Update module2/exo1/toy_notebook_en.ipynb

module2/exo1/toy_notebook_en.ipynb

@@ -12,7 +12,6 @@
    "metadata": {},
    "source": [
     "## Asking the maths library\n",
-    "\n",
     "My computer tells me that $\\pi$ is _approximatively_"
    ]
   },
@@ -39,7 +38,6 @@
    "metadata": {},
    "source": [
     "## Buffon’s needle\n",
-    "\n",
     "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
@@ -73,10 +71,7 @@
    "metadata": {},
    "source": [
     "## Using a surface fraction argument\n",
-    "\n",
-    "A method that is easier to understand and does not make use of the sin function is based on the\n",
-    "fact that if $X ∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[X^2 + Y^2 ≤ 1] = \\pi/4$ (see [\"Monte Carlo method\"\n",
-    "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 $X ∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[X^2 + Y^2 ≤ 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {
@@ -105,8 +100,10 @@
     "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",
+    "\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",
@@ -117,8 +114,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to \\pi by counting how\n",
-    "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:"
    ]
   },
   {