Some Problems and Conjectures About Voloshin Triple Systems

Abstract

In this paper we give a short survey of Voloshin Triple Systems, which are Steiner Triple Systems with a vertex colouring which colour the blocks using exactly two colours. We also provide two conjectures about $VTSs$ obtained by the construction $v⟶2v+1$, starting from $v=3$. Finally, we point out some open problems about Steiner Systems $S(2,4,v)$ with a similar vertex colouring, which could provide similar conjectures.

1. Introduction

A hypergraph of order v is a pair $\mathcal{H}=(X,\mathcal{E})$, where X is a v-set and $\mathcal{E}$ is a family of subsets of X such that the following is true (see [1,2,3]):

(1)

$E_i\subseteq X,E_i\ne \varnothing$, for every $i\in I$, I set of indices;

(2)

${\bigcup }_{i\in I}{E}_i=X$.

The elements of $X=\{x_1,x_2,\dots ,x_v\}$ are called the vertices of $\mathcal{H}$. The elements of $\mathcal{E}$ are called the edges of $\mathcal{H}$. A hypergraph $\mathcal{H}=(X,\mathcal{E})$ is said uniform of rank k, if $\left|E\right|=k$ for every $E\in \mathcal{E}$ (see [1,2,3]). If $Y\subseteq X$, the degree $d\left(Y\right)$ of Y is the number of edges of $\mathcal{H}$ containing Y (see [2]).

If $\mathcal{H}=(X,\mathcal{E})$ is a hypergraph, a colouring of $\mathcal{H}$ is a mapping $f:X⟶C$, where $C=\{1,2,\dots ,k\}$ is a set of colours, such that $\forall E\in \mathcal{E},\exists x,y\in X:f\left(x\right)\ne f\left(y\right).$

The chromatic number of $\mathcal{H}$ is defined as the minimum number $\chi \left(H\right)$ of colours needed to define a colouring of $\mathcal{H}$. A strong colouring of $\mathcal{H}$ is a mapping $f:X⟶C$, such that $\forall E\in \mathcal{E},\forall x,y\in X,x\ne y\to f\left(x\right)\ne f\left(y\right).$

The strong chromatic number of $\mathcal{H}$ is defined as the minimum number ${\chi }_f\left(H\right)$ of colours needed to define a strong colouring of $\mathcal{H}$.

In any type of colouring defined in a hypergraph $\mathcal{H}=(X,\mathcal{E})$, a colouring class of $\mathcal{H}$ is a set of all the vertices to which the same colour is assigned. Clearly, the family of all the colouring classes is a partition of the X.

In 1992, V. Voloshin introduced the concept of a mixed hypergraph (see [4,5] and also [6,7,8,9]), giving new directions for research in the colourings of hypergraphs.

The field of mixed hypergraphs is a well-established subarea of combinatorics and graph theory that has been actively studied since at least the early 1990s with the foundational work of Vitaly Voloshin. Since then, hundreds of research articles, surveys, and related publications have appeared in journals, conferences, proceedings, and monographs.

To date, there are more than 300 publications in mixed hypergraphs, see https://spectrum.troy.edu/voloshin/publishe.html (accessed on 6 November 2025).

The following definitions were all given by V. Voloshin.

(1)

“A mixed hypergraph is a triple $\mathcal{H}=(X;C,D)$, where X is a finite nonempty set and $C,D$ are two families of nonempty subsets of X called C-edge and D-edges respectively. The elements of X are the vertices. If $C=\varnothing ,D\ne \varnothing$, $\mathcal{H}$ is a D-hypergraph (D-type hypergraph). If $D=\varnothing ,C\ne \varnothing$, $\mathcal{H}$ is a C-hypergraph (C-type hypergraph). If $C=D$, then $\mathcal{H}$ is a bihypergraph”.

