Enumeration of tilings of quartered Aztec rectangles

We generalize a theorem of W. Jockusch and J. Propp on quartered Aztec diamonds by enumerating the tilings of quartered Aztec rectangles. We use subgraph replacement method to transform the dual graph of a quartered Aztec rectangle to the dual graph of a quartered lozenge hexagon, and then use Lindstr\"{o}m-Gessel-Viennot methodology to find the number of tilings of a quartered lozenge hexagon.


Introduction
A lattice divides the plane into fundamental regions. A (lattice) region is a finite connected union of fundamental regions of that lattice. A tile is the union of any two fundamental regions sharing an edge. A tiling of the region R is a covering of R by tiles with no gaps or overlaps. Denote by T(R) the number of tilings of the region R.
In general, the tiles of a region R can carry weights. The weight of a tiling is defined to be the product of the weights of all constituent tiles. The operation T(R) is now defined to be the sum of the weights of all tilings in R, and is called the tiling generating function of R. If R does not have any tiling, we let T(R) := 0.
Denote by H a,b,c the hexagon of sides a, b, c, a, b, c (in cyclic order, starting from the northwestern side) in the triangular lattice. Divide the hexagon H m,2(n−k)+1,m , where k = ⌊ m+1 2 ⌋, into four equal parts by its vertical and horizontal symmetry axes (see Figure 1.1 for an example). We consider the portion of of the hexagon that consists of unit triangles lying completely inside the upper right quarter. Remove the a 1 -st, the a 2 -nd, . . . , and the a k -th up-pointing unit triangles from the bottom of the portion. Denote by QH m,n (a 1 , a 2 , . . . , a k ) the resulting region. See the region restricted in the bold contour in Figure 1.1 for an example with k = 7, m = 13, n = 12, a 1 = 2, a 2 = 3, a 3 = 5, a 4 = 7, a 5 = 8, a 6 = 10, a 7 = 12; and Figure 1.2(a) shows an example, for k = 6, m = 12, n = 11, a 1 = 2, a 2 = 3, a 3 = 5, a 4 = 6, a 5 = 8, a 6 = 11.
Denote the resulting region by QH 2k,n (a 1 , a 2 , . . . , a k ).We have also a similar variant of QH 2k−1,n (a 1 , a 2 , . . . , a k ) defined as follows. We assign all k − 1 vertical rhombus on the left side of QH 2k−1,n (a 1 , a 2 , . . . , a k ) a weight 1/2, and remove the leftmost up-pointing unit triangle from the bottom of the region. Next, we remove the a 1 -st, the a 2 -nd, . . . , and the a k−1 -th up-pointing unit triangles from the bottom of the resulting region. The new region is denoted by QH 2k−1,n (a 1 , a 2 , . . . , a k−1 ) (illustrated in Figure 1.2(b)).
We call the four regions above holey quartered hexagons. The number of tilings of a holey quartered hexagon is given by the following theorem.
Hereafter, the empty products (like 1≤i<j≤k (a j − a i ) for k = 1) equal 1 by convention.
Theorem 1.1. For any 1 ≤ k < n and 1 ≤ a 1 < a 2 < . . . < a k ≤ n Consider a (2m+1)×(2n+1) rectangular chessboard and suppose that the corners are black. The graph whose vertices are the white unit squares of the chessboard, and whose edges connect diagonally adjacent white unit squares, is called the Aztec rectangle graph of order (m, n), and is denoted by AR m,n (see Figure 3.11(a) for an example).
A perfect matching of a graph G is a collection of edges such that each vertex of G is adjacent to exactly one selected edge. Denote by M(G) the number of perfect matchings of G.
It has been proved that the number of perfect matchings of an Aztec rectangle graph with holes (i.e. the vertices that are removed) on one side is given by a simple product formula (see [1], (4.4); [8], Lemmas 1, 2; or [6] Lemma 2).
We now consider a new situation in which we allow the holes appear on two adjacent sides of an Aztec rectangle graph. We define two new families of holey Aztec rectangle graphs in the next two paragraphs.
