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Let $\omega =e^{2\pi i/3}$. The positive-exponent quantum Fourier transform gives $|\xi ⟩=3^{-1/2}{\sum }_{y=0}^2{\omega }^{2y}|y⟩$. Reindexing the cyclic shift operator givesThus $|\xi ⟩$ is an eigenstate with eigenvalue $\omega$, with this sign fixed by the printed quantum Fourier transform convention.

Prepare the first two qutrits in $({\mathrm{QFT}}_3|0⟩{)}^{\otimes 2}$ and the answer qutrit in $|\xi ⟩$. The modular-addition quantum oracle applies $S^{f(x_1,x_2)}$ to the answer. Its eigenvalue produces quantum phase kickback, leavingApply ${\mathrm{QFT}}_3^{-1}$ to each input qutrit, then measure them in the computational basis. The result isThe preparation, inverse transforms and measurements are independent of $f$, and there is exactly one use of $U_f$. This is qutrit linear-function identification, the ternary version of Bernstein-Vazirani phase kickback.

Put $M=2^m$ and prepare $m$ control qubits in $|0{⟩}^{\otimes m}$ and the target in the supplied eigenstate $|\psi ⟩$. Apply a Hadamard gate to every control qubit. Let $k={\sum }_{j=0}^{m-1}{2}^j{k}_j$ denote the integer encoded in that quantum register.

For every $j$, apply a controlled unitary gate $U^{2^j}$ with control $k_j$ and the same target. The quantum phase kickback yieldsApplying ${\mathrm{QFT}}_M^{-1}$ to the controls therefore produces $|x⟩$ exactly. A computational-basis measurement returns $x$ with certainty, and the algorithm outputsThis is exact quantum phase estimation; exactness uses the finite dyadic phase promise, not a rounding argument.

Only a controlled-U gate is supplied as a primitive. Its $2^j$ repetitions implement controlled-$U^{2^j}$, so the total number of black-box uses is ${\sum }_{j=0}^{m-1}{2}^j=2^m-1$. Access to powers as unit-cost primitives would be a different quantum query complexity model. This distinction is the cost of exact phase estimation on a dyadic spectrum.

This is the Deutsch-Jozsa test with an arbitrary uniform-state unitary, and does not require $N$ to be a power of two. Choose a known unitary operator $F$ such that $F|0⟩=|s⟩=N^{-1/2}{\sum }_i|i⟩$. Prepare the target qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. One Boolean-oracle call gives quantum phase kickback:Apply $F^{†}$ to the index register. The amplitude on $|0⟩$ isIt equals $+1$ or $-1$ for the two constant strings, and equals zero for a balanced string. Consequently a single query decides the promised problem exactly: measure the index register and report constant for outcome zero, balanced for any other outcome. This is the Deutsch-Jozsa algorithm with the available exact state-preparation operation.

Maintain a known list of the still-active indices, initially all $N=2K$ positions. On a list of even size $M$, perform the three-step construction with $M$ replacing $N$. A known reversible relabeling prepares and queries the corresponding original indices, so quantum phase kickback still needs only one call to $O_{\mathbf{x}}$. The full unitary extension for that current size is independent of the remaining unknown values. It is allowed even when $M$ is not a power of two.

If the measurement gives $(0,0)$, its amplitude is the imbalance divided by $M$. A balanced active string has zero such amplitude, so an observed zero pair certifies that the active string is unbalanced. Return unbalanced immediately. No probability-of-error estimate is needed: an impossible outcome never occurs in the balanced case.

Otherwise the measured pair corresponds to two opposite bits. Delete those two indices and repeat. Each deletion removes exactly one zero and one one, preserving the difference between their counts. Thus the active string is balanced if and only if the original one is balanced. If all indices are removed, return balanced. This is opposite-pair elimination for exact quantum balance testing; it never requires determining which member of a deleted pair is zero.

Each query either terminates with a valid imbalance certificate or reduces the active length by two. After at most $K$ opposite-pair outcomes the active list is empty. The result is certain on every input, with worst-case query countThe same argument includes the $M=2$ last step: equal bits give $(0,0)$ with certainty, while opposite bits give the sole nonzero pair with certainty. There is no additional final query. The measurement probabilities are normalized because $(M-2w{)}^2+4w(M-w)=M^2$.

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 $a=0$.

