Closest Supporting Hyperplane to Points How do I find the hyperplane (line) that touches a cloud of 2D points from one side? I used $w^Tx+b=0$ based line definition and implemented a gradient-descent based routine that would start the support line from a certain side of points, away from them. The loss function is $w^Tx_i+b$ for all points, then I take log sum. The use of hyperplanes is similar as in SVM but in this case there are no -1 or +1 labels, there are only points I am trying to approach as close as possible. I am hoping to do this from 4 sides of a point cloud to get 4 approximate / defining lines of this point cloud. 

My solution was fine for some carefully selected step sizes, and iteration numbers, but the procedure can overshoot, trying to find another minima.

I am curious if there is a better way of defining this loss function. My implementation based on automatic differentiation package autograd, seen below:

import pandas as pd

def plot_sep(w):
    Q = np.array([[0, -1],[1, 0]])
    x = np.linspace(-20,20,3000)
    w2 = np.dot(Q,w[:2])
    m = w2[1]/w2[0]
    y = m*x + (-w[2]/w[1])
    plt.plot(x,y,'.')

df = pd.read_csv('in.csv')
df['1'] = 1.
df2 = np.array(df)

#w1_init = np.array([0.,1.,-6.]);eta  = 1e-3;iters = 30 # good
w1_init = np.array([0.,1.,-30.]);eta  = 1e-2;iters = 40 # bad

plt.plot(df2[:,0],df2[:,1],'.')
plot_sep(w1_init)
plt.xlim(0,15);plt.ylim(-10,40)
plt.savefig('out1.png')

import autograd.numpy as np
from autograd import elementwise_grad
from autograd import grad

def compute_loss(w1):
    tmp = np.dot(df2,w1)
    tmp2 = np.log(np.dot(tmp,tmp))
    return tmp2

gradient = grad(compute_loss)

w1 = np.copy(w1_init)

for i in range(iters):  
    loss = compute_loss(w1)
    print "iteration %d: loss %f" % (i, loss) 
    dw1 = gradient(w1)
    w1 += -eta*dw1
print "iteration %d: loss %f" % (i, loss)
print w1

plt.figure()
plt.plot(df2[:,0],df2[:,1],'.')
plot_sep(w1)
plt.xlim(0,15);plt.ylim(-10,40)
plt.savefig('out2.png')


Data

x,y
6.85483870968,11.875
8.06451612903,12.3958333333
7.37903225806,12.34375
8.18548387097,12.34375
8.83064516129,12.6041666667
9.43548387097,12.96875
10.0,13.0729166667
10.5241935484,13.1770833333
11.0483870968,13.2291666667
6.97580645161,10.9895833333
6.97580645161,10.4166666667
8.46774193548,10.15625
7.98387096774,10.15625
9.1935483871,10.15625
9.79838709677,10.15625
10.6048387097,10.0
11.1290322581,10.1041666667
11.1290322581,10.5208333333
10.9274193548,11.0416666667
10.9274193548,11.40625
10.9274193548,11.7708333333
10.8870967742,12.4479166667
10.0,12.7083333333
9.07258064516,11.9270833333
8.75,11.9270833333
7.86290322581,11.8229166667
7.33870967742,11.09375
7.9435483871,11.3541666667
9.15322580645,11.5104166667
9.39516129032,11.5104166667
8.50806451613,10.8854166667
9.47580645161,10.78125
9.91935483871,10.78125
10.1612903226,10.8333333333
10.1612903226,11.9270833333
9.91935483871,12.03125
9.83870967742,12.03125
9.63709677419,11.9270833333
10.564516129,11.3020833333
10.564516129,10.6770833333
9.11290322581,10.5208333333
8.02419354839,10.625
I pondered this question a little more, there might not be an optimal loss function.. But here is an idea which could utilize SVMs hyperplane seperation ability. 

I take the point cloud, if I need an approximating line from the bottom, copy points below with user defined gap, and I flip the points, as in mirror image along the x-axis. I give both point groups opposing labels, then feed the whole thing to SVM. Ideally I should get something like this:



The top line is the line I am looking for. 

Code

import pandas as pd
points1 = np.array(pd.read_csv('quadri.csv'))
plt.plot(points1[:,0], points1[:,1], 'o')
points2 = np.copy(points1)
points2[:,1] -= 7.
xmin = np.min(points2[:,0]); xmax = np.max(points2[:,0])
points2[:,0] = (xmax-points2[:,0])+xmin
points = np.vstack((points1,points2))
plt.plot(points[:,0], points[:,1], 'o')
plt.xlim(4,14); plt.ylim(0,15)