Fractional and jFold Coloring of the Plane
Discrete Comput Geom
Fractional and j Fold Coloring of the Plane
Jarosław Grytczuk 0 1
Konstanty JunoszaSzaniawski 0 1
Joanna Sokół 0 1
Krzysztof We˛sek 0 1
Jarosław Grytczuk 0 1
Editor in Charge: János Pach
0 Faculty of Mathematics and Information Science, Warsaw University of Technology , Warsaw , Poland
1 Faculty of Mathematics and Computer Science, Jagiellonian University , Kraków , Poland
We present results referring to the HadwigerNelson problem which asks for the minimum number of colors needed to color the plane with no two points at distance 1 having the same color. Exoo considered a more general problem concerning graphs G[a,b] with R2 as the vertex set and two vertices adjacent if their distance is in the interval [a, b]. Exoo conjectured χ (G[a,b]) = 7 for sufficiently small but positive difference between a and b. We partially answer this conjecture by proving that χ (G[a,b]) 5 for b > a. A j fold coloring of a graph G = (V , E ) is an assignment of j elemental sets of colors to the vertices of G, in such a way that the sets assigned to any two adjacent vertices are disjoint. The fractional chromatic number χ f (G) is the infimum of fractions k/j for j fold coloring of G using k colors. We tgheenefrraaclitzioenaamlceothloordinbgyoHf ogcrahpbhesrgGa[nad,bO], 'oDbotaninneinlg(wahboopurnodveddeptheantdGen[t1,o1n] ab . We also 4.36) for present few specific and two general methods for j fold coloring of G[a,b] for small j , in particular for G[1,1] and G[1,2]. The j fold coloring for small j has strong practical

motivation especially in scheduling theory, while graph G[
1,2
] is often used to model
hidden conflicts in radio networks.
Mathematics Subject Classification
05C15 · 05C10 · 05C62
1 Introduction
1.1 Mathematical Context
The famous Hadwiger–Nelson problem asks for the minimum number of colors
required to color the Euclidean plane R2 in such a way that no two points at
distance 1 from each other have the same color. The question can be equivalently stated
in the graph theory language: Recall that a coloring of a graph G = (V , E ) is a
function c : V → K (where K is a set of colors) such that every two adjacent vertices
x , y satisfy c(x ) = c(y). The chromatic number χ (G) of G is the minimal cardinality
of the set of colors to color G. Therefore, the Hadwiger–Nelson problem is about
determining the chromatic number of the graph on the set of vertices R2 with vertices
in distance one adjacent—it is called unit distance graph and in this article will be
denoted by G[
1,1
] (the notation will be justified shortly).
The problem was originally proposed by Edward Nelson in 1950 and was made
publicly known by Hugo Hadwiger [
4
]. Pioneers on the topic observed the following
bounds: Nelson first showed that at least 4 colors are needed (see the proof by Moser
and Moser [
7
] which uses the socalled Moser spindle) and John Isbell was first to
prove that 7 colors are enough (this result was published by Hadwiger [
4
]). Somehow
surprisingly, the aforementioned bounds remain unchanged since their origin in 1950s,
i.e. for more than 60 years nobody has found anything sharper than 4 χ (G) 7—as
long as we consider the full generality. Nevertheless, advanced studies on the question,
its subproblems and other related topics provide some understanding. For example,
if we consider only measurable colorings (i.e. with measurable colors) then at least 5
colors are necessary [
2
] and if coloring of the plane consists of regions bounded by
Jordan curves then at least 6 colors are required [
12
]. Generally, across the decades
the Hadwiger–Nelson problem inspired many interesting results in the touchpoint
of combinatorics and geometry, a vast number of challenging problems and various
applications [
10
].
One of the possible ways to generalize the first question was presented by Exoo [
1
]
(see also [
6
]). He considered graphs on the set of vertices R2 with vertices in distance
in the interval [a, b], we denote such graphs by G[a,b]. How many colors would be
enough to color such a graph, depending on a and b? Are there some important ranges
of those parameters?
