OpenLLM-France/wikimedia huggingface dataset:
# Written Chinese/Calligraphy
## Strokes

Chinese characters are made up of strokes, in the same way that English words are made up of letters. These strokes are easily learned but not easily mastered; Chinese Calligraphy is an art in Chinese culture. The image below shows all strokes that all Chinese characters are made up of, and how they look in two different writing media: the strokes on the left are written with an ink brush, and the strokes on the right are written with an ordinary ball-point pen.
The following diagram shows you how to write these strokes. It's quite straight-forward: just start at the red dot, and trace the strokes as shown with your pen. Repeat this until you can write these strokes at ease, with the exact shape as shown.
File:Write chinese strokes.gif

## Character formations

The Chinese characters should be written at the correct scale; doing otherwise results in characters that are not only imprecise but incorrect. Chinese characters are just like English letters in this respect: the vertical line in the English letter "d", for example, would never be three times the height of the circle.
Chinese characters can be thought of as contained in invisible blocks, and further divided in several ways in proportion to these blocks. Look at the following diagram for details.
These formations can be in any proportion, depending on the complexity and amount of strokes in different places in the characters. Be aware of this feature, for scale is one of the keys to writing nice Chinese characters. The following diagram illustrates some of the examples.
These formations can also be included in one another to form other characters. The following diagram illustrates this feature with the same set of characters.
# Trigonometry/Solving triangles by half-angle formulae

In this section, we present alternative ways of solving triangles by using half-angle formulae.
Given a triangle with sides a, b and c, define
    s = $^1$⁄$_2$(a+b+c).
Note that:
    a+b-c = 2s-2c = 2(s-c)
and similarly for a and b.
We have from the cosine theorem
    $\displaystyle \cos(A)={{b^{2}+c^{2}-a^{2}} \over {2bc}}$

## Sin(A/2)

    $\displaystyle 2\sin ^{2}\left({\frac {A}{2}}\right)=1-\cos(A)=1-{{b^{2}+c^{2}-a^{2}} \over {2bc}}={{(a+b-c)(a-b+c)} \over {2bc}}={{2(s-b)(s-c)} \over {bc}}$
So
    $\displaystyle \sin \left({\frac {A}{2}}\right)={\sqrt {{(s-b)(s-c)} \over {bc}}}$.
By symmetry, there are similar expressions involving the angles B and C.
Note that in this expression and all the others for half angles, the positive square root is always taken. This is because a half-angle of a triangle must always be less than a right angle.

## Cos(A/2) and tan(A/2)

    $\displaystyle 2\cos ^{2}\left({\frac {A}{2}}\right)=1+\cos(A)=1+{{b^{2}+c^{2}-a^{2}} \over {2bc}}={{(a+b+c)(b+c-a)} \over {2bc}}={{2s(s-a)} \over {bc}}$
So
    $\displaystyle \cos \left({\frac {A}{2}}\right)={\sqrt {{s(s-a)} \over {bc}}}$.
    $\displaystyle \tan \left({\frac {A}{2}}\right)={\sin({\frac {A}{2}}) \over \cos({\frac {A}{2}})}={\sqrt {{(s-b)(s-c)} \over {s(s-a)}}}$.
Again, by symmetry there are similar expressions involving the angles B and C.

## Sin(A) and Heron's formula

A formula for sin(A) can be found using either of the following identities:
    $\displaystyle \sin(A)=2\sin \left({\frac {A}{2}}\right)\cos \left({\frac {A}{2}}\right)$
    $\displaystyle \sin(A)={\sqrt {(1+\cos(A))(1-\cos(A))}}$
These both lead to
    $\displaystyle \sin(A)={\frac {2}{bc}}{\sqrt {s(s-a)(s-b)(s-c)}}$
The positive square root is always used, since A cannot exceed 180º. Again, by symmetry there are similar expressions involving the angles B and C. These expressions provide an alternative proof of the sine theorem.
Since the area of a triangle
    $\displaystyle \Delta ={\frac {1}{2}}bc\sin(A)$,
    $\displaystyle \Delta ={\sqrt {s(s-a)(s-b)(s-c)}}$
which is Heron's formula.
# Trigonometry/Solving Triangles Given SAS
## Given Two Sides and the Included Angle (SAS)

