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```					Computer Graphics

Chapter 3
Graphics Output Primitives

Andreas Savva
Basic Raster Algorithms
for 2D Primitives
Description of pictures
Specified by a set of intensities for 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 to
describe a scene in terms of basic geometric structures
referred to as output primitives.
•   Points
•   Straight lines
•   Circles
•   Splines curves and surfaces
•   Polygon color areas
•   Character strings
2
•   Etc.
Implementing Application programs

Description of objects in terms of primitives and
attributes and converts them to the pixels on the
screen.
Primitives – what is to be generated
Attributes – how primitives are to be generated

3
(0,0)

Points                          CRT
(maxx,maxy)

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 the
frame-buffer at (x, y).
 getpixel(x, y) – retrieves the current frame-buffer
intensity setting at position (x, y).
4
Lines
Analog devises, such as a random-scan display or a
vector 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.

5
Digital devices display 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 stairstep appearance
(the “jaggies”).
Particularly noticeable on systems with low resolution.
To smooth raster lines, pixel intensities along the line paths must

6
Line Drawing Algorithms

Cartesian equation:
y = mx + c           y2
where                         y1

m – slope                         x1   x2
c – y-intercept

y2  y1 y
m        
x2  x1 x

7
Slope
+ve          -ve

if |m| = 1               45°

 = 45°
45°

if |m|  1
-45° <  < 45°             °
°

if |m|  1
45° <  < 90° or     °

-90° <  < -45°                       °
8
|m| = 1
y=x
m=1
c=0   y
x     y   8                                   
0     0   7                               
1     1   6                           
2     2   5                       
3     3   4                   
4     4
3               
5     5
2           
6     6   1       
7     7   0   
8     8                                           x
0   1   2   3   4   5   6   7   8

9
|m|  1
y=½x+1
m=½
y
c=1
8
x    y    round(y)
7
0    1       1
1   1.5      2       6
2    2       2       5                                
3   2.5      3       4                        
4    3       3       3                
5   3.5      4       2        
6    4       4       1   
7   4.5      5       0
8    5       5           0   1   2   3   4   5   6   7   8   x

10
|m|  1
y = 3x - 2
m=3
y
c = -2
8
x    y   round(y)
0   -2      -2      7               
1    1       1      6
2    4       4      5
3    7       7      4           
4   10      10      3
5   13      13      2
6   16      16      1       
7   19      19      0
8   22      22                                              x
0   1   2   3   4   5   6   7   8
outside
11
The Digital Differential Analyzer
(DDA) Algorithm
y
m      means that for a unit (1) change in x there is
x         m-change in y.

x     y           Do not use
i.e. y = 3x + 1   m=3        0     1
y = 3x + 1
to calculate y.
1     4             Use m
2     7
3    10
4    13
5    16

x 1
        means that for a unit (1) change in y there is
y m       1/m change in x.                                    12
The DDA Method
Uses differential equation of the line : m
If 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.




         
                  
                    
                          
13
step through in x              step through in y
The DDA Method

Desired line

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

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

(xi,round(yi))

14
if slope m  0

if |m|  1             if |m|  1
xi+1 = xi + 1          yi+1 = yi + 1
yi+1 = yi + m          xi+1 = xi + 1/m
Right
Right

Left

15
Left
Proceeding from right-endpoint to left-endpoint
if slope m  0

if |m|  1                  if |m|  1
xi+1 = xi - 1               yi+1 = yi - 1
yi+1 = yi - m               xi+1 = xi - 1/m
Right
Right

Left

16
Left
if slope m < 0

if |m|  1                 if |m|  1
xi+1 = xi + 1              yi+1 = yi - 1
yi+1 = yi + m              xi+1 = xi - 1/m
Left
Left

Right

Right   17
Proceeding from right-endpoint to left-endpoint
if slope m  0

if |m|  1                  if |m|  1
xi+1 = xi - 1               yi+1 = yi + 1
yi+1 = yi - m               xi+1 = xi + 1/m
Left
Left

