Chapter 22. Application to discrete geometry
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138
22. Application to discrete geometry
solved by Gallai, and popularized by Erd˝
os through his article in the
American Mathematical Monthly [31].
Theorem 22.2. Any set of m noncollinear points in the plane determine at least m lines.
The idea of the following proof goes back to de Bruijn and Erd˝os [16].
Proof. Let p1 , . . . , pm be the given m points. Suppose that they
determine n lines, and let L = {l1 , l2 , . . . , ln } be the set of the lines.
Let Ci ⊂ L be the subset of lines passing through pi , and let H =
{C1 , . . . , Cm } ⊂ 2L . Then, for all i = j, Ci ∩Cj contains just one line,
that is, the line connecting two points pi and pj . Thus Ci ∩ Cj  = 1,
and by the Fisher’s inequality we get H = m ≤ L = n.
Chv´
atal posed a generalization of this problem in a metric space.
We say that three points x, y, and z in a metric space with metric d
are collinear1 if they satisfy
d(x, z) = d(x, y) + d(y, z).
For example, a graph can be viewed as a metric space by the usual
shortestpath metric. Then we can deﬁne a line in the graph in the
sense above. Such lines behave rather diﬀerently from the usual lines
in Euclidean space. Nevertheless, Chen and Chv´
atal conjectured the
following.
Conjecture 22.3 (Chen–Chv´
atal [18]). In any metric space, m noncollinear points determine at least m lines.
22.2. Chromatic number of the unitdistance
graph
A family of subsets H ⊂ 2[n] is called tavoiding if
H ∩ H  = t
for all H, H ∈ H. Applying Theorem 21.8 to the case
n = 4p − 1, s = p − 1,
we get the following.
1
This is called collinearity by betweeness.
L = {0, . . . , p − 2},
22.2. Chromatic number of the unitdistance graph
Corollary 22.4 ([67]). Let p be a prime. If a family H ⊂
(p − 1)avoiding, then
H ≤
139
[4p−1]
2p−1
is
4p − 1
4p − 1
4p − 1
+
+ ··· +
.
0
1
p−1
4p−1
Exercise 22.5. Show that 4p−1
+ 4p−1
+ · · · + 4p−1
0
1
p−1 < 2 p−1 .
s
(Hint: Use induction on s to deduce that i=0 ni < 2 ns for n ≥ 3s.)
Now we present a geometric application of the above result. The
distanced graph in ndimensional Euclidean space has vertex set Rn
and two points are adjacent if their Euclidean distance is d. If d =
1 then it is called the unitdistance graph and denoted by Gn . In
symbols:
• V (Gn ) = Rn ,
• E(Gn ) = {{x, y} : x − y = 1},
where x = x21 + · · · + x2n for x = (x1 , . . . , xn ) ∈ Rn . Note that
distanced graphs in a ﬁxed dimension are isomorphic to each other
for all d > 0.
It is diﬃcult to determine the chromatic number of Gn , and even
for the planar case (n = 2) we only know that
5 ≤ χ(G2 ) ≤ 7.
For the general case Larman and Rogers [86] proved that
χ(Gn ) < (3 + o(1))n .
On the other hand, Frankl and Wilson [67] obtained the following
lower bound using Corollary 22.4.
Theorem 22.6. Let p be a prime, and let n = 4p − 1. Then we have
χ(Gn ) > 1.1n .
√
Proof. Let k = 2p − 1 and d = 2p. As usual we identify 2[n] and
Ω = {0, 1}n . We are going to ﬁnd a large distanced structure in the
ndimensional cube Ωn . To this end we look at the points having
exactly k 1’s in the coordinate, that is, we look at subsets in [n]
k . If
satisfy
two subsets X, Y ∈ [n]
k
X ∩ Y  = p − 1,
140
22. Application to discrete geometry
then the corresponding characteristic vectors satisfy
x − y = d,
where we used x − y
2
= X Y  = 2(k − X ∩ Y ) = 2p.
