\section{Rectangular boxes}\label{sec:boxes}
We prove Theorem~\ref{thm:all-boxes}. A prescribed-family corner lemma
and a uniform lower obstruction come first, followed by the thin-box
upper bounds and the remaining shapes. The only certified input is
Theorem~\ref{thm:main}; the prescribed-family lemma is independent of it.

After ordering the sides as $a\le b\le c$, the upper bound has three
exhaustive regimes:
\begin{center}
\begin{tabular}{ll}
\toprule
Shape & Main mechanism\\
\midrule
$a=2,\ b=3$ & Exceptional-defect control and retargeting\\
$a=b=2$ & Concentration and retargeting along a rod\\
$a+b\ge6$ & Weighted compatibility and reservations\\
\bottomrule
\end{tabular}
\end{center}
For the quantitative arguments, write
$m=a-2$, $t=b-2$, $M=c-2$, and $\tau=m+t+M$. Thus
$L=8+4\tau$ is the frame size, and $N$ is the number of available copies.
A source's \emph{defect} is the number of target positions it misses.
The parameter $d$ is the Hall cutoff; a bichromatic pair is called rare
when at most $d-1$ sources possess it. Its missing family then has at least
$N-d+1$ copies. The main bounds are:
\begin{center}
\small
\begin{tabular}{llll}
\toprule
Regime & $N=L$ & $d$ & Exceptional Hall sets\\
\midrule
$2\times3\times c$, $c\ge4$ & $12+4M$ & $2M+6$ & Size $d+1$; Theorem~\ref{thm:23c-sharp}\\
$2\times2\times c$, $c\ge5$ & $8+4M$ & $2M+4$ & Sizes $d+1,d+2$; Theorem~\ref{thm:sharp-rods-upper}\\
$a+b\ge6$ & $8+4\tau$ & $\tau+M+6$ & Reservations; Theorem~\ref{thm:boxes-wide}\\
\bottomrule
\end{tabular}
\end{center}
In the wide regime every source's defect is at most $d$. The last column
lists the exceptional Hall-set sizes that require additional arguments in
the thin regimes; these are distinct from individual source defects and are handled by
concentration and retargeting. The smaller boundary shapes are treated
separately, with spare sources where necessary.
We establish the thin regimes first, then the general weighted bound.
In each case an initial corner matching verifies part of Hall's inequalities;
the final matching may rearrange those placements.

\begin{lemma}[Rare pairs]\label{lem:rare-pairs}
In a collection of five-colour cubes, any family of bichromatic pairs each
missing from more than half the source copies consists of at most two
pairwise disjoint pairs.
\end{lemma}
\begin{proof}
The missing families of two such pairs intersect. A source missing both
pairs forces them to be disjoint, by Lemma~\ref{lem:cube-incidence}.
There are at most two disjoint pairs in a five-colour palette.
\end{proof}

\subsection{Corners from a missing-pair family}
Retain the family $\mathcal H_{\alpha\beta}$ from Lemma~\ref{lem:fixed-missing-pair},
with its opposite-repeat varieties and their $2\times2$ corner array.
Write $K(W)$ for the cyclic corner types of $W$.

\begin{lemma}[Corner projections onto a missing-pair family]\label{lem:thin-projections}
Fix a bichromatic pair $\alpha\beta$.  Let $C^+$ and $C^-$ be the two sets of six
tricoloured corner types shared by the three adjacent-repeat varieties of
the respective cyclic orientations in $\mathcal H_{\alpha\beta}$.  An
opposite-repeat variety in $\mathcal H_{\alpha\beta}$ has four corners in each set.
For a source $W\notin\mathcal H_{\alpha\beta}$, put
\[
 p(W)=|K(W)\cap C^+|,\qquad m(W)=|K(W)\cap C^-|.
\]
Then
\[
 (p(W),m(W))\in\{(4,0),(2,1),(1,2),(2,2),(0,4)\}.
\]
\end{lemma}

\begin{proof}
Relabel the poles as $0,1$ and the ring colours as $2,3,4$.  Choose the
positive orientation so that
\[
 C^+=\{023,034,042,124,132,143\},\qquad
 C^-=\{024,032,043,123,134,142\}.
\]
Triples denote cyclic classes; face order is $(+x,-x,+y,-y,+z,-z)$.  The following table gives representatives of
all geometric cases outside $\mathcal H_{01}$, up to a colour permutation
preserving the pole pair and a simultaneous reflection, which may exchange
$C^+$ and $C^-$.  Such a reflection is only a symmetry of this incidence
calculation, not an allowed assembly move.
\[
\begin{array}{c|cc}
 W&p(W)&m(W)\\\hline
 (0,0,1,2,3,4)&2&2\\
 (0,1,0,2,3,4)&0&4\\
 (0,2,0,3,1,4)&2&1\\
 (0,2,1,2,3,4)&0&4\\
 (0,2,1,3,2,4)&2&1\\
 (0,2,1,3,4,4)&2&2
\end{array}
\]
To see that these cases exhaust the possibilities, first suppose a pole
colour is repeated.  Its two faces are either opposite, or adjacent; in
the latter case the other pole is opposite one repeated face or adjacent
to both.  If a ring colour is repeated, the two pole faces must be
adjacent, since otherwise $W$ belongs to $\mathcal H_{01}$.  Fix them at
$x^+$ and $y^+$.  Among the remaining faces $x^-,y^-,z^+,z^-$, an opposite
repeated pair is $z^+,z^-$.  An adjacent repeated pair is either $x^-,y^-$,
or one of $x^-,y^-$ together with one of $z^+,z^-$.  These are the final
three rows.

