Lines in the plane,

slopes,

and Euler’s formula

by Tal Harel

Part 1:Lines in the plane and

decompositions of graphs• Sylvester, 1893: “Prove that it is not possible to arrange

any finite number of real points so that a right line through every two of them shall pass through a third, unless they all lie in the same right line”

• Proved by Gallai.

Theorem 1: In any configuration of n points in the plane,

not all on a line, there is a line which contains exactly two

of the points.

■Proof: Let P be the given set of points and consider the

set L of all lines which pass through at least two points of

P. Among all pairs (P, l) with P not on l, choose a pair

(P0,l0) such that P0 has the smallest distance to l0, with Q

being the point on l0 closest to P0 (that is, on the line

through P0 vertical to l0).

Claim: This line l0 does it!

If not, then l0 contains at least 3 point of P, and thus 2 of

them, say P1 and P2, lie on the same side of Q.

It follows that the distance of P1 to the line l1 determined

by P0 and P2 is smaller than the distance of P0 to l0, and

this contradicts our choice for l0 and P0. ♠

• Proof used metric axioms (short distance) and order axioms (point lies between two points on a line) of the real plane.

• Do we really need these properties beyond the usual incidence axioms of points and lines?

3

5 4

61 2

7

Fano plane

- No 2-point line

- Therefore (Sylverster-Gallai), cannot be embedded in a real plane

• Shown by Coxeter: the order axioms will suffice for a proof of the Sylverster-Gallai theorem.

• One can devise a proof that does not use any metric properties.

• We will see a proof of the Sylverster-Gallai theorem using Euler’s formula.

• Later…

Theorem 2: Let P be a set of n ≥ 3 points in the plane, notall on a line. Then the set of L lines passing through atleast two points contains at least n lines.

■Proof:– n = 3: Nothing to show.– Induction on n - We assume correctness for n; and

need to prove correctness for n + 1:– Let |P| = n + 1. There exists a line containing

exactly two points p and q of P.– Consider the set P’ = P – {q} and the set L’ of lines

determined by P’:

• If the points of P’ do not all lie on a single line, then by induction: |L’| ≥ n and hence |L| ≥ n + 1 because of the additional line in L.

0l L

0l

• If the points in P’ are all on a single line, then we have the “pencil”:

Which results in precisely n + 1 lines. ♠

. . .

. . .p 3p 4p 1np

q

Theorem 3: Let X be a set of n ≥ 3 elements, and let

A1, … , Am be proper subsets of X, such that every pair of

elements of X is contained in precisely one set Ai.

Then m ≥ n holds.

■Proof: (Motzkin \ Conway)

For let rx be the number of sets Ai containing x.

(By the assumptions: 2 ≤ rx ≤ m)

If , then rx ≥ |Ai| because the |Ai| sets containing x

and an element of Ai must be distinct. Suppose m < n,

then m|Ai| < n rx and thus m(n - |Ai|) > n(m – rx) for

, and we come to a contradiction: ♠

x X

ix A

ix A

: :

1 1 1 11 1

i i i i ix X x X A x A A x x A Ax in n m r mm n A

• How does all of this relate to graph theory?• Think of the following statement:

“If we decompose a complete graph Kn into m cliques different from Kn, such that every edge is in a unique clique, then m ≥ n.

• Let X correspond to the vertex set of Kn and the sets Ai to the vertex sets of the cliques, then the statements are identical

• Next, we want to decompose Kn into complete bipartite graphs such that again every edge is in exactly one of these graphs.

• There is an easy way to do this:

• Number the vertices from 1 to n.• Take the complete bipartite graph joining 1 to all other

vertices. We obtain the graph K1,n-1 which is called a star

• Join 2 to vertices 3 to n, resulting in a star K1,n-2.• We go on, and decompose Kn into stars

K1,n-1, K1,n-2, … , K1,1. The decomposition uses n – 1 complete bipartite graphs.

