\section{Support and extremal corner obstructions}\label{sec:support}
Let $M(k)$ be the least number of cubes guaranteeing a corner solution when
the collection contains at most $k$ varieties. We determine these thresholds
and classify all fourteen-cube obstructions into 56 symmetry classes. Write $K(X)$ for the set of the eight distinct cyclic corner types of $X$.

\paragraph{Directed corners.}
Write $V=\col$ and encode a tricoloured cyclic triple $(i,j,k)$ by the complementary
ordered pair $ab$ for which $(a,b,i,j,k)$ is an even permutation of $V$.
Cyclically rotating the triple preserves $ab$; reversing it gives $ba$.
For example, $013\mapsto42$ and $031\mapsto24$.
Write $b(r,p)$ for the bicoloured corner $rrp$.
The tricoloured corners of an adjacent-repeat variety form a cut
\[
 C(S)=S\times(V\setminus S),\qquad |S|\in\{2,3\}.
\]
Its smaller side is a pair $P$, its repeated colour is $r\notin P$, and
its other corners are $b(r,p)$ for $p\in P$. The pair, repeated colour and
sign determine a unique adjacent-repeat variety. Call $C(P)$ positive and
$C(V\setminus P)$ negative. For example, $(0,1,0,2,3,4)$ has cut $C(34)$
and bicoloured corners $b(0,3),b(0,4)$. Colour permutations give all adjacent
varieties; odd permutations also reverse the arcs. A string such as $34$
in $C(34)$ denotes $\{3,4\}$.
An opposite-repeat variety with singleton opposite pairs $P,Q$ has arcs
$(P\times Q)\cup(Q\times P)$, a bidirected $K_{2,2}$ omitting its repeated
colour.

The identity
\[
 C(S)\cap C(T)=(S\cap T)\times\bigl(V\setminus(S\cup T)\bigr)
\]
gives the following tricoloured intersection sizes for smaller pairs $P,Q$:
\[
\begin{array}{c|ccc}
 |P\cap Q|&2&1&0\\\hline
 \text{same sign}&6&2&0\\
 \text{opposite signs}&0&1&4
\end{array}
\]
There are additionally $|P\cap Q|$ shared bicoloured corners when the
repeated colours agree, and none otherwise. In particular, a singleton
full intersection is tricoloured and joins opposite signs with different
repeated colours. By Lemma~\ref{lem:cube-incidence}, a pair involving an
opposite-repeat variety shares at least two corners; two opposite varieties
sharing exactly two corners share one bidirected edge.

\begin{lemma}\label{lem:disjoint-pair}
Four copies of $A$, together with any four sources individually
corner-disjoint from $A$, can build an opposite-repeat target.
\end{lemma}
\begin{proof}
The anchor and all four other sources are adjacent-repeat. Reflect the entire instance
if necessary so that $A$ has cut $C(P)$, $|P|=2$, and put $R=V\setminus P$.
Each disjoint source has cut $C(R)$ or $C(Q)$ for a pair $Q\subset R$.
Choose $Q$ maximizing frequency among all three pair-cut classes,
including those of frequency zero. Four $A$ copies fill
$P\times Q$ of the opposite target with singleton pairs $P,Q$.
On $Q\times P$, sources of classes $C(Q),C(R)$ are unrestricted; each other
pair class fits one two-position row. Neither row-only class occurs more
than twice, by maximality of the chosen class. Place them in their rows
and fill the remaining positions with the unrestricted sources.
\end{proof}

Consequently, eleven sources containing seven copies of one variety solve:
one of the other four sources sharing a corner completes the seven-copy
target; otherwise all four are disjoint and the lemma applies.

