Number of Dominating Sets in Cylindric Square Grid Graphs

Abstract

A dominating set of a graph is a subset D of the vertices such that every vertex not in D is adjacent to some vertex of D. In this paper, we introduce several variants of dominating sets in the square grid, periodic square grid and cylindric square grid by considering translation symmetry. We provide their exact enumerations in terms of domination polynomials. We also analyze the asymptotic behavior of the growth rates of their cardinality.

1 Introduction

Let G be a simple graph whose vertex set and edge set are denoted by V(G) and E(G). A dominating set of G is a subset D of V such that every vertex not in D is adjacent to at least one vertex of D. The domination number $\gamma (G)$ is the cardinality of a smallest dominating set. For a survey on this subject and a comprehensive bibliography, we refer the reader to read [9, 10].

The basic problem of enumerative combinatorics is that of counting the number of elements of a finite set. The domination polynomial of G is the generating function of its dominating sets with respect to the number of vertices, which is defined as

$$\begin{aligned} P(G,z) = \sum _{i=\gamma (G)}^{|V|} d(G,i) z^i, \end{aligned}$$

where d(G, i) is the number of dominating sets with cardinality i in G and $\gamma (G)$ is the domination number of G.

The closed forms for the domination polynomials of some simple classes of graphs were first introduced in [2]. An abundances of recurrence of relations for domination polynomials for arbitrary graphs is found in [13]. In particular, Theorem 5.11 in [13] is a more general result of Theorem 1 regarding the domination polynomial of the square grid in our result. So Theorem 1 can be proved in an alternative way. In this paper, instead of counting all dominating sets, we count dominating sets up to translation symmetry of a cylinder graph.

The Cartesian product $G = G_1 \square \, G_2$ of graphs $G_1$ and $G_2$ is the graph with $V(G) = V(G_1) \times V(G_2)$ and $((u_1,u_2),(v_1,v_2)) \in E(G)$ whenever $u_1=v_1$ and $(u_2,v_2) \in E(G_2)$ or $u_2=v_2$ and $(u_1,v_1) \in E(G_1)$. Let $p_n$, $p_{\infty }$, and $c_n$ denote a path of length $n-1$, a path of infinite length, and a cycle of length n. We denote by $G_{m \times n}= p_m \square \, p_n$ (a square grid), $G_{m \times {\infty }}= p_m \square \, p_{\infty }$ (an infinite strip square grid), and $G^{\mathrm{c}}_{m \times {\overline{n}}}= p_m \square \, c_n$ (a cylindric square grid). Throughout this paper, we use the notation ${\overline{n}}$ instead of n to denote periodicity.

The domination number of particular Cartesian products of paths and cycles has been deeply studied. Many efforts have been made to obtain their exact domination numbers [1, 3, 7, 11, 12, 14] and lower and upper bounds on their domination numbers [4,5,6, 8].

The author has developed the state matrix recursion method to enumerate knot mosaics [18] and two well-known lattice statistical models in the square grid, which are monomer–dimer coverings [16] and independent vertex sets [15, 17].

In this paper, we develop and apply this method to solve the enumeration problem for several variants of dominating sets in the following square grids: $G_{m \times n}$, $G_{m \times {\infty }}$ and $G^{\mathrm{c}}_{m \times {\overline{n}}}$. In contrast to $G_{m \times n}$, there are infinitely many dominating sets in $G_{m \times {\infty }}$. In this case, we focus only on (periodic) dominating sets having vertically periodic patterns whose periodic patch has fixed finite size so that the number of these dominating sets is bounded. Furthermore, $G^{\mathrm{c}}_{m \times {\overline{n}}}$ has translation symmetry due to the topology of the cylinder, so we need to consider the classes of dominating sets considered equivalent under action of a symmetry group, called cylindric dominating sets. Using the periodic dominating sets in $G_{m \times {\infty }}$, we count all cylindric dominating sets in $G^{\mathrm{c}}_{m \times {\overline{n}}}$. More precisely, we provide recurrence relations of state matrices to give their domination polynomials.

In the last section, we also analyze the asymptotic behavior of the growth rates of their cardinality.

2 Terminology and Theorems

2.1 Dominating Sets in $G_{m \times n}$

We briefly denote the domination polynomial of $G_{m \times n}$ as $P_{m \times n}(z) = P(G_{m \times n},z)$. Note that $P_{m \times n}(1)$ indicates the number of dominating sets in $G_{m \times n}$. An example of a dominating set in $G_{m \times n}$ is drawn in Fig. 1.

Fig. 1

A dominating set in $G_{m \times n}$

Hereafter, ${\mathbb {O}}_k$ is the $4^k \times 4^k$ zero-matrix and

$$\begin{aligned} L_m = \begin{bmatrix} 0&1&0&1 \end{bmatrix}^{\otimes m} \text { and }\ R_m = \big (\begin{bmatrix} 1&0&0&1 \end{bmatrix}^{\otimes m} \big )^t, \end{aligned}$$

where $M^{\otimes m}$ is the m-fold tensor product of a matrix M.

