}-n_5\left(v\right)\times {\displaystyle \frac{1}{2}}-m_5\left(v\right)\times {\displaystyle \frac{1}{4}}\hfill \\ \hfill & =-{\displaystyle \frac{2}{3}}+\left(2d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{5}{3}}-m_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}-n_5\left(v\right)\times {\displaystyle \frac{1}{2}}-m_5\left(v\right)\times {\displaystyle \frac{1}{4}}\right).\hfill \end{array}$$

CASE 5.1: Suppose $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)=d\left(v\right)=6$.

By Lemmas 4 (iii) and (iv), we have $n_5\left(v\right)=m_5\left(v\right)=0.$ It follows from a calculation in CASE 4.1 that

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& =-{\displaystyle \frac{2}{3}}+({\displaystyle \frac{4}{3}}\times d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{3}})\hfill \\ \hfill & ={\displaystyle \frac{4}{3}}\times d\left(v\right)-{\displaystyle \frac{20}{3}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{3}}\hfill \\ \hfill & \ge {\displaystyle \frac{4}{3}}\times 6-{\displaystyle \frac{20}{3}}-3\times {\displaystyle \frac{1}{3}}\hfill \\ \hfill & >0.\hfill \end{array}$$

CASE 5.2: Let $0<2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)<d\left(v\right)=6$.

It follows from a calculation in CASE 4.2 that

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& \ge -{\displaystyle \frac{2}{3}}+({\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{6}})\hfill \\ \hfill & ={\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{37}{6}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{6}}\hfill \\ \hfill & \ge {\displaystyle \frac{5}{4}}\times 6-{\displaystyle \frac{37}{6}}-3\times {\displaystyle \frac{1}{6}}\hfill \\ \hfill & >0.\hfill \end{array}$$

CASE 6: Let v be a $7^{+}$-vertex with $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)>0$.

Then v transfers at most ${\displaystyle \frac{7}{3}}$ of its charge to each incident $4^{-}$-face. By (R3), we compute ${\theta }^{*}\left(v\right)$ as follows:

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& =2d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{7}{3}}-m_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}-n_5\left(v\right)\times {\displaystyle \frac{1}{2}}-m_5\left(v\right)\times {\displaystyle \frac{1}{4}}\hfill \\ \hfill & =-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}+\left(2d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{5}{3}}-m_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}-n_5\left(v\right)\times {\displaystyle \frac{1}{2}}-m_5\left(v\right)\times {\displaystyle \frac{1}{4}}\right).\hfill \end{array}$$

CASE 6.1: Suppose $2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)=d\left(v\right)$.

By Lemmas 4 (iii) and (iv), we have $n_5\left(v\right)=m_5\left(v\right)=0$. It follows from a calculation in CASE 4.1 that

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& =-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}+\left({\displaystyle \frac{4}{3}}\times d\left(v\right)-6-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{3}}\right)\hfill \\ \hfill & ={\displaystyle \frac{4}{3}}\times d\left(v\right)-6-n_{4^{-}}\left(v\right)\times 1.\hfill \end{array}$$

-

If v is a 7-vertex, then $d\left(v\right)=7$ and $n_{4^{-}}\left(v\right)\le 3$. Thus,

$${\theta }^{*}\left(v\right)\ge {\displaystyle \frac{4}{3}}\times 7-3-6>0.$$

-

If v is an $8^{+}$-vertex, then we use $n_{4^{-}}\left(v\right)\le {\displaystyle \frac{d\left(v\right)}{2}}$ to obtain

$${\theta }^{*}\left(v\right)\ge {\displaystyle \frac{4}{3}}\times d\left(v\right)-6-{\displaystyle \frac{d\left(v\right)}{2}}>0.$$

CASE 6.2: Suppose $0<2\times n_{4^{-}}\left(v\right)+m_{4^{-}}\left(v\right)<d\left(v\right)$.

