Computational Geometry Seminar Lecture 5

1. Computational Geometry Seminar Lecture 5 Circle Packings and Lipton-Tarjan Theorem Maor Hizkiev

2. Overview Koebe Representation Theorem Lipton-Tarjan Theorem

3. Definition Convex Body - 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.

4. C1 C2 C4 V1 V2 V4 V3 C3 Definition Contact Graph – Given a packing C of convex bodies, the vertices are associated with the members of C. 2 vertices are connected  the corresponding members touch each other. Note: if the bodies are circles, then the connections are straight edges

5. Koebe's Theorem A finite graph is a contact graph of a circle packing in the plane  it is planar We will use this theorem, in order to prove Lipton-Tarjan separator theorem for planar graphs.

6. In formal Given any planar graph G, with vertex set V(G) = {v1, … , vn } and edge set E(G) = {e1, … , en }, We can find a packing of n (not necessarily congruent) circular disks C = {C1,…,Cn} in the plane with the property that Ci and Cj touch each other  vi,vj E(G) for 1  i  n V2 V1 V4 V3

7. proof It is sufficient to prove the theorem for maximal planar graphs, AKA triangulation. <= Assume that the resulting graph can be represented by circular discs, satisfying the conditions in the theorem, then, erasing the discs corresponding to the new vertices, we obtain a good representation of the original graph, because the discs are non-overlapping. V2 V1 V4 Note: in this direction we do not demand a maximal graph. V3

8. proof => • Let G be a fixed triangulated graph with vertex set V(G) = {v1, … , vn }, edge set E and face set F. • By Euler’s formula • |V| - |E| + |F| = 2 • but since 3|F| = 2|E| (triangulated graph), we get • |F| = 2|V| - 4= 2n - 4

9. ri rj rk • Let r = (r1,…,rn) be any vector of n positive reals such that r1 + … + rn = 1. For each face vivjvk of G, consider a triangle cut-out of cardboard, whose vertices are the centers of 3 mutually tangent discs of radii ri,rj and rk.

10. vi Definition Let r(vi) denote the sum of the angles at vi of all triangles that have vi as one of their vertices. In case we are (extremely) lucky, than r(vi) = 2 for every vertex G not in the exterior triangle. In this case, the triangles will perfectly fit together in the plane, and we obtain a good representation of G by discs of radii r1,…,rn NOTE: the above statement is not as trivial as it seems.

11. In the case we are not lucky The triangles overlap each other, thus, is less than 2 r(vi) vi In any case we get that Since we have |F| faces m r(vi) = |F| = (2n – 4) i=1 Recall that |F|= 2n - 4

12. Definition Let S  Rn denote the (n – 1)-dimensional simplex defined by S = { r = (r1,…,rn) | ri > 0 for all i, and  ri = 1 } And let H = { x = (x1,…,xn) |  xi = (2n – 4) } Consider the continuous mapping f:S  H. f(r) = (r(v1),…,r(vn))

13. Assume WLOG that v1,v2,v3 are the vertices of the exterior face. It is sufficient to show that, for example, x* = (2/3, 2/3,2/3, 2, 2, …, 2) Lies in the image of the map f. x* implies that the circles of the exterior face, will have equal sizes. v1,v2,v3 donates 2/3 to the sum, since each regular triangle donates /3 to the sum, plus /3 from the exterior triangle.

14. ri rj rk Claim A: f:S  H is a one-to-one function. Proof: We’ll pick 2 distinct points r,r’ S. Let I denote the set of indices i which satisfy the inequality ri < r’i. Obviously I   and I  {1,…,n}. Consider a triangle vivjvk determined by the centers of 3 mutually touching discs with radii of ri,rj,rk. We have now 4 possibilities:

15. Case 1 & 2 Case 1: All vertices of the triangle does not belong to I. Case 2: All vertices of the triangle belong to I. r(vi) In both cases, the sum does not change. iI

16. Case 3 Case 3: two vertices belongs to I, and one doesn’t. If we increase ri and rj but decrease or leave unchanged the other radii so that the discs remain tangent, then the angles of the triangle at vi will decrease. Note: such a triangle must exist, since I  {1,…,n} and I  .

