A quantitative Lov\'asz criterion for Property B

TL;DR

This paper provides an exact quantitative criterion based on edge intersections for determining non-2-colorability in hypergraphs, extending Lovász's classic criterion and characterizing extremal cases.

Contribution

It introduces a precise quantitative version of Lovász's criterion for hypergraph 2-colorability and characterizes all extremal hypergraphs achieving this bound.

Findings

Established an exact count of intersecting edge pairs for non-2-colorable hypergraphs.

Characterized all extremal hypergraphs meeting the criterion.

Combined combinatorial and probabilistic methods in the proof.

Abstract

A well known observation of Lov\'asz is that if a hypergraph is not $2$-colorable, then at least one pair of its edges intersect at a single vertex. %This very simple criterion turned out to be extremly useful . In this short paper we consider the quantitative version of Lov\'asz's criterion. That is, we ask how many pairs of edges intersecting at a single vertex, should belong to a non $2$-colorable $n$-uniform hypergraph? Our main result is an {\em exact} answer to this question, which further characterizes all the extremal hypergraphs. The proof combines Bollob\'as's two families theorem with Pluhar's randomized coloring algorithm.

Equations

$$\frac{(n-1)!(n-1)!}{(2n-1)!}=\frac{1}{n\left(\genfrac{}{}{0ex}{}{2n-1}{n}\right)}$$

$$\pi ((X\cup X^{\prime })\setminus y)<\pi (y)<\pi (Y\setminus y)$$

$$\sum _{i\in I}\frac{1}{\left(\genfrac{}{}{0ex}{}{p-|B_i|}{|A_i|}\right)}\le 1,$$

$$\sum _{i=1}^t\frac{1}{\left(\genfrac{}{}{0ex}{}{p-|B_i|}{|A_i|}\right)}=\sum _{i=1}^t\frac{1}{\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)}\ge 1,$$

$$\pi (X\setminus y)<\pi (y)<\pi ((X^{\prime }\setminus y^{\prime })\setminus X)<\pi (y^{\prime })<\pi ((Y\cup Y^{\prime })\setminus (X\cup X^{\prime })).$$

Full text

1 Introduction

A hypergraph ${\cal H}=(V,E)$ consists of a vertex set $V$ and a set of edges $E$ where each $X\in E$ is a subset of $V$. If all edges of ${\cal H}$ have size $n$ then ${\cal H}$ is called an $n$-uniform hypergraph, or $n$-graph for short. A hypergraph is $2$-colorable if one can assign each vertex $v\in V$ one of two colors, say $Red$/$Blue$, so that each $X\in E$ contains vertices of both colors. Miller [6], and later Erdős in various papers, referred to this property as Property B, after F. Bernstein [2] who introduced it in 1907. Since deciding if a hypergraph is $2$-colorable is $NP$-hard one cannot hope to find a simple characterization of all $2$-colorable hypergraphs. Instead, one looks for general sufficient/necessary conditions for having this property. For example, a famous result of Seymour [8] states that if ${\cal H}$ is not $2$-colorable then $|E|\geq|V|$. Probably the most well studied question of this type asks for the smallest number of edges in an $n$-graph that is not $2$-colorable. The study of this quantity, denote $m(n)$, was popularized by Erdős, see [1] for a comprehensive treatment. Despite much effort by many researchers, even the asymptotic value of $m(n)$ has not been determined yet.

A pair of edges $X,Y\in E({\cal H})$ is simple if $|X\cap Y|=1$. Let $m_{2}({\cal H})$ denote the number of ordered simple pairs of edges of ${\cal H}$. A well known observation of Lovász [5] states that if ${\cal H}$ is not $2$-colorable then $m_{2}({\cal H})>0$. Despite its simplicity, this observation underlies the best known bounds for $m(n)$, see [4, 7]. It is natural to ask if one can obtain a quantitative version of Lovász’s observation, that is, estimate how small can $m_{2}({\cal H})$ be in an $n$-graph not satisfying property $B$? Our main result in this paper states that (somewhat surprisingly), one can give an exact answer to the above extremal question as well as characterize the extremal $n$-graphs.

