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On the balanced upper chromatic number of finite projective
planes
Zoltán L. Blázsik∗Aart Blokhuis†Štefko Miklavič‡
Zoltán Lóránt Nagy∗Tamás Szőnyi∗,§
May 26, 2020
Abstract
In this paper, we study vertex colorings of hypergraphs in which all color class sizes differ by at
most one (balanced colorings) and each hyperedge contains at least two vertices of the same color
(rainbow-free colorings). For any hypergraph H, the maximum number kfor which there is a bal-
anced rainbow-free k-coloring of His called the balanced upper chromatic number of the hypergraph.
We confirm the conjecture of Araujo-Pardo, Kiss and Montejano by determining the balanced upper
chromatic number of the desarguesian projective plane PG(2, q)for all q. In addition, we determine
asymptotically the balanced upper chromatic number of several families of non-desarguesian projec-
tive planes and also provide a general lower bound for arbitrary projective planes using probabilistic
methods which determines the parameter up to a multiplicative constant.
1 Introduction
In recent years the notion of a proper strict coloring of hypergraphs was investigated in several papers
by Voloshin, Bacsó, Tuza and others, including [1], [2], [3] and [4]. In this work, instead of studying the
upper chromatic number we will focus on improving the known estimates of the balanced upper chromatic
number of such hypergraphs which arise from projective planes.
Let Hdenote a hypergraph with vertex set V(|V|=v) and (hyper)edge set E. A strict N-coloring C
of His a coloring of the vertices using exactly Ncolors; in other words, the collection C={C1, . . . , CN}
of color classes is a partition of V. Given a coloring C, we define the mapping ϕC:V→ {1,2, . . . , N}
by ϕC(P) = iif and only if P∈Ci. We call the numbers 1, . . . , N colors and the sets C1, . . . , CNcolor
classes. We call an edge H∈Erainbow (with respect to C)if no two points of Hhave the same color;
that is, |H∩Ci| ≤ 1for all 1≤i≤N. The upper chromatic number of the hypergraph H, denoted by
¯χ(H), is the maximal number Nfor which Hadmits a strict N-coloring without rainbow edges. Let us
call such a coloring proper or rainbow-free. A balanced coloring is a coloring in which the cardinality of
any two color classes differs by at most one. The balanced upper chromatic number of a hypergraph H,
denoted by χb(H), is the largest integer Nsuch that Hadmits a proper strict balanced N-coloring.
In the following sections we will focus on hypergraphs which arise from a projective plane Π(of order
q). The vertices are the points of the plane and the edges correspond to the lines of the plane. In 2015,
Araujo-Pardo, Kiss and Montejano proved the following results.
∗MTA–ELTE Geometric and Algebraic Combinatorics Research Group, 1117 Budapest, Pázmány P. stny. 1/C, Hungary
†Eindhoven University of Technology, Eindhoven, The Netherlands
‡Andrej Marušič Institute, University of Primorska, Koper, Slovenia
§ELTE Eötvös Loránd University, Budapest, Hungary
1
arXiv:2005.12011v1 [math.CO] 25 May 2020
Result 1.1 ([5]).All balanced rainbow-free colorings of any projective plane of order qsatisfy that each
color class contains at least three points. Thus
χb(Πq)≤q2+q+ 1
3.
Result 1.2 ([5]).For every cyclic projective plane Πqwe have
χb(Πq)≥q2+q+ 1
6.
If the difference set defining Πqin Zq2+q+1 contains {0,1,3}then
χb(Πq) = q2+q+ 1
3.
We will use the last observation to determine the balanced upper chromatic number of the desarguesian
projective plane PG(2, q)in the second section.
In the third section, we will use some well-known representations (such as affine and relative difference
sets, planar functions) of projective planes of order q(including non-desarguesians) in order to present
general lower bounds on the balanced upper chromatic number. We managed to reach the correct order
of magnitude for the remaining two cases, too. Moreover, we prove a sharp result if q≡0 (mod 3).
