1. Chapter 3 Graphics Output Primitives
Computer Graphics
Chapter 3
Graphics Output Primitives
Andreas Savva

2. 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
Etc.

3. 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

4. Points
(0,0)
(maxx,maxy)
CRT
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).

5. 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.

6. 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 be adjusted.

7. Line Drawing Algorithms
Cartesian equation:
y = mx + c
where
m – slope
c – y-intercept x1 y1 x2 y2

8. Slope +ve -ve if |m| = 1  = 45° if |m|  1 -45° <  < 45°° °
if |m|  1
45° <  < 90° or
-90° <  < -45° ° °

9. |m| = 1
y = x
m = 1
c = 0 y x y 1 2 3 4 5 6 7 8 8  7 6 5 4 3 2 1 x

10. |m|  1 y = ½ x + 1 m = ½ c = 1 y x x y round(y) 1 1.5 2 3 2.5 4 5 3.58 7 6 5  4 3 2 1 x y
round(y) 1
1.5 2 3
2.5 4 5
3.5 6 7
4.5 8 x

11. |m|  1 y = 3x - 2 m = 3 c = -2 y x x y round(y) -2 1 2 4 3 7 10 5 138 7  6 5 4 3 2 1 x y
round(y) -2 1 2 4 3 7 10 5 13 6 16 19 8 22 x
outside

12. The Digital Differential Analyzer (DDA) Algorithm
means that for a unit (1) change in x there is
m-change in y.
Do not use
y = 3x + 1
to calculate y.
Use m x y 1 4 2 7 3 10 13 5 16
i.e. y = 3x m = 3
means that for a unit (1) change in y there is
1/m change in x.

13. 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.  
step through in x
step through in y

14. The DDA Method Desired line (xi+1,round(yi+m)) (xi+1,yi+m) (xi,yi)

15. if slope m  0 if |m|  1 xi+1 = xi + 1 yi+1 = yi + m if |m|  1
xi+1 = xi + 1/m
Right
Right
Left
Left

16. Proceeding from right-endpoint to left-endpoint 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
Right
Right
Left
Left

17. if slope m < 0 if |m|  1 xi+1 = xi + 1 yi+1 = yi + m if |m|  1
xi+1 = xi - 1/m
Left
Left
Right
Right

18. Proceeding from right-endpoint to left-endpoint 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
Left
Right
Right

19. Example (DDA) y x x y round(y) 1 4/3 2 5/3 3 4 7/3 5 8/3 6 7 10/3 81
4/3 2
5/3 3 4
7/3 5
8/3 6 7
10/3 8
11/3 8 7 6 5 4  3 2 1 x

20. Example (DDA) y x y x round(x) 8 7 1/3 6 2/3 1 5 4 4/3 3 5/3 2 7/3 8/37
1/3 6
2/3 1 5 4
4/3 3
5/3 2
7/3
8/3 8  7 6 5 4 3 2 1 x

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

22. Example Draw a line from point (2,1) to (12,6)7 6 5 4 3 2 1 8 9 10 11 12

23. Bresenham Line Algorithm A more efficient approach
Basis of the algorithm:
From start position decide A or B next A B
Start position

24. 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

25. 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.

26. Example: x1 x2 x3 x4 x5 x0 y0 y1 y2 y3 y5 At x1 : At x2 : At x3 :
Start pixel at (x0,y1) x1 x2 x3 x4 x5 x0 y0 y1 y2 y3 y5
At x1 :
s1 = dy t1 = dx - dy
d1 = (si - ti) = dy - (dx - dy) = 2dy - dx
but 2dy  dx  di  0  y stays the same
hence next pixel is at (x1,y1)
True line
4dy
3dy
At x2 :
s2 = 2dy t2 = dx - 2dy
d2 = (s2 – t2) = 2dy - (dx - 2dy) = 4dy - dx
Suppose d2  0  y is incremented
hence next pixel is at (x2,y2)
2dy dy
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)

27. In General For a line with gradient ≤ 1 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.

28. Example – Draw a line from (20,10) to (30,18)
dx = 10
dy = 8
initial decision d0 = 2dy – dx = 6
Also 2dy = 16, 2(dy – dx) = -4
i di (xi+1,yi+1)
(21,11)
(22,12)
(23,12)
(24,13)
(25,14)
(26,15)
(27,16)
(28,16)
(29,17)
(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

29. 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;
y++;
d += twoDyMinusDx;
setPixel(x, y);

30. 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.

31. Parallel Line Algorithms
Take advantage of multiple processors.
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
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

32. Parallel Line Algorithms (continue)
i.e. x = 15 , np = 4 processors
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

33. 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

34. 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.

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

36. Circle Equations Polar form x = rCos y = rSin (r = radius of circle)
P=(rCos, rSin)
rSin)
rCos) x y r

37. Drawing a circle
 = 0°
while ( < 360°)
x = rCos
y = rSin
setPixel(x,y)
 =  + 1°
end while
Disadvantages
To find a complete circle  varies from 0° to 360°
The calculation of trigonometric functions is very slow.

