aa r X i v : . [ m a t h . C O ] J a n STATE MATRIX RECURSION METHOD ANDMONOMER–DIMER PROBLEM
SEUNGSANG OH
Abstract.
The exact enumeration of pure dimer coverings on the squarelattice was obtained by Kasteleyn, Temperley and Fisher in 1961. Inthis paper, we consider the monomer–dimer covering problem (allowingmultiple monomers) which is an outstanding unsolved problem in latticestatistics. We have developed the state matrix recursion method thatallows us to compute the number of monomer–dimer coverings and toknow the partition function with monomer and dimer activities. Thismethod proceeds with a recurrence relation of so-called state matricesof large size. The enumeration problem of pure dimer coverings anddimer coverings with single boundary monomer is revisited in partitionfunction forms. We also provide the number of dimer coverings withmultiple vacant sites. The related Hosoya index and the asymptotic be-havior of its growth rate are considered. Lastly, we apply this methodto the enumeration study of domino tilings of Aztec diamonds and moregeneralized regions, so-called Aztec octagons and multi-deficient Aztecoctagons. Introduction
The monomer–dimer problem is one of simplicity of definition, but famousunsolved problem, and has a long and glorious history. The monomer–dimersystem has been used as a model of a physical system [4, 12], but primarilyit is interesting as the matching counting problem in combinatorics [15].While it is known that it does not exhibit a phase transition [6], there havebeen only limited closed-form results.It gained momentum in 1961 when Kasteleyn [11] and Temperley andFisher [3, 27] found the exact solution of the enumeration of pure dimercoverings (i.e., no monomers). Pure dimer coverings are often consideredas perfect matchings or domino tilings. The total number of pure dimercoverings in the m × n square lattice with even mn is m Y j =1 n Y k =1 (cid:12)(cid:12)(cid:12)(cid:12) πjm + 1 ) + 2 i cos( πkn + 1 ) (cid:12)(cid:12)(cid:12)(cid:12) . In 1974, Temperley [26] found an intriguing bijection between spanningtrees of the m × n square lattice and pure dimer coverings in the (2 m +1) × (2 n +1) square lattice with a corner removed. This offers an alternate approach tothe vertex vacancy problem. Recently, Tzeng and Wu [28] used Temperley Mathematics Subject Classification 2010: 05A15, 05B45, 05B50, 82B20, 82D60.This work was supported by the National Research Foundation of Korea(NRF) grantfunded by the Korea government(MSIP) (No. NRF-2017R1A2B2007216). bijection to enumerate dimer coverings with a fixed single monomer on theboundary.The purpose of this paper is to introduce a method for the enumeration ofmonomer–dimer coverings (allowing multiple monomers), that is called the state matrix recursion method . More precisely, it provides a recursive for-mula of state matrices to give the partition function with respect to monomerand dimer activities. A typical example of a monomer–dimer covering inthe m × n square lattice is drawn in Figure 1. In Section 2, we state severalmonomer–dimer problems that are considered in this paper. Figure 1.
A monomer–dimer covering in Z × The state matrix recursion method is divided into three stages; • Stage 1. Conversion to the mosaic system • Stage 2. State matrix recursion formula • Stage 3. Analyzing the state matrixIn Sections 3 ∼
5, we formulate the method and show the main result at theend. Section 6 is devoted to the study of the asymptotic behavior of thegrowth rate of the Hosoya index of the m × n square lattice. In Section 7,the dimer covering problem with multiple vacant sites is handled.As an application of this method, we also consider the domino tilingproblem of the Aztec diamond and its variant regions. The Aztec diamondtheorem from the excellent article of Elkies, Kuperberg, Larsen and Propp [1]states that the Aztec diamond of order n can be tiled by dominos in exactly2 n ( n +1) / ways. A simple proof of this theorem can be found in [2]. Anaugmented Aztec diamond of order n looks much like the Aztec diamondof order n , except that there are three long columns in the middle insteadof two. Compare left two regions in Figure 2. The number of dominotilings of the augmented Aztec diamond of order n was found by Sachs andZernitz [25] as P nk =0 (cid:0) nk (cid:1) · (cid:0) n + kk (cid:1) , known as the Delannoy numbers. Noticethat the former number is much larger than the later. The enumerationproblem of domino tilings of a region is known to be very sensitive to itsboundary condition [16, 17]. Dozens of interesting patterns related to theAztec diamond allowing some squares removed have been deeply studiedand a survey of these works was proposed by Propp [23]. For example, seethe rightmost figure showing a domino tiling of a 4-by-5 Aztec rectanglewith its central square removed. TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 3
Figure 2.
