Bresenham's line drawing algorithm & Mid Point Circle algorithm
Computer Graphics
Team MembersGroup-I Assignment Topic : BRESENHARAM'S ALGORITHM (ELIPSE Drawing)
Group's representative: TANGUTURU SAI KRISHNA
S.No. BITS ID NAME Official Email ID Personal Email ID
1 2011HW69898TANGUTURU SAI KRISHNA [email protected] [email protected]
2 2011HW69900RAYAPU MOSES [email protected] [email protected]
3 2011HW69932 SHENBAGAMOORTHY [email protected] [email protected]
4 2011HW69913ANURUPA K C [email protected] [email protected]
5 2011HW69909ARUNJUNAISELVAM P [email protected] [email protected]
6 2011HW69569PRANOB JYOTI KALITA [email protected] [email protected]
7 2011HW69893TINNALURI V N PRASANTH [email protected] [email protected]
8 2011HW69904KONDALA SUMATHI [email protected] [email protected]
9 2011HW69896DASIKA KRISHNA [email protected] [email protected]
Lines
3
Analog devises, such as a random-scan display or a vector plotter, display a straight line smoothly from one endpoint to another. Linearly varying horizontal and vertical deflection voltages are generated that are proportional to the required changes in the x and y directions to produce the smooth line.
Digital devices display a straight line by plotting discrete coordinate points along the line path which are calculated from the equation of the line.
Screen locations are referenced with integer values, so plotted positions may only approximate actual line positions between two specific endpoints.A computed line position of (10.48, 20.51) will be converted to pixel position (10, 21). This rounding of coordinate values to integers causes lines to be displayed with a stairstep appearance (the “jaggies”).Particularly noticeable on systems with low resolution.To smooth raster lines, pixel intensities along the line paths must be adjusted.
4
Line Drawing Algorithms
Cartesian equation:y = mx + c
wherem – slopec – y-intercept
x
y
xx
yym
12
12
5
x1
y1
x2
y2
Slope
6
if |m| = 1 = 45°
45°45°
+ve -ve
°°
°
°
if |m| 1-45° < < 45°
if |m| 145° < < 90° or-90° < < -45°
x y
0 01 12 2
3 34 4
5 5
6 6
7 78 8
7
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8
y = xm = 1c = 0 y
x
|m| = 1
x y round(y)
0 1 11 1.5 2
2 2 23 2.5 3
4 3 35 3.5 4
6 4 47 4.5 5
8 5 5
8
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8
y = ½ x + 1m = ½c = 1 y
x
|m| 1
9
|m| 1
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8
y = 3x - 2m = 3c = -2 y
x
x y round(y)
0 -2 -21 1 12 4 43 7 74 10 10
5 13 13
6 16 16
7 19 19
8 22 22
outside
Bresenham Line AlgorithmA more efficient approach
Basis of the algorithm:
From start position decide A or B next
A
B
10
Start position
Bresenham Line Algorithm
11
For a given value of xone pixel lies at distance ti above the line, andone pixel lies at distance si below the line
True line
si
ti
Bresenham Line Algorithm
12
Decision parameter
di = (si - ti)
If di 0, then closest pixel is below true line (si smaller)
If di 0, then closest pixel is above true line (ti smaller)
We must calculate the new values for di as we move along the line.
Example:
13
)2or 5.0 (i.e. 0.5 (slope)gradientLet dxdydx
dy
3dy2dy
dy
True line
Start pixel at (x0,y1)
4dy
At x1 : s1 = dy t1 = dx - dyd1 = (si - ti) = dy - (dx - dy) = 2dy - dxbut 2dy dx di 0 y stays the samehence next pixel is at (x1,y1)
At x2 : s2 = 2dy t2 = dx - 2dyd2 = (s2 – t2) = 2dy - (dx - 2dy) = 4dy - dxSuppose d2 0 y is incrementedhence next pixel is at (x2,y2)
At x3 : s3 = 3dy - dx t2 = 2dx - 3dyd3 = (s2 – t3) = 6dy - 3dx 0so y stays the samehence next pixel is at (x3,y2)
x1 x2 x3x4 x5
x0
y0
y1
y2
y3
y5
In General
14
For a line with gradient ≤ 1d0 = 2dy – dxif di 0 then yi+1 = yi
di+1 = di + 2dyif di ≥ 0 then yi+1 = yi + 1
di+1 = di + 2(dy – dx)xi+1 = xi + 1
For a line with gradient 1d0 = 2dx – dy
if di 0 then xi+1 = xidi+1 = di + 2dx
if di ≥ 0 thenxi+1 = xi + 1
di+1 = di + 2(dx – dy)
yi+1 = yi + 1
Note: For |m| ≤ 1 the constants 2dy and 2(dy-dx) can be calculated once,so the arithmetic will involve only integer addition and subtraction.