Label the leftmost vertices of the Aztec rectangle graph AR m,n from bottom to top by 1, 2, . . . , m, and label the bottommost vertices of the graph from left to right by 1, 2, . . . , n. Remove all the leftmost vertices having even labels, and remove all bottommost vertices, except for the a 1 -st, the a 2 -nd, . . . , and the a k -th vertices, where 1 ≤ k ≤ n. Denote by QE 1 m,n (a 1 , a 2 , . . . , a k ) the resulting graph ( Figure  1.3(a) shows an example with m = 7, n = 10, a 1 = 2, a 2 = 3, a 3 = 6, a 4 = 9; the white circles indicate the vertices removed).
Next, we also start with the Aztec rectangle graph AR m,n , then remove all leftmost vertices having odd labels (as oppose to removing the ones with even labels as in the previous paragraph), and remove all bottommost vertices from AR m,n , except for the a 1 -st, the a 2 -nd, . . . , and the a l -th vertices, where 1 ≤ l ≤ n. Denote by QE 2 m,n (a 1 , a 2 , . . . , a l ) the resulting graph (an example of the graph is shown in Figure  1 The numbers of perfect matchings of the two holey Aztec rectangle graphs are given by the theorem stated below. Theorem 1.2. For any 1 ≤ k < n and 1 ≤ a 1 < a 2 < . . . < a k ≤ n M(QE 1 2k,n (a 1 , a 2 , . . . , a k )) = M(QE 1 2k−1,n (a 1 , a 2 , . . . , a k )) 2k+1,n (a 1 , a 2 , . . . , a k )) = M(QE 2 2k,n (a 1 , a 2 , . . . , a k )) Note that the two holey Aztec rectangle graphs are bipartite graphs, so the numbers of vertices in their two vertex classes must be the same if they admit perfect matchings. One readily sees that the "balancing condition" between two vertex classes of QE 1 m,n (a 1 , a 2 , . . . , a k ) holds if and only if k = ⌊ m+1 2 ⌋, and the balancing condition of QE 2 m,n (a 1 , a 2 , . . . , a l ) holds if and only if l = ⌊ m 2 ⌋. We consider next two variants of the holey Aztec rectangle graphs above in the next paragraph.
First, we remove all the leftmost vertices from the graph AR m,n+1 , label the leftmost vertices of the resulting graph from bottom to top by 1, 2, . . . , m + 1, and label the bottommost vertices of the resulting graph, except for the first one, from left to right by 1, 2, . . . , n. Second, we remove the first vertex and all vertices with even labels on the left side of the graph, and remove all vertices in the bottom, except for the ones with labels a 1 , a 2 , . . . , a k , where 1 ≤ k ≤ n. Denote by QE    vertices with even labels), we get the graph QE 2 m,n (a 1 , a 2 , . . . , a l ), where 1 ≤ l ≤ n (see Figure 1.4(b) for an example).

Proof of Theorem 1.1
Similar to the case of regions with weighted tiles, we can generalize the definition of the operation M(G) to the case of weighted graph G as follows. The weight of a perfect matching is defined to be the product of the weights of all constituent edges. The operation M(G) is now defined to be the sum of the weights of all perfect matchings in G, and is called the matching generating function of G. If G does not have any perfect matching, we let M(G) := 0.
The dual graph of a region R is the graph whose vertices are the fundamental regions in R and whose edges connect precisely two fundamental regions sharing an edge. Each edge of the dual graph carries the weight of the corresponding tile of the region. The number of tilings (resp., the tiling generating function) of a region and the number of perfect matchings (resp., the matching generating function) of its dual graph are equal by a well-known bijection.
A graph G with symmetric axis; (b) the resulting graph after the cutting procedure.
Next, we consider a useful Factorization Theorem due to Ciucu [1], which we will employ in the proof of Theorem 1.1.
