Each c_if call linear in number of circuit Clbits #7249
Description
Information
- Qiskit Terra version: main
- Python version: any
- Operating system: any
What is the current behavior?
Any call to Instruction.c_if is linear in the number of Clbit references in classical registers in the circuit. This is because of some lines in InstructionSet that unroll all registers on every call: https://github.com/Qiskit/qiskit-terra/blob/49110f9c596fc509f189f38cfbdbb3100407bff2/qiskit/circuit/instructionset.py#L61-L64
Steps to reproduce the problem
In [1]: from qiskit.circuit import QuantumCircuit
...: import numpy as np
...: import scipy.stats
...:
...: def build(n):
...: qc = QuantumCircuit(1, n)
...: instructions = qc.x(0)
...: def payload():
...: instructions.c_if(0, 0)
...: return payload
...:
...: times = []
...: ns = [100, 300, 1000, 3000, 10000]
...: for n in ns:
...: payload = build(n)
...: out = %timeit -o payload()
...: times.append(out.average)
...: scipy.stats.linregress(np.log(ns), np.log(times)).slope
10.8 µs ± 101 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
28.2 µs ± 922 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
94 µs ± 640 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
276 µs ± 5.04 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
919 µs ± 7.14 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Out[1]: 0.9710414298958975What is the expected behavior?
This should be a constant-time operation.
Suggested solutions
This can easily be fixed by the same methods as #7246.