f6

v1

v2v3

v4

v5

v6

v10v9

v8

v7

e1e2

e3

e4e5 e6

e7 e8e9

e10

e11

e12e13

e14

f1

f2f3

f4f5

Preliminaries

Planar straight line graphA planar straight line graph (PSLG) is a planar embeddingof 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 edgesuch 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.

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 edgeV2 Terminus (destination) of the edge; implies an orientationF1 Face to the left of edge, relative to V1V2 orientationF2 Face to the right of edge, relative to V1V2 orientationP1 Index of entry for first edge encountered after edge V1V2,

when proceeding counterclockwise around V1P2 Index of entry for first edge after edge V1V2,

when proceeding counterclockwise around V2

V1 V2 F1 F2 P1 P2

e1 v1 v2 f1 f2 e2 e3

e2 v4 v1 f1 f3 ? e4

e3 v2 v3 f4 f2 e5 ?

v3

v1

v2

v4

e3

e5e1

e4

e2

f2

f3

f1

f4

Example (both PSLG and DCEL are partial)

1

2 3

4

5

67

8 9

1

2

3

4

56

7 8

9

10 11

1213

F1

F2

F3F4

F5

F6

Edge V1 V2 LeftF RightF PredE NextE-------------------------------------------------1 1 2 F6 F1 7 132 2 3 F6 F2 1 43 3 4 F6 F3 2 54 3 9 F3 F2 3 125 4 6 F5 F3 8 116 6 7 F5 F4 5 107 1 5 F5 F6 9 88 4 5 F6 F5 3 79 1 7 F1 F5 1 610 7 8 F1 F4 9 1211 6 9 F4 F3 6 412 9 8 F4 F2 11 1313 2 8 F2 F1 2 10

Preliminaries

Supplementary data structuresIf the PSLG has N vertices, M edges and F faces then we knowN - M +F = 2 by Euler’s formula. DCEL can be describedby six arrays V1[1:M], V2[1:M], LeftF[1:M] Right[1:M], PredE[1:M] and NextE[1:M]. 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 HV[1:N] with one entry for each vertex;entry HV[i] denotes the minimum numbered edge that is incident on vertex vi and is the first row or edge index in the DCELwhere vi is in V1 or V2 column. Thus for our example in thepreceding slide HV=(1,1,2,3,7,5,6,10,4]. Similarly,define array HF[1:F] where F= M-N+2 ,with one entry for each face;HF[i] is the minimum numbered edge of all the edges that make the face HF[i] and is the first row or edge index in the DCELwhere 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 eachby scanning DCEL.

DCEL operationsProcedure 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 entriesfor those edges in array A [1: 3N-6] since M<= 3N-6.

Preliminaries

1 procedure EdgesIncident(j) /* VERTEX in text */2 begin3 a = HV[j] /* Get first DCEL entry for vj, a is index. */4 a0 = a /* Save starting index. */5 A[1] = a6 i = 2 /* i is index for A */7 if (V1[a] = j) then /* vertex j is the origin */8 a =PredE[a] /* Go on to next incident edge. */9 else /* vertex j is the destination vertex of edge a */10 a =NextE[a] /* Go on to next incident edge. */11 endif12 while (a a0) do /* Back to starting edge? */13 A[i] = a14 if (V1[a] = j) then15 a = PredE[a] /* Go on to next incident edge. */16 else17 a = NextE[a] /* Go on to next incident edge. */18 endif19 i = i + 120 endwhile21 end

a

NextE[a]

PredE[a]V1[a]

V2[a]

Preliminaries

DCEL notes

If we make the following changes, it will become a procedureFaceincident[j] giving all the edges that constitute a face:

3: a = HF[j]

And sustitute HV by HF and V1 by F1. A will have a sizeA[1:N].Both EdgesIncident and Faceincident requires time proportional to the number ofincident 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 Euler’s formula(2) e 3v - 6(3) f 2/3e(4) f 2v - 4

where v = number of vertices = Ne = number of edges=Mf = number of faces=F

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.

Preliminaries

Vector algebraAn ordered pair (x, y) can be a point in the plane,or a vector.

(x, y)

0

Vector additionGiven 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 withvertices 0, a, b, and a + b.

a

b

a + b

Preliminaries

Scalar multiplicationMultiplication 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 < 0, the direction is reversed.

b

2b

-b

0

Preliminaries

Vector subtractionGiven 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).

a

b

b - a

Vector lengthLength of vector a = (xa, ya) is defined as |a| = sqrt(xa

2 + ya2).

Vector Translation

b

a

-a

b-a

p

q

o

Let a =op and b =oq. Then, b-a is a translationof the vector pq at the origin o. Thus, two linesegments having same length and directionare translates of each other and can be identified with the same canonical linesegment originating at the origin o.

Preliminaries

Vector directionThe 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 < 360.

a

a

Given two vectors a and b, the angle between them ab

is measured counterclockwise starting at vector a.

b

ab

Preliminaries

Trigonometry reminderDefinition of sine and cosine based on unit circle.

x = cos y = sin 0 < < 180 y > 0 sin > 0180 < < 360 y < 0 sin < 0

(x, y)

Unit circle x2 + y2 = 1

0

90

180

270

Preliminaries

LEFT-RIGHT-ABOVE-BELOWA geometric primitive operation: triangle orientationGiven 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 .

p0

p1

p2

a

b

p0

p1

p2

a

b

ab

ab

ab < 180positive orientation

ab < 360negative orientation

+

-

p0, p1, p2 form acounter-clockwisecycle

p0, p1, p2 form aclockwise cycle

We introduce the value Q = b - a (note that Q ab).

Case Range of Q = b - a sin Q Orientation of p0p1p2

1 -360 < Q < -180 + +2 -180 < Q < 0 - -3 0 < Q < 180 + +4 180 < Q < 360 - -

(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?

Preliminaries

Observe from the table that sin(Q) has the same sign as theorientation of p0p1p2.

sin(Q) = sin(b - a) by definition of Q= sin b cos a - cos b sin a by trig identity

We know thatcos 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, sosign(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.

Preliminaries

Another way of reaching the same expression is with thedeterminant 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 TestIf 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 D=0, then the three points arecollinear.

p0 p1

p2

p2 left

right

Area Interpretation

The value of the determinant D is twicethe signed area of the triangle p0p1p2 . Thesigned area is positive if p0p1p2 form a counter clockwisesequence, it is negative if this sequence is clockwise. If thearea is zero, then D is 0.

Generalizaion

The three points p0 p1 and p2 form a plane in3-d with a positive normal. A fourth point p3 is on theupside if p3 falls on the positive side of the plane andon 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 1:3(p.17) to Section 1:5 (p.35),in O’Rourke text).

Preliminaries

Parametric equation of a lineWe 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 }

p1

p0

> 1

= 1

= 0

< 0

0 < < 1

Note, in the previous slide,we implicitlyassumed 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 canbe 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 dimensionthat 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.

Preliminaries

Point-Line classificationWe now consider the geometric primitive operation ofclassifying a point w.r.t. a line (both in the plane).A directed line segment partitions the plane into 7non-overlapping regions. The possibilities are shown below.The problem, given p0, p1, and p2, is to determine whichregion p2 lies in.

p0

p1

terminus

origin

beyond

right

left

behind

between

Preliminaries

Summary of geometric primitivesWe have seen the following primitives:1. Triangle orientation or2. Left/Right test3. Point-line classification

Others O(1) time operations:1. Point-on-plane test2. Segment-segment intersection3. Segment-triangle intersection

All require constant time (if d is fixed).