# Polyominoes and Polyiamonds as Fundamental Domains of Isohedral Tilings with Rotational Symmetry

^{1}

^{2}

^{3}

^{*}

## Abstract

**:**

**p3**,

**p31m**,

**p4**,

**p4g**, and

**p6**. There are no isohedral tilings with

**p3m1**,

**p4m**, or

**p6m**symmetry groups that have polyominoes or polyiamonds as fundamental domains. We display the algorithms’ output and give enumeration tables for small values of n. This expands earlier works [1,2] and is a companion to [3].

**MSC**52C20, 05B50, 68U05

## 1. Introduction

**p3**,

**p4**, or

**p6**symmetry groups. In this article, we consider the expanded task of producing polyomino and polyiamond tiles that generate isohedral tilings of types

**p3**,

**p3m1**,

**p31m**,

**p4**,

**p4m**,

**p4g**,

**p6**or

**p6m**and for which the tiles are fundamental domains of the tiling. Recently an extension of this study was published as a separate paper [3] in which we carried out these investigations for symmetry groups of types

**pmm**,

**pmg**,

**pgg**, and

**cmm**.

**Lemma**

**1.**

**Theorem**

**1.**

**p3**,

**p31m**,

**p3m1**,

**p6**, or

**p6m**isohedral tilings by polyominoes. There are no

**p4**,

**p4g**, or

**p4m**isohedral tilings by polyiamonds.

**p3**,

**p31m**,

**p3m1**,

**p6**, or

**p6m**all have 3-fold centers, and by Lemma 1, if the tiles are fundamental domains, these centers must lie on the boundaries of the tiles. But if the tiles are polyominoes, a 120${}^{\circ}$ rotation about such a center cannot map a tile fully onto another tile (since unit squares will not be mapped to unit squares). Thus such a tiling is impossible. Similarly, a 4-fold center is impossible in an isohedral tiling by polyiamonds (since a 90${}^{\circ}$ rotation cannot map unit triangles to unit triangles), so there can be no such tilings of types

**p4**,

**p4g**, or

**p4m**.

**Theorem**

**2.**

## 2. p4

#### 2.1. Creating Polyominoes as Fundamental Domains for **p4** Symmetry Groups

**p4**group G will be built from these unit squares. By Lemma 1, the 4-fold rotation centers for G must be located on the boundary of the polyomino tiles that are fundamental domains for G. A 4-fold center can only occur at a lattice point that is a “corner” of the polyomino tile, that is, only one unit square of the polyomino contains that point.

**p4**lattice of rotation centers; rotations about the two placed black and white centers generate a

**p4**group G.

**p4**group G is given by [1,2]

**p4**group G. We denote the equivalence class of a unit square e as $C\left(e\right)$.

**p4**group G and a given n by following Procedure 1 below, using these definitions:

- T is a set of unit squares; $B\left(T\right)$ is a set of unit squares that are edge-adjacent to the squares in T; ${\mathcal{T}}_{n}$ is a set of n-ominoes.
- When T is the empty set, we define $B(\varnothing )$ as the set of four squares around the origin (the four unit squares at the lower left in Figure 1.)
- We define $E\left(T\right)$, the Boolean function of T, which is true if $\#T=n$ and a white circle is on the boundary of T. Otherwise $E\left(T\right)$ is false.
- ${B}^{\prime}\left(T\right)=\left\{e\right|e\in B\left(T\right),C\left(e\right)\ne C\left(f\right),\mathrm{for}\mathrm{all}f\in T\}$. This is the set of all unit squares that are edge-adjacent to those in T, but not equivalent to any unit squares in T.

