Commit 2bab2628 authored by Konrad Hinsen's avatar Konrad Hinsen

Copie des modèles qu'on fournit pour module2/exo1

parent c5cc3a54
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<h1 class="title">On the computation of pi</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org9717861">1. Asking the maths library</a></li>
<li><a href="#orgf7d8991">2. Buffon's needle</a></li>
<li><a href="#orgb7b38f0">3. Using a surface fraction argument</a></li>
</ul>
</div>
</div>
<div id="outline-container-org9717861" class="outline-2">
<h2 id="org9717861"><span class="section-number-2">1</span> Asking the maths library</h2>
<div class="outline-text-2" id="text-1">
<p>
My computer tells me that \(\pi\) is <i>approximatively</i>
</p>
<div class="org-src-container">
<pre class="src src-R">pi
</pre>
</div>
<pre class="example">
[1] 3.141593
</pre>
</div>
</div>
<div id="outline-container-orgf7d8991" class="outline-2">
<h2 id="orgf7d8991"><span class="section-number-2">2</span> Buffon's needle</h2>
<div class="outline-text-2" id="text-2">
<p>
Applying the method of <a href="https://en.wikipedia.org/wiki/Buffon%27s_needle_problem">Buffon's needle</a>, we get the <b>approximation</b>
</p>
<div class="org-src-container">
<pre class="src src-R">set.seed(42)
N = 100000
x = runif(N)
theta = pi/2*runif(N)
2/(mean(x+sin(theta)&gt;1))
</pre>
</div>
<pre class="example">
[1] 3.14327
</pre>
</div>
</div>
<div id="outline-container-orgb7b38f0" class="outline-2">
<h2 id="orgb7b38f0"><span class="section-number-2">3</span> Using a surface fraction argument</h2>
<div class="outline-text-2" id="text-3">
<p>
A method that is easier to understand and does not make use of the \(\sin\) function is based on the fact that if \(X\sim U(0,1)\) and \(Y\sim U(0,1)\), then \(P[X^2+Y^2\leq 1] = \pi/4\) (see <a href="https://en.wikipedia.org/wiki/Monte_Carlo_method">"Monte Carlo method" on Wikipedia</a>). The following code uses this approach:
</p>
<div class="org-src-container">
<pre class="src src-R">set.seed(42)
N = 1000
df = data.frame(X = runif(N), Y = runif(N))
df$Accept = (df$X**2 + df$Y**2 &lt;=1)
<span style="color: #BFEBBF;">library</span>(ggplot2)
ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
</pre>
</div>
<div class="figure">
<p><img src="figure_pi_mc1.png" alt="figure_pi_mc1.png" />
</p>
</div>
<p>
It is then straightforward to obtain a (not really good) approximation to \(\pi\) by counting how many times, on average, \(X^2 + Y^2\) is smaller than 1:
</p>
<div class="org-src-container">
<pre class="src src-R">4*mean(df$Accept)
</pre>
</div>
<pre class="example">
[1] 3.156
</pre>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Author: Konrad Hinsen</p>
<p class="date">Created: 2019-03-28 Thu 11:13</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
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<h1 class="title">À propos du calcul de \(\pi\)</h1>
<div id="table-of-contents">
<h2>Table des matières</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#orga63dd54">1. En demandant à la lib maths</a></li>
<li><a href="#org23d5348">2. En utilisant la méthode des aiguilles de Buffon</a></li>
<li><a href="#org0097db4">3. Avec un argument "fréquentiel" de surface</a></li>
</ul>
</div>
</div>
<div id="outline-container-orga63dd54" class="outline-2">
<h2 id="orga63dd54"><span class="section-number-2">1</span> En demandant à la lib maths</h2>
<div class="outline-text-2" id="text-1">
<p>
Mon ordinateur m'indique que \(\pi\) vaut <i>approximativement</i>
</p>
<div class="org-src-container">
<pre class="src src-R">pi
</pre>
</div>
<pre class="example">
[1] 3.141593
</pre>
</div>
</div>
<div id="outline-container-org23d5348" class="outline-2">
<h2 id="org23d5348"><span class="section-number-2">2</span> En utilisant la méthode des aiguilles de Buffon</h2>
<div class="outline-text-2" id="text-2">
<p>
Mais calculé avec la <b>méthode</b> des <a href="https://fr.wikipedia.org/wiki/Aiguille_de_Buffon">aiguilles de Buffon</a>, on obtiendrait
comme <b>approximation</b> :
</p>
<div class="org-src-container">
<pre class="src src-R">set.seed(42)
N = 100000
x = runif(N)
theta = pi/2*runif(N)
2/(mean(x+sin(theta)&gt;1))
</pre>
</div>
<pre class="example">
[1] 3.14327
</pre>
</div>
</div>
<div id="outline-container-org0097db4" class="outline-2">
<h2 id="org0097db4"><span class="section-number-2">3</span> Avec un argument "fréquentiel" de surface</h2>
<div class="outline-text-2" id="text-3">
<p>
Sinon, une méthode plus simple à comprendre et ne faisant pas
intervenir d'appel à la fonction sinus se base sur le fait que si \(X\sim
