The Steiner distance problem for large vertex subsets in the hypercube

TL;DR

This paper investigates the asymptotic behavior of the Steiner k-diameter in large hypercubes, providing new lower bounds through probabilistic methods to understand the complexity of large vertex subset distances.

Contribution

The paper introduces a new lower bound for the Steiner k-diameter in hypercubes using probabilistic techniques, advancing understanding of large subset distances.

Findings

Established asymptotic behavior of Steiner k-diameter for large k

Developed a probabilistic method for lower bounds

Enhanced understanding of hypercube vertex subset distances

Abstract

We find the asymptotic behavior of the Steiner k-diameter of the $n$-cube if $k$ is large. Our main contribution is the lower bound, which utilizes the probabilistic method.

Equations

$$d(S)\le |S|+\frac{2^n}{n}(1+o(1)).$$

$$d(S)\ge |S|+\frac{|S{|}^2}{n{2}^n}-\frac{(n+1)}{2}.$$

$$d(S\cup \bar{S})\le |E(T)\cup E(\bar{T})|+n.$$

$$2|S|-1$$

$$=|E(T)|+|E(\bar{T})|-|E(T)\cap E(\bar{T})|+n$$

$$=2d(S)-|E(T)\cap E(\bar{T})|+n,$$

$$2d(S)-|E(T)\cap E(\bar{T})|\ge 2|S|-(n+1).$$

$$2d(S)-|E({\lambda }_1(T))\cap E({\lambda }_2(\bar{T}))|\ge 2|S|-(n+1),$$

$$\alpha :$$

$${\beta }_{i,j}:$$

$$\underset{{\lambda }_1,{\lambda }_2}{\max }\{|E({\lambda }_1(T)\cap {\lambda }_2(\bar{T}))|\}\ge \mathbb{E}(X).$$

$$\mathbb{E}(X)$$

$$=\sum _{f\in E(Q_n)}\frac{|E(T)|}{n{2}^{n-1}}\cdot \frac{|E(\bar{T})|}{n{2}^{n-1}}$$

$$=\frac{|E(T){|}^2}{n{2}^{n-1}}$$

$$=\frac{d(S{)}^2}{n{2}^{n-1}},$$

$$\underset{{\lambda }_1,{\lambda }_2}{\max }\{|E({\lambda }_1(T))\cap E({\lambda }_2(\bar{T}))\}\ge \frac{d(S{)}^2}{n{2}^{n-1}}.$$

$$2d(S)-\frac{d(S{)}^2}{n{2}^{n-1}}\ge 2|S|-(n+1).$$

$$2(|S|+x)-\frac{(|S|+x{)}^2}{n{2}^{n-1}}$$

$$2|S|+2x-\frac{|S{|}^2+2|S|x+x^2}{n{2}^{n-1}}$$

$$2x-\frac{2|S|x}{n{2}^{n-1}}+2|S|-\frac{|S{|}^2+x^2}{n{2}^{n-1}}$$

$$2x(1-\frac{|S|}{n{2}^{n-1}})$$

$$x$$

$$k+\frac{k^2}{n{2}^n}-\frac{n+1}{2}\le d(S)\le sdia{m}_k(Q_n)\le k+\frac{2^n}{n}(1+o(1)).$$

Full text

The last two authors were supported in part by the National Science Foundation contract DMS-1600811.

1. Introduction

For a connected graph $G$ of order at least 2 and $S\subseteq V(G)$, the Steiner distance $d(S)$ among the vertices of $S$ is the minimum size among all connected subgraphs whose vertex sets contain $S$. Necessarily, such a minimum subgraph must be a tree and such a tree is called a Steiner tree. The Steiner distance was introduced by G. Chartrand, O.R. Oellermann, S. Tian and H.B. Zou [2], and it has turned into a well-studied parameter of graphs. Tao Jiang, Zevi Miller, and Dan Pritikin [6] studied how large the Steiner distance of $k$ vertices can be in the $n$-dimensional hypercube $Q_{n}$ as $n\rightarrow\infty$, while Zevi Miller and Dan Pritikin [5] gave near tight bounds for the Steiner distance of a layer, i.e. vertices with the same number of 1’s, in the $n$-dimensional hypercube $Q_{n}$ as $n\rightarrow\infty$. For a given $2\leq k\leq n$, the Steiner $k$-diameter of the $n$-cube, $sdiam_{k}(Q_{n})$, is the maximum Steiner distance among all $k$ subsets of $V(Q_{n})$.

