Bounds on the Obstacle Number

Transcription

1 Bounds on the Obstacle Number OCICS Seminar Tommy Reddad Supervisors: Pat Morin, Vida Dujmović, Prosenjit Bose Computational Geometry Lab Carleton University October 25, 2013

2 Overview Obstacle number is a new (abstract) graph parameter. Introduction Examples, motivation Alpert et al. lower bound Improvements (Mukkamala et al., Dujmović & Morin) Future work

3 Introduction Definition Obstacle representation of a graph Points see each other adjacent Points don t see each other not adjacent

4 Introduction Definition Obstacle representation of a graph Points see each other adjacent Points don t see each other not adjacent Example (Petersen graph)

5 Introduction Definition Obstacle representation of a graph Points see each other adjacent Points don t see each other not adjacent Example (Petersen graph)

6 Introduction Definition Obstacle representation of a graph Points see each other adjacent Points don t see each other not adjacent Example (Petersen graph)

7 Introduction Definition Obstacle number of a graph G, obs(g): smallest number of obstacles to represent G. Example (Complete graph) obs(g) = 0 G = K n

8 Introduction Definition Obstacle number of a graph G, obs(g): smallest number of obstacles to represent G. Example (Complete graph) obs(g) = 0 G = K n Example (Petersen graph)

9 Introduction Definition Obstacle number of a graph G, obs(g): smallest number of obstacles to represent G. Example (Complete graph) obs(g) = 0 G = K n Example (Petersen graph)

10 Introduction Definition Obstacle number of a graph G, obs(g): smallest number of obstacles to represent G. Example (Complete graph) obs(g) = 0 G = K n Example (Petersen graph)

11 Examples Example (Trees)

12 Examples Example (Trees) T any tree: obs(t ) 1

13 Examples Example (Trees) T any tree: obs(t ) 1 Example (n n grid)

14 Examples Example (Trees) T any tree: obs(t ) 1 Example (n n grid) obs(g) = 1

15 Introduction Definition Worst-case obstacle number: w(n) = max{obs(g) : G has n vertices} Question: Is w(n) 1? Is w(n) O(1)?

16 Introduction Definition Worst-case obstacle number: w(n) = max{obs(g) : G has n vertices} Question: Is w(n) 1? Is w(n) O(1)? Theorem obs(g) ( n 2) E(G) Corollary w(n) O(n 2 ) Proof.

17 A first lower bound Definition G k,m : K k, and for each 2m + 2 vertices of K k, an additional vertex adjacent to each. Example (G 5,1 ) x 1 x 2 x 5 x 3 x 4

18 A first lower bound Definition G k,m : K k, and for each 2m + 2 vertices of K k, an additional vertex adjacent to each. Example (G 5,1 ) x 1 v {1,2,3,5} x 1 v {1,2,4,5} x 2 x 5 x 2 x 5 v {1,2,3,4} v {1,3,4,5} x 3 x 4 x 3 x 4 v {2,3,4,5}

19 A first lower bound Theorem (Alpert et al.) For large k, obs(g k,m ) m. Theorem (Erdős-Szekeres, Happy Ending Theorem) For any n, in a large enough point set n convex points. Example For 5 points 4 convex points

20 A first lower bound Proof. 1. Choose k so that 4m + 4 vertices P K k are convex

21 A first lower bound Proof. 1. Choose k so that 4m + 4 vertices P K k are convex 2. 2-color P in blue and red

22 A first lower bound Proof. 1. Choose k so that 4m + 4 vertices P K k are convex 2. 2-color P in blue and red 3. v adjacent to all blue and no red

23 A first lower bound Proof. 1. Choose k so that 4m + 4 vertices P K k are convex 2. 2-color P in blue and red 3. v adjacent to all blue and no red Case 1: v conv(p) Case 2: v conv(p) v v Both require at least m obstacles obs(g k,m ) m.

24 A first lower bound Best known choice for k: k = ( 8m+3 4m+2) + 1 so: V (G k,m ) = (( 8m+3 ) 4m+2) + 1 2m ( ) 8m m + 2 Corollary ( ) log n w(n) Ω log log n

25 A second lower bound Theorem (Mukkamala et al.) At most 2 O(hn log2 n) graphs G have obs(g) h. Proof. Idea: 1. View obstacles as polygons

26 A second lower bound Theorem (Mukkamala et al.) At most 2 O(hn log2 n) graphs G have obs(g) h. Proof. Idea: 1. View obstacles as polygons 2. Encode an obstacle representation 3. Count all possible encodings

27 A third lower bound Theorem (Mukkamala et al.) ( ) n w(n) Ω log n Proof. Idea: (probabilistic argument) 1. Choose large enough random graph G n,1/2 2. Disjoint slabs G i have obs(g i ) 2 w.h.p. 3. Combine slabs for lower bound w.h.p.

28 A final lower bound Theorem (Dujmović & Morin) ( ) n w(n) Ω (log log n) 2 Proof. Idea: Similar to previous argument, but more refined. 1. Choose large enough random graph G n,1/2 ( ) 2. Disjoint slabs G i have obs(g i ) Ω k w.h.p. log 2 k 3. Combine slabs for lower bound w.h.p.

29 Summary Table: Lower bounds for w(n) Authors Year Bound Alpert et al Ω Mukkamala et al Ω ( n ( log n log log n log 2 n ( Mukkamala et al Ω n ( Dujmović & Morin 2013 Ω ) ) ) log n ) n (log log n) 2

30 Future work Non-trivial upper bounds for w(n), refined lower bounds Graph families: Planar graphs: 2 wp (n) 4n 8 3 Graphs of bounded treewidth Bipartite graphs: w b (n) Explicit (bipartite) graphs with high obstacle number... Algorithms, complexity analysis...

31 References Hannah Alpert, Christina Koch, and Joshua D. Laison. Obstacle numbers of graphs. Discrete & Computational Geometry, 44(1): , Vida Dujmović and Pat Morin. On obstacle numbers. ArXiv e-prints, August Padmini Mukkamala, János Pach, and Deniz Sarıöz. Graphs with large obstacle numbers. In Dimitrios M. Thilikos, editor, Graph Theoretic Concepts in Computer Science, volume 6410 of Lecture Notes in Computer Science, pages Springer Berlin Heidelberg, Padmini Mukkamala, János Pach, and Dömötör Pálvölgyi. Lower bounds on the obstacle number of graphs. The Electronic Journal of Combinatorics, 19(2), Geza Tóth and Pavel Valtr. The Erdős-Szekeres theorem: upper bounds and related results. In Combinatorial and Computational Geometry, pages Press, 2004.