Abstract. Consider the following problem: how many collinear triples of points must a transversal of Z n × Z n have? This question is connected with venerable issues in discrete geometry. We show that the answer, for n prime, is between (n − 1)/4 and (n − 1)/2, and consider an analogous question for collinear quadruples. We conjecture that the upper bound is the truth and suggest several other interesting problems in this area

Joshua N Cooper

József Solymosi

CiteSeerX

Annals of Combinatorics 9 (2005) 169-175
0218-0006/05/020169-7
DOI 10.1007/s00026-005-0248-4
c© Birkhäuser Verlag, Basel, 2005
Annals of Combinatorics
Collinear Points in Permutations
Joshua N. Cooper1 and József Solymosi2
1Department of Mathematics, Courant Institute of Mathematical Sciences, New York University
New York, USA
cooper@cims.nyu.edu
2Department of Mathematics, University of British Columbia, 1984 Mathematics Road, Vancou-
ver, BC, Canada
solymosi@math.ubc.ca
Received August 29, 2004
AMS Subject Classification: 51E15, 11T99
Abstract. Consider the following problem: how many collinear triples of points must a transver-
sal of Zn ×Zn have? This question is connected with venerable issues in discrete geometry. We
show that the answer, for n prime, is between (n−1)/4 and (n−1)/2, and consider an analogous
question for collinear quadruples. We conjecture that the upper bound is the truth and suggest
several other interesting problems in this area.
Keywords: finite field, affine geometry, collinearity, transversal
1. Introduction
In [4], Erdős offered a construction concerning the “Heilbronn Problem”. What is the
smallest A so that, for any choice of n points in the unit square, some triangle formed
by three of the points has area at most A? His elegant construction of a point-set with
large minimum-area triangle (∼ n−2) is as follows: take the smallest prime p ≥ n, and
let the set of points be {p−1(x, x2 (mod p)) : x ∈ Zp}. (If necessary, throw out a few
points so that there are n left.) It is easy to see that this set has no three collinear points,
and therefore any three points form a nondegenerate lattice triangle – which must have
area at least p−2/2 ∼ n−2.
Another area in which collinear triples of points on a lattice arise is in connection
with the so-called “no-three-in-line” problem, dating back at least to 1917 ([1]). Is it
possible to choose 2n points on the n-by-n grid so that no three are collinear? Clearly,
if this is the case, then 2n is best possible. Guy and Kelly ([2]) conjecture that, for
sufficiently large n, not only is it true that every set of 2n points has a collinear triple,
but that it is possible to avoid collinear triples in a set of size (α− ε)n and impossible
to avoid them in a set of size (α+ ε)n, where α = (2π2/3)1/3 ≈ 1.874 and ε > 0.
169
170 J.N. Cooper and J. Solymosi
In this note, we address the question of when it is possible to avoid collinear triples
modulo n, particularly in the case of transversals (i.e., graphs of permutations) and when
n is prime.
2. Results
Suppose that S is a subset of Zn ×Zn. We say that a set of points X ⊂ S is “collinear”
if there are parameters a, b, c ∈ Zn so that each (i, j) ∈ X lies on the line {(x, y) ∈
Zn ×Zn : ax+by = c}. If n is prime, the “slope” of a such line is defined to be ab−1 if
b 6= 0 and ∞ otherwise. Clearly, each pair of points is collinear, and we say that slope
of a pair of points is the slope of the line containing them. (It is easy to see that this is
well-defined.)
If f : Zn → Zn is some function, then we say that the “graph” of f is the set
{(x, f (x)) : x ∈ Zn}. Often we will say that a set of points of a function are collinear,
where really we mean that a set of points of its graph are.
We are interested in the number of collinear triples in subsets S of Zn ×Zn, and
conditions guaranteeing that there is at least one. Note that, when n is prime, if S is
a subset with no collinear triples and |S| = m, then (m− 1)(n− 1) + 1 ≤ n2. Indeed,
