## 1. Project Overview & Quickstart (0xAX/linux-initialization-10)
# linux-initialization-10
Open-source repository 0xAX/linux-initialization-10
### Repository Details
- **Repository:** [0xAX/linux-initialization-10](https://github.com/0xAX/linux-initialization-10)
- **Primary Language:** Code
*Note: High-volume repository documentation is actively indexed and synchronized by YakaAI.*
## 2. Official Technical Reference & Guides (0xAX/0xAX.github.io)
## File: docs/content/content.md
# Exercises
:::note
The solutions for the following exercises should be considered together with my modest mathetmatical knowledge. As I am not a professional mathematician, I can not give you 100% guarantee that all of this is correct. It is only my try. If you see the mistake, please feel free to open issue or create a pull request.
:::
## Exercise 3.3-1
> Show that if $f(n)$ and $g(n)$ are monotonically increasing functions, then so are the functions $f(n) + g(n)$ and $f(g(n))$, and if $f(n)$ and $g(n)$ are in addition nonnegative, then $f(n) * g(n)$ is monotonically increasing.
First of all let's remember what is the `monotonically increasing` functions:
> A function $f(n)$ is **monotonically increasing** if $m \le n$ implies $f(m) \le f(n)$.
In accordance with the conditions of the exercise, the both our functions $f(n)$ and $g(n)$ are monotonically increasing functions. Let's consider the function $h(n) = f(n) + g(n)$. We have to prove that it is also monotonically increasing function. If we will take two numbers $m$ and $n$ where $m \le n$, we will have: $f(m) \le f(n)$ and $g(m) \le g(n)$ by the definition of monotonically increasing functions. It implies that $f(m) + g(m) \le f(n) + g(n)$.
In the same way since the $g(m) \le g(n)$ and the $f(n)$ function is monotonically increasing, the $f(g(m)) \le f(g(n))$.
If in the same time our both functions are nonnegative, we can safely multiple them and according to the facts above it also should be monotonically increasing function.
We may have even simple proof using [coq](https://coq.inria.fr/) proof assistant:
```coq
Require Import Coq.Arith.Le.
Parameter f : nat -> nat.
Parameter g : nat -> nat.
Axiom f_monotonic : forall x y : nat, x <= y -> f x <= f y.
Axiom g_monotonic : forall x y : nat, x <= y -> g x <= g y.
Theorem sum_of_monotonic_funs : forall x y : nat, x <= y -> f x + g x <= f y + g y.
Proof.
intros x y H.
apply Nat.add_le_mono.
- apply f_monotonic. assumption.
- apply g_monotonic. assumption.
Qed.
Theorem application_of_monotonic_funs : forall x y : nat, x <= y -> f(g(x)) <= f(g(y)).
Proof.
intros x y H.
apply f_monotonic.
apply g_monotonic.
assumption.
Qed.
```
## Exercise 3.3-2
> Prove that $\lfloor \alpha * n \rfloor + \lceil (1 - \alpha)n \rceil = n$ for any integer $n$ and real number \alpha in the range of $0 \le \alpha \le 1.
The cases when $\alpha = 0$ or $\alpha = 1$ are trivial.
If $\alpha = 0$ we will have:
$$
\lfloor \alpha \cdot n \rfloor + \lceil (1 - a) \cdot n \rceil = \lfloor 0 \rfloor + \lceil (1 - 0) \cdot n \rceil = 0 + n = n
$$
if $\alpha = 1$ we will have:
$$
\lfloor \alpha \cdot n \rfloor + \lceil (1 - a) \cdot n \rceil = \lfloor 1 \cdot n \rfloor + \lceil (1 - 1) \cdot n \rceil = n + 0 = n
$$
If $0 < \alpha < 1$ we may use the equations that is given in the book: $-\lceil x \rceil = \lfloor -x \rfloor$ and $-\lfloor x \rfloor = \lceil -x \rceil$. And in addition the fact that the integer number $n$ and real number $x$ we may have \lfloor n + x \rfloor = n + \lfloor x \rfloor. So let's take a look at the given expression:
$$
\lfloor \alpha * n \rfloor + \lceil (1 - \alpha)n \rceil = \lfloor \alpha * n \rfloor + \lceil n - n \cdot \alpha \rceil = \lfloor \alpha * n \rfloor + n + \lceil (- (\alpha \cdot n)) \rceil = \lfloor \alpha \cdot n \rfloor + n - \lfloor \alpha \cdot n \rfloor = n
$$
In addition we can see that the growth of both functions are equal:
```python
import math
import matplotlib.pyplot as plt
x = []
for i in range(0, 1001):
x.append(i)
y1 = []
for i in x:
y1.append(i)
y2 = []
for i in x:
y2.append(math.floor(0.3 * i) + math.ceil((1 - 0.3) * i))
plt.plot(x, y1, linewidth=6, color='red', label ='f(n) = n')
plt.plot(x, y2, '-.', linewidth=6, color='green', label ='f(n) = ⌊αn⌋ + ⌈(1 - α)n⌉')
plt.ylim(ymin=0)
plt.xlim(xmin=0)
plt.xlabel("X")
plt.ylabel("Y")
plt.legend()
plt.title('f(n) = ⌊αn⌋ + ⌈(1 - α)n⌉ and f(n) = n')
plt.show()
```
The result we may see:
## Exercise 3.3-3
> Use equation (3.14) or other means to show that $(n + o(n))^k \in \theta(n^k)$ for any real constant $k$. Conclude that $\lfloor n \rfloor^k \in \theta(n^k)$.
