On Fri, Jul 23, 2010 at 9:07 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Bill T. wrote:
<< Interleavings of two translate to words reperesemting simple curves on a punctured torus. I think each one is also valid when extended in this case. The lex-min cyclic permutations are associated <--> their slope. Tjere are O(n^2) slopes admitting O(n) cyclic permutations.
-- is there a good logical analysis of the 3D simplification base = slope for interleavings of 3? You can think of in terms of the cynical subdivision, as seen from the origin.
I foresee a whole branch of ipod-spell-correction–inspired mathematics. Cynical permutations, cynical field extensions,...
That seems a little jaundiced to me. But I was naively thinking of the torus also, perhaps in a different way:
Given three real numbers a,b,c > 0, linear independent of each other and arbitrary reals a_0, ab_0, c_0, consider the line in R^3 [thought of as tiled by unit cubes]
F(t) = (at + a_0, bt + b_0, ct + c_0))
Then F(t) enters successive unit cubes via faces perpendicular to the x_1, x_2, or x_3 axis in a specific order, which yields a sequence of 1's, 2's and 3's.
Then: Which sequences of 1's, 2's, 3's are possible from any ordered triple of a,b,c as above?
Or in general, which sequences of 1's through n's are possible ?
(Or is this equivalent to the original question?)
This is definitely the same thing as the original question, except that Dan is using "linearly independent" to cover Allen's requirement that the arithmetic progressions be non-intersecting -- what we really need in both cases is the requirement that ordering be well-defined. In 2-d, I think I understand why it's OK to assume your line passes through the origin, as long as you consider the sequence of lines-crossed up to cyclic permutation. As in Dan's model, you get all of Allen's sequences if you take arbitrary slopes and arbitrary starting positions (in the unit square). For sequences of length n, you only look at a part of the line that crosses n coordinate-axis-parallel lines. Now if you restrict your attention to the family of lines that pass through those same n lines in that same order, then the codimension-1 boundary must consist of lines that fail to be in the set because they touch an integer point, and the codimension-2 part are lines that pass through two points, one at each end of the line segment. Call one of these the origin, then the other is an integer point whose coordinates sum to n. You still need to justify uniqueness, and there's some checking to do around words that are powers of shorter words, but I think it all comes out in the wash. But in 3-d it's trickier. The events that we're recording in order involve our arbitrary line passing through unit squares parallel to one of the coordinate planes. The "wiggling the lines" argument, though, only seems to guarantee that your boundary cases touch a parallel-to-the-coordinate-axis line, not that you can get a boundary case to touch a vertex. Anyone with clarity on the 3-d picture, please enlighten us by (doing a better job than I just did of) explaining it. --Michael -- Forewarned is worth an octopus in the bush.