By Lemma 4, we have $n_{4^{-}}\left(v\right)\le {\displaystyle \frac{d\left(v\right)}{2}}$, $n_5\left(v\right)\le d\left(v\right)-2n_{4^{-}}\left(v\right)-m_{4^{-}}\left(v\right)-1$, and $m_5\left(v\right)\le d\left(v\right)-2n_{4^{-}}\left(v\right)-m_{4^{-}}\left(v\right)$. It follows from a calculation in CASE 4.2 that

$$\begin{array}{cc}\hfill {\theta }^{*}\left(v\right)& \ge -n_{4^{-}}\left(v\right)\times {\displaystyle \frac{2}{3}}+\left({\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{1}{6}}\right)\hfill \\ \hfill & ={\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}-n_{4^{-}}\left(v\right)\times {\displaystyle \frac{5}{6}}\hfill \\ \hfill & \ge {\displaystyle \frac{5}{4}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}-{\displaystyle \frac{d\left(v\right)}{2}}\times {\displaystyle \frac{5}{6}}\hfill \\ \hfill & ={\displaystyle \frac{5}{6}}\times d\left(v\right)-{\displaystyle \frac{11}{2}}\hfill \\ \hfill & >0\mathrm{for}\mathrm{each}d\left(v\right)\ge 7.\hfill \end{array}$$

CASE 7: Let f be a 3-face.

If f is incident to a terrible 3-vertex, then it must also be incident to a $5^{+}$-vertex, according to Lemma 3 (i). As a result, f can have at most two incident terrible 3-vertices.

Let k be the number of incident non-terrible 3-vertices of $f.$

If f has no incident terrible 3-vertices, then ${\theta }^{*}\left(f\right)=\theta \left(f\right)+k\times {\displaystyle \frac{1}{3}}+(3-k)\times 1+k\times {\displaystyle \frac{2}{3}}=0$ by (R1), (R2), and (R3).

If f has exactly one incident terrible 3-vertex, then ${\theta }^{*}\left(f\right)=\theta \left(f\right)+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{5}{3}}+(1-k)\times 1+k\times {\displaystyle \frac{1}{3}}+k\times {\displaystyle \frac{2}{3}}=0$ by (R1), (R2), and (R3).

If f has two incident terrible 3-vertices, then ${\theta }^{*}\left(f\right)=\theta \left(f\right)+2\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{7}{3}}=0$ by (R1) and (R3).

CASE 8: Let f be a 4-face. We use Lemma 3 (i) to obtain the following results. If f has no incident $4^{+}$-vertices, then f has four non-terrible 3-vertices. If f has exactly one incident 4-vertex, then f has three non-terrible 3-vertices. Thus, ${\theta }^{*}\left(f\right)=\theta \left(f\right)+4\times {\displaystyle \frac{2}{3}}>0$ by (R3).

If f has an incident $5^{+}$-vertex, then f has at least one non-terrible 3-vertex by Lemma 3 (i). Thus ${\theta }^{*}\left(f\right)\ge \theta \left(f\right)+{\displaystyle \frac{5}{3}}+{\displaystyle \frac{2}{3}}>0$ by (R3).

If f has at least two incident $4^{+}$-vertices, then ${\theta }^{*}\left(f\right)\ge \theta \left(f\right)+2\times {\displaystyle \frac{5}{3}}>0$ by (R2) and (R3).

CASE 9: Let f be a 5-face.

If f has no incident $4^{+}$-vertices, then f has five non-terrible 3-vertices by Lemma 3 (i). Thus, ${\theta }^{*}\left(f\right)=\theta \left(f\right)+5\times {\displaystyle \frac{1}{4}}>0$ by (R3).

If f has exactly one incident $4^{+}$-vertex, then f has at least two non-terrible 3-vertices by Lemma 3 (i). Thus ${\theta }^{*}\left(f\right)\ge \theta \left(f\right)+{\displaystyle \frac{1}{2}}+2\times {\displaystyle \frac{1}{4}}=0$ by (R2) and (R3).

