Prepare the singlet state

Both qubits start in $\lvert0\rangle$; qubit $0$ is the most significant bit, so $\lvert ab\rangle$ has qubit $0$ in state $\lvert a\rangle$. Build a circuit with at most four gates from $H$, $X$, $Z$ and $\mathrm{CNOT}$ (no measurements) whose output is $\lvert\Psi^-\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert10\rangle\big)=\big(0,\ \tfrac1{\sqrt2},\ -\tfrac1{\sqrt2},\ 0\big)^T.$ Two checks are run. is-singlet computes $F=\lvert\langle\Psi^-\vert\psi\rangle\rvert^2$ and passes when $F\ge1-10^{-6}$; a global phase does not change $F$. anti-correlated compares the Born probabilities $(p_{00},p_{01},p_{10},p_{11})$ with $(0,\tfrac12,\tfrac12,0)$ to within $10^{-6}$; it passes for both $\lvert\Psi^+\rangle$ and $\lvert\Psi^-\rangle$, because the two states differ only in the relative sign of $\lvert10\rangle$.