Can a structure be defined by the radii alone?

Factual Questions
1

This is a math question is suppose.

I am wondering if you have a cluster of balls and every ball is connected to every other by means of a rod of a fixed length, is the cluster defined uniquely by that? What I mean with this is, lets say i take the cluster apart and give you the rods and the balls and you assemble it again, is there only one solution, do you always get the same structure back as in the beginning or are there multiple possibilities?

2

If the cluster is not mirror-symmetric then there will always be at least two solutions: the original and its mirror-image.

Aside from that I think it should be possible to prove there are no more solutions by induction. That is, take the first three balls and the rods joining them; they form a unique triangle. Then add new balls to the cluster one by one; it should be possible to show that up to symmetry each ball can only go in a unique position.

But I haven’t worked out the details, and may be wrong.

3

If the assumption is that every ball is identical, I believe the cluster would be unique (except for its mirror image).

My logic is:

  1. Say you have a set of balls and instructions to connect them (your rods, in other words).
  2. Consider any two balls, with one connection between them. There is obviously only one way to connect the two.
  3. Add a third ball, with two additional connection rods. The three balls will form a unique triangle.
  4. Add a fourth ball (and three more connections). There will, in general*, be two places where the ends of the rods could intersect (making two mirror image structures). This is the same as saying “three spheres having known center locations and known radii will intersect at two points, in general*, if they intersect at all”
  5. If you add an extra ball to a non-planar* structure already containing at least four, then the ends of any four connection rods will only intersect in one place, giving a unique structure.

*The exception in step four is when all four balls are co-planar, in which case there is only one possible place to locate the ball. In that case, the first non-co-planar ball added (whether it be no. 5, no. 6, ar whatever) will have two possible locations, making mirror-image structures.

Although not rigorous, I think this demonstrates the gist of the arguement.

4

If every ball is connected to and equidistant from every other ball, there’s a limited number of solutions.

  • one ball, no rods
  • two balls, one rod
  • three balls, three rods (triangle)
  • four balls, six rods (triangular pyramid)

I’m guessing that you really mean something different.

5

Forgot about the mirror image, but thats ok, it doesn’t matter.
The balls are not equidistant, but the radii are predefined.

Thanks a lot for your help.

6

There’s an infinite number of solutions to the revised problem statement.

Take each ball as having four neighbours. There’s an infinite range in angles separating the neighbours, from cubic to “squashed flat”.

7

I don’t think so, because each ball is not only connected to its neighbours but to all other balls with a predefined length.
But your answer is exactly what I am wondering about, do I have to define the angles too or is it enough if only the radii are defined.

8

I give up.

If that’s the problem, I’ve already given you the answer.

If the problem is something else, explain it properly.

9

Maybe you should read up on packing:

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

10

Sorry, I know I’m not very good at explaining. Packing is not the problem here. But I think your answer for equidistant balls holds too if they are not equidistant.
Thanks for your help

11

I think the terminology problem here is the word “radii” in the OP, which implies radii of the balls, which is not the intention. How’s this for a rephrasing:

Suppose we have a 3D wireframe structure with n vertices, where each possible pair of vertices are connected by a straight edge. If we are given the n*(n-1) edge distances between each pair of vertices, will those distances define a unique wireframe structure?

12

That’s exactly what I mean, zut. Thanks for clarifying.

13

No, there are only finitely many solutions if the balls are all equidistant, as Demostylus said, whereas there are an infinite number of possible initial clusters when the balls aren’t equidistant. (Although it seemed clear to me that the balls were not meant to be equidistant in the OP.)

That said I think zut’s argument is correct: given the edge lengths, the cluster should be determined uniquely up to symmetry.

14

If that’s the problem, then it’s fully defined apart from translation, rotation and reflection. Taking any vertex as a reference, and any arbitrary direction as another reference, the inter-vertex distances can be transformed into a unique set of vertex coordinates.

15

Wait a second. Which of the two is the case:

We are given a pile of n balls and N rods to assemble into a shape as described, or
We are given a pile of n balls and N rods to assemble into a shape as described, and each of the rods is labelled something like “This rod goes between balls A and D”?

If it’s the second one, I agree. If it’s the first one, I don’t see any reason why the final structure has to be unique. It would probably be really difficult to come up with a counterexample, though.

16

I’m reasonably certain that the second case is the one intended. However, for the first case, a non-unique example would be four balls and six rods, where the rod lengths are all “close” to each other, but not identical. For example, rod lengths of 20, 21, 22, 23, 24, and 25 could be assembled so that the resulting tetrahedron has numerous different combinations of different face triangles.

17

Actually I was thinking about the first case.
But I think the different tetrahedons in your counterexample can be rotated and/or mirrored to fit on each other all the time.

18

Ah, OK then. The arguements I was making above apply to the second case, where the ends of each rod are labeled as to which vertex they connect to.

For the tetrahegonal counterexample, let me explain a little more:

Suppose you have six rods of length 20, 21, 22, 23, 24, and 25. If you connect these six rods, they will form a tetrahegon, having four triangular sides. There are numerous ways to connect these rods which are neither rotations nor reflections of each other. Two such ways have the following triangles for sides:
CASE 1: 20-21-22, 20-23-24, 21-23-25, and 22-24-25
CASE 2: 23-24-25, 21-22-23, 20-21-24, and 20-22-25
Note that the shapes of the triangles are different between the two cases, so the two cases can’t be reflections or rotations of each other. Multiple additional cases can be constructed; I’m too lazy to figure out how many in total.

19

ok, I thought about it and I think it is clearly defined, but only for more than 4 balls. The number of radii is

N=n/2*(n-1) where n is the number of atoms.

and the number of coordinates in 3-dimensional space is

k = 3*n

N > k only for n > 4. This means the cluster is only defined by the radii alone for clusters with more than 4 balls.

20

zut: Ah, very good. Your counterexample is extremely clear. Thanks!

missing_link: I agree with your values for N and k, but I don’t see what that has to do with anything.

Of course, I can’t come up with a counterexample with n = 5.