Given two sides and the angle included by the two given sides, we can apply the Law of Cosines to find the missing side.
* Spelling out the detail
* Let's say the sides we know are $b$ and $c$ and the included angle is $\alpha$
* We want to find the missing side $a$
* We know:
* $b^{2}+c^{2}-2bc\cos(\alpha )=a^{2}$
* We have $b,c$ and $\alpha$ . We can therefore work out the missing side $a$ as
* $a={\sqrt {b^{2}+c^{2}-2bc\cos(\alpha )}}$
* Don't be confused by the different letters we are using here to the original statement of the law of cosines! Think of the law of cosines as being a relation between two sides and the included angle, to the remaining side. We have many choices as to how we label the sides and angles.
At this point we will know a side and an opposite angle. That's where the law of sines comes in. It says that in any triangle the ratio of the length of a side to the sine of the opposite angle is the same for all three sides. We can therefore use the law of sines to find the sine of the angle opposite each side that we know. Then, knowing the sine of the angle we can work out the angle itself.
* Spelling out the detail
* We know the ratio:
* ${\frac {\sin(\alpha )}{a}}$
* is the same for the other angles and sides. So in particular:
* $\sin(\gamma )={\frac {c}{a}}\sin(\alpha )$
* and
* $\sin(\beta )={\frac {b}{a}}\sin(\alpha )$
* We know the quantities on the right. We can calculate the quantities on the left from these. Take the inverse sines and we're done.
* Not quite.
* We need to check that the angles $\alpha ,\beta ,\gamma$ add up to 180°.
* You might be horrified sometime to find that they don't. What can go wrong? See the next box.
* Ambiguity of inverse sin
* What is $\sin(80^{\circ })$?
* What is $\sin(100^{\circ })$?
* (You are expected to use a calculator for these).
* Do you see the problem?
* The inverse sine function 'makes the wrong choice' when doing the inverse sine of an obtuse angle. So, in solving an ASA triangle, you need to spot if one of the angles is obtuse. If the inverse sine for it tells you an acute angle you can then correct it to the equivalent obtuse angle.
* $\sin(70^{\circ })=\sin(110^{\circ })$
* $\sin(60^{\circ })=\sin(120^{\circ })$
* $\sin(20^{\circ })=\sin(160^{\circ })$
* The angles with the same sine add up to 180°, so you can correct the angle that came out acute that should be obtuse by subtracting it from 180°. Alternatively, since at most one of the angles in a triangle can be obtuse, the one opposite the longest side, calculate the other angle instead. Then use the fact that the angles sum to 180° to get the remaining one.
As before, the area can be found from Heron's Formula.
| previous module: "Solving Triangles Given ASA" | • Table of Contents | next module: "Worked Example: Area of a Roof" |
# Trigonometry/Solving Triangles Given ASA

One of the most common applications of trigonometry is solving triangles—finding missing sides and/or angles given some information about a triangle. The area of the triangle may also be required.
We can reconstruct the triangle given:
* ASA Angle-Side-Angle (see illustration below).
* AAS Angle-Angle-Side (see later on this page).
* SAS Side-Angle-Side (see next page).
* SSA Side-Side-Angle (later in this book). In the SSA case we may have one, two or no possible solutions.

