incidences and many faces via cuttings sivanne goldfarb 5.6.07

Incidences and Many Faces

via cuttingsSivanne Goldfarb

5.6.07

2

Outline

The Many Faces Problem

Point – Line incidences

Point – Circle incidences

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.

}{ ....2,1 mpppP

}{ ....2,1 nlllL

),( LPK

4


• Assumption:– Each lies in unique cell.

• Problem:

– Establish an upper bound for ),( LPK

ip

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 – )(),( 21

nmnOLPK


)(),( 32

32

nnmOLPK

6

Step 1 – Construct a 1/r cuttingPhase 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.

LR

1 2

3

20

8

22

lines in R lines in L\R

7

Step 1 – Construct a 1/r cuttingPhase 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(r2) trapezoids. Each trapezoid is crossed by at most lines.

1 2

3

20

8

22


r

n

r

n

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 , , . |Pi|=mi

– Li - a subset of lines that have a non empty intersection with i .|Li|=ni

• Note that

k

ii mm

1

ki 1 ii PP

lines in R

lines in L\R

P20 L20

20

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(Li) bounding the mi cells containing

points of Pi.

• Summing the weaker bound in all trapezoids gives: .

)())((2

1

1

21

nrr

nmOnnmO

k

iiii

10

Step 2 – Divide and conquer

• What happens if the cell which contains pj intersects the boundary

of i?

Solution: count all edges of this cell.


12

11

Step 3 – Complexity of zones

• Inner zone of i in A(Li ) – the collection of for all cells c in A(Li ) which

intersect the boundary of i.

ic

• Complexity of inner zone – the total number of edges bounding its cells.

• The complexity of the inner zone of iis O(ni) – we will not prove this.

• Complexity of all inner zones – )()*()( 2

1

nrOr

nrOnO

k

ii

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

)()(),(1

21

21

nrr

nmOnnmOLPK

k

iiii

)( 31

32

nmr

)(),( 32

32

nmOLPK 21

nm

21

nm

)(}||,|||),(max{),( 32

32

nnmnLmPLPKnmK

)(),( 32

32

nnmnmK

13

Outline

The Many faces Problem



14

Point – Line Incidences

• Let P be a set of m points

• Let L be a set of n lines


• 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.

)(),( 32

32

nmnmOLPI

)(),(

)(),(

21

21

mnmOLPI

nmnOLPI

15


• 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 ,

– Li - subset of lines that have a non empty intersection with .

– R – as in the Many Faces proof.

• For each trapezoid we get

i ki 1 ii PP ii mP ||

i ii nL ||

)( 2rOK

r

n

)()(),(2

1

21

r

n

r

nmOnnmOLPI iiiiiii

16


• 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!

)(1

k

ii mOm

)(),(2

1

1

nrr

nmOLPI

k

iii

1

2


17


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.

1

2 lines in R lines in L\R

l crosses both trapezoids.

RLl \

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


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

1

2


19

Point – Line Incidences3. 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.

1

2

2


20


r

n)( 2rOK

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 )(2 2 nrOr

nr

1

2

2


21


• We have shown that

• Choose for

• from weaker bound we get

• from weaker bound we get

• Putting all bounds together we get

)(),(2

1

nrmr

nmOLPI

)( 31

32

nmr nr 1

21

nm 1r

2nm nr

)(),( 32

32

nmnmOLPI

)()(),( 21

nOnmnOLPI

)()(),( 21

mOmnmOLPI

22

Outline

The Many faces Problem



23

Point – Circle Incidences

• Let P be a set of m points.

• Let C be a set of n circles.


• The idea is the same as earlier, we will use a weaker bound (again, can be

derived from the forbidden complete graphs’ results):

)(),( 54

53

mnnmOCPI

)(),(

)(),(

21

32

mnmOCPI

nmnOCPI

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.

CR

circles in R circles in C\R

25


• 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(r2)

• As earlier, at most circles cross each trapezoid. r

n

26


• Notations:

– For each trapezoid i, , ,

– Ci- subset of circles that have a non empty intersection with i. |Ci|=ni

– R – As earlier

• The number of incidences within the trapezoids:

k

i

k

iiiii

k

iii nrrmnO

r

n

r

nmOnnmOCPI

1

32

32

1

32

32

1, )()()()(

ki 1 ii PP ii mP ||

27


• 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 )(8 2 nrOr

nr

28


• There are at most m incidences between circles of R and points which are not vertices of A(R)

)(),( 32

32

nrmrmnOCPI

• Putting everything together we get

29


• Choose for and we get

• for we get

• for we get

• Putting it all together we get

)( 51

53

nmr 231

nmn )(),( 54

53

nmOCPI

31

nm )()(),( 32

nOnmnOCPI

2nm )()(),( 21

mOmnmOCPI

)(),( 54

53

mnnmOCPI

30

Questions?

incidences and many faces via cuttings sivanne goldfarb 5.6.07

Documents