Some Partitionings of Complete Designs

TL;DR

This paper introduces a novel way to partition the set of all 3-element subsets of a v-set, where v is congruent to 2 modulo 4, using simple trades, contributing to combinatorial design theory.

Contribution

It presents a new partitioning method of complete 3-uniform hypergraphs into simple trades for specific v values, advancing combinatorial design techniques.

Findings

New partitioning scheme for 3-subsets

Applicable for v ≡ 2 (mod 4)

Enhances understanding of combinatorial trades

Abstract

Let $v\geq6$ be an integer with $v\equiv2 \pmod 4$. In this paper, we introduce a new partitioning of the set of all $3$-subsets of a $v$-set into some simple trades.

Equations

$$\left(\genfrac{}{}{0ex}{}{X}{i}\right):=\{A\subseteq X:|A|=i\},0\le i\le v.$$

$$W_{tk}(v)(T,K):=\left\{\begin{array}[]{r@{\qquad}l}1&{\rm if}\ T\subseteq K\\ 0&{\rm otherwise}\end{array}\right.\quad T\in{\binom{X}{t}},\ K\in{\binom{X}{k}}.$$

$$W_{tk}f=\lambda 1.$$

$$\left(\genfrac{}{}{0ex}{}{a_0{a}_1{a}_2}{b_0{b}_1{b}_2}\right):=(a_0-b_0)(a_1-b_1)(a_2-b_2),$$

$$T(\mathcal{C}):=\sum _{(a_i{b}_i,a_j{b}_j)\in \mathcal{C}}{a}_i{b}_i(a_j+b_j)-a_j{b}_j(a_i+b_i),$$

$$\left(\begin{array}{ccc}{a}_0& a_1& a_2\\ b_0& b_1& b_2\end{array}\right)\text{+}T(a_0{b}_0,a_1{b}_1,a_2{b}_2,a_0{b}_0)$$

$$\begin{split}b_{2}&=a_{2}+1,\\ b_{1}&=\min\Big{(}\big{\{}x\in X\setminus\{a_{0},a_{1},a_{2},b_{2}\}\,\mid x>a_{1}\big{\}}\Big{)},\\ b_{0}&=\min\Big{(}\big{\{}x\in X\setminus\{a_{0},a_{1},a_{2},b_{2},b_{1}\}\,\mid x>a_{0}\big{\}}\Big{)}.\end{split}$$

$${\mathfrak{T}}_{a_0{a}_1{a}_2}:=\left(\genfrac{}{}{0ex}{}{a_0{a}_1{a}_2}{b_0{b}_1{b}_2}\right).$$

$${\mathcal{T}}_{{\alpha }_0{\alpha }_1{\alpha }_2}:=\left(\begin{array}{ccc}{a}_{{\alpha }_0}& a_{{\alpha }_1}& a_{{\alpha }_2}\\ b_{{\alpha }_0}& b_{{\alpha }_1}& b_{{\alpha }_2}\end{array}\right)+\left(\begin{array}{ccc}{a}_{{\alpha }_0}& a_{{\alpha }_1}& a_{{\alpha }_2}\\ b_{{\alpha }_1}& b_{{\alpha }_2}& b_{{\alpha }_0}\end{array}\right)-\left(\begin{array}{ccc}{a}_{{\alpha }_0}& a_{{\alpha }_1}& a_{{\alpha }_2}\\ b_{{\alpha }_2}& b_{{\alpha }_0}& b_{{\alpha }_1}\end{array}\right).$$

$$\sum _{\begin{array}{c}{\alpha }_0{\alpha }_1{\alpha }_2\subseteq \{0,\dots ,2n\}\end{array}}{\mathcal{T}}_{{\alpha }_0{\alpha }_1{\alpha }_2}\times (-1{)}^{{\alpha }_0+{\alpha }_1+{\alpha }_2}.$$

$${\mathcal{T}}_{123}-{\mathcal{T}}_{124}+{\mathcal{T}}_{125}+{\mathcal{T}}_{134}-{\mathcal{T}}_{135}+{\mathcal{T}}_{145}-{\mathcal{T}}_{234}+{\mathcal{T}}_{235}-{\mathcal{T}}_{245}+{\mathcal{T}}_{345}.$$

$$\setstacktabbedgap{2.7pt}\mathcal{T}_{235}=\parenMatrixstack{2&35\cr 7810}+\parenMatrixstack{2&35\cr 8107}-\parenMatrixstack{2&35\cr 1078}\cdot$$

