Computer Graphics

1
Computer Graphics

• Lecture 3
• Line Circle Drawing

2
Towards the Ideal Line

• We can only do a discrete approximation
• Illuminate pixels as close to the true path as
  possible, consider bi-level display only
• Pixels are either lit or not lit

3
What is an ideal line

• Must appear straight and continuous
• Only possible axis-aligned and 45o lines
• Must interpolate both defining end points
• Must have uniform density and intensity
• Consistent within a line and over all lines
• What about antialiasing?
• Must be efficient, drawn quickly
• Lots of them are required!!!

4
Simple Line
Based on slope-intercept algorithm from algebra
y mx b Simple approach
increment x, solve for y Floating point
arithmetic required
5
Does it Work?
It seems to work okay for lines with a slope of
1 or less, but doesnt work well for lines with
slope greater than 1 lines become more
discontinuous in appearance and we must add more
than 1 pixel per column to make it
work. Solution? - use symmetry.
6
Modify algorithm per octant
OR, increment along x-axis if dyltdx else
increment along y-axis
7
DDA algorithm

• DDA Digital Differential Analyser
• finite differences
• Treat line as parametric equation in t

Start point - End point -
8
DDA Algorithm

• Start at t 0
• At each step, increment t by dt
• Choose appropriate value for dt
• Ensure no pixels are missed
• Implies and
• Set dt to maximum of dx and dy

9
DDA algorithm
line(int x1, int y1, int x2, int y2) float
x,y int dx x2-x1, dy y2-y1 int n
max(abs(dx),abs(dy)) float dt n, dxdt dx/dt,
dydt dy/dt x x1 y y1 while( n-- )
point(round(x),round(y)) x dxdt y
dydt
n - range of t.
10
DDA algorithm

• Still need a lot of floating point arithmetic.
• 2 rounds and 2 adds per pixel.
• Is there a simpler way ?
• Can we use only integer arithmetic ?
• Easier to implement in hardware.

11
Observation on lines.
while( n-- ) draw(x,y) move right if( below
line ) move up
12
Testing for the side of a line.

• Need a test to determine which side of a line a
  pixel lies.
• Write the line in implicit form

• If (blt0) Flt0 for points above the line, Fgt0 for
  points below.

13
Testing for the side of a line.

• Need to find coefficients a,b,c.
• Recall explicit, slope-intercept form
• So

14
Decision variable.
Lets assume dy/dx lt 0.5 (we can use
symmetry) Evaluate F at point M Referred to as
decision variable
NE
M
E
Previous Pixel (xp,yp)
Choices for Current pixel
Choices for Next pixel
15
Decision variable.
Evaluate d for next pixel, Depends on whether E
or NE Is chosen If E chosen
NE
M
E
Previous Pixel (xp,yp)
Choices for Next pixel
Choices for Current pixel
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Decision variable.
If NE was chosen
M
NE
E
Previous Pixel (xp,yp)
Choices for Next pixel
Choices for Current pixel
17
Summary of mid-point algorithm

• Choose between 2 pixels at each step based upon
  the sign of a decision variable.
• Update the decision variable based upon which
  pixel is chosen.
• Start point is simply first endpoint (x1,y1).
• Need to calculate the initial value for d

18
Initial value of d.
Start point is (x1,y1)
But (x1,y1) is a point on the line, so F(x1,y1) 0
Conventional to multiply by 2 to remove fraction
? doesnt effect sign.
19
Midpoint algorithm

• void MidpointLine(int x1,y1,x2,y2)
• int dxx2-x1
• int dyy2-y1
• int d2dy-dx
• int increE2dy
• int incrNE2(dy-dx)
• xx1
• yy1
• WritePixel(x,y)

while (x lt x2) if (dlt 0) dincrE x
else dincrNE x y WritePixel
(x,y)
20
Circle drawing.

• Can also use Bresenham to draw circles.
• Use 8-fold symmetry

E
M
SE
Previous Pixel
Choices for Next pixel
Choices for Current pixel
21
Circle drawing.

• Implicit form for a circle is

• Functions are linear equations in terms of
  (xp,yp)
• Termed point of evaluation

22
Summary of line drawing so far.

• Explicit form of line
• Inefficient, difficult to control.
• Parametric form of line.
• Express line in terms of parameter t
• DDA algorithm
• Implicit form of line
• Only need to test for side of line.
• Bresenham algorithm.
• Can also draw circles.

22
23
Problems with Bresenham algorithm

• Pixels are drawn as a single line ? unequal line
  intensity with change in angle.

Pixel density ?2.n pixels/mm

• Can draw lines in darker colours according to
  line direction.
• Better solution antialiasing !
• (eg. Gupta-Sproull algorithm)

Pixel density n pixels/mm
24
Gupta-Sproull algorithm.

• Calculate the distance of the line and the pixel
  center
• Adjust the colour according to the distance

25
Gupta-Sproull algorithm.
Calculate distance using features of mid-point
algorithm
dy
NE
dx
Angle ?
v
M
D
E
26
Gupta-Sproull algorithm (cont)

• Recall from the midpoint algorithm
• So
• For pixel E
• So

27
Gupta-Sproull algorithm (cont)

• From previous slide
• From the midpoint computation,
• So

28
Gupta-Sproull algorithm (cont)

• From the midpoint algorithm, we had the decision
  variable (remember?)
• Going back to our previous equation

29
Gupta-Sproull algorithm (cont)

• So,
• And the denominator is constant
• Since we are blurring the line, we also need to
  compute the distances to points yp 1 and yp 1
  

30
Gupta-Sproull algorithm (cont)

• If the NE pixel had been chosen

31
Gupta-Sproull algorithm (cont)

• Compute midpoint line algorithm, with the
  following alterations
• At each iteration of the algorithm
• If the E pixel is chosen, set numerator d dx
• If the NE pixel is chosen, set numerator d dx
• Update d as in the regular algorithm
• Compute D numerator/denominator
• Color the current pixel according to D
• Compute Dupper (2dx-2vdx)/denominator
• Compute Dlower (2dx2vdx)/denominator
• Color upper and lower accordingly