Graphics Lecture Note

1
Part II Line Drawing
2
Scan Conversion for Lines

• Lets assume 1-pixel-thick line.
• If slope 1, one pixel in a column.
• If slope 1, one pixel in a row.
• Lets take slope 1 case only.
• If slope 1, we can reverse the roles of x and
  y.
• Without loss of generality, we will focus on the
  case 0
  lines can be handled separately.

y
?
x
3
A Brute Force Algorithm
LineDrawing1(x1,y1,x2,y2) // a line specified
by two integer end points m (y2 - y1) / (x2 -
x1) // slope m for the line equation B y1
- mx1 // compute the y-intercept B using
m for (xx1 x SetPixel(x, round(y))
(x2,y2)
ymxB
(x1,y1)
B
Example
(5,3)
(4,2.33)
(9,4)
3
y(1/3)x1
(3,2)
(4,2)
2
2
(5,2.66)
(3,2)
1
3
4
5
m and B computation
1st iteration
3rd iteration
2nd iteration
4
A Brute Force Algorithm (contd)

• LineDrawing1 is inefficient because, for each
  iteration, we need
• ? floating-point multiplication
• ? floating-point addition, and
• ? round operation.
• LineDrawing1(x1,y1,x2,y2)
• m (y2 - y1) / (x2 - x1)
• B y1 - mx1
• for (xx1 x
• y mx B
• SetPixel(x, round(y))
• Lets try to drop ? multiplication.

? addition
? multiplication
? round operation
5
An Improved Algorithm

• Note that, when x is increased by 1, y is
  increased by m. So, we dont have to use the line
  equation y mx B to update y value.
• LineDrawing2(x1,y1,x2,y2)
• y y1
• m (y2 - y1) / (x2 - x1)
• for (xx1 x
• SetPixel(x, round(y))
• y y m
• We now need ? floating-point addition and ? round
  operation.
• Not bad, but any better algorithm?
• Yes, there is an integer-addition-only version.
  Its Bresenhams algorithm.

6
Bresenhams Line Drawing Algorithm

• In the process of rasterization from (x1,y1) to
  (x2,y2), suppose that a pixel (xk, yk) is drawn.
• Obviously, the x-position of the next pixel is
  xk1.
• 0either yk or yk1.
• Which one to choose? E(xk1, yk) or NE(xk1,
  yk1)?

NE
E
xk
xk1
7
Bresenhams Line Drawing Algorithm (contd)

• Obviously, which one to choose between E and NE
  depends on which pixel is closer to the ideal
  line.
• If d1 d2, choose NE. Otherwise choose E. (If d1
  d2, we simply choose E. Of course, we could
  choose NE. It doesnt matter.)
• Let the ideal lines y-coordinate at xk1 be y
  m(xk1) B. Then,
• In the sense that the sign of d1-2 determines
  between E and NE, d1-2 is called the decision
  variable.

NE
yk1
d2
(xk1, y)
d1
yk
E
(xk, yk)
d1 y yk m(xk1) B yk d2 yk1 - y
yk1 - m(xk1) - B d1-2 d1 - d2 2m(xk1) 2B
2yk 1
8
Bresenhams Line Drawing Algorithm (contd)

• Lets replace the slope m by dy/dx(y2-y1)/(x2-x1)
  .
• Important is the sign of the decision variable
  d1-2, not its value. So multiply d1-2 by dx( 0)
  to make a new version of the variable D.
• D and d1-2 play the same role!!!
• If D 0, choose NE.
• Otherwise, choose E.
• In D, all except xk and yk are constants. So,
  rewrite D as follows
• Observe that, after choosing the pixel (xk, yk),
  we compute D for the next step by inserting (xk,
  yk) into Ds formula.

d1-2 2m(xk1) 2B 2yk 1 2(dy/dx)(xk1)
2B 2yk 1
D 2dy(xk1) 2Bdx 2ykdx dx 2xkdy
2ykdx 2dy 2Bdx dx
If D 0, choose NE
(xk, yk)
Otherwise, choose E
D 2xkdy 2ykdx C, where C 2dy 2Bdx dx
9
Bresenhams Line Drawing Algorithm (contd)

• Suppose that we chose a pixel (xk, yk) at x xk.
  