17. r(vi) > r’ (vi) i  I i  I Case 4 Case 4: one vertex belongs to I, and the other 2, doesn’t. If we increase ri but decrease or leave unchanged the other 2 radii so that the discs remain tangent, then the angle of the triangle at vi will decrease.  f(r)  f(r’)

18. lim r(vi) = |F(I)| r  s i  I Let s = (s1,…,sn) be a boundary point of the simplex S, and now let I denote the set of all indices for which si = 0. (0,0,1) (1,0,0) (0,0.5,0.5) (0,1,0) • Observe that if r tends to s, then in each triangle which has at least one vertex belonging to {vi | i  I}, the sum of the angles at these vertices tends to  ,hence, Where F(I) denotes the set of faces of G with at least one vertex in {vi | iI}.

19. For any fixed rS and for any nonempty proper subset I  {1,…,n}, there exists a point sBdS such that si = 0 for all iI and si > ri for all iI. • If we move r toward s along a line, then, iIr(vi) will increase. (from claim A) • Taking our former result, we get that iIr(vi) < |F(I)| In other words, the image of the map f:S  H lies in the (n-1)-dimensional convex polytope P* determined by the relations i=1xi = (2n – 4) iIxi < |F(I)| For all proper IP({1,…,n})\{} 2 conditions on a vector of H, just like linear programming

20. n x*i = (2n - 4) i = 1 • From the fact that f:SP* is a one-to-one mapping, and all accumulation points of f(r), as r tends to BdS, lie in P* (because we proved that ) , we get the following. lim r(vi) = |F(I)| r  s i  I Claim B: f:SP* is a surjective mapping, that is, f(S) = P*. • To complete the proof, we still have to show that the point x*=(x1*,…,xn*) • (defined earlier as x* = (2/3, 2/3,2/3, 2, 2, …, 2)) • belongs to P*. • Clearly,

21. For all proper IP({1,…,n})\{}, If |I| = n – 1 or n – 2, then all faces of G have at least one vertex belonging to {vi | i I}, i.e. |F(I)| = |F| = 2n-4 Therefore, in this cases, x*i < (2n – 4) = |F(I)| i  I

22. Claim C: for any subset I  {1,…,n} with 1|I|n – 3, G has more than 2|I| faces that have at least one vertex belonging to {vi | i I}. For any such I, x*i 2|I| < |F(I)| i  I In conclusion, Thus, x* satisfies all the relations defining P*, that is, x*  P* = f(S), as required.

23. Lipton-Tarjan Separator Theorem We all know the divide-and-conquer paradigm: the idea is to divide a problem into 2 smaller subproblems of the same type, which can be solved recursively, and to combine the results of the subproblems to obtain a solution to the original problem. The Lipton-Tarjan theorem, shows that any planar graph can be separated into two much smaller components by the removal of relatively few vertices.

24. In formal Let G be a planar graph with n vertices. Then the vertex set of G can be partitioned into three parts, A, B and C, such that |A|, |B|  3n/4, |C| < 2n, and no vertex in A is adjacent to any vertex in B. Note: the Lipton and Tarjan bounded |A| and |B| by 2n/3. We will prove a weaker result.

25. Half-Space x + y = 1 x + y  1

26. Helly's theorem • Suppose that • X1,…,Xn • is a finite collection of convex subsets of Rd,wheren>d. If the intersection of everyd+ 1of these sets is nonempty, then the whole collection has a nonempty intersection; that is, •  Xj   n j=1

27. Lemma for any n-element set P Rd, there exists a point q  Rd with the property that any half-space that does not contain q covers at most dn/(d+1) elements of P. (such a point q is called a centerpoint of P). For example, let’s examine the case where d = 2. n=9 q

28. Lemma's Proof Let H be the family of all half-spaces that cover more than dn/(d+1) elements of P. we have to show that the intersection of all members of H is non-empty. By Helly’s theorem, it is sufficient to prove that any d + 1 half-spaces H1,…,Hd+1H have a point in common. This is truly the case, otherwise: d+1 |P|   |(Rd – Hi)  P|  (d + 1)(n - dn/(d+1) - 1) < n i=1 Which is a contradiction.

