Pycuda - Adapt existing code and Kernel code to perform a high number of 3x3 matrix inversion

-1

Following a previous question ( Pycuda - Best way to perform a high number of small matrix inversion - CUBLAS or MAGMA ), considering the inversion of 4x4 matrix, I would like to do the same but with 3x3 matrix. As @Robert Crovella said, this change implies a complete rewrite.

Given the code shown below, I tried to test some things like putting zeros instead of values but this method doesn't seem to work.

Here is the code working for a high number of 4x4 matrix inversion

$ cat t10.py
import numpy as np
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 float 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
    inpd = np.array(inp, dtype=np.float32)
    hpatd = np.array(hpat, dtype=np.uint32)
    output = np.empty((n*16), dtype= np.float32)
    # 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
#example/test case
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, 1.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 1.0)
n = 2
result = gpuinv4x4(inp, n)
print(result.reshape(2,4,4))
$ python t10.py

[[-3.   -0.5   1.5   1.  ]
  [ 1.    0.25 -0.25 -0.5 ]
  [ 3.    0.25 -1.25 -0.5 ]
  [-3.   -0.    1.    1.  ]]

[[ 1.    0.    0.    0.  ]
  [ 0.    1.    0.    0.  ]
  [ 0.    0.    1.    0.  ]
  [ 0.    0.    0.    1.  ]]

I expect the same behavior, except that I am not longer working with 4x4 matrix but with 3x3 matrix.

If someone could help me to adapt this code above to work with 3x3 matrix inversion, this would be nice.

UPDATE 1 : Here the modifications that I made.

I have modified the dimension and used direct formula from the link given by @Robert Crovella (https://ardoris.wordpress.com/2008/07/18/general-formula-for-the-inverse-of-a-3x3-matrix/) . Below the code modified :

import numpy as np
import pycuda.driver as cuda
from pycuda.compiler import SourceModule
import pycuda.autoinit


 # kernel of 3x3 inversion
kernel_3x3 = SourceModule("""

    // in-place is acceptable i.e. out == in)
    // T = float or double only

typedef float T;
__global__ void inv3x3(const T * __restrict__ in, T * __restrict__ out, const size_t n, const unsigned * __restrict__ pat){

  size_t ix = threadIdx.x;
  size_t idx = ix + blockDim.x*blockIdx.x;
  if (ix < n*9){
    T det = in[0+idx]*(in[4+idx]*in[8+idx]-in[7+idx]*in[5+idx]) - in[1+idx]*(in[3+idx]*in[8+idx]-in[6+idx]*in[5+idx]) + in[2+idx]*(in[3+idx]*in[7+idx]-in[6+idx]*in[4+idx]);
    out[0+idx] = (in[4+idx]*in[8+idx]-in[7+idx]*in[5+idx])/det;
    out[1+idx] = (in[2+idx]*in[7+idx]-in[1+idx]*in[8+idx])/det;
    out[2+idx] = (in[1+idx]*in[5+idx]-in[2+idx]*in[4+idx])/det;
    out[3+idx] = (in[6+idx]*in[5+idx]-in[3+idx]*in[8+idx])/det;
    out[4+idx] = (in[0+idx]*in[8+idx]-in[2+idx]*in[6+idx])/det;
    out[5+idx] = (in[2+idx]*in[3+idx]-in[0+idx]*in[5+idx])/det;
    out[6+idx] = (in[3+idx]*in[7+idx]-in[4+idx]*in[6+idx])/det;
    out[7+idx] = (in[1+idx]*in[6+idx]-in[0+idx]*in[7+idx])/det;
    out[8+idx] = (in[0+idx]*in[4+idx]-in[1+idx]*in[3+idx])/det;
    __syncwarp(); 
  }
}

