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For a Boolean quantum oracle whose function is $g(x,y)=a_y\cdot x$, place its target in $|-⟩$. Conjugating the oracle by a Walsh-Hadamard transform on the $x$ register translates that register by $a_y$. This is Bernstein-Vazirani phase kickback without measuring its output, so the index $y$ may remain in an arbitrary quantum superposition. A second application undoes the translation. A phase evaluation between the two calls therefore transfers a hidden-label-dependent phase to the index register while uncomputation returns the decoding register to zero.

Use the bit-query Boolean quantum oracle $O_x|i,b,z⟩=|i,b\oplus x_i,z⟩$, where $z$ denotes workspace. Initially every probability amplitude is independent of $x$, hence is a constant polynomial. An input-independent unitary gate only forms linear combinations of probability amplitudes, so it does not increase their polynomial degree.

If $a_{i,b,z}(x)$ is an amplitude before a query, its new value isOne query increases polynomial degree by at most one. Inductively, after $T$ queries each amplitude has polynomial degree at most $T$. The acceptance probability is a sum of squared absolute values of the accepting amplitudes, so it is a real polynomial of degree at most $2T$. Intermediate quantum measurements and classical adaptation can be retained coherently, or handled by summing unnormalized branch probabilities; either approach yields the same degree bound.

An exact algorithm has acceptance probability precisely $f(x)$ at every Boolean input. The multilinear reduction on the Boolean cube replaces positive powers of each $x_i$ by $x_i$, preserving these values without increasing polynomial degree. Therefore the polynomial method for quantum query lower bounds givesHere $\mathrm{deg}(f)$ is the degree of the unique multilinear polynomial representing the Boolean function, and $Q_E$ is its exact quantum query complexity. Uniqueness follows, for example, by evaluating successively on the indicator vectors of subsets: the value on a subset determines its coefficient once all smaller-subset coefficients are known.

First use the allowed one-query algorithm to learn $p=x_1\oplus x_2$ exactly. This restricted two-index Boolean quantum oracle is realized by relabelling indices $1,2$ in a single query to the original input. Measure its output and choose the second query classically.

If $p=0$, the first two bits agree and their common value is the majority, irrespective of $x_3$. Query $x_1$ and output it. If $p=1$, the first two bits cancel in the vote, so query $x_3$ and output it. ThusThe exact two-query majority algorithm uses one parity query and one ordinary bit query on every branch, and is correct on all eight inputs. Together with the lower bound this proves $Q_E(\mathrm{MAJ})=2$.

The linear decoding must be done coherently, so its output is available inside the next Boolean quantum oracle call. Use registers $Z,X$ of $n$ qubits and a shared phase ancilla qubit in $|-⟩$. Define the Bernstein-Vazirani decoding controlled by a quantum registerThe ancilla qubit is implicit. For every computational index $x$, the Walsh-Hadamard transform calculation givesIn particular $W_g^2=I$ on these states. Initialize $Z$ to $|0⟩$ and $X$ to $H^{\otimes n}|0⟩$. The three query stages areThe middle equality uses the matching index $z=a_x$, not a classical guess of that string. The second $W_g$ performs uncomputation, removing the hidden-string register without losing its phase on $X$. Apply $H^{\otimes n}$ to $X$ to obtainThis requires precisely two queries to $U_g$ and one to $U_f$, and only $O(n)$ additional fixed quantum gates. No measurement of $a_x$ is made: such a measurement would spoil the required coherence. Preparing all registers and the phase ancilla qubit uses only the initially available zero states.

Prepare an additional qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$ by applying $X$ and then the Hadamard gate to $|0⟩$. Apply $H^{\otimes n}$ to the data register. The Boolean quantum oracle then produces quantum phase kickback:A final Walsh-Hadamard transform gives an amplitude for $z$ equal toTo see this identity, the sum factors over the bits; any position at which $a$ and $z$ differ contributes $1-1=0$. This is Bernstein-Vazirani phase kickback, and its output isThere is exactly one oracle query, $O(n)$ fixed quantum gates, and no probabilistic intermediate step. The ancilla qubit can be left in $|A⟩=|-⟩$ or reset to $|0⟩$ using its known inverse preparation. The construction includes the case $a=0$.