(2)

A k-colouring of a mixed hypergraph $\mathcal{H}=(X;C,D)$ is a mapping $f:X⟶\{1,2,\dots ,k\}$ such that

$$\forall E\in C,\exists x,y\in X,x\ne y:f\left(x\right)=f\left(y\right),$$

$$\forall E\in D,\exists x,y\in X:f\left(x\right)\ne f\left(y\right).$$

(3)

A mixed hypergraph $\mathcal{H}$ is k-colourable if there exists a k-colouring of $\mathcal{H}$, using at most k colours. A mixed hypergraph $\mathcal{H}$ is uncolourable if it does not admit any colouring. A strict k-colourings is a proper colouring using exactly k colours.

(4)

The minimum number of colours in a colouring of $\mathcal{H}$ is called the lower chromatic number of $\mathcal{H}$ and is denoted by $\underline{\chi }\left(\mathcal{H}\right)$. The maximum number of colours in a colouring of $\mathcal{H}$ is called the upper chromatic number of $\mathcal{H}$ and is denoted by $\overline{\chi }\left(\mathcal{H}\right)$.

If $k,v$ are positive integers, $1\le k\le v$, $\left|X\right|=v$, then $r_k$ denotes the number of colouring classes which partition X, such that the colouring constraint is satisfied on each C-edge and on each D-edge. Exactly, $r_k$ is the number of different k-colourings of $\mathcal{H}$, ignoring permutations of colours.

(5)

The vector $R\left(\mathcal{H}\right)=(r_1,r_2,\dots ,r_v)$ is called the chromatic spectrum of $\mathcal{H}$.

(6)

The set of values k such that $\mathcal{H}$ has a strict k-colouring is called the feasible set of $\mathcal{H}$, denoted by $S\left(\mathcal{H}\right)$. In other words, $S\left(\mathcal{H}\right)$ is the set of indices k such that $r_k>0$.

(7)

A mixed hypergraph $\mathcal{H}$ has a gap at k if $S\left(\mathcal{H}\right)$ contains values which are larger and smaller than k, but omits k. The chromatic spectrum $R\left(\mathcal{H}\right)$ is called continuous or gap free if $S\left(\mathcal{H}\right)$ has no gaps. Otherwise, it is called broken.

A Steiner system $S(h,k,v)$ is a pair $\mathrm{\Sigma }=(X,\mathcal{B})$, where X is a v-set and $\mathcal{B}$ is a family of k-subsets of X such that every h-subset of X is contained in exactly one member of $\mathcal{B}$ [2,10,11]. From definition it follows that $1\le h\le k\le v$, $h,k,v\in N$. Observe that using hypergraph theory terminology, a Steiner system is a hypergraph $\mathrm{\Sigma }=(X,\mathcal{B})$ of order v, uniform of rank k, such that every h-subset Y of X has degree $d\left(Y\right)=1$ [2]. In Design Theory the edges of $\mathrm{\Sigma }$ are also called blocks. A Steiner Triple System (STS) is a system $S(2,3,v)$, and an $STS\left(v\right)$ exists if and only if $v\equiv 1$ or 3, mod 6. A system $S(2,2,v)$ is the complete graph ${\mathcal{K}}_v$.

In a Steiner system $S(h,k,v)$$\mathrm{\Sigma }=(X,\mathcal{B})$, a parallel class is a set of blocks that partition X. $\mathrm{\Sigma }$ is called resolvable if $\mathcal{B}$ can be partitioned into parallel classes and the partition is called resolution. A resolvable $STS\left(v\right)$ is also known as a Kirkman Triple System of order v, or $KTS\left(v\right)$. A $KTS\left(v\right)$ exists if and only if $v\equiv 3$ mod 6. A system $S(2,2,v)$ is the complete graph ${\mathcal{K}}_v$. A parallel class in an $S(2,2,v)$ is called factor and a resolution is a factorisation.

