DP-colorings of uniform hypergraphs and splittings of Boolean hypercube into faces

TL;DR

This paper explores the relationship between DP-colorings of uniform hypergraphs and hypercube face coverings, establishing tight bounds for all odd uniformities and connecting combinatorial hypergraph properties with geometric hypercube structures.

Contribution

It extends previous bounds on non-2-DP-colorable hypergraphs, proving the bounds are tight for all odd uniformities, and links hypergraph colorings to hypercube face coverings.

Findings

Bound on edges is achieved for all odd k ≥ 3.

Established a connection between hypergraph DP-colorings and hypercube face coverings.

Proved tight bounds for non-2-DP-colorable hypergraphs for all odd k.

Abstract

We develop a connection between DP-colorings of $k$-uniform hypergraphs of order $n$ and coverings of $n$-dimensional Boolean hypercube by pairs of antipodal $(n-k)$-dimensional faces. Bernshteyn and Kostochka established that the lower bound on edges in a non-2-DP-colorable $k$-uniform hypergraph is equal to $2^{k-1}$ for odd $k$ and $2^{k-1}+1$ for even $k$. They proved that these bounds are tight for $k=3,4$. In this paper, we prove that the bound is achieved for all odd $k\geq 3$.

Equations

$$m=n_2-k_2-(n_3-t)+(n_3-t)n_1+t(n_1-k_1)=n_2+(n_1-1)n_3-(k_2+(k_1-1)t).$$

Full text

Keywords: hypergraph coloring, DP-coloring, covering of hypercube by faces, splitting of hypercube.

MSC 05C15, 05C65, 05C35, 51E05

1 Introduction

Let $Q_{2}^{n}$ be an $n$-dimensional Boolean hypercube. We consider coverings and splittings of $Q_{2}^{n}$ into faces. A $k$-coverings of $Q_{2}^{n}$ is a set of $(n-k)$-dimensional axis-aligned planes or $(n-k)$-faces such that a union of the faces is equal to $Q_{2}^{n}$. Two $m$-faces are called parallel if they have the same directions and a pair of parallel faces is called antipodal if for each vertex from one face there exists an antipodal vertex in another face. It is clear that each $k$-covering of $Q_{2}^{n}$ consists of $2^{k}$ or more $(n-k)$-faces. If a $k$-covering $C$ of $Q_{2}^{n}$ consists of exactly $2^{k}$ $(n-k)$-faces then $C$ is a $k$-splitting of $Q_{2}^{n}$ into $(n-k)$-faces. If $n-k=1$ then such splitting is equivalent to a perfect matching in the Boolean hypercube. A $k$-covering of $Q_{2}^{n}$ is called an antipodal $k$-splitting if it consists of exactly $2^{k}$ $(n-k)$-faces and it does not contain pairs of parallel non-antipodal faces.

The concept of DP-coloring was developed by Dvo${\rm\breve{r}\acute{a}}$k and Postle [3] in order to generalize the notation of a proper coloring. In [1] Bernshteyn and Kostochka considered a problem to estimate the minimum number of edges in non-$2$-DP-colorable $k$-uniform hypergraphs. The existence of a non-$2$-DP-colorable $k$-uniform hypergraph with $e$ edges and $n$ vertices is equivalent to the existence of a covering of $Q_{2}^{n}$ by $e$ pairs of antipodal $(n-k)$-faces. If the hypergraph has no multiple edges then the definition of DP-coloring implies that this covering does not contain pairs of parallel non-antipodal faces. If $e=2^{k-1}$ then a non-$2$-DP-colorable $k$-uniform hypergraph with $e$ edges generates an antipodal $k$-splitting and vice versa. The connection between $2$-colorings of hypergraphs and coverings of the hypercube will be stated with more details in Section 3.

