The Problem of Coloring the Plane
Problem of Coloring the Plane
Tommy Jensen
Department of Mathematics
Kyungpook National University
May 29, 2010
Tommy Jensen
The Problem of Coloring the Plane
Outline
1
The Problem of Coloring the Plane
Tommy Jensen
The Problem of Coloring the Plane
The Problem
What is the smallest number of colors sufficient for coloring the
plane in such a way that no two points of the same color are
unit distance apart?
Edward Nelson and Hugo Hadwiger, ∼ 1950
Tommy Jensen
The Problem of Coloring the Plane
Edward Nelson, 1950
Tommy Jensen
The Problem of Coloring the Plane
Edward Nelson, 2003
Tommy Jensen
The Problem of Coloring the Plane
Hugo Hadwiger, 1973
Tommy Jensen
The Problem of Coloring the Plane
A Theorem of Hadwiger
Theorem (Hadwiger 1945)
Let S1 , S2 , S3 , S4 , S5 be congruent closed sets of points in the
plane R2 such that
S1 ∪ S2 ∪ S3 ∪ S4 ∪ S5 = R2 .
Then at least one of the sets S1 , S2 , S3 , S4 , S5 contains two
points x, y of distance dR2 (x, y ) = 1.
Congruent: can be mapped onto each other by a combination
of translations, rotations and reflections.
Closed: contains all of its limit points; it is the complement of an
open set.
Tommy Jensen
The Problem of Coloring the Plane
Different Formulation of the Problem
Hadwiger-Nelson Problem
What is the smallest number k such that there exist sets
S1 , S2 , . . . , Sk of points in the plane R2 such that
S1 ∪ S2 ∪ · · · ∪ Sk = R2 ,
and none of the sets S1 , S2 , . . . , Sk contains two points x, y of
unit distance dR2 (x, y ) = 1?
This number is called the chromatic number of the plane and it
is written as χ(R2 ).
Tommy Jensen
The Problem of Coloring the Plane
What do we know about χ(R2 )?
Theorem (Nelson 1950, L. Moser 1960)
χ(R2 ) > 3
The proof is:
The Moser spindle
Tommy Jensen
Wooden spindles
The Problem of Coloring the Plane
The Upper Bound
Theorem (Isbell 1950, Hadwiger 1961)
χ(R2 ) ≤ 7
With proof:
Tommy Jensen
The Problem of Coloring the Plane
The Upper Bound
Theorem (Hadwiger 1961)
There exist congruent closed sets S1 , S2 , S3 , S4 , S5 , S6 , S7 of
points in the plane R2 , avoiding unit distance, such that
S1 ∪ S2 ∪ S3 ∪ S4 ∪ S5 ∪ S6 ∪ S7 = R2 .
With the same proof:
Tommy Jensen
The Problem of Coloring the Plane
Since 1950 we know
Theorem
4 ≤ χ(R2 ) ≤ 7
The Golomb Graph
Tommy Jensen
The Problem of Coloring the Plane
Tommy Jensen
The Problem of Coloring the Plane
Polychromatic Number
Definition
The polychromatic number χp (R2 ) of the plane is the smallest
number of colors with which it is possible to color all points, so
that for each color there exists a positive number d such that no
two points of that color have distance d.
It is obvious that χp (R2 ) ≤ χ(R2 ).
Theorem (Raiskii, 1970)
4 ≤ χp (R2 ) ≤ 6.
Dmitrij Raiskii, a high-school student in Moscow at the time.
Tommy Jensen
The Problem of Coloring the Plane
How about the Line R?
We can color the real line R with two colors.
− − − • − − − −− • − − − −− • − − − −− • − − − −− • − −−
No two points of the same color have unit distance.
χ(R) = 2
Tommy Jensen
The Problem of Coloring the Plane
The Chromatic Number of Z2
We can color the integer points of R2 with two colors.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
χ(Z2 ) = 2
Tommy Jensen
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Problem of Coloring the Plane
The Chromatic Number of Q2
Rational points are dense in R2 . Hence the chromatic number
χ(Q2 ) should be close to the chromatic number χ(R2 ), right?
Which rational points have unit distance from (0, 0)?
• ( ba , dc )
Tommy Jensen
The Problem of Coloring the Plane
Assume that ( ba , dc ) is a rational point on the unit circle.
We can assume a, b, c, d > 0, and that not both of a, b are
even, and not both of c, d are even.
Then
c
a
( )2 + ( )2 = 1 ⇔ a2 d 2 + b2 c 2 = b2 d 2 .
b
d
Can b be odd and d even? NO: b2 c 2 would be odd, and
a2 d 2 , b2 d 2 both even.
Similarly b cannot be even and d odd.
Can b and d both be even?
Tommy Jensen
The Problem of Coloring the Plane
Suppose that ( ba , dc ) is a rational point on the unit circle, both of
b, d are even, and both of a, c are odd.
