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CAD/CAM

Rasterisation

Scan conversion of lines

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What you will learn today!!

Convert vector straight lines to raster images

to be displayed on a raster terminal utilizing

the pixel information

Understand the problems associated with

displaying vectorial information on a raster

terminal

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Contents

In todays lecture well have a look at:

Vector and Raster Graphics

Bresenhams line drawing algorithm

Line drawing algorithm comparisons

Polygon fill algorithms

Anti-aliasing

Summary of raster drawing algorithms

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Representations in graphics

Vector Graphics

Image is represented by continuous geometric

objects: lines, curves, etc.

Raster Graphics

Image is represented as an rectangular grid ofcoloured squares

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Vector graphics

Graphics objects: geometry + colour

Complexity ~ O(number of objects)

Geometric transformation possible withoutloss of information (zoom, rotate, )

Diagrams, schemes, ...

Examples: PowerPoint, CorelDraw, ...

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Raster graphics

Generic

Image processing techniques

Geometric Transformation: loss of information

Complexity ~ O(number of pixels)

Jagged edges, anti-aliasing Realistic images, textures, ...

Examples: Paint, PhotoShop, ...

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Conversion

Vector graphics

Rasterization, Pattern recognition

Scan conversion

Raster graphics

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Rasterization

Rasterization (scan conversion)

Determine which pixels that are inside primitivespecified by a set of vertices

Produces a set of fragments Fragments have a location (pixel location) and

other attributes such color and texturecoordinates that are determined by interpolating

values at vertices Pixel colors determined later using color,

texture, and other vertex properties

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Line Drawing

Assuming:

Clipped (to fall within the window)

2D screen coordinates

Each pixel covers a square region

where is its center?

Rounding (X, Y) implies center at (0.0, 0.0).

Truncating implies center at (0.5, 0.5).

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DDA

DDA or Digital Differential Analyser is one of

the first algorithms developed for rasterising

the vectorial information.

The equation of a straight line is given by

Y = m X + C

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Problem

DDA = for each x plot pixel at closest y

Problems for steep lines

12

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Using Symmetry

Use for 1 m 0

For m > 1, swap role of x and y

For each y, plot closest x

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The Solution

Move across thex axis in unit intervals and ateach step choose between two differenty

coordinates

2 3 4 5

2

4

3

5 For example, fromposition (2, 3) we have

to choose between (3,

3) and (3, 4)

We would like the

point that is closer to

the original line

(xi,yi)

(xi+1,yi)

(xi+1,yi+1)

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They coordinate on the mathematical line atxi+1

is:

Deriving The Bresenham Line Algorithm

At sample positionxi+1 the verticalseparations from the

mathematical line arelabelled d2 and d1

bxmy i )1(

y

yi

yi+1

xi+1

d1

d2

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So, d2 and d1 are given as follows:

and:

We can use these to make a simple decision

about which pixel is closer to the mathematical

line

Deriving The Bresenham Line Algorithm (cont)

iyyd 1

ii ybxm

)1(

yyd i )1(2

bxmy ii )1(1

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This simple decision is based on the differencebetween the two pixel positions:

Lets substitute m with y/xwhere x and

y are the differences between the end-points:


122)1(221 byxmdd ii

)122)1(2()( 21

byxx

yxddx ii

)12(222 bxyyxxy ii

cyxxy ii 22

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So, a decision parameterpi for the ith stepalong a line is given by:

The sign of the decision parameterpi is thesame as that ofd1d2

Ifpi is negative, then we choose the lowerpixel, otherwise we choose the upper pixel


cyxxy

ddxp

ii

i

22

)(21

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Remember coordinate changes occur along

thex axis in unit steps so we can doeverything with integer calculations

At step i+1 the decision parameter is given as:

Subtractingpi from this we get:


cyxxyp iii 111 22

)(2)(2111 iiiiii yyxxxypp

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But,xi+1 is the same asxi+1 so:

whereyi+1 - yi is either 0 or 1 depending onthe sign ofpi

The first decision parameter p0 is evaluated at

(x0, y0) is given as:


)(2211 iiii yyxypp

xyp 20

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The Bresenham Line Algorithm

BRESENHAMS LINE DRAWING ALGORITHM(for |m| < 1.0)

1. Input the two line end-points, storing the left end-point in

(x0, y0)

2. Plot the point (x0, y0)

3. Calculate the constants x, y, 2y, and (2y - 2x) andget the first value for the decision parameter as:

4. At eachxi along the line, starting at i = 0, perform the

following test. Ifpi < 0, the next point to plot is

(xi+1, yi) and:

xyp 20

C

ypp ii

21

21

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The Bresenham Line Algorithm (cont)

The algorithm and derivation above assumes

slopes are less than 1. for other slopes we

need to adjust the algorithm slightly

Otherwise, the next point to plot is (xi+1, yi+1) and:

5. Repeat step 4 (x 1) timesxyC

xypp ii

222

221

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Bresenham Example

Lets plot the line from (20, 10) to (30, 18)

First off calculate all of the constants:

x: 10

y: 8

C1 =2y: 16

C2 =2y - 2x: -4

Calculate the initial decision parameterp0:

p0= 2yx = 6

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Bresenham Example (cont)

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

i pi (xi+1,yi+1)

0

1

2

3

4

56

7

8

9

6

2

-2

14

10

62

-2

14

10

(20, 10)

(21, 11)

(22, 12)

(23, 12)

(24, 13)

(25, 14)(26, 15)

(27, 16)

(28, 16)

(29, 17)

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Bresenham Exercise

Go through the steps of the Bresenham line

drawing algorithm for a line going from

(21,12) to (29,16)

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Bresenham Exercise (cont)

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

i pi (xi+1,yi+1)

0

1

2

3

4

5

6

7

8

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Bresenham Line Algorithm Summary

The Bresenham line algorithm has the followingadvantages:

An fast incremental algorithm

Uses only integer calculations Comparing this to the DDA algorithm, DDA has

the following problems:

Accumulation of round-off errors can make the

pixelated line drift away from what was intended The rounding operations and floating point arithmetic

involved are time consuming

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Anti-Aliasing

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Resolution

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Scan-Line Polygon Fill Algorithm

2

4

6

8

10 Scan Line

02 4 6 8 10 12 14 16

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Scan-Line Polygon Fill Algorithm

The basic scan-line algorithm is as follows:

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 that

lie interior to the polygon

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Scan-Line Polygon Fill Algorithm (cont)

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What next?

Co-ordinate Systems and

Transformations

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