CG-lecture07

# CG-lecture07 - Introduction to Computer Graphics Lecture 7...

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Unformatted text preview: Introduction to Computer Graphics Lecture 7 Rasterization Rasterization The Rendering Pipeline Transform Illuminate Transform Clip Project Rasterize Model & Camera Parameters Rendering Pipeline Framebuffer Display Does rasterization need programming? For earlier 2D graphics system based on Dos For system, programming may be needed. system, In win32 system or later, programming for In raterization is included in GL. The normal 2D or 3D graphics applications do not need programming for this. programming Rasterizing 1. Lines 1. Fill Areas Towards the Ideal Line We can only do a discrete approximation Illuminate pixels as close to the true path as Illuminate possible, consider bi-level display only possible, • Pixels are either lit or not lit What is an ideal line Must interpolate both defining end points Must appear straight and continuous • Only possible axis-aligned and 45o lines lines Must have uniform density and intensity • Consistent within a line and over all lines Must be efficient, drawn quickly • Lots of them are required!!! Lines 1. DDA Algorithm 1. Bresenham Algorithm (mid-point) Simple Line Based on slope-intercept algorithm from algebra: y = mx + b Simple approach: increment x, solve for y Floating point arithmetic required Does it Work? It seems to work okay for lines with a slope of 1 or less, but doesn’t work well for lines with slope greater than 1 – lines become more discontinuous in appearance and we must add more than 1 pixel per column to make it work. Solution? - use symmetry. Modify algorithm per octant OR, increment along x-axis if dy<dx else increment along y-axis DDA algorithm DDA = Digital Differential Analyser • finite differences Treat line as parametric equation in t : x(t ) = x1 + t ( x2 − x1 ) Start point End point - ( x1 , y1 ) ( x2 , y 2 ) y (t ) = y1 + t ( y2 − y1 ) 0 ≤ t ≤1 DDA Algorithm Start at t = 0 Start At each step, increment t by dt At x new = xold Choose appropriate value for n y = y Choose new old Ensure no pixels are missed: x2 − x1 dx = dt ≤ 1 n ( x 2 − x1 ) + dt n ( y 2 − y1 ) + dt n y2 − y1 dy = dt ≤ 1 n Set dt=1 and set n to maximum of (x2-x1) and (y2-y1) Set (x (y DDA algorithm line(int x1, int y1, int x2, int y2) { float x,y; int dx = x2-x1, dy = y2-y1; int n = max(abs(dx),abs(dy)); float dx = dx/n, dy = dy/n; x = x1; y = y1; while( n-- ) { point(round(x),round(y)); x += dx; y += dy; } } n - range of t. DDA algorithm Still need a lot of floating point arithmetic. • 2 ‘round’s and 2 adds per pixel. Is there a simpler way ? Can we use only integer arithmetic ? • Easier to implement in hardware. Lines 1. DDA Algorithm 1. Bresenham Algorithm (mid-point) Observation on lines. while( n-- ) { draw(x,y); move right; if( below line ) move up; } Testing for the side of a line. Need a test to determine which side of a line a Need pixel lies. pixel Write the line in implicit form: F ( x, y ) = ax + by + c = 0 • Easy to prove F<0 for points above the line, F>0 for points below. Testing for the side of a line. y c (a,b) F(x,y)=0 x Testing for the side of a line. Need to find coefficients a,b,c. Recall explicit, slope-intercept form : F ( x, y ) = ax + by + c = 0 dy y = mx + b and so y = x +b dx So: F ( x, y ) = dy ⋅ x − dx ⋅ y + c = 0 Decision variable. Evaluate F at point M Referred to as decision variable 1 d = F ( x p +1, y p + ) 2 NE M E Previous Pixel (xp,yp) Choices for Current pixel Choices for Next pixel Decision variable. Evaluated for next pixel, Depends on whether E or NE Is chosen : If E chosen : d new 1 1 = F ( x p + 2, y p + ) = a ( x p + 2) + b( y p + ) + c 2 2 But recall : NE M E Previous Pixel (xp,yp) Choices for Current pixel Choices for Next pixel 1 d old = F ( x p +1, y p + ) 2 1 = a ( x p +1) + b( y p + ) + c 2 So : d new = d old + a = d old + dy Decision variable. If NE was chosen : 3 3 d new = F ( x p + 2, y p + ) = a ( x p + 2) + b( y p + ) + c 2 2 M NE d new = d old + a + b = d old + dy − dx E Previous Pixel (xp,yp) Choices for Current pixel So : Choices for Next pixel Summary of mid-point algorithm Choose between 2 pixels at each step based upon Choose sign of decision variable. sign Update decision variable based upon which pixel Update is chosen. is Start point is simply first endpoint (x1,y1). Need to calculate initial value for d Initial value of d. Start point is (x1,y1) d start 1 1 = F ( x1 +1, y1 + ) = a ( x1 +1) + b( y1 + ) + c 2 2 b = ax1 + by1 + c + a + 2 b = F ( x1 , y1 ) + a + 2 But (x1,y1) is a point on the line, so F(x1,y1) =0 d start = dy − dx / 2 Conventional to multiply by 2 to remove fraction ⇒ doesn’t effect sign. Bresenham algorithm void MidpointLine(int void x1,y1,x2,y2) x1,y1,x2,y2) { int dx=x2-x1; int dy=y2-y1; while (x < x2) { if (d<= 0) { d+=incrE; x++ } else { d+=incrNE; x++; y++; } WritePixel(x,y); int d=2*dy-dx; int increE=2*dy; int incrNE=2*(dy-dx); x=x1; y=y1; WritePixel(x,y); } } Bresenham was not the end! 2-step algorithm by Xiaolin Wu: (see Graphics Gems 1, by Brian Wyvill) Treat line drawing as an automaton , or finite state machine, ie. looking at next two pixels of a line, easy to see that only a finite set of possibilities exist. The 2-step algorithm exploits symmetry by simultaneously drawing from both ends towards the midpoint. Two-step Algorithm Possible positions of next two pixels dependent on slope – current pixel in blue: Slope between 0 and ½ Slope between ½ and 1 Slope between 1 and 2 Slope greater than 2 Circle drawing. Can also use Bresenham to draw circles. Use 8-fold symmetry E M SE Previous Pixel Choices for Current pixel Choices for Next pixel Circle drawing. Implicit form for a circle is: f ( x, y ) = ( x − xc ) + ( y − yc ) − r 2 = 0 2 2 If SE is chosen d new = d old + ( 2 x p − 2 y p + 5) + (2 yc − xc ) If E is chosen d new = d old + ( 2 x p + 3) • Functions are linear equations in terms of (xp,yp) –Termed point of evaluation Circle drawing. d old 1 = f ( x p + 1, y p − ) 2 E: 1 1 2 d new = d ( x p + 2, y p − ) = ( x p + 2 − xc ) + ( y p − − yc ) 2 − r 2 2 2 1 2 = ( x p + 1 − xc ) + 2( x p + 1 − xc ) + 1 + ( y p − − yc ) 2 − r 2 2 = d old + 2( x p − xc ) + 3 Circle drawing. SE : d new 3 = ( x p + 2 − xc ) + ( y p − − yc ) 2 − r 2 2 2 1 1 2 = ( x p + 1 − xc ) + 2( x p + 1 − xc ) + 1 + ( y p − − yc ) − 2( y p − − yc ) + 1 − r 2 2 2 = d old + 2 x p + 2 − xc + 1 − 2 y p + 1 + 2 yc + 1 2 = d old + (2 x p − 2 y p + 5) + (2 yc − xc ) Problems with Bresenham algorithm Pixels are drawn as a single line ⇒ unequal line Pixels intensity with change in angle. intensity Pixel density = n/√2 pixels/mm Can draw lines in darker colours according to line direction. - Better solution : antialiasing ! Pixel density = n pixels/mm Summary of line drawing so far. Explicit form of line • Inefficient, difficult to control. Parametric form of line. • Express line in terms of parameter t Express • DDA algorithm Implicit form of line • Only need to test for ‘side’ of line. • Bresenham algorithm. • Can also draw circles. Rasterizing 1. Lines 1. Fill Areas Region Filling Seed Fill Approaches Seed • works at the pixel level • suitable for interactive painting apllications Scanline Fill Approaches • works at the polygon level • better performance Seed Fill Algorithms: Connectedness 4-connected region: From a given pixel, the region that 4-connected you can get to by a series of 4 way moves (N, S, E and W) you 8-connected region: From a given pixel, the region that 8-connected you can get to by a series of 8 way moves (N, S, E, W, NE, NW, SE, and SW) 4­connected 8­connected Boundary Fill Algorithm Start at a point inside a region Paint the interior outward to the Paint edge edge The edge must be specified in a The single color single Fill the 4-connected or 8connected region 4-connected fill is faster, but can 4-connected have problems: have Boundary Fill Algorithm (cont.) void BoundaryFill4(int x, int y, color newcolor, color edgecolor) { int current; current = ReadPixel(x, y); if(current != edgecolor && current != newcolor) { WritePixel (x, y, newcolor); BoundaryFill4(x+1, y, newcolor, edgecolor); BoundaryFill4(x-1, y, newcolor, edgecolor); BoundaryFill4(x, y+1, newcolor, edgecolor); BoundaryFill4(x, y-1, newcolor, edgecolor); } Polygon Representation Scanline Fill Algorithm Intersect scanline with polygon edges Fill between pairs of intersections Basic algorithm: For y = ymin to ymax 1) intersect scanline y with each edge 2) sort intersections by increasing x [p0,p1,p2,p3] 3) fill pairwise (p0->p1, p2->p3, …) Scanline Fill Algorithm Spacial Handling (cont.) Intersection is an edge end point Intersection points: (p0, p1, p2) ??? ->(p0,p1,p1,p2) so we can still fill pairwise Case 1 Spacial Handling (cont.) ->In fact, if we compute the intersection of the scanline with edge e1 and e2 separately, we will get the intersection point p1 twice. Keep both of the p1. Spacial Handling (cont.) However, in this case we don’t want to count p1 twice (p0,p1,p1,p2,p3), otherwise we will fill pixels between p1 and p2, which is wrong. Case 2 Spacial Handling (cont.) Summary: If the intersection is the ymin of the edge’s endpoint, count it. Otherwise, don’t. ...
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