B1: Copy Oracle

Editorial

The oracle takes single qubit computational basis states $x$ and $y$ and outputs $y' = x \oplus y$. Its truth table is as follows:

Observing the truth table carefully, we see that $y'$ differs from $y$ only when $x=1$. This means effecting $y \rightarrow y'$ will flip the bit when $x$ is "active". Such an operation can be achieved by using a controlled-$X$ gate ¹, with $x$ as the control bit and $y$ as the target bit.

Thus, the problem can be solved using the controlled-$X$ gate.

Sample Code

Below is a sample program:

from qiskit import QuantumCircuit, QuantumRegister
 
 
def solve() -> QuantumCircuit:
    x, y = QuantumRegister(1), QuantumRegister(1)
    qc = QuantumCircuit(x, y)
    
    qc.cx(x, y)
 
    return qc