CP411 polygon Lecture 6 1. Scan conversion algorithm for circle 2. Draw polygons 3. Antialiasing

CP411 Computer Graphics 2 Draw a circle_ straight algorithm #include const float DEG2RAD = /180; void drawCircle(float radius, float x, float y) { glBegin(GL_LINE_LOOP); for (int i=0; i < 360; i++) { float degInRad = i*DEG2RAD; glVertex2f(x+cos(degInRad)*radius, y+sin(degInRad)*radius); } glEnd }

Midpoint point algorithm ● See class note CP411 polygon

Polygon ● What is a polygon? ■ A geometric shape bounded by a closed polyline formed by a sequence of vertices. ● Why choose polygon as a primitive ? ■ In interactive graphics, polygons rule the CG world ■ Two main reasons: ○ Can represent any surface with arbitrary accuracy  Splines, mathematical functions, etc. ○ Mathematical simplicity lends itself to simple, regular rendering algorithms  Such algorithms embed well in hardware

CP411 polygon Convex / Concave Polygons ● A two-dimensional shape is convex if and only if every line segment connecting two points on the boundary is entirely contained ● OpenGL can only draw convex polygon ■ How to recognize a convex polygon ?

CP411 polygon OpenGL polygon command ● glBegin (GL_POLYGON) specify vertices of the polygon glend ● glPolygonMode( GLenum face, GLenum mode ) controls the interpretation of polygons for rasterization ■ face Specifies the polygons that mode applies to. Must be GL_FRONT for front-facing polygons, GL_BACK for back-facing polygons, or GL_FRONT_AND_BACK for front- and back-facing polygons. ■ mode Specifies the way polygons will be rasterized. Accepted values are GL_POINT, GL_LINE, and GL_FILL. The default is GL_FILL for both front- and back-facing polygons.

CP411 polygon Rasterizing Polygons ● Triangle is the minimal unit of a polygon ■ All polygons can be broken up into triangles ○ Convex, concave, complex ■ Triangles are guaranteed to be: ○ Planar ○ Convex ● We often draw polygons by breaking them into triangles

CP411 polygon Triangles ● Two basic methods for drawing triangles: ■ Edge walking ○ Use a line drawing algorithm to compute edges (edge setup) ○ Fill in pixels between edges ○ Finicky, lots of special cases ○ Touches only pixels involved in triangle ■ Edge equations

CP411 polygon Edge Walking ● Basic idea: ■ Draw edges vertically ■ Fill in horizontal spans for each scanline ■ Interpolate colors down edges ■ At each scanline, interpolate edge colors across span

CP411 polygon Edge Walking: Notes ● Order vertices in x and y ■ 3 cases: break left, break right, no break ● Walk down left and right edges ■ Fill each span ■ Until breakpoint or bottom vertex is reached ● Advantage: can be made very fast ● Disadvantages: ■ Lots of finicky special cases ■ Need to pay attention to fractional offsets

CP411 polygon Edge Walking: Notes ● Fractional offsets: ● Also: beware gaps between adjacent edges

CP411 polygon Edge Equations ● An edge equation is simply the equation of the line containing that edge ■ What is the equation of a 2D line? ■ Ax + By + C = 0 ■ Given a point (x,y), what does plugging x & y into this equation tell us? ■ Whether the point is: ○ On the line: Ax + By + C = 0 ○ “Above” the line: Ax + By + C > 0 ○ “Below” the line: Ax + By + C < 0

Edge Equations ● Edge equations thus define two half-spaces: CP411 polygon

Edge Equations ● And a triangle can be defined as the intersection of three positive half-spaces: A 1 x + B 1 y + C 1 < 0 A 2 x + B 2 y + C 2 < 0 A 3 x + B 3 y + C 3 < 0 A 1 x + B 1 y + C 1 > 0 A 3 x + B 3 y + C 3 > 0 A 2 x + B 2 y + C 2 > 0 CP411 polygon

Edge Equations ● So…simply turn on those pixels for which all edge equations evaluate to > 0: CP411 polygon

Using Edge Equations ● Which pixels: compute min,max bounding box ● Edge equations: compute from vertices ● Orientation: ensure area is positive (why?) CP411 polygon

Computing Edge Equations ● Want to calculate A, B, C for each edge from (x i, y i ) and (x j, y j ) ● Treat it as a linear system: Ax 1 + By 1 + C = 0 Ax 2 + By 2 + C = 0 ● Notice: two equations, three unknowns ● Does this make sense? What can we solve? ● Goal: solve for A & B in terms of C CP411 polygon

Computing Edge Equations ● Set up the linear system: ● Multiply both sides by matrix inverse: ● Let C = x 0 y 1 - x 1 y 0 for convenience ■ Then A = y 0 - y 1 and B = x 1 - x 0 CP411 polygon

Computing Edge Equations: Numerical Issues ● We can avoid the problem in this case by using our definitions of A and B: A = y 0 - y 1 B = x 1 - x 0 C = x 0 y 1 - x 1 y 0 Thus: C = -Ax 0 - By 0 or C = -Ax 1 - By 1 CP411 polygon

