lecture 1.pptx

Computer Graphics (CSE-457)
Lecture-1
1
Muhammad Ibrahim Khan, PhD
Professor, Dept. of CSE
Chittagong University of Engineering & Technology (CUET)

2
Pixel
In digital image processing, a pixel, or pel or picture element is a
physical point in an image.
It is the smallest addressable element in a display device; so it is t
he smallest controllable element of a picture represented on the sc
reen.
The address of a pixel corresponds to its physical coordinates.
Each pixel is a sample of an original image; more samples typicall
y provide more accurate representations of the original. So picture
quality is directly proportional to the picture resolution.

3
Pixel…

4
Scan Conversion
Rasterisation (or rasterization) is the task of taking an
image described in a vector graphics format (shapes) and
converting it into a raster image (pixels or dots) for outp
ut on a video display or printer, or for storage in a bitma
p file format.
This is also known as scan conversion.

5
Scan Conversion of a Point
• A point (x, y) within an image area, scan
converted to a pixel at location (x’, y’).
• x’ = Floor(x) and y’ = Floor(y).
• All points satisfying
and are mapped to
pixel (x’, y’).
• Point P1(1.7, 0.8) is represented by pixel
(1, 0) and points
are both represented by pixel (2, 1).
1




 x
x
x
1




 y
y
y
(2.8,1.9)
and
)
3
.
1
,
2
.
2
( 3
2 P
P

6
Scan Conversion of a Point…
• Another approach is to align the integer values
in the co-ordinate system for (x, y) with the
pixel co-ordinates.
• Here x’ = Floor(x + 0.5) and y’ = Floor(y + 0.5)
• Points both are now represented
by pixel (2, 1) and by pixel (3, 2).
2
1 and P
P
3
P

7
Line drawing algorithm
Need algorithm to figure out which intermediate pixels a
re on line path
Pixel (x, y) values constrained to integer values
Actual computed intermediate line values may be floats
Rounding may be required. Computed point
(10.48, 20.51) rounded to (10, 21)
Rounded pixel value is off actual line path (jaggy!!)
Sloped lines end up having jaggies
Vertical, horizontal lines, no jaggies

8
Line Drawing Algorithm
Line: (3,2) -> (9,6)
? Which intermediate
pixels to turn on?

9
Line Drawing Algorithm…
Slope-intercept line equation
y = mx + b
Given two end points (x0,y0), (x1, y1), how to compute
m and b?
(x0,y0)
(x1,y1)
dx
dy
0
1
0
1
x
x
y
y
dx
dy
m



 0
*
0 x
m
y
b 


10
Line Drawing Algorithm…
Numerical example of finding slope m:
(Ax, Ay) = (23, 41), (Bx, By) = (125, 96)
5392
.
0
102
55
23
125
41
96








Ax
Bx
Ay
By
m

11
Digital Differential Analyzer (DDA): Line Drawing
Algorithm
(x0,y0)
(x1,y1)
dx
dy
Walk through the line, starting at (x0,y0)
Constrain x, y increments to values in [0,1] range
Case a: x is incrementing faster (m < 1)
Step in x=1 increments, compute and round y
Case b: y is incrementing faster (m > 1)
Step in y=1 increments, compute and round x

12
DDA Line Drawing Algorithm (Case a: m < 1)
(x0, y0)
x = x0 + 1 y = y0 + 1 * m
Illuminate pixel (x, round(y))
x = x + 1 y = y + 1 * m
…
Until x == x1
(x1,y1)
x = x0 y = y0
m
y
y k
k 

1

13
DDA Line Drawing Algorithm (Case b: m > 1)
y = y0 + 1 x = x0 + 1 * 1/m
Illuminate pixel (round(x), y)
y = y + 1 x = x + 1 /m
…
Until y == y1
x = x0 y = y0
(x1,y1)
(x0,y0)
m
x
x k
k
1
1 



14
DDA Line Drawing Algorithm Pseudocode
compute m;
if m < 1:
{
float y = y0; // initial value
for(int x = x0;x <= x1; x++, y += m)
setPixel(x, round(y));
}
else // m > 1
{
float x = x0; // initial value
for(int y = y0;y <= y1; y++, x += 1/m)
setPixel(round(x), y);
}
Note: setPixel(x, y) writes current color into pixel in column x and row y in fram
e buffer