• Can we do better, i.e less graphs?• No, according to Graham and Pollak, which leads us to:

Theorem 4: If Kn is decomposed into complete bipartite subgraphs H1, … , Hm, then m ≥ n – 1.

■Proof: Let the vertex set of Kn be {1, … , n}, and let Lj, Rj

be the defining vertex sets of the complete bipartite graph

Hj, j = 1, …, m.

To every vertex i we associate a variable xi. Since

H1, …, Hm decompose Kn, we find: (1)

Now, we suppose the theory is false: m < n – 1.

1 k k

m

i j a bi j k a L b R

x x x x

Then the system of linear equations:

Has fewer equations than variables, hence there exists a

non-trivial solution c1, … , cn. From (1) we infer:

But this implies:

A contradiction, and the proof is complete ♠

1 0,

0k

n

aa L

x x

x

1, ,k m

0i ji j

c c

2 2 21

1 1

0 2 0n n

n i i j ii i j i

c c c c c c

Part 2:The slope problem

Theorem: (Scott, 1970)

If n ≥ 3 points in the plane do not lie on one single line, then

they determine at least n – 1 different slopes, where

equality is possible only if n is odd and n ≥ 5.

■Proof: (a big one)

Has 6 steps:

(1) Show for n = 2m

(2) Define permutation model

(3) Crossing moves

(4) Touching and ordinary moves

(5) Facts about the moves

(6) Putting it all together

1

2

3

4

5 6

(1)

(2)

(3)

(4)

(5)

(6)

(1)

The case n = 3 is trivial.

For any set of n odd points: n = 2m + 1 ≥ 5 (m ≥ 2) (Not all on a line) we can find a subset of n – 1 = 2m points, not all on a line, which already determines n – 1 slopes

Thus, it suffices to show that every even set of n = 2m

points in the plane (m ≥ 2) determines at least n slopes.

So for the following stages, we only consider configurations of n = 2m points in the plane that determines t ≥ 2 different slopes.

(1)

(2)

(3)

(4)

(5)

(6)

(2) We construct a periodic sequence of permutations:

- Start with some direction on the plane that is not one of the configuration’s slopes

- Number the points 1, …, n in the order in which they appear in the 1-dimensional projection in this direction

- The first permutation π0 = 123…n represents the order of the points for our starting direction.

- The direction moves counterclockwise. This changes the projection and the permutation. Changes in the order of the projected points appear exactly when the direction passes one of the configuration slopes

1

2

3

4

5 6

(1)

(2)

(3)

(4)

(5)

(6)

- Changes in the permutations are far from random or arbitrary: By performing a 180 degrees rotation of the direction, we obtain a sequence of permutations

π0 -> π1 -> π2 -> … -> πt - 1 -> πt

Which has the following special properties:• The sequence starts with π0 = 123…n and ends with

πt = n…321.• The length t of the sequence is the number of slopes

of the point configuration.• In the course of the sequence, every pair i < j is

switched exactly once. This means that on the way from π0 to πt, only increasing substrings are reversed.

• Every move consists in the reversal of one or more disjoint increasing substrings (one or more lines in the direction we pass)

By continuing the circular motion around the

configuration, one can view the sequence as a part of a

two-way infinite, periodic sequence of permutations:

… -> π-1 -> π0 -> … -> πt -> πt+1 -> … -> π2t -> …

- πi+t is the reverse of πi for all i.

- πi+2t = πi for all .

We will see that every sequence with the above properties (and t ≥ 2) must have length t ≥ n.

i

(1)

(2)

(3)

(4)

(5)

(6)

(1)

(2)

(3)

(4)

(5)

(6)

(3) We divide each permutation into a left half and a right half of equal size m = n / 2, we count the letters (number labels) that cross the imaginary barrier in the middle.

We call πi -> πi + 1 a crossing move if one of the substrings it reverses does involve letters from both sides of the barrier.

