Output primitives are the basic geometric structures which facilitate or describe a scene/picture.

Example of these include:points, lines, curves (circles, conics etc), surfaces, fill colour, character string etc.

Introduction

In order to draw the primitive objects, one has to first scan convert the object.

Scan convert: Refers to the operation of finding out the location of pixels and then setting the values of corresponding bits, in the graphic memory, to the desired intensity code.

Digital Differential Analyzer Algorithm

(DDA)

The Digital Differential Analyzer (DDA) Algorithm

5

mx

y

means that for a unit (1) change in x

there is m-change in y.

my

x 1

means that for a unit (1) change in y

there is 1/m change in x.

6

The DDA MethodUses differential equation of the line : mIf slope |m| 1 then increment x in steps of 1 pixel and find corresponding y-values.If slope |m| 1 then increment y in steps of 1 pixel and find corresponding x-values.

step through in x step through in y

DDA- Line with positive Slope

A line with positive slope (Left to Right) If m <= 1, sample at unit x intervals (Δx = 1) and compute successive y values as: yk+1 = yk + m. If m >1, sample at unit y intervals (Δy =1) andcompute successive x values as: xk+1 = xk + 1/m.

A line with positive slope (Right to Left) If m <= 1, sample at unit x intervals (Δx =-1)and compute successive y values as: yk+1 = yk – m If m >1, sample at unit y intervals (Δy = -1) and compute successive x values as: xk+1 = xk - 1/m

A line with negative slope (Left to Right) If |m| <= 1, sample at unit x intervals (Δx = 1) and compute successive y values as: yk+1 = yk + m If |m| >1, sample at unit y intervals (Δy =-1)and compute successive x values as: xk+1 = xk - 1/m

DDA- Line with negative Slope

A line with negative slope (Right to Left)If |m| <=1, sample at unit x intervals (Δx =1) and compute successive y values as: yk+1 = yk - m If |m| >1, sample at unit y intervals (Δy =1)and compute successive x values as: xk+1 = xk + 1/m

1) Accept the end point co-ordinates of the line segment AB ie A(x1,x2) and B(x2,y2). 2) Calculate dx and dy.

3) If abs(dx)>=abs(dy) then , Step=dx else Step=dy.

Dda algorithm

4) Let x increment = dx/Step ; Let y increment = dy/Step.

5) Display the pixels at standing portion put pixel (x,y,white).

6) Compute the next co-ordinate position along the line path xk+1 = xk + x increment

yk+1 = yk + y increment

Put Pixel(xk+1 , yk+1 ,white).

7) If xk+1=x2 OR / AND yk+1=y2

then STOP else go to Step (4).

Bresenham’s Line Algorithm

Bresenham’s Line Algorithm

1) In Bresenham’s Line Drawing Algorithm the pixel positions along a line path are obtained by determining the pixel i.e nearer the line at each step. 2) It is an efficient raster line generation algorithm.

1) Accept the end point co-ordinates of the line segment AB ie A(x1,x2) and B(x2,y2).

2) Calculate dx and dy.

3) If abs(dy)<=abs(dx) ie slope |m|<=1 (a) Compute initial decision parameter Po= 2abs(dy)-abs(dx).

(b) At each xk position perform the following

if Pk <0 i.e d1< d2 then,

x increment=dx/abs(dx) AND y increment=0

Pk+1= Pk + 2abs(dy)

else,

Pk >0 ie d1>d2

X increment =dx/abs(dx) Y increment =dy/abs(dy)

Pk+1= Pk +2abs(dy)-2abs(dx).

4) If abs(dy) > abs(dx) ie slope |m|>1

(a) Compute initial decision parameter Po= 2abs(dx)-abs(dy) (b) At each xk position,perform the following if Pk < 0 i.e d1 < d2 then

x increment=0 y increment=dy/abs(dy)

Pk+1= Pk + 2abs(dx)

Else, Pk >0 ie d1>d2

X increment=dx/abs(dx) Y increment=dy/abs(dy) Pk+1= Pk +2abs(dx)-2abs(dy).

5) Calculate : xnext = xk + x increment

ynext = yk + y increment.

Display (xk+1, yk+1, white).

6) Repeat Step (3) to (5) until xk+1=x2 AND /OR yk+1 =y2

7)STOP.

Midpoint Circle Algorithm

Mid-point Circle Algorithm

A method for direct distance comparison isto test the halfway position between twopixels to determine if this midpoint is insideor outside the circle boundary. This method is more easily applied toother conics, and for an integer circleradius.

1) Accept the radius r and center (xc, yc).The point of the circumference of a circle with center as origin (x0, y0) =(0,r).

2) Calculate the initial decision parameter as Po =5/4 –r .

3) At each xk position starting at k=0.Perform

the following test.

If Pk <0 then,

Xk+1= xk +1 and yk+1= yk

Pk+1 = Pk +2xk +3

Otherwise, Pk >0

Xk+1= xk +1 and Yk+1= yk -1

Pk+1= Pk +2( xk - yk ) +5

4) Determine the symmetric points in other 7 octants.

5) Translate each calculated pixel position by T ( xk , yk ) and display the pixel

X= xk+1 + xc AND Y= yk+1 + yc

Put pixel ( x,y, white).

6) Repeat step (3) to step (5) until x>=y.

7)STOP.

BRESENHAM’S CIRCLE ALGORITHM

1) Accept the radius r and center (xc, yc).The point of the circumference of a circle with center as origin (x0, y0) =(0,r).

2) Calculate the initial decision parameter as PO =3 – 2r.

3) At each xk position starting at k=0.Perform

the following test.

If Pk <0 then,

Xk+1= xk +1 and yk+1= yk

Pk+1 = Pk +4xk +6.

Otherwise, Pk > 0

Xk+1= xk +1 AND Yk+1= yk -1

Pk+1= Pk +4( xk - yk ) +10

4) Determine the symmetric points in other 7 octants.

5) Translate each calculated pixel position by T ( xk , yk ) and display the pixel

X= xk+1 + xc AND Y= yk+1 + yc

Put pixel ( x, y, white)6) Repeat step (3) to step (5) until x>=y 7) STOP.

The DDA algorithm is accomplished by taking unit step in one direction and calculating other.

The Bresenham’s Algorithm finds the closest integer co-ordinate to the actual part.

DDA algorithm uses floating point calculation. Bresenham’s uses integer point calculation. DDA is more accurate compared to

Bresenhams.

CONCLUSION

Circles and ellipses can be efficiently and accurately scan converted using midpoint methods.

Bresenham’s line algorithm and the midpoint line algorithm methods are the most efficient.

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