The other path for research leading from the Hadwiger–Nelson problem concerns
different models of graphs coloring. In the majority of this article, we investigate
fractional colorings, in some sense a generalization of the classic coloring: every
vertex gets a j elemental set of colors from the set of colors (of size k) and the sets for
adjacent vertices have to be disjoint. The ‘quality’ of such a coloring is measured by
the fraction k/j and the fractional chromatic number is the infimum of such fractions
(one can also ask for the best fraction with a fixed j ). It can be seen as in this case every
single color is just a 1/j size part of a ‘complete color’, so we divide ‘complete colors’
from the classic coloring to somehow save a bit by combining partitioned colors in
a clever way. It turns out that in fact we can save much using fractional coloring:
fractional chromatic number is always lower or equal to chromatic number of a given
graph, but the difference can be arbitrarily large.
The fractional chromatic number of the graph G[
1,1
] has been studied in the
literature. The best upper bound is due to Hochbeg and O’Donnell [5] (based on an idea
by Fisher and Ullman [
3
] of looking for a dense subset of the plane which avoids
unit distance) and the best lower bound (using a finite subgraph of the plane) can be
found in the book of Scheinerman and Ullman [
9
] (along with a good explanation of
the upper bound): 3.555 χ f (G[
1,1
]) 4.36. Note that the upper bound is much
smaller than the upper bound for χ (G[
1,1
]).
1.2 Practical Motivation
Coloring of such geometrical graphs has also some significant practical motivations
in telecommunication. We will briefly describe an example. More on this topic can be
found in the paper of Walczak and Wojciechowski [
11
].
Consider the following problem: We are given a set of transmitters with equal ranges
placed in some area (assume that ranges are equal to 1)—some of them are in each
other range, and some of them are not. If two transmitters are in each other’s range,
we assume that they can ‘quickly’ agree on their communication (there are algorithms
for it). If two transmitters are not in each other’s range but have a common neighbor
C , then it is possible that A and B would try to transmit to C in the same time—in
this case C cannot listen to both messages. Hence we have a conflict which cannot be
solved by a direct communication between A and B. If two transmitters A and B are
not in each other’s range and do not have a common neighbor, then there is no conflict.
The problem is to assign timeslots for transmitters (in an equitable way) such that no
two conflicted transmitters share a time interval.
How can we use graph coloring in this problem? We can create the graph G of
conflicts for this network of transmitters: vertices correspond to transmitters and two
vertices are adjacent if the corresponding transmitters X , Y are at distance greater than
1 and have a common transmitter in their respective ranges (note that it is possible
only if X and Y are at distance in the set (1, 2)). Hence, using a kcoloring of G, such
that each color corresponds to one timeslot is one of the possible ways of constructing
a proper transmissionschedule. By the definition of coloring, the produced schedule
does not contain any pair of conflicted transmitters sharing their time of transmission
and every transmitter gets the same amount of time in one cycle of transmission.
The length of the schedule is k. On the other hand, we can make use of a fractional
coloring of G. Since we demand that in one cycle of transmission every transmitter
gets a unit of time, then every color in a fractional coloring corresponds to an interval
of length 1/j of the time unit. This model guarantees avoiding of conflicts as well as
equability of the schedule (if we used k colors then the length of the schedule is k/j ).
The relation between considered coloring models implies that fractional coloring can
possibly produce a shorter schedule in comparison to classic coloring, which (in real
world) means: we can save a bit of time. However, the technical constraints imply that
too big values of j are not accepted—too fragmented schedule is not practical. This
issue plays an important role in our work—we devote a part of this article to fractional
colorings with ‘small j ’.
Now, consider a graph on the set of vertices R2 with vertices at distance in the set
[
1, 2
] adjacent, which will be denoted in this article as G[
1,2
]. Clearly, every conflict
graph of a transmitters network as described above is a subgraph of G[
1,2
]. Therefore,
any coloring or fractional coloring of infinite G[
1,2
] induce, respectively, a coloring or
a fractional coloring of a given finite conflict graph. Hence, we get a universal scheme
for any network. Additionally, this universal scheme works also if the transmitters are
placed on moving vehicles (thus changing the conflict graph). This is an indisputable
advantage over using standard coloring algorithms for a given conflict graph.
