Saturating sets in projective planes and hypergraph covers

Abstract

Let Πq\Pi_q be an arbitrary finite projective plane of order q. A subset S of its points is called saturating if any point outside S is collinear with a pair of points from S. Applying probabilistic tools we improve the upper bound on the smallest possible size of the saturating set to 3qln⁡q\sqrt{3q\ln{q}}. The same result is presented using an algorithmic approach as well, which points out the connection with the transversal number of uniform multiple intersecting hypergraphs.

Full text

Saturating sets in projective planes and hypergraph
covers
Zolt´an L´or´ant Nagy∗
MTA–ELTE Geometric and Algebraic Combinatorics Research Group
H–1117 Budapest, P´azm´any P. s´et´any 1/C, Hungary
nagyzoli@cs.elte.hu
Abstract
Let Πqbe an arbitrary finite projective plane of order q. A subset Sof its points
is called saturating if any point outside Sis collinear with a pair of points from S.
Applying probabilistic tools we improve the upper bound on the smallest possible
size of the saturating set to d√3qln qe+d(√q+ 1)/2e. The same result is presented
using an algorithmic approach as well, which points out the connection with the
transversal number of uniform multiple intersecting hypergraphs.
Keywords: projective plane, saturating set, dense set, transversal, blocking set,
complete arcs, hypergraph cover
1 Introduction
Let Πqbe an arbitrary finite projective plane of order q, and P,Ldenote the point and
line set of the plane, respectively. A subset Sof its points is called a saturating set if any
point outside Sis collinear with two points in S. In other words, the secants of the point
set cover the whole plane.
The size of saturating (or sometimes also called dense, saturated, determined) sets has
been widely investigated, see [14] for a recent survey. The importance of such sets relies on
connections to covering codes [8], algebraic curves over finite fields [13], sumset theory [15]
and complete arcs [14]. In fact, there is a one-to-one correspondence between saturating
sets in PG(2, q) of cardinality |S|and linear q-ary covering codes of codimension 3 and
covering radius 2, of length |S|. Here we only refer the reader to [7, 8] for the details.
∗The author is supported by the Hungarian Research Grant (OTKA) No. K 120154 and by the J´anos
Bolyai Research Scholarship of the Hungarian Academy of Sciences
1

Notation 1.1. For any points Pand Q, the set hP, Qidenotes the points of the line
determined by Pand Q. In general, if Sand Tare point sets, then the set hS, T idenotes
the points of the lines determined by a point from Sand a point from T. For convenience,
if one of the sets consists of a single point, we use Pinstead of {P}.
The set hS, Sidenotes the points of the lines determined by a distinct point pair from S.
Clearly, the notation rewrites the concept of saturating sets to the following property on
a set S:hS, Si=P(Πq).
An arc in the projective plane is a set of points such that no three points of the set are
on a line, and arcs which can not be extended to a larger arc are called complete arcs.
Observe that complete arcs are special separating sets. Due to the pioneer work of Ughi,
Sz˝onyi, Boros, Giulietti, Davydov, Marcugini, Pambianco, Kim and Vu [6, 9, 14, 16, 19]
several bounds are known for particular finite projective planes concerning maximal arcs
and saturating sets.
It is well known (and mentioned first in [19] concerning saturating sets) that the following
proposition must hold:
Proposition 1.2. |S|>√2q+ 1 for any saturating set Sin Πq.
This estimate is also known as the Lunelli-Sce bound for complete arcs.
If qis a square and the plane is Desarguesian, then the existence of saturating sets with
the same order of magnitude up to a constant factor is known due to Boros, Sz˝onyi and
Tichler [6].
Proposition 1.3. [6, 19] The union of three non-concurrent Baer sublines in a Baer
subplane from a Desarguesian plane of square order is a saturating set of size 3√q.
In fact, this can be even generalized since any 2-blocking set of a Baer-subplane, which
meets every line in at least 2 points, provides a saturating set of the ground plane. Thus
if qis a 4th power, this bound can be even improved to roughly 2√qdue to Davydov
et al. [10] and Kiss et al. [17]. Similar order of magnitude can be achieved if qis a 6th
power [10].
Proposition 1.4. [10, 17] The union of the point set of two disjoint Baer subplane of a
Baer subplane from a Desarguesian plane is a saturating set, of size 2√q+ 2 4
√q+ 2.
(The existence of Baer-subplanes in Baer subplanes of PG(2, q) tacitly implies that qis a
4th power.)
However, in the general case when the plane is not necessarily Desarguesian or the order
is arbitrary, we only have much weaker results. Following the footprints of Boros, Sz˝onyi
and Tichler [6], Bartoli, Davydov, Giulietti, Marcugini and Pambianco [3] obtained an
estimate on the minimal size of a saturating set in Πq.
2