It is known (see [1]) that for any even $k$ each $k$-uniform hypergraph with $2^{k-1}$ edges has 2-DP-coloring. Bernshteyn and Kostochka conjectured that for any odd $k\geq 3$ there exists a non-2-DP-coloring $k$-uniform hypergraph with $2^{k-1}$ edges. The main result of this paper is a construction of antipodal $k$-splittings 111 Earlier, I mistakenly claimed that such an example does not exist, i.e., each $5$-uniform hypergraph with $16$ edges is 2-DP-colorable (see [1]). for any odd $k\geq 3$ and, consequently, a proof of existence of non-2-DP-colorable $k$-uniform hypergraphs with $2^{k-1}$ edges.

2 Splittings of a hypercube

We denote a $(n-k)$-face of $Q_{2}^{n}$ by a $n$-tuple $(a_{1},a_{2},\dots,a_{n})$ of symbols $0,1,*$ where the symbol $*$ is used $n-k$ times. In more details, $(a_{1},\dots,a_{n})=\{(x_{1},\dots,x_{n}):x_{i}=a_{i}\ \mbox{if}\ a_{i}=0\ \mbox{or}\ a_{i}=1\}$.

If $A=\{(a_{1},\dots,a_{n})\}$ is an antipodal $k$-splitting then $A_{\tau}=\{(a_{\tau 1},\dots,a_{\tau n})\}$ is an antipodal $k$-splitting for any permutation $\tau$. Let us agree to $*\oplus 0=*\oplus 1=*$. We define Boolean addition of $n$-tuples to act coordinate-wise. Then for any $(n-k)$-face $a$ and any $b\in Q^{n}_{2}$ a sum $a\oplus b$ is a $(n-k)$-face of $Q^{n}_{2}$. It is clear that if $A$ is an antipodal $k$-splitting then $A\oplus b=\{a\oplus b:a\in A\}$ is an antipodal $k$-splitting for each $b\in Q^{n}_{2}$. We will refer to the aboved operations as isometries of a Boolean hypercube. $A$ and $A^{\prime}$ are called equivalent antipodal $k$-splittings if $A^{\prime}$ is obtained from $A$ by an isometry.

**Proposition 1. **If there exists an antipodal $k$-covering of $Q^{n}_{2}$ then there exists an antipodal $k$-covering of $Q^{n+1}_{2}$ with the same cardinality.

Proof. If $A$ is an antipodal $k$-covering of $Q^{n}_{2}$ then $B=\{(a_{1},\dots,a_{n},*):(a_{1},\dots,a_{n})\in A\}$ is an antipodal $k$-covering of $Q^{n+1}_{2}$. $\square$

Let $T$ be a subset of $\{1,\dots,n\}$. An antipodal $k$-splitting $A$ in $Q^{n}_{2}$ is called $t$-balanced on $T$ if any $a\in A$ has $t$ elements [math] and $1$ ($|T|-t$ asterisks) in coordinates from $T$.

**Proposition 2. ** If there exist an antipodal $k_{1}$-splitting of $Q^{n_{1}}_{2}$ and an antipodal $k_{2}$-splitting of $Q^{n_{2}}_{2}$ which is $t$-balanced on $T$, $|T|=n_{3}\leq n_{2}$, then there exists an antipodal $(k_{2}+(k_{1}-1)t)$-splitting of $Q^{n_{2}+(n_{1}-1)n_{3}}_{2}$.

Proof. Let $A$ be an antipodal $k_{2}$-splitting of $Q^{n_{2}}_{2}$ and $B=B_{0}\cup B_{1}$ be an antipodal $k_{1}$-splitting of $Q^{n_{1}}_{2}$ where sets $B_{0}$ and $B_{1}$ do not contain parallel $(n_{1}-k_{1})$-faces. Consider a $(n_{2}-k_{2})$-face $(a_{1},\dots,a_{n_{2}})\in A$. For $i\in T$, if $a_{i}=0$ we replace $a_{i}$ by arbitrary $b\in B_{0}$; if $a_{i}=1$ then we replace $a_{i}$ by arbitrary $b\in B_{1}$; if $a_{i}=*$ then we replace $a_{i}$ by $\underbrace{(*,\dots,*)}_{n_{1}}$. So, we obtain a set $C$ of $|A|(|B|/2)^{t}=2^{k_{2}}\cdot 2^{(k_{1}-1)t}$ tuples corresponding to $m$-faces in $Q^{n_{2}+(n_{1}-1)n_{3}}_{2}$, where

