rjnoe@ihlts.UUCP (Roger Noe) (09/12/84)
Everyone's familiar with the ordinary six-sided die, the regular cube.
Many are familiar with dice in the shape of the other four regular polyhedra,
the tetrahedron (4 triangular faces), octahedron (8 triangular faces),
dodecahedron (12 pentagonal faces) and the icosahedron (20 triangular faces).
If we assume all these are fair dice (i.e. absolutely constant density of
the solid material which makes up the dies and perfectly sharp edges), the
notion of symmetry dictates that each face of one of these dice is equally
likely to turn up (or down) as is any of the other faces of the same die.
But what about fair polyhedra which are not quite regular? I'm not up on
all of the terminology in this area, so I would appreciate others clarifying
some of this for me. Here's what I'm aiming at:
1. Each of the faces of these "semi-regular" polyhedra is a regular
polygon but not all the faces are the same. For instance, we
could have a polyhedron with squares and equilateral triangles
making up the faces.
2. The polyhdedra are all convex, so that the relationship
faces + vertices = edges + 2
applies.
3. Each vertex is equivalent to each of the other vertices in terms
of the number of edges incident and the pattern of faces around
it (allowing for rotation).
Symmetry again would indicate that all the faces of one type on the polyhedron
would have equal probabilities of ending on bottom if the die is rolled. I am
unsure of how to go about finding this probability. It would seem to have
something to do with the area of the face and the dihedral angles about that
face. I would expect it to be independent of the speed at which the die is
rolled. Anyone have any information on this topic or any ideas how to go about
analytically determining the probabilities of rolling one face or another?
--
"It's only by NOT taking the human race seriously that I retain what
fragments of my once considerable mental powers I still possess."
Roger Noe ihnp4!ihlts!rjnoehawk@oliven.UUCP (Rick) (09/18/84)
>Anyone know of any others?
Check your friendly neighborhood soccer ball. It is composed of a handfull of
pentagons surrounded by hexagons.
rick
--
[hplabs|zehntel|fortune|ios|tolerant|allegra|tymix]!oliveb!oliven!hawkriks@mako.UUCP (Rik Smoody) (09/18/84)
> I do not believe that it would be easy (or even possible) to > construct a regular polyhedral die with unequal, though regular faces. > However, fair dice can be made as long as all the FACES are identical > to each other, although they may not be regular polygons. The vertices > and edges need not be all identical. (I do not know if there are any further > requirements than this and a uniform material) An example of this is the 10 > ......... > > Mike Moroney > ..!decvax!decwrl!rhea!jon!moroney The requirements are more stringent than that, and here's a counter example of an unfair die, of uniform material, all faces equal. Start with an octohedron. Grab it by two opposing vertices and pull it apart and give it a half twist. Now connect each of the vertices along the rift to the two closest ones in the other half. I find it hard to draw slanty lines in text. This results in 8 new equilateral triangles connecting the old square bases of the pyramidal caps. Do this same process again. Glue those triangles onto one of the caps, and pull the other one away 'just far enough' to fill in with 8 more new triangles. (The two caps will now be aligned again, and connected by a lattice which has a square in the middle which is 45 degrees out of line. You can keep this process up for a while, if you are not already convinced that the likelihood that this die land on one of the original triangles is exceedingly small and best accomplished with glue! Rik Smoody Tektronix
lab@qubix.UUCP (Q-Bick) (09/19/84)
> >Anyone know of any others? > Check your friendly neighborhood soccer ball. It is composed of a handfull of > pentagons surrounded by hexagons. > rick @ oliven From a discussion I've had with Roger Noe: If the pentagons and hexagons were flat, the ball would favor the pentagons, because if on a pentagon face vs. a hexagon face: 1. The Center of Gravity is closer to the surface 2. The angular movement needed to upset the die is greater. When the ball becomes a sphere, these differences are removed. -- The Ice Floe of Larry Bickford {amd,decwrl,sun,idi,ittvax}!qubix!lab You can't settle the issue until you've settled how to settle the issue.
davidp@elecvax.OZ (David Martin Ward Powers) (09/20/84)
There is a dodecahedron with each of the twelve sides a rhombus - an equilateral diamond. This is the shape of the space occupied by a sphere in a regular shifting closepack, I believe. (The shape of packed bubbles?) This would be fine as a dice since it each face is identical and bears the same relationship to its neighbours. David Powers UNSW, Australia ..!decvax!mulga!davidp:elecvax
td@alice.UUCP (Tom Duff) (09/21/84)
In fact, there is a fairly large (infinite!) set of these fair polyhedral dice. For example, the duals of the Archemedian (semi-regular) polyhedra are all fair dice. The dual d(P) of a polyhedron P has a vertex at the center of each face of P, a face centered on to each vertex of P, and edges `perpendicular' to the edges of P (i.e. there is an edge joining two vertices of d(P) if the corresponding two faces of P share an edge.) It is not hard to prove that if P is Archemedian, then there is a rotation mapping d(P) onto itself which maps any face onto any other face. I.e., you can't tell one face from another by looking at it. Write numbers on the faces and you have a fair die. I believe that the rhombic dodecahedron mentioned in elecvax.328 is one of these Archemedian duals, although offhand I can't think which.
