There really IS a polyhedral d1

[jack]

In retrospect, it's obvious. That is the best d1 (one-sided die).

ETA: We went to a little difficulty defining a d1, but it was unspokenly assumed you couldn't build one without cheating a little, eg. ingeniously using a sphere... But this works.

It satisfies all the requirements for a dice. It's solid, uniform, all the faces and edges are straight. It's equally fair rolled on any surface provided with random initial orientations. And if you paint a "1" on the only stable side, it always stops on it!

That's better than we managed for d7! :)

Comments

[cartesiandaemon.livejournal.com]

Yes, I was wondering something like that.

It doesn't even matter if it lands, just so long as its not stable on any face -- but I'm fairly sure you can't construct a geometrically astable polyhedron. A sphere or frictionless-elastic-collision-thing is the nearest.

the traditional cat with buttered toast strapped to its back :-)

That works.

DM: OK, roll for sanity failure. Fetch your favourite dice and-- Thomas, what's that?
Thomas: My d0.
DM: Your d0?
Thomas: I don't want to roll a sanity failure.
DM: Well, ok. On your head be it.
Thomas: OK. Ow! Ow! Ow!
D0: Yowl!
Thomas: Agh!
D0: Yowl! Miaow! Yowl! Hiss!
Thomas: Ready...?
D0: Splat.

[simont]

Indeed, you can't construct a convex polyhedron with no stable face: consider expanding a sphere gradually out from the centre of mass until it touches a face for the first time. It must be tangent to that face, and hence if the solid rests on that face then the CoG is directly above the point of tangency and the face is stable.

If the polyhedron isn't convex, then that proof doesn't work because the first face encountered by the expanding sphere needn't be tangent to it. And indeed, non-convex polyhedra obviously do exist which can't come to rest on any of their faces; any of the standard stellations of the dodecahedron or icosahedron is a good example. In that situation, replace the polyhedron by its convex hull (equivalent for stability purposes), and you find a stable orientation if not a literal face.

Another way to look at it is to observe that gravity will attempt to get the die's centre of mass as low as possible, so a face can only be unstable if there's some other orientation available with the CoM lower. So a polyhedron can only be astable if there's no face which places the CoM lowest – i.e. there must be an infinite sequence of faces each of which makes the CoM lower than the previous one. (You probably could construct such a shape, come to think of it; but (a) it wouldn't be a polyhedron in the strict sense any more, and (b) Bolzano-Weierstrass would dictate that there was some subsequence of those faces which converged to a limiting orientation, which would then necessarily be stable.)

[simont]

(Oops. Replace the polyhedron by its convex hull but with non-uniform density if necessary to put the CoM where it previously was, I meant. Of course the expanding-sphere proof works equally well if you've arranged for the CoM to be somewhere odd; just expand your sphere out from wherever it happens to be.)

[cartesiandaemon.livejournal.com]

That sounds right. Is there any chance we could call a stellated polyhedron a d0? It doesn't land on a face... :)