Fair Dice and Dartboards

[lpetrich]

We are all familiar with cubical dice and we sometimes use coins as dice, and many players of tabletop role-playing games use dice with a variety of different shapes. For n faces, they call their dice d<n>. Thus, an ordinary cubical die is d6 and a coin is d2.

I will now derive what possible sorts of fair dice there are, a fair die being one whose faces act alike. Here are some names for such dice and similar polyhedra:

•  Isohedral figure - face-transitive (faces look alike)
•  Isotoxal figure - edge-transitive (edges look alike)
•  Isogonal figure - vertex-transitive (vertices look alike)

So I will be concerned with isohedral dice.

The arguments that I will be using I will generalize to find plane tilings. Continuing with gaming metaphors, these are isohedral dartboards, fair ones with lookalike faces.  List of convex uniform tilings has a list of them.

For a hyperbolic surface, one can also construct tilings,  Uniform tilings in hyperbolic plane. These ones have the remaining parameter values, including faces with infinite numbers of vertices and vertices with an infinite number of faces. The hyperbolic plane is usually displayed as a disk, with most of the plane scrunched up near the edge of the disk. So one puts one end of an infinite-vertex face at the edge and one also puts an infinite-face vertex there.


There is an interesting interrelationship that some polyhedra have: duality. The vertices, edges, and faces of some polyhedron map onto the faces, edges, and vertices of its dual. Among the Platonic solids, the cube and octahedron are duals of each other, and the dodecahedron and icosahedron are also duals of each other. The tetrahedron is self-dual.

 

[lpetrich]

To formalize some of this discussion, let us consider Euler's theorem about polyhedra: V - E + F = 2, for V vertices, E edges, and F faces. In general, V - E + F = X, where X is the "Euler characteristic" of the surface that they are on. X is a "topological invariant", meaning that it is independent of distortion.

More generally, for a surface (2D space), X = 2(1- g), where g is the "genus" of the surface. It is the number of holes in it, the number of cuts needed to give it spherical topology. A donut and a coffee cup each have one hole, meaning one cut needed to give them spherical topology. Thus giving them topologically equivalent shapes. Two spheres together have genus -1, since one has to undo a cut to give them spherical topology. Thus:

g = (sum of how many holes in each shape) - (how many shapes) + 1

For a complicated surface or set of surfaces, g can be very large, and in such cases, one can define a genus density, genus per unit volume. Genus density has been used a lot for summarizing the shapes of density contours of cosmology simulations.


This may not seem very useful for infinite surfaces, but one can get estimates for finite subsets of them. Consider a torus or doughnut shape. Its genus is 1, meaning that its X value is 0. One can cut a torus and flatten out its surface, giving a rectangle with periodic boundary conditions. If you try to depart from one edge, you return at the opposite edge.

One can infinitely tile such a rectangle, and as a result, X = 0 for a flat plane. It is 2 for spherical topology, and a sphere has positive curvature, and that suggests that it is less than 0 for a unit cell of a hyperbolic-plane tiling.

 

[bilby]

Interestingly, many d10 are actually dodecahedrons, designed to be so unfair that the odds of coming to rest on the two 'small' faces is effectively zero, and with the other ten faces equal in area and shape. Some are 'smoothed' so that they have rounded vertices, and have only ten flat faces.

For safety reasons, many d4 have the points 'clipped' so they, similarly, are in fact octohedral, despite being nominal tetrahedrons, but have ~0 probability of coming to rest on any of the four 'points', so their behaviour remains that of tetrahedra. Again, some are sold with rounded edges, and have only four flat faces.

 

[lpetrich]

Here is the simplest case: regular polyhedra. Their faces are all regular polygons. I will designate them with (number of vertices per face, number of faces per vertex).

I first consider triangles. Their vertex angle is 60d. Of these, I will first consider three triangles per vertex, (3,3). Their vertex angles add up to 180d, giving an angle deficit of 180d. This means that the triangles must be folded toward each other, and that gives us the regular tetrahedron.

Repeating with four triangles (3,4) gives an angle deficit of 120d, and thus the regular octahedron. With five triangles (3,5) gives 60d and the regular icosahedron. With six triangles (3,6), the angle deficit disappears and we get the triangular plane tiling. With seven triangles (3,7), we get an angle excess of 60d.

To see what is happening, we must flatten out at the vertices. That makes the triangle angles 120d, 90d, 72d, 60d, and 360/7d ~ 54.83d. The sum of the angles in those triangles becomes 360d, 270d, 216d, 180d, and 1080/7d ~ 154.29d. So the first three have spherical geometry, the fourth one planar geometry, and the fifth one hyperbolic geometry. So an angle deficit gives spherical geometry, and an angle excess hyperbolic geometry.