## Given ASA (Side and angles at each end of that side)

* Given two angles, we can find the third angle (since the sum of the measures of the three angles in a triangle is $180^{\circ }$ .
The missing angle, $\theta$ is given by:
    $\theta =180^{\circ }-(\alpha +\beta )$
* Knowing all three angles and one side, we can use the Law of Sines to find the missing sides.
In the illustration if the side we are given is the base and has length $c$ , then the side opposite the angle $\alpha$ has length $a$ given by:
    $a=c\cdot {\frac {\sin(\alpha )}{\sin(\theta )}}$
* Check This
* Check that this follows from the law of sines.
* Now is also a good time to check that you can derive the law of sines for yourself. If given a problem triangle like this one in an exam, you might be asked both to find its missing sides and to derive the law of sines.
* The area may then be found from Heron's Formula, or more easily by the formulae given in Law of Sines.

## Given AAS (one side and two other angles)

Again we can work out the missing angle, since they sum to $180^{\circ }$ . From there on we have the same information as after the first step in the ASA case.

## Exercises

* Exercise 1: Easy angles I
* A triangle $\Delta ABC$ has:
* Side ${\overline {BC}}$ of 30m
* $\angle ABC=30^{\circ }$ and
* $\angle BCA=45^{\circ }$
* Draw a rough diagram.
* What are the lengths of the remaining sides, and what is the missing angle?
* Exercise 2: Easy angles II
* A triangle $\displaystyle \Delta ABC$ has:
* Side ${\overline {BC}}$ of 30m
* $\angle BCA=30^{\circ }$ and
* $\angle CAB=45^{\circ }$
* Draw a rough diagram.
* What are the lengths of the remaining sides, and what is the missing angle?
* Exercise 3: Similar/Congruent?
* Are these two triangles in Exercise 1 and Exercise 2:
* Similar to each other?
* Congruent to each other?
| previous module: "Law of Cosines" | • Table of Contents | next module: "Solving Triangles Given SAS" |
# Trigonometry/Soh-Cah-Toa
## The Mnemonic

    $Soh-Cah-Toa$
$\displaystyle {\begin{array}{ccc}\sin(A)&=&\displaystyle {\frac {\text{opposite side}}{\rm {hypotenuse}}}\\\\\cos(A)&=&\displaystyle {\frac {\text{adjacent side}}{\rm {hypotenuse}}}\\\\\tan(A)&=&\displaystyle {\frac {\text{opposite side}}{\text{adjacent side}}}\end{array}}$
Soh-Cah-Toa is a mnemonic for remembering how to use sines, cosines, and tangents to compute lengths in right angle triangles.

# Trigonometric definitions

This triangle has sides a and b. The angle between them, C, is a right angle. The third side, c, is the hypotenuse. Side a is opposite angle A, and side b is adjacent to angle A.
Applying the definitions of the functions, we have:
    $\sin(A)={\frac {a}{c}}$ or opposite side over hypotenuse
    $\cos(A)={\frac {b}{c}}$ or adjacent side over hypotenuse
    $\tan(A)={\frac {a}{b}}$ or opposite side over adjacent side
This can be memorized using the mnemonic 'SOH-CAH-TOA' (sin = opposite over hypothenuse, cosine = adjacent over hypotenuse, tangent = opposite over adjacent).

## Video Link

These links from the Khan Academy are useful for this topic.
* Basic Trigonometry Soh-Cah-Toa (sin and cosine)
* Basic Trigonometry II Soh-Cah-Toa (starting with tan)

## Exercises: (Draw a diagram!)

* Exercise: Find sin cos and tan
* A right triangle has side a = 3, b = 4, and c = 5.
* Calculate the following:
* $\sin(A),\cos(A),\tan(A)$
* Exercise: Find a side
* A different right triangle has side $c=6$ and $\sin(A)=0.5$ .
* Calculate side a.
| previous module: "Cosine and Sine" | • Table of Contents | next module: "Sine Squared plus Cosine Squared" |
# World Stamp Catalogue/Sports
## English

Please see some images of postage stamps dedicated to sports, in the category:Sports on stamps.

