1. Incidences and Many Faces via cuttings Sivanne Goldfarb 5.6.07

2. 2 Outline The Many Faces Problem Point – Line incidences Point – Circle incidences

3. 3 The Many Faces Problem Input : –A set of m points –A set L of n lines Notations : –A(L) – the arrangement of the lines of L. Decomposition of the plane into cells. –Combinatorial complexity of a single face = The number of edges belonging to its boundary. – - The combined combinatorial complexity of faces which contain points from P.

4. 4 The Many Faces Problem Assumption: –Each lies in unique cell. Problem : –Establish an upper bound for

5. 5 We will show that The proof consists of 4 steps. The idea of the proof: Partition the plane into sub regions, apply a weaker bound in each sub region and sum up the bounds. We will use the weaker upper bound derived from the forbidden complete graphs’ results – The Many Faces Problem

6. 6 Step 1 – Construct a 1/r cutting Phase 1 Choose a subset of size r. Each cell is decomposed into trapezoids by drawing a maximal vertical line segment through each vertex of A(R ). We stop drawing a vertical line when it reaches a new line from R. lines in R lines in L\R

7. 7 Step 1 – Construct a 1/r cutting Phase 2 A trapezoid is called “heavy” if more than lines intersect it. Handle “heavy” trapezoids by further cutting them as was shown in the last lecture. As proven before, we have constructed a 1/r - cutting which consists of K = O(r 2 ) trapezoids. Each trapezoid is crossed by at most lines. lines in R lines in L\R

8. 8 Step 1 – Construct a 1/r cutting Notations: –R – for simplicity, the lines chosen for the cutting in both phases. Notice, R isn’t of size r any more. –For each trapezoid  i,,. |P i |=m i –L i - a subset of lines that have a non empty intersection with  i. |L i |=n i Note that lines in R lines in L\R P20 L20

9. 9 Step 2 – Divide and conquer Using the 1\r - cutting of the arrangement A(L), we solve a sub problem in each trapezoid. In each  i, count the edges of A(L i ) bounding the m i cells containing points of P i. Summing the weaker bound in all trapezoids gives:.

10. 10 Step 2 – Divide and conquer What happens if the cell which contains p j intersects the boundary of  i ? Solution: count all edges of this cell. lines in R lines in L\R

11. 11 Step 3 – Complexity of zones Inner zone of  i in A(L i ) – the collection of for all cells c in A(L i ) which intersect the boundary of  i. Complexity of inner zone – the total number of edges bounding its cells. The complexity of the inner zone of  i  is O(n i ) – we will not prove this. Complexity of all inner zones –

12. 12 Step 4 – wrapping it up Summing the weaker bound in each trapezoid and taking inner zones into consideration gives: Choose r to be, so it balances the two terms above, and we have - this is meaningful if According to our weaker upper bound, if then K(P, L) = O(n). We recall that in the past we proved that Finally we get

13. 13 Outline The Many faces Problem Point – Line incidences Point – Circle incidences

14. 14 Point – Line Incidences Let P be a set of m points Let L be a set of n lines We will show that As in the Many Faces problem, we will use the weaker bound derived from the forbidden complete graphs’ results: The idea of the proof (again): Partition the plane into sub regions, apply the weaker bound in each sub region and sum up the bounds.

15. 15 Point – Line Incidences We construct a 1/r - cutting as in the Many Faces proof. Decompose the plane into vertical trapezoids with pairwise disjoint interiors. Each trapezoid is crossed by at most lines of L. Notations: –For each trapezoid, –L i - subset of lines that have a non empty intersection with. –R – as in the Many Faces proof. For each trapezoid we get

16. 16 Point – Line Incidences We recall that (ignoring points on the boundary of trapezoids) and we get We still need to consider points that lie on the boundary of the trapezoids! lines in R lines in L\R

17. 17 Point – Line Incidences 1.Let p be a point which lies on an edge of the cutting (but not on a vertex of the cutting). Let be a line which passes through p. p lies on the boundary of at most two trapezoids. lines in R lines in L\R l crosses both trapezoids. We assign p to one of the trapezoids and its incidences with lines from L\R will be counted within the sub-problem associated with that trapezoid.

18. 18 Point – Line Incidences 2.Let p be a point as in the previous case. p is incident to at most one line in R There at most O(m) incidences of this kind lines in R lines in L\R

19. 19 Point – Line Incidences 3.Let p be a point which is a vertex of the cutting and l a line incident to p. l either crosses or bounds some adjacent trapezoid  i l can cross the boundary of a trapezoid in at most two points. lines in R lines in L\R

20. 20 Point – Line Incidences We charge the incident (p, l) to the pair (l,  i ) Recall that each trapezoid is crossed by at most lines and that The number of incidences involving vertices of the cutting is at most lines in R lines in L\R

21. 21 Point – Line Incidences We have shown that Choose for   from weaker bound we get Putting all bounds together we get

22. 22 Outline The Many faces Problem Point – Line incidences Point – Circle incidences

23. 23 Point – Circle Incidences Let P be a set of m points. Let C be a set of n circles. We will show that The idea is the same as earlier, we will use a weaker bound (again, can be derived from the forbidden complete graphs’ results):

24. 24 Point – Circle Incidences Construct a 1/r – cutting of the arrangement A(C). The construction is quite similar to the previous one, we will describe the changes in “phase 1”. Choose a subset of size r. Triangulate A(R) : –Draw vertical line segments through the left and rightmost points of all circles in R. –Draw a maximal vertical line segment through every vertex of A(R) and stop when it reaches another circle from R. circles in R circles in C\R

25. 25 Point – Circle Incidences A “trapezoid” is bounded by 4 edges: –2 vertical edges to the left and the right –2 circle edges at the top and the bottom. The total number of trapezoids is K = O(r 2 ) As earlier, at most circles cross each trapezoid.

26. 26 Point – Circle Incidences Notations: –For each trapezoid  i,,, –C i - subset of circles that have a non empty intersection with  i. |C i |=n i –R – As earlier The number of incidences within the trapezoids:

27. 27 Point – Circle Incidences We also have to consider incidences between points which lie on the boundaries of trapezoids and all n given circles. Each circle c intersects the boundaries of a trapezoid at most eight times – twice for each edge of the trapezoid. The number of incidences of this type is at most

28. 28 Point – Circle Incidences There are at most m incidences between circles of R and points which are not vertices of A(R) Putting everything together we get

29. 29 Point – Circle Incidences Choose for and we get for we get Putting it all together we get

30. 30 Questions ?