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Problem Statement: We are given a $2-1$ function $f:\{0,1\}^{n}\to\{0,1\}^{n}$ such that: there is a secret string $s\in\{0,1\}^{n}$ such that: $f(x)=f(x\oplus s)$. Challenge: find $s$.

Simon's algorithm says:

  1. Set up a random superposition $$\frac{1}{\sqrt{2}}|r\rangle + \frac{1}{\sqrt{2}}|r\oplus s\rangle$$

  2. Fourier sample to get a random $y$: $$y.s=0\ (\text{mod 2})$$

  3. Repeat steps $n-1$ times to generate $n-1$ linear equations in $s$.

Solve for $s$.

I don't understand steps (1) and (2). Why and how exactly do we set up the random superposition? How does it help? Also, in step (2), what does the dot operator (in $y.s$) stand for? Bit-wise multiplication?

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  • $\begingroup$ Probably a silly question, but what is $r$ in Step 1? Is it a random string? $\endgroup$
    – hola
    Sep 1, 2020 at 4:05
  • $\begingroup$ In the algorithm, where does the description of $f$ come in? I mean, $f$ not seem to be included in the algorithm? $\endgroup$
    – hola
    Sep 1, 2020 at 4:06

1 Answer 1

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$$ \newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}\newcommand{\proj}[1]{\left|#1\right\rangle\left\langle#1\right|} $$ Much of the functionality here is the same as the Bernstien-Vazirani algorithm, if that helps. The following is more or less copy and pasted from some lecture notes I prepared at some point. It explains it in a slightly different way to the direction you're coming at it from, but hopefully gets you going in the right direction.

The circuit is, in principle, the same as for the Bernstein-Vazirani Algorithm, except that since the output of the function evaluation is $n$ bits, the second register, which is used for the reversible function evaluation, is also $n$ bits, enter image description here This function evaluation is represented by the controlled-controlled...-controlled-U gate, acting as $\ket{x}\ket{y}\mapsto\ket{x}\ket{y\oplus f(x)}$.

We start with the first set of Hadamards producing an equally weighted superposition of all strings of $n$ bits, $$ \ket{0}\ket{0}\rightarrow\frac{1}{\sqrt{2^n}}\sum_{x\in\{0,1\}^n}\ket{x}\ket{0}. $$ This is followed by the function evaluation, $$ \rightarrow\frac{1}{\sqrt{2^n}}\sum_{x\in\{0,1\}^n}\ket{x}\ket{f(x)}. $$ At this point, you could measure the second register. This will return a random value of $f(r)$, so that the overall state of the system is in $$ \frac{1}{\sqrt{2}}(\ket{r}+\ket{r\oplus s})\ket{f(r)}. $$ So, the point is that you don't have to actively prepare different values of $r$; they will be selected for you by the measurement. We'll have to make a bit of an argument later on about how likely it is that we get new information each time we make such a measurement, but it'll all work out.

Actually, I don't usually think about performing the measurement at this point; it's unnecessary (but would allow you to avoid the density matrix formalism in the following calculation). Instead, calculate the action of the final set of Hadamards, yielding the final output state $$ \frac{1}{2^n}\sum_{x,z}(-1)^{x\cdot z}\ket{z}\ket{f(x)}. $$ Here, $$ x\cdot z= x_1z_1\oplus x_2z_2\oplus x_3z_3 \oplus\ldots \oplus x_nz_n, $$ where $x_k$ is the $k^{th}$ bit of $x$.

Now we collect unique values of $f(x)$, $$ \frac{1}{2^n}\sum_{z,f(x)}\left((-1)^{x\cdot z}+(-1)^{(x\oplus s)\cdot z}\right)\ket{z}\ket{f(x)}. $$ One can therefore verify that the output of the algorithm is $$ \frac{1}{2^{n-1}}\sum_{z: s\cdot z=0}(-1)^{x\cdot z}\ket{z}\sum_{f(x)}\ket{f(x)}. $$ The state of the first register clearly contains the information about $s$, but we need to extract it. Measuring the second register is not necessary (we could do it, but it doesn't help). Instead, let's trace out the second register, so we get $$ \rho=\frac{1}{2^{2n-2}}\sum_{z: s\cdot z=0}\sum_{y: s\cdot y=0}\sum_x(-1)^{x\cdot(y\oplus z)}\ket{y}\bra{z}. $$ By performing the sum over $x$, we are left with $$ \rho=\frac{1}{2^{n-1}}\sum_{s\cdot z=0}\proj{z}. $$ Measurement of the first register yields a binary string $z$ where $s\cdot z=0$. If we had $n-1$ such examples which are linearly independent, we would be able to determine $s$. This requires repeated application of the algorithm to find enough strings (this is the 'Fourier Sampling' part).

We must now justify that a linear number of applications is sufficient to find enough strings with high probability. In the absolute worst case, when we have found $n-2$ vectors, we must find the one remaining bit of information when there are still $2^{n-1}$ strings $s\cdot z=0$ to sample. These must constitute $1-2^{n-2}/2^{n-1}=\frac12$ of the space. Hence, the average number of trials to find one of these vectors is given by $$ \frac{1}{2}\sum_{n=0}^\infty\frac{n+1}{2^n}, $$ which is readily evaluated using the following identities, \begin{eqnarray} \sum_{n=0}^\infty r^n&=&\frac{1}{1-r} \nonumber\\ \frac{d}{dr}r\sum_{n=0}^\infty r^n&=&\sum_{n=0}^\infty (n+1)r^n. \nonumber \end{eqnarray} On average, we always find another linearly independent vector by making $2$ samples, i.e. within $2n$ steps, we have a high probability of determining $s$. It is this final part of the argument, requiring some classical post-processing that differentiates Simon's algorithm from the Bernstein-Vazirani algorithm. Essentially, the difference comes from not knowing the eigenvectors of $U$ in advance. If we did, we could prepare the second register in a fixed state. Instead, it is prepared in a superposition of different eigenstates (which coincides with a nice state to prepare) and we have to rely on a certain amount of randomness to sample the elements we need.

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  • $\begingroup$ +1 for the answer, but how are you drawing your circuit? It renders quite nicely. Is this a LaTeX package? $\endgroup$ Nov 11, 2021 at 14:10
  • $\begingroup$ @MarkS Yes, it's a latex package I wrote called quantikz $\endgroup$
    – DaftWullie
    Nov 11, 2021 at 14:50
  • $\begingroup$ @DaftWullie what is the Fourier sampling part. Once we measure the first register. We are as it is left with some states such that $s.z_i=0$. What is the purpose of repeating? Since repeating would give the same results? $\endgroup$
    – Upstart
    Apr 9, 2023 at 15:17
  • $\begingroup$ @Upstart Measuring once gives you a random value of $z$ for which you know $s\cdot z=0$. Repeating and measuring again makes another random selection, $z'$ such that $s\cdot z'=0$. Almost certainly, $z$ and $z'$ are linearly independent and so you get more information about what $s$ must be. Eventually, you get enough information to be able to understand what $s$ is. $\endgroup$
    – DaftWullie
    Apr 12, 2023 at 6:45
  • $\begingroup$ @DaftWullie In the actual implementation on qiskit, when we prepare superpositions, and say we do not measure the second register, and only measure the first we get all the $z_i$ we need , why is there a need to repeat.? $\endgroup$
    – Upstart
    Apr 12, 2023 at 11:06

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