A Simon's algorithm with secret string b = 01, IBM Quantum experience gives a different result from whay I calculate

Question

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:

Here is the problem: After the first Hadamard operator, if the input $x = left|10rightrangle$, the   $f(x)$ result (That is the measurement of $q2$, $q3$, which is the last two bits of $left|y_1y_2y_3y_4rightrangle$ in Computational basis states) according to my circuit, should be $left|11rightrangle$, the same as the input $x = left|11rightrangle$. And the output of $y1, y2$ should match the condition:$$bcdot{y} = 0,(,mod,2,)$$Hence, one of the output I believe is $left|1011rightrangle$, but the calculator gives me a wired (at least I think it is wired) result. What are the issues?

KAJ226 · Answer

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.

So you have the possibility of observing the state $|00rangle$ or $|01rangle$. But we know that $|00rangle$ is trivial so without much checking $|01rangle$ is what you are looking for...
Here is the code to generate the above circuit and histogram plot in Qiskit:
from qiskit import QuantumRegister, ClassicalRegister, QuantumCircuit
from numpy import pi
from qiskit import QuantumCircuit, BasicAer, execute
from qiskit.visualization import plot_histogram
%matplotlib inline

qreg_q = QuantumRegister(4, 'q')
creg_c = ClassicalRegister(2, 'c')
circuit = QuantumCircuit(qreg_q, creg_c)

circuit.h(qreg_q[0])
circuit.h(qreg_q[1])
circuit.cx(qreg_q[0], qreg_q[2])
circuit.cx(qreg_q[0], qreg_q[3])
circuit.barrier(range(4))
circuit.h(qreg_q[0])
circuit.h(qreg_q[1])
circuit.barrier(range(4))
circuit.measure(qreg_q[0], creg_c[0])
circuit.measure(qreg_q[1], creg_c[1])
circuit.draw( 'mpl',style={'name': 'bw'}, scale = 1.5, plot_barriers = False)

backend = BasicAer.get_backend('qasm_simulator')
job = execute(circuit, backend, shots = 20000)
plot_histogram(job.result().get_counts(), color='black', title="Result")

A quick note: If I use the statevector_simulator option then I will either get $|00rangle$ or $|11rangle$ as it just a one-shot simulator.

A Simon's algorithm with secret string b = 01, IBM Quantum experience gives a different result from whay I calculate

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