Computer Graphics

University of Science and TechnologyFaculty of Computer Science and

Information TechnologyComputer Science –Department

4th- Year

Computer GraphicsChapter(3)

Scan Conversion Algorithms (Output primitives)Dia Eldein Mustafa Ahmed

2018

Overview

• Primitives and Attributes.

• Raster Graphics (Rasterization).

• Why Scan Conversion?

• Algorithms for Scan Conversion:

Lines

Circles

Ellipses

Filling

Polygons

2Computer Graphics - Chapter (3) Diaa Eldein Mustafa Ahmed-2018

2D Output Primitives and Attributes

PRIMITIVESƒPoints

Lines

Circles

Ellipses

Curves

Filled Area

Text

ATTRIBUTESƒColor , size Line Type, Width and Color

Fill Styles: Hollow, Solid and Patterned

Text Font, Color, Size and Style TextTextText

Text


• Points

• A point is shown by illuminating a pixel on the screen


Each point is represented as a Pixel with (x,y ,color)


• Lines

• A line segment is

completely defined in terms

of its two endpoints.

• A line segment is thus

defined as:

Line_Seg = { (x1, y1), (x2, y2) }



(x1, y1)

(x2, y2)

• A line is produced by means of illuminating a set of intermediary pixels between the two endpoints.

• Lines is digitized into

a set of discrete integer positions that approximate the actual line path. x2


x

y

y1

y2

x1


Example: A computed line position of (10.48, 20.51) is converted to pixel position (10, 21).

• The rounding of coordinate values to integer causes all but horizonatal and vertical lines to be displayed with a stair step appearance “the jaggies”.



Rasterization

Process of converting an image described in a Vectorgraphics format (shapes), into a Raster image (pixels or dots).

Rasterization


Rasterization Problem: Given only the two end points, how to compute the intermediate pixels, so that the set of pixels closely approximate the ideal line.

Scan Conversion

• Also known as rasterization.

• In our programs objects or primitives (points , line , circle ,curves , any 2D or 3D shape) are represented symbolically.

– 3D coordinates representing an object’s position

– Attributes representing color, motion, etc.

• But, the display device is a 2D array of pixels (unless you’re doing holographic displays).


Scan Conversion is the process of representing continuous graphics object as a collection of discrete pixels

Properties of an Ideal Line

• Must appear straight and continuous

– Only possible axis-aligned and 45 degree lines

• Must interpolate both defining end points.

• Must have uniform density and intensity

– Consistent within a line and over all lines

• Must be efficient, drawn quickly .

– Lots of them are required!!!

• Lines should terminate accurately.

• Lines should be drawn rapidly.

– Efficient algorithms. مربط الفرس10

Computer Graphics - Chapter (3) Diaa Eldein Mustafa Ahmed-2018

The lines of this objectappear continuous

However, they aremade of pixels

11

Properties of an Ideal Line


• How does a machine or computer draw a line defined by a mathematical representation (e.g. using endpoints only)?

• How does the continuous line description get converted to the discrete pixel representation?

• Important part of Rasterization process is to convert the line segment into a set of pixels in the display space

Line Drawing Algorithms


(X1, Y1)

(X2, Y2)

(Xn, Yn)

Display Space

Lined Drawing Algorithms

• Slope -Intercept method (incremental method)

• Digital Differential (DDA) Algorithm

• Midpoint algorithm

• Parametric Equation

• Bresenham’s Line Algorithm

Computer Graphics - Chapter (3) Diaa Eldein Mustafa Ahmed-2018 13

1. A line in Computer graphics is a portion of straight line that extends indefinitely in opposite direction.

2. It is defined by its two end points P1(x1,y1) and P2(x2,y2)..

3. Its density should be independent of line length.

the slope intercept equation for a line:

where, m = Slope of the line

b = the y intercept of a line

Line Drawing AlgorithmsSlope -Intercept method

)1.........(bmxy


The two endpoints of a line segment are specified at positions P1(x1,y1) and P2(x2,y2).


x1 x2

x

y

P1(x1,y1)

P2(x2,y2)

b

O

y2

y2


We can determine the value for slope m & bintercept as :

)2.......(..........)

