Let’s compute the (3000,3) tensor \(c\) whose entries \(c_i^u\) are given by:
\(x\) is a (3000,3) tensor, with entries \(x_i^u\).
\(y\) is a (5000,3) tensor, with entries \(y_j^u\).
\(a\) is a (5000,1) tensor, with entries \(a_j\).
\(p\) is a scalar, encoded as a vector of size (1,).
import matplotlib.pyplot as plt import numpy as np from pykeops.numpy import Genred
Declare random inputs:
M = 3000 N = 5000 dtype = 'float32' # May be 'float32' or 'float64' x = np.random.randn(M,3).astype(dtype) y = np.random.randn(N,3).astype(dtype) a = np.random.randn(N,1).astype(dtype) p = np.random.randn(1).astype(dtype)
Define a custom formula¶
formula = 'Square(p-a)*Exp(x+y)' variables = ['x = Vi(3)', # First arg : i-variable, of size 3 'y = Vj(3)', # Second arg : j-variable, of size 3 'a = Vj(1)', # Third arg : j-variable, of size 1 (scalar) 'p = Pm(1)'] # Fourth arg : Parameter, of size 1 (scalar)
Our sum reduction is performed over the index \(j\),
i.e. on the axis
1 of the kernel matrix.
The output c is an \(x\)-variable indexed by \(i\).
my_routine = Genred(formula, variables, reduction_op='Sum', axis=1, dtype=dtype) c = my_routine(x, y, a, p, backend='auto')
The equivalent code in NumPy:
c_np = ((p - a.T)[:,np.newaxis] **2 * np.exp(x.T[:,:,np.newaxis] + y.T[:,np.newaxis,:]) ).sum(2).T # Plot the results next to each other: for i in range(3): plt.subplot(3, 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() ; plt.show()
/home/bcharlier/keops/pykeops/examples/numpy/plot_generic_syntax_numpy.py:74: UserWarning: Matplotlib is currently using agg, which is a non-GUI backend, so cannot show the figure. plt.tight_layout() ; plt.show()
Compute the gradient¶
Now, let’s compute the gradient of \(c\) with respect to \(y\). Since \(c\) is not scalar valued, its “gradient” \(\partial c\) should be understood as the adjoint of the differential operator, i.e. as the linear operator that:
takes as input a new tensor \(e\) with the shape of \(c\)
outputs a tensor \(g\) with the shape of \(y\)
such that for all variation \(\delta y\) of \(y\) we have:
Backpropagation is all about computing the tensor \(g=\partial c . e\) efficiently, for arbitrary values of \(e\):
# Declare a new tensor of shape (M,3) used as the input of the gradient operator. # It can be understood as a "gradient with respect to the output c" # and is thus called "grad_output" in the documentation of PyTorch. e = np.random.randn(M, 3).astype(dtype)
KeOps provides an autodiff engine for formulas. Unfortunately though, as NumPy does not provide any support for backpropagation, we need to specify some informations by hand and add the gradient operator around the formula:
Grad(formula , variable_to_differentiate, input_of_the_gradient)
formula_grad = 'Grad(' + formula + ', y, e)' # This new formula makes use of a new variable (the input tensor e) variables_grad = variables + ['e = Vi(3)'] # Fifth arg: an i-variable of size 3... Just like "c"! # The summation is done with respect to the 'i' index (axis=0) in order to get a 'j'-variable my_grad = Genred(formula_grad, variables_grad, reduction_op='Sum', axis=0, dtype=dtype) g = my_grad(x, y, a, p, e)
To generate an equivalent code in numpy, we must compute explicitly the adjoint of the differential (a.k.a. the derivative). To do so, let see \(c^i_u\) as a function of \(y_j\):
and implement the formula:
g_np = ((p - a.T)[:, np.newaxis, :] **2 * np.exp(x.T[:, :, np.newaxis] \ + y.T[:, np.newaxis, :]) * e.T[:, :, np.newaxis]).sum(1).T # Plot the results next to each other: for i in range(3): plt.subplot(3, 1, i+1) plt.plot(g[:40,i], '-', label='KeOps') plt.plot(g_np[:40,i], '--', label='NumPy') plt.legend(loc='lower right') plt.tight_layout() ; plt.show()
/home/bcharlier/keops/pykeops/examples/numpy/plot_generic_syntax_numpy.py:134: UserWarning: Matplotlib is currently using agg, which is a non-GUI backend, so cannot show the figure. plt.tight_layout() ; plt.show()
Total running time of the script: ( 0 minutes 1.658 seconds)