$$\setstacktabbedgap{2.7pt}\parenMatrixstack{3&46\cr 879}+\parenMatrixstack{2&510\cr 639}+\parenMatrixstack{1&35\cr 246}+\parenMatrixstack{2&38\cr 594}+\parenMatrixstack{3&710\cr 612}+\parenMatrixstack{1&29\cr 1034}+\parenMatrixstack{4&78\cr 156}+\parenMatrixstack{1&27\cr 438}\\ +\parenMatrixstack{4&79\cr 2610}+\parenMatrixstack{3&510\cr 189}+\parenMatrixstack{3&68\cr 275}+\parenMatrixstack{1&79\cr 5104}+\parenMatrixstack{8&910\cr 756}+\parenMatrixstack{1&710\cr 825}+\parenMatrixstack{6&810\cr 491}\cdot$$

$$h=\sum _{\begin{array}{c}{\alpha }_0{\alpha }_1{\alpha }_2\subseteq \{0,\dots ,2n\}\end{array}}\left(\begin{array}{ccc}{a}_{{\alpha }_0}& a_{{\alpha }_1}& a_{{\alpha }_2}\\ b_{{\alpha }_0}& b_{{\alpha }_1}& b_{{\alpha }_2}\end{array}\right)+T(\mathcal{C})$$

$$\left(\genfrac{}{}{0ex}{}{2n+1}{3}\right)\times 4+\left(\genfrac{}{}{0ex}{}{2n+1}{2}\right)\times 2=\frac{\left(\genfrac{}{}{0ex}{}{4n+2}{3}\right)}{2},$$

$$h_1=\sum _{\begin{array}{c}{\beta }_0{\beta }_1{\beta }_2\in \mathrm{STS}(2n+1)\end{array}}T(a_{{\beta }_0}{b}_{{\beta }_0},a_{{\beta }_1}{b}_{{\beta }_1},a_{{\beta }_2}{b}_{{\beta }_2},a_{{\beta }_0}{b}_{{\beta }_0})$$

$$h_2=$$

$$+T(a_{2n-2}{b}_{2n-2},a_{2n-1}{b}_{2n-1},a_{2n}{b}_{2n},a_{2n-2}{b}_{2n-2})$$

$$+\sum _{\begin{array}{c}{a}_i{a}_j\in S\end{array}}T(a_{2n-1}{b}_{2n-1},a_i{b}_i,a_{2n}{b}_{2n},a_j{b}_j,a_{2n-1}{b}_{2n-1})$$

$$\frac{1}{3}\left(\genfrac{}{}{0ex}{}{2n+1}{2}\right)\times 6=2\left(\genfrac{}{}{0ex}{}{2n+1}{2}\right),$$

$$\left(\frac{1}{3}\left(\genfrac{}{}{0ex}{}{2n-1}{2}\right)+1\right)\times 6+(n-1)\times 8=2\left(\genfrac{}{}{0ex}{}{2n+1}{2}\right),$$

$$h=\sum _{\begin{array}{c}{\alpha }_0{\alpha }_1{\alpha }_2\subseteq \{0,\dots ,6\}\end{array}}\left(\begin{array}{ccc}{a}_{{\alpha }_0}& a_{{\alpha }_1}& a_{{\alpha }_2}\\ b_{{\alpha }_0}& b_{{\alpha }_1}& b_{{\alpha }_2}\end{array}\right)+\sum _{\begin{array}{c}{\beta }_0{\beta }_1{\beta }_2\in \mathrm{STS}(7)\end{array}}T(a_{{\beta }_0}{b}_{{\beta }_0},a_{{\beta }_1}{b}_{{\beta }_1},a_{{\beta }_2}{b}_{{\beta }_2},a_{{\beta }_0}{b}_{{\beta }_0})$$

$$35(\mathrm{minimal}\mathrm{trades})+7(\mathrm{trades}\mathrm{of}\mathrm{volume}6)=182.$$

$$h=$$

$$+T(a_8{b}_8,a_9{b}_9,a_{10}{b}_{10},a_8{b}_8)$$

$$+\sum _{\begin{array}{c}{a}_i{a}_j\in S\end{array}}T(a_9{b}_9,a_i{b}_i,a_{10}{b}_{10},a_j{b}_j,a_9{b}_9)$$

$$165(\mathrm{minimal}\mathrm{trades})+13(\mathrm{trades}\mathrm{of}\mathrm{volume}6)+4(\mathrm{trades}\mathrm{of}\mathrm{volume}8)=770.$$