**Procedure**

**1.**

- Fix n (from the list in (2) above). Begin with ${\mathcal{T}}_{n}$ empty.
- Make T empty. Make ${U}_{T}={B}^{\prime}\left(T\right)=B(\varnothing )$. Make $k=0$. Make ${S}_{k}=\left\{(T,{U}_{T})\right\}$.
- For $(T,{U}_{T})$ in ${S}_{k}$, if ${U}_{T}\ne \varnothing $, choose an element e (a unit square) of ${U}_{T}$. Remove e from ${U}_{T}$ and save the pair $(T,{U}_{T})$ in ${S}_{k}$.
- Increase k by 1. Add e (from step 3) to T to create a new $k+1$-omino T, and let ${U}_{T}={B}^{\prime}\left(T\right)$ for the new T. Add the new pair $(T,{U}_{T})$ to ${S}_{k}$.
- If $T=\varnothing $ and ${U}_{T}=\varnothing $, quit the procedure.
- If $E\left(T\right)$ is true, an n-omino tile is completed and add T to ${\mathcal{T}}_{n}$ provided there is no equivalent tile in ${\mathcal{T}}_{n}$. We regard two n-ominoes equivalent if conditions (a) and (b) below are satisfied.
- (a)
- The tiles are congruent (including mirror reflection).
- (b)
- When the tiles are superimposed, the positions of m-fold rotational centers on the boundaries of the tiles are the same. If there are several m-fold rotation centers for the same m, they can be permuted appropriately before comparison.

- If ${U}_{T}=\varnothing $, decrease k by 1.
- Go back to step 3.

**p4**tiling by using the black and white circles as 4-fold rotation centers. Figure 3 shows the corresponding isohedral tilings produced by n-ominoes in Figure 2 for $n\le 5$.

#### 2.2. Symmetries of Tiles

**p4**group G generated by 4-fold rotations about the black and white rotation centers. By Theorem 2, in every case where this occurs, the n-omino that generates the tiling must have reflection and/or rotation symmetry.

- Select a polyomino that has rotation and/or reflection symmetry, and examine its tiling $\mathcal{T}$ generated by G.
- Look at all vertices and centers of unit squares in a polyomino in $\mathcal{T}$ except for the original 4-fold centers we have chosen (black and white centers), and determine whether or not $\mathcal{T}$ is invariant under a 4-fold rotation about such a point. If so, then it is a new 4-fold center for $\mathcal{T}$, and this symmetry is not in G. For example, in Figure 3, tilings (5-2) and (5-2-2), while generated differently, are the same tiling and have 4-fold centers at every vertex where 4 tiles meet. Tiling (5-3) has 4-fold rotation centers at the centers of the “cross” tiles and at every vertex where 4 tiles meet. These tilings have full symmetry groups of type
**p4**, with G as a proper subgroup. - Other new symmetry elements of $\mathcal{T}$ can be sought by using Chart 2 in [6]. If the line joining the black and white 4-fold centers we placed is an axis of mirror reflection for $\mathcal{T}$, then $\mathcal{T}$ is type
**p4m**. For example, in Figure 3, tilings (1-1) and (4-3) are type**p4m**. If this line is not a mirror reflection axis, but the line connecting two adjacent (nearest) 2-fold centers for $\mathcal{T}$ is a mirror reflection axis for $\mathcal{T}$, then $\mathcal{T}$ is type**p4g**. For example, in Figure 3, tiling (2-1) is type**p4g**.

## 3. p4g

#### 3.1. Creating Polyominoes as Fundamental Domains for **p4g** Symmetry Groups

**p4g**symmetry group contains 4-fold rotations, reflections, and glide-reflections; the subgroup generated by its 4-fold rotations is type

**p4**. Thus we begin as in Section 2 with a lattice of unit squares and place a 4-fold rotation center (a black circle) at a lattice point and call this the origin. Orthogonal unit vectors $\mathbf{u}$ and $\mathbf{v}$ are then placed at the origin (see Figure 4(a)). Next, we place a reflection axis nearest to the origin; this must lie along the edges of unit squares according to Lemma 1, so we place it at $x\mathbf{u}$ or $x\mathbf{v}$, where x is a positive integer. The placement of the origin and choice of x determine the whole

**p4g**lattice of rotation centers, reflection axes and glide-reflection axes since the

**p4g**group G is generated by 4-fold rotations about the origin and reflections in the placed axis (see Figure 4(b)).

**p4g**tilings by using the origin as a 4-fold rotation center to fill out the square bounded by the reflection axes nearest the origin, then reflecting this square in its edges. Figure 6 shows the corresponding isohedral tilings produced by polyominoes in Figure 5.

