co.combinatorics - Replacing logician-constructive with combinatorist-constructive?

Logicians interpret the word "constructive" in a very well-defined way: they take it to mean, more or less, "computability". Taking constructivity seriously and working in a world where everything must be constructive, leads to intuitionistic logic, which has been a very productive and fascinating subfield of logic.On the other hand, combinatorists use "constructive" in a different sense. They use it to mean "better than brute force". For example, Ramsey's theorem is non-constructive from the POV of a combinatorist, since its proof offers no me...Read more

co.combinatorics - A puzzle about finding three points $(x,y)$, $(x,z)$ and $(y,z)$ in a subset of a square.

I was asked (by myself) to give a proof of the following seemingly simple geometric statement, but after thinking a little I now suspect it could be less elementary than I thought (or am I being silly?). Does anybody know it, and can give an answer or a reference to it? Of course, I'm quite sure it should fit within a larger theory in combinatorics or in probability, but an elementary answer would be appreciated. Let $S$ be a (say open) subset of a square $[0,1]^2$ with Lebesgue measure $|S|>1/2$. Then, there exists a rectangle with a ver...Read more

co.combinatorics - Counting discrete functions

For $1\leq k \leq n+1$, consider the set $S_k$ of functions $f:\{1,\ldots,n\} \rightarrow \{1,\ldots,n+1\}$, with the property that $|f^{-1}\{1,\ldots,k\}| < k$. Note that $|S_1|=n^n$, and $|S_{n+1}| = (n+1)^n$. Is it true that $|S_k|$ grows with $k$? It's not hard to come up with the formula$$ |S_k| = \sum_{j=0}^{k-1} {n \choose j} k^j (n+1-k)^{n-j}$$and if you plot this it is at least apparent that the conjecture should be true.Another vague observation is that, if you divide by $(n+1)^n$ and take limit as $n\rightarrow \infty$, keeping $k...Read more

co.combinatorics - How to deconstruct a sum of intersecting upsets

A set system $\mathcal{U}\subset P([n])$ isan upset if $B\supset A \in \mathcal{U}$ implies $B\in \mathcal{U}$,intersecting if $A,B\in\mathcal{U}$ implies $A\cap B \ne \emptyset$.Note that a nonnegative linear combination of (indicator functions of) upsets is just an increasing function $f:P([n])\to[0,\infty)$, and it is a straightforward task to decompose any such $f$ back into its constituent upsets: just remove an appropriate multiple of $\mathcal{U}_1 = \lbrace A: f(A)>0 \rbrace$ and repeat.However, given a nonnegative linear combination...Read more

co.combinatorics - faces in the discrete cube

This arose from a question Gil Kalai asked about a problem I posed involving the Fourier transform on the discrete cube. Maybe it is more tractable. I'm afraid I'm not sure how to do this kind of computation.A $k$-dimensional face of the discrete cube $\{0,1\}^n$ is a set of the form: all vertices which take prescribed values (either $0$ or $1$) on some given $n-k$ coordinates and are otherwise arbitrary.The question is: does a typical subset of $\{0,1\}^n$ approximately contain a face of dimension greater than $.6n$?We are interested in the li...Read more

co.combinatorics - obtaining all vectors of given length and with with $+-1$ entries from a given one

We can "travel" on all the vector space $V =GF(2)^n$ by doing the following(a) choose a primitive polynomial $P(t)$ of degree $n$ over $GF(2)$.(b) change vector $ X = (x_1, \ldots,x_{n-1}) \in V$ into vector $Y = (y_1, \ldots, y_{n-1}) \in V$.(c) repeat until $V$ is exhausted (2^n times)where$y_1+y_2z+ \cdots + y_nz^{n-1} = z(x_1+x_2z+ \cdots + x_nz^{n-1})$and $z$ is a zero of $P$, i.e., $P(z)=0.$I want to do the same with integral vectors containing only 1 and -1I.e.: "travel" on all possible vectors $(r_1, \ldots, r_{n-1})$with $r_i^2=1$ How ...Read more

co.combinatorics - Random Vornoi Diagrams (particular measures)

This is my second question about Random Voronoi diagrams, in my first question was given some excellent advice but i was not clear in explaining what i was looking for. I'm interested to know whether there are any known results about the following measures in 2-dimensional (preferably) random Vornoi diagrams. I think i should state also that i'm working in a circular disk so if the setting of any currently known results is different would the nature of the results change if i was to change the setting?The measures are a) The variance of the num...Read more

co.combinatorics - Cluster Variables for non-convex n-gons

Most of the lectures and lecture notes on Cluster Algebras (at least from Combinatorial point of view) start with mutations of the diagonals of a convex n-gon (mostly the pentagon) as the illustration of creating the new cluster variables using the Ptolemy's theorem, which also manifests other fundamental properties, such as positivity and Laurent phenomenon. (For example this http://arxiv.org/abs/1212.6263.)Although I imagine one could address this question via the generalization of cluster algebras to other oriented surfaces with boundary, I ...Read more

co.combinatorics - Hypergraph coloring problem motivated by legal billards racks

MotivationThere are several rules about what makes a rack legal for a game of eight-ball: the top ball has to be a solid, the eight-ball is in the middle, the two bottom vertices have to be one solid and one stripe, etc. But a rule I learned when I first learned to play pool was that there should be no three balls which are pairwise touching each other that are all stripes or all solids. Apparently this is not an actual rule for professional eight-ball. Nevertheless, it is an interesting restriction.ProblemSuppose we color the points in a trian...Read more

co.combinatorics - H-generalized join graph

The $A$-join of a set of graphs $\{ G_a \}_{a \in A}$ as‎‎the graph $H$ with vertex and edge sets‎‎\begin{eqnarray*}‎‎V(H) &=& \{(x,y) \ | \ x \in V(A) \ \& \ y \in V(G_x) \},\\‎‎E(H) &=& \{ (x,y)(x^\prime,y^\prime) \ | \ xx^\prime \in E(A) \‎\text{or} \ x = x^\prime \ \& \ yy^\prime \in E(G_x)\}‎.‎\end{eqnarray*}‎‎This graph is obtained by replacing each vertex $a \in V(A)$‎‎by the graph $G_a$ and inserting either all or none of the‎‎possible edges between vertices of $G_a$ and $G_b$ depending‎‎on ...Read more