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The chromatic number of the plane is at least 5

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We present a family of finite unit-distance graphs in the plane that are not 4-colourable, thereby improving the lower bound of the Hadwiger-Nelson problem. The smallest such graph that we have so far discovered has 1567 vertices.
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THE CHROMATIC NUMBER OF THE PLANE IS AT LEAST 5
AUBREY D.N.J. DE GREY
Abstract. We present a family of finite unit-distance graphs in the plane
that are not 4-colourable, thereby improving the lower bound of the Hadwiger-
Nelson problem. The smallest such graph that we have so far discovered has
1581 vertices.
.
1. Introduction
How many colours are needed to colour the plane so that no two points at
distance exactly 1 from each other are the same colour? This quantity, termed the
chromatic number of the plane or CNP, was first discussed (though not in print) by
Nelson in 1950 (see [Soi]). Since that year it has been known that at least four and
at most seven colours are needed. The lower bound was also noted by Nelson (see
[Soi]) and arises because there exist 4-chromatic finite graphs that can be drawn in
the plane with each edge being a straight line of unit length, the smallest of which
is the 7-vertex Moser spindle [MM] (see Figure 7, left panel). The upper bound
arises because, as first observed by Isbell also in 1950 (see [Soi]), congruent regular
hexagons tiling the plane can be assigned seven colours in a pattern that separates
all same-coloured pairs of tiles by more than their diameter. The question of the
chromatic number of the plane is termed the Hadwiger-Nelson problem, because
of the contributions of Nelson just mentioned and because the 7-colouring of the
hexagonal tiling was first discussed (though in another context) by Hadwiger in 1945
[Had]. The rich history of this problem and related ones is wonderfully documented
in [Soi]. Since 1950, no improvement has been made to either bound.
We now give a brief overview of our construction. The graphs described in this
report are obtained as follows:
(1) We note that the 7-vertex, 12-edge unit-distance graph Hconsisting of the
centre and vertices of a regular hexagon of side-length 1 can be coloured
with at most four colours in four essentally distinct ways (that is, up to ro-
tation, reflection and colour transposition). Two of these colourings contain
a monochromatic triple of vertices and two do not.
(2) We construct a unit-distance graph Lthat contains 52 copies of Hand show
that, in all 4-colourings of L, at least one copy of Hcontains a monochro-
matic triple.
(3) We construct a unit-distance graph Mthat contains a copy of Hand show
that there is no 4-colouring of Min which that Hcontains a monochromatic
triple. Thus, the unit-distance graph Ncreated by arranging 52 copies of
Mso that their counterparts of Hform a copy of Lis not 4-colourable.
This completes our demonstration that the CNP is at least 5.
Date: April 7, 2018.
1
arXiv:1804.02385v2 [math.CO] 11 Apr 2018
2 AUBREY D.N.J. DE GREY
Figure 1. The essentially distinct ways to colour Hwith at most
four colours.
(4) We identify smaller non-4-colourable unit-distance graphs, first by identi-
fying vertices in Nwhose deletion preserves the absence of a 4-colouring,
and then by more elaborate methods.
2. The 4-colourings of H
Figure 1 shows the four essentially distinct (i.e., up to rotation, reflection and
colour transposition) colourings of Hwith at most four colours. The top two
colourings possess, and the bottom two lack, a triple of vertices all the same colour.
3. Construction and colourings of L
3.1. A 31-vertex graph Jassembled from 13 copies of H.The graph J,
shown in Figure 2, contains a copy of Hin its centre, six copies centred at distance
1 from its centre, and six copies centred at distance 3 from its centre.
3.2. The 4-colourings of Jin which no copy of Hcontains a monochro-
matic triple. Figure 3 shows six essentially distinct such colourings up to rotation,
reflection, colour transposition and the colours of the vertices coloured black in Fig-
ure 3 (which will not concern us hereafter). That these are the only such colourings
can be checked by grouping the possibilities according to whether the central copy
of Hhas two (the top row in Figure 3) or no (bottom row) monochromatic pairs of
vertices at distance 3 and, in the case where it has none, whether all the copies
of Hcentred at distance 1 from the centre also have none (bottom left colouring)
THE CHROMATIC NUMBER OF THE PLANE IS AT LEAST 5 3
Figure 2. The graph J, containing 31 vertices and 13 copies of H.
or some have two (last two colourings). These last two colourings are the rea-
son we need the vertices coloured black; though those vertices can be coloured in
many ways, it turns out that if they were deleted then there would be additional
4-colourings of the remaining graph in which none of the seven remaining copies of
Hcontained a monochromatic triple, and some of those new colourings lack a key
property shared by all those in Figure 3, which we now discuss.