Let $K_{2n-1}^{n}$ denote the complete $n$-graph on $2n-1$ vertices. It is easy to see that $K_{2n-1}^{n}$ is not $2$-colorable and that $m_{2}(K_{2n-1}^{n})=n\cdot\binom{2n-1}{n}$. We first observe that this simple upper bound is tight.

Proposition 1.1**.**

If an $n$-graph is not $2$-colorable then $m_{2}({\cal H})\geq n\cdot\binom{2n-1}{n}$.

As with any extremal problem, one would like to know which graphs or hypergraphs are extremal with respect to this problem. For example, Turán’s theorem states that among all $n$-vertex graphs not containing a complete $t$-vertex subgraph, there is only one graph maximizing the number of edges. In the setting of our problem, it is easy to see that $K_{2n-1}^{n}$ is not the only non $2$-colorable $n$-graph satisfying $m_{2}({\cal H})=n\cdot\binom{2n-1}{n}$, since one can take a copy of $K_{2n-1}^{n}$ and add to it more vertices and edges without increasing the number of simple pairs. Our main result in this paper characterizes the extremal $n$-graphs, by showing that this is in fact the only way to construct an $n$-graph meeting the bound of Proposition 1.1.

Theorem 1**.**

If a non $2$-colorable $n$-graph ${\cal H}$ satisfies $m_{2}({\cal H})=n\cdot\binom{2n-1}{n}$ then it contains a copy of $K_{2n-1}^{n}$.

While the proof of Proposition 1.1 is implicit in Pluhar’s [7] argument for bounding $m(n)$, the proof of Theorem 1 is more intricate, relying on Bollobás’s two families theorem [3] as well as on a refined analysis of Pluhar’s randomized algorithm for $2$-coloring $n$-graphs.

2 Proof of Proposition 1.1

In this section we describe several preliminary observations regarding a coloring algorithm introduced in [7], and use them to derive Proposition 1.1. The algorithm is the following:

Algorithm $\text{Col}({\cal H},\pi)$. The input is a hypergraph ${\cal H}=(V,E)$ and an ordering $\pi:V\mapsto\{1,\ldots,|V|\}$ (that is, $\pi$ is a bijection). The output is a $2$-coloring of $V$ (not necessarily a proper one). The algorithm runs in $|V|$ steps, where in each time step $1\leq i\leq|V|$, the vertex $\pi^{-1}(i)$ is being colored $Blue$ if this does not form any monochromatic $Blue$ edge. Otherwise, $\pi^{-1}(i)$ is colored $Red$.

We now state an important property of $\text{Col}({\cal H},\pi)$. For two disjoint subsets $X,Y\subseteq V$, we use the notation $\pi(X)<\pi(Y)$ whenever $\max_{x\in X}\pi(x)<\min_{y\in Y}\pi(y)$, that is, the elements of $X$ precede all the elements of $Y$ in the ordering $\pi$. Suppose $(X,Y)$ is a simple pair of edges in ${\cal H}$ with111Here, and in what follows, we slightly abuse notation by writing $y$ instead of the more appropriate $\{y\}$. $X\cap Y=y$. We say that $\pi$ separates $(X,Y)$ if $\pi(X\setminus y)<\pi(y)<\pi(Y\setminus y)$.

Claim 2.1**.**

If $\text{Col}({\cal H},\pi)$ fails to properly color ${\cal H}$ then $\pi$ separates at least one pair of simple edges.

Proof.

We first observe that (by definition) for every ordering $\pi$, the algorithm $\text{Col}(H,\pi)$ does not produce monochromatic $Blue$ edges. Suppose then it produced a $Red$ edge $Y\in E$. Let $y$ be the first vertex of $Y$ according to the ordering $\pi$. If $y$ was colored red, then there must have been an edge $X$ so that $y\in X$, and all other vertices of $X$ were already colored $Blue$ (otherwise the algorithm would color $y$ $Blue$). This means $(X,Y)$ is simple and that $\pi$ separates it. ∎

Note that the claim above already shows that if ${\cal H}$ is not 2-colorable then $m_{2}({\cal H})>0$. For the proof of Proposition 1.1 we will also need the following simple fact.

