gvc04 line draw

8/7/2019 GVC04 Line Draw

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GVC-432

Lecture - 4

Ref: Donald Hearn &M. Pauline Baker ,

Computer GraphicsFoley, van Dam, Feiner & Hughes,

Computer Graphics Principles & Practice

Dr Pavan ChakrabortyIIIT-Allahabad

Indian Institute of Information Technology - Allahabad


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Coordinate Systems

Pixel Coordinate System- rows and columns

Rectilinear

Usually for graphics, westart at top left corner andwork our way across anddown

Same as raster orientation

Array[row][column]

Row major used in Cand

C+ ( last index moves

fastest inmemory)

Not all languages do it this

way - eg Fortran uses

columnmajor (first indexmoves fastest)



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Cartesian

Coordinates Often we use theCartesian coordinateconvention ie x,y

coordinates (or x,y,z in3D) and map this toour display

Usually columncorresponds to x, and-row corresponds to y

This works if we know themaxmin values.

Common values are eg

640 columns x 320 rows

Or1024x768 or better

Aspect ratio is the ratio oftheseeg 4:3 - Chosen to suitthe common display deviceseg TV screens ormonitors

x

y


y


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Drawing Space or Canvas

Coordinate Systems

Drawing Primitives

Library of utilities eg drawDot( int x, int y);

Or drawLine( x1, y1, x2, y2 );

Usually we have PrimitiveModels for coordinatespaces and colours

We do not want to write ourapplication programsworrying about pixelresolutions

Wemay have libraries thatallow us to do so, but oftenthey will support moregeneral coordinates

Eg real space normalisedcoordinates [0.0,1.0]



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Colours in Brief

Red Green Blue is not theOnly colourmodel althoughstill most common

We specify separate RGBvalues foreach pixel

They map to intensities

All colours can beexpressed as combination

of these

Sometimes also an alphaor transparency value

These will conveniently packinto a computer word of 4

bytes, one byte foreach

entity

1 Byte gives us 256 values- hencenumbers of colours

Need not use this

resolution

Can also use Look-up

tables to save memory



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Graphics Libraries and

packages What is a graphics

system?

A package orLibrary that

links to a Language or

environment and lets us

write programs that are

independent of thegraphics hardware and

devices

Java Development Kit and

Java 2D and Java 3D

libraries

GL and OpenGL

(Graphics Library), VOGL

X11, DirectX, PHIGS,

and lots of others



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Summary Graphics has a varied

history

Very technology driven

Good advances in recentyears with adequate

memory and processing

power

Primitives and library

layers approach is very

common

Note devices and memorymodel

Colourmodels and drawingspaces are important ideas

for our programs



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Towards the Ideal Line

We can only do a discrete approximation

Illuminate pixels as close to the true path

as possible, consider bi-level display only

Pixels areeither lit ornot lit


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What is an ideal line Must appear straight and continuous

Only possible axis-aligned and 45o

lines

Must interpolate both defining end points

Must have uniform density and intensity Consistent within a line and over all lines

What about antialiasing?

Must beefficient, drawnquickly Lots of them are required!!!



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Simple Line

Based onslope-intercept

algorithm from algebra:

y = mx + b

Simple approach:

increment x, solve for y

Floating point arithmetic

required



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LINE-DRAWING ALGORITHMS


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Does it Work?It seems to work okay for

lines with a slope of 1 or

less,

but doesnt work well for

lines with slope greater than1 lines becomemore

discontinuous in appearance

and wemust add more than

1 pixel per column to makeitwork.

Solution? - usesymmetry.



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Modify algorithm per

octant

OR, increment along x-axis if dy


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DDA algorithm

DDA = Digital Differential Analyser

finite differences

Treat line as parametric equationin t :

)()(

)()(

121

121

yytyty

xxtxtx

!

!),(

),(

22

11

yx

yxStart point -

End point -



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DDA Algorithm Startat t= 0

At each step, incrementtbydt

Choose appropriate value fordt

Ensure no pixels are missed:

Implies: and

Setdtto maximum ofdx and dy

)()(

)()(

121

121

yytyty

xxtxtx

!

!

dt

dyyy

dt

dxxx

oldnew

oldnew

!

!

1dt

dx1

dt

dy



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DDA AlgorithmThe digitaldifferentialanalyzer(DDA)

myyxmkk!!(e

110.10.0 myyxm kk !!(e 110.10.0

m

xxymkk

110.1 1 !!(

mxxymkk

110.1 1 !!("

myyxm kk !!(u" 110.10.0

Equations are based on the assumption that lines are to be processed

from the left endpoint to the right endpoint.


