Dimer Coverings on Random Polyomino Chains

1 Introduction

The study of counting dimer coverings (or perfect matchings) on random graphs has a long history [1–9]. In Zdeborová and Mézard [9], the authors studied dimer coverings on random regular and Erdös–Rényi random graphs by means of the cavity method. Gutman et al. [3–5] studied the number of dimer coverings on random hexagonal chains, while Chen and Zhang [1] studied the same problem for random phenylene chains. More recently, Ren et al. [8] studied dimer coverings on a kind of random multiple chains of planar honeycomb lattices by the transfer matrix approach. In this paper, we shall study dimer coverings on random polyomino chains. Note that, just like the hexagonal system, the polyomino system is one of the hottest studying topics in statistical mechanics and chemistry.

Now, let us recall some basic concepts and results. In statistical physics, a dimer represents a diatomic molecule. The dimer model was first considered by Roberts [10], then by Fowler and Rushbrooke [11], and was introduced in order to describe the absorption of diatomic molecules on crystal surface. In graph theoretic terms, a dimer is a molecule that can be placed on a graph G such that it covers an edge and the two incident vertices. A dimer arrangement is a set of dimers placed on G such that no vertex is covered by more than one dimer. A dimer arrangement that covers all vertices in G is called a dimer covering or a perfect matching in terms of graph theory. The dimer model is a classical statistical mechanics model dealing with the set of all dimer coverings of a graph [12].

A polyomino system is a finite two-connected plane graph such that each interior face (also called a cell) is surrounded by a regular square of length 1. A polyomino chain is a chain polyomino system in which the joining of the centres of its adjacent cells forms a path c₁c₂…c_n, where c_i is the centre of the ith cell and denoted by R_n (see Fig. 1 for an illustration example). M(R_n) represents the number of dimer coverings on R_n. And in Zhang and Zhang [13], the authors proved that M(R_n) is always nonzero. What is more, for some special polyomino chains, the explicit expressions for M(R_n) are known [14], e.g., the linear chain, denoted by L_n (see Fig. 2a),

Figure 1:

A polyomino chain with eight cells and the corresponding path.

Figure 2:

(a) The linear chain L₇; (b) The zigzag chain Z₇.

M(Ln) = 5 + 3510(1 + 52)n + 5 − 3510(1 − 52)n,

and the zigzag chain, denoted by Z_n (see Fig. 2),

M(Zn) = n + 1.

Here we concentrate on the number of dimer coverings on random polyomino chains. In Section 2, we give a detailed description of the ensemble of random polyomino chains we shall consider. In Section 3, first, by the law of total expectation in probability theory, we establish a recurrence relation satisfied by the expectation of the number of dimer coverings in this ensemble. Then by solving the recurrence relation, we obtain the explicit expression for the expectation. From this expression, it is easy to see that, with only one exception, the average of the logarithm of this expectation (i.e., the annealed entropy of the dimer coverings) is asymptotically nonzero when n→ ∞.

2 The Ensemble of Random Polyomino Chains

Note that from the definition of polyomino chains, each polyomino chain R_n can be formed inductively. Starting from one cell, then in each step we attach a new cell to the previous one. More clearly, given a polyomino chain R_n with n squares, the squares in R_n are fixed by the path c₁c₂…c_n mentioned previously. And let R_i(i= 1, …, n) be the part of R_n with the first i squares. Then R_i is obtained from R_i–1 by attaching to it the ith square c_i. We see that, from i≥ 3, there are two ways to attach an edge of c_i to an edge e of the end square in R_i–1: either e is incident with two vertices of degree 2 in R_i–1, which is called the H-type attachment (see Fig. 3a) or e is incident with a vertex of degree 2 and the other vertex of degree 3 in R_i–1, which is called the V-type attachment (see Fig. 3b). So, for each polyomino chain in R_n(n≥ 3), a string of n– 2 symbols of H and V can be associated, indicating the attachment type of squares from 3 to n, for example, the linear chain L_n associates with H…H and the zigzag chain Z_n associates with V…V. Note that a polyomino chain is not uniquely determined by its HV sequence (see Fig. 4). But by symmetry, different polynomino chains corresponding to the same HV sequence have the same number of perfect matchings. So let

Figure 3:

(a) The H-type attachment; (b) The V-type attachment.

Figure 4:

Two polyomino chains corresponding to R_n=HVHV.

ℜn = {u1u2…un − 2|ui∈{H, V}, i = 1, …, n − 2}, (n ≥ 3).

We define an ensemble of random polyomino chains (ℜ_n, P) as follows:

1. for n= 1, 2, let ℜ₁={R₁}, ℜ₂={R₂}, that is, the probability spaces are trivial with only one element, respectively;

2. for n ≥ 3, the probability distribution

P(u1u2…un − 2) = pm(1 − p)n − 2 − m,

where m is the number of times that H appears in this sequence and p∈[0,1] is a constant. That is, the probability of the H-type attachment is p and that of the V-type attachment is 1 – p.

