diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 422697ba0ce8997e1e1a1afae180864f5f2f01aa..afaa59514057ab355cbbbd48a63d12bc7aa7bfad 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/module2/exo1/toy_notebook_en.ipynb @@ -27,16 +27,13 @@ "hidePrompt": true }, "source": [ - "My computer tells me that $\\pi$ is approximatively" + "My computer tells me that $\\pi$ is *approximatively*" ] }, { "cell_type": "code", - "execution_count": 2, - "metadata": { - "hideCode": true, - "hidePrompt": true - }, + "execution_count": 10, + "metadata": {}, "outputs": [ { "name": "stdout", @@ -58,34 +55,21 @@ "hidePrompt": true }, "source": [ - "## Buffon's needle" + "## Buffon's needle\n", + "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": 6, - "metadata": { - "hideCode": true, - "hidePrompt": true - }, - "outputs": [ - { - "data": { - "text/plain": [ - "3.128911138923655" - ] - }, - "execution_count": 6, - "metadata": {}, - "output_type": "execute_result" - } - ], + "execution_count": null, + "metadata": {}, + "outputs": [], "source": [ "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", - "theta = np.random.uniform(size=N, low=0, high =pi/2)\n", + "x = np.random.uniform(size=N, low=0, high=1)\n", + "theta = np.random.uniform(size=N, low=0, high=pi/2)\n", "2/(sum((x+np.sin(theta))>1)/N)" ] }, @@ -96,17 +80,8 @@ "hidePrompt": true }, "source": [ - "## Using a surface fraction argument" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "hideCode": true, - "hidePrompt": true - }, - "source": [ - "A method that is easier to understand and does not make the use of the sin function is based on the fact that if $X ~ U(0,1)$ and $Y ~ U(0,1)$, then $P[{x}^{2}+{y}^{2} = \\pi/4]$ (see [\"Monte Carlo method\" on wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:" + "## Using a surface fraction argument\n", + "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:" ] }, { @@ -131,16 +106,18 @@ } ], "source": [ - "%matplotlib inline\n", + "%matplotlib inline \n", "import matplotlib.pyplot as plt\n", "\n", - "np.random.seed(seed =42)\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", - "accept=(x*x+y*y)<=1\n", - "reject=np.logical_not(accept)\n", - "fig, ax=plt.subplots(1)\n", + "x = np.random.uniform(size=N, low=0, high=1)\n", + "y = np.random.uniform(size=N, low=0, high=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')"