#### 3.2. Symmetries of Tiles

**p4g**group G that generated them are tilings with square polyomino tiles. In fact, this is always true.

**Theorem**

**3.**

**p4g**group that generated it and T is a fundamental domain for the tiling, except in the case when the shape of T is square. In that case, the tiling has symmetry group

**p4m**and T is not a fundamental domain.

**p4g**group G acting on a polyomino T that is a fundamental domain for G, constructed as described in Section 3.1. If G is a proper subgroup of ${G}^{\prime}$, then by Theorem 2, there is a 2-fold or 4-fold rotation or reflection symmetry g in ${G}^{\prime}$ that is also a symmetry of T. By its construction, one corner of T is a 4-fold center for G at the origin; without loss of generality, we may assume that T contains the unit square shown in black in Figure 7.

## 4. **p4m**

**p4m**group are shown in Figure 8. Black circles are 4-fold centers, white circles are 2-fold centers, and solid lines are mirror reflection axes. The shaded region is a fundamental domain for the

**p4m**group. A polyomino that is a fundamental domain for the tiling must have its edges on the reflection axes, by Lemma 1. But this is clearly impossible.

**Theorem**

**4.**

**p4m**isohedral tilings having polyominoes as fundamental domains.

## 5. p3

#### 5.1. Creating Polyiamonds as Fundamental Domains for **p3** Symmetry Groups

**p3**isohedral tiling, we begin with a lattice of unit equilateral triangles. By Lemma 1, the 3-fold rotation centers for a

**p3**symmetry group must be located on the boundary of the tile, and in addition, these centers must be located at a lattice point that is a “corner” of the tile at which no more than two unit triangles meet. So first we place a 3-fold rotation center, a black circle, at a lattice point and call this the origin, then place vectors $\mathbf{u}$ and $\mathbf{v}$ at the origin along edges of a unit triangle. Next we place a second 3-fold rotation center, a white circle, at $x\mathbf{u}+y\mathbf{v}$, where x and y are nonnegative integers, not both zero. See Figure 9(a). These two choices of 3-fold rotation centers determine the whole

**p3**lattice of rotation centers; 3-fold rotations about the black and white centers generate the whole

**p3**symmetry group.

**p3**group G generated by the black and white 3-fold centers: a simple rhombic shape, and a polyiamond tile. Note that both fundamental domains contain the black and white 3-fold centers. The other two circles (shown in gray) that are on the boundaries of these fundamental domains are also 3-fold centers for G.

- T is a set of unit triangles; $B\left(T\right)$ is a set of unit triangles that are edge-adjacent to the triangles in T; ${\mathcal{T}}_{n}$ is a set of n-iamonds.
- When T is the empty set, we define $B(\varnothing )$ as the set of six triangles around the origin in Figure 9(a).
- We define $E\left(T\right)$, the Boolean function of T, which is true if $\#T=n$ and a white circle is on the boundary of T . Otherwise $E\left(T\right)$ is false.
- ${B}^{\prime}\left(T\right)=\left\{e\right|e\in B\left(T\right),C\left(e\right)\ne C\left(f\right),\mathrm{for}\mathrm{all}f\in T\}$. This is the set of all unit triangles that are edge-adjacent to those in T, but not equivalent to any unit triangles in T.

**p3**tiling by using the black and white circles as 3-fold rotation centers. Figure 11 shows the corresponding isohedral tilings produced by n-iamonds in Figure 10.

**p3**symmetry. However, here we analyze the symmetries of those tilings, and lay the groundwork for our investigation in Section 5.1 of tilings by polyiamonds that are fundamental domains for symmetry groups of types

**p31m**and

**p3m1**. In our older article [1], we constructed

**p3**isohedral tilings by polyiamonds that were built up from units that were ${60}^{\circ}$ rhombuses, and so the list in [1] of such

**p3**tilings is a proper subset of our present results.

#### 5.2. Symmetries of Tiles

**p3**group G generated by 3-fold rotations about the black and white rotation centers. By Theorem 2, when this occurs, the n-iamond that generates the tiling must have reflection and/or 3-fold rotation symmetry. The full symmetry group of the tiling could be of any of these types:

**p3**,

**p3m1**,

**p31m**,

**p6**, or

**p6m**. The tilings in Figure 11 that have additional symmetries, and the n-iamonds in Figure 10 that generate them are indicated by parentheses in their labels. We outline below how to identify these polyiamonds.