## Russian

См. изображения некоторых почтовых марок, посвящённых спортивной тематике, в категории:Sports on stamps. Эта филателистическая тема более детально описана в статье Спорт в филателии.
# World Stamp Catalogue/Stamp valuation
## Catalogue

Stamps are collectibles and are exchanged and traded extensively in many countries around the world.
Most postage stamps are very common either in condition used or mint (not used) condition. Fine used and mint unhinged are the two conditions most likely to be collected.
It may be hard to comprehend, but stamp catalogs exist that identify nearly every postage stamp ever created in the entire world. A collector with an interest in just one country or region may prefer a stamp catalog that describes only the stamps in the specific collecting area. For beginner collectors, a pocket catalog that identifies each stamp and lists them in chronological order may be all that is needed.
Yet it is very difficult to devine a valuation scheme in one or several currencies which reflect the true market value of any particular collectible. Therefore, the following scheme is based on scarcity (and therefore value) and a basic coding is used in this catalogue:
| Code: | Number of stamps issued or believed to exist: |
| A | 10s |
| B | 100s |
| C | 1,000s |
| D | 10,000s |
| E | 100,000s |
| F | 1,000,000s |
# World Stamp Catalogue/Stamps on Gandhi

Post the issuance of stamps on his 150th birth anniversary in 2019, Mahatma Gandhi has become the individual whose is featured in most number of postage stamps in the world surpassing Winston Churchill, who had earlier achieved this by surpassing Christopher Columbus.
# World Stamp Catalogue/Stamp Inclusion policy

In this catalogue, the term "stamp" should be interpreted as any government-issued label that are designed and issued for use in payment of postal, revenue , penalty or related charges.
Several present day postal administrations have outsource their stamp issuing activities to agencies who design, print and distribute stamps. For examples, many countries in African regions outsource/contracting their postal to Stamperija for their postage stamps needs. This is important to consider that this is not government issued stamps and it is left up to contributor's discretion whether to include on this catalogue or not.
Other stamps were issued by governments in exile, pressure groups, political parties and organizations. These stamps are often, but not always, propaganda labels that were never actually used to pay for postal charges.
Finally, there are a subset of stamps known as "Cinderella Stamp", which are basically decorative stickers that resembles postage stamps but did not have values.
* Cinderella Stamps - 1914 Christmas Seal
* Penny Black is world's first postage stamp
# World Stamp Catalogue/Stamps on stamps
## English

Please see some images of postage stamps, design of which includes images of other stamps, in the category:Stamps on stamps. This philatelic topic is described in more detail in Stamps on stamps.

## Russian

См. изображения некоторых почтовых марок, рисунок которых включает изображения других почтовых марок, в категории:Stamps on stamps. Эта филателистическая тема более детально описана в статье Марка на марке.
# World Stamp Catalogue/Stellaland

Definitives
| Date Issued | Denom. | Color | Perf. | Cat. No. Mi/Sc/Gi | Design | Designer/ Engraver | notes |
| 1884-02 | 1D. | Red | 12 | SG 1 Sc 1 | Coat of Arms | Typography | |
| 1884-02 | 1D. | Red | 12 | Sc 1a | Coat of Arms | Typography | Horizontal pair imperforate |
| 1884-02 | 1D. | Red | 12 | Sc 1b | Coat of Arms | Typography | Vertical pair imperforate |
| 1884-02 | 3D. | Orange | 12 | SG 2 Sc 2 | Coat of Arms | Typography | |
| 1884-02 | 3D. | Orange | 12 | Sc 2a | Coat of Arms | Typography | Horizontal pair imperforate |
| 1884-02 | 3D. | Orange | 12 | Sc 2b | Coat of Arms | Typography | Vertical pair imperforate |
| 1884-02 | 4D. | Blue | 12 | SG 3 Sc 3 | Coat of Arms | Typography | |
| 1884-02 | 4D. | Blue | 12 | Sc 4a | Coat of Arms | Typography | Horizontal pair imperforate |
| 1884-02 | 4D. | Blue | 12 | Sc 4b | Coat of Arms | Typography | Vertical pair imperforate |
| 1884-02 | 6D. | Mauve | 12 | SG 4 Sc 4 | Coat of Arms | Typography | |
| 1884-02 | 6D. | Mauve | 12 | Sc 4a | Coat of Arms | Typography | Horizontal pair imperforate |
| 1884-02 | 6D. | Mauve | 12 | Sc 4b | Coat of Arms | Typography | Vertical pair imperforate |
| 1884-02 | 1Sh. | Green | 12 | SG 5 Sc 5 | Coat of Arms | Typography | |
Definitive Handstamped TWEE
| Date Issued | Denom. | Paper. | Color | Perf. | Type Die/Wmk | Subject | Cat. No. Mi/Sc/Gi | Design | No. Issued | Designer/ Engraver | notes |
| 1885 | 2D. on 4D. | | Blue | 12 | | | SG 6 Sc 6 | Coat of Arms | | Typography | The Scott catalog questions its validity. |
# XML - Managing Data Exchange/XSLT and Style Sheets/Answers

