Deutsch–Jozsa with one oracle query

The register has three qubits, all starting in $\lvert0\rangle$: qubits $0$ and $1$ hold the input $x=x_0x_1$ and qubit $2$ is the ancilla $y$. Qubit $0$ is the most significant bit, so the basis state $\lvert x_0,x_1,y\rangle$ has index $4x_0+2x_1+y$.

The circuit already contains the oracle placeholder $U_f$. In the circuit editor it is the statement Uf on its own line; keep it exactly once and type gates before and after it. You may use $H$, $X$ and $Z$, at most $8$ operations in total including Uf, and no measurements.

The grader substitutes four oracles into Uf and computes the Born probabilities $(p_{00},p_{01},p_{10},p_{11})$ of the input qubits $(x_0,x_1)$, summed over the ancilla:

• constant-cases: for $f\equiv0$ ($U_f=I_8$) and for $f\equiv1$ ($U_f=I\otimes I\otimes X$) you must obtain $(1,0,0,0)$, i.e. the outcome $00$ with certainty;
• balanced-cases: for $f=x_0\oplus x_1$ ($U_f=\mathrm{CNOT}_{1\to2}\,\mathrm{CNOT}_{0\to2}$) you must obtain $(0,0,0,1)$, and for $f=x_0$ ($U_f=\mathrm{CNOT}_{0\to2}$) you must obtain $(0,0,1,0)$.

Every probability is compared within $10^{-6}$. In general the Deutsch–Jozsa circuit returns the string $z$ with probability $\big\lvert2^{-n}\sum_x(-1)^{f(x)+x\cdot z}\big\rvert^2$, which for a linear $f(x)=s\cdot x$ is $1$ exactly at $z=s$.