Here’s a brief summary of the progress made in the previous thread:

– The smallest known unit-distance graph now has 553 vertices and 2722 edges.

– Terry Tao made some progress on understanding the colorings of .

There are several interesting ideas captured in Marijn’s paper. For example, his Table 1 appears to be amenable to analysis under the probabilistic method.

There has also been some work to find 6-colorings of the plane. Along these lines, I wanted to reiterate a point made by Matthew Kahle:

**Theorem (Thomassen).** Consider any infinite connected plane graph such that each connected component of (or “face” of ) has diameter less than 1, is homeomorphic to a disk, and is bounded by a cycle in . Next, consider any proper coloring of such that each color class is the union of faces of (and part of their boundaries) such that the distance between any two of these faces is greater than 1. All such colorings require at least 7 colors.

That is, “nice” colorings will not improve the current upper bound. Regardless, as Philip suggested, it would be interesting to collect some records along these lines:

**Question.** What is the radius of the largest disk in that can be 6-colored?

Here’s a brief summary of the progress made in the previous thread:

– The smallest known unit-distance graph of chromatic number 5 now has 610 vertices and 3000 edges.

– We now have several small unit-distance graphs with no 4-coloring in which a particular vertex is bi-chromatic, and these results are human-provable.

– We now have upper bounds on the chromatic number of for .

Progress has slowed considerably since the end of the spring semester (presumably due to summer travel, etc.). I wanted to highlight a couple of questions to help keep the conversation going:

**Question.** Is Tamas’s result that every 4-coloring of is 8-periodic human-provable?

Terry Tao provided a comment along these lines. Actually, a human proof of can be obtained from something much weaker:

**Question.** Is it easy to check whether an infinite k-colorable graph has finitely many k-colorings?

Indeed, Boris has an easy proof of the following: If there exists a unit-distance graph with infinite Euclidean diameter and finitely many k-colorings, then the chromatic number of the plane is at least k+1. As such, a human proof of reduces to human-proving that has finitely many 4-colorings.

**Question.** What sort of lower bounds can we get on CNP if we use a version of Lovasz that’s higher on the SOS hierarchy?

We know that the (complementary) Lovasz number of the plane is about 3.48. Applying this semidefinite program to Marijn’s graph on 874 vertices gives a lower bound of about 3.37. Can we beat 3.48 by running an SOS program on one of our 5-chromatic graphs? What if we applied ideas from Oliveira’s thesis to compute with the entire plane?

]]>What follows is a brief summary of the progress made in the previous thread:

– We now have a 5-chromatic unit-distance graph with 633 vertices and 3166 edges. This is the current record holder.

– We now have a human-verifiable proof that there exists a finite unit-distance graph with a vertex such that there is no 5-coloring of in which is bichromatic (see this and that). This result is “in between” and , since the latter implies the result, which in turn implies the former. Domotor suggests that this might be used to prove .

– We’ve collected several probabilistic inequalities, but we’re still hunting for a human-verifiable contradiction. We have a short computer proof of , and we suspect there’s a human-verifiable proof within reach. Presumably, the proof will involve new ideas that can be leveraged for additional inequalities.

– is 5-chromatic, but so is (see this). Details are being collected in the wiki. In order to build up to a 6-chromatic ring, we might need to analyze Jannis Harder’s ring .

– Tamás Hubai identified all 4-colorings of with the help of a computer. These (finitely many!) colorings are invariant under translation by 8 and rotation by . Presumably, this will help us identify useful claims about the colorings of that are provable by hand.

]]>What follows is a summary of some of the main directions of this project:

**— Lower bounds from the probabilistic method —**

Terry Tao suggested a new approach that might lead to a human proof of :

Suppose there exists a 4-coloring , and apply a “uniform random” Euclidean isometry to and permutation to to produce a random 4-coloring (this can be made rigorous since is amenable). Then for each , we can estimate probabilities of the form by considering all proper 4-colorings of . Different can then produce inconsistent probability estimates, resulting in the desired contradiction.