Each entry follows by reading the eight corners of the displayed cube.
The projections of the nine family
varieties themselves are $(6,0),(0,6)$, and $(4,4)$, as follows directly
from their ring descriptions.
\end{proof}

\begin{lemma}[Eight cubes with a common missing pair]\label{lem:eight-common-missing}
Eight cubes from $\mathcal H_{\alpha\beta}$, together with any three additional
five-colour cubes, have a corner solution whose target belongs to
$\mathcal H_{\alpha\beta}$.
\end{lemma}

\begin{proof}
Let $O,P,Q$ count the opposite-repeat, positive
adjacent-repeat, and negative adjacent-repeat copies among the chosen eight.
Thus $O+P+Q=8$; $C^+$ and $C^-$ are as in Lemma~\ref{lem:thin-projections}.

If $O+\min(P,Q)\ge4$, the $2\times2$ Hall argument gives an
opposite-repeat corner solution from these eight copies alone. Explicitly,
choose $O_r$ whose opposite-repeat multiplicity is maximal. Each other
opposite-repeat multiplicity is at most $O/2$: it is no larger than the
chosen multiplicity, and their sum is at most $O$. Among these eight family
sources, the two positions in a cell have identical neighbourhoods. Any
deficient set can therefore be enlarged to a union of whole cells without
enlarging its neighbourhood. Each cell has
at least $O/2+\min(P,Q)\ge2$ neighbours, each row has at least
$8-O/2\ge4$, and each column has at least $O+\min(P,Q)\ge4$.
Every other nonempty union of cells has all eight copies as neighbours.

Otherwise exchange the signs so that $O+Q\le3$. Put
$\kappa=O+Q=8-P$, so $P\ge5$ and $0\le \kappa\le3$.
Some positive variety $A_r^+$ occurs at least twice. If its eight
corners have a matching among all eleven copies, we are done. Otherwise,
choose a deficient Hall set $D$ of its corners. The two copies of
$A_r^+$ fit every corner, so $|D|\ge3$ and $D$ contains a tricoloured
corner. All $P$ positive copies are therefore neighbours of $D$.
Let $R$ be its other neighbours and put $U=C^+\setminus D$, $u=|U|$.
Since $|D|+u\le8$,
\[
 P+|R|<|D|\le8-u,\qquad |R|+u\le \kappa-1.
\]
Remove the $P$ positive copies and the copies in $R$. Every surviving
copy has all its positive corners in $U$, and $u\le2$.

Every family opposite-repeat variety has four corners in $C^+$, so no
such chosen copy remains. At least
\[
 \kappa-|R|\ge u+1
\]
of the chosen nonpositive copies remain, and all of them are negative
adjacent-repeat copies. Each fits every corner in $C^-$. The remaining
pool has at least
\[
 \kappa+3-|R|\ge4+u
\]
copies.

Lemma~\ref{lem:thin-projections} gives the following implications for every source:
\[
\begin{array}{c|ccc}
 |K(W)\cap C^+|\le u &u=0&u=1&u=2\\\hline
 |K(W)\cap C^-|\ge&4&2&1.
\end{array}
\]
Indeed, the family varieties have projection sizes $(6,0),(0,6),(4,4)$,
and every variety outside the family has one of
$(4,0),(2,1),(1,2),(2,2),(0,4)$.

Partition $C^-$ into three pairs, indexed by the unordered pairs of ring
colours. The negative half of each family opposite-repeat target is the
union of two of these pairs. We will match one such four-position half
using the remaining pool, then use four positive copies for its positive
half.

Let $j$ be the number of negative-family copies in the remaining pool;
thus $j\ge u+1$. If $j\ge4$, the conclusion is immediate. If $j=3$,
another remaining copy fits at least one negative corner; choose a target
whose negative half contains it and fill the other three positions with
negative-family copies.

If $j=2$, then $u\le1$. Two other remaining copies each fit at least two
negative corners, so they can be assigned distinct ones. Any two negative
corners lie in the union of two of the three pairs. Choose the corresponding
opposite-repeat target and fill its other two negative positions with the
two negative-family copies.

Finally, if $j=1$, then $u=0$, and there are at least three other copies,
each fitting at least four of the six negative corners. Choose one of
these copies $W$. Some union of two of the three pairs contains at least
three of its negative corners. In that four-position half, every other
copy fits at least two positions. Hence $W$ and any two other copies can
be matched to three distinct positions: two copies have at least two
neighbours, and all three together have at least three. The one
negative-family copy fills the fourth position.

