On small $n$-uniform hypergraphs with positive discrepancy

TL;DR

This paper investigates the minimal size of n-uniform hypergraphs that lack a zero-discrepancy two-coloring, refining existing bounds by establishing an upper bound proportional to the logarithm of the smallest non-divisor of n.

Contribution

The authors improve the upper bound on the minimum number of edges in such hypergraphs, relating it to the logarithm of the smallest non-divisor of n, advancing understanding of discrepancy in hypergraphs.

Findings

Established an upper bound: f(n) ≤ c log snd(n)

Refined previous bounds on hypergraph discrepancy

Connected hypergraph properties to number-theoretic functions

Abstract

A two-coloring of the vertices $V$ of the hypergraph $H=(V, E)$ by red and blue has discrepancy $d$ if $d$ is the largest difference between the number of red and blue points in any edge. Let $f(n)$ be the fewest number of edges in an $n$-uniform hypergraph without a coloring with discrepancy $0$. Erd\H{o}s and S\'os asked: is $f(n)$ unbounded? N. Alon, D. J. Kleitman, C. Pomerance, M. Saks and P. Seymour proved upper and lower bounds in terms of the smallest non-divisor ($\mbox{snd}$) of $n$. We refine the upper bound as follows: $$f (n) \leq c \log \mbox{snd}\ {n}.$$

Equations

$$f(n)\le c\log \mathrm{snd}n.$$

$$c_1\frac{\log \mathrm{snd}(n/2)}{\log \log \mathrm{snd}(n/2)}\le f(n)\le c_2\frac{{\log }^3\mathrm{snd}(n/2)}{\log \log \mathrm{snd}(n/2)},$$

$$f(n)\le c\log \mathrm{snd}(n).$$

$$f(n)\le c\log \log n.$$

$$A\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}n\\ n+t\end{array}\right).$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_2\cup B_3\cup B_4\cup B_5$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_2\cup B_3\cup B_4\cup B_6$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_2\cup B_3\cup B_4\cup B_7$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_2\cup B_3\cup B_4\cup B_8$$

$$A_1\cup A_2\cup A_3\cup B_2\cup B_3\cup B_4\cup B_5\cup B_8$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_3\cup B_4\cup B_5\cup B_8$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_2\cup B_4\cup B_5\cup B_8$$

$$A_1\cup A_2\cup A_3\cup B_1\cup B_2\cup B_3\cup B_5\cup B_8$$

$$A_1\cup C\cup B_2\cup B_3\cup B_4\cup B_5\cup B_6\cup B_7\cup B_8$$

$$A_2\cup C\cup B_2\cup B_3\cup B_4\cup B_5\cup B_6\cup B_7\cup B_8$$

$$A_3\cup C\cup B_2\cup B_3\cup B_4\cup B_5\cup B_6\cup B_7\cup B_8.$$

$$q-(2^m-1)=\sum _{i=0}^{m-1}{\epsilon }_i{2}^i\text{mbox}forsome{\epsilon }_i\in \{0,1\},$$

$$q=\sum _{i=0}^{m-1}{\eta }_i{2}^i,\text{mbox}where{\eta }_i=1+{\epsilon }_i\in \{1,2\}.$$

$$v_0=({\eta }_0,\dots ,{\eta }_{m-1}),$$

$$v_i=({\eta }_0,\dots ,{\eta }_{i-2},{\eta }_{i-1}+2,{\eta }_i-1,{\eta }_{i+1},\dots ,{\eta }_{m-1})\text{mbox}fori=1,\dots ,m-1,\text{mbox}i.e.$$

$$v_{i,k}=\{\begin{array}{l}{\eta }_k,k\ne i,i-1\\ {\eta }_k-1,k=i\\ {\eta }_k+2,k=i-1.\end{array}$$

$$⟨v_i,u⟩=q;i=0,\dots ,m-1.$$

$$x_i:=2^i{x}_0\text{mbox}fori=1,\dots ,m-2;x_{m-1}:=2^{m-1}{x}_0-\delta ,$$

$$x_0:=\frac{n+2^{m-1}}{q};x_1:=2x_0-1;x_i:=2^{i-1}{x}_1\text{mbox}fori=2,\dots ,m-2;x_{m-1}:=2^{m-2}{x}_1-1.$$

$$\bigcup _{0\le i\le m-1}\bigcup _{\begin{array}{c}1\le j\le {\eta }_i\text{mbox}fori\ne k\\ j\in R\text{mbox}fori=k\end{array}}{A}_i^j$$