• For the next step x xk1, we should compute the
  decision variable by inserting (xk, yk) into Ds
  formula. Lets call it Dcurrent.
• Dcurrent is used to determine between E and NE.
  (In this example, just assume that E happens to
  be chosen.)
• After choosing between E and NE at x xk1, we
  also have to compute the new decision variable
  Dnew for the next step x xk2.
• Recall that D is computed using the just-chosen
  pixels coordinate. The value of Dnew depends on
  whether we choose E or NE.
• Importantly, Dnew can be incrementally computed
  from Dcurrent.

Dcurrent
Dold
Dnew
(1) Choose a pixel using Dold computed at x
xk-1
(3) At x xk1, choose a pixel using Dcurrent
yk1
NE
(4) Compute Dnew for x xk2, and keep going
(2) Compute Dcurrent for x xk1
yk
E
xk
xk1
xk2
xk-1
10
Bresenhams Line Drawing Algorithm (contd)

• Recall D 2xkdy 2ykdx C.
• At xk1, between E and NE, suppose we choose
  E(xk1, yk).
• Then, Dnew 2(xk1)dy 2ykdx C.
• Dcurrent 2xkdy 2ykdx C. So, Dnew Dcurrent
  2dy.
• In contrast, suppose we choose NE(xk1, yk1).
• Then, Dnew 2(xk1)dy 2(yk1)dx C. Dnew
  Dcurrent 2dy2dx.
• In summary, if Dcurrent 0, choose NE and let
  Dnew be Dcurrent 2dy2dx Otherwise, choose E
  and let Dnew be Dcurrent 2dy.
• It is very important to note that both 2dy and
  2dy-2dx are integers. So, if Dcurrent is
  guaranteed to be an integer, we are doing just
  integer additions all over the repeated steps.
• Whether Dcurrent is an integer depends on Dold,
  etc. Ultimately, if the first variable (named
  Dstart) is an integer, the world is perfect!!

11
Bresenhams Line Drawing Algorithm (contd)

• It is obvious that the first pixel to be drawn is
  (x1,y1).
• After choosing (x1,y1), compute the first
  decision variable as follows
• Note that y1 mx1B (dy/dx)x1B.
• Both dx and dy are integers, and so is Dstart.
• So, Bresenhams line drawing algorithm works
  using integer additions only. What a gorgeous
  algorithm!!!

Dstart 2x1dy 2y1dx 2dy 2Bdx dx
12
Pseudo Code of Bresenhams Algorithm

• LineDrawing3(x1,y1,x2,y2)
• x x1
• y y1
• dx (x2 - x1)
• dy (y2 - y1)
• D (2dy - dx)
• delta1 2dy
• delta2 2dy - 2dx
• SetPixel(x, y)
• for (xx11 x
• if (D0)
• y y 1
• D D delta2
• else
• D D delta1
• SetPixel(x, y)

Why dont you simulate LineDrawing1 and
LineDrawing3 and compare the CPU times?
13
Sampling, Aliasing, and Anti-aliasing

• A line consists of ? points. Such a continuous
  line is sampled with a 2D array of finite number
  of discrete points. Therefore, jaggies or
  staircasing is inevitable.
• Information distortion due to undersampling is
  called aliasing, which is a ubiquitous problem in
  signal processing and computer graphics.
• We need anti-aliasing. Intuitively, anti-aliasing
  is a technique for reducing jaggies.
• The most popular anti-aliasing technique is
  supersampling. (Recall that aliasing is caused by
  undersampling.)

14
Supersampling for Line Drawing

• Supersampling sampling an object at a higher
  resolution, and combining the results to
  reconstruct the object at a lower resolution.
• We have 4-level intensities as the max of
  subpixels of a pixel is 3.

divide each pixel Into e.g. 9 sub-pixels,
and apply Brensenhams with the sub-pixels
count the number of intensified sub-pixels
1
3
3
pixel intensity is set in proportion to the
number of sub-pixels
3
2
3
15
Antialiased Line Example

• The stairstep effect is smoothed by displaying a
  somewhat blurred line path in the vicinity of the
  stairsteps (between horizontal lines).
• Anti-aliasing is needed not only for lines, but
  also for polygon edges, textures, etc.
• Anti-aliasing is more complicated for
  polygons/textures than for lines.