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I was reading about Flexible Representation of Quantum Images (FRQI) encoding in Qiskit textbook. It says that, given $$\{\theta_0, \theta_1, ..., \theta_{4^{n}-1}\} \quad (\theta_i \in [0,\pi/2])$$ we can transform $\mid{0}\rangle^{\otimes2n+1}$ into FRQI state $\mid{I(\theta)}\rangle$ by applying unitary operator $\mathcal{R}\mathcal{H}$, with $\mathcal{H}= I \otimes H^{\otimes 2n}$ and $\mathcal{R}= \prod_{i=0}^{4^n-1}R_i$. where $$\mid{I(\theta)}\rangle= \frac{1}{2^n}\sum_{i=0}^{4^n-1}(cos \theta_i|{0}\rangle+sin \theta_i|{1}\rangle)\otimes |{i}\rangle$$

And $R_i$ is a unitary matrix named controlled rotation, given by $$R_i= I \otimes \sum_{j=0, j \neq i}^{4^n-1} |{j}\rangle \langle {j}| + R_y(2\theta_i) \otimes |{i}\rangle \langle {i}\mid$$ I would like to implement the controlled rotation $R_i$ or $\mathcal{R}$ using $C^{2n}R_y$ (controlled $R_y$) and $X$ gate.

Edit: I think I have made some progress. We can write, $$ R_i= I^{\otimes 2n+1}+(R_y(2\theta_i)-I) \otimes |{i}\rangle \langle {i}\mid$$ $$ C^{2n}R_y(2\theta_i)= I^{\otimes 2n+1}+(R_y(2\theta_i)-I) \otimes |{4^n-1}\rangle \langle {4^n-1}\mid $$ Please review my results and suggest me how to approach further.

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If you already have controlled$^{2n}-R_Y$, this is fairly straightforward. This means that what you already have is $$ C^{2n} R_y(2\theta_i)=I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |1\rangle\langle 1|^{\otimes 2n} $$ and you are aiming to make $$ R_i=I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |i\rangle\langle i|. $$ Now, let $X_x$ be a tensor product of Pauli $X$ matrices that have the action $$ X_x|i\rangle=|1\rangle^{\otimes 2n}\rangle. $$ In other words, if $y$ is the binary representation of $i$, then $$ x=y\oplus 11\ldots 1=\bar y. $$ Then \begin{align*} X_x C^{2n} R_y(2\theta_i)X_x&=X_x(I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |1\rangle\langle 1|^{\otimes 2n})X_x\\ &=I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |i\rangle\langle i|\\ &=R_i. \end{align*}

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