We assume that we want to scan-convert the line shown in the following fig, where 0<m<1.
We start with pixel P1(x1,y1) . Once
a pixel is chosen at any step, the
next pixel is either the one to its
right or the one to its right and up
due to the limit on m. The line is
best approximated by those pixels
that fall the least distance from its
true path between two considered
points.
The coordinates of the last chosen
pixel upon entering step i are (xi,yi). Our task is to choose the next one between the bottom
pixel S and top pixel T.
If S is chosen, we have
xi+1 = xi+1
and
yi+1 = yi
If T is chosen, we have
xi+1 = xi+1
and
yi+1 = yi +1
The actual y coordinates of the line at x = xi+1 is,
y = m xi+1+b = m(xi+1)+b ………………………..(i)
The distance from S to the actual line in the y direction is,
Prepared by Jesmin Akhter, Assistant professor, IIT, JU
Reference: Computer Graphics (Schaum's Outline Series)
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s = y- yi
The distance from T to the actual line in the y direction is,
t = (yi+1)-y
Considering the difference between two distance values :s-t,
If s-t < 0
s < t and the closest pixel is S.
If s-t > 0
s > t and the closest pixel is T.
We also choose T when
s-t = 0
The difference is –
s-t
= (y-yi)-[(yi+1)-y]
= y-yi-yi-1+y
= 2y-2yi-1
= 2(y-yi)-1
= 2 [ { m(xi + 1) + b } - yi) ] - 1 ……………. [from (i)]
= 2 m(xi + 1) + 2b - 2yi - 1
= 2 (∆y/∆x)(xi + 1) + 2b - 2yi – 1 [ Since, m = ∆y/∆x ]
Decision variable ,
di
= ∆x(s - t)
= ∆x { 2 (∆y/∆x)(xi + 1) + 2b - 2yi – 1}
= 2∆y(xi + 1) + ∆x (2b - 2yi – 1)
= 2∆yxi - 2∆xyi + 2∆y + 2b∆x - ∆x
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di
= 2∆yxi - 2∆xyi + c …………………..(ii)
Similarly, we can write the decision variable di +1 for next step i+1 as,
di+1 = 2∆yxi+1 - 2∆xyi+1 + c …………….(iii)
[ (iii) – (ii) ] =>
di+1 - di = 2∆y( xi+1- xi) - 2∆x( yi+1 – yi)
= 2∆y( xi + 1- xi) - 2∆x( yi+1 – yi) [ since xi+1 = xi + 1 ]
= 2∆ y- 2∆x( yi+1 – yi)
So, di+1 = di + 2∆y - 2∆x( yi+1 – yi)
If the chosen pixel is the top pixel T (meaning that di ≥ 0) then yi+1 = yi + 1 and so,
= di + 2∆y - 2∆x( yi + 1 – yi )
di+1
= di + 2∆y - 2∆x
= di + 2(∆y - ∆x)
On the other hand, if the chosen pixel is the bottom pixel S (meaning that di < 0) then yi+1 = yi
and so,
= di + 2∆y - 2∆x( yi – yi )
di+1
= di + 2∆y
= di + 2(∆y - ∆x)
Hence, we have
di+1 =
di + 2(∆y - ∆x)
di ≥ 0
di + 2∆y
di < 0
Finally, we calculate d1 ,the base case value for this recursive formula, from the original
definition of the decision variable di :
di
= ∆x [ 2m (x1 + 1) + 2b - 2y1 – 1]
= ∆x [ 2( mx1 + b - y1 ) + 2m – 1]
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= ∆x [ 2*0 + 2m – 1]
= 2(∆y / ∆x)* ∆x - ∆x
=2∆y - ∆x
In summary, Bresenham’s algorithm for scan-converting a line from p1(x1,y1) to p2(x2,y2) with x1
< x2 and 0 < m <1 can be stated as follows :
 Void Bresenham( )
 {
 Line 1: dx=x2-x1;
 Line 2: dy=y2-y1;
 Line 3: dT=2*(dy-dx);
 Line 4: dS=2*dy;
 Line 5: d=(2*dy)-dx;
 Line 6: putpixel(x1,y1);
 Line 7: while(x1<x2)
 Line 8: {
 Line 9: x1++;
 Line 10: if(d<0)
 Line 11: {
 Line 12: d=d+dS;
 Line 13: putpixel(x1,y1);
 Line 14: }
 Line 15: else
 Line 16: {
 Line 17: y1++;
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 Line 18: d=d+dT;
 Line 19: putpixel(x1,y1);
 Line 20: }
 Line 21: }
 Line 22: putpixel(x2,y2);
 Line 23: }
Scan-converting a circle using Bresenham’s
algorithm works as follows :
From fig, it can be noticed that, if points are
generated from 900 to 450 , each new point
closest to the true circle can be found by
taking either of two actions :
1. Move in the x direction one unit
or
2. Move in the x direction one unit
and move in the negative y direction one unit.