In the literature there are many constructions to obtain an $STS$, starting from a given $STS\left(v\right)$. Among them, there is the well-known construction, indicated by $v⟶2v+1$, which gives an $STS(2v+1)$ starting from an $STS\left(v\right)$. There are other constructions of type $v⟶3v$.

It is also possible to consider $S(h,k,v)$ as mixed hypergraphs, considering some blocks as C-edges and the others as D-edges.

There are very interesting problems when we consider $STS\left(v\right)s$ in which the blocks are both C-edges and D-edges. Until today, such a system has been called Bi-Steiner Triple Systems and has been denoted by $BSTS\left(v\right)$. From now on we will call these system Voloshin Triple Systems, giving them the name of their originator.

A Voloshin Triple System of order v [$VTS\left(v\right)$] is a Steiner Triple System $\mathrm{\Sigma }=(X,\mathcal{B},\varphi )$, where X is the set of vertices, $\mathcal{B}$ the set of blocks, and $\varphi$ is a vertex colouring of $\mathrm{\Sigma }$ such that $\forall B\in \mathcal{B},\left|\varphi \right(B\left)\right|=2,$ in other words,

$$\forall B=x,y,z\in \mathcal{B},\exists x,y\in B:\varphi \left(x\right)=\varphi \left(y\right),\exists y,z\in B:\varphi \left(y\right)\ne \varphi \left(z\right).$$

Consequently, the lower chromatic number and the upper chromatic number of $\mathrm{\Sigma }$ are defined.

Voloshin Triple Systems, as they are described in this article, are very significant relative to the classical $STS$s, because they have exceptional combinatorial properties. For example, consider the following important sentence:

$$“\mathit{an} \mathit{STS}\left(\mathit{15}\right) \mathit{is} \mathit{colourable} \mathit{if} \mathit{and} \mathit{only} \mathit{if} \mathit{it} \mathit{is} \mathit{a} \mathit{VTS}\left(\mathit{15}\right)”.$$

In this paper we examine $VTS$s, as bicolourings of $STS$ considered as mixed hypegraphs, and in particular, we consider the so-called extended colourings, for which interesting results can be found in [12,13,14,15].

2. Construction v ⟶ 2v + 1

It is well-known that it is always possible to construct an $STS(2v+1)$ starting from an $STS\left(v\right)$ [2,10,11].

Theorem 1. 

If $\mathrm{\Sigma }=(X,\mathcal{B})$ is an $STS\left(v\right)$, then there exists an $STS(2v+1)$ embedding Σ.

Proof. 

Let $\mathrm{\Sigma }=(X,\mathcal{B})$ be an STS(v) defined on $X=\{x_1,x_2,\dots ,x_v\}$.

• Further, let

-

$Y=\{y_1,y_2,\dots ,y_{v+1}\}$ be a set of cardinality $v+1$ (even number) such that $X\cap Y=\varnothing$;

-

$\mathcal{F}=\{F_1,F_2,\dots ,F_v\}$ be a factorisation of the complete graph $K_{v+1}$ defined in Y;

-

$\phi :X⟶\mathcal{F}$ be any bijection from X into $\mathcal{F}$.

Define the hypergraph ${\mathrm{\Sigma }}^{\prime }=(X^{\prime },{\mathcal{B}}^{\prime })$ as follows: $X^{\prime }=X\cup Y$, ${\mathcal{B}}^{\prime }=\mathcal{B}\cup \mathrm{\Gamma }$, where $\mathrm{\Gamma }=\left\{\right\{x,y,z\}:x\in X,y,z\in \phi (x\left)\right\}.$

It can be verified that ${\mathrm{\Sigma }}^{\prime }$ is an $STS(2v+1)$ (see [2,10,11]). □

3. Results About VTS

Regarding $VTSs$ (or $BSTS$), the following results have been proved in [16].

Theorem 2. 

Let $\mathrm{\Sigma }=(X,\mathcal{B})$ be a $VTS$ of order v.