[TABLE]

It is not difficult to verify that 1) all tuples of $C$ are disjoint; 2) $C$ is a covering of $Q^{n_{2}+(n_{1}-1)n_{3}}_{2}$ by counting cardinality of $\cup C$ and, consequently, $C$ is $(k_{2}+(k_{1}-1)t)$-splitting; 3) $C$ contains pairs of antipodal faces because $A$ and $B$ contain pairs of antipodal faces; 4) $C$ does not contain parallel non-antipodal faces because $A$ and $B$ do not contain such faces. $\square$

**Corollary 1. ** If there exist an antipodal $k_{1}$-splitting of $Q^{n_{1}}_{2}$ and an antipodal $k_{2}$-splitting of $Q^{n_{2}}_{2}$ then there exists an antipodal $k_{1}k_{2}$-splitting of $Q^{n_{1}n_{2}}_{2}$.

The following antipodal $3$-splitting of $Q^{4}_{2}$ correspond to the well-known antipodal perfect matching in $Q^{4}_{2}$. We will denote it by $E_{3}$.

$\begin{matrix}*&0&0&0,\quad&*&1&1&1,\\ 0&*&0&1,\quad&1&*&1&0,\\ 0&1&*&0,\quad&1&0&*&1,\\ 0&0&1&*,\quad&1&1&0&*.\\ \end{matrix}$

There exist two antipodal $5$-splittings of $Q^{8}_{2}$.

$\begin{matrix}00&1*&00&**,\quad&11&0*&11&**,\\ 00&*1&10&**,\quad&11&*0&01&**,\\ 0*&01&00&**,\quad&1*&10&11&**,\\ *0&10&01&**,\quad&*1&01&10&**,\\ 01&1*&0*&0*,\quad&10&0*&1*&1*,\\ 00&*0&1*&1*,\quad&11&*1&0*&0*,\\ 0*&10&1*&0*,\quad&1*&01&0*&1*,\\ *0&00&0*&1*,\quad&*1&11&1*&0*,\\ 01&0*&*1&0*,\quad&10&1*&*0&1*,\\ 10&*1&*0&0*,\quad&01&*0&*1&1*,\\ 1*&10&*0&0*,\quad&0*&01&*1&1*,\\ *1&00&*0&*0,\quad&*0&11&*1&*1,\\ *0&00&**&00,\quad&*1&11&**&11,\\ *0&0*&*1&01,\quad&*1&1*&*0&10,\\ *0&*1&*1&00,\quad&*1&*0&*0&11,\\ **&00&*0&01,\quad&**&11&*1&10;\\ \end{matrix}$

$\begin{matrix}0*&*0&*0&10,\quad&1*&*1&*1&01,\\ 0*&*1&1*&10,\quad&1*&*0&0*&01,\\ 0*&*1&00&*0,\quad&1*&*0&11&*1,\\ 0*&*0&00&0*,\quad&1*&*1&11&1*,\\ *1&0*&*1&00,\quad&*0&1*&*0&11,\\ *1&1*&1*&00,\quad&*0&0*&0*&11,\\ *0&0*&01&*0,\quad&*1&1*&10&*1,\\ *1&0*&10&0*,\quad&*0&1*&01&1*,\\ *1&*0&*1&10,\quad&*0&*1&*0&01,\\ *0&*1&1*&00,\quad&*1&*0&0*&11,\\ *0&*0&11&*0,\quad&*1&*1&00&*1,\\ *0&*0&10&0*,\quad&*1&*1&01&1*,\\ 1*&1*&*0&10,\quad&0*&0*&*1&01,\\ 1*&1*&0*&00,\quad&0*&0*&1*&11,\\ 1*&0*&00&*0,\quad&0*&1*&11&*1,\\ 0*&1*&01&0*,\quad&1*&0*&10&1*.\\ \end{matrix}$

