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It is known that both algorithms use the same gates: $H^{\oplus n}U_fH^{\oplus n}$.

After the circuit, the qubits are in the state $\sum_y \left( \sum_x (-1)^{f(x)+xy} \right) |y\rangle $.

In DJ's algorithm, one measures the amplitude of $|0\rangle^{\oplus n}$. It can only be $0$ or $\pm 1$, and the function is balanced iff it is 0 (for any balanced or constant function).

In Bernstein-Vazirani, one measures every qubit in its $Z$ basis and deduces the bits of the dot-product function.

It seems that one can apply Bernstein-Vazirani to any balanced or constant function (not just a dot-product function) to get a bit vector, and state that the function is constant iff the vector is zero.

What seems strange to me is that all the vector returned by BV for a DJ input is not necessarily all zero or all one, so I don't understand the criterion.

How are the measurements of each qubit and of the state $|0\rangle^{\oplus n}$ related?

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    $\begingroup$ Welcome on the site! It is "Deutsch-Jozsa". $\endgroup$ – peterh May 5 at 8:18
  • $\begingroup$ @peterh thanks, I see you have keen eyes :) $\endgroup$ – Labo May 5 at 10:32
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Finally, I found the answer myself there.

The only interesting thing is the amplitude of $|0\rangle^{\oplus n}$. If the function is constant, it is $\pm 1$ and if the function is balanced, it is $0$.

Hence, in the first case we are sure to measure all qubits in the $|0\rangle$ state and in the second case, we cannot measure all of them in this state (else we would have the state $|0\rangle^{\oplus n}$ that has amplitude $0$).

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  • $\begingroup$ Hi, Labo. Welcome to Quantum Computing SE! Could you please summarize the answer in that link over here? In case the link gets obsolete we'll no longer be able to view it. $\endgroup$ – Sanchayan Dutta May 5 at 12:24
  • $\begingroup$ @SanchayanDutta Thanks ^^ Do you see anything that is not covered by my answer? $\endgroup$ – Labo May 5 at 18:55
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    $\begingroup$ I didn't get time to read through the question and answer, sorry. If all the necessary details in that page are covered in your answer, that's great. I was just making sure that we won't miss out on anything in case the link breaks. :) $\endgroup$ – Sanchayan Dutta May 5 at 18:58

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