$$\bigcup _{0\le i\le m-1}\bigcup _{\begin{array}{c}1\le j\le {\eta }_i\text{mbox}fori\ne k,k-1\\ 1\le j\le {\eta }_i+2\text{mbox}fori=k-1\\ 1\le j\le {\eta }_i-1\text{mbox}fori=k\end{array}}{A}_i^j,$$

$$⟨v_i,d⟩=0\text{mbox}fori=0,1,\dots ,m-2\text{mbox}and⟨v_{m-1},d⟩=s$$

$$d_i=2^i{d}_0\text{mbox}fori=0,\dots ,m-2;d_{m-1}=2^{m-1}{d}_0-s;0=\sum {\eta }_i{d}_i=d_{0}q-{\eta }_{m-1}s,$$

$$|E(H)|=\left(\genfrac{}{}{0ex}{}{2r-1}{r}\right)f([n/r]{)}^r=O(\ln \mathrm{snd}[n/r]{)}^r\le O(\ln \ln n{)}^r.$$

Full text

N. Alon, D. J. Kleitman, C. Pomerance, M. Saks and P. Seymour [1] proved upper and lower bounds in terms of the smallest non-divisor ($\operatorname{snd}$) of $n$ (see (1)). We refine the upper bound as follows:

[TABLE]

Keywords: hypergraph colorings, hypergraph discrepancy, prescribed matrix determinant.

1 Introduction

A hypergraph is a pair $(V,E)$, where $V$ is a finite set whose elements are called vertices and $E$ is a family of subsets of $V$, called edges. A hypergraph is $n$-uniform if every edge has size $n$. A vertex $2$-coloring of a hypergraph $(V,E)$ is a map $\pi:V\rightarrow\{1,2\}$.

The discrepancy of a coloring is the maximum over all edges of the difference between the number of vertices of two colors in the edge. The discrepancy of a hypergraph is the minimum discrepancy of a coloring of this hypergraph. The general discrepancy theory is set out in [2, 6, 4].

Let $f(n)$ be the minimal number of edges in an $n$-uniform hypergraph (all edges have size $n$) having positive discrepancy. Obviously, if $2\nmid n$ then $f(n)=1$; if $2|n$ but $4\nmid n$ then $f(n)=3$. Erdős and Sős asked whether $f(n)$ is bounded or not. N. Alon, D. J. Kleitman, C. Pomerance, M. Saks and P. Seymour [1] proved the following Theorem, showing in particular that $f(n)$ is unbounded.

Theorem 1.1**.**

Let $n$ be an integer such that $4\,|\,n$. Then

[TABLE]

where $\operatorname{snd}(x)$ stands for the least positive integer that does not divide $x$.

To prove the upper bound they introduced several quantities. Let $\mathcal{M}$ denote the set of all matrices $M$ with entries in $\{0,1\}$ such that the equation $Mx=e$ has exactly one non-negative solution (here $e$ stands for the vector with all entries equal to $1$). This unique solution is denoted $x^{M}$. Let $z(M)$ be the least integer such that $z(M)x^{M}$ is integer and let $y^{M}=z(M)x^{M}$. For each positive integer $n$, let $t(n)$ be the least $r$ such that there exists a matrix $M\in\mathcal{M}$ with $r$ rows such that $z(M)=n$ (obviously, $t(n)\leq n+1$ because $z(J_{n+1}-I_{n+1})=n$, where $J_{n+1}$ is the $(n+1)\times(n+1)$ matrix with unit entries; $I_{n+1}$ is the $(n+1)\times(n+1)$ identity matrix). The upper bound in (1) follows from the inequality $f(n)\leq t(m)$ for such $m$ that $\left\lfloor\frac{n}{m}\right\rfloor$ is odd.

Then N. Alon and V. H. Vũ [3] showed that $t(m)\leq(2+o(1))\frac{\log m}{\log\log m}$ for infinitely many $m$. However they marked that trueness of inequality $t(m)\leq c\log m$ for arbitrary $m$ is not clear.

Our main result is the following

Theorem 1.2**.**

Let $n$ be a positive integer number. Then

[TABLE]

for some constant $c>0$.

Corollary 1.3**.**

Let $n$ be a positive integer number. Then

[TABLE]

for some constant $c>0$.

The construction of the hypergraph with positive discrepancy which yields Theorem 1.2 uses a matrix with determinant $\operatorname{snd}(n)$ and small entries satisfying some additional technical properties. Before coming to a general construction we give an example with a specific $2\times 2$ matrix which shows the vague idea.

