splitting a face
DESCRIPTION
Splitting a Face. Point . two new faces one new vertex six new half-edges. edge already split. new vertex. The Algorithm. ConstructArrangement ( L ) Compute a bounding box B ( L ) to contain all vertices of A ( L ) // Initialize DCEL for B ( L ) // O (1) for i ←1 to n - PowerPoint PPT PresentationTRANSCRIPT
Splitting a Face
Point
il
f
edge already split new
vertex il
2f
1f
two new faces one new vertexsix new half-edges
The Algorithm
Point )( 2nO
il
Face Splitting
Point Split faces in intersected by .1iA il
- Insertion of takes time linear in total complexity of faces intersected by the line.
il
ilil
il
il
Zone
Point
l
once
three times
A vertex may be counted up to four times.
fourtimes
twice
Time of Arrangement Construction
Point Proof By induction.
Time to insert all lines, and thus to construct line arrangement:
n
i
nOiO1
2 )()(
Solution to the Discrepancy Problem
Point
83)( hs
41)( hContinuous measure:
Discrete measure:
Minimize |)()(| hhs
Exactly one point brute-force method
At least two points apply duality
)( 2nO
How to Use Duality?
Point
A line through ≥ 2 sample points
Reduction
Point :an # lines above a vertex
:bn # lines below the vertex
:on # lines through the vertex
Sufficient to compute 2 of 3 numbers (with sum n).
is known from DCEL.:on
Need only compute the level of every vertex in A(S*). an
Levels of Vertices in an Arrangement
Point
0
2
1
3
1
3
4
32
3
2
level of a point = # lines strictly above it.
Counting Levels of Vertices
Point
1v
5v4v
3v2v
l121 ,...,, nvvv
Counting Levels of Vertices
Point
1vl
11v
1v
Along the line the level changesonly at a vertex .iv
A line crossing comes eitherfrom above or from below (relativeto the current traversal position)
iv
a) from above level( ) = level( ) – 1 iv1iv
b) from below level( ) = level( ) + 1 iv1iv
coming from above
coming from below
01
1 3
100
2 2no change of levelbetween vertices
Running Times
Point
)( 2nOLevels of all vertices in a line arrangement can be computed in time.
Discrete measures computable in time. )( 2nO
)(nOLevels of vertices along a line computable in time.
Duality in Higher Dimensions
Point ),...,,( 21 dpppp point
hyperplane dddd pxpxpxpxp 112211* ...:
hyperplane dddd axaxaxaxh 112211 ...:
point ),...,,(: 21*
daaaah
Inversion
Point ),( yx ppp point point ),,( 22yxyx ppppp
p
p
22 yxz
The point is lifted to theunit paraboloid.
x
y
z
Image of a Circle
Point
222 )()(: rbyaxC
22 yxz
Image of a point on the circle C has z-coordinate
22222 rbaayax
22222: rbaayaxzP
C plane
is the intersection of the planeP with the unit paraboloid.
C
Inside/Outside Below/Above
Point
222 )()(: rbyaxC
22222: rbaayaxzP
C
p
p
lies inside C iff is below . p p C
q
q