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This seems really easy? Let $A \subseteq X^2$ be the set of all $(x,y)$ such that $y \in A_x$, and let $A^r$ be the reflection of this across the main diagonal. You are asking whether $A\cup A^r = X^2 …

There is no such set $S$. Suppose for a contradiction that there was. By rescaling the coordinates, we can assume all coefficients of points in $S$ are positive integers. Now construct a set $S'$ as f …

If N is large enough, then the answer is no, since this maze can't be created: ._._._._._._. ._._| ._._. ._. |_. ._| ._. |_. ._. ._. | | ._. ._. ._. ._._| | | |_._. ._. ._. ._._| |_._._._._._. …

Not an answer, but there is an algebraic reformulation of the problem which is equivalent when $n$ is a prime. First, we assume that at the end of the process all the frogs wish to end up on leaf $0$ …

The paper Maximizing a Monotone Submodular Function subject to a Matroid Constraint gives a $(1-1/e)$-approximation algorithm for a generalized version of your second question. Even in the case where …

The orientation-preserving symmetry group $G_n$ of the $n$-dimensional cube is an index two subgroup of the full symmetry group, which is $S_n\times\{\pm 1\}^n$. By the Polya-Burnside counting theorem …

Here is a proof that for any fixed dimension $d$, there is a computable $n_0(d)$ such that for all $n\ge n_0(d)$ we can place the numbers $1, ..., n^d$ in a $d$-dimensional cube of side length $n$ suc …

It's not so hard to calculate the probability that no two points are adjacent: We may as well assume that the first vertex is chosen for us. So let's ignore it and unroll the rest of the polygon into …

I proved over here that statement ii) holds when we make the stronger assumption that $|N(u) \cap N(v)|$ is exactly $1$ for every $u, v$. Most of the argument probably does not generalize, but at lea …

Watson's nice paper "Sums of eight values of a cubic polynomial" (http://jlms.oxfordjournals.org/content/s1-27/2/217.full.pdf) shows that we may take $k = 8$.

Problem Let $* = \{0\}$ be the one matchstick nim game, let $*2 = \{0,*\}$ be the two matchstick nim game, let $*3 = \{0,*,*2\} = *2+*$ be the three matchstick nim game, let $g = \{0, *2+*3, *2+*2+*2 …

I recently managed to convert this problem into a cellular automata, and the answer to the second question appears to be no, making this question uninteresting. However, I think some people might appr …

Here's a method for sampling two commuting elements of a group $G$, given that you can sample a random element of $G$ and that you can also sample a random conjugacy class of $G$ (and assuming as well …

For the first amoeba you mentioned, I think your equations should be $e^{-kx}\pm e^{−ky}=\pm e^{-k}$, not $e^{-kx}\pm e^{−ky}=\pm e^{0}$. For the main question, I think you should be using an equatio …

This isn't intended to be a complete answer, just a formalization of the most "obvious" idea. A natural approach is to try to maximize the information gained with each test. If we are currently at the …