2 Example

Example 2.1**.**

Let us consider the matrix $A=\begin{pmatrix}3&5\\ 1&8\end{pmatrix}$ and suppose that $n$ is not divisible on $\det A=19$. Consider the system

[TABLE]

The solution of the system is $a=(3n-5t)/19$, $b=(2n+3t)/19$, which is integral if and only if $t=12n$ (mod 19) i. e. $t$ has prescribed residue modulo 19. Since $n$ is not divisible on $19$, $t$ is not equal to zero modulo $19$. So one can choose $-19<t<19$ such that $t$ has prescribed residue modulo 19 and $t$ is odd. Also, assume that $n/8>t>-2n/3$ which is certainly true if $n>200$. Then $a$ and $b$ are positive and also $b>t$ and $a$, $b$ tend to infinity simultaneously with $n$.

Let us construct an $n$-uniform hypergraph $H$ with positive discrepancy. Consider disjoint vertex sets $A_{1},A_{2},A_{3}$ of size $a$ and $B_{1},\dots,B_{8}$ of size $b$. If $t<0$ then consider a vertex set $T$ of size $|t|$ and set $C:=B_{1}\cup T$; if $t>0$ let $T$ be a $t$-vertex subset of $B_{1}$ and define $C:=B_{1}\setminus T$. The edges of $H$ are listed:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Obviously, if $H$ has a coloring with discrepancy [math], then $d(B_{5})=d(B_{6})$, where $d(X)$ is the difference between blue and red vertices in $X$, because the second edge can be reached by replacing $B_{5}$ on $B_{6}$ in the first edge. Similarly one can deduce that $d(A_{i})=d(A_{j})$ and $d(B_{i})=d(B_{j})$ for all pairs $i$, $j$. So one can put $k:=d(A_{i})$, $l:=d(B_{i})$. Because of the first edge we have $3k+5l=0$. Obviously, $k$ and $l$ are odd numbers, so the minimal solution is $k=5$, $l=-3$ (or $k=-5$, $l=3$ which is the same because of red-blue symmetry). But then the last edge gives $|k+8l|\leq|t|$ which contradicts with $|k+8l|\geq 19>|t|$.

So we got an example if $19\,\nmid\,n$ and $n>200$ of an $n$-uniform hypergraph with $11$ edges and positive discrepancy.

The number of edges in this example equals $11=3+8$, the sum of maximal entries in the columns of $A$. This is essentially (up to multiplicative constant) the general property of our construction.

3 Proofs

Proof of Theorem 1.2.

Let us denote $\operatorname{snd}(n)$ by $q$. We should construct a hypergraph with at most $c\log q$ edges and positive discrepancy. Take $m$ such that $2^{m}-1\leq q\leq 2^{m+1}-2$. Then

[TABLE]

therefore

[TABLE]

Consider $m$ vectors in $\mathbb{Z}^{m}$:

[TABLE]

[TABLE]

[TABLE]

Note that the vector $u=(1,2,\dots,2^{m-1})$ satisfies a system of linear equations

[TABLE]

Assume that $q$ is odd. Choose odd $\delta\in(-q,q)$ such that $x_{0}:=\frac{n+\eta_{m-1}\delta}{q}$ is integer. Define

[TABLE]

then the vector $x=(x_{0},\dots,x_{m-1})$ satisfies $\langle v_{i},x\rangle=n$ for $i=0,\dots,m-2$, $\langle v_{m-1},x\rangle=n+\delta$.

In the case $q=2^{m}\geq 8$ we have $n\equiv 2^{m-1}\pmod{q}$ and $\eta_{0}=2,\eta_{1}=\dots=\eta_{m-1}=1$.

Choose $x=(x_{0},\dots,x_{m-1})$ so that $\langle v_{1},x\rangle=\langle v_{m-1},x\rangle=n+1$ and $\langle v_{i},x\rangle=n$ for $i=0,2,3,\dots,m-2$. The solution is given by

[TABLE]

Now let us construct a hypergraph in the following way: for $i=0,\dots m-1$ let us take $4$ sets $A_{i}^{j}$ ($j=1,\dots,4$) of vertices of size $x_{i}$ such that all $4m$ sets $A_{i}^{j}$ are disjoint. Let the edge $e_{0}$ be the union of $A_{i}^{j}$ over $0\leq i\leq m-1$ and $1\leq j\leq\eta_{i}$. By the choice of $x_{i}$ and $\eta_{i}$ we have $|e_{0}|=n$. Then we add an edge

[TABLE]

for every $k$ and for every $R\subset[4]$ such that $|R|=\eta_{k}$. Clearly there are at most $6m$ such edges. Let us say that they form the first collection of edges. Finally, for every $1\leq k\leq m-1$ we add the edge

[TABLE]

which form the second collection of edges.

Summing up we have hypergraph with at most $7m$ edges; at most $2$ of them have size not equal to $n$. Let us correct these edges in the simplest way: if an edge has size less than $n$ then we add arbitrary vertices; if an edge has size greater than $n$ then we exclude arbitrary vertices.

