CS335 Fall 2007 Graphics and Multimedia. 2D Drawings: Lines

Transcription

1 CS335 Fall 007 Grahics and Multimedia D Drawings: Lines

2 Primitive Drawing Oerations Digital Concets of Drawing in Raster Arras PIXEL is a single arra element at x, - No smaller drawing unit exists Px, 0 0

3 Scan Conversion Pixel Value = bit, 8 bit, 4 bit,. Pixel based Geometr = address into hardware look u table to determine disla color, intensit. Points, lines, circles, conics, curves, slines, olgons, shaded olgons, text fonts & icons. all basicall reduce to scan conversion roblem: FIND Digital algorithms for continuous geometric concets

4 Drawing Lines From x, to x, Equation: = mx + b m = -/x-x b = m*x x x comute -- m, b for x=x to x, ste = m*x + b ixel x, round, color loo

5 comute m assume m < DDA Algorithm =, x = x Digital Differential Analzer [DDA avoids the multile b sloe m] Equation: x k+ = x k + k+ = k + m if sloe m > k+ = k + x k+ = x k + /m k+ m> k+ k ixelroundx, round, color for x=x to x, ste x = x + = + m ixelx, round, color end m> x k x k+ x k+ x k+3

6 DDA Algorithm Digital Differential Analzer [avoid the float multile b sloe m] sloe m, floats Problems:. Necessar to erform float addition. Necessar to have float reresentation 3. Necessar to erform -- round Float reresentations/oerations are more exensive than integer oerations. We would like to avoid them if ossible.

7 Drawing a Fast Line The Midoint Algorithm Bresenham s Algorithm Scan-converts lines Method is incremental Uses onl integer arithmetic Assume sloe of line to be drawn is 0 m Lower-left endoint: Uer-right endoint: x 0, 0 x,

8 The Midoint Algorithm x, NE E M Ke idea: at each ste, decide which ixel E or NE is closest to the line. Choose that ixel and draw it.

9 The Imlicit Line Equation If M lies above the line, draw ixel E If M lies below the line, draw ixel NE NE M x, E

10 Imlicit Lines F x, = ax + b + c = 0 Let d dx = = x x 0 0 m = d dx The sloe-intercet form gives = d dx x + B

11 Imlicit Lines Now we can convert the sloe-intercet form to the imlicit form via a few substitutions and maniulations: dx = d x + dx B 0 = d x dx + dx B Which is in the form 0 = ax + b + c b letting a = d b = dx c = Bdx

12 Imlicit Lines Now we can verif an imortant roert of this reresentation: 0, 0, 0, < > = x F x F x F for oints on the line for oints below the line for oints above the line Examle:

13 Imlicit Lines Conclusion: the sign of Fx, reveals the location of x, with resect to the line reresented b F Idea: Check the midoint at each iteration to determine which ixel, E or NE, is closest to the line. Use the imlicit equation of the line F to check

14 The Midoint Test NE M = x, + + x, E M The sign of FM gives the answer as to which ixel, E or NE, to draw

15 Formulating an Algorithm Let d be a decision variable which makes the midoint test. Then the test to decide which ixel to draw is just if d > 0 then // the midoint is below the line // the line asses closer to the uer ixel draw the NE ixel else // the midoint is above the line // the line asses closer to the lower ixel draw the E ixel

16 Making the Algorithm Incremental NE M M 3 M M M M 3 = x = x = x +, + +,, x, E x +, x +,

17 Incremental Formulation d first = F M = F x +, + = a x + + b + c + If we choose E after this comutation, then we must comute F M next. What is the relationshi between d first = F M and F M

18 Incremental Formulation Notice that d first = a x + + b + + c dnext = a x + + b + + since d next = F M which is just so that we can comute dnext = d first + dnext = d first + a d c

19 Choosing E vs. NE When we comute F M and draw the E ixel, F M can be obtained from a single addition oeration via the equation F M = F M + d What if we are drawing the NE ixel instead of the E ixel? d first = a x + + b + + c 3 dnext = a x + + b + + d next which gives this incremental comutation of F M = d first + d dx c 3

20 A Starting Point What about a starting oint? The first midoint to be comuted is F x0 +, 0 + Thus the starting value for the test variable is NE M d start = a x + + b c x, E How can we simlif this?

21 Efficiencies 0 0 d = a x + + b + + start = = ax + b + c + a F + b x0, 0 + a b c Now observe that since the oint F x 0, 0 is on the line: F M = a + = d dx start b /

22 Integer Arithmetic We can maintain and test the value of F M to make it integral d start = F M = start d dx If we move E, the increment to maintain the midoint is multilied b : dnext = d first + d If we move NE, the increment to maintain the midoint is multilied b : d next = d first + d dx

23 The Algorithm MidointLine x0, 0, x, // assumes sloe is between 0 and // and that x0, 0 is leftmost oint dx = x-x0; d = -0; d = d-dx; incre = d; incrne = d-dx; x = x0; = 0; WritePixelx,;

24 Algorithm Con t while x < x do if d <= 0 d = d + incre x = x + else d = d + incrne x = x + = + WritePixel x, end end

25 Summar of Concets Imlicit reresentation of a line Incremental algorithm for more efficient comutation Integer algorithm b introducing harmless scale factor Logic must be added to handle all line orientations