Let G be a weighted planar bipartite graph that is symmetric about a vertical line ℓ. Assume that the set of vertices lying on ℓ is a cut set of G (i.e., the removal of these vertices disconnects G). One readily sees that the number of vertices of G on ℓ must be even if G has perfect matchings, let w(G) be half of this number. Let a 1 , b 1 , a 2 , b 2 , . . . , a w(G) , b w(G) be the vertices lying on ℓ, as they occur from top to bottom. Let us color vertices of G by black or white so that any two adjacent vertices have opposite colors. Without loss of generality, we assume that a 1 is always colored white. Delete all edges on the left of ℓ at all white a i 's and black b j 's, and delete all edges on the right of ℓ at all black a i 's and white b j 's. Reduce the weight of each edge lying on ℓ by half; leave all other weights unchanged. Since the set of vertices of G on ℓ is a cut set, the graph obtained from the above procedure has two disconnected parts, one on the left of ℓ and one on the right of ℓ, denoted by G + and G − respectively (see Figure 2.1). Then Ciucu's Factorization Theorem [1] implies that Next, we quote a result on the number of tilings of a semi-hexagon due to Helfgott and Gessel [6]. A semi-hexagon of sides a, b, a, a + b is the portion of a hexagon of sides a, b, a, a, b, a (in cyclic order, starting from the northwestern side) in the triangular lattice that stays above the horizontal symmetric axis of the hexagon. We are interested in the number of tilings of the semi-hexagon sides a, b, a, a + b, where the s 1 -st, the s 2 -nd, . . . , and the s a -th up-pointing unit triangles in the base have been removed, denoted by SH a,b (s 1 , s 2 , . . . , s a ) (the removed unit triangles were called dents in [6]). The following determinant identity has been proved by Krattenthaler [9].
Proof of Theorem 1.1. Write for short R 1 := QH 2k−1,n (a 1 , a 2 , . . . , a k ). We use a standard bijection mapping each tiling µ of the region R 1 in the triangular lattice to a k-tuple of non-intersection lattice paths taking steps west or north on the square grid Z 2 .
Label the middles of the left sides of up-pointing unit triangles along the left boundary of R 1 from bottom to top by v 1 , v 2 , . . . , v k . Label the middles of the left sides of up-pointing unit triangles, which are removed from the bottom of the region, from left to right by u 1 , u 2 , . . . , u k (see Consider now a rhombus r 1 of µ whose one side contains u i , for some arbitrary but fixed 1 ≤ i ≤ k. Denote by w 1 the middle of the side of r 1 opposite the side containing u i . Let r 2 be other rhombus of µ that has a side containing w 1 . Denote by w 2 the middle point of the side of r 2 opposite the side containing w 1 . Continue our rhombi selecting process by picking a new rhombus r 3 of µ that has a side containing w 2 . This process gives a path of rhombi growing upward, and ending in a rhombus containing one of the v j 's (see the paths of shaded rhombi in Figure 2.2(b)). We can identify this path of rhombi with the linear path Consider next the obtuse (120 0 angle) coordinate system whose origin at v 1 and whose x-axis contains all the points u i 's (see Figure 1.2(a)). The linear path connecting u i and v j is a lattice path in this coordinate. Normalize this coordinate system and rotating it in standard position, we get a lattice path on square grid Z 2 (see Figure 1.2(c)). It is easy to see that v j has coordinate (j − 1, 2j − 2) and u i has coordinate (a i − 1, 0), for any 1 ≤ i, j ≤ k .
We obtain this way a k-tuple P of lattice paths (one path for each 1 ≤ i ≤ k), and they cannot touch each other (since the corresponding paths of rhombi are disjoint). One readily sees that the correspondence µ → P is a bijection between the set of tilings of R 1 and the set of k-tuples P of non-intersecting lattice paths starting at u 1 , . . . , u k , and ending at v 1 , . . . , v k . By Lindström-Gessel-Viennot theorem ( [12], Lemma 1; [14] Theorem 1.2), the number of such k-tuples P of non-intersection lattice paths is given by the determinant of the k × k matrix A whose (i, j)-entry is the number of lattice paths from Bijection between tilings of R 2 and families of nonintersecting paths.
