The left wall \(\{ (1,2),(1,3)\} \) is explicitly among the dominos of \(A_n\), is vertical, lies in column \(1\), and covers rows \(2\)–\(3\).

The left wall of \(B_n\) is \(\{ (1,1),(1,2)\} \), which is vertical, lies in column \(1\), and covers rows \(1\)–\(2\).

\(C\) consists of three horizontal dominos \(\{ (1,y),(2,y)\} \) for \(y = 1, 2, 3\). None is vertical.

Let \(k\) be a column with \(1 \le k {\lt} n\). If \(k\) is even, say \(k = 2j\), then the middle domino \(\{ (2j, 2), (2j+1, 2)\} \) spans columns \(2j\) and \(2j+1\), so there is no fault at column \(k\). If \(k\) is odd, say \(k = 2j+1\), then the basement domino \(\{ (2j+1, 1), (2j+2, 1)\} \) spans columns \(2j+1\) and \(2j+2\), so there is no fault at column \(k\).

Since \(B_n\) is the reflection of \(A_n\) (Lemma A.56), and reflection preserves faultfreeness (Lemma A.53), \(B_n\) is faultfree because \(A_n\) is (Lemma A.48).

The only potential fault in \(R_{2,3}\) is at column \(1\). But the domino \(\{ (1,1),(2,1)\} \) spans columns \(1\) and \(2\), so there is no fault.

The reflected dominos are pairwise disjoint (since the cell map is injective) and their union covers \(R_{n,3}\) (since the cell map is a bijection on \(R_{n,3}\)).

The cell map \((x, y) \mapsto (x, 4-y)\) does not change the \(x\)-coordinate. Hence \(\min \) and \(\max \) of the \(x\)-coordinates of the cells are preserved.

Reflection only changes row coordinates, not column coordinates. Therefore the minimum and maximum column of each domino are preserved (Lemma A.52), and faults (which depend only on column spans) are preserved.

The reflection swaps rows \(1 \leftrightarrow 3\) and fixes row \(2\). A vertical domino covering rows \(2\)–\(3\) in column \(c\) is sent to one covering rows \(1\)–\(2\) in column \(c\), and vice versa.

The underlying cell reflection \((x, y) \mapsto (x, 4 - y)\) is an involution on cells with \(1 \le y \le 3\).

By direct computation of the reflected domino sets. The basement dominos of \(A_n\) (in row \(1\)) reflect to the “basement” dominos of \(B_n\) (in row \(3\)), the left wall \(\{ (1,2),(1,3)\} \) reflects to \(\{ (1,1),(1,2)\} \), and similarly for the right wall, middle, and top/bottom dominos.

Reflection acts on the cells of each domino, so if \(T_1\) and \(T_2\) are equal, so are their reflections.

\(R_{n,3}\) has \(3n\) cells, and each domino covers exactly \(2\) cells. Hence \(3n\) must be even. Since \(3\) is odd, \(n\) must be even.

A vertical domino in column \(1\) covers two adjacent rows. With only \(3\) rows available, the two adjacent rows must be \(\{ 1,2\} \) or \(\{ 2,3\} \).

The vertical domino covers \((1,2)\) and \((1,3)\). The remaining cell \((1,1)\) in column \(1\) must be covered by some domino. It cannot be vertical (the cells above are taken). It cannot extend left (there is no column \(0\)). So it must be horizontal extending right: \(\{ (1,1),(2,1)\} \).

By strong induction on \(i\).

Base case (\(i = 1\)): We must prove that \(T\) contains the basement domino \(\left\{ \left( 1,1\right) ,\ \left( 2,1\right) \right\} \), the middle domino \(\left\{ \left( 2,2\right) ,\ \left( 3,2\right) \right\} \) and the top domino \(\left\{ \left( 2,3\right) ,\ \left( 3,3\right) \right\} \). For the basement domino, we have already proved this (Lemma A.60). For the middle and the top domino, we argue as follows: If \(T\) had a vertical domino in the \(2\)-nd column, then this domino would cover the top two squares of that column (since the bottom square is already covered by the basement domino \(\left\{ \left( 1,1\right) ,\ \left( 2,1\right) \right\} \)), and thus \(T\) would have a fault between the \(2\)-nd and \(3\)-rd columns (since the basement domino \(\left\{ \left( 1,1\right) ,\ \left( 2,1\right) \right\} \) also ends at the \(2\)-nd column), which would contradict the faultfreeness of \(T\). Hence, \(T\) has no vertical domino in the \(2\)-nd column. Thus, the top two squares of the \(2\)-nd column of \(T\) must be covered by horizontal dominos. These horizontal dominos must both protrude into the \(3\)-rd column (since the corresponding squares in the \(1\)-st column are already covered by the left wall), and thus must be the middle domino \(\left\{ \left( 2,2\right) ,\ \left( 3,2\right) \right\} \) and the top domino \(\left\{ \left( 2,3\right) ,\ \left( 3,3\right) \right\} \).

