The Zone Theorem The Cutting Lemma Revisited

1. The Zone TheoremThe Cutting Lemma Revisited Tom Jurgenson

2. The Zone Theorem

3. Definitions reminders • Is a sub-space of d-1 dimensions. • Is a partition of into relatively open convex sets. • Are 0/1/(d-1)-dimension faces (respectively) in . • Are the connected components of also called d-faces

4. Example d=2 (Plane) • Vertices are points, edges are segments, and facets are also segments since 2-1=1. • Cells are 2-dimesional connected components in • The blue lines are the Hyperplanes in H • Their arrangement creates: • Vertices – in red • Edges\Facets – blue segments(excluding red) • Cells – light blue areas.

5. The Zone Theorem - Motivation • Motivation – we would like to bound the number of faces that can see any part of a given hyperplane. For that we’ll need to define what ‘see’ means, and define the set of objects (faces) that ‘see’ g (the zone).

6. Definition for ‘sees’ • A face F can seehyperplane g of H (set of hyperplanes), if there are points such that the open segment xy does not intersect any hyperplane of H. • Important Note: It does not matter which point x we choose, either all of them can see g or none can. • In this example, the red line represents a segment between the thick line – g, and the face – the cell. The blue line exemplifies the fact that no segment connects that cell with the hyperplane g.

7. Zone Definition • Zone – The zone of hyperplane g, is the set of the faces of the arrangement of H that can see g. • Here is the zone (dark grey area) for hyperplane g (the thick line) and hyperplane set H (all the thin lines). Each face (vertex, edge and cell) in the zone of g can see g.

8. The Zone Theorem The number of faces in the zone of any hyperplane in an arrangement of n hyperplanes in is with the constant of proportionality depending on d. we will prove this theorem in an inductive manner

9. Base case, d=2 (The plane) • Let H be a set of n lines in the plane in general position. We consider the zone of line g. • In this case the faces would be: vertices (0-faces), edges (1-faces), and the cells are convex polygons that may or may not be bounded(1-faces). • The bound for the number of faces in the zone in this case should be

10. Bound for Edges is enough • The number of vertices is proportional to the number of edges since in a convex polygon the number of vertices is equal to the number of edges. And in open cells, the number of edges is at most twice the number of vertices. • It is clear that the number of polygons is bounded by the number of vertices by using Euler’s formula: F=v-e+2. Therefore proportional to the number of edges.

11. Bound for Edges • Let g be a horizontal line. • We’ll count the number of edges that seeg that are also above g. • The number of edges that intersect g that are also above g is bounded by n. Since each line (hyperplane in H) out of the n lines can only intersect g in one point. • The other edges are disjoint from g, and in the next slides we are going to find a bound for these edges.

12. Bound for the disjoint edges • Let uv be an edge that sees g but is disjoint from it. • Let be the line containing uv. And let a be the intersection of g and h. • Since u is a vertex - it is the intersection of two lines – so let be the other line that passes through u. Let b be the intersection of l and g.

13. Right Edge of Line l • Definition: uv is right edge of l if b is on the right of a. uv is left edge of l if b is on the left of a. • We’ll show that for each such there is at most one right edge. • Let’s assume uv is a right edge for l.

14. Only one right edge • Assume by contradiction that there are two right edges for l. Let uv and xy be the two right edges, and assume that uv lies below xy. • By definition xy should see some point of g. • However, every point to the left of a on g is obscured by the line l. And every point to the right of a on g is obscured by the line h. • This contradiction shows that only one right edge can exist for each line.

15. Only one right edge (2)

16. Conclusion for d=2 • Each line in H has at most one right edge, and by symmetry also at most one left edge. There is also an edge that intersectsg. The same can be argued for edges below g, for the lower right and lower left edges, and an edge that intersects. Altogether there are at most 6 edges for each line in H in the zone. • There are n lines in H so the number of edges is at most 6n. Linear in n. • We already argued that the total number of faces is linear in the number of edges therefore the number of faces in the zone when d=2 is: as needed.

17. Inductive Step from d-1 to d • Assume that the total number of faces of a zone in is • Mark as the max. number of (d-1) faces in the zone in an arrangement of nhyperplanes in . • Let H be an arrangement of hyperplanes and g a base hyperplanes.t is attained for them.

