Output PrimitivesPoints and LinesLine Drawing AlgorithmsDDA AlgorithmBresenham’s Line AlgorithmMidpoint Circle AlgorithmMidpoint Ellipse AlgorithmFilled Area Primitives

Points and Lines

• Point is the fundamental element of picture representation.

• It is the position in the plan defined as either pair or triplets of number depending upon the dimension.

• Two points represent line or edge and 3 or more points a polygon.

• Curved lines are represented by the short straight lines.

Line Drawing Algorithms• Slope-Intercept Equation

• Slope

• Intercept

• Interval Calculation

bxmy .

12

12

xx

yym

11 .xmyb

xmy . m

yx

Line Drawing Algorithm

DDA Algorithm

Digital Differential Analyzer– Sample the line at unit intervals in one coordinate– Determine the corresponding integer values

nearest the line path in another co-ordinate

DDA Algorithm (left to right)

• Slope

• For |m|<1 (|Δy|< |Δx|)– Sample line at unit interval in x co-ordinate

• For |m|>1 (|Δy|> |Δx|)– Sample line at unit interval in y co-ordinate

x

y

xx

yym

kk

kk

1

1

myy kk 1

mxx kk

11

11 kk xxx

11 kk yyy

DDA Algorithm (right to left)

• Slope

• For |m|<1 (|Δy|< |Δx|)– Sample line at unit interval in x co-ordinate

• For |m|>1 (|Δy|> |Δx|)– Sample line at unit interval in y co-ordinate

myy kk 1

mxx kk

11

11 kk xxx

11 kk yyy

x

y

xx

yym

kk

kk

1

1

DDA Algorithm1. Input the two line endpoints and store the left endpoint in (x0,y0)2. Plot first point (x0,y0)3. Calculate constants Δx, Δy4. If |Δx| > |Δy| steps = |Δx| else steps = |Δy|5. Calculate XInc = |Δx| / steps and YInc = |Δy| / steps 6. At each xk along the line, starting at k=0, Plot the next pixel at (xk + XInc, yk +

YInc)7. Repeat step 6 steps times

Pseudo CodeVoid lineDDA(int xa, int ya, int xb, int yb){

int dx = xb – xa, dy = yb – ya, steps, k;float xIncrement, yIncrement, x = xa, y = ya;

if( abs (dx) > abs (dy) ) steps = abs (dx);else steps = abs (dy);xIncrement = dx / (float) steps;yIncrement = dy / (float) steps;setPixel (ROUND (x), ROUND (y));for (k=0; k<steps; k++){

x += xIncrement;y += yIncrement;setPixel (ROUND(x), ROUND(y));

}}

• Use DDA algorithm for rasterizing line (0,0) to (6,6).

• Use DDA algorithm for rasterizing line (0,0) to (4,6).

Bresenham’s Line Algorithm

• Uses only incremental integer calculations

• Which pixel to draw ?– (11,11) or (11,12) ?– (51,50) or (51,49) ?– Answered by Bresenham

• For |m|<1– Start from left end point (x0,y0) step to each

successive column (x samples) and plot the pixel whose scan line y value is closest to the line path.

– After (xk,yk) the choice could be (xk+1,yk) or (xk+1,yk+1)

Then

And

Difference between separations

bxmy k )1(

kyyd 1kk ybxm )1(

yyd k )1(2

bxmy kk )1(1

122)1(221 byxmdd kk

Defining decision parameter [1]

Sign of pk is same as that of d1-d2 for Δx>0 (left to right sampling)

For Recursive calculation, initially

cyxxy kk .2.2

Constant=2Δy + Δx(2b-1) Which is independent of pixel position

cyxxyp kkk 111 .2.2

)(2)(2 111 kkkkkk yyxxxypp

c eliminated here

)(22 11 kkkk yyxypp

because xk+1 = xk + 1

yk+1-yk = 0 if pk < 0yk+1-yk = 1 if pk ≥ 0

xyp 20 Substitute b = y0 – m.x0

and m = Δy/Δx in [1]

)( 21 ddxpk

Algorithm Steps (|m|<1)1. Input the two line endpoints and store the left endpoint in (x0,y0)2. Plot first point (x0,y0)3. Calculate constants Δx, Δy, 2Δy and 2 Δy- 2Δx, and obtain p0 = 2Δy – Δx4. At each xk along the line, starting at k=0, perform the following test:

If pk<0, the next point plot is (xk+1,yk) and Pk+1 = pk + 2Δy

Otherwise, the next point to plot is (xk + 1, yk+1) andPk+1 = pk + 2Δy - 2Δx

5. Repeat step 4 Δx times

What’s the advantage?

• Answer: involves only the calculation of constants Δx, Δy, 2Δy and 2Δy- 2Δx once and integer addition and subtraction in each steps

ExampleEndpoints (20,10) and (30,18)Slope m = 0.8Δx = 10, Δy = 8P0 = 2Δy - Δx = 6

2Δy = 16, 2Δy-2Δx = -4

Plot (x0,y0) = (20,10)

• Use Bresenham’s algorithm for rasterizing the line (5,5) to (13,9)

• Digitize a line with endpoints (15,18) and (10,15) using BLA.

