Perfect matchings of trimmed Aztec rectangles Tri Lai

We consider several new families of subgraphs of the square grid whose matchings are enumerated by powers of several small prime numbers: 2, 3, 5, and 11. Our graphs are obtained by trimming two opposite corners of an Aztec rectangle. The result yields a proof of a conjecture posed by Ciucu. In addition, we reveal a hidden connection between our graphs and the hexagonal dungeons introduced by Blum.


Introduction and main results
A perfect matching of a graph G is a collection of edges of G so that each vertex is covered by precisely one edge in the collection.A perfect matching is sometimes called 1-factor (in graph theory) or dimmer covering (in statistical mechanics).The field of enumeration of perfect matchings date back to the early 1960's when Kasteleyn [8] and Temperley and Fisher [29] found the number of tilings of a rectangle on the square grid.In 1992, Elkies, Kuperberg, Larsen, and Propp [6,7] proved a striking pattern for the number of perfect matchings of an Aztec diamond graph (see the definition below), which is a power of 2. This result opened two-and-a-half flourishing decades of the enumeration of perfect matchings.
Consider a √ 2m × √ 2n rectangular contour rotated 45 • and translated so that its vertices are centers of unit squares on the square grid (see the dotted contour on the left picture of Figure 1.1).The m × n Aztec rectangle (graph) 1 AR m,n is the subgraph of the square grid induced by the vertices inside or on the boundary of the rectangular contour (see the graph restricted by the left bold contour in Figure 1.1 for the Aztec rectangle AR 6,8 ).The Aztec rectangle AR m,n is called the Aztec diamond of order n when m = n.The augmented Aztec rectangle AA m,n is obtained by stretching the Aztec rectangle AR m,n one unit horizontally, i.e. adding one square to the left of each row in the Aztec rectangle (see the graph restricted by the black contour on the right of Figure 1.1; the added squares are shaded ones).We can still define the Aztec rectangle and augmented Aztec rectangle on sub-grids or weighted versions of the square grid.The graph AA n,n is called the augmented Aztec diamond of order n.
It has been shown in [6,7] that the Aztec diamond of order n (i.e. the Aztec rectangle AR n,n ) has 2 n(n+1)/2 perfect matchings, and it is easy to see that AR m,n has no perfect matching when m = n.Sachs and Zernitz [26] proved that the augmented Aztec diamond of order n has D(n, n) perfect matchings, where the Delannoy number D(m, n), for m, n 0, is the number of lattice paths on Z 2 from the vertex (0, 0) to the vertex (m, n) using north, northeast and east steps (see e.g.Exercise 6.49 in [28]).Dana Randall later gave a simple combinatorial proof for the result.More general, the number of perfect matchings of the augmented Aztec rectangle AA m,n is given by the D(m, n), for any m, n.
Various families of subgraphs of the square grid have been investigated, focusing on the ones having perfect matchings enumerated by perfect powers (see e.g., [1], [6], [13], [23], [30], and the references therein).However, in most cases, the graphs are either highly symmetric or some variants of the Aztec rectangles.In this paper, we introduced six new families of subgraphs that are not Aztec-rectangle-like.However, their perfect matchings are still enumerated by perfect powers (see the graphs a,b,c and F a,b,c , for i = 1, 2, 3, in Section 2).
Here is a simple observation that inspires our main results.Viewing a standard brick lattice as a sub-grid of the square grid, we consider an augmented Aztec rectangle on the standard brick lattice, where the north and the south corners have been trimmed (see Figure 1.2(a)).One readily sees that the resulting graph can be deformed into the honeycomb graph whose perfect matchings are enumerated by MacMahon's formula [10] (see Figure 1.2(b)).
Next, we consider a new brick lattice B pictured in Figure 1.3.In particular, the lattice B is obtained by gluing copies of a cross pattern, which is restricted in a dotted diamond of side 2 √ 2. Let a and b be two non-negative integers.Consider a (2b+2a−2)×(2b+4a−2) augmented Aztec rectangle on the new lattice so that its east corner is the east corner of a cross pattern.Similar to the case of the standard brick lattice, we trim the north and south corners of the augmented Aztec rectangle both from right to left at levels (2a − 1) above and (4a − 1) below the eastern corner, respectively.The only difference here is that we are using two zigzag cuts containing alternatively "bumps" and "holes" of size 2, as opposed to straight cuts in the case of standard brick lattice (see Figure 1.4).Denote by T R a,b the resulting graph.The number of perfect matchings of T R a,b is given by the theorem stated below.
We note that the number of perfect matchings of T R a,b in Theorem 1.1 does not depend on b.
We now consider a 2m × 2n Aztec rectangle (m n) on the lattice B whose east corner is also the east corner of a cross pattern.Similar to the graph T R a,b , we also trim the electronic journal of combinatorics 24(4) (2017), #P4.19  the upper and lower corners of the Aztec rectangle by two zigzag cuts.The only difference here is that we are trimming the north corner from right to left and the south corner from left to right.Assume that h 1 is the distance between the top of the Aztec rectangle and the upper cut, and that h 2 is the distance between the bottom of the graph and the lower