One can repeat this analysis for squares, pentagons, hexagons, etc., and one finds these angle deficits for (# verts, # faces):

(3,3) 180d - tetrahedron
(3,4) 120d - octahedron
(3,5) 60d - icosahedron
(3,6) 0d - triangular tiling
(3,7) -60d - (hyperbolic)
(4,3) 90d - cube V=8 E=12 F=6
(4,4) 0d - square tiling
(4,5) -90d - (hyperbolic)
(5,3) 36d - dodecahedron
(5,4) -72d - (hyperbolic)
(6,3) 0d - hexagonal tiling
(6,4) -120d - (hyperbolic)
(7,3) -25.71d - (hyperbolic)
All the others are hyperbolic.

To see why all but a small finite set are hyperbolic, let us consider the unit cell of the polyhedra, the smallest region that gives those polyhedra when repeated and reflected. That unit cell is a triangle with vertices (center of a face, center of an edge, vertex). For shape (n,r), each vertex is divided among 2r unit cells, each edge among 4 unit cells, and each face among 2n unit cells, giving
V = 1/(2r)
E = 1/4
F = 1/(2n)
and
X(u.c.) = (1/2)*(1/r + 1/n - 1/2)

The sum of the unit-cell triangle's angles is 180d/r + 90d + 180d/n = 180d*(1/r + 1/n + 1/2). Its excess over 180d is 180d*(1/r+1/n - 1/2) = 360*X. Thus,
X(u.c.) > 0 -- spherical geometry
X(u.c.) = 0 -- plane geometry
X(u.c) < 0 -- hyperbolic geometry

It is easy to show that only a small set of n and r values >= 3 will give X(u.c.) >= 0.

For spherical geometry, the total X must be 2, and the number of unit cells needed is 2/X(u.c.) or
4/(1/r + 1/n - 1/2)
with
{V, E, F} = 2/(1/r + 1/n - 1/2) * {1/r, 1/2, 1/n}

Thus, we find (V,E,F) for the five Platonic solids, the completely-regular dice:
(3,3) - (4,6,4) - tetrahedron
(3,4) - (6,12,8) - octahedron
(3,5) - (12,30,20) - icosahedron
(4,3) - (8,12,6) - cube
(5,3) - (20,30,12) - dodecahedron

We also have these plane tilings, the completely-regular dartboards:
(3,6) - triangular
(4,4) - square
(6,3) - hexagonal

Finally, I note that shape (r,n) is the dual of shape (n,r) in all cases: spherical (Platonic-solid), planar, and hyperbolic.

 

[bilby]

Here is the simplest case: regular polyhedra. Their faces are all regular polygons. I will designate them with (number of vertices per face, number of faces per vertex).

I first consider triangles. Their vertex angle is 60d. Of these, I will first consider three triangles per vertex, (3,3). Their vertex angles add up to 180d, giving an angle deficit of 180d. This means that the triangles must be folded toward each other, and that gives us the regular tetrahedron.

Repeating with four triangles (3,4) gives an angle deficit of 120d, and thus the regular octahedron. With five triangles (3,5) gives 60d and the regular icosahedron. With six triangles (3,6), the angle deficit disappears and we get the triangular plane tiling. With seven triangles (3,7), we get an angle excess of 60d.

To see what is happening, we must flatten out at the vertices. That makes the triangle angles 120d, 90d, 72d, 60d, and 360/7d ~ 54.83d. The sum of the angles in those triangles becomes 360d, 270d, 216d, 180d, and 1080/7d ~ 154.29d. So the first three have spherical geometry, the fourth one planar geometry, and the fifth one hyperbolic geometry. So an angle deficit gives spherical geometry, and an angle excess hyperbolic geometry.

One can repeat this analysis for squares, pentagons, hexagons, etc., and one finds these angle deficits for (# verts, # faces):

(3,3) 180d - tetrahedron
(3,4) 120d - octahedron
(3,5) 60d - icosahedron
(3,6) 0d - triangular tiling
(3,7) -60d - (hyperbolic)
(4,3) 90d - cube V=8 E=12 F=6
(4,4) 0d - square tiling
(4,5) -90d - (hyperbolic)
(5,3) 36d - dodecahedron
(5,4) -72d - (hyperbolic)
(6,3) 0d - hexagonal tiling
(6,4) -120d - (hyperbolic)
(7,3) -25.71d - (hyperbolic)
All the others are hyperbolic.

