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\mode<presentation>
\usecolortheme[rgb={0.8, 0.2, 0}]{structure}
\usefonttheme[onlysmall]{structurebold}
\setbeamertemplate{navigation symbols}{}
%\setbeamertemplate{footline}[frame number]
\usepackage{tikz}
\usetikzlibrary{arrows,shapes,backgrounds,decorations,mindmap}
\mode
<all>
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\mode<presentation>
\useoutertheme{smoothbars}
\useinnertheme[shadow=true]{rounded}
\usecolortheme{orchid}
\usecolortheme{whale}
\usecolortheme[rgb={0.7, 0.2, 0}]{structure} % (darker red)
\useoutertheme{shadow}
\usefonttheme[onlysmall]{structurebold}
\setbeamercolor{title}{use=structure,fg=white,bg=structure.fg}
\setbeamerfont{block title}{size={}}
\mode
<all>
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#!/bin/sh
doconce clean
rm -rf *.pdf *.tex ipynb*.tar.gz *.html ._*.html *~ reveal.js Trash README.txt
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#!/bin/sh
set -x
function system {
"$@"
if [ $? -ne 0 ]; then
echo "make.sh: unsuccessful command $@"
echo "abort!"
exit 1
fi
}
if [ $# -eq 0 ]; then
echo 'bash make.sh slides1|slides2'
exit 1
fi
name=$1
rm -f *.tar.gz
opt="--encoding=utf-8"
# Note: Makefile examples contain constructions like ${PROG} which
# looks like Mako constructions, but they are not. Use --no_mako
# to turn off Mako processing.
opt="--no_mako"
rm -f *.aux
html=${name}-reveal
system doconce format html $name --pygments_html_style=perldoc --keep_pygments_html_bg --html_links_in_new_window --html_output=$html $opt
system doconce slides_html $html reveal --html_slide_theme=beige
# Plain HTML documents
html=${name}-solarized
system doconce format html $name --pygments_html_style=perldoc --html_style=solarized3 --html_links_in_new_window --html_output=$html $opt
system doconce split_html $html.html --method=space10
html=${name}
system doconce format html $name --pygments_html_style=default --html_style=bloodish --html_links_in_new_window --html_output=$html $opt
system doconce split_html $html.html --method=space10
# Bootstrap style
html=${name}-bs
system doconce format html $name --html_style=bootstrap --pygments_html_style=default --html_admon=bootstrap_panel --html_output=$html $opt
#system doconce split_html $html.html --method=split --pagination --nav_button=bottom
# IPython notebook
system doconce format ipynb $name $opt
# LaTeX Beamer slides
beamertheme=red_plain
system doconce format pdflatex $name --latex_title_layout=beamer --latex_table_format=footnotesize $opt
system doconce ptex2tex $name envir=minted
# Add special packages
doconce subst "% Add user's preamble" "\g<1>\n\\usepackage{simplewick}" $name.tex
system doconce slides_beamer $name --beamer_slide_theme=$beamertheme
system pdflatex -shell-escape ${name}
system pdflatex -shell-escape ${name}
cp $name.pdf ${name}-beamer.pdf
cp $name.tex ${name}-beamer.tex
# Handouts
system doconce format pdflatex $name --latex_title_layout=beamer --latex_table_format=footnotesize $opt
system doconce ptex2tex $name envir=minted
# Add special packages
doconce subst "% Add user's preamble" "\g<1>\n\\usepackage{simplewick}" $name.tex
system doconce slides_beamer $name --beamer_slide_theme=red_shadow --handout
system pdflatex -shell-escape $name
pdflatex -shell-escape $name
pdflatex -shell-escape $name
pdfnup --nup 2x3 --frame true --delta "1cm 1cm" --scale 0.9 --outfile ${name}-beamer-handouts2x3.pdf ${name}.pdf
rm -f ${name}.pdf
# Ordinary plain LaTeX document
rm -f *.aux # important after beamer
