The document outlines various shape drawing algorithms, including methods for drawing straight lines, circles, and ellipses, primarily focusing on Bresenham's algorithm, the midpoint circle algorithm, and techniques for filling polygons. It discusses the mathematical foundations and computational methods used in raster systems, emphasizing the importance of accuracy and efficiency in rendering shapes. Additionally, it covers polygon filling techniques and the organization of data for hardware implementation in graphics systems.

1. March 2010 1
Shape Drawing Algorithms
Shmuel Wimer
Bar Ilan Univ., Engineering Faculty

2. March 2010 2
Display
Window
maxy
miny
minx
maxx
World –coordinate limits
y
x0
1
2
3
4
5
0 1 2 3 4 5
Screen –coordinates
(4,2) pixel
coordinate

3. March 2010 3
We’d like to display straight line on the screen.
On raster systems, lines are plotted with pixels which have vertical
and horizontal finite resolution.
The outcome is a jaggy line.
In analog displays, such as vector-scan, a smooth line can be
displayed, as line equation is translated into deflection voltages.
Line Drawing Algorithms

4. March 2010 4
straight line equation: y mx b= +
0x
0y
endx
endy
end 0
0 0
end 0
;
y y
m b y mx
x x
−
= = −
−
Straight line drawing uses increments or and finds the
counterpart. If | | 1 , and if | | 1 .
x y
m y m x m x y m
δ δ
δ δ δ δ≤ = > =
In raster systems lines are plotted with pixels and step sizes in the
horizontal and vertical directions must be sampled at nearest pixel
positions.

5. March 2010 5
1If 1, sampling at 1 obtains: and pixel
coordinate is obtained by rounding to nearest interger.
k km x y y mδ +≤ = = +
samples the line in unit
steps in one coordinate and the nearest integer value is
calculated in the other coordinate.
Digital differntial analyzer (DDA)
1Similarly, if 1, 1 and 1 .k km y x x mδ +> = = +
DDA avoids mutltiplication, but involves floating point
addition. A round-off error is accumulated, drifting pixels
position from the original line.

6. March 2010 6
is accurate and efficient, using only
incremental integer calculations. Assume 1. We step to
successive columns ( 1).
m
xδ
<
=
Bresenham's algorithm
0 0
1
We start at pixel ( ,y ). Assume that pixel ( ,y ) is decided
to be displayed. Next pixel is either ( +1,y ) or ( +1,y ).
k k
k k k k
x x
x x +
1Decision on or is
made according to which
one is closer to the real
of the line.
k ky y
y
+
ky
y
1ky +
1kx +

7. March 2010 7
( )is sampled at next pixel. In real line: 1 .kx y m x b= + +
( )lowerDistance from lower pixel: 1 .k k kd y y m x b y= − = + + −
( ) ( )From upper pixel: 1 1 1 .upper k k kd y y y m x b= + − = + − + −
( )
lower upper
Compare distances. Set pixel
according to smaller distance.
2 1 2 2 1.k k
d d
m x y b
− =
+ − + −
ky
y
1ky +
1kx +
lowerd
upperd

8. March 2010 8
. Multiplying by 0 doesn't cahnge the sign
of difference. The decision parameter is defined by:k
m y x x
p
= ∆ ∆ ∆ >
( ) ( )lower upper
constant
2 2 2 2 1 .k k kp x d d x y y x y x b= ∆ − = ∆ − ∆ + ∆ + ∆ −
1442443
If 0 we plot the lower pixel; otherwise upper is drawn.kp <

9. March 2010 9
( ) ( )
( )
1 1 1
1
2 2
0 if 0
2 2 2 .
2 if 0
k k k k k k
k
k k
k
p p y x x x y y
p
y x y y y
x p
+ + +
+
− = ∆ − − ∆ − =
>
∆ − ∆ − = ∆ − 
∆ ≤
( )1 1 1
constant
The decision can be computed recursively as follows:
2 2 2 2 1 .
k
k k k
p
p y x x y y x b+ + += ∆ × − ∆ × + ∆ + ∆ −
1442443

10. March 2010 10
Circle Drawing Algorithms
( ),x y
θ
cx
cy
r
( ) ( )2 2 2
c cx x y y r− + − =
We could use above equation to
step along the x axis from xc- r
to xc+ r and calculate y positions
of pixels.
( )
22
c cy y r x x= ± − −

11. March 2010 11
This requires considerable amount of computation (square root)
The drawing appearance is poor since the vertical spacing
between drawn pixels is uneven.
Appearance can be improved by changing the role of x and y
whenever the absolute value of the slope of circle is crossing 1.
But this requires more computation.
Polar coordinates can be used with
uniform stepping of θ. This spaces
pixels evenly along circumference.
cos
sin
c
c
x x r
y y r
θ
θ
= +
= +
Time consuming due to trigonometric computations.

