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I try to design the circuit for $b = 11$ and succeeded in running. Therefore, I start to think of the circuit for different secret string $b = 01$. The circuit I made is down below: The IBM Quantum circuit I made

Here is the problem: After the first Hadamard operator, if the input $x = \left|10\right\rangle$, the $f(x)$ result (That is the measurement of $q2$, $q3$, which is the last two bits of $\left|y_1y_2y_3y_4\right\rangle$ in Computational basis states) according to my circuit, should be $\left|11\right\rangle$, the same as the input $x = \left|11\right\rangle$. And the output of $y1, y2$ should match the condition:$$b\cdot{y} = 0\,(\,mod\,2\,)$$Hence, one of the output I believe is $\left|1011\right\rangle$, but the calculator gives me a wired (at least I think it is wired) result. What are the issues?

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Measurements are only allowed at the end of the quantum circuit for current machines like those of IBM. Also for Simon's algorithm we don't care about the output of the second register. Thus only first register is measured.

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  • $\begingroup$ The measurement in the first register should be 10 or 00, since b = 01 right? $\endgroup$ – Kian Gao Oct 18 at 9:55

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