Theorem 1

The domination polynomial $P_{m \times n}(z)$ of the square grid $G_{m \times n}$ is the unique entry of the $1 \times 1$ matrix

$$\begin{aligned} L_m \cdot (A_m)^n \cdot R_m, \end{aligned}$$

where $A_m$ is a $4^m \times 4^m$ matrix recursively defined by

$$\begin{aligned} &A_{k+1} = \begin{bmatrix} {\mathbb {O}}_k &{} A_k &{} {\mathbb {O}}_k &{} A_k \\ zB_k &{} {\mathbb {O}}_k &{} zB_k &{} {\mathbb {O}}_k \\ zB_k &{} {\mathbb {O}}_k &{} zB_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} A_k &{} {\mathbb {O}}_k &{} C_k \end{bmatrix} , \ B_{k+1} = \begin{bmatrix} {\mathbb {O}}_k &{} A_k &{} {\mathbb {O}}_k &{} A_k \\ zB_k &{} {\mathbb {O}}_k &{} zB_k &{} {\mathbb {O}}_k \\ zB_k &{} {\mathbb {O}}_k &{} zB_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} A_k &{} {\mathbb {O}}_k &{} A_k \end{bmatrix} \\ & \quad {and }\,\, C_{k+1} = \begin{bmatrix} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \\ zB_k &{} {\mathbb {O}}_k &{} zB_k &{} {\mathbb {O}}_k \\ zB_k &{} {\mathbb {O}}_k &{} zB_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \end{bmatrix} , \end{aligned}$$

for $k=0, \dots , m - 1$, starting with $A_0 = B_0 = \begin{bmatrix} 1 \end{bmatrix}$ and $C_0 = \begin{bmatrix} 0 \end{bmatrix}$.

2.2 Periodic Dominating Sets in $G_{m \times {\infty }}$

Even though there are infinitely many dominating sets in an infinite strip square grid, the number of dominating sets having periodic patterns whose periodic patch has fixed finite size is bounded. To formulate more precisely, we first consider a rectangular square grid $G_{m \times n}$ as the standard periodic patch, and create an infinite strip square grid $G_{m \times {\infty }}$ by vertically translating copies of $G_{m \times n}$ and pasting them. Without loss of generality, we assume that the y-coordinates of the vertices of the standard periodic patch $G_{m \times n}$ are $\{1,2,\dots ,n\}$ in $p_{\infty }$.

An n-periodic dominating set in $G_{m \times {\infty }}$ is a dominating set in $G_{m \times {\infty }}$ such that it can also be obtained by vertically translating copies (modulo n steps up or down) of a vertex set in the standard periodic patch $G_{m \times n}$. The domination polynomial for n-periodic dominating sets in $G_{m \times {\infty }}$ is denoted as $P^{\mathrm{p}}_{m \times {\overline{n}}}(z)$.

The reader must be aware that two n-periodic dominating sets are considered to be different if they look different in the standard periodic patch, even though one can be obtained from the other by translating the set of vertices vertically in $G_{m \times {\infty }}$. Note that $P^{\mathrm{p}}_{m \times {\overline{n}}}(1)$ indicates the number of n-periodic dominating sets in $G_{m \times {\infty }}$.

An example of an n-periodic dominating set in $G_{m \times {\infty }}$ is drawn in Fig. 2. Note that an n-periodic dominating set may have a shorter fundamental period $n'$. Here, $\mathrm{tr} A$ is the trace of a square matrix A.

Fig. 2

A 5-periodic dominating set in $G_{6 \times {\infty }}$

Theorem 2

The domination polynomial for n-periodic dominating sets in the infinite strip square grid $G_{m \times {\infty }}$ is

$$\begin{aligned} P^{\mathrm{p}}_{m \times {\overline{n}}}(z) = \mathrm{tr} (A_m)^n, \end{aligned}$$

where $A_m$ is obtained in Theorem 1.

2.3 Cylindric Dominating Sets in $G^{\mathrm{c}}_{m \times {\overline{n}}}$

A cylindric dominating set in $G^{\mathrm{c}}_{m \times {\overline{n}}}$ is a dominating set in $G^{\mathrm{c}}_{m \times {\overline{n}}}$ up to cyclic rotations along the cycle $c_n$. The equivalence relation under cyclic rotations is due to the topology of the cylinder. Thus, cylindric dominating sets are obtained from periodic dominating sets modulo cyclic rotations. The domination polynomial for cylindric dominating sets in $G^{\mathrm{c}}_{m \times {\overline{n}}}$ is denoted as $P^{\mathrm{c}}_{m \times {\overline{n}}}(z)$.

Note that $P^{\mathrm{c}}_{m \times {\overline{n}}}(1)$ indicates the number of cylindric dominating sets. An example of a cylindric dominating set is drawn in Fig. 3.

