fix

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,
+   "execution_count": null,
    "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
+   "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')"