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nerding out a bit with haml

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making a custom filter for LaTeX eqnarray divs
cat-town
Dan Buch 13 years ago
parent 18b6b315f0
commit 4361300838
6 changed files with 217 additions and 310 deletions
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2
math-replay/Makefile
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@ -3,7 +3,7 @@ ALL += $(patsubst %.tex,%.pdf,$(shell find . -name '*.tex'))
%.html:%.haml
haml $^ $@
haml -Ilib -rhaml/filters-ext $^ $@
%.pdf:%.tex
texi2pdf $^ -o $@ && \

63
math-replay/algebra/linear-equations-2.haml
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@ -7,10 +7,7 @@
%link{:rel => 'stylesheet', :type => 'text/css', :href => './style.css'}
%body
%h2 Solve for $x$:
#solve-for-x-01.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
6x + 8 & = & 5 \\
-8 + 6x + 8 & = & 5 - 8 \\
6x & = & 5 - 8 \\
@ -19,110 +16,68 @@
x & = & -3 \times \frac{1}{6} \\
x & = & -3 \over 6 \\
x & = & \frac{-1}{2} \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-02.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
10x - 2 & = & 8 \\
10x - 2 + 2 & = & 8 + 2 \\
10x & = & 10 \\
10x \over 10 & = & 10 \over 10 \\
x & = & 10 \over 10 \\
x & = & 1 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-03.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
6x - 4 & = & 3 \\
6x - 4 + 4 & = & 3 + 4 \\
6x & = & 7 \\
6x \over 6 & = & 7 \over 6 \\
x & = & 7 \over 6 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-04.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
10x + 7 & = & 8 \\
10x + 7 - 7 & = & 8 - 7 \\
10x & = & 1 \\
10x \over 10 & = & 1 \over 10 \\
x & = & 1 \over 10 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-05.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
10x + 3 & = & 5 \\
10x + 3 - 3 & = & 5 - 3 \\
10x & = & 2 \\
10x \over 10 & = & 2 \over 10 \\
x & = & 1 \over 5 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-06.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
4x + 9 & = & 10 \\
4x + 9 - 9 & = & 10 - 9 \\
4x & = & 1 \\
4x \over 4 & = & 1 \over 4 \\
x & = & 1 \over 4 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-07.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
7x - 5 & = & 9 \\
7x - 5 + 5 & = & 9 + 5 \\
7x & = & 14 \\
7x \over 7 & = & 14 \over 7 \\
x & = & 2 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-08.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
3x + 9 & = & 9 \\
3x + 9 - 9 & = & 9 - 9 \\
3x & = & 0 \\
3x \over 3 & = & 0 \over 3 \\
x & = & 0 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x-09.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
3x + 10 & = & 7 \\
3x + 10 - 10 & = & 7 - 10 \\
3x & = & -3 \\
3x \over 3 & = & -3 \over 3 \\
x & = & -1 \\
\end{eqnarray}
$$

