Add stabilizer simulation example utilizing stim backend

[king-p3nguin]

Resolve #211

How the example works:

1. Creates random Clifford circuit with mid-circuit measurement. Mid-circuit measurement is inserted only when post-selection to zero state is possible.
2. Convert TensorCircuit's circuit to a Stim circuit via qir and extract the stabilizer tableau.
3. Calculate the entanglement entropy from the stabilizer tableau.
4. Validate the correctness of the entanglement entropy using TensorCircuit's function.

Output:

 ┌───┐┌───┐┌───┐┌───┐ ┌───┐┌───┐┌───┐┌───┐┌───┐ ┌───┐┌───┐ q_0: ─────┤ H ├┤ H ├┤ S ├┤ Z ├───────────────────────────────────────────┤ X ├┤ H ├┤ X ├┤ Z ├┤ Y ├───────────────┤ X ├┤ H ├───────────── ├───┤├───┤└───┘└───┘┌───┐ ┌───┐┌───┐ ┌─┐┌───┐└─┬─┘└───┘└─┬─┘├───┤├───┤┌───┐ └─┬─┘└───┘┌─┐ q_1: ─────┤ X ├┤ H ├──────────┤ X ├──■──┤ H ├┤ Y ├───────────────┤M├┤ H ├──┼─────────■──┤ X ├┤ Z ├┤ X ├────────────┼───────┤M├────────── ├───┤├───┤ ┌───┐└─┬─┘ │ ├───┤└───┘┌───┐┌───┐ └╥┘└───┘ │ ┌───┐├───┤└┬─┬┘└───┘┌───┐┌───┐ │ └╥┘ q_2: ──■──┤ X ├┤ X ├──■──┤ S ├──┼────┼──┤ S ├──■──┤ X ├┤ Y ├──────╫────────┼───────┤ X ├┤ H ├─┤M├───■──┤ S ├┤ H ├──┼────────╫─────────── │ ├───┤└─┬─┘┌─┴─┐├───┤ │ │ ├───┤ │ ├───┤├───┤┌───┐ ║ │ └─┬─┘└───┘ └╥┘ ┌─┴─┐└───┘└───┘ │ ┌───┐ ║ ┌───┐┌───┐ q_3: ──┼──┤ H ├──┼──┤ X ├┤ H ├──┼────┼──┤ Z ├──┼──┤ Y ├┤ Z ├┤ H ├─╫────────┼─────────┼─────────╫──┤ X ├───────■────┼──┤ H ├─╫─┤ X ├┤ S ├ │ ├───┤ │ ├───┤└───┘ │ │ ├───┤ │ ├───┤├───┤├───┤ ║ ┌───┐ │ ┌───┐ │ ║ └───┘ │ │ └───┘ ║ └─┬─┘└───┘ q_4: ──┼──┤ X ├──■──┤ S ├───────■────┼──┤ Y ├──┼──┤ Z ├┤ Y ├┤ S ├─╫─┤ Y ├──┼──┤ H ├──┼─────────╫──────────────┼────┼────────╫───┼─────── │ ├───┤┌───┐└───┘ ┌─┴─┐├───┤ │ └┬─┬┘└───┘└───┘ ║ ├───┤ │ ├───┤ │ ┌───┐ ║ ┌───┐ ┌─┐ │ │ ┌───┐ ║ │ q_5: ──┼──┤ S ├┤ X ├───────────────┤ X ├┤ H ├──┼───┤M├────────■───╫─┤ Z ├──┼──┤ Z ├──┼──┤ Z ├──╫──┤ Z ├─┤M├───┼────■──┤ Z ├─╫───┼─────── │ ├───┤├───┤┌───┐┌───┐┌───┐├───┤├───┤ │ └╥┘ ┌───┐ │ ║ ├───┤ │ └───┘ │ └───┘ ║ └───┘ └╥┘ │ └───┘ ║ │ q_6: ──┼──┤ X ├┤ Z ├┤ S ├┤ X ├┤ Z ├┤ H ├┤ Z ├──┼────╫──┤ Z ├──┼───╫─┤ Y ├──┼─────────┼─────────╫─────────╫────┼─────────────╫───┼─────── │ └─┬─┘├───┤├───┤├───┤├───┤└───┘└───┘ │ ║ └───┘ │ ║ └───┘ │ ┌───┐ │ ┌───┐ ║ ║ ┌─┴─┐┌───┐┌───┐ ║ │ q_7: ──┼────┼──┤ H ├┤ X ├┤ X ├┤ Y ├────────────┼────╫─────────┼───╫───■────┼──┤ H ├──■──┤ X ├──╫─────────╫──┤ X ├┤ X ├┤ H ├─╫───┼─────── ┌─┴─┐ │ ├───┤├───┤├───┤└───┘ ┌─┴─┐ ║ ┌───┐ │ ║ ┌─┴─┐ │ ├───┤ ┌─┐ ├───┤ ║ ┌───┐ ║ ├───┤└───┘└───┘ ║ │ q_8: ┤ X ├──┼──┤ Z ├┤ Y ├┤ Z ├───────────────┤ X ├──╫──┤ Z ├──┼───╫─┤ X ├──■──┤ H ├─┤M├─┤ H ├──╫──┤ Z ├──╫──┤ H ├───────────╫───┼─────── └───┘ │ ├───┤├───┤├───┤ └───┘ ║ └───┘┌─┴─┐ ║ ├───┤┌───┐├───┤ └╥┘ ├───┤ ║ └───┘ ║ └───┘ ║ │ ┌───┐ q_9: ───────■──┤ H ├┤ Y ├┤ H ├──────────────────────╫───────┤ X ├─╫─┤ S ├┤ Z ├┤ H ├──╫──┤ S ├──╫─────────╫──────────────────╫───■──┤ X ├ └───┘└───┘└───┘ ║ └───┘ ║ └───┘└───┘└───┘ ║ └───┘ ║ ║ ║ └───┘ c: 10/═══════════════════════════════════════════════╩═════════════╩══════════════════╩═════════╩═════════╩══════════════════╩═══════════ 0 1 2 3 4 5 Stim Entanglement Entropy: 1.3862943611198906 TensorCircuit Entanglement Entropy: 1.3862943618593357 

