it is necessary that all its distance ratios be square roots

of rational numbers. For a regular N-gon, if the side-length is 1,

then the distance between two vertices H hops apart along the polygon

is

sin(H*pi/N) * csc(pi/N)

by using the fact that circumradius=(1/2)*csc(pi/N).

So therefore, if for any two H values the ratio of the

(sin(H*pi/N) * csc(pi/N))^2

is irrational,

then no N-gon is embeddable on an integer lattice in any dimension.

When H=1 and 2 this formula simplifies and the criterion becomes

that

4*cos(pi/N)^2

be irrational. And this indeed is irrational for every N>0 besides 1,2,3,4,6.

J. H. Conway, C. Radin, L. Sadun:

On Angles Whose Squared Trigonometric Functions Are Rational,

Discrete & Computational Geometry 22 (1999) 321-332.

And sure enough, regular triangles and hexagons are embeddable in the 3D integer lattice

on planes x+y+z=constant, despite their non-embeddability in the plane,

and in the plane N=4 is embeddable, and on the line N=1 and 2.

So this completely settles the question for all N in all dimensions.

–Warren D. Smith

=Charlie Sitler

212 755 6666

]]>If black always names a busy beaver number he never named before,

then black wins. Otherwise white wins.

Obviously, black has a forced win but not if he has to

use computable strategies.

Also note: This game can be made to have finite length,

in the sense that Yedidia and Aaronson recently proved

that a turing machine with some exact number of states (I forget

their number, but it was about 8000; they constructed the machine

fully explicitly) cannot be proved in ZFC not to

halt; and it halts if and only if ZF is inconsistent. Which tells

us that BusyBeaver(n) is uncomputable for every n>8000.

Which means black can force a win in a finite length version of this game,

e.g. with 10000 turns — but only using uncomputable strategies. ]]>

Lewis makes a fundamental use of the singleton operator, and as we explain in the paper, to my way of thinking this makes his theory basically bi-interpretable with membership-based set theory. So it can be a perfectly fine foundation of mathematics. I view the analysis of our mereology paper in part as explaining why Lewis needed to add the singleton operator to have a satisfactory theory. If you don’t have it, then the theory is decidable. With it, the theory is basically equivalent to set theory.

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