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Suppose a known list of $2^m$ distinct $n$-bit strings $a_j$ is available and computational basis transpositions on $n+m$ qubits have polynomial-size implementations. Start with the uniform superposition state of index labels $j$, leaving the data zero. First swap $|0^n,j⟩$ with $|a_j,j⟩$ for each nonzero $a_j$. Next, for each $j\ne 0$, swap $|a_j,j⟩$ with $|a_j,0^m⟩$. Distinct index labels separate the first-stage pairs, and distinct data strings separate the second-stage pairs. Thus no later transposition disturbs an earlier term. At most $2^{m+1}-1$ transpositions prepare the uniform data state and return the quantum ancilla to zero. For $m=O(\log n)$ the quantum circuit has polynomial size. This assumes the list is available; a membership predicate alone does not supply it.

For a known finite set $B$, prepare the uniform superposition state $|s_B⟩$ and implement its reflection operator $2|s_B⟩⟨s_B|-I$. If exactly $k$ basis states in $B$ are marked, the same two-dimensional Grover rotation angle argument uses $O(\sqrt{|B|/k})$ phase queries. Preparation and reflection are known operations with no calls to the unknown predicate. Their gate cost must be accounted for separately. The success guarantee requires $k>0$, and choosing the optimal iteration number uses a known $k$.

Use three-way qutrit sign-flip discrimination. Prepare the qutrit in the uniform superposition stateFor an unknown label $j\in \{0,1,2\}$, the supplied unitary operator is $U_j=I-2|j⟩⟨j|$. Apply it once and then apply the known inversion about the meanSince $D^{†}=D$ and $D^2=I$, this is an allowed unitary transformation. Using $⟨s|j⟩=1/\sqrt{3}$ givesPerform a computational-basis measurement and guess that its outcome is the unknown label. The Born rule givesEach incorrect outcome has probability $1/27$, verifying normalization. The success probability is $25/27$ separately for every label, so no assumption about their prior probabilities is needed.

Write $N=|A|$. For $N>1$, the unmarked computational basis vectors in $A$ give the normalized vectorIts norm is one and $⟨a|\omega ⟩=0$, so these two vectors form an orthonormal basis for $\mathcal{V}$. Splitting the uniform superposition state into its marked and unmarked terms givesThe nonnegative coefficients select the specified angle in $[0,\pi /2]$.

There is one degenerate boundary case: if $N=1$, then $|{\psi }_A⟩=|a⟩$ and $\mathcal{V}$ is one-dimensional. No vector in $\mathcal{V}$ can complete $|a⟩$ to a two-element orthonormal basis. Thus the printed two-dimensional description needs $N>1$. For $N=1$ the search itself is already solved, with $\theta =\pi /2$ and no query.

List the known distinct strings in $A$ as $a_0,\dots ,a_{N-1}$, where $N=2^m$, and use an $m$-qubit index quantum ancilla. Apply Hadamard gates to that register to prepareFirst, for each $j$ with $a_j\ne 0^n$, apply the available unitary operator that transposes $|0^n,j⟩$ and $|a_j,j⟩$. The pairs are disjoint because they have distinct index labels. Skip a transposition when the two vectors coincide. This producesSecond, for each $j\ne 0$, transpose $|a_j,j⟩$ and $|a_j,0^m⟩$. These pairs are disjoint because the data strings $a_j$ are distinct. No pair contains the term with $j=0$, and no later transposition moves an earlier output. The final state is exactlyThis clean subset superposition preparation uses at most $N+(N-1)=2N-1$ of the given transpositions and $m$ Hadamard gates. Since $m=O({\log }_{2}n)$, $N$ is polynomial in $n$; multiplying this count by the assumed polynomial size of each transposition gives a polynomial-size quantum circuit. The index quantum ancilla returns to zero, so the data register is a pure uniform superposition state, rather than a mixture obtained by discarding its index. The construction uses an available enumeration of the known subset; a membership oracle alone would not justify that enumeration.

If there is no marked entry, the Boolean quantum oracle is the identity. The uniform superposition state is a positive eigenstate of the Grover diffusion operator:Thus every iteration leaves the full state $|s⟩\otimes |-⟩$ unchanged. The measured search bits have the uniform distributionOnly the first $n$ qubits are measured in the printed circuit. If the untouched ancilla were also measured in the computational basis, its bit would be an independent fair bit, giving probability $1/(2N)$ for each joint outcome.

Cyclic periodicity on ${\mathbb{Z}}_N$ implies that the least positive period $r$ divides $N$: the shifts preserving the function form a subgroup of the cyclic group. Injectivity within one period says that distinct residue classes modulo $r$ have distinct function values. Write $L=N/r$.

