The 4-girth-thickness of the complete graph

TL;DR

This paper introduces the concept of 4-girth-thickness for graphs, specifically calculating it for complete graphs, revealing a formula that applies to all but two specific cases.

Contribution

It defines the 4-girth-thickness of graphs and determines its exact value for all complete graphs except for two special cases.

Findings

The 4-girth-thickness of K_n is eil((n+2)/4)or n ,10.

eil((n+2)/4)pplies to all complete graphs except K_6 and K_{10}.

The 4-girth-thickness of K_6 is 3.

Abstract

In this paper, we define the $4$-girth-thickness $\theta(4,G)$ of a graph $G$ as the minimum number of planar subgraphs of girth at least $4$ whose union is $G$. We obtain the $4$-girth-thickness of the arbitrary complete graph $K_n$ getting that $\theta(4,K_n)=\left\lceil \frac{n+2}{4}\right\rceil$ for $n\not=6,10$ and $\theta(4,K_6)=3$.

Figures

Equations

$$\lceil \frac{n(n-1)}{2(2n-4)}\rceil =\lceil \frac{n+1}{4}+\frac{1}{2n-4}\rceil =\lceil \frac{n+2}{4}\rceil$$

$$v_1{v}_2,v_2{v}_{2k},v_{2k}{v}_3,v_3{v}_{2k-1},\dots ,v_{k+2}{v}_{k+1}.$$

$$v_{1+i-1}{v}_{2+i-1},v_{2+i-1}{v}_{2k+i-1},v_{2k+i-1}{v}_{3+i-1},v_{3+i-1}{v}_{2k-1+i-1},\dots ,v_{k+2+i-1}{v}_{k+1+i-1},$$

$$v_i^{\prime }{v}_{i+1}^{\prime },v_{i+1}^{\prime }{v}_{i-1}^{\prime },v_{i-1}^{\prime }{v}_{i+2}^{\prime },v_{i+2}^{\prime }{v}_{i-2}^{\prime },\dots ,v_{i+k+1}^{\prime }{v}_{i+k}^{\prime },$$

$$v_i{v}_{i+1}^{\prime },v_i^{\prime }{v}_{i+1},v_{i+1}{v}_{i-1}^{\prime },v_{i+1}^{\prime }{v}_{i-1},v_{i-1}{v}_{i+2}^{\prime },v_{i-1}^{\prime }{v}_{i+2},\dots ,v_{i+k+1}{v}_{i+k}^{\prime },v_{i+k+1}^{\prime }{v}_{i+k}$$

$$k+1\le \theta (4,K_{4k-1})\le \theta (4,K_{4k})\le k+1.$$

$$\{xy,x{v}_1,x{v}_2^{\prime },x{v}_3,x{v}_4^{\prime },\dots ,x{v}_{2k-1},x{v}_{2k}^{\prime },y{v}_1^{\prime },y{v}_2,y{v}_3^{\prime },y{v}_4,\dots ,y{v}_{2k-1}^{\prime },y{v}_{2k}\}.$$

$$\{x{v}_{i+\lceil \frac{3k}{2}\rceil -1}^{\prime },x{v}_{i+\lceil \frac{3k}{2}\rceil },y{v}_{i+\lceil \frac{3k}{2}\rceil -1},y{v}_{i+\lceil \frac{3k}{2}\rceil }^{\prime }\}$$

$$\{y{v}_{i+\lceil \frac{3k}{2}\rceil -1}^{\prime },y{v}_{i+\lceil \frac{3k}{2}\rceil },x{v}_{i+\lceil \frac{3k}{2}\rceil -1},x{v}_{i+\lceil \frac{3k}{2}\rceil }^{\prime }\}.$$

$$\{x{v}_{i+\lceil \frac{k}{2}\rceil -1}^{\prime },x{v}_{i+\lceil \frac{k}{2}\rceil },y{v}_{i+\lceil \frac{k}{2}\rceil -1},y{v}_{i+\lceil \frac{k}{2}\rceil }^{\prime }\}$$

$$\{y{v}_{i+\lceil \frac{k}{2}\rceil -1}^{\prime },y{v}_{i+\lceil \frac{k}{2}\rceil },x{v}_{i+\lceil \frac{k}{2}\rceil -1},x{v}_{i+\lceil \frac{k}{2}\rceil }^{\prime }\}.$$

