KernelSolve reduction

Let’s see how to solve discrete deconvolution problems using the conjugate gradient solver provided by numpy.KernelSolve.


Standard imports:

import numpy as np
import time
import matplotlib.pyplot as plt

from pykeops.numpy import KernelSolve
from pykeops.numpy.utils import IsGpuAvailable

Define our dataset:

N  = 5000 if IsGpuAvailable() else 500  # Number of points
D  = 2      # Dimension of the ambient space
Dv = 2      # Dimension of the vectors (= number of linear problems to solve)
sigma = .1  # Radius of our RBF kernel

dtype = 'float32'
x = np.random.rand(N, D).astype(dtype)
b = np.random.rand(N, Dv).astype(dtype)
g = np.array([ .5 / sigma**2]).astype(dtype) # Parameter of the Gaussian RBF kernel

KeOps kernel

Define a Gaussian RBF kernel:

formula = 'Exp(- g * SqDist(x,y)) * b'
aliases = ['x = Vi(' + str(D) + ')',   # First arg:  i-variable of size D
           'y = Vj(' + str(D) + ')',   # Second arg: j-variable of size D
           'b = Vj(' + str(Dv) + ')',  # Third arg:  j-variable of size Dv
           'g = Pm(1)']                # Fourth arg: scalar parameter

Define the inverse kernel operation, with a ridge regularization alpha:

alpha = 0.01
Kinv = KernelSolve(formula, aliases, "b", axis=1, dtype=dtype)


This operator uses a conjugate gradient solver and assumes that formula defines a symmetric, positive and definite linear reduction with respect to the alias "b" specified trough the third argument.

Apply our solver on arbitrary point clouds:

#Warmup of gpu
Kinv(x, x, b, g, alpha=alpha)

print("Solving a Gaussian linear system, with {} points in dimension {}.".format(N,D))
start = time.time()
c = Kinv(x, x, b, g, alpha=alpha)
end = time.time()
print('Timing (KeOps implementation):', round(end - start, 5), 's')


Solving a Gaussian linear system, with 5000 points in dimension 2.
Timing (KeOps implementation): 0.40102 s

Compare with a straightforward Numpy implementation:

start = time.time()
K_xx = alpha * np.eye(N) + np.exp( - g * np.sum( (x[:,None,:] - x[None,:,:]) **2, axis=2) )
c_np = np.linalg.solve( K_xx, b)
end = time.time()
print('Timing (Numpy implementation):', round(end - start, 5), 's')
print("Relative error = ", np.linalg.norm(c - c_np) / np.linalg.norm(c_np))

# Plot the results next to each other:
for i in range(Dv):
    plt.subplot(Dv, 1, i+1)
    plt.plot(   c[:40,i],  '-', label='KeOps')
    plt.plot(c_np[:40,i], '--', label='NumPy')
    plt.legend(loc='lower right')
plt.tight_layout() ;


Timing (Numpy implementation): 5.14924 s
Relative error =  8.327648884523687e-05
/home/bcharlier/keops/pykeops/examples/numpy/ UserWarning: Matplotlib is currently using agg, which is a non-GUI backend, so cannot show the figure.
  plt.tight_layout() ;

Total running time of the script: ( 0 minutes 17.180 seconds)

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