Edge Equations ● So…we can find edge equation from two verts. ● Given three corners C 0, C 1, C 0 of a triangle, what are our three edges? ● How do we make sure the half-spaces defined by the edge equations all share the same sign on the interior of the triangle? ● Be consistent (Ex: [C 0 C 1 ], [C 1 C 2 ], [C 2 C 0 ]) ● How do we make sure that sign is positive? ● A: Test, and flip if needed (A= -A, B= -B, C= -C) CP411 polygon

Edge Equations: Code ● Basic structure of code: ■ Setup: compute edge equations, bounding box ■ (Outer loop) For each scanline in bounding box... ■ (Inner loop) …check each pixel on scanline, evaluating edge equations and drawing the pixel if all three are positive CP411 polygon

findBoundingBox(&xmin, &xmax, &ymin, &ymax); setupEdges (&a0,&b0,&c0,&a1,&b1,&c1,&a2,&b2,&c2); /* Optimize this: */ for (int y = yMin; y <= yMax; y++) { for (int x = xMin; x <= xMax; x++) { float e0 = a0*x + b0*y + c0; float e1 = a1*x + b1*y + c1; float e2 = a2*x + b2*y + c2; if (e0 > 0 && e1 > 0 && e2 > 0) setPixel(x,y); } CP411 polygon

General Polygon Rasterization ● Consider the following polygon: ● How do we know whether a given pixel on the scanline is inside or outside the polygon? A B C D E F CP411 polygon

General Polygon Rasterization ● Does it still work? A B C D E F G I H CP411 polygon

General Polygon Rasterization ● Basic idea: use a parity test for each scanline edgeCnt = 0; for each pixel on scanline (l to r) if (oldpixel->newpixel crosses edge) edgeCnt ++; // draw the pixel if edgeCnt odd if (edgeCnt % 2) setPixel(pixel); CP411 polygon

Faster Polygon Rasterization ● How can we optimize the code? for each scanline edgeCnt = 0; for each pixel on scanline (l to r) if (oldpixel->newpixel crosses edge) edgeCnt ++; // draw the pixel if edgeCnt odd if (edgeCnt % 2) setPixel(pixel); ● Big cost: testing pixels against each edge ● Solution: active edge table (AET) CP411 polygon

Active Edge Table ● Idea: ■ Edges intersecting a given scanline are likely to intersect the next scanline ■ Within a scanline, the order of edge intersections doesn’t change much from scanline to scanline CP411 polygon

Active Edge Table ● Algorithm: ■ Sort all edges by their minimum y coord ■ Starting at bottom, add edges with Y min = 0 to AET ■ For each scanline: ○ Sort edges in AET by x intersection ○ Walk from left to right, setting pixels by parity rule ○ Increment scanline ○ Retire edges with Y max < Y ○ Add edges with Y min > Y ○ Recalculate edge intersections and resort (how?) ■ Stop when Y > Y max for last edges CP411 polygon

Edge Equations: Speed Hacks ● Some speed hacks for the inner loop: int xflag = 0; for (int x = xMin; x <= xMax; x++) { if (e0|e1|e2 > 0) { setPixel(x,y); xflag++; } else if (xflag != 0) break; e0 += a0; e1 += a1; e2 += a2; } ■ Incremental update of edge equation values ■ Early termination (why does this work?) ■ Faster test of equation values CP411 polygon

Antialiasing ● What is aliasing? The jaggies appeared in the drawings of lines are called aliasing. It is due to the approximation of a continuous line with discrete pixels. ● Aliasing effects: ■ Distant objects may disappear entirely ■ Objects can blink on and off in animations ● Antialiasing techniques involve some form of blurring to reduce contrast, smoothen image ● Three antialiasing techniques: ■ Prefiltering ■ Postfiltering ■ Supersampling CP411 polygon

Prefiltering ● Basic idea: ■ compute area of polygon coverage ■ use proportional intensity value ● Example: if polygon covers ¼ of the pixel ■ use ¼ polygon color ■ add it to ¾ of adjacent region color ● Cons: computing pixel coverage can be time consuming CP411 polygon

Prefiltering ● Conceive that a line is 1 pixel wide which covers certain pixel squares. ● A simple drawing method sets the pixels in C, G, H and L to the current color. ● Suppose that the overlapping proportion of a certain pixel is , the new pixel_color is then set to   current_color + (1-  )  existing_color 

CP411 polygon glEnable(GL_BLEND); glEnable(GL_LINE_SMOOTH); glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA); glHint(GL_LINE_SMOOTH_HINT, GL_DONT_CARE); Without antialiasing With antialiasing

Supersampling ● Useful if we can compute color of any (x,y) value on the screen ■ Increase frequency of sampling ■ Instead of (x,y) samples in increments of 1 ■ Sample (x,y) in fractional (e.g. ½) increments ■ Find average of samples ● Example: Double sampling = increments of ½ = 9 color values averaged for each pixe CP411 polygon

Postfiltering ● Supersampling uses average ■ Gives all samples equal importance ● Post-filtering: use weighting (different levels of importance) ● Compute pixel value as weighted average ■ Samples close to pixel center given more weight CP411 polygon