fix some point s ∈ S; then each line through s contains n− 1 points of Zn ×Zn, plus
s itself. If we count each line passing through s and some other t ∈ S, then the total
number of points is (m−1)(n−1)+1, since the lines are pairwise disjoint except for s.
Therefore the inequality holds, and moreover, m ≤ n+2. If some point s = (s1, s2) has
the property that (t, s2) 6∈ S for any t 6= s1, then (m−1)(n−1)+n ≤ n2, and therefore
m ≤ n+1. Finally, if s has the property that (t1, s2) 6∈ S and (s1, t2) 6∈ S for any t1 6= s1,
t2 6= s2, then this count yields (m−1)(n−1)+n+(n−1)≤ n2, i.e., m ≤ n. Therefore,
in general, (1) a set with no collinear triples can have at most n+2 points, (2) a set with
no collinear triples which has some column (or row) containing only one point can have
at most n + 1 points, and (3) a set with no collinear triples which has a point lying in
an otherwise empty column and row can have at most n points. It is interesting to ask,
then, how many collinear triples must a permutation of Zn have? Define Ψ(n) to be the
minimum number of collinear triples in any permutation of Zn.
2.1. Permutations
Theorem 2.1. Suppose σ is a permutation of Zn for n > 2 prime. Then σ contains at
least one collinear triple of points, i.e., Ψ(n) > 0.
Proof. Suppose not. Note that for every pair of points in the graph of σ, the slope of
that pair must be in 1, . . . , n− 1. Partition the
(n
2
)
pairs into classes according to their
slopes. Since there are n−1 classes, at least n/2 pairs lie in some class. Each of these
pairs lies on some line of the same slope, and no two of them lie on the same line, since
then we would a collinear triple. Therefore, we have at least n/2 lines, each of which
contains two points. However, n is odd, so there are in fact at least (n + 1)/2 disjoint
lines containing two points. This amounts to n+1 points in total, a contradiction.
By the argument above, it is clear that the
(n
2
)
pairs lie in at most (n− 1)2/2 =
(n
2
)
− (n− 1)/2 lines. Therefore, a natural question is to ask, if K pairs are assigned
Collinear Points in Permutations 171
to L families, what is the minimum number of triples occurring entirely within some
family? That is, suppose G is a graph with K edges, and E(G) is partitioned into L
“lines” E1, . . . , EL. Denote by Vi the set of vertices spanned by Ei. What is the least
possible value of
L
∑
i=1
(|Vi|
3
)
,
over all G and partitions {Ei}? Call this quantity T (K, L). If some G and a partition
of its edges achieves this bound, then the edges which belong to a given line must span
the minimum number of vertices. That is, if |Ei| = mi, then
|Vi| =
⌈
1+
√
1+8mi
2
⌉
=def τ(mi),
because this is the smallest number k so that
(k
2
)
≥ mi. Therefore, T (K, L) is the least
possible value of
Trip(m1, . . . , mL) =def
L
∑
i=1
(
τ(mi)
3
)
over all L-tuples m1, . . . , mL of nonnegative integers whose sum is K.
Write ρ(t) for t −2bt/2c, i.e., the parity of t.
Proposition 2.2. For K ≤ 3L, T (K, L) = max{d(K −L)/2e , 0}.
Proof. If K ≤ L, it is clear that we may set mi = 1 for 1 ≤ i ≤ K and mi = 0 oth-
erwise, so that T (K, L) = 0. Therefore, suppose L < K ≤ 3L. We make the follow-
ing claim: there is some partition K = m1 + · · ·+ mL minimizing Trip(m1, . . . , mL)
which has b(K −L)/2c indices i so that mi = 3, ρ(K − L) indices i so that mi = 2,
K−3b(K −L)/2c−2ρ(K−L) indices i so that mi = 1, and the other mi = 0. We prove
the claim by induction on K. The base case K = L we dealt with above. If K = L + 1
or K = L + 2 then we may set m1 = K −L + 1 and mi = 1 for 2 ≤ i ≤ L, resulting in
Trip(m1, . . . , mL) = 1, which is clearly best possible. Suppose, then, that K ≥ L+3.
By the pigeonhole principle, we may assume that m1 > 1 without loss of generality.
Furthermore, since τ(2) = τ(3) and increasing m1 can only decrease
Trip(m2, . . . , mL) = Trip(m1, . . . , mL)−
(
τ(m1)
3
)
,
we have that m1 ≥ 3 unless K ≤ 2, i.e., we are in the base case. If m1 = 3, then
Trip(m1, . . . , mL) = 1+Trip(m2, . . . ,mL) = 1+