• Digitize a line with endpoints (15,15) and (10,18) using BLA.

Algorithm Steps (|m|>1)1. Input the two line endpoints and store the left endpoint in (x0,y0)2. Plot first point (x0,y0)3. Calculate constants Δx, Δy, 2Δx and 2 Δx- 2Δy, and obtain p0 = 2Δx – Δy4. At each xk along the line, starting at k=0, perform the following test:

If pk<0, the next point plot is (xk, yk+1) and Pk+1 = pk + 2Δx

Otherwise, the next point to plot is (xk + 1, yk+1) andPk+1 = pk + 2Δx - 2Δy

5. Repeat step 4 Δx times

• Use BLA algorithm for rasterizing line (0,0) to (4,6).

Circle-Generating Algorithms (Basic Foundations)

• Circle Equation:

• Points along circumference could be calculated by stepping along x-axis:

222 )()( ryyxx cc

22 )( xxryy cc

Problem (in above method)

• Computational complexity• Spacing:– Non-uniform spacing of

plotted pixels

Adjustments (To fix problems)• Spacing problem (2 ways):

– Interchange the role of x and y whenever the absolute value of the slope of the circle tangent > 1

– Use polar co-ordinate:

• Equally spaced points are plotted along the circumference with fixed angular step size.

• step size chosen for θ depends on the application and display device.• Computation Problem:

– Use symmetry of circle; i.e calculate for one octant and use symmetry for others.

sin

cos

ryy

rxx

c

c

Circle Symmetry

Bresenham’s Algorithm Could Be Adapted ??

• Yes• How ?– Setting decision parameter for finding the closest

pixel to the circumference• And what to do For Non-linear equation of

circle ?– Comparison of squares of the pixel separation

distance avoids square root calculations

Midpoint Circle Algorithm

• Circle function defined as:

• Any point (x,y) satisfies following conditions

boundarycircletheoutsideisyxif

boundarycircletheonisyxif

boundarycircletheinsideisyxif

yxfcircle),(,0

),(,0

),(,0

),(

222),( ryxyxfcircle

• Decision parameter is the circle function; evaluated as:

222 )2

1()1(

)2

1,1(

ryx

yxfp

kk

kkcirclek

1)()()1(2

)2

1(]1)1[(

)2

1,1(

122

11

221

2

111

kkkkkkk

kk

kkcirclek

yyyyxpp

ryx

yxfp

yk+1 = yk if pk<0yk+1 = yk-1 otherwise

• ThusPk+1 = Pk + 2xk+1+1 if pk<0Pk+1 = Pk + 2xk+1+1-2yk+1 otherwise

• Also incremental evaluation of 2xk+1 and 2yk+1

2xk+1 = 2xk + 22yk+1 = 2yk – 2 if pk >0

• At start position (x0,y0) = (0,r)2x0 = 0 and 2y0 = 2r

• Initial decision parameter

• For r specified as an integer, round p0 to

P0 = 1-r

(because all increments are integers)

r

rr

rfp circle

4

5)2

1(1

)2

1,1(

22

0

Algorithm1. Input radius r and circle center (xc, yc) and obtain the first point on the circumference of

a circle centered on the origin as(x0,y0) = (0,r)

2. Calculate the initial value of the decision parameter asP0 = 5/4 – r

3. At each xk position, starting at k = 0, perform the following test:If pk < 0, the next point along the circle centered on (0,0) is (xk+1,yk) and

Pk+1 = pk + 2xk+1 + 1Otherwise, the next point along the circle is (xk+1,yK-1) and

Pk+1 = pk + 2xk+1 + 1 -2yk+1

Where 2xk+1 = 2xk + 2 and 2yk+1 = 2yk-24. Determine the symmetry points in the other seven octants.5. Move each calculated pixel position (x,y) onto the circular path

centered on (xc,yc) and plot the co-ordinate values:x = x + xc, y = y+yc

6. Repeat steps 3 through 5 until x ≥ y

• Given radius =10 use mid-point circle drawing algorithm to determine the pixel in the first quadrant.

Solution:P0 = 1-r = -9(x0 , y0)=(0,10)2x0 =02y0 =20

• Digitize the circle (x-2)^2+(y-3)^2=25 in first quadrant.

