Geodetic convexity and Kneser graphs

Let $n,k$ be positive integers. The Kneser graph $K(2n+k,n)$ is the graph $G=(V,E)$ such that $V=\{S\subseteq\{1,\ldots,2n+k\}:|S|=n\}$ and there is an edge $uv\in E$ whenever $u\cap v=\emptyset$. Kneser graphs have a rich combinatorial structure [9, 13], and there are many studies on this class involving colorings, independent sets, and products of graphs (see [1, 4, 11]). In addition, the $P_{3}$-hull number of a Kneser graph has been investigated in [10], where the authors determine the exact value of the $P_{3}$-hull number of $K(2n+k,n)$ for $k>1$, and provide lower and upper bounds for $k=1$. Similarly, the authors in [12] have recently studied the $q$-analogues of Kneser graphs under the same convexity and parameter. To the best of the authors’ knowledge, however, no studies on geodetic convexity parameters are known for Kneser graphs. This work investigates two of the most addressed parameters in graph convexity, the geodetic number and the geodetic hull number, in the context of Kneser graphs. In particular, we determine the exact value of these parameters for Kneser graphs of diameter two. This turns out to be a relevant question because, for a fixed $n\geq 2$, almost all graphs in the family ${\mathscr{K}}_{n}=\{K(2n+k,n):k\geq 1\}$ have diameter two (those for which $k\geq n-1$). We prove that the geodetic hull number of a Kneser graph of diameter two is two, except for $K(5,2)$, $K(6,2)$, and $K(8,2)$, which have geodetic hull number three.

The geodetic and geodetic hull numbers have been studied for several graph classes, e.g. [2, 3, 5, 14]. For general graphs, determining the geodetic or the geodetic hull number is NP-hard [6, 8]; in view of these negative results, analyzing the behavior of these parameters in classes of graphs with an interesting structure, such as the class of Kneser graphs, is a natural research direction. Another objective of this work is to contribute to the knowledge on Kneser graphs. For instance, we characterize endpoints of diametral paths in Kneser graphs, and use this characterization as a step towards the determination of the geodetic number.

The remainder of this section provides all the necessary background. Section 2 gives a characterization of vertices that are endpoints of a diametral path in $K(2n+k,n)$, in terms of their intersection size. This characterization is used as a tool for proving Theorem 4 on the geodetic number of $K(2n+k,n)$. Sections 3 and 4 present the main results on the geodetic number and the geodetic hull number of $K(2n+k,n)$, respectively.

Let $G=(V,E)$ be a finite graph. A path between $u,v\in V$ is a sequence of distinct vertices $u=x_{1},x_{2},\ldots,x_{p}=v$ such that $x_{i}x_{i+1}\in E$ for $1\leq i\leq p-1$, and its length is the number of edges therein. A shortest path between $u,v\in V$ is a path with minimum length. The distance between two vertices $u,v\in V$, denoted by $\mathit{dist}(u,v)$, is the length of a shortest path between them. The diameter of $G$ is defined as $\mathit{diam}(G)=\max_{u,v\in V}\mathit{dist}(u,v)$. A diametral path is a shortest path whose length is $\mathit{diam}(G)$. We denote the open neighborhood of a vertex $x$ by $N(x)$. For $S\subseteq V$, we define $N(S)=\cup_{x\in S}N(x)$.

An $uv$-geodesic is a shortest path between $u$ and $v$. The geodetic interval $I[u,v]$ is the set of all vertices belonging to some $uv$-geodesic. For a set $W\subseteq V(G)$, the geodetic interval $I[W]$ is defined as $I[W]=\cup_{u,v\in W}I[u,v]$. The set $W$ is geodetically convex (or $g$-convex) if $I[W]=W$, and a geodetic set of $G$ if $I[W]=V(G)$. The geodetic number $\mathit{gn}(G)$ of $G$ is the size of minimum geodetic set of $G$.