Suppose that our hypergraph has discrepancy [math], so it has a proper coloring $\pi$. For every set $A_{i}^{j}$ denote by $d(A_{i}^{j})$ the difference between the numbers of red and blue vertices of $\pi$ in $A_{i}^{j}$. Obviously, $d(A_{i}^{j_{1}})=d(A_{i}^{j_{2}})$ because there are edges $e_{1}$, $e_{2}$ from the first collection such that $e_{2}$ can be obtained from $e_{1}$ by the replacement of $A_{i}^{j_{1}}$ to $A_{i}^{j_{2}}$. So we may write $d_{i}$ instead of $d(A_{i}^{j})$.

If $q$ is odd then the vector $d=(d_{0},\dots,d_{m-1})$ satisfies

[TABLE]

for some odd $s\in(-q,q)$. Considering consequent differences of this equations we get

[TABLE]

but this fails modulo $q$. A contradiction. In the case $q=2^{m}$ we get a similar contradiction, as $(2^{m-1}-1)\pm 1$ is not divisible by $2^{m}$.

Thus we get a hypergraph on at most $7m=O(\log q)$ edges with positive discrepancy, the claim is proven. ∎

4 Discussion

• •

In fact, during the proof we have constructed a matrix of size of $O(\log k)$ with bounded integer coefficients and with determinant $k:=\operatorname{snd}(n)$. By Hadamard inequality, the determinant $k$ of $m\times m$ matrix with bounded coefficients satisfies $k=O(\sqrt{m})^{m}$, thus $\log k=O(m\log m)$, $m\geq\text{const}\,\cdot\log k/\log\log k$. We suppose that actually a matrix of size $O(\log k/\log\log k)$ with bounded integer coefficients and determinant $k$ always exists; and moreover, it may be chosen satisfying additional properties which allow to replace the main estimate with $f(n)\leq c\log\operatorname{snd}(n)/\log\log\operatorname{snd}(n)$ (which asymptotically coincides with the lower bound).

• •

It turns out, that for a fixed value of $q=\operatorname{snd}(n)$ and some values of $n$ modulo $q$, a hypergraph, constructions of above type have the discrepancy separated from zero. In particular, in Example 2.1 the choice $n\in\{\pm 4,\pm 7\}$ modulo 19 leads to the discrepancy 6.

• •

For fixed $r$ and large enough $n$ using Theorem 1.2 one can construct an $n$-uniform hypergraph with discrepancy at least $r$ and $O(\ln\operatorname{snd}{[n/r]})^{r}$ edges (here $[x]$ stands for the nearest integer to $x$), as follows: let $H_{0}$ be a hypergraph realizing $f([n/r])$, $H_{1},\dots,H_{2r-1}$ be vertex-disjoint copies of $H$. Let $V:=V(H_{1})\sqcup\dots\sqcup V(H_{2r-1})$, $E:=\{\sqcup e_{i}\ |\ i\in A\subset[2r-1],|A|=r\}$. By the construction, every $H_{i}$ has discrepancy at least $2$; so by pigeonhole principle $(V,E)$ has discrepancy at least $2r$. Define $l:=r[n/r]-n$. Finally, if $l>0$, then exclude arbitrary $l$ vertices from every edge $e\in E$; else add arbitrary $l$ vertices to every edge $e\in E$; denote the result by $H$. By definition $l\leq r$, so the discrepancy of $H$ is at least $r$. Since $|E(H_{i})|=f([n/r])$, we have

[TABLE]

• •

A. Raigorodskii independently asked the same question in a more general form: he introduced the quantity $m_{k}(n)$ that is the minimal number of edges in a hypergraph without a vertex 2-coloring such that every edge has at least $k$ blue vertices and at least $k$ red vertices. So $m_{k}(n)$ is the minimal number of a edges in a hypergraph with discrepancy at least $n-2k+2$, in particular $f(n)=m_{n/2}(n)$ for even $n$.

For the history and the best known bounds on $m_{k}(n)$ see [7]. Note that our result replaces the bound $m_{k}(2k+r)=O(\ln k)^{r+1}$ [5] with $m_{k}(2k+r)=O(\ln\ln k)^{r+1}$ for a constant $r$. It worth noting, that in the case $n=O(k)$ the behavior of $m_{k}(n)$ is completely unclear.

Acknowledgements. The work was supported by the Russian Scientific Foundation grant 16-11-10014. The authors are grateful to A. Raigorodskii for the introduction to the problem, to N. Alon for directing our attention to the paper [1] and fruitful discussions and to N. Rastegaev for a very careful reading of the draft of the paper.