Computational Geometry Seminar Lecture 4

1. Computational Geometry Seminar Lecture 4 More on straight-line embeddings Gennadiy Korol

2. Today • Restricted point set drawing. • Preliminaries. • Outerplanar graphs. • Restricted outerplanar graph drawing. • Angular resolution in graph drawing. • Lower bound on angular resolution for general graphs. • Lower bound on angular resolution for general planar graphs.

3. Restricted point set drawing • Until now we were allowed to define the position of each point in order to create the proper straight-line non crossing embedding of a graph. • Even if G is planar, there exist point sets that do not admit a straight-line embedding. Input: K4 Input point set P:

4. Restricted point set drawing • Interesting open question: Given a planar graph G and an arbitrary point set P in general position on the plane, does G have a straight-line non crossing embedding in P? • This problem for general planar graph is believed to be NP-complete. • Even then the progress have been made when G is restricted to a subclass of planar graphs.

5. Convex and non convex sets • A set S in a vector space over R is called convex if the line segment joining any pair of points of S lies entirely in S. Convex set: Non convex set:

6. Convex hull • Let P be the non-empty finite point set in the plane. Convex hull of P is the boundary of the minimal convex set that contains P. • If points of P are not collinear convex hull of P is a closed polygonal chain. Rubber band analogy: v1 v2 v5 v4 v3

7. Outerplanar graphs • A graph is called outerplanar if it has an embedding in the plane such that its vertices lie on a fixed circle and the edges lie inside of the disk of the circle and do not intersect (except for the end points). All finite/countable trees are outerplanar: Intuitively:

8. Outerplanar graphs • Equivalently graph is outerplanar if there is some face that includes every vertex. • Every planar non crossing embedding of a tree will produce exactly one face (otherwise there is a cycle and graph is not a tree). • Thus every tree is outerplanar. All finite/countable trees are outerplanar: 1 2 Cycle!

9. Outerplanar graphs • Obviously every outerplanar graph is planar. • Not every planar graph is outerplanar, for instance K4 is the smallest non-outerplanar graph known. K4 is not outerplanar.

10. Reminder: minors • Graph H is a minor of G if it can be obtained from G by repeatedly contracting or removing edges, and by deleting isolated nodes. H is a minor of G: Delete isolated nodes Contracting edges Removing edges H: G: G1: G2: G3=H:

11. K4 K2,3 Outer-planarity criterion • A beautiful analogue to Kuratowski's theorem can be found for outer-planar graphs: • A graph G is outer-planar if and only if it contains neither K4 nor K2,3 as a minor. Not outer-planar graph:

12. Maximal outerplanar graphs • Outerplanar graph is called maximal if addition of a new edge invalidates its outer-planarity. • Can be looked upon as a triangulation of every interior face of the graph. G: G2: G3: G3is maximal outer-planar

13. Drawing outerplanar graphs • Theorem: Every outerplanar graph G on n vertices admits a non crossing straight-line drawing on any set P of n points in general position on the plane as its vertex set. • Fasten your seat belts, here comes the proof!

14. Proof of the theorem Proof: • We are going to prove a stronger statement. • Given any two consecutive vertices v1, v2 on the outer face of G and any two consecutive vertices p1, p2 of the convex hull of P, there is a legal drawing in which v1 is represented by p1 and v2 is represented by p2.

15. Proof of the theorem Proof: • Without loss of generality we can assume that G is maximal outer-planar graph. • Base case: n = 3, the statement is true as we can always embed a triangle into any 3 points on a plane such that v1 is represented with p1 andv2 is represented with p2. • Assume that the assertion is true for all outerplanar graphs with less than n vertices.