Theorem 1.3. For q≡2 (mod 3), let Πqbe a projective plane of order qrepresented by an affine
difference set. Then
χb(Πq)≥q2+ 2
3.
Theorem 1.4. For q≡0 (mod 3), let Πqbe a projective plane of order qrepresented by a planar function
(or relative difference set). Then
χb(Πq) = q2+q+ 1
3=q2+q
3.
If q6≡ 0 (mod 3), we manage to give a coloring of the affine plane of order qrepresented by a suitable
planar function, thus we get a lower bound on the balanced upper chromatic number. Furthermore, as a
consequence we have a coloring of the corresponding projective plane which means that we have a lower
bound on the balanced upper chromatic number of any projective plane of order qrepresented by a planar
function.
Theorem 1.5. For q6≡ 0 (mod 3) and p > 5, let Aqand Πqbe the affine and projective plane of order
q=phrepresented by a planar function, respectively. Then
χb(Aq)≥
(1−1
p)q2
3if p= 3k+ 1,
(1−2
p)q2
3if p= 3k+ 2;
χb(Πq)≥
q2+q−1−q2
p
3if p= 3k+ 1,
q2+q+1−2q2
p
3if p= 3k+ 2, h odd,
q2+q−1−2q2
p
3if p= 3k+ 2, h even.
The fourth section is dedicated to a probabilistic argument which will give us a general lower bound
on the balanced upper chromatic number of projective planes.
Theorem 1.6. Let Πqbe an arbitrary projective plane of order q > 133. Then its balanced upper
chromatic number can be bounded from below as
χb(Πq)≥q2+q−16
10 .
2
2 Difference sets containing {0,1,3}
We recall the definition of a difference set.
Definition 2.1. Let Gbe a group of order v. A (v , k, λ)-difference set is a subset D⊂Gof size ksuch
that every nonidentity (nonzero) element of Gcan be expressed as d1d−1
2(or d1−d2if we use additive
notation) of elements d1, d2∈Din exactly λways.
Singer [6] proved PG(2, q)admits a regular cyclic collineation group and thus can be represented by a
(q2+q+ 1, q + 1,1)-difference set in a cyclic (hence abelian) group. For more details, see [7].
We start with a proof [5] of the fact mentioned above that χb(Πq) = bq2+q+1
3cif Πqcomes from a
difference set containing {0,1,3}.
Proof. Let every Ciin the partition consist of three consecutive integers, with the possible exception of
C1having 4(this happens if q= 0,2mod 3). It is clear that this coloring is balanced with the above
number of colors. To show that no line is rainbow we note that every line contains a (unique) triple
{j, j + 1, j + 3}. This triple is contained in the union of two consecutive Ci’s, so by the pigeonhole
principle two of them have the same color.
We obtain a planar difference set by starting with a primitive cubic polynomial p(x) = x3−ax2−bx−c
over GF(q)and now define the field GF(q3) = GF(q)[x]/(p(x)). Every monomial xinow reduces to a
degree (at most) 2 polynomial c2x2+c1x+c0≡(c2, c1, c0)∈GF(q)3. The exponents i, with 0≤i≤q2+q
for which xilies in a two-dimensional subspace now give a difference set. If we take the subspace c2= 0,
and if a= 0, so p(x)is of the form x3−bx −c, then our difference set will contain 0,1and 3.
By a result S.D. Cohen [8] we know that a primitive polynomial with this property exists for all q6= 4.
As a consequence, we get
Proposition 2.2.
χb(PG(2, q)) = q2+q+ 1
3.
Note that the case q= 4 has already been covered in [5].
3 Improving the lower bound on χb(Πq)for certain classes of non-
desarguesian planes
We recall the proof of Theorem 2.3. in [5]. For 0≤i≤q2+q+1
3−1define the color classes as Ci=
ni, i +q2+q+1
3, i +2(q2+q+1)
3o. Since each line contains a (unique) pair of points with difference q2+q+1
3,
having therefor the same color, there are no rainbow lines. Together with Result 1.1 this gives χb(Πq) =
q2+q+1
3, if q≡1 (mod 3).