Domino tilings of various Aztec regionsIn Section 8, we study the domino tilings of the most generalized regionamong Aztec diamond variants, called an Aztec octagon, obtained from therectangular grid with four triangular corners (not necessary to be congruent)removed as drawn in Figure 12.As another interesting application, this method provides a recursive matrix-relation producing the exact number of independent vertex sets on the squarelattice in papers [21, 22].2.
Monomer–dimer problems
Let Z m × n denote the m × n rectangular grid on the square lattice. A dimer is an edge connecting two nearest vertices. Horizontal and verticaldimers are considered as x -dimers and y -dimers, respectively. Dimers mustbe placed so that no vertex belongs to more than one dimer. An unoccupiedvertex is called a monomer . The partition function of Z m × n with monomerand dimer activities, assigned weights v, x, y to monomers, x -dimers and y -dimers respectively, is defined by G m × n ( v, x, y ) = X v n v x n x y n y with respect to the numbers n v , n x , n y of monomers, x -dimers and y -dimersrespectively, where the summation is taken over all monomer–dimer cover-ings. Note that n v + 2( n x + n y ) = mn in each term.Based upon the state matrix recursion method, we present a recursivematrix-relation producing this partition function. Hereafter O k denotes the2 k × k zero-matrix. Theorem 1.
The partition function is G m × n ( v, x, y ) = (1 , -entry of ( A m ) n , where A m is the m × m matrix recursively defined by A k = vA k − + x (cid:20) A k − O k − O k − O k − (cid:21) A k − yA k − O k − In this theorem we may replace the recursive relation by A k = A k − ⊗ (cid:20) v y (cid:21) + A k − ⊗ x in tensor product form. This will be explained after the proof of Lemma 5. S. OH for k = 2 , . . . , m , with seed matrices A = (cid:2) (cid:3) and A = (cid:20) v y (cid:21) . Theorem 1 presents a number of important consequences as follows. First,we can derive the matching polynomial for Z m × n m Z m × n ( z ) = G m × n (1 , z, z )each of whose coefficient of z k indicates the number of k -edge matchings.Second, G m × n (1 , ,
1) gives the number of monomer–dimer coverings,known as the Hosoya index of Z m × n . The Hosoya index [9] and the Merrifield-Simmons index [16, 17] of a graph are two prominent examples of topologicalindices which are used in mathematical chemistry for quantifying molecular-graph based structure descriptors. The sequence of G n × n (1 , , m = n ,grows in a quadratic exponential rate. We focus on the asymptotic behaviorof the growth rate per vertex. Let δ = lim m,n →∞ ( G m × n (1 , , mn , provided that it exists. The existence of that limit was proved in [10]. Atwo-dimensional application of the Fekete’s lemma shows again the existenceof the limit. The following theorem will be proved in Section 6. Theorem 2.