- Select a polyiamond T that has rotation and/or reflection symmetry, and examine its tiling $\mathcal{T}$ generated by G.
- Look at all vertices and centers of unit triangles in T except for those centers of rotation in G (black, white and gray centers), and determine whether or not any can be new 3-fold centers for $\mathcal{T}$. If a new 3-fold center is found, then the full symmetry group ${G}^{\prime}$ of $\mathcal{T}$ contains G as a proper subgroup and T is not a fundamental domain for $\mathcal{T}$. If a new 3-fold center is a 6-fold center, ${G}^{\prime}$ is type
**p6m**if there is a reflection axis joining it to any nearest 3-fold center; otherwise, ${G}^{\prime}$ is type**p6**. Tiling (6-4) in Figure 11 is type**p6m**, and has a 6-fold center at the center of the hexagonal 6-iamond. If there are new 3-fold centers but none are 6-fold centers, then ${G}^{\prime}$ is type**p3**. - If the tiling contains only 3-fold centers for G, we seek new symmetry elements by using Chart 2 in [6].
- If some 3-fold center for G is a 6-fold center, the tiling $\mathcal{T}$ will have
**p6m**symmetry if the line joining a 6-fold center to a nearest 3-fold center is a reflection axis for $\mathcal{T}$, otherwise it will have**p6**symmetry. In Figure 11, tilings (2-1) and (8-4) have**p6m**symmetry, and tiling (6-1) has**p6**symmetry. - Otherwise, if the line connecting a pair of adjacent 3-fold centers of the same kind (black-black, white-white, or gray-gray) is a reflection axis, $\mathcal{T}$ has
**p31m**symmetry. In Figure 11, tiling (6-2) has**p31m**symmetry. We will show in Section 7 that it is impossible for $\mathcal{T}$ to have**p3m1**symmetry.

## 6. **p31m**

#### 6.1. Creating Polyiamonds as Fundamental Domains for **p31m** Symmetry Groups

**p31m**symmetry group contains 3-fold rotations, reflections, and glide-reflections; the subgroup generated by its 3-fold rotations is type

**p3**. To build an n-iamond tile that is a fundamental domain for a

**p31m**isohedral tiling, we first place a 3-fold rotation center, a black circle, at a lattice point and call this the origin. Then we place vectors $\mathbf{u}$ and $\mathbf{v}$ at the origin along edges of a unit triangle (see Figure 12(a)).

**p31m**lattice of rotation centers, reflection axes and glide-reflection axes since the

**p31m**group G is generated by 3-fold rotations about the origin and reflections in one of the three reflection axes.

**p31m**tilings obtained by performing a 3-fold rotation about the origin (black circle) to fill out the triangle bounded by reflection axes (Figure 12(b)), then reflecting this triangle in its edges.

#### 6.2. Symmetries of Tiles

**Theorem**

**5.**

**p31m**group that generated it and T is a fundamental domain for the tiling.

**p31m**group G acting on a polyiamond T that is a fundamental domain for G, as described in Section 6.1. If G is a proper subgroup of ${G}^{\prime}$, then by Theorem 2, there is a reflection symmetry or 3-fold rotation symmetry in ${G}^{\prime}$ that is also a symmetry of T. T can be obtained from the shaded fundamental domain shown in Figure 12(b) by removing unit triangles and adding unit triangles equivalent to these (by a ${120}^{\circ}$ rotation about the origin), always keeping the new tile homeomorphic to a disc and bounded by reflection axes as in Figure 12(b). This process shows that one edge of T that lies on a reflection axis will have length at least $\lfloor (3x+1)/2\rfloor \left|\mathbf{u}\right|$, and at most one other edge of T (lying on an adjacent reflection axis of G) will have this same length (see Figure 13). From this, it follows that T cannot have 3-fold rotation symmetry.

## 7. p3m1

**p3m1**symmetry group is shown in Figure 15; black, white, and grey circles denote three inequivalent 3-fold rotation centers. Here, unlike the

**p31m**case, all 3-fold centers lie on reflection axes. The shaded region bounded by reflection axes is a fundamental domain for the

**p3m1**group that generates the tiling by reflections in those axes. By Lemma 1, this is the only polyiamond tile possible having the area of a fundamental domain. But the full symmetry group of this tiling is type

