Package: tomboy-latex Version: 0.5-1 Severity: normal Hi!
With larger notes tomboy (with LaTeX-plugin enabled) becomes really slow. I can often type whole sentences without any feedback from tomboy (no cursor movement, no charachters appearing, and tomboys windows don't redraw if I drag another window over it). When I'm finished typing and wait for about 30-60 seconds, the typed stuff will appear all at once. My impression is that there is some O(n^2) behaviour there, where n is either the total size of the note or the number of equations in it. A sample note is attached. Thanks, Jö. -- System Information: Debian Release: squeeze/sid APT prefers testing-proposed-updates APT policy: (500, 'testing-proposed-updates'), (500, 'testing') Architecture: i386 (i686) Kernel: Linux 2.6.28-1-686-bigmem (SMP w/2 CPU cores) Locale: LANG=de_DE.UTF-8, LC_CTYPE=de_DE.UTF-8 (charmap=UTF-8) Shell: /bin/sh linked to /bin/bash Versions of packages tomboy-latex depends on: ii imagemagick 7:6.4.5.4.dfsg1-1 image manipulation programs ii libgtk2.0-cil 2.12.8-2 CLI binding for the GTK+ toolkit 2 ii libmono-corlib2.0-cil 2.0.1-6 Mono core library (for CLI 2.0) ii libmono-system2.0-cil 2.0.1-6 Mono System libraries (for CLI 2.0 ii texlive-base-bin 2007.dfsg.2-6 TeX Live: Essential binaries ii texlive-latex-base 2007.dfsg.2-4 TeX Live: Basic LaTeX packages ii tomboy 0.14.2-1 desktop note taking program using tomboy-latex recommends no packages. tomboy-latex suggests no packages. -- no debconf information -- In the beginning the Universe was created. This has made a lot of people very angry and been widely regarded as a bad move. -- Douglas Adams
<?xml version="1.0" encoding="utf-8"?> <note version="0.3" xmlns:link="http://beatniksoftware.com/tomboy/link"; xmlns:size="http://beatniksoftware.com/tomboy/size"; xmlns="http://beatniksoftware.com/tomboy";> <title>Edge Basis and Transformation</title> <text xml:space="preserve"><note-content version="0.1">Edge Basis and Transformation <size:huge>The Basis</size:huge> For the edge basis we have the following properties, which fully determine them: <list><list-item dir="ltr">Each shape function \[\psi_i\] is linear. </list-item><list-item dir="ltr">Each shape function maps from \[\mathbb R^n\] to \[\mathbb R^n\], where \[n\] is the dimension. </list-item><list-item dir="ltr">For each shape function holds \[\psi_i(\mathbf x)\cdot\mathbf t^j=\delta_{ij}\quad\forall\mathbf x\in\Gamma^j\], where \[\Gamma^j\] denotes edge \[j\] and \[\mathbf t^j\] denotes a tangential unit vector to that edge.</list-item></list> We use the following numbering of edges and vertices: <monospace>| edge # | vertex 0 # | vertex 1 # | | 0 | 0 | 1 | | 1 | 0 | 2 | | 2 | 1 | 2 |</monospace> From that follow the following definitions for the tangential unit vectors \[\mathbf t^i\] and edge lengthes \[\ell^i\], where \[x^i\] are the coordinates of vertex \[i\]: <list><list-item dir="ltr">\[\mathbf t^0=(\mathbf x^1-\mathbf x^0)/\ell^0\], \[\ell^0=||\mathbf x^1-\mathbf x^0||\] </list-item><list-item dir="ltr">\[\mathbf t^1=(\mathbf x^2-\mathbf x^0)/\ell^1\], \[\ell^1=||\mathbf x^2-\mathbf x^0||\] </list-item><list-item dir="ltr">\[\mathbf t^2=(\mathbf x^2-\mathbf x^1)/\ell^2\], \[\ell^2=||\mathbf x^2-\mathbf x^1||\]</list-item></list> Since the shape functions are linear, we write them in the following way: <list><list-item dir="ltr">\[\psi^i(\mathbf x)=\mathrm A^i\mathbf x+\mathbf a^i\]</list-item></list> Evaluating the shape functions at the three vertices and taking the scalar product with the two adjacent tangential vectors \[\psi^i(\mathbf x^k)\cdot\mathbf t^j=\delta_{ij}\quad\forall k\in\{j_0,j_1\}\] leaves us with the following six conditions: <list><list-item dir="ltr">\[\psi^i(\mathbf x^0)\cdot\mathbf t^0=\delta_{i0}\quad\Longrightarrow\quad(\mathrm A^i\mathbf x^0+\mathbf a^i)\cdot(\mathbf x^1-\mathbf x^0)=\delta_{i0}\ell^0\] </list-item><list-item dir="ltr">\[\psi^i(\mathbf x^1)\cdot\mathbf t^0=\delta_{i0}\quad\Longrightarrow\quad(\mathrm A^i\mathbf x^1+\mathbf a^i)\cdot(\mathbf x^1-\mathbf x^0)=\delta_{i0}\ell^0\] </list-item><list-item dir="ltr">\[\psi^i(\mathbf x^1)\cdot\mathbf t^2=\delta_{i2}\quad\Longrightarrow\quad(\mathrm A^i\mathbf x^1+\mathbf a^i)\cdot(\mathbf x^2-\mathbf x^1)=\delta_{i2}\ell^2\] </list-item><list-item dir="ltr">\[\psi^i(\mathbf x^2)\cdot\mathbf t^2=\delta_{i2}\quad\Longrightarrow\quad(\mathrm A^i\mathbf x^2+\mathbf a^i)\cdot(\mathbf x^2-\mathbf x^1)=\delta_{i2}\ell^2\] </list-item><list-item dir="ltr">\[\psi^i(\mathbf x^2)\cdot\mathbf t^1=\delta_{i1}\quad\Longrightarrow\quad(\mathrm A^i\mathbf x^2+\mathbf a^i)\cdot(\mathbf x^2-\mathbf x^0)=\delta_{i1}\ell^1\] </list-item><list-item dir="ltr">\[\psi^i(\mathbf x^0)\cdot\mathbf t^1=\delta_{i1}\quad\Longrightarrow\quad(\mathrm A^i\mathbf x^0+\mathbf a^i)\cdot(\mathbf x^2-\mathbf x^0)=\delta_{i1}\ell1\]</list-item></list> By pairwise combination we can transform these conditions into <list><list-item dir="ltr">\[[\mathrm A^i(\mathbf x^1-\mathbf x^0)]\cdot(\mathbf x^1-\mathbf x^0)=0\quad\Longleftrightarrow\quad(\mathrm A^i\mathbf t^0)\cdot\mathbf t^0=0\] </list-item><list-item dir="ltr">\[[\mathrm A^i(\mathbf x^2-\mathbf x^0)]\cdot(\mathbf x^2-\mathbf x^0)=0\quad\Longleftrightarrow\quad(\mathrm A^i\mathbf t^1)\cdot\mathbf t^1=0\] </list-item><list-item dir="ltr">\[[\mathrm A^i(\mathbf x^2-\mathbf x^1)]\cdot(\mathbf x^2-\mathbf x^1)=0\quad\Longleftrightarrow\quad(\mathrm A^i\mathbf t^2)\cdot\mathbf t^2=0\]</list-item></list> and <list><list-item dir="ltr">\[(\mathrm A^i\mathbf x^0+\mathbf a^i)\cdot(\mathbf x^2-\mathbf x^1)=\delta_{i1}\ell^1-\delta_{i0}\ell^0\quad\Longleftrightarrow\quad(\mathrm A^i\mathbf x^0+\mathbf a^i)\cdot\mathbf t^2=(\delta_{i1}\ell^1-\delta_{i0}\ell^0)/\ell^2\] </list-item><list-item dir="ltr">\[(\mathrm A^i\mathbf x^1+\mathbf a^i)\cdot(\mathbf x^2-\mathbf x^0)=\delta_{i2}\ell^2+\delta_{i0}\ell^0\quad\Longleftrightarrow\quad(\mathrm A^i\mathbf x^1+\mathbf a^i)\cdot\mathbf t^1=(\delta_{i2}\ell^2+\delta_{i0}\ell^0)/\ell^1\] </list-item><list-item dir="ltr">\[(\mathrm A^i\mathbf x^2+\mathbf a^i)\cdot(\mathbf x^1-\mathbf x^0)=\delta_{i1}\ell^1-\delta_{i2}\ell^2\quad\Longleftrightarrow\quad(\mathrm A^i\mathbf x^2+\mathbf a^i)\cdot\mathbf t^0=(\delta_{i1}\ell^1-\delta_{i2}\ell^2)/\ell^0\]</list-item></list> Mayhaps these will be useful later. <size:huge>Coordinate Transformation </size:huge> For transformation we use a mapping \[\mathbf x=g(\mathbf{\hat x})\]. We say that we map from <italic>reference coordinates</italic> \[\mathbf{\hat x}\] into <italic>world coordinates</italic> \[\mathbf x\]. Entities with the hat (\[\hat{}\]) are in reference coordinates, entities without the hat are in world