cut. Denote by
m,n the resulting trimmed Aztec rectangle.We notice that an edge of the zigzag cut is in the graph T A h 1 ,h 2 m,n if and only if it is also a lattice edge in B. See Figure 1.5 for an example of the T A-graph; the dotted edges in the lower zigzag cut indicate the ones not in the T A-graph.We also have a variant of T A h 1 ,h 2 m,n by applying the same process to the (2m − 1) × (2n − 1) Aztec rectangle whose east-most edge is the west-most edge of a cross pattern on B. Denote by T B h 1 ,h 2 m,n the resulting graph (see Figure 1.6).Ciucu [2] conjectured that Conjecture 1.2.The numbers of perfect matchings of T A h 1 ,h 2 m,n and T B h 1 ,h 2 m,n have only prime factors less than 13 in their prime factorizations.
We prove Ciucu's conjecture by giving explicit formulas for the numbers of perfect matchings of T A h 1 ,h 2 m,n and T B h 1 ,h 2 m,n as follows.Given three integer numbers a, b, c, we define five functions g(a, b, c), q(a, b, c), α(a, b, c), β(a, b, c), and t(a, b) by setting the electronic journal of combinatorics 24(4) (2017), #P4.19 and if h 1 and h 2 are even; if h 1 is odd and h 2 is even; 0 otherwise. (1.5) The number of perfect matchings of the trimmed Aztec rectangle T A h 1 ,h 2 m,n equals α(a, b, c)2 g(a,b,c+1) 3 τ (h 1 ,h 2 ) 5 g(a,b,c) 11 q(a,b,c) , ( ) ) It is easy to see that if a bipartite graph admits a perfect matching, then the numbers of vertices in its two vertex classes are the same.A bipartite graph satisfying the later balancing condition is called a balanced graph.
m,n and T B h 1 ,h 2 m,n are not balanced.Thus, their numbers of perfect matchings are 0.
The rest of this paper is organized as follows.In Section 2, we introduce six new families of subgraphs of the lattice B and present explicit formulas for the numbers of perfect matchings of these graphs (see Theorem 2.1).The theorem is the key result of our paper; and we will prove it in the next three sections.In Section 3, we prove several recurrences for the numbers of perfect matchings of the six families of graphs by using Kuo's graphical condensation method [9].Then, in Section 4, we show that the formulas in Theorem 2.1 satisfy the same recurrences obtained in Section 3.This yields an inductive proof of Theorem 2.1, which is presented in Section 5. Section 6 is devoted to the proofs of Theorems 1.1 and 1.3 by using the result in Theorem 2.1.Finally, in Section 7, we investigate a hidden connection between the graph in Theorem 1.1 and the hexagonal dungeon introduced by Blum in his well-known conjecture (see [23], Problem 25).It is worth noticing that Blum's conjecture was confirmed by Ciucu and the author in [5].

