Testing planarity part 1

1. Testing planaritypart 1 Thomas van Dijk

2. Preface • Appendix of Planar Graph Drawing • Quite hard to read • So we’ll try to explain it, not just tell you about it

3. Preface • “There are a number of efficient algorithms for planarity testing, which are unfortunately all difficult to implement.” • (www.mathworld.com)

4. Preface • “There are a number of efficient algorithms for planarity testing, which are unfortunately all difficult to implement.” • (www.mathworld.com) • Implementation in Mathematica was actually bugged until version 4.2.1 (!) • (Also mathworld.com)

5. What is planarity testing? • Whether a graph can be drawn on the plane without edge crossings • Equivalently, whether a “planar embedding” of the graph exists

6. Embedding • Adjacency-list representation of the graph • Order in the list defines clockwise order of the edges

7. Embedding • Adjacency-list representation of the graph • Order in the list defines clockwise order of the edges 1 2 2 4 3 4 3 1 3 4 1 2 4 3 1 2

8. Embedding • Adjacency-list representation of the graph • Order in the list defines clockwise order of the edges 2 4 3 1 1 2 2 4 3 4 3 1 3 4 1 2 4 3 1 2 3 3 4 1 2 4 3 1 2 1 4 2

9. Overview • “Vertex addition algorithm” • First presented by Lempel et al (’67) • Improved to linear time byBooth and Lüker (’76) • We present this second algorithm

10. Main idea • Start with one vertex • Add vertices and their edges one by one • make sure we don’t break planarity • If you can’t add the next vertex, graph is not planar • add the vertices in an order such that if we fail, we that know that using any other order wouldn’t have worked either

11. We need to know about • Some general observations • st-Numbering • Bush forms / PQ-Trees

12. O(m) is O(n) for planar graphs • Remember from Euler’s Theorem we have thatm ≤ 3n-6for simple plane graphs • Algorithm can just reject graphs with too many edges. • So from now on, O(m) is O(n)

13. Biconnected components • Definition:A graph G is biconnected ifffor every two distinct vertices there exist two internally disjoint paths between them • Also iff G contains no cut-vertices • i.e.: cannot be made unconnected by removing a single vertex

14. Biconnected components • Theorem:A graph is planar iffall its biconnected components are planar • Proof:Induction on the number of biconnected components

15. Biconnected components • If the graph is planar, clearly its biconnected components are planar • A graph is planar iff its connected components are planar • A connected graph with no biconnected components is a tree; trees are planar

16. Biconnected components • If a graph has one biconnected component, then all vertices not in it are in trees. So the graph is planar iff the biconnected component is planar

17. Biconnected components • Induction step: • The next biconnected component is connected to the rest of the graph by a cut-vertex v.

18. Biconnected components • Induction step: • The next biconnected component is connected to the rest of the graph by a cut-vertex v. • Since the biconnected component itself is planar, an embedding with v on the exterior exists. (As Hans showed us.)

19. Biconnected components • The biconnected components of a graph can be found in linear time • (If you are interested, an exercise in Cormen et al explains it) • So from now on, we can assume graphs are biconnected.

20. st-Numbering • From the ‘main idea’ slide:“Add the vertices in an order such that if we fail, we that know using any other order wouldn’t have worked either”

21. 8 1 st-Numbering • Special nodes: source s (“1”) and sink t (“n”) • “1” and “n” adjacent

22. 8 1 st-Numbering • Special nodes: source s (“1”) and sink t (“n”) • “1” and “n” adjacent • j V-{s,t}: i, k Adj(j)st(i)<st(j)<st(k)

23. 8 1 7 2 st-Numbering • Special nodes: source s (“1”) and sink t (“n”) • “1” and “n” adjacent • j V-{s,t}: i, k Adj(j)st(i)<st(j)<st(k) • So 2 must be next to 1and n-1 must be next to n

24. 8 5 1 7 2 3 6 4 st-Numbering • Special nodes: source s (“1”) and sink t (“n”) • “1” and “n” adjacent • j V-{s,t}: i, k Adj(j)st(i)<st(j)<st(k) • So 2 must be next to 1and n-1 must be next to n

25. 2 1 4 6 3 5 11 12 9 7 10 8 5 4 1 3 2 Some st-numbering examples

26. st-Numbering • Theorem:Every biconnected graph has an st-numbering • Proof:We give an algorithm and show that it works if the graph is biconnected • Our graph is biconnected, so anst-numbering exists.

27. Finding an st-numbering • First step, standard DFS. • Can visit children in any order • Calculate and store per vertex: • Parent • “Depth first search number:” DFN • Low-point: the lowest DFN among nodes that can be reached by a path using any amount of tree edges followed by 1 back edge

28. G

29. G DFS

30. DFN 6 G 5 4 3 2 DFS 1

31. DFN LOW 6 3 G 5 3 4 1 3 1 2 1 DFS 1 1

32. Finding an st-numbering • Push the vertices onto a stack and pop them off • Four rules for what to push and pop • Assign increasing numbers to the vertices when they are popped off for the last time • Example follows

33. Initial setup • The node with DFN 2 is called the ‘source’ • The node with DFN 1 is called the ‘sink’ • Stack with sourceon top of sink • Source and sinkare “old” b a d c ac e f

34. Rule “2” • There is a “new” tree edge: follow it and take the path that leads to its low point • Push the vertices on the path ontothe stack • Mark all as old. b a abc d c e f

35. No matches for vertex a • There are now no rule matches for a. • Pop it off the stack and give it the next available number • This node is ready b 1 d c bc e f

36. Rule “3” • There is a “new” back edge into this node: from the other end, go back up tree edges to an “old” vertex • Push the vertices onthe path ontothe stack • Mark all as old bfedc b 1 d c e f

37. No matches for vertex b • So the vertex is ready and gets its number fedc 2 1 d c e f

38. No matches for vertex f • So the vertex is ready and gets its number edc 2 1 d c e 3

39. Et cetera… 2 1 5 6 4 3

40. Correctness • All vertices are numbered • The numbering is indeed an st-numbering

41. Algorithm recap • For linear complexity, we need thatO(m) = O(n). • For correctness, we need that the graph is biconnected

42. After the break … • Ron will tell you about: • bush forms and PQ-trees • how it all fits together into a planarity testing algorithm

43. Any questions?

44. Any questions? • http://www.planarity.net/