Now suppose that we can color [n]
so that any two points with
k
distance d get diﬀerent colors. Then a family of subsets having the
same color is (p − 1)avoiding, and, by Corollary 22.4 with Exern
. So in order to
cise 22.5, the size of the family is less than 2 p−1
color the distanced graph, or equivalently to color G2 , we need at
n
n
/(2 p−1
) > 1.1p colors2 .
least 2p−1
With a little bit more eﬀort one can show that χ(Gn ) > 1.2n for
suﬃciently large n ∈ N; see [67].
2
n
1.1 .
n
2p−1
/(2
n
p−1
)=
1 (3p)···(2p+1)
2 (2p−1)···p
=
3p (3p−1)···(2p+1)
2p (2p−1)···(p+1)
> (3/2)p = (3/2)
n+1
4
>
Chapter 23
Upper bounds using
inclusion matrices
For two hypergraphs F and G we can deﬁne an F × G matrix
M whose (F, G)entry is determined by a given function m(F, G).
Obviously rank M ≤ min{F, G}. If the rows of M are linearly
independent, then we get rank M = F ≤ G. In this chapter we
present some applications of this inequality.
23.1. Bounds for sindependent families
[n]
For 0 ≤ i ≤ k ≤ n, F ⊂ [n]
k , and G ⊂
i , deﬁne the inclusion
matrix M (F, G) as follows. This is an F × G matrix whose (F, G)entry m(F, G), where F ∈ F and G ∈ G, is deﬁned by
m(F, G) =
For F ⊂
(23.1)
[n]
k
1
if F ⊃ G,
0
if F ⊃ G.
and 0 ≤ j ≤ i ≤ k, simple counting yields
M (F,
[n]
i
)M (
[n]
i
,
[n]
j
)=
k−j
i−j
M (F,
[n]
j
In fact, the (F, J)entry of (23.1), where F ∈ F and J ∈
#{I ∈
[n]
i
).
[n]
j
, counts
: J ⊂ I ⊂ F }.
141
142
23. Upper bounds using inclusion matrices
For a matrix M , the column space is a vector space spanned by its
column vectors. Let colsp M denote the column space of M . Then
(23.1) also shows the following.
Lemma 23.1. Let 0 ≤ j ≤ i ≤ k and F ⊂
colsp M (F,
[n]
j
[n]
k
) ⊂ colsp M (F,
. Then
[n]
i
).
[n]
We say that F ⊂ [n]
k is sindependent if the rows of M (F, s )
are linearly independent, that is, the inclusion matrix has full rowrank. In this case, F ≤ ns immediately follows.
Lemma 23.2. Let p be a prime and let f ∈ Q[x] be a polynomial of
degree s. Let F ⊂ [n]
k . Suppose that F satisﬁes
(23.2)
f (F ∩ F )
≡ 0 (mod p)
if F = F ,
≡ 0 (mod p)
if F = F
for F, F ∈ F. Then F is sindependent.
Proof. Since f is of degree s, it can be uniquely represented as
s
f (x) =
αi
i=0
x
,
i
where αi ∈ Q. Let Mi = M (F, [n]
i ) be the inclusion matrix. Then
T
the (F, F )entry of Mi Mi counts the number of ielement subsets

. Let
contained in both F and F , that is, F ∩F
i
s
(23.3)
αi Mi MiT .
A=
i=0
Then the (F, F )entry of A is f (F ∩F ), and A (mod p) is a diagonal
matrix with no zero diagonal entries. So A is nonsingular because
det A ≡ 0 (mod p), and rank A = F. By Lemma 23.1 it follows that
colsp Mi ⊂ colsp Ms for 0 ≤ i ≤ s. This, together with (23.3), yields
colsp A ⊂ colsp Ms , and hence rank A ≤ rank Ms . Thus Ms has full
rowrank, that is, F is sindependent.
Theorem 23.3 (Frankl–Wilson [67]). Let p be a prime and let L ⊂
[0, p − 1] = {0, 1, . . . , p − 1} with L = s. Suppose that F ⊂ [n]
k