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Modular exponentiation computes the remainder $a^n\mathrm{mod}m$ for an integer base $a$, a nonnegative integer exponent $n$, and, here, an integer modulus $m\ge 2$. It applies exponentiation by squaring while reducing after each multiplication, including each squaring. The result is the unique representative in $\{0,\dots ,m-1\}$. There is no requirement that the base be coprime to the modulus when the exponent is nonnegative.^[ref]

Take $a=2$, $n=13=(1101{)}_2$, and $m=17$. Initialize $r=2$; the remaining digits giveHence $2^{13}\mathrm{mod}17=15$. The final operation multiplies $16$ by $2$, since the preceding square has already been reduced. The computation never needs to construct $2^{13}=8192$.

In general reduce the base first, $x=a\mathrm{mod}m$. For $n>0$, initialize $r=x$ and process the remaining binary digits with $r\leftarrow r^2\mathrm{mod}m$, followed, for a digit one, by $r\leftarrow rx\mathrm{mod}m$. The loop invariant iswhere $p$ is the processed prefix value. The congruence is preserved because modular arithmetic respects multiplication: if $u\equiv v(\mathrm{mod}m)$, then $u^2\equiv v^2$ and $ux\equiv va(\mathrm{mod}m)$. Thus changing representatives never changes the final congruence class.

Both factors in every product are less than $m$, so the unreduced product is less than $m^2$, and the stored result is again less than $m$. The same bounds hold for a gigantic exponent, such as $n=2^{1000}$: there are $1000$ squarings, each on bounded-size residues. For base two, the corresponding ordinary power would have $2^{1000}+1$ binary digits. The benefit is avoiding the huge ordinary power altogether, rather than constructing it and taking one final remainder. Initial reduction is essential when the supplied base is much larger than the modulus.

For $n=0$, return $1$ under the same empty-product convention as exponentiation by squaring. For $n=-k<0$, require $\gcd (a,m)=1$, find the modular multiplicative inverse $a^{-1}\mathrm{mod}m$, and compute $(a^{-1}{)}^k\mathrm{mod}m$; the extended Euclidean algorithm can find that inverse. Its cost is separate from the positive-exponent binary powering multiplication count. For example, $2^{-1}\equiv 9(\mathrm{mod}17)$, while $2^{-1}\mathrm{mod}4$ is undefined because $2$ and $4$ are not coprime.

A reversible computation gives the coherent quantum modular exponentiation used in quantum order finding and discrete-logarithm Fourier sampling.

Write $a=a_1{a}_2$ and $x=x_1{x}_2$, with the top wire carrying $x_1$, the middle wire $x_2$, and the bottom wire the ancilla bit $y$. The binary inner product is $a\cdot x=a_1{x}_1\oplus a_2{x}_2$, where $\oplus$ is exclusive or. Let $C_{j\to y}$ denote a CNOT gate controlled by wire $j$ and targeting $y$. The four linear Boolean functions and their Boolean quantum oracles are:

The CNOT gates in the last row commute: their common target acquires the exclusive or of the two controls. Each input wire is preserved, and applying any of these reversible circuits twice restores $y$.

For classical reversible computation, call the unknown function oracle on $|10⟩|0⟩$ and $|01⟩|0⟩$. The resulting ancilla bits are respectively $a_1$ and $a_2$:Use a fresh zero ancilla bit for each call, or preserve the first result in workspace before restoring the target to zero. Thus two classical calls determine the two-bit string.

