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Dominic Zimmer 2023-06-06 14:19:39 +02:00
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<head><meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Untitled</title><script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script>
<title>The Art of Memory Loss</title><script src="https://cdnjs.cloudflare.com/ajax/libs/require.js/2.1.10/require.min.js"></script>
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<p>Wir können alle Übergangswahrscheinlichkeiten $P(s \to s')$ elegant als Matrix $M$ notieren:</p>
<p>Wir können alle Übergangswahrscheinlichkeiten $P(s \to s')$ als Matrix $M$ notieren:</p>
$$
M =
\begin{pmatrix} m_{LL} &amp; m_{LR} \\ m_{RL} &amp; m_{RR}
@ -15564,6 +15564,259 @@ M =
\begin{pmatrix} 1 - p &amp; p \\ q &amp; 1-q
\end{pmatrix}
$$
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<p>Wir schreiben die Wahrscheinlichkeiten, dass der Frosch zum Zeitpunkt $t$ in $L$ oder $R$ ist, als Vektoren.</p>
$$
X_0 = \begin{pmatrix}1 \\ 0\end{pmatrix}
$$
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<p>Ein Zeitschritt $t=0 \to t=1$ verändert die Verteilung gemäß $M$:</p>
$$
X_0 = \begin{pmatrix}1 \\ 0\end{pmatrix} \to \begin{pmatrix}1-p \\ p \end{pmatrix} = X_1
$$
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<h1 id="Matrixmultiplikation">Matrixmultiplikation<a class="anchor-link" href="#Matrixmultiplikation">&#182;</a></h1>
</div>
</div>
</div>
</div>
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<p>Den Folgezustand von $X = \begin{pmatrix}L &amp; R\end{pmatrix}$ mit Übergangsmatrix $M$ berechnen wir als</p>
$$
\begin{align*}
\begin{pmatrix}L' \\ R'\end{pmatrix}^T &amp;= \begin{pmatrix}L &amp; R\end{pmatrix} \cdot M \\
&amp;= \begin{pmatrix}L &amp; R\end{pmatrix} \cdot \begin{pmatrix} \color{green}{1 - p} &amp; \color{orange}{p} \\ \color{green}{q} &amp; \color{orange}{1-q} \end{pmatrix} \\
=&amp; \begin{pmatrix}L \cdot \color{green}{(1 - p)} + R \cdot \color{green}{q} \\
L \cdot \color{orange}{p} + R \cdot \color{orange}{(1-q)}
\end{pmatrix}^T
\end{align*}
$$
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<blockquote>
<p>Man berechnet die neuen Einträge als &quot;Zeile mal Spalte&quot;</p>
</blockquote>
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</div>
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<div class="jp-InputPrompt jp-InputArea-prompt">In&nbsp;[27]:</div>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">symbols</span> <span class="k">as</span> <span class="n">S</span><span class="p">,</span> <span class="n">Matrix</span> <span class="k">as</span> <span class="n">Mat</span>
<span class="n">p</span><span class="p">,</span> <span class="n">q</span> <span class="o">=</span> <span class="n">S</span><span class="p">(</span><span class="s2">&quot;p q&quot;</span><span class="p">)</span>
<span class="n">M</span> <span class="o">=</span> <span class="n">Mat</span><span class="p">([[</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">,</span> <span class="n">p</span><span class="p">],</span> <span class="p">[</span><span class="n">q</span><span class="p">,</span> <span class="mi">1</span><span class="o">-</span><span class="n">q</span><span class="p">]])</span>
</pre></div>
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<h2 id="Beispiele">Beispiele<a class="anchor-link" href="#Beispiele">&#182;</a></h2>$$
\begin{pmatrix}1 &amp; 0\end{pmatrix}\cdot M = \begin{pmatrix}1-p &amp; p\end{pmatrix}
$$
</div>
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$$
\begin{align*}
\begin{pmatrix}1 \\ 1\end{pmatrix}^T\cdot M &amp;= \begin{pmatrix} (1-p) + q \\ p + (1-q)\end{pmatrix}^T \\
&amp; = \begin{pmatrix} (1-p) + q \\ p +(1-q)\end{pmatrix}^T \\
\end{align*}
$$
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<div class="jp-InputPrompt jp-InputArea-prompt">In&nbsp;[29]:</div>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">v1</span> <span class="o">=</span> <span class="n">Mat</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">]])</span>
<span class="n">v1</span> <span class="o">*</span> <span class="n">M</span>
</pre></div>
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$\displaystyle \left[\begin{matrix}1 - p &amp; p\end{matrix}\right]$
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">v2</span> <span class="o">=</span> <span class="n">Mat</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">]])</span>
<span class="n">v2</span> <span class="o">*</span> <span class="n">M</span>
</pre></div>
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$\displaystyle \left[\begin{matrix}- p + q + 1 &amp; p - q + 1\end{matrix}\right]$
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<h1 id="Fr%C3%B6sche-brauchen-Zufall">Fr&#246;sche brauchen Zufall<a class="anchor-link" href="#Fr%C3%B6sche-brauchen-Zufall">&#182;</a></h1><div class="alert alert-info">
<p>Viele Leute werfen Münzen in den Teich der Frösche.