The family of $g$-convex sets of graph $G$ define the geodetic convexity associated with $G$. In general, a convexity associated with a graph $G$ consists of a collection $\mathcal{C}$ of subsets of $V(G)$, called convex sets, such that: (a) $\emptyset,V(G)\in\mathcal{C}$; (b) $\mathcal{C}$ is closed under intersections. Just like the geodetic convexity is defined over shortest paths, other graph convexities can be defined using different path systems, such as the monophonic convexity [7], associated with induced paths.

The geodetic hull of a set $W\subseteq V(G)$, denoted by $H[W]$, is the minimum $g$-convex set containing $W$. Also, $W$ is a geodetic hull set of $G$ if $H[W]=V(G)$. The geodetic hull number, denoted by $\mathit{ghn}\left(G\right)$, is the size of a minimum geodetic hull set of $G$. For an integer $k\geq 0$, we define $I^{k}[W]$ as follows: $I^{0}[W]=W$ and $I^{k}[W]=I[I^{k-1}[W]]$. It is not difficult to see that $H[W]=I^{k}[W]$ for some $k\geq 0$; in fact, $k$ can be taken as the minimum index for which $H[W]=I^{k}[W]$.

Let $n$ and $k$ be positive integers, and let $[n]=\{1,\ldots,n\}$ and $[2n+k]^{n}=\{S\subseteq[2n+k]:|S|=n\}$. The Kneser graph $K(2n+k,n)$ is the graph $G=(V,E)$ such that $V=[2n+k]^{n}$ and there is an edge between two vertices $u,v\in[2n+k]^{n}$ whenever $u\cap v=\emptyset$ (see [13]). Therefore, the Kneser graph $K(2n+k,n)$ contains $\binom{2n+k}{n}$ vertices and is a $\binom{n+k}{n}$-regular graph. The Kneser graph $K(5,2)$ is the well-known Petersen graph.

In [15] the authors show that $\mathit{diam}(K(2n+k,n))=\lceil(n-1)/k\rceil+1$. Additionally, for any $u,v\in V(K(2n+k,n))$ with $|u\cap v|=s$, they show that:

Observe that if $n\geq 2$ and $k\geq n-1$ then ${\mathit{d}iam}(K(2n+k,n))=2$. Thus, the graphs of diameter two form a infinite subfamily of ${\mathscr{K}}_{n}=\{K(2n+k,n):k\geq 1\}$.

The theorem below gives a necessary and sufficient condition for two vertices in the Kneser graph $K(2n+k,n)$ to be endpoints of a diametral path.

If $k\geq n-1$, we have $\mathit{diam}(K(2n+k,n))=2$ and $s\geq 1$, that is, $u$ and $v$ are not adjacent. Therefore, $\mathit{dist}(u,v)=2=\mathit{diam}(K(2n+k,n))$.

Now, consider $n-1=2\alpha k+\beta k+\gamma$ for $\alpha\in\mathbb{N}_{0}$, $\beta\in\{0,1\}$, and $0\leq\gamma<k$. For $s=\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1+\Delta$ with $0\leq\Delta\leq H(n,k)$, we analyze two cases:

In this case, $n=2\alpha k+1$, $s=(\alpha-1)k+1+\Delta$, and $\mathit{diam}(K(2n+k,n))=\lceil(n-1)/k\rceil+1=2\alpha+1$.

If $k=1$, we have $H(n,k)=\Delta=0$. By using Eq. (1), it follows that

Let us assume now $k\geq 2$. Since $n\bmod k=1$, we have $H(n,k)=k-1$, and, thus, $0\leq\Delta\leq k-1$. Then, by Eq. (1), it follows that

In this case, we have $s=\alpha k+1+\Delta$.

