Number of bits not preserved in tk_to_qiskit conversion with RangePredicate

[CalMacCQ]

Whilst looking into #375 I found that the number of bits was not preserved in the tk_to_qiskit conversion. I think this may well be the cause of the KeyError issue.

from pytket import Circuit, OpType
from pytket.circuit.logic_exp import reg_eq
from pytket.extensions.qiskit import tk_to_qiskit


main_circ = Circuit(2, 2, "main_circ")

creg_0 = main_circ.add_c_register("creg_0", 2)

main_circ.add_gate(OpType.CH,[0, 1], condition = reg_eq(creg_0, 3))

main_circ.measure_all()
print(main_circ.n_bits) # gives 5

But if I convert to a qiskit QuantumCircuit...

qc = tk_to_qiskit(main_circ)
print(qc.num_clbits) # gives 4

We only get 4 bits in the converted qiskit circuit. it seems the tk_SCRATCH_BIT[0] bit is lost in the conversion.

[yao-cqc]

I think this is expected. In pytket, a scratch bit is required to store the evaluation of a range predicate, for example, if c == 2 tk_scratch_bit[0] = 1, and then condition a gate on that scratch bit, i.e., if tk_scratch_bit[0]==1 H(1). When converting such a conditional gate to qiskit, we combine these two instructions into one and remove the separate evaluation, resulting in if c == 2 H(1). Since tk_scratch_bit is no longer used, we don't add it to the qiskit circuit.

qc[0].condition

prints

(ClassicalRegister(2, 'creg_0'), 2)

suggests that the CH gate has the correct condition

[CalMacCQ]

Closing as I think this is expected behaviour.