COT 5520 Computational Geometry

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(No Transcript)
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Preliminaries
Planar straight line graph A planar straight line
graph (PSLG) is a planar embedding of a planar
graph G (V, E) with 1. each vertex v ? V
mapped to a distinct point in the plane, 2. and
each edge e ? E mapped to a segment between
the points for the endpoint vertices of the
edge such that no two segments intersect, except
at their endpoints.
edge (14) vertex (10) face (6)
Observe that PSLG is defined by mapping a
mathematical object (planar graph) to a geometric
object (PSLG). That mapping introduces the notion
of coordinates or location, which was not present
in the graph (despite its planarity). We will
see later that PSLGs will be useful objects. For
now we focus on a data structure to represent a
PSLG.
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Preliminaries
Doubly connected edge list (DCEL) The DCEL data
structure represents a PSLG. It has one entry
(edge node in Preparata) for each edge in the
PSLG. Each entry has 6 fields V1 Origin of the
edge V2 Terminus (destination) of the edge
implies an orientation F1 Face to the left of
edge, relative to V1V2 orientation F2 Face to the
right of edge, relative to V1V2
orientation P1 Index of entry for first edge
encountered after edge V1V2, when proceeding
counterclockwise around V1 P2 Index of entry for
first edge after edge V1V2, when proceeding
counterclockwise around V2
v3
e3
f4
f2
e5
v2
e1
v1
e4
f1
e2
f3
v4
Example (both PSLG and DCEL are partial)
4
2
2
3
F2
13
8
9
4
12
F6
11
1
3
10
F3
F1
F4
6
7
6
9
5
F5
1
4
7
8
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Edge V1 V2 LeftF RightF PredE
NextE --------------------------------------------
----- 1 1 2 F6 F1 7
13 2 2 3 F6 F2 1
4 3 3 4 F6 F3 2
5 4 3 9 F3 F2 3
12 5 4 6 F5 F3 8
11 6 6 7 F5 F4 5
10 7 1 5 F5 F6
9 8 8 4 5 F6 F5
3 7 9 1 7 F1 F5
1 6 10 7 8 F1 F4
9 12 11 6 9 F4 F3
6 4 12 9 8 F4 F2
11 13 13 2 8 F2 F1
2 10
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Preliminaries
Supplementary data structures If the PSLG has N
vertices, M edges and F faces then we know N - M
F 2 by Eulers formula. DCEL can be
described by six arrays V11M, V21M,
LeftF1M Right1M, PredE1M and
NextE1M. Since both the number of faces and
edges are bounded by a linear function of N, we
need O(N) storage for all these arrays. Define
array HV1N with one entry for each
vertex entry HVi denotes the minimum numbered
edge that is incident on vertex vi and is the
first row or edge index in the DCEL where vi is
in V1 or V2 column. Thus for our example in
the preceding slide HV(1,1,2,3,7,5,6,10,4.
Similarly, define array HF1F where F M-N2
,with one entry for each face HFi is the
minimum numbered edge of all the edges that make
the face HFi and is the first row or edge
index in the DCEL where Fii is in LeftF or
RightF column. For our example,
HF(1,2,3,6,5,1). . Both HV and HF can be
filled in O(N) time each by scanning DCEL. DCEL
operations Procedure EdgesIncident (VERTEX in
Preparata) use a DCEL to report the edges
incident to a vertex vj in a PSLG. The edges
incident to vj are given as indexes to the DCEL
entries for those edges in array A 1 3N-6
since Mlt 3N-6.
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Preliminaries
1 procedure EdgesIncident(j) / VERTEX in text
/ 2 begin 3 a HVj / Get first DCEL entry
for vj, a is index. / 4 a0 a / Save
starting index. / 5 A1 a 6 i 2 / i is
index for A / 7 if (V1a j) then /
vertex j is the origin / 8 a PredEa / Go
on to next incident edge. / 9 else / vertex
j is the destination vertex of edge a / 10 a
NextEa / Go on to next incident edge.
