= Solution
Translate the output <alphabet> by $-\min(z_1,z_2)$ and put $d=|z_2-z_1|\geq1$. The reachable outputs become
$$
\{0,1,2\}\cup\{d,d+1,d+2\},\qquad |\mathcal Y|=3+\min(d,3).
$$
For $d=1$ the two sets overlap in two points; for $d=2$ in one; for $d\geq3$ they are disjoint. In particular \b[$|\mathcal Y|\geq4$], and choosing noise values $0,1$ attains four.
With rows indexed by inputs $0,1,2$ and columns by outputs $0,1,2,3$, <independence> and uniform two-point noise give
$$
M=\begin{pmatrix}
1/2&1/2&0&0\\
0&1/2&1/2&0\\
0&0&1/2&1/2
\end{pmatrix}.
$$
Every conditional output distribution has <information entropy> one <bit>, so $I(X;Y)=H(Y)-1\leq\log_2 4-1=1$. The input <probability mass function> $(1/2,0,1/2)$ makes all four outputs equiprobable, attaining that bound. Thus
$$
\boxed{C=1\ \text{bit per use},\qquad p_X=(1/2,0,1/2).}
$$
Using a zero-probability \i[input] in a channel-capacity maximization is legitimate; it is unrelated to the strictly positive source weights needed for the tree-shape claims in Question 2.
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