At the beginning the `Use equation (3.14)` confused me pretty much. As the equation mentioned in the exercise text is $1 + x \le e^x$. I have spent enough time thinking about it but did not find a way how to apply this equation to the exercise. Luckily I remembered about the [Errata for Introduction to Alogorithms](https://mitp-content-server.mit.edu/books/content/sectbyfn/books_pres_0/11599/e4-bugs.html) and according to it, it turns out that it is a typo. It must be `Use the equation 3.13`. This equation says: $n^b \in o(a^n)$ or that any [exponential function](https://en.wikipedia.org/wiki/Exponential_function) than any [polynomial function](https://en.wikipedia.org/wiki/Polynomial). Maybe this note will help you, but it did not help me. So I decided to look at this using the most straightforward way. The $(n + o(n))^k$ is:
$$
(n + o(n))^k = \sum_{i = 0}^{k} \binom{k}{i} n^{k-i} \cdot f(n)^{i}
$$
The term with the highest power of $n$ is $n^k$ which is $\theta(n^k)$. For all other terms we will have $n^{k-i}$ and $f(n)^i$. As our $f(n) \in o(n)$ by definition, it means that with the growth of $n$ our $f(n)$ will tend to $0$ and as a result will bring insignificant growth to $n^{k-i}$.
## Exercise 3.3-4
> Proove the following:
>
> *a.* Equation (3.21)
>
> *b.* Equations (3.26)-(3.28)
>
> *c.* $log(\theta(n)) \in \theta(log(n))$
### Equation (3.21)
The equation *3.21* is:
$$
a^{log_{b}(c)} = c^{log_{b}(a)}
$$
Let's try to prove it:
$$
a^{log_{b}(c)} = c^{log_{b}(a)} = с^{\frac{log_{c}(a)}{log_{c}(b)}} = \sqrt[log_{c}(b)]{c^{log_{c}(a)}} = a^{\frac{1}{log_{c}(b)}} = a^{log_{b}{c}}
$$
### Equations (3.26)
The equation *3.26* is:
$$
n! \in o(n^n)
$$
We can try to prove it using [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation):
$$
n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n \cdot (1 + \theta(\frac{1}{n}))
$$
The definition of $o$ is:
$$
\begin{align*}
& o(g(n)) = f(n) : \text{ for any positive constant } c > 0, \text{ there exist a constant } \\
& \qquad \qquad \ \ \ n_{0} \text{ such that } 0 \le f(n) < cg(n) \text{ for all } n >= n_{0}
\end{align*}
$$
Which says us that $f(n)$ grows significantly slower than $g(n)$. So let's see what we can get out of it:
$$
\lim_{n\to\infty} \frac{n^n}{\sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n \cdot (1 + \theta(\frac{1}{n}))} = \lim_{n\to\infty} \frac{e^n}{\sqrt{2 \cdot \pi \cdot n} \cdot (1 + \theta(\frac{1}{n}))} = 2 \cdot \pi \lim_{n\to\infty} \frac{e^n}{\sqrt{\cdot n} \cdot (1 + \theta(\frac{1}{n}))} = \lim_{n\to\infty} \frac{e^n}{\sqrt{n}}
$$
Now using the [L'Hôpital's rule](https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule):
$$
\lim_{n\to\infty} (\frac{e^n}{\sqrt{n}})’ = \lim_{n\to\infty} \frac{e^n}{\frac{1}{2 \cdot \sqrt{n}}} = \lim_{n\to\infty} e^n \cdot \sqrt{n} = \infty
$$
### Equation (3.27)
The equation *3.27* is:
$$
n! \in \omega(2^n)
$$
We can do the same what we did in the previous exercise - to apply **Stirling's approximation** and see what we can get. But before this, let's try to remember the definition of the $\omega$:
