问题
I am learning CUDA with trying to solve some standard problems. As a example, I am solving the diffusion equation in two dimensions with the following code. But my results are different than the standard results and I am not able to figure that out.
//kernel definition
__global__ void diffusionSolver(double* A, double * old,int n_x,int n_y)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
int j = blockIdx.y * blockDim.y + threadIdx.y;
if(i*(n_x-i-1)*j*(n_y-j-1)!=0)
A[i+n_y*j] = A[i+n_y*j] + (old[i-1+n_y*j]+old[i+1+n_y*j]+
old[i+(j-1)*n_y]+old[i+(j+1)*n_y] -4*old[i+n_y*j])/40;
}
int main()
{
int i,j ,M;
M = n_y ;
phi = (double *) malloc( n_x*n_y* sizeof(double));
phi_old = (double *) malloc( n_x*n_y* sizeof(double));
dummy = (double *) malloc( n_x*n_y* sizeof(double));
int iterationMax =10;
//phase initialization
for(j=0;j<n_y ;j++)
{
for(i=0;i<n_x;i++)
{
if((.4*n_x-i)*(.6*n_x-i)<0)
phi[i+M*j] = -1;
else
phi[i+M*j] = 1;
phi_old[i+M*j] = phi[i+M*j];
}
}
double *dev_phi;
cudaMalloc((void **) &dev_phi, n_x*n_y*sizeof(double));
dim3 threadsPerBlock(100,10);
dim3 numBlocks(n_x*n_y / threadsPerBlock.x, n_x*n_y / threadsPerBlock.y);
//start iterating
for(int z=0; z<iterationMax; z++)
{
//copy array on host to device
cudaMemcpy(dev_phi, phi, n_x*n_y*sizeof(double),
cudaMemcpyHostToDevice);
//call kernel
diffusionSolver<<<numBlocks, threadsPerBlock>>>(dev_phi, phi_old,n_x,n_y);
//get updated array back on host
cudaMemcpy(phi, dev_phi,n_x*n_y*sizeof(double), cudaMemcpyDeviceToHost);
//old values will be assigned new values
for(j=0;j<n_y ;j++)
{
for(i=0;i<n_x;i++)
{
phi_old[i+n_y*j] = phi[i+n_y*j];
}
}
}
return 0;
}
Can someone tell me if there is anything wrong in this process? Any help will be greatly appreciated.
回答1:
One big mistake you have is that phi_old is passed to the kernel and used by the kernel but this is a host pointer.
Malloc a dev_phi_old using cudaMalloc. Set it to default value and copy it to the GPU first time before entering the z loop.
回答2:
talonmies, brano and huseyin have already pointed out some mistakes of your code.
Diffusion (heat) equation is one of the classical example of partial differential equations solvable with CUDA. There is also a thorough example in Chapter 7 of the CUDA by Example book.
As a reference to future Users, I'm providing below a full worked example including both, CPU and GPU codes. Instead of swapping pointers, as suggested by talonmies, I'm just condensing two Jacobi iterations in a single loop.
#include <iostream>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include "Utilities.cuh"
#define BLOCK_SIZE_X 16
#define BLOCK_SIZE_Y 16
/***********************************/
/* JACOBI ITERATION FUNCTION - GPU */
/***********************************/
__global__ void Jacobi_Iterator_GPU(const float * __restrict__ T_old, float * __restrict__ T_new, const int NX, const int NY)
{
const int i = blockIdx.x * blockDim.x + threadIdx.x ;
const int j = blockIdx.y * blockDim.y + threadIdx.y ;
// N
int P = i + j*NX; // node (i,j) |
int N = i + (j+1)*NX; // node (i,j+1) |
int S = i + (j-1)*NX; // node (i,j-1) W ---- P ---- E
int E = (i+1) + j*NX; // node (i+1,j) |
int W = (i-1) + j*NX; // node (i-1,j) |
// S
// --- Only update "interior" (not boundary) node points
if (i>0 && i<NX-1 && j>0 && j<NY-1) T_new[P] = 0.25 * (T_old[E] + T_old[W] + T_old[N] + T_old[S]);
}
/***********************************/
/* JACOBI ITERATION FUNCTION - CPU */
/***********************************/
void Jacobi_Iterator_CPU(float * __restrict T, float * __restrict T_new, const int NX, const int NY, const int MAX_ITER)
{
for(int iter=0; iter<MAX_ITER; iter=iter+2)
{