Let $L=2^n$ and use an $n$-qubit phase register. Begin with $|0^n⟩|v⟩$ and apply $H^{\otimes n}$ to the phase register, creating $L^{-1/2}{\sum }_{m=0}^{L-1}|m⟩|v⟩$. Controlled powers implementFor phase qubit $j$, counted from the most significant bit, the controlled power is $U^{2^{n-j}}$. It can be built from $2^{n-j}$ calls to the supplied controlled-$U$ gate. By quantum phase kickback, the phase-register state isApply the inverse quantum Fourier transform to obtain the exact quantum phase estimation mappingA computational basis measurement of the first register determines $y$ with certainty, hence $\varphi =y/L$ and the eigenvalue $e^{2\pi i\varphi }$. Exactness follows from the promised dyadic phase; no approximation or continued-fraction reconstruction is needed.

With only controlled-$U$ available as a query, the repeated-power construction uses $L-1=2^n-1$ oracle calls. The other Fourier-transform circuitry has polynomial size in $n$ in the ideal phase-gate model. The cost of exact phase estimation on a dyadic spectrum is therefore not polynomial in $n$ in this primitive-query model unless powered queries have additional implementations. The task does not require such a polynomial bound.

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.

Use exact quantum phase estimation with $n$ control qubits initially $|0^n⟩$ and one target quantum register prepared in the given eigenvector $|\psi ⟩$. Apply Hadamard gates to the control qubits. If control bit $x_j$ has significance $2^j$, apply a controlled unitary gate $U^{2^j}$ from that qubit to the target. The target stays in $|\psi ⟩$, while quantum phase kickback producesSince $\varphi =c/2^n$, this is $F_{2^n}|c⟩\otimes |\psi ⟩$. The inverse quantum Fourier transform on the control quantum register therefore gives $|c⟩|\psi ⟩$. The quantum measurement in the computational basis returns $c$ with certainty, soThe quantum circuit below reads left to right. The upper wires are control qubits, with $q_0$ least significant; the vertical ellipsis represents the intervening control wires and powers. The inverse quantum Fourier transform acts on the entire control quantum register, including any bit-order permutations.

A supplied controlled-$U$ can implement controlled-$U^{2^j}$ by $2^j$ repetitions with the same control. Hence this exact quantum circuit uses $2^n-1$ calls to the supplied controlled-$U$, unless controlled powers are additionally available cheaply. The cost of exact phase estimation on a dyadic spectrum is not automatically polynomial in $n$; the question asks for an exact algorithm, not that stronger complexity guarantee. Exactness also uses the promised dyadic eigenphase, ideal gates, and the supplied inverse quantum Fourier transform.

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 $U_f$.

When $k=N/4$,One Grover search algorithm iteration therefore givesThe selective sign change $I_G$ is implemented with one application of the Boolean quantum oracle: prepare its answer qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$ and use quantum phase kickback, obtaining $(-1{)}^{f(x)}$ on each $|x⟩$. The remaining operations $H_n$, $I_0$, and the final computational-basis measurement are independent of $f$. Measurement of $|{\psi }_G⟩$ returns some $x\in G$ with probability one, so exactly one query to $U_f$ suffices.

One classical query at any chosen $x_0$ returns only $v=s{x}_0+t$. Every possible slope $s\in {\mathbb{Z}}_3$ is compatible with that answer by choosing $t=v-s{x}_0$, so no one-query classical algorithm can determine $s$ with certainty.

For the quantum algorithm, let $\omega =e^{2\pi i/3}$ and preparePart (b) shows that the single query applies quantum phase kickback:The unknown intercept contributes only a global phase. Applying ${\mathrm{QFT}}_3^{-1}$ to the first register produces $|s⟩$, whose computational-basis measurement determines $s$ with certainty after one query.

Use the available unitaries to preparein the first register andin the answer qubit. The quantum phase kickback identity givesChoose a unitary $F$ with $F|0⟩=|s⟩$, apply $F^{-1}$ to the first register, and measure it. Its amplitude at $|0⟩$ isThis equals $+1$ or $-1$ for a constant string and zero for a balanced string. Thus outcome $0$ means constant, while every other outcome means balanced, with certainty after one oracle query. This is the Deutsch-Jozsa test with an arbitrary uniform-state unitary.

The known circuit from part (b) lets us apply $U_{\mathcal{I}}$ without querying the faulty oracle. Their composition satisfiesBecause $a=00\cdots 01$, only the last answer qubit is flipped in the exceptional case.

Prepare the answer register asSince $X|-⟩=-|-⟩$, quantum phase kickback gives, for an arbitrary search-register state,The answer register is unchanged and factors out. The induced operation on the first register is thereforeThis is the marked-state phase oracle from a single faulty identity-oracle query.