Induction step (\(j \to j+1\)): By the induction hypothesis, \(T\) contains the basement domino \(\left\{ \left( 2j-1,\ 1\right) ,\ \left( 2j,\ 1\right) \right\} \), the middle domino \(\left\{ \left( 2j,\ 2\right) ,\ \left( 2j+1,\ 2\right) \right\} \) and the top domino \(\left\{ \left( 2j,\ 3\right) ,\ \left( 2j+1,\ 3\right) \right\} \). We must now show that \(T\) also contains the basement domino \(\left\{ \left( 2j+1,\ 1\right) ,\ \left( 2j+2,\ 1\right) \right\} \), the middle domino \(\left\{ \left( 2j+2,\ 2\right) ,\ \left( 2j+3,\ 2\right) \right\} \) and the top domino \(\left\{ \left( 2j+2,\ 3\right) ,\ \left( 2j+3,\ 3\right) \right\} \).

The square \(\left( 2j+1,\ 1\right) \) cannot be covered by a vertical domino in \(T\), since this vertical domino would collide with the middle domino \(\left\{ \left( 2j,\ 2\right) ,\ \left( 2j+1,\ 2\right) \right\} \) (which we already know to belong to \(T\)). Thus, this square must be covered by a horizontal domino. This horizontal domino cannot protrude into the \(\left( 2j\right) \)-th column (since it would then collide with the basement domino \(\left\{ \left( 2j-1,\ 1\right) ,\ \left( 2j,\ 1\right) \right\} \), which we already know to belong to \(T\)), and thus must be the basement domino \(\left\{ \left( 2j+1,\ 1\right) ,\ \left( 2j+2,\ 1\right) \right\} \).

If \(T\) had a vertical domino in the \(\left( 2j+2\right) \)-nd column, then this domino would cover the top two squares of that column (since the bottom square is already covered by the basement domino \(\left\{ \left( 2j+1,\ 1\right) ,\ \left( 2j+2,\ 1\right) \right\} \)), and thus \(T\) would have a fault between the \(\left( 2j+2\right) \)-nd and \(\left( 2j+3\right) \)-rd columns (since the basement domino \(\left\{ \left( 2j+1,\ 1\right) ,\ \left( 2j+2,\ 1\right) \right\} \) also ends at the \(\left( 2j+2\right) \)-nd column), which would contradict the faultfreeness of \(T\). Hence, \(T\) has no vertical domino in the \(\left( 2j+2\right) \)-nd column. Thus, the top two squares of the \(\left( 2j+2\right) \)-nd column of \(T\) must be covered by horizontal dominos. These horizontal dominos must both protrude into the \(\left( 2j+3\right) \)-rd column (since the corresponding squares in the \(\left( 2j+1\right) \)-st column are already covered by the middle domino \(\left\{ \left( 2j,\ 2\right) ,\ \left( 2j+1,\ 2\right) \right\} \) and the top domino \(\left\{ \left( 2j,\ 3\right) ,\ \left( 2j+1,\ 3\right) \right\} \)), and thus must be the middle domino \(\left\{ \left( 2j+2,\ 2\right) ,\ \left( 2j+3,\ 2\right) \right\} \) and the top domino \(\left\{ \left( 2j+2,\ 3\right) ,\ \left( 2j+3,\ 3\right) \right\} \).

Since \(T\) is a tiling of \(R_{n,3}\) by dominos, the total # of squares of \(R_{n,3}\) must be even (since each domino covers exactly \(2\) squares). But this total # is \(3n\). Thus, \(3n\) must be even, so that \(n\) must be even.

(Alternatively, assuming \(n\) is odd, one can derive a contradiction using Claim 1 applied to \(i = (n-1)/2\), which forces the cell \((n,1)\) to be covered by a domino that collides with existing ones.)