In the following subsections we are going to investigate other representations and improve some of the
bounds.
3.1 Using affine difference sets if q≡2 (mod 3)
Our aim in this case is to use affine difference sets and the corresponding representation of affine planes
and then add the ideal points to the construction and color them in a suitable way.
Definition 3.1. Let Gbe a group of order q2−1, and let Nbe a normal subgroup of order q−1of G.
Aq-subset Dof Gis called an affine difference set of order qif {d1d−1
2:d16=d2∈D}=G\N.
3
An affine difference set Dgives rise to an affine plane (and hence to a projective plane Πq(D)) as
follows: Points of the plane are the elements of G, together with a special point O(the origin), lines
through Oare the cosets of N, the remaining lines are of the form Dg,g∈G. We refer the reader to [9],
[10] and [11] for further information about affine difference sets.
Proof of Theorem 1.3. We are going to define a coloring of the points in the orbit of size q2−1, and then
give a suitable coloring of the origin and the ideal points of the projective closure. Similarly as above
define the color classes as the right cosets of a subgroup T={1, t, t2}for a fixed element tof order three,
so the color classes are of the form Cg={g, tg, t2g}. Note that |N|is not divisible by 3, so t6∈ N. Since
every element of G\Nin particular tis of the form d1d−1
2exactly once, this means that there will be
two points with the same color in every line which avoids the origin.
There are two things left to do: the first one is to color the origin and the points of the ideal line in
order to make sure that neither the lines through the origin nor the ideal line are monochromatic. This
can be done in a greedy way. The origin, together with three ideal points get a new color, the remaining
q−2ideal points Pget the color of one of the points on the line OP in such a way that no color is used
twice (so altogether at most four times).
Observe that we now indeed get a balanced coloring of the projective plane and there are q2−1
3+ 1 =
q2+2
3color classes such that exactly q−1of them have 4 elements, the others have only 3 and there are
no monochromatic lines.
The value of the above result is questionable, since all known examples of such planes are desarguesian.
3.2 Using planar functions if q≡0 (mod 3)
If q= 3hfor some h≥1then we will use a representation of a projective plane Πq(f)based on planar
functions.
Definition 3.2. A function f: GF(q)→GF(q)is a planar function if the equation f(x+a)−f(x) = b
has a unique solution in xfor every a6= 0 and every b∈GF(q).
A planar function gives rise to an affine plane, and hence a projective plane as follows. The point
set will be the same as in AG(2, q), the vertical lines with their ideal point remain the same but we will
replace every non-vertical line with a translate of the graph of f. Note that parallel lines correspond
to translations of fthat differ by a vertical translation, and the ideal point of these translates can be
defined according to this.
We will also assume that f(0) = f(1) = 0. Moreover, let H={−1,0,1}be a 3-element subgroup in
(GF(q),+),q≡0 (mod 3). Thus the cosets of the subgroup generated by Hand the vertical line through
the origin give us a partition of the point set into q
3stripes.
Proof of Theorem 1.4. The main idea in our coloring is to color in each of these stripes the 3 points
which have the same second coordinate with the same color but do it in such a way that the points with
different second coordinates must have pairwise different colors. Let us choose a representative system
of the cosets: {x0, x1, . . . , xq
3−1}. Let us denote the stripe which contains xjwith Sj. Therefore Sjwill
correspond to the points {(xj−1, y),(xj, y ),(xj+ 1, y)}for y∈GF(q).
To begin with let us color the points {(x0−1,0),(x0,0),(x0+ 1,0)}from S0and the ideal point of the
vertical lines with the same color (let us call it C00). We can continue the coloring in S0by coloring the
triples {(x0−1, y),(x0, y),(x0+ 1, y )}with color C0y. Similarly for any j= 1,..., q
3−1color the points
{(xj−1, y),(xj, y),(xj+ 1, y )}from Sjwith color Cjy. Notice that for any jand ythe color Cjy must
be pairwise different.