Claim 2.2**.**

A random permutation separates any given simple pair with probability $1/n\binom{2n-1}{n}$.

Proof.

Let $(X,Y)$ be a simple pair, and let $X\cap Y=y$. A permutation $\pi$ separates $(X,Y)$ if and only if $\pi(X\setminus y)<\pi(y)<\pi(Y\setminus y)$, and this happens with probability exactly

[TABLE]

as desired. ∎

The above claims suffice for proving Proposition 1.1.

Proof (of Proposition 1.1):.

Assume $m_{2}({\cal H})<n\binom{2n-1}{n}$. Suppose we pick a uniformly random $\pi$. Then by the union bound and Claim 2.2, we infer that with positive probability $\pi$ does not separate any simple pair edges. Hence, there is a $\pi$ not separating any simple pair. Claim 2.1 then implies that $\text{Col}({\cal H},\pi)$ will produce a legal $2$-coloring of ${\cal H}$. ∎

3 Proof of Theorem 1

For the rest of this section fix some non $2$-colorable $n$-graph ${\cal H}=(V,E)$ satisfying $m_{2}({\cal H})=n\binom{2n-1}{n}$. We need to show that ${\cal H}$ contains a copy of $K^{n}_{2n-1}$. We start with a few preliminary claims regarding ${\cal H}$.

First, we show that no $\pi$ separates more than one simple pair.

Claim 3.1**.**

Every ordering $\pi$ separates at most one simple pair.

Proof.

Suppose $\pi$ separates two simple pairs. By Claim 2.2, the assumption on $m_{2}(\cal H)$, and by linearity of expectation, the expected number of simple pairs separated by a random permutation is exactly $1$. Hence, if $\pi$ separates $2$ simple pairs, then there must exist a permutation $\sigma$ which separates less than $1$, and therefore [math], simple pairs. Therefore, by Claim 2.1 we obtain that $\text{Col}({\cal H},\sigma)$ produces a legal $2$-coloring of ${\cal H}$, which is a contradiction to the assumption that $\mathcal{H}$ is not $2$-colorable. ∎

Claim 3.2**.**

If $(X,Y)$ and $(X^{\prime},Y)$ are simple pairs, then $X\cap Y\neq X^{\prime}\cap Y$.

Proof.

We observe that if $X\cap Y=X^{\prime}\cap Y=y$, then there is a $\pi$ that separates both $(X,Y)$ and $(X^{\prime},Y)$, and this will contradict Claim 3.1. Indeed, if $(X,Y)$ and $(X^{\prime},Y)$ are simple pairs and $X\cap Y=X^{\prime}\cap Y=y$, then $(X\cup X^{\prime})\setminus y$ and $Y$ are disjoint. Therefore, any $\pi$ satisfying

[TABLE]

separates $(X,Y)$ and $(X^{\prime},Y)$. This completes the proof. ∎

In addition to the above observations about ${\cal H}$, the last ingredient we will need is the following theorem of Bollobás [3].

Lemma 3.3**.**

Let $I$ be an index set. For all $i\in I$, let $A_{i}$ and $B_{i}$ be subsets of a set $V$ of $p$ elements satisfying the following conditions:

1. i.

$A_{i}\cap B_{i}=\emptyset$* for all $i\in I$, and* 2. ii.

$A_{j}\nsubseteq A_{i}\cup B_{i}$* for all $i\neq j\in I$.*

Then, we have

[TABLE]

with equality if and only if $B_{i}=B$ for all $i\in I$ and the sets $A_{i}$ are all the $q$-tuples of the set $P\setminus B$ for some value of $q$.

