Drawing graphs using a small number of obstacles 1

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Drawing graphs using a small number of obstacles Martin Balko, Josef Cibulka, and Pavel Valtr Charles University in Prague, Czech Republic Part of the research was

1 Drawing graphs using a small number of obstacles Martin Balko, Josef Cibulka, and Pavel Valtr Charles University in Prague, Czech Republic Part of the research was conducted during the workshop Homonolo 204 supported by the European Science Foundation as a part of the EuroGIGA collaborative research program (Graphs in Geometry and Algorithms).

2 Obstacle numbers

3 Obstacle numbers In a drawing of a graph, vertices are points in R 2 and edges are line segments.

4 Obstacle numbers In a drawing of a graph, vertices are points in R 2 and edges are line segments.

5 Obstacle numbers In a drawing of a graph, vertices are points in R 2 and edges are line segments. An obstacle is a simple polygon in the plane.

6 Obstacle numbers In a drawing of a graph, vertices are points in R 2 and edges are line segments. An obstacle is a simple polygon in the plane.

7 Obstacle numbers In a drawing of a graph, vertices are points in R 2 and edges are line segments. An obstacle is a simple polygon in the plane. The obstacle number obs(g) of a graph G is the minimum number of obstacles in a drawing of G where two vertices are connected by an edge iff the corresponding line segment avoids all the obstacles.

8 Obstacle numbers In a drawing of a graph, vertices are points in R 2 and edges are line segments. An obstacle is a simple polygon in the plane. The obstacle number obs(g) of a graph G is the minimum number of obstacles in a drawing of G where two vertices are connected by an edge iff the corresponding line segment avoids all the obstacles.

9 Examples

10 Examples

11 Examples obs(e n ) =

12 Examples T obs(e n ) =

13 Examples T obs(e n ) = obs(t ) =

14 Examples T obs(e n ) = obs(t ) =

15 Examples T obs(e n ) = obs(t ) =

16 Examples T obs(e n ) = obs(t ) = obs(p 4 P 5 ) = ( Fabrizio Frati)

17 Previous results

18 Previous results Let obs(n) be the max G obs(g) over all n-vertex graphs G.

19 Previous results Let obs(n) be the max G obs(g) over all n-vertex graphs G. Is obs(n) O(n)? (Alpert, Koch, Laison, 200)

20 Previous results Let obs(n) be the max G obs(g) over all n-vertex graphs G. Is obs(n) O(n)? (Alpert, Koch, Laison, 200) Lower bounds: obs(n) Ω(n/ log 2 n) (Mukkamala, Pach, Sarıöz, 200). obs(n) Ω(n/ log n) (Mukkamala, Pach, Pálvölgyi, 20). obs(n) Ω(n/(log log n) 2 ) (Dujmović, Morin, 203).

21 Previous results Let obs(n) be the max G obs(g) over all n-vertex graphs G. Is obs(n) O(n)? (Alpert, Koch, Laison, 200) Lower bounds: obs(n) Ω(n/ log 2 n) (Mukkamala, Pach, Sarıöz, 200). obs(n) Ω(n/ log n) (Mukkamala, Pach, Pálvölgyi, 20). obs(n) Ω(n/(log log n) 2 ) (Dujmović, Morin, 203). Only the trivial upper bound obs(n) ( n 2).

22 Previous results Let obs(n) be the max G obs(g) over all n-vertex graphs G. Is obs(n) O(n)? (Alpert, Koch, Laison, 200) Lower bounds: obs(n) Ω(n/ log 2 n) (Mukkamala, Pach, Sarıöz, 200). obs(n) Ω(n/ log n) (Mukkamala, Pach, Pálvölgyi, 20). obs(n) Ω(n/(log log n) 2 ) (Dujmović, Morin, 203). Only the trivial upper bound obs(n) ( n 2). G

23 Previous results Let obs(n) be the max G obs(g) over all n-vertex graphs G. Is obs(n) O(n)? (Alpert, Koch, Laison, 200) Lower bounds: obs(n) Ω(n/ log 2 n) (Mukkamala, Pach, Sarıöz, 200). obs(n) Ω(n/ log n) (Mukkamala, Pach, Pálvölgyi, 20). obs(n) Ω(n/(log log n) 2 ) (Dujmović, Morin, 203). Only the trivial upper bound obs(n) ( n 2). G obs(g) ( n 2) E(G)

24 Our bounds

25 Our bounds Conjecture (Mukkamala, Pach, Pálvölgyi, 20) The parameter obs(n) is around n 2.

26 Our bounds Conjecture (Mukkamala, Pach, Pálvölgyi, 20) The parameter obs(n) is around n 2. Theorem For every positive integer n, we have obs(n) n log n n +.

27 Our bounds Conjecture (Mukkamala, Pach, Pálvölgyi, 20) The parameter obs(n) is around n 2. Theorem For every positive integer n, we have obs(n) n log n n +. We can answer the question of Alpert et al. provided χ(g) is bounded.

