cg out put primitives

8/11/2019 Cg Out Put Primitives

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Computer Graphics

Chapter 3Graphics Output Primitives

Andreas Savva


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Basic Raster Algorithms

for 2D Primitives

Description of picturesSpecified by a set of intensitiesfor the pixel positions.

Describe it as a set of complex objects, such as trees,furniture and walls, positioned at specific coordinate

locations within the scene.Graphics programming packages provide functions todescribe a scene in terms of basic geometric structuresreferred to as output primitives.

Points

Straight lines

Circles

Splines curves and surfaces

Polygon color areas

Character strings

Etc.


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Implementing Application programs

Description of objects in terms of primitives and

attributes and converts them to the pixels on the

screen.Primitiveswhat is to be generated

Attributeshow primitives are to be generated


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Points

The electron beam is turned on to illuminate the phosphor

at the selected location (x, y) where

0 x maxx

0 y maxy

setpixel(x, y, intensity)loads an intensity value into theframe-buffer at (x, y).

getpixel(x, y)retrieves the current frame-buffer

intensity setting at position (x, y).

(0,0)

(maxx,maxy)

CRT


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Lines

Analog devises, such as a random-scan display or avector plotter, display a straight line smoothly from one

endpoint to another. Linearly varying horizontal and

vertical deflection voltages are generated that are

proportional to the required changes in the x and y

directions to produce the smooth line.


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Digital devicesdisplay a straight line by plotting discrete

coordinate points along the line path which are calculated

from the equation of the line.

Screen locations are referenced with integer values, so plotted

positions may only approximate actual line positions between two

specific endpoints.

A computed line position of (10.48, 20.51) will be converted to

pixel position (10, 21). This rounding of coordinate values to

integers causes lines to be displayed with a stairstepappearance(the jaggies).

Particularly noticeable on systems with low resolution.

To smooth raster lines, pixel intensities along the line paths must

be adjusted.


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Line Drawing Algorithms

Cartesian equation:

y = mx + c

wheremslope

cy-intercept

x

y

xx

yym

12

12

x1

y1

x2

y2


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Slope

if |m| = 1

= 45

45

45

+ve -ve

if |m| 1

-45< < 45

if |m| 1

45< < 90 or

-90< < -45


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8

7

6

5

4

3

2 1

0

0 1 2 3 4 5 6 7 8

y = xm = 1c = 0 y

x

x y

0 0

1 12 2

3 3

4 4

5 5

6 6

7 7

8 8

|m| = 1


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8

7

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7 8

y = x+ 1m =

c = 1y

x

x y round(y)

0 1 1

1 1.5 2

2 2 2

3 2.5 3

4 3 3

5 3.5 4

6 4 4

7 4.5 5

8 5 5

|m| 1


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|m| 1

8

7

6

5

4

3

21

0

0 1 2 3 4 5 6 7 8

y = 3x- 2m = 3

c = -2y

x

x y round(y)

0 -2 -2

1 1 1

2 4 4

3 7 7

4 10 10

5 13 13

6 16 16

7 19 19

8 22 22

outside


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The Digital Differential Analyzer

(DDA) Algorithm

mx

y

x y0 1

1 4

2 7

3 10

4 135 16

means that for a unit (1) change inxthere is

m-changeiny.

i.e. y= 3x+ 1 m= 3

my

x 1

means that for a unit (1) change inythere is

1/mchangeinx.

Do not usey= 3x + 1

to calculatey.

Use m


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The DDA Method

Uses differential equation of the line : m

If slope |m| 1then incrementxin steps of 1 pixeland find correspondingy-values.

If slope |m| 1then incrementyin steps of 1 pixel

and find correspondingx-values.

step through inx step through iny


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The DDA Method

Desired line

(xi,round(yi))

(xi,yi)

(xi+1,round(yi+m))

(xi+1,yi+m)


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if slope m0

if |m| 1

xi+1=xi+ 1

yi+1=yi+ m

if |m| 1

yi+1=yi+ 1

xi+1=xi+ 1/m

Left

Right

Left

Right


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Proceeding from right-endpoint to left-endpoint

if slope m0

if |m| 1

xi+1=xi- 1

yi+1=yi- m

if |m| 1

yi+1=yi- 1

xi+1=xi- 1/m

Left

Right

Left

Right


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if slope m< 0

if |m| 1

xi+1=xi+ 1

yi+1=yi+ m

if |m| 1

yi+1=yi- 1

xi+1=xi- 1/m

Left

Right

Left

Right


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Proceeding from right-endpoint to left-endpoint

if slope m0

if |m| 1

xi+1=xi- 1

yi+1=yi- m

if |m| 1

yi+1=yi+ 1

xi+1=xi+ 1/m

Left

Right

Left

Right


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Example (DDA)