\begin{samepage}
\begin{theorem}[Support thresholds]\label{thm:support}
For every integer $k\ge1$,
\[
 M(k)=\begin{cases}8&k=1,\\13&k=2,\\15&k\ge3.\end{cases}
\]
Three is therefore the minimum number of varieties in a fourteen-cube obstruction.
\end{theorem}
\end{samepage}
\begin{proof}
The one-variety case is immediate. Thirteen sources in two varieties include
seven of one variety, so the preceding consequence gives the upper bound.
For the lower bound take six copies each of the original $B,C$ from
Lemma~\ref{lem:lower}. Their corner union supports only $B,C$: removing
$A$ from that lemma's three-variety union cannot introduce another target,
and $A$ has five corners absent from both. The sole shared corner is $042$,
so either remaining target requires seven copies of its own variety.
Thus $6B+6C$ fails. For $k\ge3$, Theorem~\ref{thm:main} and
Lemma~\ref{lem:lower} give the matching bounds.
\end{proof}

\begin{theorem}[Multiplicity bound]\label{thm:multiplicity-five}
Every fourteen-cube obstruction has maximum multiplicity at most five.
\end{theorem}
\begin{proof}
The seven-copy case follows above. Suppose there are exactly six copies
of $A$ and eight other sources. Two others fitting distinct $A$ corners
complete $A$; four disjoint others invoke Lemma~\ref{lem:disjoint-pair}.
In a failure, therefore, at most three others are disjoint and at least
five meet $A$, all in the same singleton corner $k$.

The singleton intersection implies that $A$ is adjacent-repeat, since
an opposite-repeat source shares at least two corners with every variety.
Normalize the cut of $A$ to $C(P)$ with $P=\{p,p'\}$, and put $R=V\setminus P$.
The intersection table makes $k=pq$ for some $q\in R$ and forces every
meeting source to have cut
\[
 C(T),\qquad T=\{p\}\cup(R\setminus\{q\}).
\]
There are at most three varieties with this common cut. Reserve two
identical $B$ copies for its bicoloured corners, set aside any three further
common-cut copies, and reserve one $A$ copy for $k$. A disjoint source's
intersection with $C(T)$ is
\[
\begin{array}{c|c}
 \text{source cut}&\text{intersection with }C(T)\\\hline
 C(R)&(R\setminus\{q\})\times\{p'\}\\
 C(R\setminus\{q\})&(R\setminus\{q\})\times\{p',q\}\\
 C(\{q,r\}),\ r\in R\setminus\{q\}&\{rp'\}
\end{array}
\]
and always avoids $k$. Exactly three of the eight non-$A$ copies remain.
Each meets $C(T)\setminus\{k\}$: a common-cut copy fits all five positions,
and the displayed intersections for disjoint sources are nonempty.
If two remaining copies fit distinct corners there, use them and fill the
other three positions with the set-aside common-cut copies. Together with
the two reserved $B$ copies and the $A$ at $k$, this completes $B$.

Otherwise the three nonempty compatibility sets are one common singleton.
Indeed, a set containing two corners can be paired at distinct corners
with any other nonempty set; two different singleton sets also give such
a pair. None of the three remaining copies therefore has the common cut.
They are all disjoint from $A$, and the table identifies their shared singleton
as $rp'$. They hence have cut $C(Q)$ with $Q=\{q,r\}$. Retarget to the opposite
variety with singleton pairs $P,Q$: four $A$ copies fill $P\times Q$;
three disjoint sources and one $B$ at $rp'$ fill $Q\times P$.
Every case gives an assembly, independently of the fifteen-cube theorem.
\end{proof}

For classification, equivalence permits common colour relabelling and
simultaneous reflection of the entire collection. Reflection is not an
additional move for an individual source. In directed notation this is
arbitrary vertex relabelling and independent global arc reversal.