Fig. 3

A cylindric dominating set in $G^{\mathrm{c}}_{6 \times {\overline{5}}}$

Theorem 3

The domination polynomial for cylindric dominating sets in the cylindric square grid $G^{\mathrm{c}}_{m \times {\overline{n}}}$ is

$$\begin{aligned} P^{\mathrm{c}}_{m \times {\overline{n}}}(z) = \sum _{p | n} \frac{1}{p} \, S_{m \times p}(z^{\frac{n}{p}}), \end{aligned}$$

where the polynomial $S_{m \times p}(z)$ is recursively defined by

$$\begin{aligned} S_{m \times p}(z) = P^{\mathrm{p}}_{m \times {\overline{p}}}(z) - \!\!\! \sum _{r | p, \, r < p} \!\!\! S_{m \times r}(z^{\frac{p}{r}}). \end{aligned}$$

It is of interest to note that the numbers of $P_{m \times n}(1)$ and $P^{\mathrm{p}}_{m \times {\overline{n}}}(1)$ are registered in Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS) [19] as sequences A218354 and A286514. The following is a list of the numbers of $P^{\mathrm{c}}_{m \times {\overline{n}}}(1)$.

Table 1 $P^{\mathrm{c}}_{m \times {\overline{n}}}(1)$

2.4 Asymptotic Behavior of the Growth Rates

In the following we are interested in the asymptotic behavior of the growth rates per cite of the number of dominating sets. Consider the following limit, if it exists,

$$\begin{aligned} \lim _{m, \, n \rightarrow \infty } (P_{m \times n}(1))^{\frac{1}{mn}} = \kappa , \end{aligned}$$

which is called the dominating entropy constant of the square grid.

Theorem 4

The double limit $\kappa$ exists. More precisely,

$$\begin{aligned} \kappa = \sup _{m, \, n \ge 1} (P_{m \times n}(1))^{\frac{1}{mn}} = \inf _{m, \, n \ge 2} (P_{m \times n}(1))^{\frac{1}{(m-1)(n-1)}}. \end{aligned}$$

Furthermore, periodic dominating sets and cylindric dominating sets have the same growth rate $\kappa$, i.e.,

$$\begin{aligned} \lim _{m, \, n \rightarrow \infty } (P^{\mathrm{p}}_{m \times {\overline{n}}}(1))^{\frac{1}{mn}} = \lim _{m, \, n \rightarrow \infty } (P^{\mathrm{c}}_{m \times {\overline{n}}}(1))^{\frac{1}{mn}} = \kappa . \end{aligned}$$

3 Dominating Sets

This section is devoted to prove Theorem 1 by means of the state matrix recursion method developed by the author. This method proceeds in three stages. In the first stage we construct the mosaic system for dominating sets in the square grid. The second stage is devoted to find recursive matrix-relations producing the desired domination polynomial. In the third stage we analyze the obtained state matrix to provide the desired enumeration.

3.1 Stage 1: Conversion to the Dominating Mosaic System

The definition of the mosaic system in the paper [16] has been slightly modified in this paper. The mosaic tiles for dominating sets in $G_{m \times n}$ are distinguished into two types as illustrated in Fig. 4. There are 15 mosaic tiles of Type 1 (no dot inside), each of whose side edges is labeled with a letter ‘a’ or ‘d’, except a tile labeled with only letter d on its four sides. This exception is very crucial to satisfy the definition of domination and will be explained in the proof of Proposition 5. Furthermore there are 16 mosaic tiles of Type 2 (a dot in the center), each of whose side edges is labeled with a letter ‘b’ or ‘c’. The dot eventually indicates a vertex in a dominating set. In conclusion, we have a total of 31 mosaic tiles.

Fig. 4

Two types of mosaic tiles

An $m \times n$-mosaic is an $m \times n$ rectangular array $M = (m_{ij})$, where $m_{ij}$ is the mosaic tile placed at the ith column from ‘left’ to ‘right’ and the jth row from ‘bottom’ to ‘top’. To form a dominating set, each tile in a mosaic must match other adjacent tiles while meeting the following conditions:

• (Adjacency rule) Abutting edges of adjacent mosaic tiles are labeled with any of the following pairs of letters: a|b, c|c, d|d.

• (Boundary state requirement) All boundary edges are labeled with letters b and d only.

A mosaic is said to be suitably adjacent if any pair of mosaic tiles sharing an edge satisfies the adjacency rule. A suitably adjacent $m \times n$-mosaic is called a dominating $m \times n$-mosaic if it additionally satisfies the boundary state requirement. The key role in this method is the following one-to-one conversion.

Proposition 5

There is a one-to-one correspondence between dominating sets in $G_{m \times n}$ and dominating $m \times n$-mosaics. Obviously, the number of vertices in a dominating set is equal to the number of mosaic tiles of Type 2 in the corresponding dominating $m \times n$-mosaic.

Proof

As illustrated in Fig. 5, every dominating set D in $G_{m \times n}$ can be converted into an $m \times n$-mosaic which satisfies above two rules. Each vertex in $G_{m \times n}$ is converted into a mosaic tile of Type 2 or Type 1, whether or not the vertex is contained in D. If the mosaic tile of one side of an abutting edge is of Type 1 and the mosaic tile of the other side is of Type 2, then the corresponding labeled letters of these two mosaic tiles on the abutting edge are a and b. If the mosaic tiles of both sides are of Type 1 (or Type 2), then the corresponding labeled letters are d’s (resp. c’s). This conversion naturally satisfies the adjacency rule. Notice that the mosaic tile labeled with only letter d’s on its all four sides, which was initially excluded from the set of mosaic tiles, does not appear due to the definition of domination.