Right

Right   18
Example (DDA)
0  m  1

y  1 x 1
3                xi 1  xi  1
 yi 1  yi  1
               3
x    y     round(y)
y
0    1        1       8
1   4/3       1       7
2   5/3       2       6
5
3    2        2
4                                    
4   7/3       2
3                          
5   8/3       3
2             
6    3        3       1    
7   10/3      3       0
8   11/3      4           0   1    2   3   4   5   6   7   8   x   19
Example (DDA)
m  1

y  3 x  8            yi 1  yi  1
 xi 1  xi  ( 1 )
                 3

y   x     round(x)
y
8   0        0       8   
7   1/3      0       7   
6   2/3      1       6          
5          
5   1        1
4          
4   4/3      1
3               
3   5/3      2
2               
2   2        2       1               
1   7/3      2       0                    
0   8/3      3              0   1     2   3   4   5    6    7   8   x   20
void LineDDA(int x0, int y0, int x1, int y1)
{
int dx = x1 – x0, dy = y1 – y0, steps;
if (abs(dx)>abs(dy)) steps = abs(dx);
else steps = abs(dy);
// one of these will be 1 or -1
double xIncrement = (double)dx / (double )steps;
double yIncrement = (double)dy / (double )steps;
double x = x0;
double y = y0;
setPixel(round(x), round(y));
for (int i=0; i<steps; i++) {
{
x += xIncrement;
y += yIncrement;
setPixel(round(x), round(y));
}
}

Note: The DDA algorithm is faster than the direct use of y = mx + c.
It eliminates multiplication; only one addition.                 21
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

22
Bresenham Line Algorithm
A more efficient approach

Basis of the algorithm:

A
B

Start position

From start position decide A or B next
23
Bresenham Line Algorithm

True line

ti
si

For a given value of x
one pixel lies at distance ti above the line, and
one pixel lies at distance si below the line
24
Bresenham Line Algorithm

Decision parameter

di = (si - ti)
If di  0, then closest pixel is below true line (si smaller)
If di  0, then closest pixel is above true line (ti smaller)

We must calculate the new values for di as we move
along the line.
25
Example:
dy
(slope) 0.5 (i.e.
Let gradient                                                 0.5 or 2dy  dx)
dx
Start pixel at (x0,y1)

y5                                                          At x1 :
s1 = dy         t1 = dx - dy
d1 = (si - ti) = dy - (dx - dy) = 2dy - dx
y3                                                          but 2dy  dx  di  0  y stays the same
hence next pixel is at (x1,y1)
y2
3dy        4dy          At x2 :
2dy                                s2 = 2dy        t2 = dx - 2dy
y1             dy
d2 = (s2 – t2) = 2dy - (dx - 2dy) = 4dy - dx
Suppose d2  0  y is incremented
y0                                                          hence next pixel is at (x2,y2)
x0   x1        x2         x3         x4         x5
At x3 :
s3 = 3dy - dx      t2 = 2dx - 3dy
d3 = (s2 – t3) = 6dy - 3dx  0
so y stays the same
hence next pixel is at (x3,y2)
26
In General                For a line with gradient ≤ 1
d0 = 2dy – dx
if di  0 then yi+1 = yi
di+1 = di + 2dy
if di ≥ 0 then yi+1 = yi + 1
di+1 = di + 2(dy – dx)
xi+1 = xi + 1
For a line with gradient  1
d0 = 2dx – dy
if di  0 then xi+1 = xi
di+1 = di + 2dx
if di ≥ 0 then xi+1 = xi + 1
di+1 = di + 2(dx – dy)
yi+1 = yi + 1
Note: For |m| ≤ 1 the constants 2dy and 2(dy-dx) can be calculated once,
so the arithmetic will involve only integer addition and subtraction.    27
Example – Draw a line from (20,10) to (30,18)
(30,18)
dx = 10
dy = 8
initial decision d0 = 2dy – dx = 6
Also 2dy = 16, 2(dy – dx) = -4
(20,10)
i   di      (xi+1,yi+1)
19
0    6      (21,11)
18
1    2      (22,12)
2   -2      (23,12)       17

