OUTPUT PRIMITIVES

Set pixel(x,y intensity)

Get pixel(x,y)

LINE DRAWING ALGORITHM

Cartesian slope intercept equation for a straight line

Y=mx+b;

m=slope and b as the y-intercept

m=y2-y1/x2-x1 ------1

b=y1-m.x1 ------2

y =m x -------3

x = y/m -------4

For lines with slope magnitudes < 1, x is set proportional to small horizontal deflection voltage and corresponding y from 3

For slopes with magnitude > 1 y is set proportional to small vertical deflection voltage and corresponding horitonal deflection voltage set proportional to x from 4

DAA ALGORITHM

Digital differential analyzer is based on calculating x or y

We sample the line or unit intervals in one coordinate and determine corresponding interger values nearest the line path for the other coordinate

Positive slope

If slope <=1 x=1 and compute yk +1=yk+m

K=1 is increased by I until the final endpoint is reached

If slope >1 , y=1 and compute

Xk+1=Xk+1/m (left end point to right end point)

If this processing is reversed, starting endpoint is at right

X=-1

Yk+1=yk-m

Slope >1, y=-1 and Xk+1=Xk -1/m

DAA is faster method , disadvantage and round off error can cause calculated pixel positions to drift away from the true line path

Procedure line DAA(xa,ya,xb,yb=interger)

Var

dx,dy,steps,k:=interger;

xinc,yinc,x,y:=real

Begin

dx=xb-xa

dy=yb-ya

If abs(dx)>abs(dy) then steps =abs(x)

else steps=abs(dy)

xinc=dx/steps

yinc=dy/steps

x=xa

y=ya

Setpixel(round(x),round(y),1);

for k=1 to steps do

x=x+xinc;

y=y+yinc;

Setpixel(round(x),round(y),1)

End

End;(line DAA)

Bresenham’s line algorithm

Developed by bresenham .Used to draw line, adapted to display circles and curves.

Specified line path

13

12

11

10

10 11 12 13

Vertical axes show scan-line position and horizontal axes identify pixel columns.

Sampling at unit 11 interval, decide which of two possible pixel position is closer to the line path at each sample step.

Next position (11,11) or (11,12)

Answered by Bresenham’s line algorithm by testing the sign of the integer parameter and difference between the separations of the two pixel positions from the actual line path

Positive slope <1

Current pixel position(xk,yk),next pixel would be at (xk+1,yk) and (xk+1,yk+1)

At xk+1

Yk+1 d2

Y d1

yk

xk+1

the y coordinate on the mathematical line at pixel column position xk+1 is

y=m(xk+1)+b 1

then

d1=y-yk

=m(xk+1)+b-yk 2 from 1

And d2=(yk+1)-y

=yk+1-m(xk+1)-b

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

A decision parameter pk is obtained by finding m= y/ x

Pk= x(d1-d2)

=2 y.xk-2 x.yk+c

Sign of pk is sign of (d1-d2) since x>0

Parameter c is a constant

C=2 y+ x(2b-1)

If d1<d2 , we plot the lower pixel pk is negative. Otherwise we plot the upper pixel

Pk+1=2 y.xk+1- 2 x yk+1+c

Pk+1-pk=2 y(xk+1-xk)-2 x(yk+1-yk)

But xk+1=xk+1 so that

Pk+1=pk+2 y-2 x(yk+1-yk)

The first parameter p0 is evaluated starting from pixel position (x0 yo) and

m= x/ y

b=(y0-mx0)

p0=2 y- x

|m|<1

1)Input the two line end points and store the left endpoint(x0 y0)

2) Load (x0 y0) into the frame buffer (ie) load the first point

3)Calculate constants x, y 2 y and 2 y-2 x and obtain the starting decision parameter

P0 =2 y- x

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) and pk+1=pk+2 y

Otherwise the next plot point is(xk+1,yk+1)

Pk+1=pk+2 y-2 x

Repeat steps and x times (20 ,10) and (30,18) x=10 y=8

P0=2 y- x = 6

2 y=16 2 y- x=-4

K pk

6