Swap the jth and the (n − j + 1)th columns, for any 1 ≤ j ≤ k, in the matrix on the right hand side of (2.4), we get a new matrix Apply Lemma 2.2, with C = 1 and X i = a i and A j = n − j, to the matrix B, we obtain Therefore, by (2.4), (2.5), and (2.6), we have and (1.1) follows. Next, we prove (1.2) by the same method. We also have a bijection between the set of tilings of R 2 := QH 2k,n (a 1 , a 2 , . . . , a k ) and the set of k-tuples of non-intersecting lattice path connecting u 1 , . . . , u k and v 1 , . . . , v k , the only difference here is that the obtuse coordinate system is now selected so that v 1 has coordinate (0, 1) (as oppose to having coordinate (0, 0) in the proof of (1.1)). Figure 2.3 illustrates an example corresponding to the region in Figure 1.2(a). One readily sees that u i has also coordinate (a i − 1, 0), and v j has now coordinate (j − 1, 2j − 1) in the new coordinate system, for 1 ≤ i, j ≤ k. Again, by Lindström-Gessel-Viennot Theorem the number of tilings of R 2 is given by the determinant of the k × k matrix D whose Factor out 1 (2j−1)! from the jth column, for any 1 ≤ j ≤ k, of the matrix D, we obtain Factor out a i from the ith row of the matrix on the right hand side of (2.8), for any Swap the jth and the (n − j + 1)th columns, for any 1 ≤ j ≤ k, of the matrix on the right hand side of (2.9), we get a new matrix Apply Lemma 2.2, with C = 0 and X i = a i and A j = n − j, to the matrix E, we have Therefore, by (2.8), (2.9), (2.11), and (2.12), we ontain Apply the Factorization Theorem to the dual graph G of the semi-hexagon SH 2k,2n (S), We get G − is isomorphic to the dual graph of the region QH 2k,n (a 1 , . . . , a k ); and after removing all forced edges on the top of G + , we get a graph isomorphic to the dual graph of QH 2k−1,n (a 1 , . . . , a k ) (see Figure 2.4(a) for an example with k = 3, n = 7, a 1 = 2, a 2 = 4, a 3 = 6). Therefore, we obtain (2.14) T(SH 2k,2n (S)) = 2 k T(QH 2k−1,n (a 1 , . . . , a k )) T(QH 2k,n (a 1 , . . . , a k )).
By (1.1) and (2.14), together with Lemma 2.1, we have Proof. Denote by T C(a, a, 2b) the number of transposed complementary plane partitions contained in the box a × a × 2b. There is a bijection between the latter plane partitions and the tilings of the hexagon H a,2b,a that are invariant under refection across the vertical symmetry axis ℓ of the region (one can see [1], Theorem 6.1, for more details). Therefore, T C(a, a, 2b) is equal to the number of tilings of the subregion R of H a,2b,a that consists of unit triangles on the left of ℓ. Denote by R ′ the region obtained from R by removing the rows of horizontal rhombi that are forced on the top and on the bottom of R (see The following two families of graphs will play a special role in the proof of Theorem 1.2. Remove all the leftmost and bottommost vertices of the Aztec rectangle graph AR m,n , and denote by OR m,n the resulting graph (is sometimes called an odd Aztec rectangle graph). Apply the procedure in the definition of QE 1 m,n (a 1 , . . . , a k ) and QE 2 m,n (a 1 , . . . , a l ) to the graph OR m,n (instead of the Aztec rectangle AR m,n ). In particular, we label all the leftmost (resp., bottommost) vertices of OR m,n from bottom to top (resp., from left to right) by 1, 2, . . . , n (resp., by 1, 2, . . . , m). Remove all vertices with even labels on the left side, and remove the a 1 -st, the a 2 -nd, . . . , and the a l -th vertices from the bottom, where 1 ≤ l ≤ n. Denote by QO 1 m,n (a 1 , a 2 , . . . , a l ) the resulting graph (see Figure 3(a) for an example with m = 7, n = 10, a 1 = 1, a 2 = 2, a 3 = 4, a 4 = 6, a 5 = 7, a 6 = 9, a 7 = 10).
Similarly, we remove all vertices with odd labels from the left, and the a 1 -st, the a 2 -nd, . . . , and the a k -th vertices from the bottom of OR m,n , where 1 ≤ k ≤ n. We get the graph QO 2 m,n (a 1 , a 2 , . . . , a k ) (see Figure 3(b) for an example with m = 7, n = 10, a 1 = 2, a 2 = 7, a 3 = 5, a 4 = 7, a 5 = 8, a 6 = 10). We use the notations QO 1 m,n (∅) and QO 2 m,n (∅) if no bottom vertex has been removed. Note that the balancing conditions of the two graphs QO 1 m,n (a 1 , a 2 , . . . , a l ) and QO 2 m,n (a 1 , a 2 , . . . , a k ) are l = ⌊ m 2 ⌋ and k = ⌊ m−1 2 ⌋, respectively. The numbers of perfect matchings of the two families of graphs are given by the following theorem, that will be proved at the same time as Theorem 1.2.
2k,n (a 1 , a 2 , . . . , a k )) = M(QO 2 2k−1,n (a 1 , a 2 , . . . , a k )) An edge in a graph G is called a forced edge, if it is in every perfect matching of G. Let G be a weighted graph with weight function wt on its edges, and G ′ is obtained from G by removing forced edges e 1 , . . . , e k , and removing the vertices incident to those edges. Then one clearly has From now on, whenever we remove some forced edges, we remove also the vertices incident to them. We have the following fact by considering forced edges.