One may rephrase the existing proofs of in this probabilistic language (see the wiki). Each of these proofs finds a contradiction by demonstrating that a certain probability equals exactly 0. Doing so requires a large vertex set (e.g., de Grey’s or Exoo and Ismailescu’s ), thereby insisting on computer assistance. However, it suffices to show that the probability in question is small, meaning we might get away with a smaller (human-analyzable?) vertex set.

This line of inquiry has focused on estimating probabilities of the form . Of course, and . A spindling argument gives whenever . Jannis observed by computer assistance, contradicting this upper bound, and so in principle, it suffices to consider probabilities of this form. Many more bounds are derived in the wiki.

**Question.** Given , how efficiently can one identify all of the implied linear inequalities on , where ?

**Question.** What are “good” vertex sets to compute inequalities from?

By combining answers to these questions, we can build up a rich database of probability estimates that will hopefully lead to a contradiction from small vertex sets.

**— Computational lower bounds —**

As mentioned above, Jannis found that for every 4-coloring of , 8/3 necessarily receives the name color as 0. This implies by a spindling argument. We would like to replicate this approach to obtain new lower bounds.

**Question.** Consider all proper 5-colorings of Tamás’s in which the neighborhood of the origin avoids colors 1 and 2. Is there any vertex (other than the origin) that is colored either 1 or 2 in all such colorings?

If such a vertex exists (and the vertex’s distance from the origin has the property that is an irrational multiple of ), then the measurable chromatic number of the plane is at least 6. Presumably, this question can be answered using Jannis’s technique.

Philip asked a related question in the previous thread:

**Question.** Define , where

Does 5/3 receive the same color as 0 in every 5-coloring of ?

These two points are always monochromatic for homomorphic colorings, and if this behavior persists for general colorings, then a spindling argument implies .

**— Homomorphic coloring —**

Taking inspiration from de Grey’s , we consider rings of the form , where and is the sequence of positive Loeschian numbers.

So far, this pursuit has been fruitful: is 2-chromatic, is 3-chromatic, and is 4-chromatic (see this). Furthermore, we have a few finite subsets of that are 5-chromatic (see this and that).

**Question.** Is 5-chromatic?

I expect the answer to this question is “yes” by homomorphic coloring. The first step is to identify the ring generated by and its multiplicative inverses. This step will also help us to identify appropriate generators for Minkowski sums, enabling additional computational observations (à la Jannis).

**Question.** Does enjoy a homomorphic 5-coloring?

Presumably, the answer is “no.” To answer this question, one must first identify the ring , which in turn would identify appropriate generators to allow a computer-assisted proof of .

]]>Currently, our smallest 5-chromatic unit-distance graph has 803 vertices and 4144 edges. Much of the other progress has been in finding new 4- and 5-chromatic graphs, and hunting for features that make the chromatic number so large. I wanted to outline the role that algebraic structure appears to play in our progress to date:

Denote . Many of our constructions can be viewed as finite subsets of rings:

- , known as the Eisenstein integers and the hexagonal lattice, is 3-chromatic.
- contains 4-chromatic graphs such as the Moser spindle, Golomb’s graph (see this), and de Grey’s M. The entire ring appears to be 4-colorable by homomorphic coloring.
- contains de Grey’s original 5-chromatic graph N.
- contains 5-chromatic graphs due to Marijn and Tamás.
- contains a 5-chromatic graph due to Jannis.

Let denote the set of complex numbers of unit modulus. Instead of , several comments in the previous thread consider another ring generated by , where and . The inclusion of is helpful since it appears in :

Furthermore, by including the inverses of and in , these members are distinguished as units in the ring, making them easier to work with algebraically. In general, we want to keep track of which members of our ring have unit modulus, since these generate the edges in our Cayley graph.

**Question.** Given a subring of , how do we identify the members of ?

**Question.** Is it the case that ?

By design, any ring containing and a complex number of squared modulus will also contain . Tamás identified yet another member of , and the corresponding edges are critical to the Cayley graph being 5-chromatic.

