diff options
Diffstat (limited to 'content/know/concept/simons-algorithm/index.pdc')
| -rw-r--r-- | content/know/concept/simons-algorithm/index.pdc | 192 |
1 files changed, 192 insertions, 0 deletions
diff --git a/content/know/concept/simons-algorithm/index.pdc b/content/know/concept/simons-algorithm/index.pdc new file mode 100644 index 0000000..a65b3f0 --- /dev/null +++ b/content/know/concept/simons-algorithm/index.pdc @@ -0,0 +1,192 @@ +--- +title: "Simon's algorithm" +firstLetter: "S" +publishDate: 2021-05-01 +categories: +- Quantum information +- Algorithms + +date: 2021-04-13T10:29:02+02:00 +draft: false +markup: pandoc +--- + +# Simon's algorithm + +**Simon's algorithm** was the first proof that quantum computers +are able to solve some problems *exponentially* faster +than classical computers. +In the same spirit as +the [Deutsch-Jozsa algorithm](/know/concept/deutsch-jozsa-algorithm/) +and the [Bernstein-Vazirani algorithm](/know/concept/bernstein-vazirani-algorithm/), +the problem it solves, known as **Simon's problem**, +is of no practical use, +but nevertheless Simon's algorithm is an important landmark. + +Simon's problem is this: +we are given a "black box" function $f(x)$ +that takes an $n$-bit input $x$ +and returns an $n$-bit output. +We are promised that there exists an $s$ such that for all $x_1$ and $x_2$: + +$$\begin{aligned} + f(x_1) + = f(x_2) + \quad \Leftrightarrow \quad + x_2 = s \oplus x_1 +\end{aligned}$$ + +In other words, regardless of what $f(x)$ does behind the scenes, +its output is the same for inputs $x_1$ and $x_2$ +if and only if $x_2 = s \oplus x_1$, +or, equivalently, $x_1 = s \oplus x_2$. + +The goal is to find the $n$-bit number $s$, using as few calls to $f$ as possible. +There are two cases: +if $s = 0$, then $f$ is one-to-one, since $x_2 = 0 \oplus x_1 = x_1$. +Otherwise, if $s \neq 0$, then $f$ is two-to-one by definition: +for every $x_1$ there exists exactly one $x_2$ such that $x_2 = s \oplus x_1$. + +A classical computer solves this by randomly guessing inputs, +until it finds two that give the same output, +and then $s = x_1 \oplus x_2$. +For $n$-bit numbers, this takes $\mathcal{O}(\sqrt{2^n})$ guesses +(the square root is due to the birthday paradox). + +A quantum computer needs to query $f$ only $\mathcal{O}(n)$ times, +although the exact number varies due to the algorithm's probabilistic nature. +It uses the following circuit: + +<a href="simons-circuit.png" style="width:52%;display:block;margin:auto;"> +<img src="simons-circuit.png"> +</a> + +The XOR oracle $U_f$ implements $f$, +and has the following action for $n$-bit $a$ and $b$: + +$$\begin{aligned} + \ket{a} \ket{b} + \quad \to \boxed{U_f} \to \quad + \ket{a} \ket{b \oplus f(a)} +\end{aligned}$$ + +Starting from the state $\ket{0}^{\otimes 2 n}$, +we apply the [Hadamard gate](/know/concept/quantum-gate/) $H$ +to each of the first $n$ qubits: + +$$\begin{aligned} + \ket{0}^{\otimes n} \ket{0}^{\otimes n} + \quad \to \boxed{H^{\otimes n}} \to \quad + \ket{+}^{\otimes n} \ket{0}^{\otimes n} + = \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \ket{x} \ket{0}^{\otimes n} +\end{aligned}$$ + +Where $\ket{x}$ is shorthand for $\ket{x}_1 \cdots \ket{x}_n$. +In other words, we now have an equal superposition of all possible inputs $x$, +with a constant $\ket{0}^{\otimes n}$ beside it. +We give this to the oracle $U_f$: + +$$\begin{aligned} + \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \ket{x} \ket{0}^{\otimes n} + \quad \to \boxed{U_f} \to \quad + \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \ket{x} \ket{f(x)} +\end{aligned}$$ + +Then we apply $H^{\otimes n}$ to the first $n$ qubits again, +which, thanks to the definition of the Hadamard transform, +yields the following, +where $x \cdot y$ is the bitwise dot product: + +$$\begin{aligned} + \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \ket{x} \ket{f(x)} + \quad \to \boxed{H^{\otimes n}} \to \quad + &\frac{1}{2^n} \sum_{x = 0}^{2^n - 1} \bigg( \sum_{y = 0}^{2^n - 1} (-1)^{x \cdot y} \ket{y} \bigg) \ket{f(x)} +% \\ +% = &\sum_{y = 0}^{2^n - 1} \ket{y} \bigg( \frac{1}{2^n} \sum_{x = 0}^{2^n - 1} (-1)^{x \cdot y} \ket{f(x)} \bigg) +\end{aligned}$$ + + +Next, we measure all qubits. +The order in which we do this does not matter, +but, for clarity, let us measure the last $n$ qubits first, +yielding $\ket{f(x_1)}$ for some $x_1$. +Doing this leaves the $2n$ qubits in the following state, +where $f(x_1) = f(x_2)$ and $x_2 = s \oplus x_1$: + +$$\begin{alignedat}{2} + &\mathrm{if} \: s = 0: \qquad + &&\frac{1}{\sqrt{2^{n}}} \sum_{y = 0}^{2^n - 1} (-1)^{x_1 \cdot y} \ket{y} \ket{f(x_1)} + \\ + &\mathrm{if} \: s \neq 0: \qquad + &&\frac{1}{\sqrt{2^{n+1}}} \sum_{y = 0}^{2^n - 1} \Big( (-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \Big) \ket{y} \ket{f(x_1)} +\end{alignedat}$$ + +If $s = 0$, we get an equiprobable superposition of all $y$. +So, when we measure the first $n$ qubits, the result is a uniformly random number, +regardless of the phase $(-1)^{x_1 \cdot y}$. + +If $s \neq 0$, the situation is more interesting, +because we can only measure $y$-values where: + +$$\begin{aligned} + (-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \neq 0 +\end{aligned}$$ + +Since $x_2 = s \oplus x_1$ by definition, +we can rewrite this as follows: + +$$\begin{aligned} + (-1)^{x_1 \cdot y} + (-1)^{x_1 \cdot y \oplus s \cdot y} + = (-1)^{x_1 \cdot y} + (-1)^{x_1 \cdot y} (-1)^{s \cdot y} + \neq 0 +\end{aligned}$$ + +Clearly, the expression can only be nonzero if $s \cdot y$ is even. +In other words, when we measure the first $n$ qubits, +we get a random $y$-value, +for which $s \cdot y$ is guaranteed to be even. + +In both cases $s = 0$ and $s \neq 0$, +we measure a $y$-value that satisfies the equation: + +$$\begin{aligned} + s \cdot y = 0 \:\:(\bmod 2) +\end{aligned}$$ + +This tells us something about $s$, albeit not much. +But if we run Simon's algorithm $N$ times, +we get various $y$-values $y_1, ..., y_N$, +from which we can build a system of linear equations: + +$$\begin{aligned} + s \cdot y_1 &= 0 \:\:(\bmod 2) + \\ + s \cdot y_2 &= 0 \:\:(\bmod 2) + \\ + &\:\:\vdots + \\ + s \cdot y_N &= 0 \:\:(\bmod 2) +\end{aligned}$$ + +This can be solved efficiently by a classical computer. +In the best-case scenario, all those $y$-values would be linearly independent +(when regarded as vectors of bits), +in which case only $N = n - 1$ equations would be necessary. +Simon's algorithm is therefore $\mathcal{O}(n)$. + +It may feel like "cheating" to use a classical computer at the end. +Remember that the point of this algorithm is to limit the number of oracle queries, +which we did successfully. +Querying an oracle might be a very expensive operation, +so that is a big improvement! +That said, Simon's algorithm currently has no known practical uses. + + + +## References +1. J.S. Neergaard-Nielsen, + *Quantum information: lectures notes*, + 2021, unpublished. +2. S. Aaronson, + *Introduction to quantum information science: lecture notes*, + 2018, unpublished. |