In every case we obtain the negative half of an opposite-repeat target
in $\mathcal H_{\alpha\beta}$. Four of the $P\ge5$ positive copies fill its
positive half. The positive copies used here are disjoint from the surviving pool. This proves the lemma.
\end{proof}


\subsection{A uniform lower bound}

\begin{samepage}
\begin{lemma}[A three-variety obstruction]\label{lem:thin-three-variety-obstruction}
Let $a,b,c\ge2$ and put $\tau=a+b+c-6$. In face order
$(+x,-x,+y,-y,+z,-z)$, define
\[
 U=(0,1,0,2,3,4),\qquad
 V=(0,1,1,2,3,4),\qquad
 W=(0,3,1,4,2,3).
\]
The collection
\[
 U+(\tau+5)V+(\tau+6)W
\]
of $2\tau+12$ cubes has no $a\times b\times c$ frame using all five
colours. In particular, $U+7V+8W$ fails for both a $2\times3\times3$
frame and a $2\times2\times4$ box.
\end{lemma}
\end{samepage}

\begin{proof}
The corner types are
\[
\begin{array}{c|l}
 U&003,004,013,024,032,041,123,142\\
 V&113,114,013,024,032,041,123,142\\
 W&133,334,012,024,031,043,132,234.
\end{array}
\]
Triples denote cyclic classes. The only two-colour types in their union
are $003,004$, $113,114$, and $133,334$. An adjacent-repeat target must
therefore repeat colour $0$, $1$, or $3$. Its two repeated-colour corners
force the colours on the opposite faces at the ends of its monochromatic
edge to be, respectively, $\{3,4\}$, $\{3,4\}$, or $\{1,4\}$.
The two orders of the remaining colours give $U,V,W$, respectively, and
their mirrors. The mirrors of $U$ and $V$ require $023$, which is absent
from the union; the mirror of $W$ requires $021$, also absent.

An opposite-repeat target requires both cyclic orientations of four
distinct unordered tricoloured triples. The union above contains both
orientations only for $013$ and $123$. Thus no opposite-repeat target
is possible. The only corner-supported targets are $U,V,W$.

The corners $003$ and $004$ of target $U$ each require a $U$ copy, but only
one is available.

For target $V$, seven of its corner types are absent from $W$; the only
shared type is $024$. In addition, exactly one geometric edge in each
coordinate direction has a type absent from $W$: these are $14,14,11$.
Consequently, in any orientation of an $a\times b\times c$ frame,
$7+\tau$ positions require a $U$ or $V$ source. There are only $\tau+6$
such copies.

For target $W$, seven corner types are absent from both $U$ and $V$.
Again, one geometric edge in each direction has a private type, now
$34,33,34$. Hence $7+\tau$ positions require a $W$ source, whereas only
$\tau+6$ are available.

Every possible target therefore fails
(Figure~\ref{fig:uniform-obstruction}).
\end{proof}

\begin{figure}[ht]
\centering
\input{uniform-obstruction}
\caption{The uniform obstruction $U+(\tau+5)V+(\tau+6)W$, where $\tau=a+b+c-6$.
Each net has $+x$ at its centre, and corner types are read
counterclockwise from outside. Only $U$, $V$ and $W$ can be targets.
Their pools are $U$ alone, $U$ or $V$, and $W$ alone; faded corner types
also occur on a cube outside the pool. For $V$ and $W$, one edge in each
direction has a private type, and these edges have $\tau$ noncorner positions
in total.}
\label{fig:uniform-obstruction}
\end{figure}

\begin{proposition}\label{cor:all-lower-bounds}
For every $a,b,c\ge2$,
\[
 F_5(a,b,c)\ge\max\{15,\;4(a+b+c)-16,\;2(a+b+c)+1\}.
\]
\end{proposition}
\begin{proof}
The three terms follow from the fourteen-cube corner obstruction (Lemma~\ref{lem:lower}),
the number of frame positions, and
Lemma~\ref{lem:thin-three-variety-obstruction}, respectively.
\end{proof}

\subsection{Choosing a thin target}

\begin{lemma}[Choosing a target when one side has length two]\label{lem:thin-target}
Let $2\le b\le c$ and suppose $N\ge15$ five-colour sources are available.
There is a corner-solvable target and an orientation for a $2\times b\times c$
frame in which every noncorner edge position has a bichromatic requirement
possessed by at least $\lceil N/2\rceil$ sources. An adjacent-repeat target's
monochromatic edge is parallel to a side of length two. If $b=2$ and the
target is opposite-repeat, its repeated-colour axis can also be placed along $c$.
\end{lemma}
\begin{proof}
Put $h=\lceil N/2\rceil$ and call a bichromatic pair rare if fewer than $h$
sources possess it. Its missing family has at least
$N-h+1=\lfloor N/2\rfloor+1>N/2$ members, hence at least eight.
The missing families of any two rare pairs intersect, so
Lemma~\ref{lem:rare-pairs} gives at most two rare pairs, which are disjoint.

If no pair is rare, the fifteen-cube theorem supplies a corner target.
Put its monochromatic edge, if any, parallel to a side of length two,
where it contributes no noncorner positions. All actual edge requirements
are then bichromatic and have at least $h$ neighbours. For an opposite
rod target, put the repeated-colour axis along $c$.

