Some conditions on 5-cycles that make planar graphs 4-choosable

TL;DR

This paper identifies specific structural conditions on planar graphs involving 5-cycles that guarantee the graph is 4-choosable, expanding understanding of list coloring in planar graphs.

Contribution

It introduces new conditions on 5-cycles that ensure planar graphs are 4-choosable, broadening previous results in graph coloring theory.

Findings

Planar graphs with certain 5-cycle conditions are 4-choosable

Graphs where 5-cycles are not adjacent to 3-cycles are 4-choosable

Structural restrictions on 5-cycles influence list coloring properties

Abstract

Consider two conditions on a graph: (1) each 5-cycle is not a subgraph of 5-wheel and does not share exactly one edge with 3-cycle, and (2) each 5-cycle is not adjacent to two 3-cycles and is not adjacent to a 4-cycle with chord. We show that if a planar graph $G$ satisfies one of the these conditions, then $G$ is 4-choosable. This yields that if each 5-cycle of a planar graph $G$ is not adjacent a 3-cycle, then $G$ is 4-choosable.

Equations

$$\sum _{v\in V(G)}\mu (v)+\sum _{f\in F(G)}\mu (f)=-12.$$

Full text

Some conditions on 5-cycles that make planar graphs 4-choosable

Pongpat Sittitrai

*Department of Mathematics, Faculty of Science, Khon Kaen University, 40002, Thailand

E-mail address: pongpat@kkumail.com*

  

Kittikorn Nakprasit 111Corresponding Author

*Department of Mathematics, Faculty of Science, Khon Kaen University, 40002, Thailand

E-mail address: kitnak@hotmail.com*

Abstract

Consider two conditions on a graph: (1) each 5-cycle is not a subgraph of 5-wheel and does not share exactly one edge with 3-cycle, and (2) each 5-cycle is not adjacent to two 3-cycles and is not adjacent to a 4-cycle with chord. We show that if a planar graph $G$ satisfies one of the these conditions, then $G$ is 4-choosable. This yields that if each 5-cycle of a planar graph $G$ is not adjacent a 3-cycle, then $G$ is 4-choosable.

1 Introduction

The concept of list coloring was introduced by Vizing [13] and by Erdős, Rubin, and Taylor [4], independently. An assignment $L$ for a graph $G$ assigns a list $L(v)$ for each vertex $v.$ If $|L(v)|=k$ for each vertex $v,$ then we call $L$ a $k$-assignment. A graph $G$ is $L$-colorable if we can color $G$ with each vertex has a color from its list and no two adjacent vertices receive the same color. If $G$ is $L$-colorable for any $k$-assignment $L,$ then we say $G$ is $k$-choosable.

Thomassen [12] proved that every planar graph is 5-choosable. Voight [14] and Mirzakhani [11] presented the examples of non 4-choosable graphs. Additionally, Gutner [7] showed that the problem of determining whether a planar graph is 4-choosable is NP-hard. This leads to the interest of finding some nice sufficient conditions for planar graphs to be 4-choosable. If a planar graph has no 3-cycles, then it has a vertex of degree at most 3. Thus it is 4-choosable. More sufficient conditions are founded, for examples, it was shown that a planar graph is 4-choosable if it has no 4-cycles [10], 5-cycles [9, 16], 6-cycles [6], 7-cycles [5], intersecting 3-cycles [17], intersecting 5-cycles [8], 3-cycles adjacent to 4-cycles [2, 3], 5-cycles simultaneously adjacent to 3-cycles and 4-cycles.

In this paper, we prove the following theorems.

Theorem 1

If each 5-cycle of a planar graph $G$ is not a subgraph of 5-wheel and does not share exactly one edge with 3-cycle, then $G$ is 4-choosable.

Theorem 2

If each 5-cycle of a planar graph $G$ is not adjacent to two 3-cycles and is not adjacent to a 4-cycle with chord, then $G$ is 4-choosable.

The following corollary is an easy consequence.

Corollary 3

If each 5-cycle of a planar graph $G$ is not adjacent a 3-cycle, then $G$ is 4-choosable.

2 Preliminaries

First, we introduce some notations and definitions. A $k$-vertex (face) is a vertex (face) of degree $k,$ a $k^{+}$-vertex (face) is a vertex (face) of degree at least $k,$ and a $k^{-}$-vertex (face) is a vertex (face) of degree at most $k.$ A $(d_{1},d_{2},\dots,d_{k})$-face $f$ is a face of degree $k$ where all vertices on $f$ have degree $d_{1},d_{2},\dots,d_{k}$. A $(d_{1},d_{2},\dots,d_{k})$-vertex $v$ is a vertex of degree $k$ where all faces incident to $v$ have degree $d_{1},d_{2},\dots,d_{k}$.