Stylesheet Chapter => XSLT and Style Sheets
Stylesheet Exercises => Exercises
# Basic XPath Syntax

Expressions that start with a forward slash "/" are called absolute expressions. They start at the root of the document. All other expressions are relative to the current position within an XML document.
Expressions are created by creating a list of step expressions of the form
You can think of the predicate as a filter or conditional expression that service like a WHERE clause in SQL.

## Sample XML file

Many of the examples use a "books" example such as the following:
http://raw.github.com/dmccreary/learn-xquery/master/data/books.xml
In general the books file has the following structure:

## Basic XPath Expressions

The root document node
Note that the forward slash returns the document root, not the full books element.
The root node that contains all the books:
All book elements:
The first version is with an absolute path. The second uses a relative path - book elements at any level of the file.
Note that the first expression is faster in unindexed XML but within indexed native XML databases the second is faster.
A count of the number of books:
All the book titles:
The second book in the collection:
The title of the second book:
The third author of the second book
All books with the format "wikibook":
Get a list of all the publishers
Get a distinct list of the publishers (duplicates removed)
Books that have at least one price over 30

## XPath abbreviations

. represents the current node
.. represents the nearest parent node
@ represents the attribute delimiter
$ represents the variable delimiter
[n] represents the n-th child of the current node
ancestor::div represents the set of parent div nodes
normalize-space(firstname)="Paul" matches Paul regardless of whitespace delimiters
boolean(string($myvar) ) checks for empty strings
/ represents the absolute path of the root node
@* represents all attributes of the current node
-Return all values using a union of attributes, node names, and text values:
@*|node|text
-Return all of a node's siblings using a union of the preceding-sibling and following-sibling axes:
preceding-sibling::node | following-sibling::node
-Return the adjacent sibling of a specific type
//div/following-sibling::h3
-Check string value of current node
[. = "Matthew Bob"]
-Node identity can be checked using the count function to see if the intersection of two node-sets of the same length equals the length of either of the node sets(or in the case of a single node set whether it is equal to 1). For example, the following query returns TRUE in this case because both nodes are the same:
count(/bk:books | /bk:books/bk:book[1]/parent::*) = 1
# XML - Managing Data Exchange/XSL from chapters

* C.3
* C.4
* C.5
* C.6
# XPath
## Table of Contents
### Basic Syntax

* Basic Syntax basic syntax using simple books example

### Equivalent Expressions

* CSS Equivalents CSS and JQuery equivalent expressions
* SQL Equivalents SQL equivalent expressions

### References

* References SQL equivalent expressions

### See also

* XML - Managing Data Exchange/XPath
* XQuery/XPath examples
# XML - Managing Data Exchange/XSLT and Style Sheets/Exercises

