问题
I am facing an issue of accuracy about my code which performs a number (128, 256, 512) of 4x4 matrix inversions. When I use the original version, i.e the numpy function np.linalg.inv or np.linalg.pinv, everything works fine.
Unfortunately, with the CUDA code below, I get nan and inf values into inverted matrix.
To be more explicit, I take this matrix to invert :
2.120771107884677649e+09 0.000000000000000000e+00 0.000000000000000000e+00 0.000000000000000000e+00
0.000000000000000000e+00 3.557266600921528288e+27 3.557266600921528041e+07 3.557266600921528320e+17
0.000000000000000000e+00 3.557266600921528041e+07 3.557266600921528288e+27 3.557266600921528041e+07
0.000000000000000000e+00 3.557266600921528320e+17 3.557266600921528041e+07 1.778633300460764144e+27
If I use classical numpy "inv", I get for the following inverted 3x3 matrix :
4.715266047722758306e-10 0.000000000000000000e+00 0.000000000000000000e+00 0.000000000000000000e+00
0.000000000000000000e+00 2.811147187396482366e-28 -2.811147186834252285e-48 -5.622294374792964645e-38
0.000000000000000000e+00 -2.811147186834252285e-48 2.811147187396482366e-28 -5.622294374230735768e-48
0.000000000000000000e+00 -5.622294374792964645e-38 -5.622294374230735768e-48 5.622294374792964732e-28
To check the validity of this inverse matrix, I have multiplied it by original matrix and the result is the identity matrix.
But with CUDA GPU inversion, I get after the inversion this matrix :
0.000000000000000000e+00 0.000000000000000000e+00 0.000000000000000000e+00 0.000000000000000000e+00
0.000000000000000000e+00 0.000000000000000000e+00 0.000000000000000000e+00 0.000000000000000000e+00
-inf -inf -9.373764907941219970e-01 -inf
inf nan -inf nan
So, I woul like to increase the precision into my CUDA kernel or python code to avoid these nanand inf values.
Here is the CUDA kernel code and calling part of my main code (I have commented the classical method with numpy inv function :
# Create arrayFullCross_vec array
arrayFullCross_vec = np.zeros((dimBlocks,dimBlocks,integ_prec,integ_prec))
# Create arrayFullCross_vec array
invCrossMatrix_gpu = np.zeros((dimBlocks*dimBlocks*integ_prec**2))
# Create arrayFullCross_vec array
invCrossMatrix = np.zeros((dimBlocks,dimBlocks,integ_prec,integ_prec))
# Build observables covariance matrix
arrayFullCross_vec = buildObsCovarianceMatrix4_vec(k_ref, mu_ref, ir)
"""
# Compute integrand from covariance matrix
for r_p in range(integ_prec):
for s_p in range(integ_prec):
# original version (without GPU)
invCrossMatrix[:,:,r_p,s_p] = np.linalg.inv(arrayFullCross_vec[:,:,r_p,s_p])
"""
# GPU version
invCrossMatrix_gpu = gpuinv4x4(arrayFullCross_vec.flatten(),integ_prec**2)
invCrossMatrix = invCrossMatrix_gpu.reshape(dimBlocks,dimBlocks,integ_prec,integ_prec)
"""
and here the CUDA kernel code and gpuinv4x4 function :
kernel = SourceModule("""
__device__ unsigned getoff(unsigned &off){
unsigned ret = off & 0x0F;
off = off >> 4;
return ret;
}
const int block_size = 256;
const unsigned tmsk = 0xFFFFFFFF;
// in-place is acceptable i.e. out == in)
// T = double or double only
typedef double T;
__global__ void inv4x4(const T * __restrict__ in, T * __restrict__ out, const size_t n, const unsigned * __restrict__ pat){
__shared__ T si[block_size];
size_t idx = threadIdx.x+blockDim.x*blockIdx.x;
if (idx < n*16){
si[threadIdx.x] = in[idx];
unsigned lane = threadIdx.x & 15;
unsigned sibase = threadIdx.x & 0x03F0;
__syncwarp();
unsigned off = pat[lane];
T a,b;
a = si[sibase + getoff(off)];
a *= si[sibase + getoff(off)];
a *= si[sibase + getoff(off)];
if (!getoff(off)) a = -a;
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
off = pat[lane+16];
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
off = pat[lane+32];