28 Our bounds Conjecture (Mukkamala, Pach, Pálvölgyi, 20) The parameter obs(n) is around n 2. Theorem For every positive integer n, we have obs(n) n log n n +. We can answer the question of Alpert et al. provided χ(g) is bounded. Theorem (in the journal version) For every positive integer n and every n-vertex graph G, we have where k := min{χ(g), χ(g)}. obs(g) (n )( log k + ),

29 Our bounds Conjecture (Mukkamala, Pach, Pálvölgyi, 20) The parameter obs(n) is around n 2. Theorem For every positive integer n, we have obs(n) n log n n +. We can answer the question of Alpert et al. provided χ(g) is bounded. Theorem (in the journal version) For every positive integer n and every n-vertex graph G, we have where k := min{χ(g), χ(g)}. obs(g) (n )( log k + ), The bounds apply even if the obstacles are required to be convex.

30 Dilated drawings

31 Dilated drawings Regular drawing of K 4,4

32 Dilated drawings st level

33 Dilated drawings 2nd level

34 Dilated drawings 3rd level

35 Dilated drawings 4th level

36 Dilated drawings 5th level

37 Dilated drawings 6th level

38 Dilated drawings 7th level

39 Dilated drawings

40 Dilated drawings d 3 d 3 d 2 d 2 d d d < d 2 < d 3

41 Dilated drawings d 3 d 3 d 2 d 2 d d Levels form caps.

42 Dilated drawings d 3 d 3 d 2 d 2 d d Levels form caps. Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D.

43 Dilated drawings d 3 d 3 d 2 d 2 d d

44 Proof of the key observation Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D.

45 Proof of the key observation Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D. Part (a): p i+2 q j p i+ q j+ q j+2 p i

46 Proof of the key observation Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D. Part (a): p i+2 q j p i+ q j+ q j+2 p i

47 Proof of the key observation Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D. Part (a): p i+2 q j p i+2 +p i 2 p i+ q j+2 +q j 2 q j+ q j+2 p i

48 Proof of the key observation Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D. Part (a): p i+2 q j p i+2 +p i 2 p i+ r i r i+ q j+2 +q j 2 q j+ q j+2 p i

49 Proof of the key observation Key observation (a) If d < < d n, then all levels of a drawing D of K n,n form caps. (b) If D is ε-dilated, that is, we also have d n < ( + ε)d, and ε > 0 is small, then all edges of every cap are incident to the same face of D. Part (a): p i+2 q j p i+2 +p i 2 p i+ r i r i+ q j+2 +q j 2 q j+ q j+2 p i Part (b) follows from the fact that ε-dilated drawings converge to the regular drawing as ε 0

50 Application I: Obstacle numbers of bipartite graphs

51 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

52 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

53 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

54 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

55 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

56 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

57 Application I: Obstacle numbers of bipartite graphs Proposition For all m, n N and every bipartite graph G K m,n, we have obs(g), obs(g) m + n.

58 Application I: Obstacle numbers of general graphs

59 Application I: Obstacle numbers of general graphs For a general graph G, we use divide-and-conquer approach and iterate O(log V (G) ) times.

60 Application I: Obstacle numbers of general graphs For a general graph G, we use divide-and-conquer approach and iterate O(log V (G) ) times. We use a variant of the Horton set as the vertex set in the drawing.

61 Application I: Obstacle numbers of general graphs For a general graph G, we use divide-and-conquer approach and iterate O(log V (G) ) times. We use a variant of the Horton set as the vertex set in the drawing.

62 Application I: Obstacle numbers of general graphs For a general graph G, we use divide-and-conquer approach and iterate O(log V (G) ) times. We use a variant of the Horton set as the vertex set in the drawing.

63 Application I: Obstacle numbers of general graphs For a general graph G, we use divide-and-conquer approach and iterate O(log V (G) ) times. We use a variant of the Horton set as the vertex set in the drawing.

64 Application I: Obstacle numbers of general graphs For a general graph G, we use divide-and-conquer approach and iterate O(log V (G) ) times. We use a variant of the Horton set as the vertex set in the drawing.

65 Application II: Number of graphs with low obstacle number

66 Application II: Number of graphs with low obstacle number Let g(h, n) be the number of (labeled) n-vertex graphs G with obs(g) h.

67 Application II: Number of graphs with low obstacle number Let g(h, n) be the number of (labeled) n-vertex graphs G with obs(g) h. Upper bounds: g(h, n) 2 o(n2) (Pach, Sarıöz, 20). g(h, n) 2 O(hn log2 n) (Mukkamala, Pach, Sarıöz, 200).

68 Application II: Number of graphs with low obstacle number Let g(h, n) be the number of (labeled) n-vertex graphs G with obs(g) h. Upper bounds: g(h, n) 2 o(n2) (Pach, Sarıöz, 20). g(h, n) 2 O(hn log2 n) (Mukkamala, Pach, Sarıöz, 200). Better upper bound on g(h, n) gives better lower bound on obs(n).

69 Application II: Number of graphs with low obstacle number Let g(h, n) be the number of (labeled) n-vertex graphs G with obs(g) h. Upper bounds: g(h, n) 2 o(n2) (Pach, Sarıöz, 20). g(h, n) 2 O(hn log2 n) (Mukkamala, Pach, Sarıöz, 200). Better upper bound on g(h, n) gives better lower bound on obs(n). In 203, Dujmović and Morin conjectured g(h, n) 2 f (n) o(h) where f (n) O(n log 2 n).