(1) 

There exists in Σ exactly one colouring class having odd cardinality;

(2) 

If $v\le 2^h-1$, then $\overline{\chi }\left(\mathrm{\Sigma }\right)\le h$;

(3) 

If $v\le 2^h-1$ and $\overline{\chi }\left(\mathrm{\Sigma }\right)=h$, then $v=2^h-1$ and the colouring classes has cardinality: $2^0,2^1,2^2,\dots ,2^{h-1}$;

(4) 

If $v>3$, Σ is obtained from a $VTS\left(3\right)$ by repeated constructions $v⟶2v+1$.

Further, the following characterisation is proved in [15]:

Theorem 3. 

Let $\mathrm{\Sigma }=(X,\mathcal{B})$ be a $VTS$ of order v. It is $\overline{\chi }\left(\mathrm{\Sigma }\right)=h$ if and only if Σ is obtained from a $VTS\left(3\right)$ by repeated constructions $v⟶2v+1$.

4. Some Results and Two Conjectures

Following the results of the previous section, in [16], using computer-assisted search, the following was proved:

Theorem 4. 

If $\mathrm{\Sigma }=(X,\mathcal{B})$ is a $VTS$ of order $v=2^h-1$, obtained by a sequence of $v⟶2v+1$ constructions, starting from a $VTS\left(3\right)$, then $\underline{\chi }\left(\mathrm{\Sigma }\right)=\overline{\chi }\left(\mathrm{\Sigma }\right)=h$, for every $h<10$.

The following result proves that the situation described in the previous Theorem does not always occur [17].

Theorem 5. 

A $VTS\left(15\right)$ is colourable only if it can be obtained by a sequence $v\to 2v+1$ starting from $VTS\left(3\right)$ and it is: $\underline{\chi }=\overline{\chi }=4$.

Many other important results have been found by the authors in [17]. We cite some of them.

Theorem 6. 

(1) 

If a $VTS\left(19\right)$ is colourable, it can be coloured only by one of the colourings with a colouring class of size $(1,4,4,10),(1,2,8,8),(4,6,9)$;

(2) 

There exist $VTS\left(19\right)$ having $\underline{\chi }=\overline{\chi }=3$, $VTS\left(19\right)$ having $\underline{\chi }=\overline{\chi }=4$, $VTS\left(19\right)$ having $\underline{\chi }=3$ and $\overline{\chi }=4$, and $VTS$ uncolourable;

(3) 

The smallest order for which there exists a $VTS$ with $\underline{\chi }<\overline{\chi }$ is $v=19$.

After these results, we have formulated the following two conjectures [18], subject of a main lecture in the conference International Symposium on Graphs, Designs and Application, Messina, 30 September–4 October 2003. These conjectures are very difficult to prove or disprove, and they go further in deep exploration of the structure of VTSs.

(1)

Conjecture [M. Gionfriddo 2002]:

For all $VTS\left(v\right)$ of order $v=2^h-1$, obtained by a sequence of $v⟶2v+1$, starting from a $VTS\left(3\right)$, it is: $\underline{\chi }\left(\mathrm{\Sigma }\right)=\overline{\chi }\left(\mathrm{\Sigma }\right)=h.$

(2)

Conjecture [M. Gionfriddo 2003]:

All colourable $VTSs$ do not have a broken chromatic spectrum.

Observe that in Conjecture 2002, the initial system is a $VTS\left(3\right)$, which has only one block, necessarily colourable with two colours: one associated with one vertex and one associated with the remaining two vertices. Therefore, if $v=3⟶\underline{\chi }=\overline{\chi }=2$, and the cardinality colouring classes are $2^0,2^1$.