Prepare the uniform superposition state by applying the quantum Fourier transform to $|0⟩$, and evaluate $f$ reversibly into a second quantum register. This is efficient because the given classical Boolean circuit can be converted into a reversible circuit, with its workspace cleaned by uncomputation. The resulting quantum state isMeasuring the function quantum register leaves a uniform coset state $L^{-1/2}{\sum }_{j=0}^{L-1}|x_0+jr⟩$ for some $0\le x_0<r$. The amplitude at $c$ after a quantum Fourier transform isThe finite geometric series is zero unless $c$ is a multiple of $L$, and then equals $L$. Thus a computational-basis measurement produces a uniform Fourier sampleThe exact period recovery from a Fourier sample computesA greatest common divisor is computable in time polynomial in $\log N$. The candidate is always a divisor of the true period, and equals it precisely when $s$ is coprime to $r$.

For the success flag, compute $f(\hat{r}\mathrm{mod}N)$ and compare with $f(0)$. If $\hat{r}<r$, injectivity within the first period makes these unequal. If $\hat{r}=r$, periodicity makes them equal, including $r=N$. Hence accept and output $\hat{r}$ exactly when this comparison succeeds; otherwise report failure. This is heralded exact quantum period finding when the period divides the register size, with no false acceptance. The case $r=1$ succeeds even on the sample $c=0$.

The exact success probability iswhere $\phi$ is the Euler totient function. For $r\mid N$, the prime-product formula gives $\phi (r)/r\ge \phi (N)/N$. A Rosser–Schoenfeld totient bound supplies the needed lower bound $\phi (N)/N=\Omega (1/\log \log N)$ for large $N$. Every single attempt, including verification, has the assumed polynomial-in-$\log N$ gate cost. Repeating $O(\log \log N)$ times gives constant success probability with a certified result whenever one is found.

The printed $O$ in the totient hint should be a lower-bound statement, $\Omega$. Literally, $\phi (N)=O(N/\log \log N)$ is false: at primes $N$, $\phi (N)=N-1$. Likewise the exact single-sample success probability need not be $O(1/\log \log N)$; for prime period it tends to one. The useful intended guarantee is the displayed inverse-log-log lower bound, not a universal upper bound. If an algorithm satisfying the literal success upper bound is desired, one can additionally accept an otherwise certified run only when $k=\lceil \log \log N\rceil$ independent fair bits are all zero. This retains a detectable failure flag and polynomial runtime, and has success at most $2^{-k}=O(1/\log \log N)$, but deliberately weakens the useful guarantee. The standard unsuppressed algorithm above gives the stronger intended performance. Small $N$ can be handled directly. Primary evidence for the totient lower bound is Theorem 15, equations (3.41)–(3.42), in denisevellachemla.eu/Rosser-Schoenfeld-1962.pdf .

A good item must exist, so assume $1\le k<N$; if $k=0$, the requested search is impossible, while $k=N$ makes every output good. Put $p=k/N$ and $\theta =\arcsin \sqrt{p}$. Normalize the good and bad components of the uniform superposition state asThe Boolean phase oracle negates $|g⟩$ and fixes $|b⟩$. The Grover diffusion operator is $-I_{{\psi }_0}=2|{\psi }_0⟩⟨{\psi }_0|-I$. Its implementation is independent of $f$: conjugate the reflection about $|0^n⟩$ by Hadamard gates. One Grover search algorithm iteration isHence, by the Grover rotation angle formula,Choose the nonnegative integer $j$ nearest to $\pi /(4\theta )-1/2$. Its final angle differs from $\pi /2$ by at most $\theta$, so a computational-basis measurement succeeds with probability at least ${\cos }^2\theta =1-p$. The allowed small-density regime includes $p\le 1/3$, giving the requested $2/3$ bound. Each iteration uses one Boolean phase oracle query, and $j=O(1/\sqrt{p})=O(\sqrt{N/k})$.

For $k=N/4$, $\theta =\pi /6$ and one iteration reaches $3\theta =\pi /2$ exactly. Thus one query gives a good outcome with certainty:This is exact Grover search on four entries applied to a marked fraction of one quarter, not only to a four-element register.

Use the positive-exponent quantum Fourier transform convention $F_N|x⟩=N^{-1/2}{\sum }_t{e}^{2\pi ixt/N}|t⟩$. The period of a function here is the least positive translation leaving the function unchanged on the cyclic group ${\mathbb{Z}}_N$. The injective promise means $f(x)=f(x^{\prime })$ exactly when $x\equiv x^{\prime }(\mathrm{mod}r)$. In particular $r\mid N$: the subgroup of period translations is generated by $r$, and wrapping the argument by $N$ preserves its value. This is exact quantum period finding on a finite cyclic group, rather than the truncated, nondividing-period version requiring continued fractions.

Prepare a uniform superposition state in the first quantum register using $F_N|0⟩$, and perform a reversible computation of the function into a second quantum register, cleaning its workspace by uncomputation:A quantum measurement in the computational basis of the second quantum register leaves a periodic coset state in the first:For the quantum Fourier transform of this periodic coset state, the finite geometric series vanishes unless the measured label is $t=mN/r$. On those labels its normalized amplitude is $r^{-1/2}{e}^{2\pi iam/r}$. Thus $m$ has the uniform distribution on a finite set on $\{0,\dots ,r-1\}$, irrespective of $a$.