Full text

1 Introduction

Integers $t,k$, and $v$ with $0\leq t\leq k\leq v-t$ are considered. Let $X=\{0,1,\dots,v-1\}$ be a linearly ordered $v$-set, and

[TABLE]

The elements of $\binom{X}{k}$ are called blocks. For the sake of brevity, sometimes a set $\{a_{0},a_{1},\dots,a_{i}\}$ is denoted by the string $a_{0}a_{1}\dots a_{i}$.

The inclusion matrix $W_{tk}(v)$ (known as Wilson Matrix) is defined to be a $\binom{v}{t}$ by $\binom{v}{k}$ $(0,1)$-matrix whose rows and columns are indexed by (and referred to) the members of $\binom{X}{t}$ and $\binom{X}{k}$, respectively, where

[TABLE]

For the sake of convenience, sometimes we use $W_{tk}$ or just a bare $W$ for $W_{tk}(v)$.

Now, suppose that the elements of $\binom{X}{k}$ are lexicographically ordered. A $\mathbb{Z}$ -collection of the elements of $\binom{X}{k}$ is a function $f:\binom{X}{k}\to\mathbb{Z}$ with the vector representation \big{(}f(A_{1}),\dots,f(A_{\binom{v}{k}})\big{)}^{T}.

It is well known that $W_{tk}$ is a full rank matrix over $\mathbb{Q}$ [4]. As a linear operator, $W_{tk}$ acts on a $\mathbb{Z}$ -collection of blocks, and algebraically counts the number of times that any member of $\binom{X}{t}$ appears in the blocks of the collection.

Let 1 be the all 1 vector, and let $\lambda$ be a nonnegative integer. We call the following equation the fundamental equation of design theory:

[TABLE]

For $\lambda>0$, every nonnegative integral solution of Equation (* ‣ 1) is called a $t$-$(v,k,\lambda)$ design. Also, a simple $2$-$(v,3,1)$ design is called a Steiner triple system, denoted by ${\rm STS}(v)$. It is well known that an ${\rm STS}(v)$ exists if and only if $v\equiv 1,3\pmod{6}$. For more on Steiner triple systems, see [3].

For $\lambda=0$, every integral solution of Equation (* ‣ 1) is called a ${\rm T}(t,k,v)$ trade. Let $T$ be a ${\rm T}(t,k,v)$ trade. Clearly $\rm T$ has some negative and positive entries. Therefore, we can assume that $\rm T=T_{0}-T_{1}$, where $\rm T_{0}$ and $\rm T_{1}$ are two nonnegative $\mathbb{Z}$ -collections. Now, $W_{tk}\ {\rm{T}}_{0}=W_{tk}\ {\rm{T}}_{1}$ certifies that every element of $\binom{X}{t}$ appears equally often in $\rm T_{0}$ and $\rm T_{1}$. This means that $\rm T_{0}$ and $\rm T_{1}$, which are called the two legs of T, are mutually $t$-wise balanced. The number of blocks in $T_{0}$ and $T_{1}$ are equal and is called the volume of $T$.

The foundation of a trade is the set of $s$ elements from $X$ appearing in the trade. A nonzero T$(t,k,v)$ trade whose both volume and foundation size take the minimum values is called a minimal trade. It is known that the volume and foundation size of a T$(t,k,v)$ trade are at least $2^{t}$ and $k+t+1$, respectively [6]. A trade is called simple if it contains no repeated blocks.

In what follows, we present two kinds of simple trades.

• $\bullet$

Let $a_{0}a_{1}a_{2}$ and $b_{0}b_{1}b_{2}$ be two disjoint blocks. Then

[TABLE]

is a minimal ${\rm T}(2,3,v)$ trade.

• $\bullet$

Suppose that $a_{0}a_{1}\dots a_{n-1}$ and $b_{0}b_{1}\dots b_{n-1}$، are two disjoint subsets of $X$, and $3\leq n\leq\frac{v}{2}$. Let $G$ be a graph with vertices $\{a_{0}b_{0},a_{1}b_{1},\dots,a_{n-1}b_{n-1}\}$. For every cycle $\mathcal{C}$ of $G$

[TABLE]

is a ${\rm T}(2,3,v)$ trade.