Six new families of graphs
In this section, we consider six new families of graphs whose matching enumerations will be employed in our proofs of Theorems 1.1 and 1.3 in Section 6.
Pick the center V s of a cross pattern on the lattice B. Let a, b, c, d, e, x be six nonnegative integers.We create a six-sided contour starting from V s as follows.We go 2a √ 2 units southeast from V s , then 2b √ 2 units northeast, 4c units west, 2d √ 2 units northwest, and 2e √ 2 units southwest.We adjust e so that the ending point V e of the fifth side are on the same level as V s .Finally, we close the contour by going x units west or east, based on whether V e is on the east or the west of V s (see Figures 2.1(a) and (b)).We denote by C (1) (a, b, c) the resulting contour.We will show in the next three paragraphs that our contour depends on only three parameter a, b, c.
The above choice of e requires that a + e = b + d.
If V e is x units on the east of V s (i.e. a > c + d), then the closure of the contour yields 4a = x + 4d + 4c, so x = 4(a − c − d).Similarly, in the case when V e is x units on the west of V s (i.e. a < c + d), we get x = 4(c + d − a).Thus, in all cases, we must have x = 4f , where f = |a − c − d|.
We now consider the subgraph G a,b,c of the lattice B induced by the vertices inside the contour (see the graphs restricted by the solid boundaries in  Next, along each horizontal side of the contour, we apply a zigzag cut as in the definition of the trimmed Aztec rectangles in the previous section (illustrated by bold zigzag lines in Figures 2.1(c) and (d); the shaded vertices indicate the vertices of G a,b,c , which have been removed by the trimming process).In particular, for the c-side 2 , we perform the zigzag 2 From now on, we usually call the first, the second, . . ., the sixth sides of the contour the a-, the b-, . . ., and the f -sides.The same terminology will be used for the sides of the graph induced by vertices inside the contour.
the electronic journal of combinatorics 24(4) (2017), #P4.19 cut from right to left and stop when having no room on this side for the next step of the cut.Similarly, we cut along f -side from left to right and also stop when cannot reach further.Denote by F Recall that if a bipartite graph G admits a perfect matching, then the numbers of vertices in the two vertex classes of G must be the same; and we say that the graph G is balanced.One readily sees that the balance of F 3) It means that d, e, f depend on a, b, c.This explains why our graph F a,b,c is indeed determined by a, b, c (and so is the contour Next, we consider a variant A a,b,c of the graph F    In the definitions of the next families of graphs, we always assume the constraints (2.1), (2.2), (2.3), and b 2.
The proof of Theorem 2.1 will be given in the next three sections.In particular, Sections 3 and 4 show that the expressions on the left and right hand sides of each of the equalities (2.4)-(2.9)both satisfy the same recurrences.Then we will give an inductive proof for the theorem in Section 5.