Compute the greatest common divisor by the Euclidean algorithm, and set $q=N/\gcd (t,N)$, with $\gcd (0,N)=N$. Since $t=mN/r$,The exact period recovery from a Fourier sample is therefore certified by testing $f(q\mathrm{mod}N)=f(0)$. Every candidate $q$ divides $r$ and lies between $1$ and $r$, so the injective promise makes that equality equivalent to $q=r$. This includes $r=N$, where the test uses input $0$, and $r=1$, which always succeeds. The test does not assert that an arbitrary equality of function values certifies an arbitrary period; the Fourier-derived divisor restriction is essential.

The concise result isHere $\phi$ is the Euler totient function, with $\phi (1)=1$. The totient lower bound from the Mertens product gives $\phi (r)/r=\Omega (1/\max \{1,\log \log N\})$ for $r\le N$. Each trial uses polynomial time quantum Fourier transforms, function evaluations and Euclidean algorithm arithmetic in $\log N$. Repeating $O(\max \{1,\log \log N\})$ independent trials gives constant success probability; the herald accepts only correct answers.

The printed asymptotic bound has the wrong direction for this argument. In Big O notation, the stated upper bound on the number of coprime integers supplies no success lower bound and is itself false uniformly: for prime $N$, $\phi (N)=N-1$, not $O(N/\log \log N)$. Likewise the single-trial success need not be $O(1/\log \log N)$, since it tends to one for $r=N$ prime. The meaningful corrected guarantee is the lower bound above. If an upper-bounded success rate is literally required, one may deliberately retain an accepted result with an independent probability proportional to $1/\max \{1,\log \log N\}$; this discards useful successes and is unnecessary for efficient recovery.

Interpret a function collision as two distinct inputs with equal outputs. Without distinctness the request is vacuous, since an input paired with itself always qualifies. Set $m=\lceil N^{1/3}\rceil$ (handling small $N$ by direct queries), choose a known set $A$ of size $m$, and query all its inputs. Store a classical table of their outputs and corresponding inputs. If this table contains a function collision, return it immediately.

Otherwise the table contains $m$ distinct outputs. Because $g$ is exactly two-to-one, each $a\in A$ has exactly one partner in $B$, the known complement of $A$. None lies in $A$, and different table outputs have different partners. Thus exactly $m$ inputs in $B$, of size $L=N-m$, satisfy the known-table membership predicateBuild a marked-state phase oracle for this predicate by a compute-phase-uncompute construction: use $U_g$ on a clean answer quantum register, reversibly compare its output against the classical table, apply the conditional minus sign, undo the comparison, and apply $U_g$ again. Since the answer operation is bitwise exclusive or, $U_g^{-1}=U_g$. All workspace returns to zero, and each phase query costs two calls to $U_g$.

Perform known-subset Grover search in the span of the computational basis states labelled by $B$, starting in their uniform superposition state. Reflection about this known state uses no $U_g$ queries; choosing $A$ as an initial interval makes $B$ a known interval that can also be encoded by its own $L$ labels. Its marked fraction is $m/L$, so the previous Grover rotation angle analysis uses $O(\sqrt{L/m})$ phase queries. On measuring $b$, one further $U_g$ query checks the output and retrieves its partner $a$ from the table. The inputs are distinct because the two sets are disjoint.

Consequently the quantum query complexity isFor large $N$, rounding the optimal Grover search algorithm iteration count gives success at least $1-m/(N-m)>1/2$. An internal table function collision already gives certain success. The balance $m\asymp \sqrt{N/m}$ explains the cube-root choice. This is the Brassard–Høyer–Tapp collision algorithm: its table uses $O(m\log N)$ classical bits. The bound counts calls to $U_g$, as requested; it does not charge table lookup and its reversible implementation as constant-time gates, or claim an overall polynomial time in $\log N$.

An injective self-map of the finite set ${\mathbb{Z}}_N$ is a bijection. For sufficiently large $N$, exactlyoutputs satisfy the integer comparison, so there are exactly $m$ good inputs. Prepare the uniform superposition state $|\psi ⟩=N^{-1/2}{\sum }_x|x⟩$, whose good probability isUse the phase oracle from the previous part and the reflection $2|\psi ⟩⟨\psi |-I$. The amplitude amplification theorem requires $O(p^{-1/2})$ iterations; each uses two $U_f$ queries, while the state preparation and its reflection are independent of $f$. With the nearest-integer iteration count from that theorem, the failure probability is at most $p$, tending to zero. Therefore the success probability eventually exceeds $0.9$, withThis is a bound on quantum query complexity; implementation costs of the known gates are not counted as oracle queries.