In computational complexity theory and quantum computing, Simon's problem is a computational problem that is proven to be solved exponentially faster on a quantum computer than on a classical (that is, traditional) computer. The quantum algorithm solving Simon's problem, usually called Simon's algorithm, served as the inspiration for Shor's algorithm.^[1] Both problems are special cases of the abelian hidden subgroup problem, which is now known to have efficient quantum algorithms.

The problem is set in the model of decision tree complexity or query complexity and was conceived by Daniel R. Simon in 1994.^[2] Simon exhibited a quantum algorithm that solves Simon's problem exponentially faster with exponentially fewer queries than the best probabilistic (or deterministic) classical algorithm. In particular, Simon's algorithm uses a linear number of queries and any classical probabilistic algorithm must use an exponential number of queries.

This problem yields an oracle separation between the complexity classes BPP (bounded-error classical query complexity) and BQP (bounded-error quantum query complexity).^[3] This is the same separation that the Bernstein–Vazirani algorithm achieves, and different from the separation provided by the Deutsch–Jozsa algorithm, which separates P and EQP. Unlike the Bernstein–Vazirani algorithm, Simon's algorithm's separation is exponential.

Because this problem assumes the existence of a highly-structured "black box" oracle to achieve its speedup, this problem has little practical value.^[4] However, without such an oracle, exponential speedups cannot easily be proven, since this would prove that P is different from PSPACE.

Problem description

Given a function (implemented by a black box or oracle) ${\displaystyle f:\{0,1\}^{n}\rightarrow \{0,1\}^{n}}$ with the promise that, for some unknown ${\displaystyle s\in \{0,1\}^{n}}$, for all ${\displaystyle x,y\in \{0,1\}^{n}}$,

${\displaystyle f(x)=f(y)}$ if and only if ${\displaystyle x\oplus y\in \{0^{n},s\}}$,

where ${\displaystyle \oplus }$ denotes bitwise XOR. The goal is to identify ${\displaystyle s}$ by making as few queries to ${\displaystyle f(x)}$ as possible. Note that

${\displaystyle a\oplus b=0^{n}}$ if and only if ${\displaystyle a=b}$

Furthermore, for some ${\displaystyle x}$ and ${\displaystyle s}$ in ${\displaystyle x\oplus y=s}$, ${\displaystyle y}$ is unique (not equal to ${\displaystyle x}$) if and only if ${\displaystyle s\neq 0^{n}}$. This means that ${\displaystyle f}$ is two-to-one when ${\displaystyle s\neq 0^{n}}$, and one-to-one when ${\displaystyle s=0^{n}}$. It is also the case that ${\displaystyle x\oplus y=s}$ implies ${\displaystyle y=s\oplus x}$, meaning that

$${\displaystyle f(x)=f(y)=f(x\oplus {s})}$$

${\displaystyle f}$

The associated decision problem formulation of Simon's problem is to distinguish when ${\displaystyle s=0^{n}}$ (${\displaystyle f}$ is one-to-one), and when ${\displaystyle s\neq 0^{n}}$ (${\displaystyle f}$ is two-to-one).

Example

The following function is an example of a function that satisfies the required property for ${\displaystyle n=3}$:

${\displaystyle x}$	${\displaystyle f(x)}$
000	101
001	010
010	000
011	110
100	000
101	110
110	101
111	010

In this case, ${\displaystyle s=110}$ (i.e. the solution). Every output of ${\displaystyle f}$ occurs twice, and the two input strings corresponding to any one given output have bitwise XOR equal to ${\displaystyle s=110}$.

For example, the input strings ${\displaystyle 010}$ and ${\displaystyle 100}$ are both mapped (by ${\displaystyle f}$) to the same output string ${\displaystyle 000}$. That is, ${\displaystyle {\displaystyle f(010)=000}}$ and ${\displaystyle {\displaystyle f(100)=000}}$. Applying XOR to 010 and 100 obtains 110, that is ${\displaystyle {\displaystyle 010\oplus 100=110=s}.}$

${\displaystyle s=110}$ can also be verified using input strings 001 and 111 that are both mapped (by f) to the same output string 010. Applying XOR to 001 and 111 obtains 110, that is ${\displaystyle 001\oplus 111=110=s}$. This gives the same solution ${\displaystyle s=110}$ as before.

In this example the function f is indeed a two-to-one function where ${\displaystyle {\displaystyle s\neq 0^{n}}}$.

Problem hardness

Intuitively, this is a hard problem to solve in a "classical" way, even if one uses randomness and accepts a small probability of error. The intuition behind the hardness is reasonably simple: if you want to solve the problem classically, you need to find two different inputs ${\displaystyle x}$ and ${\displaystyle y}$ for which ${\displaystyle f(x)=f(y)}$. There is not necessarily any structure in the function ${\displaystyle f}$Zdroj:https://en.wikipedia.org?pojem=Simon's_problem
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