Scan Conversion II Revision 1.4 2/13/07 Copyright 2006, Dr. Zachary Wartell, UNCC, All Rights...

Scan Conversion II

Revision 1.42/13/07

Copyright 2006, Dr. Zachary Wartell, UNCC, All Rights Reserved

Polygons

A polygon is a many-sided planar figure composed of vertices and edges. A polygon is bounded (finite area) and closed (includes boundary).

Vertices are represented by points (x,y).

Edges are represented as line segments which connect two points, (x1,y1) and (x2,y2).

P = { (xi , yi ) } i=1,n

E3

(x3,y3)

E2

(x2,y2)E1(x1,y1)

Larry Hodges

Polygons: Complex vs Simple

Zachary Wartell

• A simple polygon – edges only intersect a vertices, no coincident vertices

• A complex polygon – edges intersect and/or coincident vertices

A

B

C

C

EF

G

H

A B,E

C

D

F

A

B

C

A B

C

D

E

Simple Polygons: Convex and Concave

Convex Polygon - For any two points P1, P2 inside the polygon, all points on the line segment which connects P1 and P2 are inside the polygon.

– All points P = uP1 + (1-u)P2, u in [0,1] are inside the polygon provided that P1 and P2 are inside the polygon.

Concave Polygon - A polygon which is not convex.

Larry Hodges, Zachary Wartell

Inside vs Outside Polygon: Odd-Parity Test


-Is point P inside polygon?

-Count # times a ray from P intersects edge of polygon

-odd count implies inside

-When crossing a vertex, if the vertex’s two edges are on same side of line then count it twice, else count it once.

Count onceCount twice

oror

Filling Concave Polygons

Fill the polygon 1 scanline at a time

Determine which pixels on each scanline are inside the polygon and set those pixels to the appropriate value.

Look only for those pixels at which changes occur.


Scan-Line Algorithm


For each scan-line:

-Find the intersections of the scan line with all edges of the polygon.

-Sort the intersections by increasing x-coordinate.

-Fill in all pixels between pairs of intersections.

Possible Problems

1. Horizontal edges ==> Ignore

2. Vertices ==> If local max or min, then count twice, else count once. (This is implemented by shortening one edge by one pixel.)

3. Calculating intersections is slow.

2

4

6

8

For scan-line number 7 the sorted list of x-coordinates is (1,3,7,9)

Therefore fill pixels with x-coordinates 1-3 and 7-9.

Edge Coherence


-Not all polygon edges intersect each scanline.

-Many edges intersected by scanline yk will also be intersected by scanline yk+1

-Nearby pixels on given edge and span have similar coordinates, attributes and equation components

“scan-line aligned” trapezoids

Zachary Wartell

T1

T2

T3

T4

T5T6

-break up problem into simpler ”scan-line aligned” trapezoids- only two side edges: left, right

left edgeof T2

right edgeof T2

Edge Table

Zachary Wartell

T1

T2

T3

T4

T5T6

-sort vertices by y coordinate building Edge Table (ET)

ymax

y=0A

B

CD

F

E

G

H

yD,C

yG

yE

yB

yH

yA AH AB

HG

BC

EF ED

GF

ET

not needed

Scan AET with AEL

Zachary Wartell

T1

T2

T3

T4

T5T6

-scan up through ET and track Active Edge Table (AET). At each scan-line fill in span of pixel for each edge pair.

ymax

y=0A

CD

F

E

G

H

yG

yE

yB

yH

yA AH AB

HG

BC

EF ED

GF

ET AETk

(k = 1..10)

[(AH,AB)]

[(HG,AB)]

[(HG,BC)]

[(HG,EF),(DE,BC)]

[(GF,EF),(DE,BC)]

[(DE,BC)]

B

How do we fill trapezoids?

Zachary Wartell

-Need to generate pixel coordinates of left and right edges of edge pair-- (lx,ly) and (rx,ry)-Need to fill pixel in span between current

A

CD

F

E

G

H

B

A

CD

F

E

G

H

B

rx0, lx0

y0

rx2rx3

lx1,

lx2,

lx3

H

G

y1

y2

y3

rx1

Rasterize Edge

Zachary Wartell

-From y=mx + b we have: yk+1= yk+1,

xk+1= xk+ 1 / m = xk+ Δx / Δy - fractions yuck!

-So split into integer and fraction and look at fraction numerator: xk+1= xk+ (Δx div Δy) + (Δx mod Δy) / Δy

integer fraction

xk = xik + xnk / Δ y, xi is integer component xn is x’s fraction’s numerator

xk=0

yk=0

yk

yk+1

xk xk+1 …

…

B

Rasterize Edge

Larry Hodges, Zachary WartellLarry Hodges, Zachary Wartell

-Base Case: xi = x0 , xn = 0-Iteration Case: xi is integer component, xn is fraction numerator

xi := xi + (Δx div Δy)xn := xn + (Δx mod Δy)if xn ≥ Δy then

xi := xi + 1 xn := xn – Δyendif

A=(6,1)

B=(13,4)

Δx=7Δy=3Δx div Δy= 2Δx mod Δy= 1 k y xi xn x (=xi+xn/Δy)

0 1 6 0 61 2 8 1 8 1/32 3 10 2 10 2/33 4 13 0 13 0/3

(8,2)(10,3)