**p6m**, which has a fundamental domain with area 1/6 that of the shaded tile. Thus,

**Theorem**

**6.**

**p3m1**isohedral tilings having polyiamonds as fundamental domains.

**p3m1**isohedral tiling.

## 8. p6

#### 8.1. Creating Polyiamonds as Fundamental Domains for **p6** Symmetry Groups

**p6**symmetry group G contains 6-fold, 3-fold, and 2-fold rotations; the subgroup generated by its 3-fold rotations is type

**p3**. Thus we begin as in Section 5.1, with a lattice of equilateral triangles, to build n-iamond tiles that are fundamental domains for G. By Lemma 1, 6-fold and 3-fold centers for G are located on the boundaries of tiles, and we have seen that 3-fold centers can only occur at lattice points that are “corners” of the tiles where at most two unit triangles meet. Clearly 6-fold centers can only occur at lattice points that are “corners” where just one unit triangle in the tile meets that point.

**p6**lattice of rotation centers; rotations about the two chosen black and white centers generate G.

**p6**group is given by [2]

**p6**group classifies all unit triangles into n equivalence classes and we denote the equivalence class of a unit triangle e as $C\left(e\right)$. We construct a set ${\mathcal{T}}_{n}$ of n-iamonds that are fundamental domains for the given

**p6**group as in Section 5.1, with n chosen from the list in (10). Figure 17 shows the set of inequivalent n-iamonds in ${\mathcal{T}}_{n}$ for $n\le 9$. From each of these n-iamonds we obtain the associated

**p6**tiling by using the black circles as 6-fold rotation centers and white circles as 3-fold rotation centers. Figure 18 shows the corresponding isohedral tilings produced by n-iamonds in Figure 17.

**p6**symmetry. However, in our next Section 8.2 we analyze the symmetries of those tilings. In our older article [1], we presented a list of polyiamond tiles for

**p6**isohedral tilings constructed from ${60}^{\circ}$ rhombic units. Every tile in that list can be obtained by performing a half-turn about the 2-fold center on the boundary of a tile in our list here in Figure 17.

#### 8.2. Symmetries of Tiles

**p6**group G that generated them. These tilings and their polyiamond tiles in Figure 17 are indicated by parentheses in their labels. We can identify these polyiamonds as follows.

- Select a polyiamond that has rotation and/or reflection symmetry, and examine its tiling $\mathcal{T}$ generated by G.
- If the line connecting adjacent 6-fold and 3-fold centers (black to white) is a reflection axis for $\mathcal{T}$, the tiling has
**p6m**symmetry. Tilings (1-1) and (4-1) in Figure 18 have**p6m**symmetry. These tilings also have 3-fold centers for the tiling at the centers of the polyiamond tiles. - Otherwise, look at all vertices and centers of unit triangles in a polyiamond in $\mathcal{T}$ except for those black and white rotation centers in the
**p6**group we have generated and determine whether or not they can be new 3-fold centers for $\mathcal{T}$. If new 3-fold centers are found, the full symmetry group of the tiling contains G as a proper subgroup. Tiling (7-1) in Figure 18 has a larger**p6**symmetry group than G since there are 3-fold rotation centers for $\mathcal{T}$ at the centers of the rotor-like polyiamond tiles. Note that the same tiles appear in tiling 7-1-2, but in this tiling, the centers of the tiles are not 3-fold centers of rotation for $\mathcal{T}$.

## 9. **p6m**

**p6m**group. Black, white, and gray circles are the 6-, 3-, and 2-fold centers, respectively. Solid lines are reflection axes. The shaded ${30}^{\circ}$- ${60}^{\circ}$- ${90}^{\circ}$-triangle is a fundamental domain. A polyiamond that is a fundamental domain for the tiling must have its edges on the reflection axes, by Lemma 1. But this is clearly impossible. Thus,

**Theorem**

**7.**

**p6m**isohedral tilings having polyiamonds as fundamental domains.

## 10. Enumeration Tables

**p3**,

**p31m**,

**p3m1**,

**p4**,

**p4m**,

**p4g**,

**p6**, or

**p6m**that generate an isohedral tiling having the n-omino or n-iamond tiles as fundamental domain. Tiles are equivalent only if they generate the same tiling by the action of the group G when each tile is marked with an asymmetric motif. For example, the tiles 5-2 and 5-2-2 in Figure 3 are congruent and their corresponding tilings are the same, but the placement of their 4-fold centers is different, and so if the tiles are marked with an asymmetric motif, they generate different isohedral tilings.

