How to implement finding the coordinates of points from distance matrix in python based on gram-matrix?

Question

I want to study a bus stop optimization problem. However, I'm now stuck in how to convert the distance matrix to the real coordinates of points.

I have browsed a lot resouce and known using the formula：M(i, j) = 0.5(D(1, j)^2 + D(i, 1)^2 - D(i, j)^2)* to solve the problem enter link description here. I'm not good at math and I just want to implement to it.

Firstly, I try to understand the principle of Mathematics and here is my solution. enter link description here.

Then, I want to implment the algorithm using python for the following example. Here is my matrix which represents different distance for each bus-station. I want to transfer it to the coordinates of points.

Here is my code for implmenting:

import csv
import numpy as np
import math

class csv_util():

    def generate_coordinate_point(self):
        '''transfer the distance matrix to the real coordinate points'''
        sqrt_result = 2*math.sqrt(2)

        matrix = np.array([[0,2,2,sqrt_result],[2,0,sqrt_result,2],[2,sqrt_result,0,2],[sqrt_result,2,2,0]])

        gram_matrix = self.calculate_gram_matrix(matrix)

        a, b = np.linalg.eig(gram_matrix)

        #b = b.astype(np.int16)
        a = a.astype(np.int16)


        eigen_vector = format(b)

        length = a.size
        tmp_matrix = np.zeros(length * length)
        random_point_matrix = tmp_matrix.reshape(length, length)

        for item1 in range(length):
            random_point_matrix[item1][item1] = a[item1]

        print("the eigen-value is: " + format(random_point_matrix))
        print("the eigen-vector is: " + eigen_vector)

        new_matrix = (np.sqrt(random_point_matrix))*b

        print("the coordinate points: "+format(new_matrix))



    def calculate_gram_matrix(self,matrix):
        '''get the gram matrix for transfer to the coordinate points'''

        length = matrix[0].size
        tmp_matrix = np.zeros(length*length)
        gram_matrix = tmp_matrix.reshape(length,length)

        for item1 in range(length):
            for item2 in range(length):
                gram_matrix[item1][item2] = (math.pow(matrix[0][item2],2)+math.pow(matrix[0][item1],2)-math.pow(matrix[item1][item2],2))/2
                if gram_matrix[item1][item2]<0.1 and gram_matrix[item1][item2]>-0.1:
                    gram_matrix[item1][item2] = 0

        return gram_matrix

However, the result of final matrix isn't correct. The result like this:

the eigen-value is: [[12.  0.  0.  0.]
 [ 0.  0.  0.  0.]
 [ 0.  0.  4.  0.]
 [ 0.  0.  0.  0.]]
-------------
the eigen-vector is: [[ 0.00000000e+00  0.00000000e+00  0.00000000e+00  1.00000000e+00]
 [ 4.08248290e-01 -5.77350269e-01 -7.07106781e-01  0.00000000e+00]
 [ 4.08248290e-01 -5.77350269e-01  7.07106781e-01  0.00000000e+00]
 [ 8.16496581e-01  5.77350269e-01  1.57009246e-16  0.00000000e+00]]
-------------
the coordinate points: [[ 0.          0.          0.          0.        ]
 [ 0.         -0.         -0.          0.        ]
 [ 0.         -0.          1.41421356  0.        ]
 [ 0.          0.          0.          0.        ]]

The final points like this: [0,0],[-0.0,-0.0],[-0.0,1.414421],[0.0,0.0]. They cannot be satisfied with the distance matrix in this example. Please help me how to get the correct points. Thanks!

Answer 1

The construction of the Gram matrix of dot products associated to the distance matrix and its further factorization is generally a great method, which also allows you to infer the dimension of the coordinate realization of the distance matrix. However, if in your case the realization is planar (two dimensional), then I think it is (arguably) easier (and possibly faster) to just approach it a bit more geometrically (again, you should be certain that the distance matrix is for points in 2D):

import numpy as np
import math

def x_coord_of_point(D, j):
    return ( D[0,j]**2 + D[0,1]**2 - D[1,j]**2 ) / ( 2*D[0,1] )

def coords_of_point(D, j):
    x = x_coord_of_point(D, j)
    return np.array([x, math.sqrt( D[0,j]**2 - x**2 )])
    
def calculate_positions(D):
    (m, n) = D.shape
    P = np.zeros( (n, 2) )
    tr = ( min(min(D[2,0:2]), min(D[2,3:n])) / 2)**2
    P[1,0] = D[0,1]
    P[2,:] = coords_of_point(D, 2)
    for j in range(3,n):
        P[j,:] = coords_of_point(D, j) 
        if abs( np.dot(P[j,:] - P[2,:], P[j,:] - P[2,:]) - D[2,j]**2 ) > tr:
            P[j,1] = - P[j,1]
    return P 
    
sqrt_result = 2*math.sqrt(2)
D = np.array([[0, 2, 2, sqrt_result],
              [2, 0, sqrt_result, 2], 
              [2, sqrt_result, 0, 2], 
              [sqrt_result, 2, 2, 0]])

P = calculate_positions(D)
print(P)
     

You may want to add some checks and improvements to make sure that the vectors P[1,:] and P[2,:] are not aligned, which is equivalent to checking that

abs( P[1,0]*P[2,1] - P[1,1]*P[2,0] ) < 0.0001 (or some more appropriate threshold)

If they are, simply implement a while loop until you find the first vector P[j0, :] that is not aligned with P[1,0]. The role of this first vector P[j0,:] unaligned with the initial vector P[1,:] allows you to have a useful if clause in the function vector(D). I did not include it to avoid obscuring the idea of the code.