Solution:P0=(1-r)= -4

(x0 , y0)=(0,5)2x0 =02y0 =10

Ellipse Generating Algorithms

• Equation of ellipse:

• F1(x1,y1), F2(x2,y2)

• General Equation

• Simplified Form

• In polar co-ordinate

constantyyxxyyxx 22

22

21

21 )()()()(

constantdd 21

022 FEyDxCxyByAx

1

22

y

c

x

c

r

yy

r

xx

sincosyc

xcryyrxx

• Ellipse function222222),( yxyellipse rryrxryxf x

boundaryellipsetheoutsideisyxifboundaryellipsetheonisyxifboundaryellipsetheinsideisyxif

yxfellipse),(,0),(,0),(,0

),(

• From ellipse tangent slope:

• At boundary region (slope = -1):

• Start from (0,ry), take x samples to boundary between 1 and 2

• Switch to sample y from boundary between 1 and 2 (i.e whenever )

yr

xr

dx

dy

x

y

2

2

2

2

yrxr xy22 22

yrxr xy22 22

Slope=-1

• In the region 1

• For next sample

• Thus increment

222222 )2

1()1(

)2

1,1(1

yxkxky

kkellipsek

rryrxr

yxfp

22

1222

1

2221

222

111

2

1

2

1)1(211

)2

1(1)1(

)2

1,1(1

kkxykykk

yxkxky

kkellipsek

yyrrxrpp

rryrxr

yxfp

01,2201,2

122

12

21

2

kkxyky

kyky

pifyrrxrpifrxr

increment

•For increment calculation; Initially:

•Incrementally:Update x by adding 2ry

2 to first equation and update y by subtracting 2rx

2 to second equation

yxx

y

rryrxr

22

2

2202

• Initial value

2220

222

22

0

4

11

2

12

1,11

xyxy

yxyxy

yellipse

rrrrp

rrrrr

rfp

(0,ry)

• In the region 2

• For next sample

• Initially

222222 )1()2

1(

)1,2

1(2

yxkxky

kkellipsek

rryrxr

yxfp

22

1222

1

22222

12

111

2

1

2

1)1(222

112

1

)1,2

1(2

kkyxkxkk

yxkxky

kkellipsek

xxrryrpp

rryrxr

yxfp

2220

22

02

0

000

)1(2

12

1,2

12

yxxy

ellipse

rryrxrp

yxfp

For simplification calculation of p20 can be done by selecting pixel positions in counter clockwise order starting at (rx,0) and unit samples to positive y direction until the boundary between two regions

Algorithm1. Input rx,ry, and the 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 value of the decision parameter in region 1 as

3. At each xk position in region 1, starting at k = 0, perform the following test: If p1k < 0, the next point along the ellipse centered on (0,0) is (xk+1,yk) and

Otherwise, the next point along the ellipse is (xk+1,yk-1) and

With

and continue until

2220 4

11 xyxy rrrrp

21

21 211 ykykk rxrpp

21

21

21 2211 ykxkykk ryrxrpp

221

2221

2 222,222 xkxkxykyky ryryrrxrxr

yrxr xy22 22

4. Calculate the initial value of decision parameter in region 2 using the last point (x0,y0) calculated in region 1 as

5. At each yk position in region 2, starting at k = 0, perform the following test: If p2k>0, the next point along the ellipse centered on (0,0) is (xk,yk-1) and

Otherwise the next point along the ellipse is (xk+1,yk-1) and

Using the same incremental calculations for x and y as in region 1.6. Determine the 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 co-ordinate

values:X = x + xc, y = y+ yc

8. Repeat the steps for region 1 until

2220

22

02

0 )1(2

12 yxxy rryrxrp

21

21 222 xkxkk ryrpp

21

21

21 2222 xkxkykk ryrxrpp

yrxr xy22 22

• Digitize the ellipse with parameter rx =8 and ry =6 in first quadrant.

Solution:

Filled Area Primitives

• Two basic approaches for area filling:– By determining overlap intervals for scan lines– Start from given interior position and filling

outwards

Scan Line Polygon Filled Algorithm

• Intersection points of scan line with the polygon edge are calculated.

• Points are sorted from left to right.• Corresponding frame buffer position between

each intersection pair are set by specified color.

• A scan line pass through vertex, intersect two polygon edges.

• Scan line y’– Intersect even number of edges– 2 pairs of intersection points correctly find the

interior span• Scan line y– Intersect an odd number of edges(5)– Must count the vertex intersection as only one

point

How to distinguish these cases?

• Scan line y– Intersecting edges are on the opposite side

• Scan line y’– Intersecting edges are on the same side.

• By tracing around the boundary,– If the endpoint y values of two consecutive edges

monotonically increases or decreases count middle vertex as single

Implementation

• In determining edge intersection, we can set up incremental set up calculation using fact that slope of edge is constant.

Inside Outside test

• Odd Even rule:– Odd edge means interior

• Non zero winding number rule– Non zero winding number means interior

Scan line fill for Curved Area

• Require more works than polygon filling due to non-linear boundary.

• We calculate the scan line intersection and fill all the interior points.

• Symmetries between the quadrant can be used to reduce the calculation.

Boundary fill algorithm

• Fill the region starting from the interior point until the appropriate color boundary is reached.

• Two ways:– 4 connected– 8 connected

4 connected

8 connected

Using 4 connected flood fill algorithm