Assume that (xi,yi) are the coordinates of the last scan-converted pixel upon entering step i .
Our task is to choose the next one between the bottom pixel S and top pixel T.
Let, the distance from origin to pixel T squared minus the distance to the true circle squared =
D(T)
Then, let the distance from origin to pixel S squared minus the distance to the true circle
squared = D(S)
As the coordinates of
T = (xi + 1,yi) and
S = (xi + 1,yi - 1), the following expressions can be developed :
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D(T) = (xi + 1)2 + yi2 – r2
D(S) = (xi + 1)2 + (yi -1)2 – r2
Since D(T) will always be positive and D(S) will always be negative, a decision variable di may
be defined as follows :
di = D(T) + D(S)
=(xi + 1)2 + yi2 – r2 + (xi + 1)2 + (yi -1)2 – r2
=2(xi + 1)2 + yi2 +(yi -1)2 –2 r2
So,
di =2(xi + 1)2 + yi2 +(yi -1)2 –2 r2 …………………………………………..(i)
When di < 0 , we have |D(T)| < |D(S)| and pixel T is chosen.
When di ≥ 0 , we have |D(T)| ≥ |D(S)| and pixel S is chosen.
We can also write the decision variable di+1 for the next step i+1 :
di+1 = 2(xi+1 + 1)2 + yi+12 +(yi+1 -1)2 –2 r2 ………………………….(ii)
Hence, [ (ii) – (i) ] =>
di+1 - di = 2(xi+1 + 1)2 + yi+12 +(yi+1 -1)2 –2 r2 - 2(xi + 1)2 - yi2 -(yi -1)2 +2 r2
= 2(xi + 1 + 1)2 + yi+12 +(yi+1 -1)2 - 2(xi + 1)2 - yi2 - (yi -1)2
[ since xi+1 = xi + 1 ]
= 2[ (xi + 1)2 + 2(xi + 1) + 1 ] + yi+12 + yi+12 -2yi+1 +1-2(xi + 1)2 - yi2 - yi 2 +
2 yi -1
= 2(xi + 1)2 + 4(xi + 1) + 2 + 2 yi+12 -2yi+1 -2(xi + 1)2 - yi2 - yi 2 + 2 yi
= 4xi + 4 + 2 + 2 yi+12 -2yi+1 - 2yi 2 + 2 yi
= 4xi + 6 + 2( yi+12- yi 2 ) – 2(yi+1 - yi)
So,
di+1 = di + 4xi + 2( yi+12- yi 2 ) – 2(yi+1 - yi) + 6
If T is the chosen pixel (meaning that di < 0) then yi+1 = yi and so,
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di+1 = di + 4xi + 2( yi2- yi 2 ) – 2(yi - yi) + 6
= di + 4xi + 6
On the other hand, if the chosen pixel is the bottom pixel S (meaning that di ≥ 0) then yi+1 = yi 1 and so,
di+1 = di + 4xi + 2[( yi -1) 2- yi 2 ) – 2[(yi - 1- yi)] + 6
= di + 4xi + 2[( yi 2- 2 yi + 1- yi 2 ) +2 + 6
= di + 4xi -4yi + 2 + 2 + 6
= di + 4(xi -yi) + 10
Hence, we have,
di+1 =
di + 4(xi -yi) + 10
di ≥ 0
di + 4xi + 6
di < 0
Finally, we set (0,r) to be the starting pixel coordinates and compute the base case value d1 for
this recursive formula from the original definition of d i :
d1 =2(0 + 1)2 + r2 +(r -1)2 –2r2
= 2 + r2 + r2-2r +1 -2r2
= 3 – 2r
We can now summarize the algorithm for generating all the pixel coordinates in the 900 to 450
octant that are needed when scan-converting a circle of radius r.