Example – Draw a line from (20,10) to (30,18)
19
18
17
16
15
14
13
12
11
10
20 21 22 23 24 25 26 27 28 29 30 31 32
15
(20,10)
(30,18)dx = 10dy = 8
initial decision d0 = 2dy – dx = 6Also 2dy = 16, 2(dy – dx) = -4
i di (xi+1,yi+1)
0 6 (21,11) 1 2 (22,12) 2 -2 (23,12) 3 14 (24,13) 4 10 (25,14) 5 6 (26,15) 6 2 (27,16) 7 -2 (28,16) 8 14 (29,17) 9 10 (30,18)
16
void LineBres(int x0, int y0, int x1, int y1) // line for |m| < 1{ int dx = abs(x1 – x0), dy = abs(y1 – y0); int d = 2 * dy – dx, twoDy = 2 * dy, twoDyMinusDx = 2 * (dy – dx); int x, y;
if (x0 > x1) { // determines which point to use as start, which as end
x = x1; y = y1; x1 = x0; } else { x = x0; y = y0; } setPixel(x,y);
while (x < x1) { x++; if (d < 0) d += twoDy; else { y++; d += twoDyMinusDx; } setPixel(x, y); }}
Special cases
17
Special cases can be handled separatelyHorizontal lines (y = 0)Vertical lines (x = 0)Diagonal lines (|x| = |y|)
directly into the frame-buffer without processing them through the line-plotting algorithms.
Circle Equations
18
Polar formx = rCosy = rSin (r = radius of circle)
P=(rCos, rSin)
rSin)
rCos)x
y
r
Drawing a circle
19
DisadvantagesTo find a complete circle varies from 0° to 360°The calculation of trigonometric functions is very
slow.
= 0°while ( < 360°)
x = rCosy = rSinsetPixel(x,y) = + 1°
end while
Cartesian formUse Pythagoras theorem
x2 + y2 = r2
20
x
ry
y
x x
y
r 2 2,P x r x
Circle algorithms
21
Step through x-axis to determine y-values
Disadvantages:– Not all pixel filled in– Square root function is very slow
Circle Algorithms
22
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)
Bresenham’s Circle Algorithm
General PrincipleThe circle function:
and
2 2 2( , )circlef x y x y r
23
Consider only 45° ≤ ≤ 90°
if (x,y) is inside the circle boundary
if (x,y) is on the circle boundary
if (x,y) is outside the circle boundary
0
( , ) 0
0circlef x y
Bresenham’s Circle Algorithm
24
p1 p3
p2
D(si)D(ti)
After point p1, do we choose p2 or p3?
yi
yi - 1
xi xi + 1
r
Bresenham’s Circle Algorithm
25
Define: D(si) = distance of p3 from circle
D(ti) = distance of p2 from circle
i.e. D(si) = (xi + 1)2 + yi2 – r2 [always +ve]
D(ti) = (xi + 1)2 + (yi – 1)2 – r2 [always -ve]
Decision Parameter pi = D(si) + D(ti)
so if pi < 0 then the circle is closer to p3 (point above)
if pi ≥ 0 then the circle is closer to p2 (point below)
The Algorithm
26
x0 = 0y0 = rp0 = [12 + r2 – r2] + [12 + (r-1)2 – r2] = 3 – 2r
if pi < 0 thenyi+1 = yi
pi+1 = pi + 4xi + 6
else if pi ≥ 0 thenyi+1 = yi – 1pi+1 = pi + 4(xi – yi) + 10
Stop when xi ≥ yi and determine symmetry points in the other octants
xi+1 = xi + 1
Example
10
9
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 9 1027
i pi xi, yi
0 -17 (0, 10)
1 -11 (1, 10)
2 -1 (2, 10)
3 13 (3, 10)
4 -5 (4, 9)
5 15 (5, 9)
6 9 (6, 8)
7 (7,7)
r = 10
p0 = 3 – 2r = -17
Initial point (x0, y0) = (0, 10)
Midpoint Circle Algorithm
yi yi-1
xi xi+1 xi+2
28
Midpoint
x2 + y2 – r2 = 0
Assuming that we have just plotted the pixels at (xi , yi).
Which is next? (xi+1, yi) OR (xi+1, yi – 1).
- The one that is closer to the circle.
Midpoint Circle AlgorithmThe decision parameter is the circle at the midpoint
between the pixels yi and yi – 1.
If pi < 0, the midpoint is inside the circle and the pixel yi is closer to the circle boundary.
If pi ≥ 0, the midpoint is outside the circle and the pixel yi - 1 is closer to the circle boundary.
