fix

parent e2bbc063
...@@ -27,16 +27,13 @@ ...@@ -27,16 +27,13 @@
"hidePrompt": true "hidePrompt": true
}, },
"source": [ "source": [
"My computer tells me that $\\pi$ is approximatively" "My computer tells me that $\\pi$ is *approximatively*"
] ]
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 2, "execution_count": 10,
"metadata": { "metadata": {},
"hideCode": true,
"hidePrompt": true
},
"outputs": [ "outputs": [
{ {
"name": "stdout", "name": "stdout",
...@@ -58,34 +55,21 @@ ...@@ -58,34 +55,21 @@
"hidePrompt": true "hidePrompt": true
}, },
"source": [ "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", "cell_type": "code",
"execution_count": 6, "execution_count": null,
"metadata": { "metadata": {},
"hideCode": true, "outputs": [],
"hidePrompt": true
},
"outputs": [
{
"data": {
"text/plain": [
"3.128911138923655"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [ "source": [
"import numpy as np\n", "import numpy as np\n",
"np.random.seed(seed=42)\n", "np.random.seed(seed=42)\n",
"N = 10000\n", "N = 10000\n",
"x = np.random.uniform(size=N, low=0, high =1)\n", "x = np.random.uniform(size=N, low=0, high=1)\n",
"theta = np.random.uniform(size=N, low=0, high =pi/2)\n", "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
"2/(sum((x+np.sin(theta))>1)/N)" "2/(sum((x+np.sin(theta))>1)/N)"
] ]
}, },
...@@ -96,17 +80,8 @@ ...@@ -96,17 +80,8 @@
"hidePrompt": true "hidePrompt": true
}, },
"source": [ "source": [
"## Using a surface fraction argument" "## 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:"
},
{
"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:"
] ]
}, },
{ {
...@@ -131,16 +106,18 @@ ...@@ -131,16 +106,18 @@
} }
], ],
"source": [ "source": [
"%matplotlib inline\n", "%matplotlib inline \n",
"import matplotlib.pyplot as plt\n", "import matplotlib.pyplot as plt\n",
"\n", "\n",
"np.random.seed(seed =42)\n", "np.random.seed(seed=42)\n",
"N = 1000\n", "N = 1000\n",
"x=np.random.uniform(size=N, low=0, high=1)\n", "x = np.random.uniform(size=N, low=0, high=1)\n",
"y=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", "\n",
"reject=np.logical_not(accept)\n", "accept = (x*x+y*y) <= 1\n",
"fig, ax=plt.subplots(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[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.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
"ax.set_aspect('equal')" "ax.set_aspect('equal')"
......
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