The aspect of the colourings in Figure 3 that will be our focus below is that
they feature only three essentially distinct colourings of the vertices at distance 2
from the centre, which we shall hereafter term the linking vertices. Either (a) the
linking vertices are all the same colour as the centre (left-hand cases in Figure 3),
or (b) four consecutive ones (when enumerated going clockwise around the centre)
are the same colour as the centre and the other two are a second colour (middle
cases), or (c) two opposite ones are the same colour as the centre and all the other
four are a second colour (right-hand cases). Hereafter we refer to an opposite pair
of linking vertices as a linking diagonal.
4 AUBREY D.N.J. DE GREY
Figure 3. The essentially distinct colourings of Jin which no
copy of Hcontains a monochromatic triple.
3.3. A 61-vertex graph Kassembled from two copies of J.We now construct
Kby arranging two copies of Jwith their centres coincident and with their relative
rotation chosen so that corresponding linking vertices are at distance 1 from each
other; see Figure 4.
We note that, in any 4-colouring of Kin which none of the 26 copies of Hcontains
a monochromatic triple, both copies of Jmust have their linking vertices coloured
according to option (c) above. Further, in option (c) each of the three linking
diagonals of Jis monochromatic. Thus, in all 4-colourings of Kwhere no copy of
Hcontains a monochromatic triple, all six linking diagonals are monochromatic.
3.4. A 121-vertex graph Lassembled from two copies of K.Finally, we
construct Lby arranging two copies of Kwith their counterparts of one linking
vertex coincident and their counterparts of the opposite linking vertex distance 1
apart (see Figure 5). These linking vertices are labelled.
The property noted above for Ktrivially implies that in no 4-colouring of Ldo
all of its 52 constituent copies of Hlack a monochromatic triple.
4. Construction and colourings of M
In seeking graphs that can serve as Min our construction, we focus on graphs
that contain a high density of Moser spindles. The motivation for exploring such
graphs is that a spindle contains two pairs of vertices distance 3 apart, and these
pairs cannot both be monochromatic. Intuitively, therefore, a graph containing a
high density of interlocking spindles might be constrained to have its monochro-
matic 3-apart vertex pairs distributed rather uniformly (in some sense) in any
THE CHROMATIC NUMBER OF THE PLANE IS AT LEAST 5 5
Figure 4. The graph K, containing 61 vertices and 26 copies of H.
4-colouring. Since such graphs typically also contain regular hexagons of side-
length 1, one might be optimistic that they could contain some such hexagon that
does not contain a monochromatic triple in any 4-colouring of the overall graph,
since such a triple is always an equilateral triangle of edge 3 and thus consti-
tutes a locally high density, i.e. a departure from the aforementioned uniformity,
of monochromatic 3-apart vertex pairs.
4.1. Graphs with high edge density and spindle density. In seeking graphs
with high spindle density, we begin by noting two attractive features of the 9-vertex
unit-distance graph Tthat is obtained by adding two particular vertices to a spindle
(see Figure 6, left). These vertices, Pand Q:
(1) form an equilateral triangle with the tip Xof the spindle
(2) lie on (the extension of ) the line forming the base Y Z of the spindle
6 AUBREY D.N.J. DE GREY
Figure 5. The graph L, containing 121 vertices and 52 copies of H.
We can thus construct a 15-vertex unit-distance graph U(Figure 6, right) that
contains three spindles and possesses rotational and reflectional triangular symme-
try. This symmetry suggests that graphs formed by combining translations and
60-degree rotations of Umight have particulatly high edge and spindle density.
4.2. Construction of a graph that serves as M.The speculation just men-
tioned turns out to be true; for example, we have found a 97-vertex graph containing
78 spindles (not shown). Therefore, a custom program (outlined in the next section)
was written to test graphs of this form for possession of a 4-colouring in which the
central Hcontains a monochromatic triple. Disappointingly, we could not find a
graph of this form that enforces sufficient uniformity of the distribution of 3-apart
monochromatic vertex pairs to deliver the property we require for M, even though
we checked examples with well over 1000 vertices.
However, this approach can be extended. Graphs arising from the construction
described thus far can have spindles in only six different orientations, with edges
falling into just three equivalence classes based on triangular symmetry; see Figure 7
(left). A natural elaboration is to add such classes based on the relative orientations
of spindles that share a lot of vertices, such as in Figure 7 (middle). The maximum
possible degree of a vertex in a graph constructed from these vectors is thereby
increased from 18 to 30; see Figure 7 (right), which we denote as graph V.