Although for this coloring there will be no rainbow translates of f, almost all of the vertical lines are
rainbow. That is the reason why we will modify this coloring a little bit. For every j= 1,...,q
3−1
4
delete Cj0and Cj1. For every j= 0,...,q
3−1denote the ideal point of the translate of fwhich goes
through (xj−1,0) and (xj,0) with Pj1; through (xj,0) and (xj+ 1,0) with Pj2; through (xj−1,0) and
(xj+ 1,0) with Pj3. Introduce new color classes for every j= 1,..., q
3−1:
Cjα ={(xj−1,0),(xj−1,1), Pj1}
Cjβ ={(xj,0),(xj,1), Pj2}
Cjγ ={(xj+ 1,0),(xj+ 1,1), Pj3}.
With this modification we certainly achieved that now every vertical line has two points with the same
color. Moreover, by coloring the appropriate ideal points with these new colors we achieved that there are
no rainbow parabolas. But on the ideal line all of the points have pairwise different colors so far. Notice
that we did not color P01,P02 and P03 yet. If we color these 3 points with a new color then this will take
care of the ideal line, too. One can see that there is only one color class C00 which has 4 elements hence
we used exactly q2+q
3color classes.
Figure 1: Modified coloring if q≡0 (mod 3)
Remark 3.3. In [12, 13] Dembowski, Ostrom, Coulter, Matthews showed that there are planar functions
so that the represented geometry is not desarguesian.
5
3.3 Using planar functions if q6≡ 0 (mod 3)
We discuss here two further constructions which give a bit weaker results for the desarguesian projective
plane but it completes the constructions for any projective plane which can be represented with planar
functions (which is a strictly larger class). Moreover, these constructions give us a lower bound on the
balanced upper chromatic number of the affine planes represented by planar functions, too.
Proof of Theorem 1.5. Let fbe a planar function with f(0) = f(1) = 0. Without loss of generality we
can assume that f(2) = 1 (otherwise we can divide every value of fwith the value of f(2)). If p= 3k+ 1
is a prime, q=ph, then we will color the affine plane of order prepresented by f. Every color class will
be on two consecutive horizontal lines y=cand y=c+ 1.
Point (0,0) has color 1, points (1,0) and (2,0) have color 2, point (3,0) has color 3and this pattern
is repeated until (3k−3,0) (which is a single point). The last color class has 3 consecutive points,
(3k−2,0),(3k−1,0),(3k, 0). On the line y= 1, the same pattern appears but everything is shifted by
x→x+ 3 so that the single color classes have pairwise different new colors, and the pairs with the same
color inherit their color from the single element of the previous line. More precisely, the points (1,1) and
(2,1) get color 1, then point (3,1) gets a new color, then (4,1),(4,2) get the color of (3,0), etc. At the
end, (3k, 1) gets a new color, and finally (0,1) gets color 1, too. With this coloring, on each horizontal
line there will be three consecutive points having the same color (these points are in a color class of size
4), the remaining color classes will have size 3. After psteps, we get back to the coloring on line y= 0.
Notice that this coloring also make the vertical lines rainbow-free.
If p= 3k+ 2 and p > 5is a prime, q=ph, then the pattern changes a little bit. We need to finish
sooner the alternating sequence of single and double classes, namely at (3k−6,0), and then close with
two 3 element classes separated with a single element with a new color. We include the examples for
p= 7 and 11 in Figure 2.
2 13 13 13 4 14 14
12 13 11 11 14 12 12
10 11 9 9 9 12 10
8 8 9 7 7 10 8
6 6 7 5 5 5 8
1 1 1 5 3 3 6
1 2 2 3 4 4 4
(a) p= 7
2 31 31 31 4 32 32 32 6 33 33
30 31 28 28 28 32 29 29 33 30 30
27 27 28 25 25 29 26 26 26 30 27
24 24 25 22 22 22 26 23 23 23 27
19 19 19 22 20 20 20 23 21 21 24
19 16 16 16 20 17 17 21 18 18 18
15 16 13 13 17 14 14 14 18 15 15
12 13 10 10 10 14 11 11 11 15 12
9 9 10 7 7 7 11 8 8 12 9
1 1 1 7 3 3 8 5 5 5 9
1 2 2 3 4 4 4 5 6 6 6
(b) p= 11
Figure 2: Colorings of affine planes represented by planar functions
If h > 1then we can extend these colorings of the similar p×pgrids and get a balanced coloring of
the affine plane defined by the planar function fin both cases.