The double limit δ exists. More precisely, δ = sup m,n ≥ ( G m × n (1 , , mn . Third, known as the pure dimer problem for even mn , G m × n (0 , x, y ) isthe partition function of Z m × n only with dimer activity, assigned weights x, y to x -dimers and y -dimers respectively. Remark that, instead of theform G m × n (0 , ,
1) of the number of pure dimer coverings, a better closedform of this number was already founded as mentioned in the introduction.Hammersley [5] showed that the following limit exists and from the exactresults [11, 27], we know thatlim n →∞ ( G n × n (0 , , n = e Cπ = 1 . · · · , where C is the Catalan’s constant.Fourth, the coefficient of the degree 1 term v of G m × n ( v, ,
1) indicatesthe number of dimer coverings with a single vacancy (non-fixed and on/offthe boundary) for odd m and n . But, more interesting models are dimercoverings with a fixed single vacancy on the boundary [13, 28, 29]. A fixedsingle boundary monomer covering, say, in Z m × n for odd m and n is amonomer–dimer covering with exactly one fixed monomer on the boundary,having odd-numbered x - and y -coordinates. It is known that the numberof dimer coverings with fixed single boundary monomer does not depend onthe location of the fixed monomer [28]. Theorem 3.
Let G sm × n ( v, x, y ) be the (2 , -entry of ( A m ) n in Theorem 1for odd m and n . Then G sm × n (0 , , is the number of fixed single boundarymonomer coverings in Z m × n . TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 5
Note that, instead of the (2,1)-entry, we may use any ( i, j )-entry of ( A m ) n for { i, j } = { , k +1 } and k = 0 , , , . . . , m − S be a set of verticesin Z m × n , called a fixed monomer set as in Figure 3. In this case, we onlyconsider the number of monomer–dimer coverings instead of the partitionfunction with monomer and dimer activities, by assigning 1 to the weights v , x and y . g m × n ( S ) denotes the number of distinct monomer–dimer coveringswhich have monomers exactly at the sites of S . Here ( k, i ) indicates thevertex placed at the k th column from left to right and the i th row frombottom to top. Theorem 4.
For a given fixed monomer set S in Z m × n , g m × n ( S ) = (1 , -entry of n Y i =1 A m,i , where A m,i is defined by the recurrence relations, for k = 1 , . . . , m ,if the vertex ( k, i ) is contained in S , A k,i = (cid:20) A k − ,i O k − O k − O k − (cid:21) and B k,i = O k or if the vertex ( k, i ) is not contained in S , A k,i = (cid:20) B k − ,i A k − ,i A k − ,i O k − (cid:21) and B k,i = (cid:20) A k − ,i O k − O k − O k − (cid:21) with seed matrices A ,i = (cid:2) (cid:3) and B ,i = (cid:2) (cid:3) . fixed monomer Figure 3.
A monomer–dimer covering with fixed monomers3.
Stage 1. Conversion to the monomer–dimer mosaic system
This stage is dedicated to the installation of the mosaic system for monomer–dimer coverings on the square lattice. Mosaic system is introduced byLomonaco and Kauffman [14] to give a precise and workable definition ofquantum knots. This definition is intended to represent an actual physicalquantum system.Recently, the author et al . have developed a state matrix argument forknot mosaic enumeration [7, 8, 18, 19, 20]. We follow the notation and
S. OH terminology used in [19] with much modification to adjust to the dimersystem.Five symbols T , T , T , T and T illustrated in Figure 4 are called mosaictiles (for monomer–dimer coverings on the square lattice). Their side edgesare labeled with two letters a and b as follows: letter a if it is not touched bya thick arc on the tile, and letter b for otherwise. In the original definition ofknot mosaic theory, eleven symbols were used to represent a knot diagram. T T T T T a ba aa abaaa aa ba a aa aab
Figure 4.
Five mosaic tiles labeled with two lettersFor positive integers m and n , an m × n -mosaic is an m × n rectangulararray M = ( M ij ) of those tiles, where M ij denotes the mosaic tile placed atthe i th column from left to right and the j th row from bottom to top. Weare exclusively interested in mosaics whose tiles match each other properlyto represent monomer–dimer coverings. This requires the followings: • (Adjacency rule) Abutting edges of adjacent mosaic tiles in a mosaic arelabeled with the same letter. • (Boundary state requirement) All boundary edges in a mosaic are labeledwith letter a .As illustrated in Figure 5, every monomer–dimer covering in Z m × n can beconverted into an m × n -mosaic which satisfies the two rules. In this mosaic,a dot in each T indicates a monomer, and T and T (or, T and T ) canbe adjoined along the edges labeled b to produce a dimer. Note that thestatements of the adjacency rule and boundary state requirement vary indifferent lattice models. n m Figure 5.
