1. Triangle Scan Conversion

2. 2 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Rasterization Rasterization (scan conversion) –Determine which pixels that are inside primitive specified by a set of vertices –Produces a set of fragments –Fragments have a location (pixel location) and other attributes such color and texture coordinates that are determined by interpolating values at vertices Pixel colors determined later using color, texture, and other vertex properties

3. Triangle Area Filling Algorithms Why do we care about triangles? Edge Equations Edge Walking

5. Do something easier! Instead of polygons, let’s do something easy! TRIANGLES! Why? 1) All polygons can be broken into triangles 2) Easy to specify 3) Always convex 4) Going to 3D is MUCH easier

6. Polygons can be broken down Triangulate - Dividing a polygon into triangles. Is it always possible? Why?

7. Any object can be broken down into polygons

8. Specifying a model For polygons, we had to worry about connectivity AND vertices. How would you specify a triangle? (What is the minimum you need to draw one?) –Only vertices (x 1,y 1 ) (x 2,y 2 ) (x 3,y 3 ) –No ambiguity –Line equations A 1 x 1 +B 1 y 1 +C 1 =0A 2 x 2 +B 2 y 2 +C 2 =0A 3 x 3 +B 3 y 3 +C 3 =0

9. Triangles are always convex What is a convex shape? An object is convex if and only if any line segment connecting two points on its boundary is contained entirely within the object or one of its boundaries. Think about scan lines again!

10. Scan Converting a Triangle Recap what we are trying to do Two main ways to rasterize a triangle –Edge Equations A 1 x 1 +B 1 y 1 +C 1 =0 A 2 x 2 +B 2 y 2 +C 2 =0 A 3 x 3 +B 3 y 3 +C 3 =0 –Edge Walking

11. Types of Triangles What determines the spans? Can you think of an easy way to compute spans? What is the special vertex here?

12. Edge Walking 1. Sort vertices in y and then x 2. Determine the middle vertex 3. Walk down edges from P 0 4. Compute spans P0P0 P1P1 P2P2

13. Edge Walking Pros and Cons Pros Fast Easy to implement in hardware Cons Special Cases Interpolation can be tricky

14. Color Interpolating P0P0 P1P1 P2P2 (?, ?, ?)

15. Edge Equations A 1 x 1 +B 1 y 1 +C 1 =0 A 2 x 2 +B 2 y 2 +C 2 =0 A 3 x 3 +B 3 y 3 +C 3 =0 How do you go from: x 1, y 1 - x 2, y 2 to A 1 x 1 +B 1 y 1 +C 1 ? P0P0 P1P1 P2P2

16. Given 2 points, compute A,B,C C = x 0 y 1 – x 1 y 0 A = y 0 – y 1 B = x 1 – x 0

17. Edge Equations What does the edge equation mean? A 1 x 1 +B 1 y 1 +C 1 =0 Pt1[2,1], Pt2[6,11] A=-10, B=4, C=16 What is the value of the equation for the: –gray part –yellow part –the boundary line What happens when we reverse P 0 and P 1? P0P0 P1P1

18. Combining all edge equations 1) Determine edge equations for all three edges 2) Find out if we should reverse the edges 3) Create a bounding box 4) Test all pixels in the bounding box whether they too reside on the same side P2P2 P1P1 P0P0

19. Edge Equations: Interpolating Color Given redness at the 3 vertices, set up the linear system of equations: The solution works out to:

20. Edge Equations: Interpolating Color Notice that the columns in the matrix are exactly the coefficients of the edge equations! So the setup cost per parameter is basically a matrix multiply Per-pixel cost (the inner loop) cost equates to tracking another edge equation value (which is?) –A: 1 add

21. Pros and Cons of Edge Equations Pros If you have the right hardware (PixelPlanes) then it is very fast Fast tests Easy to interpolate colors Cons Can be expensive if you don’t have hardware 50% efficient

22. Recap P0P0 P1P1 P2P2 P1P1 P2P2 P0P0

23. 23 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Scan Conversion of Line Segments Start with line segment in window coordinates with integer values for endpoints Assume implementation has a write_pixel function y = mx + h

24. 24 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 DDA Algorithm Digital Differential Analyzer –DDA was a mechanical device for numerical solution of differential equations –Line y=mx+ h satisfies differential equation where m =  y/  x = y 2 -y 1 /x 2 -x 1 Along scan line  x = 1 at each iteration of the loop For(x=x1; x<=x2,ix++) { y+=m; write_pixel(x, round(y), line_color) }

25. 25 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Problem DDA = for each x plot pixel at closest y –Problems for steep lines

26. 26 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Using Symmetry Use for 1  m  0 For m > 1, swap role of x and y –For each y, plot closest x

27. 27 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Bresenham’s Algorithm DDA requires one floating point addition per step We can eliminate all fp through Bresenham’s algorithm Consider only 1  m  0 –Other cases by symmetry Assume pixel centers are at half integers If we start at a pixel that has been written, there are only two candidates for the next pixel to be written into the frame buffer

28. 28 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Candidate Pixels 1  m  0 last pixel candidates

29. 29 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Decision Variable - d =  x(b-a) d is an integer d > 0 use upper pixel d < 0 use lower pixel

30. 30 Angel: Interactive Computer Graphics 5E © Addison-Wesley 2009 Incremental Form More efficient if we look at d k, the value of the decision variable at x = k d k+1 = d k –2  y, if d k <0 d k+1 = d k –2(  y-  x), otherwise For each x, we need do only an integer addition and a test Single instruction on graphics chips