path: root/source/know/concept/shors-algorithm/index.md

---
title: "Shor's algorithm"
sort_title: "Shor's algorithm"
date: 2021-04-13
categories:
- Quantum information
- Cryptography
- Algorithms
layout: "concept"
---

**Shor's algorithm** was the first truly useful quantum algorithm.
It can solve important problems,
most notably integer factorization,
much more efficiently than any classical algorithm.
It weakens widely-used cryptographic schemes,
such as RSA and [Diffie-Hellman](/know/concept/diffie-hellman-key-exchange/).

In essence, Shor's algorithm's revolutionary achievement
is that it can efficiently find the periods $$s_1, ..., s_A$$
of a function $$f(x_1, ..., x_A)$$ on a discrete finite field, where:

$$\begin{aligned}
    f(x_1, ..., x_A)
    = f(x_1 + s_1, ..., x_A + s_A)
\end{aligned}$$

This is a so-called *hidden subgroup problem* for a *finite Abelian group*.
With minimal modifications,
Shor's algorithm can solve practically every such problem.



## Integer factorization

Originally, Shor's algorithm was designed to factorize an integer $$N$$,
in which case the goal is to find the period $$s$$ of
the modular exponentiation function $$f$$ (for reasons explained later):

$$\begin{aligned}
    f(x)
    = a^x \bmod N
\end{aligned}$$

For a given $$a$$ and $$N$$.
The period $$s$$ is the smallest integer satisfying $$f(x) = f(x+s)$$.
To do this, the following $$2q$$-qubit quantum circuit is used,
with $$q$$ chosen so that $$N^2 \le 2^q < 2 N^2$$:

{% include image.html file="shors-circuit.png" width="70%" alt="Shor's circuit" %}

Here, $$\mathrm{QFT}_q$$ refers to the $$q$$-qubit
[quantum Fourier transform](/know/concept/quantum-fourier-transform/),
and the oracle $$U_f$$ calculates $$f(x)$$ for predetermined values of $$a$$ and $$N$$.
It is an XOR oracle, working as follows:

$$\begin{aligned}
    \Ket{x} \Ket{y}
    \quad \to \boxed{U_f(a, N)} \to \quad
    \Ket{x} \Ket{y \oplus f(x)}
\end{aligned}$$

Execution starts by applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
to the first $$q$$ qubits, yielding:

$$\begin{aligned}
    \Ket{0}^{\otimes q} \Ket{0}^{\otimes q}
    \quad \to \boxed{H^{\otimes q}} \to \quad
    \Ket{+}^{\otimes q} \Ket{0}^{\otimes q}
    = \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
\end{aligned}$$

Where $$Q = 2^q$$, and $$\Ket{x}$$ is the computational basis state $$\Ket{x_1} \cdots \Ket{x_q}$$.
Moving on to $$U_f$$:

$$\begin{aligned}
    \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
    \quad \to \boxed{U_f(a, N)} \to \quad
    \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{f(x)}
\end{aligned}$$

Then we measure $$f(x)$$, causing it collapse as follows,
for an unknown arbitrary value of $$x_0$$:

$$\begin{aligned}
    f(x_0) = f(x_0 + s) = f(x_0 + 2s) = \cdots = f(x_0 + (L-1) s)
\end{aligned}$$

Due to [entanglement](/know/concept/quantum-entanglement/),
the unmeasured (top $$q$$) qubits change state into a superposition:

$$\begin{aligned}
    \frac{1}{\sqrt{L}} \sum_{\ell = 0}^{L - 1} \Ket{x_0 + \ell s}
\end{aligned}$$

Clearly, there is a periodic structure here,
but we cannot measure it directly,
because we do not know the value of $$x_0$$,
which, to make matters worse, changes every time we run the algorithm.
This is where the QFT comes in, which outputs the following state:

$$\begin{aligned}
    \frac{1}{\sqrt{QL}} \sum_{k = 0}^{Q - 1} \bigg( \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg) \Ket{k}
\end{aligned}$$

Where $$\omega_Q$$ is a $$Q$$th root of unity.
Measuring this state yields a $$\Ket{k}$$, with a probability $$P(k)$$:

$$\begin{aligned}
    P(k)
    = \frac{1}{QL} \bigg| \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg|^2
    = \frac{1}{QL} \bigg| \omega_Q^{x_0 k} \sum_{\ell = 0}^{L - 1} \omega_Q^{\ell s k} \bigg|^2
    = \frac{1}{QL} \bigg| \sum_{\ell = 0}^{L - 1} \omega_Q^{\ell s k} \bigg|^2
\end{aligned}$$

The last step holds because $$|\omega_Q| = 1$$.
Surprisingly, this implies that we did not need
to perform the measurement of $$f(x)$$ earlier!
This makes sense: the period $$s$$ does not depend on $$x_0$$,
so why would we need an implicit $$x_0$$ to determine $$s$$?