If $v=7$, the $VTS\left(7\right)$ is obtained by a system $VTS\left(3\right)$ having only a block $\{1^{\prime },2^{\prime },3^{\prime }\}$, a factorisation $\mathcal{F}=\{F_1,F_2,F_3\}$, where $F_1=\{\{1,2\},\{3,4\}\},F_2=\{\{1,3\},\{2,4\}\},F_3=\{\{1,4\},\{2,3\}\}$, and associating $i^{\prime }$ with $F_i$ (this does not harm the generality) provides the following blocks:

$$\{1^{\prime },1,2\},\{2^{\prime },1,3\},\{3^{\prime },1,4\},$$

$$\{1^{\prime },3,4\},\{2^{\prime },2,4\},\{3^{\prime },2,3\}.$$

It can be verified that, necessarily, this $VTS\left(7\right)$ can be coloured only giving a colour C, different from $A,B$, to all the vertices $1,2,3,4$. It follows that

$$v=7⟶\underline{\chi }=\overline{\chi }=3,\mathrm{colouring}\mathrm{classe}\mathrm{cardinalities}:2^0,2^1,2^2.$$

Similarly, it can be verified that

$$v=15⟶\underline{\chi }=\overline{\chi }=4,\mathrm{colouring}\mathrm{classe}\mathrm{cardinalities}:2^0,2^1,2^2,2^3.$$

The Conjecture 2002 states that

$$v=2^h-1⟶\underline{\chi }=\overline{\chi }=h,\mathrm{colouring}\mathrm{classe}\mathrm{cardinalities}:2^0,2^1,2^2,\dots ,2^{h-1}.$$

5. Some Particular Cases

It is clear that the system obtained by the construction $v⟶2v+1$ overall depends on the choice of the factorisation defined in Y, $\left|Y\right|=v+1$. We see a case where this choice allows us to construct systems that verify Conjecture 2002. In what follows, given a vertex x, the symbolism $x⟶A$ will mean that the colour A is assigned to the vertex x, while $x⟶(A,B)$ will mean that the vertex x can be coloured only by A or by B, and finally, $(x,y,\dots ,z)⟶A$ will mean that all the vertices $(x,y,\dots ,z)$ can be coloured only by A. Following the symbolism and terminology of Section 2, consider the construction $v⟶2v+1$, where

(1)

CASE v = 3:

-

$X=\{x_1,x_2,x_3\},Y=\{1,2,3,4\}$, $X\cap Y=\varnothing$;

-

$\mathrm{\Sigma }=(X,\mathcal{B})$ is a $VTS$ of order $v=3$, where $\mathcal{B}=\left\{B\right\},B=\{x_1,x_2,x_3\}$;

-

f is a vertex colouring defined in $\mathrm{\Sigma }$ such that $f\left(x_1\right)=A,f\left(x_2\right)=f\left(x_3\right)=B$, where $A,B$ are colours;

-

$\mathcal{F}=\{F_1,F_2,F_3\}$ is a factorisation defined in Y;

-

$\varphi :X⟶\mathcal{F}$ is a bijection such that $\varphi \left(x_i\right)=F_i$ for every $i=1,2,3$;

-

${\mathrm{\Sigma }}^{\prime }=(X^{\prime },{\mathcal{B}}^{\prime })$ is a $VTS\left(7\right)$ obtained by construction $v⟶2v+1$.

It follows that

Theorem 7. 

If $\mathcal{F}=\{F_1,F_2,F_3\}$ is a factorisation defined on Y as follows:

$$F_1=\{\{1,2\},\{3,4\}\},$$

$$F_2=\{\{1,3\},\{2,4\}\},$$

$$F_3=\{\{1,4\},\{2,3\}\},$$

then $\underline{\chi }\left({\mathrm{\Sigma }}^{\prime }\right)$ =$\overline{\chi }\left({\mathrm{\Sigma }}^{\prime }\right)=3$ and the colouring classes have cardinality $2^0,2^1,2^2.$

Proof. 