system doconce format pdflatex $name --minted_latex_style=trac --latex_admon=paragraph $opt
system doconce ptex2tex $name envir=minted
# Add special packages
doconce subst "% Add user's preamble" "\g<1>\n\\usepackage{simplewick}" $name.tex
doconce replace 'section{' 'section*{' $name.tex
pdflatex -shell-escape $name
pdflatex -shell-escape $name
mv -f $name.pdf ${name}-minted.pdf
cp $name.tex ${name}-plain-minted.tex
# Publish
dest=../../pub
if [ ! -d $dest/$name ]; then
mkdir $dest/$name
mkdir $dest/$name/pdf
mkdir $dest/$name/html
mkdir $dest/$name/ipynb
fi
cp ${name}*.pdf $dest/$name/pdf
cp -r ${name}*.html ._${name}*.html reveal.js $dest/$name/html
# Figures: cannot just copy link, need to physically copy the files
if [ -d fig-${name} ]; then
if [ ! -d $dest/$name/html/fig-$name ]; then
mkdir $dest/$name/html/fig-$name
fi
cp -r fig-${name}/* $dest/$name/html/fig-$name
fi
cp ${name}.ipynb $dest/$name/ipynb
ipynb_tarfile=ipynb-${name}-src.tar.gz
if [ ! -f ${ipynb_tarfile} ]; then
cat > README.txt <<EOF
This IPython notebook ${name}.ipynb does not require any additional
programs.
EOF
tar czf ${ipynb_tarfile} README.txt
fi
cp ${ipynb_tarfile} $dest/$name/ipynb
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TITLE: Data Analysis and Machine Learning Lectures: Cubic Splines and Gradient Methods
AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
DATE: today
!split
===== Cubic Splines =====
!bblock
Cubic spline interpolation is among one of the most used
methods for interpolating between data points where the arguments
are organized as ascending series. In the library program we supply
such a function, based on the so-called cubic spline method to be
described below.
A spline function consists of polynomial pieces defined on
subintervals. The different subintervals are connected via
various continuity relations.
Assume we have at our disposal $n+1$ points $x_0, x_1, \dots x_n$
arranged so that $x_0 < x_1 < x_2 < \dots x_{n-1} < x_n$ (such points are called
knots). A spline function $s$ of degree $k$ with $n+1$ knots is defined
as follows
* On every subinterval $[x_{i-1},x_i)$ *s* is a polynomial of degree $\le k$.
* $s$ has $k-1$ continuous derivatives in the whole interval $[x_0,x_n]$.
!eblock
!split
===== Splines =====
!bblock
As an example, consider a spline function of degree $k=1$ defined as follows
!bt
\[
s(x)=\begin{bmatrix} s_0(x)=a_0x+b_0 & x\in [x_0, x_1) \\
s_1(x)=a_1x+b_1 & x\in [x_1, x_2) \\
\dots & \dots \\
s_{n-1}(x)=a_{n-1}x+b_{n-1} & x\in
[x_{n-1}, x_n] \end{bmatrix}.
\]
!et
In this case the polynomial consists of series of straight lines
connected to each other at every endpoint. The number of continuous
derivatives is then $k-1=0$, as expected when we deal with straight lines.
Such a polynomial is quite easy to construct given
$n+1$ points $x_0, x_1, \dots x_n$ and their corresponding
function values.
!eblock
!split
===== Splines =====
!bblock
The most commonly used spline function is the one with $k=3$, the so-called
cubic spline function.
Assume that we have in adddition to the $n+1$ knots a series of
functions values $y_0=f(x_0), y_1=f(x_1), \dots y_n=f(x_n)$.
By definition, the polynomials $s_{i-1}$ and $s_i$
are thence supposed to interpolate the same point $i$, that is
!bt
\[
s_{i-1}(x_i)= y_i = s_i(x_i),
\]
!et
with $1 \le i \le n-1$. In total we have $n$ polynomials of the
type
!bt
\[
s_i(x)=a_{i0}+a_{i1}x+a_{i2}x^2+a_{i2}x^3,
\]
!et
yielding $4n$ coefficients to determine.