A trio is a graphs consist of a vertex set of five elements, namely $\{x,y,z,u,v,w\},$ and an edge set $\{xy,xu,xv,yv,yw,uv,vw\}$ (see Fig. 1). We call $s$ on a 3-cycle $f$ a good vertex of $f$ if $f$ is not in any trio, a bad vertex of $f$ if $f$ is in a trio and is the only 3-cycle on that trio that $s$ is belong to, a worst vertex of $f$ if $s$ is on $f$ and all other 3-cycles, otherwise $s$ is called a worse vertex of $f.$ We call a face $f$ is a bad (worse, or worst, respectively) face of a vertex $v$ if $v$ is a bad (worse, or worst, respectively) vertex of $f.$

Lemma 4

*Suppose $G$ is a minimal non 4-choosable graph. Then each of the followings holds.

(1) Each vertex degree is at least 4.

(2) $G$ does not contain a $(4,4,4,4)$-face.

(3) $G$ does not contain a subgraph $H$ as in Fig. 2(1).*

Proof.   Let $L$ be a 4-list assignment of $G.$

(1) This is a well-known fact, see e.g. [10].

(2) Suppose $G$ contains a $(4,4,4,4)$-face $F.$ By minimality of $G,$ the graph $G-V(F)$ has an $L$-coloring. A list of available colors of each vertex in $F$ has size at least 2. Thus we can extend $L$-coloring to $G,$ a contradiction.

(3) Suppose $G$ contains a subgraph $H$ as in Fig. 2. By minimality of $G,$ the graph $G-V(H)$ has an $L$-coloring. The lower bound on size of each residual list is as in Fig. 2(2). Next, choose colors for $x$ and $u.$ We can extend an $L$-coloring to $G$ unless every residual list is identical and has size 2 but one can recolor $x$ and $u$ to avoid this situation. This completes the proof. $\Box$

Next, we introduce a famous theorem established by Alon and Tarsi [1]. This theorem shows the connection between the list coloring and an orientation of a graph. A digraph $D$ consists of a vertex set $V(D)$ and an arc set $A(D).$ For an arc $a=(u,v)\in A(D),$ we say $a$ has a tail $u$ and a head $v.$ The indegree $d^{-}_{D}(v)$ of a vertex $v$ is the number of arcs with head $v,$ and the outdegree $d^{+}_{D}(v)$ of a vertex $v$ is the number of arcs with tail $v.$ A subgraph $H$ of $D$ is called Eulerian if $d^{-}_{H}(v)=d^{+}_{H}(v)$ for each vertex $v$ of $H.$ In this context, $H$ can be disconnected or even edgeless. A digraph $H$ is even (odd) if it has even (odd) number of arcs. Let $EE(D)$ denote the number of even spanning Eulerian subgraphs of $D,$ and $EO(D)$ denote the number of odd spanning Eulerian subgraphs of $D.$

Theorem 5

[1]** Let $D$ be a digraph. Let $L$ be a list assignment with $|L(v)|=d^{+}_{D}(v)+1$ for each vertex $v$ of $D.$ If $EE(D)\neq EO(D),$ then $D$ is $L$-colorable.

Lemma 6

A minimal non 4-choosable graph $G$ does not contain a subgraph isomorphic to one of the configurations in Fig. 3.

Proof.   Let $L$ be a 4-list assignment of $G.$ Suppose $G$ contains a subgraph $H$ isomorphic to one of the configurations in Fig. 3. By minimality of $G,$ the graph $G-V(H)$ has an $L$-coloring. An orientation and availability of colors for every vertex are shown in Fig. 4. Since $EE(G_{1})=2>EO(G_{1})=1,EE(G_{2})=3>EO(G_{2})=1,$ and $EE(G_{3})=2>EO(G_{3})=1,$ each $G_{i}$ satisfies Theorem 5. Thus we can extend an $L$-coloring to $G,$ a contradiction. $\Box$

3 Main results and proof

Theorem 7

If each 5-cycle of a planar graph $G$ is not a subgraph of 5-wheel and does not share exactly one edge with 3-cycle, then $G$ is 4-choosable.

Theorem 8

If each 5-cycle of a planar graph $G$ is not adjacent to two 3-cycles and is not adjacent to $\theta(1,2,2),$ then $G$ is 4-choosable.

The following corollary is an easy consequence.