These colorings use pand 2p4-element classes in every p×pgrid, respectively. In total there are q2
p2
such grids which means that in the affine plane of order qthe number of the 4-element color classes are
6
q2
p,if p= 3k+ 1 and 2q2
p,if p= 3k+ 2.
We can use these constructions in order to give a balanced coloring of the projective plane, too. Since
the affine lines are rainbow-free, we can arbitrarily color the points of the ideal line with 3 and 4 element
color classes. Obviously we use the most colors if we use as many 3 element color classes on the ideal
line as we can. If p= 3k+ 1 then q+ 1 ≡2 (mod 3) therefore in the ideal line there must be at least
two 4-element color classes. If p= 3k+ 2 then the remainder of q+ 1 when divided by 3 depends on the
parity of h. If his odd then q+ 1 ≡0 (mod 3), and if even then q+ 1 ≡2 (mod 3). In the following
table we calculated the number of color classes of size 3 and 4 in every possible setup.
p= 3k+ 1 p= 3k+ 2,hodd p= 3k+ 2,heven
on the
ideal line
in the
affine plane
on the
ideal line
in the
affine plane
on the
ideal line
in the
affine plane
the number of
3-element color classes q−7
3
q2
3·p−4
p
q+1
3
q2
3·p−8
p
q−7
3
q2
3·p−8
p
the number of
4-element color classes 2q2
p02q2
p22q2
p
The construction for p= 3k+2 does not work for p < 11. What can we say about p= 5? Surprisingly,
it turned out that for p= 5 there is no balanced coloring of the affine plane of order 5 represented by
a planar function fwith color classes of size 3 and 4. Moreover, none exists if there is at least one
color class of size 4. These claims can be shown by a rather long case analysis which we choose to omit.
However, by a computer search we found out that there exist a coloring such that all but one vertical
line and one other line are rainbow, but we couldn’t correct these errors by coloring the ideal points in
order to get a balanced coloring of the projective plane of order 5 represented by a planar function f.
It is straightforward to find a balanced coloring of the affine plane of order 5 with color classes of size 5
which can be generalized to get a balanced coloring for any affine and projective plane of order q= 5h
with roughly q2
5color classes. In Figure 3, we included the above mentioned colorings of the affine plane
of order 5.
6277 6
4 7 5 6 4
5 3 3 4 5
3 1 1 5 1
12 2 2 3
(a) „almost” good color-
ing for p= 5
4 5 4 4 4
3 3 4 3 3
2 2 2 3 2
1 1 1 1 2
1 5 5 5 5
(b) with 5 element color
classes
Figure 3: Colorings of affine planes of order 5 represented by f
4 General lower bound with a probabilistic approach
In this section, we prove a general lower bound for all projective planes. In order to prove Theorem 1.6,
we need a technical lemma which appeared in the paper of Nagy ([14], Lemma 3.4).
7
Lemma 4.1. Denote Qk
i=1 1−i
nby An(k). Then
An(k)<exp −k(k+ 2)
2n−k−2∆(n, k),
where ∆(n, k)is the product of error terms √n−1
√n−k−1,1 + k2
12(n−k−1)2kand 1−(k+2)2
12n2
k(k+2)
2n−k−2.
We continue the preparation with a well known bound on the number of incidences I(P,L)between
a point set Pand a line set L
Lemma 4.2 (Incidence bound, see [15]).
I(P,L)≤min n|P|p|L| +|L|,|L|p|P| +|P|o.