12
2 2 212
( 1, )
( 1) ( )
i circle i i
i i
p f x y
x y r
29
Decision ParametersDecision Parameters are obtained using
incremental calculations
OR
where yi+1 is either yi or yi-1 depending on the sign of pi
11 1 1 2
2 2 211 2
( 1, )
( 2) ( )
i circle i i
i i
p f x y
x y r
2 2 21 1 12( 1) ( ) ( ) 1i i i i i i ip p x y y y y
30
Note:xi+1 = xi +1
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) andpi+1 = pi + 2xi+1 + 1 – 2yi+1
where 2xi+1 = 2xi + 2 and 2yi+1 = 2yi – 2
4. Determine symmetry points in the other octants5. 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 10 2 2 4(1, ) 1 ( )circlep f r r r r
31
move circle origin at (0,0) byx = x – xc and y = y – yc
Example
10
9
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 9 1032
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)
33
Midpoint functionvoid plotpoints(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 circle(int r){
int x = 0, y = r;plotpoints(x,y);int p = 1 – r;while (x<y) {
x++;if (p<0) p += 2*x + 1;else {
y--;p += 2*(x-y) + 1;
}plotpoints(x,y);
}}
34
Ellipse-Generating AlgorithmsEllipse – A modified circle whose radius varies from a
maximum value in one direction (major axis) to a minimum value in the perpendicular direction (minor axis).
P=(x,y)F1
F2
d1
d2
The sum of the two distances d1 and d2, between the fixed positions F1 and F2 (called the foci of the ellipse) to any point P on the ellipse, is the same value, i.e.
d1 + d2 = constant
35
Ellipse PropertiesExpressing distances d1 and d2 in terms of the focal
coordinates F1 = (x1, x2) and F2 = (x2, y2), we have:
Cartesian coordinates:
Polar coordinates:
2 2 2 21 1 2 2( ) ( ) ( ) ( ) constantx x y y x x y y
ry
rx
22
1c c
x y
x x y y
r r
cos
sinc x
c y
x x r
y y r
36
Ellipse AlgorithmsSymmetry between quadrantsNot symmetric between the two octants of a quadrantThus, we must calculate pixel positions along the
elliptical arc through one quadrant and then we obtain positions in the remaining 3 quadrants by symmetry
(x, y)(-x, y)
(x, -y)(-x, -y)
rx
ry
37
Ellipse Algorithms
Decision parameter:
2 2 2 2 2 2( , )ellipse y x x yf x y r x r y r r
1
Slope = -1
rx
ry 2
0 if ( , ) is inside the ellipse
( , ) 0 if ( , ) is on the ellipse
0 if ( , ) is outside the ellipseellipse
x y
f x y x y
x y
2
2
2
2y
x
r xdySlope
dx r y
38
Ellipse Algorithms
Starting at (0, ry) we take unit steps in the x direction until we reach the boundary between region 1 and region 2. Then we take unit steps in the y direction over the remainder of the curve in the first quadrant.
At the boundary
therefore, we move out of region 1 whenever
1Slope = -1
rx
ry 2
2 21 2 2y x
dyr x r y
dx
2 22 2y xr x r y
39
Midpoint Ellipse Algorithm
yi yi-1
xi xi+1 xi+2
Midpoint
Assuming that we have just plotted the pixels at (xi , yi).
The next position is determined by:12
2 2 2 2 2 212
1 ( 1, )
( 1) ( )
i ellipse i i
y i x i x y
p f x y
r x r y r r
If p1i < 0 the midpoint is inside the ellipse yi is closer
If p1i ≥ 0 the midpoint is outside the ellipse yi – 1 is closer
40
Decision Parameter (Region 1)
At the next position [xi+1 + 1 = xi + 2]
OR
where yi+1 = yi
or yi+1 = yi – 1
11 1 1 2
2 2 2 2 2 211 2
1 ( 1, )
( 2) ( )
i ellipse i i
y i x i x y
p f x y
r x r y r r
2 2 2 2 2 21 11 1 2 21 1 2 ( 1) ( ) ( )i i y i y x i ip p r x r r y y
41
Decision Parameter (Region 1)Decision parameters are incremented by:
Use only addition and subtraction by obtaining
At initial position (0, ry)
2 21
2 2 21 1
2 if 1 0
2 2 if 1 0y i y i
y i y x i i
r x r pincrement
r x r r y p
2 22 and 2y xr x r y
2
2 2
2 2 2 2 21 10 2 2
2 2 214
2 02 2
1 (1, ) ( )
y
x x y
ellipse y y x y x y
y x y x
r xr y r r
p f r r r r r rr r r r
42
Region 2
Over region 2, step in the negative y direction and midpoint is taken between horizontal pixels at each step.
