Download ppt - 5.1 Si23_03 SI23 Introduction to Computer Graphics Lecture 5 – Drawing A Line

5.1Si23_03

SI23Introduction to Computer

Graphics

SI23Introduction to Computer

Graphics

Lecture 5 – Drawing A Line

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Course OutlineCourse Outline

ImageDisplay

URLGIMP

colour

Vector graphics– Line drawing– Area filling

Graphical interaction

SVGViewer

2D vectorgraphics

URLlines.,areas

graphicsalgorithms

interaction

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Line DrawingLine Drawing

Line drawing is a fundamental operation in computer graphics

Demanding applications require very fast drawing speeds - say, 1 million lines per second

Hence the algorithm for converting a line to a set of pixels will be implemented in hardware - and efficiency is vital

– Want to use integer arithmetic– Want to use additions rather than

multiplications

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Line Drawing - The Problem

Line Drawing - The Problem

To draw a line from one point to another on a display

- which pixels should be illuminated?

Suppose pt1 = (0,1)and pt2 = (5,4)

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Where to draw the line?

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Line Drawing - By HandLine Drawing - By Hand

In a simple example, one can do by eye

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... but we would likean algorithm!

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Equation of a LineEquation of a Line

Line equation is:

y = m x + c

If line joins (x1,y1) to (x2,y2), then

m = (y2-y1)/(x2-x1) and c = y1 - m x1

Suppose pt1 = (0,1)and pt2 = (5,4)

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Then m = (4-1)/(5-0) ie 0.6and c = 1

Thus: y = 0.6 x + 1

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Drawing a Line From Its Equation

Drawing a Line From Its Equation

We can use the equation to draw the pixels

y = 0.6 x + 1

x = 0 -> y =1 y =1x = 1 -> y = 1.6 y = 2x = 2 -> y = 2.2 y = 2x = 3 -> y = 2.8 y = 3x = 4 -> y = 3.4 y = 3

Calculate y for x = 0,1,2,3,4,5and round to nearest integer

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The Line DrawnThe Line Drawn

y = 0.6 x + 1

x = 0 -> y =1 y =1x = 1 -> y = 1.6 y = 2x = 2 -> y = 2.2 y = 2x = 3 -> y = 2.8 y = 3x = 4 -> y = 3.4 y = 3

Calculate y for x = 0,1,2,3,4,5and round to nearest integer

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This gives us (after 1 multiplication, 1 addition,and 1 rounding operation per step):

How do we make more efficient?

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DDA AlgorithmDDA Algorithm

We can do this more efficiently by simply incrementing y at each stage

y* = y + q where q = (y2-y1)/(x2-x1)

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12345 y = 1.0 y = 1

y = 1.0 + 0.6 = 1.6 y = 2y = 1.6 + 0.6 = 2.2 y = 2y = 2.2 + 0.6 = 2.8 y = 3y = 2.8 + 0.6 = 3.4 y = 3

Here q = 0.6

One addition, one rounding

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Slope of LineSlope of Line

We have assumed so far that slope of line (m) lies between -1 and 1

When slope of line is greater than 1 in absolute value, then we increment y by 1 at each step, and x by:

d = (x2-x1)/(y2-y1)

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Exercise: calculate thepixels to draw line from(1,0) to (4,5)

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EfficiencyEfficiency

The DDA (Digital Differential Analyser)algorithm just described has the problem:– floating point operations (real numbers)

are expensive - add and round at each step

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Yet really all we need todecide is whether to movehorizontally or diagonally (for a line of slope < 1).

Can we do this in integerarithmetic?

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Bresenham’s AlgorithmBresenham’s Algorithm

One of the classic algorithms in computer graphics is the Bresenham line drawing algorithm

– Published in 1965– Still widely used– Excellent example

of an efficient algorithm Jack Bresenham

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Bresenham’s Algorithm - First Step

Bresenham’s Algorithm - First Step

Look in detail at first step:

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We have choice of amove to (1,1) or (1,2)

Exact y-value at x=1 is:y* = 0.6 + 1 = 1.6

Since this is closer to 2than 1, we choose (1,2)

0

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2d2

d1

Decision based on d1>d2, ie d1-d2>0

0.6

0.4

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In general, suppose we are at (xk,yk) - slope of line, m, is between 0 and 1