Proof. The proofs of the first two equalities are illustrated in Figures 3.1(a) and (b) respectively. The last two equalities can be obtained similarly.
Next, we will employ several basic preliminary results stated below.        (c) The statement of part (b) is also true when K and K are the graphs indicated in Figure 3.4(b) (in this case G ′ has two new vertices C and D, they are adjacent only to one another and to B and A, respectively). By the above fundamental lemmas, we have the following fact.
Lemma 3.7. For any 1 ≤ k < n and 1 ≤ a 1 < a 2 < . . . < a k ≤ n n (a 1 , . . . , a k )),  Proof. The proofs of (3.7) and (3.8) are essentially the same, so we present only the proof of (3.7). The proof of (3.7) is illustrated in Figure 3.6, for the case k = 3, n = 6, a 1 = 1, a 2 = 3, and a 3 = 6. First, apply Vertex-splitting Lemma 3.3 to all vertices of G := QE 1 2k−1,n (a 1 , . . . , a k ) which are on the dotted lines as in Figure 3.6(a). In particular, the vertices on the vertical dotted lines are separated horizontally, and the vertices on the horizontal dotted lines are separated vertically. We get the solid graph in Figure 3.6(b). Second, apply the suitable replacements in Spider Lemma 3.5 at the positions of the (2k−1)n−n+1 diamond and n−1 partial diamonds in the solid graph; the diamonds and partial diamonds with legs will be replaced by the 4cycle with the dotted edges. We get the graph in the Figure 3.6(c), the dotted edges are weighted by 1/2. Third, remove all forced edges (the solid edges in 3.6(c)), and apply the Star Lemma 3.4 with factor t = 2 to the resulting graph at all (2k −1)n−k black vertices shown in Figure 3.8(c). We get the graph G ′ := QO 2 2k−1,n (a 1 , . . . , a k ). By Lemmas 3.3, 3.4 and 3.5, we obtain which implies (3.7).
Remark 1. One can have another proof of the equality (3.7) by using the Complementation Theorem [2]. The cellular completion of QE 1 2k−1,n (a 1 , . . . , a k ) is the graph G obtained from QO 2 2k−1,n (a 1 , . . . , a k ) by assigning all edges a weight 1/2. Moreover, it is easy to see that each perfect matching of G ′ consists of (2k − 1)n − k edges of weight 1/2, so each perfect matching of G has weight 2 −(2k−1)n+k . Thus, (3.7) follows.
The connected sum G#G ′ of two disjoint graphs G and G ′ along the ordered sets of vertices {v 1 , . . . , v n } ⊂ V (G) and {v ′ 1 , . . . , v ′ n } ⊂ V (G ′ ) is the graph obtained from G and G ′ by identifying vertices v i and v ′ i , for i = 1, . . . , n. Lemma 3.8. Let G be a graph , and let {v 1 , v 2 , . . . , v n } be an ordered subset of its vertex set.  (a) Assume that K is the graph obtained from QE 2 2q,n ([n]) by appending n vertical edges to n vertices in its bottom. Then  In the two equalities (3.10) and (3.11), the connected sum acts on G along {v 1 , v 2 , . . . , v n } and acts on the other two summands along their bottom vertices (ordered from left to right).
Proof. Since the proofs of parts (a) and (b) are essentially the same, the proof of part(b) is omitted.
