Build strongly entangling layers

Complete the TypeScript function

stronglyEntanglingLayers(weights: number[][][], n: number): Operation[]

that returns the operations of $L$ strongly entangling layers on $n$ qubits, in time order.

Shapes. weights has shape $(L,n,3)$: weights[l][i] = [phi, theta, omega] holds the three angles of wire $i$ in layer $\ell$. The circuit has $3nL$ rotations and, for $n>1$, $nL$ CNOTs. Qubit $0$ is the most significant bit and $R_P(\alpha)=e^{-i\alpha P/2}$.

Layer $\ell$ ($\ell=0,\dots,L-1$):

1. on every wire $i=0,\dots,n-1$ the rotation $\mathrm{Rot}(\phi,\theta,\omega)=R_Z(\omega)\,R_Y(\theta)\,R_Z(\phi),$ written as the three operations { kind: 'gate', gate: 'RZ', targets: [i], params: [phi] }, then 'RY' with [theta], then 'RZ' with [omega];
2. if $n>1$, with the range $r_\ell=(\ell\bmod(n-1))+1\in\{1,\dots,n-1\},$ the CNOTs $\mathrm{CNOT}\big(i,(i+r_\ell)\bmod n\big)$ for $i=0,1,\dots,n-1$ in this order, each written { kind: 'gate', gate: 'CNOT', targets: [control, target] }. For $n=1$ there are no CNOTs.

Grading. For every test the grader builds the circuit { numQubits: n, ops } from your list, computes its $2^n\times2^n$ unitary $U$ and compares it with the reference unitary $T$ through the process fidelity $\lvert\mathrm{Tr}(T^\dagger U)\rvert/2^n\ge1-10^{-9}$, so a different but equivalent gate sequence also passes. The weights of the hidden tests are $w_{\ell,i,k}=2\pi\,u$ with $u$ drawn from mulberry32(seed) in the order $\ell$, $i$, $k$. The tests are grouped into three checks:

• rot-symmetric (hidden): one layer with $\phi=\omega$ on every wire, $n=2$ (seed $11$) and $n=3$ (seed $12$);
• single-layer: the visible test zero-weights-L1-n2 and hidden random tests of shapes $(1,2,3)$ (seed $1$) and $(1,3,3)$ (seed $4$);
• multi-layer (hidden): shapes $(2,3,3)$ (seed $2$) and $(3,4,3)$ (seed $3$).

Visible test. weights = [[[0, 0, 0], [0, 0, 0]]] and n = 2: all rotations are the identity and $r_0=1$, so the circuit is $\mathrm{CNOT}(0,1)$ followed by $\mathrm{CNOT}(1,0)$. On basis labels $(x_0,x_1)\mapsto(x_0,x_0\oplus x_1)\mapsto(x_1,x_0\oplus x_1)$, hence $T=\mathrm{CNOT}_{1\to0}\,\mathrm{CNOT}_{0\to1}=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&0&0&1\\0&1&0&0\end{pmatrix},$ with $\lvert00\rangle\mapsto\lvert00\rangle$, $\lvert01\rangle\mapsto\lvert11\rangle$, $\lvert10\rangle\mapsto\lvert01\rangle$, $\lvert11\rangle\mapsto\lvert10\rangle$.