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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

> I am trying to put N dots equispaced on a spherical surface

>

> For N=2 the answer is clearly `Poles apart` - distance pi times R

> For `The Platonic Solids` there are exact answers

>

> But HOW do we handle N=17 etc and how do we get there?

> The `most evenly-spaced configuration` and the VARIATION in the nearest-neighbour gt circle distances.

>

> For N=millions once again the answer should become computable and be yery near that for a hexagonal array in 2-d

> But how near?

>

> Many thanks

> John

>

> Might this work:-

> Place n dots at random lat/long

> For each dot in turn, Find the ONE nearest-neighbour that is FARTHEST

> Find a new position for it that reduces that distance

> Then find the ONE nearest-neighbour that is NEAREST and move it farther away

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

Did you find any of this useful?

http://markcordell.blogspot.com/2011/11 … ts-on.html

This site suggests that it might not be possible except for only certain N's.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

Yes, Bobbym, very helpful

Many ingenious ideas, but no solution

Spheres COMTAINED WITHIN a sphere

Points on the SURFACE of a sphere

In BOTH cases there are two oft-suggested ideas:-

A kind of "crystal structure" as when we "grow" close-spaced spirals of increasing latitude and longitude.

A kind of "random" structure in which dots are placed more likely the bigger the area that dot gets placed in.

The "random" has very much more variation in distances (and risks overlaps)

I suspect there IS a computer approx method

Start with the 8 dots of a cube vertices

or the 12 midpoints of its edges.

We now want to add ONE more dot.

The obvious place for it is equidistant from as many other dots as possible.

For example for the N=12 case we get n=13 by placing that one extra dot midway in between 4 other closest dots

We mow have UNevenly-spaced dots.

So we move the nearest dots to the new dot a small distance d away from the new dot.

And for each of those dots we move their nearest-meighbours away by d/2

And so on.

The computer will happily work away on this all night

Reminds me of R V Southwell's brilliant "Relaxation of Constraints"

The general plan is "Always reduce the biggest errors" and keep going!

The site that says "May be impossible" is a mere "pure" maths site!

I simply want the solution that gives for closest neighbours the minimum distance and how much these minimum distances vary

For N as small as 13 this can be listed exactly and fully

For N millions we will have to think carefully how we WEIGHT the importance of nearest-dots agains those progressively farther away!

ALL ideas welcome. that is what I care about - new IDEAS!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

For millions of dots I would use a random distribution of points. One that had the same expected number of points for each area of the sphere. As N approaches infinity that should mean they will get closer and closer to equidistant.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

Good idea!

By the way I like your "In mathematics, you don't understand things. You just get used to them." but for me it is "Just try them out!"

OK, so for a certain sphere, 1256200 points, the surface points nearest to one point vary depending on what we mean by nearest

Of ALL points only 2 (one pair) are nearest of all.

But for every point there are:-

A nearest point

A next nearest point

and so on

For example for point A its nearest dot is distance 3.12 away

The next nearest are 3.2,3.33,3.7 and 4

Is that "good"

What would be better?

For point B the nearest neighbours were

3.13, 3.19,3.3,3.73, 3.9

Is B better that A?

Might it pay to skip point B (or A) and replace it with a "better" point

How do we judge?

HOW do we begin to puzzle this out?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

For that many points I would use this method:

http://mathworld.wolfram.com/SpherePointPicking.html

the points would be randomly placed on the surface. You do not have to decide anything. The randomness will spread the points out in nearly an optimal way.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

testing why it refuses to allow me to reply

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

I am not following you. Are you having trouble posting?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

I was trying "Quick Reply" and after it had already acknowledged I was signed on

Quick Reply is denioed with idiot message endinmg "Say when"

I was trying to send you a reference illustrating the vast difference between random and uniform

Random means irregular!

The ref is now lost and I have spent an hour trying to recover it - sorry abiut that.

To DENY crystalline's claim to be more uniform we need a way to MEASURE uniformity of spacing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

To do better you would have to define the criterion of measure of space.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

Yes indeed - we must say what we decide we mean by "evenly spaced"

For the cases N=2,3,4,6,8,12,20,30 it is plain enough

We mean maximally spaced but in groiups as large as possible the distances to be equal

For example for N=8 we want every dot to be equal distance from every other of 3 other dots and equal spaced but more distant from three other dots.

For N=4,5,7,,9,10,11 etc we can make up similar rules

But for all N>4 we need to say what weight we give to the dots closest compared to those farther away. In that way we decide which dot-arrangement is best for each N in turn.

This is where I am stuck and need your ideas and suggestions

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi Mrwhy;

In mathematics or any other problem solving field we rarely have the choice of picking what we want. Very often it is what we are capable of doing that decides what we can pick. Pick some very, very simple criterion and use that.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

So for N=2,3,4,6,8.12,20,30 there is ONE best pattern of dots for each N

But for all other N there are as many "best srrangements" as the number of ways we care to define "best even placement" - how many neighbours are at WHICH distance.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

But for all other N there are as many "best srrangements" as the number of ways we care to define "best even placement" - how many neighbours are at WHICH distance

That may be true, there are certainly numerous ways. You can pick one that you can solve, or is computationally feasible.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

The simplest idea so far is this:-

For N>2 points, each point has a neighbour that is nearer than of all the others.

Lets make this distance as great as we can without going outside the sphere.

VERY SMALL increments are best!

We now REPEAT this for every single point in turn.

Now each point ends up max-spaced from its nearest neighbour

This is ONE variety of evenly-spaced.

But is it the only one?

To SEE this we need a way to visualise the pattern of points (on a flat screen)

For example for N=5 there is one pattern (of several?)

When we add a point we now have N=6 points.

If we could SEE how the points moved when we added that point it would help us begin to understand "what is happening" and why!

PLease suggest some ideas of how to visualise points scattered on or inside a sphere.

Many thanks

*Last edited by Mrwhy (2013-04-19 19:17:16)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

You would need a 3D plotter.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

bobbym wrote:

Hi;

You would need a 3D plotter.

Is there a good way to simulate a 3-d plotter on a computer screen?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

I am not sure I am following you. To show those points you will have to get a 3D plotting program.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

A good 3-d plotter would give you a model of the dots - a model you could hold in your hands and twist about to see really what was where.

So what are the BEST ways of simulating "twist about to see really what was where" on a flat computer screen?.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,444

Hi;

There are programs to do that. You could maybe find one and then it will take your points that are in (x,y,z) and render them on your screen. They also rotate, dilate, translate, generally anything you can imagine.

Here is an example with 500 random points on a sphere.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

Many thanks bobbym

I will search for those progs in out coder section.

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