Non-crossing geometric spanning trees with bounded degree and monochromatic leaves on bicolored point sets

TL;DR

This paper proves the existence of non-crossing geometric spanning trees on bicolored point sets with degree bounds on red points and all blue points as leaves, under certain size constraints.

Contribution

It introduces a method to construct non-crossing spanning trees with degree bounds and monochromatic leaves on bicolored point sets, extending previous geometric graph results.

Findings

Existence of such trees under specified conditions

Construction method for non-crossing spanning trees with degree constraints

Application to geometric graph theory problems

Abstract

Let $R$ and $B$ be a set of red points and a set of blue points in the plane, respectively, such that $R\cup B$ is in general position, and let $f:R \to \{2,3,4, \ldots \}$ be a function. We show that if $2\le |B|\le \sum_{x\in R}(f(x)-2) + 2$, then there exists a non-crossing geometric spanning tree $T$ on $R\cup B$ such that $2\le \operatorname{deg}_T(x)\le f(x)$ for every $x\in R$ and the set of leaves of $T$ is $B$, where every edge of $T$ is a straight-line segment.

Equations

$$\deg_{Q}(x)=f(x)\quad\mbox{for all $x\in R$,~{}~{}and~{}~{} }\deg_{Q}(y)=1\quad\mbox{for all $y\in B$.}$$

$$\sum _{x\in R}f(x)+\sum _{y\in B}1=\sum _{x\in R}(f(x)-2)+2|R|+|B|$$

$$=$$

Full text

1 Introduction and related work

Let $R$ be a set of red points and $B$ be a set of blue points in the plane. We always assume that $R$ and $B$ are disjoint and $R\cup B$ is in general position (i.e, no three points of $R\cup B$ are collinear). Several works [1, 8] have considered problems on non-crossing geometric spanning trees and geometric graphs (where edges are straight-line segments) on $R\cup B$. See also the survey [7].

1.1 Contributions of this work

For a tree $T$ and a vertex $v$ of $T$, let us denote by $\deg_{T}(v)$ the degree of $v$ in $T$. A vertex of $T$ with degree one is called a leaf of $T$, and the set of leaves of $T$ will be denoted by $Leaf(T)$. Further, let us denote by $|T|$ the order of a tree $T$ (i.e., its number of vertices) and by $|X|$ the cardinality of a set $X$.

In this paper, given $R$ and $B$ in the plane as above, we aim for non-crossing geometric spanning trees $T$ on $R\cup B$ such that $Leaf(T)=B$. We prove the following theorem:

Theorem 1

Assume that $R$ and $B$ are given in the plane and a function $f:R\to\{2,3,4,\ldots\}$ is given. If $2\leq|B|\leq\sum_{x\in R}(f(x)-2)+2$, then there exists a non-crossing geometric spanning tree $T$ on $R\cup B$ such that $Leaf(T)=B$ and $2\leq\deg_{T}(x)\leq f(x)$ for every $x\in R$. Moreover, if $|B|=\sum_{x\in R}(f(x)-2)+2$, then $T$ satisfies that $\deg_{T}(x)=f(x)$ for every $x\in R$ (see (2) and (3) of Figure 1).

By setting $f(x)=k$ for every $x\in R$ in the above theorem, we obtain the following corollary:

Corollary 2

Let $k\geq 2$ be an integer. Assume that $R$ and $B$ are given in the plane. If $2\leq|B|\leq(k-2)|R|+2$, then there exists a non-crossing geometric spanning tree $T$ on $R\cup B$ such that $Leaf(T)=B$ and the maximum degree of $T$ is at most $k$. Moreover, if $|B|=(k-2)|R|+2$, then $T$ satisfies that $\deg_{T}(x)=k$ for every $x\in R$.

Observation 3

For $k=2$, the above corollary says that if $|B|=2$ and $|R|\geq 1$, then there exists a non-crossing geometric path which passes through all the points of $R\cup B$ and whose endvertices are the two blue points.

1.2 Related work

Ikebe et al. [4] proved that, given one red point $R=\{r\}$ and a set $B$ of blue points in the plane, any rooted tree $T$ with root $w$ of order $|B|+1$ can be straight-line embedded on $\{r\}\cup B$ in such a way that $w$ is mapped to $r$ and no crossings arise. Kaneko and Kano [6] proved that, given two red points $R=\{r_{1},r_{2}\}$ and a set $B$ of blue points in the plane, together with two rooted trees $T_{1}$ with root $w_{1}$ and $T_{2}$ with root $w_{2}$, if $|T_{1}|+|T_{2}|=|B|+2$, then $T_{1}\cup T_{2}$ can be straight-line embedded on $\{r_{1},r_{2}\}\cup B$, without crossings, in such a way that $w_{1}$ and $w_{2}$ are mapped to $r_{1}$ and $r_{2}$, respectively.