In this note we give natural upper bounds for the Steiner distance of a large vertex set in the hypercube. It turns out that even the second order term in this estimate is close to tight. With these bounds, we determine $sdiam_{k}(Q_{n})$ asymptotically for large $k$.

2. Upper Bound

For the upper bound, we utilize connected dominating sets of $Q_{n}$. A set $S\subset V(Q_{n})$ is a dominating set of $Q_{n}$ if every vertex of $Q_{n}$ is either an element of $S$ or has a neighbor in $S$. The minimum size of all dominating sets is called the domination number of $Q_{n}$ and is denoted $\gamma(Q_{n})$. The connected domination number, denoted by $\gamma_{c}(Q_{n})$, is minimum size of all connected dominating sets.

In 1988, Kabatyanskii and Panchenko [4] showed $\lim_{n\rightarrow\infty}\frac{\gamma(Q_{n})}{2^{n}/n}=1$. In an upcoming paper, Griggs [3] utilizes this result to show that $\lim_{n\rightarrow\infty}\frac{\gamma_{c}(Q_{n})}{2^{n}/n}=1$. We use this last result to develop an upper bound for the Steiner diameter of subsets of $V(Q_{n})$.

Lemma 1**.**

Suppose that $S\subset V(Q_{n})$. Then,

[TABLE]

Proof.

Begin with a minimum connected dominating set of $Q_{n}$. Simply connect each of the elements of $S$ to this connected dominating set. The resulting subgraph spans $S$ and contains at most $|S|+\gamma_{c}(Q_{n})-1$ edges. Using [3], we then have that $d(S)\leq|S|+\frac{2^{n}}{n}(1+o(1)).$ ∎

3. Lower Bound

To bound the Steiner distance of large vertex subsets of $Q_{n}$ from below, we partition the vertices of the hypercube into two sets. Identifying each vertex of $Q_{n}$ into a binary string of length $n$, we let vertices with an even number of 1’s make up the set of even vertices and denote this set by $\mathcal{E}_{n}$. Similarly, we let the vertices with an odd number of 1’s make up the set of odd vertices and denote this set by $\mathcal{O}_{n}$. We refer to changing the value of the $i$’th entry of a binary string $v=v_{0}\cdots v_{i}\cdots v_{n}$ as “flipping” the $i$’th entry of $v$. Given an entry $v_{i}$, we let $\bar{v}_{i}=1-v_{i}$. That is, $\bar{v_{i}}$ is the flipped value of $v_{i}$. For the proof of Theorem 2, we use probabilistic methods similar to those found in [1].

Theorem 2**.**

Suppose that $S\subset\mathcal{E}_{n}$, i.e., each vertex in $S$ contains an even number of 1’s. Then,

[TABLE]

Proof.

Suppose that $S$ is a subset of the set of even vertices of $Q_{n}$. Let $\bar{S}$ be some subset of the odd vertices which is the image of $S$ under some automorphism of $Q_{n}$. That is, $S\subset\mathcal{E}_{n}$, $\bar{S}\subset\mathcal{O}_{n}$, and $\bar{S}=\gamma(S)$ for some $\gamma\in\operatorname{Aut}(Q_{n})$. Such a subset exists. Indeed, consider the set of all vertices in $S$ with the first entry flipped. Since $S$ and $\bar{S}$ are isomorphic, we have that $d(S)=d(\bar{S})$. Suppose that $T=(V(T),E(T))$ and $\bar{T}=(V(\bar{T}),E(\bar{T}))$ are Steiner trees of $S$ and $\bar{S}$, respectively. Naively, we have that $d(S\cup\bar{S})\geq 2|S|-1$. Furthermore, connecting $T$ and $\bar{T}$ with at most $n$ edges yields a subgraph of $Q_{n}$ connecting $S\cup\bar{S}$. Hence,

[TABLE]

Putting these inqualities together and applying the principle of inclusion and exclusion, we have

[TABLE]

which implies that

[TABLE]

Note that if $\lambda_{1}$ and $\lambda_{2}$ are automorphisms of $Q_{n}$ which preserve the parity of their inputs, then the inequality above extends to

[TABLE]

where $\lambda_{i}(T)$ and $\lambda_{i}(\bar{T})$ are the images of $T$ and $\bar{T}$ under $\lambda_{i}$ for $1\leq i\leq 2$.