2 Preliminaries
First, we need to formally define fractional and classic coloring of graphs.
Definition 1 A coloring of a graph G = (V , E ) with k colors (or kcoloring) is
an assignment of colors {1, 2, . . . , k} to the vertices of G such that no two adjacent
vertices have the same color. The smallest number of colors needed to color a graph
G is called chromatic number and denoted by χ (G).
Definition 2 Let Pj (k) be a family of all j elemental subsets of {1, 2, . . . , k}. A j
fold coloring of graph G = (V , E ) is an assignment of j element sets of colors to the
vertices of G, i.e. f : V −→ Pj (k), such that for any two adjacent vertices v, w ∈ V
we have f (v) ∩ f (w) = ∅. The smallest number of colors k needed for a j fold
coloring of a graph G is called the j fold chromatic number and denoted by χ j (G).
The fractional chromatic number is defined to be
χ f (G) := ji∈nNf
χ j (G)
j
= jl→im∞
χ j (G)
j
.
Since a 1fold coloring of a graph is just a classic coloring χ f (G) χ (G).
Now, we will present two possible, equivalent notions of graphs on the Euclidean
plane we are considering in this work. The first is due to Exoo [
1
]. The second is
introduced in this paper and is more convenient for our work, except Sect. 3.
Definition 3 An εunit distance graph, denoted by Gε is a graph whose vertices are
all points of the plane, in which two points are adjacent if their distance d satisfies
1 − ε d 1 + ε, i.e.
Gε =
R2, {{x , y} ⊂ R2  1 − ε
Definition 4 A graph G[a,b] is a graph whose vertices are all points of the plane
V = R2, in which two points are adjacent if their distance d satisfies a d b.
G[a,b] =
R2, {{x , y} ⊂ R2  a
G[1,b] since G[a,b] =∼ G[1,b/a].
Note that every Gε graph can be defined as G[1−ε,1+ε] and every G[a,b] graph can
b−a . Additionally, it is enough to consider graphs
be defined as Gε graph with ε = b+a
3 Coloring of G[a,b]
The classic Hadwiger–Nelson problem is considered to be very difficult, in particular
giving a better general lower bound than 4. In his article Exoo [
1
] looked for values of
ε such that we can determine the chromatic number of Gε and he succeed in finding
some such values. He proved that for 0.134756 . . . < ε < 0.138998 . . . the exact
value of χ (Gε) is 7, and for ε > 0.008533 . . . we get χ (Gε) 5.
His work (including computational experiments) suggested that for small enough
ε the exact value of the chromatic number of Gε is 7.
Conjecture 1 [
1
] For any ε > 0 we have χ (Gε) = 7.
We will give a partial answer to this conjecture. It appears that the pure positivity
of ε allow us to establish a lower bound of 5, strictly better than 4. We will use a result
by Nielsen [
8
] and for that we need some additional notions.
Given a coloring of the plane F , a triangle T = x yz is a monochromatic limit
triangle if there is a monochromatic set {x1, y1, z1, x2, y2, z2, . . .} such that xn −→ x ,
yn −→ y, zn −→ z and each of the triangles Tn = xn yn zn is similar to T .
We will say that triangles x yz and x y z are εclose if x − x < ε, y − y < ε
and z − z < ε.
Theorem 1 [
8
] Let F be a twocoloring of the plane and let T be a triangle. Then F
admits a monochromatic limit triangle congruent to T .
Theorem 2 For any ε > 0 we have χ (Gε)
5.
Proof Let ε > 0 and suppose that χ (Gε) 4. Let F be a 4coloring of Gε and let
F be a 2coloring of the plane such that each point of color 1 or 2 in F is white in
F and each point of color 3 or 4 in F is black in F . Let T = x yz be an equilateral
triangle with a side length 1.
From Theorem 1 F admits a monochromatic limit triangle congruent to T (lets say
it is white). So from the definition of monochromatic limit triangle there exists n such
that the triangle Tn = xn yn zn is 2ε close to T . The side lengths of Tn are within the
interval [1 − ε, 1 + ε], hence the sides of Tn are edges of Gε. Since all vertices of Tn are
white in F then each of them is colored with 1 or 2 in F . Since there are three vertices
and two colors there is a monochromatic edge, which contradicts the assumption that
F is a 4coloring of Gε.