12(

)12

(

x

y

xx

yym

)3.(..........*xmybii

We can substitute in equation (3) either P1(x1,y1) or P2(x2,y2) to compute b .


)4....(..........1

*1 xmyb


1- Input end points P1(x1,y1) ,P2(x2 ,y2)

2- Compute

3- Compute

4 –Compute the slope

3- Compute the intercept

4- If (x2 > x1)

dx = 1

else

dx = -1

5- while (x1 != x2)

y = m * x1 + b

PlotPixel(x1, round(y))

x1 += dx

6- EndComputer Graphics - Chapter (3) Diaa Eldein Mustafa Ahmed-2018 17

What’s the problem?If the slope m> 1, the line is not continuous, pixels are skipped


xxx12

yyy12

x

ym

xmyb1

*1

P2(x2,y2)

P1(x1,y1)

P1(x1,y1)

P2(x2,y2)

Example 1: The endpoints of line are(1,3) & (6,18). Compute each value of y as x steps from 1 to 6 and plot the result.

Solution :



51612

xxx

1531812

yyy

35

15

x

ym

112

dxxx01*33

1*

1 xmyb

The challenge is: to find a way to calculate the next x,y position by previous one as quickly as possible.


(x1, round(y))Y= m * x1 + bX1

(1,3)=3*1+0=31

(2,6)=3*2+0=62

(3,9)=3*3+0=93

(4,12)=3*4+0=124

(5,15)=3*5+0=155

(6,18)=3*6+0=186

xx dx 11011



P1(1,3)

P2(2,6)

P3(3,9)

P4(4,12)

P5(5,15)

P6(6,18)

1

2

3

X

Y

The algorithm performs a floating-point multiplication for every step in x.This method therefore requires an enormous number of floating-pointmultiplications, and is therefore more expensive

• Exercises :

1)Draw line given by end points (12,10) and (21, 13).

2) Draw and sketch the line given by the end points

(-3,8) and (-6 ,-11).


21


In the Slope -Intercept method algorithm, we increment x and find the y value corresponding to the new x using the line equation.

The result is a real number that is rounded to the nearest integer.

The Digital differential analyzer (DDA) algorithm is anincremental scan-conversion method.

Such an approach is characterized by performingcalculations at each step using results from thepreceding step.



DDA -Algorithm


• The differential equation of a straight line is given by:


DDA -Algorithm

)1(

1

1

12

12 .....)(

)(

)(

)(

xx

yy

xx

yy

x

y

ii

iim

yi 1

xi xi 1 xi 2xy

i

yi 2

y

y

x

)2(1 ...........

)(

xm yy ii

)3(1

............xmyyii

)4(1

............xmyyii


(Case a: m < 1) distance on x-axis is greater than on y-axis( )

and from (4)

Then the straight line equation model is given by (5) & (6):


DDA -Algorithm

)5(1 .......1 ii xxx

)6(1 ...............myy ii

yx

(Case a: m < 1) or Δy<Δx

• In this case we use equations (7) with increment in x by1 and increment y by m

(7)

• That means y is independent of x , but depend on the slope m.



DDA -Algorithm

myy

xx

ii

ii

1

1 1



DDA -Algorithm

(Case b: m > 1) or Δy>Δx

And from (4)

or

ii xxm

1

1

)8(1 .......1 ii yyy

)9(1 .).........(1 ii xxm

)10(1 .......1

mxx ii

(Case b: m > 1) or Δy>Δx

In this case we use equations (8) and (10) to construct the straight line equation model (11) , with increment in y by 1 and increment x by .