Let $\Gamma=\langle\alpha,\beta_{i,j}:1\leq 0<j\leq n-1\rangle$ be the subgroup of the group of automorphisms of $Q_{n}$ generated by the automorphisms

[TABLE]

In words, $\alpha$ shifts each entry of its input to the left by 1 (modulo $n$), while $\beta_{i,j}$ flips only the values of the $i$’th and $j$’th entries of its input. Note that each element of $\Gamma$ preserves the parity of its input. We now verify the following claim:

Claim: For any two edges $e_{1},e_{2}\in E(Q_{n})$, there exists a unique element of $\lambda\in\Gamma$ such that $\lambda(e_{1})=e_{2}$.

Suppose that $e_{1}=ab$ and $e_{2}=uv$ where $a$ and $u$ are even vertices while $b$ and $v$ are odd vertices. Without loss of generality, we may assume that $a=\mathbf{0}$, the vertex of all zeros. This implies that the string $b$ contains a single 1. We shall first prove existence of an automorphism $\lambda\in\Gamma$ mapping $e_{1}$ to $e_{2}$.

Since $u\in\mathcal{E}_{n}$, using a composition of automorphisms of the form $b_{i,j}$ we may map $uv$ to $\mathbf{0}\hat{v}$, where $\hat{v}$ has a single 1. Then, using some power of the automorphism $\alpha$, we may then map the edge $\mathbf{0}\hat{v}$ to the edge $\mathbf{0}b=e_{1}$. Let $\lambda$ be these composition of autmorphisms in $\Gamma$.

To show that this automorphism is unique, we show that $|\Gamma|=n2^{n-1}$. Since $\alpha\circ\beta_{ij}=\beta_{i-1,j-1}\circ\alpha$ (where the indexes are taken modulo $n$), any $\lambda\in\Gamma$ can be described as first applying an appropriate power of $\alpha$ and then flipping an even number of digits. As we have $n$ choices for the power of $\alpha$ and $2^{n-1}$ choices for the subset of digits we flip, $|\Gamma|=n2^{n-1}$. Since $Q_{n}$ has $n2^{n-1}$ edges, and any $\lambda\in\Gamma$ maps the edge $\mathbf{0}b$ to one of these in such a way that [math] gets mapped to the endvertex in $\mathcal{E}_{n}$, and all edges of $Q_{n}$ will be the image of $\mathbf{0}b$ under some $\lambda\in\Gamma$, the claim follows.

We now consider the experiment of selecting elements $\lambda_{1},\lambda_{2}\in\Gamma$ independently with uniform probability, and applying them to $T$ and $\bar{T}$, respectively. Consider the random variable $X=|E(\lambda_{1}(T))\cap E(\lambda_{2}(\bar{T}))|$. For the expected value of $X$, $\mathbb{E}(X)$, we have that

[TABLE]

Using our claim, we observe that

[TABLE]

which implies

[TABLE]

Using $\lambda_{1}$ and $\lambda_{2}$ which achieve this maximum and applying inequality (1), we see that

[TABLE]

We are going to bootstrap the calculation above. Assume without loss of generality that $d(S)=|S|+x$ for some $x>0$. So,

[TABLE]

and the result is proven. ∎

With these results in hand, we can determine the asymptotic growth of $sdiam_{k}(Q_{n})$ as for large $k$. In particular, we can determine the first and second order terms if $k=\Omega(2^{n})$, while we can determine the first order term if $2^{n}/n=o(k)$.

Corollary 3**.**

If $k=k(n)$, then

1. (1)

If $k=\Omega(2^{n})$, then $sdiam_{k}(Q_{n})=k+\Theta(2^{n}/n)$, and 2. (2)

If ${2^{n}/n}=o(k)$, then $\lim\limits_{n\rightarrow\infty}\frac{sdiam_{k}(Q_{n})}{k}=1$.

Proof.

If $k\leq 2^{n-1}$, let $S\subset V(Q_{n})$ be a subset of the even vertices of size $k$. If $k>2^{n-1}$, let $S$ contain all even vertices and choose the remaining odd vertices randomly. Applying the bounds determined in Lemma 1 and Theorem 2, we see that

[TABLE]

If $k=\Omega(2^{n})$, $sdiam_{k}(Q_{n})$ is bounded above and below by $k+\Theta(2^{n}/n).$ Alternatively, if $2^{n}/n=o(k)$, we have $sdiam(Q_{n})=k(1+o(1))$, giving $\lim_{n\rightarrow\infty}\frac{sdiam_{k}(Q_{n})}{k}=1$. ∎