Once has been determined, one may hunt for homomorphic colorings of .

**Question.** Does admit a homomorphic 5-coloring?

**Question.** Does there exist such that does not admit a homomorphic 5-coloring?

In theory, we could use the absence of homomorphic colorings to flag the possible nonexistence of a coloring, and then run a SAT solver on a 3-fold (say) Minkowski sum of an appropriate finite subset of to establish nonexistence. (Warning: This may lead to many false positives considering the entire plane is not homomorphically colorable.)

]]>What follows is a summary of the progress made thus far:

**Simplifying de Grey’s graph.** There have been three strides along these lines. One is to decrease the size of the graph by iteratively removing vertices that are not necessary to be 5-chromatic. This approach has led to our current record holder, which has 826 vertices and 4273 edges. Another approach has been to hunt for more symmetric 5-chromatic graphs, in the hopes that such graphs might be easier to analyze by hand (e.g., see this and that). Finally, others have been hunting for extremely small alternatives to de Grey’s L and M graphs that are amenable to by-hand analysis, in the hopes that these would eventually lead to a 5-chromatic graph (e.g., see this and that).

**Fundamental limits.** In 1998, Pritikin proved that every graph with at most 12 vertices is 4-colorable, and every graph with at most 6197 vertices is 6-colorable. Pritikin’s bounds are obtained by coloring the plane with k colors and an additional “wild” color such that points of unit distance are both allowed to receive the wild color. If the wild color occupies a small fraction p of the plane, then an exercise in the probabilistic method gives that any fixed unit-distance graph with n vertices enjoys an embedding in the plane that avoids the wild color (and is therefore k-colorable) provided n<1/p. Pritikin’s coloring for the k=4 case cannot be improved without improving on the densest known subset of the plane that avoids unit distances (originally due to Croft in 1967). See this MO thread for additional information. As such, improving the bound in the k=4 case might require a new technique, whereas the k=5 and 6 cases might be amenable to optimization.

**Larger k.** Can we use de Grey’s graph to construct unit-distance graphs that are not 5-colorable? To answer this question, we first need to understand how 5-colorings of de Grey’s graph force small collections of vertices to be colored. Varga and Nazgand provide some thoughts along these lines. Even if we can’t stitch together a 6-chromatic unit-distance graph with these ideas, we might be able to apply them to prove that the measurable chromatic number of the plane is at least 6.

**Vertices with algebraic structure.** It appears as though the coordinates of our smallest 5-chromatic graph lie in (see this). If we view the plane as the complex plane, what is the smallest ring that admits a 5-chromatic single-distance graph? Every single-distance graph in the Eistenstein integers and Gaussian integers is 2-chromatic (see this). David Speyer suggests looking at next.

The Hadwiger–Nelson problem is that of determining the chromatic number of the plane (CNP), defined as the minimum number of colours that must be assigned to the points of the plane in order to prevent any two points unit distance apart from being the same colour. It was first posed in 1950 and was reduced the following year to a finite graph problem; in particular, CNP is at least k if and only if there exists a finite graph of chromatic number k that can be drawn in the plane with each edge being a straight line of unit length. Such graphs are termed unit-distance graphs. It is easy to find 4-chromatic unit-distance graphs, the smallest being the 7-vertex Moser spindle.

In April 2018, Aubrey posted to the arXiv a report [deG2018] of a family of 5-chromatic unit-distance graphs. However, the smallest such graph that he discussed has 1581 vertices, and its lack of a 4-colouring requires checking for the nonexistence of particular types of 4-colourings of subgraphs of it that have almost 400 vertices, which requires a computer search.

This has led to interest in the search for simpler examples. Along these lines, our goals for this project are quite varied:

**Goal 1:** Find progressively smaller 5-chromatic unit-distance graphs. The current record is 1577, and recent insights promise to lead to further progress soon.