Note that any isometry of the hypercube exchange only the order of columns and symbols [math] and $1$ in any fixed column. Consequently, the above two antipodal $5$-splittings are nonequivalent. The last antipodal $5$-splitting is $1$-balanced on the sets $\{1,2\}$ and $\{3,4\}$. We will denote it by $E_{5}$.

**Theorem 1. ** There exists an antipodal $k$-splitting for every odd $k\geq 3$.

Proof. Let us use antipodal $5$-splitting $E_{5}$ and $3$-splitting $E_{3}$ in the construction from Proposition 2 with $k_{1}=3,k_{2}=5$, $n_{1}=4,n_{2}=8$, $T=\{1,2\}$, $t=1$. We obtain an antipodal $7$-splittings $E_{7}$. Replacing $E_{3}$ by $E_{5}$ in this construction, we obtain an antipodal $9$-splittings $E_{9}$. It is clear that we can construct an antipodal $(2s+1)$-splittings $E_{2s+1}$ from $(2s-3)$-splittings $E_{2s-3}$ and $1$-balanced $5$-splittings $E_{5}$ by Proposition 2. Consequently, we prove the theorem by induction. $\square$

The proof of the following statement can be found in [1]. Here it is rewritten in notations of this article.

Proposition 3[1] ([1]). If $k$ is even then an antipodal $k$-splitting of $Q^{n}_{2}$ does not exist.

Proof. Let $A$ be an antipodal $k$-splitting and $k$ is even. Let us consider an $(n-k)$-face $a\in A$, the $(n-k)$-face $\overline{a}\in A$ antipodal to $a$ and a $k$-face $a^{\bot}$ orthogonal (dual) to $a$, i.e., positions of asterisks in $a$ and $a^{\bot}$ are complementary. By the definitions, we obtain that $x=a\cap a^{\bot}$ and $\widetilde{x}=\overline{a}\cap a^{\bot}$ are antipodal vertices within the face $a^{\bot}$. For example, $a=(0,1,1,0,*,*)$, $a^{\bot}=(*,*,*,*,1,0)$, $x=(0,1,1,0,1,0)$, $\widetilde{x}=(1,0,0,1,1,0)$. The vertices $x$ and $\widetilde{x}$ have the same parity because $k$ is even. But for all other $b\in A$ we obtain that $b\cap a^{\bot}$ has the same number of even-weighted and odd-weighted vertices. Since $A$ is a splitting, the set $\{b\cap a^{\bot}:b\in A\}$ is a splitting of $a^{\bot}$ as well. Because the numbers of even-weighted and odd-weighted vertices in $a^{\bot}$ are equal, we have a contradiction. $\square$

**Proposition 4. ** For any $k$-splitting $A$ of $Q^{n}_{2}$ ($n>k>0$) and for any direction of faces the number of $(n-k)$-faces of this direction in $A$ is even.

Proof. Suppose $a\in A$ and $A$ contains $m$ $(n-k)$-faces of the same direction as $a$. Consider a face $a^{\bot}$. If $b\in A$ has the same direction as $a$ then $|b\cap a^{\bot}|=1$, otherwise the number $|b\cap a^{\bot}|$ is even. Since $|a^{\bot}|=2^{k}=\sum_{b\in A}|b\cap a^{\bot}|$ and all terms except $m$ are even, $m$ is even. $\square$

A splitting of hypercube into $(n-k)$-faces is a special case of A-designs. In [6] there are given constructions of A-design with additional properties, for example, with no adjacent parallel faces. An antipodal $k$-splitting does not exist if $k$ even. Nevertheless we can find a $k$-splitting of $Q^{n}_{2}$ with a pair of parallel $(n-k)$-faces on almost maximal distance.