/ 11 endif 12 while (a ? a0) do / Back to
starting edge? / 13 Ai a 14 if (V1a
j) then 15 a PredEa / Go on to next
incident edge. / 16 else 17 a NextEa /
Go on to next incident edge. / 18 endif 19 i
i 1 20 endwhile 21 end
NextEa
V2a
a
V1a
PredEa
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Preliminaries
DCEL notes If we make the following changes, it
will become a procedure Faceincidentj giving
all the edges that constitute a face 3 a
HFj And sustitute HV by HF and V1 by F1. A
will have a size A1N. Both EdgesIncident and
Faceincident requires time proportional to the
number of incident edges reported. How does that
relate to N, the number of vertices in the
PSLG? We have these facts about planar graphs
(and thus PSLGs) (1) v - e f 2 Eulers
formula (2) e ? 3v - 6 (3) f ? 2/3e (4) f ? 2v -
4 where v number of vertices N e number
of edgesM f number of facesF v ? O(N) by
definition ? e ? O(N) by (2) ? EdgesIncident
requires time ? O(N) and ? DCEL requires storage
? O(N), one entry per edge.
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Preliminaries
Vector algebra An ordered pair (x, y) can be a
point in the plane, or a vector.
Vector addition Given vectors a (xa, ya) and b
(xb, yb), vector addition is defined as a b
(xa xb, ya yb). Geometrically, vectors a and
b determine a parallelogram with vertices 0, a,
b, and a b.
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Preliminaries
Scalar multiplication Multiplication of vector b
by a scalar (a real number) t. Scalar
multiplication is defined as tb (txb, tyb). The
vector length is scaled by t. If t lt 0, the
direction is reversed.
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Preliminaries
Vector subtraction Given vectors a (xa, ya) and
b (xb, yb), vector subtraction is defined as b
- a b (-1)a, carried out as b - a (xb - xa,
yb - ya).
b
b - a
a
Vector length Length of vector a (xa, ya) is
defined as a sqrt(xa2 ya2).
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Vector Translation
q
b
p
a
b-a
o
-a
Let a op and b oq. Then, b-a is a
translation of the vector pq at the origin o.
Thus, two line segments having same length and
direction are translates of each other and can
be identified with the same canonical
line segment originating at the origin o.
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Preliminaries
Vector direction The direction of vector a is
described by its polar angle ?a, the angle the
vector makes with the positive x axis. Measured
in counterclockwise rotation, starting at the
positive x axis. Values are in the range 0 ? ?a lt
360.
b
a
?ab
?a
Given two vectors a and b, the angle between them
?ab is measured counterclockwise starting at
vector a.
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Preliminaries
Trigonometry reminder Definition of sine and
cosine based on unit circle.
90
Unit circle x2 y2 1
(x, y)
0
180
270
x cos ? y sin ? 0 lt ? lt 180 ? y gt 0 ? sin ?
gt 0 180 lt ? lt 360 ? y lt 0 ? sin ? lt 0
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Preliminaries
LEFT-RIGHT-ABOVE-BELOW A geometric primitive
operation triangle orientation Given three
non-collinear points p0, p1, p2, the triangle
??p0p1p2 is positively oriented if p2 lies to the
left of p0p1, and negatively oriented if p2 lies
to the right of p0p1. Let vector a p1 - p0 and
vector b p2 - p0 .
p2
b
?????ab lt 180 positive orientation
p0, p1, p2 form a counter-clockwise cycle

a
?ab
p1
p0
???????ab lt 360 negative orientation
-
?ab
p0, p1, p2 form a clockwise cycle
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We introduce the value Q ?b - ?a (note that Q ?
?ab).
(Equality holds if a and b are collinear.) With
this information, we could compute Q ?b -
?a and then use the table to give the orientation
of ??p0p1p2. But computing ?a and ?b require
expensive trig functions. Can we do better?