Conversion of the monomer–dimer coveringdrawn in Figure 1 to a monomer–dimer m × n -mosaic TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 7
A mosaic is said to be suitably adjacent if any pair of mosaic tiles sharingan edge satisfies the adjacency rule. A suitably adjacent m × n -mosaic is calleda monomer–dimer m × n -mosaic if it additionally satisfies the boundary staterequirement. Key role is the following one-to-one conversion. One-to-one conversion.
There is a one-to-one correspondence betweenmonomer–dimer coverings in Z m × n and monomer–dimer m × n -mosaics. Stage 2. State matrix recursion formula
Now we introduce two types of state matrices for suitably adjacent mo-saics to produce the partition function G m × n ( v, x, y ).4.1. States and state polynomials.
Let p ≤ m and q ≤ n be positiveintegers, and consider a suitably adjacent p × q -mosaic M . A state is a fi-nite sequence of two letters a and b . The b -state s b ( M ) ( t -state s t ( M )) isthe state of length p obtained by reading off letters on the bottom (top,respectively) boundary edges of M from right to left, and the l -state s l ( M )( r -state s r ( M )) is the state of length q on the left (right, respectively) bound-ary edges from top to bottom as shown in Figure 6. State aa · · · a is calledtrivial. s l r s baa bab aabaaa bb t s b s aa Figure 6.
A suitably adjacent 5 × s r ( M ) = aba , s b ( M ) = ababa , s t ( M ) = baaba ,and s l ( M ) = baa Given a triple h s r , s b , s t i of r -, b - and t -states, we associate the statepolynomial : P h s r ,s b ,s t i ( v, x, y ) = X k ( n v , n x , n y ) v n v x n x y n y , where k ( n v , n x , n y ) equals the number of all suitably adjacent p × q -mosaics M , having n v , n x , n y numbers of T , T , T mosaic tiles, respectively, suchthat s r ( M ) = s r , s b ( M ) = s b , s t ( M ) = s t and trivial s l ( M ) = aa · · · a .Mosaic tiles T , T , T are respectively related to a monomer, an x -dimer’sright part and a y -dimer’s top part. The last triviality condition of s l ( M )is necessary for the left boundary state requirement. See Figure 7 for anexplicit example. S. OH a aaa bb a aaa ba
Figure 7.
Twelve suitably adjacent 3 × P h baa , aba , aab i ( v, x, y ) = v y + 2 v xy + 3 v y + 2 v xy + 3 v y + y Bar state matrices.
Consider suitably adjacent p × bar mosaics . Bar mosaics of length p have possibly 2 p kinds of b - and t -states, especially called bar states . We arrange all bar states in thelexicographic order. For 1 ≤ i ≤ p , let ǫ pi denote the i th bar state in thisorder. Bar state matrix X p ( X = A, B ) for the set of suitably adjacent barmosaics of length p is a 2 p × p matrix ( m ij ) given by m ij = P h x ,ǫ pi ,ǫ pj i ( v, x, y ) , where x = a , b , respectively. We remark that information on suitably adja-cent bar mosaics with trivial l -state is completely encoded in two bar statematrices A p and B p . Lemma 5.
Bar state matrices A p and B p are obtained by the recurrencerelations: A k = (cid:20) vA k − + xB k − A k − yA k − O k − (cid:21) and B k = (cid:20) A k − O k − O k − O k − (cid:21) with seed matrices A = (cid:20) v y (cid:21) and B = (cid:20) (cid:21) . Note that we may start with matrices A = (cid:2) (cid:3) and B = (cid:2) (cid:3) instead of A and B . Proof.
