Distance example (#31)

* [docs] Draft for distance example

* Change color and line style of the distance line

examples/plot_distance.py

@@ -0,0 +1,103 @@
+"""
+Distance between splines
+------------------------
+
+In this example, we compute the euclidean distance between two splines.
+"""
+
+import matplotlib.pyplot as plt
+import numpy as np
+import scipy
+import splinebox
+
+# %%
+# For simplicity we create a function that can plot the two splines and the
+# distance between the splines at parameter values t_min and s_min.
+
+
+def plot_splines(spline1, spline2, t_min=None, s_min=None):
+    vals1 = spline1.eval(np.linspace(0, spline1.M, 1000))
+    vals2 = spline2.eval(np.linspace(0, spline2.M, 1000))
+    knots1 = spline1.knots
+    knots2 = spline2.knots
+
+    plt.plot(vals1[:, 1], vals1[:, 0])
+    plt.plot(vals2[:, 1], vals2[:, 0])
+    plt.scatter(knots1[:, 1], knots1[:, 0])
+    plt.scatter(knots2[:, 1], knots2[:, 0])
+
+    if t_min is not None and s_min is not None:
+        point1 = spline1.eval(t_min)
+        point2 = spline2.eval(s_min)
+        plt.plot([point1[1], point2[1]], [point1[0], point2[0]], color="k", linestyle="--")
+
+    plt.gca().set_aspect("equal", "box")
+
+    plt.show()
+
+
+# %%
+# We start by constructing two arbitrary closed splines.
+
+basis_function = splinebox.basis_functions.B3()
+M = 5
+spline1 = splinebox.spline_curves.Spline(M=M, basis_function=basis_function, closed=True)
+spline2 = splinebox.spline_curves.Spline(M=M, basis_function=basis_function, closed=True)
+
+spline1.control_points = np.array(
+    [
+        [1, 2],
+        [2, 2],
+        [3, 2.5],
+        [2.2, 3],
+        [1.3, 2.2],
+    ]
+)
+
+spline2.control_points = np.array(
+    [
+        [2, -2],
+        [2.5, -1],
+        [2, -1.5],
+        [1.5, -1],
+        [1, -2],
+    ]
+)
+
+# %%
+# Plot the splines
+plot_splines(spline1, spline2)
+
+# %%
+# To get an initial guess of the spline parameter pair with the
+# smallest distance, we perform a brute force search.
+# Note: This can be made more accurate by increasing the number
+# of parameters we interogate.
+
+t = np.linspace(0, spline1.M, 5)
+s = np.linspace(0, spline2.M, 5)
+vals1 = spline1.eval(t)
+vals2 = spline2.eval(s)
+distance_vectors = vals1[:, np.newaxis] - vals2[np.newaxis, :]
+distances = np.linalg.norm(distance_vectors, axis=-1)
+indices = np.unravel_index(np.argmin(distances), distances.shape)
+t_min = t[indices[0]]
+s_min = s[indices[1]]
+
+plot_splines(spline1, spline2, t_min, s_min)
+
+# %%
+# To further refine the estimate we can
+# run an optimization.
+
+
+def distance(parameters):
+    val1 = spline1.eval(parameters[0])
+    val2 = spline2.eval(parameters[1])
+    return np.linalg.norm(val1 - val2)
+
+
+result = scipy.optimize.minimize(distance, np.array([t_min, s_min]), bounds=((0, spline1.M), (0, spline2.M)))
+t_min, s_min = result.x
+
+plot_splines(spline1, spline2, t_min, s_min)