CS465 Fall 2006
Homework 7 Solution
-------------------

1.1

Using the given image width and height, nx = 12/2 = 6, ny = 8/2 = 4 so M_vp = 

[6 0 0 5.5]
[0 4 0 3.5]
[0 0 1 0  ]
[0 0 0 1  ]


1.2

We know n and f. r and l (b and t) can be derived by transforming 2
eye-space vertices into pixel space and solving the system of equations for
the first (second) homogenized coordinate. This gives

l=-3
r=3
b=-2
t=2

and the matrix M_proj = 

[-5/3    0     0     0   ]
[  0   -2.5    0     0   ]
[  0     0   -9/7  -80/7 ]
[  0     0     1     0   ]


1.3

This simply involves multiplying M_vp*M_proj*V for each vertex, and
dividing by w.

(1.75, 0.75, 1/7)
(8.25, 0.75, -4/7)
(5, 5.75, -5/7)



2.1

Barycentric coordinates: alpha, beta, gamma
Canonical depth: z'
Texture coordinates: u/w, v/w, 1/w


2.2

The point has relative height of 2.25 pixels, and the top vertex has height
5. Thus Gamma = 2.25/5 = 0.45. 

From symmetry, alpha = beta = (1.0 - 0.45)/2 = 0.275.


2.3

Barycentric coords from 2.2: 0.275, 0.275, 0.45
z' = 0.275*(1/7) + 0.275*(-4/7) + 0.45*(-5/7) = -123/280 = -0.439285 

w = original depth of point; so 1/w at vertices = -1/8, -1/16, -1/20
    and u/w at vertices = 0, -1/16, -1/40
	v/w at vertices = 0, 0, -1/20

u/w = 0.275*0 + 0.275*(-1/16) + 0.45*(-1/40) = -91/3200 = -0.02844
v/w = 0.275*0 + 0.275*0 + 0.45*(-1/20) = -9/400 = -0.0225
1/w = 0.275*(-1/8) + 0.275*(-1/16) + 0.45*(-1/20) = -237/3200 = -0.07406


2.4 

u = (u/w) / (1/w) = 91/237 = 0.38397
v = (v/w) / (1/w) = 72/237 = 0.30380

This result can be verified by reprojecting the pixel into world space,
finding barycentric coordinates there, and interpolating u,v.