THE CHROMATIC NUMBER OF THE PLANE IS AT LEAST 5 7
Figure 6. The graphs Tand U.
Figure 7. The vector classes present in one (left), or three tightly
linked (middle), Moser spindles; the graph V(right).
Happily, this turns out to suffice. The 301-vertex graph Wshown in Figure
8 (left) is built by centring copies of Vat all vertices of Vand then deleting all
vertices that lie further than 3 from its centre. The 1345-vertex graph shown in
Figure 8 (right) is then built by centring copies of Wat all seven vertices of H.
Our program did not find any 4-colouring of this latter graph in which the central
Hcontains a monochromatic triple, so it can serve as our M. Arranging 52 copies
of it so that their counterparts of Hform a copy of Lgives a 20425-vertex unit-
distance graph Nwhich, therefore, is not 4-colourable. (We do not show a picture
of N, because it is visually impenetrable and we have anyway discovered smaller
examples, discussed below. Even Wand Mare too large for their pictures to be
explanatory; we include them just in case it is interesting to see the general shape.)
4.3. Testing 4-colourability of edge-dense, spindle-dense graphs. A vertex
count of 1345 exceeds the size of general graphs whose chromatic number can be
determined computationally in reasonable time by simplistic search, and a count
of 20425 may not be tractable even with state-of-the-art algorithms and hardware.
It was thus necessary to develop a custom program to test graphs for possession of
the property required for our graph M.
8 AUBREY D.N.J. DE GREY
Figure 8. The graphs W(left) and M(right).
Because we are only asking whether a specific number of colours is or is not
sufficient, and also because of the high density of edges and spindles in our tar-
get graphs, the required test turns out to be computationally far cheaper than a
general determination of chromatic number of comparable-sized graphs. It can be
performed rapidly by a simple depth-first search optimised only slightly, as follows:
(1) Since our question is whether there is any 4-colouring of Min which the
central Hcontains a monochromatic triple, we begin by specifying the
colourings of the vertices of that central H, which we term the initialising
vertices. Since Mhas the same symmetry as H, we need only check the
two essentially distinct triple-containing colourings of those vertices.
(2) We order the remaining vertices according to a hierarchy (most significant
first) of parameters (all decreasing): how many spindles they are part of,
their degree, and how many unit triangles they are part of.
(3) We colour the next not-yet-coloured vertex (initially vertex 8) with the first
colour that we have not already tried for it (initially colour 1).
(4) We check each not-yet-coloured neighbour (if any) of the just-coloured ver-
tex to see how many colours are still permissible for it. If any such vertex
already has neighbours of all four colours, we will need to backtrack (see
step 6 below). If any has neighbours of exactly three colours, we assign the
remaining (forced) colour to it.
(5) If we do not need to backtrack, but we did colour some vertices in step 4,
we repeat step 4 for each such vertex.
(6) If we need to backtrack, we uncolour everything that we just coloured in
the most recent iteration of steps 3 and (any resulting iterations of) 4.
(7) If we just did an uncolouring and the just-uncoloured vertex that was
coloured at step 3 has no colours that have not yet been tried, we repeat
step 6 for the next-most-recent iteration of steps 3 and 4 unless we have
already backtracked all the way to the vertices of H. Otherwise, if there
are still some uncoloured vertices we return to step 3.
(8) We terminate when we get here, i.e. when either all vertices are coloured
or we have backtracked all the way down to the vertices of H.
THE CHROMATIC NUMBER OF THE PLANE IS AT LEAST 5 9
This algorithm was implemented in Mathematica 11 on a standard MacBook
Air and terminated in only a few minutes for our candidate M, without finding a
4-colouring starting from either of the triple-containing colourings of its central H.
Essentially, the speed arises because the fixing of only 20 or so colours at step 3
typically lets almost all remaining colours be forced at step 4.
5. Identification of smaller solutions
One may naturally wonder how large the smallest non-4-colourable unit-distance
graph is. Moreover, if substantially smaller graphsare found, they may be verifiable
independently by standard algorithms such as SAT solvers. Accordingly, in this
section we outline some promising lines of attack for identifying non-4-colourable
unit-distance graphs with fewer than 20425 vertices.
