# Solving positive definite linear systems¶

This benchmark compares the performances of KeOps versus Numpy and Pytorch on a inverse matrix operation. It uses the functions torch.KernelSolve (see also here) and numpy.KernelSolve (see also here).

In a nutshell, given $$x \in\mathbb R^{N\times D}$$ and $$b \in \mathbb R^{N\times D_v}$$, we compute $$a \in \mathbb R^{N\times D_v}$$ so that

$b = (\alpha\operatorname{Id} + K_{x,x}) a \quad \Leftrightarrow \quad a = (\alpha\operatorname{Id}+ K_{x,x})^{-1} b$

where $$K_{x,x} = \Big[\exp(-\|x_i -x_j\|^2 / \sigma^2)\Big]_{i,j=1}^N$$. The method is based on a conjugate gradient scheme. The benchmark tests various values of $$N \in [10, \cdots,10^6]$$.

## Setup¶

Standard imports:

import importlib
import os
import time

import numpy as np
import torch
from matplotlib import pyplot as plt

from scipy.sparse import diags
from scipy.sparse.linalg import aslinearoperator, cg
from scipy.sparse.linalg.interface import IdentityOperator

from pykeops.numpy import KernelSolve as KernelSolve_np, LazyTensor
from pykeops.torch import KernelSolve
from pykeops.torch.utils import squared_distances

use_cuda = torch.cuda.is_available()


Benchmark specifications:

D  = 3  # Let's do this in 3D
Dv = 1  # Dimension of the vectors (= number of linear problems to solve)
MAXTIME = 10 if use_cuda else 1   # Max number of seconds before we break the loop
REDTIME = 5  if use_cuda else .2  # Decrease the number of runs if computations take longer than 2s...

# Number of samples that we'll loop upon
NS = [10, 20, 50,
100, 200, 500,
1000, 2000, 5000,
10000, 20000, 50000,
100000, 200000, 500000,
1000000
]


Create some random input data:

def generate_samples(N, device, lang):
"""Create point clouds sampled non-uniformly on a sphere of diameter 1."""
if lang == 'torch':
if device == 'cuda':
torch.cuda.manual_seed_all(1234)
else:
torch.manual_seed(1234)

x = torch.rand(N, D, device=device)
b = torch.randn(N, Dv, device=device)
gamma = torch.ones(1, device=device) * .5 / .01 ** 2  # kernel bandwidth
alpha = torch.ones(1, device=device) * 0.8  # regularization
else:
np.random.seed(1234)

x  = np.random.rand(N, D).astype('float32')
b  = np.random.randn(N, Dv).astype('float32')
gamma = (np.ones(1) * 1 / .01 ** 2).astype('float32')   # kernel bandwidth
alpha = (np.ones(1) * 0.8).astype('float32')  # regularization

return x, b, gamma, alpha


## KeOps kernel¶

Define a Gaussian RBF kernel:

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


Note

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

Define the Kernel solver, with a ridge regularization alpha:

def Kinv_keops(x, b, gamma, alpha):
Kinv = KernelSolve(formula, aliases, "a", axis=1)
res = Kinv(x, x, b, gamma, alpha=alpha)
return res

def Kinv_keops_numpy(x, b, gamma, alpha):
Kinv = KernelSolve_np(formula, aliases, "a", axis=1, dtype='float32')
res = Kinv(x, x, b, gamma, alpha=alpha)
return res

def Kinv_scipy(x, b, gamma, alpha):
x_i, y_j = LazyTensor( gamma * x[:, None, :]), LazyTensor( gamma * x[None, :, :])
K_ij = (- ((x_i - y_j) ** 2).sum(2)).exp()
A = aslinearoperator(diags(alpha * np.ones(x.shape))) +  aslinearoperator(K_ij)
A.dtype = np.dtype('float32')
res = cg(A, b)
return res


Define the same Kernel solver, using a tensorized implementation:

def Kinv_pytorch(x, b, gamma, alpha):
K_xx = alpha * torch.eye(x.shape, device=x.get_device()) + torch.exp( - squared_distances(x, x) * gamma)
res = torch.solve(b, K_xx)
return res

def Kinv_numpy(x, b, gamma, alpha):
K_xx = alpha * np.eye(x.shape) + np.exp( - gamma * np.sum( (x[:,None,:] - x[None,:,:]) **2, axis=2) )
res = np.linalg.solve(K_xx, b)
return res


## Benchmarking loops¶

def benchmark(Routine, dev, N, loops=10, lang='torch') :
"""Times a routine on an N-by-N problem."""