To see why all but a small finite set are hyperbolic, let us consider the unit cell of the polyhedra, the smallest region that gives those polyhedra when repeated and reflected. That unit cell is a triangle with vertices (center of a face, center of an edge, vertex). For shape (n,r), each vertex is divided among 2r unit cells, each edge among 4 unit cells, and each face among 2n unit cells, giving
V = 1/(2r)
E = 1/4
F = 1/(2n)
and
X(u.c.) = (1/2)*(1/r + 1/n - 1/2)

The sum of the unit-cell triangle's angles is 180d/r + 90d + 180d/n = 180d*(1/r + 1/n + 1/2). Its excess over 180d is 180d*(1/r+1/n - 1/2) = 360*X. Thus,
X(u.c.) > 0 -- spherical geometry
X(u.c.) = 0 -- plane geometry
X(u.c) < 0 -- hyperbolic geometry

It is easy to show that only a small set of n and r values >= 3 will give X(u.c.) >= 0.

For spherical geometry, the total X must be 2, and the number of unit cells needed is 2/X(u.c.) or
4/(1/r + 1/n - 1/2)
with
{V, E, F} = 2/(1/r + 1/n - 1/2) * {1/r, 1/2, 1/n}

Thus, we find (V,E,F) for the five Platonic solids, the completely-regular dice:
(3,3) - (4,6,4) - tetrahedron
(3,4) - (6,12,8) - octahedron
(3,5) - (12,30,20) - icosahedron
(4,3) - (8,12,6) - cube
(5,3) - (20,30,12) - dodecahedron

We also have these plane tilings, the completely-regular dartboards:
(3,6) - triangular
(4,4) - square
(6,3) - hexagonal

Finally, I note that shape (r,n) is the dual of shape (n,r) in all cases: spherical (Platonic-solid), planar, and hyperbolic.

The platonic solids won't have sex with you; Planar figures just lie there; And hyperbolae are out there but crazy.

Sorry, this should probably be in 'Only reply when you are drunk'.

 

[lpetrich]

Interestingly, many d10 are actually dodecahedrons, designed to be so unfair that the odds of coming to rest on the two 'small' faces is effectively zero, and with the other ten faces equal in area and shape. Some are 'smoothed' so that they have rounded vertices, and have only ten flat faces.

Thus making a shape called a 5-trapezohedron.

One might also use a d20 as a d10 by using only the ones digit, or a d10*d2 by using both digits, the d2 coming from the tens digits being either even or odd.

For safety reasons, many d4 have the points 'clipped' so they, similarly, are in fact octohedral, despite being nominal tetrahedrons, but have ~0 probability of coming to rest on any of the four 'points', so their behaviour remains that of tetrahedra. Again, some are sold with rounded edges, and have only four flat faces.

Like all the cubic dice that are commonly manufactured -- they have rounded edges and rounded vertices, making them something like truncated cuboctahedra.

 

[lpetrich]

Properties of Dice by Klaus Æ. Mogensen has a proof of all the kinds of isohedral, fair dice that there are, and I will extend that proof to isohedral, fair dartboards.

KAeM calls the number of faces at a vertex that vertex's rank, and different vertices may have different ranks.

For finding a shape's overall geometry, one must find the shape's Euler characteristic, and a convenient unit cell for isohedral shapes is their faces. Each face has n vertices, and vertex k has rank r(k). Counting sharing of vertices by neighboring faces, each face has n/2 edges and 1/r(k) of vertex k. This gives us the Euler characteristic per face:

Xf = (sum of 1/r over the vertices) - (n/2) + 1

Xf > 0 is the spherical case, and F = 2/Xf (dice)
Xf = 0 is the flat case (dartboards)
Xf < 0 is the hyperbolic case


Since each r >= 3, Xf has an upper limit of (1 - n/6), meaning that for polyhedra, n = 3, 4, and 5, and that n = 6 has only one non-hyperbolic case: all r's equal to 3. For n > 6, every shape is hyperbolic.

This gives us a solution: the hexagonal plane tiling, the honeycomb dartboard.

-

We use r >= 3 because r = 2 is a degenerate case. This can be seen from merging the first two terms of Xf:

Xf = (sum over vertices of (1/r - 1/2)) + 1

If r = 2, then it does not contribute. Any such vertex can be removed and its two edges merged to make one edge.

 

[lpetrich]

The rest of the non-hyperbolic solutions are for the faces being triangles, quadrangles, pentagons, and hexagons, with polyhedra only for the first three. They are semiregular, so while their faces look alike to within reflection, they may have non-lookalike vertices and edges. Since we have already done lookalike vertices, we must continue to non-lookalike ones, ones that come in different types.

KAeM addresses their representation in the faces by finding the possible type content of each face's vertices, or corners as he calls them.