Proposition 1.5 (Bartoli et al.).[3] min |S| ≤ p4(q+ 1) ln (q+ 1) + 2, if Sis a satu-
rating set in Πq.
Our main theorem improves the constant term to √3.
Theorem 1.6. min |S| ≤ (√3 + o(1))√qln q, if Sis a saturating set in Πq.
In fact, we prove the exact result min |S|≤d√3qln qe+d(√q+ 1)/2e.
Computer searches suggest that for Galois planes, this bound is still not sharp and the
correct order of magnitude is probably O(√q), see [9]. However, we will see in Section 4
that our estimate might be sharp in general, if we cannot built on any algebraic structure
of the plane. We focus mainly on the order of magnitude of min |S|, partly because it has
been determined for qsmall in P G(2, q), see [4, 5] and the references therein. In particular,
Bartoli et al. showed an upper bound slightly above (√3)√qln qfor the smallest size of
complete arcs in P G(2, q) under a certain probabilistic conjecture, and their computer
search results suggest that this is indeed the right order of magnitude [4, 5]. These results
also support the assumption that Theorem 1.6 might be sharp for general planes.
The paper is built up as follows. In Section 2, we present the first proof of Theorem
1.6, applying a refined version of the first moment method, where an almost suitable
structure is proved via random argument, which can be fixed. This idea appeared first in
Erd˝os’ work, who proved this way the existence of dense, complete bipartite graph free
and dense, even cycle free graphs, see e.g. in [2] (Chapter 6). In Section 3, we proceed by
showing that an advanced greedy-type algorithm also provides a saturating set of this size.
Finally, in Section 4 we analyze the above approaches and point out their connection to
hypergraph cover problems, which suggests that probably in general (non-desarguesian)
projective planes it would be hard to improve the order of magnitude. Note that similar
connection between geometric problems and transversals in hypergraph appeared before
several times [1]. We finish with a number of open problems.
2 Probabilistic argument - First proof
Proof of Theorem 1.6. The key idea is the following: we put in every point to our future
saturating set S∗with a given probability p:= p(q) determined later on, independently
from the other points. Then we complete it to obtain a saturating set via Lemma 2.1.
Lemma 2.1. Consider a set of points S∗and the corresponding set Rcontaining those
points which are not determined by the lines of S∗. One can add at most d|R|/2epoints
to S∗so that the resulting set is a saturating set.
Proof. Pair up the points of Rand for each pair {x, x0}, choose the intersection of hx, si
and hx,0s0ifor two different points s, s0in S∗. Clearly, the addition of the intersection
makes the points x, x0determined.
3