**Goal 2:** Reduce (ideally to zero) the reliance on computer assistance for the proof. Computer assistance was leveraged in [deG2018] to analyze a subgraph of size 397. This has so far been decreased to 278, though the corresponding full graph is actually larger than Aubrey’s original one.

**Goal 3:** Apply these simpler graphs to inform progress in related areas. For example:

- Find a 6-chromatic unit-distance graph.
- Improve the corresponding bound in higher dimensions.
- Improve the current record of 105/29 for the fractional chromatic number of the plane.
- Find the smallest unit-distance graph of a given minimum degree (excluding, in some natural way, boring cases like Cartesian products of a graph with a hypercube).

Following guidance from Terry Tao, we are setting a deadline for this project at Apr 15, 2019.

See this page for a newcomer’s guide to the initial approach of this project. In the short term, we are working to make further progress on Goals 1 and 2 by stitching together newly identified graphs that can substitute for the building-blocks L and M in [deG2018] and then passing them through shrinking routines of the sort described in section 5 of [deG2018].

To properly document progress on these goals, it might be helpful for participants to post links to files that contain coordinates of points in the plane and/or the resulting unit-distance graph. I launched a Dropbox folder that can serve as a repository of such files. Participants who wish to contribute a file to this folder should email me so I can grant editing privileges or post the file myself.

]]>The core idea of de Grey’s proof is to select a finite point set H in the plane and a coloring property P, and then demonstrate two incongruous statements:

(a) For every proper k-coloring of the plane, there exists a copy of H whose inherited coloring satisfies P.

(b) For every proper k-coloring of the plane, every copy of H inherits a coloring that doesn’t satisfy P.

Indeed, if both statements hold, then there is no proper k-coloring of the plane. To prove (a) and (b), we pass to finite unit-distance graphs in the plane. Explicitly, to prove (a), one finds a finite point set L such that for every proper k-coloring of L, there exists a copy of H in L whose inherited coloring satisfies P. Similarly, to prove (b), one finds a finite superset M of H such that every proper k-coloring of M forces the inherited coloring of H to not satisfy P.

For example, for k=3, we may take H to be two points of distance , and let P be the property that the vertices receive different colors. Then letting L be the vertices of a – – triangle, we see that for every proper 3-coloring of L, the two vertices of unit distance must receive different colors, and so one of these must have a different color from the third vertex that’s away. This copy of H necessarily satisfies P. Next, let M be the diamond graph of four vertices forming two equilateral triangles of unit side length. Here, the end points have distance , forming H, and any 3-coloring of M will force the points in H to have the same color, i.e., H’s inherited coloring fails to satisfy P.

Given L and M, one may produce a finite unit-distance graph that is not k-colorable. Indeed, extend each copy of H in L to a copy of M. Performing this operation to the L and M described above produces the Moser spindle.

To tackle k=4, de Grey takes H to be the 7 points in the hexagonal lattice of norm at most 1. Observe that the 6 points of norm 1 can be decomposed as the vertices of two triangles of side length . He takes P to be the property that one of these triples receives the same color. To prove (a), he constructs L by overlaying portions of 4 copies of the hexagonal lattice, resulting in a graph of 121 vertices. With this choice of L, the proof of (a) does not require computer assistance, though a computer makes the proof less mundane. To prove (b), he constructs M by aligning several copies of the Moser spindle, resulting in 1345-vertices. For this choice of M, the proof of (b) requires computer assistance.

As in the 3-colorable case, one may combine these graphs to produce a finite unit-distance graph that is not 4-colorable. In de Grey’s paper, the resulting graph N has 20425 vertices (presumably, there are many duplicated vertices here, considering L contains 52 copies of H, each of which are extended to a copy of M, i.e., 121+52*1345 = 70061 vertices with double-counting). de Grey then proceeds to “shrink” N by removing unnecessary vertices, or replacing point subsets with smaller ones that still preclude 4-colorability. These moves are necessarily computer assisted. The result is a 1581-vertex graph that can be independently verified as not 4-colorable with the help of a SAT solver (in fact, this point set is a superset of the 1577 points identified in the second update here).

