From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 .../concept/bernstein-vazirani-algorithm/index.md  | 34 +++++++++++-----------
 1 file changed, 17 insertions(+), 17 deletions(-)

(limited to 'source/know/concept/bernstein-vazirani-algorithm')

diff --git a/source/know/concept/bernstein-vazirani-algorithm/index.md b/source/know/concept/bernstein-vazirani-algorithm/index.md
index af49841..f91c0ba 100644
--- a/source/know/concept/bernstein-vazirani-algorithm/index.md
+++ b/source/know/concept/bernstein-vazirani-algorithm/index.md
@@ -17,10 +17,10 @@ It is extremely similar to the
 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,
+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 $s$:
+of $$x$$ with an unknown $$N$$-bit string $$s$$:
 
 $$\begin{aligned}
     f(x)
@@ -28,10 +28,10 @@ $$\begin{aligned}
     = (s_1 x_1 + s_2 x_2 + \:...\: + s_N x_N) \:\:(\bmod \: 2)
 \end{aligned}$$
 
-The goal is to find $s$.
+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 $n \in \{ 0, ..., N \!-\! 1\}$.
+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:
@@ -40,9 +40,9 @@ It uses the following circuit:
 <img src="bernstein-vazirani-circuit.png" style="width:52%">
 </a>
 
-Where $U_f$ is a phase oracle,
+Where $$U_f$$ is a phase oracle,
 whose action is defined as follows,
-where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$:
+where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$:
 
 $$\begin{aligned}
     \Ket{x}
@@ -51,14 +51,14 @@ $$\begin{aligned}
     = (-1)^{s \cdot x} \Ket{x}
 \end{aligned}$$
 
-That is, it introduces a phase flip based on the value of $f(x)$.
+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.
+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/) $H$
+Starting from the state $$\Ket{0}^{\otimes N}$$,
+applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
 to all qubits yields:
 
 $$\begin{aligned}
@@ -68,7 +68,7 @@ $$\begin{aligned}
     = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x}
 \end{aligned}$$
 
-This is an equal superposition of all candidates $\Ket{x}$,
+This is an equal superposition of all candidates $$\Ket{x}$$,
 which we feed to the oracle:
 
 $$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Then, thanks to the definition of the Hadamard transform,
-a final set of $H$-gates leads us to:
+a final set of $$H$$-gates leads us to:
 
 $$\begin{aligned}
     \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \Ket{x}
@@ -87,9 +87,9 @@ $$\begin{aligned}
     = \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 $s$.
+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 $$s$$.
 
 
 
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