Clearly if Xis random variable counting the number of points in S∗, we have E(X) =
p(q2+q+ 1). Then if Yis random variable counting the number of points not in hS∗, S∗i
we get
E(Y) = (q2+q+ 1)P(a given point is not determined )
= (q2+q+ 1)(1 −p)(q2+q+1) p
1−p+1 + qp
1−pq+1!.(2.1)
Indeed, in order to obtain that a given point is not determined, either a point is chosen,
thus no further points may appear; or it is not chosen and on every line on the point at
most one point is chosen. The expression we get this way simplifies to (2.1).
It is easy to see that (q2+q+ 1)(1 −p)(q2+q+1) p
1−pdoes not contribute to the main
term of E(Y), so we omit it.
We want to choose the value of p:= p(q) in such a way that the sum E(X) + d1
2E(Y)eis
minimized, which would provide a suitable saturating set via the first moment method,
in view of Lemma 2.1. This yields p≈1
2exp 2p
p−2(q2+q+ 1)exp (q+ 1) ·2pq
1−p
pq
1−p+2 if we
apply the approximation forms via Taylor’s theorem
(1+1/x)x+0.5=e·1 + 1
12x2−O1
x3 (x→ ∞)⇔1+z≈exp 2z
z+ 2(1+O(z3)) (z→0),
for z=pq
1−p, as 1 + z2
12 −O(z3)2z
z+2 = (1 + O(z3)) if z1, and
(1−1/x)x−0.5=e−1·1−1
12x2−O1
x3 (x→ ∞)⇔1−p≈exp 2p
p−2(1−O(p3)) (p→0).
From this, one can derive
1
2E(Y) = 1
2(q2+q+ 1) exp 2p
p−2(q2+q+ 1)·∆q2+q+1
1exp (q+ 1) ·2pq
1−p
pq
1−p+ 2!·∆q+1
2,
where ∆1= (1 −O(p3)) and ∆2= (1 + O(z3)) denote the error terms with z=pq
1−p.
4

By simplifying the main term in E(Y), we get
1
2(q2+q+ 1) ·exp 2p
p−2(q2+q+ 1)exp (q+ 1) ·2pq
1−p
pq
1−p+ 2!=
1
2(q2+q+ 1) ·exp 2p
p−2+ 2p(q2+q)·(1
p−2+1
pq + 2(1 −p))≈
1
2(q2+q+ 1) ·exp 2p
p−2+ 2pq(q+ 1) ·−p(q−1)
4≈
1
2(q2+q+ 1) ·exp −1
2q(q2−1)p2,
(2.2)
after we omit the smaller order terms during the approximation.
The calculation above leads to the choice p=√3√qlnq
q2+q+1 , which in fact implies that the main
term in E(Y) (2.2) equals to O(√q), and it is easy to check that the same holds for the
error terms coming from ∆1and ∆2. This in turn provides the existence of a saturating
set of size
E(X) + d1
2E(Y)e= (1 + o(1))p3qln q.
3 Algorithmic approach - Second proof
Below we will present an algorithm to choose the point set of a saturating set S. We start
with an empty set S0in the beginning, and increase its cardinality by adding one point
in each step.
Notation 3.1. In the ith step, we denote the current set Sithat will be completed to a
saturating set, Didenotes the points of the plane outside Siwhich are determined by Si
and Ridenotes the set of points not in Si∪Di. For a point set H,σ(H)denotes the set
of lines skew to H.
The benefit of a point b(P)(in step i) is the amount of points from Riwhich would become
determined by the point set Si∪ {P}, that is, b(P) = |hP, Sii ∩ Ri|.
To obtain Di+1 from Di, we would like to add a point to Siwhich has the largest benefit.
Consider all the lines skew to Si, and choose one of them for which the intersection with
Riis minimal. Adding up the benefits of the points of this line `∗, we get
X
P∈`∗
b(P) = |Ri∩`∗|+i· |Ri\`∗|.(3.1)
Indeed, the double counting counts any point P∈Ri∩`∗exactly once as these points
become determined only if we choose Pitself from `∗. On the other hand, any point Q
5