Recurrences of M A
Eric Kuo (re)proved the Aztec diamond theorem by Elkies, Kuperberg, Larsen, and Propp (see [6,7]) by using a method called "graphical condensation" [9].The key of his proof is the following combinatorial interpretation of the Desnanot-Jacobi identity in linear algebra (also mentioned as Dodgson condensation; see e.g.[11], pp.136-149).We refer the reader to [3,4,5,14,15,16,17,18,19,20,21,22,24,25,27] for various recent applications of this powerful method.Theorem 3.1 (Kuo's Condensation Theorem [9]).Let G be a planar bipartite graph, and V 1 and V 2 the two vertex classes with |V 1 | = |V 2 |.Assume in addition that x, y, z and t are four vertices appearing in a cyclic order on a face of G so that x, z ∈ V 1 and y, t ∈ V 2 .Then In this section, we use the Kuo's Condensation Theorem to prove that the numbers of perfect matchings of the six families of graphs a,b,c 's, for i = 1, 2, 3, satisfy the following six recurrences (with some constraints).
We use the notations (a, b, c) and (a, b, c) for general functions from Z 3 to Z.We consider the following six recurrences: and We notice that if ≡ in the recurrence (R6), we get the recurrence (R3).
the electronic journal of combinatorics 24(4) (2017), #P4.19 a,b,c ) all satisfy the recurrence (R1), i.e. we have and Proof.First, we prove the equality (3.2), for i = 1.Apply Kuo's Condensation Theorem 3.1 to the graphs G := A a,b,c with the four vertices u, v, w, t chosen as in Figure 3.1(a) (for a = b = 8 and c = 3).In particular, we pick u on the west corner, v and w on the south corner, and t on the east corner of the graph.
, A , A , A   The similarity between the statement for the A-and the statement for the B-graphs allows us to prove only first one.This part can be treated like part (a) by applying Kuo's Theorem 3.1 to the graphs A   For i = 1, 2, 3 denote by Φ i (a, b, c) the product on the right hand sides of the equalities (2.4)-(2.6) in Theorem 2.1, respectively, and Ψ i (a, b, c) the product on the right hand sides of equalities (2.7)-(2.9),respectively.In the next section, we will show that these functions satisfy the same recurrences (R1)-(R6).
4 Recurrences for the functions Φ i (a, b, c) and Ψ i (a, b, c) Lemma 4.1.For any integers a, b, c, and i = 1, 2, 3, the functions Φ i (a, b, c) and Ψ i (a, b, c) all satisfy the recurrence (R1).
Proof.We need to show that and for i = 1, 2, 3. We first consider the case of even b.There are six subcases to distinguish, based on the values of a − c (mod 6).We show in details here the subcase when a − c ≡ 0 (mod 6) (the other five subcases can be obtained in a perfectly analogous manner).
If a − c ≡ 0 (mod 6), by the definition of functions g(a, b, c) and q(a, b, c), we can cancel out almost all the exponents of 2, 5 and 11 on both sides of the equalities (4.1) and (4.2).The equality (4.1) becomes and the equality (4.2) becomes By definition of the functions α(a, b, c) and β(a, b, c), we have here α(a, b, c Then the above equalities are equivalent to the following obvious equalities respectively.The remaining case of odd b turns out to follow from the case of even b.Indeed, for j = 0, 1, . . ., 5, verification of the case of odd b and a − c ≡ j (mod 6) is the same as the verification of the case of even b and a − c ≡ 3 + j (mod 6).Ψ i (a, b, c) also satisfy the recurrence (R2), i.e. and the electronic journal of combinatorics 24(4) (2017), #P4.19 for i = 1, 2, 3.
Proof.(a) Arguing the same as the proof of Lemma 4.1, we only need to consider the case of even b.When b is even, we have also six subcases to distinguish, depending on the values of a − c (mod 6).Again, we only show here the verification for the subcase a − c ≡ 0 (mod 6), and the other subcases can be obtained similarly.
Assume now that a − c is a multiple of 6.Similar to Lemma 4.1, we can cancel out almost all the exponents of 2, 5 and 11 on two sides of the equalities (4.5) and (4.6).These equalities are simplified to and respectively.By the definition of α(a, b, c) and β(a, b, c), one can verify easily the above equalities.
(b) We only show that Φ 1 (a, b, c) and Ψ 1 (a, b, c) satisfy (R3), as the second statement can be proven similarly.
By definition of functions Φ 1 and Ψ 1 , one readily verifies that and ) and for i = 1, 2, 3.
Proof.(a) This part can be treated similarly to Lemma 4.1 and Lemma 4.2(a).(b) From part (a), we only need to prove that for i = 1, 2, 3.However, this follows directly from the definition of the functions Φ i 's and Ψ i 's.
the electronic journal of combinatorics 24(4) (2017), #P4.19 5 Proof of Theorem 2.1 We will prove Theorem 2.1 in the same fashion as the proof of Theorem 3.1 in [5].
Proof of Theorem 2.1.We define the a function P (a, b, c) by setting In particular, P (a, b, c) is always even.We call P (a, b, c) the perimeter of our six graphs a,b,c 's and a,b,c 's.We need to show that for i = 1, 2, 3, by induction of the value the perimeter P (a, b, c) of the six graphs.
For the base cases, one can verify easily (5.1) with the help of vaxmacs, a software written by David Wilson3 , for all the triples (a,b,c) satisfying at least one of the following conditions: For the induction step, we assume that (5.1) holds for all graphs A a,b,c → A a,b,c → A a,b,c → F a,b,c → F  1, one readily verifies that (b, a, f ) ∈ D 3 ; and if c = 0, it is easy to see that (b, a, f ) ∈ D 1 .Then, by the cases treated above, we obtain also (5.15).