We believe that there is still room to prove a better bound.
The best known way to find an upper bound for the fractional chromatic number of
G[
1,1
] is by looking for a set in the plane with highest possible density but without
vertices in distance equal to 1. The densest such set was presented by Hochbeg and
O’Donnell [5]. They presented a proof of the upper bound by a limit argument. In
the following theorem we generalize their construction for G[1,b]. We give a complex
description of the method for any b: not only the upper bound, but also an explicit
specification of a sequence of fractional colorings “converging” to the upper bound.
This sequence give a finite fractional coloring with “quality” as close to the upper
bound as needed.
Theorem 3 If b ≥ 1 then χ f (G[1,b]) √33 · b+√x1−x2 where x is the root of
bx = π6 − arc sin(x ). Moreover there exists a sequence of ( 2(bn+1) − 1)2fold
colorings with n2 colors for n 1.
Proof Note: In order to keep this proof shorter we omit some details of calculations.
We will define a set S in the plane with high density but without vertices in distance in
[1, b] in the following way. Let set A be an intersection of a disk of unit diameter and
a hexagon with common center point as in Fig. 1. The shape of A changes when we
change the size of the hexagon. In the fractional coloring of G[1,b] we choose the set
A in such a way that the ratio between the length of the circular arc y and the segment
x is equal to b (it will be explained later).
From the construction we get the following equalities: x = sin( α2 ), y = β2 ,
α + β = π3 . Therefore we obtain
π
y = 6 − arcsin(x ).
Then we build S by placing copies of A on the plane with distance b like in Fig. 2.
Assume that two neighboring components of S are centered at (0, 0) and (s, 0) which
determines the location of all components of S. Let n be a positive integer. We define
a tiling with hexagons of width s/n in such a way that the left side of some hexagon
is a part of the left vertical segment of the border of A.
For 0 i, j < n let Si, j = S + (si /n)(1, 0) + (s j/n)(1/2, √3/2). We define
hnfold coloring of the plane as follows. We assign color (i, j ) to all points in hexagons
4 Fractional Chromatic Number of G[a,b]
(1)
y
x
s
s
b
s
fully contained in Si, j . Now we will estimate the number hn of hexagons that are
fully contained in one copy of A, let say A0. Let A0n be a rescaled copy o A with
t(hπe−sa6marecscienn(txe)r+as6xA√01an−d xd2ia)m,aentderth(e1a−rea2 onsf).thTehheexaraegaonofisA(s0n/ins)p2(n√=3/412()1.O−bs2enrv)e
s 2
that if a hexagon of the tiling has nonempty intersection with A0n then it is fully
contained in A0. Therefore we can bound hn from below:
hn
=
(s/n)2(√3/2) =
√3 √n1 − x 2 − 2 2(bx + x 1 − x 2).
b +
(s/n)2(√3/2)
pn
41 (1 − 2 ns )2(π − 6 arcsin(x ) + 6x √1 − x 2)
On the other hand we conclude that every hexagon is contained in hn of Si, j (since we
can get all the hexagons by shifting one of them by (si /n)(1, 0)+(s j/n)(1/2, √3/2)),
hence every point in a hexagon has hn of n2 colors assigned.
With n tending to infinity the set of points colored 1 tends to S and our sequence
of hnfold colorings gives us an upper bound for the fractional chromatic number of
G[1,b]:
χ f (G[1,b])
nl→im∞ 2 √13 b+√√n1−x2 − 2 2 π − 6 arcsin(x ) + 6x √1 − x 2
2 3
= π − 6 arcsin(x ) + 6x √1 − x 2 · nl→im∞
n2
b+√1−x2 − 2 2
n
n2
2√3(b + √1 − x 2)2
= π − 6 arcsin(x ) + 6 x √1 − x 2
2√3(b + √1 − x 2)2
= π − 6 arcsin(x ) + 3 sin(2 arcsin(x ))
.