of Rioutside `∗will be determined by adding a point from `∗∩hQ, Sii. This latter point
set is of cardinality ias Qwas not determined before.
Lemma 3.2.
min
`∈σ(Si)|Ri∩`| ≤ |Ri|
q.
Proof. There are at least (q2+q+ 1) −iq −1 = q(q+ 1 −i) skew lines to any set of i
points. Every point of Riis appearing on exactly (q+ 1 −i) lines from σ(Si), hence
X
`∈σ(Si)|Ri∩`|=|Ri|(q+ 1 −i),
thus the statement follows.
Proposition 3.3.
|Ri+1|≤|Ri|1−i
q+ 2
if we add the point having the largest benefit to Si.
Proof. We have b(P)≥1
q+1 (|Ri∩`∗|+i· |Ri\`∗|) for the point Phaving the largest
benefit on the skew line `∗of minimal intersection with Ri, in view of Equation 3.1.
Observe that |Ri∩`∗| ≤ |Ri|
q, according to Lemma 3.2. By adding this point Pto Si, we
get
|Ri+1|=|Ri|−b(P)≤ |Ri|− |Ri|
q+i(|Ri| − |Ri|
q)
q+ 1 ≤ |Ri|1−i(q−1)
q(q+ 1)<|Ri|1−i
q+ 2.
Lemma 3.4.
k:=d√3qln qe
Y
i=1 1−i
q+ 2< q−3/2.
Proof. Denote Qk
i=1 1−i
q+2 =(q+1)!
(q+2)k(q−k+1)! by Aq(k). We apply Stirling’s approxima-
tion
n!
t!<√2πn n
en
√2πt t
et
for t<nwith t=q−k+ 1 and n=q+ 1. This implies
Aq(k)<p(q+ 1)
p(q−k+ 1) ·(q+ 1)q+1
(q+ 2)k·(q−k+ 1)q−k+1 ·ek.
One again we use the Taylor-form for the approximation (1+1/x)x+0.5< e·1 + 1
12x2(x > 1),
6

to obtain
1 + k
q−k+ 1q−k+1
=1 + k
q−k+ 1(q−k+1
k+1/2)k1 + k
q−k+ 1−k/2
<
ek·1 + k2
12(q−k+ 1)2kq−k+ 1
q+ 1 k/2
.
(3.2)
Applying the approximation also for (1 −1/x)x−0.5< e−1·1−1
12x2(x→ ∞), we can
simplify
q−k+ 1
q+ 1 k/2q+ 1
q+ 2k
=1−(q+ 2)(k+ 2) + 1
(q+ 2)2k/2
<1−(k+ 2)
(q+ 2) k/2
=
1−(k+ 2)
(q+ 2) (q+2
k+2 −0.5)·k/2·(q+2
k+2 −0.5)−1
<e−1·1−(k+ 2)2
12(q+ 2)2k(k+2)
2q−k+2
.
(3.3)
In total, if we take into consideration both (3.2) and (3.3), we get that
Aq(k)< exp −k(k+ 2)
2q−k+ 2∆3(k),
where ∆3(k) is the product of error terms √(q+1)
√(q−k+1) ,1 + k2
12(q−k+1)2kand
1−(k+2)2
12(q+2)2k(k+2)
2q−k+2 .
If one plugs in k=d√3qln qe, careful calculations on the error term imply that
Aq(k)< e−((1+)3
2ln q),
where =(q)>0.
By applying Proposition 3.3 successively k=d√3qln qetimes it follows from Lemma 3.4
that we end up with at most √q+ 1 points remaining in Rk. But note that these points
can be covered by adding a further d(√q+ 1)/2epoints to the saturating set, via Lemma
2.1, hence this algorithm provides a saturating set of size d√3qln qe+d(√q+ 1)/2e.
4 Connections with hypergraph coverings, applica-
tions and open problems
To improve further the bound of Theorem 1.6 in the main term, one must ensure at least
one of the followings.
7