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Line 1
Line 2
Line 3
Line 4
Line 5
:
:
:
:
:
x = 0;
y = r;
d = 3 – 2r;
while ( x ≤ y)
{
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Line 6 :
Line 7 :
Line 8 :
Line 9 :
Line 10 :
Line 11 :
Line 12 :
Line 13 :
Line 14 :
Line 15 :
setPixel(x,y);
if(d < 0)
d=d + 4x + 6;
else
{
d = d + 4 ( x – y ) + 10;
y- -;
}
x++;
}
Scan-converting a circle using Midpoint Circle algorithm works as follows :
Assume that (xi,yi) are the coordinates of the last
scan-converted pixel upon entering step i . Our task
is to choose the next one between the bottom pixel S
and top pixel T.
Our decision parameter is the circle function
evaluated at the midpoint between these two pixels.
Considering the coordinates of the point halfway
between pixel T and pixel S (xi +1, yi-1/2) known as
midpoint, we find our decision parameter :
pi = fcircle (xi +1, yi-1/2) = (xi +1)2 + (yi -1/2)2 – r2……………..(i)
If pi < 0 , this midpoint is inside the circle and we choose pixel T.
Otherwise, the mid position is outside or on the circle boundary, and we choose pixel S.
Similarly, the decision parameter for next step i+1 is
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Pi+1 = fcircle(xi+1+1, yi+1-1/2)
= [xi+1+1]2 + (yi+1 -1/2)2 –r2………………………………..(ii)
Hence, [ (ii) – (i) ] =>
pi+1 - pi = (xi+1 + 1)2 + (yi+1 -1/2)2 – r2 - (xi + 1)2 - (yi -1/2)2 + r2
= (xi + 1 + 1)2 + (yi+1 -1/2)2 - (xi + 1)2 - (yi -1/2)2 [ since xi+1 = xi + 1 ]
= (xi + 1)2 + 2(xi + 1) + 1 + yi+12 - yi+1 +1/4 - (xi + 1)2 - yi 2 + yi -1/4
= 2(xi + 1) + 1 + yi+12 - yi+1 - yi 2 + yi
= 2(xi + 1) + 1 + (yi+12- yi 2 ) – ( yi+1 - yi )
So,
pi+1 = pi + 2(xi + 1) + 1 + (yi+12- yi 2 ) – ( yi+1 - yi )
If T is the chosen pixel (meaning that pi < 0) then yi+1 = yi and so,
pi+1
= pi + 2(xi + 1) + 1 + (yi2- yi 2 ) – ( yi - yi )
= pi + 2(xi + 1) + 1
= pi + 2xi + 3
On the other hand, if the chosen pixel is the bottom pixel S (meaning that p i ≥ 0) then yi+1 = yi -1
and so,
pi+1 = pi + 2(xi + 1) + 1 + [(yi-1)2- yi 2 )] – (yi-1- yi )
= pi + 2(xi + 1) + 2+( yi 2 - 2 yi +1 - yi 2)
= pi + 2(xi + 1) + 2 - 2 yi +1
= pi + 2(xi + 1) +1 - 2 yi + 2
= pi + 2xi + 2 +1 – 2 yi + 2
= pi + 2(xi – yi) + 5
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Hence, we have,
pi+1 =
pi + 2xi + 3
pi < 0
pi + 2(xi – yi) + 5
pi ≥ 0
Finally, we compute the initial value for the decision parameter using the original definition of
pi and (0,r) :
Pi = (0 + 1) 2 + ( r - ½ ) 2 - r2
= 1+ r2 - r + ¼ - r2
= 5/4 - r
This is not really integer computation. However, when r is an integer, we can simply set,
P1 = 1- r.
We can now summarize the algorithm for generating all the pixel coordinates in the 90 0 to 450
octant :
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Line 1 : Int x=0,y=r,p=1-r;
Line 2 : While(x<=y)
Line 3 : {
Line 4 : setPixel(x,y);
Line 5 : If(p<0)
Line 6 : p=p+2x+3;
Line 7 : Else
Line 8 : {
Line 9 : P=p+2(x-y)+5;
Line 10 : y--;
Line 11 : }
Line 12 : x++;
Line 1 3: }