We use induction on k . A straightforward observation on four mosaictiles T , T , T and T establishes the lemma for k = 1. For example, (2 , A is P h a ,ǫ ,ǫ i ( v, x, y ) = P h a , b , a i ( v, x, y ) = y since only mosaic tile T satisfies this requirement.Assume that A k − and B k − satisfy the statement. For one case, weconsider A k . Partition this matrix of size 2 k × k into four block subma-trices of size 2 ( k − × ( k − , and consider the 11-submatrix of A k , i.e., the(1 , × i, j )-entry ofthis 11-submatrix is the state polynomial P h a , a ǫ k − i , a ǫ k − j i ( v, x, y ) where a ǫ k − i (similarly a ǫ k − j ) is a bar state of length k obtained by concatenating twostates a and ǫ k − i . A suitably adjacent k × TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 9 triple h a , a ǫ k − i , a ǫ k − j i has two choices T and T for the rightmost mosaictile, and so its second rightmost tile must have r -state a or b , respectively,by the adjacency rule. By considering the contribution of the rightmost tiles T and T to the state polynomial, one easily gets P h a , a ǫ k − i , a ǫ k − j i ( v, x, y ) = v (cid:0) ( i, j )-entry of A k − (cid:1) + x (cid:0) ( i, j )-entry of B k − (cid:1) . Thus the 11-submatrix of A k is vA k − + xB k − . See Figure 8. aa aaa T aa aab T Figure 8.
Expanding a bar mosaicAll the other cases have no or unique choice for the rightmost mosaic tileand the same argument gives Table 1 presenting all possible eight cases asdesired. (cid:3)
Submatrix for h s r , s b , s t i Rightmost tile Submatrix A k h a , a ·· , a ··i T , T vA k − + xB k − h a , a ·· , b ··i T A k − h a , b ·· , a ··i T yA k − h a , b ·· , b ··i None O k − B k h b , a ·· , a ··i T A k − h b , a ·· , b ··i None O k − h b , b ·· , a ··i None O k − h b , b ·· , b ··i None O k − Table 1.
Eight submatrices of A k and B k Remark that we may replace the recursive relation in Lemma 5 by A k = A k − ⊗ (cid:20) v y (cid:21) + B k − ⊗ (cid:20) x
00 0 (cid:21) and B k = A k − ⊗ (cid:20) (cid:21) in tensor product form. This will be done by re-defining b - and t -statesso as reading off m -tuple of states on the bottom and top, respectively,boundary edges from left to right (the reverse direction). Now follow thesame argument as in the above proof.4.3. State matrices.
State matrix A m × q for the set of suitably adjacent m × q -mosaics is a 2 m × m matrix ( a ij ) given by a ij = P h a ··· a ,ǫ mi ,ǫ mj i ( v, x, y ) . The trivial state condition of s r is necessary for the right boundary staterequirement. We get state matrix A m × n by simply multiplying the bar statematrix n times. Lemma 6.
State matrix A m × n is obtained by A m × n = ( A m ) n . Proof.
We use induction on n . For n = 1, A m × = A m since A m × countssuitably adjacent m × r -state a .Assume that A m × ( k − = ( A m ) k − . Let M m × k be a suitably adjacent m × k -mosaic with trivial l - and r -states. Also let M m × ( k − and M m × bethe suitably adjacent m × ( k − m × k − t -state of M m × ( k − and the b -state of M m × must coincide as shown in Figure 9. mr - th among 2 choices b i - thj - th b a aaa b aaa a b s aa baa aaab t s a M m ( k -1) + M m + Figure 9.
Expanding M m × ( k − to M m × k Let A m × k = ( a ij ), A m × ( k − = ( a ′ ij ) and A m × = ( a ′′ ij ). Note that a ij is the state polynomial for the set of suitably adjacent m × k -mosaics M which admit splittings into M m × ( k − and M m × satisfying s b ( M ) = s b ( M m × ( k − ) = ǫ mi , s t ( M ) = s t ( M m × ) = ǫ mj , and s t ( M m × ( k − ) = s b ( M m × ) = ǫ mr (1 ≤ r ≤ m ). Obviously, all l - and r -states of them must be trivial.Thus, a ij = m X r =1 a ′ ir · a ′′ rj . This implies A m × k = A m × ( k − · A m × = ( A m ) k , and the induction step is finished (cid:3) Stage 3. Analyzing the state matrix
We analyze state matrix A m × n = ( A m ) n to find the partition function G m × n ( v, x, y ). Proof of Theorem 1.
