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Enumerative combinatorics, graph theory, order theory, posets, matroids, designs and other discrete structures. It also includes algebraic, analytic and probabilistic combinatorics.
6
votes
0
answers
108
views
What is the largest subgraph of the Kneser graph which has a small chromatic number?
While trying to characterize constraint satisfaction problems which can be solved by the Linear Programming relaxation, I've run into a few perplexing puzzles related to the existence of certain order …
4
votes
Accepted
Creating mazes with colored tiles
If N is large enough, then the answer is no, since this maze can't be created:
._._._._._._.
._._| ._._. ._. |_.
._| ._. |_. ._. ._. |
| ._. ._. ._. ._._| |
| |_._. ._. ._. ._._|
|_._._._._._. …
7
votes
Accepted
Free median algebras and maximal linked systems
The answer is yes. The reason is rather simple: since the median operation is a special case of a majority operation, that is, an operation satisfying the identities
$\forall x,y,\ m(x,x,y) = m(x,y,x) …
2
votes
Identifying a subset with as few tests as possible
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 …
2
votes
A variation of longest paths in directed acyclic graph
It’s NP-hard to approximate this within any constant factor, by a reduction from 2-CSP (that it is hard to approximate 2-CSP follows from the PCP theorem and parallel repetition). Suppose I have an in …
1
vote
A variant of set cover problem reformulated
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 …
6
votes
A puzzle with some jumping frogs
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 $0$ …
4
votes
Algorithm to generate random commuting permutations
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 …
3
votes
What is the optimal size in the finite axiom of symmetry?
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 …
10
votes
2
answers
442
views
Iterated sumset inequalities in cancellative semigroups
This question is motivated by the following well-known theorems:
Thm (Plünnecke): If $A$ is a finite nonempty subset of an abelian group, then for every $n$ we have $|A^n| \le \frac{|AA|^n}{|A|^n}|A| …
6
votes
Accepted
Waring problem for binomial coefficients (generalization of Gauss' Eureka Theorem)
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 $k = 8$.
15
votes
Accepted
Placing numbers $1,2,\ldots,n^3$ in a cube so that numbers of any two adjacent unit subcube ...
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 …
4
votes
How many types of jigsaw puzzle pieces in n dimensions?
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 …
28
votes
Accepted
"Circular" domination in ${\mathbb R}^4$
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 …
4
votes
Accepted
Can anyone analyze this misere game?
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 …