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\section*{Chapter 8 Solutions}
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8.17 Let case 0 be the case with no black or no white vertices,
case 1 be the case with 1 black or one white vertex, case 2 be the
case with two adjacent white vertices and and case 3 be the case with
two diagonally opposite white vertices. Coloring consective vertices
as 0 or 1, the 16 cases are

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0000 0 all white\\
0001 1 one white\\
0010 1 one white\\
0011 2 two consecutive white\\
0100 1 one white\\
0101 3 diagonally opposite white\\
0110 2 two consecutive white\\
0111 1 one black\\
1000 1 one white\\
1001 2 two consecutive white\\
1010 3 diagonally opposite white\\
1011 1 one black\\
1100 2 two consecutive white\\
1101 1 one black\\
1110 1 one black\\
1111 0 all black\\

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12.5 Using the case numbers from page 336, for each case we use
symmetry arguments. For example, for case 1, the black vertex can
be any of the 8 vertices and we can swtich white and black, yielding
16 possibilities. For case 6, the adjacent blacks can be along any of the
12 edges but there are 2 places we can place the third black and not 
have all three blacks on the same face, which yields 24 cases, but then
we can switch black and white, yielding 48 possibilities.
The numbers of each case are

\begin{verbatim}
0   2
1  16
2  24
3  24
4   8
5  48
6  48
7  16
8   6
9   8
10  6
11 24
12 24
13  2
\end{verbatim}

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12.9 If the values of the function $f(x,y,z)$ are given at the 
corners of a unit cube with a vertex at the origin and lying in
the positive octant, then trilinear interpolation approximates $f$
inside the cube by the function
$$g(x,y,z)=(1-x)(1-y)(1-z)f(0,0,0)+x(1-y)(1-z)f(1,0,0)$$
$$+(1-x)y(1-z)f(0,1,0)+(1-x)(1-y)zf(0,0,1)+xy(1-z)f(1,1,0)$$
$$+x(1-y)zf(1,0,1)+(1-x)yzf(0,1,1)+xyzf(1,1,1).$$
Clearly where the cube is located is unimportant and the values of $f$ can
be replaced by samples at the vertices of any parallelepiped.

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