Bernstein–Vazirani algorithm

The Bernstein–Vazirani algorithm, which solves the Bernstein–Vazirani problem is a quantum algorithm invented by Ethan Bernstein and Umesh Vazirani in 1992. It's a restricted version of the Deutsch–Jozsa algorithm where instead of distinguishing between two different classes of functions, it tries to learn a string encoded in a function. The Bernstein–Vazirani algorithm was designed to prove an oracle separation between complexity classes BQP and BPP.

Problem statement

Given an oracle that implements a function $f\colon \{0,1\}^{n}\rightarrow \{0,1\}$ in which $f(x)$ is promised to be the dot product between $x$ and a secret string $s\in \{0,1\}^{n}$ modulo 2, $f(x)=x\cdot s=x_{1}s_{1}+x_{2}s_{2}+\cdots +x_{n}s_{n}$, find $s$.

Algorithm

Classically, the most efficient method to find the secret string is by evaluating the function $n$ times where $x=2^{i}$, $i\in \{0,1,...,n-1\}$

${\begin{aligned}f(1000\cdots 0_{n})&=s_{1}\\f(0100\cdots 0_{n})&=s_{2}\\f(0010\cdots 0_{n})&=s_{3}\\&\,\,\,\vdots \\f(0000\cdots 1_{n})&=s_{n}\\\end{aligned}}$

In contrast to the classical solution which needs at least $n$ queries of the function to find $s$, only one query is needed using quantum computing. The quantum algorithm is as follows:

Apply a Hadamard transform to the $n$ qubit state $|0\rangle ^{\otimes n}$ to get

${\frac {1}{\sqrt {2^{n}}}}\sum _{x=0}^{2^{n}-1}|x\rangle .$

Next, apply the oracle $U_{f}$ which transforms $|x\rangle \to (-1)^{f(x)}|x\rangle$. This can be simulated through the standard oracle that transforms $|b\rangle |x\rangle \to |b\oplus f(x)\rangle |x\rangle$ by applying this oracle to ${\frac {|0\rangle -|1\rangle }{\sqrt {2}}}|x\rangle$. ($\oplus$ denotes addition mod two.) This transforms the superposition into

${\frac {1}{\sqrt {2^{n}}}}\sum _{x=0}^{2^{n}-1}(-1)^{f(x)}|x\rangle .$

Another Hadamard transform is applied to each qubit which makes it so that for qubits where $s_{i}=1$, its state is converted from $|-\rangle$ to $|1\rangle$ and for qubits where $s_{i}=0$, its state is converted from $|+\rangle$ to $|0\rangle$. To obtain $s$, a measurement in the standard basis ($\{|0\rangle ,|1\rangle \}$) is performed on the qubits.

Graphically, the algorithm may be represented by the following diagram, where $H^{\otimes n}$ denotes the Hadamard transform on $n$ qubits:

$|0\rangle ^{n}\xrightarrow {H^{\otimes n}} {\frac {1}{\sqrt {2^{n}}}}\sum _{x\in \{0,1\}^{n}}|x\rangle \xrightarrow {U_{f}} {\frac {1}{\sqrt {2^{n}}}}\sum _{x\in \{0,1\}^{n}}(-1)^{f(x)}|x\rangle \xrightarrow {H^{\otimes n}} {\frac {1}{2^{n}}}\sum _{x,y\in \{0,1\}^{n}}(-1)^{f(x)+x\cdot y}|y\rangle =|s\rangle$

The reason that the last state is $|s\rangle$ is because, for a particular $y$,

${\frac {1}{2^{n}}}\sum _{x\in \{0,1\}^{n}}(-1)^{f(x)+x\cdot y}={\frac {1}{2^{n}}}\sum _{x\in \{0,1\}^{n}}(-1)^{x\cdot s+x\cdot y}={\frac {1}{2^{n}}}\sum _{x\in \{0,1\}^{n}}(-1)^{x\cdot (s\oplus y)}=1{\text{ if }}s\oplus y={\vec {0}},\,0{\text{ otherwise}}.$

Since $s\oplus y={\vec {0}}$ is only true when $s=y$, this means that the only non-zero amplitude is on $|s\rangle$. So, measuring the output of the circuit in the computational basis yields the secret string $s$.

See also

• Hidden Linear Function problem