Note that this also follows from the upper bound of the Zarankiewicz number z(m, n, 2,2) [16], which
denotes then maximum number of 1s in an m×nmatrix which does not contain an all-1 2 ×2submatrix.
Proof of Theorem 1.6. We show the existence of a suitable coloring with color classes of size 10 or 11 by
the combination of a probabilistic argument and an application of the point-line incidence bound Lemma
4.2 together with Hall’s marriage theorem.
(Step 1.) Take an arbitrary point Qof the plane and t=dcqelines incident to Q, where the parameter
c∈(0,1) is determined later on. We choose uniformly at random a pair of points from each `\Qof these
lines `incident to Q, and we assign a distinct color to each pair.
(Step 2.) Write tq = 9s+r, where r∈ {0,1,...,8}. Next we take a random coloring of the non-colored
points of the set St
i=1 `i\Qso that apart from rcolor classes of size 10, each color is used 9times. We
say that a color resolves a line if the line contains at least two points from that color class. And let’s call
a line resolved if it contains two points from the same color class.
The probability that a line not incident to Qis not resolved by this random coloring is less than
1·1−8
tq −1·1−2·8
tq −2· · · · 1−(t−1) ·8
tq −(t−1)< Atq/8(t−1).
Hence we may apply Lemma 4.1 to obtain that expected number of not resolved lines which are not
incident to Qis at most
E(not resolved lines not on Q)< q2·Atq/8(t−1) < q2·exp −t2−1
2tq/8−t−1∆(tq/8, t −1),
by the linearity of expectation.
Here the right hand side can be bounded from above as
q2·exp −t2−1
tq/4−t−1∆(tq/8, t −1) < q2exp −4t
qexp −t2−t+q/4
q/4(tq/4−t−1) ∆(tq/8, t −1).
If q > 133 holds and t=dcqeis chosen appropriately, a careful calculation of the Taylor series of the
error terms proves that the expected value can be bounded above by the main term
E(not resolved lines not on Q)< q2exp −4t
q< q2exp(−4c).(1)
To perform Step 3, let us take a coloring as above with less than q2exp(−4c)lines not resolved, beside
the lines through Q. In order to resolve these lines as well, our aim is to assign distinct not-colored points
8
Pfto each of these lines fwith Pf∈f, and choose a color for each assigned point from the colors used
already on f. In this step we also require that every color must be used at most once. Finally in Step 4,
we have to color the remaining uncolored points in such a way that all the lines through Qare resolved
and every color class is of size 10 or 11. Here we might apply new colors as well.
(Step 3.) First, we have to find a matching between the uncolored (q−t+ 1)q+ 1 points and the set of
not resolved lines which are not incident to Q, that covers the set of the lines in view. To resolve at the
end all the remaining lines as well (i.e., those that passes through Q), we extend this incidence graph by
adding two copies of not resolved lines through Qand joining them to the points incident to them. We
apply Hall’s theorem twice combined with Lemma 4.2 to prove the existence of the covering matching in
view. The incident points chosen in this step to the lines are called the assigned points.
Suppose that we have a set of not resolved lines X, and a set Yof uncolored points of cardinality less
than |X|incident to them. Lemma 4.2 implies that I(X, Y )<|X|p|X| − 1 + |X|on the one hand, and
we also know that I(X, Y )≥(q−t+ 1)|X|. This is in turn a contradiction if
p|X| − 1<(q−t),thus if q2·exp(−4c) + 2(1 −c)q < q2(1 −c)2−2(1 −c)q, (2)
where we took into account that Xis of size at most q2exp(−4c) + 2(1 −c)q, and the error term which
may occur while we are considering the ceiling in t=dcqe.