So, what does the above probability $$P(k)$$ work out to?
There are two cases:

$$\begin{alignedat}{2}
    &\mathrm{if} \: \omega_Q^{sk} = 1: \qquad
    &&P(k) = \frac{1}{QL} |L|^2 = \frac{L}{Q}
    \\
    &\mathrm{if} \: \omega_Q^{sk} \neq 1: \qquad
    &&P(k) = \frac{1}{QL} \Bigg| \frac{1 - \omega_Q^{sk L}}{1 - \omega_Q^{sk}} \Bigg|^2
\end{alignedat}$$

Where the latter case was evaluated as a geometric series.
The condition $$\omega_Q^{sk}\!=\!1$$ is equivalent to asking
if $$sk$$ is a multiple of $$Q$$, i.e. if $$sk = cQ$$, for an integer $$c$$.

Recall that $$L$$ is the number of times that $$s$$ fits in $$Q$$,
so $$L\!=\!\lfloor Q / s \rfloor$$.
Assuming $$Q/s$$ is an integer, then $$L\!=\!Q/s$$ and $$Q\!=\!s L$$,
which tells us that
$$\omega_Q^{sk}\!=\!\omega_{s L}^{s k}\!=\!\omega_L^k$$.
This implies that if $$k$$ is a multiple of $$L$$ (i.e. $$k\!=\!c L$$),
then $$\omega_L^k\!=\!1$$, so $$P(k) = L / Q$$,
which is exactly what we got earlier!

In other words, the condition $$\omega_Q^{sk}\!=\!1$$
is equivalent to $$Q/s$$ being an integer.
In that case, we have that $$Q\!=\!sL$$,
which we substitute into $$P(k)$$ from earlier:

$$\begin{aligned}
    \mathrm{if} \: (Q/s) \in \mathbb{N}: \qquad
    P(k)
    = \frac{L}{Q}
    = \frac{1}{s}
\end{aligned}$$

And because $$k$$ is a multiple of $$L$$,
and $$L$$ fits $$s$$ times in $$Q$$,
there must be exactly $$s$$ values of $$k$$ that satisfy $$P(k) = 1/s$$.
Therefore the probability of all other $$k$$-values is zero!
This becomes clearer when you look at the sum used to calculate $$P(k)$$:
if $$Q\!=\!sL$$, then it sums $$\omega_L^{\ell k}$$ over $$\ell$$,
leading to perfect destructive interference for the "bad" $$k$$-values,
leaving only the "good" ones.

**So, to summarize: if** $$Q/s$$ **is an integer**,
then measuring only yields $$k$$-values that are multiples of $$L\!=\!Q/s$$.
Running Shor's algorithm several times then gives
several $$k$$-values separated by $$L$$.
That tells us what $$L$$ is, and we already know $$Q$$,
so we *finally* find the period $$s = Q/L$$.

That begs the question: what if $$Q/s$$ is not an integer?
We cannot *check* this, since $$s$$ is unknown!
Instead, we rewrite the probability $$P(k)$$ as follows:

$$\begin{aligned}
    \mathrm{if} \: (Q/s) \not\in \mathbb{N}: \qquad
    P(k)
    = \frac{1}{QL} \Bigg| \frac{1 - \omega_Q^{sk L}}{1 - \omega_Q^{sk}} \Bigg|^2
    = \frac{1}{QL} \Bigg| \frac{\sin(\pi s k L / Q)}{\sin(\pi s k / Q)} \Bigg|^2
\end{aligned}$$

This function peaks if $$s k$$ is close to a multiple of $$Q$$, i.e. $$s k \approx c Q$$,
which we rearrange:

$$\begin{aligned}
    \frac{k}{Q} \approx \frac{c}{s}
\end{aligned}$$

We know the left-hand side,
and, from the definition of $$f(x)$$,
clearly $$s \le N$$.
We chose $$Q \sim N^2$$,
so $$s$$ is quite small,
and consequently $$c$$ is too, since $$k < Q$$.