coordines. The mapping \[g\] is affine linear and can thus be written as <list><list-item dir="ltr">\[g(\mathbf{\hat x})=\mathrm B\mathbf{\hat x}+\mathbf b\]</list-item></list> The conversion of the vertex coordinates is simple: <list><list-item dir="ltr">\[\mathbf x^i=\mathrm B\mathbf{\hat x}^i+\mathbf b\]</list-item></list> The tangential vectors and edge lengthes are a bit more complicated: <list><list-item dir="ltr">\[\mathbf t^0=\frac{\mathbf x^1-\mathbf x^0}{||\mathbf x^1-\mathbf x^0||}=\frac{\mathrm B(\mathbf{\hat x}^1-\mathbf{\hat x}^0)}{||\mathrm B(\mathbf{\hat x}^1-\mathbf{\hat x}^0)||}=\frac{\mathrm B\mathbf{\hat t}^0}{||\mathrm B\mathbf{\hat t}^0||}\], \[\ell^0=||\mathbf x^1-\mathbf x^0||=||B(\mathbf{\hat x}^1-\mathbf{\hat x}^0)||\] </list-item><list-item dir="ltr">\[\mathbf t^1=\frac{\mathbf x^2-\mathbf x^0}{||\mathbf x^2-\mathbf x^0||}=\frac{\mathrm B(\mathbf{\hat x}^2-\mathbf{\hat x}^0)}{||\mathrm B(\mathbf{\hat x}^2-\mathbf{\hat x}^0)||}=\frac{\mathrm B\mathbf{\hat t}^1}{||\mathrm B\mathbf{\hat t}^1||}\], \[\ell^1=||\mathbf x^2-\mathbf x^0||=||B(\mathbf{\hat x}^2-\mathbf{\hat x}^0)||\] </list-item><list-item dir="ltr">\[\mathbf t^2=\frac{\mathbf x^2-\mathbf x^1}{||\mathbf x^2-\mathbf x^1||}=\frac{\mathrm B(\mathbf{\hat x}^2-\mathbf{\hat x}^1)}{||\mathrm B(\mathbf{\hat x}^2-\mathbf{\hat x}^1)||}=\frac{\mathrm B\mathbf{\hat t}^2}{||\mathrm B\mathbf{\hat t}^2||}\],\[\ell^2=||\mathbf x^2-\mathbf x^1||=||B(\mathbf{\hat x}^2-\mathbf{\hat x}^1)||\]</list-item></list> If we require that \[\hat\psi^i\] fullfills the same conditions in reference coordinates as \[\psi^i\] fullfills in world coordinates, how what is the transformation \[\psi^i(g(\mathbf{\hat x}))=L(\hat\psi^i(\mathbf{\hat x}))\]? Let us assume that \[L\] is linear and independent of \[\mathbf x\] and \[\mathbf{\hat x}\]: <list><list-item dir="ltr">\[\mathbf y=L(\mathbf{\hat y})=\mathrm C\mathbf{\hat y}+\mathbf c\]</list-item></list> How do \[\mathrm C\] and \[\mathbf c\] relate to \[\mathrm B\] and \[\mathbf b\]? We have the three vertex coordinates times two directions in each coordinate to evaluate \[\psi^0\] and \[\hat\psi^0\], that should give us six conditions with which we should be able to fix the six coefficients of \[\mathrm C\] and \[\mathbf c\] (ideally, \[\mathbf c=0\]). OK, evaluating just one basis function is probably not enough, since I already know what to expect there. Instead, I probably have to evaluate a function \[\mathbf y(\mathbf x)\] (\[\mathbf{\hat y}(\mathbf{\hat x})\]) to learn whats going on. <list><list-item dir="ltr">Let \[\mathbf{\hat y}=\sum_iy^i\hat\psi^i\] and \[\mathbf y=\sum_iy^i\psi^i\] with the same coeffitients \[y^i\].</list-item></list> Now we require \[\mathbf y(\mathbf x^i)=\mathrm C\mathbf{\hat y}(\mathbf{\hat x}^i)+\mathbf c\]. We will not test this with cartesian components </note-content></text> <last-change-date>2009-06-13T13:45:31.8804240+02:00</last-change-date> <last-metadata-change-date>2009-06-13T13:45:46.5230680+02:00</last-metadata-change-date> <create-date>2009-06-13T07:02:50.7525010+02:00</create-date> <cursor-position>31</cursor-position> <width>582</width> <height>489</height> <x>425</x> <y>89</y> <tags> <tag>system:notebook:Edge Basis and Transformation</tag> </tags> <open-on-startup>False</open-on-startup> </note>
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