3   14      (24,13)       16
4   10      (25,14)       15
5    6      (26,15)       14
6    2      (27,16)       13
7   -2      (28,16)
12
8   14      (29,17)
11
9   10      (30,18)
10
20   21   22   23   24   25   26   27   28   29   30   31   32   28
void LineBres(int x0, int y0, int x1, int y1) // line for |m| < 1
{
int dx = abs(x1 – x0), dy = abs(y1 – y0);
int d = 2 * dy – dx, twoDy = 2 * dy, twoDyMinusDx = 2 * (dy – dx);
int x, y;
if (x0 > x1) {   // determines which point to use as start, which as end
x = 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);
}                                                                          29
}
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.

30
Parallel Line Algorithms
Given np processors, subdivide the line path into np Bresenham
segments.
For a line with slope 0  m  1 and leftpoint (x0,y0) the distance to
the right endpoint (left endpoint for next segment) is

x  n p  1
x p 
np
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 kth partition is
xk = x0 + kxp
31
Parallel Line Algorithms (continue)
i.e. x = 15 ,     np = 4 processors

15  4  1 18
x p               4
4       4

Starting x-values at x0, x0 + 4, x0 + 8, x0 + 12

yp = mxp
At the kth segment, the starting y-coordinates is
yk = y0 + round(kyp)
Also, the initial decision parameter for Bresenham’s
algorithm at the start of the kth subinterval is:
pk = (kxp)(2y) – round(kyp)(2x) + 2y – x
32
Anti-aliasing for straight lines
Lines generated can have jagged or stair-step appearance,
one aspect of phenomenon called aliasing, caused by fact
that pixels are integer coordinate points.

Use anti-aliasing routines to smooth out display of a line
by adjusting pixels intensities along the line path

33
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 is I, then the intensity per unit
length of line A is I, whereas for line B it is only I/2; this
discrepancy is easily detected by the viewer.
34
Circle Generating Algorithms

 Circles and ellipses are common components in
many pictures.
 Circle generation routines are often included in
packages.

35
Circle Equations
 Polar form
x = rCos
y = rSin         (r = radius of circle)
y

P=(rCos, rSin)
r        rSin)

x
rCos)

36
Drawing a circle
 = 0°
while ( < 360°)
x = rCos
y = rSin
setPixel(x,y)
 =  + 1°
end while

 To find a complete circle  varies from 0° to
360°
 The calculation of trigonometric functions
is very slow.                                   37
Cartesian form
 Use Pythagoras theorem
x2 + y2 = r2   r           y
x
y


P  x, r 2  x 2   
r       y
x            x

38
Circle algorithms
 Step through x-axis to determine y-values

– Not all pixel filled in                   39
– Square root function is very slow
Circle Algorithms
 Use 8-fold symmetry and only compute pixel
positions for the 45° sector.

(-x, y)          (x, y)

(-y, x)              45°             (y, x)

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

(-x, -y)         (x, -y)             40
Bresenham’s Circle Algorithm
Consider only
45° ≤  ≤ 90°

General Principle
 The circle function:

f circle ( x, y )  x 2  y 2  r 2

 and
 0 if (x,y) is inside the circle boundary

f circle ( x, y )   0 if (x,y) is on the circle boundary
 0 if (x,y) is outside the circle boundary

41
Bresenham’s Circle Algorithm
p1       p3
yi
D(si)
yi - 1                      D(ti)
p2

r

xi    xi + 1

After point p1, do we choose p2 or p3?
42
Bresenham’s Circle Algorithm
Define: D(si) = distance of p3 from circle
D(ti) = distance of p2 from circle

i.e. D(si) = (xi + 1)2 + yi2 – r2       [always +ve]
D(ti) = (xi + 1)2 + (yi – 1)2 – r2 [always -ve]

 Decision Parameter pi = D(si) + D(ti)
so if pi < 0 then the circle is closer to p3 (point above)
if pi ≥ 0 then the circle is closer to p2 (point below)