The vertices of the only block of $\mathrm{\Sigma }$ have colours $f\left(x_1\right)=A,f\left(x_2\right)=B,f\left(x_3\right)=B.$ The blocks of ${\mathrm{\Sigma }}^{\prime }$ are $\{x_1,x_2,x_3\}$ and

$$\{x_1,1,2\},\{x_2,1,3\},\{x_3,1,4\},$$

$$\{x_1,3,4\},\{x_2,2,4\},\{x_3,2,3\}.$$

Note that it is not possible to colour two distinct vertices of Y with two different colours $C,D$, such that $\{A,B\}\cap \{C,D\}=\varnothing$. Therefore, if we define an extension of the colouring f to $X\cup Y$ by setting $f\left(1\right)=f\left(2\right)=f\left(3\right)=f\left(4\right)=C$ (C colour different from $A,B$), then we have a colouring of ${\mathrm{\Sigma }}^{\prime }$ with three colours. Hence, $\overline{\chi }=3$ and the three colouring classes have the cardinality $2^0,2^1,2^2$.

Now, we see that it is not possible to colour the vertices of Y using only the colours $A,B$.

Indeed, suppose that $f\left(Y\right)=\{A,B\}$. Necessarily, a pair $\{x,y\}\in F_1$ is coloured by $A,B$. Let $1⟶A$, $2⟶B$. Therefore, from $F_2$ it is $4⟶\left(A\right)$, and from $F_3$ it is $3⟶A$. But this implies $\{x_1,3,4\}⟶A$, with a monochromatic block. Therefore, $\underline{\chi }=3$.

Now, we see that, in any case, the colouring classes have cardinality $2^0,2^1,2^2$.

Let 1 be a vertex of Y of colour C. From $F_1,F_2,F_3$, respectively, it follows $2⟶(A,C)$, $3⟶(B,C)$, $4⟶(B,C)$. Therefore,

(1)

If $2⟶A$, then from $F_2$ it is $4⟶B$ and from $F_3$ $3⟶B$. Finally, $C_A=\{x_1,2\},$$C_C=\left\{1\right\}$.

(2)

If $2⟶C$, then $(3,4)⟶B$ or $(3,4)⟶C$, and in both cases the colouring classes have cardinality $2^0,2^1,2^2$.

□

(2)

CASE v = 7:

-

$X=\{x_1,x_2,\dots ,x_7\},Y=\{1,2,\dots ,8\}$, $X\cap Y=\varnothing$;

-

$\mathrm{\Sigma }=(X,\mathcal{B})$ is a $VTS$ of order $v=7$;

-

f is a vertex colouring defined in X such that $f\left(x_1\right)=A,f\left(x_2\right)=f\left(x_3\right)=B,f\left(x_4\right)=f\left(x_5\right)=f\left(x_6\right)=f\left(x_7\right)=C$;

-

$\mathcal{F}=\{F_1,F_2,\dots ,F_7\}$ is a factorisation defined in Y;

-

$\varphi :X⟶\mathcal{F}$ is a bijection such that $\varphi \left(x_i\right)=F_i$ for every $i=1,2,\dots ,8$;

-

${\mathrm{\Sigma }}^{\prime }=(X^{\prime },{\mathcal{B}}^{\prime })$ is the $VTS\left(15\right)$ obtained by construction $v⟶2v+1$.

Theorem 8. 

If $\mathcal{F}=\{F_1,F_2,\dots ,F_7\}$ is a factorisation on Y defined as follows:

$$F_1:\{1,2\},\{3,4\},\{5,6\},\{7,8\}$$

$$F_2:\{1,3\},\{2,4\},\{5,7\},\{6,8\}$$

$$F_3:\{1,4\},\{2,3\},\{5,8\},\{6,7\}$$

$$F_4:\{1,5\},\{2,6\},\{3,7\},\{4,8\}$$

$$F_5:\{1,6\},\{2,7\},\{3,8\},\{4,5\}$$

$$F_6:\{1,7\},\{2,8\},\{3,5\},\{4,6\}$$

$$F_7:\{1,8\},\{2,5\},\{3,6\},\{4,7\}$$

and $f\left(x_1\right)=A,f\left(x_2\right)=f\left(x_3\right)=B,f\left(x_4\right)=f\left(x_5\right)=f\left(x_6\right)=f\left(x_7\right)=C,$ then $\underline{\chi }\left({\mathrm{\Sigma }}^{\prime }\right)=\overline{\chi }\left({\mathrm{\Sigma }}^{\prime }\right)=4.$

Proof. 