70 Application II: Number of graphs with low obstacle number Let g(h, n) be the number of (labeled) n-vertex graphs G with obs(g) h. Upper bounds: g(h, n) 2 o(n2) (Pach, Sarıöz, 20). g(h, n) 2 O(hn log2 n) (Mukkamala, Pach, Sarıöz, 200). Better upper bound on g(h, n) gives better lower bound on obs(n). In 203, Dujmović and Morin conjectured g(h, n) 2 f (n) o(h) where f (n) O(n log 2 n). Theorem For all n, h N with h < n, we have g(h, n) 2 Ω(hn).

71 Application III: Complexity of faces in arrangements of segments

72 Application III: Complexity of faces in arrangements of segments An arrangement of segments

73 Application III: Complexity of faces in arrangements of segments Faces of the arrangement.

74 Application III: Complexity of faces in arrangements of segments F The complexity of F is 4.

75 Application III: Complexity of faces in arrangements of segments F 2 The complexity of F 2 is 7.

76 Application III: Complexity of faces in arrangements of segments F 2 The complexity of F 2 is 7. Question (Arkin et al., 995) What is the complexity of M faces in an arrangement of line segments with n endpoints?

77 Application III: Complexity of faces in arrangements of segments F 2 The complexity of F 2 is 7. Question (Arkin et al., 995) What is the complexity of M faces in an arrangement of line segments with n endpoints? Upper bound: O(min{nM log n, n 4/3 M 2/3 + n 2 log M}) (Aronov et al., 992, and Arkin et al., 995)

78 Application III: Complexity of faces in arrangements of segments F 2 The complexity of F 2 is 7. Question (Arkin et al., 995) What is the complexity of M faces in an arrangement of line segments with n endpoints? Upper bound: O(min{nM log n, n 4/3 M 2/3 + n 2 log M}) (Aronov et al., 992, and Arkin et al., 995) Theorem Lower bound: Ω(min{nM, n 4/3 M 2/3 }).

79 Application III: Complexity of faces in arrangements of segments F 2 The complexity of F 2 is 7. Question (Arkin et al., 995) What is the complexity of M faces in an arrangement of line segments with n endpoints? Upper bound: O(min{nM log n, n 4/3 M 2/3 + n 2 log M}) (Aronov et al., 992, and Arkin et al., 995) Theorem Lower bound: Ω(min{nM, n 4/3 M 2/3 }). Tight for M n log 3/2 n.

80 Work in progress: Point-line incidences

81 Work in progress: Point-line incidences Theorem (Szemerédi, Trotter, 983) The number of incidences between M points and N lines is at most O(M 2/3 N 2/3 + M + N).

82 Work in progress: Point-line incidences Theorem (Szemerédi, Trotter, 983) The number of incidences between M points and N lines is at most O(M 2/3 N 2/3 + M + N). Best upper bound is 2.44 M 2/3 N 2/3 + M + N (Ackerman, 204).

83 Work in progress: Point-line incidences Theorem (Szemerédi, Trotter, 983) The number of incidences between M points and N lines is at most O(M 2/3 N 2/3 + M + N). Best upper bound is 2.44 M 2/3 N 2/3 + M + N (Ackerman, 204). Lower bounds: 0.42 M 2/3 N 2/3 + M + N (Erdős, 946, Edelsbrunner, Welzl, 986) 0.63 M 2/3 N 2/3 + M + N (Elekes, 2002)

84 Work in progress: Point-line incidences Theorem (Szemerédi, Trotter, 983) The number of incidences between M points and N lines is at most O(M 2/3 N 2/3 + M + N). Best upper bound is 2.44 M 2/3 N 2/3 + M + N (Ackerman, 204). Lower bounds: 0.42 M 2/3 N 2/3 + M + N (Erdős, 946, Edelsbrunner, Welzl, 986) 0.63 M 2/3 N 2/3 + M + N (Elekes, 2002) Theorem The number of incidences between M points and N lines is at least. M 2/3 N 2/3 + M + N.

85 Work in progress: Point-line incidences Theorem (Szemerédi, Trotter, 983) The number of incidences between M points and N lines is at most O(M 2/3 N 2/3 + M + N). Best upper bound is 2.44 M 2/3 N 2/3 + M + N (Ackerman, 204). Lower bounds: 0.42 M 2/3 N 2/3 + M + N (Erdős, 946, Edelsbrunner, Welzl, 986) 0.63 M 2/3 N 2/3 + M + N (Elekes, 2002) Theorem The number of incidences between M points and N lines is at least. M 2/3 N 2/3 + M + N. Thank you.

Lower bounds on the obstacle number of graphs

Lower bounds on the obstacle number of graphs Padmini Mukkamala, János Pach, and Dömötör Pálvölgyi Rutgers, The State University of New Jersey Ecole Polytechnique Fédérale de Lausanne Eötvös University