15
DDA Example (Case a: m < 1)
Suppose we want to dra
w a line starting at pixel
(2,3) and ending at pixel
(12,8).
What are the values of t
he variables x and y at e
ach timestep?
What are the pixels colo
red, according to the D
DA algorithm?
t x y R(x) R(y)
0 2 3 2 3
1 3 3.5 3 4
2 4 4 4 4
3 5 4.5 5 5
4 6 5 6 5
5 7 5.5 7 6
6 8 6 8 6
7 9 6.5 9 7
8 10 7 10 7
9 11 7.5 11 8
10 12 8 12 8

16
DDA Algorithm Drawbacks
DDA is the simplest line drawing algorithm
Not very efficient
Floating point operations and rounding operations are expensiv
e.

17
The Bresenham Line Algorithm
The Bresenham algorithm is another incremental scan co
nversion algorithm
The big advantage of this algorithm is that it uses only in
teger calculations: integer addition, subtraction and multi
plication by 2, which can be accomplished by a simple a
rithmetic shift operation.

18
The Big Idea
Move across the x axis in unit intervals and at each st
ep choose between two different y coordinates
2 3 4 5
2
4
3
5
For example, from
position (2, 3) we have to
choose between (3, 3) and
(3, 4)
We would like the point
that is closer to the
original line
(xk, yk)
(xk+1, yk)
(xk+1, yk+1)

19
Deriving The Bresenham Line Algorithm
The y coordinate on the mathematical line at xk+1
is:
At sample position xk+1
the vertical separations
from the mathematical line
are labelled dupper and
dlower
b
x
m
y k 

 )
1
(
y
yk
yk+1
xk+1
dlower
dupper

20
Deriving The Bresenham Line Algorithm…
So, dupper and dlower are given as follows:
and:
We can use these to make a simple decision about which
pixel is closer to the mathematical line
y
y
d k
upper 

 )
1
(
b
x
m
y k
k 



 )
1
(
1
k
lower y
y
d 

k
k y
b
x
m 


 )
1
(

21
This simple decision is based on the difference between the
two pixel positions:
Let’s substitute m with ∆y/∆x where ∆x and
∆y are the differences between the end-points:
1
2
2
)
1
(
2 




 b
y
x
m
d
d k
k
upper
lower
)
1
2
2
)
1
(
2
(
)
( 








 b
y
x
x
y
x
d
d
x k
k
upper
lower
)
1
2
(
2
2
2 









 b
x
y
y
x
x
y k
k
c
y
x
x
y k
k 





 2
2

22
So, a decision parameter pk for the kth step along a li
ne is given by:
The sign of the decision parameter pk is the same as
that of dlower – dupper
If pk is negative, then we choose the lower pixel,
otherwise we choose the upper pixel
c
y
x
x
y
d
d
x
p
k
k
upper
lower
k










2
2
)
(

23
Remember coordinate changes occur along the x axis
in unit steps so we can do everything with integer cal
culations
At step k+1 the decision parameter is given as:
Subtracting pk from this we get:
c
y
x
x
y
p k
k
k 





 

 1
1
1 2
2
)
(
2
)
(
2 1
1
1 k
k
k
k
k
k y
y
x
x
x
y
p
p 





 



24
But, xk+1 is the same as xk+1 so:
where yk+1 - yk is either 0 or 1 depending on the sign
of pk
The first decision parameter p0 is evaluated at
(x0, y0) is given as:
)
(
2
2 1
1 k
k
k
k y
y
x
y
p
p 




 

x
y
p 


 2
0

25
The Bresenham Line Algorithm…
BRESENHAM’S LINE DRAWING ALGORITHM
(for |m| < 1.0)
1. Input the two line end-points, storing the left end-point in (x0, y0)
2. Plot the point (x0, y0)
3. Calculate the constants Δx, Δy, 2Δy, and (2Δy - 2Δx) and get
the first value for the decision parameter as:
4. At each xk along the line, starting at k = 0, perform the following
test. If pk < 0, the next point to plot is
(xk+1, yk) and:
x
y
p 


 2
0
y
p
p k
k 


 2
1

26
The Bresenham Line Algorithm…
Otherwise, the next point to plot is (xk+1, yk+1) and:
5. Repeat step 4 (Δx – 1) times
x
y
p
p k
k 




