Obviously, not much confidence in anything beyond 5, but it looks like I picked good opening values. I'll need to study Tom's and APG's proposals and get them coded up, but I don't doubt them at all. I'm not that sure on 2D, either, but the initial terms will be easier. On Mon, Jun 20, 2016 at 4:51 PM, <rcs@xmission.com> wrote:
Has the 2D question beem addressed? --Rich
---
Quoting apgoucher@gmx.com:
I think I can increase 16/50 to 16/51, but I'll need to verify that all points are distinct:
Operate on a sphere S of unit radius, and let ABC be a fixed equilateral triangle on S, all of whose edges have length pi/3 (i.e. OABC is a regular tetrahedron where O is the centre of S).
Now we want points A', B' such that AA'B'B is a spherical 'rhombus' (all edges are unit length). There is a set of such solutions parametrised by S^1. Let C' be the third point such that A'B'C' is an equilateral triangle oriented in the *opposite* way to ABC. By the intermediate value theorem, there is some choice of parameter which makes CC' = 1.
Then OABC and OA'B'C' are regular tetrahedra sharing a vertex (the origin) with AA' = BB' = CC' = 1.
Take the Minkowski sum of these two tetrahedra; this yields 16 points with 51 edges of unit length.
-- APG.
-----Original message----- Sent: Monday, 20 June 2016 at 21:44:03 From: "Tom Rokicki" <rokicki@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Maximal 3D unit distance graphs. I can bump your 12/31 value to 12/32.
Start with your 6/12 octohedron; place it flat on a surface with two of the faces parallel to the surface.
Copy the octohedron exactly one unit off the surface and add edges between the two pairs; we are now at 12 nodes and 24+6 or 30 edges, and the top octohedron has two degrees of freedom (fixing the bottom octohedron).
Move the top octohedron so that one of its vertices forms the apex of a tetrahedron from the base triangle of the bottom octohedron; this lets you add two more edges from that vertex, and also two more edges from the corresponding upper plane, for a total of 32 edges.
If this isn't clear I can draw a diagram.
On Mon, Jun 20, 2016 at 7:52 AM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
In 3D space, with n distinct points, what is the maximal number of unit
distance lengths?
I'm trying to build up a sequence for OEIS, but I'm only positive I've got the right values on two or three of the entries. Here's what I have so far.
V -- E -- figure 4 -- 6 -- tetrahedron 5 -- 9 -- triangular bipyramid 6 -- 12 -- octahedron 7 -- 15 -- pentagonal bipyramid 8 -- 18 -- snub disphenoid or Raiskii spindle 9 -- 21 -- triaugmented triangular prism 10 -- 25 -- Nechustan spindle 11 -- 28 -- Augmented Nechustan spindle 12 -- 31 -- Double Pacman spindle 13 -- 36 -- Cuboctahedron + center 14 -- 40 -- Cuboctahedron + center + pyramid 15 -- 45 -- Icosahedron + 3 internal points 16 -- 50 -- Icosahedron + 4 internal points
http://math.stackexchange.com/questions/1830194/maximal-unit-lengths-in-3d-w... has a bit more info.
Ed Pegg Jr
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