check_code_snippets.py Documentation

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1. Project Overview & Quickstart (0xAX/check_code_snippets.py)

check_code_snippets.py

Open-source repository 0xAX/check_code_snippets.py

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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 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) = ⌊αnβŒ‹ + ⌈(1 - Ξ±)nβŒ‰')

plt.ylim(ymin=0)
plt.xlim(xmin=0)
plt.xlabel("X")
plt.ylabel("Y")
plt.legend()
plt.title('f(n) = ⌊αnβŒ‹ + ⌈(1 - Ξ±)nβŒ‰ 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 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 than any polynomial function. 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:

$$
n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n \cdot (1 + \theta(\frac{1}{n}))
$$

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'HΓ΄pital's 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 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, the first thing that comes to mind from this book is Stirling's 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:

$$
log^*(n) =
\begin{cases}
0 & \qquad if n \le 1; \
1 + log^*(log(n)) & \qquad if n > 1 \
\end{cases}
$$

So basically iterated logarithm is a number of times we need to apply logarithm 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, 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

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 and i18n.

Anything unclear or buggy in this tutorial? Please report it!

What's next?

File: docs/tutorial-basics/create-a-blog-post.md

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: SΓ©bastien 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.

File: docs/tutorial-basics/create-a-document.md

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.

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

Create a Page

Add Markdown or React files to src/pages to create a standalone page:

  • src/pages/index.js β†’ localhost:3000/
  • src/pages/foo.md β†’ localhost:3000/foo
  • src/pages/foo/bar.js β†’ 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 (
    <Layout>
      <h1>My React page</h1>
      <p>This is a React page</p>
    </Layout>
  );
}

A new page is now available at 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.

File: docs/tutorial-basics/deploy-your-site.md

Deploy your site

Docusaurus is a static-site-generator (also called Jamstack).

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/.

You can now deploy the build folder almost anywhere easily, for free or very small cost (read the Deployment Guide).

File: docs/tutorial-basics/markdown-features.mdx

Markdown Features

Docusaurus supports Markdown and a few additional features.

Front Matter

Markdown documents have metadata at the top called Front Matter:

text title="my-doc.md"
// highlight-start
---
id: my-doc-id
title: My document title
description: My document description
slug: /my-custom-url
---
// highlight-end

## Markdown heading

Markdown text with [links](./hello.md)

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.

Images

Regular Markdown images are supported.

You can use absolute paths to reference images in the static directory (static/img/docusaurus.png):

md
![Docusaurus logo](/img/docusaurus.png)

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
![Docusaurus logo](./img/docusaurus.png)

Code Blocks

Markdown code blocks are supported with Syntax highlighting.

md
```jsx title="src/components/HelloDocusaurus.js"
function HelloDocusaurus() {
  return <h1>Hello, Docusaurus!</h1>;
}
```
jsx title="src/components/hellodocusaurus.js"
function HelloDocusaurus() {
  return <h1>Hello, Docusaurus!</h1>;
}

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 can make your documentation more interactive and allows using any React components inside Markdown:

jsx
export const Highlight = ({children, color}) => (
  <span
    style={{
      backgroundColor: color,
      borderRadius: '20px',
      color: '#fff',
      padding: '10px',
      cursor: 'pointer',
    }}
    onClick={() => {
      alert(`You clicked the color ${color} with label ${children}`)
    }}>
    {children}
  </span>
);

This is <Highlight color="#25c2a0">Docusaurus green</Highlight> !

This is <Highlight color="#1877F2">Facebook blue</Highlight> !

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

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

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/ 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

File: docs/intro.md

Tutorial Intro

Let's discover Docusaurus in less than 5 minutes.

Getting Started

Get started by creating a new site.

Or try Docusaurus immediately with docusaurus.new.

What you'll need

  • Node.js version 18.0 or above:
    • When installing Node.js, you are recommended to check all checkboxes related to dependencies.

Generate a new site

Generate a new Docusaurus site using the classic template.

The classic template will automatically be added to your project after you run the command:

bash
npm init docusaurus@latest my-website classic

You can type this command into Command Prompt, Powershell, Terminal, or any other integrated terminal of your code editor.

The command also installs all necessary dependencies you need to run Docusaurus.

Start your site

Run the development server:

bash
cd my-website
npm run start

The cd command changes the directory you're working with. In order to work with your newly created Docusaurus site, you'll need to navigate the terminal there.

The npm run start command builds your website locally and serves it through a development server, ready for you to view at http://localhost:3000/.

Open docs/intro.md (this page) and edit some lines: the site reloads automatically and displays your changes.

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