Mid-Point Cirle Drawing Algorithm

NEHA KAURAV
HIMANSHI GUPTA
CLASS:
BCA 3rd SEM.

• A circle is all points in the same plane that lie at an equal distance from a
center point. The circle is only composed of the points on the border.
• The distance between the midpoint and the circle border is called the
radius. A line segment that has the endpoints on the circle and passes
through the midpoint is called the diameter. The diameter is twice the size
of the radius. A line segment that has its endpoints on the circular border
but does not pass through the midpoint is called a chord.

Circle Algorithms
4
• Use 8-fold symmetry and only compute pixel
positions for the 45° sector.
45°
(x, y)
(y, x)
(-x, y)
(y, -x)
(x, -y)(-x, -y)
(-y, x)
(-y, -x)

• Circle function:
Fcircle(x,y)=x2+y2-r2
Fcircle(x,y)>0,then (x,y) lies outside the circle.
Fcircle(x,y)<0,then (x,y) lies inside the circle
Fcircle(x,y)=0,then (x,y) is on the circle boundary.
.
Fk<0
Yk+1=yk
Next pixel=(xk+1,yk)
Midpoint inside the circle
.
Fk>=0
Yk+1=yk-1
Next pixel=(xk+1,yk-1)
Midpoint above the
circle

• The decision parameter is the circle at the midpoint
between the pixels ykand yk – 1.
f=fcircle(xk+1,yk-1/2)
=(xk+1)2+(yk-1/2)2-r2
We know that
Xk+1=xk+1,
Fk=f(xk+1,yk-1/2)
Fk=(xk+1)2+(yk-1/2)2-r2 (i)
Fk+1=f(xk+2,yk+1-1/2)
Fk+1=(xk+2)2+(yk+1-1/2)2-r2 (ii)
Now subtracting eqn (ii) by (i)
Fk+1-fk=2(xk+1)+(y2
k+1-y2
k)-(yk+1-yk)+1

So,
If fk<0: yk+1=yk
Fk+1=fk+2xk+1+1
If fk>=0: yk+1=yk-1
Fk+1=fk+2xk+1-2yk+1+1
For the initial points, (x0,y0)=(0,r)
F0=fcircle(1,r-1/2)
=1+(r-1/2)2-r
=5/4-r
=1-r

The Algorithm1. Initial values:- point(0,r)
x0 = 0
y0 = r
2. Initial decision parameter
3. At each xi position, starting at i = 0, perform the
following test: if pi < 0, the next point is (xi + 1, yi) and
pi+1 = pi + 2xi+1 + 1
If pi ≥ 0, the next point is (xi+1, yi-1) and
pi+1 = pi + 2xi+1 + 1 – 2yi+1
where 2xi+1 = 2xi + 2 and 2yi+1 = 2yi – 2
4. Determine symmetry points in the other octants
5. Move pixel positions (x,y) onto the circular path
centered on (xc, yc) and plot the coordinates: x = x + xc,
y = y + yc
6. Repeat 3 – 5 until x >= y
2 2 51 1
0 2 2 4(1, ) 1 ( )circlep f r r r r       
8

void plotpoints(int xcenter,int ycenter,int x,int y)
{
setpixel(xcenter+x, ycenter+y);
setpixel(xcenter+x, ycenter-y);
setpixel(xcenter-x, ycenter+y);
setpixel(xcenter-x, ycenter-y);
setpixel(xcenter+y, ycenter+x);
setpixel(xcenter+y, ycenter-x);
setpixel(xcenter-y, ycenter+x);
setpixel(xcenter-y, ycenter-x);
}
void midpoint(int xcenter,int ycenter,int radius)
{
int x = 0, y = radius;
int f = 1 – radius;
plotpoints(xcenter,ycenter,x,y);
while (x<y) {
x++;
if (f<0) f += 2*x + 1;
else {
y--;
f+= 2*(x-y) + 1;
}
plotpoints(xcenter,ycenter,x,y);
}
}

Example
10    
9  
8 
7 
6 
5 
4 
3 
2 
1 
0 
0 1 2 3 4 5 6 7 8 9 10
10
i pi xi+1,
yi+1
2xi+
1
2yi+
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)
r = 10
p0 = 1 – r = -9 (if r is integer round p0 = 5/4 – r to integer)
Initial point (x0, y0) = (0, 10)

Mid-Point Cirle Drawing Algorithm

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