[king-p3nguin]

Hi @AbdullahKazi500, I modified your code to calculate the correct entanglement entropy and support mid-circuit measurement correctly, and I would not mind splitting the bounty with you. Please let me know what you think.

[AbdullahKazi500]

Hi @king-p3nguin I appreciate that you informed me and I am happy that you were able to solve it with my code @refraction-ray we can split the bounty I am happy with that, and any decision taken by the organizer
thank you once again

[refraction-ray]

Thanks for the contribution, the PR now looks better. However, I can still find the inconsistence of the entropy calculation with num_qubits = 10, depth = 6, where stim gives EE value 0 while tc gives 0.693. Could you debug the issue of this inconsistency? It would also be better to generalize the code a bit to support any partition for EE calculation instead of half chain now as num_qubits//2

Attached is the circuit yielding inconsistent results

[AbdullahKazi500]

Hi @refraction-ray and @king-p3nguin I have updated the code here Stim Entropy in the new patch I have got the results
Stim Entanglement Entropy: 0.6931471805599453
TensorCircuit Entanglement Entropy: 0.6931471813752873

The cut variable is used to define the partition size and can be adjusted as needed.
The get_bipartite_binary_matrix function is updated to take cut as a parameter.
Ensured that postselection logic is correctly applied in both TensorCircuit and Stim.
The simulate_stim_circuit_with_mid_measurement function now correctly handles postselections.

[king-p3nguin]

@refraction-ray I was able to fix the bug in the example. I have tested the code with num_qubits = 10, depth=1~30, cut = [i for i in range(num_qubits // 3)].

• tc to stim circuit conversion via qir does not seem a good idea because when converted to qir, all the mid-circuit measurements are pushed to the end of the circuit, so I just created a list of operations.
• np.isclose() is used to check the probability since even if the probability is close to 1.0, p < 1.0 can be True due to floating-point error.
• I also modified the code to accept tracing out any qubits.

[refraction-ray]

Thanks for your contribution. I'll assign the issue to both of you first and then further testing the code :)

all the mid-circuit measurements are pushed to the end of the circuit, so I just created a list of operations

That's why we need c.measurement_instruction() for each mid-circuit measurement to record, and merge_qir to ensure the order of gates and mid-circuit measurements

[refraction-ray]

@all-contributors please add @AbdullahKazi500 for example

[allcontributors]

@refraction-ray

I've put up a pull request to add @AbdullahKazi500! 🎉