He considers various symmetries, and I can apply group theory to this issue. The symmetries of a regular n-gon is the group Dih, and it has subgroups Cyc(m) and Dih(m), where m evenly divides n and can be 1 or n. Here are some trivial cases:
Dih, Cyc: all alike
Cyc(1), the identity group: all different
Here are the nontrivial cases for the various numbers of face vertices:

Triangle:
Dih(1): abb

Quadrangle:
Cyc(2): abab
Dih(2): aaaa, abab
Dih(1): aabb, abcb

Pentagon:
Dih(1): abccb

Hexagon (only non-hyperbolic one a flat-plane tiling):
Cyc(3): abcabc
Cyc(2): ababab
Dih(3): abbabb
Dih(2): aaaaaa
Dih(1): abcdcb, aabccb

All the others only appear in hyperbolic tilings, and I won't try to analyze those. Since we want polyhedra, we look at only triangles, quadrangles, and pentagons, and in them, non-lookalike corners only appear as singlets and doublets.

KAeM continues with some relationships between vertices and corners.

The first one is that any vertex that contains only one kind of corner has these constraints on that corner's neighbors. They have the same type, the vertex's rank is even, or both. The second constraint is for when the two neighbors have different types. Then the vertex has neighboring vertices whose corners alternate between those two types as one goes around.

The second one is that when a vertex contains more than one type of corner, the types occur in proportion to their multiplicity in the face, doublets twice as often as singlets.

These will be enough to determine the semregular isohedral polyhedra and flat-plane tilings.

 

[lpetrich]

Looking in triangles, we first consider the doublet-singlet case, aab. If the doublets and singlets are in different types of vertices, then the doublets must have even rank, since each doublet corner has the other one and the singlet as neighbors. The ranks of each triangle's vertices are
2s, 2s, r

and the triangles form pyramids with the bases being r-gons of doublet corners and the peaks being singlet corners. Let us see what overall geometry one gets by calculating Xf:
Xf = 1/s + 1/r - 1/2

This is twice the X(u.c.) that we found for regular-polygon shapes, and the solutions that we found there carry over into this case. We find regular polyhedra and tilings that have pyramids on them, and an additional case. That is for s = 2, a case that yields a degenerate regular polyhedron: two polygons back to back and sharing all their vertices and edges. This gives a bipyramid, two pyramids with their bases matching each other.

The regular octahedron is a 4-bipyramid, and (2r)-bipyramids can be deformed while still being isohedral. They can have their base vertices alternately inward and outward or else alternately in opposite directions along the peak-to-peak axis.

So we have these non-hyperbolic cases:
(4,4,r) r-bipyramid - V=(r,2)=r+2, E=(r,2r)=3r, F=2r
(6,6,3) pyramid tetrahedron - V=(4,4)=8, E=(6,12)=18, F=12
(6,6,4) pyramid cube - V=(8,6)=14, E=(12,24)=36, F=24
(6,6,5) pyramid dodecahedron - V=(20,12)=32, E=(30,60)=90, F=60
(6,6,6) pyramid hexagonal tiling - triangular tiling
(8,8,3) pyramid octahedron - V=(6,8)=14, E=(12,24)=36, F=24
(8,8,4) pyramid square tiling
(10,10,3) pyramid icosahedron - V=(12,20)=32, E=(30,60)=90, F=60
(12,12,3) pyramid triangular tiling

The remaining case is all corners having different types. That makes all vertices have even rank. The largest value that the smallest rank can have without making a hyperbolic shape is 6, and the only non-hyperbolic case is (6,6,6) -- the triangular tiling. So we set one of the ranks to 4 and let the others vary: (4,2r,2s).

To find the overall shape, we find Xf again:
Xf = (1/2) * (1/r + 1/s - 1/2)

Our old friend the regular-polyhedron X equation. So we get super pyramids on the faces of the regular polyhedra and tilings. From their peaks, edges extend to their polygons' vertices and edge centers.

(4,4,2r) r-bi-super-pyramid - (2r)-bipyramid - V=(r,r,2)=2(r+1), E=(2r,4r)=6r, F=4r
(4,6,6) super-pyramid tetrahedron - pyramid cube - V=(4,6,4)=14, E=(12,24)=36, F=24
(4,6,8) super-pyramid cube / octahedron - V=(6,12,8)=26, E=(24,48)=72, F=48
(4,6,10) super-pyramid dodecahedron / icosahedron - V=(12,30,20)=62, E=(60,120)=120, F=120
(4,6,12) super-pyramid triangular/hexagonal tiling
(4,8,8) super-pyramid square tiling - pyramid square tiling


So we have one infinite family of polyhedra, seven additional polyhedra, and three flat-plane tilings.

 

[steve_bank]

Imagine a grate with rectangular openings the width of a needle plus a little more/ For a random toss what is the probability of the needle falling through the grate? I do not remember the solution.

If surfaces are not flat then you would seem to have a continuous probability distribution instead of discrete.

Way back in the early 80s I sat up one night tossing dice and flipping coins to prove it to myself. Also pulling smaples of colored balls from a box.

 

[lpetrich]

Turning to quadrangles, we find that the largest minimum rank that gives a non-hyperbolic shape is 4, and that the only non-hyperbolic shape with that minimum rank has all ranks equal to 4: (4,4,4,4), the square tiling.

