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All-pairs Euclidean distance

Date: 2019-05-10, Author: Nenad Markuš

Given two arrays of \(d\)-dimensional vectors, \(\mathbf{A}\in\mathbb{R}^{M\times d}\) and \(\mathbf{B}\in\mathbb{R}^{N\times d}\), we are interested in efficient vectorized code for obtaining a matrix \(\mathbf{D}\in\mathbb{R}^{M\times N}\) such that

\[(\mathbf{D})_{ij} = \vert\vert\mathbf{a}_i - \mathbf{b}_j\vert\vert_2\]

where \((\mathbf{D})_{ij}\) is the \(ij\)th entry of \(\mathbf{D}\), \(\mathbf{a}_i\) is the \(i\)th row of \(\mathbf{A}\) and \(\mathbf{b}_j\) is the \(j\)th row of \(\mathbf{B}\). In other words, we want to compute the Euclidean distance between all vectors in \(\mathbf{A}\) and all vectors in \(\mathbf{B}\). The following numpy code does exactly this:

def all_pairs_euclid_naive(A, B):
	D = numpy.zeros((A.shape[0], B.shape[0]), dtype=numpy.float32)
	for i in range(0, D.shape[0]):
		for j in range(0, D.shape[1]):
			D[i, j] = numpy.linalg.norm(A[i, :] - B[j, :])
	return D

Unfortunately, this code is really inefficient. To rectify the issue, we need to write a vectorized version in which we avoid the explicit usage of loops. To arrive at a solution, we first expand the formula for the Euclidean distance:

\[\vert\vert\mathbf{a}_i - \mathbf{b}_j\vert\vert_2= (\mathbf{a}_i - \mathbf{b}_j)^T(\mathbf{a}_i - \mathbf{b}_j)= \mathbf{a}_i^T\mathbf{a}_i - 2\mathbf{a}_i^T\mathbf{b}_j + \mathbf{b}_j^T\mathbf{b}_j\]

This leads us to the following equation for \(\mathbf{D}\):

\[\mathbf{D} = \sqrt{\mathbf{S}_A - 2\cdot\mathbf{A}\cdot\mathbf{B}^T + \mathbf{S}_B}\]

where \(\mathbf{S}_A\in\mathbb{R}^{M\times N}\) is such that \((\mathbf{S}_A)_{ij}=\mathbf{a}_i^T\mathbf{a}_i\) and \(\mathbf{S}_B\in\mathbb{R}^{M\times N}\) is such that \((\mathbf{S}_B)_{ij}=\mathbf{b}_j^T\mathbf{b}_j\). The square root is taken elementwise.

Each of the three above terms can be obtain with vectorized code.

The first term is obtained for each \(i\) by squaring the entries of \(\mathbf{A}\), summing along the second dimension after that and repeating the obtained result \(N\) times to obtain an \(M\times N\) matrix. The second term can be computed with the standard matrix-matrix multiplication routine. The third term is obtained in a simmilar manner to the first term.

Without further ado, here is the numpy code:

def all_pairs_euclid_numpy(A, B):
	sqrA = numpy.broadcast_to(numpy.sum(numpy.power(A, 2), 1).reshape(A.shape[0], 1), (A.shape[0], B.shape[0]))
	sqrB = numpy.broadcast_to(numpy.sum(numpy.power(B, 2), 1).reshape(B.shape[0], 1), (B.shape[0], A.shape[0])).transpose()
	return numpy.sqrt(
		sqrA - 2*numpy.matmul(A, B.transpose()) + sqrB

And the following implementation uses Pytorch:

def all_pairs_euclid_torch(A, B):
	sqrA = torch.sum(torch.pow(A, 2), 1, keepdim=True).expand(A.shape[0], B.shape[0])
	sqrB = torch.sum(torch.pow(B, 2), 1, keepdim=True).expand(B.shape[0], A.shape[0]).t()
	return torch.sqrt(
		sqrA - 2*torch.mm(A, B.t()) + sqrB

Note that the above code can be executed on the GPU as well.

By setting \(M=2048\), \(N=2048\) and \(d=2048\), we obtain the following timings for the modern 40-thread Intel CPU and the Nvidia 1080Ti GPU:

Implementation Time [ms]
naive 29500
numpy 160
pytorch 20
pytorch (cuda) 15

You can get the time measurements for your hardware by running the following script.