Wir sollen einen Frosch aussuchen, der der neue Schatzmeister werden soll.
Wie können wir aus den 7 Fröschen <strong>fair</strong> einen auswählen?</p>
</div>
<p><strong>Bedingungen</strong>:</p>
<ul>
<li>Jeder Frosch wird mit $\frac{1}{7}$ ausgewählt</li>
<li>Zufall nur durch die Münzen</li>
</ul>
</div>
</div>
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<div class=" highlight hl-ipython3"><pre><span></span>
</pre></div>
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<ul>
<li>die Berechnung von ganzfrüh aus der Tabelle nenen wir Matrixmultiplikation.</li>
<li>MM captures the idea that &quot;</li>
<li>denn, wir können Lineare Algebra machen. $v_0^T\cdot{}M = v_1$</li>
</ul>
</div>
</div>
</div>
@ -15572,18 +15825,15 @@ $$
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<div class="jp-InputPrompt jp-InputArea-prompt">In&nbsp;[4]:</div>
<div class="jp-InputPrompt jp-InputArea-prompt">In&nbsp;[26]:</div>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="c1">## Frösche brauchen Zufall</span>
<span class="p">:::</span><span class="n">info</span>
<span class="n">Viele</span> <span class="n">Leute</span> <span class="n">werfen</span> <span class="n">Münzen</span> <span class="ow">in</span> <span class="n">den</span> <span class="n">Teich</span> <span class="n">der</span> <span class="n">Frösche</span><span class="o">.</span> <span class="n">Wir</span> <span class="n">sollen</span> <span class="n">einen</span> <span class="n">Frosch</span> <span class="n">aussuchen</span><span class="p">,</span> <span class="n">der</span> <span class="n">der</span> <span class="n">neue</span> <span class="n">Schatzmeister</span> <span class="n">werden</span> <span class="n">soll</span><span class="o">.</span> <span class="n">Wie</span> <span class="n">können</span> <span class="n">wir</span> <span class="n">aus</span> <span class="n">den</span> <span class="mi">7</span> <span class="n">Fröschen</span> <span class="o">**</span><span class="n">fair</span><span class="o">**</span> <span class="n">einen</span> auswählen<span class="o">?</span>
<span class="p">:::</span>
<span class="o">**</span><span class="n">Bedingungen</span><span class="o">**</span><span class="p">:</span>
<span class="o">-</span> <span class="n">Jeder</span> <span class="n">Frosch</span> <span class="n">wird</span> <span class="n">mit</span> <span class="err">$</span>\<span class="n">frac</span><span class="p">{</span><span class="mi">1</span><span class="p">}{</span><span class="mi">7</span><span class="p">}</span><span class="err">$</span> <span class="n">ausgewählt</span>
<span class="o">-</span> <span class="n">Zufall</span> <span class="n">nur</span> <span class="n">durch</span> <span class="n">die</span> <span class="n">Münzen</span>
<span class="o">---</span>
@ -15668,7 +15918,7 @@ $$
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<span class="ansi-cyan-fg"> Cell </span><span class="ansi-green-fg">In[4], line 3</span>
<span class="ansi-cyan-fg"> Cell </span><span class="ansi-green-fg">In[26], line 3</span>
<span class="ansi-red-fg"> :::info</span>
^
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@ -130,13 +130,7 @@
}
},
"source": [
"## Frösche Simulieren\n",
"\n",
"<div class=\"alert alert-success\">\n",
" \n",
"Wie oft ist der Frosch im Schnitt in $l$, wie oft in $r$?\n",
"\n",
"</div>"
"## Frösche Simulieren"
]
},
{
@ -162,25 +156,13 @@
"# Los geht's!"
