我正在写一个GPU上的图像恢复算法,详情在
Cuda: least square solving , poor in speed
求解线性系统的QR分解法
Ax=b
工作原理如下
min||Ax-b|| ---> ||QRx-b|| ---> ||(Q^T)QRx-(Q^T)b|| ---> ||Rx-(Q^T)b||
其中 R
是上三角矩阵。由此产生的上三角线性系统很容易求解。
我想用CULA工具来实现这个方法。 CULA 例程 GEQRF
计算 QR 因式分解。手册说:
On exit, the elements on and above the diagonal of the array contain the
min(M,N)-by-N
upper trapezoidal matrixR
(R
is upper triangular ifm >= n)
; the elements below the diagonal, with the arrayTAU
, represent the orthogonal/unitary matrixQ
as a product ofmin(m,n)
elementary reflectors.
我无法弄清楚 Q
的存储位置,算法对我来说似乎太复杂了。你能给点建议吗?
谢谢!
最佳答案
截至 2015 年 2 月,CUDA 7.0(现在为候选版本)提供了新的 cuSOLVER 库,包括计算矩阵的 QR 分解的可能性。这与 cuBLAS 库结合使用,能够根据 cuSOLVER 用户指南附录 C 中阐述的指南求解线性系统。
您必须遵循三个步骤:
1) geqrf
:它通过返回A
的上三角部分的上三角矩阵R
来计算矩阵的QR分解和矩阵 Q
以 Householder 向量的形式存储在 A
的下三角部分,而 Householder 向量的比例因子由 TAU< 返回
参数;
2) ormqr
:它通过覆盖C
返回Q
和矩阵C
的乘积;
3) trsm
:求解上三角线性系统。
下面,我将提供这些例程用法的完整示例。
#include "cuda_runtime.h"
#include "device_launch_paraMeters.h"
#include<iostream>
#include<fstream>
#include<iomanip>
#include<stdlib.h>
#include<stdio.h>
#include<assert.h>
#include <cusolverDn.h>
#include <cublas_v2.h>
#include <cuda_runtime_api.h>
#include "Utilities.cuh"
#include "TimingGPU.cuh"
#define BLOCK_SIZE 32
#define prec_save 10
/***************/
/* COPY KERNEL */
/***************/
__global__ void copy_kernel(const double * __restrict d_in, double * __restrict d_out, const int M, const int N) {
const int i = blockIdx.x * blockDim.x + threadIdx.x;
const int j = blockIdx.y * blockDim.y + threadIdx.y;
if ((i < N) && (j < N)) d_out[j * N + i] = d_in[j * M + i];
}
/****************************************************/
/* LOAD INDIVIDUAL REAL MATRIX FROM txt FILE TO CPU */
/****************************************************/
// --- Load individual real matrix from txt file
template <class T>
void loadCPUrealtxt(T * __restrict h_out, const char *filename, const int M) {
std::ifstream infile;
infile.open(filename);
for (int i = 0; i < M; i++) {
double temp;
infile >> temp;
h_out[i] = (T)temp;
}
infile.close();
}
/************************************/
/* SAVE REAL ARRAY FROM GPU TO FILE */
/************************************/
template <class T>
void saveGPUrealtxt(const T * d_in, const char *filename, const int M) {
T *h_in = (T *)malloc(M * sizeof(T));
gpuErrchk(cudaMemcpy(h_in, d_in, M * sizeof(T), cudaMemcpyDeviceToHost));
std::ofstream outfile;
outfile.open(filename);
for (int i = 0; i < M; i++) outfile << std::setprecision(prec_save) << h_in[i] << "\n";
outfile.close();
}
/********/
/* MAIN */
/********/
int main(){
// --- Extension of Appendix C.1 of cuSOLVER library User's Guide
// --- See also http://www.netlib.org/lapack/lug/node40.html
// --- ASSUMPTION Nrows >= Ncols
const int Nrows = 500;
const int Ncols = 500;
TimingGPU timerGPU;
double timingQR, timingSolve;
// --- cuSOLVE input/output parameters/arrays
int work_size = 0;
int *devInfo; gpuErrchk(cudaMalloc(&devInfo, sizeof(int)));
// --- CUDA solver initialization
cusolverDnHandle_t solver_handle;
cusolveSafeCall(cusolverDnCreate(&solver_handle));
// --- CUBLAS initialization
cublasHandle_t cublas_handle;
cublasSafeCall(cublasCreate(&cublas_handle));
/***********************/
/* SETTING THE PROBLEM */
/***********************/
// --- Setting the host, Nrows x Ncols matrix
double *h_A = (double *)malloc(Nrows * Ncols * sizeof(double));
loadCPUrealtxt(h_A, "D:\\Project\\solveNonSquareLinearSystemQRCUDA\\solveNonSquareLinearSystemQRCUDA\\testMatrix.txt", Nrows * Ncols);
// --- Setting the device matrix and moving the host matrix to the device
double *d_A; gpuErrchk(cudaMalloc(&d_A, Nrows * Ncols * sizeof(double)));
gpuErrchk(cudaMemcpy(d_A, h_A, Nrows * Ncols * sizeof(double), cudaMemcpyHostToDevice));
// --- Initializing the data matrix C (Of course, this step could be done by a kernel function directly on the device).
