---
title: "Bernstein-Vazirani algorithm"
sort_title: "Bernstein-Vazirani algorithm"
date: 2021-05-01
categories:
- Quantum information
- Algorithms
layout: "concept"
---

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%">
</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/quantum-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.