1. The odd-distance graph
Moshe Rosenfeld
University of Washington, Tacoma
&
Hanoi University of Science 。
11/20/2018

2. iff |u-v|  D Geometric Graphs
By geometric graphs we mean graphs G(M,D) where M is a metric space and D is a subset of R+.
V(G)  M, (u,v)  E(G)
iff |u-v|  D 。
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3. Probably the most famous geometric graph is the unit distance graph G(R2, 1).
This problem asks for the smallest number of colors needed to color the points of the plane R2 so that two points at distance 1 apart receive different colors. 。
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4. He called it the alternative four color problem.
Ed Nelson, a graduate student at the University of Chicago noted in 1950 that at least four colors are needed to color this graph.
He called it the alternative four color problem.
John Isbell in the same year proved that seven colors suffice.
And thus begun its march to the hall of fame.
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5. How does a problem become famous? What made this problem famous?
It is 63 years old.
Very simple to understand
At least five mathematicians were “credited” with it: Edward Nelson, Paul ErdÖs, Hugo Hadwiger, Leo Moser and Martin Gardener
The lower and upper bounds (4,7) were established in 1950.
No progress since!
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6. What made it famous?

7. The attractions of this problem are many.
It is very easy to understand.
It is finite in nature, all we need to find is a finite 7 chromatic graph that can be embedded in G(R2 , 1).
Subgraphs of G(R2,1) with large girth.
Many variations, such as choosability, circular chromatic number, color sets missing one color, other set axioms…
G(Rd , 1), G(R2 , D),
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8. It can get “ugly.” This is a graph of order 45, girth 5,
                                                                                                                                                                               
It can get “ugly.” This is a graph of order 45, girth 5,
chromatic number 4 embedded in G(R2 ,1)
Different directions, many opportunities, sometimes getting ugly need computers.

9. The two triangles form a 3-color transfer.
This is Hadwiger’s 7-coloring of G(R2 ,1)
Inside is Moser’s spindle showing that at
least four colors are needed.
The two triangles form a 3-color transfer.
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10. If s is the length of the side of the hexagon then:
||x – y||  2s if x,y are in the same hexagon.
||x – y||  7s if x,y have same color but different hexagons.
As long as 1/ 7 < s < ½ we get a coloring of the unit distance graph by 7 colors.
This means also that:
(G(R2, 1/ 7 < s < ½) )  7
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11. This brings us to the odd-distance graph: G(R2 , {1,3,5,7,…})
It follows that if D  R is bounded and min(D)   > 0 then (G(R2 , D) is finite.
What if D is not bounded?
This brings us to the odd-distance graph: G(R2 , {1,3,5,7,…}) 。
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12. Odd distances in R2
The six distances determined by 4 points in R2 cannot be all odd.
Putnam 1992
R. Graham, B. Rothschild, E. Strauss the maximum number of points in Rd such that all distances among them is an odd integer is d+1 unless d = k. In these dimensions we can have d + 2 points.

13. An elementary classroom proof
X2 Y2
X1 Y1
X3 Y3 o
Rank of this matrix is < 3 so determinant is o.

15. The odd-distance graph
In 1994 in Boca Raton I asked Paul Erdös and Herbert Wilf whether R2 can be colored in a finite number of colors so that two points at odd integral distance have distinct colors? (obvious lower bound 4)
Erdös also asked what is the maximum number of odd integral distances among n points in R2
Curious: in 1946 Erdos asked to determine the maximum number of unit distances determined by n points in the plane (still open). In graph terms, asks for the largest size among all subgraphs of order n.

16. Density, typical ErdÖs questions.
Given n points in R2, how many distances can be 1? (ErdÖs, 1946).
How many distances can be odd?
How many times can the largest distance occur among n points in R2?
A “biological proof.”
All largest distance segments must intersect. 2. The maximum number of mutually intersecting segments among n points no 3 on a line is n.

17. Clearly, the maximum number of odd distances among n points in R2 is  t4(n) (Turán’s number)
L. Piepemeyer proved that Kn,n,n can be embedded in R2 so that all edges have odd integral distance.