**p6**and $n=7$, there are 20 tiles in Figure 17, with corresponding isohedral tilings in Figure 18, so ${N}_{7}=20$. Since tiling 7-1 of Figure 18 has parentheses around its label, ${S}_{7}=19$. From Figure 17, we can see that four pairs of tiles are congruent: 7-1 and 7-1-2; 7-3 and 7-3-2; 7-5 and 7-5-2; 7-12 and 7-12-2. Thus ${N}_{7}^{\prime}=20-4=16$. Among the 19 tiles counted for ${S}_{7}$, there are also 16 non-congruent tiles, so ${S}_{7}^{\prime}=16$.

## 11. Summary

**p3**,

**p31m**,

**p4**,

**p4g**, and

**p6**. We have shown that there are no isohedral tilings with symmetry groups of types

**p3m1**,

**p4m**, or

**p6m**that have polyominoes or polyiamonds as fundamental domains. For symmetry groups of types

**p3**,

**p31m**,

**p4**,

**p4g**, and

**p6**we used the backtracking Procedure 1 to obtain a set ${\mathcal{T}}_{n}$ of n-omino or n-iamond tiles where each tile produced one isohedral tiling, generated by a given symmetry group G of one of these five types. We can denote ${\mathcal{T}}_{n}\left(G\right)$ as the set ${\mathcal{T}}_{n}$ for that symmetry group G and ${\mathcal{T}}_{n}^{*}\left(G\right)$ the corresponding set of isohedral tilings.

**p3m1**symmetry group. In Table 1, Table 2 and Table 3 of Section 10, we used the notation ${N}_{n}=\#{\mathcal{T}}_{n}^{*}\left(G\right)$ and ${S}_{n}=\#{\mathcal{S}}_{n}^{*}\left(G\right)$.

## Acknowledgement

## References

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**Figure 1.**A lattice of unit squares with black and white 4-fold rotation centers that generate a

**p4**group G. At left, a simple triangular fundamental domain (shaded) for G; at right, a polyomino fundamental domain (shaded) for G. Here $x=1$, $y=3$; the area of each fundamental domain is 5 square units.

**Figure 2.**List of n-ominoes produced by the procedure in Section 2.1 for $n\le 8$ [2]. The tiles are fundamental domains for the

**p4**group used to construct them. The labels indicate n followed by the tile number for that n. Parentheses indicate that the tiles produce tilings having more symmetries than the

**p4**group that generates the tilings.

**Figure 4.**(

**a**) A lattice of unit squares with a 4-fold rotation center (black circle, the origin) and reflection axis (thick line) at $x\mathbf{u}$ that generate a

**p4g**symmetry group G. The shaded triangle is a fundamental domain for G; (

**b**) The lattice of all symmetry elements for G: 4-fold centers are black circles, 2-fold centers are white circles, reflection axes are thick solid lines, glide-reflection axes are dotted lines. The shaded square region is a polyomino fundamental domain for G. Here $x=2$; the area of each fundamental domain for G is 4.

**Figure 5.**List of n-ominoes produced as described in Section 3.1 for $n\le 9$; these are fundamental domains for the

**p4g**group used to construct them. The labels indicate n followed by the tile number for that n. Parentheses indicate that the tiles produce tilings having more symmetries than the

**p4g**group that generates the tilings.

**Figure 7.**(

**a**)–(

**c**) The derivation of polyomino fundamental domains T for a

**p4g**group G by removing unit squares from and adding equivalent unit squares to the original $x\times x$ square polyomino fundamental domain in (

**a**). Here $x=2$; (

**d**) The polyomino T in (

**b**) has reflection symmetry with mirror axis M, but this is not a symmetry of the tiling $\mathcal{T}$ having T as fundamental domain, since ${T}^{\prime}$, the image of T by a ${90}^{\circ}$ rotation about the origin, is mapped by a reflection in M to a tile that crosses a reflection axis for G.

**Figure 8.**The rotation and reflection symmetry elements of a

**p4m**group G. Black circles are 4-fold centers, white circles are 2-fold centers, and lines are reflection axes. (Glide-reflection axes are not shown.) The shaded area is a fundamental domain for G.