// --- Notice that, in this case, only the first column of C contains actual data, the others being empty (zeroed). However, cuBLAS trsm
// has the capability of solving triangular linear systems with multiple right hand sides.
double *h_C = (double *)calloc(Nrows * Nrows, sizeof(double));
loadCPUrealtxt(h_C, "D:\\Project\\solveNonSquareLinearSystemQRCUDA\\solveNonSquareLinearSystemQRCUDA\\testVector.txt", Nrows);
double *d_C; gpuErrchk(cudaMalloc(&d_C, Nrows * Nrows * sizeof(double)));
gpuErrchk(cudaMemcpy(d_C, h_C, Nrows * Nrows * sizeof(double), cudaMemcpyHostToDevice));
/**********************************/
/* COMPUTING THE QR DECOMPOSITION */
/**********************************/
timerGPU.StartCounter();
// --- CUDA QR GEQRF preliminary operations
double *d_TAU; gpuErrchk(cudaMalloc((void**)&d_TAU, min(Nrows, Ncols) * sizeof(double)));
cusolveSafeCall(cusolverDnDgeqrf_bufferSize(solver_handle, Nrows, Ncols, d_A, Nrows, &work_size));
double *work; gpuErrchk(cudaMalloc(&work, work_size * sizeof(double)));
// --- CUDA GEQRF execution: The matrix R is overwritten in upper triangular part of A, including diagonal
// elements. The matrix Q is not formed explicitly, instead, a sequence of householder vectors are
// stored in lower triangular part of A.
cusolveSafeCall(cusolverDnDgeqrf(solver_handle, Nrows, Ncols, d_A, Nrows, d_TAU, work, work_size, devInfo));
int devInfo_h = 0; gpuErrchk(cudaMemcpy(&devInfo_h, devInfo, sizeof(int), cudaMemcpyDeviceToHost));
if (devInfo_h != 0) std::cout << "Unsuccessful gerf execution\n\n";
timingQR = timerGPU.GetCounter();
printf("Timing for QR calculation = %f [ms]\n", timingQR);
/*****************************/
/* SOLVING THE LINEAR SYSTEM */
/*****************************/
timerGPU.StartCounter();
// --- CUDA ORMQR execution: Computes the multiplication Q^T * C and stores it in d_C
cusolveSafeCall(cusolverDnDormqr(solver_handle, CUBLAS_SIDE_LEFT, CUBLAS_OP_T, Nrows, Ncols, min(Nrows, Ncols), d_A, Nrows, d_TAU, d_C, Nrows, work, work_size, devInfo));
// --- Reducing the linear system size
double *d_R; gpuErrchk(cudaMalloc(&d_R, Ncols * Ncols * sizeof(double)));
double *d_B; gpuErrchk(cudaMalloc(&d_B, Ncols * sizeof(double)));
dim3 Grid(iDivUp(Ncols, BLOCK_SIZE), iDivUp(Ncols, BLOCK_SIZE));
dim3 Block(BLOCK_SIZE, BLOCK_SIZE);
copy_kernel << <Grid, Block >> >(d_A, d_R, Nrows, Ncols);
gpuErrchk(cudaMemcpy(d_B, d_C, Ncols * sizeof(double), cudaMemcpyDeviceToDevice));
// --- Solving an upper triangular linear system - compute x = R \ Q^T * B
const double alpha = 1.;
cublasSafeCall(cublasDtrsm(cublas_handle, CUBLAS_SIDE_LEFT, CUBLAS_FILL_MODE_UPPER, CUBLAS_OP_N,
CUBLAS_DIAG_NON_UNIT, Ncols, 1, &alpha, d_R, Ncols, d_B, Ncols));
timingSolve = timerGPU.GetCounter();
printf("Timing for solution of the linear system = %f [ms]\n", timingSolve);
printf("Overall timing = %f [ms]\n", timingQR + timingSolve);
/************************/
/* CHECKING THE RESULTS */
/************************/
// --- The upper triangular part of A contains the elements of R. Showing this.