$$\{x{v}_{i+\lceil \frac{3k}{2}\rceil +1},x{v}_{i+\lceil \frac{3k}{2}\rceil }^{\prime },y{v}_{i+\lceil \frac{3k}{2}\rceil +1}^{\prime },y{v}_{i+\lceil \frac{3k}{2}\rceil }\}$$

$$\{y{v}_{i+\lceil \frac{3k}{2}\rceil +1},y{v}_{i+\lceil \frac{3k}{2}\rceil }^{\prime },x{v}_{i+\lceil \frac{3k}{2}\rceil +1}^{\prime },x{v}_{i+\lceil \frac{3k}{2}\rceil }\}.$$

$$\{x{v}_{i+\lceil \frac{k}{2}\rceil },x{v}_{i+\lceil \frac{k}{2}\rceil -1}^{\prime },y{v}_{i+\lceil \frac{k}{2}\rceil }^{\prime },y{v}_{i+\lceil \frac{k}{2}\rceil -1}\}$$

$$\{y{v}_{i+\lceil \frac{k}{2}\rceil },y{v}_{i+\lceil \frac{k}{2}\rceil -1}^{\prime },x{v}_{i+\lceil \frac{k}{2}\rceil }^{\prime },x{v}_{i+\lceil \frac{k}{2}\rceil -1}\}.$$

$$\{x{v}_1^{\prime },x{v}_2,x{v}_3^{\prime },x{v}_3,\dots ,x{v}_{2k-1}^{\prime },x{v}_{2k},y{v}_1,y{v}_2^{\prime },y{v}_3,y{v}_3^{\prime },\dots ,y{v}_{2k-1},y{v}_{2k}^{\prime }\}.$$

$$k+1\le \theta (4,K_{4k+1})\le \theta (4,K_{4k+2})\le k+1.$$

Full text

Keywords: Thickness, planar decomposition, girth, complete graph.

2010 Mathematics Subject Classification: 05C10.

1 Introduction

A finite graph $G$ is planar if it can be embedded in the plane without any two of its edges crossing. A planar graph of order $n$ and girth $g$ has size at most $\frac{g}{g-2}(n-2)$ (see [6]), and an acyclic graph of order $n$ has size at most $n-1$, in this case, we define its girth as $\infty$. The thickness $\theta(G)$ of a graph $G$ is the minimum number of planar subgraphs whose union is $G$; i.e. the minimum number of planar subgraphs into which the edges of $G$ can be partitioned.

The thickness was introduced by Tutte [11] in 1963. Since then, exact results have been obtained when $G$ is a complete graph [1, 3, 4], a complete multipartite graph [5, 12, 13] or a hypercube [9]. Also, some generalizations of the thickness for the complete graph $K_{n}$ have been studied such that the outerthickness $\theta_{o}$, defined similarly but with outerplanar instead of planar [8], and the $S$-thickness $\theta_{S}$, considering the thickness on a surfaces $S$ instead of the plane [2]. See also the survey [10].

We define the $g$-girth-thickness $\theta(g,G)$ of a graph $G$ as the minimum number of planar subgraphs of girth at least $g$ whose union is $G$. Note that the $3$-girth-thickness $\theta(3,G)$ is the usual thickness and the $\infty$-girth-thickness $\theta(\infty,G)$ is the arboricity number, i.e. the minimum number of acyclic subgraphs into which $E(G)$ can be partitioned. In this paper, we obtain the $4$-girth-thickness of an arbitrary complete graph of order $n\not=10$.

2 The exact value of $\theta(4,K_{n})$ for $n\not=10$

Since the complete graph $K_{n}$ has size $\binom{n}{2}$ and a planar graph of order $n$ and girth at least $4$ has size at most $2(n-2)$ for $n\geq 3$ and $n-1$ for $n\in\{1,2\}$ then the $4$-girth-thickness of $K_{n}$ is at least

[TABLE]

for $n\geq 3$ and also $\left\lceil\frac{n+2}{4}\right\rceil$ for $n\in\{1,2\}$, we have the following theorem.

Theorem 2.1**.**

The $4$-girth-thickness $\theta(4,K_{n})$ of $K_{n}$ equals $\left\lceil\frac{n+2}{4}\right\rceil$ for $n\not=6,10$ and $\theta(4,K_{6})=3$.

Proof.

Figure 1 displays equality for $n\leq 5$.