31

1

131 1

10

1

ii

ii

yyxx

m

xy

x y round(y)

0 1 11 4/3 1

2 5/3 2

3 2 2

4 7/3 25 8/3 3

6 3 3

7 10/3 3

8 11/3 4

87

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7 8

y

x


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Example (DDA)

)(1

1

83

31

1

1

ii

ii

xxyy

m

xy

y x round(x)

8 0 07 1/3 0

6 2/3 1

5 1 1

4 4/3 13 5/3 2

2 2 2

1 7/3 2

0 8/3 3

8 7

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7 8

y

x


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voidLineDDA(intx0, inty0, intx1, inty1){

intdx = x1x0, dy = y1y0, steps;

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

elsesteps = abs(dy);// one of these will be 1 or -1doublexIncrement = (double)dx / (double)steps;doubleyIncrement = (double)dy / (double)steps;

doublex = x0;

doubley = y0;setPixel(round(x), round(y));

for(inti=0; i


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Example

Draw a line from point (2,1) to (12,6)

Draw a line from point (1,6) to (11,0)

7

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7 8 9 10 11 12


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Bresenham Line AlgorithmA more efficient approach

Basis of the algorithm:

From start position decide Aor Bnext

A

B

Start position


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Bresenham Line Algorithm

For a given value ofxone pixel lies at distance tiabove the line, andone pixel lies at distancesibelow the line

True line

siti


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Bresenham Line Algorithm

Decision parameter

di= (siti)

If di0, then closest pixel is below true line (sismaller)

If di0, then closest pixel is above true line (tismaller)

We must calculate the new values for dias we move

along the line.


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3dy2dy

dy

Example:

)2or5.0(i.e.0.5(slope)gradientLet dxdydx

dy

Start pixel at (x0,y1)

4dy

At x1:s1= dy t1= dxdy

d1= (siti) = dy- (dxdy) = 2dy- dx

but 2dydxdi0ystays the same

hence next pixel is at (x1,y1)

At x2:s2= 2dy t2= dx- 2dy

d2= (s2t2) = 2dy- (dx- 2dy) = 4dy- dx

Supposed2 0yis incremented


At x3:s3= 3dy - dx t2= 2dx- 3dy

d3= (s2t3) = 6dy- 3dx0

soystays the same


x1 x2 x3 x4 x5x0y0

y1

y2

y3

y5


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In General For a line with gradient 1d0= 2dydx

if di0 then yi+1=yi

di+1= di+ 2dyif di0 then yi+1=yi+ 1

di+1= di+ 2(dydx)

xi+1=xi+ 1

For a line with gradient 1d0= 2dxdy

if di0 then xi+1=xidi+1= di+ 2dx

if di0 then xi+1=xi+ 1di+1= di+ 2(dxdy)

yi+1=yi+ 1

Note: For |m| 1 the constants 2dyand 2(dy-dx) can be calculated once,

so the arithmetic will involve only integer addition and subtraction.

E l


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Example Draw a line from (20,10) to (30,18)

19

18

17

16

15

14

13

12

11

10

20 21 22 23 24 25 26 27 28 29 30 31 32

(20,10)

(30,18)dx= 10

dy= 8

initial decision d0= 2dydx= 6Also 2dy= 16, 2(dydx) = -4

i di (xi+1,yi+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)

id Li B (i t 0 i t 0 i t 1 i t 1) // li f | | 1


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voidLineBres(intx0, inty0, intx1, inty1) // line for |m| < 1{

intdx = abs(x1x0), dy = abs(y1y0);intd = 2 * dydx, twoDy = 2 * dy, twoDyMinusDx = 2 * (dydx);intx, y;

if(x0 > x1) { // determines which point to use as start, which as endx = x1;y = y1;x1 = x0;

}else{

x = x0;y = y0;}setPixel(x,y);

while (x < x1) {x++;if (d < 0) d += twoDy;else{

y++;d += twoDyMinusDx;

}setPixel(x, y);

}}


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Special cases

Special cases can be handled separately

Horizontal lines (y = 0)

Vertical lines (x = 0)

Diagonal lines (|x| = |y|)

directly into the frame-buffer without

processing them through the line-plotting

algorithms.