Otherwise choose a rare pair $\alpha\beta$. Apply
Lemma~\ref{lem:eight-common-missing} to eight sources from
$\mathcal H_{\alpha\beta}$ and any three others. The resulting corner
solution has a target in that family. Put its singleton $\alpha\beta$
axis along a side of length two. Every actual edge requirement joins a
pole colour to a ring colour, so it avoids $\alpha\beta$ and the other
possible rare pair, whose colours are both ring colours. The possible
monochromatic edge is parallel to the short axis. In an opposite rod
target the repeated-colour axis is perpendicular to the singleton axis,
so it can simultaneously be put along $c$.
\end{proof}

\subsection{The smallest thin box}

\begin{lemma}[Compatibility for a thin frame]\label{lem:thin-compatibility}
Choose an orientation of a five-colour target for a $2\times2\times3$
frame such that its four noncorner edge positions are bichromatic.
Every source cube fits at least four of the twelve positions.
\end{lemma}

\begin{proof}
If either source or target is opposite-repeat,
Lemma~\ref{lem:cube-incidence} gives two shared corners.
The four parallel bichromatic edges have either four distinct types or
two types occurring twice each. In the latter case both contain the
target's repeated colour. The source's missing pairs are disjoint, so
it fits at least two of these edges, giving four compatible positions.

If both varieties are adjacent-repeat, the four parallel edge types are
distinct and transverse to the monochromatic edge. The source fits either
all four, or at least three and a corner by
Lemma~\ref{lem:transverse-corner}. Again it fits four positions.
\end{proof}

\begin{theorem}[The $2\times2\times3$ box]\label{thm:223}
We have
\[
 F_5(2,2,3)=15.
\]
In particular, any fifteen five-colour cubes contain twelve that form a
$2\times2\times3$ box with monochromatic exterior faces using all five
colours.
\end{theorem}

\begin{proof}
The fourteen-cube corner obstruction (Lemma~\ref{lem:lower}) gives the lower bound: the eight
corner cubes of any box would themselves give a corner solution.

For the upper bound, apply Lemma~\ref{lem:thin-target} to fifteen sources.
The target has a corner matching, and each of its four noncorner edge
positions has a bichromatic requirement with at least eight neighbours.

For this target and orientation, every source fits at least
four of the twelve positions, by Lemma~\ref{lem:thin-compatibility}.
Thus each source misses at most eight positions.

Apply Lemma~\ref{lem:partial-matching} with the corners as $P_0$,
$L=12$, $N=15$ and $d=8$. It supplies a matching of all twelve positions
to distinct source copies. The final matching may rearrange the preliminary
corner placements.
\end{proof}

\subsection[Sharp thresholds for 2 by 3 by c frames]{Sharp thresholds for $2\times3\times c$ frames}

\begin{lemma}[Compatibility for $2\times3\times c$]\label{lem:23c-compatibility}
Let $M=c-2\ge2$ and put $d=2M+6$. Orient a five-colour target for
a $2\times3\times c$ frame so that its monochromatic edge, if any,
is parallel to the side of length two. An adjacent-repeat source misses
at most $d$ positions, and an opposite-repeat source misses at most
$d+1$ positions. If a source misses $d+1$ positions, it is the unique
opposite-repeat variety lacking the two bichromatic types appearing
among those missed edge positions.
\end{lemma}

\begin{proof}
The nonzero edge weights, counting noncorner positions, are $1,M$.
By Lemma~\ref{lem:cube-incidence},
for an opposite-repeat source,
at most six corners are incompatible. Its two missing bichromatic pairs
are disjoint, so at most three geometric edges are incompatible, with
at most two parallel to either nonzero direction. If their counts are
$k_1,k_M$, the defect is at most
\[
 6+k_1+Mk_M\le6+3+2(M-1)=2M+7=d+1.
\]
Equality requires six incompatible corners, two incompatible long
edges, and one incompatible edge of weight $1$. These three geometric
edges display exactly two bichromatic types: there are at least two
because any target type occurs at most twice, and at most two because
the source lacks only two types. They are disjoint and determine the
opposite-repeat source variety uniquely.

An adjacent-repeat source lacks only one bichromatic type. Against an
opposite-repeat target, at most six corners and edge positions of total
weight $2M$ are incompatible, giving defect at most $d$.
Against an adjacent-repeat target, an incompatible edge lies transverse
to the monochromatic edge, so Lemma~\ref{lem:transverse-corner}
gives a shared corner.
The absent type contributes weight at most $M+1$: a doubled type has
one edge in each nonzero direction. The defect is therefore at most
\[
 7+(M+1)=M+8\le2M+6=d.
\]
If no actual edge is incompatible, the defect is at most eight, which
is also at most $d$.
\end{proof}

\begin{lemma}[An opposite-repeat target along the short axis]
\label{lem:23c-opposite-short}
Let $c\ge3$. For a $2\times3\times c$ frame, orient an opposite-repeat target with
its repeated-colour axis along the side of length two. Every source
misses at most $2(c-2)+6$ positions.
\end{lemma}