In Section 2, we construct a simple ${\rm T}(2,3,v)$ trade with volume $\binom{v}{3}/2$ by using disjoint trades. For example, Let $X=\{a_{0},a_{1},a_{2},b_{0},b_{1},b_{2}\}$, and $(a_{0}b_{0},\,a_{1}b_{1},\,a_{2}b_{2},\,a_{0}b_{0})$ be a cycle of graph $G$ with vertices $\{a_{0}b_{0},a_{1}b_{1},a_{2}b_{2}\}$, then

[TABLE]

is a simple ${\rm T}(2,3,6)$ trade with volume $10$.

A simple $\rm{T}(t,k,v)$ trade with volume $\binom{v}{k}/2$ is called a $(t,k,v)$-halving. The following conjecture (known as halving conjecture) is due to Alan Hartman [5]:

Conjecture. There exists a $(t,k,v)$-halving if and only if for $0\leq i\leq t$, $\binom{v-i}{k-i}\equiv 0\pmod{2}$.

Clearly, every $(t,k,v)$-halving can be written as a linear combination of some trades with smaller volumes. In what follows, we briefly describe three methods available in the literature for constructing a $(2,3,v)$-halving. All methods employ the linear combination of trades.

• $\circ$

AK algorithm [7]. For a block $a_{0}a_{1}a_{2}$, we choose the block $b_{0}b_{1}b_{2}$ such that

[TABLE]

Then we define the trade

[TABLE]

Ajoodani and Khosrovshahi proved that the following algorithm produces a $(2,3,v)$-halving.

Although, through this algorithm the halving is constructed directly, but it does not reveal the trade-like structure of the halving.

• $\circ$

Standard recursive method [1]. By reordering the columns of $W$, we can write the reduced row echelon form of $W$ as \big{(}C\mid{\rm I}\big{)}.

Then \big{(}\frac{\rm~{}I}{-C}\big{)} is called the standard basis for the kernel of $W$ and denoted by $\mathbb{S}_{tk}(v)$.

Naturally, every $(t,k,v)$-halving is a linear combination of the columns of $\mathbb{S}_{tk}(v)$. The complicated structure of $\mathbb{S}_{tk}(v)$ does not reveal much information about the structure of $(t,k,v)$-halving and in this regard the following two conjectures remain open [1, 10].

• –

The elements of every row of $\mathbb{S}_{tk}(v)$ have the same sign.

• –

For $t>1$, the matrix $\mathbb{S}_{tk}(v)$ contains a nowhere zero row.

Nevertheless, in [2], by carefully studying $\mathbb{S}_{23}(v)$, a $(1,-1)$-vector $\eta$ is constructed recursively such that $\mathbb{S}_{23}(v)\eta=h$, and $h$ is a $(2,3,v)$-halving.

In this method $\binom{v}{3}-\binom{v}{2}$ trades are used which are not necessarily simple and every two trade are not disjoint.

• $\circ$

V10 Method [10]. Let $v=4n+2$ and $X$ is partitioned into two subsets $a_{0}a_{1}\dots a_{2n}$ and $b_{0}b_{1}\dots b_{2n}$. If $\alpha_{0}\alpha_{1}\alpha_{2}$ is a subset of $\{0,1,\dots,2n\}$, then we define the trade

[TABLE]

It is easy to show that $\mathcal{T}_{\alpha_{0}\alpha_{1}\alpha_{2}}$ is a simple trade with volume $10$ and foundation size $6$. The following summation gives a $(2,3,v)$-halving

[TABLE]

Example. The following summation is a $(2,3,10)$-halving

[TABLE]

For instance

[TABLE]

In this method, the linear combination of $\binom{\frac{v}{2}}{3}$ trades with volume $10$ and foundation size $6$ is used to construct a $(2,3,v)$-halving. Every two trades are disjoint or have exactly $4$ blocks in common.

2 Partition method

Recently, Sauskan and Tarannikov partitioned $\binom{\{1,\dots,10\}}{3}$ into $15$ disjoint minimal trades and subsequently by augmenting these trades they obtained a $(2,3,10)$-halving [9]. Then, for $v=4(2n)+2$, they constructed a $(2,3,v)$-halving by utilizing these trades and the method of combining $t$-designs. Originally, this method was used to obtain some infinite families of $t$-designs [8].

