3/26/20 CMPS 3130/6130 Computational Geometry - PowerPoint Presentation

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3/26/20

CMPS 3130/6130 Computational Geometry

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CMPS 3130/6130 Computational GeometrySpring 2020

ArrangementsCarola Wenk

Arrangement of Lines

Let be a set of lines in . Then is called the arrangement of . It is defined as the planar subdivision induced by all lines in .

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CMPS 3130/6130 Computational Geometry

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is

simple

if no three lines meet in one point, and no two lines are parallel.

Lines. Not line segments.

Vertices, edges, faces.

Arrangement Complexity

The complexity of is its #vertices + #edges + #faces.

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CMPS 3130/6130 Computational Geometry

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Because it’s a planar subdivision.

#vertices

#edges

#faces

We have

lines, and in the worst case every pair

intersects (vertex).

On one line we can have at most

vertices,

so at most

edges total.

Consider an incremental construction, just for counting purposes right now.

Let . Line splits a face of in two. This creates i additional faces (since has at most i edges in see above). #faces in

Arrangement Construction

Input: A set of lines in Output: The arrangement stored in a DCEL

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CMPS 3130/6130 Computational Geometry

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1) Sweep-line construction:Takes time.2) Incremental construction:Insert one line after the other. Again let . Construct_arrangement(){ whole plane for i=1 to n{ insert into }

by threading through face by face and splitting edges and faces accordingly (using the DCEL!).

Runtime:

Using “zone theorem”

Zone Theorem

Zone Theorem:

Let be an arrangement of lines and let be another line. The zone of in is the planar subdivision consisting of all faces, edges, and vertices intersected by . The complexity of the zone of in is .

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CMPS 3130/6130 Computational Geometry

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How can the zone have complexity

if

has complexity

?

Zone Theorem Proof

Assume is horizontal. Also assume is simple and as no horizontal edges

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CMPS 3130/6130 Computational Geometry

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right-bounding edges

left-bounding edges

Goal:

Prove that # left-bounding edges in the zone is

, using induction.

Base:

Step:

Let

be a set of

lines and

its arrangement.

line that has the

rightmost

intersection with

vertex on

above

, closest to

vertex on

below

, closest to

Zone Theorem Proof

3/26/20

CMPS 3130/6130 Computational Geometry

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Goal:

Prove that # left-bounding edges in the zone is , using induction.

Step: Let be a set of lines and its arrangement.

line that has the

rightmost

intersection with

vertex on

above

, closest to

vertex on

below

, closest to

Think about



# left-bounding edges in is by inductive hypothesis.

Now insert into :

is a new edge, and two edges were split into two.3 new edgesNo more new edges: Region is not in but is the only part of above that could contribute with left-bounding edges.In total, the zone has edges

For

in