First, let us assume $k=1$. Then, $\gamma=\Delta=H(n,k)=0$, which implies $\beta=1$ and $\mathit{diam}(K(2n+k,n))=\lceil(2\alpha+1)/1\rceil+1=2\alpha+2$. By using Eq. (1) and the fact that $n=2\alpha k+\beta k+\gamma+1$, we have that:

Assume now $k\geq 2$. Note that $n\bmod k=(\gamma+1)\bmod k$. Substituting in Eq. (3):

From the above equation, $\gamma-H(n,k)=1-k$ if $\gamma=0$, and $\gamma-H(n,k)=1$ if $1\leq\gamma\leq k-1$. Then, we have $1\leq\gamma-\Delta\leq k-1$ when $\gamma\neq 0$. We analyze the possible cases in Eq. (1):

• –

  If $\beta=1$ and $\gamma\neq 0$, $\mathit{dist}(u,v)=\min\{2\alpha+4,2\alpha+3\}=2\alpha+3$ and $\mathit{diam}(K(2n+k,n))=\lceil(2\alpha k+\beta k+\gamma)/k\rceil+1=2\alpha+3$;

• –

  If $\beta=0$ and $\gamma\neq 0$, $\mathit{dist}(u,v)=\min\{2\alpha+2,2\alpha+3\}=2\alpha+2=\mathit{diam}(K(2n+k,n))$;

• –

  If $\beta=1$ and $\gamma=0$, $\mathit{dist}(u,v)=\min\{2\alpha+2,2\alpha+3\}=2\alpha+2$ and $\mathit{diam}(K(2n+k,n))=\lceil(2\alpha k+\beta k)/k\rceil+1=2\alpha+2$.

This concludes Case 2 and the first part of the proof.

Conversely, suppose $\mathit{dist}(u,v)=\mathit{diam}(K(2n+k,n))$, and assume by contradiction that there is an integer $\varepsilon\geq 1$ such that

Assuming $n-1=2\alpha k+\beta k+\gamma$ for $\alpha\in\mathbb{N}_{0}$, $\beta\in\{0,1\}$ again, we analyze the two possible cases.

We divide the proof of Case 1 in two subcases, analyzing possible values of $\beta$ and $\gamma$.

This case implies $n=2\alpha k+1$ and $s=(\alpha-1)k+1-\varepsilon$. Thus, substituting in Eq. (1),

However, $\mathit{diam}(K(2n+k,n))=\lceil(n-1)/k\rceil+1=\lceil 2\alpha k/k\rceil+1=2\alpha+1$.

This case implies $n=2\alpha k+\beta k+\gamma+1$ and $s=\alpha k+1-\varepsilon$. Substituting in Eq. (1),

However, for $\beta=1$ or $\gamma\neq 0$, $2\alpha+2\leq\mathit{diam}(K(2n+k,n))\leq 2\alpha+3$. Therefore, both Cases 1.1 and 1.2 lead to contradictions, and this concludes Case 1.

Again, we analyze the possible values of $\beta$ and $\gamma$.

This case implies $n=2\alpha k+1$ and $s=(\alpha-1)k+1+H(n,k)+\varepsilon$.

Assume now $k\geq 2$. Then, recall that $H(n,k)=k-1$, since $n\bmod k=1$. Substituting in Eq. (1), we have that

In this case, $n=2\alpha k+\beta k+\gamma+1$ and $s=\alpha k+1+H(n,k)+\varepsilon$.

If $k=1$, we have $H(n,k)=\gamma=0$, $\beta=1$, and $\mathit{diam}(K(2n+k,n))=2\alpha+2$. Thus, by using Eq. (1), we have that

Finally, assume $k\geq 2$. Again, by Eq. (1):

Recall from Case 2 in the first part of the proof that $\gamma-H(n,k)=1$, for all $1\leq\gamma\leq k-1$. By confronting Eq. (5) with the possibilities for $\beta$ and $\gamma$, we have:

• –

  If $\beta=1$ and $\gamma\neq 0$,

  	$\displaystyle\mathit{dist}(u,v)$	$\displaystyle=\min\{2\lceil\alpha+1+(1-\varepsilon)/k\rceil,2\lceil\alpha+(1+H(n,k)+\varepsilon)/k\rceil+1\}$
  		$\displaystyle=2\lceil\alpha+1+(1-\varepsilon)/k\rceil\leq 2\alpha+2<\mathit{diam}(K(2n+k,n))=2\alpha+3.$