The (1 , A m × n is the state polynomial forthe set of suitably adjacent m × n -mosaics associated to the triple h a · · · a , ǫ m , ǫ m i = h a · · · a , a · · · a , a · · · a i , so having trivial r -, b -, t - and l -states. According to the boundary staterequirement, monomer–dimer coverings in Z m × n are converted into suitablyadjacent m × n -mosaics M with trivial r -, b -, t - and l -states as the left picture TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 11 in Figure 10. Thus this state polynomial represents the partition function G m × n ( v, x, y ). In short, we get G m × n ( v, x, y ) = (1,1)-entry of A m × n . This combined with Lemmas 5 and 6 completes the proof. Note that the tworecurrence relations in Lemma 5 easily merge into one recurrence relationas in Theorem 1. (cid:3) a G ( v,x,y )m + n G (0,1,1)m + ns aaa abaa aaaaaa aaa aaa aaaaa aa aaa aa Figure 10.
Examples of monomer–dimer coverings relatedto (1 , , A m × n Proof of Theorem 3. G sm × n ( v, x, y ) which is the (2 , A m ) n is thestate polynomial associated to the triple h a · · · a , ǫ m , ǫ m i , so having thesecond b -state and trivial t -, l - and r -states. Since the second b -state is aa · · · ab , M , must be mosaic tile T and we may consider it as a fixed sin-gle monomer. Now G sm × n (0 , ,
1) is the number of pure dimer coverings witha single monomer at M , , as desired. It is well-known that this number isindependent of location of the monomer, provided that it places at boundarysites with odd-numbered x - and y -coordinates [28]. Therefore, instead of the(2,1)-entry, we may use any ( i, j )-entry of ( A m ) n for { i, j } = { , k +1 } and k = 0 , , , . . . , m − (cid:3) Growth constant of the Hosoya index
We will need the following result called Fekete’s lemma whose conse-quences are many and deep. In this paper we state and prove its two-variatemultiplicative version with generalization.
Lemma 7 (Generalized Fekete’s Lemma) . Let { a m,n } m, n ∈ N be a doublesequence with a m,n ≥ , and k be a nonnegative integer.If the sequence satisfies a m ,n · a m ,n ≤ a m + m + k,n and a m,n · a m,n ≤ a m,n + n + k for all m , m , m , n , n and n , then lim m,n →∞ ( a m,n ) mn = sup m,n ≥ ( a m,n ) m + k )( n + k ) , provided that the supremum exists.Instead, if it satisfies a m + m ,n ≤ a m + k,n · a m ,n and a m,n + n ≤ a m,n + k · a m,n , then lim m,n →∞ ( a m,n ) mn = inf m,n>k ( a m,n ) m − k )( n − k ) . Remark that in this paper we only use the supermultiplicative inequalitypart with k = 0. The other parts will be used in on-going papers. Proof.
Let S = sup m,n ( a m,n ) m + k )( n + k ) and let B be any number less than S .Choose any positive integers i and j satisfying B < ( a i,j ) i + k )( j + k ) . Forsufficiently large integers m and n , there are integers p m and q m (simillary p n and q n for n and j ) such that m = p m ( i + k ) + q m and 0 ≤ q m < i + k bythe division algorithm. By the supermultiplicative inequalities prescribed inthe lemma,( a m,n ) mn ≥ ( a i,n ) pmmn ≥ ( a i,j ) pmpnmn = ( a i,j ) i + k )( j + k ) (cid:16) pm ( i + k ) m (cid:17)(cid:16) pn ( j + k ) n (cid:17) . Since p m ( i + k ) m , p n ( j + k ) n → m, n → ∞ , we have B < ( a i,j ) i + k )( j + k ) ≤ lim m, n →∞ ( a m,n ) mn ≤ S. This provides the desired limit. The submultiplicative inequality part of theproof can be proved in similar way. (cid:3)
Proof of Theorem 2.