!eblock
!split
===== Splines =====
!bblock
Every subinterval provides in addition the $2n$ conditions
!bt
\[
y_i = s(x_i),
\]
!et
and
!bt
\[
s(x_{i+1})= y_{i+1},
\]
!et
to be fulfilled. If we also assume that $s'$ and $s''$ are continuous,
then
!bt
\[
s'_{i-1}(x_i)= s'_i(x_i),
\]
!et
yields $n-1$ conditions. Similarly,
!bt
\[
s''_{i-1}(x_i)= s''_i(x_i),
\]
!et
results in additional $n-1$ conditions. In total we have $4n$ coefficients
and $4n-2$ equations to determine them, leaving us with $2$ degrees of
freedom to be determined.
!eblock
!split
===== Splines =====
!bblock
Using the last equation we define two values for the second derivative, namely
!bt
\[
s''_{i}(x_i)= f_i,
\]
!et
and
!bt
\[
s''_{i}(x_{i+1})= f_{i+1},
\]
!et
and setting up a straight line between $f_i$ and $f_{i+1}$ we have
!bt
\[
s_i''(x) = \frac{f_i}{x_{i+1}-x_i}(x_{i+1}-x)+
\frac{f_{i+1}}{x_{i+1}-x_i}(x-x_i),
\]
!et
and integrating twice one obtains
!bt
\[
s_i(x) = \frac{f_i}{6(x_{i+1}-x_i)}(x_{i+1}-x)^3+
\frac{f_{i+1}}{6(x_{i+1}-x_i)}(x-x_i)^3
+c(x-x_i)+d(x_{i+1}-x).
\]
!et
!eblock
!split
===== Splines =====
!bblock
Using the conditions $s_i(x_i)=y_i$ and $s_i(x_{i+1})=y_{i+1}$
we can in turn determine the constants $c$ and $d$ resulting in
!bt
\begin{align}
s_i(x) =&\frac{f_i}{6(x_{i+1}-x_i)}(x_{i+1}-x)^3+
\frac{f_{i+1}}{6(x_{i+1}-x_i)}(x-x_i)^3 \nonumber \\
+&(\frac{y_{i+1}}{x_{i+1}-x_i}-\frac{f_{i+1}(x_{i+1}-x_i)}{6})
(x-x_i)+
(\frac{y_{i}}{x_{i+1}-x_i}-\frac{f_{i}(x_{i+1}-x_i)}{6})
(x_{i+1}-x).
\end{align}
!et
!eblock
!split
===== Splines =====
!bblock
How to determine the values of the second
derivatives $f_{i}$ and $f_{i+1}$? We use the continuity assumption
of the first derivatives
!bt
\[
s'_{i-1}(x_i)= s'_i(x_i),
\]
!et
and set $x=x_i$. Defining $h_i=x_{i+1}-x_i$ we obtain finally
the following expression
!bt
\[
h_{i-1}f_{i-1}+2(h_{i}+h_{i-1})f_i+h_if_{i+1}=
\frac{6}{h_i}(y_{i+1}-y_i)-\frac{6}{h_{i-1}}(y_{i}-y_{i-1}),
\]
!et
and introducing the shorthands $u_i=2(h_{i}+h_{i-1})$,
$v_i=\frac{6}{h_i}(y_{i+1}-y_i)-\frac{6}{h_{i-1}}(y_{i}-y_{i-1})$,
we can reformulate the problem as a set of linear equations to be
solved through e.g., Gaussian elemination
!eblock
!split
===== Splines =====
!bblock
Gaussian elimination
!bt
\[
\begin{bmatrix} u_1 & h_1 &0 &\dots & & & & \\
h_1 & u_2 & h_2 &0 &\dots & & & \\
0 & h_2 & u_3 & h_3 &0 &\dots & & \\
\dots& & \dots &\dots &\dots &\dots &\dots & \\
&\dots & & &0 &h_{n-3} &u_{n-2} &h_{n-2} \\
& && & &0 &h_{n-2} &u_{n-1} \end{bmatrix}
\begin{bmatrix} f_1 \\
f_2 \\
f_3\\
\dots \\
f_{n-2} \\
f_{n-1} \end{bmatrix} =
\begin{bmatrix} v_1 \\
v_2 \\
v_3\\
\dots \\
v_{n-2}\\
v_{n-1} \end{bmatrix}.