**Proposition 5. ** For even $k\geq 4$ there exits a $k$-splitting of $Q^{n}_{2}$ with the distance $k-1$ between every pair of $(n-k)$-faces of the same direction.

Proof. As mentioned above, for any permutation $\tau$ and any $k$-splitting $A$ a set $A_{\tau}$ is a $k$-splitting. Consider $k$-splitting $(E_{5})_{\tau}$ for $\tau=\begin{pmatrix}1234&5678\\ 5678&1234\\ \end{pmatrix}$. Since the first four coordinates of faces from $E_{5}$ contain two symbols $*$ and the last four coordinates of faces from $E_{5}$ contain one symbol $*$, we conclude that $(E_{5})_{\tau}$ does not contain faces of the same direction as $E_{5}$. Denote a set $E_{5}0=\{a0:a\in E_{5}\}$. So a distance between parallel faces in a $6$-splitting $E_{5}0\cup(E_{5})_{\tau}1$ is maximal minus $1$, i.e., it is equal to $5$. By the same way, we can prove that $E_{2s+1}**$ and $(E_{2s+1}**)_{\sigma}$, where $\sigma$ permute two last pairs of coordinates, do not contain faces of the same direction. Then $E_{2s+1}**0\cup(E_{2s+1}**)_{\sigma}1$ is the required $(2s+2)$-splitting. $\square$

Now we will find a minimal dimension of a hypercube containing a $k$-splitting with at most two $(n-k)$-faces of any fixed direction.

**Proposition 6. ** There exits a $k$-splitting of $Q^{n}_{2}$ with at most two $(n-k)$-faces of each direction if $n-2k+2\geq 0$, $0<k<n$.

Proof. By induction on $n$. Any $1$-splitting of $Q^{n}_{2}$ consists of two parallel faces. For $n=3,k=2$ it is easy to verify this statement directly. The case $n=4$, $k=3$ corresponds to $E_{3}$. Let $B$ be a $k$-splitting of $Q^{n}_{2}$ with at most two $(n-k)$-faces of each direction. By Proposition 2 $B$ contains two or zero faces of each direction. Let $B=B_{0}\cup B_{1}$ where sets $B_{0}$ and $B_{1}$ do not contain parallel $(n-k)$-faces. Consider the set $A=\{b0*,b1*:b\in B_{0}\}\cup\{b*1,b*0:b\in B_{1}\}$. By the construction, the set $A$ is a $(k+1)$-splitting of $Q^{n+2}_{2}$ with at most two $(n-k+1)$-faces of each direction. Besides, the set $C=B*$ is a $k$-splitting of $Q^{n+1}_{2}$. $\square$

3 2-DP-colorings

Let $G$ be an $r$-uniform hypergraph on $n$ vertices. For each $e\in E(G)$ we consider two antipodal 2-colorings $\varphi_{e}:e\rightarrow\{0,1\}$ and $\overline{\varphi_{e}}=\varphi_{e}\oplus 1$. Let $\Phi$ be a collection of $\varphi_{e}$, $e\in E(G)$. We say that a 2-coloring $f:V(G)\rightarrow\{0,1\}$ avoids $\Phi$ if $f|_{e}\neq\varphi_{e}$ and $f|_{e}\neq\overline{\varphi}_{e}$ for each $e\in E(G)$.

A hypergraph $G$ is called a proper 2-colorable if there exists a $2$-coloring $f$ avoiding $\Phi_{0}$, where $\Phi_{0}$ consists of constant maps. A hypergraph $G$ is called 2-DP-colorable if for every $\Phi$ there exists a 2-coloring $f$ avoiding $\Phi$.