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
T det = si[sibase + (lane>>2)]*a;
det += __shfl_down_sync(tmsk, det, 4, 16); // first add
det += __shfl_down_sync(tmsk, det, 8, 16); // second add
det = __shfl_sync(tmsk, det, 0, 16); // broadcast
out[idx] = a / det;
}
}
""")
# python function for inverting 4x4 matrices
# n should be an even number
def gpuinv4x4(inp, n):
# internal constants not to be modified
hpat = ( 0x0EB51FA5, 0x1EB10FA1, 0x0E711F61, 0x1A710B61, 0x1EB40FA4, 0x0EB01FA0, 0x1E700F60, 0x0A701B60, 0x0DB41F94, 0x1DB00F90, 0x0D701F50, 0x19700B50, 0x1DA40E94, 0x0DA01E90, 0x1D600E50, 0x09601A50, 0x1E790F69, 0x0E391F29, 0x1E350F25, 0x0A351B25, 0x0E781F68, 0x1E380F28, 0x0E341F24, 0x1A340B24, 0x1D780F58, 0x0D381F18, 0x1D340F14, 0x09341B14, 0x0D681E58, 0x1D280E18, 0x0D241E14, 0x19240A14, 0x0A7D1B6D, 0x1A3D0B2D, 0x063D172D, 0x16390729, 0x1A7C0B6C, 0x0A3C1B2C, 0x163C072C, 0x06381728, 0x097C1B5C, 0x193C0B1C, 0x053C171C, 0x15380718, 0x196C0A5C, 0x092C1A1C, 0x152C061C, 0x05281618)
# Convert parameters into numpy array
# float32
"""
inpd = np.array(inp, dtype=np.float32)
hpatd = np.array(hpat, dtype=np.uint32)
output = np.empty((n*16), dtype= np.float32)
"""
# float64
"""
inpd = np.array(inp, dtype=np.float64)
hpatd = np.array(hpat, dtype=np.uint32)
output = np.empty((n*16), dtype= np.float64)
"""
# float128
inpd = np.array(inp, dtype=np.float128)
hpatd = np.array(hpat, dtype=np.uint32)
output = np.empty((n*16), dtype= np.float128)
# Get kernel function
matinv4x4 = kernel.get_function("inv4x4")
# Define block, grid and compute
blockDim = (256,1,1) # do not change
gridDim = ((n/16)+1,1,1)
# Kernel function
matinv4x4 (
cuda.In(inpd), cuda.Out(output), np.uint64(n), cuda.In(hpatd),
block=blockDim, grid=gridDim)
return output
As you can see, I tried to increase accuracy of inverting operation by replacing np.float32 by np.float64 or np.float128 but problem remains.
I have also replaced typedef float T; by typedef double T;but without success.
Anyone could help me to perform the right inversion of these matrices and mostly avoid the 'nan' and 'inf' values ? I think this is a real issue of precision but I can't find how to circumvent this problem.
回答1:
This question has previous related questions here and (to a lesser extent) here. It's unclear to me why the title of the question refers to 3x3, the bolded text in the question refers to to 3x3, but the presented problem is a 4x4 matrix inverse (and as stated to OP previously, this code can only be used to invert 4x4 matrices). I will proceed under the assumption that the example case is the desired case.
According to my testing, the only thing that is necessary is to take the previous answer code and convert it to use double (or, in pycuda, float64) rather than float (or in pycuda, float32). I think this should be evident because the example matrix values exceed the range of a float32 type. Here is a worked example:
$ cat t10.py
import numpy as np
# import matplotlib.pyplot as plt
import pycuda.driver as cuda
from pycuda.compiler import SourceModule
import pycuda.autoinit
# kernel
kernel = SourceModule("""
__device__ unsigned getoff(unsigned &off){
unsigned ret = off & 0x0F;
off = off >> 4;
return ret;
}
const int block_size = 256;
const unsigned tmsk = 0xFFFFFFFF;
// in-place is acceptable i.e. out == in)
// T = float or double only
typedef double T; // *** change this typedef to convert between float and double
__global__ void inv4x4(const T * __restrict__ in, T * __restrict__ out, const size_t n, const unsigned * __restrict__ pat){
__shared__ T si[block_size];
size_t idx = threadIdx.x+blockDim.x*blockIdx.x;
if (idx < n*16){
si[threadIdx.x] = in[idx];
unsigned lane = threadIdx.x & 15;
