Scan conversion, pixel

1. Scan Conversion
A. Samal

2. Scan Conversion
• Last step in the graphics pipeline
• Efficiency is a central issue
• Common primitives
– Lines
– Polygons
– Circles
• Hardware implementations preferable

3. Lines
• Compute the coordinate of the pixels that lie on a
line on a 2D raster
• Defined by the coordinates of the end points
• Criteria
– Thin
– As close to ideal as possible
– 1 pixel wide
• If -1<= slope <= 1 exactly 1 pixel in one column
• Else exactly 1 pixel in one row
– Constant brightness, independent of the length &
orientation
– Should be drawn fast

4. Scan Conversion of Lines

5. Basic Incremental Algorithm
b
mx
y
line
a
of
Equation 

:
0
1
0
1
x
x
y
y
x
y
m






1
1 

 m
Assume
y
than
faster
increases
x
So,
)
, 0
y
(x
:
point
Starting 0
)
, 1
1 y
(x
:
point
Ending

6. Problem
• for each x plot pixel at closest y
–Problems for steep lines

7. Basic Incremental Algorithm
)
y
,round
(x i
i )
(
)
,y
(x i
i
m)
,y
(x i
i 
1
)
m
y
,round
(x i
i )
(
1 

Desired Line

8. Basic Incremental Algorithm
• Brute force algorithm
• Inefficient
• Each iteration
– Floating point multiply
– Floating point addition
– Rounding
{
)
,
,
,
( 1
1 y
x
y
x
Line 0
0
;
1
1
0
0
x
x
y
y
m



){
;
; 1 


 i
i
0
i x
x
x
x
for(x
);
( b
mx
Round
y i
i 

}
}
);
,
( i
i y
x
drawpixel

9. Basic Incremental Algorithm
• Eliminate the
multiplication
• Incremental
computation
• Digital Differential
Analyzer (DDA)
b
mx
y i
i 
 
 1
1
}
}
;
));
(
,
(
){
;
;
;
;
{
);
,
,
,
(
1
1
1
1
1
m
y
y
y
Round
x
drawpixel
x
x
x
x
for(x
y
y
x
x
y
y
m
y
x
y
x
Line
i
i
i
i
i
i
0
i
0
i
0
0
0
0










m
y
y
then
x
If i
i 


 1
1
b
x
x
m i 


 )
(
x
m
yi 


b
mx
y i
i 


10. Basic Incremental Algorithm
• If the magnitude of the slope is more than 1 change
the role of x and y in the algorithm
– y increases faster than x
• Simple but inefficient
– Still there are floating point operations
• Bresenham’s algorithm
– Uses only integer arithmetic
– Avoids round function
– Extended to drawing circles

11. Midpoint Line Algorithm
1
1 

 m
Assume
(NE)
NorthEast
(E)
East
choices
Two :
right
upper
:
y
x
left
lower
:
y
x
)
,
(
)
,
(
1
1
0
0
)
,
(
: p
p y
x
pixel
Previous
next?
be
should
pixel
Which
E
NE
M
Previous
Pixel
Current Pixel
Choices
Next Pixel
Choices
)
,
( p
p y
x
P 
Q

12. Midpoint Line Algorithm
y
x
pixel
Previous p
p )
,
(
:
• Compute the distance between
– Q and E
– Q and NE
• See which side of the line M lies
• M lies below the line
– Choose NE
• M lies above the line
– Choose E
• M lies on the line
– Choose either
1

 p
x
x
with
line
the
of
on
intersecti
of
Point
:
Q
E
NE
M
Previous
Pixel
Current Pixel
Choices
Next Pixe
Choices
)
,
( p
p y
x
P 
Q

13. Midpoint Line Algorithm
lies?
point
a
line
the
of
side
which
know
we
do
How
)
.(
);
(
);
( 0
1
0
1
0
1 x
x
B
c
x
x
dx
b
y
y
dy
a 









line
the
of
tion
Representa
Implicit
0
c
by
ax
y)
F(x, 



line
the
of
tion
Representa
Explicit
B
x
dx
dy
B
mx
y 



0
.
.
. 