First of all, observe that it is not possible that two vertices of Y can be coloured by two distinct colours $D,E$ such that $\{A,B\}\cap \{C,D\}=\varnothing$. Indeed, if $x,y\in Y$, $\{x,y\}\in F_j\in \mathcal{F}$, the block $\{x_j,x,y\}$ [where $f\left(x_j\right)\in \{A,B\},f\left(x\right)=C,f\left(y\right)=D$] would have three vertices with three distinct colours.

Therefore, since it is possible to colour all the vertices of Y by a new colour D, it follows that $\overline{\chi }\left({\mathrm{\Sigma }}^{\prime }\right)=4$.

Now, suppose that $f\left(Y\right)\subseteq \{A,B,C\}$. If ${\mathcal{C}}_A,{\mathcal{C}}_B,{\mathcal{C}}_C$ are the colouring classes in Y, associated, respectively, with the colours $A,B,C$, then $|{\mathcal{C}}_A|+|{\mathcal{C}}_B|+|{\mathcal{C}}_C|=|Y|=8$.

Prove that

$$|{\mathcal{C}}_A|\le 4,|{\mathcal{C}}_B|\le 4,|{\mathcal{C}}_C|\le 4.$$

Indeed, if for some $U\in \{A,B,C\}$ it were $|{\mathcal{C}}_U|\ge 5$, there would be a factor $F_j$ with $f\left(x_j\right)=U$, containing a pair of vertices $\{x,y\}$ both coloured with U. But this is not possible due to the existence of the monochromatic block $\{x_j,x,y\}$.

(1)

Let $|{\mathcal{C}}_C|=4$. Necessarily, since all the pair of vertices both coloured by C are contained in $F_1\cup F_2\cup F_3$, it can be only ${\mathcal{C}}_C=\{1,2,3,4\}$ or $\{5,6,7,8\}$. Without loss of generality, we can suppose that ${\mathcal{C}}_C=\{1,2,3,4\}$ and $\{5,6,7,8\}\subseteq {\mathcal{C}}_A\cup {\mathcal{C}}_B$, and, since at least a vertex $x\in \{5,6,7,8\}$ must be coloured by A (or by B), let $5⟶A$ (or B).

This implies $6⟶B$, otherwise $\{x_1,5,6\}⟶A$. Hence, from the existence of the blocks $\{x_2,6,8\},\{x_3,6,7\}$, it follows that, respectively, $8⟶A,7⟶A$. But this is not possible due the existence of the monochromatic block $\{x_1,7,8\}$.

(2)

Let $|{\mathcal{C}}_C|=3$. To avoid monochromatic blocks with vertices all coloured by C, without loss of generality we suppose $1⟶C,2⟶C,3⟶C$. This from $F_1$ implies $4⟶A$ and from $F_2$ (or $F_3$) implies $4⟶B$. But this is not possible.

(3)

Let $|{\mathcal{C}}_C|=2$. Similarly to the previous cases, we can suppose that the two vertices coloured by C belong to $\{1,2,3,4\}$. If $(1,3)⟶C$, then from $F_1$ it follows that $4⟶A$ and from $F_3$ it follows that $4⟶B$, and this is not possible. If $(1,2)⟶C$, then from $F_2$ and $F_3$ it follows that $(3,4)⟶B$. Hence, from $F_4,F_5,F_6,F_7$: $(5,6,7,8)⟶B$. But this implies the existence of the following monochromatic blocks $\{x_2,5,7\},\{x_2,6,8\},\{x_3,5,8\},\{x_3,6,7\}$, and this is not acceptable.

