= Solution
Use <two-bit remote state preparation on a graph-state path>. Alice performs the first two measurements from the preceding <measurement-based quantum computation>: angle $\alpha$ on her first qubit, giving $r$, then angle $(-1)^r\beta$ on her second, giving $s$. She sends the pair $(r,s)$ to Bob, using exactly two <classical communication> bits.
The branch calculation already gives Bob's state as $e^{ir\beta}X^sZ^r|\chi\rangle$, where $|\chi\rangle=U(\beta)U(\alpha)|+\rangle$. Bob applies $X^s$ first and $Z^r$ second, so his correction operator is $Z^rX^s$. Since $X^2=Z^2=I$,
$$
(Z^rX^s)(X^sZ^r)|\chi\rangle=|\chi\rangle.
$$
Thus
$$
\boxed{\text{Alice sends }(r,s);\qquad\text{Bob applies }Z^rX^s.}
$$
The final state is exactly the desired state up to the irrelevant global phase. No postselection is used: all four measurement branches have probability $1/4$ and are corrected. Bob does not need the angles themselves; Alice uses them in her measurement choices, while the two bits specify his Pauli correction.
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