Line drawing algorithm and antialiasing techniques

AMITY UNIVERSITY, HARYANA

COMPUTER GRAPHICS

Ankit GargAssistant Professor

Amity University, Haryana

Module 1& 2• Module I: Introduction to Graphics and Graphics Hardware System• Application of computer graphics, Video Display Devices, Raster Scan Display,

Random Scan Display, Input Devices, Graphic Software and graphics standards, Numerical based on Raster and Random scan display, Frame buffer, Display processor.

• Module II: Output Primitives and Clipping operations• Algorithms for drawing 2D Primitives lines (DDA and Bresenham‘s line

algorithm), circles (Bresenham‘s and midpoint circle algorithm), ellipses (midpoint ellipse algorithm), Antialiasing (Supersampling method and Area sampling method), Line clipping (cohen-sutherland algorithm), Curve clipping algorithm, and polygon clipping with Sutherland Hodgeman algorithm, Area fill algorithms for various graphics primitives: Scanline fill algorithm, boundary fill algorithm, flood fill algorithm, Polygon representation, various method of Polygon Inside test: Even-Odd method, winding number method, Character generation techniques.

Module-IILine Drawing Algorithms

Line Drawing Algorithm

• DDA Line drawing algorithm• Breshenham’s Line drawing algorithm (Also

called Antialiasing Algorithm)• These Line drawing algorithms are used for

scan conversion of graphic object i.e. Line.

Line Drawing Algorithm

• Scan conversion is a process to represent graphics objects as a collection of pixels.

• Various algorithm and mathematical computations are available for this purpose.

• Rasterization-Process of determining which pixels give the best approximation to a desired line on the screen.

Lines with slope magnitude |m| <1, ∆y = m ∆x

∆x> ∆y

Lines with slope magnitude |m| >1, ∆x = ∆y /m

∆y> ∆x

P1

P2

P1

P2

Digital Differential Algorithm(DDA)• Scan conversion algorithm based on

calculating either ∆x or ∆y.• Sample the line at unit intervals in one

coordinate and determine the corresponding integer values nearest the line path for the other coordinate.

Case 1 Line with positive slope less than or equal to 1,

we sample at unit x intervals (∆x =1) and determine the successive y value as

Yk+1= Yk + m ∆y=m ∆x Yk+1- Yk =m ∆x Yk+1- Yk =m (∆x =1)

• k takes integer values starting from 1 and increases by 1 until the final point is reached.

• m can be any number between 0 and 1.• Calculated y values must be rounded off to

the nearest integer.• For lines with positive slope greater than 1,

sample at unit y intervals (∆y = 1) and calculate each successive x value as

xk+1= xk + 1/m

StepsP1 ( xa,ya) and P2 (xb,yb) are the two end points.

2. Horizontal and vertical differences between the endpoint positions are computed & assigned to two parameters namely dx and dy.

3. The difference with greater magnitude determines the ‘value’ of increments to be done. That means the number of times sampling has to be done. This value is assigned to a parameter called ‘steps’.

4. Starting from the 1st pixel ,we determine the offset needed at each step to generate the next pixel position along the line path. Call the offset value as xincrement and yincrement.The starting points are xa and ya.

Assign x =xa and y=yax= x+ xincr & y= y+ yincr

5. Loop through the process steps times, till the last point xb, yb is reached.

P1 (xa,ya)

P2 (xb,yb)

DDA Pseudo-code// assume that slope is gentleDDA(float x0, float x1, float y0, float y1) { float x, y; float xinc, yinc; int numsteps;

numsteps = Round(x1) – Round(x0); xinc = (x1 – x0) / numsteps; yinc = (y1 – y0) / numsteps; x = x0; y = y0; putpixel(Round(x),Round(y)); for (int i=0; i<numsteps; i++) { x += xinc; y += yinc; putpixel(Round(x),Round(y)); }

}

DDA Example• Suppose we want to draw a line starting

at pixel (2,3) and ending at pixel (12,8).numsteps = 12 – 2 = 10xinc = 10/10 = 1.0 X1-X0= 10yinc = 5/10 = 0.5 Y1- y0= 5

Iteration x y Round(x) Round(y)

0 2 3 2 3

1 3 3.5 3 4

2 4 4 4 4

3 5 4.5 5 5

4 6 5 6 5

5 7 5.5 7 6

6 8 6 8 6

7 9 6.5 9 7

8 10 7 10 7

9 11 7.5 11 8

10 12 8 12 8

// assume that slope is gentle M<=1DDA(float x0, float x1, float y0, float y1) { float x, y; float xinc, yinc; int numsteps;

numsteps = Round(x1) – Round(x0); xinc = (x1 – x0) / numsteps; yinc = (y1 – y0) / numsteps; x = x0; y = y0; putpixel(Round(x),Round(y)); for (int i=0; i<numsteps; i++) { x += xinc; y += yinc; putpixel(Round(x),Round(y)); }}

DDA Example• Suppose we want to draw a line starting

at pixel (15,4) and ending at pixel (20,17).numsteps = xinc = X1-X0=yinc = Y1- y0=

Iteration x y Round(x) Round(y)

0

1

2

3

4

5

6

7

8

9

10

// assume that slope is gentle M>1DDA(float x0, float x1, float y0, float y1) { float x, y; float xinc, yinc; int numsteps;

numsteps = Round(y1) – Round(y0); xinc = (x1 – x0) / numsteps; yinc = (y1 – y0) / numsteps; x = x0; y = y0; putpixel(Round(x),Round(y)); for (int i=0; i<numsteps; i++) { x += xinc; y += yinc; putpixel(Round(x),Round(y)); }}