\]
!et
Note that this is a set of tridiagonal equations and can be solved
through only $O(n)$ operations.
!eblock
!split
===== Splines =====
!bblock
The functions supplied in the program library are *spline* and *splint*.
In order to use cubic spline interpolation you need first to call
!bc cppcod
spline(double x[], double y[], int n, double yp1, double yp2, double y2[])
!ec
This function takes as
input $x[0,..,n - 1]$ and $y[0,..,n - 1]$ containing a tabulation
$y_i = f(x_i)$ with $x_0 < x_1 < .. < x_{n - 1}$
together with the
first derivatives of $f(x)$ at $x_0$ and $x_{n-1}$, respectively. Then the
function returns $y2[0,..,n-1]$ which contains the second derivatives of
$f(x_i)$ at each point $x_i$. $n$ is the number of points.
This function provides the cubic spline interpolation for all subintervals
and is called only once.
!eblock
!split
===== Splines =====
!bblock
Thereafter, if you wish to make various interpolations, you need to call the function
!bc cppcod
splint(double x[], double y[], double y2a[], int n, double x, double *y)
!ec
which takes as input
the tabulated values $x[0,..,n - 1]$ and $y[0,..,n - 1]$ and the output
y2a[0,..,n - 1] from *spline*. It returns the value $y$ corresponding
to the point $x$.
!eblock
!split
===== Conjugate gradient (CG) method =====
!bblock
The success of the CG method for finding solutions of non-linear problems is based
on the theory of conjugate gradients for linear systems of equations. It belongs
to the class of iterative methods for solving problems from linear algebra of the type
!bt
\begin{equation*}
\hat{A}\hat{x} = \hat{b}.
\end{equation*}
!et
In the iterative process we end up with a problem like
!bt
\begin{equation*}
\hat{r}= \hat{b}-\hat{A}\hat{x},
\end{equation*}
!et
where $\hat{r}$ is the so-called residual or error in the iterative process.
When we have found the exact solution, $\hat{r}=0$.
!eblock
!split
===== Conjugate gradient method =====
!bblock
The residual is zero when we reach the minimum of the quadratic equation
!bt
\begin{equation*}
P(\hat{x})=\frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T\hat{b},
\end{equation*}
!et
with the constraint that the matrix $\hat{A}$ is positive definite and symmetric.
If we search for a minimum of the quantum mechanical variance, then the matrix
$\hat{A}$, which is called the Hessian, is given by the second-derivative of the function we want to minimize. This quantity is always positive definite. In our case this corresponds normally to the second derivative of the energy.
!eblock
!split
===== Conjugate gradient method, Newton's method first =====
!bblock
We seek the minimum of the energy or the variance as function of various variational parameters.
In our case we have thus a function $f$ whose minimum we are seeking.
In Newton's method we set $\nabla f = 0$ and we can thus compute the next iteration point
!bt
\begin{equation*}
\hat{x}-\hat{x}_i=\hat{A}^{-1}\nabla f(\hat{x}_i).
\end{equation*}
!et
Subtracting this equation from that of $\hat{x}_{i+1}$ we have
!bt
\begin{equation*}
\hat{x}_{i+1}-\hat{x}_i=\hat{A}^{-1}(\nabla f(\hat{x}_{i+1})-\nabla f(\hat{x}_i)).
\end{equation*}
!et
!eblock
!split
===== Simple example and demonstration =====
!bblock
The function $f$ can be either the energy or the variance. If we choose the energy then we have
!bt
\begin{equation*}
\hat{\alpha}_{i+1}-\hat{\alpha}_i=\hat{A}^{-1}(\nabla E(\hat{\alpha}_{i+1})-\nabla E(\hat{\alpha}_i)).