So to get a polyhedron, at least one rank must be 3. This causes problems for symmetry aabb. For a vertex containing a and b, it must contain them in equal numbers, giving it an even rank. For a vertex containing only a or only be, that corner will have neighbors a and b, giving it an even rank. Thus, rank 3 is not possible. This leaves symmetries abab, abcb, and all different.


Turning to abab, that symmetry indicates rhombus faces. For corner ranks (r,s,r,s), we find
Xf = 2*(1/r + 1/s - 1/2)

which contains our old friend the regular-polyhedron X value. So one constructs a rhombic polyhedron by taking a regular one, and around each edge center, connecting the vertices and face centers. So we have:

(3,3,3,3) rhombi on tetrahedron - cube - V=(3,3)=6, E=12, F=6
(3,4,3,4) rhombi on cube / octahedron - rhombic dodecahedron - V=(6,8)=14, E=24, F=12
(3,5,3,5) rhombi on dodecahedron / icosahedron - rhombic triacontahedron - V=(12,20)=32, E=60, F=30
(3,6,3,6) rhombi on tri / hex tiling
(4,4,4,4) rhombi on square tiling - square tiling

The rhombic dodecahedron can be deformed with the 3-face vertices alternately inward and outward.


The next symmetry is abcb, kite symmetry. Its corner ranks are (r,t,s,t).

If the doublet and one of the singlets are part of the same vertex, then that vertex's rank must be a multiple of 3, giving us (r,3s,3s,3s). For s > 1, this is hyperbolic, so we look at (r,3,3,3). This one is a trapezohedron. Its faces are arranged much like the bipyramid's faces, but how they meet at the shape's equator is different. In each hemisphere, the second corner of each one is in the same vertex as the fourth corner of is neighbor in that hemisphere and the third corner of its neighbor in the opposite hemisphere. It has Xf = 1/r, giving this solution:

(r,3,3,3) r-trapezohedron - V=(2r,2)=2(r+1), E=(2r,2r)=4r, F=2r

The 3-trapezohedron is the cube. For all r, a trapezohedron can be deformed by moving the third corner of each face relative to the other two.

If the doublet and the singlet are in separate vertices, then the doublet's rank must be even. If it is 6, then the only non-hyperbolic solution is (3,6,3,6), the rhombus-on-tri/hex-grid solution. So we work with it being 4. For (r,4,s,4), its Xf value is
Xf = 1/r + 1/s - 1/2

the regular-polyhedron value again. So the kites are arranged as follows: the first vertex of each face is on a regular-polyhedron vertex, the third one on a neighboring face center, and the second and forth ones on the centers of the neighboring edges. The non-hyperbolic solutions are
(3,4,3,4) - kites on tetrahedron - rhombic dodecahedron - V=(4,6,4)=14, E=(12,12)=24, F=12
(3,4,4,4) - kites on cube / octahedron - V=(6,12,8)=26, E=(24,24)=48, F=24
(3,4,5,4) - kites on dodecahedron / icosahedron - V=(12,30,20)=62, E=(60,60)=120, F=60
(3,4,6,4) - kites on tri/hex grid
(4,4,4,4) - kites on square grid - square grid

The kites on cube/octa one can be deformed by rotating the 3-vertices, with neighbors being rotated in opposite directions and the regular-polyhedron-edge vertices being moved along those edges.

The all-vertices-different case gives us no new solutions.

So we have one infinite family of polyhedra, four additional ones, and two flat-plane tilings.

 

[lpetrich]

The remaining polyhedral-die face shapes are pentagons.

The first symmetry to consider is 2 doublets, 1 singlet: abccb. If b and c are in the same type of vertex, then their rank must be even, and thus at least 4. This case is always hyperbolic. If one doublet and the singlet, a, are part of the same vertex type, then that type has a rank that is a multiple of 3. If that rank is at least 6, then the shape is hyperbolic. So we consider that rank being 3.

The case abaab gives a dodecahedron, with all five vertices having rank 3, and the case aabba with ranks (3,3,4,4,3) is a flat-plane tiling that looks something like interleaved picket fences. The pentagons look like rectangles with triangles attached at one edge, and one row of them is made by attaching them on the sides next to the triangle. That row looks something like a picket fence. Its neighbors are upside-down ones, one attached to the bottom edges, and the other being shifted and attached to the top edges, the triangle-part edges.

When both doublets and the singlet are each part of different vertex types, the doublet ranks must be even and thus at least 4, and there are thus no non-hyperbolic solutions.

So we consider all singlets. If four of them are part of the same vertex type, then that type must have even rank, at least 4. No non-hyperbolic solutions. If three of them are part of one type and the rest of another type, then the first type has a multiple-of-3 rank and the second type even rank. The only non-hyperbolic solution is the picket-fence solution, (3,3,4,4,3). For three vertices with one corner type and the remaining two vertices with one corner type each, we get solutions (r,3,s,3,3) and (r,s,3,3,3).