moroney@jon.DEC (09/21/84)
>>Anyone know of any others? >Check your friendly neighborhood soccer ball. It is composed of a handfull of >pentagons surrounded by hexagons. >rick No, I was looking for polyhedrons with all edges and FACES identical. I doubt very much a soccer ball would make a fair die, because of the different shapes and areas of the faces. Mike Moroney ..!decvax!decwrl!rhea!jon!moroney
moroney@jon.DEC (09/21/84)
>> However, fair dice can be made as long as all the FACES are identical >> to each other, although they may not be regular polygons. The vertices >> ......... >The requirements are more stringent than that, and here's a counter >example of an unfair die, of uniform material, all faces equal. >Start with an octohedron. Grab it by two opposing vertices >and pull it apart and give it a half twist. Now connect each >of the vertices along the rift to the two closest ones in the other >half. I find it hard to draw slanty lines in text. >This results in 8 new equilateral triangles connecting the old >square bases of the pyramidal caps. >Do this same process again. .......the likelihood that this die land on one of the original triangles is exceedingly small... You are right.. There is another important requirement - the angle between any 2 faces must be the same, or each face must have an identical environment around it to all the other faces. So far the only polyhedra which meet these that I know of are the 5 regular polyhedra, the "30" in role-playing games (Anyone know the official name for this?) and the 12-sided rhomboid I mentioned earlier. Mike Moroney ..!decvax!decwrl!rhea!jon!moroney
moroney@jon.DEC (09/22/84)
>In fact, there is a fairly large (infinite!) set of these fair polyhedral dice. >For example, the duals of the Archemedian (semi-regular) polyhedra are all >fair dice. The dual d(P) of a polyhedron P has a vertex at the center of >each face of P, a face centered on to each vertex of P, and edges `perpendicular' >to the edges of P (i.e. there is an edge joining two vertices of d(P) if the >corresponding two faces of P share an edge.) >It is not hard to prove that if P is Archemedian, then there is a rotation >mapping d(P) onto itself which maps any face onto any other face. I.e., you >can't tell one face from another by looking at it. Write numbers on the faces >and you have a fair die. Only one problem with that, if you take a dual of an Archemedian, what you get is another Archemedian. The dual of a cube is an octahedron, the dual of an octahedron is a cube, dodecahedron <--> icosahedron, and the dual of a tetrahedron is another tetrahedron. So you don't gain any more fair dice by taking duals of Archemedians. There is a permutation of polyhedra which is similar to taking duals I will call it the edge-dual. It is somewhat hard to visualize and explain, but here goes. The dual can be visualized by taking a polyhedron and filing down all the vertices evenly until you just reach the point where none of the original faces remain. Similarly, the edge dual can be formed by taking a polyhedron and filing down all the edges evenly until you just reach the point where none of the original faces remain. You will then have a figure which has one face for each of the original edges and one vertex for each of the original vertices and faces. When this is done on an Archemedian, you get only 2 new polyhedra. When you do this to a cube or an octahedron, you get the rhombic dodecahedron. When you do this to a (pentagonal) dodecahedron or an icosahedron you get the "30" I mentioned in a previous article. (probably called a rhombic {whatevertheprefixfor30is}ahedron) The edge dual of a tetrahedron is a cube standing on a corner. (Try visualizing that process!) >I believe that the rhombic dodecahedron mentioned in elecvax.328 is one of these >Archemedian duals, although offhand I can't think which. The edge-dual of a cube or octahedron. Mike Moroney ..!decvax!decwrl!rhea!jon!moroney
td@alice.UUCP (Tom Duff) (09/23/84)
I've been receiving letters telling me that the dual of an Archemedian polyhedron is another Archemedian polyhedron, and now I see the same blather in net.math. In reply: NO! The dual of a *regular* (Platonic, NOT Archemedian) polyhedron is a regular polyhedron. The dual of an Archemedian (SEMI-regular -- don't you people *read* things before running off at the mouth?) polyhedron generally doesn't have regular faces, although all it's faces are all identical. Archemdedian polyhedra have regular faces (not necessarily all with the same number of edges) and the same configuration of faces at each vertex (e.g. triangle-square-triangle-square.) They are characterized by having a symmetry group under which any vertex may be transformed into any other vertex. These are the Archemedian polyhedra: Vertex Name Configuration 3 3 3 Tetrahedron 4 4 4 Cube 3 3 3 3 Octahedron 5 5 5 Dodecahedron 3 3 3 3 3 Icosahedron 3 6 6 Truncated Tetrahedron 3 8 8 Truncated Cube 3 10 10 Truncated dodecahedron 3 4 3 4 Cuboctahedron 3 5 3 5 Icosidodecahedron 5 6 6 Truncated Icosahedron 4 6 6 Truncated Octahedron 4 6 8 Truncated Cuboctahedron 4 6 10 Truncated icosidodecahedron 3 4 4 4 Rhombicuboctahedron 3 4 5 4 Rhombicosidodecahedron (My favorite name) 4 3 3 3 3 Snub Cuboctahedron (My favorite shape) 5 3 3 3 3 Snub Icosidodecahedron 4 4 n N-gonal Prism, n>=3 3 3 3 n N-gonal Antiprism, n>=3 43333 and 53333 exist in left- and right-handed versions. Also, cube==square prism, and octahedron==triangular antiprism. As I stated in my previous message, the dual of any of these is a fair die, because the dual's symmetry group transforms any faces into any other face. Of course, there are other fair dice than these. For example, take two n-gonal pyramids and stick them back to back. Lo and behold, a fair die with 2n faces appears. I don't know a general construction for fair dice with odd numbers of faces, or even if they must exist for all 2n+1>=5.