Since the first column has no vertical domino, each of the three cells \((1,1)\), \((1,2)\), \((1,3)\) must be covered by a horizontal domino extending into column \(2\). These three dominos cover all of column \(2\) as well. If \(n {\gt} 2\), no domino would span columns \(2\) and \(3\), creating a fault at column \(2\), which contradicts faultfreeness. For \(n = 1\), the rectangle \(R_{1,3}\) has \(3\) cells, which is odd, so no domino tiling exists.

It clearly suffices to prove parts (a), (b) and (c). We begin with the easiest part, which is (c):

(c) Clearly, \(C\) is a faultfree domino tiling of \(R_{2,3}\) that contains no vertical domino in the first column. It remains to show that \(C\) is the only such tiling.

Indeed, let \(T\) be any faultfree domino tiling of \(R_{2,3}\) that contains no vertical domino in the first column. Thus, its first column must be filled with three horizontal dominos, which all must protrude into the second column and thus cover that second column as well. If \(T\) had any further column, then \(T\) would have a fault between its \(2\)-nd and its \(3\)-rd column, which is impossible for a faultfree tiling. Thus, \(T\) must consist only of the three horizontal dominos already mentioned. In other words, \(T=C\).

(a) It is straightforward to check that the tilings \(A_{2},A_{4},A_{6},A_{8},\ldots \) are faultfree (indeed, the basement dominos prevent faults between the \(\left( 2i-1\right) \)-st and \(\left( 2i\right) \)-th columns, whereas the top dominos prevent faults between the \(\left( 2i\right) \)-th and \(\left( 2i+1\right) \)-st columns). Thus, they are faultfree domino tilings of a height-\(3\) rectangle that contain a vertical domino in the top two squares of the first column. It remains to show that they are the only such tilings.

Indeed, let \(T\) be any faultfree domino tiling of a height-\(3\) rectangle that contains a vertical domino in the top two squares of the first column. Let \(n\) be the width of this rectangle (so that the rectangle is \(R_{n,3}\)). We shall show that \(n\) is even, and that \(T=A_{n}\).

We know that \(T\) contains a vertical domino in the top two squares of the first column. In other words, \(T\) contains the left wall. The remaining square in the first column is the square \(\left( 1,1\right) \), and it must thus be covered by the basement domino \(\left\{ \left( 1,1\right) ,\ \left( 2,1\right) \right\} \) (since no other domino would fit). Hence, \(T\) contains the basement domino \(\left\{ \left( 1,1\right) ,\ \left( 2,1\right) \right\} \).

Now, Claim 2 (Lemma A.62) shows that \(n\) is even, so that \(n/2\) is a positive integer. Furthermore, we know that the tiling \(T\) contains the left wall, the first \(n-1\) basement dominos (by Claim 1, Lemma A.61), all the middle dominos and all the top dominos (also by Claim 1). This leaves only the four squares \(\left( n-1,\ 1\right) \), \(\left( n,\ 1\right) \), \(\left( n,\ 2\right) \) and \(\left( n,\ 3\right) \) unaccounted for, but there is only one way to tile them: namely, with the last basement domino \(\left\{ \left( n-1,\ 1\right) ,\ \left( n,\ 1\right) \right\} \) and the right wall \(\left\{ \left( n,2\right) ,\ \left( n,3\right) \right\} \). Thus, \(T\) must contain all the basement dominos, all the middle dominos, all the top dominos and both walls (left and right). Since these dominos cover all the squares of \(R_{n,3}\), this entails that \(T\) consists precisely of these dominos. In other words, \(T=A_{n}\).

(b) Proposition A.64 (b) is just Proposition A.64 (a) reflected across the horizontal axis of symmetry of \(R_{n,3}\). More precisely, if \(T\) is a faultfree tiling with a bottom vertical in column \(1\), then its reflection \(T'\) is faultfree (Lemma A.53) with a top vertical in column \(1\) (Lemma A.54). By part (a), \(T' = A_n\) for some even \(n \ge 2\). Since \(B_n\) is the reflection of \(A_n\) (Lemma A.56) and reflection preserves equality of tilings (Lemma A.57), reflecting back gives \(T = B_n\).

The overall classification follows by Lemma A.59: any faultfree tiling of \(R_{n,3}\) with \(n \ge 1\) falls into exactly one of the three cases (top vertical, bottom vertical, or no vertical in column \(1\)).