Now let us explain why did we choose the ratio between the length of the circular
arc—y and the segment—x to be b. To find minimal value we take the derivative of
the upper bound for χ f (G[1,b]) presented above and check when it equals 0.
√
4√3x b +
1 − x 2 (6bx + 6 arcsin(x ) − π )
√1 − x 2 (−6 arcsin(x ) + 3 sin (2 arcsin(x )) + π )2 = 0
⇐⇒ 6bx + 6 arcsin(x ) − π = 0.
Applying equality 1 we obtain
and hence
To complete the proof we transform the formula
using the previous equality and by doing so we get
6bx − 6y = 0
y = bx .
2√3(b + √1 − x 2)2
π − 6 arcsin(x ) + 3 sin(2 arcsin(x ))
√
3 b +
3 ·
√
x
1 − x 2
as the upper bound for the fractional chromatic number of graph G[1,b].
5 j Fold Coloring of the Plane
Note that the upper bounds for the fractional chromatic numbers of G[1,b] are
established by presenting an infinite sequence of j fold colorings. These colorings give
results close to the upper bound only for very big j . However, as it was mentioned in
the introduction, for the practical reasons it is often the case that only j fold coloring
with small j are valuable to consider. Furthermore, this consideration leads also to a
purely mathematical question: How fast, in terms of j , can we get close to the infinite
limit  the upper bound for the fractional chromatic number? In this section, we give
some insight into this matter.
f (G [1,b ])
140
120
100
80
60
40
20
2
4
In this subsection, we concentrate on G[
1,1
].
Theorem 4 There are twofold coloring of G[
1,1
] with 12 colors and threefold coloring
of G[
1,1
] with 16 colors.
Proof For our colorings we are going to use classic covering of the plane with hexagons
with side length 1/2. Obviously in a hexagon of side length 1/2 there are pairs of point
in a distance 1, so when we say we color a hexagon we mean the interior of the hexagon
plus its right border and its three vertices: the upper one and the two on the right (see
Fig. 4).
Firstly we show our twofold coloring of G[
1,1
] with 12 colors. We are going to
use the hexagon grid twice with the set of colors {1, 2, . . . , 12}. We create the first
layer by simply giving each row of hexagons three numbers and use it periodically so
that if one row has colors from the set {1, 2, 3} the next gets colors {4, 5, 6} and so
on. Having this coloring as our model we get the second layer by moving the colored
grid by a vector [3√3/4, −3/2]. See Fig. 5 with the first layer of colors and marked
placement of color 1 in the layers. The distances between two hexagons of the same
√
color are: vertically 2, horizontally √3 and diagonally 5 8 3 ≈ 1.08, so they are all
bigger than 1.
To create a threefold coloring of G[
1,1
] with 16 colors we again start with the
classic covering of the plane with hexagons with side length 1/2. We are going to use
the hexagon grid 3 times with the set of colors {1, 2, 3, . . . , 16}. We create the first
layer by simply giving each row of hexagons four numbers and use it periodically so
Fig. 5 Twofold coloring of
G[
1,1
] with 12 colors
1
7
1
4
10
8
2
5
1
11
3
9
6
12
1
1
that if one row has colors from the set {1, 2, 3, 4} the next gets colors {5, 6, 7, 8} and
so on. Having this coloring as our model we get second and third layers by moving the
colored grid by a vector [√3, −1]. See Fig. 6 with the first layer of colors and marked
placement of color 1 in other layers. The distances between two hexagons of the same
color are at least one (exactly one in case of two hexagons from different layers).
Theorem 5 There is a sevenfold coloring of G[
1,1
] with 37 colors i.e. 377 ≈ 5.285.
Proof To create sevenfold coloring of G[
1,1
] we start with a hexagon grid with hexagon
√1 . We color with the first color some hexagons in the pattern
side’s length s = 2 7
√
presented as shadowed on Fig. 7. Then we shift our pattern by a vector [ 2√37 , 0] and
color it with second color (bold contour in Fig. 7). Repeating this action 37 times we
have 7 colors for each hexagon and there are 37 colors total. In our base coloring figure
made by 7 hexagons the largest distance between two points is 1 so choosing half of
the border to be colored and the other not is enough to make sure there are no two
point in distance 1 in one figure. The distance between two such figures is
1
1
5
7
8
2
9
6
13
3
1
10
1
4
11
1
12
1
14
15
16
1
1
In this section we give two general methods for building j fold colorings (for small
j ) for graphs G[1,b].