 dx
B
y
dx
x
dy

14. Midpoint Line Algorithm
lies?
point
a
line
the
of
side
which
know
we
do
How
)
2
1
,
1
(
) 

 p
p x
x
F
F(M
of
sign
the
Check








line
the
below
points
for
line
the
above
points
for
line
the
on
points
for
y
x
F
0
0
0
)
,
(
line
the
below
or
above
lies
Q
if
test
To

15. Midpoint Line Algorithm
d
variable
decison
a
Define
NE
or
E
Choose
d 
 0
)
2
1
,
1 

 p
p y
F(x
d
E
Choose
Q)
M
d

 above
is
(
0
NE
Choose
Q)
below
is
(M
d

 0
E
NE
M
Previous
Pixel
Current Pixel
Choices
Next Pixel
Choices
)
,
( p
p y
x
P 
Q

16. Midpoint Line Algorithm
value
the
in
change
the
on
based
Add
directly
F(M)
of
value
the
compute
not
Do

• What happens at the next grid point?
– Depends on whether we choose N or NE
• If E is chosen
c
y
b
x
a
y
x
F
d p
p
p
p
new 






 )
2
1
(
)
2
(
)
2
1
,
2
(
c
y
b
x
a
d p
p
old 



 )
2
1
(
)
1
(
a
d
d old
new 
 dy
a
d
d old
new
E 




E)
choose
we
after
d
in
increment
E (
 E
N
E
M
Previo
us
Pixel
Current
Pixel
Choices
Next Pixel
Choices
)
,
( p
p y
x
P 
Q

17. Midpoint Line Algorithm
• If NE is chosen
c
y
b
x
a
y
x
F
d p
p
p
p
new 






 )
2
3
(
)
2
(
)
2
3
,
2
(
NE)
choose
we
after
d
in
increment
NE (

c
y
b
x
a
d p
p
old 



 )
2
1
(
)
1
(
b
a
d
d old
new 


dx
dy
b
a
d
d old
new
NE 






value
the
in
change
the
on
based
Add
directly
F(M)
of
value
the
compute
not
Do

E
N
E
M
Previo
us
Pixel
Current
Pixel
Choices
Next Pixel
Choices
)
,
( p
p y
x
P 
Q

18. Midpoint Line Algorithm
• Summary
– Choose between the two pixels (E/NE) based on
the sign of the decision variable d=F(M)
– Update the decision variable by adding the change
along that direction
– Repeat this process until the end point is reached

19. Midpoint Line Algorithm
2
dx
dy 

d
variable
decision
the
for
value
Initial
)
2
1
,
1 


 0
0
start y
F(x
F(M)
d
c
y
b
x
a 0
0 



 )
2
1
(
)
1
(
2
b
a
c
by
ax 0
0 




line
the
on
lies
)
,y
(x
since
b
a 0
0
2



20. Midpoint Line Algorithm
• Fractional arithmetic still problematic
• Should be avoided if possible
• Redefine F(x,y) by multiplying by 2
• Thus F(x,y) = 2(ax+by+c)
• Replace all constants and the decision variable
by 2
• Removes all multiplications; only additions left

21. Midpoint Line Algorithm
c)
by
(ax
F(x,y) 

 2
dx
dy
b
a
dstart 


 2
2
)
(
*
2 dx
dy
NE 


dy
E .
2



22. Midpoint Line Algorithm
MidpointLine(x0,y0,x1,y1,color)
{
int dx,dy,dE,dNE,d,x,y;
dx=x1-x0; dy=y1-y0;
d=2*dy-dx;
dE=2*dy; dNE=2*(dy-dx);
x=x0;y=y0;
DrawPixel(x,y,color);
for(x=x0;x<x1; x++) {
if (d<=0) d+=dE ;
else {d+=dNE; y++;}
DrawPixel(x,y,color);
}

23. Bresenham’s Algorithm
An example
4 5 6 7 8 9
7
8
9
10
11
12
(x0 , y0 )  (5,8)
(x4 , y4 )  (9,11)
x  4
y  3
incrE  2y  6
incrNE  2(y  x)  2
d0
 2y  x  2
 first choice is NE

24. Bresenham’s Algorithm
An example
4 5 6 7 8 9
7
8
9
10
11
12
d1  d0  incrNE  2  2  0
second choice is E

25. Bresenham’s Algorithm
An example
4 5 6 7 8 9
7
8
9
10
11
12
d2  d1  incrE  0  6  6
third choice is NE