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This is a Quantum Fourier Transform and the Wiki page provides a generic decomposition: Can you find your two qubit version?

There is A Polynomial Quantum Algorithm for Approximating the Jones Polynomial For simple knots, you can do that with 2 qubits...

Not an answer yet, but to long for comment: Alice and Bob share two entangled pairs. Now look only at Alice's. The last line gives the state that only one of'em has: \frac1{\sqrt 2}(|00〉+|11〉) =\...

Instead of saving the information of a single comparison into a y qubit, you can use a controlled adder. Instead of $n$ y qubits you would only need $\log_2 n$.

Let's start with the Bell state $\frac1{\sqrt2}(|01\rangle+|10\rangle)$. Let's further assume that the entire universe (or at least the parts of relevance) are pervaded by a static magnetic field ...

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The simple answer is: yes...
3 out of 4 states are inverted, that leads to the following situation before you do the Grover Diffusion, which mirror around the red line: This will bring the all states to zeros, but not $|11\... View answer 2 votes In Elementary gates for quantum computation it is conjectured that based on dimension counting, ... a lower bound on the number of two-bit gates required to produce an arbitrary$n\$-bit unitary ...