If f has at least two incident $4^{+}$-vertices, then ${\theta }^{*}\left(f\right)\ge \theta \left(f\right)+2\times {\displaystyle \frac{1}{2}}=0$ by (R2) and (R3).

CASE 10: Let f be a $7^{+}$-face.

Note that a 7-face is incident to at most six 3-vertices that are also incident to a 3-face; otherwise, there are two adjacent 3-faces, which contradicts Condition (i) of G.

By (R4), ${\theta }^{*}\left(f\right)=\theta \left(f\right)-6\times {\displaystyle \frac{1}{6}}=0$ if f is a 7-face, and ${\theta }^{*}\left(f\right)=\theta \left(f\right)-d\left(v\right)\times {\displaystyle \frac{1}{6}}={\displaystyle \frac{5}{6}}\times d\left(v\right)-6>0$ if f is a $8^{+}$-face.

From all the cases considered, we obtain ${\sum }_{x\in V\left(G\right)\cup F\left(G\right)}{\theta }^{*}\left(x\right)>0,$ a contradiction. This completes the proof.

4. Concluding Remarks and Open Problem

Before we show that our main result in Theorem 3 implies Theorems 1 and 2, two following lemmas are required.

Lemma 5.

Let G be a graph with no chordal $7^{-}$-cycles where G has no $2^{-}$-vertices. Given f and g are adjacent faces in G, the following statements hold.

(i)

If f is a 3-face, then g is not an m-face where $m\in \{3,\dots ,6\}$.

(ii)

If f is a 4-face, then g is not an m-face where $m\in \{3,\dots ,5\}$.

Proof. 

(i) Suppose to the contrary that f is a 3-face and g is $6^{-}$-face.

-

Let $b\left(g\right)$ be bounded by one cycle (a $d\left(g\right)$-cycle).

If $b\left(f\right)$ and $b\left(g\right)$ share exactly two vertices, then $b\left(f\right)\cup b\left(g\right)$ is a ($d\left(f\right)+d\left(g\right)-2$)-cycle with a chord where $4\le d\left(f\right)+d\left(g\right)-2\le 7$, a contradiction.

If $b\left(f\right)$ and $b\left(g\right)$ share exactly three vertices, then $b\left(g\right)$ is a $d\left(g\right)$-cycle with a chord where $d\left(g\right)\le 6$, a contradiction.

-

Let $b\left(g\right)$ be bounded by at least two cycles. Since G has no 2-vertices, we only consider the case g is a 6-face where $b\left(g\right)$ is intersecting two 3-cycles $C_1$ and $C_2$.

If $b\left(f\right)$ and $b\left(g\right)$ share exactly two vertices, then $b\left(f\right)\cup C_1$ or $b\left(f\right)\cup C_2$ is a 4-cycle with a chord, a contradiction.

If $b\left(f\right)$ and $b\left(g\right)$ share exactly three vertices, then $b\left(g\right)=C_1$ or $b\left(g\right)=C_2$. Consequently, G contains a 2-vertex, a contradiction.

(ii) Suppose to the contrary that f is a 4-face and g is $5^{-}$-face. Note that $b\left(g\right)$ is bounded by one cycle (a $d\left(g\right)$-cycle).

If $b\left(f\right)$ and $b\left(g\right)$ share exactly two vertices, then $b\left(f\right)\cup b\left(g\right)$ is a ($d\left(f\right)+d\left(g\right)-2$)-cycle with a chord where $d\left(f\right)+d\left(g\right)-2\le 7$, a contradiction.

If $b\left(f\right)$ and $b\left(g\right)$ share at least three vertices, then let $b\left(f\right)=v_1{v}_2{v}_3{v}_4$ and $b\left(g\right)=v_1{v}_2{u}_1\dots u_{d\left(g\right)-2}$. By symmetry, we only consider the case $v_3=u_i$ where $i\in \{1,2,\dots ,d(g)-2\}.$ If $i=1$, then $v_2$ is a 2-vertex which contradicts Lemma 1, otherwise $b\left(g\right)$ is a $d\left(g\right)$-cycle with a chord which is also a contradiction. □

Lemma 6.