18. Motivation: number of blue facets • We color one hyperplane in H red, let it be . The rest n-1hyperplanes in H we color blue. • A Blue Facet is a facet that lies in a blue hyperplane. • Since every facet has a chance of becoming blue (choose any blue hyperplane out of n blue hyperplanes), the expected number of blue facets is • Now we bound the expected number of blue facets in a different way and use it to estimate • Example in the next slide…

19. Blue and Red Facets

20. Bound for Expected Blue Facets • Consider the arrangement of blue hyperplanes only. Since there are n-1 such hyperplanes the number of blue facets is bounded by • In the next stage we add the red hyperplaneh and see how it affects the number of blue facets in the zone of g.

21. The increase in number of blue facets caused by h • The red hyperplaneh may split an existing blue facet F into two new facets – F1 and F2. If both new facets can see g then we get an increase in the number of facets. • Claim: is visible from (we’ll prove it later) • If we intersect all blue hyperplanes with g and with h we get a d-1 dimensional arrangement in which is a facet in a zone of the d-2 dimensional hyperplane . Using the induction assumption we get such “splits”. The total number of facets becomes:

22. Split Example

23. Proof for the above claim • Reminder: We wanted to show the following claim: is visible from • Proof: Let C be a cell of the zone in the arrangement of blue hyperplanes having F on the boundary. • Because F1 and F2 see g we get that sees and sees • The interior of is contained in C and we get that the intersection of with the hyperplaneh contains a segment witnessing the visibility of from

24. Split Example (II)

25. Conclusion for facets • On the one hand we calculated the expected number of blue facets: • On the other hand we found an upper bound for the number of blue facets: • We combine the two and get:

26. Conclusion for facets (II) • And as we can see: as needed. • Next we are going to discuss d-k faces and not just d-1 faces (or facets)

27. Expected number of d-k faces • Denote the max. possible number of j-faces in the zone for an arrangement of nhyperplanes of dimension d. • Let H be an arrangement of n hyperplanes where is attained. • Again a random hyperplaneh in H is colored red and the rest blue. • Reminder: a d-k face is the intersection of k hyperplanes. • A d-k face is blue if its relative interior is disjoint from the red hyperplaneh.

28. Expected number of d-k faces (2) • The probability for a d-k face to be blue is : [For the selection of first blue hyperplane, then the second on so on until we select the k-thhyperplane] • As in the facet case we get that the expected number of blue d-k faces in the required zone is:

29. Upper bound for d-k faces • By adding the red hyperplaneh the number of blue d-k faces can increase by at most by the inductive hypothesis. • Using the same arguments in lower dimensions as the facets case we get to the conclusion that: • And in the next slide we’ll show:

30. Algebraic proof

31. Bound for edges and vertices • For the case k=d-1 (edges) by using this method we only get the bound: • So the number of edges and vertices must be bound separately: • Vertices: Each vertex is contained in some3-face of the zone. Within such 3-face, the number of vertices is at most 3 times the number of 2-faces, because the 3-face is a 3D convex polyhedron. Since H is simple each 2-face is contained in a bounded number of 3-face. It follows that the total number of vertices is at most proportional to

32. Final details • Edges are proportional to vertices. • In conclusion, for every k=1…d-2 we get • For vertices and edges we get • In conclusion we get that the zone has

33. Zones in other arrangements • The max. complexity of a zone can be investigated for objects other than hyperplanes, and that leads to many related problems that we won’t discuss here.

34. Cutting Lemma Revisited

35. General Plan • First, we’ll prove the cutting lemma for the planar case with a tight bound. We already saw a proof for a bound that was not tight that uses random sampling, and a different proof that is tight but does not use random sampling. • The following proof is both tight and uses random sampling in a way that makes it easy to be generalized to higher dimensions. • But first we’ll do a review for some cutting-lemma related definitions.

36. Review of Definitions (I) • A generalized triangle is the intersection of 3 half planes. • A cutting of the plane into generalized triangles, is the subdivision of the plane to a disjoint set generalized triangles that their union covers the entire plane.