In other words, Condition (2) yields a suitable assignment of distinct not colored points for the not
resolved lines. To assign distinct colors for these points from their respective lines skew to Q, we apply
Hall’s theorem again and suppose to the contrary that there is a set Xof not resolved lines skew to Qon
which less than |X|colors were used. Hence the total number of colored points on these lines is at most
10|X|as each color can appear at most 10 times. However, the number of incidences between the colored
points of these lines could not exceed |X|p10|X|+ 10|X|according to Lemma 4.2, while this incidence
number is t|X|=dcqe|X|. Thus the got a contradiction to our assumption if
|X|p10|X|+ 10|X|<dcqe|X|i.e., if q2·exp(−4c)<(cq −10)2
10 .(3)
If both Condition (2) and (3) hold then we are able to resolve all the lines skew to Q. In order to choose
the optimal constant c, we may suppose that these upper bounds are close to each other (asymptotically)
i.e. we choose csuch that the values of q2(1 −c)2and c2q2
10 are almost the same.
If we pick that constant cto be c= 0.77, it is easy to verify that both Condition (2) and (3) hold
when q > 133.
(Step 4.) We finish are proof by coloring the remaining uncolored points such that all the lines
containing Qare resolved and each color is used 10 or 11 times. To guarantee the resolving property, we
introduce new colors and take 5pair of assigned points from each distinct 5lines of the q−t+ 1 not yet
resolved ones. This makes further color classes of size 10, with less than 10 assigned points left uncolored.
These leftout assigned points finally get yet a new color, and this color class is completed to have size 10
by putting in arbitrary uncolored points.
Up to this point, we already resolved all the lines but we have color classes of size 9,10 and possibly 11
as well. To end up with a balanced coloring we try to complete the classes of size 9to have size 10 by
coloring the remaining uncolored points. This is doable since the number of uncolored points is at least
(q−t+ 1)q+ 1 −q2·exp(−4c)which is more than the number N < 1
9tq of color classes of size 9at this
point.
Thus we obtained a balanced coloring of almost all the points which resolves every line. Finally, we
partition the remaining uncolored points to color classes of size 10 and put the remaining at most 9points
into distinct formerly created color classes of size 10. This provides at most 17 classes of size 11 and
further classes of size 10, which completes the proof.
9
Remark 4.3. For 11 ≤q≤133, we are able to verify by a computer aided search that there exists a
balanced coloring for an arbitrary projective plane of order qwith color classes of size 11 and 12, namely
the number of colors needed is at most q2+q−18
11 . One should repeat the steps in the proof of Theorem 1.6
but use the concrete expected value instead of Condition (1) and use the stronger inequalities in Condition
(2) and (3). Finally, suppose that the number of color classes is at most q2+q−18
11 and q≤10. Then we
have less than 11 colors, thus every line has a pair of monochromatic points by the pigeon-hole principle.
Concluding remarks
We showed that for certain non-desarguesian planes one can construct a rainbow-free coloring with color
classes of size 3 and 4. It would definitely be interesting to find more classes of projective planes with
this property. Some nice construction for potential planes can be found in the paper by Pott [17].
Probably, the most natural extension to our problem is to consider higher dimensional projective
spaces. In this case, we can consider subspaces of fixed dimension kin PG(n, q), and try to determine the
balanced upper chromatic number. Some initial results in this direction can be found in Araujo-Pardo,
Kiss, Montejano [5].
One can also extend the problem to the case when more color class sizes are allowed. For example,
one can consider rainbow-free coloring with color classes of size at most k, and determine the maximum
number of colors under this condition.
Acknowledgement
The first author was supported by the ÚNKP-18-3 New National Excellence Program of the
Ministry of Human Capacities. In the first part of this research, the first, third and fifth authors gratefully
acknowledge the support of the bilateral Slovenian–Hungarian Joint Research Project no. NN 114614 (in
Hungary) and N1-0140 (in Slovenia). In the second part of this research, these three authors were
supported by the Slovenian–Hungarian Bilateral project Graph colouring and finite geometry (NKM-
95/2019000206) of the two Academies. The third author was supported by research project N1-0140 of
the Slovenian Research Agency. The fourth author is also supported by the Hungarian Research Grants
(NKFI) No. K 120154 and SNN 132625 and by the János Bolyai Scholarship of the Hungarian Academy
of Sciences. The fifth author was also supported by research project J1-9110 of ARRS.
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