For $n$ input bits, query the standard basis strings $e_j$, since $a\cdot e_j=a_j$. This gives an upper bound of $n$ calls. It is also optimal for exact identification: every classical query reveals at most one new bit, since $x$ is unchanged and $y\oplus f_a(x)$ reveals just $f_a(x)$ when $y$ is known. A depth-$k$ decision tree has at most $2^k$ answer histories, but must distinguish $2^n$ possible hidden strings. Hence $k\ge n$, including adaptive strategies. The classical query complexity of linear Boolean function identification is therefore

Allow clean workspace bits, as in standard reversible computation. Boolean algebra expresses any Boolean function as XORs of ANDs and constants. A Toffoli gate maps $(a,b,c)$ to $(a,b,c\oplus ab)$, so an initially zero target computes an AND without erasing the inputs. A CNOT gate accumulates XORs, and a NOT gate supplies complemented bits or constants. Intermediate products can be computed into workspace, XORed into an output, and then uncomputed by reversing the workspace computation. This realizes the reversible embedding $(x,y)\mapsto (x,y\oplus f(x))$ for any Boolean function and, with reversible clean-up, universal reversible classical computation.

The Fredkin transformation has control $a$ and targets $b,c$. A network of three gates, shown in i, iswhere the sequence is executed left to right. After the first gate $c_1=c\oplus b$; after the second $b_1=b\oplus a(c\oplus b)$; and the last gives $c_2=c\oplus b\oplus b_1$. If $a=0$, $(b_1,c_2)=(b,c)$; if $a=1$, $(b_1,c_2)=(c,b)$. Thus the circuit is exactly a Fredkin gate.

The workspace qualification matters for the unrestricted universality claim. On four or more wires, every embedded NOT, CNOT and Toffoli gate is an even permutation of all computational states. Without ancillary wires they cannot implement, for example, a permutation swapping just $0000$ and $0001$ while fixing the other four-bit strings. Clean ancillas remove this obstruction on the designated input subspace.

Write $Q=2^n$, $t=(p-1)(q-1)$ and $a=t/Q$. The easy starting state is uniform over all $n$-bit labels, so its good fraction is $a$, not $t/N$. For distinct primes one has $a\ge 1/4$. If both primes are odd, $t/N\ge (1-1/3)(1-1/5)=8/15$ and $N/Q\ge 1/2$, giving $a\ge 4/15$. If one prime is two, the other is an odd prime $q$; since $2q$ has $n$ bits, $q>2^{n-2}$, and $t=q-1\ge 2^{n-2}=Q/4$.

The uniform coprime state for a semiprime can now be prepared by one exact Grover rotation. SetPrepareUse the good subspace spanned by $|k⟩|1⟩$ with $1\le k<N$ and $\gcd (k,N)=1$. Its squared overlap with $|\Psi ⟩$ is exactly $a\lambda =1/4$. The Euclidean algorithm supplies a reversible computation of the membership predicate, including the range check; condition a sign flip on membership and the extra qubit being one, then uncompute the workspace. This implements $S_G=I-2{\Pi }_G$.

The starting-state reflection $D_{\Psi }=2|\Psi ⟩⟨\Psi |-I$ uses the inverse of its known preparation and a reflection on the all-zero state. Applying $D_{\Psi }{S}_G$ once invokes exact amplitude amplification at $\theta =\pi /6$, and givesThe extra qubit factors off, and the arithmetic workspace is returned to zero. Although the stated range includes $N$, that label is not coprime to itself, so the predicate $k<N$ is equivalent and avoids admitting labels outside the intended range.

The supplied $p,q$ determine $t$ in polynomial time; no factoring procedure is needed. Binary arithmetic, range comparison, the reversible greatest common divisor, the two starting-state reflections and the controlled sign flip all have polynomial-size circuits. Thus the construction takes polynomial time in $n$ in the ideal model allowing the specified one-qubit state preparation. Exactness uses the rotation with known amplitudes $\sqrt{\lambda },\sqrt{1-\lambda }$; with a fixed finite approximate gate library one obtains arbitrary accuracy with precision overhead, rather than an automatic promise of exact state preparation. In the ideal model the preparation succeeds with certainty, without rejection sampling. If desired, the extra $|1⟩$ can be reset by a Pauli X gate.

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.

Quantum modular exponentiation is the reversible computation that mapsIt can be implemented in polynomial time in the bit lengths of $x$ and $N$ by repeated squaring and controlled modular multiplication.