In [9], for constructing $(2,3,10)$-halving, no algorithm has been presented. Here, we construct $(2,3,10)$-halvings based on a hill climbing process.

Example 1**.**

*A ${(2,3,10)}$-halving. *

[TABLE]

We note that a $(2,3,v)$-halving exists if and only if $v=4n+2$. Clearly, by augmetation of minmal trades for $v=4(2n-1)+2$, one can not construct a $(2,3,v)$-halving. In what follows, we describe our partitioning mehtod to construct $(2,3,v)$-halvings.

Theorem 2.1**.**

For positive integer $n$, let $v=4n+2$ and suppose that $X$ is partitioned into two subsets $a_{0}a_{1}\dots a_{2n}$ and $b_{0}b_{1}\dots b_{2n}$. Let $\mathcal{C}$ be the Eulerian cycle of complete graph $G$ with vertices $\{a_{0}b_{0},a_{1}b_{1},\dots,a_{2n}b_{2n}\}$, then

[TABLE]

is a $(2,3,4n+2)$-halving.

Proof.

It is easy to check that all the blocks of $h$ are disjoint. Then by the following relation

[TABLE]

clearly $h$ is a $(2,3,4n+2)$-halving. ∎

In the following Theorem, we partition all blocks of $T(\mathcal{C})$ of Theorem 2.1 into trades with volume $6$ and volume $8$.

Theorem 2.2**.**

For positive integer $n$, let $v=4n+2$ and suppose that $X$ is partitioned into two subsets $a_{0}a_{1}\dots a_{2n}$ and $b_{0}b_{1}\dots b_{2n}$. Then

• $(i)$

Let $\,2n+1{\equiv}1,3\pmod{6}$. Consider an ${\rm STS}(2n+1)$ with elements from $\{a_{0},a_{1},\dots,a_{2n}\}$. Then

[TABLE]

is a ${\rm T}(2,3,4n+2)$ trade with volume $2\binom{2n+1}{2}$.

• $(ii)$

Let $\,2n+1{\equiv}5\pmod{6}$. Consider ${\rm STS}(2n-1)$ with elements from $\{a_{0},\dots,a_{2n-2}\}$. If $S$ is a partition of $\{a_{0},\dots,a_{2n-3}\}$ into $2$-subsets, then

[TABLE]

is a ${\rm T}(2,3,4n+2)$ trade with volume $2\binom{2n+1}{2}$.

Proof.

• $(i)$

Since the number of blocks of ${\rm STS}(2n+1)$ is $\frac{1}{3}\binom{2n+1}{2}$, then $h_{1}$ is the augmentation of $\frac{1}{3}\binom{2n+1}{2}$ trades with volume $6$. Now, by the relation

[TABLE]

$h_{1}$ is a ${\rm T}(2,3,4n+2)$ trade.

• $(ii)$

Since the number of blocks of ${\rm STS}(2n-1)$ is $\frac{1}{3}\binom{2n-1}{2}$, then $h_{2}$ is the augmentation of $\frac{1}{3}\binom{2n-1}{2}+1$ trades of volume $6$ and $n-1$ trades with of $8$. Now, by the following relation

[TABLE]

$h_{2}$ is a ${\rm T}(2,3,4n+2)$ trade.

∎

3 Some Examples

**1. $\mathbf{(2,3,14)}$-halving. ** Let $a_{i}=i$ and $b_{i}=i+7$, where $0\leq i\leq 6$. Consider ${\rm STS}(7)=\{013,026,045,124,156,235,346\}.$ Then

[TABLE]

is a $(2,3,14)$-halving. Therefore, the trade structure of $(2,3,22)$-halving is the following:

[TABLE]

**2. $\mathbf{(2,3,22)}$-halving. ** Let $a_{i}=i$ and $b_{i}=i+11$, where $0\leq i\leq 10$. Consider ${\rm STS}(9)=\{012,036,048,057,138,147,156,237,246,258,345,678\}$ and $S=\{01,23,45,67\}$. Then

[TABLE]

is a $(2,3,22)$-halving. Therefore, the trade structure of $(2,3,22)$-halving is the following:

[TABLE]

Acknowledgements

The authors thank Professor Denis Krotov of Sobolev Institute of Mathematics for mentioning the paper by Sauskan and Tarannikov [9].

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