$$
\begin{align*}
& \omega(g(n)) = f(n) : \text{ for any positive constant } c > 0, \text{ there exist a constant } \\
& \qquad \qquad \ \ \ n_{0} \text{ such that } 0 \le cg(n) < f(n) \text{ for all } n >= n_{0}
\end{align*}
$$
Or in other words $g(n)$ grows significantly slower than $f(n)$. So let's see:
$$
\lim_{n\to\infty} \frac{2^n}{\sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n \cdot (1 + \theta(\frac{1}{n}))} = \lim_{n\to\infty} \frac{2^n}{(\frac{n}{e})^n} = \lim_{n\to\infty} \frac{(2 \cdot e)^n}{n^n}
$$
Since $n^n$ grows much faster than $(2 \cdot e)^n$ the limit will strive to $0 => n! \omega(n^n)$.
### Equations (3.28)
The equation *3.28* is:
$$
log(n!) \in \theta(n \cdot log(n))
$$
Going over the same way, using the **Stirling's approximation** we will get:
$$
log(n)! = log(\sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^{n} \cdot (1 + \theta(\frac{1}{n}))) = log(\sqrt{2 \cdot \pi \cdot n }) + log((\frac{n}{e})^n) = \theta(sqrt_{n}) + n \cdot log(\frac{n}{e})
$$
In the last expression the leading term is $n \cdot log(\frac{n}{e})$ or $nlog(n) - n(log(e))$. Skipping the least significant terms we will get what we need to prove: $log(n!) \in \theta(n \cdot log(n))$.
### Equation (c)
According to the definition of $\theta$ we need to show that:
$$
log(c_{1} \cdot n) \le log(f(n)) \le log(c_{2} \cdot n)
$$
We can consider $f(n)$ as a [linear function](https://en.wikipedia.org/wiki/Linear_function) so we can rewrite our inequation as:
$$
log(c_{1} \cdot n) \le log(f(a \cdot n + b)) \le log(c_{2} \cdot n)
$$
In other words, we need to find such constants $c_{1}$ and $c_{2}$ and $n \ge n_{0}$ after which this inequation will be true. For this case we can consider $c_{1} = \frac{1}{n}$ and $c_{2} \le n$. Let's take a look at the left part of the inequation:
$$
\begin{align*}
& log(c_{1} \cdot n) \le log(f(a \cdot n + b)) \\
& log(c_{1}) + log(n) \le log(f(a \cdot n + b)) \\
& log(\frac{1}{n}) + log(n) \le log(a \cdot n + b) \\
\end{align*}
$$
The left part gives $0$ for any $n$.
Now we consider the right side of inequation with $c_{2} \le n$ or even the case $c_{2} = n$:
$$
\begin{align*}
& log(f(a \cdot n + n)) \le log(c_{2} \cdot n) \\
& log(f(a \cdot n + n)) \le log(c_{2}) + log(n) \\
& log(f(a \cdot n + n)) \le log(n) + log(n) \\
& log(f(a \cdot n + n)) \le 2 \cdot log(n) \text{ since } a \text{ and } b \text{ are constants }
\end{align*}
$$
## Exercise 3.3-5
> Is the function $\lceil log_{2}(n) \rceil!$ polynomially bounded? Is the function $\lceil log_{2}(log_{2}(n)) \rceil!$ polynomially bounded?
From the chapter *Polynomials* we know that `polynomially bounded` means - $f(n) \in O(n^{k})$. The definition of the `O` is:
$$
\begin{align*}
& O(g(n)) = f(n) : \text{ there exist positive constants c and } n_{0} \text{ such that } \\
& \qquad \qquad \ \ \ 0 <= f(n) <= cg(n) \text{ for all } n >= n_{0}
\end{align*}
$$
So we basically should prove (or refute) that $\lceil log(n) \rceil! \in O(n^{k})$ and $\lceil log_{2}(log_{2}(n)) \rceil! \in O(n^{k})$.