Given this breakthrough, it is natural to wonder whether the techniques can be extended to prove that the plane is not 5-colorable. Considering the Moser spindle was a fundamental gadget in de Grey’s proof, it seems like an appropriate first step to hunt for the smallest 5-chromatic unit-distance graph (or even a 5-chromatic unit-distance graph whose lack of 4-colorability is more easily verified). de Grey has proposed a polymath project to this effect.

The last few days have seen a bit of activity in various directions along these lines, which I’ll summarize below:

- We found a modification to L that has 97 vertices and 40 copies of H. (The proof of (a) with this new L still doesn’t require a computer.) Aubrey de Grey has observed additional symmetries in this modification, suggesting that further reductions might be made.
- Tamás Hubai identified various simplifications that can be made to M. He also pointed out that we could let H be the vertices of an equilateral triangle of side length (which looks like a natural generalization of the k=3 proof above), though it would require more copies of M in the pre-shrunk graph N.
- Marijn Heule recast whether M forces H to not satisfy P as a SAT problem, and can apply DRAT-trim to produce a proof of unsatisfiability. The vertices that are inconsequential to this proof are removable. Randomly permuting clauses produces a different proof, suggesting a randomized approach to iteratively remove vertices. Apparently, this shrinks M considerably, but the final result is yet to come

On the side, I have been discussing with collaborators about other methods of simplification, which I summarize below:

- It would be nice if one could prove that a given graph is not 4-colorable by producing a Lovasz certificate, that is, a weighting of the vertices and a weighting of the edges such that the resulting weighted adjacency matrix with vertex weights on the diagonal is positive semidefinite with unit trace. If the sum of these matrix entries is larger than 4, then this would certify that the graph is not 4-colorable. Soledad Villar coded up a projected gradient descent solver to hunt for such a certificate (exploiting the sparsity of the weighted adjacency matrix), and so far, she reports the sum of the entries as 3.9. Time will tell if this approach brings fruit, but Fernando de Oliveira Filho points out that this will do no better than the Lovasz number of the entire plane’s unit-distance graph. Does this produce a new lower bound on the Lovasz number and fractional chromatic number of the plane?
- At the moment, it seems that the fundamental barrier to simplifying N is the construction of M that proves (b). It seems that de Grey’s approach in his paper was to start with H, then L, then M. As an alternative, one might consider searching through candidates for M to identify how subsets of points are forced to be colored, and then use that information to inform the choice of H and L. Boris Alexeev has indicated to me that this approach should be amenable to computer search.

Last night, I worked with Boris Alexeev to get a handle on de Grey’s final graph construction. He describes two collections of vertices and . Here, is a subset of , but for simplicity, we will take to be all of , which we will denote . This set is described as the orbit of 39 points under the action of the group generated by rotation by 60 degrees and reflection about the -axis (warning: the 5th point in de Grey’s writeup has a parenthesis mismatch). The resulting set has size 397 (there are duplicates to remove). Now we take two copies of (one is actually and the other is ), except we rotate the second copy by . This will produce an edge between and its rotated version (it will also produce 5 additional edges due to the rotational symmetry of ). The resulting set of points has size 793 (only the origin was double-counted). Now we shift all of these points to the right by to produce a set we’ll call . Take a second copy of and rotate by . This introduces an edge between and its rotated version (this corresponds to the edge between B and C in Figure 5 of de Grey’s paper). The resulting set of points has size 1585 (again, only the origin was double-counted). Note that this is 18 vertices more than reported by de Grey because our two copies of each have 9 additional vertices for simplicity.

For the sake of reproducibility, the following files might be useful:

- The 1585-vertex graph described above in DIMACS format
- A naive translation of 4-colorability of this graph into a SAT problem in DIMACS format
- The vertices of this graph in explicit Sage notation

Boris started running the above SAT instance last night using a few of popular SAT solvers, and the first to return UNSATISFIABLE was Glucose 4.0 (which is based on MiniSat), after 5.5 total hours on 16 cores (i.e., an m4.4xlarge instance on Amazon EC2), amounting to 20 minutes of clock time. It also takes less than a second to 5-color this graph.