DDA Example• Suppose we want to draw a line starting

at pixel (6,3) and ending at pixel (8,10).numsteps = ?xinc = ? X1-X0= yinc = ? Y1- y0=

Iteration x y Round(x) Round(y)

0

1

2

3

4

5

6

7

8

9

10

// assume that slope is gentle M>1DDA(float x0, float x1, float y0, float y1) { float x, y; float xinc, yinc; int numsteps;

numsteps = Round(y1) – Round(y0); xinc = (x1 – x0) / numsteps; yinc = (y1 – y0) / numsteps; x = x0; y = y0; putpixel(Round(x),Round(y)); for (int i=0; i<numsteps; i++) { x += xinc; y += yinc; putpixel(Round(x),Round(y)); }}

DDA Example• Suppose we want to draw a line starting

at pixel (2,3) and ending at pixel (5,12).numsteps = ?xinc = ? X1-X0= yinc = ? Y1- y0=

Iteration x y Round(x) Round(y)

0

1

2

3

4

5

6

7

8

9

10

DDA Algorithm (continued)

This DDA algorithm suffers from two reasons. Floating point calculations and rounding operations are expensive and time consuming.• Due to limited precision and Rounding operations calculated points will not

be displayed at its original point.

Y_inc

X_inc

Bresenham’s Algorithm

• Uses only integer calculations

• Requires less time to generate line due to elimination of floating point calculations.

Bresenham’s Algorithm m<=1 1. Input the two line endpoints and store left endpoint as (x0,y0)2. Pre-calculate the values dx, dy3. Color pixel (x0,y0)4. Let p0 = 2dy –dx, Here P0 is decision parameter which tells us which next pixel will glow.5. At each xk along the line, starting with k=0:

6. Repeat Step-5 Until we does not reach up to final point.

Remember! The algorithm and derivation above assumes slopes are less than or equal to 1. for other slopes we need to adjust the algorithm slightly.

If pk<0, then the next point to plot is (xk + 1,yk),and pk+1 = pk + 2dy

Otherwise, the next point to plot is (xk + 1, yk + 1),and pk+1 = pk + 2dy – 2dx

Bresenham’s Algorithm Example Where m<=1

• Suppose we want to draw a line starting at pixel (2,3) and ending at pixel (12,8).

dx = 12 – 2 = 10dy = 8 – 3 = 5p0 = 2dy – dx = 0

t p P(x) P(y)

0 0 2 3

1 -10 3 4

2 0 4 4

3 -10 5 5

4 0 6 5

5 -10 7 6

6 0 8 6

7 -10 9 7

8 0 10 7

9 -10 11 8

10 0 12 8

2dy = 102dy – 2dx = -10

Algorithm1. Input the two line endpoints and

store left endpoint as (x0,y0)2. Pre-calculate the values dx, dy,

2dy and 2dy -dx3. Color pixel (x0,y0)4. Let p0 = 2dy –dx5. At each xk along the line,

starting with k=0:

6. Repeat Step-4 dx times

If pk<0, then the next point to plot is (xk + 1,yk),and pk+1 = pk + 2dy (Down pixel will be selected)Otherwise, the next point to plot is (xk + 1, yk + 1),and pk+1 = pk + 2dy – 2dx (Upper pixel will be selected)

Anti-aliasing and filtering techniques

Anti-aliasing• Anti-aliasing is a method of fooling the eye

that a jagged edge is really smooth.• Due to low resolution aliasing effect will

occur, which can be removed by increasing the screen resolution.

Circle after applying antialiasingJagged edges due to aliasing

Some more Examples

Reducing Aliasing• By increasing Resolution

The aliasing effect can be minimized by increasing resolution of the raster display.

Disadvantage of improving resolution

• More memory requirement (Size of frame buffer will become Large) • More scan conversion time

Anti-aliasing Methods

• Super-sampling method or post filtering

• Area sampling or Pre filtering

Super-sampling Method (Cont….)

• In this method every individual pixel is subdivided in to sub-pixel.

• In this method we count the number of pixel which are overlapped by the object.

• The intensity value of a pixel is the average of the intensity values of all the sampled sub-pixels with in that pixel.

• In this method every pixel on the screen have different intensity.

Super-sampling for a line object having Non-Zero width.

Super-sampling Method (Cont….)

• Pixel at upper right corner is assigned 7/9 because seven of its nine-sub pixels are inside the object area.

• Suppose the color of object is RED(1,0,0), and the background color is light yellow (.5,.5,.5).

• At what intensity the pixel will glow? (1 X 7/9 +.5 X 2/9, 0X7/9+ 0.5 X 2/9, 0X7/9+.5X2/9)

R G BBlending of background color and object color will occur only in area of pixel where object overlaps.

Question-

1. What will be the intensity of center pixel?Answer- (1 X 1+.5X0, 0X1+.5X0, 0X1+.5X0)2. What will be the intensity of lower right side pixel?Answer- (1X1/9 + .5X8/9, 0X1/9+.5X8/9, 0X1/9+.5X8/9)

Intensity Variation on pixels after Super sampling method

Write Formula for Blending of Colors for following Conditions1. Object Background is (.5,.5,.5)2. Object Color is (1,1,0)Ans: Write Formula for Blending of Colors for following Conditions1. Object Background is (.5,.5,.5)2. Object Color is (1,1,1)

Area Sampling Method • This figure shows how line with a non-Zero width have

different intensity value at each pixel on the screen• In this Area sampling method intensity value is

determined according to area covered by the object.

90%

15%

65%

50%

Thank you!!