1. Scan Conversion or Rasterization
Drawing lines, circles, and etc. on a grid implicitly involves approximation.
The general process: Scan Conversion or Rasterization
Ideally, the following properties should be considered
smooth
continuous
pass through specified points
uniform brightness
efficient
Computer Graphics, KKU. Lecture 6

2. Line Drawing and Scan Conversion
There are three possible choices which are potentially useful.
Explicit: y = f(x)
y = m (x - x0) + y0 where m = dy/dx
Parametric: x = f(t), y = f(t)
x = x0 + t(x1 - x0), t in [0,1]
y = y0 + t(y1 - y0)
Implicit: f(x, y) = 0
F(x,y) = (x-x0)dy - (y-y0)dx
if F(x,y) = 0 then (x,y) is on line
F(x,y) > 0 then (x,y) is below line
F(x,y) < 0 then (x,y) is above line
Computer Graphics, KKU. Lecture 6

3. Line Drawing - Algorithm 1
A Straightforward Implementation
DrawLine(int x1,int y1, int x2,int y2, int color) {
float y;
int x;
for (x=x1; x<=x2; x++)
y = y1 + (x-x1)*(y2-y1)/(x2-x1)
WritePixel(x, Round(y), color ); }
Computer Graphics, KKU. Lecture 6

4. Line Drawing - Algorithm 2
A Better Implementation
DrawLine(int x1,int y1,int x2,int y2, int color) {
float m,y;
int dx,dy,x;
dx = x2 - x1;
dy = y2 - y1;
m = dy/dx;
y = y ;
for (x=x1; x<=x2; x++)
WritePixel(x, Floor(y), color );
y = y + m; }
Computer Graphics, KKU. Lecture 6

5. Line Drawing Algorithm Comparison
Advantages over Algorithm 1
eliminates multiplication
improves speed
Disadvantages
round-off error builds up
get pixel drift
rounding and floating point arithmetic still time consuming
works well only for |m| < 1
need to loop in y for |m| > 1
need to handle special cases
Computer Graphics, KKU. Lecture 6

6. Line Drawing - Midpoint Algorithm
The Midpoint or Bresenham’s Algorithm
The midpoint algorithm is even better than the above algorithm in that it uses only integer calculations. It treats line drawing as a sequence of decisions. For each pixel that is drawn the next pixel will be either N or NE, as shown below.
Computer Graphics, KKU. Lecture 6

7. 168 471 Computer Graphics, KKU. Lecture 6
Midpoint Algorithm
The midpoint algorithm makes use of the implicit definition of the line, F(x,y) =0. The N/NE decisions are made as follows.
d = F(xp + 1, yp + 0.5)
if d < 0 line below midpoint choose E
if d > 0 line above midpoint choose NE
if E is chosen
dnew = F(xp + 2, yp + 0.5)
dnew- dold = F(xp + 2, yp + 0.5) F(xp + 1, yp + 0.5)
Delta = d new -d old = dy
Computer Graphics, KKU. Lecture 6

8. 168 471 Computer Graphics, KKU. Lecture 6
Midpoint Algorithm
If NE is chosen
dnew = F(xp+2, yp+1.5)
Delta = dy-dx
Initialization
dstart = F(x0+1, y0+0.5) = (x0+1-x0)dy - (y0+0.5-y0)dx = dy-dx/2
Integer only algorithm
F’(x,y) = 2 F(x,y) ; d’ = 2d
d’start = 2dy - dx
Delta’ = 2Delta
Computer Graphics, KKU. Lecture 6

9. Midpoint Algorithm for x1 < x2 and slope <= 1
DrawLine(int x1, int y1, int x2, int y2, int color) {
int dx, dy, d, incE, incNE, x, y;
dx = x2 - x1;
dy = y2 - y1;
d = 2*dy - dx;
incE = 2*dy;
incNE = 2*(dy - dx);
y = y1;
for (x=x1; x<=x2; x++) {
WritePixel(x, y, color);
if (d>0) {
d = d + incNE;
y = y + 1;
} else {
d = d + incE; }
Computer Graphics, KKU. Lecture 6

10. General Bressenham’s Algorithm
To generalize lines with arbitrary slopes
consider symmetry between various octants and quadrants
for m > 1, interchange roles of x and y, that is step in y direction, and decide whether the x value is above or below the line.
If m > 1, and right endpoint is the first point, both x and y decrease. To ensure uniqueness, independent of direction, always choose upper (or lower) point if the line go through the mid-point.
Handle special cases without invoking the algorithm: horizontal, vertical and diagonal lines
Computer Graphics, KKU. Lecture 6

11. Scan Converting Circles
Explicit: y = f(x)
Usually, we draw a quarter circle by incrementing x from 0 to R in unit steps and solving for +y for each step.
Parametric:
- by stepping the angle from 0 to 90
- avoids large gaps but still insufficient.
Implicit: f(x) = x2+y2-R2
If f(x,y) = 0 then it is on the circle.
f(x,y) > 0 then it is outside the circle.
f(x,y) < 0 then it is inside the circle.
Computer Graphics, KKU. Lecture 6

12. 168 471 Computer Graphics, KKU. Lecture 6
Eight-way Symmetry
Computer Graphics, KKU. Lecture 6

13. Midpoint Circle Algorithm
dold = F(xp+1, yp+0.5)
If dold < 0, E is chosen
dnew = F(xp+2, yp-0.5)
= dold+(2xp+3)
DeltaE = 2xp+3
If dold >= 0, SE is chosen
dnew = F(xp+2, yp-1.5) = dold+(2xp-2yp+5)
DeltaSE = 2xp-2yp+5
Initialization
dinit = 5/4 – R = 1 - R
Computer Graphics, KKU. Lecture 6

14. Midpoint Circle Algorithm (cont.)
Computer Graphics, KKU. Lecture 6

15. Scan Converting Ellipses
2a is the length of the major axis along the x axis.
2b is the length of the minor axis along the y axis.
The midpoint can also be applied to ellipses.
For simplicity, we draw only the arc of the ellipse that lies in the first quadrant, the other three quadrants can be drawn by symmetry
Computer Graphics, KKU. Lecture 6

16. Scan Converting Ellipses: Algorithm
j > i in region 1
j < i in region 2
Firstly we divide the quadrant into two regions
Boundary between the two regions is
the point at which the curve has a slope of -1
the point at which the gradient vector has the i and j components of equal magnitude
Computer Graphics, KKU. Lecture 6

17. Ellipses: Algorithm (cont.)
At the next midpoint, if a2(yp-0.5)<=b2(xp+1), we switch region 1=>2
In region 1, choices are E and SE
Initial condition: dinit = b2+a2(-b+0.25)
For a move to E, dnew = dold+DeltaE with DeltaE = b2(2xp+3)
For a move to SE, dnew = dold+DeltaSE with DeltaSE = b2(2xp+3)+a2(-2yp+2)
In region 2, choices are S and SE
Initial condition: dinit = b2(xp+0.5)2+a2((y-1)2-b2)
For a move to S, dnew = dold+Deltas with Deltas = a2(-2yp+3)
For a move to SE, dnew = dold+DeltaSE with DeltaSE = b2(2xp+2)+a2(-2yp+3)
Stop in region 2 when the y value is zero.
Computer Graphics, KKU. Lecture 6

18. Midpoint Ellipse Algorithm
Computer Graphics, KKU. Lecture 6

19. Midpoint Ellipse Algorithm (cont.)
Computer Graphics, KKU. Lecture 6