I would be interested to find alternative certificates of this theorem. The Lovász number of a graph’s complement is famously sandwiched between the clique number and the chromatic number of the original graph. A short MATLAB script (running CVX) demonstrates that the Lovász number of the Moser spindle‘s complement is , forcing the chromatic number (integer) to be at least 4. (**EDIT:** My original MATLAB script neglected to take the graph complement.) In general, semidefinite programs are slow for large problem instances, but Lovász might be faster than SAT solvers in this setting. Furthermore, to certify that the 1585-vertex graph above is not 4-colorable, it suffices to find a dual-feasible point with value strictly larger than 4. Perhaps finding such a point is a reasonable pursuit. (This is not unlike the pursuit of fast certificates here and here.)

**UPDATE:** Aubrey suggested that we consider what happens when we remove the remaining 18 vertices. This amounts to removing 9 vertices from two of the copies of in the above construction. Instead, we performed a more conservative test by only removing 9 vertices from one of the copies of . However, we found that doing so resulted in a graph that is 4-colorable (and in fact, the coloring is identified in just a few seconds). Just to be sure, we also tried removing rotated versions of these 9 vertices in case our interpretation of was different from Aubrey’s intended construction, but every choice we made produced a graph that is (quickly) 4-colorable. As such, we cannot verify the 1567-vertex construction (or even a 1576-vertex construction). Regardless, Aubrey’s main result is apparently valid with the 1585-vertex construction described above.

**UPDATE:** After the previous update, we wondered which vertices could be removed from the 1585-vertex graph without the graph becoming 4-colorable. After an exhaustive search, we found that exactly 8 vertices are individually removable, namely, the vertices indexed by 3, 4, 23, 24, 948, 949, 952, and 953 in our files. Furthermore, we found that removing all 8 of these vertices produces a 1577-vertex graph that is also not 4-colorable. As such, this graph appears to be the unique minimal 5-chromatic induced subgraph of the 1585-vertex graph above. Therefore, for the proposed polymath project, the pursuit of a smaller unit-distance graph with chromatic number at least 5 will necessarily require a different approach than hunting for subgraphs of this graph.

**UPDATE:** To avoid any possibility of confusion, I wanted to point out that of the files I linked to above, our methodology first generated the file that contains exact coordinates of the vertices. (We derived this from an interpretation of Aubrey’s construction, extrapolating from the explicit coordinates that he lists at the end of his paper.) Next, we identified likely edges by numerical approximation. Finally, we used exact arithmetic to check that each of these likely edges honestly corresponds to unit distance. This last step took about an hour of running Sage on a standard laptop.

What follows is the list of speakers and links to their slides. (I anticipate referencing these slides quite a bit in the near future.) Thanks to all who participated!

Henry Cohn — Why are packing problems much easier in some cases than others?

Oleg R. Musin — Densest sphere packing in four dimensions

Kasso Okoudjou — New results in minimizing the p-frame potentials

Wei-Hsuan Yu — New bounds for equiangular lines and spherical two-distance sets

Mark Magsino — Constructing Tight Gabor Frames Using CAZAC Sequences

Steven T. Flammia — Zauner’s Conjecture and Algebraic Number Theory

Marcus Appleby — A number-theoretic technique for constructing exact sets of complex equiangular lines

Blake C. Stacey — Symmetric Informationally Complete quantum measurements: Where sphere packing meets quantum information

Desai Cheng — The solution to the frame Quantum Detection Problem

Matthew Fickus — Equiangular tight frames and combinatorial designs

Hanmeng Zhan — Graph covers and equiangular frames

James P. Solazzo — Complex Two-Graphs

John I. Haas — Optimally packed fusion frames via symmetric and quasi-symmetric block designs

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