Bah – so I realised that the volume of these cells is 1/2, so the determinant method gives 64 colours for the vectors mentioned (my claim that {+/-2,+/-2,0,0} gives 64 was nonsense). I have now found a better vector set – {{5/2, 1/2, 1/2, 1/2}, {2, -1, -1, -1}, {3/2, -3/2, -1/2, 3/2}, {3/2, -3/2, 3/2, -1/2}} – but it still doesn’t get to 54, only 56. T’m pretty sure that that’s the best we can do with this tesselation, unfortunately.

]]>I’ve been revisiting the area of upper bounds in higher dimensions, largely as a result of a very cool realisation by Jaan that the known 15-colouring of R^3 uses distinct sets of five colours repeated every three layers of cells, with each layer being assigned colours in the manner also used by Soifer nearly 30 years ago to tile the plane with six colours when one colour has a different distance excluded than the other five. (See Figure 6.5 of his book.) I’ve been focusing on R^4, and I am really new to 4D geometry, but I am fairly convinced by now that the tesselation of R^4 with the “24-cell”:

https://en.wikipedia.org/wiki/24-cell_honeycomb

should be colorable with fewer than 54 colours. When each cell is centred either on a unit grid point or on that grid shifted by 1/2 in all four directions, the cells have diameter Sqrt[2], and it seems that cells with centres Sqrt[7] apart are Sqrt[5/2] apart, and thus that the number of cells around a given cell that cannot be its colour is only 328 (I think), as compared with 370 for the permutohedron. That’s as far as I’ve got, but it seems like such a big difference that it would be astonishing if it doesn’t translate into a better colouring. My first (manual) attempt at an efficient sublattice is the vectors {{1/2,3/2,3/2,3/2},{-3/2,-1/2,3/2,3/2},{-3/2,3/2,-1/2,3/2},{-3/2,3/2,3/2,-1/2}} – what number does that give? Four of the inter-centre distances in a minimal simplex based on these vectors are Sqrt[7] and the other six are Sqrt[8], which seems pretty promising given that we get a 64-colouring just by naively using the permutations of {+/-2,+/-2,0,0}, where everything is Sqrt[8].

I’ll try to work more on this myself soon, but I won’t have much time for a while – I’m hoping others can help.

]]>I’m delighted to say that our article is out! I have placed a copy of the whole issue here:

https://www.dropbox.com/s/8zmhm4ouhol8zco/xxx-4.pdf?dl=0

because it also contains the latest solo report from Jaan, a very nice demonstration of improved multi-fold chromatic numbers in the plane. I’m sure he would be happy to engage in discussion of that work here.

]]>It was an open problem whether a UDG on a sphere of non-special radius can have a superlinear number of edges, and we also discussed it somewhere during this polymath. These are the two relevant papers:

P. Erdos, D. Hickerson, and J. Pach, A problem of Leo Moser about repeated distances on the sphere, Amer. Math. Monthly 96 (1989), 569–575.

https://doi.org/10.1080/00029890.1989.11972243

Here it is shown that a UDG can have n log*n edges.

Swanepoel, Konrad and Valtr, P. (2004) The unit distance problem on spheres. In: Pach, János, (ed.) Towards a Theory of Geometric Graphs. Contemporary mathematics (342). American Mathematical Society, Budapest, Hungary, pp. 273-280. ISBN 9780821834848

Here it is shown that a UDG can have n sqrt(log n) edges.

Can you even have dense UDG graphs on a sphere? I think the limitation is far more than “no hexagons”. If I understand correctly, the core pieces of planar 5-chromatic graphs are all based on a lattice in Z[w1,w3], where every lattice point is automatically part of dozens of Moser spindles. But a sphere doesn’t have this underlying lattice. If you start on the equator then [move east 100 miles] and [move north 100 miles], where you end up depends on the order of those two steps. Without this commutative property, making dense graphs seems impossible.

You can still make a Moser spindle on the sphere. But can you combine multiple Moser spindles in a useful way? Can you even combine triangles in a useful way?

An intuitive way to think about this: Take some nice dense graph on the plane, and think about it as some super-rigid structure. Try to warp it to a sphere. You can imagine that most of the edges will “snap”. Only after removing a big fraction of the edges will the graph be flexible enough. Note that this flexibility is critical if you change your radius just a smidgen. (Is there a name for this kind of flexibility?)

A tangent for triangles on a sphere — If you add 6 triangles around a point, you’ll get a hexagon-ish graph where the “first” and “last” points on the perimeter are close but don’t touch. Keep adding triangles. For certain sphere radii, your triangles will have a rational angle and repeat after T triangles (e.g. 6000000001 triangles) (a 3-chromatic graph). When T=3,4,5 you can chain together these graphs to form a tetrahedron, octohedron, icosahedron. With T=11 and two revolutions, do you get a similar structure? I’m pretty sure not because this would give a perfectly isomorphic distribution of a lot of points on the sphere, which isn’t possible. Anyways, it might be interesting if for large spheres it mattered if the triangle angles are rational or not.

]]>Because then we could look for 5-chromatic graphs that do not have such graphs as subgraphs. The idea of using 1.3 as a width is just because we know that we can get to 5-chromatic without having the easy non-sphere-drawable graphs (such as the hexagon).

Of course the real first step should be to check whether Jaan’s latest graph is already sphere-drawable! If it isn’t, we can probably find smallish subgraphs of it that are not.

]]>such

]]>You want them suc that they canNOT be drawn on a sphere? Why?

]]>Thanks Jaan.

I did try other distributions of the figures but in the end this was the one that minimises the sum of the distances between a figire and the text that references it and also has each figure on a different page, so I think it is the best.

I can’t figure out how to remove the page numbers. I copied the entire LaTeX from your “green article” down to “maketitle” but they didn’t disappear. What is the trick?

You are quite right about declaring victory, of course. I think the discovery of a human (or even human-verifiable :-)) proof of CNP!=4 was such a big milestone that it makes sense to call it a victory, but for sure we should continue to work on the many remaining dragons!

Perhaps a good one to look at now is the lower bound for the surface of a large sphere. We could start by looking for UDGs that fit inside a strip of width 1.3 and cannot be drawn on a sphere.

]]>At the moment, I am unable to improve the result. And this is somewhat good, because it creates the feeling that we are stuck in some fundamental limitation, and now it is interesting to understand what it is connected with.

The article turned out to be nice both in form and in content. I understood the system in the arrangement of the pictures. The only thing I do not like is that the latter climb onto the bibliography. Can’t we try to move everything towards the beginning? However, I’m not sure that it will be better this way. Do not forget to remove the page numbers (according to the journal’s requirements, they should be invisible).

Btw, I don’t think we should declare victory.