43
The Algorithm
x0 = 0
y0 = r
p0 = [12 + r2 – r2] + [12 + (r-1)2 – r2] = 3 – 2r

if pi < 0 then
yi+1 = yi
pi+1 = pi + 4xi + 6
xi+1 = xi + 1
else if pi ≥ 0 then
yi+1 = yi – 1
pi+1 = pi + 4(xi – yi) + 10

 Stop when xi ≥ yi and determine symmetry
44
points in the other octants
Example
r = 10
p0 = 3 – 2r = -17
Initial point (x0, y0) = (0, 10)
10      
i     pi   xi, yi       9                    
0    -17    (0, 10)      8                           
7                               
1    -11    (1, 10)
6                                   
2      -1   (2, 10)
5                                       
3      13   (3, 10)      4                                       
4      -5   (4, 9)       3                                           
5      15   (5, 9)       2                                           
1                                           
6      9    (6, 8)
0                                           
7           (7,7)
0   1   2   3   4   5   6   7   8   9   10
45
Exercises

 Draw the circle with r = 12 using the
Bresenham algorithm.

 Draw the circle with r = 14 and center at
(15, 10).

46
Decision Parameters

 Prove that if pi < 0 and yi+1 = yi then
pi+1 = pi + 4xi + 6

 Prove that if pi ≥ 0 and yi+1 = yi – 1 then
pi+1 = pi + 4(xi – yi) + 10

47
 Only involves integer addition, subtraction
and multiplication
 There is no need for squares, square roots
and trigonometric functions

48
Midpoint Circle Algorithm
Midpoint
yi    
yi-1              x2 + y2 – r2 = 0

xi xi+1 xi+2

Assuming that we have just plotted the pixels at (xi , yi).
Which is next? (xi+1, yi) OR (xi+1, yi – 1).
- The one that is closer to the circle.

49
Midpoint Circle Algorithm
 The decision parameter is the circle at the midpoint
between the pixels yi and yi – 1.

pi  f circle ( xi  1, yi  1 )
2

 ( xi  1)2  ( yi  1 )2  r 2
2

 If pi < 0, the midpoint is inside the circle and the pixel
yi is closer to the circle boundary.
 If pi ≥ 0, the midpoint is outside the circle and the pixel
yi - 1 is closer to the circle boundary.

50
Decision Parameters
 Decision Parameters are obtained using
incremental calculations
Note:
pi 1  f circle ( xi 1  1, yi 1  ) 1
2           xi+1 = xi +1
 ( xi  2) 2  ( yi 1  1 ) 2  r 2
2

OR

pi 1  pi  2( xi  1) 2  ( yi21  yi2 )  ( yi 1  yi )  1

where yi+1 is either yi or yi-1 depending on the sign of pi

51
1.   Initial values:- point(0,r)        The Algorithm
x0 = 0
move circle origin at (0,0) by
y0 = r                           x = x – xc and y = y – yc
2.   Initial decision parameter
p0  f circle (1, r  1 )  1  (r  1 ) 2  r 2  5  r
2              2             4

3.   At each xi position, starting at i = 0, perform the
following test: if pi < 0, the next point is (xi + 1, yi) and
pi+1 = pi + 2xi+1 + 1
If pi ≥ 0, the next point is (xi+1, yi-1) and
pi+1 = pi + 2xi+1 + 1 – 2yi+1
where 2xi+1 = 2xi + 2 and 2yi+1 = 2yi – 2
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 3 – 5 until x ≥ y                                      52
Example
r = 10
p0 = 1 – r = -9 (if r is integer round p0 = 5/4 – r to integer)
Initial point (x0, y0) = (0, 10)
10      
i    pi   xi+1, yi+1 2xi+   2yi+   9                     
1     1     8                            
0    -9      (1, 10)   2    20     7                                
1    -6      (2, 10)   4    20     6                                    
2    -1      (3, 10)   6    20     5                                        

3    6       (4, 9)    8    18
4                                        
3                                            
4    -3      (5, 9)    10   18
2                                            
5    8       (6, 8)    12   16     1                                            
6    5       (7, 7)                0                                            
0   1   2   3   4   5   6   7   8   9   10
53
Exercises

 Draw the circle with r = 12 using the
Midpoint-circle algorithm

 Draw the circle with r = 14 and center at
(15, 10).