**Figure 9.**(

**a**) A lattice of unit triangles with two 3-fold rotation centers that generate a

**p3**group G; (

**b**) A simple rhombic fundamental domain (shaded) for G; (

**c**) A polyiamond fundamental domain (shaded) for G. Here $x=2$, $y=1$; the area of each fundamental domain is 14 triangular units.

**Figure 10.**List of n-iamonds produced as described in Section 5.1 for $n\le 8$ [2]. The tiles are fundamental domains for the

**p3**group used to construct them. The labels indicate n followed by the tile number for that n. Parentheses indicate that the tiles produce tilings having more symmetries than the

**p3**group that generates the tilings.

**Figure 12.**(

**a**) A lattice of unit triangles with a 3-fold rotation center at the origin (black circle) and reflection axis (solid line); together these generate a

**p31m**group G; (

**b**) Repeated 3-fold rotations about the origin and reflections about the axis in (

**a**) generate the whole lattice of symmetry elements for G, which includes reflection axes (solid heavy lines), 3-fold rotation centers (black and white circles), and glide-reflection axes (dashed lines). A shaded triangular fundamental domain for G is shown in (

**a**), and a shaded trapezoidal polyiamond fundamental domain for G is shown in (

**b**). Here $x=2$; the area of each fundamental domain is 12 triangular units.

**Figure 13.**List of n-iamonds produced as described in Section 6.1 for $n\le 12$. The tiles are fundamental domains for the

**p31m**group used to construct them. The labels indicate n followed by the tile number for that n.

**Figure 15.**The lattice of reflection axes and 3-fold rotation centers of a

**p3m1**group (glide-reflection axes are not shown). Three inequivalent 3-fold centers are black, grey, and white circles. Lines are reflection axes. The shaded region is a fundamental domain for the

**p3m1**group generated by reflections in the axes surrounding the region.

**Figure 16.**(

**a**) A lattice of unit triangles with a black 6-fold rotation center and white 3-fold rotation center that generate a

**p6**group G; (

**b**) A triangular fundamental domain (shaded) for G; (

**c**) A polyiamond fundamental domain (shaded) for G. Here $x=2$, $y=1$; the area of each fundamental domain is 7 triangular units.

**Figure 17.**List of n-iamonds produced as described in Section 8.1 for $n\le 9$ [2]. The tiles are fundamental domains for the

**p6**group used to construct them. The labels indicate n followed by the tile number for that n. Parentheses indicate that the tiles produce tilings having more symmetries than the

**p6**group that generates the tilings.

**Figure 19.**The lattice of rotation and reflection symmetry elements for a

**p6m**group G. Solid lines are reflection axes, black, white, and gray circles are 6-, 3-, and 2-fold centers respectively. (Glide-reflection axes are not shown.) The shaded region is a fundamental domain for G.

**Table 1.**The number of isohedral tilings of types

**p3**,

**p31m**and

**p3m1**having n-ominoes or n-iamonds as fundamental domains.

**Table 2.**The number of isohedral tilings of types

**p4**,

**p4g**and

**p4m**having n-ominoes or n-iamonds as fundamental domains.

**Table 3.**The number of isohedral tilings of types

**p6**and

**p6m**having n-ominoes or n-iamonds as fundamental domains.

© 2011 by the authors; licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution license (http://creativecommons.org/licenses/by/3.0/.)

## Share and Cite

**MDPI and ACS Style**

Fukuda, H.; Kanomata, C.; Mutoh, N.; Nakamura, G.; Schattschneider, D.
Polyominoes and Polyiamonds as Fundamental Domains of Isohedral Tilings with Rotational Symmetry. *Symmetry* **2011**, *3*, 828-851.
https://doi.org/10.3390/sym3040828

**AMA Style**

Fukuda H, Kanomata C, Mutoh N, Nakamura G, Schattschneider D.
Polyominoes and Polyiamonds as Fundamental Domains of Isohedral Tilings with Rotational Symmetry. *Symmetry*. 2011; 3(4):828-851.
https://doi.org/10.3390/sym3040828

**Chicago/Turabian Style**

Fukuda, Hiroshi, Chiaki Kanomata, Nobuaki Mutoh, Gisaku Nakamura, and Doris Schattschneider.
2011. "Polyominoes and Polyiamonds as Fundamental Domains of Isohedral Tilings with Rotational Symmetry" *Symmetry* 3, no. 4: 828-851.
https://doi.org/10.3390/sym3040828