4-Dimensional Objects

[Roman]

How do you calculate the number of sides of a 4 dimensional object from the number of sides of a 3 dimensional object?

For example, a cube has 6 sides. A hypercube has 16 sides - not too difficult, because the hypercube is not that hard to 'visualize', but say a 20 sided object in 3-D is not so easy to visualize in 4-D.

I began thinking about it when I started thinking about D&D dice. Dice can apparently lead you to strange thoughts.

I am mostly looking for the above for dice, but other objects are interesting too, so how many sides would 4, 6, 8, 10, 12, 16 (not a die, but still seems interesting), 20, 30, 40, 50, 60, 80 and 100 sided 3-D objects have in 4-D? Is there any kind of formula?

From visualization (which may well be wrong), I think:

d4 would have 11 sides in 4-D
d6 would have 16 sides in 4-D

 

[-Warlord-]

11 sides on a d4? I'm going to ask my DM is I can roll for my Magic Missiles in 4-D

 

[hong]

Drink more of this. Soon, you will be visualising all sorts of things.

 

[Turanil]

Roman, you obviously have an IQ of 295, and lots of ranks in Knowledge [Superscience]. However, I would humbly point out that most people on these boards (including me) do not exceed 130 of IQ. As such, explaining a little bit about 4-D dice and hypercubes (plus what the use of this in rpg) would help, especially if some Enworlders are expected to answer your question...

 

[Kashkadir]

You could start by asking google about polychoron or polytope and take e.g. the links to wikipedia or mathworld.

 

[tarchon]

Hmm, AFAIK, a 6-sided 4d object has... 6 sides.

 

[Fieari]

I do believe he's referring to n-sided 3d objects being extruded into 4d (or possibly higher).

I think some of the difficulty arises in the method of "Extrusion" used. Consider a square, and a circle. Brought from 2d to 3d, the square becomes a cube, and generally you think that the cirlce becomes a sphere. Except that if you use the same extrusion method on the circle as the square, you'll get a cylinder instead. If you use the same extrusion method on the square as you used to get a circle into a sphere, you'll... also get a cylinder. In order to produce both a cube and a sphere, you need to use different methods.

So... what method are you using the extrude your hyperpyramid? Hyperdiamond? Hyperdecahedron? Hyperduodecahedron? Etc? The method used will alter the topography drastically, as seen in my 2d->3d example above. Unless you can define the translation, you can't define the number of sides.

 

[the Jester]

I am utterly fascinated by this thread. I wish I had something to contribute to it. :\

Edit: other than my appreciation.

 

[Nifft]

D&D Application of 4-D SmartyStuff(tm):

Q: How do you trap a 4D critter when you're just a lowly 3D critter?

A: Stick a sword through it. (Seriously -- you can't box it in, you have to "hook" it by piercing something through it, then you have to hold on hard.)

-- N

 

[Woocash]

Fieari is right. It is not easy to "convert" a polyhedron from 3d to 4d. There is no way to do so with most of platonic polyhedrons (there are no 4d objects of corresponding regularity).

The only dices we can make in 4d (or even more dimensions) without problems are d4 (regular simplex) and d6 (cube).

n-dimensional simplex has (n+1) sides, each of them being (n-1)-dimensional simplex (3-dim d4 has 4 sides being triangles, a triangle has 3 sides being lines; in higher dimensions it works the same).

n-dimensional cube has 2n sides ("top" and "bottom" in each dimension), each being (n-1)-dimensional cube (d6 has 6 sides being squares, a square has 4 sides being lines).



Edit: I just found complete formulas for a simplex and a cube in n dimensions.

n-dimensional simplex has (n+1)!/(k+1)!(n-k)! k-dimensional components (simplexes).
ie. 3-dim d4 has (3+1)!/(2+1)!(3-2)! = 24/6 = 4 2-dim (triangle) sides, (3+1)!/(1+1)!(3-1)! = 24/(2*2) = 6 1-dim (line) edges and (3+1)!/(0+1)!(3-0)! = 24/6 = 4 0-dim (point) vortices.

n-dimensional cube has (n!/k!(n-k)!)*2^(n-k) k-dimensional components (cubes).
ie. 3-dim d6 has (3!/2!(3-2)!)*2^(3-2) = (6/2)*2 = 6 2-dim (square) sides, (3!/1!(3-1)!)*2^(3-1) = (6/2)*4 = 12 1-dim (line) edges and (3!/0!(3-0)!)*2^(3-0) = (6/6)*8 = 8 0-dim (point) vortices.

Both above formulas are quite easy to prove, but I don't want to use too much mathematical language in this post .