The most direct way to find such graphs is to seek a succession of small simplifi-
cations of N. Although our colouring program is extremely fast for graphs like M,
unfortunately it is not fast enough to be practical for the more sparsely-interlinked
combinations of Mthat constitute N. Thus, we now list a selection of approaches
to shrinking Nthat are subtler than simply deleting individual vertices from it and
seeking a 4-colouring.
5.1. Shrinking M.Most simply, Ncan potentially be shrunk by identifying ver-
tices in Mwhose deletion does not introduce any 4-colourings in which its central
copy of Hcontains a monochromatic triple.
5.2. Shrinking S.Define Sto be the graph formed by arranging 13 copies of M
so that their central copies of Hform a copy of J. The property of Jthat we use
in building Lrefers only to its linking vertices (and its central vertex), not to the
copies of Hfrom which Jis built. Thus, we can delete vertices from Sif no new
4-colouring of the remaining graph arises that incorporates a new colouring of its
linking vertices, even if some such new colourings do give some constituent copy of
Ha monochromatic triple. Since Nis built from four copies of S, this translates
into a potential shrinking of N.
5.3. Relaxing the constraint on Jand thus on S.Further, it is not strictly
required that the linking vertices of J(hence of S) conform to one of the three
colouring schemes from Figure 3. In order to ensure that all the linking diagonals
of Kare monochromatic, it suffices that any of the following hold for Jand S:
(1) At least five of the linking vertices are the same colour as the centre
(2) Exactly four are, and they do not form a rectangle
(3) Exactly two are, and all three linking diagonals are monochromatic
5.4. Relaxing the construction of N.Nis constructed by joining four copies
of Sin a chain, so those four copies fall into two pairs with potentially different
constraints. Let Nbe built from two copies each of Sa and Sb, such that the
vertices labelled in Figure 5 are linking diagonals of the copies of Sa rather than
Sb. Two relaxations then arise:
(1) Only one, not all three, of the linking diagonals of S a need be monochro-
matic (though we must specify which one);
10 AUBREY D.N.J. DE GREY
(2) The possible 4-colourings of the linking vertices of Sa and S b must be
mutually incompatible when the specified linking diagonal of Sa is not
monochromatic, but that is all. Thus, for example, it is enough that in
Sa either the specified linking diagonal is monochromatic or at least one
vertex of each linking diagonal is the same colour as the centre, so long as
in Sb at least one linking diagonal has both vertices the same colour as the
centre. Both of these constraints are weaker than that given above for S.
5.5. Adding new vertices that enable the deletion of others. S(whether
generically or as Sa and Sb) contains many sets Xof four or more vertices that lie
on a circle of unit radius whose centre is not a vertex of S. Suppose that (perhaps
after shrinking it) the deletion of any vertex of Sallows the remaining graph to be
4-coloured, but that for some such set Xthere is a set Yof at least two vertices
of Swhose simultaneous deletion allows the remaining graph to be 4-coloured but
only in ways that assign all four colours to members of X. Then S(hence N) can
be shrunk by adding a single vertex joined to all the members of X(and located
at the centre of the circle on which they lie) and deleting all members of Y. In
principle this approach can of course also be applied directly to Nitself.
6. Status of shrinking N
We have so far shrunk Nby a factor of nearly 13, our current record being the
1581-vertex graph Gthat (again more for artistic than expository reasons) is shown
in Figure 9. It was verified using the same program described above, applied to its
constituent Sa and Sb with the permitted 4-colourings of linking vertices that were
given in section 5.4(2) above. Sb retains the same symmetry as Hso the initialising
vertices and colourings used for it were as before; Sa has only reflectional symmetry
about the x-axis so we used (0,0), (-2,0) and (2,0) with colourings 1,1,2 or 1,2,1 or
1,2,3. To construct it, we create Sb by placing vertices at
(0,0),(1/3,0),(1,0),(2,0),((33 3)/6,0),(1/2,1/12),(1,1/3),(3/2,3/2),
(7/6,11/6),(1/6,(12 11)/6),(5/6,(12 11)/6),
(2/3,(11 3)/6),(2/3,(3311)/6),(33/6,1/12),
((33 + 3)/6,1/3),((33 + 1)/6,(3311)/6),((33 1)/6,(3311)/6),
((33 + 1)/6,(11 3)/6),((33 1)/6,(11 3)/6),
((33 2)/6,(2311)/6),((33 4)/6,(2311)/6),
((33 + 13)/12,(11 3)/12),((33 + 11)/12,(3 + 11)/12),
((33 + 9)/12,(11 3)/4),((33 + 9)/12,(33 + 11)/12),
((33 + 7)/12,(3 + 11)/12),((33 + 7)/12,(3311)/12),
((33 + 5)/12,(5311)/12),((33 + 5)/12,(11 3)/12),
((33 + 3)/12,(311 53)/12),((33 + 3)/12,(3 + 11)/12),
((33 + 3)/12,(3311)/12),((33 + 1)/12,(11 3)/12),
((33 1)/12,(3311)/12),((33 3)/12,(11 3)/12),
((15 33)/12,(11 3)/4),((15 33)/12,(73311)/12),
((13 33)/12,(3311)/12),((11 33)/12,(11 3)/12)
THE CHROMATIC NUMBER OF THE PLANE IS AT LEAST 5 11
Figure 9. The 1581-vertex, non-4-colourable unit-distance graph G.