Let b = 2k s and d = 2` t, where k , ` > 0 and s, t are odd
numbers. We can assume ` ≤ k .
Then
a2 d 2 + b2 c 2 = b2 d 2 ⇔ 22` a2 t 2 + 22k s2 c 2 = 22k +2` s2 t 2
⇔ a2 t 2 +22k −2` s2 c 2 = 22k s2 t 2 , so ` = k must hold, since a2 t 2 is odd.
Then we have
a2 t 2 + s2 c 2 = (at)2 + (sc)2 = 22k s2 t 2 .
If at = 2n + 1 and sc = 2m + 1, then
(at)2 + (sc)2 = 4(n2 + n + m2 + m) + 2.
But 22k s2 t 2 is divisible by 4, a contradiction.
We deduce that b and d cannot both be even. Therefore they
are both odd.
Tommy Jensen
The Problem of Coloring the Plane
Odd Rationals
Definition
Call a rational number odd if it can be written as
p
q
for integers p and q where q is odd.
Definition
Call a point (r1 , r2 ) in Q2 odd if both r1 and r2 are odd rational
numbers.
Tommy Jensen
The Problem of Coloring the Plane
• ( ba , dc )
Every rational point on the unit circle is odd.
Tommy Jensen
The Problem of Coloring the Plane
Lemma 1
If (r1 , r2 ) and (q1 , q2 ) are rational points at unit distance, then
(r1 − q1 , r2 − q2 ) is odd.
Definition
Let ∼ be the relation defined on Q2 by (r1 , r2 ) ∼ (q1 , q2 ) if and
only if (r1 − q1 , r2 − q2 ) is odd.
This relation ∼ is reflexive and symmetric.
Lemma 2
∼ is an equivalence relation.
Tommy Jensen
The Problem of Coloring the Plane
Proof of Lemma 2
Assume (r1 , r2 ) ∼ (q1 , q2 ) and (q1 , q2 ) ∼ (s1 , s2 ), such that
r1 − q1 =
c1
a2
c2
a1
, r2 − q2 =
, q1 − s1 =
, q2 − s2 =
,
b1
d1
b2
d2
where b1 , d1 , b2 , d2 are odd integers.
Then
r1 − s1 =
a1 a2
a1 b2 + a2 b1
+
=
,
b1 b2
b1 b2
which is odd since b1 b2 is an odd integer.
Similarly for r2 − s2 .
Now (r1 , r2 ) ∼ (s1 , s2 ) follows.
Tommy Jensen
The Problem of Coloring the Plane
Coloring Q2
Since any two points of unit distance in Q2 belong to the same
equivalence class of ∼, it is enough to color one equivalence
class.
This coloring may then be translated to give colorings of all
other classes as well.
Definition
For every odd point ( ba , dc ) in Q 2 , where b and d are odd
integers, define:
a c
red if a + c is even,
ϕ( , ) =
blue if a + c is odd.
b d
Tommy Jensen
The Problem of Coloring the Plane
Lemma 3
No two points at unit distance are colored the same by ϕ.
Proof of Lemma 3
Assume that ( ab11 , dc11 ) and ( ab22 , dc22 ) are odd points at unit
distance, b1 , b2 , d1 , d2 are odd integers:
(a1 b2 − b1 a2 )2 d12 d22 + b12 b22 (c1 d2 − d1 c2 )2 = b12 b22 d12 d22
implies that (a1 b2 − b1 a2 ) and (c1 d2 − d1 c2 ) have different
parities; one is odd and the other is even.
Then also a1 + a2 and c1 + c2 have different parities, and
a1 + a2 + c1 + c2 is an odd integer.
It follows that a1 + c1 and a2 + c2 have different parities.
Tommy Jensen
The Problem of Coloring the Plane
The 2-Color Theorem for Q2
Using Lemmas 1–3:
Theorem (Woodall 1973)
χ(Q2 ) = 2
We see that there is far between χ(Q2 ) and χ(R2 ).
Tommy Jensen
The Problem of Coloring the Plane
A Problem in Between χ(Q2 ) and χ(R2 ).
Question
How many colors are needed to color all of R2 , so that points p1
and p2 at unit distance are colored differently whenever
p1 − p2 ∈ Q2 ?
Definition
Call this smallest number of colors χ∆Q (R2 ).
Tommy Jensen
The Problem of Coloring the Plane
The bounds for χ∆Q (R2 )
Michael Payne, an australian undergraduate student, proved
two bounds for χ∆Q (R2 ).
Upper bound for χ∆Q (R2 ) (Payne 2007)
χ∆Q (R2 ) ≤ 2
Proof of the upper bound
For each point p of R2 use the 2-coloring of Q2 , which exists by
Woodall’s theorem, to color all points p + Q2 by 2 colors.