(4)

Let $|{\mathcal{C}}_C|=1$. Suppose $1⟶C$. From $F_1,F_2,F_3$, it follows immediately that $2⟶A,3⟶B,4⟶B$. Further, from $F_5$ it follows that $7⟶A,8⟶B,5⟶B$. But the existence of the pair $\{2,8\}\in F_6$ implies that the block $\{x_6,2,8\}$ contains three vertices coloured with $A,B,C$.

(5)

Let $|{\mathcal{C}}_C|=0$. This implies (considering the factors) that $|{\mathcal{C}}_A|=|{\mathcal{C}}_B|=4$. From $F_1$ it follows that $(1,2)⟶(A,B)$ and from $F_2$ and $F_3$ it follows that $(3,4)⟶\left(A\right)$. But this means that the block $\{x_1,3,4\}$ is monochromatic, with all the vertices coloured by A.

From (1), (2), (3), (4), (5) it follows that $\underline{\chi }\left({\mathrm{\Sigma }}^{\prime }\right)=4$. □

6. New Open Problems

In this last section we consider an $S(2,4,v)$$\mathrm{\Sigma }=(X,\mathcal{B})$, in which it is defined a vertex-colouring $\varphi :⟶C$ such that

$$\forall B=\{x,y,z,t\}\in \mathcal{B},\left|\varphi \right(B\left)\right|=2\mathrm{and}\exists x,y,z\in B:\varphi \left(x\right)=\varphi \left(y\right)=\varphi \left(z\right).$$

in other words, every block is bicoloured, with three vertices having the same colour.

It is well-known, for $S(2,4,v)$, the following construction $v⟶3v+1$ (see [2]).

Observe that if $v\equiv 1$ or 4 mod 12, then $2v+1\equiv 3$ mod 6, and then there exist $KTS(2v+1)$.

Theorem 9. 

If $\mathrm{\Sigma }=(X,\mathcal{B})$ is an $S(2,4,v)$, then there exists an $S(2,4,3v+1)$ embedding Σ.

Proof. 

Let $\mathrm{\Sigma }=(X,\mathcal{B})$ be an $S(2,4,v)$ defined on $X=\{x_1,x_2,\dots ,x_v\}$.

Further, let

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$Y=\{y_1,y_2,\dots ,y_{2v+1}\}$ be a set of cardinality $2v+1$ such that $X\cap Y=\varnothing$;

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$\mathrm{\Omega }=(Y,\mathcal{C})$ be a $KTS(2v+1)$, where $\mathrm{>\Pi }=\{C_1,C_2,\dots ,C_v\}$ is a resolution of $\mathrm{\Omega }$;

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$\phi :X⟶\mathrm{\Pi }$ be any bijection from X into $\mathrm{\Pi }$.

If ${\mathrm{\Sigma }}^{\prime }=(X^{\prime },{\mathcal{B}}^{\prime })$ is a hypergraph defined as follows:

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$X^{\prime }=X\cup Y$;

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${\mathcal{B}}^{\prime }=\mathcal{B}\cup \mathrm{\Gamma }$, where $\mathrm{\Gamma }=\left\{\right\{x,y,z,t\}:x\in X,y,z,t\in \phi (x\left)\right\};$

• then it is possible to verify that ${\mathrm{\Sigma }}^{\prime }$ is an $S(2,4,3v+1)$ (see [2]). □
• Open Problems:

(1)

Determine the lower chromatic number $\underline{\chi }$ and the upper chromatic number $\overline{\chi }$ of $S(2,4,v)$, starting for $v=4$ and for systems obtained by the construction $v⟶3v+1$.

(2)

Determine the lower chromatic number $\underline{\chi }$ and the upper chromatic number $\overline{\chi }$ of $S(2,4,v)$.