12. March 2010 12
We can use large angular steps and
connect the points by straight lines.
For smoother boundary we can use
angular steps of 1/r which plots
pixels approximately one pixel
apart of each other (r∙1/r =1).
O
45
( ),x y
( ),x y−
( ),x y−
( ),x y− −
( ),y x− ( ),y x
( ),y x− − ( ),y x−
Any of the methods can use
symmetry for efficiency.
All methods involve
expensive computations.

13. March 2010 13
( )
determines the mid point between
two pixels. If it falls inside the circle, the lower pixel is plotted,
otherwise the upper does. Assume w.l.o.g that circle is centered
at 0,0 .
Midpoint Circle Algorithm
( ), can be added later.c cx y
( ) 2 2 2
circ ,f x y x y r= + −
( )circ
0 inside
, 0 boundary
0 outside
f x y
<

=
>
ky
1ky −
2kx +1kx +kx
midpoint
2 2 2
0x y r+ − =

14. March 2010 14
Successive decision parameters are obtained by incremental
calculations.
( ) ( ) ( )
2 2 2
circ 1, 1 2 1 1 2k k k k kp f x y x y r= + − = + + − −
( )
( ) ( )
Assuming that last plotted pixel was , , we need decide
whether next plotted is 1, or 1, 1 .
k k
k k k k
x y
x y x y+ + −
( )
( )
If 0 midpoint is inside the circle and pixel 1, is
closer to circle and plotted. Otherwise 1, 1 does.
k k k
k k
p x y
x y
< +
+ −
( ) ( ) ( )
( ) ( ) ( )
2 2 2
1 circ 1 1 1
2 2
1 1
1, 1 2 2 1 2
2 1 1
k k k k k
k k k k k k
p f x y x y r
p x y y y y
+ + + +
+ +
= + − = + + − −
= + + + − − − +

15. March 2010 15
1 1
1
1
1 1
The value of used in the calculation of is if 0, or
1 if 0. So the increments of is the following:
2 1 if 0
2 1 2 if 0
k k k k
k k
k k
k k
k k k
y p y p
y p p
x p
p p p
x y p
+ +
+
+
+ +
<
− ≥
+ <
∆ = − = 
+ − ≥
( )
1 1 1
1
Evaluations of 2 and 2 are also incremental 2 2 2,
2 2 2. At start position 0, these values are 0 and 2 ,
respectively.
k k k k
k k
x y x x
y y r r
+ + +
+
= +
= −
( ) ( ) ( )
( )
0 0 0 0 circ
2 2
0
is obtained from , 0, , 1, 1 2
1 1 2 5 4 . updates are made by integral
increments and decrements, so if is integer, we set 1 .
p x y r p f r
r r r p
r p r
= = −
= + − − = −
= −

16. March 2010 16
y
x
1 2 3 4 5 6 7 8 90 10
1
2
3
8
9
0
10
5
6
7
4
k pk (xk+1,yk+1) 2xk+1 2yk+1
0 - 9 (1,10) 2 20
1 - 6 (2,10) 4 20
2 - 1 (3,10) 6 20
3 6 (4,9) 8 18
4 - 3 (5,9) 10 18
5 8 (6,8) 12 16
6 5 (7,7) 14 14
0 0 0
Example: Draw a circle with 10.
Initialization: 1 10 9 ; 2 0 , 2 20
r
p x y
=
= − = − = =