The illustration of the proof of part (a) is shown in Figure 3.8. First, we apply Vertex-splitting Lemma 3.3 to all vertices of K staying on the dotted line in Figure  3.8(a). In particular, the vertices on the vertical dotted lines are separated horizontally, and the vertices on the horizontal dotted lines are separated vertically. We get the graph with solid edges in Figure 3.8(b). Second, apply the suitable replacements in Spider Lemma 3.5 at the places of (2q − 1)(n − 1) diamonds and 4q + n − 2 partial diamonds in the graph with solid line; the diamonds and partial diamonds with legs will be replaced by the 4-cycles with the dotted edges. We get the graph in the     Lemma 3.9. For 1 ≤ k < n and 1 ≤ a 1 < a 2 < . . . < a k ≤ n (3.12) M(QO 2 2k−1,n (a 1 , a 2 , . . . , a k )) = 2 k(k−1) T(QH 2k−1,n (a 1 , a 2 , . . . , a k )) Proof. Apply alternately the replacements in Lemma 3.8 parts (a) and (b), for q = 1, 2, . . . , k−1, to the dual graph of QH k,n (a 1 , a 2 , . . . , a k ) from the top. The procedure is illustrated by Figures 3.9 and 3.10, the portion above the dotted line in a graph is replaced by the portion above that line in the next graph; the graphs in Figure 3.9(d) and 3.10(a) are the same. Finally, we get the graph QO 2 2k−1,n (a 1 , a 2 , . . . , a k ), and (3.13) M(QO 2 2k−1,n (a 1 , a 2 , . . . , a k )) T(QH 2k−1,n (a 1 , a 2 , . . . , a k )) which implies (3.12).
Before presenting the proof of Theorems 1.2 and 1.3, we quote two results about the number of perfect matchings of an Aztec rectangle graph with holes on one side.  Next, we consider a variant of the lemma above (see [6], Lemma 2; or [8], Lemma 2).
Lemma 3.11. The number of perfect matchings of a m × n Aztec rectangle, where all the vertices in the bottom-most row have been removed, and where the a 1 -st, the a 2 -nd, . . . , and the a m -th vertex, have been removed from the resulting graph (see Figure 3.11(c), for and example with m = 3, n = 5, a 1 = 3, a 2 = 4,a 3 = 6), equals Denote by AR m,n (a 1 , . . . , a m ) and AR m,n (a 1 , . . . , a m ) the graphs in Lemmas 3.10 and 3.11, respectively.
Next, we combine the proofs of Theorem 1.2 and Theorem 3.1 into a single proof as follows.
The following two families of graphs will play the key role in the proof of Theorem 1.3.
We remove all the bottommost vertices of AR m,n , and label the vertices on the left side of resulting graph by 1, 2, . . . , m (from bottom to top), and label the vertices in its bottom, except for the first one, by 1, 2, . . . , n (from left to right). Second, we remove the first vertex and all vertices with even labels on the left side of the graph, and remove the vertices with labels a 1 , a 2 , . . . , a l from it bottom, for 1 ≤ l ≤ n. Denote by QO 1 m,n (a 1 , a 2 , . . . , a l ) the resulting graph (see Figure 3.13(a) for an example).
Repeat the procedure in the previous paragraph, the only change is that we remove now the vertices of odd labels on the left side in the second step, we get the graph QO 2 m,n (a 1 , a 2 , . . . , a k ), where a 1 , a 2 , . . . , a k are the labels of vertices removed from the bottom, and where 1 ≤ k ≤ n (see Figure 3.13(b) for an example). One readily sees that the balancing conditions for the two graph in the previous paragraph are l = ⌊ m 2 ⌋ and k = ⌊ m−1 2 ⌋, respectively. The numbers of perfect matchings of the above two families of graphs are given by the following theorem.

Quartered holey Aztec rectangles
The Aztec diamond of order n is defined to be the union of all the unit squares with integral corners (x, y) satisfying |x| + |y| ≤ n + 1. The Aztec diamond of order 7 is shown in Figure 4.1, and the Aztec diamond of order 8 is shown in Figure 4.2. It has been shown that the number of tilings of the Aztec diamond of order n is 2 n(n+1) [4].
An Aztec rectangle region is defined to be a portion of an Aztec diamond as the shaded regions in Figures 4.1(a) and (b), for the case of m < n and m > n respectively. One readily see that the dual graph of the Aztec rectangle region of order (m, n) is the Aztec rectangle graph of the same order. Denote by AR m,n the Aztec rectangle region of order (m, n).
Jockusch and Propp [7] considered a region, named quartered Aztec diamonds, obtained from an Aztec diamond by dividing into four parts by two zigzag cuts. Figure 4.2 illustrates the three kinds of quartered Aztec diamonds of order 8, that are denoted by R(8), K a (8) nd K na (8) (on can see the precise definition in [1]). We generalize the family of quartered Aztec diamonds to a family of regions, which we called quartered holey Aztec rectangles 1 , in the next paragraph.