// --- Only update "interior" (not boundary) node points
for(int j=1; j<NY-1; j++)
for(int i=1; i<NX-1; i++) {
float T_E = T[(i+1) + NX*j];
float T_W = T[(i-1) + NX*j];
float T_N = T[i + NX*(j+1)];
float T_S = T[i + NX*(j-1)];
T_new[i+NX*j] = 0.25*(T_E + T_W + T_N + T_S);
}
for(int j=1; j<NY-1; j++)
for(int i=1; i<NX-1; i++) {
float T_E = T_new[(i+1) + NX*j];
float T_W = T_new[(i-1) + NX*j];
float T_N = T_new[i + NX*(j+1)];
float T_S = T_new[i + NX*(j-1)];
T[i+NX*j] = 0.25*(T_E + T_W + T_N + T_S);
}
}
}
/******************************/
/* TEMPERATURE INITIALIZATION */
/******************************/
void Initialize(float * __restrict h_T, const int NX, const int NY)
{
// --- Set left wall to 1
for(int j=0; j<NY; j++) h_T[j * NX] = 1.0;
}
/********/
/* MAIN */
/********/
int main()
{
const int NX = 256; // --- Number of discretization points along the x axis
const int NY = 256; // --- Number of discretization points along the y axis
const int MAX_ITER = 1; // --- Number of Jacobi iterations
// --- CPU temperature distributions
float *h_T = (float *)calloc(NX * NY, sizeof(float));
float *h_T_old = (float *)calloc(NX * NY, sizeof(float));
Initialize(h_T, NX, NY);
Initialize(h_T_old, NX, NY);
float *h_T_GPU_result = (float *)malloc(NX * NY * sizeof(float));
// --- GPU temperature distribution
float *d_T; gpuErrchk(cudaMalloc((void**)&d_T, NX * NY * sizeof(float)));
float *d_T_old; gpuErrchk(cudaMalloc((void**)&d_T_old, NX * NY * sizeof(float)));
gpuErrchk(cudaMemcpy(d_T, h_T, NX * NY * sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_T_old, d_T, NX * NY * sizeof(float), cudaMemcpyDeviceToDevice));
// --- Grid size
dim3 dimBlock(BLOCK_SIZE_X, BLOCK_SIZE_Y);
dim3 dimGrid (iDivUp(NX, BLOCK_SIZE_X), iDivUp(NY, BLOCK_SIZE_Y));
// --- Jacobi iterations on the host
Jacobi_Iterator_CPU(h_T, h_T_old, NX, NY, MAX_ITER);
// --- Jacobi iterations on the device
for (int k=0; k<MAX_ITER; k=k+2) {
Jacobi_Iterator_GPU<<<dimGrid, dimBlock>>>(d_T, d_T_old, NX, NY); // --- Update d_T_old starting from data stored in d_T
Jacobi_Iterator_GPU<<<dimGrid, dimBlock>>>(d_T_old, d_T , NX, NY); // --- Update d_T starting from data stored in d_T_old
}
// --- Copy result from device to host
gpuErrchk(cudaMemcpy(h_T_GPU_result, d_T, NX * NY * sizeof(float), cudaMemcpyDeviceToHost));
// --- Calculate percentage root mean square error between host and device results
float sum = 0., sum_ref = 0.;
for (int j=0; j<NY; j++)
for (int i=0; i<NX; i++) {
sum = sum + (h_T_GPU_result[j * NX + i] - h_T[j * NX + i]) * (h_T_GPU_result[j * NX + i] - h_T[j * NX + i]);
sum_ref = sum_ref + h_T[j * NX + i] * h_T[j * NX + i];
}
printf("Percentage root mean square error = %f\n", 100.*sqrt(sum / sum_ref));
// --- Release host memory
free(h_T);
free(h_T_GPU_result);
// --- Release device memory
gpuErrchk(cudaFree(d_T));
gpuErrchk(cudaFree(d_T_old));
return 0;
}
The Utilities.cu and Utilities.cuh files needed to run such an example are maintained at this github page.
回答3:
Here:
A[i+n_y*j] = A[i+n_y*j] + (old[i-1+n_y*j]+old[i+1+n_y*j]+old[i+(j-1)*n_y]+old[i+(j+1)*n_y] -4*old[i+n_y*j])/40;
You are dividing by 40(integer) which can result in wrong diffusing rate. Actually can result in none-diffusing.
But A is an array of doubles.
Divide the diffuse rate by 40.0 and see if it works.
If this is from Jos-Stam's solver, it should be 4.0 not 40
Theres also another thing:
-4*old[i+n_y*j])/40;
Here you are multiplying with 4(integer). This can cause a integral-casting too!
This:
-4.0*old[i+n_y*j])/40.0;
decreases some errors.
Have a nice day.
来源：https://stackoverflow.com/questions/11994679/solving-2d-diffusion-heat-equation-with-cuda