Let G be a graph with neither 4-cycles nor 6-cycles where G has no $2^{-}$-vertices. If f and g are adjacent faces in G where f is a 3-face, then g is not an m-face where $m\in \{3,\dots ,6\}$.

Proof. 

Suppose to the contrary that f is a 3-face and g is $6^{-}$-face.

-

Let $b\left(g\right)$ be bounded by one cycle (a $d\left(g\right)$-cycle).

From the condition of G, we assume $d\left(g\right)\in \{3,5\}$. Consider $d\left(g\right)=3$, we have G contains a 4-cycle if $b\left(f\right)$ and $b\left(g\right)$ share exactly two vertices, and G contains a 2-vertex if $b\left(f\right)$ and $b\left(g\right)$ share exactly three vertices, a contradiction.

Consider $d\left(g\right)=5$, we have G contains a 6-cycle if $b\left(f\right)$ and $b\left(g\right)$ share exactly two vertices, and G contains a 4-cycle if $b\left(f\right)$ and $b\left(g\right)$ share exactly three vertices, a contradiction.

-

Let $b\left(g\right)$ be bounded by at least two cycles. By Lemma 1, we only consider the case that g is a 6-face where $b\left(g\right)$ is intersecting two 3-cycles $C_1$ and $C_2$.

If $b\left(f\right)$ and $b\left(g\right)$ share exactly two vertices, then $b\left(f\right)\cup C_1$ or $b\left(f\right)\cup C_2$ is a 4-cycle, a contradiction.

If $b\left(f\right)$ and $b\left(g\right)$ share exactly three vertices, then $b\left(g\right)=C_1$ or $b\left(g\right)=C_2$. Consequently, G has a 2-vertex. □

Now, we are in a position to show that our main result in Theorem 3 implies Theorems 1 and 2.

Theorem 4.

Every planar graph (i) with no chordal $7^{-}$-cycles, or (ii) with neither 4- nor 6-cycles, has an (${\mathcal{F}}_2,\mathcal{F}$)-partition.

Proof. 

Suppose to the contrary that there exists a minimal counterexample G. By Lemma 3 (i), G has no $2^{-}$-vertices.

(i)

Suppose G has no chordal $7^{-}$-cycles. By Lemma 5, none of its 3-faces is adjacent to a $6^{-}$-face, and none of its 4-faces is adjacent to a $5^{-}$-face. It follows from Theorem 3 that G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition.

(ii)

Suppose G has neither 4- nor 6-cycles. By Lemma 6, none of its 3-faces is adjacent to a $6^{-}$-face. Since G has no 4-cycles, it has no 4-faces. It follows from Theorem 3 that G has an (${\mathcal{F}}_2,\mathcal{F}$)-partition.

Thus, a minimal counterexample does not exist. This completes the proof. □

Note that Theorem 4 (i) implies Theorem 2, while Theorem 4 (ii) is exactly Theorem 1. Apart from the latter result, it is shown by Kang et al. [29] that each planar graph with neither 4- nor 6-cycles also has an $({\mathcal{F}}_1,\mathcal{I},\mathcal{I})$-partition. Motivated by these results, we propose the following conjecture.

Problem 1.

Every planar graph with neither 4- nor 6-cycles has an $({\mathcal{F}}_1,\mathcal{F})$-partition.

It is worth noting that vertex partition problems have broad applications in various domains. For instance, they play a key role in frequency assignment for wireless networks and in the design of fault-tolerant networks, where partitioning helps to minimize interference and maintain functionality under failures. Moreover, planar graphs themselves are widely used to model real-world structures such as circuit layouts, traffic systems, and geographic networks.