17. March 2010 17cx
cy
yr
xr
Ellipse Drawing Algorithms
1F
2F
1d
2d
( ),P x y=
( ) ( )
( ) ( )
1 2 1 2
2 2
1 1
2 2
2 2
2 2
and are foci,
0
F F d d const
x x y y
x x y y const
Ax By Cxy Dx Ey F
+ =
− + − +
− + − =
+ + + + + =
( ) ( )
2 22 2
Simplified if ellipse axes are parallel
to coordinate axes.
1
cos ; sin
c x c y
c x c y
x x r y y r
x x r y y rθ θ
− + − =
= + = +

18. March 2010 18
Unlike circles, symmetry exists only in quadrants.
Midpoint Ellipse Algorithm is working similar to circle with a few
adaptations.
( ) 2 2 2 2 2 2
ellipse , y x x yf x y r x r y r r= + −
( )ellipse
0 inside
, 0 boundary
0 outside
f x y
<

=
>
slope 1= −
yr
xr
region 1
region 2
( ) ( )2 2
2 2
2 2
Move from region 1 to region 2 when 2 2
y x
y x
dy dx r x r y
r x r y
= −
≥

19. March 2010 19
ky
1ky −
2kx +1kx +kx
midpoint
2 2 2 2 2 2
0y x x yr x r y r r+ − =
( )
( ) ( )
ellipse
2 22 2 2 2
1 1, 1 2
1 1 2
k k k
y k x k x y
p f x y
r x r y r r
= + −
= + + − −
( )
( ) ( )
1 ellipse 1 1
2 22 2 2 2
1
1 1, 1 2
2 1 2
k k k
y k x k x y
p f x y
r x r y r r
+ + +
+
= + −
= + + − −
( )
2 2
1
1 2 2 2
1 1
2 2 2
2 if 1 0
1 1
2 2 if 1 0
These terms can be calculated by addition only.
At 0, 2 0 and 2 2
y k y k
k k
y k y x k k
y y x x y
r x r p
p p p
r x r r y p
r r x r y r r
+
+
+ +
 + <
∆ = − = 
+ − ≥
= =

20. March 2010 20
( ) ( )
( )
0 0 0
2 2 2
0 ellipse
1 is obtained from , 0, ,
1 1, 1 2 4.
y
y y x y x
p x y r
p f r r r r r
=
= − = − +
( ) ( )0 0
2 is handled similarly. We could start at the end point
of region1 and proceed clockwise, but it is better to walk
counterclockwise, starting at , ,0 .
k
x
p
x y r=

21. March 2010 21
Polygon Fill Areas
Convex polygons are
easy to fill
Concave polygons are
more difficult to fill
Some graphics packages support convex polygon filling only.
How should polygon split into convex components?
We’d like to avoid any trigonometric or division operations.
We’ll use vector product a x b for convex / concave test.

22. March 2010 22
( )1 2 0z
× >E E
( )2 3 0z
× >E E
( )3 4 0z
× <E E
( )4 5 0z
× >E E
( )5 6 0z
× >E E
( )6 1 0z
× >E E
1E1V 2V
3V
5V
6V
4V
2E
3E
4E
5E
6E
x y z
i j ix iy iz
jx jy jz
E E E
E E E
× =
u u u
E E
Polygon vertices are counterclockwise
oriented.
The cross product of successive edges
is a perpendicular vector.
Z component is positive for convex
angle and negative for concave.
Only multiplications are involved.

23. March 2010 23
( )
1Define oriented edge by:
Apply cross product:
, ,
k k k
i j iy jz iz jy iz jx ix jz ix jy iy jxE E E E E E E E E E E E
+= −
× = − − −
E V V
E E
A convex corner following concave one can be eliminated by
splitting a triangle and reducing vertex degree, until all concave
corners are eliminated.
The remaining convex polygon can be further split into triangles
by a successive traversal of its vertices.

24. March 2010 24
The problem turns to filling of triangles
Found
concave
followed by
convex
Tear off
triangle
Tear off
triangle

25. March 2010 25
Inside-Outside Tests
?
Odd-even rule: Draw “infinite” ray and
count the number of crossings. Odd is
inside, even outside.
Avoid passing through vertices.
?
Nonzero winding-number rule:
Draw “infinite” directed ray. Traverse
edges and check for crossing
direction (e.g. by cross product). Add
±1 accordingly. If it sums to zero
point is outside, otherwise it is
inside.