Furthermore, the side letter of a mosaic tile of Type 1 (or Type 2) lying on the boundary of the mosaic is labeled with letter d (resp. b). This is due to the boundary state requirement. Obviously there are two options for the boundary label for a mosaic tile of Type 1: a or d. But, if we choose letter a, then, for example, the mosaic tile with labeled a’s on the left and the bottom boundaries and d’s on the right and the top boundaries can be placed at the bottom-right corner, violating the definition of domination. For the same reason as before, we choose letter b for the boundary label for a mosaic tile of Type 2.

This conversion is unique, and conversely we can obtain a dominating set D from a given dominating mosaic by directly selecting the dots in the mosaic tiles of Type 2. Due to the absence of the mosaic tile labeled with only letter d on its four sides, every vertex not in D is adjacent to at least one vertex of D. $\square$

Fig. 5

Conversion of the dominating set in Fig. 1 to the corresponding dominating $m \times n$-mosaic

3.2 Stage 2: State Matrix Recursion Formula

We consider a suitably adjacent $p \times q$-mosaic M for positive integers $p \le m$ and $q \le n$. The number of appearances of mosaic tiles of Type 2 in M is denoted by d(M).

A state is a finite sequence of four letters a, b, c, and d. The b-state $s_b(M)$ and t-state $s_t(M)$ are the states of length p obtained by reading off letters on the bottom and the top, respectively, boundary edges of M from ‘right’ to ‘left’. Similarly the l-state $s_l(M)$ and r-state $s_r(M)$ are the states of length q on the left and the right, respectively, boundary edges from ‘top’ to ‘bottom’. For examples, the four state indications of the dominating $6 \times 5$-mosaic drawn in Fig. 5 are $s_b =$ dbdddb, $s_t =$ dbbddb, $s_l =$ bdbdb, and $s_r =$ dddbd.

Given r-, b-, and t-states $s_r, s_b, s_t$, a state polynomial for the triple $\langle s_r, s_b, s_t \rangle$ is

$$\begin{aligned} Q_{\langle s_r, s_b, s_t \rangle }(z) = \sum k(d) z^d, \end{aligned}$$

where k(d) is the number of all suitably adjacent $p \times q$-mosaics M such that $d(M)=d$, $s_r(M) = s_r$, $s_b(M) = s_b$, $s_t(M) = s_t$, and $s_l(M)$ is any state of length q consisting of only two letters b and d. The last condition for $s_l(M)$ is due to the boundary state requirement on the left side of mosaics.

First, consider a suitably adjacent $p \times 1$-mosaic, which is called a bar mosaic. Bar mosaics of length p have $4^p$ kinds of b- and t-states, called bar states. Arrange the bar states in two ways as follows; the abcd-order which is just the lexicographic order and the bacd-order which is similar to the lexicographic order, but b precedes a, (for example of $p=2$, bb, ba, bc, bd, ab, aa, ..., db, da, dc, and dd). For $1 \le i \le 4^p$, let $\epsilon ^p_i$ and $\lambda ^p_i$ denote the ith bar states of length p among the abcd- and the bacd-ordered state sets, respectively.

Bar state matrix $X_p$ ($X = A, B, C, D$) for the set of suitably adjacent bar mosaics of length p is a $4^p \times 4^p$ matrix $(m_{ij})$ given by

$$\begin{aligned} m_{ij} = Q_{\langle \text {x}, \epsilon ^p_i, \lambda ^p_j \rangle }(z), \end{aligned}$$

where x $=$ a, b, c, d, respectively.

Lemma 6

Bar state matrices $A_p$, $B_p$, $C_p$, and $D_p$ are obtained by the recurrence relations;

$$\begin{aligned} A_{k+1}= & {} {\begin{bmatrix} {\mathbb {O}}_k &{} B_k + D_k &{} {\mathbb {O}}_k &{} B_k + D_k \\ {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} B_k + D_k &{} {\mathbb {O}}_k &{} B_k + D_k \end{bmatrix}}, \\ D_{k+1}= & {} {\begin{bmatrix} {\mathbb {O}}_k &{} B_k + D_k &{} {\mathbb {O}}_k &{} B_k + D_k \\ {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} B_k + D_k &{} {\mathbb {O}}_k &{} B_k \end{bmatrix}}, \\ {and }\,\, B_{k+1}= & {} C_{k+1} = {\begin{bmatrix} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \\ zA_k + zC_k &{} {\mathbb {O}}_k &{} zA_k + zC_k &{} {\mathbb {O}}_k \\ zA_k + zC_k &{} {\mathbb {O}}_k &{} zA_k + zC_k &{} {\mathbb {O}}_k \\ {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k &{} {\mathbb {O}}_k \end{bmatrix}} \end{aligned}$$

with seed matrices

$$\begin{aligned} A_1 = {\begin{bmatrix} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 1 \end{bmatrix}} , \ B_1 = C_1 = {\begin{bmatrix} 0 &{} 0 &{} 0 &{} 0 \\ z &{} 0 &{} z &{} 0 \\ z &{} 0 &{} z &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 \end{bmatrix}} {, and } D_1 = {\begin{bmatrix} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 \end{bmatrix}} . \end{aligned}$$

Note that we may start with matrices $A_0 = D_0 = \begin{bmatrix} 1 \end{bmatrix}$ and $B_0 = C_0 = \begin{bmatrix} 0 \end{bmatrix}$ instead of $A_1$, $B_1$, $C_1$, and $D_1$.