The solution (r,s,3,3,3) has r and s both even, and the only non-hyperbolic solution is the picket-fence solution, (4,4,3,3,3).

The solution (r,3,s,3,3) has this Xf value:
Xf = 1/r + 1/s - 1/2

the regular-polyhedron value yet again. So the solutions are pentagons overlaid on regular polyhedra. The first vertex of each pentagon goes onto a regular-polyhedron vertex, the third one onto a neighboring face center, and the second, fourth, and fifth ones into additional vertices, each one containing a face's second point, another face's fourth point, and another face's fifth point.

Expanding out the above solution:
(3,3,3,3,3) pentagons on tetrahedron - dodecahedron - V=(4,12,4)=20, E=(12,12,6)=30, F=12
(4,3,3,3,3) pentagons on cube / octahedron - V=(6,24,8)=38, E=(24,24,12)=60, F=24
(5,3,3,3,3) pentagons on dodecahedron / icosahedron - V=(12,60,20)=92, E=(60,60,30)=150, F=60
(6,3,3,3,3) pentagons on tri / hex tiling
(4,3,4,3,3) pentagons on square tiling

The pentagons are not mirror imaged, and in general, a solution and its mirror image will differ.

So the pentagon case has two polyhedra and three flat-plane tilings.

 

[lpetrich]

Imagine a grate with rectangular openings the width of a needle plus a little more/ For a random toss what is the probability of the needle falling through the grate? I do not remember the solution.

This is related to  Buffon's needle (Buffon's Needle Problem -- from Wolfram MathWorld). It is Buffon-Laplace Needle Problem -- from Wolfram MathWorld (On Laplace's Extension of the Buffon Needle Problem | Mathematical Association of America), and a related one is Clean Tile Problem -- from Wolfram MathWorld.

 

[lpetrich]

I should finish with the rest of KAeM's discussion of pentagon-face isohedral dice.

After his listing of solutions, he continues with two vertex types, each with two corner types, and one vertex type with one corner type. The first two vertex types have even ranks, ranks at least 4. All hyperbolic. The next one has one vertex type with two corner types and three vertex types, each with one corner type. The first one and at least two others have even ranks, at least 4. All hyperbolic. Finally, all five corners having their own vertex types. All with even ranks, at least 4. All hyperbolic.

No more dice, and no more dartboards. KAeM then gives a table of what he derived, complete with computer-graphics pictures of them.


I have omitted discussions of hyperbolic tilings, in part because I don't know of any closed-form expressions that give all the vertex positions for any tiling, expressions comparable to what exist for spherical and flat-plane geometry. I looked for algorithms for making hyperbolic-plane graphics, and the only algorithm that I found was to create a tile and then add tiles to already-present tiles.

I myself have implemented such an algorithm. I worked in coordinates where the space was one sheet of a two-sheet hyperboloid: (x,y,z) where z² = x² + y² + 1 and z > 0. Much of spherical geometry could carry over with suitable modifications, like geodesics (generalized straight lines), distances, angles, and rotation matrices. I then projected the results onto a disk, with a choice of angle-preserving (conformal) or geodesic-preserving projections. Those were easy: {x,y}/(1+z) or {x,y}/z.

 

[lpetrich]

Returning to isohedral polyhedra and tilings, fair dice and dartboards, I will now list the solutions.

Polyhedra:

• Axial: 2 infinite families (semiregular)
• Quasi-Spherical: 18
  • Platonic: 5 (regular)
  • Catalan: 13 (semiregular)

d2 is a degenerate case: two polygons back to back. Coins are the best-known d2 dice.

(ranks of vertices) name - V=vertices, E=edges, F=faces

The two infinite families are r-bipyramids and r-trapezohedra. They are d(2r) for r >= 3, at least d6.
(4,4,r) r-bipyramid - V=(r,2)=r+2, E=(r,2r)=3r, F=2r
(3,3,3,r) r-trapezohedron - V=(2r,2)=2(r+1), E=(2r,2r)=4r, F=2r

