BRESSENHAM’S MIDPOINT CIRCLE

DRAWING ALGORITHM(EXPLAINED)

WHAT IS A CIRCLE?• A circle is the set of all points in a plane that are

at a given distance from a given point, the centre.• Equivalently it is the curve traced out by a point

that moves so that its distance from a given point is constant. • A circle may also be defined as a special ellipse in

which the two foci are coincident and the eccentricity is 0.

GENERATING CIRCLESCircle equation : f(x,y) = x2 + y2 – r2

Any point on the circumference of the circle will satisfy the above equation.Hence :

>0 Point is outside the boundary of the circlef(x,y) =0 Point is on the boundary of the circle

<= Point is inside the boundary of the circle

FIRST OCTANT• For a pixel at (xk, yk) the next pixel (xk +1, yk) or at (xk+1,

yk–1) is closer to the circle.

• The pixel point Pk for a corresponding point on the circle is given as:Pk = f(xk + 1, yk – ½) = (xk + 1)2 + yk - ½)2 – r2

• The next pixel point Pk+1 is given as:Pk+1 = f(xk+1 + 1, yk+1 - ½) = [(xk + 1) + 1]2 + (yk+1 - ½ )2 – r2

(xk +1, yk)(xk+1, yk–1)

Expanding Pk and Pk+1 and subtracting them, the eq. obtained is:

Pk+1 = Pk + 2(xk+1) + (y2k+1 – y2

k) – (yk+1 – yk) + 1

(where the next Y – coordinate (yk+1) is either yk or yk-1 depending on the sign of Pk.)

Since xk+1 denotes the next point along the X – axis (going rightwards)

xk+1 = xk + 1 2(xk+1) = 2xk + 2

Since yk+1 denotes the next point along the Y – axis (going downwards)

yk+1 = yk - 1 2(yk+1) = 2yk - 2

We’ve have to begin somewhere. So let the

starting position be (x0,y0) = (0, r)

Substituting these values in the circle

equation, P0 = f(1, r – ½) = - r

Since r is a integer, neglecting the small

value of ¼, we have P0 = 1 - r

(0, r)

(0, 0)

BRESSENHAM’S ALGORITHM

1. Input radius r and centre of the circle (xc, yc), and obtain the first point on the circumference of a circle centred on the origin as

(x0, y0) = (0, r)

2. Calculate the initial value of the decision parameter as P0 = 1 – r

where r is the radius of the circle

3. At each xk position, starting at k= 0, perform the following test:

If (Pk < 0), then the NEXT POINT along the circle centred on (0, 0) is (xk+1, yk) and Pk+1 = Pk + 2xk+1 +1

Pk+1 = Pk + 2xk + 3

Otherwise, the NEXT POINT along the circle is (xk+1, yk–1) and Pk+1 = Pk + 2xk+1 + 1 – 2yk+1

Pk+1 = Pk + 2xk – 2yk + 5

2xk+1 = 2xk + 2

2yk+1 = 2yk – 2

4. Determine the Symmetry Points on the other 7 Octants

5. Move each calculated pixel position (x,y) onto the circular path

centered at (xc,yc) and plot the coordinate values as

x = x + xc and y = y + yc

6. Repeat steps 3 to 5 until x >= y

FOR EXAMPLEConsider a circle that has to be drawn with its center at (0, 0) and a

radius of 10:

Initial point on the circumference of the circle is (0, r) = (0 ,10)

Initial pixel P0 = 1 – r = 1 – 10 = -9

Therefore, x = 0, y = 10, 2x = 0, 2y = 20, P0 = -9

Applying Bressenham’s Algorithm, we tabulate the data into a table

Pk Formula Pk Xk Yk

Initial Values -9 0 10

<0 -9 + 2(0) + 3 = -6 1 10

<0 -6 + 2(1) + 3 = -1 2 10

<0 -1 + 2(2) + 3 = 6 3 10

>0 6 + 2(3) -2(10) + 5 = -3 4 9

<0 -3 + 2(4) + 3 = 8 5 9

>0 8 + 2(5) – 2(9) + 5 = 5 6 8

>0 5 + 2(6) – 2(8) + 5 = 6 7 7

>0 6 + 2(7) – 2(7) + 5 = 11 8 6

REMEMBER:

xk+1 and yk+1 is derived based on

the sign of Pk

REMEMBER:

Pk+1 derived

from Pk, xk and yk

First calculate xk+1 and yk+1 values from Pk and then Pk+1 from Pk, xk, yk

1110

9876543210 1 2 3 4 5 6 7 8 9 10 11

• This is what the output

would look like.

• Only the first Octant is

shown.

• The Remaining Octants can

be derived symmetrically by

changing the signs.

FINAL RENDER

Prepared byMrinmoy Dalal13311A0506

S.N.I.S.T.