⌈
K −3− (L−1)
2
⌉
=
⌈
K −L
2
⌉
.
172 J.N. Cooper and J. Solymosi
If, on the other hand, m1 > 3, then τ(m1) ≥ 4 and L ≥ 2. In that case,
Trip(m1, . . . , mL) =
(
τ(m1)
3
)
+Trip(m2, . . . , mL)
=
(
τ(m1)
3
)
+
⌈
K −L−m1 +1
2
⌉
≥ K −L+1
2
+
(
m1
3
)
− m1
2
>
⌈
K −L
2
⌉
,
where the second line follows from induction and the fourth from the fact that
(τ(m1)
3
)
−
m1
2 > 0 for m1 > 3. Since this contradicts the minimality of the partition, we must have
m1 = 3. Furthermore, there are
1+
⌊
K −L−2
2
⌋
=
⌊
K −L
2
⌋
indices i with mi = 3, ρ(K −L−2) = ρ(K −L) indices i with mi = 2,
K −3−3b(K −L−2)/2c−2ρ(K−L−2) = K −3b(K −L)/2c−2ρ(K−L)
indices i with mi = 1 and the rest zeroes.
We may apply this immediately to the original question by setting K =
(n
2
)
and
L = (n−1)2/2.
Corollary 2.3. For n > 2 prime, Ψ(n) ≥ d(n−1)/4e.
On the other hand, we have the following proposition.
Proposition 2.4. For n > 2 prime, Ψ(n) ≤ (n−1)/2.
Proof. Define the function f : Zn → Zn by f (x) = x−1 if x 6= 0 and f (0) = 0. This
is clearly a permutation, and we show that its graph has exactly (n− 1)/2 collinear
triples. First, suppose that 1 ≤ x < y < z ≤ n−1. It is easy to see that the condition that
(x, f (x)), (y, f (y)), and (z, f (z)) are collinear is equivalent to the statement that
( f (z)− f (x))(y− x)− ( f (y)− f (x))(z− x)
= (z−1 − x−1)(y− x)− (y−1− x−1)(z− x) = 0 (mod n).
Multiplying by xyz yields
y(x− z)(y− x)− z(x− y)(z− x) = (y− z)(x− z)(y− x) = 0,
which is impossible because x, y, and z are distinct. Therefore, if f exhibits a collinear
triple, it must have some point with abscissa 0. Without loss of generality, we may
assume that x = 0. Therefore, if 0 = x < y < z ≤ n−1,
z−1y− y−1z = 0,
so that y2 = z2, i.e., y = ±z. Since y 6= z, y = −z, and there are (p− 1)/2 unordered
triples of the type {0, y, −y}.
Collinear Points in Permutations 173
In fact, we believe the following conjecture to be true.
Conjecture 2.5. Ψ(n) = (n−1)/2 for n > 2 prime.
We have included computational data supporting this conjecture in Table 1.
Table 1: The first 18 values of Ψ(x).
x Ψ(x) x Ψ(x) x Ψ(x)
1 0 7 3 13 6
2 0 8 0 14 9
3 1 9 5 15 6
4 0 10 2 16 4
5 2 11 5 17 8
6 0 12 0 18 ≤ 16
Problem 2.6. What about n composite?
About this question, unfortunately, we can say nothing.
Note that, by the proof of Proposition 2.4, any fractional linear transformation of the
form (ax+b)/(cx+d) with c 6= 0 (along with −dc−1 7→ ac−1) gives rise to (n−1)/2
collinear triples for n > 2 prime.
Conjecture 2.7. The function
g(x) =
{
x/(x−1), if x 6= 1,
1, if x = 1.
is the lexicographic-least permutation with (n−1)/2 collinear triples for n > 2 prime.
2.2. Quadruples
The permutation from Proposition 2.4 has the property that, if we remove the point
(0, 0), the resulting graph is a collinear triple-free set of n−1 points with no two on a
single row or column – showing that Theorem 2.1 is right. A moment’s reflection also
reveals that it has no collinear 4-tuple.
There are permutations with many collinear triples which have no collinear 4-tuples,
however. Consider h(x) = x3, a function on Zn, where n > 2 is prime and congruent
to 2 mod 3. Then h is a permutation, since the unique solution of x3 = c (mod n) is
x = c(n−1)/3 (mod n). Clearly, x3 − ax− b = 0 has at most three solutions, so no line
intersects the graph of h (or h0) in four points. Furthermore, x3 − ax− b cannot have
exactly two roots in Zn unless it has a double root, since
x3 −ax−b = (x− r1)(x− r2)(x− r3)
174 J.N. Cooper and J. Solymosi
implies that r3 = br−11 r
−1
2 unless r1 = 0 or r2 = 0. If r1 = 0, then x
3 − ax− b = x3 −
x2(r2 + r3)+ xr2r3, so r3 = −r2, and either both r2 and r3 are in Zn or neither is. (The
same holds for r1 and r3 if r2 = 0.) If it has a double root, then
x3 −ax−b = (x− r1)2(x− r2) = x3 − (2r1 + r2)x2 +(r21 +2r1r2)x− r21r2,
so r2 = −2r1. Therefore, a = 3r21 and b = −2r31. This gives n− 1 lines containing
exactly two points (r1 = 0 gives a triple root).