By Claim 2, \(n\) is even and \(n \ge 2\). By Claim 1 and the base case lemma, \(T\) contains all the dominos of \(A_n\). Since both \(T\) and \(A_n\) tile the same rectangle, they have the same number of dominos, and so \(T = A_n\).

Reflect \(T\) to obtain \(T'\), which is faultfree with a top vertical in column \(1\). By part (a), \(T' = A_n\) for some even \(n \ge 2\). Since reflection is an involution, \(T\) is the reflection of \(T'\), and hence \(T\) equals the reflection of \(A_n\), which is \(B_n\).

In a \(2 \times 3\) rectangle with no vertical domino in column \(1\), every domino must be horizontal (since a vertical domino in column \(2\) would force its column-\(1\) neighbor to also be vertical in column \(1\), or would leave a cell that can only be covered by a vertical domino in column \(1\)). Hence the tiling consists of three horizontal dominos \(\{ (1,y),(2,y)\} \) for \(y = 1, 2, 3\), which is exactly \(C\).

The vertical domino has column coordinate \(c\). Since all cells of a tiling lie in \(R_{n,3}\), the column coordinate satisfies \(c \le n\).

By Claim 2 (Lemma A.62), \(n\) is even. Since \(T\) is nonempty (faultfree implies nonempty), \(n \ge 1\), and being even forces \(n \ge 2\).

The cell \((1,1) \in R_{n+1,2}\), and by the coverage condition, some domino in \(T\) must contain it.

The domino is either horizontal or vertical. If horizontal, its second cell is \((2,1)\) (cannot be \((0,1)\) since \(0 {\lt} 1\)). If vertical, its second cell is \((1,2)\) (cannot be \((1,0)\) since \(0 {\lt} 1\)).

The cell \((1,2)\) must be covered by some domino \(d'\). This \(d'\) cannot overlap with \(d\) (which occupies \((1,1)\) and \((2,1)\)). The only possibility for \(d'\) is \(\{ (1,2),(2,2)\} \).

By Lemma A.70, some domino covers \((1,1)\). By Lemma A.71, it is either vertical (giving the first case) or horizontal (giving the second case by Lemma A.72).

The prepended vertical covers column \(1\); the shifted dominos cover columns \(2\) through \(n+1\). Pairwise disjointness holds by coordinate separation.

The two horizontal dominos cover columns \(1\)–\(2\); the shifted dominos cover columns \(3\) through \(n+2\).

After removing the vertical domino, the remaining dominos all have column coordinates \(\ge 2\). Shifting left by \(1\) maps them to \(R_{n,2}\). Disjointness and coverage are preserved.

After removing the two horizontal dominos, the remaining dominos all have column coordinates \(\ge 3\). Shifting left by \(2\) maps them to \(R_{n,2}\).

If \(T\) has a vertical first column, the restrict-then-prepend roundtrip produces a tiling whose dominos are the vertical domino plus all remaining dominos shifted right then left, recovering the original. The horizontal pair case is analogous.

Take \(x\) to be the classification of \(T\) (by its first-column structure). Then the prepended image equals \(T\) by Lemma A.78.

If \(x\) corresponds to a tiling \(T\) of \(R_{n+1,2}\): prepending a vertical domino produces a tiling whose first column is vertical, so the classification recovers \(T\) after restriction. The key is that restricting after prepending a vertical domino recovers \(T\).

If \(x\) corresponds to a tiling \(T\) of \(R_{n,2}\): prepending a horizontal pair produces a tiling whose first column is not vertical, so the classification recovers \(T\) after restriction.

For the tiling conversion: the converted dominos are pairwise disjoint (since the coordinate casting preserves disjointness) and their union covers the integer-based rectangle (by the rectangle correspondence).

The map is injective (since \(\mathbb {N} \hookrightarrow \mathbb {Z}\) is injective) and surjective onto elements with positive coordinates.

Since the coordinate casting preserves the cell set (up to the injective \(\mathbb {N} \hookrightarrow \mathbb {Z}\) embedding), disjointness is preserved.

Every cell \((i,j)\) in the \(\mathbb {Z}\)-based rectangle corresponds (via Lemma A.82) to a cell in the natural-number-based rectangle, which is covered by some domino. The image of that domino under the coordinate casting covers the original cell.