SRI LAXMI Bresenham Circle Drawing Algorithm,
Jan 08, 2016
SRI LAXMI
Bresenham
Circle Drawing Algorithm,
2of39
KSL
Contents
In today’s lecture we’ll have a look at:– Bresenham’s Circle drawing algorithm– Exercise using Bresenham’s algorithm
3of39
KSL
CIRCLE
The set of points that are all at a given distance ‘r’ from a center position (Xc,Yc).
4of39
KSL
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 solving the equation for y at unit x intervals using:
222 ryx
22 xry
5of39
KSL
A Simple Circle Drawing Algorithm (cont…)
20020 220 y
20120 221 y
20220 222 y
61920 2219 y
02020 2220 y
6of39
KSL
A Simple Circle Drawing Algorithm (cont…)
However, unsurprisingly this is not a brilliant solution!Firstly, the resulting circle has large gaps where the slope approaches the verticalSecondly, 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
7of39
KSL
Eight-Way Symmetry
The first thing we can notice to make our circle drawing algorithm 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
8of39
KSL
Mid-Point Circle Algorithm
Similarly to the case with lines, there is an incremental algorithm for drawing circles – the mid-point circle algorithm
In the mid-point circle algorithm we use eight-way symmetry so only ever calculate the points for the top right eighth of a circle, and then use symmetry to get the rest of the points
The mid-point circle algorithm was developed by Jack Bresenham, who we heard about earlier.
9of39
KSL
Mid-Point Circle Algorithm (cont…)
(xk+1, yk)
(xk+1, yk-1)
(xk, yk)
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?
10of39
KSL
Mid-Point Circle Algorithm (cont…)
Let’s re-jig the equation of the circle slightly to give us:
…(1)The equation evaluates as follows:
By evaluating this function at the midpoint between the candidate pixels we can make our decision
222),( ryxyxfcirc
,0
,0
,0
),( yxfcirc
boundary circle theinside is ),( if yx
boundary circle on the is ),( if yx
boundary circle theoutside is ),( if yx
11of39
KSL
Mid-Point Circle Algorithm (cont…)
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:mid point b/w 2 points (xk+1,Yk) and (Xk+1, Yk-1) is [xk+1, yk-1/2]
...2If pk < 0 the midpoint is inside the circle and the
pixel at yk is closer to the circle
Otherwise the midpoint is outside and yk-1 is closer
222 )21()1(
)21,1(
ryx
yxfp
kk
kkcirck
12of39
KSL
Mid-Point Circle Algorithm (cont…)
To ensure things are as efficient as possible we can do all of our calculations incrementallyFirst consider: ( since Xk+1 = Xk+1)
or:
where yk+1 is either yk or yk-1 depending on the
sign of pk
22
12
111
21]1)1[(
21,1
ryx
yxfp
kk
kkcirck
1)()()1(2 122
11 kkkkkkk yyyyxpp
13of39
KSL
the initial value of Pk is given by the circle function at the position (0,r) as,
Substituting k=0,Xk=0,Yk=r in the above function results in,
222 )21()1(
)21,1(
ryx
yxfp
kk
kkcirck
14of39
KSL
Mid-Point Circle Algorithm (cont…)
The first decision variable is given as:
if r is an integer, then Po can be rounded to P0= 1 – r.
Then if pk < 0 then the next decision variable is
given as:
If pk > 0 then the decision variable is:
r
rr
rfp circ
45
)21(1
)21,1(
22
0
12 11 kkk xpp
1212 11 kkkk yxpp
15of39
KSL
The Mid-Point Circle Algorithm
MID-POINT CIRCLE ALGORITHM
• Input radius r and circle centre (xc, yc), then set the coordinates for the first point on the circumference of a circle centred on the origin as:
• Calculate the initial value of the decision parameter as:
• Perform the test, Starting with k = 0 at each position xk, perform the following test.