Each of the X collinear triples of h contains three pairs of points. No pair is con-
tained in two triples, since two collinear triples that intersect in two points form a
collinear quadruple. Furthermore, only n−1 pairs are not contained in some collinear
triple, so we have 3X + (n− 1) =
(n
2
)
, or X = (n− 1)(n− 2)/6. We may conclude
that, for each prime n > 2 congruent to 2 mod 3, there is some permutation with
(n− 1)(n− 2)/6 collinear triples, but no collinear quadruples. Unfortunately, such a
construction cannot work for p = 1 (mod 3), since no cubic permutation polynomials
exist for such p (see [3]).
Now, suppose that a permutation has X collinear triples but no collinear quadruples.
Each of the X triples contains three pairs of points and, again, each such pair appearing
in a triple appears in only one of them. Therefore, we must have 3X ≤ n(n−1)/2, i.e.,
X ≤ bn(n−1)/6c – and this holds for any n.
By the above observations, limsupn→∞ CT (n)/n
2 = 1/6. However, the question
remains what the lower bound is when n 6≡ 2 (mod 3), or when n is composite. There
is also a gap of about n/3 between the upper and lower bounds even in the case of n ≡ 2
(mod 3). We also wish to know:
Problem 2.8. What is the maximum number CT0(n) of collinear triples in a subset of
Zn ×Zn which has no collinear quadruples?
2.3. Pair Packing
We return to the argument of Proposition 2.2. Consider the following “greedy” process
for finding an assignment of K pairs to L lines. We proceed through the lines one at a
time, and when we reach Ei with K ′ pairs/edges unaccounted for, we place into it
min
{
K′,
(
τ(dK ′/(L− i+1)e)
2
)}
edges, arranged as a graph on τ(dK ′/(L− i+1)e) vertices. That is, we distribute the
remaining edges as equally as possible into the remaining lines, but round the number
of edges we place into the current line Ei up to the nearest triangular number whenever
possible. One might conjecture that this process results in the optimal configuration,
i.e, minimizing T (K, L) – but it does not. Indeed, already for K = 28 and L = 2, the
optimal configurations are (m1, m2) = (21, 5) and (20, 6), neither of which has a line
with mi = 15 =
(τ(28/2)
2
)
.
Problem 2.9. Describe those configurations which achieve T (K, L).
Collinear Points in Permutations 175
It is easy to see that, for each K and L, some optimal configuration has at most one
mi which is not a triangular number. Furthermore, if mi ≤ r for some r, then we may
move q points from the largest line m j =
(a
2
)
+q, 1 ≤ q ≤ a, to line i and in the process
change the number of total triples by
(
τ(q+ r)
3
)
−
(
a+1
3
)
+
(
a
3
)
≤
(
τ(a+ r)
3
)
−
(
a
2
)
.
Therefore, we have a contradiction if
(τ(a+r)
3
)
<
(a
2
)
. Since τ(x) ≤ 2+
√
2x,
(
τ(a+ r)
3
)
≤
(
2+
√
2(a+ r)
)3
6
.
For r large, a = 2r3/4 provides a contradiction. Therefore,
Proposition 2.10. If mini mi = r, r sufficiently large, in a configuration (m1, . . . , mL)
achieving T (K, L), then maxi mi ≤ 2r3/2.
It is easy to see that the function
g(x) =
(
(1+
√
1+8x)/2
3
)
=
x
6
(√
1+8x−3
)
is concave-up for x ≥ 0. Therefore, by Jensen’s Inequality, T (K, L) ≥ Lg(K/L). This,
in particular, implies that the number of collinear triples in a subset of Zn ×Zn of
cardinality X , n prime, is  X 3/n2 when X/n → ∞.
Surely, more than this can be said concerning Problem 2.9.
Acknowledgments. Thank you to Greg Martin, Jozef Skokan and Joel Spencer for valuable
discussions and insights.
References
1. H.E. Dudeney, Amusements in Mathematics, Nelson, Edinburgh, 1917.
2. R.K. Guy and P.A. Kelly, The no-three-in-line problem, Canad. Math. Bull. 11 (1968) 527–
531.
3. R. Lidl, and H. Niederreiter, Finite Fields, Encyclopedia of Mathematics and Its Applica-
tions, Vol. 20, Cambridge University Press, Cambridge, 1983.
4. K.F. Roth, On a problem of Heilbronn, J. London Math. Soc. 26 (1951) 198–204.


Collinear points in permutations

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Collinear points in permutations

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