Proof

The inductive proof of Lemma 5 in the paper [16] has been slightly modified and adopted here. First, for example, (4, 2)-entry of $A_1$ is

$$\begin{aligned} Q_{\langle {\texttt {a}}, \epsilon ^1_4, \lambda ^1_2 \rangle }(z) = Q_{\langle {\texttt {a}}, {\texttt {d}}, {\texttt {a}} \rangle }(z) = 1, \end{aligned}$$

since there is a unique mosaic tile M of Type 1 such that $s_r(M) =$ a, $s_b(M) =$ d, $s_t(M) =$ a, and $s_l(M)$ is any of b and d (but, in this case, only a can be used). Here, $\epsilon ^1_4 =$ d in the abcd-order and $\lambda ^1_2 =$ a in the bacd-order. The other entries of $A_1$, $B_1$, $C_1$, and $D_1$ can be obtained in the same manner.

Assume that $A_k$, $B_k$, $C_k$, and $D_k$ satisfy the statement. Consider, for example, the matrix $B_{k+1}$, which is of size $4^{k+1} \times 4^{k+1}$. Partition this matrix into sixteen block submatrices of size $4^k \times 4^k$, and consider the 31-submatrix of $B_{k+1}$, i.e., the (3, 1)-component in the $4 \times 4$ array of the sixteen blocks. According to the abcd- and the bacd-orders, the (i, j)-entry of the 31-submatrix is the state polynomial $Q_{\langle {\texttt {b}}, {\texttt {c}}\epsilon ^k_i, {\texttt {b}}\lambda ^k_j \rangle }(z)$, where c$\epsilon ^k_i$ (similarly b$\lambda ^k_j$) is a bar state of length $k + 1$ obtained by concatenating two states c and $\epsilon ^k_i$. A suitably adjacent $(k + 1) \times 1$-mosaic corresponding to this triple $\langle {\texttt {b}}, {\texttt {c}}\epsilon ^k_i, {\texttt {b}}\lambda ^k_j \rangle$ must have a rightmost mosaic tile M of Type 2 whose right, bottom, and top labels are b, c, and b. Thus, the left label of M can be any of b and c. This implies that the second rightmost mosaic tile of the $(k + 1) \times 1$-mosaic must have the right label a or c, respectively, by the adjacency rule. The contribution of the rightmost mosaic tile of Type 2 to the state polynomial gives

$$\begin{aligned} Q_{\langle {\texttt {b}}, {\texttt {c}}\epsilon ^k_i, {\texttt {b}}\lambda ^k_j \rangle }(z) = (i,j)\text {-entry of } (zA_k + zC_k). \end{aligned}$$

This implies that the 31-submatrix of $B_{k+1}$ is $zA_k + zC_k$. In the same manner, we derive the other sixty three entries as in the lemma. $\square$

Now, consider mosaics of any width. State matrix $Y_{m \times q}$ for the set of suitably adjacent $m \times q$-mosaics is a $4^m \times 4^m$ matrix $(y_{ij})$ given by

$$\begin{aligned} y_{ij} = \sum _{s_r} Q_{\langle s_r, \epsilon ^m_i, \lambda ^m_j \rangle }(z), \end{aligned}$$

where the summation is taken over $2^q$ r-states $s_r$ of length q consisting of only two letters b and d. The summation condition of $s_r$ is due to the boundary state requirement on the right side of mosaics.

Lemma 7

State matrix $Y_{m \times n}$ is obtained by

$$\begin{aligned} Y_{m \times n} = (B_m + D_m)^n. \end{aligned}$$

Proof

The inductive proof of Lemma 6 in the paper [16] has been slightly modified and adopted here. For $n=1$, $Y_{m \times 1} =B_m + D_m$, since $Y_{m \times 1}$ counts suitably adjacent $m \times 1$-mosaics with r-state b or d.

Assume that $Y_{m \times k} = (B_m + D_m)^k$. Consider a suitably adjacent $m \times (k + 1)$-mosaic $M^{m \times (k+1)}$ with r- and l-states consisting of letters b and d. Split it into two suitably adjacent $m \times k$-mosaic $M^{m \times k}$ and $m \times 1$-mosaic $M^{m \times 1}$ which is the top bar mosaic. The adjacency rule gives a certain relation between the t-state of $M^{m \times k}$ and the b-state of $M^{m \times 1}$ as shown in Fig. 6; letters a and b are changed by b and a, respectively, from one state to the other. The key point is that, for some $r = 1, \dots , 4^m$, $s_t(M^{m \times k}) = \lambda ^m_r$ and $s_b(M^{m \times 1}) = \epsilon ^m_r$.