\end{equation*}
!et
In the simple harmonic oscillator model, the gradient and the Hessian $\hat{A}$ are
!bt
\begin{equation*}
\frac{d\langle E_L[\alpha]\rangle}{d\alpha} = \alpha-\frac{1}{4\alpha^3}
\end{equation*}
!et
and a second derivative which is always positive (meaning that we find a minimum)
!bt
\begin{equation*}
\hat{A}= \frac{d^2\langle E_L[\alpha]\rangle}{d\alpha^2} = 1+\frac{3}{4\alpha^4}
\end{equation*}
!et
!eblock
!split
===== Simple example and demonstration =====
!bblock
We get then
!bt
\begin{equation*}
\alpha_{i+1}=\frac{4}{3}\alpha_i-\frac{\alpha_i^4}{3\alpha_{i+1}^3},
\end{equation*}
!et
which can be rewritten as
!bt
\begin{equation*}
\alpha_{i+1}^4-\frac{4}{3}\alpha_i\alpha_{i+1}^4+\frac{1}{3}\alpha_i^4.
\end{equation*}
!et
!eblock
!split
===== Conjugate gradient method =====
!bblock
In the CG method we define so-called conjugate directions and two vectors
$\hat{s}$ and $\hat{t}$
are said to be
conjugate if
!bt
\begin{equation*}
\hat{s}^T\hat{A}\hat{t}= 0.
\end{equation*}
!et
The philosophy of the CG method is to perform searches in various conjugate directions
of our vectors $\hat{x}_i$ obeying the above criterion, namely
!bt
\begin{equation*}
\hat{x}_i^T\hat{A}\hat{x}_j= 0.
\end{equation*}
!et
Two vectors are conjugate if they are orthogonal with respect to
this inner product. Being conjugate is a symmetric relation: if $\hat{s}$ is conjugate to $\hat{t}$, then $\hat{t}$ is conjugate to $\hat{s}$.
!eblock
!split
===== Conjugate gradient method =====
!bblock
An example is given by the eigenvectors of the matrix
!bt
\begin{equation*}
\hat{v}_i^T\hat{A}\hat{v}_j= \lambda\hat{v}_i^T\hat{v}_j,
\end{equation*}
!et
which is zero unless $i=j$.
!eblock
!split
===== Conjugate gradient method =====
!bblock
Assume now that we have a symmetric positive-definite matrix $\hat{A}$ of size
$n\times n$. At each iteration $i+1$ we obtain the conjugate direction of a vector
!bt
\begin{equation*}
\hat{x}_{i+1}=\hat{x}_{i}+\alpha_i\hat{p}_{i}.
\end{equation*}
!et
We assume that $\hat{p}_{i}$ is a sequence of $n$ mutually conjugate directions.
Then the $\hat{p}_{i}$ form a basis of $R^n$ and we can expand the solution
$ \hat{A}\hat{x} = \hat{b}$ in this basis, namely
!bt
\begin{equation*}
\hat{x} = \sum^{n}_{i=1} \alpha_i \hat{p}_i.
\end{equation*}
!et
!eblock
!split
===== Conjugate gradient method =====
!bblock
The coefficients are given by
!bt
\begin{equation*}
\mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}.
\end{equation*}
!et
Multiplying with $\hat{p}_k^T$ from the left gives
!bt
\begin{equation*}
\hat{p}_k^T \hat{A}\hat{x} = \sum^{n}_{i=1} \alpha_i\hat{p}_k^T \hat{A}\hat{p}_i= \hat{p}_k^T \hat{b},
\end{equation*}
!et
and we can define the coefficients $\alpha_k$ as
!bt
\begin{equation*}
\alpha_k = \frac{\hat{p}_k^T \hat{b}}{\hat{p}_k^T \hat{A} \hat{p}_k}
\end{equation*}
!et
!eblock
!split
===== Conjugate gradient method and iterations =====
!bblock
If we choose the conjugate vectors $\hat{p}_k$ carefully,
then we may not need all of them to obtain a good approximation to the solution
$\hat{x}$.
