Teleport a qubit

The first four operations are locked: $R_Y(1.1)$ and $R_Z(0.6)$ on qubit $0$ prepare $e^{-0.3i}\lvert\psi\rangle$, and $H_1$, $\mathrm{CNOT}_{1\to2}$ prepare the shared pair $\lvert\Phi^+\rangle_{12}=\tfrac1{\sqrt2}(\lvert00\rangle+\lvert11\rangle)$. Qubit $0$ is the most significant bit. Append at most four gates from $\{H,\mathrm{CNOT},\mathrm{CZ},X,Z\}$, at most three of them two-qubit gates, and no measurements, so that Bob's qubit $2$ ends in $\lvert\psi\rangle$. The target is $\sigma=\begin{pmatrix}\cos^2 0.55 & \cos0.55\sin0.55\,e^{-0.6i}\\ \cos0.55\sin0.55\,e^{0.6i} & \sin^2 0.55\end{pmatrix}.$ Two checks are run. bob-state computes $\rho_2=\mathrm{Tr}_{01}\lvert\Psi\rangle\langle\Psi\rvert$ and passes when $T(\rho_2,\sigma)=\tfrac12\lVert\rho_2-\sigma\rVert_1\le10^{-6}$. alice-register computes $\rho_{01}=\mathrm{Tr}_2\lvert\Psi\rangle\langle\Psi\rvert$ and passes when it is within $10^{-6}$ of $\lvert{+}{+}\rangle\langle{+}{+}\rvert$ (every entry $\tfrac14$), the state the deferred protocol leaves on Alice's side.

Same locked preparation, now with two classical bits $c_0,c_1$. Alice measures qubit $0$ into $c_0$ and qubit $1$ into $c_1$; Bob may then apply $X$ and $Z$ to qubit $2$ only as classically conditioned gates (a gate with a condition such as "execute if $c_1=1$"). The budget is at most four gates from $\{H,\mathrm{CNOT},X,Z\}$, measurements not counted, and at most one two-qubit gate, so the quantum-controlled corrections of the deferred step are no longer available; $X$ and $Z$ must both appear. The grader evaluates the circuit on the density-matrix backend, which sums all four measurement branches exactly: $\rho=\sum_{m_0,m_1}\rho_{m_0m_1}$. The check bob-state passes when $T(\mathrm{Tr}_{01}\rho,\sigma)\le10^{-6}$.