Theorem 6 There exists a nmfold coloring with ( √2b3 + 1) · n · ( √2b3 + 1) · m colors
of the graph G[1,b] i.e.
χnm (G[1,b])
nm
(2b/√3 + 1) · n n·m (2b/√3 + 1) · m .
Proof In the proof we are going to create n · m colored hexagon grids. A color of a
hexagon will be a pair of numbers, first of which will be related to the row the hexagon
is in and the second corresponds to the column.
Let W11 be a hexagon grid with hexagons with side length 1/2. Let H be one of
the hexagons from W11. For 2 j n let W1j be a hexagon grid created by moving
uncolored W11 by a vector ( j − 1)/n [√3/2, 0]. Now, lets say H is colored (1, 1) and
let’s find first hexagon H in the same row from any W1j that can also be colored (1, 1)
without creating a monochromatic edge in G[1,b] i.e at distance greater or equal to b
(see Fig. 8).
Counting the hexagons from H to H we find that there are exactly
b + 2
of them, since the distance between the centers of H and H has to be at least b + √23 .
Then we shift each of W1j by vectors (i − 1)/m [√3/4, −3/4] for 2 i m getting
grids W j . Now remembering that H is colored (1, 1) we find the first hexagon H
i
from Wi1 to have the same color (see Fig. 9). The number of hexagons between the
two is (2b/√3 + 1) · m .
Now we have three hexagons H , H , and H that can be colored with the same
color, since the distance between H and H would be the smallest when n = m and
then the centers of the three hexagons create an equilateral triangle with side length
√
at least b + 23 (because H and H are in proper distance).
The final coloring is created by giving each hexagon a pair of numbers, so that H gets
color (1, 1), the next hexagon in the same row gets (1, 2) and so on until H gets (1, 1)
again and the cycle repeats, and in the row below we get colors (2, 1), (2, 2), (2, 3) . . .
and so on until we have row with H in which we use (1, 1), (1, 2), (1, 3) . . . again.
Since we have nm hexagon grids every point gets nm colors. Finally we get nmfold
coloring of G[1,b] with ( √2b3 + 1) · n · ( √2b3 + 1) · m colors (Fig. 10).
The following result can be seen as a combination of twofold coloring approach
from Theorem 4 and the main method from Theorem 6.
H
3
1
2
1
H
3
2
3
4
5
1
H’
4
5
1
H’’
2
3
4
5
1
H’
4
5
1
H’’
Theorem 7 There exists a 2nmfold coloring with 2 (b + 1)2n (b + 1) 23m
of the graph G[1,b] i.e.
colors
2 √3(b + 1) 23n
2nm
(b + 1) 23m .
Proof Note that the proof uses similar methods to the previous one. We are going to
create 2 · n · m colored hexagon grids. A color of a hexagon will be a pair of numbers,
first of which will be related to the row the hexagon is in and the second corresponds
to the column.
Let W11 be a hexagon grid with hexagons with side length 1/2. Let H be one of
the hexagons from W11. For 2 i m let Wi1 be a hexagon grid created by moving
uncolored W11 by a vector (i − 1)/m [0, −3/2]. Now lets say H is colored (1, 1) and
let’s find first hexagon H in the same column from any Wi1 that can also be colored
(1, 1) without creating a monochromatic edge in G[1,b] i.e at distance greater or equal
to b, so the distance between the centers of H and H needs to be at least b + 1 (see
Fig. 11 for fourfold coloring with 4 · 6 = 24 colors of G[
1,1
] with n = 1 and m = 2)
Since the distance between the centers of H and H is at least b + 1 but we chose
H to be as close as possible to H , then there are (b + 1) 23m rows between H and H .
Now for 2 j m and 2 j n let Wij be a hexagon grid created by shifting Wi1 by
a vector ( j − 1)/n [√3/2, 0]. Let H be the first hexagon in the same row as H in any
of W1j such that the distance between the centers of the two is at least √3(b + 1). The
number of hexagons between H and H is √3(b + 1) √2n3 = (b + 1)2n b + √23 .
