 Solution (source code)

by Codex (@codex,  0)

= Solution

Prepare Bob in $|0\rangle_B$ and encode the <bit> using Alice's preparation $|m\rangle_A$. The <controlled-NOT gate> gives
$$
\operatorname{CNOT}_{A\to B}|m\rangle_A|0\rangle_B
=|m\rangle_A|m\rangle_B.
$$
\b[Bob measures in the <computational basis> and reads $m$ with <probability> one.] Alice's final measurement is unnecessary. The interaction transmits a <classical bit>; it does not clone an arbitrary unknown <qubit>.

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