 Solution

A constant function has $f(0)=f(1)$, whereas a balanced one has one zero and one one, so $f(0)\ne f(1)$. The four oracles require only the following circuits:Prepare $|0⟩|1⟩$, apply a Hadamard gate to each qubit, apply the oracle once, apply another Hadamard gate to the first qubit, and measure that first wire in the computational basis. Immediately before the oracle the state is $|+⟩|-⟩$. The identity $X|-⟩=-|-⟩$ gives phase kickback:The output is certainly zero for constant and one for balanced. In circuit terms the target $X$ contributes only a global sign, and a controlled $X$ with target $|-⟩$ acts as $Z$ on the control; $HZH=X$ then turns that phase into a measured bit. These identities prove the network for all four explicit oracles. The unused sign contains $f(0)$ only as a global phase, consistent with the full-identification lower bound.