(11)



DDA -Algorithm

mxx

yy

ii

ii

1

1

1

1

m

1

• Example (1) : Draw the line with end points P1(20,10) and P2(30,18)



DDA -Algorithm

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

step x y Plot(x,y)

1 20 10 (20,10)

2 21 10.8 (21,11)

3 22 11.6 (22,12)

4 23 12.4 (23,12)

5 24 13.2 (24,13)

6 25 14 (25,14)

7 26 14.8 (26,15)

8 27 15.6 (27,16)

9 28 16.4 (28,16)

10 29 17.2 (29,17)

11 30 18 (30,18)

x = 30 – 20 = 10 y = 18 – 10 = 8 Larger length = 10 = x

Yinc =m= 8 / 10 = 0.8 Xinc = 1

29


DDA -Algorithm




DDA -Algorithm

Example (2) : Draw the line with end points P1(8,12) and P2(16,28)

step x y Plot(x,y)

1 8 12 (8,12)

2 8.5 13 (9,13)

3 9 14 (9,14)

4 9.5 15 (10,15)

5 10 16 (10,16)

6 10.5 17 (11,17)

7 11 18 (11,18)

8 11.5 19 (12,19)

9 12 20 (12,20)

10 12.5 21 (13,21)

11 13 22 (13,22)

12 13.5 23 (14,23)

13 14 24 (14,24)

14 14.5 25 (15,25)

15 15 26 (15,26)

16 15.5 27 (16,27)

17 16 28 (16,28)

x = 16 – 8 = 8

y = 28 – 12 = 16

Larger length = 16 = y

Xinc =1/m= 8 / 16 =0.5

Yinc = 1



19

18

17

16

15

14

13

12

20

1715141312111098 16 18

2122

23

24

25

26

27

28


• Algorithm:

1. Input end points p(x0 , y0 ) and p(xend , yend ) .

2. Compute Δx=xend -x0.

3. Compute Δy =yend -y0.

4. Plot p(x0 , y0 )

5. Compute m= Δy/Δx6. If (m>1)

for y=y0 +1 to yendx= x0+ 1/m

plot p(round (x),y)else

for x=x0 +1 to xendy= y0 + m

plot p(x,round (y))

7. End

Line Drawing AlgorithmsDDA -Algorithm


Digital Differential

Analyzer

Line Drawing Algorithm

Bresenhams Line Drawing Algorithm

ArithmeticDDA algorithm uses floating

points i.e. Real Arithmetic.

Bresenhams algorithm uses fixed points

i.e. Integer Arithmetic.

Operations

DDA algorithm uses

multiplication and division

in its operations.

Bresenhams algorithm uses only

subtraction and addition in its

operations.

Speed

DDA algorithm is rather

slowly than Bresenhams

algorithm in line drawing

because it uses real arithmetic

(floating-point operations).

Bresenhams algorithm is faster than DDA

algorithm in line drawing because it

performs only addition and subtraction in

its calculation and uses only integer

arithmetic so it runs significantly faster.


Difference Between DDA Line Drawing Algorithm and

Bresenhams


Accuracy &

Efficiency

DDA algorithm is not as

accurate and efficient as

Bresenham algorithm.

Bresenhams algorithm is more

efficient and much accurate than DDA

algorithm.

Drawing

DDA algorithm can draw circles

and curves but that are not as

accurate as Bresenhams

algorithm.

Bresenhams algorithm can draw circles

and curves with much more accuracy than

DDA algorithm.

Round Off

DDA algorithm round off the

coordinates to integer that is

nearest to the line.

Bresenhams algorithm does not round off

but takes the incremental value in its

operation.

Expensive

DDA algorithm uses an enormous

number of floating-point

multiplications so it is expensive.

Bresenhams algorithm is less expensive

than DDA algorithm as it uses only

addition and subtraction.


Difference Between DDA Line Drawing Algorithm and

Bresenhams

Exercise :(1) Scan convert a line having end points (3,2) & (4,7)

using DDA.

(2) Using the above end points (4,7) as start point and (3,2) as end point , draw the line using DDA.

(3) Scan convert a line with end points (4,11) & (18,8)

Using DDA.



DDA -Algorithm


Bresenham’s Line Algorithm

The Bresenham’s algorithm is another incrementalscan conversion algorithm.

The big advantage of this algorithm is that, it uses onlyinteger calculations.

Moving across the x- axis in unit intervals and at eachstep choose between two different y coordinates.

