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This is a question about identifying a theorem or maybe even a subject matter. I've asked myself a question, but I don't know where to look for answers. Typing a problem statement in a search engine doesn't really work unless you know a theorem exists and the exact statement it addresses.

I'm going to try to present my thoughts/questions with hopes that someone might recognize a more specific theorem/subject. My thoughts/questions which follow are not meant to be rigorous...just trying to help convey what I'm looking for.

Part I

Given a set of unique points arbitrarily distributed over the domain $\mathbb{R}^2$, there exists a uniform grid in which all points coincide with a vertex of the grid.

(When I say "uniform grid", think checkerboard...can be oriented any angle, but all elements must be squares)

Part II

Under what conditions is Part I true or false?

(Maybe it's always true or always false)

Part III

If Part I is true, how does the edge length of the square element relate to the distribution of the points?

Again, the specific questions aren't meant to be answered... I'm trying to find references to research further.

EDIT

The set of points would be finite. Imagine dumping a bucket of marbles on the floor of an empty room. Where the marbles stop, those are your points. The grid has to be generated such that every marble coincides with a vertex of the checkboard. The boundaries do not bound the grid (they're only there to contain the marbles)

Initially I was thinking that you have to relate it to the distance between the points. In which case, I would think that all distances would have to be a multiple of the smallest. Then I was thinking, what if the distances were primes? I don't think anything would work. Then I was thinking that the only possible case which would work for all, would be if the smallest distance was the diameter of the point. But then what is the diameter of a point if its just a point...therefore, the only possible grid would be the plane itself. I don't know if that would "qualify" as a grid, but I think you get my point. So we have some potential worst case and best case scenarios...I'm looking for more rigorous theorems or proofs which help answer these questions. I just assumed they exist as the question is not that abstract.

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  • $\begingroup$ Is the given set of points supposed to be finite? The statement is clearly false if it can be infinite, since then it could be the set of points on a continuous curve. $\endgroup$
    – saulspatz
    Commented Nov 23, 2019 at 6:43
  • $\begingroup$ Do you know what specific math topic this question would belong to? I'm not even sure I'm directing this to the right people. $\endgroup$ Commented Nov 23, 2019 at 7:04
  • $\begingroup$ What is a "vertex" of a checkerboard? $\endgroup$
    – bof
    Commented Nov 23, 2019 at 7:21
  • $\begingroup$ "In geometry, a vertex (plural: vertices or vertexes) is a point where two or more curves, lines, or edges meet" $\endgroup$ Commented Nov 23, 2019 at 7:24
  • $\begingroup$ Checkerboard....chess board....whatever. It's just a uniform grid of squares. $\endgroup$ Commented Nov 23, 2019 at 7:27

3 Answers 3

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Part I is false. If $A$, $B$, $C$ lie on a square lattice, then $$\left(\frac{|AB|}{|AC|}\right)^2\in\mathbb{Q}.$$ So, a counterexample with only three points may be constructed: take a triangle with side lengths $1$, $\pi$, $\pi$, for example. It will be impossible to fit it into any square grid.

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  • $\begingroup$ I'm assuming A,B,C are points and AB, AC are lengths of a segment. Then from your example, I can say $AB = \pi$, $AC = \pi$ ... the length of two segments. Your equation gives an integer = 1... am I missing something? I agree with your case, but I think the condition might be missing something? $\endgroup$ Commented Nov 23, 2019 at 8:50
  • $\begingroup$ @ThatsRightJack You can take $|AB|=\pi$, $|AC|=1$. Actually, if any two lengths in your set of points have a ratio whose square is not an integer, they won’t be able to fit in a lattice. $\endgroup$
    – ViHdzP
    Commented Nov 23, 2019 at 14:15
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You can't draw an equilateral triangle on a square grid.

Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners?

I suspect that a random triangle (with an appropriate definition of random) can't live on a square grid.

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The set of points on a 2D smooth surface on two curvilinear parameters $(u,v)$:

$$ (x,y,z) = (f(u,v), g(u,v), h(u,v)) $$ which belongs to a specific pattern defined by these three functions with continuous partial derivatives.

Given points $ (x(u,v),y(u,v),z(u,v)) $ cannot be mapped/fitted in by a set of arbitrarily embedded scattered grid or net points (vertices) onto any patch in 3-space.

That said, specific examples can be given of non-arbitrary defined curvilinear orthogonal "square" projected planar grids in which you are interested.

In 3D conformal mapping using inversions, $90^{\circ}$ vertices get mapped onto a plane tangent to south pole in the following instances.

In Stereographic Projn latitudes and longitudes of the globe are projected in a plane forming.... polar grid or bipolar grid depending on whether chosen point of projection is from 1) the north pole or from 2) the equator.

Loxodromes/rhumb lines intersecting lat/long lines orthogonally at $45^{\circ}$ are given (before projection) by

$$ x= sech \theta \cos \theta, \,y= sech \theta \sin \theta, \, z= \tanh \theta $$

It would be interesting to find again such planar projections repeated with loxodromes/rhumb lines intersecting lat/long lines at $\pm 45^{\circ}$ as square curvilinear grids from the earlier two points (pole and equator).

Ortho Loxodromes

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