Lines in the plane, slopes, and Euler’s formula by Tal Harel

1. Lines in the plane,slopes,and Euler’s formulaby Tal Harel

2. 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.

3. 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 P0vertical to l0).

4. Claim: This line l0does it! If not, thenl0contains 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 l1determined by P0 and P2 is smaller than the distance of P0 to l0, and this contradicts our choice for l0 and P0. ♠

5. 3 Fano plane 5 4 7 2 1 6 • 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? - No 2-point line - Therefore (Sylverster-Gallai), cannot be embedded in a real plane

6. 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…

7. Theorem 2:Let P be a set of n ≥ 3 points in the plane, not all on a line. Then the set of L lines passing through at least 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.

8. . . . . . . • If the points in P’ are all on a single line, then we have the “pencil”: Which results in precisely n + 1 lines. ♠

9. 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 letrxbe the number of setsAicontainingx. (By the assumptions:2 ≤rx≤ m) If, thenrx≥ |Ai|because the|Ai|sets containing x and an element ofAimust be distinct. Supposem < n, thenm|Ai| < n rxand thusm(n - |Ai|) > n(m – rx) for , and we come to a contradiction: ♠

10. 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 Knand the sets Aito the vertex sets of the cliques, then the statements are identical • Next, we want to decompose Kninto complete bipartite graphs such that again every edge is in exactly one of these graphs. • There is an easy way to do this:

11. 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:

12. Theorem 4:If Kn is decomposed into complete bipartite subgraphs H1, … , Hm, then m ≥ n – 1. ■Proof: Let the vertex set of Knbe {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.

13. 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 ♠

14. 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.

15. 4 1 3 5 6 2 ■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

16. (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 – 1slopes Thus, it suffices to show that every even set of n = 2mpoints 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)

17. (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) 4 1 3 5 6 2

18. - 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 < jis 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) (1) (2) (3) (4) (5) (6)

19. 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. (1) (2) (3) (4) (5) (6)

20. (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 + 1a 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)

21. 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 ccrossing 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) (3) (4) (5) (6)

22. (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)

23. 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)

24. (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 dthe 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)

25. 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)

26. (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) (2) (3) (4) (5) (6)

27. This implies that its length t satisfies: which completes the proof ♠ (1) (2) (3) (4) (5) (6)

28. 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

29. ■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 graphG* 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.

30. 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.

31. 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 ♠

32. We will show some consequences of Euler’s formula, and also three “beautiful” proofs that have Euler’s formula at their core: • A proof of the Sylvester-Gallai theorem • 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.

33. First, we show some important “local” consequences of Euler’s formula: - Let G be any simple planar graph with n > 2 vertices. • G has a vertex of degree at most 5. • G has at most 3n – 6 edges. • 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.

34. ■Proof: • Every face has at least 3 sides (since G is simple), so: (fkis 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: (niis 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.

35. 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 ≥ 0and thus: 3n – 6 = 3e – 3f ≥ e from Euler’s formula. • 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.

36. So we conclude that: using the inequalities from before again. So we have e ≥ n + f, again contradicting Euler’s formula. ♠

37. 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)

38. 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…