U(0,1)\) et \(Y\sim U(0,1)\) alors \(P[X^2+Y^2\leq 1] = \pi/4\) (voir <a href="https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80">méthode de
Monte Carlo sur Wikipedia</a>). Le code suivant illustre ce fait :
</p>
<div class="org-src-container">
<pre class="src src-R">set.seed(42)
N = 1000
df = data.frame(X = runif(N), Y = runif(N))
df$Accept = (df$X**2 + df$Y**2 &lt;=1)
<span class="org-constant">library</span>(ggplot2)
ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
</pre>
</div>
<div class="figure">
<p><img src="figure_pi_mc1.png" alt="figure_pi_mc1.png" />
</p>
</div>
<p>
Il est alors aisé d'obtenir une approximation (pas terrible) de \(\pi\) en
comptant combien de fois, en moyenne, \(X^2 + Y^2\) est inférieur à 1 :
</p>
<div class="org-src-container">
<pre class="src src-R">4*mean(df$Accept)
</pre>
</div>
<pre class="example">
[1] 3.156
</pre>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Auteur: Arnaud Legrand</p>
<p class="date">Created: 2018-09-19 mer. 21:51</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
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<h1 class="title">On the computation of pi</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org00d8e4a">1. Asking the math library</a></li>
<li><a href="#org7025e2c">2. * Buffon's needle</a></li>
<li><a href="#org7d2ca3d">3. Using a surface fraction argument</a></li>
</ul>
</div>
</div>
<div id="outline-container-org00d8e4a" class="outline-2">
<h2 id="org00d8e4a"><span class="section-number-2">1</span> Asking the math library</h2>
<div class="outline-text-2" id="text-1">
<p>
My computer tells me that \(\pi\) is <i>approximatively</i>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #F0DFAF; font-weight: bold;">from</span> math <span style="color: #F0DFAF; font-weight: bold;">import</span> *
pi
</pre>
</div>
<pre class="example">
3.141592653589793
</pre>
</div>
</div>
<div id="outline-container-org7025e2c" class="outline-2">
<h2 id="org7025e2c"><span class="section-number-2">2</span> * Buffon's needle</h2>
<div class="outline-text-2" id="text-2">
<p>
Applying the method of <a href="https://en.wikipedia.org/wiki/Buffon%27s_needle_problem">Buffon's needle</a>, we get the <b>approximation</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #F0DFAF; font-weight: bold;">import</span> numpy <span style="color: #F0DFAF; font-weight: bold;">as</span> np
np.random.seed(seed=42)
<span style="color: #DFAF8F;">N</span> = 10000
<span style="color: #DFAF8F;">x</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #DFAF8F;">theta</span> = np.random.uniform(size=N, low=0, high=pi/2)
2/(<span style="color: #DCDCCC; font-weight: bold;">sum</span>((x+np.sin(theta))&gt;1)/N)
</pre>
</div>
<pre class="example">
3.12891113892
</pre>
</div>
</div>
<div id="outline-container-org7d2ca3d" class="outline-2">
<h2 id="org7d2ca3d"><span class="section-number-2">3</span> Using a surface fraction argument</h2>
<div class="outline-text-2" id="text-3">
<p>
A method that is easier to understand and does not make use of the \(\sin\) function is based on the fact that if \(X\sim U(0,1)\) and \(Y\sim U(0,1)\), then \(P[X^2+Y^2\leq 1] = \pi/4\) (see <a href="https://en.wikipedia.org/wiki/Monte_Carlo_method">"Monte Carlo method" on Wikipedia</a>). The following code uses this approach:
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #F0DFAF; font-weight: bold;">import</span> matplotlib.pyplot <span style="color: #F0DFAF; font-weight: bold;">as</span> plt
np.random.seed(seed=42)
<span style="color: #DFAF8F;">N</span> = 1000
<span style="color: #DFAF8F;">x</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #DFAF8F;">y</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #DFAF8F;">accept</span> = (x*x+y*y) &lt;= 1
<span style="color: #DFAF8F;">reject</span> = np.logical_not(accept)
<span style="color: #DFAF8F;">fig</span>, <span style="color: #DFAF8F;">ax</span> = plt.subplots(1)
ax.scatter(x[accept], y[accept], c=<span style="color: #CC9393;">'b'</span>, alpha=0.2, edgecolor=<span style="color: #BFEBBF;">None</span>)
ax.scatter(x[reject], y[reject], c=<span style="color: #CC9393;">'r'</span>, alpha=0.2, edgecolor=<span style="color: #BFEBBF;">None</span>)
ax.set_aspect(<span style="color: #CC9393;">'equal'</span>)
plt.savefig(matplot_lib_filename)
<span style="color: #F0DFAF; font-weight: bold;">print</span>(matplot_lib_filename)
</pre>
</div>
<div class="figure">
<p><img src="figure_pi_mc2.png" alt="figure_pi_mc2.png" />
</p>
</div>
<p>
It is then straightforward to obtain a (not really good) approximation to \(\pi\) by counting how many times, on average, \(X^2 + Y^2\) is smaller than 1:
</p>
<div class="org-src-container">
<pre class="src src-python">4*np.mean(accept)
</pre>
</div>
<pre class="example">
3.1120000000000001
</pre>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Author: Konrad Hinsen</p>
<p class="date">Created: 2019-03-28 Thu 11:13</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
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<div id="content">
<h1 class="title">À propos du calcul de \(\pi\)</h1>
<div id="table-of-contents">
<h2>Table des matières</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org5d62750">1. En demandant à la lib maths</a></li>
<li><a href="#org48ad359">2. En utilisant la méthode des aiguilles de Buffon</a></li>
<li><a href="#orgd097712">3. Avec un argument "fréquentiel" de surface</a></li>
</ul>
</div>
</div>
<div id="outline-container-org5d62750" class="outline-2">
<h2 id="org5d62750"><span class="section-number-2">1</span> En demandant à la lib maths</h2>
<div class="outline-text-2" id="text-1">
<p>
Mon ordinateur m'indique que \(\pi\) vaut <i>approximativement</i>:
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #F0DFAF; font-weight: bold;">from</span> math <span style="color: #F0DFAF; font-weight: bold;">import</span> *
pi
</pre>
</div>
<pre class="example">
3.141592653589793
</pre>
</div>
</div>
<div id="outline-container-org48ad359" class="outline-2">
<h2 id="org48ad359"><span class="section-number-2">2</span> En utilisant la méthode des aiguilles de Buffon</h2>
<div class="outline-text-2" id="text-2">
<p>
Mais calculé avec la <b>méthode</b> des <a href="https://fr.wikipedia.org/wiki/Aiguille_de_Buffon">aiguilles de Buffon</a>, on obtiendrait
comme <b>approximation</b> :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #F0DFAF; font-weight: bold;">import</span> numpy <span style="color: #F0DFAF; font-weight: bold;">as</span> np
np.random.seed(seed=42)
<span style="color: #DFAF8F;">N</span> = 10000
<span style="color: #DFAF8F;">x</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #DFAF8F;">theta</span> = np.random.uniform(size=N, low=0, high=pi/2)
2/(<span style="color: #DCDCCC; font-weight: bold;">sum</span>((x+np.sin(theta))&gt;1)/N)
</pre>
</div>
<pre class="example">
3.12891113892
</pre>
</div>
</div>
<div id="outline-container-orgd097712" class="outline-2">
<h2 id="orgd097712"><span class="section-number-2">3</span> Avec un argument "fréquentiel" de surface</h2>
<div class="outline-text-2" id="text-3">
<p>
Sinon, une méthode plus simple à comprendre et ne faisant pas
intervenir d'appel à la fonction sinus se base sur le fait que si \(X\sim
U(0,1)\) et \(Y\sim U(0,1)\) alors \(P[X^2+Y^2\leq 1] = \pi/4\) (voir <a href="https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80">méthode de
Monte Carlo sur Wikipedia</a>). Le code suivant illustre ce fait :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #F0DFAF; font-weight: bold;">import</span> matplotlib.pyplot <span style="color: #F0DFAF; font-weight: bold;">as</span> plt
np.random.seed(seed=42)
<span style="color: #DFAF8F;">N</span> = 1000
<span style="color: #DFAF8F;">x</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #DFAF8F;">y</span> = np.random.uniform(size=N, low=0, high=1)
<span style="color: #DFAF8F;">accept</span> = (x*x+y*y) &lt;= 1
<span style="color: #DFAF8F;">reject</span> = np.logical_not(accept)
<span style="color: #DFAF8F;">fig</span>, <span style="color: #DFAF8F;">ax</span> = plt.subplots(1)
ax.scatter(x[accept], y[accept], c=<span style="color: #CC9393;">'b'</span>, alpha=0.2, edgecolor=<span style="color: #BFEBBF;">None</span>)
ax.scatter(x[reject], y[reject], c=<span style="color: #CC9393;">'r'</span>, alpha=0.2, edgecolor=<span style="color: #BFEBBF;">None</span>)
ax.set_aspect(<span style="color: #CC9393;">'equal'</span>)
plt.savefig(matplot_lib_filename)
<span style="color: #F0DFAF; font-weight: bold;">print</span>(matplot_lib_filename)
</pre>
</div>
<div class="figure">
<p><img src="figure_pi_mc2.png" alt="figure_pi_mc2.png" />
</p>
</div>
<p>
Il est alors aisé d'obtenir une approximation (pas terrible) de \(\pi\) en
comptant combien de fois, en moyenne, \(X^2 + Y^2\) est inférieur à 1 :
</p>
<div class="org-src-container">
<pre class="src src-python">4*np.mean(accept)
</pre>
</div>
<pre class="example">
3.1120000000000001
</pre>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Auteur: Konrad Hinsen</p>
<p class="date">Created: 2019-03-28 Thu 11:06</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
</body>
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