 2
2
1
• The algorithm and derivation above assumes slopes are less
than 1. for other slopes we need to adjust the algorithm
slightly

27
Bresenham’s Line Algorithm ( Example)
using Bresenham’s Line-Drawing Algorithm, Digitize the line wit
h endpoints (20,10) and (30,18).
y = 18 – 10 = 8,
x = 30 – 20 = 10
m = y / x = 0.8
2*y = 16
2*y – 2* x = -4
plot the first point (x0, y0) = (20, 10)
p0 = 2 * y – x = 2 * 8 – 10 = 6 , so the next point is (21, 11)

28
Example (cont.)
K Pk (xk +1, yk +1) K Pk (xk +1, yk +1)
0 6 (21,11) 5 6 (26,15)
1 2 (22,12) 6 2 (27,16)
2 -2 (23,12) 7 -2 (28,16)
3 14 (24,13) 8 14 (29,17)
4 10 (25,14) 9 10 (30,18)

29
Example (cont.)

30
Bresenham’s Line Algorithm (cont.)
Notice that bresenham’s algorithm works on lines with slope in ra
nge 0 < m < 1.
We draw from left to right.
To draw lines with slope > 1, interchange the roles of x and y dire
ctions.

31
Code (0 < slope < 1)
Bresenham ( int xA, yA, xB, yB) {
int d, dx, dy, xi, yi;
int incE, incNE;
dx = xB – xA;
dy = yB – yA;
incE = dy << 1; // Q
incNE = incE – dx << 1; // Q + R
d = incE – dx; // initial d = Q + R/2
xi = xA; yi = yA;
writePixel(xi, yi);
while(xi < xB) {
xi++;
if(d < 0) // choose E
d += incE;
else { // choose NE
d += incNE;
yi++;
}
writePixel(xi, yi);
}}

32
Bresenham Line Algorithm Summary
The Bresenham line algorithm has the following advanta
ges:
An fast incremental algorithm
Uses only integer calculations
Comparing this to the DDA algorithm, DDA has the follo
wing problems:
Accumulation of round-off errors can make the pixel at end line
drift away from what was intended
The rounding operations and floating point arithmetic involved
are time consuming

33
A Simple Circle Drawing Algorithm
The equation for a circle is:
where r is the radius of the circle
So, we can write a simple circle drawing algorithm by so
lving the equation for y at unit x intervals using:
2
2
2
r
y
x 

2
2
x
r
y 



34
A Simple Circle Drawing Algorithm (cont…)
20
0
20 2
2
0 


y
20
1
20 2
2
1 


y
20
2
20 2
2
2 


y
6
19
20 2
2
19 


y
0
20
20 2
2
20 


y

35
A Simple Circle Drawing Algorithm (cont…)
However, unsurprisingly this is not a brilliant solution!
Firstly, the resulting circle has large gaps where the slop
e approaches the vertical
Secondly, the calculations are not very efficient
The square (multiply) operations
The square root operation – try really hard to avoid these!
We need a more efficient, more accurate solution

36
Eight-Way Symmetry
The first thing we can notice to make our circle drawing al
gorithm more efficient is that circles centred at (0, 0) have
eight-way symmetry
(x, y)
(y, x)
(y, -x)
(x, -y)
(-x, -y)
(-y, -x)
(-y, x)
(-x, y)
2
R

37
Mid-Point Circle Algorithm
Similarly to the case with lines, there is an incremental alg
orithm for drawing circles – the mid-point circle algorithm
In the mid-point circle algorithm we use eight-way symmet
ry so only ever calculate the points for the top right eighth
of a circle, and then use symmetry to get the rest of the poi
nts

38
Mid-Point Circle Algorithm (cont…)
Assume that we have
just plotted point (xk, yk)
The next point is a
choice between (xk+1, yk)
and (xk+1, yk-1)
We would like to choose
the point that is nearest to
the actual circle
So how do we make this choice?
(0, r)

39
Let’s re-jig the equation of the circle slightly to give us:
The equation evaluates as follows:
By evaluating this function at the midpoint between the
candidate pixels we can make our decision
2
2
2
)
,
( r
y
x
y
x
fcirc 










,
0
,
0
,
0
)
,
( y
x
fcirc
boundary
circle
the
inside
is
)
,
(
if y
x
boundary
circle
on the
is
)
,
(
if y
x
boundary
circle
the
outside
is
)
,
(
if y
x

