An arrangement of lines: A(H)

112

3

4

5

1

55 5

11

1

1111111111 3455

11111

11

12

3

4

5

Elements of the arrangement

Vertices – intersection of lines

Edges – portions of lines bounded by vertices, except when unbounded at one end

Faces – regions bounded by edges and vertices

Counts of the elements

Number of vertices (V) : n(n-1)/2

Number of edges (E): n^2 (interesting)

Number of faces (F): n^2 – n(n-1)/2 + 1 (follows from the modified Euler formula V-E+F = 1)

Computing an arrangement

We must output a data structure that encapsulates the mutual relationships of the elements (vertices, edges and faces), corresponding to the pictorial representation of an A(H) as in the first slide

EG paper only shows how to traverse A(H)

Sweepline paradigm

The standard sweepline paradigm requires sorting the O(n^2) intersection points

This would require O(n^2 log n) time

Topological sweep

Gets rid of the log n factor, by processing the intersection points without having to sort them !!

It does this at the expense of just O(n) additional space needed by lots of extra book-keeping.

A partial order

We can define the following partial order on elements of the arrangement

An element A is above element B if A is above B at every vertical line that intersects both A and B

The above relationship is acyclic The inverse of above is below

Consequences

There exists a unique element in A(H) that is not below any other and a unique element that is not above any other.

These are called respectively the top-most and the bottom-most element

Prove uniqueness from the acyclicity of the above partial order

Cuts

A cut is a sequence of edges (c1, c

2, ...,c

n),

one from each line of A(H), such that for each i (1.. n-1) there is a (unique) face f

i such that c

i

is above fi , and c

(i+1) is below f

i

c1 is below the top-most face and c

n is above

the bottom-most face

A cut – pictorially

c1c2

c3

c4

c5

Ordering the cuts

A cut C is is to the left of a cut C' if for each line l in A(H), c

i on l from C is to the left of or

identical with cj from C' on l

Thus there is a lefmost cut and a rightmost one

Important Fact

In a given cut there exists an i such that ci

and c(i+1)

have a common right end-point The so-called topological sweep exploits this

to move from the leftmost cut to the rightmost one in a series of elementary moves

In each such move, the topological line moves across such a common right end-point

We demonstrate this on the example arrangement in the next few slides

Sweeping the arrangement

The leftmost cut

Sweeping A(H) : 1st elementary move

2nd elementary move

3rd elementary move

4th elementary move

5th elementary move

6th elementary move

7th elementary move

8th elementary move

9th elementary move

10th elementary move

The rightmost cut

Representing a cut

A cut is represented by an array C[1..n], where each C[i] = ( λi, ρi , µi), representing respectively the index of the line that defines the left and right end-point of the edge c i and the index of the line on which it lies

From one cut to the next..

This is done in an elementary step An elementary step is one in which the

topological sweep moves across an intersection defined by c

i and c

(i+1) for some i in

the current cut Such an i always exists except when the cut is

the rightmost one

Implementing an elementary move

An elementary move is implemented with the help of two data structures derived from a cut C – namely the upper horizon tree T+(C) and the lower horizon tree T-(C)

Upper horizon tree

Lower horizon tree

Initializing the horizon trees

Data Structures for horizon trees

An array HTU[1..n] for the upper horizon tree, where HTL[i] = (λi, µi), where λi (µi ) is the index of the line that defines the left (right) end point of the segment from li that belongs to the upper horizon tree

λi = -1 if segment left-unbounded and µi = 0 if right-unbounded

A similar definition for HTL[1..n]

Initializing HTU (and HTL similarly)

Updating HTU (and similarly HTL)

Example HTU[1..5]

HTU[1]= (-1,2)HTU[2]= (-1,5)HTU[3]= (5,4)HTU[4]= (5,0)HTU[5]= (3,0)

Example HTL[1..5]

HTL[1]= (-1,0)HTL[2]= (-1,1)HTL[3]= (5,1)HTL[4]= (5,3)HTL[5]= (3,1)

Data Structures for horizon trees

Array M[1..n] stores the index of the line on which ci lies

Array N[1..n] stores the description of a cut; N[i] = (λi, µi), where λi (µi ) is the index of the line that defines the left-end (right-end) of ci

N[1..n] can be obtained from HTL[1..n], HTU[1..n] and M[1..n]

Example N[1..5]

M[1..5] = [1,2, 5, 3, 4] (slide 32) This gives: N[1..5] = [(-1,2), (-1,1), (3,1),

(5,4),(5,3)]

Data Structures for horizon trees

Finally, we have a stack I that stores the indices i such that ci and c(i+1) have a common right end-point.

This is obtained from N[1..n] by examining pairs of entries in N[1..n] and checking if µi =

µi+1 + 1, for i= 1, .., n-1, and stacking the i for which the above holds

Example I

I=[1,4 …….., for the example N[1..5]

Updating HTU (and similarly HTL)

Updating all the other data structures

It is easy to update M[1..n] N[i] = HTL[M[i]] ∩ HTU[M[i]] From N[i] we can update the stack I

Analysis

The analysis shows that the cost of traversing the bays associated with a fixed line l is O(n) and hence O(n2) for all n lines.

Amortized Analysis