Stylesheet Chapter => XSL and Style Sheets
Stylesheet Answers => Answers
* For Chapter 2's exercises, we created an XML document, schema, and stylesheet to define a museum entity. Change that stylesheet to transform that XML document into another XML document, instead of into HTML.
* In Chapter 4 we created an XML document, schema, and stylesheet to list restaurants, including the most popular, in a city.
** a. Create a new stylesheet that uses a named template to output the name of the city. Then use two templates with different modes to print out the restaurants, using a larger font for the most popular restaurant.
** b. Now create a stylesheet that will sort the restaurant names alphabetically, and number them using the number function and a single level of numbering.
** c. Finally, we will use the call-template and with-param elements to create a stylesheet that will count the restaurants in a city, and add the prices up, and output it to a text file. This stylesheet will first call a template named "count-restaurants" and will use with-param to pass a parameter to count the restaurants. Then it will call a template named "sum-prices" that will use with-param to pass the parameter "num" to sum the upper price for the restaurant.
Stylesheet Chapter => XSL and Style Sheets
Stylesheet Answers => Answers
# XML - Managing Data Exchange/XUL/Answers

Exercises
* Create a window that contains two text boxes that are approximately 200 pixels down the window. Use either vertical or horizontal layout. Label each text box and set default values for each text box that are different from each other. Create a button labeled ‘Swap text’. Next, create a function that when the button is clicked, will swap the text from one text box to the other.
* In this exercise, you will need to make two separate files. On one page, create a window that has two tabbed panels with different content on each. On the other page, simply make a popup menu with one menu item ‘Open’. Create a function that will redirect the browser to the tabbed page when the ‘Open’ menu option is clicked.
Question 1: Answer
XulNum1.js
* function swap
* var noDanaObj=document.getElementById('nodana');
* var onlyXulObj=document.getElementById('onlyxul');
* var noDanaStr=noDanaObj.getAttribute("value");
* var onlyXulStr=onlyXulObj.getAttribute("value");
* noDanaObj.setAttribute("value",onlyXulStr);
* onlyXulObj.setAttribute("value",noDanaStr);
XulNum1.xul
* <?xml version="1.0"?>
* <?xml-stylesheet href="chrome://global/skin/" type="text/css"?>
* <window
* id="Ex1"
* title="Excercise 1"
* orient="vertical"
* xmlns= "http://www.mozilla.org/keymaster/gatekeeper/there.is.only.xul">
* <script src="XulNum1.js" />
* <hbox width="200px" height="200px" flex = "1">
* <text value= "There is no Dana..." id="nodana" style="font-size:18pt"/>
* </hbox>
* <hbox width="200px" height="200px" flex = "1">
* <text value="There is only XUL" id="onlyxul" style="font-size:24pt"/>
* </hbox>
* <box width="200px">
* <button id="swap" label="swap" default="true" oncommand="swap;"/>
* </box>
* </window>
# YafaRay/How do I add my own scene?

If you think you have a useful/interesting scene you want to share with the rest of the world, just add it. On top of every Wikibooks page you should see a 'edit this page' link. Just click on it when you are in the 'YafaRay: Scenes with Explanations' page and give your scene an appropriate place/name. After you save the page (always preview it first, of course), you can click on your scene name and put your scene on the empty page you are linked to.
If you are not sure about anything, just look at how other pages are made.
# YafaRay

YafaRay is an open source ray tracing program available under the LGPL license. It supersedes YafaRay (Yet Another Free Raytracer). It enables the creation of images and animations to a photo-realistic quality.
Yafaray was developed in C++ and can be ported to a number of operating systems including Linux, Windows, and Mac OS.
It has a number of rendering features including: global illumination, skydome illumination, caustics, depth of field and blurry reflections.

## Contents

* User Manual
* Scenes with Explanations
* How do I add my own scene?
* Simple Preview

## Further reading

* Yafaray at the FSF directory
* libYafaRay on GitHub