Re: [math-fun] Maximal 3D unit distance graphs.
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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Here are my first values on the hex grid (I'm just adding triangles randomly at the moment). Starting from n=5: (9,12,15,18,22,27,31,35,39,44,48,52,57,62,66,71) Should anyone desire, I can post actual coordinates for any of these. On Mon, Jun 20, 2016 at 3:19 PM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
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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Looks like 12 points is at least 34 unit edges, tweaking Tom's idea with two octahedra. {{0, 0, 0}, {1, 0, 0}, {1, 1, 0}, {0, 1, 0}, {1/2, 1/2, -(1/Sqrt[2])}, {1/2, 1/2, 1/Sqrt[2]}, {5/6, 1/2, -(1/(3 Sqrt[2]))}, {11/6, 1/2, -(1/(3 Sqrt[2]))}, {11/6, 3/2, -(1/(3 Sqrt[2]))}, {5/6, 3/2, -(1/(3 Sqrt[2]))}, {4/3, 1, -((2 Sqrt[2])/3)}, {4/3, 1, Sqrt[2]/3}} On Mon, Jun 20, 2016 at 5:27 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Here are my first values on the hex grid (I'm just adding triangles randomly at the moment). Starting from n=5:
(9,12,15,18,22,27,31,35,39,44,48,52,57,62,66,71)
Should anyone desire, I can post actual coordinates for any of these.
On Mon, Jun 20, 2016 at 3:19 PM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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In 2D, I think a simple hex grid has optimal values up to a certain size. Asymptotically this yields 6n (or more precisely 6n-O(sqrt(n))). At some point a simple n-cube starts winning (said n-cube drawn by simply drawing the (n-1) cube twice, offset at distance 1 and at an appropriate angle to avoid point coincidence). This gives exactly 1/2 n log_2 n for n a power of two. Slightly better is the 3^n torus, which yields n log_3 n for n a power of 3. This can be generalized. The Cartesian product of any two unit-distance graphs is a unit-distance graph (you have to adjust the angle of one of the two graphs, and then just add the coordinates pair-wise). If we start with two graphs (v_1, e_1) and (v_2, e_2), this yields a unit-distance graph with v_1 * v_2 vertices and v_1 * e_2 + v_2 * e_1 edges. On Mon, Jun 20, 2016 at 3:27 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Here are my first values on the hex grid (I'm just adding triangles randomly at the moment). Starting from n=5:
(9,12,15,18,22,27,31,35,39,44,48,52,57,62,66,71)
Should anyone desire, I can post actual coordinates for any of these.
On Mon, Jun 20, 2016 at 3:19 PM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Silly correction:
In 2D, I think a simple hex grid has optimal values up to a certain size. Asymptotically this yields 6n (or more precisely 6n-O(sqrt(n))).
Obviously it only yields *3n* asymptotically . . . my apologies. On Mon, Jun 20, 2016 at 4:07 PM, Tom Rokicki <rokicki@gmail.com> wrote:
In 2D, I think a simple hex grid has optimal values up to a certain size. Asymptotically this yields 6n (or more precisely 6n-O(sqrt(n))).
At some point a simple n-cube starts winning (said n-cube drawn by simply drawing the (n-1) cube twice, offset at distance 1 and at an appropriate angle to avoid point coincidence). This gives exactly 1/2 n log_2 n for n a power of two.
Slightly better is the 3^n torus, which yields n log_3 n for n a power of 3.
This can be generalized. The Cartesian product of any two unit-distance graphs is a unit-distance graph (you have to adjust the angle of one of the two graphs, and then just add the coordinates pair-wise). If we start with two graphs (v_1, e_1) and (v_2, e_2), this yields a unit-distance graph with v_1 * v_2 vertices and v_1 * e_2 + v_2 * e_1 edges.
On Mon, Jun 20, 2016 at 3:27 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Here are my first values on the hex grid (I'm just adding triangles randomly at the moment). Starting from n=5:
(9,12,15,18,22,27,31,35,39,44,48,52,57,62,66,71)
Should anyone desire, I can post actual coordinates for any of these.
On Mon, Jun 20, 2016 at 3:19 PM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (4)
-
apgoucher＠gmx.com -
Ed Pegg Jr -
rcs＠xmission.com -
Tom Rokicki