CS 418 : EXAM 2 REVIEW TA: Ryan Freedman Sushma Kini Chi Zhou

LIST of TOPICS  Vertex Shader  Image and Color Spaces  Rasterization  Texture Mapping and Coordinates  Visibility

CMY Subtractive Color  Cyan, Magenta, Yellow  Color model used in pigments and reflective materials (ink,paint)  Grade school color rules Blue + Yellow = Green? Cyan + Yellow = Green  Also CMYK (blacK) C + M + Y = Brown? C + M + Y = Black (in theory) C + M + Y = Gray (in practice) C M Y

Selecting Colors HSV = Hue, Saturation, Value  1978, Alvy Ray Smith  Hue [0,360] is angle about color wheel 0  = red, 60  = yellow, 120  = green, 180  = cyan, 240  = blue, 300  = magenta  Saturation [0,1] is distance from gray S = (maxRGB – minRGB)/maxRGB  Value [0,1] is distance from black V = maxRGB HSL = Hue, Saturation, Lightness  Double cone, saturation in middle  = maxRGB – minRGB maxRGB = R  H = (G – B)/  maxRGB = G  H = 2+(B – R)/  maxRGB = B  H =4+ (R – G)/  H = (60*H) mod 360 Eric Pierce

Exercise For each row’s named color, enter its corresponding coordinates in each of the column’s color spaces and compute its luminance in the last column. Color Name(R, G, B)(C, M, Y)(H, S, V)YLuminance Blue(0, 0, 1)

Exercise For each row’s named color, enter its corresponding coordinates in each of the column’s color spaces and compute its luminance in the last column. Color Name(R, G, B)(C, M, Y)(H, S, V)YLuminance Blue(0, 0, 1)(1, 1, 0)

Exercise For each row’s named color, enter its corresponding coordinates in each of the column’s color spaces and compute its luminance in the last column. Color Name(R, G, B)(C, M, Y)(H, S, V)YLuminance Blue(0, 0, 1)(1, 1, 0)

Exercise For each row’s named color, enter its corresponding coordinates in each of the column’s color spaces and compute its luminance in the last column. Color Name(R, G, B)(C, M, Y)(H, S, V)YLuminance Blue(0, 0, 1)(1, 1, 0)(240, 1, 1).11

Rasterization Converts  lines and triangles  with floating point vertices  in viewport (screen) coordinates into  pixels  with integer coordinates  in viewport (screen) coordinates pixels centered at grid vertices, not grid cells

Line Rasterization  How to rasterize a line from (0,0) to (4,3)  Pixel (0,0) and (4,3) easy  One pixel for each integer x-coordinate  Pixel’s y-coordinate closest to line  If line equal distance between two pixels, pick on arbitrarily but consistently ? ?

Midpoint Algorithm  Which pixel should be plotted next?  East?  Northeast?  Line equation y = mx + b m = (y 1 – y 0 )/(x 1 – x 0 ) b = y 0 – mx 0 f(x,y) = mx + b – y NE E M f > 0 f < 0

Midpoint Increments f(M) = f(P) + m – ½ f(M E )= f(x+2,y+½) = m(x+2) + b – (y+½) = f(P) + 2m – ½ = f(M) + m f(M NE ) = f(x+2,y+1½) = m(x+2) + b – (y+1½) = f(P) + 2m – 1½ = f(M) + m – 1 f(1, ½)= m + b – ½ = m – ½ if line starts at origin NE E M P EE ENE NE 2 MEME M NE

Exercise For a line that extends from (0,0) to (8,2): Define a function f(x,y) that is negative above the line and positive below the line. Determine f(x,y) at the first midpoint Determine the point to rasterize (E or NE) Determine the location of the pixel drawn

Exercise For a line that extends from (0,0) to (8,2): Define a function f(x,y) that is negative above the line and positive below the line. f(x,y) = mx+b-y = (x/4)+0-y = (x/4)-y Determine f(x,y) at the first midpoint Determine the point to rasterize (E or NE) Determine the location of the pixel drawn

