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@ahelwer has provided a smart construction, and I will provide a more brute-force method as a complementary for beginners like me. It actually starts out by listing all the possible 16 moves. Let x,y=0,1,2,3, then all the possible moves are $ x=0, \quad y=(0123)\rightarrow(0123) \\ x=1, \quad y=(0123)\rightarrow(1230) \\ x=2, \quad y=(0123)\rightarrow(2301) ...


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The question comes in two parts. Firstly, for the top circuit, can you find the gates $A$, $B$ and $C$ such that $ABC=I$ and $CXBXA=R_x(\theta)$ or $R_y(\theta)$. Secondly, can you reduce either of these to the lower circuit with $AB=I$ and $XBXA=R_x(\theta)$ or $R_Y(\theta)$. The first of these is detailed in the bit of Nielsen and Chuang just before the ...


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Note: $I_k$ is unit matrix of order $k$ in the following text. First step of the algorithm is $H \otimes H \otimes I_2 \otimes I_2$ as you mentioned. A controlled gate $U$ with $n$ qubits between the control qubit and the target qubit can expressed as a matrix $$ CU_{n} = \begin{pmatrix} I_{\frac{N}{2}} & O_{\frac{N}{2}} \\ O_{\frac{N}{2}} & I_{\...


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You can use $Ry$ (y-rotation gate). Its general matrix is $$ Ry(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix} $$ Applying the gate on $|0\rangle$ state, you get a state $$ |\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle $$ Hence $\cos(\theta/2) = \frac{1}{\sqrt{3}}$ ...


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Here is an implementation of a circuit producing state $|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle)$ on IBM Q: Note that $\theta = 1.2310$ for $\mathrm{Ry}$ on $q_0$. $\theta = \frac{\pi}{4}$ and $\theta = -\frac{\pi}{4}$ for first and second $\mathrm{Ry}$ on $q_1$. The $\mathrm{Ry}$ on $q_0$ prepares qubit in superposition $|...


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Consider some simpler cases, $(j,j+1)$ for general $j$. Then you can do you want with plugging in $j=i$, $j=i+1$, $j=i+3$ and $j=i+4$ and concatenating the circuits appropriately and then simplifying. So how to do $(j,j+1)$? That is conjugate to $(0,1)$, so just consider that for now. $(0,1)$ would be NOT but controlled on making sure all the higher places ...


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A brute force solution :). You can also obtain CCH via qiskit's basic gates with help of get_controlled_circuit method. from qiskit import * from qiskit.aqua.utils.controlled_circuit import get_controlled_circuit q_reg = QuantumRegister(3, 'q') qc_h = QuantumCircuit(q_reg) qc_ch = QuantumCircuit(q_reg) qc_cch = QuantumCircuit(q_reg) qc_h.h(q_reg[0]) ...


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Summarization based on discussion with user met927: Transpiled circuit form depends on used backend - it is different for simulator and real quantum processor: On simulator, the $\mathrm{CH}$ gate is transpiled to the circuit shown above On real quantum processor, the gate is implemented with two $\mathrm{U2}$ gates and $\mathrm{CNOT}$ (i.e. like in the ...


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Assuming you've got Toffoli and single-qubit rotations, you can implement the following: This basically works because if either of the controls is not $|1\rangle$, the Toffoli does nothing and the two single-qubit unitaries cancel each other. Whereas, if both controls are $|1\rangle$, then the net gate on the target qubit is $$ (\cos\frac{\pi}{8}I+i\sin\...


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Nielsen and Chuang book may be confusing; I will try to explain. Any classical algorithm can be presented as a circuit consisting of $NOT$ and $AND$ gates; this means that if we can make quantum gates computing $NOT$ and $AND$, and make fanout, we can run any classical algorithm on quantum computer. $NOT$ is reversible and we have quantum $NOT$ gate; many ...


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Based on paper Five Two-Bit Quantum Gates are Sucient to Implement the Quantum Fredkin Gate provided by Norbert Schuch, I realized that there is a more efficient implementation in terms of number of gates. Here is a result: Matrix of CNOT acting on $|q_1\rangle$ controlled by $|q_2\rangle$ is \begin{equation} CNOT_{2}= \begin{pmatrix} 1 & 0 & 0 &...


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It's not possible to implement a Toffoli using only Fredkin gates, because Fredkin gates preserve the number of 1s in the state while Toffolis do not.


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