pi with monte carlo

app.py

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+import numpy as np
+from pyDOE2.doe_lhs import lhs
+import matplotlib.pyplot as plt
+import streamlit as st
+import mpmath
+
+# Set the precision to 50 decimal places
+mpmath.mp.dps = 50
+
+# Compute the value of pi with 50 decimal places of precision
+pi = mpmath.pi
+
+st.title("Compute Pi with Monte Carlo")
+
+st.markdown(
+    r"""
+    We know that area of a circle is $\pi r^2$. If we enclose the circle in a square of side $2r$, then the area of the square is $4r^2$. The ratio of the area of the circle to the area of the square is $\frac{\pi r^2}{4r^2} = \frac{\pi}{4}$. If we randomly sample points in the square, then the ratio of the number of points in the circle to the total number of points is also $\frac{\pi}{4}$.
+            """
+)
+
+
+# draw a circle with matplotlib
+def draw_the_plot():
+    fig, ax = plt.subplots()
+    ax.set_aspect("equal")
+    # draw a square with matplotlib
+    square = plt.Rectangle(
+        (-1, -1), 2, 2, facecolor="lightblue", edgecolor="r", linewidth=2
+    )
+    ax.add_artist(square)
+    # add a two sides arrow outside the square to indicate the length of the side
+    upper_margin = 1.05
+    ax.arrow(
+        -1,
+        upper_margin,
+        2,
+        0,
+        head_width=0.05,
+        color="k",
+        length_includes_head=True,
+        clip_on=False,
+    )
+    ax.arrow(
+        1,
+        upper_margin,
+        -2,
+        0,
+        head_width=0.05,
+        color="k",
+        length_includes_head=True,
+        clip_on=False,
+    )
+    # annotate the arrow
+    ax.annotate(
+        "$2r$",
+        xy=(0, upper_margin),
+        xytext=(0, upper_margin),
+        horizontalalignment="center",
+        verticalalignment="bottom",
+    )
+    circle = plt.Circle((0, 0), 1, facecolor="lightgreen", edgecolor="b", linewidth=2)
+    ax.add_artist(circle)
+    # draw a horizontal arrow from the center to the edge
+    ax.arrow(0, 0, 1, 0, head_width=0.05, color="k", length_includes_head=True)
+    # annotate the arrow
+    ax.annotate(
+        "$r$",
+        xy=(0.5, 0),
+        xytext=(0.5, -0.1),
+        horizontalalignment="center",
+        verticalalignment="bottom",
+    )
+
+    ax.set_xlim(-1.1, 1.1)
+    ax.set_ylim(-1.1, 1.1)
+    ax.set_xticks([])
+    ax.set_yticks([])
+    # remove the x and y axis
+    ax.spines["left"].set_visible(False)
+    ax.spines["right"].set_visible(False)
+    ax.spines["top"].set_visible(False)
+    ax.spines["bottom"].set_visible(False)
+
+    return fig, ax
+
+
+fig, ax = draw_the_plot()
+st.pyplot(fig)
+
+st.subheader("Monte Carlo Simulation")
+
+seed = st.number_input("Random seed", min_value=0, max_value=10000000, value=0, step=1)
+N = st.number_input(
+    "Number of points", min_value=1, max_value=10000000, value=10000, step=1
+)
+type = st.selectbox("Type of sampling", ["Random", "Grid"])
+
+if type == "Random":
+    samples = lhs(2, samples=N, random_state=seed) * 2 - 1
+else:
+    samples = np.linspace(-1, 1, int(np.sqrt(N)))
+    samples = np.array(np.meshgrid(samples, samples)).T.reshape(-1, 2)
+
+# compute the distance of each point from the origin
+distances = np.linalg.norm(samples, axis=1)
+
+# compute the number of points inside the circle
+inside = samples[distances < 1, :]
+print(samples.shape, inside.shape)
+
+st.markdown(
+    f"""
+    Points inside the circle: = {len(inside)}\n
+    Ratio: = {len(inside)} / {len(samples)} = {len(inside) / len(samples)}\n
+    $\pi$ = {len(inside) / len(samples)} * 4\n
+    | $\pi$ | value |
+    | --- | --- |
+    | estimated $\pi$ | {len(inside) / len(samples) * 4:.50f} |
+    | real $\pi$ | {pi} |
+    """
+)
+
+fig, ax = draw_the_plot()
+ax.scatter(samples[:, 0], samples[:, 1], s=0.1, color="r")
+ax.scatter(inside[:, 0], inside[:, 1], s=0.1, color="b")
+st.pyplot(fig)

requirements.txt

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+numpy
+matplotlib
+mpmath
+pyDOE2