Let M_n(u₁u₂…u_n–2) be the number of perfect matchings of the polyomino chains associated with this sequence, where M_n is an integer valued random variable on (ℜn, P). In the next section, we will determine the expectation E(M_n) of M_n.

3 Main Results and Proofs

In order to obtain the expectation of M_n, we first recall two basic results: one is from graph theory and the other is from probability theory.

Lemma 1 [15]. Let e= uv be an edge of a graph G. Then,

M(G) = M(G − e) + M(G − u − v),

where G– e is the subgraph obtained from G by deleting edge e and G– u– v is the subgraph obtained from G by deleting vertices u and v together with their incident edges.

Lemma 2 [16]. (Law of total expectation) If X is a random variable with E(|X|)<∞ and Y is any random variable, on the same probability space, then

E(X) = EY(EX|Y(X|Y)).

One special case states that if A₁, A₂,…, A_n is a partition of the whole outcome space, where A₁, A₂,…, A_n satisfying these events are mutually exclusive and exhaustive, then

E(X) = ∑i = 1np(Ai)E(X|Ai).

From the above Lemmas, we can obtain the following result.

Theorem 1. The expectations E(M_n) satisfy the following recurrence relation:

E(Mn) = E(Mn − 1) + p∑i = 0n − 3(1 − p)iE(Mn − 2 − i) + (1 − p)n − 2

with initial conditions E(M₁)=2 and E(M₂)=3.

Proof. The initial conditions E(M₁)=2 and E(M₂)=3 are easy to get from the definition. Now, for n≥ 3, recall that

ℜn = {u1u2…un − 2|ui∈{H, V}, i = 1,…, n − 2}.

Let

Ain = {*HV…V︸i|*∈ℜn − i − 1,}, 0 ≤ i ≤ n − 3;An − 2n = {VV…V}.

Then,

ℜn = ∪i = 0n − 2Ain,  Ain∩Ajn = ∅, (i ≠ j).

And it is easy to see that

P(Ain) = p(1 − p)i, (0 ≤ i ≤ n − 3), P(An − 2n) = (1 − p)n − 2.

So by Lemma 2, we have

(*)E(Mn) = ∑i = 0n − 2P(Ain)E(Mn|Ain) = ∑i = 0n − 3p(1 − p)iE(Mn|Ain) + (1 − p)n − 2E(Mn|An − 2n).  (*)

Now, for any given R_n, take e to be the edge in the last cell, which is incident with two vertices of degree 2 (see Fig. 5). Then by Lemma 1, we have

Figure 5:

(a) M(R_a–e)=M(R_b); (b) M(R_a – u – v)=M(R_c).

Case 1. If Rn∈A0n, it means that R_n=u₁u₂…u_n–3H. Then

M(u1u2…un−3H) = M(u1u2…un − 3) + M(u1u2…un − 4).

Therefore,

(1)E(Mn|A0n) = E(Mn − 1) + E(Mn − 2). (1)

Case 2. If Rn∈Ain(1 ≤ i ≤ n − 3), it means that Rn = u1…un − i − 3HV…V︸i, (1 ≤ i ≤ n − 3). Then,

M(u1…un − i − 3HV…V︸i) = M(u1…un − i − 3HV…V︸i − 1)  + M(u1u2…un − i − 4).

Therefore,

(2)E(Mn|Ain) = E(Mn − 1|Ai − 1n − 1) + E(Mn − 2 − i). (2)

Case 3. If Rn∈An − 2n, that is Rn = V…V︸n − 2, then

M(V…V︸n − 2) = M(V…V︸n − 3) + 1.

Therefore,

(3)E(Mn|An − 2n) = E(Mn − 1|An − 3n − 1) + 1. (3)

Substituting (1), (2) and (3) into the formula *, we have

E(Mn) = p(E(Mn − 1) + E(Mn − 2)) + ∑i = 1n − 3p(1 − p)i(E(Mn − 1|Ai − 1n − 1) + E(Mn − 2 − i))+ (1 − p)n − 2(E(Mn − 1|An − 3n − 1) + 1)= p(E(Mn − 1) + E(Mn − 2)) + ∑i = 1n − 3p(1 − p)iE(Mn − 2 − i) + (1 − p)n − 2+ (1 − p)[∑i = 1n − 3p(1 − p)i − 1E(Mn − 1|Ai − 1n − 1) + (1 − p)n − 3E(Mn − 1|An − 3n − 1)].

Note that

∑i = 1n − 3p(1 − p)i − 1E(Mn − 1|Ai − 1n − 1) + (1 − p)n − 3E(Mn − 1|An − 3n − 1)  = E(Mn − 1).