Tommy Jensen
The Problem of Coloring the Plane
Lower bound for χ∆Q (R2 ) (Payne 2007)
χ∆Q (R2 ) ≥ 3
Proof of the lower bound
Suppose that R2 is colored with only two colors red and blue.
Let R be the set of red points, B the set of blue points.
Then R2 = R ∪ B, and since the Lebesgue measure µ of R2 is
infinite, one of R and B has positive measure, say µ(B) > 0.
Tommy Jensen
The Problem of Coloring the Plane
Lemma 1
The rational points are dense on the unit circle.
Proof
The points
2
2nm
n − m2
,
n2 + m2 n2 + m2
with (m, n) ∈ Z2 \ {(0, 0)}
are rational points on the unit circle.
It is easy to check that they are dense on the unit circle.
Tommy Jensen
The Problem of Coloring the Plane
Lemma 2
Let p0 be any point in R2 and let ε > 0.
Then a rational number q ∈ Q with |q| < ε and three points
p1 , p2 , p3 ∈ R2 exist, such that
pi − pi−1 ∈ Q2 for i = 1, 2, 3,
|pi − pi−1 | = 1 for i = 1, 2, 3, and
p3 − p0 = (q, 0).
Proof
By Lemma 1 there exists (r1 , r2 ) ∈ Q2 such that |(r1 , r2 )| = 1
and |r1 − 1/2| < ε/2.
Let:
p1 = p0 + (r1 , r2 ), p2 = p0 + (r1 − 1, r2 ), p3 = p0 + (2r1 − 1, 0). Tommy Jensen
The Problem of Coloring the Plane
Because of µ(B) > 0, the Lebesgue Density Theorem implies
that for almost every point x ∈ B,
µ(B ∩ Bε (x))
→ 1 as ε → 0+ ,
µ(Bε (x))
where Bε (x) is a ball of radius ε and center x.
Tommy Jensen
The Problem of Coloring the Plane
µ(B ∩ S)
9
>
.
µ(S)
10
We can assume the sides of S parallel to the coordinate axes,
and that they have length 1. In particular µ(S) = 1.
By Lemma 2 we can find points
We choose a square S in R2 so that
(u, v ) = p0 , p1 , p2 , p3 = (u + q, v ) for some q ∈ Q with |q| <
so that (pi − pi−1 ) ∈ Q2 and |pi − pi−1 | = 1 for i = 1, 2, 3.
Let
B 0 = {(x − q, y ) : (x, y ) ∈ B},
so that B 0 is another copy of B moved a bit to the right or left.
Then µ(B 0 ∩ S) >
8
because of |q| <
10
Tommy Jensen
1
10 .
1
10 ,
The Problem of Coloring the Plane
9
8
and µ(B 0 ∩ S) >
it follows that:
10
10
7
=⇒ B ∩ B 0 ∩ S 6= ∅.
µ(B ∩ B 0 ∩ S) >
10
Choose a point (x, y ) ∈ B ∩ B 0 ∩ S.
Because of (x, y ) ∈ B 0 , the point (x + q, y ) is in B.
By choice of q there are points p0 , p1 , p2 , p3 with
From
µ(B ∩ S) >
(x, y ) = p0 , p1 , p2 , p3 = (x+q, y ), (pi − pi−1 ) ∈ Q2 , |pi − pi−1 | = 1.
Both p0 , p3 are in B. This implies both p1 , p2 are in R,
contradiction.
This proves that no 2-coloring exists.
Tommy Jensen
The Problem of Coloring the Plane
Summarizing Michael Payne’s bounds
Upper bound
In ZFC (Zermelo-Fraenkel-Choice set theory) the following is
true:
χZFC
(R2 ) = 2.
∆Q
Lower bound
In ZFS (ZF+ACℵ0 +LM set theory) the following is true:
χZFS
(R2 ) > 2.
∆Q
ZFS was introduced and proved consistent with ZF by Robert
Solovay in 1970.
LM : every subset of R is Lebesgue measurable.
Tommy Jensen
The Problem of Coloring the Plane
Generalizations of the Lower Bound
Payne, Nov. 2009
A lower bound of n + 3 can be proved for coloring Rn with
measurable color sets, so that points p1 and p2 are colored
differently whenever their distance is 1 and their difference
belongs to F n , where F is any field between Q and R,
Q ⊆ F ⊆ R.
Tommy Jensen
The Problem of Coloring the Plane
Conclusion
We would like to know the number of colors which is really
needed to color R2 .
The precise answer might well be something like:
We can use 4 colors in ZFC, but need 5,6 or 7 colors in ZFS.
Unfortunately we do not know which system of axioms is the
one which tells the truth about the real R2 .
Maybe it is ZFC, or it is ZFS, or it is a third one.
Tommy Jensen