saveGPUrealtxt(d_A, "D:\\Project\\solveNonSquareLinearSystemQRCUDA\\solveNonSquareLinearSystemQRCUDA\\d_R.txt", Nrows * Ncols);
// --- The first Nrows elements of d_C contain the result of Q^T * C
saveGPUrealtxt(d_C, "D:\\Project\\solveNonSquareLinearSystemQRCUDA\\solveNonSquareLinearSystemQRCUDA\\d_QTC.txt", Nrows);
// --- Initializing the output Q matrix (Of course, this step could be done by a kernel function directly on the device)
double *h_Q = (double *)malloc(Nrows * Nrows * sizeof(double));
for (int j = 0; j < Nrows; j++)
for (int i = 0; i < Nrows; i++)
if (j == i) h_Q[j + i*Nrows] = 1.;
else h_Q[j + i*Nrows] = 0.;
double *d_Q; gpuErrchk(cudaMalloc(&d_Q, Nrows * Nrows * sizeof(double)));
gpuErrchk(cudaMemcpy(d_Q, h_Q, Nrows * Nrows * sizeof(double), cudaMemcpyHostToDevice));
// --- Calculation of the Q matrix
cusolveSafeCall(cusolverDnDormqr(solver_handle, CUBLAS_SIDE_LEFT, CUBLAS_OP_N, Nrows, Ncols, min(Nrows, Ncols), d_A, Nrows, d_TAU, d_Q, Nrows, work, work_size, devInfo));
// --- d_Q contains the elements of Q. Showing this.
saveGPUrealtxt(d_Q, "D:\\Project\\solveNonSquareLinearSystemQRCUDA\\solveNonSquareLinearSystemQRCUDA\\d_Q.txt", Nrows * Nrows);
// --- At this point, d_C contains the elements of Q^T * C, where C is the data vector. Showing this.
// --- According to the above, only the first column of d_C makes sense.
//gpuErrchk(cudaMemcpy(h_C, d_C, Nrows * Nrows * sizeof(double), cudaMemcpyDeviceToHost));
//printf("\n\n");
//for (int j = 0; j < Nrows; j++)
// for (int i = 0; i < Nrows; i++)
// printf("C[%i, %i] = %f\n", j, i, h_C[j + i*Nrows]);
// --- Check final result
saveGPUrealtxt(d_B, "D:\\Project\\solveNonSquareLinearSystemQRCUDA\\solveNonSquareLinearSystemQRCUDA\\d_B.txt", Ncols);
cusolveSafeCall(cusolverDnDestroy(solver_handle));
return 0;
}
运行此类示例所需的 Utilities.cu
和 Utilities.cuh
文件在 github page 中维护。 . TimingGPU.cu
和 TimingGPU.cuh
文件保存在 github page 中。 .
可以通过以下Matlab代码生成数据并检查结果:
clear all
close all
clc
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% GENERATE RANDOM NON-SQUARE MATRIX WITH DESIRED CONDITION NUMBER %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% --- Credit to https://math.stackexchange.com/questions/198515/can-we-generate-random-singular-matrices-with-desired-condition-number-using-mat
Nrows = 500; % --- Number of rows
Ncols = 500; % --- Number of columns
% condNumber = 10 * sqrt(2); % --- Desired condition number
% A = randn(Nrows, Ncols);
% [U, S, V] = svd(A);
% S(S~=0) = linspace(condNumber, 1, min(Nrows, Ncols));
% A = U * S * V';
% --- Setting the problem solution
x = ones(Ncols, 1);
% y = A * x;
%
% Asave = reshape(A, Nrows * Ncols, 1);
% save testMatrix.txt Asave -ascii -double
% save testVector.txt y -ascii -double
load testMatrix.txt
load testVector.txt
A = reshape(testMatrix, Nrows, Ncols);
y = testVector;
[Q, R] = qr(A);
xMatlab = R \ (Q.' * y);
fprintf('Percentage rms of solution in Matlab %f\n', 100 * sqrt(sum(sum(abs(xMatlab - x).^2)) / sum(sum(abs(x).^2))));
fprintf('Percentage rms of Q * R - A %f\n', 100 * sqrt(sum(sum(abs(Q * R - A).^2)) / sum(sum(abs(A).^2))));
load d_R.txt
d_R = reshape(d_R, Nrows, Ncols);
d_R = d_R(1 : Ncols, :);
R = R(1 : Ncols, :);
fprintf('Percentage rms of matrix R between Matlab and CUDA %f\n', 100 * sqrt(sum(sum(abs(triu(R) - triu(d_R)).^2)) / sum(sum(abs(triu(d_R)).^2))));
load d_QTC.txt
fprintf('Percentage rms of Q^T * y - d_QTC %f\n', 100 * sqrt(sum(sum(abs(Q.' * y - d_QTC).^2)) / sum(sum(abs(d_QTC).^2))));
load d_B.txt
fprintf('Percentage rms of solution in Matlab %f\n', 100 * sqrt(sum(sum(abs(d_B - x).^2)) / sum(sum(abs(x).^2))));
请根据需要评论/取消评论行。
时间
计时(以毫秒为单位)(在 GTX960 卡上执行的测试,cc. 5.2):
Size QR decomposition Solving system Overall
100x100 0.89 1.41 2.30
200x200 5.97 3.23 9.20
500x500 17.08 21.6 38.7
关于algorithm - QR 分解求解 CUDA 中的线性系统,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22399794/