26. March 2010 26
Polygon Tables
Objects are described by sets of polygon surface facets.
1E
1V
2V
3V
5V
4V
2E
3E
4E
5E
6E
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
VERTEX
TABLE
: , ,
: , ,
: , ,
: , ,
: , ,
V x y z
V x y z
V x y z
V x y z
V x y z
1 1 2
2 2 3
3 3 1
4 3 4
5 4 5
6 5 1
EDGE
TABLE
: ,
: ,
: ,
: ,
: ,
: ,
E V V
E V V
E V V
E V V
E V V
E V V
1 1 2 3
2 3 4 5 6
SURFACE-FACET
TABLE
: , ,
: , , ,
S E E E
S E E E E

27. March 2010 27
• In hardware implementation these tables translate into
physical RW memories.
• For hardware implementations triangles are preferred,
since it makes SURFACE-FACET MEMORY
homogeneous.
• Data in memories needs to be organized to maximize
pipeline processing of facets:
– Vertices are stored such that vertex progression in VERTEX
MEMORY completes new triangle for every vertex.
– Vertex indices are their memory addresses.
• It is the SW application responsibility to triangulate
objects and load HW memories in the right order.

28. March 2010 28
Plane Equations
• Graphics system processes input data through several
steps:
– Transformation of modeling through viewing pipeline.
– Identification of visible surfaces.
– Rendering of surface facets.
• Information about the spatial orientation of surfaces is
needed
– Plane equations are in order.
– How to calculate those efficiently w/o divisions?

29. March 2010 29
General plane equation: 0
, , is any point on the plane. , , , are constansts
describing spatial propertis of the plane.
Ax By Cz D
x y z A B C D
+ + + =
( ) ( ) ( )
( ) ( ) ( )
1 1 1 2 2 2 3 3 3, , , , , and , , non co-linear ordered
counterclockwise, 1, 1,2,3.k k k
x y z x y z x y z
A D x B D y C D z k+ + = − =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
1 2 3 2 3 1 3 1 2
1 2 3 2 3 1 3 1 2
1 2 3 2 3 1 3 1 2
1 2 3 3 2 2 3 1 1 3 3 1 2 2 1
A y z z y z z y z z
B z x x z x x z x x
C x y y x y y x y y
D x y z y z x y z y z x y z y z
= − + − + −
= − + − + −
= − + − + −
= − − + − + −  

30. March 2010 30
Front and Back Polygon Faces
• Polygon surfaces enclose objects.
– Need to know the outward an inward sides of polygon.
• Need to find the side of a polygon visible by an observer.
( )
( )
( )
A point , , not on the plane satisfies: 0.
If 0 the point , , is behind the plane.
If 0 the point , , is in front of the plane.
x y z Ax By Cz D
Ax By Cz D x y z
Ax By Cz D x y z
+ + + ≠
+ + + <
+ + + >
( ) ( ) ( )2 1 3 1The normal is , , , yielding
plane coeficient. Normal is used for coloring and highlighting.
A B C V V V V= = − × −N

31. March 2010 31
Scan-Line Polygon-Fill Algorithms
• This is most time consuming operation in computer
graphics.
– Use only integer arithmetic with minimum operations
• Polygons are filled with scan-line
– The sweep is vertical
– A scan-line is the pixels of same y coordinates
Pixels between odd-even
intersections with border
are inside.
y
x
1 2 3 4

32. March 2010 32
11 12 Scan-line y
21 1
Scan-line y’
A problem may occur at vertices. Scan-line y’ is okay, but y is not.
The remedy is to traverse edges prior to filling and detect the values
of three consecutive of two successive edges. If they are monotonic
then one of the edges is shortened by one pixel at corner

33. March 2010 33
Polygon filling takes advantage of y increments by one unit.
( ),k kx y
( )1 1,k kx y+ + Scan-line 1ky +
Scan-line ky
How to calculate efficiently the intersection of an edge with scan line?
( )
( )
1
1 1
1
1
, 1k k
k k k k
k k
y y
m y y x x
x x m
+
+ +
+
−
= = + ⇒ = +
−
How to avoid division?