\begin{proof}
Every actual edge type contains the target's repeated colour. By
Lemma~\ref{lem:cube-incidence}, a source
can lack at most one such bichromatic type, because its missing pairs
are disjoint. The absent type accounts for at most two parallel geometric
edges, of total weight at most $2(c-2)$. At least two corners are shared,
so at most six further positions are missed.
\end{proof}

\begin{theorem}[The remaining $2\times3\times c$ frames]
\label{thm:23c-sharp}
\[
 F_5(2,3,3)=17,\qquad
 F_5(2,3,c)=4c+4\quad(c\ge4).
\]
\end{theorem}

\begin{proof}
First let $c=3$ and take seventeen sources. Lemma~\ref{lem:thin-target}
gives a corner target with its possible monochromatic edge along the side
of length two and at least nine neighbours for each actual edge position.

Every source misses at most nine of the sixteen frame positions. Indeed,
if source or target is opposite-repeat, at most six corners and three
actual geometric edges are incompatible. If both are adjacent-repeat,
at most two actual geometric edges are incompatible; whenever one is
incompatible, Lemma~\ref{lem:transverse-corner} supplies a shared corner. If no edge
is incompatible, at most eight positions are missed.

The chosen target has a corner matching, and every edge requirement has
at least nine neighbours. Lemma~\ref{lem:partial-matching}, with the corners
as the matched subset, completes the frame. This proves $F_5(2,3,3)\le17$; the
collection $U+7V+8W$ in Lemma~\ref{lem:thin-three-variety-obstruction}
gives equality.

Now let $c\ge4$, put $M=c-2\ge2$, and set
\[
 d=2M+6,\qquad N=4M+12=2d.
\]
There are $N=4c+4$ frame positions. Apply Lemma~\ref{lem:thin-target} to
any $N$ sources. The target has a corner matching, its possible monochromatic
edge lies along the side of length two, and every actual edge requirement
has at least $d=N/2$ neighbours. Hall's inequality
holds for corner-only sets, and for any set of at most $d$ positions
containing an edge. By Lemma~\ref{lem:23c-compatibility}, sets of more
than $d+1$ positions neighbour every source.

Suppose the frame has no matching. A deficient Hall set $D$ must therefore
have exactly $d+1$ positions. Every nonneighbour misses all of those
positions, so Lemma~\ref{lem:23c-compatibility} says that all nonneighbours
have one opposite-repeat variety $V$, determined by the two absent
bichromatic types. Since $D$ has at most $d$ neighbours among $N=2d$
copies, the inventory contains at least $d$ copies of $V$.

Retarget the frame to $V$ with its repeated-colour axis along the side
of length two. By Lemma~\ref{lem:23c-opposite-short}, every source misses
at most $d$ positions. The $d$ copies of $V$ fit every position, so the universal-copy consequence
of Lemma~\ref{lem:partial-matching} supplies a frame.

Thus $N$ sources always suffice. The number of frame positions supplies
the matching lower bound.
\end{proof}
% Analytic resolution of the 2 x 2 x c family.
% Inputs: N_5=15; the eight-family-plus-three corner lemma; basic geometry.

\subsection[Sharp thresholds for 2 by 2 by c boxes]{Sharp thresholds for $2\times2\times c$ boxes}

Put $M=c-2$. The frame of a $2\times2\times c$ box consists of eight
corners and four parallel geometric edges, each with $M$ positions. Its
size is $L=4M+8=4c$. We call an orientation of an adjacent-repeat target
\emph{transverse} if its monochromatic edge is not parallel to the long
axis. For an opposite-repeat target, its \emph{axial} orientation has
its repeated-colour axis along the long direction.

\begin{lemma}[Directional compatibility]\label{lem:rod-compatibility}
Let $M\ge2$.
\begin{enumerate}
\item For a transverse adjacent-repeat target, every adjacent-repeat
source misses at most $M+7$ positions. Every opposite-repeat source
misses at most $2M+6$ positions. An opposite-repeat source missing more
than $M+6$ positions must be one of two varieties; their missing
long-edge types are complementary pairs.
\item For an axial opposite-repeat target $V$, every source misses at
most $d=2M+4=L/2$ positions. If $M\ge3$, a source different from $V$
can miss $d$ positions only if it is one of the other two opposite-repeat
varieties with the same repeated colour. These two varieties have
complementary missing long-edge types.
\end{enumerate}
\end{lemma}

\begin{proof}
For (i), write the target in face order $(+x,-x,+y,-y,+z,-z)$ as
\[
 (r,\alpha,r,\beta,\gamma,\delta),
\]
and take the long direction to be $x$. Its four long-edge types are
\[
 r\gamma,\quad r\delta,\quad \beta\gamma,\quad \beta\delta.
\]
An adjacent-repeat source lacks at most one of these. If it lacks one,
Lemma~\ref{lem:transverse-corner} gives a shared corner, so its defect
is at most $M+7$. If it lacks none, its defect is at most eight, also
bounded by $M+7$.