device = torch.device(dev)
x, b, gamma, alpha = generate_samples(N, device, lang)

# We simply benchmark a kernel inversion
code = "a = Routine(x, b, gamma, alpha)"
exec( code, locals() ) # Warmup run, to compile and load everything
if use_cuda: torch.cuda.synchronize()

t_0 = time.perf_counter()  # Actual benchmark --------------------
for i in range(loops):
exec( code, locals() )
if use_cuda: torch.cuda.synchronize()
elapsed = time.perf_counter() - t_0  # ---------------------------

print("{:3} NxN kernel inversion, with N ={:7}: {:3}x{:3.6f}s".format(loops, N, loops, elapsed / loops))
return elapsed / loops

def bench_config(Routine, backend, dev, l) :
"""Times a routine for an increasing number of samples."""

print("Backend : {}, Device : {} -------------".format(backend, dev))

times = []
not_recorded_times = []
try :
Nloops = [100, 10, 1]
nloops = Nloops.pop(0)
for n in NS :
elapsed = benchmark(Routine, dev, n, loops=nloops, lang=l)

times.append( elapsed )
if (nloops * elapsed > MAXTIME) or (nloops * elapsed > REDTIME/nloops and len(Nloops) > 0):
nloops = Nloops.pop(0)

except RuntimeError:
print("**\nMemory overflow !")
not_recorded_times = (len(NS)-len(times)) * [np.nan]
except IndexError:
print("**\nToo slow !")
not_recorded_times = (len(NS)-len(times)) * [np.Infinity]

return times + not_recorded_times

def full_bench(title, routines) :
"""Benchmarks a collection of routines."""

backends = [ backend for (_, backend, _) in routines ]

print("Benchmarking : {} ===============================".format(title))

lines  = [ NS ]
for routine, backend, lang in routines :
lines.append(bench_config(routine, backend, "cuda" if use_cuda else "cpu", lang) )

benches = np.array(lines).T

# Creates a pyplot figure:
plt.figure(figsize=(12,8))
linestyles = ["o-", "s-", "^-", "x-", "<-"]
for i, backend in enumerate(backends):
plt.plot( benches[:,0], benches[:,i+1], linestyles[i],
linewidth=2, label='backend = "{}"'.format(backend) )

for (j, val) in enumerate( benches[:,i+1] ):
if np.isnan(val) and j > 0:
x, y = benches[j-1,0], benches[j-1,i+1]
plt.annotate('Memory overflow!',
xy=(x, 1.05*y),
horizontalalignment='center',
verticalalignment='bottom')
break
elif np.isinf(val) and j > 0:
x, y = benches[j-1,0], benches[j-1,i+1]
plt.annotate('Too slow!',
xy=(x, 1.05*y),
horizontalalignment='center',
verticalalignment='bottom')
break

plt.title('Runtimes for {} in dimension {}'.format(title, D))
plt.xlabel('Number of samples')
plt.ylabel('Seconds')
plt.yscale('log') ; plt.xscale('log')
plt.legend(loc='upper left')
plt.grid(True, which="major", linestyle="-")
plt.grid(True, which="minor", linestyle="dotted")
plt.tight_layout()

# Save as a .csv to put a nice Tikz figure in the papers:
header = "Npoints " + " ".join(backends)
os.makedirs("output", exist_ok=True)
np.savetxt("output/benchmark_kernelsolve.csv", benches,


## Run the benchmark¶

routines = [(Kinv_numpy, "NumPy", "numpy"),
(Kinv_pytorch, "PyTorch", "torch"),
(Kinv_keops_numpy, "NumPy + KeOps", "numpy"),
(Kinv_keops,   "PyTorch + KeOps", "torch"),
(Kinv_scipy,   "Scipy + KeOps", "numpy"),
]
full_bench( "Inverse radial kernel matrix", routines )

plt.show() Out:

Benchmarking : Inverse radial kernel matrix ===============================
Backend : NumPy, Device : cuda -------------
100 NxN kernel inversion, with N =     10: 100x0.000048s
100 NxN kernel inversion, with N =     20: 100x0.000101s
100 NxN kernel inversion, with N =     50: 100x0.000797s
10 NxN kernel inversion, with N =    100:  10x0.007036s
10 NxN kernel inversion, with N =    200:  10x0.057652s
1 NxN kernel inversion, with N =    500:   1x0.371515s
1 NxN kernel inversion, with N =   1000:   1x1.213036s
1 NxN kernel inversion, with N =   2000:   1x2.528340s
1 NxN kernel inversion, with N =   5000:   1x5.223090s
1 NxN kernel inversion, with N =  10000:   1x20.332538s
**
Too slow !
Backend : PyTorch, Device : cuda -------------
100 NxN kernel inversion, with N =     10: 100x0.003185s
10 NxN kernel inversion, with N =     20:  10x0.003181s
10 NxN kernel inversion, with N =     50:  10x0.003189s
10 NxN kernel inversion, with N =    100:  10x0.003251s
10 NxN kernel inversion, with N =    200:  10x0.005839s
10 NxN kernel inversion, with N =    500:  10x0.006834s
10 NxN kernel inversion, with N =   1000:  10x0.009535s
10 NxN kernel inversion, with N =   2000:  10x0.018448s
10 NxN kernel inversion, with N =   5000:  10x0.080955s
1 NxN kernel inversion, with N =  10000:   1x0.283361s
1 NxN kernel inversion, with N =  20000:   1x1.651585s
**
Memory overflow !
Backend : NumPy + KeOps, Device : cuda -------------
100 NxN kernel inversion, with N =     10: 100x0.000190s
100 NxN kernel inversion, with N =     20: 100x0.000182s
100 NxN kernel inversion, with N =     50: 100x0.000184s
100 NxN kernel inversion, with N =    100: 100x0.000280s
100 NxN kernel inversion, with N =    200: 100x0.000379s
100 NxN kernel inversion, with N =    500: 100x0.000720s
10 NxN kernel inversion, with N =   1000:  10x0.000842s
10 NxN kernel inversion, with N =   2000:  10x0.001212s
10 NxN kernel inversion, with N =   5000:  10x0.002684s
10 NxN kernel inversion, with N =  10000:  10x0.004884s
10 NxN kernel inversion, with N =  20000:  10x0.014212s
10 NxN kernel inversion, with N =  50000:  10x0.063108s
1 NxN kernel inversion, with N = 100000:   1x0.250456s
1 NxN kernel inversion, with N = 200000:   1x1.141655s
1 NxN kernel inversion, with N = 500000:   1x8.048549s
1 NxN kernel inversion, with N =1000000:   1x40.217673s
**
Too slow !
Backend : PyTorch + KeOps, Device : cuda -------------
100 NxN kernel inversion, with N =     10: 100x0.000546s
10 NxN kernel inversion, with N =     20:  10x0.000535s
10 NxN kernel inversion, with N =     50:  10x0.001110s
10 NxN kernel inversion, with N =    100:  10x0.001112s
10 NxN kernel inversion, with N =    200:  10x0.001698s
10 NxN kernel inversion, with N =    500:  10x0.002028s
10 NxN kernel inversion, with N =   1000:  10x0.002119s
10 NxN kernel inversion, with N =   2000:  10x0.003288s
10 NxN kernel inversion, with N =   5000:  10x0.004837s
10 NxN kernel inversion, with N =  10000:  10x0.007439s
10 NxN kernel inversion, with N =  20000:  10x0.016130s
10 NxN kernel inversion, with N =  50000:  10x0.072380s
1 NxN kernel inversion, with N = 100000:   1x0.298504s
1 NxN kernel inversion, with N = 200000:   1x1.415842s
1 NxN kernel inversion, with N = 500000:   1x11.282376s
**
Too slow !
Backend : Scipy + KeOps, Device : cuda -------------
100 NxN kernel inversion, with N =     10: 100x0.000917s
10 NxN kernel inversion, with N =     20:  10x0.000902s
10 NxN kernel inversion, with N =     50:  10x0.000906s
10 NxN kernel inversion, with N =    100:  10x0.000914s
10 NxN kernel inversion, with N =    200:  10x0.000948s
10 NxN kernel inversion, with N =    500:  10x0.001006s
10 NxN kernel inversion, with N =   1000:  10x0.001046s
10 NxN kernel inversion, with N =   2000:  10x0.001141s
10 NxN kernel inversion, with N =   5000:  10x0.001413s
10 NxN kernel inversion, with N =  10000:  10x0.001921s
10 NxN kernel inversion, with N =  20000:  10x0.004865s
10 NxN kernel inversion, with N =  50000:  10x0.010613s
10 NxN kernel inversion, with N = 100000:  10x0.034312s
10 NxN kernel inversion, with N = 200000:  10x0.242128s
1 NxN kernel inversion, with N = 500000:   1x1.772097s
1 NxN kernel inversion, with N =1000000:   1x7.001583s


Total running time of the script: ( 3 minutes 36.116 seconds)

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