34. March 2010 34
1Calculate interceps by: , k k
y x
x m x x
x y
+
∆ ∆
= = +
∆ ∆
Initialize a counter to 0. Then increment the counter by each time
is incremented by 1 (moving to new scal-line). Once counter meets
or exceeds , is increased by 1 and counter is decreased b
x
y
y x
∆
∆ y .y∆
This is equivalent to maintaining integer and fractional parts of
intercepts and incrementing the fractional parts until it reaches
the next integer value.

35. March 2010 35
Instead of truncation we can round to nearest pixel intersection
by comparing the counter to 2. This can be done with integer
arithmetics by adding 2 , comparing to , and then decreasing
counter
x
y
x y
∆
∆ ∆
by 2 .y∆
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
0 0 0 0
0 0 0 0 0 0 0
0 0 0
Example: 7 3. , : ,0 : , 1,3 : ,
2,6 : , 3,9 3,2 : 1, 4,5 : 1,
5,8 5,1 : 2,...
y x y count x y x y x
y x y y x y x
y y x
∆ ∆ = = +
+ + → + + + +
+ → + +
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
0 0 0 0
0 0 0 0 0 0
0 0
Example: 7 3. , : ,0 : , 1,6 : ,
2,12 2, 2 : 1, 3,4 : 1, 4,10
4, 4 : 2,...
y x y count x y x y x
y y x y x y
y x
∆ ∆ = = +
+ → + − + + + + →
+ − +

36. March 2010 36
Antialiasing
• Graphics primitives generated by raster algorithms have
jagged (stair-step) appearance, called aliasing.
– This is due to under sampling.
– Antialiasing compensates for this.
• Sampling rate could increase by higher raster resolution.
– Expensive, increase of frame buffer (memory).
– More processing (time, hardware).
– Jagging phenomena is not canceled, but only improved.
• Raster systems capable of gradual intensity of pixels can
support antialiasing by modifying pixel intensity near
boundary of primitives.

37. March 2010 37
Antialiasing by Supersampling
• Treat the screen as if it has finer grid than actual.
• Modify real pixel intensity according to the number of
covered sub-pixels
There are 4 possible intensities,
depending on the count of covered
pixels at every 3x3 superpixel
(1,1) is covered by 3 subpixels
(2,2) and ((3,2) are covered by 2
subpixels
(2,1) and ((3,3) are covered by 1
subpixels

38. March 2010 38
• Line was treated as zero width.
• In case of finite width we count the number of subpixels
inside the polygon representing the line.
Useful also for mixing foreground
and background colors in a pixel.
For instance, if foreground is red
and background is blue:
color
5 red 4 blue
Pixel
9
× + ×
=

39. March 2010 39
1
1
1
1
2
2
2 24
Central pixel is most important and has
a weight of 1/4. Others has a weight of
1/8 and 1/16.
Antialiasing by subpixel Weighting Masks
Antialiasing compensates for line-intensity differences
Diagonal lines are less instance than horizontal
and vertical ones.
Supersampling with higher weights for subpixels
close to diagonal will intensify diagonal lines

40. March 2010 40
Antialiasing Area Boundaries
Pixel intensities are adjusted along
the boundary of area.
Subdivide area of pixels and double scan-
line. Count how many subpixels are inside.
Scan-line 1
Scan-line 2
75% pixel intensity

41. March 2010 41
Percentage of pixel area within fill area by midpoint
method (Pitteway & Watkinson).
ky
0.5ky +
1ky +
kx 1kx +
y mx b= +
( )
1
Assume that , has been
plotted. The next nearest
line at is either at or at
1.
k k
k k
k
x y
x y
y
+
+
( ) ( )mid
if 0 select 1
1 0.5 ,
if 0 select
or by adding the term 1
k
k k
k
y
y y m x b y
y
m
> +
− = + + − +    <
−
( ) ( ) ( )
if 1 select 1
1 0.5 1
if 1 select
k
k k
k
m y
p m x b y m
m y
> − +
= + + − + + −    < −

42. March 2010 42
0.5kx +0.5kx − kx
0.5ky +
0.5ky −
ky
( )0.5ky m x b= − + ( )0.5ky m x b= + +
boundary line
overlap area
( )The overlap area of a pixel rectangle at , with the interior
of polygon is: area 0.5 .
We can intensify the pixel for antialiasing accordingly.
k k
k k
x y
mx b y p= + − + =