For example, as shown in the following illustration,from position (2, 3) you need to choose between (3, 3)and (3, 4) in order to elongate it to (5,5) .

You would like the point that is closer to the originalline.




• At sample position , the vertical separations from the mathematical line are labeled as and


xk 1

d upper d lower

1xk

yk

y

1yk



• You can use these to make a simple decision about which pixel is closer to the mathematical line.

• This simple decision is based on the difference between the two pixel positions.




• From the above illustration, the y coordinate on the mathematical line at is −

• So, and are given as follows −


bxmy k )1(

d upperd lower

bxmyyyd kkkupper

)1(11

ybxmyyd kkklower )1(

1kx

byxm

xmbyy

xmbyydd

kk

kkk

kkkupperlower

221)1(2

)1(22)1(

)1(22)(1



• Let us substitute m with dy/dx where dx and dy are the differences between the end-points.


byxdx

dydd kkupperlower 221)1(2

Cdxydyx

bdxdydxydyx

byxdx

dydxdddx

kk

kk

kkupperlower

)(2)(2

)12(22)(2)(2

)221)1(2()(




So, a decision parameter for the step along a line is given by −

kP thk

Cdxydyxdddxpkkupperk lower

)(2)(2)(

• The sign of the decision parameter is the same as that of .• If is negative, then choose the lower pixel, otherwise choose the upper pixel.• Remember, the coordinate changes occur along the x axis in unit steps, so you can do everything with integer calculations.

kPdd upperlower

kP

• At step , the decision parameter is given as −




1k

Cdxydyxpkkk

)(2)(2111

Subtracting from this we get −

But, is the same as . So −

kP

))(2)(2())(2)(2(111

CdxydyxCdxydyxppkkkkkk

xk 1 1xk

)(2211 yydxdypp

kkkk

• Where, is either 0 or 1 depending on the sign of .

• The first decision parameter is evaluated at is given as −

• Now, keeping in mind all the above points and calculations, here is the Bresenham’s example for slope m < 1


)(1

yykk

kP

0P ),( 00 yx



dxdyP 20

• If the next point to plot is and

• If the next point to plot is and


0kP ),1( yxkk

dyppkk

21

0kP



)1,1( kk yx

)(21

dxdyppkk

• To illustrate the algorithm, we digitize the line with endpoints (20,10) and (30,18). This line has slope of 0.8, with

Δx = 10

Δy =8

• The initial decision parameter has the value

p0 = 2Δy − Δx = 6

• and the increments for calculating successive decision parameters are

2 Δy = 16

2 Δy - 2 Δx = -4




• We plot the initial point

(x0 , y0)=(20,10) and determine

successive pixel positions

along the line path from the

decision parameter as


k pk (xk +1, yk +1)

0 6 (21,11)

1 2 (22,12)

2 -2 (23,12)

3 14 (24,13)

4 10 (25,14)

5 6 (26,15)

6 2 (27,16)

7 -2 (28,16)

8 14 (29,17)

9 10 (30,18)






17

16

15

14

13

12

11

10

18

292726252423222120 28 30

• Algorithm1. Input the line end points p(x0 , y0 ) and p(xend , yend ) .2. Plot the first point p(x0 , y0 ) .3. Calculate the constants Δx, Δy, 2Δy, and 2(Δy − Δx), and

obtain the starting value for the decision parameter asp0 = 2Δy − Δx

4. At each xk along the line, starting at k = 0 , perform the following test:

If pk < 0,the next point to plot is (xk +1, yk )andpk+1=pk+2Δy

Otherwise, the next point to plot is (xk +1, yk +1) andpk+1=pk+2Δy−2Δx

5. Repeat step 4, Δx−1 times.Computer Graphics - Chapter (3) Diaa Eldein Mustafa Ahmed-2018 49



• Exercise :

(1) Go through the steps of the Bresenham’s line drawing algorithm for a line going from (21,12) to (29,16).

(2) Using the three above methods (slope-intercept , DDA and Bresenham’s algorithm , draw the line joining the end points (3,11) and (-9,18). What are the accurate line regarding the points laying in the path of the line ?.