\paragraph{The generalized family.}
Let $A$ be the opposite variety with singleton pairs $\{1,2\},\{3,4\}$.
For $r\in\{2,3,4\}$, let $B_r$ have cut $C(01)$ and repeat $r$;
for $s\in\{1,2,4\}$, let $C_s$ have cut $C(124)$ and repeat $s$.
In the notation of Lemma~\ref{lem:lower}, $B_2=B$ and $C_1=C$.

\begin{samepage}
\begin{corollary}\label{cor:extremal-family}
With the same $A,B_r,C_s$, every collection
\[
 4A+\sum_{r\in\{2,3,4\}}x_rB_r+\sum_{s\in\{1,2,4\}}y_sC_s,
 \qquad \sum_r x_r=\sum_s y_s=5,
\]
where the counts are nonnegative integers, is an obstruction if and only if
$x_2y_2=x_4y_4=0$.
\end{corollary}
\end{samepage}
\begin{proof}
When $x_2y_2=x_4y_4=0$, the two groups have disjoint repeated-colour supports.
Their tricoloured union is contained in that of $A$, $C(01)$ and $C(124)$,
which misses the arcs $01,21,30,34$. Its bidirected edges are
$02,04,13,14,23,24$. No four-cycle passes through $0$, since its neighbours
$2,4$ have no second common neighbour; on the other four vertices the only
four-cycle is $A$. Thus $A$ is the only covered opposite-repeat target.
For an adjacent target, the bicoloured corners determine an active $B_r$ or
$C_s$, up to reversing its cut. The reversed cuts require absent arcs
$21,01$, respectively. Each remaining target has the following positions
private to its indicated source group:
\[
\begin{array}{c|l|c|c}
 \text{target}&\text{private positions}&\text{demand}&\text{supply}\\\hline
 A&24,31,32,41,42&5&4\\
 B_r&02,03,04,12\text{ and both bicoloured corners}&6&5\\
 C_s&10,20,40,43\text{ and both bicoloured corners}&6&5
\end{array}
\]
All targets therefore fail. Missing varieties within either group can only
remove suppliers.
Conversely, suppose both groups contain repeat $r\in\{2,4\}$. Build $B_r$:
reserve one $B_r$ at $b(r,1)$ and one $C_r$ at $b(r,0)$; the other four
$B$-group copies fill $02,03,04,12$, and two $A$ copies fill $13,14$.
These are eight distinct sources, giving a solution.
Taking $y_1=5$ gives a 21-member
subfamily. The general family also contains, for example,
$4A+2B_2+3B_3+2C_1+3C_4$, whose maximum multiplicity is four.
\end{proof}

\input{classification}

\begin{theorem}[Three-variety obstructions]\label{thm:three-variety-classification}
Up to this equivalence, the fourteen-cube obstructions using exactly three
varieties are precisely five collections $4A+5B_r+5C_s$. The varieties are those of Corollary~\ref{cor:extremal-family};
Class I is the original illustrated obstruction. Representatives are
\[
\begin{array}{c|ccccc}
 \text{class}&\mathrm{I}&\mathrm{II}&\mathrm{III}&\mathrm{IV}&\mathrm{V}\\\hline
 r&2&2&3&3&4\\
 s&1&4&1&2&2
\end{array}
\]
\end{theorem}
\begin{proof}
The other two forms in Theorem~\ref{thm:classification14} have six varieties.
A three-variety member of the first form must concentrate each five-copy
group on one variety, giving $4A+5B_r+5C_s$ with $r\ne s$.
Corollary~\ref{cor:extremal-family} proves failure for all seven such pairs.
A symmetry between normalized collections fixes the unique
four-copy variety $A$, hence fixes $0$ and preserves its two unordered
pairs. Without arc reversal it fixes both cuts, hence every vertex.
With reversal it exchanges the cuts and is $\sigma=(1\ 3)(2\ 4)$.
It sends $(r,s)$ to $(\sigma(s),\sigma(r))$, pairing $(2,1)$ with $(3,4)$
and $(3,2)$ with $(4,1)$, while fixing $(2,4),(3,1),(4,2)$.
This gives exactly the five inequivalent representatives displayed.
\end{proof}
This displayed proof uses the complete classification. The formal endpoint
instead has an independent proof without interval certificates
(Appendix~\ref{app:extremal}); the repository also retains the elementary
necessity argument in \path{paper/three-variety-necessity.tex}.