Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can anyone give a concrete example of n points in the unit square (for instance, n runs from 3 through a large number) that can be generated by the algorithm here or any other algorithm or any construction such that the areas of the smallest triangles, multiplied by n^2, form a sequence that is non-decreasing for large n or tends to be unbounded?

share|improve this question
2  
Example of what? Your question provides no information about what you are asking! –  Mariano Suárez-Alvarez Aug 30 '13 at 3:56
    
I'm sure that for any concrete set of $n$ points, there is an algorithm or construction that will generate that set of points, so I suppose you can take any $n$ points you like as an example. –  Gerry Myerson Aug 30 '13 at 4:34
    
@GerryMyerson the link points to an algorithm for finding sets that are counterexamples to a conjecture of Heilbronn. I agree, though, that in its current form the question should be closed. –  Yemon Choi Aug 30 '13 at 4:37
2  
If the comment above is meant to be a clarification of the question, please edit it into the question. People shouldn't have to wade through the comments to understand the question. –  Gerry Myerson Aug 30 '13 at 4:40
    
The comment mentions the unit square. If that's an important part of the problem, that should be in the body of the question, too. –  Gerry Myerson Aug 30 '13 at 10:05

1 Answer 1

Unbounded might not be too hard if I understand the question correctly, although my argument is very much vaguer than I would like. For a set $S$ of points in the unit square let $n=|S|$ and $f(S)$ be $n^2$ times the smallest area of any of the $\binom{n}{3}$ triangles determined by three of the points. You link to an article which constructs for each $n \ge 3$ an explicit (in some sense) set $T_n$ with $f(T_n) \gt c\log(n)$ for some fixed $c$. Are you asking for a sequence of points so that the sequence $a_n=f(\lbrace p_1,p_2,\cdots,p_n\rbrace)$ is increasing or at least unbounded?

Here is a possible attack using only that information. Considering the actual construction in the linked article might give better results. It seems possible that the following could work, i.e. have $\limsup(a_n)=\infty$, although it could at the same time have $\liminf(a_n)=0$: First let me comment that we can assume that the points in the set $T_n$ have the form $(\frac{a}{n^k},\frac{b}{n^k})$ for $k=4$, I don't know that that helps but it makes things seem more controlled (proof at end).

Consider the sets $T_n$ above (about which I know nothing beyond what is stated). Perhaps one can always find a sequence of indices $n_1,n_2,\cdots$ so that $T_{n_i}$ is "almost a subset" of $T_{n_{i+1}}$ in the sense that we can perturb the points of the smaller one by a sufficiently small amount (perhaps none by more than $1/2^{n_i}$) such that they coincide with a subset $T'_{n_i}$ of the larger one having $f(T'_{n_i}) \gt \frac{c}{2}\log(n_i).$ Then (skipping some routine details) there would be a sequence so that $a_n \gt \frac{c}{2}\log(n)$ for $n=n_1,n_2,\cdots.$

The question of $a_n$ increasing or at least having a positive lower bound might be harder.

My argument that the points of $T_n$ can have coordinates of the form $\frac{\cdot}{n^k}$ for $k=4$ is that moving each point to make this happen will change the area of each little triangle by no more than $\frac{1}{n^k}$. I don't know if we really need $k$ as big as $4$.

In the mean time I glanced at the article which seems to use interesting hyper-graph methods. A close reading might help with the question.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.