A crossing move has order d if it moves 2d letters across the barrier – the crossing string has exactly d letters on one side and at least d letters on the other side.

π2 = 213:564 -> 265:314 = π3 is a crossing move of order 2.

652:341 -> 654:321 is a crossing move of order 1.

(1)

(2)

(3)

(4)

(5)

(6)

In the course of the sequence π0 -> π1 -> … -> πt, each of the letters 1,2,…,n has to cross the barrier at least once. This implies that, if the orders of the c crossing moves are d1, d2,…,dc, then we have:

This also implies that we have at least 2 crossing moves, since a crossing move with 2di = n occurs only if all the points are on one line, i.e for t = 1.

Geometrically, a crossing move corresponds to the direction of a line of the configuration that has less than m points on each side.

1

2 # letters that cross the barrierc

ii

d n

(1)

(2)

(3)

(4)

(5)

(6)

(4) A touching move is a move that reverses some

string that is adjacent to the central barrier, but does not

cross it. For example: π4 = 625:314 -> 652:341 = π5 is a

touching move.

Geometrically, a touching move corresponds to the

slope of a line of the configuration that has exactly m

Points on one side, and hence at most m – 2 points on

the other side.

Moves that are neither touching nor crossing will be

called ordinary moves. For example:

π1 = 213:546 -> 213:564 = π2.

So every move is either crossing, touching or ordinary,

and we use the letters T,C,O to denote the types of

moves.

(1)

(2)

(3)

(4)

(5)

(6)

C(d) will denote a crossing move of order d.

Thus for our small example we get:

π0 – T > π1 – O > π2 – C(2) > π3 – O > π4 – T > π5 – C(1) > π6

Or even shorter we can record this sequence as:

T, O, C(2), O, T, C(1)

(1)

(2)

(3)

(4)

(5)

(6)

(5) To complete the proof, we need these 2 facts:

Between any two crossing moves, there is at least one

touching move.

Between any crossing move of order d and the next

touching move, there are at least d – 1 ordinary moves.

In fact, after a crossing move of order d the barrier is

contained in a symmetric decreasing substring of length

2d, with d letters on each side of the barrier. For the next

crossing move the central barrier must be brought into

an increasing substring of length at least 2. But only

touching moves affect whether the barrier in in an

increasing substring. This yields the first fact.

(1)

(2)

(3)

(4)

(5)

(6)

For the second fact, note that with each ordinary move

(reversing some increasing substring) the decreasing

2d-string can get shortened by only one letter on each

side. And, as long as the decreasing string has at least 4

letters, a touching move is impossible. This yields the

second fact.

(1)

(2)

(3)

(4)

(5)

(6)

(6) The T-O-C pattern of the infinite sequence of

permutations, as derived in (2), is obtained by repeating

over and over again the T-O-C pattern of length t of the

sequence π0 -> … -> πt. Thus with the facts of (5) we see

that in the infinite sequence of moves, each crossing

move of order d is embedded into a T-O-C pattern of the

type (*)

of length 1 + (d – 1) + 1 + (d – 1) = 2d.

In the infinite sequence, we may consider a finite

segment of length t that starts with a touching move.

This segment consists of substrings of the type (*), plus

possibly extra inserted T’s.

1 1

, , ,..., , ( ), , ,..., ,d d

T O O O C d O O O

(1)

(2)

(3)

(4)

(5)

(6)

This implies that its length t satisfies:

which completes the proof ♠1

2c

ii

t d n

Part 3:Three applications of Euler’s formula

Euler’s formula: (Scott, 1750)

If G is a connected plane graph with n vertices, e edges

and f faces, then:

n – e + f = 2

■Proof: Let be the edge set of a spanning tree for G,

that is, of a minimal sub graph that connects all the vertices

of G.

- Does not contain a cycle because of the minimality assumptions.