yi yi-1
xi xi+1 xi+2
Midpoint
Decision parameter:12
2 2 2 2 2 212
2 ( , 1)
( ) ( 1)
i ellipse i i
y i x i x y
p f x y
r x r y r r
If p2i > 0 the midpoint is outside the ellipse xi is closer
If p2i ≤ 0 the midpoint is inside the ellipse xi + 1 is closer
43
Decision Parameter (Region 2)
At the next position [yi+1 – 1 = yi – 2]
OR
where xi+1 = xi
or xi+1 = xi + 1
11 1 12
2 2 2 2 2 211 2
2 ( , 1)
( ) ( 2)
i ellipse i i
y i x i x y
p f x y
r x r y r r
2 2 2 2 21 11 1 2 22 2 2 ( 1) ( ) ( )i i x i x y i ip p r y r r x x
44
Decision Parameter (Region 2)
Decision parameters are incremented by:
At initial position (x0, y0) is taken at the last position selected in region 1
2 21
2 2 21 1
2 if 2 0
2 2 if 2 0x i x i
y i x i x i
r y r pincrement
r x r y r p
10 0 02
2 2 2 2 2 210 02
2 ( , 1)
( ) ( 1)
ellipse
y x x y
p f x y
r x r y r r
45
Midpoint Ellipse Algorithm1. Input rx, ry, and ellipse center (xc, yc), and obtain the first
point on an ellipse centered on the origin as(x0, y0) = (0, ry)
2. Calculate the initial parameter in region 1 as
3. At each xi position, starting at i = 0, if p1i < 0, the next point along the ellipse centered on (0, 0) is (xi + 1, yi) and
otherwise, the next point is (xi + 1, yi – 1) and
and continue until
2 2 210 41 y x y xp r r r r
2 21 11 1 2i i y i yp p r x r
2 2 21 1 11 1 2 2i i y i x i yp p r x r y r
2 22 2y xr x r y
46
Midpoint Ellipse Algorithm4. (x0, y0) is the last position calculated in region 1. Calculate
the initial parameter in region 2 as
5. At each yi position, starting at i = 0, if p2i > 0, the next point along the ellipse centered on (0, 0) is (xi, yi – 1) and
otherwise, the next point is (xi + 1, yi – 1) and
Use the same incremental calculations as in region 1. Continue until y = 0.
6. For both regions determine symmetry points in the other three quadrants.
7. Move each calculated pixel position (x, y) onto the elliptical path centered on (xc, yc) and plot the coordinate values
x = x + xc , y = y + yc
2 2 2 2 2 210 0 022 ( ) ( 1)y x x yp r x r y r r
2 21 12 2 2i i x i xp p r y r
2 2 21 1 12 2 2 2i i y i x i xp p r x r y r
47
Example
i pi xi+1, yi+1 2ry2xi+1 2rx2yi+1
0 -332 (1, 6) 72 768
1 -224 (2, 6) 144 768
2 -44 (3, 6) 216 768
3 208 (4, 5) 288 640
4 -108 (5, 5) 360 640
5 288 (6, 4) 432 512
6 244 (7, 3) 504 384
rx = 8 , ry = 6
2ry2x = 0 (with increment 2ry
2 = 72)
2rx2y = 2rx
2ry (with increment -2rx2 = -128)
Region 1 (x0, y0) = (0, 6)
2 2 210 41 332y x y xp r r r r
Move out of region 1 since2ry
2x > 2rx2y
48
Example
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8
i pi xi+1, yi+1 2ry2xi+1 2rx2yi+1
0 -151 (8, 2) 576 256
1 233 (8, 1) 576 128
2 745 (8, 0) - -
Region 2 (x0, y0) = (7, 3) (Last position in region 1)
10 22 (7 ,2) 151ellipsep f
Stop at y = 0
49
Midpoint Ellipse Functionvoid ellipse(int Rx, int Ry){
int Rx2 = Rx * Rx, Ry2 = Ry * Ry;int twoRx2 = 2 * Rx2, twoRy2 = Ry2 * Ry2;int p, x = 0, y = Ry;int px = 0, py = twoRx2 * y;
ellisePlotPoints(xcenter, ycenter, x, y);// Region 1p = round(Ry2 – (Rx2 * Ry) + (0.25 * Rx2));while (px < py) {
x++;px += twoRy2;if (p < 0) p += Ry2 + px;else {
y--;py -= twoRx2;p += Ry2 + px – py;
}ellisePlotPoints(xcenter, ycenter, x, y);
}// Region 2p = round(Ry2 * (x+0.5) * (x+0.5) + Rx2 * (y-1)*(y-1) – Rx2 * Ry2;while (y > 0) {
y--;py -= twoRx2;if (p > 0) p += Rx2 – py;else {
x++;px += twoRy2;p += Rx2 – py + px;
}ellisePlotPoints(xcenter, ycenter, x, y);
}}
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