As soon as we see [factorial function](https://en.wikipedia.org/wiki/Factorial), the first thing that comes to mind from this book is [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation). Let's remember how its definition looked in the book:
$$
n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^{n} \cdot (1 + \theta(\frac{1}{n}))
$$
First of all let's consider the $\lceil log_{2}(n) \rceil!$ function and will try to apply `Stirling's approximation` to it:
$$
\lceil log_{2}(n) \rceil! = \sqrt{2 \cdot \pi \cdot log_{2}(n)} \cdot (\frac{log_{2}(n)}{e})^{log_{2}(n)} \cdot (1 + \theta(\frac{1}{log_2{(n)}}))
$$
The leading term here is $(\frac{log_{2}(n)}{e})^{log_{2}(n)}$ so we can skip others and consider only it. By the using well known formula of logarithms $a^{log_{b}(c)} = c^{log_{b}(a)}$ we can rewrite our expression as:
$$
\left(\frac{\log_{2}n}{e}\right)^{\log_{2}n} = n^{\log_{2}\left(\frac{\log_2{n}}{e}\right)} = n^{\log_{2}(\log_{2}n - \log_{2}e)}
$$
To answer the question does it polynomially bounded, we need to compare it with $n^{k}$. As $k$ is constant (as well as $\log_{2}e$, we will have: $n^{\log_{2}(\log_{2}n - \log_{2}e)} > n^k$ with growth of $n$ which means $\lceil \log(n) \rceil!$ is not polynomially bounded.
The another and more easy way to prove that could be considering $n = 2^k$. In this case the $\lceil log_{2}(n) \rceil!$ becomes just $k!$. The factorial function is $1 \cdot 2 \cdot 2 \cdot 3 \cdot \ldots \cdot k$. From another side our polynom is $2^{mk}$. As we know the factorial function growth faster than polynom, thus the function $\lceil log_{2}(n) \rceil!$ is not polynomially bounded.
For the second function $\lceil log(log(n)) \rceil!$ we can consider that $n$ is reresented by the $2^{2^{k}}$. By the logarithm rules it becomes $k! \le 2^{2^{k}}$. The $k!$ is $1 \cdot 2 \cdot 3 \cdot \ldots \cdot k$. The $2^{2^{k}}$ is $4^1 \cdot 4^2 \cdot 4^3 \cdot \ldots \cdot 4^k$. Thus our main expression $k! \le 2^{2^{k}}$ is true which means the $\lceil log(log(n)) \rceil!$ function is polynomially bounded.
## Exercise 3.3-6
> Which is asymptotically larger: $log^\*(log(n))$ or $log(log^*(n))$?
Personally for me it was much easier to use the definition of the **iterated logarithm** from [wikipedia](https://en.wikipedia.org/wiki/Iterated_logarithm):
$$
log^\*(n) =
\begin{cases}
0 & \qquad if n \le 1; \\
1 + log^\*(log(n)) & \qquad if n > 1 \\
\end{cases}
$$
So basically **iterated logarithm** is a number of times we need to apply [logarithm](https://en.wikipedia.org/wiki/Logarithm) function to $n$ while we will not get `1`. To understand it better, we can consider the following simple examples:
$$
log(log^\*(65536)) = log(1 + log^\*(16)) = log(1 + 1 + log^\*(4)) = log(1 + 1 + 1 + log^\*(2)) + log(1 + 1 + 1 + 1) = 2
$$
and:
$$
log^\*(log(65536)) = log^\*(16) = 1 + log^\*(log(4)) = 1 + log^\*(2) = 1 + 1 + 1 = 3
$$
To understand which one function growth faster, we may divide one into another and to check the result. If the result is `0` the function that is in the dividend growth slower and vice versa. So let's consider the following expression in assumption that $n = 2^m$:
$$
\frac{log^\*(log(n))}{log(log^*(n))} = \frac{log(1 + log^\*(m))}{log^\*(m)} = \frac{log(1 + 1 + log^\*(k))}{1 + log^\*(k)} = \frac{log(1 + 1 + 1 + \ldots + 1)}{1 + 1 + 1 + \ldots 1}
$$
Obviously the function in the dividend growth slower than in divisor. This means the $log(log^*(n))$ function grows assymptotically faster than $log^\*(log(n))$ function.