We construct the dual graph G* of G: we put a vertex into

the interior of each face of G, and connect two such

vertices of G* by edges that correspond to common

boundary edges between the corresponding faces. If there

are several common boundary edges, then we draw

several connecting edges in the dual graph.

- G* may have multiple edges, even if G is simple.

T E

Consider the collection of edges in the dual graph

that corresponds to edges in E – T. The edges in T*

connect all the faces, since T does not have a cycle; but

also T* does not contain a cycle, since otherwise it would

separate some vertices of G inside the cycle from vertices

outside.

- Cannot be, because T is a spanning sub graph, and

the edges of T and of T* do not intersect.

For every tree the number of vertices is one larger than the

number of edges. To see this, choose one vertex as the

root, and direct all edges “away from the root”: this yields a

bijection between the non-root vertices and the edges, by

matching each edge with the vertex it points at.

* *T E

Applied to the tree T this yields n = eT + 1, while for the tree

T* it yields f = eT* + 1. Adding both equations we get:

n + f = (eT + 1) + (eT* + 1) = e + 2 ♠

We will show some consequences of Euler’s formula, and

also three “beautiful” proofs that have Euler’s formula at

their core:

1. A proof of the Sylvester-Gallai theorem

2. A theorem on two-colored point configurations

- Both of these use Euler’s formula alongside other arithmetic relationships between basic graph parameters.

3. Pick’s theorem about areas of elementary triangles.

First, we show some important “local” consequences of

Euler’s formula:

- Let G be any simple planar graph with n > 2 vertices.

(A) G has a vertex of degree at most 5.

(B) G has at most 3n – 6 edges.

(C) If the edges of G are two-colored, then there is a vertex of G with at most two color-changes in the cyclic order of the edges around the vertex.

Note: For each of the three statements, we assume that G is connected.

■Proof:

(A) Every face has at least 3 sides (since G is simple), so:

(fk is the number of k-faces)

f = f1 + f2 + f3 +… and: 2e = f1 + 2f2 + 3f3 + …

yield:

f = f3 + f4 + f5 + … and: 2e = 3f3 +4f4 + 5f5 + …

and thus 2e – 3f ≥ 0.

Now if each vertex has degree at least 6, then:

(ni is the number of vertices of degree i in G)

n= n0 + n1 + n2 + … and: 2e = n1 + 2n2 + 3n3 + …

imply:

n = n6 + n7 + n8 + … and: 2e = 6n6 +7n7 + 8n8 + …

and thus 2e – 6n ≥ 0.

Taking both inequalities together, we get:

6(e – n – f) = (2e – 6n) + 2(2e – 3f) ≥ 0

and thus e ≥ n + f, contradicting Euler’s formula.

(B) As in the first step of part (A), we obtain 2e – 3f ≥ 0 and

thus:

3n – 6 = 3e – 3f ≥ e

from Euler’s formula.

(C) Let c be the number of corners where color changes occur. Suppose the statement is false, then we have

c ≥ 4n corners with color changes, since at every vertex there is an even number of changes. Now every face with 2k or 2k + 1 sides has at most 2k such corners.

So we conclude that:

using the inequalities from before again.

So we have e ≥ n + f, again contradicting Euler’s formula. ♠

3 4 5 6 7 8

3 4 5 6 7

3 4 5 6 7

3 4 5 6 7

4 2 4 4 6 6 8

2 4 6 8 10

2(3 4 5 6 7 )

4( )

4 4

n c f f f f f f

f f f f f

f f f f f

f f f f f

e f

1. The Sylvester-Gallai theorem, revisited:

The Sylvester-Gallai theorem: Given any set of n ≥ 3

points in the plane, not all on one line, there is always a line

that contains exactly two of the points.

■Proof: (Sylvester-Gallai via Euler)

2. Monochromatic lines

Theorem: Given any finite configuration of “black” and “white” points in the plane, not all on one line, there is always a “monochromatic” line: a line that contains at least two points of one color and none of the other

■Proof:

skipped…