18. Equilateral triangle side 7
Rotate once
Distances: 3, 5, 8
Realizes K2,2,2

19. 3 equilateral triangles with side 72. P1
The embedding of K(3,3,3):
3 equilateral triangles with side 72.
P1, Y2, R3 sides 16, 56, P3, Y1, R2 same P2, Y3,R1 same
All other edges are: 49 (equilateral triangles), 21, 35 and 39 R1 Y2 Y1 R2 P3 P2 R3 Y3

20. What makes it work. 32 + 52 + 3.5 = 72 212 + 352 + 21.35 = 492
But also = 492
In general, we proved that a2 + b2 + a.b = 72n has a solution with gcd(a,7) = 1. Which implies that it has n different solutions {a,b}.
This allows us to rotate an equilateral triangle of side length 72n n+1 times about its center and obtain an embedding of Kn+1,n+1,n+1 in G(R2,Odd). 。
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21. Collateral benefits
As an aside, we also obtain a set of 3n points in R2, no three on a line, such that all distances among them are integers.
Haiko Harbroth was first to use this construction in search of the smallest diameter of n points in R2 with all distances integers.
We believe that our approach will help us find sets with smaller diameter.
We are also looking at sets of points in general position (no 3 on a line no 4 on a circle) with pairwise integral distances.
Current record is 7.
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22. Can the odd-distance graph compete with the unit-distance graph for fame?
Hayri Ardal (SFU)
Jano Manuch (SFU)
Moshe Rosenfeld (UWT)
Saharon Shelah (Hebrew University)
Ladislav Stacho (SFU)
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23. Some notable subgraphs of the odd-distance graph.
The integral lattice is 2-colorable
The rational points are 2-colorable.
You cannot go from one rational point to another making odd integral jumps stepping only on rational points.
Every 3-colorable graph is a subgraph of G(R2,odd).
G(R2,odd) is not k-list colorable for any integer k.
It contains Kn,n as a subgraph.

24. Theorem: The chromatic number of G(R2 , odd) is  5.
Construct a 4-color transfer.
Key: “120o Pythagorean triples”:
a2 + b2 + ab = c2 (3, 5, 7)

25. THE
ARMS SPINDLE
21 points in R2 that require 5 colors.

26. You have only four colors available.6 5 1 4 3 2 。
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We get a four color transfer!

27. Choosability
The unit distance graphs in R2 and R3 are countably choosable.
The R2 odd-distance graph is countably choosable.
The R3 odd-distance graph is not countably choosable.
#2 of course implies that the unit distance graph is countably choosable.

28. R3 is not countably choosable
Bn = (0,0,4n2 + 4n)
{(x,y,0) | x2 + y2 =1}
L((x,y,0)) = A  N,
|A| = 0
L(Bn) = {n, n+1, …}

29. The circular chromatic number of the unit distance graph is  4.
G(R2 ,odd) is countably choosable.
The circular chromatic number of the unit distance graph is  4.
The circular chromatic number of
G(R2 ,odd)  4.5 (Xuding Zhu et. al.)
Every 4-colorable graph is a subgraph of G(R3, odd)
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30. A terminal, logical inexactitude.
Peter Brass, William Moser and Janos Pach wrote in their book Research Problems in Discrete Geometry :
..the existence of a K4 is the only obstruction. That is, every finite K4 - free graph can be represented by odd distances in the plane. (p. 252)
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31. Odd wheels W5 is not a Subgraph of G(R2,odd) Nam Le Tien and M. R.
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32. Is the chromatic number of G(R2,odd) finite?
Interestingly, G(R2,odd) has no finite measurable coloring.
This follows immediately from a theorem of Furstenberg, Katznelson and Weiss [FKW] which asserts that for every Lebesgue measurable subset A  R2 with positive upper density, there exists a number r0 so that A contains a pair of points at distance r for every r > r0 .
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33. HELP!!! 1. Odd wheels are not subgraphs of G(R2,odd).
2. Are there triangle–free graphs that are not subgraphs of G(R2,odd)
3. Can every 4-coloarble graph be faithfully embedded in G(R3,odd)
4. What is the chromatic number of a circle in G(R2, odd)? 。
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34. Open Problems Find a 6-chromatic subgraph of G(R2,odd)
Find a subgraph of G(R2,odd) with circular chromatic number 5.
Find lower bounds for G(R3,odd)
If G is a subgraph of G(R2,odd) can it always be faithfully embedded in G(R2,odd)?
Just solved; every 3-chromatic graph can be faithfully embedded in the odd-distance graph 。
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35. Thank you
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