From 138e77ba49300f7cf383d67c1ded68510fc1c849 Mon Sep 17 00:00:00 2001 From: 2e07cdc5aaf5c3993089d76e376cb88f <2e07cdc5aaf5c3993089d76e376cb88f@app-learninglab.inria.fr> Date: Sun, 15 Oct 2023 10:44:05 +0000 Subject: [PATCH] solution2 --- module2/exo1/toy_notebook_en.ipynb | 30 ++++++++++++++++-------------- 1 file changed, 16 insertions(+), 14 deletions(-) diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index fe30778..4be35c1 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -4,14 +4,14 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "# 1 On the computation of $\\pi$" + "# On the computation of $\\pi$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ - "## 1.1 Asking the maths library" + "## Asking the maths library" ] }, { @@ -43,19 +43,19 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## 1.2 Buffalo's needle" + "## Buffalo's needle" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ - "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**" + "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__" ] }, { "cell_type": "code", - "execution_count": 2, + "execution_count": 5, "metadata": {}, "outputs": [ { @@ -64,13 +64,13 @@ "3.128911138923655" ] }, - "execution_count": 2, + "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ - "In [2]: import numpy as np\n", + "import numpy as np\n", "np.random.seed(seed=42)\n", "N = 10000\n", "x = np.random.uniform(size=N, low=0, high=1)\n", @@ -89,12 +89,12 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "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 $P$[$X^2$ + $Y^2$ $\\le$ 1] = $\\pi$/4 (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:" + "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 [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:" ] }, { "cell_type": "code", - "execution_count": 3, + "execution_count": 6, "metadata": {}, "outputs": [ { @@ -113,29 +113,31 @@ "source": [ "%matplotlib inline\n", "import matplotlib.pyplot as plt\n", + "\n", "np.random.seed(seed=42)\n", "N = 1000\n", "x = np.random.uniform(size=N, low=0, high=1)\n", "y = np.random.uniform(size=N, low=0, high=1)\n", - "1\n", + "\n", "accept = (x*x+y*y) <= 1\n", "reject = np.logical_not(accept)\n", + "\n", "fig, ax = plt.subplots(1)\n", "ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n", "ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n", - "ax.set_aspect('equal')\n" + "ax.set_aspect('equal')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ - "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:" + "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:" ] }, { "cell_type": "code", - "execution_count": 4, + "execution_count": 7, "metadata": {}, "outputs": [ { @@ -144,7 +146,7 @@ "3.112" ] }, - "execution_count": 4, + "execution_count": 7, "metadata": {}, "output_type": "execute_result" } -- 2.18.1