Proofs of Theorems 1.1 and 1.3
Before going to the proofs of Theorems 1.1 and 1.3, we quote the following useful lemma that was first introduced in [12] (see Lemma 3.6(a)).Lemma 6.1 (Graph Splitting Lemma).Let G be a bipartite graph, and let V 1 and V 2 be the two vertex classes..
Assume that an induced subgraph H of G satisfies following two conditions: (i) (Separating Condition) There are no edges of G connecting a vertex in It is easy to see that G 2 has a unique perfect matching (see the bold edges in Figure 6.1), and the graph G 1 and G 2 are isomorphic to A 2a,3a,2a and F 2a,3a,2a , respectively.By Theorem 2.1, we obtain 2a,3a,2a ) = 2 g(2a,3a,2a−1)+g(2a,3a,2a+1) 5 2g(2a,3a,2a) 11 2q(2a,3a,2a) = 2 a(a+1)+a(a−1) 5 2a Proof of Theorem 1.3.The proof is illustrated in Figure 6.2, for m = 5, n = 7, h 1 = 4 and h 2 = 3.Consider the rightmost subgraph G 1 of T A h 1 ,h 2 m,n , which is restricted by a dotted contour in Figure 6.2.By Graph Splitting Lemma 6.1, we obtain Next, we consider the graph G obtained from T A h 1 ,h 2 m,n − G 1 by removing horizontal forced edges (the circled ones on the right of G 1 in Figure 6.2).Applying Graph-splitting Lemma 6.1 again to the second subgraph G 2 of T A h 1 ,h 2 m,n , which is restricted by a dotted contour, we have M 2 − 2 more times the above process, we get a graph G and Apply the same process for the lower part of G.We get a graph G (the subgraph restricted by the bold contour in Figure 6.2) and where k = h 2 +1

2
, and H j is the j-th subgraph (from the left) restricted by a dotted contour in the lower part of G.
Combining (6.6) and (6.7), we deduce It is easy to see that M (G j ) = 1 if h 1 is odd, and 3 if h 1 is even, for any j = 1, 2, . . ., i. Similarly, M (H j ) = 1 if h 2 is odd, and 3 if h 2 is even, for any j = 1, 2, . . ., k.Thus, On the other hand, G is isomorphic to the graph A a,b,c , where . By (6.9) and Theorem 2.1, the equality (1.6) follows.The equality (1.7) can be proved analogously.
Next, we consider a variation of Theorem 1.3 as follows.Instead of using horizontal trimming lines as in Theorem 1.3, we consider two new stair-shaped trimming lines.The structure of each level in the new trimming lines is similar to the old ones (i.e. is a zigzag line with alternatively bumps and holes of size 2), and each two consecutive levels are connected by a "staircase" (see Figure 6.3).Assume that h 1 is the distance between the top of the Aztec rectangle and the highest level of the upper trimming line, and h 2 is the distance between the bottom of the Aztec rectangle and the lowest level of the lower trimming line.Again, by Graph Splitting Lemma 6.1(a), we can cut off small subgraphs with the same structure as that of G i and H j .We get again the final graph isomorphic to the graph A

Relation between the graph T R a,b and Hexagonal Dungeons
We consider the hexagonal dungeon HD a,2a,b introduced by Blum [23] (see detailed definition of the hexagonal dungeon in [5]).Figure 7.1 shows the hexagonal dungeon HD 2,4,6 .We consider the dual graph of HD a,2a,b (i.e. the graph whose vertices are the small right triangles in HD a,2a,b and whose edges connect precisely two triangles sharing an edge), which is denoted by G a,2a,b .The upper graph with solid edges in Figure 7.3 illustrates the dual graph of HD 2,4,6 .It has been proven by Ciucu and the author in [5] that the number of perfect matchings of G a,2a,b is given by 13 2a 2 14 a 2 2 , which is similar to the number of perfect matchings of T R a,b (10 2a 2 11 a 2 2 ).This suggests the existence of a hidden connection between T R a,b and the dual graph G a,2a,b of the hexagonal dungeon.
If we assign some weights to the edges of a graph G, then we use the notation M(G) for the sum of weights of the perfect matchings of G, where the weight of a perfect matching is the product of weights of its constituent edges.We call M(G) the matching generating function of the weighted graph G.
Next, we quote a well-known subgraph replacement trick called urban renewal, which was first discovered by Kuperberg, and its variations found by Ciucu [1].Next, we apply suitable replacement rules in Lemma 7.1 to G a,2a,b around the dotted rectangles as in the upper graph in Figure 7.3).Then we deform the resulting graph into a weighted graph G a,2a,b on the square lattice (see the lower graph in Figure 7.3; the bold edges have weight 1  2 ).We want to emphasize that even though the graphs G a,2a,b and T R a,b have the same shape, their weight assignments are different.This means that Theorem 1.1 can not be deduced from the work in [5], and vice versa.
We now want to consider a common generalization of the weight assignments in the graphs G a,2a,b and T R a,b as follows.Assume that x, y, z are three indeterminate weights.We assign weights to edges of the lattice B so that each cross pattern is weighted as in Figure 7

Figure 1 . 2 :
Figure 1.2: Obtaining a honeycomb graph from a trimmed augmented Aztec rectangle on the brick lattice.