(1) Prove the existence of a point that has significantly more benefit then the others,
for several steps.
(2) Prove a lemma, which provides a completion of almost saturating sets to obtain a
saturating set with fewer points than Lemma 2.1. That might enable us to apply
Lemma 3.4 until a smaller summation limit.
Concerning (1), observe that while avoiding lines `which are secants of Siduring the choice
for the newly added point, it is not guaranteed that the new point won’t lie eventually
on a secant of Si, so this method does not enable us to provide arcs. However, as these
tangents does not contain points for Ri, the sum of the benefits on these tangent lines
are certainly less than the sum determined in Equation 3.1. Despite all this, the known
constructions with O(√q) points on Galois planes of square order consist of point sets
having large secant sizes — but note that we heavily built on the algebraic (sub)structure.
Concerning (2), notice first that the constant √3 in the bound (and also in the background
of the first probabilistic proof) is explained by the application domain of the approaches,
as the first phase of both methods lasted until the number of not saturated points decreases
from q2to O(√q), so if one cannot exploit the structure of the not saturated points hence
cannot obtain a good bound on the variance of the distribution of the benefit values, then
Lemma 3.4 would provide an evidence that O(√q) points won’t form a saturating set in
general.
This problem is strongly connected to the theory of bounding the transversal number of
certain hypergraphs. Recall that for an r-uniform hypergraph H, the covering number or
transversal number τ(H) is the the minimum cardinality of a set of vertices that intersects
all edges of H.
Consider a point set S0in the projective plane. Let X(S0) = {x1, . . . , xm}be the set
of points not saturated by S0, that is, those points which are not incident to the lines
determined by the point pairs of S0. Assign to each xia set Hiof |S0|(q−1) + 1 points
which make xisaturated, i.e. Hi=hxi, S0i \ S0.
Note that the intersection of these sets consists of many points.
Lemma 4.1. |Hi∩Hj|=|S0|(|S0| − 1) or |Hi∩Hj|= (|S0| − 1)(|S0| − 2) + qfor every
1≤i<j≤m.
Proof. Suppose first that hxi, xji ∩ S0=∅. Then for every z6=z0∈S0, the points
hxi, zi ∩ hxj, z0iare distinct, hxi, zi ∩ hxj, z0i ∈ Hi∩Hj, and they are points outside S0,
hence |Hi∩Hj|=|S0|(|S0| − 1).
In the other case when |hxi, xji ∩ S0|= 1, a point z∗∈S0is incident to hxi, xji. Now
for every z6=z0∈S0with z6=z∗6=z0, the points hxi, zi ∩ hxj, z0i ∈ Hi∩Hjare distinct
again and not belonging to S0, while hxi, z∗i=hxj, z∗ithus hxi, z∗i \ z∗also belongs to
the intersection, hence the claim follows.
8

Lemma 2.1 may thus be altered to a much more general lemma concerning the transversal
number of t-intersecting uniform hypergraphs, applied to the sets {Hi=hxi, S0i\S0}. So
an (approximate) solution for the open problem below would imply stronger results for
the saturation problem as well.
Problem 4.2. Given F={Hi:i= 1 . . . m}an r-uniform t-intersecting set system on
an n-element ground set, prove sharp upper bounds on τ(F)in terms of n, m,tand r.
It is well known that while the trivial bound τ(F)≤r−t+ 1 can be sharp for dense
hypergraphs, the approximation of the transversal number is hard in general. Due to
Lov´asz [18], a connection is made with the fractional transversal number τ∗(F) (which is
also called fractional covering number) as τ < (1 + ln d)τ∗, where dis the maximal degree
in the hypergraph.
Note that this result yields instantly the bound τ < (1 + ln d)rm
r+(m−1)t.
By exploiting the intersection property, the greedy algorithm yields that there exists a
point in any Hithat is contained in at least 1 + tm
rsets from F. Hence the following
proposition follows from the recursion:
Proposition 4.3. Let F={Hi:i= 1 . . . m}be an r-uniform t-intersecting set system
on an n-element ground set. Then τ(F)≤ d rm
tm+rln me.
For further results, we refer to the survey of F¨uredi [12] and the paper of Alon, Kalai and
Matouˇsek and Meshulam [1]. The investigation of the transversal number of t-intersecting
hypergraphs was proposed also by F¨uredi in [11].
Multiple saturating sets (see e.g. in [3]) and their generalizations in higher dimensional
spaces are also investigated. A point set Sin PG(n, q) is saturating if any point of
PG(n, q)\Sis collinear with two points in S. The two proof techniques presented in
Sections 2 and 3 are applicable in these more general settings as well. Namely, the natural
analogue of the Lunelli-Sce bound provides the following lower bound for a saturating set
Sin PG(n, q):
Proposition 4.4. |S| ≥ √2qn−1
2.
The direct analogue of the first (probabilistic) approach shows the existence of a saturating
set in PG(n, q) of size |S| ≤ (1+o(1))p(n+ 1)qn−1ln q, which improves previously known
bounds for n= 4.
Acknowledgement
Grateful acknowledgement is due to the anonymous referees for their helpful suggestions
in order to improve the presentation of the paper.
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