## Exercise 3.3-7
> Show that golden ratio $\phi$ and its conjugate $\overline \phi$ both satisfy the equation $x^2 = x + 1$
Let's look closer at the equation:
$$
\begin{align*}
& x^2 = x + 1 \\
& x^2 - x - 1 = 0 \\
& x_{1} = \frac{1 + \sqrt{1^2 - (4 \cdot 1 \cdot -1}}{2 \cdot 1)} = \frac{1 + \sqrt(5)}{2} \\
& x_{2} = \frac{1 - \sqrt{1^2 - (4 \cdot 1 \cdot -1}}{2 \cdot 1)} = \frac{1 - \sqrt(5)}{2} \\
\end{align*}
$$
Where $x_{1}$ and $x_{2}$ are exactly $\phi$ and $\overline \phi$.
## Exercise 3.3-8
> Prove by induction that the $i_{th}$ Fibonacci number satisfies the equation $F_{i} = \frac{\phi^i - \overline \phi^i}{\sqrt{5}}$
Using [mathematical induction](https://en.wikipedia.org/wiki/Mathematical_induction), we should consider the base of induction first of all:
$$
P(1) = \frac{\frac{1 + \sqrt(5)}{2} - \frac{1 - \sqrt(5)}{2}}{\sqrt(5)} = \frac{\frac{1 + \sqrt{5} - 1 + \sqrt{5}}{2}}{\sqrt{5}} = 1
$$
As for the $i = 1$ the Fibonacci number is $1$ the base of induction is correct. For the Fibonacci numbers for $i = i - 1$ and $i = i - 2$ we will have:
$$
\begin{align*}
& F_{i} = F_{i - 1} + F_{i - 2} = \\
& \frac{\phi^{i - 1} - \overline \phi^{i - 1} + \phi^{i - 2} - \overline \phi^{i - 2}}{\sqrt{5}} = \\
& \frac{\phi^{i - 2}(\phi + 1) - \overline \phi^{i - 2}(\overline \phi + 1)}{\sqrt{5}} = \\
& \frac{\phi^{i - 2} \cdot \phi^2 - \overline \phi^{i - 2} \cdot \overline \phi^2}{\sqrt{5}} = \\
& \frac{\phi^i - \overline \phi^i}{\sqrt{5}}
\end{align*}
$$
## Exercise 3.3-9
> Show that $k \cdot log(k) \in \theta(n)$ implies $k = \theta(n / log(n))$
By the rule of **symmetery** we have $f(n) \in \theta(g(n)) \text{ if and only if } g(n) \in \theta(f(n))$. According to this rule, we may rewrite our expression as:
$$
n \in \theta(k \cdot log(k))
$$
Let's take logarithm of both parts:
$$
log(n) = log(k \cdot log(k)) = log(k) + log(log(k))
$$
The $log(log(k))$ grows significantly slower than $log(k)$ so we may omit it. Now let's devide $n$ to $log(n)$:
$$
\frac{n}{log(n)} = \frac{k \cdot log(k)}{log(k)} = k \in \theta(k)
$$
So we should that $k \in \theta(n / log(n))$.
---
## File: docs/tutorial-basics/congratulations.md
---
sidebar_position: 6
---
# Congratulations!
You have just learned the **basics of Docusaurus** and made some changes to the **initial template**.
Docusaurus has **much more to offer**!
Have **5 more minutes**? Take a look at **[versioning](../tutorial-extras/manage-docs-versions.md)** and **[i18n](../tutorial-extras/translate-your-site.md)**.
Anything **unclear** or **buggy** in this tutorial? [Please report it!](https://github.com/facebook/docusaurus/discussions/4610)
## What's next?
- Read the [official documentation](https://docusaurus.io/)
- Modify your site configuration with [`docusaurus.config.js`](https://docusaurus.io/docs/api/docusaurus-config)
- Add navbar and footer items with [`themeConfig`](https://docusaurus.io/docs/api/themes/configuration)
- Add a custom [Design and Layout](https://docusaurus.io/docs/styling-layout)
- Add a [search bar](https://docusaurus.io/docs/search)
- Find inspirations in the [Docusaurus showcase](https://docusaurus.io/showcase)
- Get involved in the [Docusaurus Community](https://docusaurus.io/community/support)
---
## File: docs/tutorial-basics/create-a-blog-post.md
---
sidebar_position: 3
---
# Create a Blog Post
Docusaurus creates a **page for each blog post**, but also a **blog index page**, a **tag system**, an **RSS** feed...