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Parallel Line Algorithms

Take advantage of multiple processors.

Given npprocessors, subdivide the line path into npBresenhamsegments.

For a line with slope 0 m1 and leftpoint (x0,y0) the distance tothe right endpoint (left endpoint for next segment) is

where

x= width of the line

xp

is computed using integer division

Numbering the segments, and the processors, as 0, 1, 2, , np-1,

starting x-coordinate for the kthpartition is

xk=x0+ kxp

p

pp

nnxx 1


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Parallel Line Algorithms (continue)

i.e. x= 15 , np= 4 processors

Startingx-values atx0,x0+ 4,x0+ 8,x0+ 12

yp= mxpAt the kthsegment, the starting y-coordinates is

yk=y0+ round(kyp)Also, the initial decision parameter for Bresenhamsalgorithm at the start of the kthsubinterval is:

pk= (kxp)(2y)round(kyp)(2x) + 2yx

44

18

4

1415

px

A i li i f i h li


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Anti-aliasing for straight lines

Lines generated can have jagged or stair-step appearance,

one aspect of phenomenon called aliasing, caused by factthat pixels are integer coordinate points.

Use anti-aliasing routines to smooth out display of a line

by adjusting pixels intensities along the line path


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Another raster effect

Line B

Line A

Both lines plotted with the same number of pixels, but

the diagonal line is longer than the horizontal line

Visual effect is diagonal line appears less thick.

If the intensity of each pixel isI, then the intensity per unit

length of line A isI, whereas for line B it is onlyI/2; this

discrepancy is easily detected by the viewer.


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Circle Generating Algorithms

Circles and ellipses are common components in

many pictures.

Circle generation routines are often included inpackages.


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Circle Equations

Polar formx= rCos

y= rSin (r = radius of circle)

P=(rCos, rSin)

rSin)

rCos)x

y

r


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Drawing a circle

Disadvantages

To find a complete circle varies from 0 to

360

The calculation of trigonometric functions

is very slow.

= 0while (< 360)

x= rCosy = rSinsetPixel(x,y)= + 1

end while


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Cartesian form

Use Pythagoras theorem

x2+ y2= r2

x

ry

y

x x

y

r

2 2,P x r x

Ci l l ith


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Circle algorithms Step throughx-axis to determiney-values

Disadvantages:Not all pixel filled in Square root function is very slow


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Circle Algorithms

Use 8-fold symmetry and only compute pixelpositions for the 45sector.

45

(x,y)

(y,x)

(-x,y)

(y, -x)

(x, -y)(-x, -y)

(-y,x)

(-y, -x)


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Bresenhams Circle Algorithm

General Principle

The circle function:

and

Consider only

45 90

2 2 2( , )circlef x y x y r

if (x,y) is inside the circle boundary

if (x,y) is on the circle boundary

if (x,y) is outside the circle boundary

0( , ) 0

0

circlef x y


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p1 p3

p2

D(si)D(ti)

After pointp1, do we choosep2orp3?

yi

yi- 1

xi xi+ 1

r

B h Ci l Al i h


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Define:D(si) = distance ofp3from circleD(ti) = distance ofp2from circle

i.e. D(si) = (x

i+ 1)2+y

i

2r2 [always +ve]

D(ti) = (xi+ 1)2+ (yi1)

2r2 [always -ve]

Decision Parameter pi= D(si) + D(ti)

so ifpi< 0 then the circle is closer top3(point above)

ifpi 0 then the circle is closer top2(point below)

The Al ith


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The Algorithmx0= 0

y0

= r

p0= [12+ r2r2] + [12+ (r-1)2r2] = 32r

i f pi< 0 then

yi+1= yipi+1= pi+ 4xi+ 6

else if pi0 then

yi+1= yi1pi+1= pi+ 4(xiyi)+ 10

Stop when xi yiand determine symmetry

points in the other octants

xi+1 = xi+ 1

Example


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Example

10

9

8

7

6

5

4

3 2

1

0

0 1 2 3 4 5 6 7 8 9 10

i pi xi, yi

0 -17 (0, 10)

1 -11 (1, 10)

2 -1 (2, 10)

3 13 (3, 10)

4 -5 (4, 9)5 15 (5, 9)

6 9 (6, 8)

7 (7,7)

r = 10

p0= 32r = -17

Initial point (x0,y0) = (0, 10)


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Exercises

Draw the circle with r= 12 using the

Bresenham algorithm.