Corollary 9

If each 5-cycle of a planar graph $G$ is not adjacent a 3-cycle, then $G$ is 4-choosable.

We use the same proof for Theorems 7 and 8. In fact the proof of the latter is easier because it does not involve a trio.

Proof.

Suppose that $G$ is a minimal counterexample. The discharging process is as follows. Let the initial charge of a vertex $v$ in $G$ be $\mu(v)=2d(v)-6$ and the initial charge of a face $f$ in $G$ be $\mu(f)=d(f)-6$. Then by Euler’s formula $|V(G)|-|E(G)|+|F(G)|=2$ and by the Handshaking lemma, we have

[TABLE]

Now, we establish a new charge $\mu^{*}(x)$ for all $x\in V(G)\cup F(G)$ by transferring charge from one element to another and the summation of new charge $\mu^{*}(x)$ remains $-12$. If the final charge $\mu^{*}(x)\geq 0$ for all $x\in V(G)\cup F(G)$, then we get a contradiction and the proof is completed.

Let $w(v\rightarrow f)$ be the charge transfered from a vertex $v$ to a face $f.$ The discharging rules are as follows.

(R1) $w(v\rightarrow f)=0.2$ if $f$ is a 5-face.

(R2) For a 4-vertex $v$, $w(v\rightarrow f)=1$ if $v$ is a good, bad, or, worse vertex of a 3-face $f,$ $w(v\rightarrow f)=2/3$ if $v$ is a worst vertex of a 3-face $f,$ $w(v\rightarrow f)=2/3$ if $v$ is a worst vertex of a 3-face $f,$ and $w(v\rightarrow f)=1/3$ if $v$ is vertex on a 4-face $f.$

(R3) For a 5-vertex $v$, $w(v\rightarrow f)=1$ if $v$ is a good or worst vertex of a 3-face $f,$ $w(v\rightarrow f)=1.5$ if $v$ is a bad vertex of a 3-face $f,$ $w(v\rightarrow f)=1.25$ if $v$ is a worse vertex of a 3-face $f,$ $w(v\rightarrow f)=1$ if $v$ is vertex on a (4,4,4,5)-face $f,$ $w(v\rightarrow f)=2/3$ if $v$ is vertex on a 4-face $f$ that is not a (4,4,4,5)-face.

(R4) For a 6+-vertex $v$, $w(v\rightarrow f)=1$ if $v$ is a good or worst vertex of a 3-face $f,$ $w(v\rightarrow f)=1.5$ if $v$ is a bad vertex of a 3-face $f,$ $w(v\rightarrow f)=1.25$ if $v$ is a worse vertex of a 3-face $f,$ $w(v\rightarrow f)=1$ if $v$ is vertex on a (4,4,4,5)-face $f,$ $w(v\rightarrow f)=2/3$ if $v$ is vertex on a 4-face $f$ that is not a (4,4,4,5)-face.

(R5) After (R1)-(R4), redistribute charge in three 3-faces in the same trio to make their charge equal.

It remains to show that resulting $\mu^{*}(x)\geq 0$ for all $x\in V(G)\cup F(G)$.

Consider a 4-vertex $v.$ If $v$ is not incident to any 3-face, then $\mu^{*}(v)\geq\mu(v)-4\cdot(1/3)=2/3.$ If $v$ is incident to exactly one 3-face, then $\mu^{*}(v)\geq\mu(v)-(1+3\cdot(1/3))=0.$ If $v$ is incident to two 3-faces, then $v$ is not a worst vertex of any face and other incident faces are $6^{+}.$ Using (R2), we obtain $\mu^{*}(v)=\mu(v)-(2\cdot 1)=0.$ If $v$ is incident to three 3-faces, then $v$ is a worst vertex of these faces and the other incident face is $6^{+}.$ Using (R2), we obtain $\mu^{*}(v)=\mu(v)-(3\cdot(2/3))=0.$