By Lemma~\ref{lem:cube-incidence}, an opposite-repeat source shares at
least two corners and lacks at most
two long-edge types, giving defect at most $2M+6$. To lack two of the
four displayed types, its two missing pairs must be one of the disjoint
matchings
\[
 \{r\gamma,\beta\delta\}\qquad\text{or}\qquad\{r\delta,\beta\gamma\}.
\]
Each matching determines exactly one opposite-repeat variety, with
repeated colour $\alpha$. The two varieties have complementary missing
long-edge types. Every other opposite-repeat source lacks at most one
long-edge type, and hence has defect at most $M+6$.

For (ii), let the repeated colour of $V$ be $r$, and put its singleton
opposite pairs on the two transverse axes. The four long-edge types are
the four edges of the resulting $K_{2,2}$ on the singleton colours.
The other two opposite-repeat varieties with repeated colour $r$ each
share four corners and two long-edge types with $V$. Their compatible
position count is $4+2M=d$, and their missing long-edge pairs are
complementary.

Every adjacent-repeat source lacks at most one long-edge type. An
opposite-repeat source with repeated colour different from $r$ also
lacks at most one: one of its two missing pairs contains $r$, whereas
no long-edge type does. By Lemma~\ref{lem:cube-incidence}, these sources share at least two
corners with $V$, so they fit at least $2+3M\ge d$ positions. If $M\ge3$, this last
inequality is strict. The target variety $V$ itself fits every position.
This proves both assertions.
\end{proof}

\begin{lemma}[Concentrated opposite-repeat copies]\label{lem:rod-concentration}
Let $M\ge3$ and $d=2M+4$. A collection of $2d=4M+8$ cubes containing
at least $d-1$ copies of one opposite-repeat variety forms a
$2\times2\times(M+2)$ box.
\end{lemma}

\begin{proof}
Let that variety be $V$ with multiplicity $v$. Use its axial target
orientation. Every source misses at most $d$ positions. If $v\ge d$, the universal-copy consequence of
Lemma~\ref{lem:partial-matching} applies, since $N=L=2d$.

Suppose $v=d-1$. The same reasoning handles all position sets except
sets $D$ of size exactly $d$. If such a set fails Hall, no non-$V$
copy neighbours it. Those copies must each miss $d$ positions, so
Lemma~\ref{lem:rod-compatibility}(ii) restricts them to the other two
opposite-repeat varieties with the same repeated colour. Since $d>8$,
$D$ contains a long-edge position. The two candidate varieties have
complementary missing long-edge types, so at most one can be absent
from the neighbourhood of $D$. All $2d-v=d+1$ non-$V$ copies are
therefore one variety $W$. Using $W$ as the axial target gives the
previous case, with at least $d$ universal copies, and completes the box.
\end{proof}

\begin{theorem}[Thin rods]\label{thm:sharp-rods-upper}
\[
 F_5(2,2,4)\le17,
 \qquad
 F_5(2,2,c)=4c\quad(c\ge5).
\]
\end{theorem}

\begin{proof}
First let $c=4$, so $M=2$ and $L=16$, and take seventeen source
copies. Lemma~\ref{lem:thin-target} gives a corner target, transverse if
adjacent-repeat and axial if opposite-repeat, with at least nine sources
for every long-edge type.
If the target is opposite-repeat, every source misses at most eight positions
by Lemma~\ref{lem:rod-compatibility}. Lemma~\ref{lem:partial-matching},
with the corners as the matched subset, completes the frame.

For an adjacent-repeat target, adjacent-repeat sources miss at most nine
positions and opposite-repeat sources at most ten. A failing Hall set
$D$ cannot consist only of corners. Since it contains an edge position,
it has at least nine neighbours; and if $|D|>10$, it neighbours every
source. Therefore a failure would have $|D|=10$ and exactly nine
neighbours. Its eight non-neighbouring source copies must all be
opposite-repeat. Each must miss two long-edge types, since otherwise
its defect is at most $M+6=8$. The two candidate varieties in
Lemma~\ref{lem:rod-compatibility}(i) have complementary missing
long-edge types. As $D$ contains an edge position, all eight copies
must be the same opposite-repeat variety $V$.

For the axial target $V$, every source misses at most eight positions.
Its eight universal copies and $N=17\ge L=16$ satisfy the universal-copy
consequence of Lemma~\ref{lem:partial-matching}, proving the upper bound.

Now let $M=c-2\ge3$, put $d=2M+4$, and take exactly $N=2d=L$
cubes. Lemma~\ref{lem:thin-target} gives a transverse or axial corner
target as appropriate, with at least $d=N/2$ sources for each long-edge type. An opposite-repeat
target is completed directly by Hall, since every source misses at
most $d$ positions.