In other words, $$c/s$$ is a "simple" fraction,
so our goal is to find a "simple" fraction
that is close to the "complicated" fraction $$k/Q$$.
For example, if $$k/Q\!=\!0.332$$,
then probably $$c/s\!=\!1/3$$.

This can be done rigorously using the **continued fractions algorithm**:
write $$k/Q$$ as a continued fraction,
until the non-integer part of the denominator becomes small enough.
This part is then neglected,
and we calculate whatever is left, to get an estimate of $$c/s$$.

Of course, $$P(k)$$ is a probability distribution,
so even though the odds are in our favour,
we might occasionally measure a misleading $$k$$-value.
Running Shor's algorithm several times "fixes" this.

**So, to summarize: if** $$Q/s$$ **is not an integer**,
the measured $$k$$-values are generally close to $$c Q / s$$ for an integer $$c$$.
By approximating $$k/Q$$ using the continued fraction algorithm,
we estimate $$c/s$$.
Repeating this procedure gives several values of $$c/s$$,
such that $$s$$ is easy to deduce
by taking the least common multiple of the denominators.

In any case, once we think we have $$s$$,
we can easily verify that $$f(x)\!=\!f(x\!+\!s)$$.
Whether $$s$$ is the *smallest* such integer depends on how lucky we are,
but fortunately, for most applications of this algorithm,
that does not actually matter,
and usually we find the smallest $$s$$ anyway.

You typically need to repeat the algorithm $$\mathcal{O}(\log{q})$$ times,
and the QFT is $$\mathcal{O}(q^2)$$.
The bottleneck is modular exponentiation $$f$$,
which is $$\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$$
and therefore worse than the QFT,
yielding a total complexity of $$\mathcal{O}(q^2 (\log{q})^2 \log{\log{q}})$$.

OK, but what does $$s$$ have to do with factorizing integers?
Well, recall that $$f$$ is given by:

$$\begin{aligned}
    f(x)
    = a^x \bmod N
\end{aligned}$$

$$N$$ is the number to factorize, and $$a$$ is a random integer *coprime* to $$N$$,
meaning $$\gcd(a, N) = 1$$.
The fact that $$s$$ is the period of $$f$$ for a certain $$a$$-value, implies that:

$$\begin{aligned}
    a^x
    = a^{x + s} \bmod N
    \quad \implies \quad
    1
    = a^s \bmod N
\end{aligned}$$

Suppose that $$s$$ is even. In that case,
we can rewrite the above equation as follows:

$$\begin{aligned}
    (a^{s/2})^2 - 1
    = 0 \bmod N
\end{aligned}$$

In other words, $$(a^{s/2})^2 \!-\! 1$$ is a multiple of $$N$$.
We then use that $$(a\!-\!b) (a\!+\!b) = a^2\!-\!b^2$$:

$$\begin{aligned}
    \big( a^{s/2} - 1 \big) \big( a^{s/2} + 1 \big)
    = 0 \bmod N
\end{aligned}$$

Because $$s$$ is even by assumption, the two factors on the left are integers,
and as just mentioned, their product is a multiple of $$N$$.
Then we only need to calculate:

$$\begin{aligned}
    \gcd\!\big( a^{s/2}\!-\!1, N \big) > 1
    \quad\:\: \mathrm{and} \quad\:\:
    \gcd\!\big( a^{s/2}\!+\!1, N \big) > 1
\end{aligned}$$

And there we have the factors of $$N$$!
The $$\gcd$$ can be calculated efficiently in $$\mathcal{O}(q^2)$$ time.

But what if $$s$$ is odd?
No problem, then we just choose a new $$a$$ coprime to $$N$$,
and keep repeating Shor's algorithm until we do find an even $$s$$.
We do the same if $$a^{s/2}\!\pm\!1$$ is itself a multiple of $$N$$.



## References
1.  J.S. Neergaard-Nielsen,
    *Quantum information: lectures notes*,
    2021, unpublished.
2.  S. Aaronson,
    *Introduction to quantum information science: lecture notes*,
    2018, unpublished.