40
Assuming we have just plotted the pixel at (xk,yk) so we
need to choose between (xk+1,yk) and (xk+1,yk-1)
Our decision variable can be defined as:
If pk < 0 the midpoint is inside the circle and and the pixel
at yk is closer to the circle
Otherwise the midpoint is outside and yk-1 is closer
2
2
2
)
2
1
(
)
1
(
)
2
1
,
1
(
r
y
x
y
x
f
p
k
k
k
k
circ
k









41
To ensure things are as efficient as possible we can do al
l of our calculations incrementally
First consider:
or:
where yk+1 is either yk or yk-1 depending on the sign of
pk
 
  2
2
1
2
1
1
1
2
1
]
1
)
1
[(
2
1
,
1
r
y
x
y
x
f
p
k
k
k
k
circ
k













1
)
(
)
(
)
1
(
2 1
2
2
1
1 






 

 k
k
k
k
k
k
k y
y
y
y
x
p
p

42
The first decision variable is given as:
Then if pk < 0 then the next decision variable is given as
:
If pk > 0 then the decision variable is:
r
r
r
r
f
p circ








4
5
)
2
1
(
1
)
2
1
,
1
(
2
2
0
1
2 1
1 

 
 k
k
k x
p
p
1
2
1
2 1
1 



 
 k
k
k
k y
x
p
p

43
Mid-point Circle Algorithm - Steps
1. Input radius r and circle center (xc, yc ). set the first point (x0 , y0 ) = (0, r ).
2. Calculate the initial value of the decision parameter as p0 = 1 – r.
(p0 = 5 /4 – r ≅ 1 – r )
3. If pk < 0,
plot (xk + 1, yk ) and pk+1 = pk + 2xk + 1 + 1,
Otherwise,
plot (xk+ 1, yk – 1 ) and pk+1 = pk + 2xk+1 + 1 – 2yk+1,
where 2xk + 1 = 2xk + 2 and 2yk + 1 = 2yk – 2.

44
4. Determine symmetry points on the other seven octants.
5. Move each calculated pixel position (x, y) onto the circular path c
entered on (xc, yc) and plot the coordinate values: x = x + xc ,
y = y + yc
6. Repeat steps 3 though 5 until x  y.
7. For all points, add the center point (xc, yc )

45
Now we drew a part from circle, to draw a complete circle, we must
plot the other points.
We have (xc + x , yc + y), the other points are:
(xc - x , yc + y)
(xc + x , yc - y)
(xc - x , yc - y)
(xc + y , yc + x)
(xc - y , yc + x)
(xc + y , yc - x)
(xc - y , yc - x)

46
Mid-point circle algorithm (Example)
Given a circle radius r = 10, demonstrate the midpoint circle algorit
hm by determining positions along the circle octant in the first quad
rant from x = 0 to x = y.
Solution:
p0 =1 – r = – 9
Plot the initial point (x0, y0 ) = (0, 10),
2x0 = 0 and 2y0 =20.
Successive decision parameter values and positions along the circle
path are calculated using the midpoint method as appear in the next
table:

47
K Pk (xk+1, yk+1) 2 xk+1 2 yk+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

48

49
Mid-point Circle Algorithm – Example (2)
Given a circle radius r = 15, demonstrate the midpoint circle algor
ithm by determining positions along the circle octant in the first q
uadrant from x = 0 to x = y.
Solution:
p0 = 1 – r = – 14
plot the initial point (x0 , y0) = (0, 15),
2x0 = 0 and 2y0 = 30.
Successive decision parameter values and positions along the circl
e path are calculated using the midpoint method as:

50
K Pk (xk+1, yk+1) 2 xk+1 2 yk+1
0 – 14 (1, 15) 2 30
1 – 11 (2, 15) 4 30
2 – 6 (3, 15) 6 30
3 1 (4, 14) 8 28
4 – 18 (5, 14) 10 28

51
K Pk (xk+1, yk+1) 2 xk+1 2 yk+1
5 – 7 (6,14) 12 28
6 6 (7,13) 14 26
7 – 5 (8,13) 16 26
8 12 (9,12) 18 24
9 7 (10,11 ) 20 22
10 6 (11,10) 22 20

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