Figure 1 . 5 :
Figure 1.5:The graph T A 4,3 5,7 is obtained by trimming the rectangle AR 10,14 (on the lattice B) by two zigzag cuts.

( 1 )
a,b,c the resulting graph.(see Figure 2.1(c) for the case a > c + d, and Figure 2.1(d) for the case a c + d).

( 1 )
a,b,c requires d = 2b − a − 2c.Moreover, to guarantee that the graph is not empty, we assume in addition b 2. In summary, we have d = 2b − a − 2c 0, (2.1)

( 1 )
a,b,c as follows.We remove all vertices along the a-, b-, d-, and e-sides of G a,b,c (illustrated by white circles in Figures 2.2 (a) and (b)).We now trim along the c-and f -sides of the resulting graph in the same way as in the definition of F (1) a,b,c (the vertices of G a,b,c , which are removed by the trimming process, are illustrated by shaded points in Figures 2.2 (a) and (b); the edges removed are shown by dotted edges).Figures 2.2 (c) and (d) give examples of the graph A (1) a,b,c in the cases a > c + d and a c + d, respectively.
(a) and (b) for examples).The contour C (2) (a, b, c) determines a new induced subgraph H a,b,c of the lattice B (i.e., H a,b,c is induced by vertices inside the contour).The graphs H 9,8,2 and H 5,8,4 are illustrated by the ones restricted by the solid boundaries in Figures 2.3 (a) and (b), respectively.Next, we remove all vertices along the a-and d-sides of H a,b,c ; and we also remove the vertices along the f -side if a > c + d (see white circles in Figures 2.

4
(c) and (d)).Similar to the graph F (1) a,b,c , we obtain the graph F (2) a,b,c by applying the zigzag cuts along two horizontal sides of the resulting graph (which are now the b-and e-sides).The graphs F are pictured in Figures 2.

3
(c) and (d), respectively; the shaded points also indicate the vertices of H a,b,c , which are removed by the trimming process.Similarly, the graph A (2) a,b,c is obtained from H a,b,c by removing vertices along the c-side, as well as vertices along the f -side when a c + d, and trimming along two horizontal sides (illustrated in Figures 2.3
(a) and (b)).We denote by K a,b,c the subgraph of the lattice B induced by the vertices inside C (3) a,b,c (see the graph restricted by the solid contours in Figures 2.4 (a) and (b)).Our last two graphs are obtained from K a,b,c by a similar removing-trimming process as follows.The graph F (3) a,b,c is obtained from K a,b,c by removing vertices along its b-, c-, and e-sides, as well as vertices along the f -side when a c+d, and trimming along two horizontal sides (along the a-side from right to left, and the d-side from left to right).The graph A (3) a,b,c is obtained similarly by removing vertices along the f -side if a > c + d, and trimming along the two horizontal sides.Figures 2.4
) with the four vertices u, v, w, t in Lemma 3.3.Substituting the above five equalities into the equality (3.1) in Kuo's Condensation Theorem, we get (3.2), for i = 1.Similarly, apply Kuo's Condensation Theorem 3.1 to the graphs A c with the four vertices u, v, w, t chosen as in Figures 3.2

