Drawing conditional measurement operations

Required prerequisites

• Consult the security policy. If reporting a security vulnerability, do not report the bug using this form. Use the process described in the policy to report the issue.
• Make sure you've read the documentation. Your issue may be addressed there.
• Search the issue tracker to verify that this hasn't already been reported. +1 or comment there if it has.
• If possible, make a PR with a failing test to give us a starting point to work on!

Describe the bug

Drawing functionality does not plot out mid circuit measurements:

@cudaq.kernel 
def kernel(): 
    
    qubits = cudaq.qvector(3)
    
    h(qubits[0])
    
    m1 = mz(qubits[0])
        
    if m1 == 1: 
        z(qubits[2])
    

In this circuit, the z gate is only applied if the measurement result is 1.

This is how conditional gates are represented in literature:

Steps to reproduce the bug

NA

Expected behavior

NA

Is this a regression? If it is, put the last known working version (or commit) here.

Not a regression

Environment

• CUDA Quantum version:
• Python version:
• C++ compiler:
• Operating system:

Suggestions

No response

Start with the Python kernel example in the issue and inspect the existing circuit-drawing entry point to determine how mid-circuit measurements and conditional gates are represented. Done means the measurement and its conditional z gate are visibly represented in the drawn circuit, with coverage for the shown case.