Geometric Graphs and Quasi-Planar Graphs

1. Geometric GraphsandQuasi-Planar Graphs Dafna Tanenzapf

2. Definition – Geometric Graph • A geometric graph is a graph drawn in the plane by straight-line segments . • There are no three points in which are collinear. • The edges of can be possibly crossing.

3. Forbidden Geometric Graphs • Given a class H of forbidden geometric graphs determine (or estimate) the maximum number of edges that a geometric graph with n vertices can have without containing a subgraph belonging to H.

4. Definition • Let be the class of all geometric graphs with vertices, consisting of pairwise disjoint edges ( ). • Example (k=2):

5. Theorem • Let be the maximum number of edges that a geometric graph with n vertices can have without containing two disjoint edges (straight-line thrackle). Then for every :

6. Theorem (Goddard and others) • Let be the maximum number of edges that a geometric graph with n vertices can have without containing three disjoint edges. Then:

7. Definitions • Edge xy is Leftmost at x if we can rotate it by (180 degrees) counter-clockwise around x without crossing any other edge of x. • A vertex x is called pointed if it has an angle between two consecutive edges that is bigger than . z y x x

8. Proof • Let G be a geometric graph with n vertices and at least 3n+1 edges. • We will show that there exist three edges which are disjoint.

9. Proof - Continue • For each pointed vertex at G, delete the leftmost edge. • We will denote the new subgraph as G1. • For each vertex at G1 delete an edge if there are no two edges from its right (begin from the leftmost edge of every vertex).

10. Proof - continue • Examples: |E|=1 |E|=2 |E|=3 |E|≥4 Number of deleted edges: 1 2 3 3 z u v z u y y z y y x x x x

11. Proof - continue • For every vertex we deleted at most 3 edges. • The remaining graph has at least one edge x0y0 (3n+1-3n=1). • In G1 there were two edges to the right of x0y0: x0y1 and x0y2. • In G1 there were two edges to the right of x0y0: y0x1 and y0x2.

12. Proof - continue • In G there was the edge x2y to the left of x2y0. • In G there was the edge y2x to the left of y2x0. • WLOG we can assume that the intersection of x0y2 and y0x2 is on the same side of y2 (if we split the plane into two parts by x0y0). • The edges y0x2, x0y1, y2x don’t intersect.

13. Conclusions • The upper bound is: • The lower bound is:

14. What’s next? • Dilworth Theorem • Pach and Torocsik Theorem

15. Definition – Partially ordered sets • A partially ordered set is a pair . • X is a set. • is a reflexive, antisymmetric and transitive binary relation on X. • are comparable if or . • If any two elements of a subset are comparable, then C is a chain. • If any two elements of a subset are incomparable, then C is an antichain.

16. Theorem (Dilworth) • Let be a finite partially ordered set. • If the maximum length of a chain is k, then X can be partitioned into k antichains. • If the maximum length of an antichain is k, then X can be partitioned into k chains.

17. Explanation (Dilworth Theorem) • If the maximum length of a chain is k, then X can be partitioned into k antichains. • For any x X, define the rank of x as the size of the longest chain whose maximal element is x. • 1≤rank(x)≤k • The set of all elements of the same rank is an antichain. • If the maximum length of an antichain is k, then X can be partitioned into k chains. • It can be shown by induction on k.

18. Definitions • Let uv and u’v’ be two edges. • For any vertex v let x(v) be the x-coordinate and let y(v) be the y-coordinate. • Suppose that x(u)<x(v) and x(u’)<x(v’). • uv precedes u’v’ (uv<<u’v’) if x(u) x(u’) and x(v) x(v’). • The edge uv lies below u’v’ if there is no vertical line l that intersects both uv and u’v’ such that: y(l∩uv) y(l∩u’v’). u’ v’ v u u’ v’ u v l

19. Theorem (Pach and Torocsik) • Let denote the maximum number of edges that a geometric graph with n vertices can have without containing k+1 pairwise disjoint edges. Then for every k,n 1:

20. Proof • Let G be a geometric graph with n vertices, containing no k+1 pairwise disjoint edges. • WLOG, no two vertices of G have the same x-coordinate. • Let uv and u’v’ be two disjoint edges of G such that uv lies below u’v’.

21. Proof - continue • We define four binary relations on E(G): • uv <1u’v’ if uv<<u’v’. • uv <2u’v’ if u’v’<<uv. • uv <3u’v’ if [x(u),x(v)] [x(u’),x(v’)]. • uv <4u’v’ if [x(u’),x(v’)] [x(u),x(v)].

22. Proof - continue

23. Proof - continue • We can conclude from the definitions that: • is a partially ordered set . • Any pair of disjoint edges is comparable by at least one of the relations .