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#include 'device. h"

void lineDDA (int xa, int ya, int xb, int yb){ int dx = xb - xa, dy = yb - ya, steps, k;

float xIncrement, yIncrement, x = xa, y = ya;

if( abs(dx) >abs(dy) ) steps = abs (dx) ;

else steps = abs dy);

xIncrement = dx /(float) steps;

yIncrement = dy / (float) steps;

setpixel (ROUNDlxl, ROUND(y) ) :

for (k=O; k


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DDA algorithmline(int x1, int y1, int x2, int y2)

{

float x,y;

int dx = x2-x1, dy = y2-y1;int n = max(abs(dx),abs(dy));

float dt = n, dxdt = dx/dt, dydt = dy/dt;

x = x1;

y = y1;

while( n-- ) {

point(round(x),round(y));x += dxdt;

y += dydt;

}

}

n - range oft.



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DDA algorithm Still need a lot of floating point arithmetic.

2 rounds and 2 adds per pixel.

Is there a simpler way ?

Can we use only integer arithmetic ?

Easier to implement in hardware.



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Observation on lines.

while( n-- )

{draw(x,y);

move right;

if( below line )

move up;

}



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Testing for the side of a

line. Need a test to determine which side of a line

a pixel lies.

Write the lineinimplicit form:

0),( !! cbyaxyxF

Easy to prove F0 for points below.



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Testing for the side of a

line.

Need to find coefficients a,b,c.

Recall explicit, slope-intercept form :

So:

0),( !! cbyaxyxF

bxdx

dyybmxy !! soand

0..),( !! cydxxdyyx



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Decision variable.

Evaluate d fornext pixel, Depends on whetherE orNE Is chosen :

IfE chosen :

cybxayxFd ppppnew !! )21()2()

21,2(

Butrecall :

cybxa

yxd

pp

ppold

!

!

)21()1(

)2

1,1(

So :

dyd

add

old

oldne

!

!

M

E

NE

Previous

Pixel

(xp,yp)

Choices for

Current pixel

Choices for

Next pixel



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Decision variable.IfNE was chosen :

cybxayxd ppppne !! )

2

3()2()

2

3,2(

So :

dxdyd

badd

old

oldnew

!

!

M

E

NE

Previous

Pixel

(xp,yp)

Choices for

Current pixel

Choices for

Next pixel



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Summary of mid-point

algorithm Choose between 2 pixels at each step based

upon sign of decision variable.

Update decision variable based upon which

pixel is chosen.

Start point is simply first endpoint(x1,y1).

Need to calculate initial value ford



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Initial value of d.

2

)2

1()1()

2

1,1(

11

1111

bacbyax

cybxayxFdstart

!

!!

But(x1,y1) is a point on the line, so F(x1,y1)=0

2/dxdydstart !

Conventionalto multiply by 2 to remove fraction doesnt effect sign.

2),( 11

bayx !

Start point is (x1,y1)



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Bresenham algorithmvoid MidpointLine(int

x1,y1,x2,y2)

{

int dx=x2-x1;int dy=y2-y1;

int d=2*dy-dx;

int increE=2*dy;

int incrNE=2*(dy-dx);

x=x1;y=y1;

WritePixel(x,y);

while (x < x2) {

if (d


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Bresenham was not the end!

2-step algorithm by XiaolinWu:

(see Graphics Gems 1, by BrianWyvill)

Treat line drawing as an automaton , orfinite statemachine, ie. looking at next two

pixels of a line, easy to see that only a finite

set of possibilities exist.

The2-step algorithmexploits symmetry bysimultaneously drawing from both ends

towards themidpoint.



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Two-step Algorithm

Possible positions ofnext two pixels dependent on slope

current pixel in blue:

Slope between 0 and

Slope between and 1

Slope between1 and 2

Slope greater than2



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Circle drawing. Can also use Bresenham to draw circles.

U

se 8-fol

d symmetr

y

Choices for

Next pixel

M

E

SE

Previous

PixelChoices for

Current pixel



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Circle drawing. Implicit form foracircle is:

)32(chosenisIf

)522(chosenisSIf

!

!

poldne

ppoldne

xdd

yxdd

Functions are linearequations in terms of(xp,yp)

Termed point of evaluation



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Problems with Bresenham

algorithm

Pixels are drawn as a single lineunequal lineintensity with changein angle.

Pixel density= n pixels/mm

Pixel density=2.n pixels/mm

Can draw lines in darkercoloursaccording to line direction.

- Better solution : antialiasing !



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Summary of line drawing

so far. Explicit form of line

Inefficient, difficult to control.

Parametric form of line. Express linein terms of parametert

DDA algorithm

Implicit form of line

Only need to test for side of line. Bresenham algorithm.

Can also draw circles.



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Sampling theorytells us aliasing is caused by frequencies being presentabovethe Nyquistlimit.

Ideal solution : band-pass filterto remove high frequencies.

Fouriertransform tells us the transform ofa band-pass filter is a sinc function.

Convolution theorytells us we can convolve witha sinc function in the spatialdomain instead.

A sinc function is an impractical filter.

Summary of aliasing.


gvc04 line draw

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