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LECTURE: 07

DDA (DIGITAL DIFFERENTIAL

 ANALYZER ALGORITHM)

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LINE DRAWING ALGORITHM-

STEPS

• Specifies (x,y) values of end pixels/points

• Need algorithm to figure out which intermediate

pixels are on line path

• Pixel (x,y) values constrained to integer valuesthough actual computed intermediate line values

may be floats

• Rounding may be required. E.g. computed point

(10.48, 20.51) rounded to (10, 21)• Rounded pixel value is off actual line path (jaggy!!)

• Sloped lines end up having jaggies

• Vertical, horizontal lines, no jaggies

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LINE DRAWING ALGORITHM

0 1 2 3 4 5 6 7 8 9 10 11 12

87

654321

Line: (3,2) -> (9,6)

?Which intermediatepie!" t# t$rn #n?

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Slope-intercept line equationy = mx + b

Given two end points (x0,y0), (x1, y1), how to compute m

and b?

(0,%0)

(1,%1)

d

d%

01

01

 x x

 y y

dx

dym

−

−

==

0*0   xm yb   −=

LINE DRAWING ALGORITHM

4

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LINE DRAWING ALGORITHM

Numerical example of finding slope m:

(Ax, Ay) = (23, 41), (Bx, By) = (125, 96)

5392.0102

55

23125

4196==

−

−=

−

−=

 Ax Bx

 Ay Bym

5

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DIGITAL DIFFERENTIAL ANALYZER (DDA): LINE

DRAWING ALGORITHM

(0,%0)

(1,%1)

d

d%

Walk through the line, starting at (x0,y0)Constrain x, y increments to values in [0,1] rangeCase a: x is incrementing faster (m < 1) Step in x=1 increments, compute and round y

Case b: y is incrementing faster (m > 1) Step in y=1 increments, compute and round x

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DDA ALGORITHM

 Y_inc

 X_inc

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DDA LINE DRAWING ALGORITHM (CASE A: M < 1)

(0, %0)

& 0 ' 1 % & %0 ' 1 m

!!$minate pie! (, r#$nd(%))

& ' 1 % & % ' 1 m

!!$minate pie! (, r#$nd(%))

*

+nti! && 1

(1,%1)

& 0 % & %0

!!$minate pie! (, r#$nd(%))

m y y k k    +=+1

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DDA LINE DRAWING ALGORITHM

(CASE B: M > 1)

% & %0 ' 1 & 0 ' 1 1m

!!$minate pie! (r#$nd(), %)

% & % ' 1 & ' 1 m

!!$minate pie! (r#$nd(), %)

*

+nti! % && %1

& 0 % & %0

!!$minate pie! (r#$nd(), %)(1,%1)

(0,%0)

m x x k k 

1

1  +=

+

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DDA PSEUDO-CODE

 // assume that slope is gentle

DDA(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;

 ColorPixel(Round(x),Round(y)); 

for (int i=0; i<numsteps; i++) {

  x += xinc;

  y += yinc;

  ColorPixel(Round(x),Round(y));

 }}

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DDA EXAMPLE

Suppose we want todraw a line startingat pixel (2,3) andending at pixel (12,8).

What are the valuesof the variables x andy at each timestep?

What are the pixelscolored, according tothe DDA algorithm?

numsteps = 12 – 2 = 10xinc = 10/10 = 1.0

yinc = 5/10 = 0.5

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t x y R(x) R(y)

0 2 3 2 3

1 3 3.5 3 42 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

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LINE DRAWING ALGORITHM

DRAWBACKS

DDA is the simplest line drawing algorithmNot very efficient

Round operation is expensive

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