To prove that $\theta(4,K_{6})=3>\left\lceil\frac{6+2}{4}\right\rceil=2$, suppose that $\theta(4,K_{6})=2$. This partition define an edge coloring of $K_{6}$ with two colors. By Ramsey’s Theorem, some part contains a triangle obtaining a contradiction for the girth $4$. Figure 2 shows a partition of $K_{6}$ into tree planar subgraphs of girth at least $4$.

For the remainder of this proof, we need to distinguish four cases, namely, when $n=4k-1$, $n=4k$, $n=4k+1$ and $n=4k+2$ for $k\geq 2$. Note that in each case, the lower bound of the $4$-girth thickness require at least $k+1$ elements. To prove our theorem, we exhibit a decomposition of $K_{4k}$ into $k+1$ planar graphs of girth at least $4$. The other three cases are based in this decomposition. The case of $n=4k-1$ follows because $K_{4k-1}$ is a subgraph of $K_{4k}$. For the case of $n=4k+2$, we add two vertices and some edges to the decomposition obtained in the case of $n=4k$. The last case follows because $K_{4k+1}$ is a subgraph of $K_{4k+2}$. In the proof, all sums are taken modulo $2k$.

Case $n=4k$. It is well-known that a complete graph of even order contains a cyclic factorization of Hamiltonian paths, see [7]. Let $G$ be a subgraph of $K_{4k}$ isomorphic to $K_{2k}$. Label its vertex set $V(G)$ as $\{v_{1},v_{2},\dots,v_{2k}\}$. Let $\mathcal{F}_{1}$ be the Hamiltonian path with edges

[TABLE]

Let $\mathcal{F}_{i}$ be the Hamiltonian path with edges

[TABLE]

where $i\in\{2,3,\dots,k\}$.

Such factorization of $G$ is the partition $\{E(\mathcal{F}_{1}),E(\mathcal{F}_{2}),\dots,E(\mathcal{F}_{k})\}$. We remark that the center of $\mathcal{F}_{i}$ has the edge $e=v_{i+\left\lceil\frac{k}{2}\right\rceil}v_{i+\left\lceil\frac{3k}{2}\right\rceil}$, see Figure 3.

Now, consider the complete subgraph $G^{\prime}$ of $K_{4k}$ such that $G^{\prime}=K_{4k}\setminus V(G)$. Label its vertex set $V(G^{\prime})$ as $\{v^{\prime}_{1},v^{\prime}_{2},\dots,v^{\prime}_{2k}\}$ and consider the factorization, similarly as before, $\{E(\mathcal{F}^{\prime}_{1}),E(\mathcal{F}^{\prime}_{2}),\dots,E(\mathcal{F}^{\prime}_{k})\}$ where $\mathcal{F}^{\prime}_{i}$ is the Hamiltonian path with edges

[TABLE]

where $i\in\{1,2,\dots,k\}$.

Next, we construct the planar subgraphs $G_{1}$, $G_{2}$,…,$G_{k-1}$ and $G_{k}$ of girth $4$, order $4k$ and size $8k-4$ (observe that $2(4k-2)=8k-4$), and also the matching $G_{k+1}$, as follows. Let $G_{i}$ be a spanning subgraph of $K_{4k}$ with edges $E(\mathcal{F}_{i})\cup E(\mathcal{F}^{\prime}_{i})$ and

[TABLE]

where $i\in\{1,2,\dots,k\}$; and let $G_{k+1}$ be a perfect matching with edges $v_{j}v^{\prime}_{j}$ for $j\in\{1,2,\dots,2k\}$. Figure 4 shows $G_{i}$ is a planar graph of girth at least $4$.

To verify that $K_{4k}=\underset{i=1}{\overset{k+1}{\bigcup}}G_{i}$: 1) If the edge $v_{i_{1}}v_{i_{2}}$ of $G$ belongs to the factor $\mathcal{F}_{i}$ then $v_{i_{1}}v_{i_{2}}$ belongs to $G_{i}$. If the edge is primed, belongs to $G^{\prime}_{i}$. 2) The edge $v_{i_{1}}v^{\prime}_{i_{2}}$ belongs to $G_{k+1}$ if and only if $i_{1}=i_{2}$, otherwise it belongs to the same graph $G_{i}$ as $v_{i_{1}}v_{i_{2}}$. Similarly in the case of $v^{\prime}_{i_{1}}v_{i_{2}}$ and the result follows. 2. 2.

Case $n=4k-1$. Since $K_{4k-1}\subset K_{4k}$, we have

[TABLE] 3. 3.