We use briefly G m × n to denote G m × n (1 , ,
1) whichis obviously at least 1 for all m , n . First, we prove the existence of thelimit of ( G m × n ) mn . The supermultiplicative inequalities G m × n · G m × n ≤ G ( m + m ) × n and similarly G m × n · G m × n ≤ G m × ( n + n ) are obvious becausewe can create a new monomer–dimer ( m + m ) × n -mosaic by simply adjoiningtwo monomer–dimer m × n - and m × n -mosaics. Since sup m, n ( G m × n ) mn ≤ (cid:3) Fixed monomers problem
Proof of Theorem 4.
Let S be a fixed monomer set. We find bar state matrix A m,i for some i th bar mosaic with fixed monomers by using relevant bar statematrix recurrence relations similar to Lemma 5 with some modifications ineach step ( k, i ) as below. We may assume that { ( p, i ) , . . . , ( q, i ) } , 1 ≤ p ≤ q ≤ m , is a subset of maximal consecutive vertices in S on the i th bar mosaicin the sense that two vertices ( p − , i ) and ( q + 1 , i ) (if they exist) are notcontained in S . See Figure 11. Note that, in this case, we only consider thenumber of monomer–dimer coverings instead of the partition function. i - th bar mosaicp - th q - thno T T, no T
T,only T Figure 11.
A set of maximal consecutive vertices of S If ( k − , i ), ( k, i ) and ( k +1 , i ) are not contained in S ,(1) A k,i = (cid:20) B k − ,i A k − ,i A k − ,i O k − (cid:21) and B k,i = (cid:20) A k − ,i O k − O k − O k − (cid:21) TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 13 because T cannot be used in this step. Also if k = p, . . . , q , i.e., ( k, i ) iscontained in S , A k,i = (cid:20) A k − ,i O k − O k − O k − (cid:21) and B k,i = O k because only T can be located at ( k, i ).In the remaining cases of k = p − q +1, A p − ,i = (cid:20) B p − ,i A p − ,i A p − ,i O p − (cid:21) and B p − ,i = O p − because T and T cannot be located at ( p − , i ) and so the 11-submatrix of B p − ,i is O p − instead of A p − ,i , and A q +1 ,i = (cid:20) O q A q,i A q,i O q (cid:21) and B q +1 ,i = (cid:20) A q,i O q O q O q (cid:21) because T and T cannot be located at ( q +1 , i ). Indeed, the equations inthese remaining cases can be replaced by Eq. (1) because A p,i and B p,i donot use B p − ,i and the 11-submatrix of A q +1 ,i is O q which is equal to B q,i .Therefore all of these recurrence relations eventually merge into the re-currence relations in Theorem 4. By applying the rest of the state matrixrecursion method, we conclude that g m × n ( S ) = (1,1)-entry of n Y i =1 A m,i , which completes the proof. (cid:3) Domino tilings in the Aztec octagon
An Aztec diamond of order n consists of all lattice squares that lie com-pletely inside the diamond shaped region { ( x, y ) : | x | + | y | ≤ n + 1 } . Anaugmented Aztec diamond of order n looks much like the Aztec diamond oforder n , except that there are three long columns in the middle instead oftwo. A domino is a 1-by-2 or 2-by-1 rectangle. There are exact enumerationsof domino tilings of these two regions and dozens of interesting variants asstated in the introduction.In this section, we study the domino tilings on the most extended versionof the Aztec diamond. An m × n - Aztec octagon of order ( p, q, r, s ), denoted by A m × n ( p, q, r, s ), is defined as the union of mn − ( p − p + q − q + r − r + s − s ) unitsquares, arranged in the m × n rectangular grid with four triangular cornerswith side lengths p − q − r − s − n canbe represented as A n × n ( n, n, n, n ) and A (2 n +1) × n ( n, n, n, n ), respectively,and the m × n rectangular region Z m × n is indeed A m × n (1 , , , α m × n ( p, q, r, s ) denote the number of domino tilings of A m × n ( p, q, r, s ). Theorem 8.
