From 075683cdf4588fe16f41d9f7b46b9720b42b2553 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 17 Jul 2024 10:01:43 +0200
Subject: Improve knowledge base

---
 .../know/concept/deutsch-jozsa-algorithm/index.md  | 22 +++++++++++-----------
 1 file changed, 11 insertions(+), 11 deletions(-)

(limited to 'source/know/concept/deutsch-jozsa-algorithm/index.md')

diff --git a/source/know/concept/deutsch-jozsa-algorithm/index.md b/source/know/concept/deutsch-jozsa-algorithm/index.md
index 44b06ad..223877a 100644
--- a/source/know/concept/deutsch-jozsa-algorithm/index.md
+++ b/source/know/concept/deutsch-jozsa-algorithm/index.md
@@ -72,8 +72,8 @@ $$\begin{aligned}
     + \frac{1}{2} \Ket{1} \Big( \Ket{0 \oplus f(1)} - \Ket{1 \oplus f(1)} \Big)
 \end{aligned}$$
 
-The parenthesized superpositions can be reduced.
-Assuming that $$f(b) = 0$$, we notice:
+The parenthesized superpositions can be reduced:
+let us suppose that $$f(b) = 0$$, then:
 
 $$\begin{aligned}
     \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
@@ -91,7 +91,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We can thus combine both cases, $$f(b) = 0$$ or $$f(b) = 1$$,
-into the following single expression:
+into the following expression:
 
 $$\begin{aligned}
     \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
@@ -106,8 +106,8 @@ $$\begin{aligned}
     \frac{1}{2} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
 \end{aligned}$$
 
-The second qubit in state $$\Ket{-}$$ is garbage; it is no longer of interest.
-The first qubit is given by:
+The second qubit in state $$\Ket{-}$$ is garbage (i.e. no longer of interest).
+The first qubit is:
 
 $$\begin{aligned}
     \frac{1}{\sqrt{2}} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big)
@@ -126,8 +126,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Depending on whether $$f$$ is constant or balanced,
-the mearurement outcome of this state will be $$\Ket{0}$$ or $$\Ket{1}$$
-with 100\% probability. We have solved the problem!
+the measurement outcome of this state will be $$\Ket{0}$$ or $$\Ket{1}$$
+with 100% probability. We have solved the problem!
 
 Note that we only consulted the oracle (i.e. applied $$U_f$$) once.
 A classical computer would need to query it twice,
@@ -146,7 +146,7 @@ This algorithm is then implemented by the following quantum circuit:
     alt="Deutsch-Jozsa circuit" %}
 
 There are $$N$$ qubits in initial state $$\Ket{0}$$, and one in $$\Ket{1}$$.
-For clarity, the oracle $$U_f$$ works like so:
+The oracle $$U_f$$ performs this action:
 
 $$\begin{aligned}
     \Ket{x_1} \Ket{x_2} \cdots \Ket{x_N} \Ket{y}
@@ -167,7 +167,7 @@ $$\begin{aligned}
 Where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$ denotes a classical binary state.
 For example, if $$x = 5 = 2^0 + 2^2$$ in the summation,
 then $$\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$$
-(from least to most significant).
+(from least to most significant digit).
 
 We give this state to the oracle,
 and, by the same logic as for the Deutsch algorithm,
@@ -217,8 +217,8 @@ we only need to measure the $$N$$ qubits once;
 $$f$$ is constant if and only if all are zero.
 
 The Deutsch-Jozsa algorithm needs only one oracle query to give an error-free result,
-whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case;
-a revolutionary discovery.
+whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case.
+A revolutionary discovery!
 
 
 ## References
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