Case $n=4k+2$ (for $k\not=2$). Let $\{G_{1},\dots,G_{k+1}\}$ be the planar decomposition of $K_{4k}$ constructed in the Case 1. We will add the two new vertices $x$ and $y$ to every planar subgraph $G_{i}$, when $1\leq i\leq k+1$, and we will add $4$ edges to each $G_{i}$, when $1\leq i\leq k$, and $4k+1$ edges to $G_{k+1}$ such that the resulting new subgraphs of $K_{4k+2}$ will be planar. Note that $\binom{4k}{2}+4k+4k+1=\binom{4k+2}{2}$.

To begin with, we define the graph $H_{k+1}$ adding the vertices $x$ and $y$ to the planar subgraph $G_{k+1}$ and the $4k+1$ edges

[TABLE]

The graph $H_{k+1}$ has girth $4$, see Figure 5.

In the following, for $1\leq i\leq k$, by adding vertices $x$ and $y$ to $G_{i}$ and adding $4$ edges to $G_{i}$, we will get a new planar graph $H_{i}$ such that $\{H_{1},\dots,H_{k+1}\}$ is a planar decomposition of $K_{4k+2}$ such that the girth of every element is $4$. To achieve it, the given edges to the graph $H_{i}$ will be $v^{\prime}_{j}x,xv_{j-1},v_{j}y,yv^{\prime}_{j-1}$, for some odd $j\in\{1,3,\dots,2k-1\}$.

According to the parity of $k$, we have two cases:

• •

Suppose $k$ odd. For odd $i\in\{1,2,\dots,k\}$, we define the graph $H_{i}$ adding the vertices $x$ and $y$ to the planar subgraph $G_{i}$ and the $4$ edges

[TABLE]

when $\left\lceil\frac{k}{2}\right\rceil$ is even, otherwise

[TABLE]

Additionally, for even $i\in\{1,2,\dots,k\}$, we define the graph $H_{i}$ adding the vertices $x$ and $y$ to the planar subgraph $G_{i}$ and the $4$ edges

[TABLE]

when $\left\lceil\frac{k}{2}\right\rceil$ is even, otherwise

[TABLE]

Note that the graph $H_{i}$ has girth $4$ for all $i$, see Figure 6.

• •

Suppose $k$ even. Similarly that the previous case, for odd $i\in\{1,2,\dots,k\}$, we define the graph $H_{i}$ adding the vertices $x$ and $y$ to the planar subgraph $G_{i}$ and the $4$ edges

[TABLE]

when $\left\lceil\frac{k}{2}\right\rceil$ is even, otherwise

[TABLE]

On the other hand, for even $i\in\{1,2,\dots,k\}$, we define the graph $H_{i}$ adding the vertices $x$ and $y$ to the planar subgraph $G_{i}$ and the $4$ edges

[TABLE]

when $\left\lceil\frac{k}{2}\right\rceil$ is even, otherwise

[TABLE]

Note that the graph $H_{i}$ has girth $4$ for all $i$, see Figure 7.

In order to verify that each edge of the set

[TABLE]

is in exactly one subgraph $H_{i}$, for $i\in\{1,\dots,k\}$, we obtain the unicyclic graph $\mathcal{F}^{*}_{i}$ identifying $v_{j}$ and $v^{\prime}_{j}$ resulting in $v_{j}$; identifying $x$ and $y$ resulting in a vertex which is contracted with one of its neighbours. The resulting edge, in dashed, is showed in Figures 6 and 7. The set of those edges are a perfect matching of $K_{2k}$ proving that the added two paths of length 2 in $G_{i}$ have end vertices $v_{j}$ and $v^{\prime}_{j-1}$, and the other $v^{\prime}_{j}$ and $v_{j-1}$. The election of the label of the center vertex is such that one path is $v_{even}xv^{\prime}_{odd}$ and $v^{\prime}_{even}yv_{odd}$ and the result follows. 4. 4.

Case $n=4k+1$ (for $k\not=2$). Since $K_{4k+1}\subset K_{4k+2}$, we have

[TABLE]

For $k=2$, Figure 8 displays a decomposition of three planar graphs of girth at least $4$ proving that $\theta(4,K_{9})=\left\lceil\frac{9+2}{4}\right\rceil=3$.

By the four cases, the theorem follows. ∎

About the case of $K_{10}$, it follows $3\leq\theta(4,K_{10})\leq 4$. We conjecture that $\theta(4,K_{10})=4$.