29. Lipton-Tarjan Proof Let S denote the unit sphere in R3 centered at (0,0,1), whose poles are N = (0,0,2) and (0,0,0). The stereographic projection :R2 S maps any point p of the xy-plane, into the intersection of the segment Np with S. the image of any circular disc under  will be a . Conversely, the pre-image of any spherical cap S, which does not contain N in its interior, is a circular disc or a half-plane in the xy-plane. spherical cap N = (0,0,2) (0,0,0)

30. How to make a stereographic projection? N = (0,0,2) For a point p, XY-plane we will cross a line from N through p, until it hits the xy-plane. p If we want to map N to the xy-plane, then we will need an infinite line.

31. Koebe’s Theorem implies that the vertices of G can be represented by non-overlapping spherical caps C1,…,Cn S, such that 2 of them touch each other  the corresponding vertices are adjacent. Pick a point pi in the interior of Ci, and set P = {p1,…pn}. Let q  R3 be the centerpoint of P, satisfying the condition of the previous lemma. Recall that we are working in R3, thus q should cover at most 3n/4 vertices. By symmetry, we can assume that q belongs to the segment connecting the center and the south poles of S (we can rotate the sphere).

32. ( ) 1 + d -1(p) 1 - d Suppose q does not coincide with (0,0,1) - the center of S. Then d = |q – (0,0,1)|  0 Let :S  S be a mapping defined by {  If p  N (p) = N if p = N Since  is the composition of an inverse stereographic projection, a dilatation of the xy-plane, and a stereographic projection, it carries spherical caps into spherical caps, and it also preserves the incidences between them. (pi) will lie in the interior of the cap (Ci), and it can shown that (0,0,1) will be a centerpoint of (P).

33. n  ri2 < 4 i=1 Now, we can assume WLOG, that (0,0,1) is a centerpoint of P. This means that any plane passing through (0,0,1) has at most 3n/4 elements of P strictly on its right side (and left side). Now we need to prove that there’s a plane that covers < 2n Let ri denote the (spherical) radius of the cap Ci. Since C1,…,Cn form a packing, and the area of Ci is at mostri2 , we have Unit sphere area size

34. Jensen's Inequality The average of a convex function is bigger than the function in the average of the points. x1 + … + xn f( ) f(x1) + … + f(xn)  n n If you use the convex function exp and x = log(a), we will get the inequality of arithmetic and geometric means n a1 + … + an a1 * … * an  n

35. ( n ) 2  ri  n i=1 n n  ri2 < 4 i=1 n  ri < 2 n i=1 By Jensen’s Inequality, we have

36. Consider now any plane passing through (0,0,1), and let u(H) denote the unit normal vector sitting at (0,0,1). The location of the endpoint of u(H), as H varies over all planes through (0,0,1) that intersect a fixed spherical cap Ci, will be a ringlike region Ri symmetric about a great circle of S. Ri = 2ri { { 2ri Ci

37. Obviously, the area of Ri satisfies A(Ri) < (2ri)(2) = 4ri Hence, there exists a point of S which is covered by at most n n  A(ri)  4ri n i=1 i=1 <  ri < 2 n = A(S) 4 i=1 Regions Ri. Equivalently, there is a plane H0 intersecting fewer than Caps of Ci. 2 n

38. Let A, B and C denote the set of those vertices of G, for which the corresponding spherical caps Ci lie entirely on one side of H0, on the other side of H0, or meet H0, respectively. It follows directly from the above properties of H0 that |A|, |B|  3n/4 and |C| < . 2 n Clearly, no vertex in A can be adjacent to any vertex in B, because two closed caps lying in two complementary half-spaces can never touch each other.