## Create your first Post
Create a file at `blog/2021-02-28-greetings.md`:
```md title="blog/2021-02-28-greetings.md"
---
slug: greetings
title: Greetings!
authors:
- name: Joel Marcey
title: Co-creator of Docusaurus 1
url: https://github.com/JoelMarcey
image_url: https://github.com/JoelMarcey.png
- name: Sébastien Lorber
title: Docusaurus maintainer
url: https://sebastienlorber.com
image_url: https://github.com/slorber.png
tags: [greetings]
---
Congratulations, you have made your first post!
Feel free to play around and edit this post as much as you like.
```
A new blog post is now available at [http://localhost:3000/blog/greetings](http://localhost:3000/blog/greetings).
---
## File: docs/tutorial-basics/create-a-document.md
---
sidebar_position: 2
---
# Create a Document
Documents are **groups of pages** connected through:
- a **sidebar**
- **previous/next navigation**
- **versioning**
## Create your first Doc
Create a Markdown file at `docs/hello.md`:
```md title="docs/hello.md"
# Hello
This is my **first Docusaurus document**!
```
A new document is now available at [http://localhost:3000/docs/hello](http://localhost:3000/docs/hello).
## Configure the Sidebar
Docusaurus automatically **creates a sidebar** from the `docs` folder.
Add metadata to customize the sidebar label and position:
```md title="docs/hello.md" {1-4}
---
sidebar_label: 'Hi!'
sidebar_position: 3
---
# Hello
This is my **first Docusaurus document**!
```
It is also possible to create your sidebar explicitly in `sidebars.js`:
```js title="sidebars.js"
export default {
tutorialSidebar: [
'intro',
// highlight-next-line
'hello',
{
type: 'category',
label: 'Tutorial',
items: ['tutorial-basics/create-a-document'],
},
],
};
```
---
## File: docs/tutorial-basics/create-a-page.md
---
sidebar_position: 1
---
# Create a Page
Add **Markdown or React** files to `src/pages` to create a **standalone page**:
- `src/pages/index.js` → `localhost:3000/`
- `src/pages/foo.md` → `localhost:3000/foo`
- `src/pages/foo/bar.js` → `localhost:3000/foo/bar`
## Create your first React Page
Create a file at `src/pages/my-react-page.js`:
```jsx title="src/pages/my-react-page.js"
import React from 'react';
import Layout from '@theme/Layout';
export default function MyReactPage() {
return (
My React page
This is a React page
);
}
```
A new page is now available at [http://localhost:3000/my-react-page](http://localhost:3000/my-react-page).
## Create your first Markdown Page
Create a file at `src/pages/my-markdown-page.md`:
```mdx title="src/pages/my-markdown-page.md"
# My Markdown page
This is a Markdown page
```
A new page is now available at [http://localhost:3000/my-markdown-page](http://localhost:3000/my-markdown-page).
---
## File: docs/tutorial-basics/deploy-your-site.md
---
sidebar_position: 5
---
# Deploy your site
Docusaurus is a **static-site-generator** (also called **[Jamstack](https://jamstack.org/)**).
It builds your site as simple **static HTML, JavaScript and CSS files**.
## Build your site
Build your site **for production**:
```bash
npm run build
```
The static files are generated in the `build` folder.
## Deploy your site
Test your production build locally:
```bash
npm run serve
```
The `build` folder is now served at [http://localhost:3000/](http://localhost:3000/).
You can now deploy the `build` folder **almost anywhere** easily, **for free** or very small cost (read the **[Deployment Guide](https://docusaurus.io/docs/deployment)**).
---
## File: docs/tutorial-basics/markdown-features.mdx
---
sidebar_position: 4
---
# Markdown Features
Docusaurus supports **[Markdown](https://daringfireball.net/projects/markdown/syntax)** and a few **additional features**.
## Front Matter
Markdown documents have metadata at the top called [Front Matter](https://jekyllrb.com/docs/front-matter/):
```text title="my-doc.md"
// highlight-start
---
id: my-doc-id
title: My document title
slug: /my-custom-url
---
// highlight-end
## Markdown heading
Markdown text with [links](./hello.md)
```
## Links
Regular Markdown links are supported, using url paths or relative file paths.
```md
Let's see how to [Create a page](/create-a-page).