unsigned sibase = threadIdx.x & 0x03F0;
__syncwarp();
unsigned off = pat[lane];
T a,b;
a = si[sibase + getoff(off)];
a *= si[sibase + getoff(off)];
a *= si[sibase + getoff(off)];
if (!getoff(off)) a = -a;
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
off = pat[lane+16];
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
off = pat[lane+32];
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
b = si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
b *= si[sibase + getoff(off)];
if (getoff(off)) a += b;
else a -=b;
T det = si[sibase + (lane>>2)]*a;
det += __shfl_down_sync(tmsk, det, 4, 16); // first add
det += __shfl_down_sync(tmsk, det, 8, 16); // second add
det = __shfl_sync(tmsk, det, 0, 16); // broadcast
out[idx] = a / det;
}
}
""")
# host code
def gpuinv4x4(inp, n):
# internal constants not to be modified
hpat = ( 0x0EB51FA5, 0x1EB10FA1, 0x0E711F61, 0x1A710B61, 0x1EB40FA4, 0x0EB01FA0, 0x1E700F60, 0x0A701B60, 0x0DB41F94, 0x1DB00F90, 0x0D701F50, 0x19700B50, 0x1DA40E94, 0x0DA01E90, 0x1D600E50, 0x09601A50, 0x1E790F69, 0x0E391F29, 0x1E350F25, 0x0A351B25, 0x0E781F68, 0x1E380F28, 0x0E341F24, 0x1A340B24, 0x1D780F58, 0x0D381F18, 0x1D340F14, 0x09341B14, 0x0D681E58, 0x1D280E18, 0x0D241E14, 0x19240A14, 0x0A7D1B6D, 0x1A3D0B2D, 0x063D172D, 0x16390729, 0x1A7C0B6C, 0x0A3C1B2C, 0x163C072C, 0x06381728, 0x097C1B5C, 0x193C0B1C, 0x053C171C, 0x15380718, 0x196C0A5C, 0x092C1A1C, 0x152C061C, 0x05281618)
# Convert parameters into numpy array
# *** change next line between float32 and float64 to match float or double
inpd = np.array(inp, dtype=np.float64)
hpatd = np.array(hpat, dtype=np.uint32)
# *** change next line between float32 and float64 to match float or double
output = np.empty((n*16), dtype= np.float64)
# Get kernel function
matinv4x4 = kernel.get_function("inv4x4")
# Define block, grid and compute
blockDim = (256,1,1) # do not change
gridDim = ((n/16)+1,1,1)
# Kernel function
matinv4x4 (
cuda.In(inpd), cuda.Out(output), np.uint64(n), cuda.In(hpatd),
block=blockDim, grid=gridDim)
return output
inp = (1.0, 1.0, 1.0, 0.0, 0.0, 3.0, 1.0, 2.0, 2.0, 3.0, 1.0, 0.0, 1.0, 0.0, 2.0, 1.0, 2.120771107884677649e+09, 0.0, 0.0, 0.0, 0.0, 3.557266600921528288e+27, 3.557266600921528041e+07, 3.557266600921528320e+17, 0.0, 3.557266600921528041e+07, 3.557266600921528288e+27, 3.557266600921528041e+07, 0.0, 3.557266600921528320e+17, 3.557266600921528041e+07, 1.778633300460764144e+27)
n = 2
result = gpuinv4x4(inp, n)
print(result)
$ python t10.py
[ -3.00000000e+00 -5.00000000e-01 1.50000000e+00 1.00000000e+00
1.00000000e+00 2.50000000e-01 -2.50000000e-01 -5.00000000e-01
3.00000000e+00 2.50000000e-01 -1.25000000e+00 -5.00000000e-01
-3.00000000e+00 -0.00000000e+00 1.00000000e+00 1.00000000e+00
4.71526605e-10 0.00000000e+00 0.00000000e+00 0.00000000e+00
0.00000000e+00 2.81114719e-28 -2.81114719e-48 -5.62229437e-38
0.00000000e+00 -2.81114719e-48 2.81114719e-28 -5.62229437e-48
0.00000000e+00 -5.62229437e-38 -5.62229437e-48 5.62229437e-28]
$
Apart from updating the input matrix to use the one supplied in this question, note that only 3 lines of code were changed from the previous answer to convert from 32-bit floating point to 64-bit floating point. Each of those 3 lines are marked by a comment that contains triple-asterisk (***) in it.
Also, if you are concerned about accuracy beyond the first 9 or so digits displayed here, I haven't investigated that. It may be that this code is not numerically suitable for every case or need.
As an aside, the code in the question also shows a float128 type in the python section. There is no native CUDA type that corresponds to that.
来源：https://stackoverflow.com/questions/55386164/how-to-perform-pycuda-4x4-matrix-inversion-with-same-accuracy-than-numpy-linalg