69
math-replay/algebra/new-definitions-1.haml
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@ -7,62 +7,37 @@
%link{:rel => 'stylesheet', :type => 'text/css', :href => './style.css'}
%body
%h2 $x `not` $y = 8x - 3y$:
.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
x `not` y & = & 8x - 3y \\
6 `not` -2 & = & 8(6) - 3(-2) \\
6 `not` -2 & = & 48 + 6 \\
6 `not` -2 & = & 54 \\
\end{eqnarray}
$$
%h2 $x `star` y = 8y + 2$:
.solution
:escaped
$$
\begin{eqnarray}
x `star` y = 8y + 2 \\
-2 `star` 5 = 8(5) + 2 \\
-2 `star` 5 = 40 + 2 \\
-2 `star` 5 = 42 \\
\end{eqnarray}
$$
:eqnarray
x `star` y & = & 8y + 2 \\
-2 `star` 5 & = & 8(5) + 2 \\
-2 `star` 5 & = & 40 + 2 \\
-2 `star` 5 & = & 42 \\
%h2 $x `diamond` y = xy + 3x - y$:
.solution
:escaped
$$
\begin{eqnarray}
x `diamond` y = xy + 3x - y \\
3 `diamond` -2 = (3)(-2) + 3(3) - (-2) \\
3 `diamond` -2 = -6 + 9 + 2 \\
3 `diamond` -2 = -6 + 11 \\
3 `diamond` -2 = 5 \\
\end{eqnarray}
$$
:eqnarray
x `diamond` y & = & xy + 3x - y \\
3 `diamond` -2 & = & (3)(-2) + 3(3) - (-2) \\
3 `diamond` -2 & = & -6 + 9 + 2 \\
3 `diamond` -2 & = & -6 + 11 \\
3 `diamond` -2 & = & 5 \\
%h2 $x `not` y = x + 3y$:
.solution
:escaped
$$
\begin{eqnarray}
x `not` y = x + 3y \\
-4 `not` 6 = -4 + 3(6) \\
-4 `not` 6 = -4 + 18 \\
-4 `not` 6 = 14 \\
\end{eqnarray}
$$
:eqnarray
x `not` y & = & x + 3y \\
-4 `not` 6 & = & -4 + 3(6) \\
-4 `not` 6 & = & -4 + 18 \\
-4 `not` 6 & = & 14 \\
%h2 $x `donut` y = 3x - 3y$:
.solution
:escaped
$$
\begin{eqnarray}
x `donut` y = 3x - 3y \\
0 `donut` 0 = 3(0) - 3(0) \\
0 `donut` 0 = 0 - 0 \\
0 `donut` 0 = 0 \\
\end{eqnarray}
$$
:eqnarray
x `donut` y & = & 3x - 3y \\
0 `donut` 0 & = & 3(0) - 3(0) \\
0 `donut` 0 & = & 0 - 0 \\
0 `donut` 0 & = & 0 \\