```
```md
Let's see how to [Create a page](./create-a-page.md).
```
**Result:** Let's see how to [Create a page](./create-a-page.md).
## Images
Regular Markdown images are supported.
You can use absolute paths to reference images in the static directory (`static/img/docusaurus.png`):
```md
```
You can reference images relative to the current file as well. This is particularly useful to colocate images close to the Markdown files using them:
```md
```
## Code Blocks
Markdown code blocks are supported with Syntax highlighting.
````md
```jsx title="src/components/HelloDocusaurus.js"
function HelloDocusaurus() {
return Hello, Docusaurus!
;
}
```
````
```jsx title="src/components/HelloDocusaurus.js"
function HelloDocusaurus() {
return Hello, Docusaurus!
;
}
```
## Admonitions
Docusaurus has a special syntax to create admonitions and callouts:
```md
:::tip My tip
Use this awesome feature option
:::
:::danger Take care
This action is dangerous
:::
```
:::tip My tip
Use this awesome feature option
:::
:::danger Take care
This action is dangerous
:::
## MDX and React Components
[MDX](https://mdxjs.com/) can make your documentation more **interactive** and allows using any **React components inside Markdown**:
```jsx
export const Highlight = ({children, color}) => (
{
alert(`You clicked the color ${color} with label ${children}`)
}}>
{children}
);
This is Docusaurus green !
This is Facebook blue !
```
export const Highlight = ({children, color}) => (
{
alert(`You clicked the color ${color} with label ${children}`);
}}>
{children}
);
This is Docusaurus green !
This is Facebook blue !
---
## File: docs/tutorial-extras/manage-docs-versions.md
---
sidebar_position: 1
---
# Manage Docs Versions
Docusaurus can manage multiple versions of your docs.
## Create a docs version
Release a version 1.0 of your project:
```bash
npm run docusaurus docs:version 1.0
```
The `docs` folder is copied into `versioned_docs/version-1.0` and `versions.json` is created.
Your docs now have 2 versions:
- `1.0` at `http://localhost:3000/docs/` for the version 1.0 docs
- `current` at `http://localhost:3000/docs/next/` for the **upcoming, unreleased docs**
## Add a Version Dropdown
To navigate seamlessly across versions, add a version dropdown.
Modify the `docusaurus.config.js` file:
```js title="docusaurus.config.js"
export default {
themeConfig: {
navbar: {
items: [
// highlight-start
{
type: 'docsVersionDropdown',
},
// highlight-end
],
},
},
};
```
The docs version dropdown appears in your navbar:
## Update an existing version
It is possible to edit versioned docs in their respective folder:
- `versioned_docs/version-1.0/hello.md` updates `http://localhost:3000/docs/hello`
- `docs/hello.md` updates `http://localhost:3000/docs/next/hello`
---
## File: docs/tutorial-extras/translate-your-site.md
---
sidebar_position: 2
---
# Translate your site
Let's translate `docs/intro.md` to French.
## Configure i18n
Modify `docusaurus.config.js` to add support for the `fr` locale:
```js title="docusaurus.config.js"
export default {
i18n: {
defaultLocale: 'en',
locales: ['en', 'fr'],
},
};
```
## Translate a doc
Copy the `docs/intro.md` file to the `i18n/fr` folder:
```bash
mkdir -p i18n/fr/docusaurus-plugin-content-docs/current/
cp docs/intro.md i18n/fr/docusaurus-plugin-content-docs/current/intro.md
```
Translate `i18n/fr/docusaurus-plugin-content-docs/current/intro.md` in French.
## Start your localized site
Start your site on the French locale:
```bash
npm run start -- --locale fr
```
Your localized site is accessible at [http://localhost:3000/fr/](http://localhost:3000/fr/) and the `Getting Started` page is translated.
:::caution
In development, you can only use one locale at a time.
:::
## Add a Locale Dropdown
To navigate seamlessly across languages, add a locale dropdown.
Modify the `docusaurus.config.js` file:
```js title="docusaurus.config.js"
export default {
themeConfig: {
navbar: {
items: [
// highlight-start
{
type: 'localeDropdown',
},
// highlight-end
],
},
},
};
```
The locale dropdown now appears in your navbar:
## Build your localized site
Build your site for a specific locale:
```bash
npm run build -- --locale fr
```
Or build your site to include all the locales at once:
```bash
npm run build
```