Consider an Aztec rectangle AR a,b . Denote by ℓ nd ℓ ′ the southeast-to-northwest and the southwest-to-northeast symmetry axes of the region. Remove some unit squares along ℓ so that their appearance is symmetric about ℓ ′ . In particular, there are four cases to distinguish as follows.
We remove 2l = 2n−2m unit squares along ℓ. Divide the region into two congruent parts by a zigzag cut with (0,2) and (2, 0) steps that passes the center of the region and runs along ℓ ′ . Divide also the region into two congruent parts by a zigzag cut passing the center of the region and running along ℓ, so that when we identify the opposite vertices on ℓ of any removed unit squares, the zigzag cut becomes a zigzag cut with (0, −2) and (2, 0) steps. Up to symmetric, we have two ways to superimpose the two zigzag cuts as in Figure 4 We remove 2l = 2m − 2n unit squares running along ℓ. Similar to Case 1, there are two ways to superimpose the two zigzag cuts as in Figure 4  We admit a removed unit square containing the center of the region. The number of removed unit square along ℓ is now 2l + 1 = 2m − 1 − 2n. Divide the region into 4 parts by superimposing the two zigzag cuts running along ℓ and ℓ ′ , so that when we identify four vertices of the center square and identify the opposite vertices on ℓ of any other removed unit squares, we get two zigzag cuts of 2-unit steps as in Cases 1 and 2. Up to symmetry, we also have two ways to superimpose the two zigzag cuts as in Figure 4.5.
We admit also a removed unit square containing the center of the region. The number of removed unit square along ℓ is now 2l + 1 = 2n+ 1 −2m. Up to symmetry we have two ways to superimpose the two zigzag cuts as in Figure 4.6.
Call four pieces obtained from the cutting procedure above quartered holey Aztec rectangles. We label all unit squares running along ℓ by 1, 2, 3, . . . away from the center (if there is a square containing the center, then that square is labeled 0). Next, we color alternately black or white all the unit squares resting on ℓ which are not removed. Going along ℓ from left to right, we assume that the first colored unit square on the right of ℓ ′ is white. Let O and E be the label sets of the white and the black unit squares, respectively. Denote by RN a,b (S), RE a,b (S), RS a,b (S) and        ). Remark 2. In the Cases 1 and 2, our procedure still works when m = n, and when there are no removed unit squares. Then our quartered holey Aztec rectangles become the quartered Aztec diamonds. This means that Theorems 1.2,1.3, 3.1,3.12 imply the formulas for the number of tilings of quartered Aztec diamonds (see [1]). The author has also a simple proof for the number of tilings of quartered Aztec diamonds [11].
We are interested in the number cyclically symmetric tilings of the holey Aztec rectangle region that are invariant under the 180 0 -rotation at the center. We call these tilings cyclically symmetric tilings. Denote by T * (R) the number of cyclically symmetric tilings of the region R.
Recall that the operation ∆ on a finite set S := {s 1 , s 2 , . . . , s k } is defined to be the product 1≤i<j≤k (s j − s i ). We define five more operations as follows.  where, as mentioned before, the empty products are equal 1 by convention. The number of cyclically symmetric tilings of a holey Aztec rectangle region is given by the following theorem.  Proof. We prove only part (a) (the proofs of the other parts are similar, and are omitted).
Consider the dual graph G of the the region AR 2m,2n (c 1 , c 2 , . . . , c n−m )), i.e. the graph AR 2m,2n with the corresponding vertices on its horizontal symmetry axis ℓ removed. Let G be the subgraph of G induced by the vertices below or on ℓ (see  . Let S be the set of vertices of G lying on ℓ. Label two vertices of S lying closest to the center of the graph by 1, label the two next closest vertices by 2 and so on. Let G be the graph obtained from G by identifying all pair of vertices having the same label (see Figure 4.7(b)). The number of cyclically symmetric tilings of the region is equal to the number of cyclically symmetric perfect matchings of its dual graph, and the latter number is exactly the number of perfect matchings of G.
Note that we can put all vertices of G, which are obtained from identifying two vertices of the same label in G, on the vertical symmetric axis of G. Thus, G admits a vertical symmetry axis, and we can apply the Factorization Theorem to G (see Figure 4.8). Since the number of holes in the Aztec rectangle graph is l = 2m − 2n, we have 2w(G) = m + (n − l) = 2m. Moreover, G − is isomorphic to QE 2 m,m (E), and G + is isomorphic to QE 1 m,n (O). Therefore, we obtain