For an adjacent-repeat target, adjacent-repeat sources miss at most
$M+7\le d$ positions; opposite-repeat sources miss at most
$2M+6=d+2$. Any failing Hall set $D$ therefore has
\[
 d+1\le |D|\le d+2.
\]
It has at most $d+1$ neighbours, leaving at least $d-1$ non-neighbouring
source copies. All are opposite-repeat. An opposite-repeat source
missing only one long-edge type has defect at most $M+6<d+1$, so
all these copies are among the two varieties in
Lemma~\ref{lem:rod-compatibility}(i). Since $D$ contains an edge,
all must be one variety $V$. Lemma~\ref{lem:rod-concentration} then
supplies a box from the original collection. Thus every collection of
$L=4c$ cubes works. The number of positions gives the matching lower
bound.
\end{proof}

\begin{proposition}[Complete rod threshold]\label{cor:sharp-rods}
Together with Theorem~\ref{thm:main}, Theorem~\ref{thm:223} and
the sixteen-cube obstruction of Lemma~\ref{lem:thin-three-variety-obstruction},
Theorem~\ref{thm:sharp-rods-upper} gives
\[
 F_5(2,2,c)=\max\{15,\,4c,\,2c+9\}\qquad(c\ge2).
\]
Equivalently, the values are $15,15,17$ for $c=2,3,4$, respectively,
and $4c$ for every $c\ge5$.
\end{proposition}
\begin{proof}
The cases $c=2,3$ are the established corner and smallest thin-box
thresholds. For $c=4$, Lemma~\ref{lem:thin-three-variety-obstruction} and the seventeen-cube
upper bound give equality. For $c\ge5$, the number of positions gives
$F_5(2,2,c)=4c$, as proved above.
\end{proof}
\subsection{Boxes whose two shortest sides sum to at least six}

Order the side lengths as $2\le a\le b\le c$ and put
\[
 m=a-2,\qquad t=b-2,\qquad M=c-2,\qquad \tau=m+t+M.
\]
The frame has $L=8+4\tau$ positions. In this section assume $m+t\ge2$.

\begin{lemma}[Weighted compatibility]\label{lem:box-weighted}
Let $2\le a\le b\le c$ with $a+b\ge6$, so that $m+t\ge2$ and $M\ge1$.
Every source misses at most $d=\tau+M+6$ positions of every target frame.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem:cube-incidence}, if the target is opposite-repeat,
at most three geometric edges are
incompatible, with at most two in any direction. Their total weight is
at most $2M+t\le \tau+M$. At most six corners are incompatible, so the
claimed bound follows.

For an opposite-repeat source, take the adjacent-repeat target
\[
 (r,\alpha,r,\beta,\gamma,\delta),
\]
whose monochromatic edge is parallel to $z$.
A doubled bichromatic type, $r\gamma$ or $r\delta$, occupies one edge in each of
the $x,y$ directions. If such a type is missing, it and the monochromatic
edge contribute one edge in each direction; the other disjoint missing
type contributes at most one further edge. The missed weight is at most
$\tau+M$. If neither missing type is doubled, two missing single types
cannot have the same direction: the single types in each direction
share a colour. Adding the monochromatic edge therefore gives at most
three edges, with at most two parallel, again of weight at most
$2M+t\le \tau+M$. There are at most six incompatible corners.

For two adjacent-repeat varieties, at most three geometric edges are
incompatible. If there are at most two, the total defect is at most
\[
 2M+8\le \tau+M+6,
\]
because $m+t\ge2$. If there are three, they are the monochromatic edge
and a doubled missing bichromatic type, one edge in each direction.
Lemma~\ref{lem:transverse-corner}
gives a shared corner. The defect is at most
$\tau+7\le \tau+M+6$, since $M\ge1$.
\end{proof}

\begin{theorem}\label{thm:boxes-wide}
If $2\le a\le b\le c$ and $a+b\ge6$, then
\[
 F_5(a,b,c)=4(a+b+c)-16.
\]
\end{theorem}
\begin{proof}
Take $N=L=8+4\tau$ source copies and put $d=\tau+M+6$. Call a bichromatic
pair rare if at most $d-1$ copies possess it. Its missing family has
at least
\[
 N-d+1=3\tau-M+3
\]
copies, exceeding $N/2$ by $m+t-1\ge1$. Lemma~\ref{lem:rare-pairs}
again gives at most two disjoint rare pairs.

By Lemma~\ref{lem:partial-matching}, a target avoiding all rare pairs
has a full frame matching whenever its corners and monochromatic edge
positions have a matching: every other position has at least $d$
neighbours, and Lemma~\ref{lem:box-weighted} bounds each source's defect
by $d$.

\emph{No rare pair.} For each colour reserve up to $m$ sources having
that colour on adjacent faces, taking all such sources if fewer exist.
The reservations are disjoint. At least $N-5m\ge7m+8\ge15$ sources
remain when $m\ge1$; when $m=0$, none are reserved and $N\ge24$.
The corner theorem supplies a target on the unreserved sources.
Orient its possible monochromatic edge along the shortest direction.
If the target repeats colour $r$ adjacently, a compatible source with
adjacent $r$-faces survived the reservation, so the $r$-reserve has its
full size $m$. It fills the monochromatic edge. The Hall observation
completes the frame.