37. Review of Definitions (II) • A cutting of the plane with a set of n of lines H – is a cutting of the plane into generalized triangles s.t the interior of each such is intersected by at most lines from H. • Note – the triangles’ edges may or may not include lines from H.

38. ½ cutting example • In the next example we have a ½ cutting. The red lines are lines in H, the black lines are the lines that create the triangulation.

39. The Cutting Lemma • For every set H of n lines in the plane and every r>1 there exists a 1/r cutting for H the size • In other words, there is a subdivision of the plane into generalized triangles s.t the interior of each triangle is intersected by at most n/r lines of H.

40. The sampler • Goal – sample line. Once the lines are sampled their intersection (possibly with some modifications) would result in a triangulation. • In the previous lecture when we wanted to select a subgroup S of H, we picked |S| lines out of H with repetitions. This procedure may result with a subgroup of size smaller than |S|. • In order to simplify the calculation, we’ll choose S by independent Bernoulli trials. We fix the probability of p=s/n, and include a line from H in S using that probability. • For this point on, let s=|S|

41. Why sampling and triangulation does not work? • In a previous lecture we already saw a sampling algorithm that outputs a triangulation that with probability close to 1 non of the triangles in the outputted triangulation is intersected by more than where is some constant (Weak Cutting Lemma) • We will demonstrate that a similar statement with a is not generally true. So our method would have to include more than just sampling in order to achieve the desired triangulation. • To see that we’ll examine a simpler case than the plane – 1-dimensional situation.

42. The 1-D case • H is a set of n points. • A generalized triangle in this case is the segment between two points. • We would like to count the longest consecutive count of unselected points of H. Let k be our target number for such a count. • In example below, n=30 and s=15 so p=0.5. The black dots are selected points (In this example there are exactly 15 selected dots, but it doesn’t have to be this way)

43. k consecutive unselected points • We’ll show that for a fixed k it is very likely that k consecutive unselected points show up in a sequence of n points, where n is sufficiently large • For simplicity assume n is divisible by k and divide our n points in to blocks the length of k. Notice that this is an even more restrictive case than a general consecutive count. • In each such a block there is a probability of for the entire block to be unselected. Therefore the probability of not obtaining any block as entirely unselected is and this probability is exponentially small for k proportional to

44. Why is it exponentially small? • p is a constant, let p=0.5. The probability becomes: • Now set k=0.5logn: • And that exponentially small as n goes to infinity.

45. Conclusion of 1D • A sequence of k that is proportional to log of n is very likely to appear using this sampling method. This is why this method is not good enough to get the 1/r cutting we require. • Of course, in the 1D case we can define a sampler that selects every n/s point. However it is not clear how to transform this selection to the plane (Where the lines are not ordered). • 

46. 2nd attempt – Two level decomposition • General plan: Instead of trying to improve the sampling algorithm to generate a 1/r cutting from the start: • we’ll take a random sample (obtained as before with probability p=s/n independent Bernoulli trials) • We just saw that this is not good enough, so we’ll take every triangle that intersects more lines than required and divide it so each new triangle will be good for our 1/r cutting. • Finally, we’ll show that the total number of triangles is

47. But first, more definitions • Let T be the collection of triangles after triangulating the random sample. • Let denote the set of lines of H that intersect the triangle and number of lines that intersect. • Let the excess of be • Note: if so the triangle can be included in the 1/r cutting. If we subdivide it to a collection of finer triangles.

48. Subdivision of triangles with excess greater than 1 • We consider the arrangement of (instead of H), and we construct a cutting for it. (How? We’ll see later...) • We intersect this triangulation with . This process may produce triangles, but also quadrilaterals, pentagons, and hexagons. Each of these convex polygons is subdivided into triangles. • Each triangle in the cutting is intersected by at most triangles, and therefore are valid for the 1/r cutting.

49. Example: Up – the process of subdividing a triangle with excess greater than 1. Left – An example for a hexagon created by combining the two triangulations.

50. But How? The suboptimal cutting lemma. • As of now, we can’t subdivide because the cutting lemma is not valid yet. So let’s use a weaker claim: • The Suboptimal Cutting Lemma for every finite collection of lines and any u>1, there exists a 1/u cutting consisting of at most: triangles, where K is a suitable constant. • We’ll prove the suboptimal cutting lemma later.