For domino tilings of an Aztec octagon A m × n ( p, q, r, s ) , α m × n ( p, q, r, s ) = ( b m ( r, s ) , b m ( q, p )) -entry of ( A m ) n , { sp { { { rq Figure 12. A × (3 , , , Figure 13.
Domino tilings of Aztec octagons without/with holes where A m is the m × m matrix recursively defined by A k = (cid:20) A k − O k − O k − O k − (cid:21) A k − A k − O k − for k = 2 , . . . , m , with seed matrices A = (cid:2) (cid:3) and A = (cid:20) (cid:21) . Here b m ( r, s ) = 23 (2 m − r +2[ r ] − m − r ) + 13 (2 s − s − s ] ) + 1 . Proof.
It is worthwhile mentioning that the enumeration of domino tilings ofthis special case A m × n (1 , , ,
1) is answered as G m × n (0 , ,
1) in Theorem 1.Domino tilings of a region (or equivalently, pure dimer coverings on therelated square lattice) is known to be very sensitive to its boundary condi-tion. As an evidence, if it has a non-trivial boundary state in some part ofwhich letters a and b appear in turn as in Figure 14, then not only boundarybut some interior squares must be covered by dominos in the unique way asthe shaded region in the figure.Using this trick, by letting the bottom and top states as in Figure 15, wecan cover the related four triangular corners by dominos in the unique wayso that the remaining set of squares is the Aztec octagon we consider. Thebottom state of length m consists of three parts; r − a and b in TATE MATRIX RECURSION METHOD AND MONOMER–DIMER PROBLEM 15 ababaa aabb a { Figure 14.
Non-trivial boundary state conditionturn ending with b , m − r − s +2 letters of only a , and followed by s − a and b in turn beginning with b . This is the b m ( r, s )th state among 2 m states, where for r, s ≥ b m ( r, s ) = 2 m − (or 2 m − ) + · · · + 2 m − r +3 + 2 m − r +1 + 2 s − + 2 s − + · · · + 2 (or 2 ) + 1 , where the choice of 2 m − or 2 m − (similarly 2 or 2 ) depends on whether r (respectively s ) is even or odd. Further denote that b m (1 ,
1) = 1 ,b m (1 , s ) = 2 s − + 2 s − + · · · + 2 (or 2 ) + 1 ,b m ( r,
1) = 2 m − (or 2 m − ) + · · · + 2 m − r +3 + 2 m − r +1 + 1 . Remember that the bottom and top states are obtained by reading off lettersfrom right to left. This number b m ( r, s ) can be re-defined as written in thetheorem. { s - p - {{ { r - q - babaaaaaba ab ababaaaaab ba Figure 15.
Suitably adjacent 12 × b m ( r, s ) , b m ( q, p ))-entry of ( A m ) n afterapplying x = y = 1 and v = 0 is the number of suitably adjacent m × n -mosaics associated to the triple h a · · · a , ǫ mb m ( r,s ) , ǫ mb m ( q,p ) i . This completes theproof. (cid:3) Lastly, we mention the enumeration problem of domino tilings of anAztec octagon with holes as the right picture in Figure 13. Let S be aset of squares in A m × n ( p, q, r, s ). Let α m × n ( S ; p, q, r, s ) denote the numberof domino tilings of A m × n ( p, q, r, s ) restricting all squares of S removed. Theorem 9.
For domino tilings of an Aztec octagon A m × n ( p, q, r, s ) with aset S of holes, α m × n ( S ; p, q, r, s ) = ( b m ( r, s ) , b m ( q, p )) -entry of n Y i =1 A m,i , where A m,i is recursively defined in Theorem 4 and b m ( r, s ) is in Theorem 8.Proof. Theorem 4 combined with Theorem 8 guarantees the theorem. (cid:3)
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