4 . 3 . 4 .
) with the selection of the vertices u, v, w, t in Lemma 3.Lemma Assume that a, b, c are three non-negative integers satisfying a 2, b 5, c 2, d := 2b − a − 2c 0, and e := 3b − 2a − 2c 0. Assume in addition that a c + d.(a) If d 1, then for i = 1, 2, 3 the numbers perfect matchings M(A (i) a,b,c ) and M(F (i) a,b,c ) all satisfy the recurrence (R4).(b) If d = 0, then for i = 1, 2, 3 the pairs of the numbers of perfect matchings M(A (i) a,b,c ), M(F (4−i) a,b,c ) and M(A (i) a,b,c ), M(F (4−i) a,b,c ) both satisfy the recurrence (R5) for ( , ).Proof.(a) We prove here the number of perfect matchings of A (i) a,b,c 's satisfies the recurrence (R4), and the corresponding statement for F (i) a,b,c 's can be obtained in the same fashion.We need to show that M A the electronic journal of combinatorics 24(4) (2017), #P4.19Similar to Lemmas 3.2 and 3.3, we apply Kuo's Theorem 3.1 to the graph A (i) a,b,c with the four vertices u, v, w, t chosen as inFigures 3.4
,c 's with the four vertices u, v, w, t selected as inFigures  3.4 (d), (e), and (f).The only difference here is that, after removing forced edges from the graph A (i) a,b,c − {u, v}, we get the graph F (4−i) c,b−1,a−1 (see the graphs restricted by the bold contours in Figures 3.4 (d), (e), and (f)), instead of the graph A (i) a−1,b−1,c as in the part (a).
,c 's and F (i) a,b,c 's having perimeter P (a, b, c) less than p, for some p 16.We will prove (5.1) for all Aand F -graphs with perimeter p.By the base cases, we only need to show (5.1) for all graphs having the triple (a, b, c) in the following domainD := {(a, b, c) ∈ Z 3 : P (a, b, c) = p, b 5, c + d 3, d 0, e 0}.(5.2)First, we partition D into four subdomains as followsD 1 := D ∩ {2 a c + d}, D 2 := D ∩ {a 1}, D 3 := D ∩ {a > c + d, e d},andD 4 := D ∩ {a > c + d, e < d}.Next, we verify that (5.1) holds in each of the above subdomains.fori = 1, 2, 3.By (5.11) and (5.12), we only need to show that M(A(i) f,e,d ) = Φ i (f, e, d) and M(F (i) f,e,d ) = Ψ i (f, e, d),(5.13)then(5.1) follows.If e 4, then the triple (f, e, d) satisfy the condition (ii) in the base cases, thus (5.13) follows.Then we can assume that e 5. We now need to verify that the triple (f, e, d) is in the domain D 1 .It is more convenient to re-write the domain D 1 with all constraints in terms of a, b, c as follows:D 1 = {(a, b, c) ∈ Z 3 : P (a, b, c) = p; 2 a 2b − a − c; b 5; 2b − a − c 3; 2b − a − 2c 0; 3b − 2a − 2c 0}.(5.14)We have in this case f = c+d−a c+d−1 2 (we are assuming that a 1 and c+d 3).Moreover, the inequality f 2e − f − d is equivalent to a 0, so (f, e, d) satisfies the second constraint of the domain D 1 .This implies from the definition of the perimeter that P (f, e, d) = 8e − 4f − 4d = 8b − 4a − 4c = p.Next, the third constraint follows from our assumption e 5.For the fourth constraint, we have 2e− f − d = 2b − a − c = c + d 3.Finally, we have 2e − f − 2d = c 0 and 3e − 2f − 2d = b 0, which imply the last two constraints.Thus, the triple (f, e, d) is indeed in D 1 .By the case treated above, we have again(5.13).It means that (5.1) has just been verified for the case (a, b, c) ∈ D 2 .The case (a, b, c) ∈ D 3 can be treated similarly to the case (a, b, c) ∈ D 1 .We also divide further D 3 into three subdomains:D 3a := D 3 ∩ {c 2},D 3b := D 3 ∩ {c = 1}, and D 3c := D 3 ∩ {c = 0}.If (a, b, c) ∈ D 3,a , D 3b , or D 3c , we use the recurrences (R1) (see Lemmas 3.2 and 4.1), (R2) (see Lemmas 3.3(a) and 4.2(a)), or (R3) and (R6) (see Lemmas 3.3(b) and 4.2(b)), respectively.Finally, we consider the case (a, b, c) ∈ D 4 (i.e., we are assuming that a > c + d).We also reflect the graphs A (i) a,b,c 's and F (i) a,b,c 's over a horizontal line, and get the reflection diagram as follows:A

Lemma 7 . 1 (
Urban renewal).Let G be a weighted graph.Assume that G has a subgraph K as one of the graphs on the left column in Figure7.2,where only white vertices can the electronic journal of combinatorics 24(4) (2017), #P4.19