• –

  if $\beta=0$ and $\gamma\neq 0$,

  	$\displaystyle\mathit{dist}(u,v)$	$\displaystyle=\min\{2\lceil\alpha+(1-\varepsilon)/k\rceil,2\lceil\alpha+(1+H(n,k)+\varepsilon)/k\rceil+1\}$
  		$\displaystyle=2\lceil\alpha+(1-\varepsilon)/k\rceil\leq 2\alpha<\mathit{diam}(K(2n+k,n))=2\alpha+2.$

• –

  If $\beta=1$ and $\gamma=0$, we have $-(H(n,k)+\varepsilon)\leq-1$. Thus,

  	$\displaystyle\mathit{dist}(u,v)$	$\displaystyle=\min\{2\lceil\alpha+1-(H(n,k)+\varepsilon)/k\rceil,2\lceil\alpha+(1+H(n,k)+\varepsilon)/k\rceil+1\}$
  		$\displaystyle=2\lceil\alpha+1-(H(n,k)+\varepsilon)/k\rceil\leq 2\alpha<\mathit{diam}(K(2n+k,n))=2\alpha+2.$

Therefore, both Cases 2.1 and 2.2 lead to contradictions. This concludes Case 2 and the proof of the theorem.  $\Box$

An interesting fact derived from Theorem 1 is:

The next theorem gives a sufficient condition for a set $S\subseteq V(K(2n+k,n))$ to be a geodetic set. Say that two vertices $u,v$ in a graph $G$ are diametrically opposed if $d(u,v)={\mathit{d}iam}(G)$.

In order to prove that $D\cup\{r\}$ is a geodetic set, we show that each $x\in{\mathscr{L}}(i)$, with $0\leq i<\lceil(n-1)/k\rceil+1$, has at least one neighbor $z\in{\mathscr{L}}(i+1)$. This is trivial for $i=0$. By Theorem 1, either $|r\cap x|<\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1$ or $|r\cap x|>\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1+H(n,k)$.

Let $y\in N(x)\cap{\mathscr{L}}(i-1)$. We analyze two cases.

In this case, by Corollary 2, $|r\cap y|<\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1$.

Let $k^{\prime}=\min\{k,\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1-|r\cap y|\}$, and let $z_{1}$ be a set formed by $k^{\prime}$ elements in $r\setminus(x\cup y)$. Observe that $z_{1}$ exists because ${\mathit{d}ist}(x,r)<{\mathit{d}iam}(K(2n+k,n))$, which in turn implies the existence of a set that has at least $k^{\prime}$ more elements in common with $r$ than $y$. Additionally, let $z_{2}$ be a set formed by $n-k^{\prime}-|r\cap y|$ elements in $y\setminus r$. It follows that $z=(r\cap y)\cup z_{1}\cup z_{2}$ is a neighbor of $x$ such that $L(z)=i+1$, and this concludes Case 1.

In this case, by Corollary 2, $|r\cap y|>\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1+H(n,k)$.

Let $k^{\prime\prime}=\min\{k,|r\cap y|-(\left(\left\lceil\frac{n-1}{2k}\right\rceil-1\right)k+1+H(n,k))\}$, and let $z_{3}$ be a set formed by $k^{\prime\prime}$ elements in $r\cap y$. In addition, let $z_{4}$ be formed by $n-k^{\prime\prime}$ elements in $[2n+k]\setminus(r\cup x)$. Again, $z_{3}$ and $z_{4}$ exist because there is a set with at least $k^{\prime\prime}$ elements that intersect $r$ less than $y$, since ${\mathit{d}ist}(x,r)<{\mathit{d}iam}(K(2n+k,n))$. To conclude Case 2, note that $z=(y\setminus z_{3})\cup z_{4}$ is a neighbor of $x$ with $L(z)=i+1$.