]
},
{
"cell_type": "markdown",
"id": "68dd379a",
"metadata": {
"slideshow": {
"slide_type": "subslide"
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"source": [
"## Frösche Simulieren"
]
},
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"cell_type": "code",
"execution_count": 2,
"execution_count": 5,
"id": "41017b8e",
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@ -189,8 +171,8 @@
"output_type": "stream",
"text": [
"100000/100000\n",
" 0.2473 of the time at 0 (24735 total)\n",
" 0.7527 of the time at 1 (75265 total)\n"
" 0.2495 of the time at 0 (24945 total)\n",
" 0.7506 of the time at 1 (75055 total)\n"
]
}
],
@ -238,7 +220,10 @@
"metadata": {
"slideshow": {
"slide_type": "subslide"
}
},
"tags": [
"hide_cell"
]
},
"source": [
"## Frösche Analysieren\n",
@ -258,7 +243,10 @@
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"tags": [
"hide_cell"
]
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"source": [
"<div class=\"alert alert-success\">\n",
@ -322,7 +310,10 @@
"metadata": {
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"<div class=\"alert alert-info\">\n",
@ -345,7 +336,10 @@
"metadata": {
"slideshow": {
"slide_type": "subslide"
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"tags": [
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"source": [
"## Frösche Analysieren\n",
@ -534,7 +528,7 @@
},
{
"cell_type": "markdown",
"id": "aa49d253",
"id": "12dde881",
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@ -556,7 +550,7 @@
},
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@ -571,15 +565,16 @@
]
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"source": [
"Wir können alle Übergangswahrscheinlichkeiten $P(s \\to s')$ elegant als Matrix $M$ notieren:\n",
"Wir können alle Übergangswahrscheinlichkeiten $P(s \\to s')$ als Matrix $M$ notieren:\n",
"\n",
"$$\n",
"M = \n",
@ -590,9 +585,245 @@
"$$"
]
},
{
"cell_type": "markdown",
"id": "4287f203",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"source": [
"Wir schreiben die Wahrscheinlichkeiten, dass der Frosch zum Zeitpunkt $t$ in $L$ oder $R$ ist, als Vektoren.\n",
"\n",
"$$\n",
"X_0 = \\begin{pmatrix}1 \\\\ 0\\end{pmatrix}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "25a506f5",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"source": [
"Ein Zeitschritt $t=0 \\to t=1$ verändert die Verteilung gemäß $M$:\n",
"\n",
"$$\n",
"X_0 = \\begin{pmatrix}1 \\\\ 0\\end{pmatrix} \\to \\begin{pmatrix}1-p \\\\ p \\end{pmatrix} = X_1\n",
"$$"
]
},
{
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"id": "0d56b780",
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"source": [
"# Matrixmultiplikation"
]
},
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"source": [
"Den Folgezustand von $X = \\begin{pmatrix}L & R\\end{pmatrix}$ mit Übergangsmatrix $M$ berechnen wir als\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"\\begin{pmatrix}L' \\\\ R'\\end{pmatrix}^T &= \\begin{pmatrix}L & R\\end{pmatrix} \\cdot M \\\\\n",
"&= \\begin{pmatrix}L & R\\end{pmatrix} \\cdot \\begin{pmatrix} \\color{green}{1 - p} & \\color{orange}{p} \\\\ \\color{green}{q} & \\color{orange}{1-q} \\end{pmatrix} \\\\\n",
"=& \\begin{pmatrix}L \\cdot \\color{green}{(1 - p)} + R \\cdot \\color{green}{q} \\\\\n",
"L \\cdot \\color{orange}{p} + R \\cdot \\color{orange}{(1-q)} \n",
"\\end{pmatrix}^T\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "782d4f7a",
"metadata": {
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},
"source": [
"> Man berechnet die neuen Einträge als \"Zeile mal Spalte\""
]
},