The quasi-spherical ones are:
d4:
(3,3,3) Tetrahedron - V=4, E=6, F=4
d6:
(3,3,3,3) Cube - V=8, E=12, F=6
d8:
(4,4,4) Octahedron - V=6, E=12, F=8
d12:
(3,3,3,3,3) Dodecahedron - V=20, E=30, F=12
(3,6,6) Pyramid tetrahedron - V=(4,4)=8, E=(6,12)=18, F=12
(3,4,3,4) Rhombi on octahedron - V=(6,8)=14, E=24, F=12
d20:
(5,5,5) Icosahedron - V=12, E=30, F=20
d24:
(3,8,8) Pyramid octahedron - V=(6,8)=14, E=(12,24)=36, F=24
(4,6,6) Pyramid cube - V=(8,6)=14, E=(12,24)=36, F=24
(3,4,4,4) - Kites on octahedron - V=(6,12,8)=26, E=(24,24)=48, F=24
(3,3,3,3,4) Pentagons on octahedron - V=(6,24,8)=38, E=(24,24,12)=60, F=24
d30:
(3,5,3,5) Rhombi on icosahedron - V=(12,20)=32, E=60, F=30
d48:
(4,6,8) Super-pyramid octahedron - V=(6,12,8)=26, E=(24,48)=72, F=48
d60:
(3,10,10) Pyramid icosahedron - V=(12,20)=32, E=(30,60)=90, F=60
(5,6,6) Pyramid dodecahedron - V=(20,12)=32, E=(30,60)=90, F=60
(3,4,5,4) Kites on icosahedron - V=(12,30,20)=62, E=(60,60)=120, F=60
(3,3,3,3,5) Pentagons on icosahedron - V=(12,60,20)=92, E=(60,60,30)=150, F=60
d120:
(4,6,10) Super-pyramid icosahedron - V=(12,30,20)=62, E=(60,120)=120, F=120

Sizes: 4, 6, 8, 12, 20, 24, 30, 48, 60, 120

 

[lpetrich]

The isohedral (face-transitive) shapes have duals: the isogonal (vertex-transitive) ones. Lookalike faces become lookalike vertices, and the different types of vertices become different types of faces. Types of edges remain unchanged.

The Platonic solids are both isohedral and isogonal, and the duals of the Catalan solids (isohedral, quasi-spherical) are the Archimedean solids (isogonal, quasi-spherical).

Here are all the isogonal ones, matched to the isohedral ones.

Axial:
Bipyramid - Prism: polyhedral barrel with the sides being rectangular
Trapezohedron - Antiprism: polyhedral barrel with the sides being alternating-direction triangles

Quasi-Spherical (Catalan - Archimedean):
Pyramid tetrahedron - Truncated tetrahedron
Pyramid cube - Truncated octahedron
Pyramid octahedron - Truncated cube
Pyramid dodecahedron - Truncated icosahedron
Pyramid icosahedron - Truncated dodecahedron
Super-pyramid octahedron - Truncated cuboctahedron
Super-pyramid icosahedron - Truncated icosidodecahedron
Rhombi on octahedron - Cuboctrahedron (maximal truncated cube/octahedron)
Rhombi on icosahedron - Icosidodecahedron (maximal truncated dodecahedron/icosahedron)
Kites on octahedron - Rhombicuboctahedron (cube/octahedron with its edges cut off)
Kites on icosahedron - Rhombicosidodecahedron (dodecahedron/icosahedron with its edges cut off)
Pentagons on octahedron - Snub cube (snub octahedron) (cuboctahedron with edges replaced by back-to-back triangles)
Pentagons on icosahedron - Snub dodecahedron (snub icosahedron) (icosidodecahedron with edges replaced by back-to-back triangles)

The truncation here is cutting off the vertices.


The Archimedean solids are not very useful as dice, though prisms and antiprisms are sometimes used as dice. Prisms are the only semiregular-polyhedron dice that can have an odd number of faces.

 

[lpetrich]

Now for dartboards. Isohedral plane tilings also have isogonal duals.

The regular-polygon ones are both isohedral and isogonal.
(6,6,6) Triangle
(3,3,3,3,3,3) Hexagon
(4,4,4,4) Square
Triangle ~ Hexagon, Square is self-dual

Catalan (isohedral) - Archimedean (isogonal)
(4,8,8) pyramid square tiling - truncated square tiling
(3,3,4,3,4) pentagons on square tiling - snub square tiling (like snub polyhedra)
(3,3,3,4,4) picket fence - alternation of rectangle strips and alternating-triangle strips (split picket fence)
(3,12,12) pyramid triangular tiling - truncated hexagonal tiling
(4,6,12) super-pyramid triangular tiling - truncated trihexagonal tiling
(3,6,3,6) rhombi on triangular tiling - trihexagonal tiling
(3,4,6,4) kites on triangular tiling - rhombitrihexagonal tiling (trihexagonal with edges cut off)
(3,3,3,3,6) pentagons on triangular tiling - snub trihexagonal tiling (like snub polyhedra)

Truncation is cutting off vertices here


One can extend all the Catalan and Archimedean constructions to hyperbolic surfaces, as one can with regular-polygon tilings. Here are some renderings of such constructions: Hyperbolic Planar Tesselations and Semi-Regular Tilings of the Plane Part 3: General Theorems

 

[lpetrich]

Fair Dice has 3D-model renderings of all the various kinds of isohedral polyhedra. Isohedra has more renderings, including lots of deformations.