Fig. 6

Expanding $M^{m \times k}$ to $M^{m \times (k+1)}$

Let $Y_{m \times (k+1)} = (y_{ij})$, $Y_{m \times k} = (y_{ij}')$, and $Y_{m \times 1} = (y_{ij}'')$. Each entry $y_{ij}$ of $Y_{m \times (k+1)}$ is the state polynomial for the set of suitably adjacent $m \times (k + 1)$-mosaics M, which admit splittings into $M^{m \times k}$ and $M^{m \times 1}$ satisfying $s_b(M) = s_b(M^{m \times k}) = \epsilon ^m_i$, $s_t(M) = s_t(M^{m \times 1}) = \lambda ^m_j$, and for some $r = 1, \dots , 4^m$, $s_t(M^{m \times k}) = \lambda ^m_r$ and $s_b(M^{m \times 1}) = \epsilon ^m_r$. Obviously, all l- and r-states of them consist of letters b and d. Because all $4^m$ kinds of states occur on these m horizontal abutting edges,

$$\begin{aligned} y_{ij} = \sum ^{4^m}_{r=1} y_{ir}' \cdot y_{rj}''. \end{aligned}$$

This gives

$$\begin{aligned} Y_{m \times (k+1)} = Y_{m \times k} \cdot Y_{m \times 1} = (B_m + D_m)^{k+1}, \end{aligned}$$

and the induction step is finished. $\square$

3.3 Stage 3: State Matrix Analyzing

Finally, we are ready to prove Theorem 1 by analyzing the state matrix $Y_{m \times n}$.

Proof of Theorem 1

The (i, j)-entry of $Y_{m \times n}$ is the state polynomial for the set of suitably adjacent $m \times n$-mosaics M with $s_b(M) = \epsilon ^m_i$, $s_t(M) = \lambda ^m_j$ and r- and l-states consisting of letters b and d. According to the boundary state requirement, dominating sets in $G_{m \times n}$ are converted into suitably adjacent $m \times n$-mosaics M with b-, t-, r-, and l-states consisting of letters b and d only. Thus, the domination polynomial $P_{m \times n}(z)$ is the sum of all entries of $Y_{m \times n}$ associated to b- and t-states which are consisting of letters b and d.

Now, we define the matrices $L_m$ and $R_m$ as in the paragraph before Theorem 1. Obviously, $L_m \cdot Y_{m \times n}$ is the $1 \times 4^m$ matrix obtained from $Y_{m \times n}$ by nullifying each $i\hbox {th}$ row where the corresponding $i\hbox {th}$ state in the abcd-order uses at least one letter of a or c, followed by summing the rest non-zero entries columnwise. Again, $(L_m \cdot Y_{m \times n}) \cdot R_m$ is the $1 \times 1$ matrix obtained from $L_m \cdot Y_{m \times n}$ by nullifying each $j\hbox {th}$ column where the corresponding jth state in the bacd-order has at least one letter of a or c, followed by summing the rest non-zero entries in the unique row. Therefore, we get $P_{m \times n}(z)$ from the unique entry of $L_m \cdot Y_{m \times n} \cdot R_m$. This fact combined with Proposition 5 and Lemmas 6 and 7 completes the proof.

Note that the recurrence relation in Lemma 6 is easily translated to that of Theorem 1 by substituting $B_k + D_k$, $A_k + C_k$, and $B_k$ into $A_k$, $B_k$, and $C_k$, respectively. $\square$

4 Periodic Dominating Sets

In this section, we prove Theorem 2.

Proof of Theorem 2

For constructing the periodic dominating mosaic system in Stage 1 of the state matrix recursion method, we use the same mosaic tiles (drawn in Fig. 4) and $m \times n$-mosaics as the dominating mosaic system. But, the adjacency rule and the boundary state requirement should be changed to satisfy the periodic condition. We say that two mosaic tiles are $\partial _v$-adjacent if they lie on the vertical opposite ends in a column.

• (Adjacency rule) Abutting edges of adjacent mosaic tiles are labeled with any of the following pairs of letters: a|b, c|c, d|d.

• (Periodic adjacency rule) Boundary edges in $\partial _v$-adjacent mosaic tiles in each column are labeled with any of the following pairs of letters: a|b, c|c, d|d.

• (Boundary state requirement) All left and right boundary edges are labeled with letters b and d only.

A mosaic is said to be suitably adjacent if any pair of mosaic tiles sharing an edge satisfies the adjacency rule. A suitably adjacent $m \times n$-mosaic is called a p.dominating $m \times {\overline{n}}$-mosaic if it additionally satisfies the periodic adjacency rule and the boundary state requirement. Then we have the following one-to-one conversion as Proposition 5.

Proposition 8

There is a one-to-one correspondence between n-periodic dominating sets in $G_{m \times {\infty }}$ and p.dominating $m \times {\overline{n}}$-mosaics. Obviously, the number of vertices in a periodic dominating set in the standard periodic patch is equal to the number of mosaic tiles of Type 2 in the corresponding p.dominating $m \times {\overline{n}}$-mosaic.

In Stage 2, exactly the same argument as in the case of the dominating mosaic system is applied to produce the same state matrix $Y_{m \times n}$. Note that the periodic adjacency rule is not considered in this stage.