Let us now show how to use Lemma 3.3 in order to derive Theorem 1. Recall that $V$ is the vertex set of ${\cal H}$ and set $p:=|V|$. Let $M(\mathcal{H})$ be a collection of simple pairs $(X,Y)$ defined as follows; out of all the simple pairs $(X,Y)$ with the same “second” set $Y$, put in $M(\mathcal{H})$ one of these pairs. Observe that by Claim 3.2 each $Y$ belongs to at most $|Y|=n$ simple pairs of the form $(X,Y)$ (i.e, with $Y$ as the second set), implying that $t:=|M(\mathcal{H})|\geq\frac{1}{n}\cdot m_{2}(\mathcal{H})=\binom{2n-1}{n}$. We now define a collection $\mathcal{F}$ consisting of pairs of subsets of $V$ as follows: For every simple pair $s:=(X,Y)\in M(\mathcal{H})$, define $A_{s}=X\setminus Y$ and $B_{s}=V\setminus(X\cup Y)$, and let $\mathcal{F}=\{(A_{s},B_{s}):\text{ }s\in M(\mathcal{H})\}$. For convenience, let us rename the pairs in ${\cal F}$ as $(A_{i},B_{i})$ with $1\leq i\leq t$.

Now we wish to show that $\mathcal{F}$ satisfies the conditions in Lemma 3.3. Observe that if it does, then since

[TABLE]

it follows by the first part of Lemma 3.3 that the last inequality is in fact an equality. Therefore, by the second part of Lemma 3.3, we conclude that all the $B_{i}$’s are the same set $B$, and the set of all the $A_{i}$’s consists of all $n-1$ subsets of a ground set of size $2n-1$. That is, let $B=B_{i}$ and $U=V\setminus B$. Then we have that $|U|=2n-1$, and that the sets $A_{i}$ are all the $n-1$ subsets of $U$. Since by construction we have that $U\setminus A_{i}\in E(\mathcal{H})$ for all $i$, we conclude that $\mathcal{H}$ restricted to the set $U$ is a copy of $K^{n}_{2n-1}$ as desired. It thus remains to show the following:

Claim 3.4**.**

$\mathcal{F}$* satisfies the conditions in Lemma 3.3*

Proof.

The first condition $A_{i}\cap B_{i}=\emptyset$ for all $i$ is trivially satisfied by construction. For the second condition, let $(A,B)$ and $(A^{\prime},B^{\prime})$ be two elements in $\mathcal{F}$ coming from simple pairs $(X,Y)$ and $(X^{\prime},Y^{\prime})$ belonging to $M({\cal H})$, respectively. Recall that by the way we defined $M(\mathcal{H})$ and $\mathcal{F}$ we have $Y\neq Y^{\prime}$. Let us use $y$ and $y^{\prime}$ to denote the unique elements in $X\cap Y$ and $X^{\prime}\cap Y^{\prime}$, respectively. We wish to show that $A\nsubseteq A^{\prime}\cup B^{\prime}$, which, by construction, is implied by $(X\setminus y)\cap Y^{\prime}\neq\emptyset$. Assuming $(X\setminus y)\cap Y^{\prime}=\emptyset$, we will derive a contradiction to Claim 3.1 by showing that there is a permutation $\pi$ separating two distinct simple pairs.

Observe that it cannot be that $y\in Y^{\prime}$. Indeed, if it was the case, then together with the assumption that $(X\setminus y)\cap Y^{\prime}=\emptyset$ we would infer that $(X,Y)$ and $(X,Y^{\prime})$ are both simple pairs intersecting at $y$ (and distinct as $Y\neq Y^{\prime}$), contradicting Claim 3.2. Assume then that $y\not\in Y^{\prime}$ (so in particular $y\neq y^{\prime}$). We claim that we can find a $\pi$ satisfying

[TABLE]

Indeed, the only thing that needs to be justified is the ability to place $y^{\prime}$ as above, which follows from the fact that $y^{\prime}\in Y^{\prime}$ and the assumption $(X\setminus y)\cap Y^{\prime}=\emptyset$ which together imply that $y^{\prime}\not\in X$. Observe that since $\pi$ first places $X\setminus y$ and then $y$, the pair $(X,Y)$ is separated by $\pi$. Such a $\pi$ clearly places $X^{\prime}\setminus y^{\prime}$ before $y^{\prime}$ and the assumption $(X\setminus y)\cap Y^{\prime}=\emptyset$ together with the fact that $y\not\in Y^{\prime}$ imply that such a $\pi$ places all of $Y^{\prime}\setminus y^{\prime}$ after $y^{\prime}$, so it separates $(X^{\prime},Y^{\prime})$ as well, giving us the desired contradiction. ∎

This completes the proof of Theorem 1.