Draw the circle with r= 14 and center at

(15, 10).


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Decision Parameters

Prove that ifpi< 0 andyi+1=yithen

pi+1=pi+ 4xi+ 6

Prove that ifpi0 andyi+1=yi1 then

pi+1

=pi

+ 4(xi

yi

) + 10


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Advantages of Bresenham circle

Only involves integer addition, subtraction

and multiplication

There is no need for squares, square rootsand trigonometric functions

Mid i Ci l Al i h


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Midpoint Circle Algorithm

yi yi-1

xi xi+1 xi+2

Midpoint

x2+y2r2= 0

Assuming that we have just plotted the pixels at (xi , yi).Which is next? (xi+1, yi)OR (xi+1, yi1).

- The one that is closer to the circle.

Mid i Ci l Al i h


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Midpoint Circle Algorithm

The decision parameter is the circle at the midpoint

between the pixelsyiandyi1.

Ifpi< 0, the midpoint is inside the circle and the pixel

yiis closer to the circle boundary. Ifpi0, the midpoint is outside the circle and the pixel

yi- 1 is closer to the circle boundary.

12

2 2 212

( 1, )

( 1) ( )

i circle i i

i i

p f x y

x y r

D i i P t


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Decision Parameters

Decision Parameters are obtained using

incremental calculations

OR

whereyi+1is eitheryioryi-1 depending on the sign ofpi

11 1 1 2

2 2 21

1 2

( 1, )

( 2) ( )

i circle i i

i i

p f x y

x y r

2 2 2

1 1 12( 1) ( ) ( ) 1i i i i i i ip p x y y y y

Note:

xi+1=xi+1

Th Al ith1 Initial values: point(0 r)


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The Algorithm1. Initial values:- point(0,r)x0= 0

y0= r

2. Initial decision parameter

3. At eachxiposition, starting at i = 0, perform the

following test: ifpi< 0, the next point is (xi+ 1,yi) and

pi+1=pi+ 2xi+1+ 1

Ifpi 0, the next point is (xi+1,yi-1) and

pi+1=pi+ 2xi+1+ 12yi+1

where 2xi+1

= 2xi+ 2 and 2y

i+1= 2y

i2

4. Determine symmetry points in the other octants

5. Move pixel positions (x,y) onto the circular path centered

on (xc, yc) and plot the coordinates:x=x+xc,y=y+yc

6. Repeat 35 untilxy

move circle origin at (0,0) byx=xxcandy=yyc

2 2 51 10 2 2 4

(1, ) 1 ( )circlep f r r r r

Example


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Example

10

9

8

7

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7 8 9 10

i pi xi+1, yi+1 2xi+1

2yi+1

0 -9 (1, 10) 2 20

1 -6 (2, 10) 4 20

2 -1 (3, 10) 6 20

3 6 (4, 9) 8 18

4 -3 (5, 9) 10 18

5 8 (6, 8) 12 16

6 5 (7, 7)

r = 10

p0= 1r = -9(if ris integer roundp0= 5/4rto integer)

Initial point (x0,y0) = (0, 10)


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Exercises

Draw the circle with r= 12 using the

Midpoint-circle algorithm

Draw the circle with r= 14 and center at

(15, 10).


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Exercises

Prove that ifpi< 0 andyi+1=yithen

pi+1=pi+ 2xi+1+ 1

Prove that ifpi0 andyi+1=yi-1 then

pi+1=pi+ 2xi+1+ 12yi+1

Midpoint function


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Midpoint functionvoid plotpoints(int x, int y){

setpixel(xcenter+x, ycenter+y);

setpixel(xcenter-x, ycenter+y);setpixel(xcenter+x, ycenter-y);setpixel(xcenter-x, ycenter-y);setpixel(xcenter+y, ycenter+x);setpixel(xcenter-y, ycenter+x);setpixel(xcenter+y, ycenter-x);setpixel(xcenter-y, ycenter-x);

}

void circle(int r){

int x = 0, y = r;plotpoints(x,y);int p = 1 r;while (x


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Ellipse-Generating Algorithms

EllipseA modified circle whose radius varies from amaximum value in one direction (major axis) to a minimumvalue in the perpendicular direction (minor axis).