54
Exercises

 Prove that if pi < 0 and yi+1 = yi then
pi+1 = pi + 2xi+1 + 1

 Prove that if pi ≥ 0 and yi+1 = yi-1 then
pi+1 = pi + 2xi+1 + 1 – 2yi+1

55
Midpoint function
void 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<y) {
x++;
if (p<0) p += 2*x + 1;
else {
y--;
p += 2*(x-y) + 1;
}
plotpoints(x,y);
}                                   56
}
Ellipse-Generating Algorithms
 Ellipse – A modified circle whose radius varies from a
maximum value in one direction (major axis) to a minimum
value in the perpendicular direction (minor axis).

d1
F1                      P=(x,y)
d2
F2

 The sum of the two distances d1 and d2, between the fixed
positions F1 and F2 (called the foci of the ellipse) to any
point P on the ellipse, is the same value, i.e.
d1 + d2 = constant
57
Ellipse Properties
 Expressing distances d1 and d2 in terms of the focal
coordinates F1 = (x1, x2) and F2 = (x2, y2), we have:
( x  x1 )2  ( y  y1 )2  ( x  x2 )2  ( y  y2 )2  constant

ry
rx

2
 x  xc   y  yc 
2

 Cartesian coordinates:          
 r    
1
 rx   y 

 Polar coordinates:            x  xc  rx cos 
y  yc  ry sin                         58
Ellipse Algorithms
 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)
ry
rx

(-x, -y)             (x, -y)
59
Ellipse Algorithms
fellipse ( x, y)  r x  r y  r r
2 2
y     x
2   2    2 2
x y

 Decision parameter:
 0 if ( x, y ) is inside the ellipse

f ellipse ( x, y )   0 if ( x, y ) is on the ellipse
 0 if ( x, y ) is outside the ellipse

Slope = -1
1
ry       2                           dy   2ry2 x
Slope      2
rx                          dx   2rx y

60
Ellipse Algorithms                                 Slope = -1
ry 1 2
rx

 Starting at (0, ry) we take unit steps in the x direction until
we reach the boundary between region 1 and region 2.
Then we take unit steps in the y direction over the
remainder of the curve in the first quadrant.
 At the boundary
dy
 1          2ry2 x  2rx2 y
dx
 therefore, we move out of region 1 whenever

2ry2 x  2rx2 y
61
Midpoint Ellipse Algorithm
Midpoint
yi    
yi-1

xi   xi+1 xi+2

Assuming that we have just plotted the pixels at (xi , yi).
The next position is determined by:
p1i  f ellipse ( xi  1, yi  1 )
2

 ry2 ( xi  1) 2  rx2 ( yi  1 ) 2  rx2 ry2
2

If p1i < 0 the midpoint is inside the ellipse  yi is closer
If p1i ≥ 0 the midpoint is outside the ellipse  yi – 1 is closer   62
Decision Parameter (Region 1)

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

p1i 1  f ellipse ( xi 1  1, yi 1  1 )
2

 ry2 ( xi  2)2  rx2 ( yi 1  1 ) 2  rx2 ry2
2

OR
p1i 1  p1i  2ry2 ( xi  1)2  ry2  rx2 ( yi 1  1 )2  ( yi  1 )2 
          2             2    

where yi+1 = yi
or    yi+1 = yi – 1
63
Decision Parameter (Region 1)
Decision parameters are incremented by:
2ry2 xi 1  ry2
                              if p1i  0
increment   2

 2ry xi 1  ry2  2rx2 yi 1 if p1i  0

Use only addition and subtraction by obtaining
2ry2 x and 2rx2 y

At initial position (0, ry)
2ry2 x  0
2rx2 y  2rx2 ry

p10  f ellipse (1, ry  1 )  ry2  rx2 (ry  1 ) 2  rx2 ry2
2                     2
 ry2  rx2 ry  1 rx2
4                                         64
Region 2
Over region 2, step in the negative y direction and midpoint is
taken between horizontal pixels at each step.

