I fiddled around getting the calculations right. The method was right, but the actual consists are rather hard to do in the head. I used pen and paper, several calculations, to the same result.

Any triangle or quadralateral, can, by rotation on its edges, tile the euclidean plane. The symmetry is 2 2 2 2 in both cases: that is, the tiling is of the nature of snubs and wrap-snubs. The edges of of the triangle or quadralateral can be of any measure that forms such a figure in the euclidean plane.

Most of the uniform polyhedra have vertex-figures of this nature, and so there is a tiling of these uniform polyhedra, with ideal vertices (horoteelons) that have the 2 2 2 2 symmetry. The ones with five or six polygons need to be addressed separately.

The following table shows the 20 spherous (compact), and 11 horous (paracompact) uniform polytopes.

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` 1 C 4,3,6 9 ID 17 tCO 24 SH `

2 Pp (u) 10 rCO 18 tID 25 rSH

3 Ap (u) 11 rID 19 sC (*) 26 tQ

4 T 3,3,6 12 tT 20 sD (*) 27 tH

5 O 3,4,4 13 tO --------------- 28 tSH

6 I (*) 14 tC 21 Q 4,4,4 29 sQ (*)

7 D 6,3,5 15 tI 22 S 3,6,3 # 30 sH (#)

8 CO 16 tD 23 H 6,3,6 31 LPC1 (*)

p,q,r This construction has a wythoff construction, as given.

(u) An infinite class, for p>5, and P3, A4.

(*) The vertex figure does not tile the euclidean plane.

S The vertex figure tiles, but leads to no apeirohedra

sH The vertex-figure tiles with vertices of two kinds.

Most of these are laminate: they have one or more classes of apeirohedra (planes made entirely of cell-faces), suitable for making laminatopes. All of the examples have reflexes in the ordinary uniform tilings, so we should use these to explain etc.

An apeirohedron is made of hedra laid without a perimeter: that is, an unbounded cover of polygons. While one might consider these in isolation, they occur in higher structures too. For example, all of the faces of a cubic fall into apeirohedra {4,4}. The oct-tet tiling has apeirotopes {3,6}. There is a tiling of triangular prisms, which has squares falling in {4,4} and the triangles in {3,6}. The tiling of hexagonal prisms has apeirohedra of the form {6,3}, but the squares form no such apeirohedron.

S = {3,6,3} forms something like the tiling of tO, which lets no apeirohedra, even though the tiling exists.

sH = s3s6s, forms something like x3x3o3o3z, a tiling of tetrahedra and truncated tetrahedra. All faces fall on apeirohedra, but you can not select a set of parallels that contain all vertices. It does not form a uniform laminate.

In this count, we ignore the wythoffs, the non-functional ones, and count the rest.

(u) 2 ( ) 16 Together, 2u+16.

Vertex FiguresAll triangles tile the plane under the orbifold 2 2 2 2, such that there are through-lines that contain lines of one type (ie a or b or c).

The quadralaterals to form through-line tilings under 2 2 2 2 are parallelograms and trapezia. In the first case, it is as if were formed from two triangles, the second case alternates the two bases.

All of the quadralateral vertex-figures are of this type.

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` `

P 3 P 3 P P P

o------o--o------o--o-- o----o----o----

\ / \ / \ str = 3 or 4 Q | | xPoQx has

--o--o------o--o------o-- | | this form

/ \ / \ / o----o

o------o--o------o--o---- P

For triangles, the triangle can be placed on the apeirohedron so that they are assymetric. The example is the oct-tet tiling, whose continuation over the {3,6} is not a mirror image. This leads to the A and B laminates, ie LA2 and LB2.

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\ // \ // \ // \// \// \//

--o------o -----o---- --o------o------o--

// \ // \ //\ /\\ /\\ /\\

Normal (A form) Reversed (B form)

LA2 = {3,4,A} LB2.

The polytopes mentioned above are all 'A' form, there are crossing apeirogons formed by the three kinds of edge. The B form is like LB2, the tiling with the close-pack vertices.

Laminate PolytopesWe now turn to laminates. One can cut along a series of apeirotopes, such that every vertex is exposed. The apeirotopes would need to be identical. Given such a set, one can create alternating layers as long as the etching is identical. That is, one can make a tiling of alternating triangular-prisms and oct-tet layers, etc.

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\ // \ // \ // A \// \// \// A

--o------o -----o---- --o------o------o--

| | | P | | | P

--o------o -----o---- --o------o------o--

// \ // \ //\ A /\\ /\\ /\\ Ar

Normal (A form) Reversed (B form)

LPA2 LPB2.

Here there are alternating layers of a symmetric form P, with a reversable set A, Ar. Note here that the A's must be equal, and the P's must be equal.

The number of LaminatesThe following table shows all possible apeirohedra, along with the symmetric (f) and assymetric (h) tilings that support this kind of layer. There are five kinds of laminates, which have these numbers.

C LC2 f(f+1)/2 symmetric, excluding XX

A LA2 h(h-1)/2 normal assymetric, excluding XX

B LB2 h(h+1)/2 reversed assymetric, including XX'

PA LPA2 fh normal extended

PB LPB2 fh reversed extended.

The sum of the last four is h(h+2f).

We then list the various apeirohedra, which are p = xPoUo and P3 = o3xPoUo. These are lines variously of alternating 3,P and of simple P.

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3. P3 T O CO ID

tT tC tD SH tQ 45

4. C CO tO Q tQ 10

4: (Pp) tCO tID tSH 39

43. A4 rCO 1

5. P5 D ID tI 6

53. A5 rID 1

6. P6 H SH 3

6: tT tO tI tCO tID tSH 72

63. A6 rSH 1

8. P8 0

8: tC tCO tD 15

10. P10 0

10: tD tID 8

12. P12 0

12: tH tSH 8

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67 142 = 209

The above table gives the 'sundries'. Specifically, 39 is found with f=5, h=3, excludes the infinite families. When we put f=5,h=4, one gets 56, meaning that there are 17 classes of infinite laminates. This includes the reversed layering of (Pp) / (Pp), which we shall remove from here, to get 16 infinite families ie 16u.

The form Pp can alternate with Pq, both normal and reversed, which leads to two bi-infinite classes. So the total of laminates is

Primitives 2u + 16

Laminates 2u^2 + 16u + 209

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Total 2u^2 + 18u + 225 or 245 separate entries.