Kaneko [5] considered sets $R$ and $B$ in the plane with $|R|=|B|$ and proved that, then, there exists a non-crossing geometric spanning tree $T$ on $R\cup B$ such that every edge of $T$ joins a red point to a blue point and the maximum degree of $T$ is at most 3 (see (1) of Figure 1). Finally, Biniaz et al. [2] considered sets $R$ and $B$ in the plane with $|B|\leq|R|$ and proved that, then, there exists a non-crossing geometric spanning tree $T$ on $R\cup B$ such that every edge of $T$ joins a red point to a blue point and the maximum degree of $T$ is at most $\max\{3,\lceil(|R|-1)/|B|\rceil+1\}$.

2 Proof of Theorem 1

In this section we prove Theorem 1.We first state the following proposition, which is a special case of Theorem 1.

Proposition 4

Assume that $R$ and $B$ are given in the plane and a function $f:R\to\{2,3,4,\ldots\}$ is given. If $|B|=\sum_{x\in R}(f(x)-2)+2$, then there exists a non-crossing geometric spanning tree $T$ on $R\cup B$ such that $Leaf(T)=B$ and $\deg_{T}(x)=f(x)$ for every $x\in R$ (see (2) and (3) of Figure 1).

In order to prove Proposition 4 we will use the following lemma:

Lemma 5** (Theorem 3.6 of [3], and Exercises 2.1.12 of [9])**

Let $n\geq 2$ be an integer, and let $d_{1},d_{2},\ldots,d_{n}$ be positive integers. If $d_{1}+d_{2}+\cdots+d_{n}=2n-2$, then there exists a tree $T$ with vertex set $\{v_{1},v_{2},\ldots,v_{n}\}$ that satisfies $\deg_{T}(v_{i})=d_{i}$ for every $1\leq i\leq n$.

Proof of Proposition 4. We fist show that there exists a geometric spanning tree $Q$ on $R\cup B$ that might have crossings but satisfies

[TABLE]

It follows from the condition of Proposition 4 that

[TABLE]

Hence by Lemma 5, there exists a geometric spanning tree $Q$ that satisfies the condition (1) but might have some crossings. Among all geometric spanning trees $Q$ satisfying (1), choose a geometric spanning tree $T$ on $R\cup B$ such that the sum $\sum_{xy\in E(T)}|xy|$ is minimum, where $|xy|$ denotes the length of the straight-line edge $x$ to $y$. We shall show that $T$ has no crossings.

The following three possible types of crossings could arise. First, that two edges $st$ and $uv$ of $T$ intersect, where $s,t,u,v$ are red points (see (1) of Figure 2). Since $T-st-uv$ consists of three components, by symmetry, we may assume that $u$ and $t$ are contained in the same component of $T-st-uv$, that is, $u$ and $t$ are connected by a path in $T-st-uv$. Then $T-st-uv+su+vt$ is another geometric spanning tree on $R\cup B$ satisfying the degree condition (1) and its total sum of edge-lengths is smaller than that of $T$. This contradicts the choice of $T$. Hence this case does not occur.

Second, that two edges $st$ and $ux$ of $T$ intersect, where $s,t,u$ are red points and $x$ is a blue point (see (2) of Figure 2). Since $T-st-ux$ consists of three components and $\{x\}$ forms one component, $u$ and $t$ are connected by a path in $T-st-ux$ or $u$ and $s$ are connected by a path in $T-st-ux$. By symmetry, we may assume that $u$ and $t$ are connected by a path in $T-st-ux$. Then $T-st-ux+su+tx$ is another geometric spanning tree on $R\cup B$ satisfying the degree condition (1) and its total sum of edge lengths is smaller than that of $T$. This is a contradiction.

Third, that two edges $sy$ and $ux$ of $T$ intersect, where $s,u$ are red points and $x,y$ are blue points (see (3) of Figure 2). Since $T-sy-ux$ consists of three components and $\{x\}$ and $\{y\}$ form two components, $s$ and $u$ are connected by a path in $T-sy-ux$. Then $T-sy-ux+sx+uy$ is another geometric spanning tree on $R\cup B$ satisfying the degree condition (1) and its total sum of edge-lengths is smaller than that of $T$. This is a contradiction.

Note that blue points being leaves implies that these three were the only possible cases for crossings and, therefore, $T$ has no crossings. Consequently, $T$ is the desired non-crossing geometric spanning tree on $R\cup B$, and Proposition 4 is proved.   $\Box$

We next prove Theorem 1 by making use of Proposition 4.

Proof of Theorem 1. We may assume that $2\leq|B|<\sum_{x\in R}(f(x)-2)+2$ since if $|B|=\sum_{x\in R}(f(x)-2)+2$, then the theorem holds by Proposition 4. It is easy to see that there exists a mapping $f^{\prime}:R\to\{2,3,4,\ldots\}$ that satisfies $f^{\prime}(x)\leq f(x)$ for all $x\in R$ and $|B|=\sum_{x\in R}(f^{\prime}(x)-2)+2$. By Proposition 4, there exists a non-crossing geometric spanning tree $T$ such that $Leaf(T)=B$ and $\deg_{T}(x)=f^{\prime}(x)$ for all $x\in R$. Hence, $T$ is the desired geometric spanning tree on $R\cup B$. Consequently Theorem 1 is proved.    $\Box$