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. Input: K 4 Input point set P:  Even if G is planar, there exist point sets that do not admit a straight-line embedding.

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. Rubber band analogy: v1v1 v2v2 v3v3 v5v5 v4v4  If points of P are not collinear convex hull of P is a closed polygonal chain.

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. All finite/countable trees are outerplanar: 1 2 Cycle!  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.

9. Outerplanar graphs  Obviously every outerplanar graph is planar.  Not every planar graph is outerplanar, for instance K 4 is the smallest non-outerplanar graph known. K 4 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: H: G: Removing edges G1:G1: Contracting edges G2:G2: Delete isolated nodes G 3 =H:

11. 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 K 4 nor K 2,3 as a minor. K4K4 K 2,3 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:G2:G3:G3:G 3 is 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 v 1, v 2 on the outer face of G and any two consecutive vertices p 1, p 2 of the convex hull of P, there is a legal drawing in which v 1 is represented by p 1 and v 2 is represented by p 2.

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 v 1 is represented with p 1 and v 2 is represented with p 2.  Assume that the assertion is true for all outerplanar graphs with less than n vertices.

16. Proof continued  Let v 1, v 2, …, v n denote the vertices of G along the outer face in counter clockwise order.  Let v i be the third vertex of the unique triangle sitting on the edge {v 1, v 2 } in E(G). Input G: v i = v1v1 v8v8 v7v7 v6v6 v5v5 v4v4 v3v3 v2v2

17. Proof continued  Let p 1, p 2 be any 2 consecutive vertices of the convex hull of P. w Input P: p1p1 p2p2

18. Proof continued  There is a point p in P - {p 1, p 2 } such that: a)Triangle contains no point of P. b)There is a line l through p and an interior point of the segment p 1 p 2 that has precisely i–2 points of P strictly on its right-hand (p 2 ) side. Input P: p1p1 p2p2 p l

19. Proof continued  First select p’ such that p 2 p’ has exactly i-3 elements of P strictly on its right-hand side. Input P: p1p1 p2p2  If triangle p 1 p 2 p’ is empty, let p := p’  Otherwise choose p to be the point in p 1 p 2 p’ such that the angle p 2 p 1 p is minimal. p’ p

20. Proof continued  Let G 1 and G 2 be the subgraphs of G induced by {v 2, v 3, …, v i } and {v i, v i+1,..., v n, v 1 } resp.  Now let H 1 and H 2 denote the points of P on the right and left half-planes of l respectively (incl p). Input G: v i = v1v1 v8v8 v7v7 v6v6 v5v5 v4v4 v3v3 v2v2 Input P: p1p1 p2p2 p l G1G1 G2G2 H1H1 H2H2

21. Proof continued  By induction hypothesis G 1 has a non crossing straight line drawing on the point set H 1.  So does G 2 on H 2.  Add the edge {v 1, v 2 } as p 1 p 2 and we’re done. □ Input G:Input P: p1p1 p2p2 p G1G1 G2G2 H1H1 H2H2 v1v1 v2v2 vivi

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 K 4 or K 2,3 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. Poor angular resolution:Much better layout:

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 K 3 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 a d-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 {D 1,..., D n } 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, D 0, …., D t-1 ) is called a wheel of length t with C as a hub if:  All D i are adjacent to C.  Each D i is adjacent to D i+1 mod t C P is a wheel of length 6:

35. Fans  Ordered disk packing P = (C, D 0, …., D t ) is called a fan of length t with C as a hub if:  All D i are adjacent to C.  Each D i is adjacent to D i+1 C P is a fan of length 4:

36. Lemma 1  Lemma 1: Let the ordered disk packing P = (C, A, D 1,…, D t, B) be a fan of length t+2 with a hub C.  Let r C, r A, r 1,…, r t, r B, respectively denote the radii of the discs in P. Assume that A and B are adjacent.  Let r = min {r A, r B, r C }.  Then each D j has radius r j >= a t r, P is a fan of length 4: B A C

37. Lemma 2  Lemma 2: Take ordered disc packing P = (C, D 0, …, D t-1 ) that is a wheel of length t with hub C.  Let the discs in P have respective radii r C, r 0, …, r t-1.  Then each disk D j has radius r j >= a t-3 r C. P is a wheel of length 6: D0D0 D5D5 D1D1 D2D2 D3D3 D4D4 C

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 K 4 :

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 a d-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 a d-3. ( r D >= a d-3 r C ) 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: r D /r C >= a d-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 a d-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 r A <= r B <= r C respectively. C A B

44. Proof, the end  If we normalize the radii by taking r C = 1, by the remarks above: a d-2 <= r A <= r B  It is easy t see that the smallest angle L in the triangle is achieved when r A = r B = a d-2  Then sin(L/2) = a d-2 / (1 + a d-2 )  From basic calculus: L/2 > sin(L/2)  Finally L > 2*a d-2 / (1 + a d-2 ) > a d-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/d 2 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 a d-2 is known (a ~ 0.15).