We want to regard the conjugate gradient method as an iterative method.
This will us to solve systems where $n$ is so large that the direct
method would take too much time.
We denote the initial guess for $\hat{x}$ as $\hat{x}_0$.
We can assume without loss of generality that
!bt
\begin{equation*}
\hat{x}_0=0,
\end{equation*}
!et
or consider the system
!bt
\begin{equation*}
\hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0,
\end{equation*}
!et
instead.
!eblock
!split
===== Conjugate gradient method =====
!bblock
One can show that the solution $\hat{x}$ is also the unique minimizer of the quadratic form
!bt
\begin{equation*}
f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n.
\end{equation*}
!et
This suggests taking the first basis vector $\hat{p}_1$
to be the gradient of $f$ at $\hat{x}=\hat{x}_0$,
which equals
!bt
\begin{equation*}
\hat{A}\hat{x}_0-\hat{b},
\end{equation*}
!et
and
$\hat{x}_0=0$ it is equal $-\hat{b}$.
The other vectors in the basis will be conjugate to the gradient,
hence the name conjugate gradient method.
!eblock
!split
===== Conjugate gradient method =====
!bblock
Let $\hat{r}_k$ be the residual at the $k$-th step:
!bt
\begin{equation*}
\hat{r}_k=\hat{b}-\hat{A}\hat{x}_k.
\end{equation*}
!et
Note that $\hat{r}_k$ is the negative gradient of $f$ at
$\hat{x}=\hat{x}_k$,
so the gradient descent method would be to move in the direction $\hat{r}_k$.
Here, we insist that the directions $\hat{p}_k$ are conjugate to each other,
so we take the direction closest to the gradient $\hat{r}_k$
under the conjugacy constraint.
This gives the following expression
!bt
\begin{equation*}
\hat{p}_{k+1}=\hat{r}_k-\frac{\hat{p}_k^T \hat{A}\hat{r}_k}{\hat{p}_k^T\hat{A}\hat{p}_k} \hat{p}_k.
\end{equation*}
!et
!eblock
!split
===== Conjugate gradient method =====
!bblock
We can also compute the residual iteratively as
!bt
\begin{equation*}
\hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1},
\end{equation*}
!et
which equals
!bt
\begin{equation*}
\hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{p}_k),
\end{equation*}
!et
or
!bt
\begin{equation*}
(\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{p}_k,
\end{equation*}
!et
which gives
!bt
\begin{equation*}
\hat{r}_{k+1}=\hat{r}_k-\hat{A}\hat{p}_{k},
\end{equation*}
!et
!eblock
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\mode<presentation>
\usecolortheme[rgb={0.8, 0.2, 0}]{structure}
\usefonttheme[onlysmall]{structurebold}
\setbeamertemplate{navigation symbols}{}
%\setbeamertemplate{footline}[frame number]
\usepackage{tikz}
\usetikzlibrary{arrows,shapes,backgrounds,decorations,mindmap}
\mode
<all>
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\mode<presentation>
\useoutertheme{smoothbars}
\useinnertheme[shadow=true]{rounded}
\usecolortheme{orchid}
\usecolortheme{whale}
\usecolortheme[rgb={0.7, 0.2, 0}]{structure} % (darker red)
\useoutertheme{shadow}
\usefonttheme[onlysmall]{structurebold}
\setbeamercolor{title}{use=structure,fg=white,bg=structure.fg}
\setbeamerfont{block title}{size={}}
\mode
<all>
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#!/bin/sh
doconce clean
rm -rf *.pdf *.tex ipynb*.tar.gz *.html ._*.html *~ reveal.js Trash README.txt
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#!/bin/sh
set -x
function system {
"$@"
if [ $? -ne 0 ]; then
echo "make.sh: unsuccessful command $@"
echo "abort!"