and at all points obtained by rotating those vectors around the origin by multiples
of 60 degrees and/or by negating their y-coordinates, giving a total of 397 vertices.
Sa is then created by deleting the vertices (1/3,0) and (1/3,0) from Sb, and the
full graph is assembled from two copies each of Sa and S b as before, using (-2,0)
and (2,0) as the specified linking diagonal of Sa. Happily, Ghas turned out to be
within the reach of standard SAT solvers, with which others have now confirmed its
chromatic number to be 5 without the need to resort to checking weaker properties
of subgraphs.
This attempt to identify smaller examples has thus far been rather cursory, so
we think it highly likely that smaller examples than Gexist.
7. Acknowledgements
I am indebted to Boris Alexeev, Rob Hochberg, Brendan McKay, Dustin Mixon,
Paul Phillips, Landon Rabern and Gordon Royle for testing various graphs with
independent code. I also thank Imre Leader and Graham Brightwell for comments
on earlier versions of the manuscript, as well as for their role in germinating my
interest in graph theory fully thirty years ago.
12 AUBREY D.N.J. DE GREY
References
1. H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae
Math. 4(1945), 238–242.
2. L. Moser and M. Moser, Solution to Problem 10, Can. Math. Bull. 4(1961), 187–189.
3. A. Soifer, The Mathematical Coloring Book, Springer, 2008, ISBN-13: 978-0387746401.
SENS Research Foundation, Mountain View, California 94041, USA; email aubrey@sens.org
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... The upper bound is a classical coloring of a regular hexagon tiling due to Isbell. The lower bound were obtained by de Grey [4] in 2018, breaking a 70 year-old record set. (another constructions are contained in [7,6,17,13,14]). ...
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We prove that for an arbitrary $\varepsilon > 0$ holds \[ \chi (\mathbb{R}^3 \times [0,\varepsilon]^6) \geq 10, \] where $\chi(M)$ stands for the chromatic number of an (infinite) graph with the vertex set $M$ and the edge set consists of pairs of monochromatic points at the distance 1 apart.
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Full-text available
We prove that for an arbitrary $\varepsilon > 0$ holds \[ \chi (\mathbb{R}^3 \times [0,\varepsilon]^6) \geq 10, \] where $\chi(M)$ stands for the chromatic number of an (infinite) graph with the vertex set $M$ and the edge set consists of pairs of monochromatic points at the distance 1 apart.
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Logic Artificial Intelligence (AI) is a subfield of AI where variables can take two defined arguments, True or False, and are arranged in clauses that follow the rules of formal logic. Several problems that span from physical systems to mathematical conjectures can be encoded into these clauses and solved by checking their satisfiability (SAT). In contrast to machine learning approaches where the results can be approximations or local minima, Logic AI delivers formal and mathematically exact solutions to those problems. In this work, we propose the use of logic AI for the design of optical quantum experiments. We show how to map into a SAT problem the experimental preparation of an arbitrary quantum state and propose a logic-based algorithm, called Klaus, to find an interpretable representation of the photonic setup that generates it. We compare the performance of Klaus with the state-of-the-art algorithm for this purpose based on continuous optimization. We also combine both logic and numeric strategies to find that the use of logic AI significantly improves the resolution of this problem, paving the path to developing more formal-based approaches in the context of quantum physics experiments.
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We prove that every finite colouring of the plane contains a monochromatic pair of points at an odd distance from each other.
Uberdeckung des euklidischen Raum durch kongruente Mengen
  • H Hadwiger
H. Hadwiger, Uberdeckung des euklidischen Raum durch kongruente Mengen, Portugaliae Math. 4 (1945), 238-242.