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Preliminaries
Observe from the table that sin(Q) has the same
sign as the orientation of ??p0p1p2. sin(Q)
sin(?b - ?a) by definition of Q sin ?b cos ?a
- cos ?b sin ?a by trig identity We know
that cos ?a xa / a sin ?a ya /
a cos ?b xb / b sin ?b yb /
b by definition of sine and cosine. Then by
substitution sin(?b - ?a) (yb / b)(xa /
a) - (xb / b)(ya / a) (1 / a b)
(ybxa - xbya). a and b are positive
constants, so sign(sin(?b - ?a)) sign(ybxa -
xbya). By definition, xa (x1 - x0) ya
(y1 - y0) xb (x2 - x0) yb (y2 -
y0) so sign(sin(?b - ?a)) sign((y2 - y0)(x1 -
x0) - (x2 - x0)(y1 - y0)). The latter expression
can be used to find the orientation of
??p0p1p2 from the coordinates of p0, p1, p2 in
constant time.
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Preliminaries
Another way of reaching the same expression is
with the determinant of the coordinates of the
points x0 y0 1 D x1 y1 1 x2 y2 1
Evaluating this determinant gives the
expression x0 y1 x2 y0 x1 y2 - x2 y1 - x0 y2
- x1 y0 which is the expansion of the final
expression previously derived. Left turn/Right
Turn Test If the sign of the determinant D is
positive then ??p0p1p2 is counter- clockwise (p2
is left of p0p1 ) and if D is negative then
??p0p1p2 is clockwise (p2 is right of p0p1 ). If
D0, then the three points are collinear.
p2
left
p0
p1
p2
right
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Area Interpretation The value of the determinant
D is twice the signed area of the triangle
??p0p1p2 . The signed area is positive if p0p1p2
form a counter clockwise sequence, it is
negative if this sequence is clockwise. If
the area is zero, then D is 0. Generalizaion The
three points p0 p1 and p2 form a plane
in 3-d with a positive normal. A fourth point p3
is on the upside if p3 falls on the positive side
of the plane and on the downside if p3 falls on
the negative side of the plane. The test for this
is with respect to the sig of the determinant
x0 y0 x0 1
x1 y1 z1 1
D x2 y2 z2 1
x3 y3 z3 1 This D represents
the signed volume of the polyhedron. The
fomulation extends to n-dimensional space.
(Further Reading Section 13(p.17) to Section
15 (p.35), in ORourke text).
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Preliminaries
Parametric equation of a line We use the
following equation of a line line ?(p0) (1
- ??(p1) , where ????? (real numbers) where p0
and p1 as usual are the points determining the
line. p0 (x0, y0) p1 (x1, y1) Substituting
gives ?(x0, y0) (1 - ??(x1, y1) Multiplying
through gives the coordinates ?x0 (1 - ??x1,
?y0 (1 - ??y1
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Note, in the previous slide,we implicitly assumed
that the points are located in a plane. The
definition is valid in d- dimension where a point
is specified by giving a vector of d real
numbers. Thus, for three dimension, the two
points can be specified as p0(x0,y0,z0)
p1(x1,y1,z1) And the equation for the line in
parametric form will look like line ?(p0)
(1 - ??(p1) , where ????? (real numbers) This
is again the set of points in three
dimension that lie on the line connecting these
two points. The discussion is valid for Euclidean
space Ed. If ? is restricted to be in the range
between 0 and 1, the equation becomes the
equation of the line segment connecting these
two points.
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Preliminaries
Point-Line classification We now consider the
geometric primitive operation of classifying a
point w.r.t. a line (both in the plane). A
directed line segment partitions the plane into
7 non-overlapping regions. The possibilities are
shown below. The problem, given p0, p1, and p2,
is to determine which region p2 lies in.
beyond
p1
left
terminus
between
p0
origin
right
behind
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Preliminaries
Summary of geometric primitives We have seen the
following primitives 1. Triangle orientation
or 2. Left/Right test 3. Point-line
classification Others O(1) time operations 1.
Point-on-plane test 2. Segment-segment
intersection 3. Segment-triangle
intersection All require constant time (if d is
fixed).