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Caveat: This is just a random thought, and I realise the question may not be well-posed.

I'm trying to figure out how fast a certain sequence of graphs would grow. Here, I use $n$ as the number of dimensions. The property (let's just call it $P$) the graphs must have is that all the nodes must be connected, and no 2 edges can intersect at any point.

Thus, for n = 1, the largest graph we can make with this property is quite trivial - it is just the graph consisting of two nodes with one edge between them.

Consider another example. For n = 2, the largest graph one can make is $K_4$, which can be embedded as e.g. the vertices and centroid of an equilateral triangle with the edges drawn as straight lines.

Let $f(n)$ be the function that returns the number of nodes the largest $n$-dimensional graph satisfying $P$ would have. So $f(1) = 2$, and $f(2)= 4$.

What would $f(3)$ be and what would the graph look like? I originally thought $f(3) = 6$, where the octahedron graph would be the largest graph satisfying $P$, but later I realised one could just add an extra node outside the tetrahedron, and the graph of those 7 nodes would satisfy $P$. Upon further reflection, I realised that in each successive dimension, there is "more room" to move around, and so as $n$ increases $f(n)$ would start to become very large.

Thus, what would the sequence $f(1), f(2), f(3), f(4)...$ look like, what would the corresponding graphs look like (is there $n$ such that the corresponding graph is not unique i.e. there are multiple graphs that satisfy $P$ and have the maximum number of vertices), and how fast does this sequence grow?

Peter Taylor
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  • Why isn't $f(2)=4$? Why can't you put a node at the center of the triangle and connect it with edges to the three corners of the triangle? – bof Nov 02 '17 at 06:54
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    In $3$ dimensions you can put nodes at $(t,t^2,t^3)$ for $t=1,2,3,\dots$, and connect them with straight line segments, so $f(3)=\infty.$ – bof Nov 02 '17 at 06:57
  • I believe you are correct. I will amend the question. – Rohin Garg Nov 02 '17 at 06:57
  • But isn't the number of nodes still 3(one at $t$, one at $t^2$,and one at $t^3$)? – Rohin Garg Nov 02 '17 at 07:01
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    $(t,t^2,t^3)$ are the coordinates of one point in space: $x=t,y=t^2,z=t^3.$ The first point is $(1,1,1),$ the second is $(2,4,8),$ the third is $(3,9,27),$ the fourth is $(4,16,64),$ and so on. – bof Nov 02 '17 at 07:04
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  • Ah, how silly of me - I didn't think of the triple that way. I guess looking at the other question helps, but It would be nice to have a more precise reason why $f(3) = \infty$. – Rohin Garg Nov 02 '17 at 07:14
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    Put the vertices, as many as you want, on the curve $x=t,y=t^2,z=t^3.$ Connect them all to one another with straight edges. If two of the edges cross, then the plane determined by those two edges intersects the curve in $4$ points. But you can show that no $4$ points on the curve are coplanar. – bof Nov 02 '17 at 07:29
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    To save me a lot of typing, let $P(t)=(t,t^2,t^3).$ Consider any four distinct points on the curve: $P(t_1),P(t_2),P(t_3),P(t_4).$ I claim that those four points can't lie on a plane. For the four points to lie on the plane $A+Bx+Cy+Dz=0,$ the quantities $A,B,C,D$ must satisfy a certain system of four linear equations: $$A+t_iB+t_i^2C+t_i^3D=0$$ for $i=1,2,3,4.$ But this linear system has no nontrivial solutions, because its matrix is a Vandermonde matrix and so is nonsingular. – bof Nov 02 '17 at 07:39
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    It is easy to see that any graph with no more than $\frak c$ vertices can be drawn with straight-line edges in $\Bbb R^3$ without crossings, because any placement of its vertices with no four at one plane generates such a drawing. – Alex Ravsky Nov 02 '17 at 09:26
  • Thanks @bof, that was really helpful. – Rohin Garg Nov 02 '17 at 17:32
  • @AlexRavsky, could you provide a link to a proof? – Rohin Garg Nov 02 '17 at 17:33
  • I don’t have such a link but the claim is obvious, because, as bof already mentioned above, if two of the edges cross then their endpoint vertices are coplanar. The required placement of vertices with no four at one plane can be easily constructed by (transfinite) induction using that the space $\Bbb R^3$ cannot be covered by finitely (or even less than $\frak c$) many planes. – Alex Ravsky Nov 02 '17 at 17:50
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    @AlexRavsky Or simply place the $\frak c$ vertices on the curve $x=t,y=t^2,z=t^3.$ – bof Nov 02 '17 at 21:40
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    @RohinGarg I've added details to this answer, I hope it's clearer now. – bof Nov 02 '17 at 23:57
  • That was great, thanks. – Rohin Garg Nov 03 '17 at 11:37

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