Warning: H&B doesn’t handle Δx > Δy

Rasterization Special Cases


-Previous slide truncated to obtain integer x coordinate, alternatively use rounding

-you end up with a fraction (comparing xn to Δy/2) and must multiply xn by 2 to get all integer operations

-Often polygons share edges, to avoid writing the same pixel twice possibly:

-horizontal edge – only drawn when on top-right edge – round up-left edge – round down, if xn = 0

Rasterization Special Cases


-Edge Shortening Trick:-Recall Odd-Parity Rule Problem:

-Implement “Count Once” case with edge shortening:


oror


oror

A

B

C

A

B

C

A

B

C

B'

A

B

C

B'

C

B

A

B

C

B

A

B'B

B'

xA,yB’,1/mAB xC,yB’,1/mCB

Splitting Concave Polygons – Concept

Concave can be split into Convex Polygons – this simplifies Edge Tableand Active Edge Table in polygon rasterizer (only one scan-line trapezoid is active at a time!)

Identify: look for interior angle greater than 180º - cross product signs will differ

X

YE1

E2

E3

E4

E5

E6

(E1 X E2)z > 0

(E2 X E3)z > 0

(E3 X E4)z < 0

(E4 X E5)z < 0

(E5 X E6)z > 0

(E6 X E1)z > 0

i j kx0 y0 z0x1 y1 z1

©Zachary Wartell - 1/26/2005

Splitting Concave Polygons

Edge Vector: Compute Edge Vectors (Ek = Vk+1- Vk) traversing polygon counter-clockwise.If cross product of consecutive edge is negative, then split polygon alongthe embedding line of first edge in cross product

Rotational: put successive edges on x-axis. If next edge has negative x coordinate split polygon on x-axis.

X

YE1

E2

E3

E4

E5

E6

X

Y E1

E2

E3

E4

E5

E6

XY

X

Y


Splitting Convex Polygon to Triangles

Triangle rasterizing is even simpler. Only 3 edges to “sort”. Only fill in 1 or 2 scan-line aligned triangles. Pretty much what all graphics hardware does.

Pick 3 vertices of convex polygon. Split off triangle. Repeat with remaining vertices. Continue until only 3 vertices left.

v0

v1

v2

v3

v4

v0

v1

v2

v3

v4

v0

v1

v2

v3

v4

A) B) C)


Filling Irregular Boundaries


boundary fill: expand and fill region until you reach boundary color

flood fill: expand and fill region while you find interior color

Boundary Fill

Interior Fill

4 vs 8 connected


Define: 4-connected versus 8-connected,its about the neighbors

4 vs 8 connected

Zachary Wartell

Fill Result: 4-connected versus 8-connected

“seed pixel”

Recursive Pixel Boundary Fillhttp://www.cosc.canterbury.ac.nz/people/mukundan/cogr/Fill.html

void boundaryFill4(int x, int y){ if(getPixel(x,y)!=borderColor && getPixel(x,y)!=fillColor ){ setPixel(x,y); boundaryFill4(x+1,y); boundaryFill4(x-1,y); boundaryFill4(x,y+1); boundaryFill4(x,y-1); } }

http://www.cosc.canterbury.ac.nz/people/mukundan/cogr/Fill.html

65

47

1

2

131

3

12

14156

141

5

5

7

4

Recursive Scan-line (boundary) fill

Zachary Wartell

A B

C D

Stack

Fill Patterns


Fill patterns can be used to put a noticeable “texture” inside a polygon. A fill pattern can be defined in a 0-based, m x n array. A pixel (x,y) is assigned the value found in:

pattern((x mod m), (y mod n))

Pattern Pattern filled polygon

Halftoning


For 1-bit (B&W) displays, fill patterns with different fill densities can be used to vary the range of intensities of a polygon. The result is a tradeoff of resolution (addressability) for a greater range of intensities and is called halftoning. The pattern in this case should be designed to avoid being noticed.

These fill patterns are chosen to minimize banding.

Dithering


Another method to increasing the number of apparent intensities on a 1-bit (B&W) display is dithering. In an ordered dither the decision to turn a pixel on or off at point (x,y) depends on the desired intensity I(x,y) at that point and on an (n by n) dither matrix Dn. The dither matrix is indexed from 0 to (n-1) along its rows and columns. Each of the integers 0 to n2 -1 appears once in the matrix. For instance when n = 4, we have D4 =

0 8 2 10

12 4 14 6

3 11 1 9

15 7 13 5

To process the point at (x,y), we first compute

i = x MOD 4, j = y MOD 4

Then if I(x,y) > D4(i, j) the point (x,y) is turned on; otherwise it is not.

Revisions


1.11. Removed slide 17 “Antialiasing Polygons” (putting anti-aliasing from H&B in separate .ppt)

1.2 1. Fixed error in slide 11

1.4– Added Slide 23 ‘Recursive Pixel Boundary Fill’– clarified diagram in slide 24 and noted subtly/error found in textbook.

Scan Conversion II Revision 1.4 2/13/07 Copyright 2006, Dr. Zachary Wartell, UNCC, All Rights...

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Transcript of Scan Conversion II Revision 1.4 2/13/07 Copyright 2006, Dr. Zachary Wartell, UNCC, All Rights...