""")

def gpuinv3x3 (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
  inpd = np.array(inp, dtype=np.float32)
  hpatd = np.array(hpat, dtype=np.uint32)
  output = np.empty((n*9), dtype= np.float32)
  # Get kernel function
  matinv3x3 = kernel_3x3.get_function("inv3x3")
  # Define block, grid and compute
  blockDim = (81,1,1) # do not change
  gridDim = ((n/9)+1,1,1)
  # Kernel function
  matinv3x3 (
      cuda.In(inpd), cuda.Out(output), np.uint64(n), cuda.In(hpatd),
      block=blockDim, grid=gridDim)
  return output
#example/test case
inp = (1.0, 1.0, 1.0, 0.0, 0.0, 3.0, 1.0, 2.0, 2.0, 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0)
n = 2
result = gpuinv3x3(inp, n)
print(result.reshape(2,3,3))

The first matrix is correctly inverted but not the second one (the identity matrix which has identity matrix as inverse) :

[[[ 2.         -0.         -1.        ]
  [-1.         -0.33333334  1.        ]
  [-0.          0.33333334 -0.        ]]

 [[ 1.         -0.5        -0.        ]
  [       -inf  1.         -1.        ]
  [        nan         nan  1.        ]]]

So, this issue doesn't seem to come from the kernel code but rather the batch size or something similar with dimensions of global 1D array (in my code, you can see the 2 3x3 matrices formatted as 1D array of 18 elements ( inp = (1.0, 1.0, 1.0, 0.0, 0.0, 3.0, 1.0, 2.0, 2.0, 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0) ).

Anyone could see what's wrong in this code ? Especially, the issue of bad inversion on the second matrix . Just a last point, an odd size of group doesn't imply issues for GPU processing ?

python
matrix
cuda
matrix-inverse
asked on Stack Overflow Mar 26, 2019 by youpilat13 • edited Mar 29, 2019 by youpilat13

1 Answer

1

This answer will closely follow my answer on the 4x4 invert question, both in terms of answer layout and calculation method/kernel design. The formulas are described here.

First, as before, we will show a CUDA C++ version with comparison to cublas:

$ cat t432.cu
#include <iostream>
#include <cublas_v2.h>
#include <cstdlib>
// 3x3 matrix inversion
// https://stackoverflow.com/questions/1148309/inverting-a-4x4-matrix
// https://ardoris.wordpress.com/2008/07/18/general-formula-for-the-inverse-of-a-3x3-matrix/
// 9 threads per matrix to invert
// 32 matrices per 288 thread block

const unsigned block_size = 288;
typedef double mt;

#define cudaCheckErrors(msg) \
    do { \
        cudaError_t __err = cudaGetLastError(); \
        if (__err != cudaSuccess) { \
            fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
                msg, cudaGetErrorString(__err), \
                __FILE__, __LINE__); \
            fprintf(stderr, "*** FAILED - ABORTING\n"); \
            exit(1); \
        } \
    } while (0)

#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL

long long dtime_usec(unsigned long long start){

  timeval tv;
  gettimeofday(&tv, 0);
  return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}

__device__ unsigned pat[9];
const unsigned hpat[9] = {0x07584, 0x08172, 0x04251, 0x08365, 0x06280, 0x05032, 0x06473, 0x07061, 0x03140};


__device__ unsigned getoff(unsigned &off){
  unsigned ret = off & 0x0F;
  off >>= 4;
  return ret;
}

// in-place is acceptable i.e. out == in)
// T = float or double only
template <typename T>
__global__ void inv3x3(const T * __restrict__ in, T * __restrict__ out, const size_t n){

  __shared__ T si[block_size];
  size_t idx = threadIdx.x+blockDim.x*blockIdx.x;
  T det = 1;
  if (idx < n*9)
    det = in[idx];
  unsigned sibase = (threadIdx.x / 9)*9;
  unsigned lane = threadIdx.x - sibase; // cheaper modulo
  si[threadIdx.x] = det;
  __syncthreads();
  unsigned off = pat[lane];
  T a  = si[sibase + getoff(off)];
  a   *= si[sibase + getoff(off)];
  T b  = si[sibase + getoff(off)];
  b   *= si[sibase + getoff(off)];
  a -= b;
  __syncthreads();
  if (lane == 0) si[sibase+3] = a;
  if (lane == 3) si[sibase+4] = a;
  if (lane == 6) si[sibase+5] = a;
  __syncthreads();
  det =  si[sibase]*si[sibase+3]+si[sibase+1]*si[sibase+4]+si[sibase+2]*si[sibase+5];
  if (idx < n*9)
    out[idx] = a / det;
}

