The document describes various output primitives in computer graphics, including points, lines, and circles. It provides details on how each primitive is represented and algorithms for drawing them, such as the digital differential analyzer (DDA) algorithm for lines and the midpoint circle algorithm. Key points covered include how points map to individual pixels, how lines are drawn by plotting discrete points, and how circles can be rendered using either Cartesian equations or parametric equations in polar coordinates.

1. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
OutputPrimitive
1
St. Francis Institute of Technology
Department of Computer Engineering
1
Subject Incharge
Rupesh Mishra
Asst. Professor
Room No. 401
rupeshmishra@sfit.ac.in
31 July 2020

2. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
2
Module II
Lecture 4
Computer Graphics: Output Primitive
St. Francis Institute of Technology
Department of Computer Engineering
1

3. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in

4. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
OutputPrimitive
• Graphics programming packages
• Describe a scene in terms of basic geometric structures (Output Primitives)
• Group output primitives into more complex structures.
• Each output primitive is specified with
• Input coordinate data
• Information about object to be displayed.
• Point, Line, Circle, Conic Section, Spline curves, polygon and Character String.

5. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Point

6. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Point
• Point plotting is accomplished by converting a single coordinate position into
appropriate operations for the output device in use.
• Setting value in frame buffer
• Moving beam of electron at the selected location (CRT)
• Point plotting instruction in display list (Random Scan)
• Converted to deflection voltages to position the electron beam.
• Frame buffer is loaded with the color codes (RGB System)

7. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Line

8. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Line
• Collection of points over a path
• Plotting discrete points between the two endpoints.
• Calculated from the equation of the line
• Frame buffer loaded with intensity value at the corresponding pixel coordinates.
• Video Controller plots the screen pixels
• Approximation of actual points
• (10.4, 20.5) converted to (10, 21)
• Stair-step appearance in low resolution

9. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
LineDrawing
• The Cartesian slope-intercept equation
for a straight line
•
•
• intercept
• Given end point of line
•
•
y = m ⋅ x + c
m : Slope
c : y
(x1, y1), (x2, y2)
m =
(y2 − y1)
(x2 − x1)
c = y1 − m ⋅ x1
Line Segment
Y intercept
x intercept
(x1, y1)
(x2, y2)
x axis
Y axis

10. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• For any given interval along a line, we
can compute the corresponding y interval
•
• Similarly
•
x Δx
Δy
Δy = m . Δx
Δx =
Δy
m
Δy
Δx

11. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Δy
Δx Δx
Δy
Δx
Δy
For angle more than 45∘
Δy
Δx
> 1
For angle less than 45∘
Δy
Δx
< 1
For angle equal 45∘
Δy
Δx
= 1

12. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
DigitalDifferentialAnalyser
• Scan-conversion line algorithm
• Calculating either or
• Sample the line at unit intervals in one coordinate
• Determine corresponding integer values nearest the line path for the other
coordinate.
• DDA algorithm works even if points are reversed
Δx Δy
DDA

13. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
(1) Compute
(2)If then
else
Δx = x2 − x1
Δy = y2 − y1
abs(Δx) > abs(Δy)
steps = abs(Δx)
steps = abs(Δy)
(3)
(4)
(5)repeat for all steps
xinc =
Δx
steps
yinc =
Δy
steps

14. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
LimitationofDDA
• Time Consuming
• Floating Point Numbers
• Roundoff Error

15. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in

16. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Bresenham's Line Algorithm

17. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Accurate and efficient raster line-generating
algorithm
• Using only incremental integer calculations
• Sampling at unit intervals
• Decide possible pixel positions closer to
the line path at each sample step.
• Weather to plot pixel at position
in the next step
• Testing the sign of an integer parameter,
whose value is proportional to the
difference between the separations of the
two pixel positions from the actual line
path.
x
(11,11)or(11,12)
10
11
12
13
14
10 11 12 13
.
. .
.

18. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
du
dl
Decisión parameter
y = mx + c
y = m(xk + 1) + c
dl = y − yk = [m(xk + 1) + c] − yk
du = yk+1 − y = yk+1 − [m(xk + 1) + c]
if(dl − du < 0) − > yk
if(dl − du > 0) − > yk+1
xk
yk+1
yk
xk+1
y

19. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
dl − du = [[m(xk + 1) + c] − yk] − [yk+1 − [m(xk + 1) + c]]
dl − du = m(xk + 1) + c − yk − yk+1 + m(xk + 1) − c
dl − du = 2m(xk + 1) − 2yk + 2c − 1
m =
Δy
Δx
dl − du =
2Δy
Δx
(xk + 1) − 2yk + 2c − 1
dl − du = Δx[
2Δy
Δx
(xk + 1) − 2yk + 2c − 1]
Δx(dl − du) = 2Δy(xk + 1) − 2Δxyk + 2Δxc − Δx
pk = Δx(dl − du) = 2Δyxk + 2Δy − 2Δxyk + 2Δxc − Δx
pk = Δx(dl − du) = 2Δyxk − 2Δxyk

20. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
pk = 2Δyxk − 2Δxyk
pknext = 2Δyxknext − 2Δxyknext
pknext − pk = 2Δyxknext − 2Δxyknext − 2Δyxk + 2Δxyk
pknext − pk = 2Δy(xknext − xk) − 2Δx(yknext − yk)
pknext − pk = 2Δy(xknext − xk) − 2Δx(yknext − yk)
if(pknext − pk) < 0
pknext = pk + 2Δy(xknext − xk) − 2Δx(yknext − yk)
pknext = pk + 2Δy
if(pknext − pk) > = 0
pknext = pk + 2Δy(xknext − xk) − 2Δx(yknext − yk)
pknext = pk + 2Δy − 2Δx

21. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
InitialDecisionParameter
p0 = 2Δyx1 + 2Δy − 2Δxy1 + 2Δxc − Δx
c = y −
Δy
Δx
x1
p0 = 2Δyx1 + 2Δy − 2Δxy1 + 2Δx[y −
Δy
Δx
x1] − Δx
p0 = 2Δyx1 + 2Δy − 2Δxy1 + 2Δxy1 − 2Δyx1 − Δx
p0 = 2Δy − Δx

22. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Algorithm breshanmsLine( )
______________________________________
x1, y1, x2, y2
x = x1
y = y1
dx = x2 − x1
dy = y2 − y1
p = 2dy − dx
while(x < = x2)
putpixel(x, y)
x + +
if(p < 0)
p = p + 2dy
else
p = 2dy − 2dx + p
y + +

23. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
x Y P
9 18 3
10 19 1
11 20 -1
12 20 7
13 21 5
14 22 3
x = x1
y = y1
dx = x2 − x1
dy = y2 − y1
p = 2dx − dy
while(x < = x2)
putpixel(x, y)
x + +
if(p < 0)
p = p + 2dy
else
p = 2dy − dx + p
y + +
(9,18) (14,22)
dx = 14 − 9 = 5
dy = 22 − 18 = 4
p = (2 * 4) − 5 = 3
pnext = 3 + (2 * 4) − (2 * 5) = 1

24. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Yaxis
15
16
17
18
19
20
21
22
23
24
25
X axis
6 7 8 9 10 11 12 13 14 15 16

25. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
x Y P
20 10
21 11 6
22 12 2
23 12 -2
24 13 14
25 14 10
26 15 6
27 16 2
28 16 -2
29 17 14
30 18 10
dx= 10
dy= 8
p=6
2dy = 16
2dy-2dx=-4

26. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
x Y P
2 3
3 4 7
4 5 11
5 6 15
6 7 19
dx= 3
dy= 5
p=7
2dy = 10
2dy-2dx=4
(2,3) and (5,8)

27. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Circle

28. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Frequently used component in
image and graph
• Properties of Circles
• A Circle is defined as the set of
points that are all at a given
distance r from a centre position
•
• Distance Relationship
Center(xc, yc)
(x − xc)2
+ (y − yc)2
= r2
• Scan conversion
• Calculate the position of points
on a circle circumference by
stepping along the x axis in unit
step
•
• Calculate
• Not a best method
• Lot of computation
• Unequal spacing
(xc − r)to(xc + r)
y = yc ± (r2
− (xc − x)2
r
x
y
(xc, yc)

29. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Calculate points along the circular boundary using polar
coordinates r and
• Circle equation in parametric polar form
•
•
• Generated with these equations using a fixed angular step size
• The step size chosen for 8 depends on the application and the
display device
• Step size at
• Plots pixel positions that are approximately one unit apart.
θ
x = xc + r ⋅ cosθ
y = yc + r ⋅ sinθ
1
r
UniformSpacing

30. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Symmetry of Circle
• Shape of the circle is similar in each quadrant
• Generate the circle section in the second quadrant
• The section of circle is symmetric across y axis
• Above section of circle to generate III and IV quadrant
• One step further
• Symmetry between octant
• A one-eighth circle sector is mapped to seven circle points
in the other octants of the plane
• Cartesian equation - multiplications and square root
• Parametric equations - Multiplications & Trigonometric
• Incremental calculation of decision parameters
xy
Computations
.
(x, y)
.
(x, − y)
.
(y, x)
.(y, − x)
.(−y, − x)
.
(−x, − y)
.(−y, x)
.
(−x, y)

31. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Midpoint Circle Algorithm

32. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Center
•
•
• For any pixel
•
•
(0,0)
x2
+ y2
= r2
x2
+ y2
− r2
= 0
(x1, y1)
x2
1 + y2
1 − r2
= ?
res = ?
• If
• Point Lies on the Circle
• If
• Point Lies inside the Circle
• If
• Point Lies outside the Circle
(res = 0)
(res < 0)
(res > 0)
CircleEquation

33. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
.
. (x1, y1)
r
.
< r
. > r
(x2, y2)
(x3, y3)
•
•
• Point Lies on the Circle
•
• Point Lies inside the Circle
•
• Point Lies outside the Circle
x2
+ y2
− r2
= 0
x2
1 + y2
1 = r2
x2
1 + y2
1 < r2
x2
1 + y2
1 > r2
(0,0)

34. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
.
.
.
.
.
.
.
.
.
.
. South Pixel
South East Pixel
. Starting Pixel (0, r)
Point − > (xk, yk)
(xk + 1,yk)
(xk + 1,yk − 1)
(xk, yk) (xk + 1,yk)
(xk + 1,yk − 1)
X always increases
Midpoint ( Decision Parameter )
.
If Mid Point is inside the circle P < 0
Select South Pixel
If Mid Point is Outside the circle P > 0
Select South East Pixel
(xk + 1,yk)
(xk + 1,yk − 1)

35. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Midpoint
midpoint = [
xk+1 + xk+1
2
,
yk + yk−1
2
]
midpoint = [
xk + 1 + xk + 1
2
,
yk + yk − 1
2
]
midpoint = [xk + 1,yk −
1
2
]
midpoint = [xk+1, yk− 1
2
]
midpoint = [xm, ym]
(xk + 1,yk)
(xk + 1,yk − 1)
.
.

36. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Decision Parameter
midpoint = [xm, ym]
(pk)
pk = x2
m + y2
m − r2
pk = x2
k+1 + y2
k− 1
2
− r2
pk+1 = (xk+1 + 1)2
+ (yk+1 −
1
2
)2
− r2
pk+1 − pk = (xk+1 + 1)2
+ (yk+1 −
1
2
)2
− r2
− x2
k+1 − y2
k−1
2
+ r2
pk+1 − pk = (xk + 1 + 1)2
+ (yk+1 −
1
2
)2
− (xk + 1)2
− y2
k−1
2
pk+1 − pk = (xk + 2)2
+ (yk+1 −
1
2
)2
− (xk + 1)2
− y2
k−1
2

37. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
•
•
•
•
•
•
•
•
pk+1 − pk = (xk + 2)2
+ (yk+1 −
1
2
)2
− (xk + 1)2
− y2
k−1
2
pk+1 − pk = (x2
k + 4 + 4xk) + (y2
k+1 +
1
4
− yk+1) − (x2
k + 1 + 2xk) − (y2
k +
1
4
− yk)
pk+1 − pk = x2
k + 4 + 4xk + y2
k+1 +
1
4
− yk+1 − x2
k − 1 − 2xk − y2
k −
1
4
+ yk
pk+1 − pk = 2xk + 3 + y2
k+1 − yk+1 − y2
k + yk
pk+1 = pk + 2xk + 3 + y2
k+1 − y2
k − yk+1 + yk
if(pk > 0)
yk+1 = ?
yk+1 = yk − 1

38. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
pk+1 = pk + 2xk + 3 + y2
k+1 − y2
k − yk+1 + yk
if(pk < 0)
yk+1 = yk
pk+1 = pk + 2xk + 3 + y2
k − y2
k − yk + yk
pk+1 = pk + 2xk + 3
if(pk > 0)
yk+1 = yk − 1
pk+1 = pk + 2xk + 3 + (yk − 1)2
− y2
k − (yk − 1) + yk
pk+1 = pk + 2xk + 3 + y2
k − 2yk + 1 − y2
k − yk + 1 + yk
pk+1 = pk + 2xk − 2yk + 5

39. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Initial Point
p0 = ?
(0,r)
p0 = x2
m + y2
m − r2
p0 = (xk + 1)2
+ (yk −
1
2
)2
− r2
p0 = (0 + 1)2
+ (r −
1
2
)2
− r2
p0 = (0 + 1)2
+ (r2
− r +
1
4
) − r2
p0 =
5
4
− r

40. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
midpointCircle(r)
1. or
2. and project on
remaining Octant
3. Loop
p0 =
5
4
− r p0 = 1 − r
plot(xi, yi)
if(pi < 0)then
pi+1 = pi + 2xi + 3
yi+1 = yi
xi+1 = xi + 1
project on remaining
Octant
Repeat until
if(pi > 0)then
pi+1 = pi + 2(xi − yi) + 5
yi+1 = yi − 1
xi+1 = xi + 1
plot(xi+1, yi+1)
(x > = y)

41. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
(1) or
(2)
(3)
(4) while
p =
5
4
− r p = 1 − r
x = 0,y = r
plot(x, y)
(x > = y)
if(p < 0)then
p = p + 2x + 3
if(p > 0)then
p = p + 2 × (x − y) + 5
y = y − 1
x = x + 1
plot(x, y)
midPointCircle(xc, yc, r)
p x Y
0 10
-9 1 10
-6 2 10
-1 3 10
6 4 9
-3 5 9
8 6 8
5 7 7
midPointCircle(0,0,10)
p = 1 − 10 = − 9
p < 0
p = (−9) + 2 × 0 + 3 = − 6

42. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
(1) or
(2)
(3)
(4) while
p =
5
4
− r p = 1 − r
x = 0,y = r
plot(x, y)
(x > = y)
if(p < 0)then
p = p + 2x + 3
if(p > 0)then
p = p + 2(x − y) + 5
y = y − 1
x = x + 1
plot(x, y)
midPointCircle(xc, yc, r)
p x Y
0 5
-4 1 5
-1 2 5
4 3 4
-3 4 4
midPointCircle(0,0,5)
p = 1 − 5 = − 4
p < 0
p = (−9) + 2 * 0 + 3 = − 6

43. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
p x Y
0 10
-9 1 10
-6 2 10
-1 3 10
6 4 9
-3 5 9
8 6 8
5 7 7
midPointCircle(3,4,10)
p = 1 − 10 = − 9
p < 0
p = (−9) + 2 * 0 + 3 = − 6
x Y
3 14
4 14
5 14
6 14
7 13
8 13
9 12
10 11

44. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
p x Y
0 8
-7 1 8
-4 2 8
1 3 7
-6 4 7
3 5 6
midPointCircle(5,5,8)
x Y
5 13
6 13
7 13
8 12
9 12
10 11

45. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Ellipse
x2
r2
x
+
y2
r2
y
= 1
(0,0) rx
ry
⋅
R1
R2
Sample x point
Sample y point
Slope = -1
Slope < -1
Slope > -1

46. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• An ellipse is an elongated circle
• An ellipse is defined as the set of
points such that the sum of the
distances from two fixed positions
(foci) is the same for all points
•
•
•
•
P = (x, y)
d1 : Distance from the First foci
d2 : Distance from the Second foci
d1 + d2 = constant

47. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Focal Coordinates
•
•
•
• squaring this equation, isolating the remaining radical
•
• Evaluation of the coefficients
• Focal Coordinates
• Dimensions of the major and minor axes
• Ellipse equations are greatly simplified if the major and minor axes are oriented to
align with the coordinate axes.
F1 = (x1, y1)
F2 = (x2, y2)
(x − x1) + (y − y1) + (x − x2) + (y − y2) = constant
Ax2
+ By2
+ Cxy + Dx + Ey + F = 0

48. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Ellipse in standard position
• Major and Minor axes oriented
parallel to the x and y axes
•
•
• The equation of the ellipse in terms
of the centre coordinates
and parameters and
•
• Center
•
• Using polar coordinates
•
•
• Symmetry considerations can be
used to further reduce computation
• Symmetric between quadrants
rx semi major axis
ry semi minor axis
(xc, yc)
rx ry
(
x − xc
rx
)2
+ (
y − yc
ry
)2
= 1
(0,0)
(
x
rx
)2
+ (
y
ry
)2
= 1
x = xc + rx cosθ
y = yc + ry sinθ
calculate pixel positions along the elliptical arc throughout one quadrant,
then we obtain positions in the remaining three quadrants by symmetry

49. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Mid Point Ellipse

50. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Similar to display raster circle
• Input
• Ellipse in standard position centred
on the origin
• Translate and Rotate to orient the
Major and Minor axes
• The midpoint ellipse method is
applied throughout the first
quadrant in two parts.
• Region I
• Region II
•rx, ry, (xc, yc)
(Slope of curve < − 1)
(Slope of curve > − 1)
ry
rx
Slope < − 1
Slope > − 1

51. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Region I
• Unit steps in the direction
• Slope of the curve less than -1
•
• Start position
• Step clockwise along the elliptical
path in the first quadrant
• Shifting from unit steps in to unit
steps in when the slope becomes
less than
• Region II
• Unit steps in the y direction
• Slop greater than -1
•
• Start position
• Step counter clockwise along the
elliptical path in the first quadrant
• Shifting from unit steps in to unit
steps in when the slope becomes
greater than
• Parallel processing
x
|slope| < 1
(0,ry)
x
y
−1
|slope| > 1
(rx,0)
y
x
−1
Calculate pixel positions in the two regions simultaneously

52. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Sequential Implementation
• Start Point
• Step along the ellipse path in
clockwise order throughout the
first quadrant.
•
• For any point
•
•
• Decision Parameter
•
• Point lies inside the ellipse
•
• Point lies on the ellipse
•
• Point lies outside the ellipse
• At each position, the next pixel
along the ellipse path is selected
• Sign of the ellipse function
evaluated at the midpoint between
the two candidate pixels.
(0,ry)
r2
y x2
+ r2
x y2
− r2
x r2
y = 0
(xk, yk)
r2
y x2
k + r2
x y2
k − r2
x r2
y = ?
res = ?
if(res < 0)
if(res = 0)
if(res > 0)

53. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Starting Point
• Take unit steps in the direction
• Continue till the boundary between
region 1 and region 2
• Then take unit steps in the
direction over the remainder of the
curve in the first quadrant.
• At each step, test the value of the
slope of the curve.
•
•
• On the boundary
•
•
• Out of region I
•
(0,ry)
x
y
d
dx
(r2
y x2
+ r2
x y2
− r2
x r2
y ) = 0
dy
dx
= −
2r2
y x
2r2
x y
dy/dx = − 1
2r2
y x = 2r2
x y
2r2
y x > = 2r2
x y
Slope

54. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
[Region I]
• Previous position
• Sampling position
• Midpoint between the two
candidate pixels
•
• Evaluating the decision parameterat
midpoint
•
•
•
• Mid point is inside the ellipse
•
•
• Mid point is on the ellipse
•
•
• Mid point is outside the ellipse
•
(xk, yk)
xk+1
(xk + 1,yk) (xk + 1,yk − 1)
Mid Point (xk + 1,yk −
1
2
)
p1k = r2
y (xk + 1)2
+ r2
x (yk −
1
2
)2
− r2
x r2
y
if (p1k < 0)
(xk + 1,yk)
if (p1k = 0)
(xk + 1,yk − 1)
if (p1k > 0)
(xk + 1,yk − 1)

55. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
p1k = r2
y (xk + 1)2
+ r2
x (yk −
1
2
)2
− r2
x r2
y
p1k+1 = r2
y (xk+1 + 1)2
+ r2
x (yk+1 −
1
2
)2
− r2
x r2
y
p1k+1 = r2
y (xk + 1 + 1)2
+ r2
x (yk+1 −
1
2
)2
− r2
x r2
y
p1k+1 − p1k = r2
y (xk + 1 + 1)2
+ r2
x (yk+1 −
1
2
)2
− r2
x r2
y − r2
y (xk + 1)2
− r2
x (yk −
1
2
)2
+ r2
x r2
y
p1k+1 − p1k = r2
y ((xk + 1)2
+ 1 + 2(xk + 1) − r2
y (x2
k + 1 + 2xk) + r2
x (y2
k+1 +
1
4
− yk+1) − r2
x (y2
k +
1
4
− yk)
p1k+1 − p1k = r2
y (x2
k + 1 + 2xk + 1 + 2xk + 2 − x2
k − 1 − 2xk) + r2
x (y2
k+1 +
1
4
− yk+1 − y2
k −
1
4
+ yk)
p1k+1 − p1k = r2
y {2(xk + 1) + 1} + r2
x (y2
k+1 − y2
k − yk+1 + yk)
p1k+1 = p1k + r2
y {2(xk + 1) + 1} + r2
x (y2
k+1 − y2
k − yk+1 + yk)

56. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
•
•
•
•
•
•
•
p1k+1 = p1k + r2
y {2(xk + 1) + 1} + r2
x (y2
k+1 − y2
k − yk+1 + yk)
if (p1k < 0)
yk+1 = yk
p1k+1 = p1k + 2r2
y xk+1 + 1 + r2
y
if (p1k > = 0)
yk+1 = yk − 1
p1k+1 = p1k + 2r2
y xk+1 + 1 + r2
y − 2r2
x yk+1

57. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
• Initial Point
•
•
•
•
•
•
(0,ry)
p1k = r2
y (xk + 1)2
+ r2
x (yk −
1
2
)2
− r2
x r2
y
p10 = r2
y (0 + 1)2
+ r2
x (ry −
1
2
)2
− r2
x r2
y
p10 = r2
y + r2
x (r2
y +
1
4
− ry) − r2
x r2
y
p10 = r2
y + r2
x r2
y +
r2
x
4
− r2
x ry − r2
x r2
y
p10 = r2
y +
r2
x
4
− r2
x ry
⋅
⋅(0,ry)
(0,0)
Ist Quadrant

58. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
(0,0) rx
ry
⋅
Sample x point
Sample y point
Slope = -1
Slope < -1
Slope > -1
⋅⋅⋅
⋅⋅⋅(xk + 1,yk − 1)(xk, yk − 1)
(xk + 1,yk)
(xk + 1,yk − 1)

59. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
[Region II]
• Previous position
• Sampling position
• Midpoint between the two
candidate pixels
•
• Evaluating the decision parameter
at midpoint
•
•
•
• Mid point is inside the ellipse
•
•
• Mid point is on the ellipse
•
•
• Mid point is outside the ellipse
•
(xk, yk)
yk+1
(xk, yk − 1) (xk + 1,yk − 1)
Mid Point (xk +
1
2
, yk − 1)
p2k = r2
y (xk +
1
2
)2
+ r2
x (yk − 1)2
− r2
x r2
y
if (p2k < 0)
(xk + 1,yk − 1)
if (p2k = 0)
(xk, yk − 1)
if (p2k > 0)
(xk, yk − 1)

60. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
p2k = r2
y (xk +
1
2
)2
+ r2
x (yk − 1)2
− r2
x r2
y
p2k+1 = r2
y (xk+1 +
1
2
)2
+ r2
x (yk+1 − 1)2
− r2
x r2
y
p2k+1 = r2
y (xk+1 +
1
2
)2
+ r2
x (yk − 1 − 1)2
− r2
x r2
y
p2k+1 − p2k = r2
y (xk+1 +
1
2
)2
+ r2
x (yk − 1 − 1)2
− r2
x r2
y − r2
y (xk +
1
2
)2
− r2
x (yk − 1)2
− r2
x r2
y
p2k+1 − p2k = r2
y (x2
k+1 +
1
4
+ xk+1) − r2
y (x2
k +
1
4
+ xk) + r2
x ((yk − 1)2
+ 1 − 2(yk − 1)) − r2
x (y2
k + 1 + 2yk)
p2k+1 − p2k = r2
y (x2
k+1 +
1
4
+ xk+1 − x2
k −
1
4
− xk) + r2
x (y2
k + 1 − 2yk + 1 − 2yk + 2 − y2
k − 1 + 2yk)
p2k+1 − p2k = r2
y (x2
k+1 + xk+1 − x2
k − xk) + r2
x (1 − 2(yk − 1))
p2k+1 = p2k + r2
y (x2
k+1 + xk+1 − x2
k − xk) + r2
x (1 − 2yk+1)

61. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
•
•
•
•
•
•
•
p2k+1 = p2k + r2
y (x2
k+1 + xk+1 − x2
k − xk) + r2
x (1 − 2yk+1)
if (p2k > 0)
xk+1 = xk
p2k+1 = p2k − 2r2
x yk+1 + r2
x
if (p2k < = 0)
xk+1 = xk + 1
p2k+1 = p2k + 2r2
y xk+1 − 2r2
x yk+1 + r2
x

62. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
InitialPoint
• Last Point of [Region I]
• Let
•
•
(x, y)
p2k = r2
y (xk +
1
2
)2
+ r2
x (yk − 1)2
− r2
x r2
y
p20 = r2
y (x +
1
2
)2
+ r2
x (y − 1)2
− r2
x r2
y

63. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
Algorithm

64. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
midPointEllipse
(1)
(2)
(3) ,
(4) while
(5)
(6)
(rx, ry, xc, yc)
(x0, y0) = (0,ry)
p10 = r2
y +
r2
x
4
− r2
x ry
dx = 2r2
y x dy = 2r2
x y
(dx < = dy)
plot (x, y)
if (p1 < 0)
x = x + 1
dx = 2r2
y x
p1 = p1 + 2r2
y x + r2
y
else
x = x + 1
y = y − 1
dx = 2r2
y x
dy = 2r2
x y
p1 = p1 + 2r2
y x − 2r2
x y + r2
y
p20 = r2
y (x +
1
2
)2
+ r2
x (y − 1)2
− r2
x r2
y
while (y > 0)
plot (x, y)
if (p2 > 0)
y = y − 1
dy = 2r2
x y
p2 = p2 − 2r2
x y + r2
x
else
x = x + 1
y = y − 1
dx = 2r2
y x
dy = 2r2
x y
p2 = p2 + 2r2
y x − 2r2
x y + r2
x

65. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
(x_k,y_k) P (x_k+1, y_k+1) dx dy
(0,6) -332 (1,6) 72 764
(1,6) -224 (2,6) 114 768
(2,6) -44 (3,6) 216 768
(3,6) 208 (4,5) 288 640
(4,5) -108 (5,5) 360 640
(5,5) 288 (6,4) 432 512
(6,4) 244 (7,3) 504 384
End of Region 1
midPointEllipse(rx : 8, ry : 6)
p10 = r2
y − r2
x ry +
1
4
r2
x
(6)2
− (8)2
⋅ 6 +
1
4
⋅ (8)2
= − 332
p1 < 0
dx = 2r2
y ⋅ xk+1
dx = 2(6)2
(1)
dy = 2r2
x ⋅ yk+1
dy = 2(8)2
(6)
p1 = p1 + 2r2
y x + r2
y
p1 = − 332 + 72 + (6)2
p14 = p13 + 2r2
y x − 2r2
x y + r2
y
p14 = 208 + (2 ⋅ (6)2
⋅ 4) − (2 ⋅ (8)2
⋅ 5) + (6)2
p1 > 0

66. Mr. Rupesh Mishra | rupeshmishra@sfit.ac.in
(x_k,y_k) P (x_k+1, y_k+1) dx dy
(6,4) 244 (7,3) 504 384
Start of Region II
(7,3) -23 (8,2) 576 256
(8,2) 361 (8,1) 576 128
(8,1) 297 (8,0) - -
midPointEllipse(rx : 8, ry : 6)
p20 = r2
y (x +
1
2
)2
+ r2
x (y − 1)2
− r2
x r2
y
(36)2
(7 +
1
2
)2
+ (6)2
(3 − 1)2
− (8)2
(6)2
p2 < 0
dx = 2r2
y ⋅ xk+1
dx = 2(6)2
(8)
dy = 2r2
x ⋅ yk+1
dy = 2(8)2
(2)
p2 = p2 + 2r2
y x − 2r2
x y + r2
x
Region II − Start Point (7,3)
p2 = − 23 + 576 − 256 + 64
p2 > 0
p2 = p2 − 2r2
x y + r2
x
p2 = 361 − 128 + 64