Prepare the two data qubits in $|s⟩=\frac{1}{2}{\sum }_{x\in B_2}|x⟩$ and a phase ancilla qubit in $|-⟩$. A single Boolean quantum oracle query flips only the marked amplitude. Its success fraction is $p=1/4$, so $\theta =\pi /6$ in amplitude amplification. The fixed diffusion reflection $D=2|s⟩⟨s|-I$ gives $\sin (3\theta )=1$ after one iteration.

Directly, after the query the marked amplitude is $-1/2$ and the other three are $1/2$. Their mean is $1/4$. The diffusion reflection replaces each amplitude $c_x$ by $2(1/4)-c_x$, yielding one at the marked input and zero elsewhere. ThusHere $U_f$ is understood with its target ancilla qubit, and $D$ acts only on the data. It is independent of $f$: $D=H^{\otimes 2}(2|00⟩⟨00|-I)H^{\otimes 2}$, with the central diagonal gate implementable using two Pauli Z gates and one Controlled-Z gate. Measuring the data in the computational basis therefore finds the unique marked string with certainty after one oracle query.

Prepare an ancillary qubit inIt can be prepared from $|0⟩$ by an $X$ operation followed by a Hadamard gate, independently of $g$. One call to the Boolean quantum oracle then gives quantum phase kickback:The data state receives a minus sign precisely on the basis vectors spanning $\mathcal{G}$, while the ancilla remains unchanged and unentangled. One oracle call realizes the reflection. By linearity it is the marked-state phase oraclewith one oracle call. The ancillary preparation and its optional inverse use only fixed operations, so no additional knowledge of $g$ is needed.

Follow the printed examples by counting positive squares, so zero is not marked. For $N\ge 2$, the marked output set isA one-to-one map from the finite set $[N]$ to itself is a permutation, so exactly $K$ inputs have outputs in $S$. In the uniform quantum superposition $|s⟩=N^{-1/2}{\sum }_x|x⟩$, the initial good probability is therefore $p=K/N$. For permutation-preimage quantum search, this known marked density under a permutation is $\Theta (N^{-1/2})$.

Define the efficiently computable predicate $h(y)$ to be one if and only if $y$ is a positive square. The assumed Boolean quantum oracle for $h$, acting on $y$ and a $|-⟩$ ancilla, implements its diagonal sign operation $D_h|y⟩=(-1{)}^{h(y)}|y⟩$. Compute the modular-addition quantum oracle into a zero output register, apply this sign operation, and uncompute:This implements $I_{\mathcal{G}}$ for $\mathcal{G}=\mathrm{span}\{|x⟩:f(x)\in S\}$, with the work registers clean. Only forward calls to $U_f$ were promised, but its inverse costs one such call: if $J|y⟩=|-y\mathrm{mod}N⟩$ on the second register, thenIndeed the output addition is changed into subtraction. Thus each marked reflection costs exactly two queries to $U_f$; all other operations used here are independent of the unknown $f$.

Use the amplitude amplification theorem with $|\psi ⟩=|s⟩$, whose reflection is allowed by the assumptions. With $\theta =\arcsin \sqrt{K/N}$, choose $m$ nearest to $\pi /(4\theta )-1/2$. Here $p\le 1/2$ for every $N\ge 2$, so a trial succeeds with probability at least $1/2$. AlsoMeasure $x$ and use one additional $U_f$ query to evaluate and check $f(x)$. On failure, restart with fresh registers. Four independent trials have failure probability at most $2^{-4}$, so the required confidence and query bound areThis is an upper bound on quantum query complexity, not a claim that arbitrary reflections or every classical gate have constant cost. Including zero as a square would change $K$ to $K+1$ without changing the asymptotic bound. With the printed positive-square interpretation, $N=1$ has no valid output and no algorithm can satisfy the success requirement; the intended asymptotic task necessarily has $N\ge 2$.

Apply a Pauli X gate and then a Hadamard gate to the third qubit, and apply Hadamard gates to the first two. This preparesOne application of the Boolean quantum oracle now gives, by quantum phase kickback,The answer qubit is unentangled and may be ignored. If all three qubits must be restored to the displayed register form, applying $H$ and then $X$ to the answer qubit maps $|-⟩$ back to $|0⟩$. Thus the first register is exactly $|f⟩$ after one use of $U_f$.