Finally, in Stage 3, we analyze this state matrix $Y_{m \times n}$. Recall that the (i, j)-entry of $Y_{m \times n}$ is the state polynomial for the set of $m \times n$-mosaics M with $s_b(M) = \epsilon ^m_i$, $s_t(M) = \lambda ^m_j$, satisfying the adjacency rule and the boundary state requirement (i.e., its r- and l-states consist of letters b and d). In the periodic case, we only consider the different rule applied to the top and the bottom boundaries of each mosaic M. According to the periodic adjacency rule, it must also have b-state $\epsilon ^m_i$ and t-state $\lambda ^m_i$ for the same ith order. The only mosaics counted in the diagonal entries of $Y_{m \times n}$ have this property. This fact combined with Proposition 8 gives

$$\begin{aligned} P^{\mathrm{p}}_{m \times {\overline{n}}}(z) = \mathrm{tr} (Y_{m \times n}), \end{aligned}$$

as we desired. $\square$

5 Cylindric Dominating Sets

In this section, we prove Theorem 3.

Proof of Theorem 3

We denote that two p.dominating mosaics M and $M'$ are equivalent if one can be obtained from the other by a cyclic rotation on the cylinder. For a p.dominating $m \times {\overline{n}}$-mosaic $M=(M_{i,j})$, [M] denotes an equivalence class of M, called a c.dominating $m \times {\overline{n}}$-mosaic. Then we have the following easy one-to-one conversion.

Proposition 9

There is a one-to-one correspondence between cylindric dominating sets in $G^{\mathrm{c}}_{m \times {\overline{n}}}$ and c.dominating $m \times {\overline{n}}$-mosaics. Obviously, the number of vertices in a cylindric dominating set is equal to the number of mosaic tiles of Type 2 in the corresponding c.dominating $m \times {\overline{n}}$-mosaic.

Let f be the quotient map from the set of p.dominating mosaics to the set of c.dominating mosaics defined by $f(M) = [M]$. We define $g_{t}(M)=(M'_{i,j})$ where $M'_{i,j} = M_{i,j-t}$ for all i, j, performing a cyclic rotation by t steps in up direction. We use the set of indices $\{ 1,2, \dots , n \}$ as a complete residue system modulo n. In the set of c.dominating mosaics, $[M] = [g_{t}(M)]$ for all t.

A p.dominating $m \times {\overline{n}}$-mosaic M is said to have a fundamental period p if p is the smallest positive integer such that $M = g_{p}(M)$. Then n must be divisible by p and $M = g_{ap}(M)$ for any integer a. Denote

$$\begin{aligned} S_{m \times p}(z) = \sum k'(d) z^d, \end{aligned}$$

where $k'(d)$ is the number of p.dominating $m \times {\overline{p}}$-mosaics with fundamental period p, which have d vertices.

Since the set of all p.dominating $m \times {\overline{n}}$-mosaics is a disjoint union of sets of all p.dominating $m \times {\overline{n}}$-mosaics with fundamental period p for all possible factors p of n, we have

$$\begin{aligned} P^{\mathrm{p}}_{m \times {\overline{n}}}(z) = \sum _{p | n} S_{m \times p}(z^{\frac{n}{p}}). \end{aligned}$$

The reason we use $S_{m \times p}(z^{\frac{n}{p}})$ instead of $S_{m \times p}(z)$ is that a p.dominating $m \times {\overline{n}}$-mosaic with fundamental period p is merely $\frac{n}{p}$ copies of a p.dominating $m \times {\overline{p}}$-mosaic with fundamental period p arrayed vertically. This guarantees that there is a bijection between p.dominating $m \times {\overline{n}}$-mosaics with fundamental period p and p.dominating $m \times {\overline{p}}$-mosaics with fundamental period p, and the formers have $\frac{n}{p}$ times many vertices than the letters.

We repeat this argument to p.dominating $m \times {\overline{p}}$-mosaics with fundamental period r with r|p to find $S_{m \times p}(z)$ recursively. Thus, we have a similar equation $P^{\mathrm{p}}_{m \times {\overline{p}}}(z) = \sum _{r | p} S_{m \times r}(z^{\frac{p}{r}})$, or

$$\begin{aligned} S_{m \times p}(z) = P^{\mathrm{p}}_{m \times {\overline{p}}}(z) - \!\!\! \sum _{r | p, \, r < p} \!\!\! S_{m \times r}(z^{\frac{p}{r}}). \end{aligned}$$

Now consider the set $f^{-1}([M])$ of the pre-image of [M] for a p.dominating $m \times {\overline{n}}$-mosaic M with fundamental period p. We will show that $f^{-1}([M])$ consists of exactly p totally different elements, i.e., all cyclic rotations $g_{a}(M)$’s for $a=0,1, \dots , p - 1$ are totally different. If not, $g_{a}(M) = g_{b}(M)$ for some integers $0 \le a < b \le p - 1$, and so $g_{b-a}(M) = g_{0}(M) = M$ for $0< b - a < p$, a contradiction. This gives

$$\begin{aligned} P^{\mathrm{c}}_{m \times {\overline{n}}}(z) = \sum _{p | n} \frac{1}{p} \, S_{m \times p}(z^{\frac{n}{p}}), \end{aligned}$$

as we desired. $\square$

6 Asymptotic Behaviors

The following two-dimensional variation of the Fekete’s lemma gives a simple proof of the existence and computational lower and upper bounds for three growth rates in Sect. 2.4.