exit 1
fi
}
if [ $# -eq 0 ]; then
echo 'bash make.sh slides1|slides2'
exit 1
fi
name=$1
rm -f *.tar.gz
opt="--encoding=utf-8"
# Note: Makefile examples contain constructions like ${PROG} which
# looks like Mako constructions, but they are not. Use --no_mako
# to turn off Mako processing.
opt="--no_mako"
rm -f *.aux
html=${name}-reveal
system doconce format html $name --pygments_html_style=perldoc --keep_pygments_html_bg --html_links_in_new_window --html_output=$html $opt
system doconce slides_html $html reveal --html_slide_theme=beige
# Plain HTML documents
html=${name}-solarized
system doconce format html $name --pygments_html_style=perldoc --html_style=solarized3 --html_links_in_new_window --html_output=$html $opt
system doconce split_html $html.html --method=space10
html=${name}
system doconce format html $name --pygments_html_style=default --html_style=bloodish --html_links_in_new_window --html_output=$html $opt
system doconce split_html $html.html --method=space10
# Bootstrap style
html=${name}-bs
system doconce format html $name --html_style=bootstrap --pygments_html_style=default --html_admon=bootstrap_panel --html_output=$html $opt
#system doconce split_html $html.html --method=split --pagination --nav_button=bottom
# IPython notebook
system doconce format ipynb $name $opt
# LaTeX Beamer slides
beamertheme=red_plain
system doconce format pdflatex $name --latex_title_layout=beamer --latex_table_format=footnotesize $opt
system doconce ptex2tex $name envir=minted
# Add special packages
doconce subst "% Add user's preamble" "\g<1>\n\\usepackage{simplewick}" $name.tex
system doconce slides_beamer $name --beamer_slide_theme=$beamertheme
system pdflatex -shell-escape ${name}
system pdflatex -shell-escape ${name}
cp $name.pdf ${name}-beamer.pdf
cp $name.tex ${name}-beamer.tex
# Handouts
system doconce format pdflatex $name --latex_title_layout=beamer --latex_table_format=footnotesize $opt
system doconce ptex2tex $name envir=minted
# Add special packages
doconce subst "% Add user's preamble" "\g<1>\n\\usepackage{simplewick}" $name.tex
system doconce slides_beamer $name --beamer_slide_theme=red_shadow --handout
system pdflatex -shell-escape $name
pdflatex -shell-escape $name
pdflatex -shell-escape $name
pdfnup --nup 2x3 --frame true --delta "1cm 1cm" --scale 0.9 --outfile ${name}-beamer-handouts2x3.pdf ${name}.pdf
rm -f ${name}.pdf
# Ordinary plain LaTeX document
rm -f *.aux # important after beamer
system doconce format pdflatex $name --minted_latex_style=trac --latex_admon=paragraph $opt
system doconce ptex2tex $name envir=minted
# Add special packages
doconce subst "% Add user's preamble" "\g<1>\n\\usepackage{simplewick}" $name.tex
doconce replace 'section{' 'section*{' $name.tex
pdflatex -shell-escape $name
pdflatex -shell-escape $name
mv -f $name.pdf ${name}-minted.pdf
cp $name.tex ${name}-plain-minted.tex
# Publish
dest=../../pub
if [ ! -d $dest/$name ]; then
mkdir $dest/$name
mkdir $dest/$name/pdf
mkdir $dest/$name/html
mkdir $dest/$name/ipynb
fi
cp ${name}*.pdf $dest/$name/pdf
cp -r ${name}*.html ._${name}*.html reveal.js $dest/$name/html
# Figures: cannot just copy link, need to physically copy the files
if [ -d fig-${name} ]; then
if [ ! -d $dest/$name/html/fig-$name ]; then
mkdir $dest/$name/html/fig-$name
fi
cp -r fig-${name}/* $dest/$name/html/fig-$name
fi
cp ${name}.ipynb $dest/$name/ipynb
ipynb_tarfile=ipynb-${name}-src.tar.gz
if [ ! -f ${ipynb_tarfile} ]; then
cat > README.txt <<EOF
This IPython notebook ${name}.ipynb does not require any additional
programs.
EOF
tar czf ${ipynb_tarfile} README.txt
fi
cp ${ipynb_tarfile} $dest/$name/ipynb