Next move will be to (xk+1, yk) or (xk+1, yk+1)

d2

d1

xk xk+1

yk

yk+1

d1 = [m(xk + 1)+c] - yk d2 = yk + 1 - [m(xk + 1)+c]

d1 - d2 =2m(xk + 1) - 2yk + 2c -1

>0 indicates (xk+1, yk+1)<0 indicates (xk+1, yk)

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d1 - d2 = 2m(xk + 1) - 2yk + 2c -1 : d1-d2 >0 diagonal

Let m = Dy/Dx and multiply both sides by Dx, settingDx(d1 – d2) = pk

pk = 2Dy xk - 2Dx yk + b (b a real constant)

Decision is still pk>0 - but can we work only in integers?

How does pk change at next step to pk+1? We knowthat xk increases by 1, and yk either increases by 1 orstays the same. So ….

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pk = 2Dy xk - 2Dx yk + b

pk+1 = pk + 2Dy - 2Dx - if y increasespk+1 = pk + 2Dy - if y stays the same

Thus we have a decision strategy that only involves Integers, no multiplication and no rounding

.. we start things off withp0 = 2Dy - Dx

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Bresenham’s Algorithm - A Summary

Bresenham’s Algorithm - A Summary

To draw from A to B with slope 0<m<1 – let (x0,y0) = A and plot (x0,y0) – calculate Dx, Dy

– calculate p0 = 2Dy - Dx– for k = 0,1,2,.. until B reached:

If pk < 0 then plot (xk+1 , yk) and set

pk+1 = pk + 2 Dy

If pk > 0 then plot (xk+1, yk+1) and set

pk+1 = pk + 2 Dy - 2 Dx

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Bresenham’s Algorithm - Example

Bresenham’s Algorithm - Example

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–(x0,y0) = (0,1)

–Dx = 5, Dy = 3

- p1 = p0 + 2Dy -2Dx

= -3 <0

hence plot (2,2)

–p0 = 2Dy - Dx = 1 > 0

–hence plot (1,2)

And so on …

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Bresenham’s Algorithm - Other Slopes and Other

Shapes

Bresenham’s Algorithm - Other Slopes and Other

Shapes

There are straightforward variations for lines with slopes:

m > 1-1 < m < 0

m < -1 There are similar algorithms for

drawing circles and ellipses efficiently - see Hearn & Baker textbook

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Other Ways of Drawing Lines

Other Ways of Drawing Lines

Over past 38 years, researchers have strived to improve on Bresenham’s incremental algorithm

– for example, think how to parallelise the algorithm

Alternative structural approach comes from recognition that horizontal (0) and diagonal (1) moves come in runs

We had:– 10101

One of the moves occurs singly, and is evenly spread within the sequence - the other occurs in up to two lengths (consecutive numbers)

– 01001000100100010– ie 1 occurs singly, 0

occurs in twos and threes– decision is now which

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AliasingAliasing

Lines on raster displays suffer from aliasing effects - especially noticeable with lines near to horizontal

This is caused because we are `sampling’ the true straight line at a discrete set of positions.Called `aliasing’ because many straight lineswill have the same raster representation.

More noticeable on screen than laser printer - why?

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Supersampling – Bresenham on Finer Grid

Supersampling – Bresenham on Finer Grid

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Each pixel isdivided into9 sub pixels.

Either 0,1,2 or 3sub-pixels can beset in any pixel.

This can be usedto set the intensitylevel of each pixel.

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SupersamplingSupersampling

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This shows the intensitylevels for the pixels inthe previous slide.

We are really usingintensity levels tocompensate for lackof resolution.

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Antialiased LineAntialiased Line

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The end result isa thicker line, with theintensity spread outover a larger area.

To the eye, the lineappears much sharperand straighter.

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d1 - d2 = 2m(xk + 1) - 2yk + 2c -1 : d1-d2 >0 diagonal

Let m = Dy/Dx and multiply both sides by Dx, settingDx(d1 – d2) = pk

pk = 2Dy xk - 2Dx yk + b (b a constant)

Decision is still pk>0 - but can we work only in integers?

pk+1 = 2Dy xk+1 - 2Dx yk+1 + b ... So

pk+1 = pk + 2Dy - 2Dx(yk+1 - yk) - where (yk+1 - yk) = 0 or 1

Thus we have a decision strategy that only involves integers.. we start things off with

p0 = 2Dy - Dx