Exercise For a line that extends from (0,0) to (8,2): Define a function f(x,y) that is negative above the line and positive below the line. f(x,y) = mx+b-y = (x/4)+0-y = (x/4)-y Determine f(x,y) at the first midpoint Midpoint = (1,1/2) f(1,1/2) = ¼ - ½ = -¼ Determine the point to rasterize (E or NE) Determine the location of the pixel drawn

Exercise For a line that extends from (0,0) to (8,2): Define a function f(x,y) that is negative above the line and positive below the line. f(x,y) = mx+b-y = (x/4)+0-y = (x/4)-y Determine f(x,y) at the first midpoint Midpoint = (1,1/2) f(1,1/2) = ¼ - ½ = -¼ Determine the point to rasterize (E or NE) Because f(m) <= 0 : E, the midpoint is above the line Determine the location of the pixel drawn E: Current pixel P = (Px,Py), E = (Px+1,Py) (0+1,0) = (1,0)

The Over Operator  How to indicate which parts of front picture are clear and which are opaque  Use alpha channel to indicate opacity [Smith]  Over operator [Porter & Duff S’84]  A over B: C A over B =  A C A + (1 –  A )  B C B  A over B =  A + (1 –  A )  B  Note that  A C A used in color equations, so store  A C A instead of C A  A over B w/premultiplied alpha C A over B = C A + (1 –  A ) C B  A over B =  A + (1 –  A )  B B A C = (  R  G  B  )

Z-Buffer  Algorithm: For each rasterized fragment (x,y) If z > zbuffer(x,y) then framebuffer(x,y) = fragment color zbuffer(x,y) = z  Question: For fragment (1,1), if zbuffer(1,1) contains the value -3. Will the point having z value -2 be displayed?

Z-Buffer  Algorithm: For each rasterized fragment (x,y) If z > zbuffer(x,y) then framebuffer(x,y) = fragment color zbuffer(x,y) = z  Question: For fragment (1,1), if zbuffer(1,1) contains the value -3. Will the point having z value -2 be displayed? Yes

Normalized View Volume x y z x y z Model Coords World Coords Viewing Coords Clip Coords Screen Coords (left,top,-near) (right,bottom,-near) (0,0,-far) glFrustum(left,right,bottom,top,near,far) (-1,1,1) z 1 x 1 y 1

Question  Consider a scene. For the eye’s view, let V be its viewing matrix, P its perspective distortion matrix and W its window-to-viewpoint matrix. Let p= [x,y,z,1] T represent the pixel position (x,y) with depth value z (Assume that multiplication by P automatically performs a homogeneous divide afterward, so that the homogeneous coordinate w = 1 always for the result of all matrix-vector products.)  (a) Let the transformation matrix A map the point p back to world coordinates, such that q= Ap is a homogeneous point in world coordinates that would be rendered onto pixel (x,y) with depth value z What is the formula for matrix A?

Question  Consider a scene. For the eye’s view, let V be its viewing matrix, P its perspective distortion matrix and W its window-to-viewpoint matrix. Let p= [x,y,z,1] T represent the pixel position (x,y) with depth value z (Assume that multiplication by P automatically performs a homogeneous divide afterward, so that the homogeneous coordinate w = 1 always for the result of all matrix-vector products.)  (a) Let the transformation matrix A map the point p back to world coordinates, such that q= Ap is a homogeneous point in world coordinates that would be rendered onto pixel (x,y) with depth value z What is the formula for matrix A? A=V -1 P -1 W -1

Normalized Perspective Distortion x y z x y z (left,top,-near) (right,bottom,-near) (0,0,-far) (-1,1,1) z 1 x 1 y 1

Question x y z (2,1,-1) (-2,-1,-1) (0,0,-3)  Find the Normalized perspective distortion transformation matrix from the data given below:

Question x y z (2,1,-1) (-2,-1,-1) (0,0,-3)