2D Output Primitives

1. 2D Output Primitives Points Lines Circles Ellipses Other curves Filling areas Text Patterns Polymarkers

2. Point plotting A coordinate position in an application program corresponds to a pixel (picture element) position on an output device Typically, a memory position in a frame buffer (display memory) corresponds to a specific display position

3. Points (pixels) The size of a display point (pixel) is a function of the diameter of the electron beam (if CRT) and the size of the phosphor spot on the display Too small points => a grainy image Too large points => a fuzzy image The minimal functionality required to perform computer graphics: setPixel(x, y, intensity) intensity = getPixel(x, y)

4. Coordinate system

5. Coordinate pixels

6. Line plotting Given the two endpoints, line plotting means calculating intermediate pixel positions (scan conversion) Requirements on a good line plotter: Straight line Ending correctly Constant thickness Thickness independent of length and direction Fast drawing

7. Incremental methods Normally, incremental methods are used (point-by-point plotting) Most usual are DDA methods (Digital Differential Analyzer), parametric Lines (curves) are generated using the differential equation of the lines (curves) For lines: y� = ?y/?x (constant) where y = m.x + b (line equation) and m = ?y/?x (line slope)

8. Technique Increment x and y by small steps proportional to the first derivatives of x and y, respectively For an ideal display: increment x and y by ??x and ??y, ? �small�, from one endpoint position to the other In practise: only integer coordinate positions are available on a device=> rounding to nearest integer position in each step necessary Choose the increments ??x and ??y = 1 !

9. Line drawing algorithms Symmetric DDA choose ? so that ?=2-n and 2n-1=max(?x,?y)<2n Simple DDA choose ?=1/max(?x,?y), i.e. unity steps in one direction Bresenham�s algorithm with an error term and unity steps in one direction

10. Example Symmetric DDA Endpoints (0,0) and (12,10) ?x =12, ?y=10, 2n-1=max(12,10)<2n 23=max(12,10)<24, i.e. n=4 and ?=1/16 Use ??x=12.1/16=3/4 and ??y= =10.1/16= 5/8 as increments in each step Initiate with fraction part=0.5 (cp. rounding to nearest integer) and start in endpoint (0,0) Results in 15 different points out of 17 generated in total

11. Example Simple DDA Endpoints (0,0) and (12,10) ?x =12, ?y=10, ?=1/max(?x,?y) ?=1/max(12,10)=1/12 Use ??x=12.1/12=1 (unity step!) and ??y= =10.1/12= 5/6 as increments in each step Initiate with fraction part=0.5 (cp. rounding to nearest integer) and start in endpoint (0,0) Results in 13 different points out of 13 generated in total

12. Plotted line Line with endpoints in (0,0) and (12,10) plotted with Symmetric DDA in the first diagram and Simple DDA and Bresenham in the second

13. Bresenham�s algorithm 1 A unit step is incremented in one direction - the direction with the biggest slope m. If -1=m=1 then unit steps in x else unit steps in y. The other coordinate is changed depending on an error term e, which will be influenced by the algorithm The error term e is defined by the distance between the exact line and a generated pixel position (see figure)

14. Error term

15. Bresenham�s algorithm 2 In each step m=?y/?x is added to e. Before this is done, the sign of e decides whether the coordinate with the smallest slope is changed or not (always unit steps in the other coordinate) Initiating e=?y/?x-0.5 gives (assume biggest slope in x): e=0 => the exact line is above the current pixel, i.e. y is incremented and e is decremented by one e<0 => the exact line is below the current pixel, i.e. y is not changed Using integers only, if e is multiplied by 2?x => no divisions and only multiplications by 2

16. The algorithm Again, assume biggest slope in x and 0=m=1 Initiate e := 2?y - ?x if e=0 then e:= e+(2?y - 2?x), xk+1 := xk+1, yk+1 := yk +1 if e<0 then e:= e + 2?y, xk+1 := xk+1

17. Example Bresenham Endpoints (0,0) and (12,10) ?x =12, ?y=10 => 0=m=1, i.e. unit steps in x (biggest slope) => 2?y = 20, 2?y-2?x = -4 Initiate error term e = 2?y-?x = 8 In each step (always incr x): if e=0 then e:=e+(2?y-2?x)= e - 4 and incr y if e<0 then e:=e+2?y= e + 20 Results in 13 different points out of 13 generated in total

18. Circle generating methods In general, assume circle midpoint in origin If not, only a simple translation is required f(x,y) = x2 + y2 - r2 = 0 (circle equation) DDA-technique Direct from the equation Polar coordinates Bresenham

19. DDA-technique First derivative of the equation: 2x + 2y . y� = 0 <=> y� = -x/y If 2n-1 = r < 2n => choose ? = 2-n xk+1 := xk + ? . yk yk+1 := yk - ? . xk

20. Direct from the equation y2 = r2 - x2 In the sector from 90� to 45�, take small steps in x from 0 to r/sqrt(2) and cal-culate y=sqrt(r2 - x2) for each x

21. Using polar coordinates x = r . cos ? y = r . sin ? From 0 to ?/4 (45�), take small steps in ? and calculate x and y for each ?

22. Symmetry of a circle Calculation of a circle point (x,y) in one octant yields the circle points for the other seven octants

23. Intro to Bresenham The best approximation will be described if the pixels choosen are those closest to the true circle. If plotting from 90� to 45� the next pixel to be plotted is either given by a) increment x by one or b) increment x by one and decrement y by one A method which is only using integers, integer additions and multiplications by 2 and 4 (cp binary shifts) is described by Bresenham

24. Deriving Bresenham

25. Bresenham continued A decision (error) variable di is then defined: di = D(Si) + D(Ti) The initial value on this variable can be derived as: d1 = 3 - 2r, where x0= 0 and yo= r

26. Bresenham continued The sign of the decision variable will then decide how to continue: if di<0 then xi:=xi-1 + 1 di+1:=di + 4xi-1 +6 else (if di=0) xi:=xi-1 +1 yi:=yi-1 - 1 di+1:=di + 4(xi-1 - yi-1) + 10

27. Ellipse properties Given two fixed positions F1 and F2, the sum of the two distances from these points to any point P on the ellipse (d1+d2) is constant.

28. Ellipse Equation With the major and minor axes aligned with the coordinate axes the ellipse is said to be in standard position:

29. One approach Take small steps in x from xc to xc+rx and calculate y:

30. Symmetry can be used Only one quadrant needs to be calculated since symmetry gives the three other quadrants

31. General curves A direct method: Given a general curve y = f(x), take small steps in x and calculate y=f(x) for each x.