{
"cell_type": "code",
"execution_count": 4,
"execution_count": 27,
"id": "e9f7e2f6",
"metadata": {
"slideshow": {
"slide_type": "skip"
}
},
"outputs": [],
"source": [
"from sympy import symbols as S, Matrix as Mat\n",
"p, q = S(\"p q\")\n",
"M = Mat([[1-p, p], [q, 1-q]])"
]
},
{
"cell_type": "markdown",
"id": "daddfd1e",
"metadata": {
"slideshow": {
"slide_type": "subslide"
}
},
"source": [
"## Beispiele\n",
"\n",
"$$\n",
"\\begin{pmatrix}1 & 0\\end{pmatrix}\\cdot M = \\begin{pmatrix}1-p & p\\end{pmatrix}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "c1a6597e",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"source": [
"$$\n",
"\\begin{align*}\n",
"\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}^T\\cdot M &= \\begin{pmatrix} (1-p) + q \\\\ p + (1-q)\\end{pmatrix}^T \\\\\n",
"& = \\begin{pmatrix} (1-p) + q \\\\ p +(1-q)\\end{pmatrix}^T \\\\\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 29,
"id": "aa2e9aca",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"outputs": [
{
"data": {
"text/latex": [
"$\\displaystyle \\left[\\begin{matrix}1 - p & p\\end{matrix}\\right]$"
],
"text/plain": [
"Matrix([[1 - p, p]])"
]
},
"execution_count": 29,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"v1 = Mat([[1, 0]])\n",
"v1 * M"
]
},
{
"cell_type": "code",
"execution_count": 25,
"id": "b31deb83",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"outputs": [
{
"data": {
"text/latex": [
"$\\displaystyle \\left[\\begin{matrix}- p + q + 1 & p - q + 1\\end{matrix}\\right]$"
],
"text/plain": [
"Matrix([[-p + q + 1, p - q + 1]])"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"v2 = Mat([[1, 1]])\n",
"v2 * M"
]
},
{
"cell_type": "markdown",
"id": "8a0fa546",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"source": [
"# Frösche brauchen Zufall\n",
"\n",
"<div class=\"alert alert-info\">\n",
"\n",
"Viele Leute werfen Münzen in den Teich der Frösche.\n",
"Wir sollen einen Frosch aussuchen, der der neue Schatzmeister werden soll. \n",
"Wie können wir aus den 7 Fröschen **fair** einen auswählen?\n",
" \n",
"</div>\n",
"\n",
"**Bedingungen**:\n",
"- Jeder Frosch wird mit $\\frac{1}{7}$ ausgewählt\n",
"- Zufall nur durch die Münzen\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "5d3f53ad",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "markdown",
"id": "9514b1c1",
"metadata": {
"slideshow": {
"slide_type": "fragment"
}
},
"source": [
"- die Berechnung von ganzfrüh aus der Tabelle nenen wir Matrixmultiplikation.\n",
"- MM captures the idea that \"\n",
"- denn, wir können Lineare Algebra machen. $v_0^T\\cdot{}M = v_1$"
]
},
{
"cell_type": "code",
"execution_count": 26,
"id": "9d30de72",
"metadata": {
"slideshow": {
@ -605,7 +836,7 @@
"evalue": "invalid syntax (2451516634.py, line 3)",
"output_type": "error",
"traceback": [
"\u001b[0;36m Cell \u001b[0;32mIn[4], line 3\u001b[0;36m\u001b[0m\n\u001b[0;31m :::info\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
"\u001b[0;36m Cell \u001b[0;32mIn[26], line 3\u001b[0;36m\u001b[0m\n\u001b[0;31m :::info\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
]
}
],
@ -614,12 +845,9 @@
"## Frösche brauchen Zufall\n",
"\n",
":::info\n",
"Viele Leute werfen Münzen in den Teich der Frösche. Wir sollen einen Frosch aussuchen, der der neue Schatzmeister werden soll. Wie können wir aus den 7 Fröschen **fair** einen auswählen?\n",
"\n",
":::\n",
"\n",
"**Bedingungen**:\n",
"- Jeder Frosch wird mit $\\frac{1}{7}$ ausgewählt\n",
"- Zufall nur durch die Münzen\n",
"\n",
"---\n",
"\n",

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