38. Cartesian form
Use Pythagoras theorem
x2 + y2 = r2 x r y y x r

39. Circle algorithms Step through x-axis to determine y-values
Disadvantages:
Not all pixel filled in
Square root function is very slow

40. Circle Algorithms
Use 8-fold symmetry and only compute pixel positions for the 45° sector.
45°
(x, y)
(y, x)
(-x, y)
(y, -x)
(x, -y)
(-x, -y)
(-y, x)
(-y, -x)

41. Bresenham’s Circle Algorithm
Consider only
45° ≤  ≤ 90°
General Principle
The circle function:
and

42. Bresenham’s Circle Algorithmp1 p3 yi
D(si)
D(ti)
yi - 1 p2 r xi
xi + 1
After point p1, do we choose p2 or p3?

43. 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 – r [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)

44. 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
else if pi ≥ 0 then
yi+1 = yi – 1
pi+1 = pi + 4(xi – yi) + 10
Stop when xi ≥ yi and determine symmetry points in the other octants
xi+1 = xi + 1

45. Example r = 10 p0 = 3 – 2r = -17 Initial point (x0, y0) = (0, 10) i pi 9 8 7 6 5 4 3 2 1 i pi
xi, yi
-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)

46. Exercises Draw the circle with r = 12 using the Bresenham algorithm.
Draw the circle with r = 14 and center at (15, 10).

47. 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

48. Advantages of Bresenham circle
Only involves integer addition, subtraction and multiplication
There is no need for squares, square roots and trigonometric functions

49. Midpoint Circle Algorithmyi 
yi-1 xi
xi+1
xi+2
Midpoint
x2 + y2 – r2 = 0
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.

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

51. Decision Parameters
Decision Parameters are obtained using incremental calculations OR
where yi+1 is either yi or yi-1 depending on the sign of pi
Note:
xi+1 = xi +1

52. move circle origin at (0,0) by
The Algorithm
Initial values:- point(0,r)
x0 = 0
y0 = r
Initial decision parameter
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 – 2yi+1
where 2xi+1 = 2xi + 2 and 2yi+1 = 2yi – 2
Determine symmetry points in the other octants
Move pixel positions (x,y) onto the circular path centered on (xc, yc) and plot the coordinates: x = x + xc, y = y + yc
Repeat 3 – 5 until x ≥ y
move circle origin at (0,0) by
x = x – xc and y = y – yc

53. 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  9 8 7 6 5 4 3 2 1 i pi
xi+1, yi+1
2xi+1
2yi+1 -9
(1, 10) 2 20 1 -6
(2, 10) 4 -1
(3, 10) 6 3
(4, 9) 8 18 -3
(5, 9) 10 5
(6, 8) 12 16
(7, 7)

54. Exercises
Draw the circle with r = 12 using the Midpoint-circle algorithm
Draw the circle with r = 14 and center at (15, 10).

55. 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 – 2yi+1

56. 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;

57. 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).
P=(x,y) F1 F2 d1 d2
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

58. Ellipse Properties
Expressing distances d1 and d2 in terms of the focal coordinates F1 = (x1, x2) and F2 = (x2, y2), we have:
Cartesian coordinates:
Polar coordinates: ry rx

59. 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

60. Ellipse Algorithms
Decision parameter: 1
Slope = -1 rx ry 2

61. Ellipse Algorithms 1
Slope = -1 rx ry 2
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
therefore, we move out of region 1 whenever

62. Midpoint Ellipse Algorithmyi 
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:
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

63. Decision Parameter (Region 1)
At the next position [xi = xi + 2] OR
where yi+1 = yi
or yi+1 = yi – 1

64. Decision Parameter (Region 1)
Decision parameters are incremented by:
Use only addition and subtraction by obtaining
At initial position (0, ry)

65. Region 2
Over region 2, step in the negative y direction and midpoint is taken between horizontal pixels at each step. yi 
yi-1 xi
xi+1
xi+2
Midpoint
Decision parameter:
If p2i > 0 the midpoint is outside the ellipse  xi is closer
If p2i ≤ 0 the midpoint is inside the ellipse  xi + 1 is closer

66. Decision Parameter (Region 2)
At the next position [yi+1 – 1 = yi – 2] OR
where xi+1 = xi
or xi+1 = xi + 1

67. Decision Parameter (Region 2)
Decision parameters are incremented by:
At initial position (x0, y0) is taken at the last position selected in region 1

68. Midpoint Ellipse Algorithm
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)
Calculate the initial parameter in region 1 as
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
otherwise, the next point is (xi + 1, yi – 1) and
and continue until

69. Midpoint Ellipse Algorithm
(x0, y0) is the last position calculated in region 1. Calculate the initial parameter in region 2 as
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
otherwise, the next point is (xi + 1, yi – 1) and
Use the same incremental calculations as in region 1. Continue until y = 0.
For both regions determine symmetry points in the other three quadrants.
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

70. Move out of region 1 since
Example
rx = 8 , ry = 6
2ry2x = 0 (with increment 2ry2 = 72)
2rx2y = 2rx2ry (with increment -2rx2 = -128)
Region 1
(x0, y0) = (0, 6) i pi
xi+1, yi+1
2ry2xi+1
2rx2yi+1
-332
(1, 6) 72
768 1
-224
(2, 6)
144 2
-44
(3, 6)
216 3
208
(4, 5)
288
640 4
-108
(5, 5)
360 5
(6, 4)
432
512 6
244
(7, 3)
504
384
Move out of region 1 since
2ry2x > 2rx2y

71. Example Region 2 (x0, y0) = (7, 3) (Last position in region 1) i pi
xi+1, yi+1
2ry2xi+1
2rx2yi+1
-151
(8, 2)
576
256 1
233
(8, 1)
128 2
745
(8, 0) -
Stop at y = 0 6  5 4 3 2 1 7 8

72. Exercises Draw the ellipse with rx = 6, ry = 8.
Draw the ellipse with rx = 14, ry = 10 and center at (15, 10).

73. 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; }
// Region 2
p = round(Ry2 * (x+0.5) * (x+0.5) + Rx2 * (y-1)*(y-1) – Rx2 * Ry2;
while (y > 0) {
if (p > 0) p += Rx2 – py;
p += Rx2 – py + px;