16. v3 v4 v2 v5 v1 v8 v6 v7 Proof continued • Let v1, v2,… , vn denote the vertices of G along the outer face in counter clockwise order. • Let vi be the third vertex of the unique triangle sitting on the edge {v1, v2} in E(G). Input G: vi =

17. Proof continued • Let p1, p2 be any 2 consecutive vertices of the convex hull of P. w Input P: p2 p1

18. Proof continued • There is a point p in P-{p1, p2} such that: a) Triangle <p1 p2p> contains no point of P. b) There is a line l through p and an interior point of the segment p1p2 that has precisely i–2 points of P strictly on its right-hand (p2) side. Input P: p2 p p1 l

19. Proof continued • First select p’ such that p2p’ has exactly i-3 elements of P strictly on its right-hand side. • If triangle p1 p2 p’ is empty, let p:=p’ • Otherwise choose p to be the point in p1 p2 p’ such that the angle p2 p1 p is minimal. Input P: p2 p p’ p1 p’ p’ p’

20. Proof continued • Let G1 and G2 be the subgraphs of G induced by {v2, v3, …, vi} and {vi, vi+1, ..., vn, v1} resp. • Now let H1 and H2 denote the points of P on the right and left half-planes of l respectively (incl p). v3 Input G: Input P: H1 p2 v4 v2 G1 p v5 vi = v1 p1 G2 l v8 v6 H2 v7

21. Proof continued • By induction hypothesis G1 has a non crossing straight line drawing on the point set H1. • So does G2 on H2. • Add the edge {v1, v2} as p1p2 and we’re done. □ Input G: Input P: H1 p2 v2 G1 p vi v1 p1 G2 H2

22. On constrained drawing • Outerplanar graphs are known to be the largest subclass of planar graphs that admits a straight line drawing on a general set of points. • Corollary: Every tree with n vertices admits a non crossing straight-line drawing on any set of n points in general position in the plane.

23. Rooted tree drawing • Theorem (Ikebe, Perles, Tamura and Shinnichi) • Every rooted tree with n vertices admits a non crossing straight-line drawing on any set P of n points in the plane in general position such that the root is embedded in an arbitrarily specified element of P. • Proof is beyond the scope of our lecture.

24. Concluding • Even when G is planar there are point sets that forbid non-crossing straight line drawing. • Outerplanar graphs have a drawing where all vertices lie on some face. • Graph is outerplanar iff it does not contain K4 or K2,3 as a minor. • Every outerplanar graph has non-crossing straight-line embedding on an arbitrary set of points in the plane. • Every tree can be drawn on any set of points while embedding the root at a specific point.

25. Moving on • Here we leave our outer-planar graph study and move on to general graph drawing questions. • Good time for your questions.

26. Angular resolution • Putting a graph on a small grid guarantees that the picture does not get too “crowded”. • Another parameter is the angular resolution of the graph G, which is the smallest angle between two adjacent edges in the best drawing of G. Much better layout: Poor angular resolution:

27. On angular resolution • Clearly if the maximum degree of the vertices in G is d, the angular resolution cannot exceed . • This is not a tight bound. For instance K3 with d=2 but with an angular resolution of π/3. • This parameter is known to be NP-hard to compute, even for d=4. But this problem is not known to be in NP.

28. On angular resolution • Theorem (Formann et al.) Every graph with maximum degree d has a straight-line drawing with resolution • Edges are allowed to cross in such representation.

29. On angular resolution • With a little care we can make sure that all vertices are embedded into distinct points of a polynomial size grid and the resolution remains roughly the same. • For planar graphs, the situation is better, a resolution of at least c*(1/d) can be achieved. • But again however, in such a representation the edges may cross!

30. On angular resolution • Theorem (Malitz, Papakostas)Every planar graph with maximum degree d admits a non-crossing straight-line drawing with resolution of at least ad-2, where a > 0 is an absolute constant. • We are going to base the proof on 3 lemmas and 1 important theorem.

31. Disk packing • Disk packing P is a set {D1,..., Dn} of closed disks of zero, finite or infinite radius on a plane. • Some disks maybe adjacent but none overlap on interior points. Disk packing P:

32. Disk packing • Disk packing induces a planar graph G. • Let F be an exterior face of G is this planar drawing. • We will say that P realizes the pair (G, F). Disk packing P induces G: F

33. Theorem 1 • Theorem 1 (Andrew and Thurston): For every triangulated planar graph G with exterior face F there exists a disk packing P that realizes (G, F). • No algorithm with polynomial time is known to find such P.

