Computer Graphics - Bresenham's line drawing algorithm & Mid Point Circle algorithm

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 saikrishna.tanguturu@wipro.com sai.tsk2008@gmail.com

2 2011HW69900RAYAPU MOSES rayapu.moses@wipro.com stalinkvd001@gmail.com

3 2011HW69932 SHENBAGAMOORTHY Ashenbagamoorthy.a83@wipro.com moorthy2626@gmail.com

4 2011HW69913ANURUPA K C anurupa.c85@wipro.com anu.rupa30@gmail.com

5 2011HW69909ARUNJUNAISELVAM P arunjunaiselvam.p95@wipro.com arunjunai.carrer@gmail.com

6 2011HW69569PRANOB JYOTI KALITA pranob.kalita@wipro.com pranob.kalita90@gmail.com

7 2011HW69893TINNALURI V N PRASANTH prasanth.tinnaluri@wipro.com naga.prasanth985@gmail.com

8 2011HW69904KONDALA SUMATHI sumathi.kondala@wipro.com sumathi.kondala@gmail.com

9 2011HW69896DASIKA KRISHNA dasika.krishna@wipro.com dasikakrishnas@gmail.com

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);

}}

Thank You

Click icon to add picture