Midpoint
yi     
yi-1

xi   xi+1 xi+2
Decision parameter:
p 2i  f ellipse ( xi  1 , yi  1)
2

 ry2 ( xi  1 ) 2  rx2 ( yi  1) 2  rx2 ry2
2

If p2i > 0 the midpoint is outside the ellipse  xi is closer
65
If p2i ≤ 0 the midpoint is inside the ellipse  xi + 1 is closer
Decision Parameter (Region 2)

At the next position [yi+1 – 1 = yi – 2]

p 2i 1  f ellipse ( xi 1  1 , yi 1  1)
2

 ry2 ( xi 1  1 )2  rx2 ( yi  2) 2  rx2 ry2
2

OR
p2i 1  p2i  2rx2 ( yi  1)  rx2  ry2 ( xi 1  1 )2  ( xi  1 )2 
          2             2    

where xi+1 = xi
or    xi+1 = xi + 1
66
Decision Parameter (Region 2)
Decision parameters are incremented by:

2rx2 yi 1  rx2
                                   if p 2i  0
increment   2

 2ry xi 1  2rx2 yi 1  rx2      if p 2i  0

At initial position (x0, y0) is taken at the last
position selected in region 1

p 20  f ellipse ( x0  1 , y0  1)
2

 ry2 ( x0  1 ) 2  rx2 ( y0  1) 2  rx2 ry2
2

67
Midpoint Ellipse Algorithm
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 1 as
p10  ry2  rx2ry  1 rx2
4

3. At each xi position, starting at i = 0, if p1i < 0, the next
point along the ellipse centered on (0, 0) is (xi + 1, yi) and
p1i 1  p1i  2ry2 xi 1  ry2

otherwise, the next point is (xi + 1, yi – 1) and
p1i 1  p1i  2ry2 xi 1  2rx2 yi 1  ry2

2ry2 x  2rx2 y
68
and continue until
Midpoint Ellipse Algorithm
4. (x0, y0) is the last position calculated in region 1. Calculate the
initial parameter in region 2 as
p20  ry2 ( x0  1 )2  rx2 ( y0 1)2  rx2ry2
2

5. At each yi position, starting at i = 0, if p2i > 0, the next point
along the ellipse centered on (0, 0) is (xi, yi – 1) and
p 2i 1  p 2i  2rx2 yi 1  rx2
otherwise, the next point is (xi + 1, yi – 1) and
p2i 1  p2i  2ry2 xi 1  2rx2 yi 1  rx2
Use the same incremental calculations as in region 1. Continue
until y = 0.
6. For both regions determine symmetry points in the other three
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                 69
Example
rx = 8 , ry = 6
2ry2x = 0          (with increment 2ry2 = 72)
2rx2y = 2rx2ry (with increment -2rx2 = -128)
Region 1
(x0, y0) = (0, 6)
p10  ry2  rx2ry  1 rx2  332
4

i     pi    xi+1, yi+1 2ry2xi+1 2rx2yi+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      Move out of region 1 since
2ry2x > 2rx2y          70
Example
Region 2
(x0, y0) = (7, 3)          (Last position in region 1)
p 20  f ellipse (7  1 , 2)  151
2

i    pi       xi+1, yi+1 2ry2xi+1 2rx2yi+1
0    -151        (8, 2)       576             256
1    233         (8, 1)       576             128
2    745         (8, 0)           -            -    Stop at y = 0

6      
5                         
4                                    
3                                        
2                                             
1                                             
0                                                                 71
0     1    2    3    4   5       6   7    8
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 and
center at (15, 10).

72
Midpoint Ellipse Function
void 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 1
p = 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 2
p = 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;
}
ellisePlotPoints(xcenter, ycenter, x, y);                      73
}
}

```
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