Lemma 10

[16, Lemma 7] Let $\{ a_{m,n} \}_{m, \, n \in \, {\mathbb {N}}}$ be a double sequence with $a_{m,n} \ge 1$, and k be a nonnegative integer. If the sequence satisfies $a_{m_1,n} \cdot a_{m_2,n} \le a_{m_1 + m_2 + k,n}$ and $a_{m,n_1} \cdot a_{m,n_2} \le a_{m,n_1 + n_2 + k}$ for all indices, then

$$\begin{aligned} \lim _{m, \, n \rightarrow \infty } (a_{m,n})^{\frac{1}{mn}} = \sup _{m, \, n \ge 1} (a_{m,n})^{\frac{1}{(m+k)(n+k)}}, \end{aligned}$$

provided that the supremum exists. On the other hand, if the sequence satisfies $a_{m_1+m_2,n} \le a_{m_1+k,n} \cdot a_{m_2,n}$ and $a_{m,n_1+n_2} \le a_{m,n_1+k} \cdot a_{m,n_2}$, then

$$\begin{aligned} \lim _{m, \, n \rightarrow \infty } (a_{m,n})^{\frac{1}{mn}} = \inf _{m, \, n > k} (a_{m,n})^{\frac{1}{(m-k)(n-k)}}. \end{aligned}$$

Now, we prove Theorem 4.

Proof of Theorem 4

Obviously, $P_{m \times n}(1)$ is at least 1 for all m, n. First, we prove the existence of the limit of their ${\frac{1}{mn}}\hbox {th}$ powers. For any two dominating $m_1 \times n$- and $m_2 \times n$-mosaics (similarly for the other index), we can always create a new dominating $(m_1 + m_2) \times n$-mosaic by simply adjoining them. This induces the supermultiplicative inequalities on both indices. Since $\sup _{m, \, n} (P_{m \times n}(1))^{\frac{1}{mn}} \le 31$, which are the number of possible mosaic tiles at each site,

$$\begin{aligned} \kappa = \lim _{m, \, n \rightarrow \infty } (P_{m \times n}(1))^{\frac{1}{mn}} = \sup _{m, \, n \ge 1} (P_{m \times n}(1))^{\frac{1}{mn}} \end{aligned}$$

by applying the first part of Lemma 10.

On the other hand, we split a dominating $(m_1 + m_2) \times n$-mosaic $M^{(m_1+m_2) \times n}$ into two suitably adjacent $m_1 \times n$- and $m_2 \times n$-mosaics, which are not dominating any more. We attach instead an extra (lightly shaded) column to the $m_1 \times n$-mosaic on its right side as illustrated in Fig. 7. More precisely, each mosaic tile in the extra column has a dot inside, except the following case; if there is a pair of horizontally adjacent mosaic tiles lying on $m_1$th and $(m_1+1)$th columns in $M^{(m_1+m_2) \times n}$ as darkly shaded in the left figure (i.e., the left mosaic tile of the pair has a dot inside, but the right mosaic tile does not have), then the mosaic tile of the pair lying in $M^{m_2 \times n}$ has a dot inside instead of the mosaic tile in the extra column having the same row as the pair. The reader easily check that the new two mosaics $M^{(m_1+1) \times n}$ and $M^{m_2 \times n}$ are also dominating and they are injectively constructed. We repeat the same argument for the other index. Now we apply the second part of Lemma 10 to get

$$\begin{aligned} \lim _{m, \, n \rightarrow \infty } (P_{m \times n}(1))^{\frac{1}{mn}} = \inf _{m, \, n \ge 2} (P_{m \times n}(1))^{\frac{1}{(m-1)(n-1)}}. \end{aligned}$$

This proves the first part of the theorem.

Fig. 7

Splitting dominating mosaics

A dominating mosaic can be uniquely considered as a p.dominating mosaic of the same size. Furthermore, we get a dominating $m \times (n + 1)$-mosaic from a p.dominating $m \times {\overline{n}}$-mosaic by simply adjoining a properly dotted extra row on the top like Figure 7 (figures with $90^{\circ }$ rotation). The same exceptional case here is applied. Therefore, we have

$$\begin{aligned} P_{m \times n}(1) \le P^{\mathrm{p}}_{m \times {\overline{n}}}(1) \le P_{m \times (n+1)}(1). \end{aligned}$$

Since at most n different p.dominating $m \times {\overline{n}}$-mosaics can be produced from a c.dominating $m \times {\overline{n}}$-mosaic by cyclic rotations, one easily deduces that

$$\begin{aligned} \frac{1}{n} \, P^{\mathrm{p}}_{m \times {\overline{n}}}(1) \le P^{\mathrm{c}}_{m \times {\overline{n}}}(1) \le P^{\mathrm{p}}_{m \times {\overline{n}}}(1). \end{aligned}$$

From the above two inequalities, we conclude that periodic dominating sets and cylindric dominating sets have the same growth rate $\kappa$. $\square$