Three-bit phase estimation of a phase gate

The register has four qubits, all starting in $\lvert0\rangle$. Qubits $0,1,2$ form the counting register $\lvert k\rangle=\lvert k_0k_1k_2\rangle$ with qubit $0$ the most significant bit, $k=4k_0+2k_1+k_2$; qubit $3$ is the target.

The circuit already contains three controlled placeholders acting on qubit $3$: U1 controlled by qubit $2$ (weight $2^0$), U2 controlled by qubit $1$ (weight $2^1$) and U4 controlled by qubit $0$ (weight $2^2$). Keep each exactly once and add gates from $\{H,X,\mathrm{CP},\mathrm{SWAP}\}$, at most $16$ operations in total including the three placeholders, and no measurements.

The grader runs four cases, substituting $U^{2^j}=P(2\pi\varphi\cdot2^j)$:

• phase-1-8: $\varphi=1/8$ ($U=T$) must give the counting distribution $P(001)=1$;
• phase-3-8: $\varphi=3/8$ must give $P(011)=1$;
• phase-5-8: $\varphi=5/8$ must give $P(101)=1$;
• phase-4-8: $\varphi=1/2$ ($U=Z$) must give $P(100)=1$.

Each check compares the full vector $(p_{000},\dots,p_{111})$ of the counting qubits, summed over qubit $3$, within $10^{-6}$. After the controlled powers the counting register holds $\tfrac1{\sqrt8}\sum_{k=0}^{7}e^{2\pi i\varphi k}\lvert k\rangle=\mathrm{QFT}_8\lvert m\rangle$ when $\varphi=m/8$, so the last stage must undo the Fourier transform.