\emph{One rare pair $\alpha\beta$.} For each of the three ring colours,
reserve up to $m$ sources in the whole collection having that colour
on adjacent faces, again taking all such sources if fewer exist.
At most $3m$ family copies are removed, so $\mathcal H_{\alpha\beta}$ retains at least
\[
 (3\tau-M+3)-3m=3t+2M+3\ge8
\]
copies, and at least eleven total sources remain unreserved. Apply
Lemma~\ref{lem:eight-common-missing} to eight unreserved family copies
and any three further unreserved sources. The resulting corner target
lies in $\mathcal H_{\alpha\beta}$. Orient its $\alpha\beta$ axis along the shortest
direction. Any monochromatic edge is parallel to this axis and has its
full reserve of $m$ copies, by the same survivor argument. The target
avoids the only rare pair, so the Hall observation applies.

\emph{Two rare pairs $e,f$.} Their common missing variety $V$ is the
unique opposite-repeat variety with $e,f$ as singleton opposite pairs.
Its multiplicity satisfies
\[
 v\ge N-2(d-1)=2(m+t)-2\ge2.
\]
If $v\ge8$, use eight copies for the corners of $V$. If $4\le v\le7$,
the family $\mathcal H_e$ has at least eleven copies and hence at least
four not of variety $V$. Each shares at least four corners with $V$,
by the ring array. Place four such copies at distinct corners and fill
the other four with $V$. In either case, $V$ has no monochromatic edge
and avoids both rare pairs; the Hall observation applies.

\emph{The final exceptional case: $v=2$ or $3$.} Since
$v\ge2(m+t)-2$ and $v\le3$, we have $m+t=2$, hence $N=2d$.
The ordering $m\le t$ also gives $m\le1$. If $n_e,n_f$
count the sources possessing the rare pairs, then
\[
 n_e,n_f\le d-1,\qquad
 v\ge N-n_e-n_f.
\]
Consequently,
\[
 n_e,n_f\ge d-2,\qquad n_f+v\ge d+1.
\]
For each ring colour relative to $e$, reserve up to $m$ sources repeating
that colour adjacently, taking all if fewer exist. As in the one-rare-pair
case, at least eight $\mathcal H_e$-sources and eleven sources in total
remain unreserved. Apply Lemma~\ref{lem:eight-common-missing} to eight
unreserved family sources and three other unreserved sources. Orient the
resulting target $T\in\mathcal H_e$ with its $e$ axis shortest. Its corners
have a matching on the unreserved sources, and any monochromatic edge has
its full reserve of $m$ copies by the same survivor argument.

Within this family, $V$ shares at least four target corners and can
miss at most two geometric edges, both parallel to the $e$ axis. For
a different opposite-repeat target these are its two $f$-edges; for
an adjacent-repeat target they are its $f$-edge and monochromatic edge.
Thus $V$ misses at most $2m+4\le6$ actual positions.

The only Hall sets not covered as before contain bichromatic edge
positions of type $f$ and no non-rare bichromatic edge. Such a set $D$
has at least $n_f$ neighbours. If $|D|>n_f\ge d-2\ge8$, every copy
of $V$ is also a neighbour, since it misses at most six positions.
These are distinct from the $n_f$ copies possessing $f$, so
\[
 |\Gamma(D)|\ge n_f+v\ge d+1.
\]
This handles $|D|\le d$; larger sets neighbour every source. Hall's
theorem completes the frame in the final case.

The number of frame positions gives the lower bound, proving equality.
All preliminary placements certify Hall inequalities; the final
matching may rearrange them.
\end{proof}

\subsection{Completion of the classification}
\begin{proof}[Proof of Theorem~\ref{thm:all-boxes}]
Order the sides as $2\le a\le b\le c$. When $a+b\ge6$,
Theorem~\ref{thm:boxes-wide} gives $F_5=L$. The remaining possibilities
are exactly $(a,b)=(2,2)$ and $(2,3)$, settled by
Proposition~\ref{cor:sharp-rods} and Theorem~\ref{thm:23c-sharp}.
Together with the corner theorem at $(2,2,2)$, these give $F_5=15$
for sums six and seven, $F_5=17$ for sum eight, and $F_5=L$
for every larger sum. Proposition~\ref{cor:all-lower-bounds} proves
sharpness and yields the stated maximum formula.

For the complete box, take $\max\{abc,F_5(a,b,c)\}$ sources and first
choose a frame. Every source has every colour, so any unused source can
fill any face-interior position after a suitable proper rotation. All
fully interior positions are unrestricted. There are enough unused
sources to fill the remaining $abc-L$ positions. Conversely, any complete
box supplies a frame and uses $abc$ sources. Hence
\[
 G_5(a,b,c)=\max\{abc,F_5(a,b,c)\}.
\]
Putting $x=a-2$, $y=b-2$, and $z=c-2$ gives
\[
 abc-L=xyz+2(xy+xz+yz)\ge0,
\]
which simplifies this identity to the claimed formula for $G_5$.

If $(a,b)=(2,2)$, this formula is $\max\{4c,15,2c+9\}$: it exceeds
the volume only for $c=2,3,4$. In every other sorted case, $b\ge3$;
then $abc\ge18\ge15$ and
\[
 abc-2(a+b+c)-1\ge 2bc-2b-2c-5\ge1.
\]
The first inequality follows by increasing $a$ from two, and the second
is minimised at $b=c=3$. There are no further exceptions.
\end{proof}