diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb index 681f5400b5500f8c510768a301982042120993c1..91b8fcf3292015b1080a2817010d27b05404da52 100644 --- a/module2/exo1/toy_notebook_en.ipynb +++ b/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:" ] }, {