P=(x,y)F1

F2

d1

d2

The sum of the two distances d1and d2, between the fixedpositionsF1andF2(called thefociof the ellipse) to anypointPon the ellipse, is the same value, i.e.

d1+ d2= constant

Ellipse Properties


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Ellipse Properties

Expressing distances d1and d2in terms of the focal

coordinatesF1= (x1,x2) andF2= (x2,y2), we have:

Cartesian coordinates:

Polar coordinates:

2 2 2 2

1 1 2 2( ) ( ) ( ) ( ) constantx x y y x x y y

ry rx

22

1c c

x y

x x y yr r

cos

sin

c x

c y

x x r

y y r


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Ellipse Algorithms

Symmetry between quadrants

Not symmetric between the two octants of a quadrant

Thus, we must calculate pixel positions along the

elliptical arc through one quadrant and then we obtain

positions in the remaining 3 quadrants by symmetry

(x,y)(-x,y)

(x, -y)(-x, -y)

rx

ry

Ellipse Algorithms


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Ellipse Algorithms

Decision parameter:

2 2 2 2 2 2( , )ellipse y x x y

f x y r x r y r r

1

Slope = -1

rx

ry 2

0 if ( , ) is inside the ellipse

( , ) 0 if ( , ) is on the ellipse0 if ( , ) is outside the ellipse

ellipse

x y

f x y x yx y

2

2

2

2

y

x

r xdySlope

dx r y

Ellipse Algorithms Slope = 1


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Ellipse Algorithms

Starting at (0, ry) we take unit steps in thexdirection until

we reach the boundary between region 1and region 2.

Then we take unit steps in theydirection over theremainder of the curve in the first quadrant.

At the boundary

therefore, we move out of region 1whenever

1

Slope = -1

rx

ry 2

2 2

1 2 2y xdy

r x r ydx

2 22 2y xr x r y

Midpoint Ellipse Algorithm


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yi yi-1

xi xi+1 xi+2

Midpoint

Assuming that we have just plotted the pixels at (xi , yi).

The next position is determined by:

12

2 2 2 2 2 212

1 ( 1, )

( 1) ( )

i ellipse i i

y i x i x y

p f x y

r x r y r r

Ifp1i< 0 the midpoint is inside the ellipse yiis closer

Ifp1i0 the midpoint is outside the ellipse yi1is closer

D i i P (R i 1)


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Decision Parameter (Region 1)

At the next position [xi+1+ 1 =xi+ 2]

OR

whereyi+1=yi

or yi+1=yi1

11 1 1 2

2 2 2 2 2 21

1 2

1 ( 1, )

( 2) ( )

i ellipse i i

y i x i x y

p f x y

r x r y r r

2 2 2 2 2 21 11 1 2 2

1 1 2 ( 1) ( ) ( )i i y i y x i ip p r x r r y y



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Decision Parameter (Region 1)Decision parameters are incremented by:

Use only addition and subtraction by obtaining

At initial position (0, ry)

2 2

1

2 2 2

1 1

2 if 1 0

2 2 if 1 0

y i y i

y i y x i i

r x r p

increment r x r r y p

2 2

2 and 2y xr x r y

2

2 2

2 2 2 2 21 10 2 2

2 2 21

4

2 0

2 2

1 (1, ) ( )

y

x x y

ellipse y y x y x y

y x y x

r x

r y r r

p f r r r r r r

r r r r

Region 2


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Region 2Overregion 2, step in the negative y direction and midpoint istaken between horizontal pixels at each step.

yi yi-1

xi xi+1 xi+2

Midpoint

Decision parameter:

12

2 2 2 2 2 212

2 ( , 1)

( ) ( 1)

i ellipse i i

y i x i x y

p f x y

r x r y r r

Ifp2i> 0 the midpoint is outside the ellipse xiis closer

Ifp2i0 the midpoint is inside the ellipse xi+ 1is closer

D i i P t (R i 2)


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At the next position [yi+11 =yi2]

OR

wherexi+1=xi

or xi+1=xi+ 1

11 1 12

2 2 2 2 2 21

1 2

2 ( , 1)

( ) ( 2)

i ellipse i i

y i x i x y

p f x y

r x r y r r

2 2 2 2 21 11 1 2 2

2 2 2 ( 1) ( ) ( )i i x i x y i ip p r y r r x x



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Decision parameters are incremented by:

At initial position (x0, y0) is taken at the last

position selected in region 1

2 2

1

2 2 2

1 1

2 if 2 0

2 2 if 2 0

x i x i

y i x i x i

r y r pincrement

r x r y r p

10 0 02

2 2 2 2 2 210 02

2 ( , 1)

( ) ( 1)

ellipse

y x x y

p f x y

r x r y r r



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p p g

1. Input rx, ry, and ellipse center (xc,yc), and obtain the first

point on an ellipse centered on the origin as

(x0,y0) = (0, ry)

2. Calculate the initial parameter in region 1as

3. At eachxiposition, starting at i= 0, ifp1i< 0, the next

point along the ellipse centered on (0, 0) is (xi + 1,yi) and

otherwise, the next point is (xi+ 1,yi1) and

and continue until

2 2 210 4

1 y x y xp r r r r

2 2

1 11 1 2i i y i yp p r x r

2 2 2

1 1 11 1 2 2i i y i x i yp p r x r y r

2 2

2 2y xr x r y



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p p g4. (x0,y0)is the last position calculated in region 1. Calculate the

initial parameter in region 2as

5. At eachyiposition, starting at i= 0, ifp2i> 0, the next point

along the ellipse centered on (0, 0) is (xi,yi1)and

otherwise, the next point is (xi+ 1,yi1)and

Use the same incremental calculations as in region 1. Continue

untily= 0.

6. For both regions determine symmetry points in the other threequadrants.

7. Move each calculated pixel position (x, y) onto the elliptical

path centered on (xc,yc) and plot the coordinate values

x= x+ xc , y= y+ yc

2 2 2 2 2 21

0 0 022 ( ) ( 1)y x x yp r x r y r r

2 2

1 12 2 2i i x i xp p r y r

2 2 2

1 1 12 2 2 2i i y i x i xp p r x r y r

Example


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p

i pi xi+1, yi+1 2ry2xi+1 2rx

2yi+1

0 -332 (1, 6) 72 768

1 -224 (2, 6) 144 768

2 -44 (3, 6) 216 768

3 208 (4, 5) 288 640

4 -108 (5, 5) 360 640

5 288 (6, 4) 432 512

6 244 (7, 3) 504 384

rx= 8 , ry= 6

2ry2x = 0 (with increment 2ry

2= 72)

2rx2y = 2rx

2ry (with increment -2rx2= -128)

Region 1

(x0, y0) = (0, 6)

2 2 210 4

1 332y x y xp r r r r

Move out of region 1since

2ry2

x> 2rx2

y

Example


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71

p

6

5

4 3

2

1

0

0 1 2 3 4 5 6 7 8

i pi xi+1, yi+1 2ry2xi+1 2rx

2yi+1

0 -151 (8, 2) 576 256

1 233 (8, 1) 576 128

2 745 (8, 0) - -

Region 2

(x0, y0) = (7, 3) (Last position inregion 1)

10 22 (7 ,2) 151ellipsep f

Stop aty= 0


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Exercises

Draw the ellipse with rx= 6, ry= 8.

Draw the ellipse with rx= 10, ry= 14.

Draw the ellipse with rx= 14, ry= 10 andcenter at (15, 10).

Midpoint Ellipse Function


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p pvoid ellipse(int Rx, int Ry){

int Rx2 = Rx * Rx, Ry2 = Ry * Ry;int twoRx2 = 2 * Rx2, twoRy2 = Ry2 * Ry2;int p, x = 0, y = Ry;

int px = 0, py = twoRx2 * y;

ellisePlotPoints(xcenter, ycenter, x, y);// Region 1p = round(Ry2 (Rx2 * Ry) + (0.25 * Rx2));while (px < py) {

x++;px += twoRy2;if (p < 0) p += Ry2 + px;

else {y--;py -= twoRx2;p += Ry2 + px py;

}ellisePlotPoints(xcenter, ycenter, x, y);

}// Region 2p = round(Ry2 * (x+0.5) * (x+0.5) + Rx2 * (y-1)*(y-1) Rx2 * Ry2;while (y > 0) {

y--;py -= twoRx2;if (p > 0) p += Rx2 py;else {

x++;px += twoRy2;p += Rx2 py + px;

}

cg out put primitives

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