size_t nr = 2048;
int main(int argc, char *argv[]){
  if (argc > 1) nr = atoi(argv[1]);


  const mt m2[] = {1.0, 1.0, 1.0, 0.0, 0.0, 3.0, 1.0, 2.0, 2.0};
  const mt i2[] = {2.0, 0.0, -1.0, -1.0, -0.33333334, 1.0, 0.0, 0.33333334,  0.0};
  const mt m1[] = {1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0};
  const mt i1[] = {1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0};

  mt *h_d, *d_d;
  h_d = (mt *)malloc(nr*9*sizeof(mt));
  cudaMalloc(&d_d, nr*9*sizeof(mt));
  cudaMemcpyToSymbol(pat, hpat, 9*sizeof(unsigned));
  for (int i = 0; i < nr/2; i++){
    memcpy(h_d+i*2*9, m1, sizeof(m1));
    memcpy(h_d+i*2*9+9, m2, sizeof(m2));}
  cudaMemcpy(d_d, h_d, nr*9*sizeof(mt), cudaMemcpyHostToDevice);
  long long t = dtime_usec(0);
  inv3x3<<<((nr*9)/block_size)+1, block_size>>>(d_d, d_d, nr);
  cudaDeviceSynchronize();
  t = dtime_usec(t);
  cudaMemcpy(h_d, d_d, nr*9*sizeof(mt), cudaMemcpyDeviceToHost);
  for (int i = 0; i < 2; i++){
    for (int j = 0; j < 9; j++) std::cout << h_d[i*9 + j] << ",";
    std::cout << std::endl;
    for (int j = 0; j < 9; j++) std::cout << ((i==0)?i1[j]:i2[j]) << ",";
    std::cout << std::endl;}
  std::cout << "kernel time: " << t << " microseconds" << std::endl;
  cudaError_t err = cudaGetLastError();
  if (err != cudaSuccess) std::cout << cudaGetErrorString(err) << std::endl;
  //cublas
  for (int i = 0; i < nr/2; i++){
    memcpy(h_d+i*2*9, m1, sizeof(m1));
    memcpy(h_d+i*2*9+9, m2, sizeof(m2));}
  cudaMemcpy(d_d, h_d, nr*9*sizeof(mt), cudaMemcpyHostToDevice);
  cublasHandle_t h;
  cublasStatus_t cs = cublasCreate(&h);
  if (cs != CUBLAS_STATUS_SUCCESS) std::cout << "cublas create error" << std::endl;
  mt **A, **Ai, *Aid, **Ap, **Aip;
  A  = (mt **)malloc(nr*sizeof(mt *));
  Ai = (mt **)malloc(nr*sizeof(mt *));
  cudaMalloc(&Aid, nr*9*sizeof(mt));
  cudaMalloc(&Ap,  nr*sizeof(mt *));
  cudaMalloc(&Aip, nr*sizeof(mt *));
  for (int i = 0; i < nr; i++) A[i]  =  d_d + 9*i;
  for (int i = 0; i < nr; i++) Ai[i] =  Aid + 9*i;
  cudaMemcpy(Ap, A, nr*sizeof(mt *), cudaMemcpyHostToDevice);
  cudaMemcpy(Aip, Ai, nr*sizeof(mt *), cudaMemcpyHostToDevice);
  int *info;
  cudaMalloc(&info, nr*sizeof(int));
  t = dtime_usec(0);
  cs = cublasDmatinvBatched(h, 3,  Ap, 3, Aip, 3, info, nr);
  if (cs != CUBLAS_STATUS_SUCCESS) std::cout << "cublas matinv error" << std::endl;
  cudaDeviceSynchronize();
  t = dtime_usec(t);
  cudaMemcpy(h_d, Aid, nr*9*sizeof(mt), cudaMemcpyDeviceToHost);
  for (int i = 0; i < 2; i++){
    for (int j = 0; j < 9; j++) std::cout << h_d[i*9 + j] << ",";
    std::cout << std::endl;
    for (int j = 0; j < 9; j++) std::cout << ((i==0)?i1[j]:i2[j]) << ",";
    std::cout << std::endl;}
  std::cout << "cublas time: " << t << " microseconds" << std::endl;
  err = cudaGetLastError();
  if (err != cudaSuccess) std::cout << cudaGetErrorString(err) << std::endl;
  return 0;
}
$ nvcc -o t432 t432.cu -lcublas
$ ./t432
1,0,0,0,1,0,0,0,1,
1,0,0,0,1,0,0,0,1,
2,-0,-1,-1,-0.333333,1,-0,0.333333,-0,
2,0,-1,-1,-0.333333,1,0,0.333333,0,
kernel time: 59 microseconds
1,0,0,0,1,0,0,0,1,
1,0,0,0,1,0,0,0,1,
2,0,-1,-1,-0.333333,1,0,0.333333,0,
2,0,-1,-1,-0.333333,1,0,0.333333,0,
cublas time: 68 microseconds
$