So we have

E(Mn) = E(Mn − 1) + p∑i = 0n − 3(1 − p)iE(Mn − 2 − i) + (1 − p)n − 2.

To solve the recurrence relation in Theorem 1, we first give the following result.

Lemma 3.

|α + βαβ0⋯001α + βαβ⋯0001α + β⋯00⋮⋮⋮⋮⋮⋮000⋯1α + β|n × n = αn + 1 − βn + 1α − β.

Proof. This result can be proved easily by the induction on n.

Theorem 2. The expectation of the number of perfect matchings in random polyomino chains is

E(Mn) = (− 1)nα − β[(3 + 2β)αn − 1 − (3 + 2α)βn − 1)],

where α = p − 2 + p2 + 4p2, β = p − 2 − p2 + 4p2.

Proof. Since E(M₁)=2, E(M₂)=3, so by Theorem 1, we have

E(Mn) − E(Mn − 1) − pE(Mn − 2) − ⋯ − p(1 − p)n − 5E(M3)  − p(1 − p)n − 4E(M2) = p(1 − p)n − 3E(M1) + (1 − p)n − 2 = (1 − p)n − 3(1 + p).

Thus we get a system of linear equations,

[1− 1− p⋯− p(1 − p)n − 5− p(1 − p)n − 401− 1⋯− p(1 − p)n − 6− p(1 − p)n − 5⋮⋮⋮⋮⋮⋮000⋯− 1− p000⋯1− 1000⋯01][E(Mn)E(Mn − 1)⋮E(M4)E(M3)E(M2)]  = [(1 − p)n − 3(1 + p)(1 − p)n − 4(1 + p)⋮(1 − p)(1 + p)1 + p3].

So, by Cramer’s rule, E(M_n) is equal to

|(1 − p)n − 3(1 + p)− 1− p− p(1 − p)⋯− p(1 − p)n − 5− p(1 − p)n − 4(1 − p)n − 4(1 + p)1− 1− p⋯− p(1 − p)n − 6− p(1 − p)n − 5⋮⋮⋮⋮⋮⋮⋮(1 − p)(1 + p)000⋯− 1− p1 + p000⋯1− 13000⋯01|(n − 1) × (n − 1).

From the second to the last row in the above determinant, multiplying each row by p – 1 and adding to its previous row, we get

E(Mn) = |0p − 21 − 2p⋯00001p − 2⋯000001⋯000⋮⋮⋮⋮⋮⋮⋮000⋯1p − 21 − 2p4p − 200⋯01p − 2300⋯001|(n − 1) × (n − 1)

= (4p − 2)( − 1)n − 1|p − 21 − 2p⋯0001p − 2⋯00001⋯000⋮⋮⋮⋮⋮⋮00⋯p − 21 − 2p000⋯1p − 21 − 2p|(n − 2) × (n − 2)

+ 3( − 1)n|p − 21 − 2p⋯0001p − 2⋯00001⋯000⋮⋮⋮⋮⋮⋮00⋯p − 21 − 2p000⋯1p − 21 − 2p00⋯01p − 2|(n − 2) × (n − 2).

Write

Dn = |p − 21 − 2p⋯0001p − 2⋯00001⋯000⋮⋮⋮⋮⋮⋮00⋯p − 21 − 2p000⋯1p − 21 − 2p00⋯01p − 2|n × n.

Then

E(Mn) = ( − 1)n3Dn − 2 + ( − 1)n − 1(4p − 2)Dn − 3.

Let α = p − 2 + p2 + 4p2, β = p − 2 − p2 + 4p2. Then α + β=p – 2, αβ=1 – 2p. So by Lemma 3,

E(Mn) = ( − 1)n3Dn − 2 + ( − 1)n − 1(4p − 2)Dn − 3= 1α − β[(− 1)n3(αn − 1 − βn − 1) + (− 1)n − 1(4p − 2)(αn − 2 − βn − 2)]= (− 1)nα − β[(3 + 2β)αn − 1 − (3 + 2α)βn − 1)].

Now the proof is completed.

Remark 1: From the expression of Theorem 2, we can easily deduce the number of dimer coverings of the linear chain L_n since

M(Ln) = limp→1E(Mn) = 5 + 3510(1 + 52)n + 5 − 3510(1 − 52)n.

Similarly, the number of dimer coverings of the zigzag chain Z_n is

M(Zn) = limp → 0E(Mn) = n + 1.

Besides p=0 and p=1, the other special case that is worth pointing out is p = 12, that is, the uniform random polyomino chains. In this case,

E(Mn) = 2(32)n − 1.

Remark 2: In statistical physics, for a random variable X, the asymptotic behaviour of the average of log E(X), namely, the annealed entropy of X, plays an important role. From Theorem 2, we see

limn→∞logE(Mn)2n + 2 = log(− β)2,

where 2n + 2 is the vertex number of polyomino chain with n squares. So except for p=0, the annealed entropy is nonzero.