34. Wheels • Ordered disk packing P = (C, D0, …., Dt-1) is called a wheel of length t with C as a hub if: • All Di are adjacent to C. • Each Di is adjacent to Di+1 mod t P is a wheel of length 6: C

35. Fans • Ordered disk packing P = (C, D0, …., Dt) is called a fan of length t with C as a hub if: • All Di are adjacent to C. • Each Di is adjacent to Di+1 P is a fan of length 4: C

36. Lemma 1 • Lemma 1:Let the ordered disk packing P = (C, A, D1,…, Dt, B) be a fan of length t+2 with a hub C. • Let rC, rA, r1,…, rt, rB, respectively denote the radii of the discs in P. Assume that A and B are adjacent. • Let r = min {rA, rB, rC}. • Then each Dj has radius rj >= at r, P is a fan of length 4: A B C

37. Lemma 2 • Lemma 2:Take ordered disc packing P = (C, D0, …, Dt-1) that is a wheel of length t with hub C. • Let the discs in P have respective radii rC, r0, …, rt-1. • Then each disk Dj has radius rj >= at-3 rC. P is a wheel of length 6: D5 D0 D4 C D1 D3 D2

38. Lemma 3 • Lemma 3:Every triangulated planar graph G with exterior face F is realized by a disk packing in which all three exterior discs have the same radius. Realizing K4:

39. Back to the theorem • Theorem (Malitz, Papakostas)Every planar graph with maximum degree d admits a non-crossing straight-line drawing with resolution of at least ad-2, where a > 0 is an absolute constant. • For proof for lemmas please refer to original paper:“On the Angular Resolution of Planar Graphs”by Seth Malitz and Achilleas Papakostas.

40. Proving the theorem • Proof: • Given any triangulated planar graph G with maximum degree d and exterior face F, we know by the Theorem 1 and Lemma 3 that there exists a disk packing P for (G, F) where all three outer disks have the same radius, say unit one. • Consider any adjacent pair of discs C, D in P and let D be the one with the smaller radius.

41. Proof, continued • If C is not one of the three outer disks then C and D are part of a wheel of length <= d with C as a hub. • By Lemma 2 the radius of D divided by that of C is at least ad-3.(rD >= ad-3 rC) C

42. Proof, continued • Now suppose that C is an exterior disc and D is an interior one. • In this case C and D are part of a fan with hub C, including 2 other exterior disks, named A and B. • Using Lemma 1: rD/rC >= ad-2 C A B

43. Proof, continued • Hence, given any two adjacent discs in P, the radius of the smaller disk divided by that of the larger is at least ad-2. • Take any triangle in the straight-line drawing of G induced by P as above. • Call discs generating this triangle A, B and C with radii rA <= rB <= rC respectively. C A B

44. Proof, the end • If we normalize the radii by taking rC= 1, by the remarks above: ad-2 <= rA <= rB • It is easy t see that the smallest angle L in the triangle is achieved when rA = rB =ad-2 • Then sin(L/2) = ad-2 / (1 + ad-2) • From basic calculus: L/2 > sin(L/2) • Finally L > 2*ad-2 / (1 + ad-2) > ad-2□ C A B

45. On angular resolution • It turned out that by using disk packing technique there is no lower bound decreasing less than exponentially fast for the angular resolution. • Proved by Papacostas and Malitz in the same paper.

46. Summing it up • Angular resolution is an important measure in graph drawing. • A lower bound of order 1/d2 exists when edges are allowed to cross. • Can preserve the resolution while drawing on the grid. • A better lower bound exists for planar graphs that is of order of 1/d. • For non crossing drawing of planar graphs no better bound than ad-2 is known (a ~ 0.15).