---
title: "Bernstein-Vazirani algorithm"
firstLetter: "B"
publishDate: 2021-05-01
categories:
- Quantum information
- Algorithms

date: 2021-04-13T10:28:44+02:00
draft: false
markup: pandoc
---

# Bernstein-Vazirani algorithm

In quantum information,
the **Bernstein-Vazirani algorithm** proves
the supremacy of quantum computers
over classical deterministic or probabilistic computers.
It is extremely similar to the
[Deutsch-Jozsa algorithm](/know/concept/deutsch-jozsa-algorithm/),
and even uses the same circuit.

It solves a very artificial problem:
we are given a "black box" function $f(x)$
that takes an $N$-bit $x$ and returns a single bit,
which we are promised is the lowest bit of the bitwise dot product
of $x$ with an unknown $N$-bit string $s$:

$$\begin{aligned}
    f(x)
    = s \cdot x \:\:(\bmod 2)
    = (s_1 x_1 + s_2 x_2 + \:...\: + s_N x_N) \:\:(\bmod 2)
\end{aligned}$$

The goal is to find $s$.
To solve this problem,
a classical computer would need to call $f(x)$ exactly $N$ times
with $x = 2^n$  for $n \in \{ 0, ..., N \!-\! 1\}$.
However, the Bernstein-Vazirani algorithm
allows a quantum computer to do it with only a single query.
It uses the following circuit:

<a href="bernstein-vazirani-circuit.png">
<img src="bernstein-vazirani-circuit.png" style="width:52%;display:block;margin:auto;">
</a>

Where $U_f$ is a phase oracle,
whose action is defined as follows,
where $\ket{x} = \ket{x_1} \cdots \ket{x_N}$:

$$\begin{aligned}
    \ket{x}
    \quad \to \boxed{U_f} \to \quad
    (-1)^{f(x)} \ket{x}
    = (-1)^{s \cdot x} \ket{x}
\end{aligned}$$

That is, it introduces a phase flip based on the value of $f(x)$.
For an example implementation of such an oracle,
see the Deutsch-Jozsa algorithm:
its circuit is identical to this one,
but describes $U_f$ in a different (but equivalent) way.

Starting from the state $\ket{0}^{\otimes N}$,
applying the [Hadamard gate](/know/concept/hadamard-gate/) $H$
to all qubits yields:

$$\begin{aligned}
    \ket{0}^{\otimes N}
    \quad \to \boxed{H^{\otimes N}} \to \quad
    \ket{+}^{\otimes N}
    = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \ket{x}
\end{aligned}$$

This is an equal superposition of all candidates $\ket{x}$,
which we feed to the oracle:

$$\begin{aligned}
    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \ket{x}
    \quad \to \boxed{U_f} \to \quad
    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \ket{x}
\end{aligned}$$

Then, thanks to the definition of the Hadamard transform,
a final set of $H$-gates leads us to:

$$\begin{aligned}
    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \ket{x}
    \quad \to \boxed{H^{\otimes N}} \to \quad
    \ket{s}
    = \ket{s_1} \cdots \ket{s_N}
\end{aligned}$$

Which, upon measurement, gives us the desired binary representation of $s$.
For comparison, the Deutsch-Jozsa algorithm only cares whether $s = 0$ or $s \neq 0$,
whereas this algorithm is interested in the exact value of $s$.



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