The distance between the centers of H and H is at least:
(b + 1)
Since the distance is at least 2b + 2 they can be colored the same color and there is
enough space between them to put another hexagon in the same color between them.
So we create new hexagon grids Vi j by shifting Wij by a vector
√
2n
3(b + 1) √
√3
3 · 4n , (b + 1)
2m
3
The final coloring is created by giving each hexagon a pair of numbers, so that H
gets color (1, 1) the next hexagon in the same row from any W1j gets (1, 2) and so
on until H gets (1, 1) again and the cycle repeats. The next row from W1j has the
first coordinate in all colors equal to 2. The set of numbers for second coordinate is
{1, 2, . . . , √3(b + 1) √2n3 }, and for the first coordinate 2 (b + 1) 23m (since there are
2 rows of hexagons between 2 hexagons from the same grid in the same column). So
we use 2 (b + 1) 23m · √3(b + 1) √2n3 colors using 2nm hexagon grids.
Table 3 Selected results in
j fold coloring of G[
1,1
] with
small j
5.3 Summary
1
H
1
2
3
5
6
1
4 H’
3
4
1
H’’
Method
j = nm
j = 2 nm
k =
j =
k/j ≈
j =
k =
k/j ≈
k =
k/j ≈
7
Theorems 6 and 7 give different results and we cannot say one is stronger than another.
Table 2 presents comparison of the two for j fold coloring of G[
1,1
] with small values
of j (k is the numbers of colors used). We bold best results for a fixed j .
Our best results in j fold coloring of G0 with small values of j using k colors are
summarized in Table 3. They follow from Theorems 4, 5, 6, 7.
For practical applications it is useful to consider j fold coloring of graph G[
1,2
],
especially with small j . Table 4 presents our results in coloring G[
1,2
] using method
from Theorem 6. It appears that in this case the method from Theorem 7 does not give
good results.
Acknowledgments We thank Professor Zbigniew Lonc, Zbigniew Walczak and Professor Jacek
Wojciechowski for introducing us to the problem.
k =
j =
100
9
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0
International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution,
and reproduction in any medium, provided you give appropriate credit to the original author(s) and the
source, provide a link to the Creative Commons license, and indicate if changes were made.
1. Exoo , G.: εUnit distance graphs . Discrete Comput. Geom . 33 , 117  123 ( 2005 )
2. Falconer , K.J.: The realization of distances in measurable subsets covering Rn . J. Comb. Theory, Ser. A 31 , 187  189 ( 1981 )
3. Fisher , D. , Ullman , D. : The fractional chromatic number of the plane . Geombinatorics 2 ( 1 ), 8  12 ( 1992 )
4. Hadwiger , H.: Ungelöste Probleme . Elem. Math. 16 , 103  104 ( 1961 )
5. Hochberg , R., O 'Donnell , P.: A large independent set in the unit distance graph . Geombinatorics 3 ( 4 ), 83  84 ( 1993 )
6. Ivanov , L.L. : On the chromatic numbers of R2 and R3 with intervals of forbidden distances . Electron . Notes Discrete Math. 29 , 159  162 ( 2007 )
7. Moser , L. , Moser , W. : Solution for problem 10 . Can . Bull. Math. 4 , 187  189 ( 1961 )
8. Nielsen , M.J.: Approximating monochromatic triangles in a twocolored plane . Acta Math. Hung . 74 ( 4 ), 279  286 ( 1997 )
9. Scheinerman , E.R. , Ullman , D.H. : Fractional Graph Theory . Wiley, New York ( 2008 )
10. Soifer , A. : The Mathematical Coloring Book . Springer, New York ( 2008 )
11. Walczak , Z. , Wojciechowski , J.M.: Transmission scheduling in packet radio networks using graph coloring algorithm, wireless and mobile communications . In: ICWMC '06 ( 2006 )
12. Woodall , D.R. : Distances realized by sets covering the plane . J. Comb. Theory, Ser. A 74 ( 4 ), 279  286 ( 1997 )