A $2$-coloring $f$ of a $k$-uniform hypergraph on $n$ vertices is one-to-one corresponding to an $n$-tuple over alphabet $\{0,1\}$ ($f\in Q^{n}_{2}$). Each $k$-hyperedge correspond to $(n-k)$-faces of $Q^{n}_{2}$ of some direction. For example, a $k$-hyperedge consisting of $i_{1}$th,…,$i_{k}$th vertices corresponds to faces $(*,\dots,*,a_{i_{1}},*,\dots,*,a_{i_{2}},*,\dots,a_{i_{k}},*\dots,*)$ where $a_{i_{j}}\in\{0,1\}$. The set $\{a_{i_{j}}\}$ corresponds to some coloring of vertices from the hyperedge. A $2$-coloring $f$ avoids $\varphi_{e}=(*,\dots,1,\dots,*,\dots,0,\dots,*)$ iff $f\not\in\varphi_{e}$. A 2-coloring $f$ avoids $\Phi$ if $f\not\in\varphi_{e}\cup\overline{\varphi}_{e}$ for each $\varphi_{e}\in\Phi$.

Consider a table of size $n\times\ell$, where every column corresponds to a $(n-k)$-face of an antipodal covering of $Q^{n}_{2}$. Let us replace in the table symbols $*$ by [math] and other symbols by $1$. By the definition, the resulting table is the incidence matrix of a non-$2$-DP-colorable $k$-uniform hypergraph with $\ell$ edges. Consequently, we have the following statement.

**Proposition 7. ** A $k$-uniform hypergraph with $\ell$ edges and $n$ vertices is non-$2$-DP-colorable if and only if its incidence matrix corresponds to a $k$-covering of $Q_{2}^{n}$ by $\ell$ pairs of antipodal $(n-k)$-faces.

Moreover, Proposition 3 implies the following statement.

**Corollary 2. ** There exists a non-$2$-DP-colorable $k$-uniform hypergraph with $2^{k-1}$ edges if and only if there exists an antipodal $k$-splitting of $Q^{n}_{2}$.

A non-$2$-DP-colorable $3$-uniform hypergraph with $4$ edges that corresponding to the antipodal $3$-splitting $E_{3}$ is presented in [1]. By Theorem 2 and Corollary 3, we obtain the following statement.

**Corollary 3. ** If $k\geq 3$ is odd there exists a non-2-DP-colorable $k$-uniform hypergraph with $2^{k-1}$ edges.

Since a union of at most $2\ell$ $(n-k)$-faces contains $\ell 2^{n-k+1}$ vertices, we obtain the following corollary.

Corollary 4[1] ([1]). Every $k$-uniform hypergraph with $\ell<2^{k-1}$ edges is $2$-DP-colorable.

By definition a proper coloring of hypergraph is not contained monochromatic edges. Consequently, a non-$2$-proper colorable hypergraph corresponds to a covering of the hypercube consisting of faces which contain vertices $\overline{0}$ or $\overline{1}$. Therefore, each $k$-uniform hypergraph with $2^{k-1}$ edges is proper $2$-colorable. Moreover, by similar arguments we obtain that every $k$-uniform hypergraph with $s^{k-1}$ or less edges is proper $s$-colorable. A better bound for the case of proper colorings is known. Cherkashin and Kozik [2] Radhakrishnan and Srinivasan [4] (for $s=2$) showed that any $k$-uniform hypergraph with $c(s)(\frac{k}{\ln k})^{\frac{s-1}{s}}s^{k-1}$ or less edges is proper $s$-colorable, where $c(s)>0$ does not depend on $k$ ($k$ is large enough). A survey of results on proper colorings of hypergraphs and related problems were found in [7]. A some Brooks’ type theorem for DP-colorings of hypergraphs is proved in [5].

4 Acknowledgments

The author is grateful to S.Avgustinovich, A.Kostochka and A.Taranenko for their attention to this work and useful discussions.