24. Proof - continue • cannot contain a chain of length k+1. • Otherwise, G has k+1 pairwise disjoint edges. • According to the previous theorem, for any i, E(G) can be partitioned into at most k anti-chains (classes), so that no two edges belonging to the same class are comparable by . • The edges can be partitioned into classes Ej , such that no two elements of Ej are comparable by any relation.

25. Proof - continue • From the second conclusion, any Ej does not contain two disjoint edges. • We saw earlier that . • Then: • To sum up:

26. Definition – Quasi Planar Graphs • A graph is called quasi planar if it can be drawn in the plane so that no three of its edges are pairwise crossing.

27. Motivation • We want to find an upper bound to the number of edges of quasi-planar graph. • Pach had shown that a quasi-planar graph with n vertices has edges. • For the general case, k-quasi-planar graph (a graph with no k pairwise crossing edges), the upper bound is: • We will prove a Theorem. • Its conclusion will be that the upper bound is: (can be shown in induction).

28. Theorem (Agarwal, Aronov, Pach, Pollack and Sharir) • If G(V,E) is a quasi-planar graph (undirected, without loops or parallel edges), then |E|=O(|V|). • We will prove the theorem for the case that G has a straight-line drawing in the plane with no three pairwise crossing edges.

29. Definition • The arrangement A(E) of E(G) is a complex set consisted of: • Nodes: N={V(G) X(G)|X(G)=crossing points}. • Segments: S={E’(G)|E’(G)=the edges between the vertices of N}. • Faces F={the faces of the graph G’=(N,S)}.

30. Definition • The complexity |f| of f F is the number of segments in S on the boundary f of f. • If a segment is in the interior of f, then it contributes 2 to |f|.

31. Lemma2 • Let G=(V,E) be a quasi-planar graph drawn in the plane. • The complexity of all f of A(E) such that: • f is a non-quadrilateral face. • f is quadrilateral face incident to at least on vertex of G. is O(|V|+|E|).

32. Definition • A graph is called overlap graph if its vertices can be represented by intervals on a line so that two vertices are adjacent if and only if the corresponding intervals overlap but neither of them contains the other.

33. Proof (Theorem) • Let G be a quasi-planar graph drawn in the plane with n vertices. • WLOG G is a connected graph. • Let G0=(V,E0) be a spanning tree of G, |E0|=n-1 • E*=E\E0. G: G0:

34. Proof - continue • Each face of A(E0) is simply connected. • By Lemma 2, the complexity of non-quadrilateral and quadrilateral faces of A(E0) incident to a point of V is O(n). • We call the remaining faces of A(E0) crossing quadrilateral.

35. Proof - continue • For each edge e E*, let Ω(e) denote the set of segments of A(E0 {e}) that are contained in e. • Every s Ω(e) is fully contained in some face f A(E0) and its two endpoints lie on f.

36. Proof - continue • For each f A(E0) let X(f) denote the set of all segments in that are contained in f. • Any two segments in X(f) cross each other if and only if their endpoints alternate along f. • We “cut” the face in some vertex and get an open interval. • Two elements of X(f) cross each other if and only if the corresponding intervals overlap.

37. Proof - continue • This defines a triangle-free overlap graph on the vertices X(f). • By Gyarfas and Kostochka, every triangle-free overlap graph can be colored by 5 colors. • Therefore, the segments of X(f) can be colored by at most 5 colors, so that no two segments with the same color cross each other.

38. Proof - continue • For each f A(E0) let H(f) denote the quasi- planar graph whose edges are X(f). • A monochromatic graph is a graph which all its edges are colored in one color. • Let f A(E0) be a face that is not crossing quadrilateral. • Let H1(f),…,H5(f) be the monochromatic subgraphs of H.

39. Proof - continue • We fix one of Hi, assume WLOG H1, and we reinterpret it to a new graph: every edge on the boundary of the face and vertices of V on the boundary will be a vertex, and all the interior segments will be the edges in the new graph. • The resulting graph H1* is planar. b b a c c a d p e p e d

40. Proof - continue • Face of H1*(f) is a digon if it is bounded by two edges of H1*(f). • Edge of H1*(f) is shielded if both of the faces incident to it are digons. • The remaining edges of H1*(f) are called exposed. b b a c c a d p e p e d

41. Proof - continue • By Euler’s formula (|E|≤3|V|-6), there are at most O(nf) exposed edges in H1*(f). • nf is the number of vertices of H1*(f). • |nf|≤2|f|.

42. Proof - continue • We repeat this analysis for every Hi(f) (2≤i≤5). • The number of edges e E* containing at least one exposed segment is . • By Lemma2, this sum is O(n). • We need to bound the number of e E* with no exposed segments (shielded edges).

43. Lemma 3 • There are no shielded edges.

44. Proof - continue • The total number of edges of E* is O(n). • The total number of edges of E is O(n).