如何在 Python 和 function 中实现计算环面上两个任意点之间的欧几里得距离

Question

Given a 10x10 grid (2d-array) filled randomly with numbers, either 0, 1 or 2 . How can I find the Euclidean distance (the l2-norm of the distance vector) between two given points considering periodic boundaries?

Let us consider an arbitrary grid point called centre . Now, I want to find the nearest grid point containing the same value as centre . I need to take periodic boundaries into account, such that the matrix/grid can be seen rather as a torus instead of a flat plane. In that case, say the centre = matrix[0,2] , and we find that there is the same number in matrix[9,2] , which would be at the southern boundary of the matrix. The Euclidean distance computed with my code would be for this example np.sqrt(0**2 + 9**2) = 9.0 . However, because of periodic boundaries, the distance should actually be 1 , because matrix[9,2] is the northern neighbour of matrix[0,2] . Hence, if periodic boundary values are implemented correctly, distances of magnitude above 8 should not exist.

So, I would be interested on how to implement in Python a function to compute the Euclidean distance between two arbitrary points on a torus by applying a wrap-around for the boundaries.

import numpy as np

matrix = np.random.randint(0,3,(10,10))
centre = matrix[0,2]

#rewrite the centre to be the number 5 (to exclude itself as shortest distance)
matrix[0,2] = 5

#find the points where entries are same as centre
same = np.where((matrix == centre) == True)
idx_row, idx_col = same

#find distances from centre to all values which are of same value 
dist = np.zeros(len(same[0]))
for i in range(0,len(same[0])):
    delta_row = same[0][i] - 0 #row coord of centre
    delta_col = same[1][i] - 2 #col coord of centre
    dist[i] = np.sqrt(delta_row**2 + delta_col**2)

#retrieve the index of the smallest distance
idx = dist.argmin() 
print('Centre value: %i. The nearest cell with same value is at (%i,%i)' 
      % (centre, same[0][idx],same[1][idx]))

Answer 1

For each axis, you can check whether the distance is shorter when you wrap around or when you don't. Consider the row axis, with rows i and j .

• When not wrapping around, the difference is abs(i - j) .
• When wrapping around, the difference is "flipped", as in 10 - abs(i - j) . In your example with i == 0 and j == 9 you can check that this correctly produces a distance of 1.

Then simply take whichever is smaller:

delta_row = same[0][i] - 0 #row coord of centre
delta_row = min(delta_row, 10 - delta_row)

And similarly for delta_column .

The final dist[i] calculation needs no changes.

Answer 2

I have a working 'sketch' of how this could work. In short, I calculate the distance 9 times, 1 for the normal distance, and 8 shifts to possibly correct for a closer 'torus' distance.

As n is getting larger, the calculation costs can go sky high as the numbers go up. But, the torus effect, is probably not needed as there is always a point nearby without 'wrap around'.

You can easily test this, because for a grid of size 1, if a point is found of distance 1/2 or closer, you know there is not a closer torus point (right?)

import numpy as np

n=10000

np.random.seed(1)

A = np.random.randint(low=0, high=10, size=(n,n))

I create 10000x10000 points, and store the location of the 1's in ONES .

ONES = np.argwhere(A == 0)

Now I define my torus distance, which is trying which of the 9 mirrors is the closest.

def distance_on_torus( point=[500,500] ):
    index_diff = [[1],[1],[0],[0],[0,1],[0,1],[0,1],[0,1]]
    coord_diff = [[-1],[1],[-1],[1],[-1,-1],[-1,1],[1,-1],[1,1]]
    
    tree = BallTree( ONES, leaf_size=5*n, metric='euclidean')
    
    dist, indi = tree.query([point],k=1, return_distance=True )

    distances = [dist[0]]

    for indici_to_shift, coord_direction in zip(index_diff, coord_diff):
        MIRROR = ONES.copy()
        for i,shift in zip(indici_to_shift,coord_direction):
            MIRROR[:,i] = MIRROR[:,i] + (shift * n)

        tree = BallTree( MIRROR, leaf_size=5*n, metric='euclidean')
        dist, indi = tree.query([point],k=1, return_distance=True )
        
        distances.append(dist[0])
        
    
    return np.min(distances)

%%time

distance_on_torus([2,3])

It is slow, the above takes 15 minutes.... For n = 1000 less than a second.

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A optimisation would be to first consider the none-torus distance, and if the minimum distance is possibly not the smallest, calculate with only the minimum set of extra 'blocks' around. This will greatly increase speed.