• Circle Generation Algorithms


Circle Drawing AlgorithmsCartesian Co-ordinate Form

The equation of a circle:

We could solve for y in terms of xP(x0 , y0)

P(x,y)

x

y

r

If the center at any point (x0 , y0 )

Equation of the circle in this form is:

then from (1) we can find that

From (2) we can compute the value of with respect to the corresponding value of .



)1(..........222 ryx

)2(..........22

xry

y

x

x

y

dx

r

P(x,y)

If the center at the origin O(0 , 0)

O(0,0)



We can draw the upper part of the circle with the equation:

And the lower part with

the equation

If we implement

a decrement in x by dx ,then

Can compute the corresponding

Values of y from (3) and (4).

y

x

dx

r

P(x,y)

P(r,0)P(-r,0)

P(0,r)

P(0,-r)

xry 22

xry 22

)3.........(22

xry

)4.......(22

xry

P(-x,y)

Algorithm:( while version)1- input the radius r2- x=r 3- dx=13- while (x > - r)

PlotPixel( x, round(y))PlotPixel( x, round(-y))

x=x-dx4- End



xry 22

Algorithm: :( for version)1-input the radius r2- x=r 3- dx=13- for x=r to - r

PlotPixel( x, round(y))PlotPixel( x, round(-y))

x=x-dx4- End

xry 22

To draw a more smooth circle (dense points) we can take small value of dx (e.g. dx=0.5 or dx=0.1


Circle Drawing AlgorithmsPolar Co-ordinate Form

x

y

d

P(x,y)=P(r , )

P(r , +d)

)2)........(sin(

)1)........(cos(

ry

rx

From the previous concepts of Polar Co-ordinates :

20

rr

In this case the circle will be drawn as one unit based on the value of ()

When =0

00*)0sin(

1*)0cos(

rry

rrrx

P (0, r)

P (r, 0)

When =90

rrry

rrx

1*)90sin(

00*)90cos(

P (r, 0)

P (0, r)


Circle Drawing AlgorithmsPolar Co-ordinate Form

Algorithm:( while version)1- input the radius r2- =3.143- C= /1804- =04- while ( < 2* )

PlotPixel( round(x), round(y))

= +15- End

)*sin(

)*cos(

Cry

Crx

Algorithm:( for version)1- input the radius r2- =3.143- C= /1804- for =0 to 2*


= +15- End

kCry

hCrx

)*sin(

)*cos(

1

Center at the origin (0,0 ) Center at the origin (h,k )

• The Cartesian equation for the ellipse shown below can be written in terms of the ellipse center coordinates and parameters rx and ry as:

• If the center at the origin

the equation will be:


Ellipse Drawing Algorithms

1

22

y

c

x

c

r

yy

r

xx

),( yxcc

rx

ry

),( yxcc

xc

yc

X

Y

1

22

yx r

y

r

x


22

1

x

c

y

c

r

xx

r

yy

c

x

cy y

r

xxry

2

12

1

x

c

y

c

r

xx

r

yy


22

1

xy r

x

r

y

2

1

x

yr

xry

2

1

xy r

x

r

y

)(0 origintheatcentertheyxif cc

• The Polar equation for the ellipse shown below can be written in terms of the ellipse center coordinates and parameters rx and ry as:

• If the center at the origin

the equation will be:



yy

xx

cry

crx

)sin(*

)cos(*

),( yx cc

rx

ryX

Y

P(r, )

)sin(*

)cos(*

y

x

ry

rx

)0,0(



Algorithm:( Cartesian)

1- input the center (Xc ,Yc)

2- input the radius rx , ry

3- for x= rx to -rx


PlotPixel( round(x), round(-y))

4- End

Algorithm:( Polar)1- input the center (Xc ,Yc)

2- input the radius rx , ry

3- =180

4- C= /180

5- for =0 to 2*

x=rx *cos(C*) + Xc

Y=ry *sin(C*) + Yc

PlotPixel( x, round(y))

6- End

yr cy rxx

yx

c

2)(1


Polygon Drawing Algorithms


Polygon Fill Algorithms

Thank You End

Questions ?


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