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{\bf Gaetano Quattrocchi}
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{\bf Colouring $4$-cycle Systems with Specified Block Colour
Patterns: the Case of Embedding $P_3$-designs}
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\font\msam=msam10 \def\square{\mathbin{\hbox{\msam\char"03}}}
        A {\it colouring} of a $4$-cycle system $(V,{\cal B})$ is a
surjective mapping $\phi : V \rightarrow \Gamma$. The elements of
$\Gamma$ are {\it colours}. If $|\Gamma|=m$, we have an
$m$-{\it colouring} of $(V,{\cal B})$. For
every $B\in{\cal B}$, let $\phi(B)=\{\phi(x) | x\in B\}$.
There are seven distinct colouring patterns in which a
$4$-cycle can be coloured: type $a$ ($\times \times \times \times$,
monochromatic),
type $b$ ($\times \times \times \square$, two-coloured of pattern
$3+1$),
type $c$ ($\times \times \square \square$, two-coloured of pattern
$2+2$),
type $d$ ($\times \square \times \square$, mixed two-colored),
type $e$ ($\times \times \square \triangle$, three-coloured of
pattern $2+1+1$),
type $f$ ($\times \square \times \triangle$, mixed three-coloured),
type $g$ ($\times \square \triangle \diamondsuit$, four-coloured
or polychromatic).

Let $S$ be a subset of $\{a,b,c,d,e,f,g\}$. An
$m$-colouring $\phi$ of $(V,{\cal B})$
is said {\it of type} $S$ if
the
type of every $4$-cycle of $\cal B$ is in $S$.
A type $S$ colouring is
said to be {\it proper} if for every type $\alpha \in S$ there is
at least one $4$-cycle of $\cal B$
having colour type $\alpha$.

We say that a $P(v,3,1)$, $(W,{\cal P})$, {\it is embedded}
in a $4$-cycle system of order $n$, $(V,{\cal B})$,
if every path $p=[a_1,a_2,a_3] \in {\cal P}$ occurs
in a $4$-cycle $(a_1,a_2,a_3,x) \in {\cal B}$
such that $x \not\in W$.


In this paper we consider
the following spectrum problem:
given an integer
$m$ and a set $S \subseteq \{b,d,f\}$, determine the set of integers $n$ such that
there exists a $4$-cycle system of order $n$
with a proper $m$-colouring of type $S$
(note that each colour class of a such coloration is the point set
of a $P_3$-design {\it embedded} in the
$4$-cycle system).




We give a complete answer to the above problem except when
$S=\{b\}$. In this case the problem is completely solved
only for $m=2$.



\bye

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