63
math-replay/algebra/solving-for-a-variable.haml
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@ -7,10 +7,7 @@
%link{:rel => 'stylesheet', :type => 'text/css', :href => './style.css'}
%body
%h2 Solve for $l$:
#solve-for-l-01.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-2lmn + 8m - 3n + 9 & = & -5m + 3n - 7 \\
-2lmn + 8m - 3n + 9 - 9 & = & -5m + 3n - 7 - 9 \\
-2lmn + 8m - 3n & = & -5m + 3n - 16 \\
@ -25,13 +22,8 @@
l & = & \frac{\frac{-5m - 16 + 6n - 8m}{-2}}{mn} \\
l & = & \frac{\frac{-16 + 6n - 13m}{-2}}{mn} \\
l & = & \frac{8 - 3n - \frac{13m}{-2}}{mn} \\
\end{eqnarray}
$$
#solve-for-l-02.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-2lmn + 8m - 3n + 9 & = & -5m + 3n - 7 \\
-2lmn + 8m - 3n + 9 - 9 & = & -5m + 3n - 7 - 9 \\
-2lmn + 8m - 3n & = & -5m + 3n + 2 \\
@ -46,13 +38,8 @@
lmn & = & \frac{-13m}{-2} - 3n - 1 \\
\frac{lmn}{mn} & = & \frac{\frac{-13m}{-2} - 3n - 1}{mn} \\
l & = & \frac{\frac{-13m}{-2} - 3n - 1}{mn} \\
\end{eqnarray}
$$
#solve-for-l-03.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-2lmn + 8m - 3n + 9 & = & -5m + 3n - 7 \\
-9 + -2lmn + 8m - 3n + 9 & = & -9 + -5m + 3n - 7 \\
-2lmn + 8m - 3n & = & -9 + -5m + 3n - 7 \\
@ -66,14 +53,9 @@
\frac{-2lmn}{-2mn} & = & \frac{-13m + 6n + -16}{-2mn} \\
l & = & \frac{-13m + 6n + -16}{-2mn} \\
l & = & \frac{13m - 6n + 16}{2mn} \\
\end{eqnarray}
$$
%h2 Solve for $x$:
#solve-for-x-01.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-2xy + 5xz - 5x + 10 & = & -6y - 2 \\
-2xy + 5xz - 5x + 10 - 10 & = & -6y - 2 - 10 \\
-2xy + 5xz - 5x & = & -6y + -2 + -10 \\
@ -85,13 +67,8 @@
-5x \over -5 & = & -6y - 12 - 5xz + 2xy \over -5x \\
x & = & -6y - 12 - 5xz + 2xy \over -5x \\
x & = & \frac{-6y}{-5x} - \frac{12}{-5x} - \frac{5xz}{-5x} + \frac{2xy}{-5x} \\
\end{eqnarray}
$$
#solve-for-x-02.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-2xy + 5xz - 5x + 10 & = & -6y - 2 \\
-2xy + 5xz - 5x + 10 - 10 & = & -6y - 2 - 10 \\
-2xy + 5xz - 5x & = & -6y + -2 + -10 \\
@ -108,13 +85,8 @@
-5x \over -5 & = & -6y - 12 \over -2y + 5z - 5 \\
x & = & -6y - 12 \over -2y + 5z - 5 \\
x & = & 6y + 12 \over 2y - 5z + 5 \\
\end{eqnarray}
$$
#solve-for-x-03.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-2xy + 5xz - 5x + 10 & = & -6y - 2 \\
-2xy + 5xz - 5x + 10 - 10 & = & -6y - 2 - 10 \\
-2xy + 5xz - 5x & = & -6y - 12 \\
@ -122,14 +94,9 @@
x \cdot (-2y + 5z - 5) \over -2y + 5z - 5 & = & -6y - 12 \over -2y + 5z - 5 \\
x & = & -6y - 12 \over -2y + 5z - 5 \\
x & = & 6y + 12 \over 2y - 5z + 5 \\
\end{eqnarray}
$$
%h2 Solve for $m$:
#solve-for-m-01.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
-8m - 4n - 8p - 5 & = & -5n - p + 6 \\
-8m - 4n - 8p - 5 + 5 & = & -5n - p + 6 + 5 \\
-8m - 4n - 8p & = & -5n - p + 11 \\
@ -143,33 +110,21 @@
m & = & -n + 11 + 7p \over -8 \\
m & = & n - 11 - 7p \over 8 \\
m & = & n - 7p - 11 \over 8 \\
\end{eqnarray}
$$
%h2 Solve for $x$ (again):
#solve-for-x2-01.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
7xy + 4xz + x + 1 & = & y + 6 \\
7xy + 4xz + x + 1 - 1 & = & y + 6 - 1 \\
7xy + 4xz + x & = & y + 5 \\
x \cdot (7y + 4z + 1) & = & y + 5 \\
x \cdot (7y + 4z + 1) \over 7y + 4z + 1 & = & y + 5 \over 7y + 4z + 1 \\
x & = & y + 5 \over 7y + 4z + 1 \\
\end{eqnarray}
$$
%h2 Solve for $v$:
#solve-for-v-01.solution
:escaped
$$
\begin{eqnarray}
:eqnarray
3vw - 2vx + 10v + 1 & = & w + 4 \\
3vw - 2vx + 10v + 1 - 1 & = & w + 4 - 1 \\
3vw - 2vx + 10v & = & w + 3 \\
v \cdot (3w - 2x + 10) & = & w + 3 \\
v \cdot (3w - 2x + 10) \over 3w - 2x + 10 & = & w + 3 \over 3w - 2x + 10 \\
v & = & w + 3 \over 3w - 2x + 10 \\
\end{eqnarray}
$$

2
math-replay/algebra/style.css
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@ -9,7 +9,7 @@ h2 {
margin-top: 40px;
}
.solution {
.eqnarray {
border-radius: 5px;
box-shadow: 2px 5px 5px #000;
padding: 10px;

22
math-replay/lib/haml/filters-ext.rb
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@ -0,0 +1,22 @@
require 'haml/filters'
require 'haml/helpers'
module Haml
module Filters
module Eqnarray
include Base
def render(text)
<<END
<div class="eqnarray">
$$
\\begin{eqnarray}
#{Haml::Helpers.html_escape(text).rstrip.gsub("\n", "\n ")}
\\end{eqnarray}
$$
</div>
END
end
end
end
end
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