Now for some real dice. Cubical ones have been around for millennia, and icosahedral ones go back to the Roman Empire. I'm sure that coins have been used as dice for as long as they have been made, but I don't know how far back one can find literary references to flipping coins -- I don't know if anyone has researched that issue. It's hard for me to find anything on how far back other die shapes go.

DiceCollector.com - What shapes do dice have? and Alea Kybos' Dice Shapes both have numerous pictures of different kinds of dice.

All of the 18 kinds of quasi-spherical isohedral dice are represented, as are the lowest members of the two infinite families of axial isohedral dice and their duals.

• 3: 3-prism
• 4: 4-prism, tetrahedron (2-antiprism with endcaps degenerate)
• 5: 5-prism
• 6: 6-prism, 3-antiprism, 3-bipyramid, cube (3-trapezohedron, 4-prism with endcaps usable)
• 7: 7-prism
• 8: 8-prism, 4-antiprism, 4-trapezohedron, octahedron (4-bipyramid, 3-antiprism with endcaps usable)
• 9: 9-prism
• 10: 10-prism, 5-antiprism, 5-bipyramid, 5-trapezohedron
• 11: 11-prism
• 12: 12-prism, 6-antiprism, 6-bipyramid, 6-trapezohedron, pyramids on tetrahedron, rhombi on octahedron, dodecahedron
• 13: 13-prism
• 14: 14-prism, 7-bipyramid, 7-trapezohedron, cuboctachedron
• 15: 15-prism
• 16: 16-prism, 8-antiprism, 8-bipyramid
• 17: 17-prism
• 18: 18-prism, 9-trapezohedron
• 19: 19-prism
• 20: 20-prism, 10-antiprism, icosahedron
• 22: 11-trapezohedron
• 24: pyramids on cube, pyramids on octahedron, kites on octahedron, pentagons on icosahedron
• 26: 13-trapezohedron, rhombicuboctahedron
• 28: 14-bipyramid
• 30: rhombi on icosahedron
• 32: 16-bipyramid, truncated icosahedron (buckyball)
• 34: 17-trapezohedron
• 36: 18-bipyramid
• 48: super pyramids on octahedron
• 50: 25-trapezohedron
• 60: pyramids on dodecahedron, pyramids on icosahedron, kites on icosahedron, pentagons on icosahedron
• 120: super pyramids on icosahedron

(edits because some of the shapes and descriptions are difficult to interpret)

 

[Kharakov]

d2 is a degenerate case: two polygons back to back. Coins are the best-known d2 dice.

Degenerate meaning trivial? I don't get it.

Irregular polygons aren't going to make "fair" coins. In fact, assuming a bit of skill on the part of the flipper of polygons, one could learn any non-poly<100gons intrinsic flip-aparity. In fact, I'm thinking that any poly<=12 would be doable, given standard coin to human size ratio.


Some coins are infinigons, such as Pennies and Nickels. Some coins are reeded, such as Dimes which are 118gons and Quarters which are 119gons, or k9gons/dogons.

I'll flip a dogon to decide. Geddit? Flip a dog on to dee-side. Nogod down under.... because the Austra-aliens (pronounced astra-aliens) are from the land down under, a quantum supercomputer consisting of quark star matter, that exists at the center of the earth. Australia is an illusion! They are experimenting with different, more fragile forms of life, that exist at the edge of quantum possibility. Well, not really. Maybe. I don't know.

(ranks of vertices) name - V=vertices, E=edges, F=faces

( face configuration) name - V... ??

 

[lpetrich]

d2 is a degenerate case: two polygons back to back. Coins are the best-known d2 dice.

Degenerate meaning trivial? I don't get it.

Degenerate in the sense of being a flat surface.

Irregular polygons aren't going to make "fair" coins. ...

I don't see how boundary shape makes any difference. What does is the symmetry. If it is reflection symmetric, then it is fair.

(ranks of vertices) name - V=vertices, E=edges, F=faces

( face configuration) name - V... ??

Indeed it is:  Vertex configuration.

It mentions some flat-plane tilings that are the infinite limits of the four infinite families of semiregular polyhedra:

Isohedral:
(4,4,oo) - oo-bipyramid: a line with evenly-spaced perpendicular lines
(3,3,3,oo) - oo-trapezohedron - a zigzag line with perpendicular half-lines coming from each vertex in that vertex's direction
Isogonal:
(4,4,oo) - oo-prism: a strip of rectangles with infinite endcaps on each side
(3,3,3,oo) - oo-antiprism: a strip of alternating-direction triangles with infinite endcaps on each side

The polygonal coin is (2ⁿ: n 2's) for n vertices, and its dual is an orange-slice shape, with n lunes.