bio | website | |
---|---|---|
location | DE | |
age | ||
visits | member for | 2 years, 5 months |
seen | Feb 16 at 21:56 | |
stats | profile views | 321 |
Jan 20 |
awarded | Yearling |
Dec 11 |
comment |
“Circular” domination in ${\mathbb R}^4$
Why is the first reduction from $S$ to $S'$ true? For example, when $(5,5,5,1),(1,1,1,5)\in S$, then there must exist a vector that dominates $(5,5,5,5)$. But $(5,5,5,5)$ cannot be built as the elementwise maximum from two of the twelve "two-coordinate" versions $(5,5,0,0),(5,0,5,0),\ldots,(0,0,1,5)$. |
Nov 26 |
comment |
A problem on chains of squares — can one find an easy combinatorial proof?
A) What is the well-known solution that uses Brouwer's fixed-point theorem? B) The discrete problem has definitely easy direct proofs, just by specializing the polygonal Jordan curve theorem. C) If the two paths don't cross, one can extend them to a crossing-free drawing of the (non-planar) complete bipartite graph $K_{3,3}$ by adding two new vertices and some curves outside the square. But I guess this does not count, as non-planarity of $K_{3,3}$ is based on Euler's polyhedral formula, and that in turn depends on the Jordan curve theorem, it seems. |
Nov 26 |
comment |
Powers of orthogonal matrices is closed
@Sebastian, What do you mean, "any real matrix", for which matrix? Was my argument not short enough? |
Nov 26 |
revised |
Simplifying triangulations of 3-manifolds
fixed grammar and spelling |
Nov 26 |
answered | Powers of orthogonal matrices is closed |
Nov 26 |
suggested | approved edit on Simplifying triangulations of 3-manifolds |
Jan 20 |
awarded | Yearling |
Oct 22 |
comment |
Tiling a rectangle with weighted cells (min-max problem)
What precisely do you mean by "sparse"? A constant number of entries in each row and column? At most $m+n$ entries in total? Or why would the total number of entries not appear in the runtime? (And edit the problem rather than posting an answer) (And the upper bound $a,b<1$ seems to be irrelevant.) |
Oct 11 |
awarded | Necromancer |
Oct 11 |
answered | Tiling A Rectangle With A Hint of Magic |
Oct 10 |
revised |
How many vertices/edges/faces at most for a convex polyhedron that tiles space?
added the original reference of the figure |
Oct 10 |
suggested | approved edit on How many vertices/edges/faces at most for a convex polyhedron that tiles space? |
Oct 8 |
revised |
The Cayley Menger Theorem and integer matrices with row sum 2
small clarification |
Oct 8 |
comment |
Vertices of a Polytope
Indeed, fullerenes are a good example. Almost all of the surface is a hexagonal lattice, and there you can simultaneously cut away vertices as long as their minimum distance is 2. Thus, you can cut away half of the vertices (almost; staying away from the pentagons). |
Oct 7 |
answered | Vertices of a Polytope |
Sep 18 |
comment |
Is there a 3d equivalent of this picture?
1. It should be pointed out that the density does not increase indefinitely inside the "black spot" but there is some minimum distance (if the points were really generated as indicated in the text) 2. Would you tolerate, for example (in the planar example) occasional vertices with 5 or 7 incident triangles, or should it be "completely regular"? |
Jun 25 |
awarded | Revival |
Jun 25 |
awarded | Excavator |
Apr 28 |
comment |
What is the expected value for this
The exponent would not change. For example, take random points in a circular disk. Fit a triangle in the disk. A constant fraction of the n points falls inside the triangle (w.h.p.), and thus you get at least $c′n^{1/3}$ points in convex position; similarly, an ellipse (affine image of a disk) fits into any triangle, showing the other direction of the inequality. Any two convex shapes are related like this (after affine transformation). |