(i) If pk < 0, the next point along the circle centred on (0, 0) is (xk+1, yk) and:
),0(),( 00 ryx
rp 45
0
12 11 kkk xpp
16of39
KSL
The Mid-Point Circle Algorithm (cont…)
(ii) If Pk >0 then the next point along the circle is (xk+1, yk-1) and:
where = 2Xk+2 and = 2Yk – 2
• Identify the symmetry points in the other seven octants
• Move (x, y) according to:
• Repeat steps 3 to 5 until x >= y
12 ky
111 212 kkkk yxpp
cxxx cyyy
12 kx
17of39
KSL
Mid-Point Circle Algorithm Example
To see the mid-point circle algorithm in action lets use it to draw a circle centred at (0,0) with radius 10
Determine the positions along the circle octant in the first quadrant from x=0 to x=y.
The intial value of the decision parameter is
P0 = 1-r = 1-10 = -9
For circle centred on the coordinate origin, the initial point is (X0,Y0)=(0,10) and initial increment terms for calculating the decision parameters are
2X0 =0 and 2Y0 = 20
18of39
KSL
K=0 and P0 = -9 (1,10) (pk<0)K=1, P1=P0+2Xk+1 => -9+2(1)+1 =-9+3=-6 (2,10)K=2, P2= P1 +2(2)+1 = -6+4+1 = -1 (3,10)K=3 P3=P2+2(3)+1 = -1+7= 6 (4,9)K=4 (Pk>0) P4= P3+2(4)+1-2(9) => 6+8+1-18 = -3 (5,9)K=5 p5= p4+2(5)+1 => -3+10+1 = 8 (6,8)K=6 p6=8+2(6)+1-2(8) => 8+12+1-16 = 5 (7,7)K=7 p7= 6 (8,6)K=8 p8=11 (9,5)K=9 p9 =20 (10,4)
111 212 kkkk yxpp
12 11 kkk xpp
19of39
KSL
Mid-Point Circle Algorithm Example (cont…)
9
7
6
5
4
3
2
1
0
8
976543210 8 10
10 k pk (xk+1,yk+1) 2xk+1 2yk+1
0
1
2
3
4
5
6
-9
-6
-1
6
-3
8
5
(1,10)
(2,10)
(3,10)
(4,9)
(5,9)
(6,8)
(7,7)
2
4
6
8
10
12
14
20
20
20
18
18
16
14
20of39
KSL
Mid-Point Circle Algorithm Exercise
Use the mid-point circle algorithm to draw the circle centred at (0,0) with radius 15
21of39
KSL
Mid-Point Circle Algorithm Example (cont…)
k pk (xk+1,yk+1) 2xk+1 2yk+1
0
1
2
3
4
5
6
7
8
9
10
11
12
9
76543210
8
976543210 8 10
10
131211 14
15
1312
14
11
16
1516
22of39
KSL
Mid-Point Circle Algorithm Summary
The key insights in the mid-point circle algorithm are:
– Eight-way symmetry can hugely reduce the work in drawing a circle
– Moving in unit steps along the x axis at each point along the circle’s edge we need to choose between two possible y coordinates
23of39
KSL
Mid-Point Circle Algorithm (cont…)
6
2 3 41
5
4
3
24of39
KSL
Mid-Point Circle Algorithm (cont…)
6
2 3 41
5
4
3
25of39
KSL
Mid-Point Circle Algorithm (cont…)
M
6
2 3 41
5
4
3
26of39
KSL
Mid-Point Circle Algorithm (cont…)
M
6
2 3 41
5
4
3
27of39
KSL
Mid-Point Circle Algorithm (cont…)
M
6
2 3 41
5
4
3
28of39
KSL
Blank Grid
29of39
KSL
Blank Grid
30of39
KSL
Blank Grid
9
7
6
5
4
3
2
1
0
8
976543210 8 10
10
31of39
KSL
Blank Grid