Consider a 5-vertex $v.$ If $v$ is a (3,4,4,4,5)- or a (4,4,4,4,5)-vertex, then at least one of its incident faces is a (4,4,5,5)-face by Lemma 6(3). This means $v$ sends charge to its incident faces at most $3+(2/3)+(1/5)$ which is less than 4. If $v$ is a (3,4,4,4,4)- or a (4,4,4,4,4)-vertex, then at least three of its incident faces is a (4,4,5,5)-face by Lemma 6(3). This means $v$ sends charge to its incident faces at most $2+3\cdot(2/3)=4.$ If $v$ is incident to at least two 5-face, then $v$ sends charge at most $2\cdot(1/5)+3$ in the case that $v$ has a bad face, and at most $2\cdot(1/5)+3$ if $v$ has no bad faces. If $v$ has two bad faces, then $v$ has three 6+-faces. Using (R3), we obtain $\mu^{*}(v)=\mu(v)-(2\cdot(1.5))=2.$ So we assume that $v$ is incident to a 6+-face and has at most one bad face. If $v$ has a worst face, a worse face, or a bad face, then $v$ is incident to at least two 6+-faces and $\mu^{*}(v)=\mu(v)-3=1$ in the case that $v$ has a worst face, or $\mu^{*}(v)\geq\mu(v)-(2\cdot(1.25)+1)>0$ in the case that $v$ has a worse face or a bad face. If $v$ has neither a worst face, a worse face, nor a bad face, then $\mu^{*}(v)\geq\mu(v)-4=0.$

If $v$ has a worst face, a worse face or a bad face, respectively, then $v$ is incident to at least two 6+-faces and $\mu^{*}(v)=\mu(v)-3=1$ or $\mu^{*}(v)\geq\mu(v)-(2\cdot(1.25)+1)>0,$ respectively.

Consider a $k$-vertex $v$ with $k\geq 6.$ Assume $v$ is incident to the faces $f_{1},\ldots,f_{k}$ in a cyclic order, where all subscripts are taken modulo $k.$ To calculate $\mu(v),$ we redistribute $w(v\rightarrow f_{i}).$ However, we still use $w(v\rightarrow f_{i})$ according to (R4) for the final charge. If $w(v\rightarrow f_{i})=1.5,$ then $v$ is a worse vertex of $f_{i}$ and $f_{i-1}$ (or $f_{i+1}$) is a 6+-face. Reduce $w(v\rightarrow f_{i})$ to 1 and transfer the remaining charge 0.5 to $f_{i-1}.$ If $w(v\rightarrow f_{i})=2,$ then $v$ is a bad vertex of $f_{i}$ whereas both $f_{i-1}$ and $f_{i+1}$) are 6+-faces. Reduce $w(v\rightarrow f_{i})$ to 1 and transfer charge 0.5 to $f_{i-1}$ and $f_{i+1}.$ According to this average, $v$ sends charge at most 1 to each of its adjacent faces. Thus $\mu^{*}(v)\geq\mu(v)-\deg(v)\geq 0.$

It is clear that $\mu^{*}(f)=\mu(f)\geq 0$ for $f$ a $6^{+}$-face $f.$ If $f$ is a 5-face, then $\mu^{*}(f)=\mu(f)+5\cdot(0.2)=0$ by (R1). Consider a 4-face $f.$ Lemma 4 yields that every vertex on $f$ has degree at least 4 and one of them has degree at least 5. If $f$ is a $(4,4,4,5)$-face, then $\mu^{*}(f)\geq\mu(f)+3(2/3)+1=0,$ otherwise $\mu^{*}(f)\geq\mu(f)+2(1/3)+2\cdot(2/3)=1$ by (R2) and (R3). If $f$ is a 3-face that is not in a trio, then $\mu^{*}(f)=\mu(f)+3(1/3)=0$ by (R2), (R3), and (R4).

According to (R5), it remains to show that the total increased charge of three 3-faces in each trio is at least 9 (or in the other words, the summation of their new charge is 0.) If the degree of a worst vertex is at least 5, then each vertex send charge at least 1 to its incident face. So we assume that a worst vertex has degree 4 and it sends charge totaling $3\cdot(2/3)=2$ to three faces in a trio. By (R2)-(R4), each of the remaining vertices in the trio sends charge at least 1 to its incident face in a trio. Note that a 6+-vertex $v$ in the trio sends charge totaling $2\cdot(1.5)=3$ to two faces in the trio if $v$ is a worse vertex, otherwise it sends charge 2 to a face in the trio according to (R4). Additionally, a 5-vertex $v$ in a trio sends charge totaling $2\cdot(1.25)=2.5$ to two faces in the trio if $v$ is a worse vertex, otherwise it sends charge 1.5 to a face in the trio according to (R4). Thus total increased charge is at least 9 if the trio has a 6+-vertex $v$ or two $5$-vertex. Use Lemma 4 and 6 to eliminate the remaining case. This completes the proof. $\Box$

It is interesting to see that whether one can reduce some restrictions in Theorems 7 or 8 but a planar graph is still 4-choosable.

Acknowledgments

The first author is supported by Development and Promotion of Science and Technology talents project (DPST).