So this is perhaps slightly faster than cublas but not much, for this 2048 matrix test case, CUDA 10.0, Tesla P100, linux.

Similar to the previous answer, here is a simplified (only 2 matrices) pycuda test case:

$ cat t14.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 >>= 4;
  return ret;
}

// in-place is acceptable i.e. out == in)
// T = float or double only
const int block_size = 288;
typedef double T; // *** can set to float or double
__global__ void inv3x3(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;
  T det = 1;
  if (idx < n*9)
    det = in[idx];
  unsigned sibase = (threadIdx.x / 9)*9;
  unsigned lane = threadIdx.x - sibase; // cheaper modulo
  si[threadIdx.x] = det;
  __syncthreads();
  unsigned off = pat[lane];
  T a  = si[sibase + getoff(off)];
  a   *= si[sibase + getoff(off)];
  T b  = si[sibase + getoff(off)];
  b   *= si[sibase + getoff(off)];
  a -= b;
  __syncthreads();
  if (lane == 0) si[sibase+3] = a;
  if (lane == 3) si[sibase+4] = a;
  if (lane == 6) si[sibase+5] = a;
  __syncthreads();
  det =  si[sibase]*si[sibase+3]+si[sibase+1]*si[sibase+4]+si[sibase+2]*si[sibase+5];
  if (idx < n*9)
    out[idx] = a / det;
}

""")
# host code
def gpuinv3x3(inp, n):
    # internal constants not to be modified
    hpat = (0x07584, 0x08172, 0x04251, 0x08365, 0x06280, 0x05032, 0x06473, 0x07061, 0x03140)
    # 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*9), dtype= np.float64)
    # Get kernel function
    matinv3x3 = kernel.get_function("inv3x3")
    # Define block, grid and compute
    blockDim = (288,1,1) # do not change
    gridDim = ((n/32)+1,1,1)
    # Kernel function
    matinv3x3 (
        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, 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0)
n = 2
result = gpuinv3x3(inp, n)
print(result.reshape(2,3,3))
$ python t14.py
[[[ 2.         -0.         -1.        ]
  [-1.         -0.33333333  1.        ]
  [-0.          0.33333333 -0.        ]]

 [[ 1.          0.          0.        ]
  [ 0.          1.          0.        ]
  [ 0.          0.          1.        ]]]
$

The above happens to be using double i.e. float64 in pycuda. Changing it to float i.e. float32 in pycuda involves changing the same 3 lines as described in this answer.

answered on Stack Overflow Mar 30, 2019 by Robert Crovella • edited Mar 30, 2019 by Robert Crovella

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